The Classification of the Finite Simple Groups, Number 10 [10] 9781470475536, 9781470475666

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M ATHEMATICAL

Surveys and Monographs Volume 40, Number 10

The Classification of the Finite Simple Groups, Number 10 Part V, Chapters 9–17: Theorem C6 and Theorem C*4, Case A Inna Capdeboscq Daniel Gorenstein Richard Lyons Ronald Solomon

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The Classification of the Finite Simple Groups, Number 10 Part V, Chapters 9–17: Theorem C6 and Theorem C*4, Case A

M ATHEMATICAL

Surveys and Monographs Volume 40, Number 10

The Classification of the Finite Simple Groups, Number 10 Part V, Chapters 9–17: Theorem C6 and Theorem C*4, Case A Inna Capdeboscq Daniel Gorenstein Richard Lyons Ronald Solomon

EDITORIAL COMMITTEE Alexander H. Barnett Michael A. Hill Bryna Kra (chair)

David Savitt Natasa Sesum Jared Wunsch

The authors gratefully acknowledge the support provided by grants from the Engineering and Physical Sciences Research Council (EP/V05547X/1) and the Simons Foundation (425816). 2020 Mathematics Subject Classification. Primary 20D05, 20D06, 20D08. For additional information and updates on this book, visit www.ams.org/bookpages/surv-40.10 The ISBN numbers for this series of books includes ISBN 978-1-4704-7553-6 (number 10) ISBN 978-1-4704-6437-0 (number 9) ISBN 978-1-4704-4189-0 (number 8) ISBN 978-0-8218-4069-6 (number 7) ISBN 978-0-8218-2777-2 (number 6) ISBN 978-0-8218-2776-5 (number 5) ISBN 978-0-8218-1379-9 (number 4) ISBN 978-0-8218-0391-2 (number 3) ISBN 978-0-8218-0390-5 (number 2) ISBN 978-0-8218-0334-9 (number 1) Library of Congress Cataloging-in-Publication Data The first volume was catalogued as follows: Gorenstein, Daniel. The classification of the finite simple groups / Daniel Gorenstein, Richard Lyons, Ronald Solomon. p. cm. (Mathematical surveys and monographs: v. 40, number 1–) Includes bibliographical references and index. ISBN 0-8218-0334-4 [number 1] 1. Finite simple groups. I. Lyons, Richard, 1945– . II. Solomon, Ronald. III. Title. IV. Series: Mathematical surveys and monographs, no. 40, pt. 1–;. QA177 .G67 1994 512.2-dc20 94-23001 CIP Copying and reprinting. Individual readers of this publication, and nonprofit libraries acting for them, are permitted to make fair use of the material, such as to copy select pages for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication is permitted only under license from the American Mathematical Society. Requests for permission to reuse portions of AMS publication content are handled by the Copyright Clearance Center. For more information, please visit www.ams.org/publications/pubpermissions. Send requests for translation rights and licensed reprints to [email protected]. c 2023 by the American Mathematical Society. All rights reserved.  The American Mathematical Society retains all rights except those granted to the United States Government. Printed in the United States of America. ∞ The paper used in this book is acid-free and falls within the guidelines 

established to ensure permanence and durability. Visit the AMS home page at https://www.ams.org/ 10 9 8 7 6 5 4 3 2 1

28 27 26 25 24 23

For Lucie, Catherine, and David

Contents Preface

xi

Chapter 9. General Group-Theoretic Lemmas 1. Fusion 2. Groups X with e(X) ≤ 3 3. Signalizer Functors 4. Balance 5. p-Groups 6. p-Components 7. Standard Components of Involution Centralizers 8. Coprime Action 9. Generation 10. Subgroups of GL4 (3) 11. Strong p-Embedding 12. Characteristic-p Representations 13. Miscellaneous

1 1 3 6 8 10 14 17 18 19 21 25 26 26

Chapter 10. Theorems C6 and C∗6 1. Introduction 2. Tp -Groups and TGp -Groups in Lop (G) 3. Centralizers with Tp -Components 4. Balance and Generation 5. Theorem 3: 3/2-Balance 6. Theorem 3: 5/2-Balance and 3-Balance 7. Nonquasisimple Pumpups of K 8. Nontrivial Signalizer Functors 9. The Subgroup M 10. K Lies in M 11. The Pumpup of K in M 12. The L-Preuniqueness of M : the Nonsimple Case 13. The L-Preuniqueness of M : the Simple Rank 2 Case 14. The Rank 1 Case: 3/2-Balance 15. The Rank 1 Case: Controlling Cores 16. The L-Preuniqueness of M : the Rank 1 Case 17. Theorem 4: Almost Strong p-Embedding, the General Case 18. The Residual Cases 19. The Case Op (M ) = 1 vii

29 29 36 39 40 43 54 59 66 77 79 88 91 96 102 111 115 123 126 129

viii

CONTENTS

Chapter 11. Theorems C4 and C∗4 : Introduction 1. Theorem C∗4 and Its Cases 2. The Stages of Theorem C∗4 (Case A) 3. Some Definitions

143 143 146 148

Chapter 12. Theorem C∗4 : Stage A1. First Steps 1. Introduction 2. Initial Remarks 3. An Exceptional Configuration Involving L3 (4) 4. An Exceptional Configuration Involving L2 (pp ) 5. Balance with Respect to P0 6. The Subgroup M , and the Second Case of Balance 7. Balance in the Remaining Cases 8. The Residual Case 9. B-Signalizers and Ipo (G): The Centralizing Case 10. The Non-Centralizing Case: A Dichotomy 11. The Case BT ∩ Ipo (G) = ∅ 12. The M12 Case 13. The L3 (4a ) Case 14. Completion of the Proof

151 151 152 153 155 161 167 171 178 179 181 186 191 195 199

Chapter 13. Theorem C∗4 : Stage A2. Nonconstrained p-Rank 3 Centralizers 1. Introduction 2. Components and p-Components 3. Theorem 1: Introduction 4. Some √ Nonsimple Cases √ for 2F4 ( 32), Sp4 (8), and 3D4 (2) 5. 2F4 ( 2), and Preliminaries √ 6. The L2 (8) and 2B2 ( 32) Cases 7. Some Cases of p-Rank 2 8. Configurations Involving M12 and HJ 9. L2 (q), q ∈ {5, 7, 17} 10. The Remaining Cases of Theorem 1 11. Theorem 2: Introduction 12. Components with Outer p-Automorphisms 13. K Is Not J3 14. From U4 (2) to 2Sp6 (2) 15. K ∼ = 2Sp6 (2) Implies G ∼ = Co3 16. The Z2 × Sp6 (2) Case

203 203 204 207 213 214 216 224 228 234 238 245 248 265 272 276 278

Chapter 14. Theorem C∗4 : Stage A3. KM -Singularities 1. Introduction 2. Centralizers of Noncyclic p-Groups 3. KM -Singularities of Rank 2 and Type A6 : The Case Hinv = ∅ 4. The Case Hinv = ∅: {2, 3}-Local Subgroups 5. p = 3; L2 (3n ) 3-Components

291 291 291 295 298 299

CONTENTS

6. 2-Subgroups of G Normalized by E33 -Subgroups 7. Subgroups H ∈ H and Involution Centralizers 8. The Residual Cases

ix

300 303 306

Chapter 15. Theorem C∗4 : Stage A4. Setups for Recognizing G 1. Introduction 2. Pumpups of a Rank 2 KM -Singularity of Type A6 3. Centralizers of 2-Central Involutions

311 311 312 317

Chapter 16. Theorem C∗4 : Stage A5. Recognition 1. Introduction 2. The Extraspecial Setup 3. The 32 U4 (3) and G2 (3) Cases: G ∼ = Suz or F3 4. The U4 (2) Setup 5. 3-Local Structure 6. G ∼ = U5 (2) 7. G ∼ = U6 (2) 8. G ∼ = D4 (2) 9. G ∼ = Co2 10. The A9 –Sp6 (2) Setup 11. 3-Local Structure 12. The Mixed Case 13. The A9 Case 14. G ∼ = A12 15. G ∼ = F5 16. The Sp6 (2) Case 17. G ∼ = Sp8 (2) 18. G ∼ = F4 (2)

333 333 333 335 343 344 346 350 356 364 380 381 385 386 388 389 396 397 406

Chapter 17. Properties of K-Groups 1. Simple Sections of G, e(G) = 3 2. Flat TGp -Groups 3. Groups of Low p-Rank, p Odd 4. Balance and Signalizers 5. Generation 6. The Balance-Generation Dichotomy 7. Cp -Groups, p odd 8. Tp -Groups and TGp -Groups 9. C2 -Groups 10. Some Familiar C2 -Groups and Their Subgroups 10.1. O8+ (2) 10.2. Sp6 (2) 10.3. U6 (2) 10.4. L6 (2) 10.5. 2 F4 (q) 10.6. L3 (4) and L3 (4n )

409 409 416 418 424 436 446 457 473 476 482 482 490 499 500 500 504

x

CONTENTS n

10.7. 2B2 (2 2 ), n ≥ 3 10.8. Sp4 (2) ∼ = Σ6 10.9. U4 (2) ∼ = P Sp4 (3) 10.10. L± 4 (3) 10.11. M12 , HJ 10.12. M22 10.13. J3 10.14. Suz 11. Some Other Familiar Groups 11.1. L2 (q), q odd 11.2. L2 (27) 11.3. (S)L3 (q) 11.4. L3 (4) and F i22 11.5. A9 11.6. A10 12. Sylow Subgroups 13. Pumpups and Subcomponents 14. Cross-Characteristic “Pumpups” 15. Connecting p-Groups 16. Small Representations 17. Elements of Order 2p 18. p-Ranks 19. p-Components 20. Miscellaneous

506 506 507 511 518 519 520 521 522 522 523 524 526 529 530 530 531 542 546 547 549 552 553 554

Bibliography

565

Index

569

Preface In the first volume of this series [I2 , Sections 14–16], we formulated seven theorems, Theorems Ci , 1 ≤ i ≤ 7, to handle our sevenfold case division for the proof of the classification theorem for the finite simple groups. The seven rows of the “Classification Grid” [I2 , Section 3] gave a coarse outline of the proofs of Theorems C1 –C7 as we foresaw them. Together with various supporting Uniqueness theorems collected in the “Uniqueness Grid” [I2 , Section 2], and together with a specified list of Background Results, Theorems C1 –C7 imply the classification theorem. Several of the necessary Uniqueness theorems, but not all, were proved in Book 4. These included all the so-called 2-Uniqueness theorems — those necessary to complete the characterization of target groups of odd type. (Roughly speaking, the simple groups of odd type are the alternating groups, and the groups of Lie type of odd characteristic, with a small finite number of additions and subtractions.) It is important to note that the completion of the so-called Odd Uniqueness theorems, which are those Uniqueness theorems needed for target groups of even type, is a major step still to be covered in future volumes. Theorem C1 is the Odd Order Theorem of Feit-Thompson, which is one of our assumed Background Results. Theorems C2 and C3 , on the Special Odd case, were proved in Book 6. In Books 5, 7, and 8, we proved Theorem C7 (the theorem for the Generic case), modulo a new Theorem C∗6 , which in turn is a mild strengthening of Theorem C6 . In particular, in conjunction with the 2-Uniqueness theorems established in Book 4 (the archetype of which is the Bender-Suzuki theorem on groups containing a strongly embedded subgroup), Books 5–8 established that a minimal counterexample G to the classification theorem is of even type, indeed “restricted” even type. In fact, with a boost from Theorem C∗6 , proved in Chapter 10 of the current volume, and Odd Uniqueness theorems, they would establish more, as we shall see presently. In Book 9, we began the special even type analysis by proving Theorem C5 , yielding among other targets some large sporadic groups, including the Monster. Also in that book we began the proofs of Theorems C6 and a variant C∗6 . In this book we first complete the proof of Theorems C6 and C∗6 , ∗∗ and a consequence Theorem C∗∗ 6 , in Chapter 10. Theorem C6 is stated as follows (Chapter 10, Theorem 1.5): Theorem. If G is of weak LTp -type, p odd, then G contains a strong p-uniqueness subgroup of component type. xi

xii

PREFACE

(See the Introductory section of Chapter 10 for the definition of weak LTp -type.) Together with Theorems C5 [V1 , p. 1] and C∗7 [III1 , p. 5] from Books 5, 7, 8, and 9, these theorems yield the following milestone: Theorem. Let G be a K-proper simple group of restricted even type. If p is an odd prime such that m2,p (G) ≥ 4, then either G ∈ K or G has a strong p-uniqueness subgroup. Modulo Odd Uniqueness theorems, then, a minimal counterexample to the classification theorem satisfies e(G) ≤ 3. In Chapter 11, we consider what cases remain for the classification theorem. Now, Theorems C5 , C6 , and C7 are distinguished from one another by their hypotheses on the p-local structure of G for some single odd prime p (let us call it the “critical” prime). In these different theorems, different hypotheses are placed on the isomorphism types of components of CG (x)/Op (CG (x)) for certain elements x ∈ G of order p (the set of all such isomorphism types is called Lop (G)). In all cases the 2-local p-rank m2,p (G) of G is assumed to be at least 4, and indeed some 2-local subgroup of G containing almost a whole Sylow 2-subgroup of G is assumed to have p-rank at least 4. On the other hand, certain generalizations (C∗6 and C∗7 ) of Theorems C6 and C7 are also proved in Books 5, 7, and 8 and Chapter 10 of the current volume. These generalizations weaken the assumption m2,p (G) ≥ 4 to mp (G) ≥ 4 for the critical prime p. Hence a minimal counterexample G to the classification theorem must have, at this point, additional properties beyond e(G) ≤ 3. Taking advantage of this, we formulate in Chapter 11 a new Theorem C∗4 to serve in the place of Theorem C4 . We then see that the proof of Theorem C∗4 divides into three main cases. First, there is the “classical” Quasithin case e(G) ≤ 2. Next is the “Bicharacteristic” case, in which e(G) = 3 but mp (G) > 3 = m2,p (G) for some odd prime p, and every group in Lop (G) is in the explicit set Cp of “Chev(p)-like” quasisimple K-groups [I2 , 12.1]. Last is the so-called “Case B,” in which e(G) = 3 and mp (G) = 3 for every odd prime p such that m2,p (G) = 3. The classical Quasithin case has been handled by a celebrated theorem of Aschbacher and Smith [ASm1], [ASm2]. One version of their result was designed to be quotable here: Theorem (Aschbacher, Smith). Let G be a K-proper simple group of even type with e(G) ≤ 2. Then G ∈ K. In Chapters 12–16 of the current volume, we handle the Bicharacteristic case, thereby proving: Theorem C∗4 (Case A). Let G be a K-proper simple group of restricted even type. Suppose that p is an odd prime such that m2,p (G) = 3 < mp (G).

PREFACE

Suppose (a) (b) (c)

xiii

also that Lop (G) ⊆ Cp . Then p = 3 and one of the following holds: G∼ = A12 ; or G∼ = Co2 , Co3 , Suz, F3 , or F5 ; or ∼ G = U5 (2), U6 (2), D4 (2), Sp8 (2), or F4 (2).

Note that in this theorem, as was also the case in Theorem C5 , there is no alternative conclusion that a strong p-uniqueness subgroup exists. This contrasts with Theorems C∗6 and C∗7 , and also with the future Theorem C∗4 (Case B). (Note also that in Definition 1.1 of Chapter 1 of Book 5 [III1 , p. 1], we have defined “strong p-uniqueness subgroup of G of component type” for any prime p which, if odd, satisfies mp (G) ≥ 4. The definition does not require m2,p (G) ≥ 4.) In summary, the theorems to be proved in future volumes are Case B of Theorem C∗4 and Odd Uniqueness theorems. The change to Theorem C∗4 has consequences for the proofs of both of these. While we originally conceived of a mainly 2-local proof of Theorem C4 , we now plan to use odd local analysis as well, following the fundamental groundbreaking papers of Aschbacher [A13], [A24]. In particular, Sections 15 and 22 of [I2 ] no longer fit our plans and can be replaced by Chapter 11 of this volume, which contains more precise details on Theorem C∗4 . Furthermore, our Theorem C∗4 will also necessitate a strengthened version of the Odd Uniqueness theorems — Theorem and Corollary U (σ), which in turn depend on Theorem M(S). Just as we have found it necessary in other volumes of this series, we again need to expand the Background Results, this time in connection with the recognition of the sporadic group Co2 . The expansion is made precise in (9B) of Chapter 16. We are happy to acknowledge the help of Michael Aschbacher, George Glauberman, Jon Hall, and Pham Huu Tiep, with whom we have had very useful communications. During the preparation of this book, the first author received support from the Engineering and Physical Sciences Research Council, and the fourth author received support from a Simons Foundation award. We gratefully acknowledge this support. We have recently passed January 1, 2023, the centenary of the birth of Danny Gorenstein. It is fitting that we extend a special tribute to his inspiration and guidance as we continue to work towards the fulfillment of his vision. Inna Capdeboscq Richard Lyons Ronald Solomon May 2023

CHAPTER 9

General Group-Theoretic Lemmas 1. Fusion As is the case throughout this series, G is our minimal counterexample to the Classification theorem. As such, G is a simple K-proper group. All groups are assumed finite unless explicitly stated otherwise. Lemma 1.1 (Thompson transfer lemma). Let T0 ≤ T ∈ Syl2 (X) with |T : T0 | = 2. Let t ∈ I2 (T ). If t ∈ [X, X] (or more generally t lies in every subgroup of X of index 2), then t has an X-conjugate in T0 that is extremal in T . Proof. See [IG , 15.16] or [IG , 15.15].



Lemma 1.2. Let x ∈ Ip (G) and P ∈ Sylp (CG (x)). Let y ∈ Ip (Z(P )) be p-central in G. If CG (x) controls the G-fusion of y in Z(P ), then x ∈ [CG (y), CG (y)]. 

Proof. See [III2 , 6.1].

Recall that for any group X and any set π of positive primes, Oπ (X) ia the largest normal π-subgroup of X, and Oπ (X) is the smallest normal subgroup N of X such that X/N is a π-group. Equivalently, Oπ (X) is the subgroup of X generated by all elements of X whose orders are divisible by primes only in π  . Here π  is the complement of π in the set of all positive primes. Also, we write Op (X), Op (X), etc., instead of O{p} (X), O{p} (X), etc. Also, if A is an element or subset of a group X, and B ⊆ X, then AB = {Ab | b ∈ B}. Lemma 1.3 (Gr¨ un). Let 1 = P ∈ Sylp (X) for some prime p and group X. Let Z ≤ Z(P ) and N = NG (Z), and assume that Z is weakly closed in P with respect to X. If X is simple, then N = Op (N ). Proof. By [IG , 16.9], N controls X-fusion in P . Then as X = Op (X),  [IG , 15.10(ii)] completes the proof. Lemma 1.4. Suppose that x ∈ G is a p-element, P ∈ Sylp (CG (x)), and W is an abelian subgroup of P that is weakly closed in P with respect to G. If x ∈ W , then xG ∩ W = xNG (W ) . 

Proof. See [III8 , 6.4]. 1

2

9. GENERAL GROUP-THEORETIC LEMMAS

∼ F  T . If X = O2 (X), Lemma 1.5. Let T ∈ Syl2 (X) and suppose Z4 = then for every involution z ∈ I2 (T ) and Tz ∈ Syl2 (CX (z)), Tz has a normal subgroup Fz ∼ = Z4 . Proof. Replacing z by a conjugate if necessary, we may assume that z is extremal in T and subject to this, if possible, [F, z] = 1. If indeed [F, z] = 1 then the result is obvious. If [F, z] = 1 then all extremal conjugates of z in T lie outside the subgroup CT (F ) of index 2, so by Lemma 1.1, z ∈ O2 (X), contrary to assumption.  Definition 1.6. For a group X and prime p, we define Sp (X) to be the set of all elementary abelian p-subgroups A of X such that A contains Ip (CX (A)). Equivalently, Sp (X) is the set of all maximal elementary abelian p-subgroups of X with respect to inclusion. Lemma 1.7. Let P ∈ Syl3 (X) with P ∼ = Z3 × (Z3 Z3 ). Let Z = Z(P ) ∩ [P, P ] and assume that Z is not weakly closed in P with respect to X. Then there exists D ∈ S3 (P ) and g ∈ NX (D) such that Z g = Z. Proof. By the Alperin-Goldschmidt theorem [IG , 16.1] there exist D ≤ P and g ∈ NX (D) such that CP (D) ≤ D and Z g = Z. If D ≤ J(P ), then D = J(P ) ∈ S3 (P ). So assume that D ≤ J(P ). Then if [D, D] = 1, Z = Z(D) ∩ [D, D]  NX (D), contradicting the existence of g. So D is abelian, and if Φ(D) = 1, then Z = Φ(D)  NX (D), contradiction. Hence D is elementary abelian, whence D ∈ S3 (P ), and the proof is complete.  If V is an elementary abelian p-group, we use the terms GL(V ) and Aut(V ) interchangeably. The set of all elementary abelian subgroups of X of order pn , n > 0, is written Epn (X). If the prime p is clear from context, the superscript “p” may be omitted. Lemma 1.8. Let X ≤ Aut(V ) where V ∼ = Ep3 , p an odd prime. Suppose that V = W × U with |W | = p, X stabilizes U , CX (W ) acts irreducibly on U , and X contains a transvection t moving W . Then X is transitive on E1 (V ) − E1 (U ). Proof. Let M = NAut(V ) (U ), so that X ≤ M and Op (M ) is transitive on E1 (V ) − E1 (U ). We may thus assume that Op (M ) ≤ X, and argue to a contradiction. But CX (W ) has equivalent actions on U and Op (M ), so it is irreducible on Op (M ). Thus, X ∩ Op (M ) = 1, and X embeds (irreducibly) in M/Op (M ) ∼ = Aut(U ). As X contains a transvection, the image of X in Aut(U ) contains SL(U ). Since p is odd, X has a central involution and therefore normalizes a unique element of E1 (V ) − E1 (U ). Given the action of CX (W ), this element must be W . But t moves W , a contradiction. The proof is complete. 

2. GROUPS X WITH e(X) ≤ 3

3

2. Groups X with e(X) ≤ 3 Recall that for any prime p and group X, mp (X) is the largest nonnegative integer such that X contains an elementary abelian subgroup of order pn . Also, for any odd prime p and group X of even order, m2,p (X) = max{mp (H) | H ≤ X, O2 (H) = 1} and e(X) = max{m2,p (X) | p is an odd prime}. Obviously if X ≤ Y , then m2,p (X) ≤ m2,p (Y ) and e(X) ≤ e(Y ). As usual, Kp is the set of (isomorphiusm classes of) quasisimple K-groups such that Op (K) = 1. Moreover, for J ∈ Kp , we also define m  p (J) = mp (J) − mp (Z(J)) and m  2,p (J) = m2,p (J) − mp (Z(J)). Lemma 2.1 (Thompson dihedral lemma). Let p be an odd prime. Suppose that the elementary abelian 2-group E of rank m > 0 acts faithfully on the p-group P . Then P contains an elementary abelian E-invariant subgroup Q such that QE = D1 × D2 × · · · × Dm , where each Di is dihedral of order 2p. Moreover, e(EP ) ≥ m − 1. 

Proof. [IG , 24.1].

Lemma 2.2. Let X be a group. Let p be an odd prime and A an elementary abelian p-subgroup of X. Suppose that CX (A) contains an elementary abelian 2-subgroup E. Suppose also that mp (CX (A)) > m2,p (X) or that Op (CX (A)) has odd order. If CX (A) is p-constrained, then there exists an EA-invariant Sylow p-subgroup P of Op p (CX (A)), and m2 (E/CE (P )) ≤ m2,p (X) − mp (A) + 1. In particular, if e(X) ≤ 3, then m2 (E/CE (P )) ≤ 4 − mp (A). Proof. If Op (CX (A)) has even order, then by a Frattini argument, m2,p (X) ≤ mp (CX (A)), contrary to assumption. So Op (CX (A)) has odd order, implying that P exists. Replacing E by a complement to CE (P ), we may assume that CE (P ) = 1. Then by Lemma 2.1, EAP contains A × D, where D is the direct product of m2 (E) dihedral groups of order 2p. Let z be an involution in one of these dihedral direct factors. Then m2,p (X) ≥ mp (CAD (z)) = mp (A) + mp (D) − 1 = mp (A) + m2 (E) − 1. The result follows.  Corollary 2.3. Let X = K z with K quasisimple, z ∈ I2 (Z(X)), and O2 (X) = 1. Suppose X acts on a p-group P , p odd, and CK (P ) = 1. Then m2,p (XP ) ≥ mp (P ∩ Z(XP )) + m2 (X) − 1. Proof. Let E ≤ X be elementary abelian of order 2m , m = m2 (X). Then z ∈ E. As X/K is cyclic, m2 (E ∩ K) ≥ m − 1. Since CK (P ) = 1, (E ∩ K)P ≥ (P ∩ Z(XP )) × D, where D is the direct product of m − 1 dihedral groups of order 2p and Op (D) ≤ P . Therefore mp (P ) ≥ mp (P ∩ Z(XP )) + m − 1. If CE (P ) = 1, then the conclusion is obvious, so suppose

4

9. GENERAL GROUP-THEORETIC LEMMAS

that CE (P ) = 1. Then EP contains (P ∩ Z(XP )) × D ∗ , where D ∗ is the direct product of m copies of D2p , and again the conclusion is obvious.  Corollary 2.4. Let X = K z with K/O2 (K) simple, K = [K, K], = 1 and z ∈ CX (K/O2 (K)). If m2 (X) ≥ 5 and e(X) ≤ 3, then K is quasisimple. z2

Proof. As K/O2 (K) is simple and K = [K, K], any proper normal subgroup of K lies in O2 (K). If the lemma fails, then F (O2 (K)) = F ∗ (O2 (K)) = 1 and CK (F (O2 (K))) ≤ O2 (K) by [IG , 3.17(iii)], and so CK (Op (K)) ≤ O2 (K) for some odd prime p. Hence m2 (AutK (Op (K))) ≥ 4. By Lemma 2.1, mp (Op (K)) ≥ 4. Hence if CKz (Op (K)) has even order, then m2,p (K z) ≥ 4 > e(K z), a contradiction. Thus, we may assume that CKz (Op (K)) ≤ O2 (K). Therefore m2 (AutKz (Op (K))) ≥ 5, so by Lemma 2.1 again, m2,p (K z) ≥ m2 (AutKz (Op (K))) − 1 ≥ 4 > e(K z), a contradiction. The lemma is proved.  Lemma 2.5. Let X be a K-group such that e(X) ≤ 3. Let p be an odd prime and let Y ≤ CX (Op p (X)/Op (X)). If Y = [Y, Y ], then Y ≤  Lp (X)Op (X). If Y = [Y, Y ] = Op (Y ), then Y ≤ Lp (X). 

Proof. Since Lp (X) = Op (Lp (X)Op (X)), it suffices to prove the first statement, and we may assume that Op (X) = 1. Let L be a component of E(X). If LY has m components, then using [IG , 16.11], we see that m2,p (X) ≥ m − 1, so m ≤ 4 as e(X) = 3. As Σm is solvable and Y = [Y, Y ], Y normalizes L. Then by the Schreier property, Y induces inner automorphisms on L, and hence on all of F ∗ (X) = E(X)Op (X). Then Y ≤ F ∗ (X) by [IG , 3.6], so Y = [Y, Y ] ≤ F ∗ (X)(∞) = E(X) = Lp (X).  Corollary 2.6. Assume that X and Y are as in Lemma 2.5. Suppose that in addition, (a) Op (X) has odd order;  (b) Y = O p (Y ) is a component of CX (z) for some z ∈ I2 (X); and (c) O2 (Y ) = 1. Then there is a p-component J of X such that Y ∼ = Y , where X = X/Op (X), and one of the following holds: (a) Y is a component of CJ (z); or z (b) J = J z , Y is a diagonal subgroup of JJ , and Y ∼ = J/Op p (J). Proof. By Lemma 2.5, Y ≤ Lp (X). Since Op (X) has odd order and O2 (Y ) = 1, Y ∼ = Y . We may then assume that Op (X) = 1, so that Y ≤ E(X). Then the desired conclusions follow from L2 -balance and the  fact that O2 (Y ) = 1. Remark 2.7. A typical application of Lemma 2.5 and Corollary 2.6 will be a kind of “cross-characteristic L-balance” to be used throughout the proof of Theorem C∗4 : Stage A2. Its purpose is to transfer structural information

2. GROUPS X WITH e(X) ≤ 3

5

from CG (z), where z ∈ I2 (G), to CG (A), where A is an elementary abelian p-subgroup of CG (z) such that (2A)

(1) A# ∩ Ipo (G) = ∅ and Op (CG (A)) has odd order; (2) E(CCG (z) (A)) has a p-component Y ∈ Co2 .

(Note that if mp (CCG (z) (A)) ≥ 3, then A# ⊆ Ipo (G) and (2A1) holds by Theorem C∗4 : Stage A1. Also if K is a component of CG (z), then with a few exceptions K ∈ Chev(2), so any component Y of CK (A) as in (2A2) lies in Chev(2) as well, by [IA , 4.9.6]). Let X = CG (A) and X = X/Op (X), so that Y   CX (z). If (2B)

m2 (Y z) is sufficiently large,

then [Op (X), Y ] = 1, since otherwise (by the Thompson dihedral lemma) e(X) ≥ m2,p (X) > 3, contrary to assumption. Corollary 2.6 then applies and yields that there is a p-component J of X such that either Y is a z component of CJ (z), or J = J z and Y is a diagonal subgroup of J J . Consequently, (2C)

either Y ↑2 J/Op p (J) or Y ∼ = J/Op p (J).

(The latter is possible whether J z = J or not.) Since A ∩ Ipo (G) = ∅, J/Op (J) ∈ Cp by the hypotheses of Theorem C∗4 : Case A and by [IA , 7.1.10]. However, Y ∈ Co2 . Hence there are only a few possibilities for p, Y , and J/Op (J) in (2C). Those satisfying Y ↑2 J/Op p (J) are listed in [V17 , 14.1], and those satisfying Y ∼ = J/Op p (J) are listed in [V17 , 14.2]. Whenever we refer to [V17 , 14.1] or [V17 , 14.2], it is to be understood that we are limiting the possibilities for Y and J/Op (J) because (2C) holds. A final remark: when J z = J, the possibilities for J/Op (J) are even more highly restricted. Indeed as e(G) ≤ 3, in that case we have z

(2D)

 2,p (J) + mp (A) ≤ m  p (J) + m  2,p (J ) + mp (Op (X)) m  p (J) + m z

≤ m2,p (JJ Op (X)) ≤ 3.

Cf. [V17 , 14.3]. Lemma 2.8. Let H be a K-group with e(H) ≤ 3. Let N  H with |N | odd and set H = H/N . If K = [K, K] ≤ H, then either [F (H), K] = 1 or K ≤ E(H). If in addition K is a component of CH (z) for some involution z ∈ H, and [F (H), K] = 1, then there exists a component L of E(H) such that K is a component of CLLz (z). Proof. Without loss, N = 1. We may assume that K = [K, K] ≤ H and [F (H), K] = 1, so that CK (E(H)) ≤ Z(K). Let L be a component of H such that [L, K] = 1. If K does not normalize L (or any component L0

6

9. GENERAL GROUP-THEORETIC LEMMAS

of H), then |LH 0 | ≥ 5 as K = [K, K] and Σ4 is solvable. But then using [IG , 16.9], we get m2,p (H) ≥ |K H | − 1 ≥ 4 > e(H), a contradiction. Thus K normalizes each component of H, inducing inner automorphisms on E(H) by the Schreier property. Therefore by [IG , 3.6], K = [K, K] ≤ E(H). As K is a component of CH (z), it is a component of LLz by L2 -balance, and the lemma follows.  Lemma 2.9. Let K be a 3-component (of itself ) such that O3 (K) ≤ O2 (K). Assume that e(K) ≤ 3, i.e., m2,r (K) ≤ 3 for all odd primes r, that K contains a Frobenius subgroup F of order 9.8, and that m2 (K) ≥ 4. Then K is quasisimple. Proof. Suppose that K is not quasisimple. Then there exists a prime r > 3 such that CK (Or (K)) ≤ O2 (K), for otherwise F ∗ (K) = F (K) ≤ Z(K) and K is abelian, contradiction. Let R = Or (K) and E ∈ E2∗ (K), so that CE (R) = 1. By the Thompson dihedral lemma, K contains the direct product H of m2 (K) copies of D2r , so 3 ≥ m2,r (K) ≥ m2 (K) − 1. Hence m2 (K) = 4. Moreover if CZ(R) (E) = 1, then m2,r (K) ≥ mr (HCZ(R) (E)) − 1 ≥ m2 (K) = 4, a contradiction. Thus CZ(R) (E) = 1. In particular, Z(R) ≤  Z(K). But then CK (Z(R)) ≤ O2 (K), and so CF (R) ≤ O3 (O2 (F )) = 1. Then by [IG , 9.12], a Sylow 2-subgroup Q of F has a free submodule on Ω1 (Z(R)), whence m2,r (K) ≥ mr (CR (Z(Q))) ≥ 4, a final contradiction.  Lemma 2.10. Let K be a 3-component of X such that O3 (K) ≤ O2 (K). Assume the following: (a) e(X) ≤ 3, i.e., m2,r (X) ≤ 3 for all odd primes r, (b) X contains a Frobenius subgroup H of order 9.8 or 8.7 such that F (H) ≤ K but F (H) ∩ O3 3 (K) = 1; and (c) There is z ∈ I2 (NX (K)) such that z inverts O3 (K) elementwise. Then K is quasisimple. Proof. Because of (c), O3 (K) is abelian. Suppose that K is not quasisimple. Then for some prime r > 3, H z acts faithfully on V := Ω1 (Or (K)), because of (b). Note that [K, z] ≤ CK (O3 (K)) ≤ O3 (K). Let t ∈ CK (z) − O3 (K) be an involution. Then mr (CV (t)) ≤ e(G) = 3 and similarly mr (CV (tz)) = 3. As CV (z) = 1, mr (V ) ≤ 6. But by [IG , 9.12], the faithful action of H on V forces mr (V ) ≥ 7, a contradiction completing the proof.  3. Signalizer Functors Throughout this section, G is our K-proper simple group (or any Kproper simple group). Let k be a positive integer (only k = 1, 2, and 3 are relevant in this volume) and p a prime, and assume that mp (G) ≥ k + 2.

3. SIGNALIZER FUNCTORS

7

Definition 3.1. For any D ∈ Epk (G) and A ∈ Epk+1 (G), ΔD = ΔG (D), Θk (G; A), and Θk+ 1 (G; A) are defined by 2

ΔD = ΔG (D) :=



Op (CG (d)),

d∈D #

(3A)

  Θk (G; A) := ΔD | D ∈ Epk (A) , and   Θk+ 1 (G; A) := [ΔD , A] | D ∈ Epk (A) ΔA . 2

We make the same definition of Θk (G; A) even if mp (G) = k + 1. It is important to note that for all g ∈ G, (ΔD )g = ΔDg , (3B)

(Θk (G; A))g = Θk (G; Ag ), and (Θk+ 1 (G; A))g = Θk+ 1 (G; Ag ). 2

2

The following theorem is the key to applying the signalizer functor method once an appropriate level of balance has been established. Theorem 3.2. Let B ∈ Epk+2 (G). Then the following conditions hold: (a) If G is k-balanced with respect to B, then (1) Θk (G; B) is a p -group; (2) Θk (G; A) = Θk (G; B) for every A ∈ Epk+1 (B); and (3) ΓB,k+1 (G) ≤ NG (Θk (G; B)); and (b) If G is k + 12 -balanced with respect to B, then (1) Θk+1 (G; B) is a p -group; (2) Θk+ 1 (G; A) = Θk+1 (G; B) for every A ∈ Epk+1 (B); and 2 (3) ΓB,k+1 (G) ≤ NG (Θk+1 (G; B)). Remark 3.3. In (a), the k-balanced situation, the group Θk (G; B) and its normalizer play the significant role. In (b), it is Θk+1 (G; B) playing the analogous role, but to remind us of the level of balance that underlies the conclusions of the theorem, it is customary to write Θk+ 1 (G; B) instead 2 of Θk+1 (G; B). Thus, k-balance leads to the p -group Θk (G; B), for both integer and half-integer values of k. Moreover, with this extended notation, the conjugation properties (3B) still hold. Proof. This results from a combination of [IG , Theorem 21.6] (the  Signalizer Functor theorem) and [IG , Propositions 21.8, 21.10]. Properties (a2) and (b2) lead immediately to a “connection” corollary. Corollary 3.4. Let k ≥ 1 be an integer or half-integer, and suppose that r is an integer with r > k. Let B1 , B2 , . . . , Bn be elements of Epr+1 (G) such that mp (Bi ∩ Bi+1 ) ≥ r for all i = 1, . . . , n − 1. Let A ∈ Epr (B1 ) and

8

9. GENERAL GROUP-THEORETIC LEMMAS

B ∈ Epr (Bn ). Suppose that G is k-balanced with respect to each Bi , i = 1, . . . , n, and let W = Θk (G; A). Then (a) W = Θk (G; Bi ) = Θk (G; B) for all i = 1, . . . , n; and (b) ΓB,r (G) ≤ NG (W ). Remark 3.5. In the situation of Corollary 3.4, we say that A is rconnected to B by the chain B1 , . . . , Bn , provided mp (Bi ∩ Bi+1 ) ≥ r for all i = 1, . . . , n − 1. We may also say that the chain A, B1 ∩ B2 , B2 ∩ B3 , . . . , Bn−1 ∩ Bn , B r-connects A to B, the key property being that these are elements of Epr (G) any two successive of which commute elementwise. These two notions of rconnectedness are equivalent. Without the parameter r, “connected” means “2-connected.” Proof. We apply part (a2) or (b2) of Theorem 3.2, whichever is appropriate, several times, with the role of (A, B) taken successively by (A, B1 ), (B1 ∩ B2 , B1 ), (B1 ∩ B2 , B2 ), (B2 ∩ B3 , B2 ), . . . , (B, Bn ). This proves (a), and (b) follows then from part (a3) or (b3) of the theorem.  4. Balance We review some terminology and theory from [IG , Section 20]. Recall that if k is a positive integer and B ∈ Epk+1 (G), G is said to be k-balanced with respect to B if and only if for every A ∈ Epk (B) and every b ∈ B # , (4A)

ΔA ∩ CG (b) ≤ Op (CG (b)).

Here, as before, ΔA = ΔG (A) :=



Op (CG (a)).

a∈A#

Moreover, G is weakly k-balanced with respect to B if and only if for every A ∈ Epk (B) and every b ∈ B # , (4B)

[ΔA ∩ CG (b), B] ≤ Op (CG (b)).

As is customary, we use the terms “balanced” and “weakly balanced” in place of “1-balanced” and “weakly 1-balanced.” We say that G is k + 12 -balanced with respect to B ∈ Epk+2 (G) if and only if G is both (k + 1)-balanced and weakly k-balanced with respect to B. These global notions have local analogues, applicable to elements of Lop (G). When the local analogues hold for certain elements of Lop (G), the global notions follow. Let K ∈ Kp and B ∈ Epk+1 (Aut(K)). Then K is locally k-balanced with respect to B if and only if for every X such that B Inn(K) ≤ X ≤ Aut(K), and every E ∈ Epk (B),we have ΔX (E) := ∩x∈E Op (CX (x)) = 1. Also, K is weakly locally balanced with respect to B ∈ Epk+1 (Aut(K)) if and only if

4. BALANCE

9

for every such X and E, [ΔX (E), B] = 1. Also, K is locally k + 12 -balanced with respect to B ∈ Epk+2 (Aut(K)) if and only if it is locally k + 1-balanced and weakly locally k-balanced with respect to B. Finally, K is locally kbalanced, weakly locally k-balanced, or locally k + 12 -balanced for p if it is locally k-balanced, weakly locally k-balanced, or locally k + 12 -balanced, respectively, with respect to all B ∈ Ep (Aut(K)) of the relevant p-ranks. Lemma 4.1. Let B ∈ Epk+1 (G). If G is not k-balanced with respect to B, then there exists b ∈ B # and a p-component J of CG (b) such that in CG (b) = CG (b)/Op (CG (b)), J is not locally k-balanced with respect to AutB (J). More precisely, there is A ∈ Ek (B) and a subgroup D ≤ ΔA ∩ CG (b) such that  = AD normalizes J and [J, D] = 1, so that if we put H = ADJ and H  ≤ Δ  (J).  H/CH (J), then D H

Proof. See [IG , 20.6].



The configuration in Lemma 4.1, consisting of b, J, B, A, and D, is called a k-obstruction. We often just say that (b, J) is a k-obstruction, and do not mention B or A. The analogue of Lemma 4.1 for k + 12 -balance has an extra twist. Failure of k + 12 -balance can be due to failure of local k + 1-balance or failure of weak local k-balance (as one would expect), or failure of local k-balance in a component that is not B-invariant. Lemma 4.2. Let B ∈ Epk+2 (G). If G is not k + 12 -balanced with respect to B, then there exists b ∈ B # and a p-component J of CG (b) such that in CG (b) = CG (b)/Op (CG (b)), J is either not locally k + 1-balanced or not locally k-balanced with respect to AutB (J). More precisely, either there is a k + 1-obstruction (b, J), or there is A ∈ Ek (B) and a subgroup D ≤ [ΔA ∩ CG (b), B] such that AD normalizes J and [J, D] = 1. In the latter case, if B normalizes J, moreover, then J is not even weakly locally kbalanced with respect to AutB (J). Definition 4.3. When b, J, A, B, and D satisfy the hypotheses (and conclusion) of Lemma 4.2, we say that they form a k + 12 -obstruction. For brevity we usually just say that (b, J) is a k + 12 obstruction. Appropriate pumpups of such obstructions inherit the obstruction property. Lemma 4.4. Let b, J, A, B, D be a k-obstruction (resp. k + 12 -obstruction) as in Lemma 4.1 (resp. Lemma 4.2). Let b1 ∈ B # and suppose that [b1 , J] = 1. Let L be a p-component of the subnormal closure of J in CG (b1 ). Then b1 , L, A, B, D1 also form a k-obstruction (resp. k + 12 -obstruction), where we put D1 = CD (b1 ) (resp. D1 = [CD (b1 ), B]). Proof. Let D = ΔA ∩ CG (b) (resp. D = [ΔA ∩ CG (b), B]). Then D = [D, b1 ]CD (b1 ) (resp. D = [D, b1 ][CD (b1 ), B]). As D normalizes J

10

9. GENERAL GROUP-THEORETIC LEMMAS

and [b1 , J] = 1, [D, b1 ] centralizes J. But [J, D] = 1, so [J, CD (b1 )] = 1 (resp. [J, [CD (b1 ), B]] = 1). Defining D1 as in the lemma, we have [J, D1 ] = 1, and  D1 ≤ CD (b1 ) ≤ ΔA ∩CG (b1 ). Now J0 := Lp (CJ (b1 )) covers J/Op (J), and LB = J0B , so D1 does not centralize L/Op (L). The proof is complete.  Lemma 4.5. Let p be an odd prime and X a K-group containing a subgroup A ∼ = Ep2 . Let W = ∩a∈A# O{2,p} (CX (a)). Then [O2 (X), W ] = 1 and W ∩ Op (X) ≤ O{2,p} (X). Proof. Without loss we may pass to X/O{2,p} (X) and assume that O{2,p} (X) = 1. Let Y = Op (X), so that O2 (Y ) = 1. Note that for each a ∈ A# , O{2,p} (CX (a)) ∩ Y ≤ O{2,p} (CY (a)) = O2 (CY (a)). First, for any a ∈ A# , [CO2 (X) (a), W ] ≤ [O2 (CX (a)), O2 (CX (a))] = 1, so [O2 (Y ), W ] ≤ [O2 (X), W ] = 1 as A is noncyclic. Next, let L be a component of Y . If a ∈ A# and [L, a] = 1, then [L, W ] ≤ [E(CX (a)), O2 (CX (a))] = 1, as desired, so we assume that CA (L) = 1. But L is a p -group and a K-group, so mp (Aut(L)) ≤ 1 by [IA , 7.1.2]. Therefore A ≤ NX (L). Fix a ∈ A − NA (L) and set L∗ = La , a product of p components of which L0 := [CL∗ (a), CL∗ (a)] is a diagonal. In particular O 2 (L0 ) =  1. As E(Y ) = L∗ . W ≤ O2 (CY (a)), [L0 , W ] = 1. In particular W normalizes L0 Since [W, a] = 1 and W is a p -group, W normalizes every component of L∗ , and in particular normalizes L. As [L0 , W ] = 1 and L0 projects onto L/Z(L), [L, W ] = 1. We have proved that [F ∗ (Y ), W ] = 1. Therefore W ≤ Z(F ∗ (Y )) so W is a 2-group. But W has odd order so W = 1. The proof is complete.  5. p-Groups Lemma 5.1. Let H be a group with S = E ∗ S0  H, where S is of symplectic type with E extraspecial and S0 either cyclic of order ≥ 4 or of maximal class and of order > 8. Let C0 be the largest cyclic subgroup of S0 . Then C0 char S, C0  H and H/CH (C0 ) is abelian. Proof. As Aut(C0 ) is abelian, it suffices to prove that C0  H. If C0 = S0 , then C0 = Z(S)  H. So we may assume that S0 is of maximal class and |C0 | = 2m ≥ 8. Then [S, S] = [E, E][S0 , S0 ] = Φ(C0 ) ∼ = Z2m−1 .  Hence, CS ([S, S]) = EC0 , and C0 = Z(CS ([S, S])) char S, so C0  H. Lemma 5.2. Let R ≤ GL(V ) for some finite-dimensional vector space V over F2 . Suppose that [V, R, R] = 0. Then R is an elementary abelian 2-group. Proof. Let 1 = x ∈ R. Then V (x2 − 1) = V (x − 1)2 = [V, x, x] = 0 so  x2 = 1. Hence R is elementary abelian.

5. p-GROUPS

11

Lemma 5.3. Let P be a 2-group, N  P , and u ∈ P . Suppose that N ≤ M  P with |M : N | = 2. Let R0 = {[x, u] | x ∈ P }, and suppose that N ⊆ R0 ∩ M = M . Then R0 ∩ M = N . Proof. This follows easily from the identity [xy, u] = [x, u]y [y, u], with y chosen so that [y, u] ∈ M − N and [x, u] ranging over N .  If X is any group and p any prime, the sectional p-rank rp (X) of X is the largest integer n such that X has subgroups N  M such that M/N is elementary abelian of order pn . The following lemma is elementary. We omit the proof. Lemma 5.4. Let p be a prime and X a group. Then (a) If Z ≤ Y ≤ X, then rp (Z) ≤ rp (Y ); and (b) If Z  Y ≤ X, then rp (Y ) ≤ rp (Z) + rp (Y /Z). Lemma 5.5. Let Q = Qy u be a 2-group with Qy ≤ Q and u ∈ I2 (Q) − Qy . Suppose that |CQy (u)| ≤ 4. Then r2 (Q) ≤ 5. Proof. Let R = [Qy , u] and R0 = {[x, u] | x ∈ Qy } ⊆ R. Then u inverts every element of R0 . Also |R| ≥ |R0 | = |Qy : CQy (u)| ≥ |Qy |/4. If |Qy : R| = 4, then equality holds throughout, and in particular R0 = R. Thus R is abelian and Ω1 (R) = CR (u) has order at most 4, so R has rank 2. As |Q : Qy | = 2 and |Qy : R| = 4, the lemma follows in this case. As R < Qy , we may thus assume that |Qy : R| = 2 = |R|/|R0 |. Choose a chief series 1 < T1 < T2 < · · · < Q of Q passing through R, and let n be the greatest integer such that Tn ⊆ R0 . Since R0 = R, Tn+1 ≤ R. Then Tn is abelian and inverted by y. It follows from Lemma 5.3 that R0 ∩ Tn+1 = R0 ∩ Tn . This implies that CQy /Tn (u) covers CQy /Tn+1 (u), so |CQy /Tn+1 (u)| = 12 |CQy /Tn (u)| ≤ 12 |CQy (u)| ≤ 2. Hence Q/Tn+1 has maximal class, whence r2 (Q) ≤ r2 (Tn ) + r2 (Tn+1 /Tn ) + r2 (Q/Tn+1 ) ≤ 2 + 1 + 2 = 5.  Lemma 5.6. Let P be a 3-group with m3 (P ) ≥ 4. Let U  P with ∼ U = E32 . Then there exists A  P such that [U, A] = 1 and A ∼ = E34 . Proof. By Konvisser’s theorem [IG , 10.17], there is A0  P such that  A0 ∼ = E34 . Either A0 or U CA0 (U ) then satisfies the requirement. We need a few results about p-groups of low rank and low nilpotence class. Lemma 5.7. Let p be an odd prime and X a group such that X = N H, where N = F ∗ (X) ∼ = p1+2 and H ∼ = SL2 (p). Then mp (X) = 3.

12

9. GENERAL GROUP-THEORETIC LEMMAS 

Proof. One such group X is Op (M ), where M is a certain parabolic subgroup of P Sp4 (p), and mp (M ) = 3 [IA , 3.3.3]. Therefore it is enough to show that the image of H in Aut(N ) is uniquely determined up to conjugacy.  But Inn(N ) ∼ = N/Z(N ) ∼ = Ep2 and Op (Aut(N )) is an extension of Inn(N ) by SL2 (p), acting naturally. Thus the image of H in Aut(N ) is the centralizer  of an involution in Op (Aut(N )), which has one class of involutions. The result follows.  Lemma 5.8. Let R be a p-group, p odd, of exponent p and class at most 2. If mp (R) ≤ 2 or 3, then |R| ≤ p3 or p5 , respectively, and in case of equality, R is extraspecial. Proof. Let A = CR (A)  R be abelian; there exists such an A by [G1, 3.12]. Then A ∼ = Epn , n ≤ 3, and R/A embeds in Aut(A). Since R has class at most 2, R/A acts quadratically on A. Thus, according as n = 2 or 3, |R/A| ≤ p or p2 . Hence |R| ≤ p3 or p5 , respectively. Suppose that |R| = p3 or p5 , respectively. If |Z(R)| = p, then R is extraspecial, as desired. If |Z(R)| > p2 , then mp (R) > 3, so |Z(R)| = p2 . Hence |R| = p5 . If there is z ∈ Z(R) − Φ(R), then R = z × R1 for some R1 which must be extraspecial, an impossibility. So Φ(R) = Z(R). Let P = P/Φ(P ) = x1  × x2  × x3  and zi = [xi , x3 ] ∈ Z(R). Then Z(R) = z1 , z2 , for otherwise CR (x3 ) > Z x3  and mp (R) > 3. Hence 1 = [x1 , x2 ] = z1a z2b = [xa1 xb2 , x3 ] for some a and b (mod p). Without loss, a ≡ 0 (mod p). There is c such that ac ≡ 1 (mod p). Then bc a [x1 xbc 2 , x2 ] = [x1 , x2 ] = [x1 x2 , x3 ],  −a in R is isomorphic to Ep4 , a final contraso the preimage of x1 xbc 2 , x2 x3 diction. 



Lemma 5.9. Let P be a p-group of odd exponent p. Let Z be a subgroup of Z(P ) and suppose that P/Z ∼ = Ep5 . Then mp (P ) ≥ 4. Proof. By assumption P has class at most 2. If mp (P ) ≤ 3, then |P | = p5 by Lemma 5.8, so Z = 1 and mp (P ) = 5, contradiction. The lemma follows.  Lemma 5.10. Let P be a p-group of class at most 2 and exponent p, p an odd prime. Suppose that P has sectional rank at least 4, but rank at most 3. Then |Z(P )| = p. Proof. Clearly |P | > p4 . Then Lemma 5.8 implies that P is extraspecial, whence the result.  We conclude this section with several results about connectedness in p-groups, mostly for odd primes p. See Remark 3.5 for relevant definitions. ∼ Ep2 . Then Lemma 5.11. Let P be a p-group, and U  P with U = mp (CP (U )) = mp (P ). Proof. See [IG , 10.20(ii)].



5. p-GROUPS

13

Lemma 5.12. Let P be a p-group, p odd. If Q  P and mp (Q) ≥ 2, then there is U ≤ Q such that Ep2 ∼ = U  P. Proof. See [IG , 10.11].



Lemma 5.13. Let P = 1 be a p-group, p odd. If mp (CP (t)) ≥ 3 for all t ∈ Ip (P ), then P is connected. Proof. Since mp (P ) ≥ 3, Lemma 5.12 provides an Ep2 -subgroup U  P . Let V ≤ P with V ∼ = Ep2 ; we show that V is connected to U , which proves the lemma. It suffices to show that V ≤ A ≤ P for some A ∼ = Ep3 , for then the chain V, CA (U ), U connects V to U . Let z ∈ Ip (Z(P )). If z ∈ V , we take A = z V . If z ∈ V , then V = z, t for some t ∈ Ip (P ). Then t ∈ B ∼ = Ep3 for some B ≤ P by assumption, and we can take A ≤ V B = z B.  Lemma 5.14. Suppose that X is a p-group, p odd. Assume that X = X1 × · · · × Xn × X0 , n ≥ 1, where mp (Xi ) = 2 for each i = 1, . . . , n, and X0 = 1. If mp (X) ≥ 5, then all elements of Ep5 (X) are 4-connected. If mp (X) = 4, then all elements of Ep4 (X) are 3-connected. Proof. For each i = 1, . . . , n, let Ui  Xi with Ui ∼ = Ep2 . If mp (X0 ) > 1 (which must be the case if n = 1), we let U0  X0 with U0 ∼ = Ep2 ; otherwise let U0 = Ω1 (X0 ) = 1. Let U ∗ = U1 · · · Un U0 . Then U ∗ is elementary abelian of rank at least 4. Let D ∈ Ep5 (X) if mp (X) ≥ 5; otherwise let D ∈ Ep4 (X). Let Di be the projection of D on Xi for each i, and D ∗ = D1 · · · Dn D0 , so that D ≤ D ∗ . Successively replacing each Di by CDi (Ui )Ui , we obtain a chain of subgroups (mp (D ∗ )−1)-connecting D ∗ to U ∗ . As mp (D ∗ ) ≥ mp (D), the lemma follows.  Lemma 5.15. Let X = X1 ×X2 ×X3 be a p-group, p odd, with mp (Xi ) ≥ 2 for i = 1, 2. Then any two elements of Ep5 (X) are 4-connected in X. p Proof. Let Ui  Xi with Ui ∼ = Ep2 , i = 1, 2. Then for any A ∈ E5 (X), the chain A, CA (U1 )U1 , CA (U1 U2 )U1 U2 , U1 U2 4-connects A to U1 U2 , yielding the lemma. 

Lemma 5.16. Let P be a p-group, p odd, with mp (P ) ≥ 8. Then all elements of Ep8 (P ) are 4-connected in P . Proof. Let A be a maximal normal elementary abelian subgroup of P . By [IA , 10.16], 8 ≤ mp (P ) ≤ mp (A) + mp (Aut(A)), which forces mp (A) ≥ 4. Hence P has a normal subgroup B ∼ = Ep4 . Then mp (Aut(B)) = mp (GL4 (p)) = 4 [IA , 3.3.3]. Hence if A1 , A2 ∈ Ep8 (P ), then the chains  Ai , CAi (B)B, B 4-connect the Ai to B. The result follows. Lemma 5.17. Suppose that p is an odd prime, K   X, K is simple of p-rank ≥ 2, and mp (CX (K)) ≥ 2. Let S ∈ Sylp (X). Then all elements of Ep5 (S) are 4-connected.

14

9. GENERAL GROUP-THEORETIC LEMMAS

Proof. Replacing S by a conjugate if need be, we may assume that CS (K) ∈ Sylp (CX (K)). Fix Ep2 -subgroups U, V  NS (K) such that U ≤ K and V ≤ CS (K). We show that any A ∈ Ep5 (S) is 4-connected to U V . Write A = NA (K) × A1 for some A1 ; we argue by induction on |A1 |. If A1 = 1, so that A ≤ NS (K), then the chain A, CA (U ), U CA (U V ), U V demonstrates the desired connectedness. So assume that A1 = 1. Write A1 = B  1 × a  for some a = 1 and hyperplane B1 of A1 . Set A0 = NA (K)B1 , K ∗ = K A , and J = E(CK ∗ (B1 )). Then J = J1 × Jp where each Ji ∼ = K, and J1 , . . . , Jp are cycled by a. We assume without loss that J1 is the direct factor in this decomposition that projects onto K. Moreover, each Ji is A0 invariant, as A is abelian. For each i = 1, . . . , p, choose zi ∈ Ip (CS∩Ji (A0 )) and set A∗ = A0 z1 . As Z(K) = 1 and z1 ∈ A (because z1a ∈ J1a = J1 ), A and A∗ ∈ Ep5 (S) are 4-connected via A0 . But z1 ∈ J1 normalizes K, so NA∗ (K) = NA0 (K) z1  = NA (K) z1  > NA (K). Hence by induction, A∗ is 4-connected to U V . The proof is complete.  Lemma 5.18. Let X be a group with O3 (X) = 1, and suppose that two components of E(X) are L, of 3-rank 1, and I ∼ = L2 (27). Let P ∈ Syl3 (X). Then any element of E34 (P ) is 3-connected to P ∩ I ∼ = E33 .   3 (P ). Without loss X = L∗ I ∗ A, where L∗ = LA Proof. Let A ∈ E 4  and I ∗ = I A . As L ∼ I, [L∗ , I ∗ ] = 1. Let Q = Ω1 (P ∩ L∗ ) and R = = ∗ Ω1 (P ∩ I ), both elementary abelian groups. Then Z := CR (A) = 1 and by Lemma 5.12, there is U  P with U ≤ Q and U ∼ = E32 . Then the chain  A, CA (U ), ZU, P ∩ I 3-connects A to P ∩ I, as required. 6. p-Components Lemma 6.1. Let K be a p-component of the group X for some prime p. Then the following hold: (a) If N  K and K = Op (K)N , then K = N ; and (b) K is quasisimple if and only if CK (Op (K)) ≤ Op p (K); and (c) If e(X) ≤ 3, Op (K) has odd order, and m2 (K/Op (K)) > 4, then K is quasisimple. 

Proof. Part (a) holds as K = Op (K). In (b), the implication in the forward direction is trivial, and the reverse implication is an application of (a) to N = CK (Op (K)). Finally, (c) is an application of Corollary 2.4 with z = 1.  Recall [IG , 5.15] that for any group X and prime p, (6A)

L∗p (X) = Op (X)Lp (X)Op (CX (P ))for any P ∈ Sylp (Lp (X)).

Lemma 6.2 (L∗p -Balance). Let P be a p-group acting on the K-group X. Then L∗p (CX (P )) ≤ L∗p (X). Proof. See [IG , 5.18].



6. p-COMPONENTS

15

Lemma 6.3. Let (x, K) ∈ ILp (G) for some prime p, and assume that K is not quasisimple. Let D ∈ Ep (CG (x)) be noncyclic, and assume that I := Lp (CK (D)) = 1. Then there is a hyperplane F of D such that [ΔF , I] = 1. Proof. Since K is not quasisimple, CK (Op (K)) ≤ Op p (K) by Lemma 6.1, whence W := [Op (CG (x)), I] = 1. By [IG , 4.3(i)], W = [W, I]. Hence by Lemma 9.2, there is a hyperplane F of D such that X := [CW (F ), I] = 1. Again X = [X, I], so it suffices to show that X ≤ ΔF . Of course f = Cf /Op (Cf ). X ≤ CG (F ). Let f ∈ F # , set Cf = CG (f ), and set C  By Lp -balance, I ≤ L := Lp (CG (f )), and indeed I is a product of pcomponents of Lp (CL (D)). Using Lp -balance and the Bp -property, we see  is semisimple. by an inductive argument (on the rank of D) that I = Lp (I)     Hence X = [X, I] = 1 as X is a p -group. Therefore, X ≤ Op (Cf ). As f ∈ F # was arbitrary, X ≤ ΔF . The proof is complete.  Lemma 6.4. Let (a0 , J0 ) ∈ ILop (G) be Ipo -rigid, p odd. Assume that mp (J0 ) ≥ 2. Then there exists an Ipo -terminal and Ipo -rigid long pumpup (x∗ , K ∗ ) of (a, J) such that K ∗ /Op (K ∗ ) ∼ = J0 /Op (J0 ), mp (C(x∗ , K ∗ )) ≥ mp (C(a0 , J0 )), and in case of equality, |C(x∗ , K ∗ )|p ≥ |C(a0 , J0 )|p . Proof. Among all long pumpups of (a0 , J0 ) (including (a0 , J0 ) itself), choose (x∗ , K ∗ ) to maximize mp (C(x∗ , K ∗ )), and, subject to this, to maximize |C(x∗ , K ∗ )|p . (Thus (x∗ , K ∗ ) is Gilman-maximal; see [IG , 6.16].) Then the proof of the slightly different result [IG , 1.10, p. 13] may be inserted here with (x∗ , K ∗ ) in the role of (x, K) there, and with only the following change: on lines 11 and 12, the justification should be changed to the rigidity hypothesis (which was not assumed there). This produces a complete proof of our lemma.  Lemma 6.5. Let (a, J) ∈ ILo3 (G) with m := m3 (C(a, J)) ≥ 2, J/O3 (J) simple, and m3 (J/O3 (J)) = m3 (Aut(J/O3 (J))) = 2. Suppose that whenever (x, I), (y, H) ∈ ILo3 (G), (x, I) < (y, H), and I/O3 (I) ∼ = J/O3 (J), ∗ ∗ o ∼ then H/O3 (H) = J/O3 (J). Then there is (x , K ) ∈ IL3 (G) which is onestep rigid and Gilman-maximal and such that K ∗ /O3 (K ∗ ) ∼ = J/O3 (J) and m3 (C(x∗ , K ∗ )) ≥ m. Proof. There exists a pumpup chain from (a, J) to a Gilman-maximal pair (an , Jn ): (a, J) = (a0 , J0 ) < (a1 , J1 ) < · · · < (an , Jn ). Inductively we show that m3 (NCG (ai ) (Ji )) ≥ 4 and Ji /O3 (Ji ) ∼ = J/O3 (J), i = 0, . . . , n. This holds by assumption if i = 0. Suppose that it holds for i. If (ai , Ji ) < (ai+1 , Ji+1 ) is a diagonal pumpup, it obviously holds for i + 1. In the case of a trivial or vertical pumpup, we have m3 (C(ai , Ji )) ≥ 4 − m3 (Aut(Ji /O3 (Ji ))) ≥ 2, so m3 (NCG (ai+1 ) (Ji+1 )) ≥ m3 (CG (ai+1 ) ∩ NCG (ai ) (Ji )) ≥ 2 + m3 (Ji ) ≥ 4. Hence ai+1 ∈ I3o (G), so by assumption

16

9. GENERAL GROUP-THEORETIC LEMMAS

Ji+1 /O3 (Ji+1 ) ∼ = J/O3 (J). This completes the induction. Finally, by definition of Gilman-maximality, m3 (C(an , Jn )) ≥ m3 (C(a, J)) ≥ m, so the lemma holds.  Lemma 6.6. Let (x, K) ∈ ILp (G) and Q ∈ Sylp (C(x, K)). For every y ∈ Ip (Q), let Ky be the subnormal closure of Lp (CK (y)) in CG (y). Let V ∈ Ep2 (Q) and w ∈ Ip (Q) with [V, w] = 1. (a) If Kv ≤ E(CG (v)) for all v ∈ V # , then Kw ≤ E(CG (w)); and (b) If Kv is a trivial pumpup of K for all v ∈ V # , then Kw is not a diagonal pumpup of K. Proof. Set K0 = Lp (CK ( w V )), so that for all y ∈ Ip ( w V ), Ky is the subnormal closure of K0 in CG (y). In (a), set Wy = Op (CG (y)) for all such y. Our hypothesis is that [Wv , K0 ] = 1for all v ∈ V # , and it suffices to prove that [Ww , K0 ] = 1, as Kw = K0Kw . Thus it is enough to show that Xv := [CWw (v), K0 ] = 1 for all v ∈ V # . As CWw (v) is K0 -invariant, Xv = [Xv , K0 ] by [IG , 4.3(i)]. Then since K0 ≤ Kv   CG (v), Xv ≤ Kv . If Kv is a diagonal pumpup of K0 , then K0 projects onto each simple direct factor of Kv /Z(Kv ), and so as Xv is K0 -invariant, Wv ≤ Z(Kv ). The same conclusion holds by the Bp -property [IA , 7.1.3] if Kv is quasisimple. In either case, Xv = [Xv , K0 ] = 1. Thus (a) holds. In (b), suppose that Kw is a diagonal pumpup of K. Then by [IG , 6.19], V /CV (Kw /Op (Kw )) is cyclic, so there exists 1 = v ∈ CV (Kw /Op (Kw )). But then the subnormal closure Kv of K0 in CG (v) covers Kw /Op (Kw ). This contradicts the assumption that Kv is a trivial pumpup of K. Thus, (b) holds as well.  Lemma 6.7. Let X be a K-group and p an odd prime. Suppose that O (X) = 1. Let P ∈ Sylp (G) and U  P with U ∼ = Ep2 . Then there exists E ∈ Ep∗ (P ) such that U ≤ E and E normalizes all components of E(X). p

Proof. Choose F ∈ Ep∗ (P ) normalizing as many components of X as possible. Suppose F does not normalize some component K of X. By [IG , 8.7], |K F | = p = 3 and the isomorphism type of K is such that Aut(K/Z(K)) = Inndiag(K/Z(K)) has 3-rank 2. Hence   m3 (F/CF ( K F )) ≤ 1 + m3 (Aut(K/Z(K))) = 3.   Choose x1 , x2 , x3 of order 3, one in each component L of K F but outside Z(L). Then   F0 := CF ( K F ) x1 , x2 , x3  satisfies m3 (F0 ) ≥ m3 (F ), and F0 fixes more components of X than F does. This contradicts our choice of F . Hence F fixes all components of X. So does U , because U = z, u with z ∈ Z(P ) and |[P, u]| ≤ p (and p is odd).  Hence CF (U )U satisfies the requirements of the lemma.

7. STANDARD COMPONENTS OF INVOLUTION CENTRALIZERS

17

7. Standard Components of Involution Centralizers Let (z, K) ∈ ILo2 (G), that is z is an involution of G and K is a 2component of CG (z). Recall that K is said to be terminal (in G) if and only if K is quasisimple and for every involution y ∈ CG (K), K is a component of CG (y). Furthermore, K is standard (in G) if and only if it is terminal in G and for all g ∈ G, [K, K g ] = 1. In our K-proper group G of even type, the notions of terminal and standard coincide. Lemma 7.1. Let (z, K) ∈ ILo2 (G), with G a K-proper simple group of even type. Then K is standard in G if and only if it is terminal in G. Proof. One direction is trivial. Since G is of even type, K ∈ C2 . Hence by definition of C2 [V11 , 3.1], K ∼ = SL2 (q), q odd, or 2A7 . Thus, m2 (K) > 1,  and the required equivalence is a consequence of [II3 , Corollary PU4 ]. The following is the usual setup when a standard component has been found. (1) (z, K) ∈ ILo2 (G) with K ∈ K; (2) K is standard in G; (7A) (3) M = NG (K); (4) Q ∈ Syl2 (CG (K)). Lemma 7.2. Assume (7A). Then ΓQ,1 (K) ≤ M . Proof. See [IG , 18.7].



Lemma 7.3. Assume (7A). Let g ∈ G − M and V = Qg ∩ M . If V = 1, then ΓV,1 (K) < K. In particular V acts faithfully on K. Proof. See [IG , 18.8].



It can be easier to work in Aut(K) than in M , and the following variant of Lemma 7.3 permits this. Lemma 7.4. Assume (7A). Let g ∈ G − M , V = Qg ∩ M , and W = AutV (K). If V = 1, then ΓW,1 (K) < K. Proof. Let w ∈ W ≤ Aut(K) be induced by v ∈ V . Then CK (v) =  CK (w). It follows that ΓW,1 (K) = ΓV,1 (K) < K, by Lemma 7.3. Note that if 1 = V0 ≤ V and W0 = AutV0 (K), then NK (W0 ) does not necessarily normalize V0 , so the prime superscript in Lemma 7.4 cannot in general be removed. Next, a quasisimple group K is said to be outer well-generated (or, for brevity, outer generated) for the prime p if and only if whenever the noncyclic elementary abelian p-group A acts on K, either ΓA,1 (K) = K or AutA (K) ≤ Inn(K).

18

9. GENERAL GROUP-THEORETIC LEMMAS

Lemma 7.5. Suppose that K is quasisimple and p is a prime dividing |K|. Then K is outer well-generated for p provided each Ep2 -subgroup B of Aut(K) satisfies either ΓB,1 (K) = K or B ≤ Inn(K). This follows from the definition of outer well-generation just as Lemma 7.4 followed from Lemma 7.3. Lemma 7.6. Assume (7A). Suppose also that (a) K is outer well-generated for the prime 2; (b) The full preimage in K of any elementary abelian 2-subgroup of K/O2 (Z(K)) is also elementary abelian; and (c) For some x ∈ G − M , m2 (Qx ∩ M ) ≥ 2. Then there is g ∈ G − M such that Qg ≤ M . Proof. See [IG , 18.11]. Note that if K is simple, or Z(K) = O2 (K), then (b) is vacuously true. Also, the conclusion implies that ΓQg ,1 (K) < K  and ΓAutQg (K),1 (K) < K, by Lemmas 7.3 and 7.4. Lemma 7.7. Assume (7A). Let g ∈ G−M , set V = Qg ∩M , and suppose −1 −1 that u ∈ V # . If [CK (u), V ] has even order then Qg ≤ C(z, K)g ≤ M (and g −1 ∈ G − M ). Proof. Let z be an involution in [CK (u), V ]. Since K is standard in G, ΓQ,1 (G) ≤ M . Therefore CK (u) ≤ K ∩ M g . Also V ≤ Qg ≤ CG (K g )  M g , so z ∈ CG (K g ). As K  M , z ∈ CK (K g ). But [Q, K] = 1 so Q ≤ C(z, K) ≤ CM (z) ≤ M g since z ∈ CK (K g ). The lemma follows.  8. Coprime Action Lemma 8.1. If T is a 2-group and mp (Aut(T )) ≥ 2, where p ≥ 5 and p is prime, then m2 (T ) ≥ 3. Proof. Suppose false and let T be a minimal counterexample. By the critical subgroup lemma [IG , 11.11], we may assume that Φ(T ) ≤ Z(T ) = Ω1 (Z(T )), and of course that m2 (Z(T )) ≤ 2. We claim that m2 (T /Φ(T )) ≥ 8. Since p ≥ 5, this is clear unless possibly p = 7 and m2 (T /Φ(T )) = 6. But in the latter case, there is x ∈ I7 (Aut(T )) such that for Tx = [T, x], Tx Z(T )/Z(T ) ∼ = E23 . Then for any hyperplane Z0 of Z(T ), Tx Z(T )/Z0 cannot be extraspecial, so it is abelian by the action of x. As Z0 is arbitrary, Tx is abelian of rank at least 3, contradiction. This proves our claim. Now note that for any t ∈ T , |T : CT (t)| = |[T, t]| ≤ |Z(T )| = 4. We choose ti ∈ T , 1 ≤ i ≤ 3, as follows. First, t1 ∈ T −Z(T ) is arbitrary; then t2 ∈ CT (t1 )− Z(T ) t1 ; then |CT ( t1 , t2 )/Z(T )| ≥ 24 and we choose t3 ∈ CT ( t1 , t2 ) − Z(T ) t1 , t2 . Thus t1 , t2 , t3  is abelian of rank 3, a final contradiction.  Lemma 8.2. Let p be an odd prime. Let Y ∼ = Z pp −1 act nontrivially on p−1

the group R of order pp+1 . Then R = CR (Y ) × [R, Y ] with |CR (Y )| = p and [R, Y ] ∼ = Epp .

9. GENERATION

19

Proof. For any prime r dividing |Y |, ordr (p) = p, so every noncentral chief factor of Y R in R has order pp . As [R, Y ] = 1, there is one such chief factor, and one central chief factor within R. If R is nonabelian, it follows that |[R, R]| = p and hence R is extraspecial, by the action of Y . But then mp (R/[R, R]) = p is odd, which is impossible. Therefore R is abelian and the lemma follows.  Lemma 8.3. Let A = a ∼ = Z3 act on the 2-group T . Suppose that ∼ CT (a)  T and T /CT (a) = E22 . Then [T, a] ∼ = E22 or Q8 . Proof. We may assume that CT (a) = 1. Let C be a critical subgroup of T . Then [C, a] = 1 so [T, a] = [C, a] has class at most 2. We may assume that T = [T, a], which implies that T /[T, T ] ∼ = E22 . Hence T is extraspecial and the result follows.  Lemma 8.4. Let the p-group B act on the p -group Y . Let q be a prime distinct from p and let X be a B-invariant q-subgroup of Y . Then X ≤ S for some B-invariant Sylow q-subgroup of Y . 

Proof. See [IG , 11.21].

Lemma 8.5. Let B ∼ = E3n act on the 2-group T with CT (B) = 1 and CB (T ) = 1. Then BT contains the direct product of n copies of A4 . Proof. Suppose false and let BT be a minimal counterexample. By the critical subgroup lemma [IG , 11.11], Φ(T ) ≤ Z(T ) and Z(T ) is elementary abelian. By Maschke’s theorem applied to the action of B on T /Φ(T ), and the fact that CT (B) = 1, there exists T0 ≤ T such that Φ(T ) ≤ T0 , T0 is B-invariant, and |T : T0 | = 4. If n = 1, then CZ(T0 ) (B) = 1 so BT ∼ = A4 , by minimality. So we may assume that n > 1. By minimality, CB (T0 ) = 1 (as obviously CT0 (B) = 1). Let b ∈ CB (T0 )# and set T1 = [T, b]. Then as T0  T B, [T0 , T, b] = 1. But [b, T0 ] = 1 so [T1 , T0 ] = 1 by the Three Subgroups lemma. By the case n = 1, T1 b contains some T2 B ∼ = A4 . As [T0 , b] = 1, T2 covers T /T0 , so T = T2 × T0 and T2 = T1 is B-invariant. Now  the result follows by induction in T0 CB (T1 ). 9. Generation Lemma 9.1. Suppose that the p-group B acts on the p -group X. Then X = [X, B]CX (B). Proof. Note that [X, B]  X. Therefore using Lemma 8.4, we can reduce to the case that X is a q-group for some prime q = p, and then the  proof is finished by [IG , 11.4]. Lemma 9.2. If the noncyclic elementary abelian p-group B = 1 acts on the p -group W , then W = CW (A) | A ≤ B, |B : A| = p. Moreover if W is a K-group, then for any D ≤ B, [W, D] = [CW (A), D] | A ≤ B, |B : A| = p. Proof. See [IG , 11.23] and [IA , 7.1.2].



20

9. GENERAL GROUP-THEORETIC LEMMAS

Recall that if k is a positive integer and A an elementary abelian psubgroup of X, then for any Y ≤ X or any Y acted upon by A,   ΓA,k (Y ) := NY (B) | B ∈ Epk (A)   ΓA,k (Y ) := CY (B) | B ∈ Epk (A) . We adopt the convention that (9A)

If k ≤ 0, then ΓA,k (Y ) = Y .

Lemma 9.3. Let K be a p-component  B of X, where Op (X) = 1. Let B ∈ ∗ with mp (B) = n. Set K = K , A = NB (K), and mp (A) = m. Let s be a positive integer. Then the following hold: (a) If K = ΓA,m−s (K) or s ≥ m, then K ∗ = ΓB,n−s (K ∗ ); and (b) If K = ΓA,s (K), then K ∗ = ΓB,s (K ∗ ). Ep (X)



Proof. See [GL1, I, (22-4)]. In the next three lemmas we consider the following configuration: 

(9B)

(1) p is a prime and X = Op (E(X)); (2) A = x, w ∼ = Ep2 acts faithfully on X and permutes its components transitively; (3) K is a component of X and K = K w ; and (4) J = E(CX (w)).

  Lemma 9.4. Assume (9B). Set J ∗ = J A and I = E(CJ ∗ (w)). Then I = [I, x]. Proof. If J ∗ has p2 components, then x cycles the p components of I, and so [I, x] = I. Otherwise I is a single component and there is some a ∈ NA (K). Since [a, w] = 1 and [J ∗ , a] = 1, a must act nontrivially on K. But I projects surjectively on K/Z(K), so [I, a] = I. Hence we may assume that x = a. Now x has a generator of the form x = awi for some 1 < i < p. Then [I, x] = [I, a] = I, as claimed.   w  Lemma 9.5. Assume (9B), with X = K . Then it is not the case # that J is a p-component of CX (a) for each a ∈ A . Proof. Suppose J   CX (a) for all a ∈ A# . As A is noncyclic, some a ∈ A# normalizes K. Since a acts nontrivially on K and K is quasisimple, a acts nontrivially on K/Z(K). Let z = aw. Choose g ∈ K such that g a ≡ g p−1 p−1 ∈ J, but g 1+z+···+z ∈ J. As the second (mod Z(K)). Then g 1+w+···+w of these lies in E(CX (z)) (see [IG , 3.27(v)]), we have a contradiction. This completes the proof. 

10. SUBGROUPS OF GL4 (3)

21

  Lemma 9.6. Assume (9B). Then X = Lp (CX (a)) | a ∈ A# . 

Proof. See [IG , 3.28(ii)]. 10. Subgroups of GL4 (3) The setup for this section is the following: (1) D = x×DL with x ∼ = Z3n , n ≥ 1, and DL ∼ = E33 ; also x0  = Ω1 ( x); (2) N ≤ Aut(D) and Nx0 := NN ( x0 ) normalizes DL and x; (10A) (3) Cx0 := CN (x0 ), and AutCx0 (DL ) has a normal subgroup NL ∼ = Σ4 or Σ4 × Z2 of index at most 2; (4) Z := N ∩ Z(Aut(D)); and (5) X := x0 N ⊆ E1 (D).

NL acts irreducibly and monomially on DL . Thus there are three minimal O2 (O2 (NL ))-invariant subgroups xi , i = 1, 2, 3, of DL , and they are permuted by Nx0 since O2 (O2 (NL )) = O2 (O2 (Nx0 )). We label the Nx0 -orbits on E1 (D) according to their lengths, as follows. For this paragraph, let i and j run over arbitrary but distinct  values  in ±1 [3], and the x x {1, 2, 3}. Then{x0 } isorbit [1]; the xi  form orbit form 0 i  

±1 form orbit [6], and the x0 x±1 form orbit [12]; orbit [61 ]; the xi x±1 j i xj   ±1 ±1 and the x1 x2 x3 form orbit [4]. The remaining eight subgroups x0 y, where y ∈ [4], form either a single orbit [8] or the union  orbits  −1[41 ]  of two −1 = x0 y . and [42 ], depending on whether x0 y is Nx0 -conjugate to x0 y Given two Nx0 -orbits O and O on E1 (D), we write O ∼ O if and only if O and O are fused by N . If N happens to be irreducible on Ω1 (D), then D = Ω1 (D) and it will turn out that N preserves a nondegenerate quadratic form on D, up to sign. When D = Ω1 (D), for any sign  = ±1, consider the nondegenerate symmetric bilinear form on D defined by

(10B) f (x0 ) = , O  (D)

f (xi , xi ) = 1, i > 0,

f (xi , xj ) = 0, 0 ≤ i < j ≤ 3,

GO (D))

(resp. be the subgroup of GL(D) preserving f and let (resp. up to sign). Also set Ω (D) = [O (D), O (D)] = O2 (GO (D)). Then Ω− (D) ∼ = A6 and O+ (D) is an extension of Ω+ (D) ∼ = SL2 (3) ∗ SL2 (3) by D8 . We have (1) { y ∈ E1 (D) | f− (y, y) = 0} = [4] ∪ [61 ]; (2) { y ∈ E1 (D) | f+ (y, y) = 0} = [4] ∪ [12]; (10C) (3) { y ∈ E1 (D) | f+ (y, y) = 1} = [1] ∪ [3] ∪ Y, Y = [41 ] ∪ [42 ] or [8]. The Weyl groups of type D4 , B4 , and F4 are embeddable in O+ (D). Namely, in each case the root lattice Λ satisfies (α, β) ∈ 12 Z for all α, β ∈ Λ, and the determinant of the form reduces to 1 (mod 3), so the Weyl group

22

9. GENERAL GROUP-THEORETIC LEMMAS

∼ Λ/3Λ, preserving f+ , the mod 3 reduction of the form ( , ). acts on D = Hence we may regard W (D4 ) ≤ W (B4 ) ≤ W (F4 ) = O+ (D), the equality being true by orders. We define W (D4 )∗ = W (B4 ) ∩ SL(D), so that W (D4 ) and W (D4 )∗ are the two subgroups of W (B4 ) of index 2 and having a Σ3 quotient. We now state our main result of this section. Proposition 10.1. Assume (10A). Then one of the following holds: (a) N normalizes x0 ; (b) D ∼ = E34 , N normalizes DL , and X = E1 (D) − E1 (DL ). N has a normal subgroup R ∼ = E33 which acts regularly on X and trivially on DL , and R ∼ = DL as F3 NL -modules; or (c) D ∼ = E34 , N is irreducible on D, [1] ∼ [4], and one of the following holds: (1) N ≤ GO − (D), [4] ∼ [61 ], N = N o × Z with N o ∼ = Σ5 , and |X | = 5; (2) N ≤ GO − (D), [4] ∼ [61 ], N contains Ω− (D) ∼ = A6 , and |X | = 15 or 30; (3) N ≤ O + (D), [4] ∼ [12], N is conjugate to W (D4 ) or W (D4 )∗ in GL(D), |X | = 4, and X = [1] ∪ [3]; (4) N ≤ O + (D), [4] ∼ [12], N is conjugate to W (C4 ) in GL(D), |X | = 4 or 8, and X = [1] ∪ [3] or [1] ∪ [3] ∪ [4i ] for some i = 1 or 2; or (5) O2,3 (O+ (D)) ≤ N ≤ GO+ (D), [1] ∪ [3] ⊆ X , [4] ∼ [12], [6] ∼ [61 ], and |X | = 12 or 24. Proof. We may assume that (a) fails, whence D ∼ = E34 . Suppose that N is reducible on D. As NL is irreducible on DL and centralizes x0 , N normalizes DL . Let N0 be the stabilizer in GL(D) of the chain D > DL > 1. Then N0 ∼ = DL , and as NL centralizes x0 , N0 ∼ = DL as F3 NL -modules. Let R = N ∩ N0 , which is NL -invariant. If R = 1, it follows that R = N0 , which is transitive on E1 (D) − E1 (DL ). Hence N = RNx0 and (b) follows in this case. So assume for a contradiction that R = 1. Then N embeds in Aut(DL ) × Aut( x0 ). Now AutNx0 (DL ) has a normal absolutely irreducible Σ4 -subgroup, so | AutNx0 (DL )| ≤ 24 .3. Then as |N : Nx0 | = |X | ≤ 33 , | AutN (DL )| ≤ 24 .34 < |[Aut(DL ), Aut(DL )]|. As all proper subgroups of L3 (3) are solvable, O2 (N ) = O2 (Nx0 ) ∼ = A4 , so x0  = CD (O2 (Nx0 )) is N -invariant. This is a contradiction since (a) fails. Now we may assume that N is irreducible on D. Let O be the N -orbit containing [4]. We show that (10D)

O = [4] ∪ [61 ] or O = [4] ∪ [12].

Let u ∈ [4] and Qu ∈ Syl3 (CN (u)). Then Qu contains some Ru ∈ Syl3 (Nx0 ). We have CD (Ru ) = x0 , u, and u = CD (Ru ) ∩ [D, Ru ]; in particular x0 , u and u are NN (Ru )-invariant. Suppose that ug ∈ CD (Ru ) for some g ∈ N ; we claim that u = ug . If Qu = Ru , then Ru ∈ Syl3 (CN (ug ))

10. SUBGROUPS OF GL4 (3)

23

and so we can modify g by an element of CN (ug ) if necessary and assume that g ∈ NN (Ru ). But then g normalizes u = CD (Ru ) ∩ [D, Ru ], as claimed. So we may assume that Qu > Ru , whence ug ∈ x0 . Moreover, NQu (Ru ) shears x0 to u, so ug ∈ x0 , u − u. Thus our claim holds. As a result, [4] ∼ [1], [41 ], [42 ], or [8]. Thus, O ⊆ [3] ∪ [4] ∪ [6] ∪ [61 ] ∪ [12]. Of course |O| divides |N |, which divides | Aut(D)|; and O ⊆ E1 (DL ), by irreducibility of N . These conditions imply that if (10D) fails, then O = [4] ∪ [61 ] ∪ [6] or [4] ∪ [61 ] ∪ [3]. In the latter case, 13 divides |N | and |X |, so N has an orbit on E1 (D) − O − X of length 1 or 14. As 14 does not divide | Aut(D)|, (a) must hold, contrary to assumption. If, on the other hand, O = [4] ∪ [61 ] ∪ [6], then choose g ∈ N g g g such that DL = DL . Then |O ∩ (DL ∪ DL )| ≥ 10 + 10 −4 so O ⊆ DL ∪ D L . In g g ±1 particular, [61 ] ⊆ DL . But [61 ] = x0 xi | i = 1, 2, 3 = D, so DL = D, which is absurd. Hence (10D) is established. Then O is the set of isotropic 1-spaces in D with respect to f for some sign , by (10C). We will prove the following lemma below. Lemma 10.2. Let V be an n-dimensional vector space over F3 , n ≥ 4, and let f and g be two nondegenerate symmetric bilinear forms on V . If for all v ∈ V , f (v, v) = 0 if and only if g(v, v) = 0, then g = ±f . The lemma implies that N ≤ GO  (D) for some sign . Set A = O2 (NL ) ∼ = A4 , so that CD (A) = x0 . Since N is irreducible, O2 (A)  N , and so O2 (N ) > A. Suppose first that  = −, so F ∗ (GO− (D)) ∼ = Z2 ×A6 and O2 (GO− (D)) ∼ = 2 A6 . Then O (N ) ∼ = A5 or A6 . In the first case, NAut(Ω− (D)) (O2 (N )) lies in a Σ6 -subgroup, so N ≤ O− (D). Moreover, as NL ∩ O− (D) contains Σ4 , N/Z ∼ = Σ5 and conclusion (c1) follows. In the second case, by Witt’s lemma, Ω− (D) is transitive on the sets of those v ∈ E1 (D) with a given nonzero value of f− (v, v). Each such set has 15 members, and this implies that conclusion (c2) holds. Now suppose that  = +, and set U = O2 (O+ (D)) ∼ = Q8 ∗ Q8 , so that GO+ (D)/U ∼ = Σ3 Z2 . Thus O2 (A) ≤ U and [g, U ] = U for any g ∈ I3 (A). If U ≤ N , then g ∈ Syl3 (N ) as O3 (N ) = 1, and so O2 (A) = [O2 (N ), O2,3 (N )]  N , contradicting what we saw above. Thus, U ≤ N . In particular, Z(Aut(D)) ≤ Nx0 and |Nx0 | = 2b .3, b = 4 or 5. In the case g ∈ Syl3 (N ), N = U NN ( g) ≤ U S g, where S ∈ Syl2 (NGO+ (D) ( g)). Now in the central product O2 (GO+ (D)) ∼ = SL2 (3) ∗ SL2 (3), g is on the diagonal, so U S g is conjugate to W (B4 ), and in particular lies in O+ (D) and has order 27 .3. Thus f+ (v, v) = 1 for all v ∈ X , so X ⊆ [1] ∪ [3] ∪ Y as in (10C3). As N contains Σ4 , |U S|2 /|N |2 ≤ 2, whence N is conjugate to W (D4 ), W (D4 )∗ , or W (B4 ). But |X | = |N : Nx0 | is a power of 2, so |X | ≥ 4. Thus, in the W (D4 ) and W (D4 )∗ cases, |X | = 26−b = 4. All the statements in conclusions (c3) and (c4) then follow.

24

9. GENERAL GROUP-THEORETIC LEMMAS

Finally, in the case g ∈ Syl3 (N ), O2,3 (O+ (D)) = U T ≤ N , where E32 ∼ = D8 . Then both X = T ∈ Syl3 (O+ (D)). We have GO+ (D)/U T ∼ + and the orbit O = x0 N ∩O (D) have cardinalities of the form 2a .3. Using (10C3) we see that |O | = 12, so |X | = 12 or 24. In the latter case, clearly [6] ∪ [61 ] ⊆ X . In the former case, [6] ∪ [61 ], as the set of all v ∈ E1 (D) − X such that f+ (v, v) = 0, is N -invariant. Since N is irreducible, [6] ∼ [61 ] in this case as well. Hence (c5) holds, and the proposition is proved.  Proof of Lemma 10.2. Let A = {v ∈ V | f (v, v) = 0} and set λ(v) = f (v, v)/g(v, v) for all v ∈ A. We must prove that λ is constant on A. For v, w ∈ A, define v ∼ w ⇐⇒ f (v + δw, v + δw) = 0 for both δ = 1 and δ = −1. Thus v ∼ w ⇐⇒ g(v + δw, v + δw) = 0 for both δ’s. Then v ∼ w implies f (v, v)+f (w, w) = g(v, v)+g(w, w) = 0, so λ(v) = λ(w). It therefore suffices to show that any v, w ∈ A can be connected by a chain v = v0 , v1 , . . . , vm = w such that vi−1 ∼ vi for all i = 1, . . . , m. As n ≥ 4, any v, w ∈ A are orthogonal to a common x ∈ A, so we may assume that v ⊥ w. Again as n ≥ 4, v, w⊥ contains some x such that f (x, x) = −f (v, v). Then v ∼ x ∼ w, completing the proof.  Corollary 10.3. In Proposition 10.1, suppose that N contains a reflection. Then in (c2), N has a Σ6 -subgroup. In (c3), N is Aut(D)-conjugate to W (D4 ). In (c5), there are three possibilities: N is Aut(D)-conjugate to a group containing W (D4 ) with index 3, or to W (F4 ) or GO+ (D). Finally, if N has a unique conjugacy class of reflections, then in (c2), N∼ = Σ6 or GO− (D); (c4) cannot occur; nor can W (F4 ) occur in (c5). Proof. Let y ∈ N be a reflection. In (c2), O− (D) = Z(O− (D)) × ∼ = Z2 × Σ6 . In (c3), W (D4 )∗ lies in SL(D) so contains no reflection. And in (c5), let H = GO+ (D)/O2,3 (GO+ (D)) ∼ = D8 . Then y interchanges the two Q8 factors of O2 (GO+ (D)), so its image y ∈ H is a noncentral involution. The image N of N in H is therefore y, y Z(H), or H, yielding the three stated possibilities for N . The last sentence follows for (c4) and (c5) as B4 and F4 have two root lengths. In (c2), if Z = 1, then F ∗ (N ) has two classes of reflections, which must be fused, hence N ∼ = GO − (D). If − − ∼ ∼  Z = 1, then N = Σ6 as GO (D)/E(GO (D)) = D8 . Corollary 10.4. In Proposition 10.1c, if N ∼ = GO (D), then the N orbits on E1 (D) are O and E1 (D) − O. Here O = [4] ∪ [12] if  = + and O = [4] ∪ [61 ] if  = −. (Ω− (D) y)

Proof. As |D| = 34 , the orbits of GO (D) on E1 (D) are two in number: the singular 1-spaces and the nonsingular 1-spaces. The corollary follows from (10D).  Corollary 10.5. Suppose (x0 , L) ∈ ILo3 (G) with L/O3 (L) simple. Let D = x0  × (D ∩ L) ∈ E3∗ (LC(x0 , L)) with D ∩ L ∈ E3∗ (L), m3 (D ∩ L) = 3, and m3 (C(x0 , L)) = 1. Suppose that (10A) applies with N = AutG (D) and x = x0 , and conclusion (b) of Proposition 10.1 holds. Let R  AutG (D) and

11. STRONG p-EMBEDDING

25

DL be as in the proposition, and let T be the preimage of R in NG (D) := NG (D)/O3 (NG (D)). Then T has a NL -invariant subgroup T0 mapping onto R but with T0 ∩ D = DL ≤ Z(T0 ). Moreover, T0 ∼ = E36 or Z9 × Z9 × Z9 , or T0 is special of complexion 33+3 . In the last case, no maximal subgroup of T0 is abelian. Proof. Note that D ∈ Syl3 (CG (D)) so CG (D) = D × O3 (CG (D)). Set ND = NG (D) and N D = ND /O3 (CG (D)), so that N D /D ∼ = N . The action of NL on T /D is irreducible, so T /D is elementary abelian. Set D = N D /DL . Now N D /D  ∼  = T ∼ N = R = D L as modules for F3 NL , and D .  ∼ D = Z3 . A Sylow 3-subgroup of NL acts freely on DL , hence on N    Therefore we may write ND = T0 × D as NL -module, for some T0 ≤ T . Then T0 /DL ∼ = D L as irreducible NL -modules. The irreducibility implies that T0 is homocyclic abelian or special. Also if T1 is an abelian maximal subgroup of T0 , then T1g = T1 for some g ∈ NL , so T1 ∩ T1g ≤ Z(T1 T1g ) = Z(T0 ) and  hence T0 is abelian. The proof is complete. 11. Strong p-Embedding Lemma 11.1. Let M be a strongly p-embedded subgroup of X for some prime p and group X. If X is simple, then M = Op (M ). Proof. Let P ∈ Sylp (M ), so that P = 1 and NX (Q) ≤ M < G for all 1 = Q ≤ P by assumption. In particular P ∈ Sylp (X). Then by the AlperinGoldschmidt theorem [IG , 16.1], M controls X-fusion in P . As X = Op (X),  the result follows by the Focal Subgroup theorem (see [IG , 15.10(ii)]). The next proof involves the set U(G; M ; p), which can be characterized as follows [II1 , 1.1, (1B)], for any subgroup M of G and any prime p. (11A)

U(G; M ; p) = {x ∈ Ip (M ) | for any g ∈ G, xg ∈ M ⇐⇒ g ∈ M }

Lemma 11.2. Suppose that G is of even type. Assume that (a) For some M ≤ G, M ∩ M g is a 3 -group for all g ∈ G − M ; (b) M contains CG (z) for some 2-central involution z of G; and (c) For every involution t ∈ M , 3 divides |CM (t)|. If m3 (M ) > 2, then M = G. Proof. Let S ∈ Syl2 (CG (z)), so that S ≤ M and S ∈ Syl2 (G). We show that z ∈ U(G; M ; 2). Namely, if z g ∈ M for some g ∈ G, then CG (z g ) ≤ M g by (b). By (c), 3 divides |M ∩ M g |, so g ∈ M by (a). It follows by [II1 , 1.7] and the fact that z ∈ Z(S) that M controls G-fusion in S. By Holt’s theorem (see [II2 , Theorem SF, p. 24]), M = G or G ∼ = A9 or G is a simple Bender group. In the last case, m3 (G) ≤ 2 [IA , 4.10.3], contrary to assumption. Finally let P ∈ Syl3 (M ) ⊆ Syl3 (G). By (a), ΓP,1 (G) ≤ M . But if G ∼ = A9 , then P contains disjoint 3-cycles a and b, and G = CG (a), CG (b) ≤ ΓP,1 (G) ≤ M , so M = G. The proof is complete. 

26

9. GENERAL GROUP-THEORETIC LEMMAS

12. Characteristic-p Representations We shall need a few basic facts about modular representations of finite groups, due to Brauer. We refer the reader to [Is2, Chapter 15] for the definition of Brauer characters. Lemma 12.1. Let X be a group, p a prime divisor of |X|, and IBr(X) the set of Brauer characters of the p-modular absolutely irreducible representations of X. Then the following hold: (a) Brauer characters of inequivalent absolutely irreducible representations are distinct and linearly independent; and (b) |IBr(G)| is the number of conjugacy classes of p -elements in G. Proof. See [Is2, 15.5, 15.10, 15.11].



13. Miscellaneous Lemma 13.1. Let R and S be 3-subgroups of H = GL9 (4) such that S≥R∼ = 31+4 and m3 (S) = 3. Then S = R. Proof. It suffices to assume that |S : R| = 3 and derive a contradiction. First, for any V ≤ R such that V ∼ = E33 , V has 9 distinct irreducible constituents on the natural H-module, since R ∼ = 31+4 . Hence CH (V ) is ∼ diagonalizable and CH (V ) = E39 . As m3 (S) = 3, V = CS (V ). Set Z = Z(R) and S = S/Z, and let Y be the preimage of Z(S)∩R =: Y . Then Y is noncyclic. If m3 (Y ) = 2, then Y ∼ = 31+2 , T := CR (Y )  S, and T > 1. Hence Y ∩ T = 1, which is absurd. Thus, m3 (Y ) = 3 and we may take V ≤ Y . But then V  S and the group S/V = S/CS (V ) of order 33 maps injectively into the group CAut(V ) (V /Z) ∩ CAut(V ) (Z) of order 9, an impossibility.  Lemma 13.2. Suppose that X = AB, where A ≤ X and B < X. If A normalizes a nonidentity subgroup Z of B, then X is not simple.         Proof. 1 = Z X = Z AB = Z B ≤ B < X. But Z X  X.  Lemma 13.3. Let p be an odd prime and Q  H = Y Q ≤ X with Q an s-group, s = p, and Y ∼ = Z(pp −1)/(p−1) . Suppose that either

A 0 ; or (a) X = GL2p (p) and a generator y of Y has the form y = 0 A (b) X = GLp (p). Then Y centralizes Q. Proof. For any y0 ∈ Y # , we see by [IA , 4.8.2] that (13A)

CX (y0 ) = CX (Y ).

Let F = Q1 /Q2 be a Y -chief factor in Q, and suppose that Y centralizes Q2 but not Q1 . Choose Q1 minimal such, so that either Q1 is elementary abelian, or Q1 has class 2 and exponent s or 4, by the critical subgroup

13. MISCELLANEOUS

27

lemma [IG , 11.11]. If s divides |Y |, then s > 2 and Sylow s-subgroups of X are abelian (of rank at most 2). Hence Os (Y ) centralizes Q, whence Y centralizes Q by (13A). So we may assume that s is coprime to |Y |. Then Y acts irreducibly and nontrivially on Q1 /Q2 . By (13A), Y acts fixed-pointfreely on Q1 /Q2 . If 1 < [Q1 , Q1 ] < Q2 , then Q2 /[Q1 , Q1 ] = CQ1 /[Q1 ,Q1 ] (Y ) has a Y -invariant complement in Q1 /[Q1 , Q1 ], contradicting the minimality of Q1 . So either Q1 is elementary abelian or Q2 = [Q1 , Q1 ] ≤ Z(Q1 ), Q2 is elementary abelian, and Q2 ≤ Z(Q1 ). If Q1 is elementary abelian, then Q1 Y is a Frobenius group and as s = p, 2p ≥ |Y | by [IA , 9.12]. But 2p < |Y |, so this is impossible. Thus Q1 is not abelian, whence Q2 = 1. The minimal dimension of a faithful characteristic p representation of Q1 is then at least that of an extraspecial group Q1 /Z for some hyperplane Z of Q2 , which is at least |Q1 /Q2 |1/2 > |Y |1/2 > 1 1 p 2 (p−1) . Therefore 2p ≥ |Q1 /Q2 |1/2 > p 2 (p−1) . It follows that p = 3, whence |Q1 /Q2 | = 42 or 52 . Hence 13 = |Y | divides |O4± (2)| or GL2 (5), which is a contradiction. The lemma follows.  Lemma 13.4 (B. H. Neumann). Let P be a (finite) group and let α ∈ I3 (Aut(P )) with CP (α) = 1. Then P is nilpotent of class at most 2. Proof. We follow Neumann [N1]. First, the mapping x → x−1 xα is bijective on P , since if x−1 xα = y −1 y α , then (yx−1 ) = (yx−1 )α so yx−1 = 1. 2 Then for any x ∈ G, x = y −1 y α for some y, and so xxα xα = 1. Similarly, 2 2 x → xα x−1 is bijective, leading to xα xα x = 1. Hence xxα = (x−1 )α = xα x. Next, (x−1 y)(x−1 y)α = (y −1 x)α = (y −1 )α xα , 2

2

2

x−1 y(x−1 )α y α = yy α x−1 (x−1 )α , y −1 x−1 yx = y α x−1 (x−1 )α (y −1 )α xα x = y α xα (y −1 )α (x−1 )α , 2

2

[y, x] = [(y −1 )α , (x−1 )α ]. 2

Applying the last equation twice more in succession, [y, x] = [(y −1 )α , (x−1 )α ] = [y α , xα ] = [y −1 , x−1 ], 2

2

so yx commutes with xy. But yx and xy are a typical pair of conjugate elements, so every element of P commutes with all its P -conjugates. Equivalently, as [y, x] = (x−1 )y x, [y, x, x] = 1 for all x, y ∈ P . Now since P is clearly a 3 -group, a result of F. W. Levi implies that P is nilpotent of class at most 2 [Hu1, III.6.5]. 

CHAPTER 10

Theorems C6 and C∗6 1. Introduction In this chapter we complete the proofs, begun in [V7 ], of Theorems C6 and C∗6 . We also state and prove an important consequence, which we call Theorem C∗∗ 6 , which should have been stated in [V7 ]. Theorem C6 concerns the case in which our K-proper simple group G is of LTp -type for some odd prime p, and Theorem C∗6 is a mild generalization designed to assist in the generic case (Theorem C7 ) for G. Theorem C∗∗ 6 is precisely the remark that in Theorem C6 , the assumption m2,p (G) ≥ 4 can be replaced by the weaker condition mp (G) ≥ 4. This is important for our future work on the case e(G) = 3. For Theorem C6 , as a group of LTp -type, G and an odd prime p satisfy the following conditions by assumption:

(1A)

(1) G is of restricted even type; (2) p ∈ σ(G); (3) Every element of Lop (G) is either a Cp -group or a Tp -group; and (4) Some element of Lop (G) is a Tp -group.

For Theorem C∗∗ 6 , we weaken only condition (2) and say that G is of weak LTp -type if and only if

(1B)

(1) G is of restricted even type; (2) mp (G) ≥ 4; (3) Every element of Lop (G) is either a Cp -group or a Tp -group; and (4) Some element of Lop (G) is a Tp -group.

Recall that G is of even type if and only if m2 (G) ≥ 3, and for all involutions z ∈ G, O2 (CG (z)) = 1 and every component of CG (z) is a C2 group. In this chapter, we shall occasionally use the even type hypothesis, but we shall not need the minor technical strengthening to “restricted” even type. That strengthening will occasionally be used in subsequent chapters of this book, however. Recall also that σ(G) is the set of odd primes r such that m2,r (G) ≥ 4. 29

10. THEOREMS C6 AND C∗6

30

Thus condition (1A2) means that some 2-local subgroup H of G satisfies mp (H) ≥ 4. Furthermore, for odd p, Ipo (G) = {x|x ∈ Ip (G), mp (CG (x)) ≥ 4} and Lop (G) is the subset of Lp (G) consisting of all the components of CG (x)/Op (CG (x)) as x ranges over the elements of Ipo (G). The set of (quasisimple) Cp -groups is listed in [V3 , Sec. 1] and also [V11 , 3.1]. It consists of the groups in Kp which lie in Chev(p), together with a finite number of additional “characteristic p-like” groups (and with a very small number of subtractions). The critical property of Cp -groups for this chapter, for odd primes p, is that they are balanced for the prime p, with a single exception, namely, L2 (27) for p = 3. We only occasionally use other properties of these groups. The set Tp of (quasisimple) Tp -groups, for p an odd prime, is similarly enumerated in [I2 , Sec. 13] and [V7 , Sec. 1]. It is of central importance in this chapter, as is a stratification of Tp into five subsets Tpi , 1 ≤ i ≤ 5. Rather than repeat the enumeration of Tp , we give the stratification, which is nontrivial only for p = 3 and p = 5. Thus, Tp is the disjoint union of Tpi , i = 1, . . . , 5, where Tpi is defined as follows. Here  = ±1, and Kp , as always, is the set of all (isomorphism types of) quasisimple K-groups K such that Op (K) = 1. Definition 1.1. T31 = {SL3 (q) | q ≡  (mod 3)} ∪ {3A6 , 3A7 , 3M22 , HJ, G2 (8)}; and Tp1 = ∅ if p > 3; T32 = {L3 (q) | q ≡  (mod 9)}; and Tp2 = ∅ if p > 3; T33 = {L3 (q) | q ≡ 4 or 7 (mod 9), q > 4} ∪ {A7 , M12 , M22 }, and Tp3 = ∅ if p > 3; T34 = {L3 (4)} and T54 = {F i22 }, and Tp4 = ∅ if p > 5; Tp5 = {K ∈ Kp − Cp | mp (K) = 1}. We shall prove Theorem 1.2 (Theorem C6 ). If G is of LTp -type, p odd, then G contains a strong p-uniqueness subgroup. The proof of Theorem C6 is based on the construction of a p-component preuniqueness subgroup M that satisfies the conditions of [II3 , Theorem PU2 ] (apart from one exceptional subcase involving Tp4 ). We give more details on the proof below. Some remarks on the set Tp , p odd: The bulk of the set is the subset 5 Tp of all groups of p-rank 1, excepting those groups of p-rank 1 which lie 5

in Cp , namely L2 (p), Ap , and for p = 3 or 5, L2 (8) or 2B2 (2 2 ), respectively. Furthermore, mp (K) ≤ 2 for all Tp -groups (“p-thin” groups) [V17 , 8.1]. Also by [IA , 6.1.4], if K is a nonsimple Tp -group, then p = 3 and K ∈ T31 . The

1. INTRODUCTION

31

groups L3 (4) and F i22 in Tp4 are exceptional in that they have strongly pembedded subgroups which normalize a nontrivial p -subgroup in Aut(K), a very rare occurrence among groups in Kp . One distinction between the groups L3 (q) in T32 and those in T33 is this: if K ∈ T32 , then |K|3 > 32 and there is y ∈ I3 (K) such that CK (y) has a component isomorphic to SL2 (q), whereas if K ∈ T33 , then |K|3 = 32 and CK (y) is solvable for every y ∈ I3 (K) (see [IA , 4.8.2,4.8.4]). Furthermore, by [V17 , 8.3,7.13], if K is a Cp -group or a Tpi -group and J is a component of CK (y) for some element y of order p acting on K, or J is a quotient of K, then J/Op (J) is a Cp -group or a Tpj -group for some i ≤ j; and if K is a Cp -group then so is J/Op (J). We define dT,p = min{i | Lop (G) ∩ Tpi = ∅}.

(1C)

It follows immediately that in Theorem C6 , the set of elements of Lop (G) that d

are Tp T,p -groups is closed under pumping up within the set of all elements of Lop (G). (This condition will be relaxed in Theorem C∗6 .) We refer to the integer dT,p as the Tp -depth of G. (Cf. the Gp -depth, analogously defined in [III3 , Def. 5.1].) In this chapter the prime p will have been fixed, and we always write d for dT,p . To state the first step in the proof of Theorem C6 , we introduce the following additional terminology. If x ∈ Ipo (G) and K is a p-component of CG (x) with (x, K) Ipo -terminal in G and K/Op (K) a Tp -group, we call (x, K) an Ipo -terminal Tp -pair. Moreover, if K/Op (K) is a Tpi -group, and we wish to specify the integer i, we refer to (x, K) as an Ipo -terminal Tpi -pair. We also recall the notion of Ipo -rigidity for a pair (x, K) ∈ ILop (G) [IG , 6.11]. It stipulates that for any long pumpup (x, K) 3, then every element of TGp (G) lies in G5p ∪G6p , and in particular, has p-rank 2 if it lies in Chev. In the exceptional situation, because of the slightly different definition of d0 (Definition 1.7), we need a supplementary result to Theorem 1∗ . We restate this result, Proposition 1.8, and prove it here. Proposition 2.5. In the exceptional situation, d0 is the least d > 3  such that T3d (G) = ∅, and there exists an I3o -terminal T3d0 -pair, which is Ipo -rigid if d0 = 4. Proof. Since p = 3 and d0 = 4 or 5 in the exceptional situation, we see that what needs to be proved is that (in the exceptional situation) if there exists (y, L) ∈ ILo3 (G) such that L/O3 (L) ∼ = L3 (4), then there exists such a pair which is both I3o -terminal and I3o -rigid. By the definitions of d0 and the exceptional situation, every element of Lo3 (G) is a C3 -group, isomorphic to A7 or L3 (4), a group of 3-rank 1, or a TG3 -group, necessarily flat by Proposition 2.1. By [V17 , 13.18], and as |A7 | < |L3 (4)|, none of these is a vertical pumpup of L3 (4), and since the covering group SL3 (4) is also excluded, (y, L) is already I3o -rigid, as all its long pumpups must be. But by [IG , 6.22] (see also the definition [IG , 6.9]), some such long pumpup is Ipo -terminal in G, so the proposition follows.  In many circumstances we have control over mp (C(y, L)) for any pumpup (y, L) of (x, K). Proposition 2.6. Let (y, L) be a pumpup of (x, K), with y ∈ Q. Suppose that (a) or (b) holds: (a) d0 = 1a, 3 or 4; or (b) y ∈ Z(Q). Then mp (C(y, L)) ≤ mp (C(x, K)) (= mp (Q)), and in case of equality, |C(y, L)|p ≤ |C(x, K)|p . Proof. This is trivial if y ∈ Z(Q), for then Q ∈ Sylp (C(y, L)) by definition of Ipo -terminality [IG , 6.9]. If d0 = 1a, then by (1I), x ∈ O3 3 (K),  whence x ∈ O 3 (CK (y)) ≤ L and then x ∈ O3 3 (L) as (x, K) is Ipo -terminal. Hence for some Q∗ ∈ Sylp (C(y, L)), x ∈ Z(Q∗ ). It follows that Q∗ ≤ C(x, K), which implies the desired result. Finally, suppose that d0 = 3 or 4. Then as (x, K) ∈ TJop (G), (x, K) is Ipo rigid. It follows that (y, L) is Ipo -rigid, as are all its long pumpups. By [V9 , 6.4], (y, L) has an Ipo -terminal long pumpup (z, I) such that mp (C(z, I)) ≥ mp (C(y, L) ≥ mp (C(x, K)), and in case of equality, |C(z, I)|p ≥ |C(y, L)|p > |C(x, K)|p . This violates the maximality properties held by (x, K) as a  member of TJop (G), and completes the proof.

3. CENTRALIZERS WITH Tp -COMPONENTS

39

3. Centralizers with Tp -Components In this section we collect a few basic results that will be needed throughout the proof of Theorem C∗6 . These results concern the structure of CG (y) for arbitrary elements y of Ipo (G) such that CG (y) contains a p-component J with J/Op (J) a Tp -group. y = Cy /Op (Cy ), let We fix such a y and J, set Cy = CG (y) and C  We assume, as we may, that V was V ∈ Sylp (Cy ), and set W = CV (J). chosen so that W ∈ Sylp (C(y, J)). Note that mp (V ) ≥ 4. By [IG , 10.22(iii)], we obtain Lemma 3.1. If p = 3 and mp (W ) ≥ 4, then W is connected (whence mp (CW (w)) ≥ 3 for every w ∈ I3 (W )). By “connected” we always mean “2-connected” (see Remark 3.5 or [IG , 10.18]). Definition 3.2. Let (y, J) ∈ ILop (G). Then N (y, J) = NCG (y) (J). We next prove Lemma 3.3. The following conditions hold: (a) mp (Cy ) = mp (V ) = mp (N (y, J)); and (b) Either J is V -invariant or W is connected of rank ≥ 3. Proof. By [V17 , 19.1], mp (N (y, J)) = mp (Cy ), so (a)  holds. ∗ ∗ V Suppose that J is not V -invariant and set J = J , so that J is the product of n p-components J = J1 , J2 , . . . , Jn , transitively permuted by V with n = pr for some r ≥ 1. Set L = J2 J3 · · · Jn , so that L is invariant under W and W ∩ L ∈ Sylp (L) (inasmuch as W ∈ Sylp (C(y, J))). By / Op p (Ji ), 2 ≤ i ≤ n. Since [IG , 16.11], there is wi ∈ Ip (W ∩ Ji ) with wi ∈ n − 1 ≥ p − 1 ≥ 2, mp ( y, wi |2 ≤ i ≤ n) ≥ 3, so mp (W ) ≥ 3.  = 1. The product Consider first the case that J is simple, whence Z(L) of all p-components of C(y, J) then contains some w ∈ Ip (Z(W )) with w ∈ y. Therefore Z(W ) is noncyclic. But now W is trivially connected, so (b) holds in this case. Thus we can assume that J is nonsimple. As J ∈ Tp , p = 3 and |Z(Ji )| = 3, 1 ≤ i ≤ n. But now if m3 (W ) ≥ 4, then W is connected by Lemma 3.1 and again (b) holds, so we can also assume that m3 (W ) ≤ 3. As there exist  to be cyclic and xi ∈ I3 (Ji ) − O3 3 (Ji ) for each i = 2, . . . , n, this forces Z(J)  = J2 ∗ J3 . In particular, W leaves both J2 and J3 invariant. By [V9 , 5.12] L there exists E32 ∼ = Ui ≤ W ∩ Ji with Ui  W , i = 2 and 3. Set U = U2 U3 . Then U is elementary abelian and since U2 ≤ J3 , |U | ≥ 33 . To prove connectivity, we need only show that m3 (CW (t)) ≥ 3 for every t ∈ I3 (W ), by [IG , 10.22(i)]. Clearly this is true for t ∈ U , so we can assume t∈ / U , in which case it will suffice to prove that |CU (t)| ≥ 32 . But t, like W , centralizes U/U ∩ O3 3 (J2 J3 ) ∼ = E32 , so the desired statement holds by [IG , 9.16]. Thus (b) holds in this case as well and the lemma is proved. 

10. THEOREMS C6 AND C∗6

40

We next prove / Ipo (G), then the following Lemma 3.4. If there is w ∈ Ip (W ) with w ∈ conditions hold: (a) mp (CW (w)) = 2; (b) J is V -invariant; (c) W contains an Ep2 -subgroup U such that U # ⊆ Ipo (G) and U w ∼ = p1+2 ; and  (d) If J is nonsimple (whence p = 3), then  y  = Z(J). Proof. As usual, there is v ∈ Ip (V ∩ J) with v ∈ / Op p (J), so v ∈ / W . Since W centralizes v, it follows that mp (CV (w)) > mp (CW (w)). But mp (CG (w)) ≤ 3 as w ∈ / Ipo (G), so mp (CW (w)) ≤ 2. On the other hand,

w = y as y ∈ Ipo (G), so mp (CW (w)) = 2, proving (a). But now (b) follows from (a) and Lemma 3.3b. Since J is V -invariant, W  V . Hence again by [V9 , 5.12], there is Ep2 ∼ = U ≤ W with U  V , and by [V9 , 5.11], mp (CV (U )) = mp (V ) ≥ 4. Thus mp (CG (u)) ≥ 4 for each u ∈ U # and consequently U # ⊆ Ipo (G). In particular, w ∈ / U , so by (a), w does not centralize U . Thus U w ∼ = p1+2 , proving (c).  =  z Finally, suppose J is nonsimple, whence as J ∈ Tp , p = 3 and Z(J) for some z ∈ I3 (Z(V )). Then w ∈ / z, y, otherwise w ∈ Z(V ), whence w ∈ I3o (G), contradiction. But as mp (CW (w)) ≤ 2 by (a), z, y = y = z, proving (d).  Finally, we need the following more specialized result. Lemma 3.5. If mp (W ) ≤ 3, then mp (V ) ≤ 6. Proof. By Lemma 3.3a, there is A ∈ E∗ (V ) leaving J invariant. Set  so that mp (A0 ) ≤ mp (W ) ≤ 3. But A/A0 ≤ Aut(J),  so A0 = CA (J), mp (A/A0 ) ≤ 3 by [V17 , 8.6]. Thus mp (A) ≤ 6.  4. Balance and Generation As remarked in the Introduction, the proof of Theorem C∗6 is based largely on signalizer functor theory, which as usual depends for its execution on balance and generational properties of the elements of Lop (G). Here we collect the various balance and generation properties we shall need for the analysis. We defer until Section 9 the corresponding general results concerning signalizer functor theory, which are more easily described in the context in which they will be needed. The results on balance are a direct consequence of the corresponding local properties of the elements of Lop (G), which by hypothesis are either Cp -,Tp -, or TGp (G)-groups. See [V9 , Sec. 4] or [IG , Sec. 20] for the definitions of local k-balance, weak local k-balance, and local k + 12 -balance, for integer values of k ≥ 1. We have the following result by [V17 , 4.8].

4. BALANCE AND GENERATION

41

Lemma 4.1. Let L ∈ Cp ∪ Tp ∪ TGp (G). Then the following conditions hold: (a) L is locally 3-balanced and locally 5/2-balanced for p; (b) Either L is locally 2-balanced for p, or p = 3 and L/Z(L) ∼ = L3 (q), ∼ q ≡  (mod 3),  = ±1, or p = 3 and L = A11 , or p = 5 and L∼ = F i22 ; and (c) If L is a Cp -group, then either L is locally balanced for p, or p = 3 and L ∼ = L2 (27). The relativized forms of these results give more precise information [V17 , 4.8]. Lemma 4.2. Let L ∈ Cp ∪ Tp ∪ TGp (G). Let H be a subgroup of Aut(L) containing Inn(L), and let A be an Ep2 -subgroup of H. Then the following conditions hold: (a) If L is not locally 2-balanced (in H) with respect to A, then A ≤ Inndiag(L) (and A ≤ Inn(L) if L ∼ = A11 or F i22 with p = 3 or 5, respectively); and (b) If L is locally 2-balanced, but not locally 3/2-balanced (in H) with respect to A, then one of the following holds: (1) Some element of A# induces a nontrivial field, graph, or graphfield automorphism on L ∈ Chev − Chev(p); (2) p = 3, A induces inner-diagonal automorphisms on L/Z(L) ∼ = L3 (q), q ≡  (mod 3),  = ±1, and A is the image of a nonabelian 3-subgroup of GL3 (q); (3) p = 3, A induces inner-diagonal automorphisms on L/Z(L) ∼ = L6 (q), q ≡  (mod 3),  = ±1, and A is the image of a nonabelian 3-subgroup of GL6 (q); or (4) p = 3 and L ∼ = L2 (27), A7 , 3A7 , A10 , M12 , M22 , 3M22 , or HJ. We note that M12 contains two conjugacy classes of E32 -subgroups, with respect to one of which it is locally 3/2-balanced and to the other not locally 3/2-balanced (see [IA , 7.5.5b, 5.3b]). Combining Lemmas 4.1 and 4.2 with the general balance results of [IG , 20.4, 20.6, 20.7], we obtain the following four global balance properties of G. Proposition 4.3. G is 3-balanced with respect to each element of E5 (G). Note that if B ∈ Ep4 (G), then B # ⊆ Ipo (G). Proposition 4.4. If B ∈ Ep4 (G) and G is not 5/2-balanced with respect to B, then there is b ∈ B # and a p-component J of CG (b) such that the following conditions hold (we write J = J/Op (J)): (a) Either p = 3 with J /Z(J) ∼ = L3 (q), q ≡  (mod 3),  = ±1, or ∼ F i22 , or p = 3 and J = ∼ A11 ; p = 5 and J =

42

10. THEOREMS C6 AND C∗6

(b) There is Ep2 ∼ = D ≤ B leaving J invariant and inducing faithful inner-diagonal or inner automorphisms on J; (c) [ΔD ∩ CG (b), B] normalizes, but does not centralize J (whence J is not locally 2-balanced with respect to D); and (d) B does not leave J invariant and mp (CB (J)) ≤ mp (B) − 3. Proposition 4.5. If B ∈ Ep4 (G) and G is not 2-balanced with respect to B, then there is b ∈ B # and a component J of CG (b) such that the following conditions hold (again with J = J/Op (J)): (a) p and J are as in Proposition 4.4a; (b) There is Ep2 ∼ = D ≤ B normalizing J and inducing faithful innerdiagonal or inner automorphisms on J; (c) ΔD ∩ CG (b) normalizes but does not centralize J, so that J is not locally 2-balanced with respect to D; and (d) If d0 = 5, then p = 3 and J ∼ = A11 . Proposition 4.5d depends on the facts that all the groups in Proposition 4.4a except A11 are Tpi -groups for some i < 5; and in the exceptional situation, that A7 is not among them. Proposition 4.6. If B ∈ Ep3 (G) with B # ⊆ Ipo (G) and G is not 3/2balanced with respect to B, then there is b ∈ B # and a p-component J of CG (b) such that the following conditions hold (again with J = J/Op (J)): (a) One of the following holds: (1) p and J are as in Proposition 4.4a but J is not a TGp -group; (2) mp (J) = 1; (3) p = 3 and J ∼ = L2 (27), (3)A7 , M12 , (3)M22 , or HJ, all with m3 (CAut(J) (AutB (J))) = 2; or (4) J ∈ TGp (G); (b) Either the conclusions of Proposition 4.5 hold, or there is d ∈ B # leaving J invariant such that the following hold: (1) [Op (CG (d)) ∩ CG (b), B] normalizes, but does not centralize J (whence J is not weakly locally balanced with respect to d); (2) Either d induces a nontrivial inner or inner-diagonal automorphism on J, or p = 3 and J ∼ = L2 (27); (3) mp (CB (J)) ≤ m(B) − 2; and (4) If B leaves J invariant and either mp (J) = 1 or J is a TGp group, then some element of B # induces a nontrivial field, graph, or graph-field automorphism on J, or p = 3 and J /Z(J) ∼ = A10 or L6 (q), q ≡  (mod 3). If B, b, and J are as in Proposition 4.4, 4.5, or 4.6, we refer to the pair (b, J) as a 5/2-obstruction, 2-obstruction, or 3/2-obstruction for B, respectively. Now we turn to generation. Primarily we require local results. Recall [IA , 7.2.1] that for any group X and any elementary abelian p-group B of

5. THEOREM 3: 3/2-BALANCE

43

rank ≥ k acting on X, ΓB,k (X) = CX (D) | D ∈ Ek (B) . In the next four lemmas, we fix a simple K-group L, set H = Aut(L), and assume that mp (Aut(L)) ≥ 2. We also fix Ep2 ∼ = A ≤ H. First, [V17 , 11.3.3, 5.9] yield: Lemma 4.7. If L is a simple Tp -group, then (a) One of the following holds: (1) L = ΓA,1 (L); (2) p = 3, L ∼ = M12 , and L is locally balanced (in H) with respect to A; or (3) p = 3 and L ∼ = F i22 , with A ≤ L in = L3 (4), or p = 5 and L ∼ either case; and (b) If L is a Tp2 -group (whence p = 3 and L ∼ q ≡  (mod 9),  = L3 (q),   = ±1), then we can choose A in L so that L = L3 (CL (a)) | a ∈ A# . Since M12 , L3 (4), and F i22 each have automorphism groups of p-rank 2 [IA , 5.6.1, 4.10.3a], as an immediate corollary we obtain Lemma 4.8. If L is a simple Tp -group and B ∼ = Ep3 acts on L, then L = ΓB,1 (L). Likewise by [V17 , 5.12], we have: Lemma 4.9. Suppose that L1 ∈ TGp (G), L = L1 /Z(L1 ), and B ∼ = Epm acts on L, m ≥ 2. If ΓB,1 (L) < L, then m = 2 and one of the following holds: (a) p = 3, L ∼ = An , 10 ≤ n ≤ 12, and E acts on L like an E32 -subgroup of L with a regular orbit on n letters; (b) L ∼ = A4p and E acts on L like an Ep2 -subgroup of L with all orbits on 4p letters of length p; or (c) p = 3 and L ∼ = 3ON . = Sp8 (2) or L1 ∼ In particular, if mp (B) ≥ 3, then L = ΓB,1 (L). Finally [V17 , 11.2.2] yields: Lemma 4.10. If p = 3 and L ∼ = L2 (27), then either A ≤ L or L =

ΓA,1 (L).

5. Theorem 3: 3/2-Balance We introduce notation in connection with our Ipo -terminal Tpd0 -pair (1F) (5A)

(x, K) ∈ TJop (G) ⊆ TJp (G).

This notation will be fixed for the remainder of the proof of Theorem C∗6 .

10. THEOREMS C6 AND C∗6

44

We let (5B)

(1) (2) (3) (4)

CG (x) = CG (x)/Op (CG (x)), T ∈ Sylp (CG (x)), T1 = NT (K), P = T ∩ K, Q = CT (K), and R = P Q.

Thus P ∈ Sylp (K), and we assume, as we may, that our choice was made so that (5C) T1 ∈ Sylp (N (x, K)), so Q ∈ Sylp (C(x, K)), mp (Q) ≥ 2, and Q  T1 . We have x ∈ Z(Q) and P centralizes Q. Also R is the subgroup of T consisting of the elements of T inducing inner automorphisms on K. Note that as x ∈ Ipo (G), and by Lemma 3.1, we have (5D)

mp (T1 ) = mp (T ) ≥ 4.

For each t ∈ Ip (Q), we set (5E)

Kt = the pumpup of K in CG (t), so that provided t ∈ Ipo (G), we have Kt /Op (Kt ) ∼ = K/Op (K) as K is Ipo -terminal in G. Recall that if K/Op (K) is not simple (whence p = 3 and d0 = 1a), we have chosen (x, K) so that x ∈ K (1I). As noted above, all this notation is fixed for the rest of this chapter. In this section and the next, for each of the possible values of d0 , we shall define a nonempty set B(T ) of subsets of Ep (T ), all of the same rank, and then verify a useful level of balance for G with respect to any B ∈ B(T ). The elements of B(T ) will all lie in T1 and hence will leave K invariant. In this section, we treat the case d0 = 1a, whence p = 3 and K is nonsimple. We let K = K1 , K2 , . . . , Km be all the 3-components of CG (x) such that K i is nonsimple, 1 ≤ i ≤ m, and we put K ∗ = K1 K2 · · · Km . In particular, K ∗  CG (x). Also as m3 (Ki ) = 2 and Ki contains an element of order 3 not in O3 3 (Ki ), 1 ≤ i ≤ m [IG , 16.11], we have m3 (K ∗ ) ≥ m + 1. Furthermore, as T1 leaves K = K1 invariant, T1 leaves K2 · · · Km invariant (assuming m > 1). We set Pi = T ∩ Ki = T1 ∩ Ki , so that Pi ∈ Syl3 (Ki ) and [Pi , Pj ] = 1, 1 ≤ i < j ≤ m, with P = P1 . Furthermore, Pi ≤ Q for 2 ≤ i ≤ m. For technical reasons, the case m3 (K ∗ ) ≤ 3 is slightly exceptional, so we define B(T ) in that case first. In particular, m ≤ 2 and T leaves K invariant in this case, so T = T1 . Since P = P1  T1 = T , there is E32 ∼ = B1 ≤ P with B1  T by [V9 , 5.12]. Likewise Q  T and if m = 2, then also P2  T . According as m = 1 or 2, again by [V9 , 5.12], there is thus E32 ∼ = U ≤ Q or

5. THEOREM 3: 3/2-BALANCE

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P2 with x ∈ U and U  T . [If m = 2, then x ∈ P2 as m3 (K ∗ ) ≤ 3.] Since m3 (K ∗ ) ≤ 3, it follows in either case that B1 U ∼ = E27 . Thus if m3 (K ∗ ) ≤ 3, we define B(T ) to be the set of all elements B ∈ E3 (T1 ) of the form B = B1 U for such normal subgroups B1 and U of T = T1 . In particular, B ∼ = E27 and U ≤ Q. On the other hand, if m3 (K ∗ ) ≥ 4, we set P ∗ = P2 P3 · · · Pm x (whence ∗ P  T1 ) and let U be a normal E32 -subgroup of T1 with x ∈ U and U ≤ P ∗ . Again U ≤ Q. This time we define B0 (T ) to be the set of all elements B ∈ E3 (T1 ) of the form B = B1 B2 · · · Bm , where Bi is any E32 -subgroup of Pi , 1 ≤ i ≤ m; and we define B(T ) to be the set of all B ∈ B0 (T ) such that U ≤ B for some such normal subgroup U of T1 . [Clearly we can choose the set of Bi , 2 ≤ i ≤ m, so that U ≤ B.] Since m3 (Bi ) = m3 (Pi ) = m3 (Ki ) = 2, 2 ≤ i ≤ m, we have m3 (B) = m3 (K ∗ ) ≥ 4. We easily obtain in all cases (when d0 = 1a): Lemma 5.1. If B ∈ B(T ), then B ≤ A for some A ∈ E4 (T1 ). In particular, B # ⊆ I3o (G). Proof. If m3 (B) ≥ 4, the lemma obviously holds with B = A, so assume m3 (B) = 3, whence m3 (K ∗ ) ≤ 3 and T = T1 . We have B = B1 U in this case with B1 and U normal in T . Set T0 = CT (U ). By [V9 , 5.11], m3 (T0 ) = m3 (T ). Also B1 ≤ T0 , whence B1  T0 . But then again by [V9 , 5.11], m3 (CT0 (B1 )) = m3 (T0 ) and consequently m3 (CT (B)) = m3 (CT (B1 U )) = m3 (T0 ) = m3 (T ) ≥ 4 by (5D). Thus, B ≤ A for some A ∈ E4 (CT (B)), and the lemma holds in this case as well.  Note that it is immediate from the definition of B(T ) that for each B ∈ B(T ), we have (5F)

|B : CB (K)| = 3.

We shall prove Proposition 5.2. Suppose that d0 = 1a. If B ∈ B(T ), then G is 3/2balanced with respect to B. Moreover, if m3 (K ∗ ) ≥ 4 and B ∈ B0 (T ), then G is 3/2-balanced with respect to B. We assume false for some B ∈ B(T )∪B0 (T ) and argue to a contradiction in a sequence of lemmas. Since d0 = 1, we are in the normal situation. We first set up some notation. If m3 (K ∗ ) ≥ 4, then B ∈ B0 (T ), so B = B1 B2 · · · Bm , and we set E = x B2 · · · Bm , so that E ≤ Q and m3 (Q) ≥ 3. Suppose, on the other hand, that m3 (K ∗ ) ≤ 3, whence B = B1 U ∈ B(T ). If m3 (Q) ≥ 3, then m3 (CQ (U )) ≥ 3 by [V9 , 5.11] and we let E ∈ E3 (CQ (U )). Next consider the case m3 (Q) = 2. In particular, T = T1 . Since m3 (T ) ≥ 4, [V17 , 8.7a] implies that K ∼ = SL3 (q), q ≡  (mod 3),  = ±1. If some t ∈ I3 (CT (U )) induces a nontrivial field automorphism on K, then by [V17 , 8.7b], we can choose t to centralize B1 , and for such a choice of t, we set E = U t (∼ = E27 ). Finally, if no such t exists, we set E = U . Note that E

10. THEOREMS C6 AND C∗6

46

centralizes B1 in all cases. Moreover, (5G)

E # ⊆ I3o (G).

Indeed, B1 ≤ E, so m3 (BE) > m3 (E) and (5G) holds unless possibly m3 (E) = 2, in which case E = U  T . As m3 (T ) = m3 (CG (x)) ≥ 4, m3 (CT (U )) ≥ 4, establishing (5G). We need the following property of E. Lemma 5.3. The following conditions hold: (a) For any e ∈ E # such that Ie := E(CK (e)) = 1, Ie has a trivial pumpup Ke in CG (e). In particular, B1 ≤ Ke and x ∈ O3 3 (Ke ); and (b) For any e ∈ E # such that Ie = 1, x centralizes every component L1 of CG (e)/O3 (CG (e)) which is not isomorphic to D4− (2), D5 (2), or E6 (2), and which satisfies at least one of the following for some b ∈ CG (E x) with b3 = 1: (1) Let L1 = L1 /O3 (L1 ). Then L3 (CL1 (b)) has a component J 1 such that J 1 /Z(J 1 ) is not locally balanced with respect to p = 3; (2) L3 (CL1 (b)) has a component J 1 with J 1 /Z(J 1 ) 2, q1 ≡ 1 (mod 3), 1 = ±1. Proof. If e ∈ Q, then as K is 3-terminal, the lemma is clear, so we can assume that e ∈ / Q, whence m3 (Q) = 2, E = U e , K ∼ = SL3 (q), q ≡  (mod 3),  = ±1, and e induces a nontrivial field automorphism on K by definition of E. Suppose that q > 8, whence I e ∼ = SL3 (q 1/3 ) with B1 ≤ Ie and x = Z(I e ). To prove (a), we must show that Ie covers Ke /O3 (Ke ). Note first that as x ∈ Ie ≤ Ke , Ke is not the product of three 3components of CG (e) cycled by x, so by L3 -balance, Ke is a 3-component e = Ce /O3 (Ce ) and I = L3 (CI (U )), so that of CG (e). Set Ce = CG (e), C e I = Ie . We know that for any u ∈ U # , K covers Ku /O3 (Ku ). Hence if we set Ku,e = L3 (CKu (e)), it follows that Ie and hence I covers Ku,e /O3 (Ku,e ) for each u ∈ U # . Thus for each u ∈ U # , I has a trivial pumpup in Cu,e and hence in CCe (u), which implies that I = Ie is a component of CCe (u) and hence of CK e (u) for each u ∈ U # . But now we conclude from [III11 , 1.16]  e , and (a) is proved. that I = Ie = K Suppose then that q = 8, so that x ∈ CK (e) ∼ = SU3 (2), a solvable 3component of CCe (x). We may assume that x ∈ O3 3 (Ce ), for otherwise x centralizes every component of Ce /O3 (Ce ). Then by solvable L3 -balance,  L := O 3 (CK (e)) lies in a 3-component Y of Ce . In particular, x ∈ Y − O3 3 (Y ). Let L1 , L1 , b and J 1 be as in the statement of the lemma and assume for a contradiction that [L1 , x] = 1. Then L1 = Y and by (5G), L1 ∈ CTG3 .

5. THEOREM 3: 3/2-BALANCE

47

∼ D − (2), D5 (2), or E6 (2), By [V17 , 13.17], applied to X = Ce , L1 = 4 contrary to assumption. The proof is complete.  Suppose now that the proposition fails and let (b, J) be a 3/2-obstruction for B. In particular, Proposition 4.6 gives the possibilities for J/O3 (J) and also implies that (5H)

m3 (B/CB (J/O3 (J))) ≥ 2.

We use (5H) to prove Lemma 5.4. x does not centralize J/O3 (J). Proof. Suppose false and let H be the pumpup of L3 (CJ (x)) in CG (x). Hence if we set B0 = CB (H/O3 (H)), then as H covers J/O3 (J), it follows from (5H) that (5I)

m3 (B/B0 ) ≥ 2.

This immediately yields that H does not centralize K ∗ . Indeed, if [H, K ∗ ] = 1, then if m3 (K ∗ ) ≥ 4, B = B1 B2 · · · Bm centralizes H, contrary to (5I). If m3 (K ∗ ) ≤ 3, on the other hand, then B = B1 U with B1 ∩U = x. Since then B1 centralizes H, it follows that m3 (B/B0 ) ≤ m3 (B/B1 ) = 1, again contradicting (5I). Hence by L3 -balance, H is either a single b-invariant 3-component of K ∗ or the product of three such 3-components cycled by b. However, the latter possibility is excluded here as b ∈ B leaves each Ki invariant, 1 ≤ i ≤ m. We conclude that H = Kk for some k, 1 ≤ k ≤ m. But then Bj centralizes H for all j = k, 1 ≤ j ≤ m, and U centralizes H if m = 1. Also x centralizes H. Thus each such Bj (or U ) and x lie in B0 , and it is immediate then that m3 (B/B0 ) = 1, again contradicting (5I).  We shall reduce to the case E = U . We define Je = L3 (CJ E  (e)) for all e ∈ E # and begin with the following lemma. Lemma 5.5. Suppose that E = U . Then there exists e ∈ E # such that the following conditions hold: (a) q = 8 and e induces a nontrivial field automorphism on K/O3 (K); (b) m3 (Q) = 2; (c) The subnormal closure I1 of Je in CG (e) is a 3-component such that x ∈ O3 3 (CK (e)) ≤ I1 and I1 /O3 (I1 ) ∼ = D4− (2), D5 (2), or E6 (2); and (d) No component of Je /O3 3 (Je ) is locally balanced for p = 3. Proof. Since E = U , m3 (E) ≥ 3. Set J = J/O3 (J), and suppose first that  = 1, (5J) E normalizes J and CE (J)

48

10. THEOREMS C6 AND C∗6

so that (5K)

 E embeds in C := CAut(J)  (AutB (J)),

 ≥ m3 (C) ≥ 3. and in particular, m3 (Aut(J)) # We set E1 = {e ∈ E | Ie := L3 (CK (e)) = 1}. Let D be a subgroup # of E of maximal order such  that CE (K) ≤ D and D ⊆ E1 , and set # J0 = L3 (CJ (e)) | e ∈ D . By Lemma 5.3a, x centralizes J0 /O3 (J0 ). As x does not centralize J by Lemma 5.4, J0 does not cover J. Note that m3 (D) ≥ m3 (CE (K)) ≥ 2, and strict inequality holds unless some e ∈ E induces a nontrivial field automorphism on K. As (b, J) is a 3/2-obstruction for B, one of the situations in Proposition 4.6a must occur. If situation (1) occurs, then as m3 (C) ≥ 3 and A11 ∈ TG3 ,  J)  ∼ it follows from [V17 , 4.15] that J/Z( = L3 (q13 ) for some q1 > 2, q1 ≡   Let (mod 3), and some e2 ∈ E induces a nontrivial field automorphism on J.  E2 be the set of all such e2 ; then by [IA , 7.3.8], J is covered by Je | e ∈ E2 . Here Je /O3 3 (Je ) ∼ = L3 (q1 ) for each e ∈ E2 , and in particular Je /O3 (Je ) is not locally balanced for p = 3, so e satisfies (d). It then follows with Lemma 5.4 that [Je /O3 (Je ), x] = 1 for some e ∈ E2 . By Lemma 5.3a, e ∈ E1 . Hence q = 8 and (a) holds. Let I1 be the subnormal closure of Je in CG (e), so that x does not centralize I1 /O3 (I1 ). By solvable L3 -balance [IG , 13.8], x ∈ I1 − O3 3 (I1 ) and I1 is a single 3-component, as [b, x] = 1. Now by Lemma 5.4b, conclusion (c) of this lemma holds. Since e ∈ E − E1 , the defnition of E implies that m3 (Q) = 2, so the lemma holds in this case. Since m3 (C) ≥ 3, neither (2) nor (3) of Proposition 4.6a can hold. Hence by that proposition we may assume that J ∈ TG3 (G). Again since m3 (C) ≥ 3 and J is not locally balanced for p = 3, the only sporadic, alternating, or Chev(3) possibilities for J are An , n = 10 or 11, by [V17 , 4.25], with E acting as a group generated by three mutually disjoint 3-cycles. In these cases, letting E3 be the set of these 3-cycles, we have J covered by Je | e ∈ E3 , with Je /O3 (Je ) ∼ = An−3 for each e ∈ E3 . As An−3 is not locally balanced for p = 3, we may apply part (a) or (b1) of Lemma 5.3 again to obtain the conclusions of this lemma. Suppose finally that J ∈ Chev(r), r = 3. Again, x centralizes J0 /O3 (J0 ). However, J ∈ TG3 (G) is not locally balanced with respect to B, and in view  Hence of (5K), we can apply [V17 , 6.10] and conclude that J0 covers J.  x] = 1, contradicting Lemma 5.4. [J, Therefore (5J) must fail, whence  = ∅. E0 := (E − NE (J)) ∪ CE (J)  whence by Lemma 5.4, there is e ∈ E0 By [IG , 3.28(ii)], Je | e ∈ E0  covers J, such that [Je /O3 (Je ), x] = 1. Lemma 5.3a then implies that e ∈ E1 , whence again (a) and (b) hold. Now Lemma 5.3b1 implies (c). Also, (d) holds since

5. THEOREM 3: 3/2-BALANCE

49

 J),  completing the every component of Je /O3 3 (Je ) is isomorphic to J/Z( proof of the lemma.  Now we can reduce to the case E = U . Lemma 5.6. We have E = U and m3 (Q) = 2. Proof. Suppose false, so that m3 (E) ≥ 3 by definition of E, and continue the above argument. We assume that e ∈ E has been chosen as in Lemma 5.5, if possible so that I1 ∼ = E6 (2).  ∼ Suppose in fact that I1 = E6 (2). Then O3 (CI1 (x)) is an extension of CK (e) ∗ M ∼ = SU3 (2) ∗ SL3 (4) by a subgroup of order 3 inducing outer diagonal automorphisms on each of the two central factors. By L3 -balance, [M, K/O3 (K)] = 1. As m3 (Q) = 2, it follows easily that m3 (C(e, I1 )) = 1. Let R ∈ Syl3 (CG (e)) with CR (x) ∈ Syl3 (CG ( e, x)). Then R = (R ∩ I1 ) × CR (I1 /O3 (I1 )) and Ω1 (Z(R)) = x, e. Since x = [R, R] ∩ Ω1 (Z(R)), either eG ∩ Ω1 (Z(R)) = E1 (Z(R)) − { x} (if R ∈ Syl3 (G)) or x  NG (R) with R ∈ Syl3 (G). In either case, x is weakly closed in x, e. Note that since C(x, K) has rank 2 and contains M , K  CG (x). Therefore e ∈ [CG (x), CG (x)]. By a transfer lemma [III2 , 6.1], e ∈ [G, G], contradicting the simplicity of G. Therefore, I1 ∼ = D4− (2) or D5 (2). Since x, B, and E all commute, and x ∈ I1 − O3 3 (I1 ), B normalizes I1 . Hence B normalizes all 3-components of CI1 (b), by [V17 , 13.21]. In particular, B normalizes Je . As Je /O3 (Je ) 1. Otherwise, by [V17 , 8.8], H/O3 (H) ∼ = SL3 (q  ), q  ≡  (mod 3),  = ±1, or 3A6 , 3A7 , or 3M22 . Since T ∈ Syl3 (G), we can assume without loss that T ∗ = CT (t) ∈ Syl3 (CG (t)). If the lemma fails,   ∗ is the product of 3-components H = H1 , H2 , . . . , Hn then H ∗ = H T for some n ≥ 3. But as x ∈ Z(T ), x ∈ Z(T ∗ ) and hence x centralizes each Hi /O3 (Hi ), 1 ≤ i ≤ n, by [V17 , 12.1]. But then if we set Ii = L3 (CHi (x)), Ii covers Hi /O3 (Hi ), 1 ≤ i ≤ n. However, by L3 -balance and as m3 (Aut(K)(∞) ) = m3 (K) = 2, all but at most one Hi centralizes K. Since n ≥ 3, it follows therefore that m3 (Q) = m3 (C(x, K)) ≥ 3, contrary  to m3 (Q) = 2.

5. THEOREM 3: 3/2-BALANCE

51

Since m3 (Q) = 2 and m3 (T ) = 4, it follows from [V17 , 8.7a] that K∼ (5M) = SL (q), q ≡  (mod 3),  = ±1. 3

Furthermore, by Lemma 5.1, (5N)

B ≤ A for some A ∈ E34 (T ) = E3∗ (T ).

However, as m3 (E) = 2, no element of A induces a nontrivial field automorphism on K by definition of E. Hence by [V17 , 11.3.4a], there is E27 ∼ = F ≤ A with E = U ≤ F such that if we set I = L3 (CK (F )), then (1) I/O3 (I) ∼ = L2 (q); (2) B0 = B1 ∩ I ∼ (5O) = Z3 (with B0 = x); and (3) F = B. On the other hand, by Lemmas 5.7a and 5.9, F leaves J invariant. Since m3 (J) ≥ 2, it follows therefore from [V17 , 6.11] that there is f ∈ F # such that if we set H = L3 (CJ (f )), then (5P)

(1) m3 (H) ≥ 2; and (2) x acts nontrivially on H/O3 (H).

We let N be the pumpup of H in CG (f ), so that each 3-component of N has 3-rank ≥ 2. Hence by Lemma 5.9, N is not the product of three 3-components cycled by b, so by L3 -balance, N is a single b-invariant 3f = Cf /O3 (Cf ). Then component of CG (f ). We set Cf = CG (f ) and C  ∈ CTG3 . Moreover, H is a 3-component of CN (b), so x acts nontrivially N  by (5P2). But x ∈ O3 3 (CG (e)) for every e ∈ E # by Lemma 5.3, so on N certainly f ∈ / E. Thus we also have (5Q)

f ∈ F − E.

In addition, as I ≤ CG (F ), I ≤ Cf and by L3 -balance, either I ≤ N or  . We treat these two possibilities separately, first proving I centralizes N . Lemma 5.10. I centralizes N Proof. Suppose false, in which case I ≤ N . But F leaves N invariant by Lemma 5.9, so I is a 3-component of CN (F ) as I is a 3-component of CK (F ) and hence of CG (F ). It follows therefore from [V17 , 5.13] that for e , where Ne = L3 (CN (e)). Howsome e ∈ E − x, x acts nontrivially on N ever, as x ∈ O3 3 (CG (e)), it follows, as usual, that x centralizes Ne /O3 (Ne ), e , contradiction.  whence x centralizes N We now repeat this last argument as follows. We have that B0 f  cen (as B0 ≤ I). Furthermore, by [V17 , 11.3.4b], there is f0 ∈ tralizes N B0 f  − f  such that if we set I0 = L3 (CK (f0 )), then (1) I0 /O3 (I0 ) ∼ = L2 (q); and (5R) (2) I, I0  covers K/O3 (K).

10. THEOREMS C6 AND C∗6

52

We let N0 be the pumpup of N in Cf0 := CG (f0 ), so that x does not centralize N0 /O3 (N0 ) and N0 is a single A-invariant 3-component of Cf0 , again by Lemma 5.9. Hence repeating the proof of Lemma 5.10 with N0 , I0 , f0 , and F0 := f0  E in place of N , I, f , and F , we conclude: Lemma 5.11. I0 centralizes N0 /O3 (N0 ). With this information, we can prove Lemma 5.12. C(x, K) contains a Sylow 3-subgroup of N x. Proof. Let D be a Sylow 3-subgroup of N x containing x and centralizing B0 f . Also let D0 be a subgroup of D maximal subject to centralizing K/O3 (K). Since x ∈ D ∩ K, D0 centralizes x. Thus D0 ≤ C(x, K). We argue that D0 = D. Set D1 = ND (D0 ). We need only show that D1 centralizes K/O3 (K), for then D1 = D0 and hence D = D0 , as required. Set K0 = L3 (CK (D0 )), so that K0 covers K/O3 (K) and K0 is a 3component of CG (D0 ) as x ∈ D0 . Furthermore, as D ≤ N x, D central = I and D1 leaves L  so if we set L = L3 (CIO (C ) (D0 )), then L izes I, f 3  But as D centralizes B0 f  and hence invariant with D1 centralizing L. f0 , likewise D ≤ N0 x (N0 is defined following (5R)). Hence if we put L0 = L3 (CI0 O3 (Cf0 ) (D0 )), then L0 covers I0 O3 (Cf0 )/O3 (Cf0 ) and D1 leaves L0 invariant with D1 centralizing L0 /O3 (L0 ).  On the other hand, by (5P), L, L0 =K, soif we set CD0 = CG (D0 )  0 . But then as D1 D = CD /O3 (CD ), it follows that L,  L 0 = K and C 0

0

0

 0 , so D1 centralizes acts on CD0 , our argument yields that D1 centralizes K  K/O3 (K), as required, and the lemma is proved. Let then D be a Sylow 3-subgroup of N x with D ≤ C(x, K), so that x ∈ Z(D). We also immediately obtain  . Moreover, Lemma 5.13. We have D ≤ N with D acting faithfully on N  ) = 2.  is simple and m3 (N N  , then as x does not centralize Proof. If some t ∈ I3 (D) centralizes N , x ∈ N / O3 3 (CG (t)). But t ∈ C(x, K), so by the 3-terminality of K, x ∈  . In particular, N  O3 3 (CG (t)), contradiction. Thus D acts faithfully on N is simple (of 3-rank ≥ 2). Furthermore, as m3 (Q) = 2, m3 (D) = 2 by the preceding lemma. Since x ∈ Z(D) and m3 (N ) ≥ 2, it follows that x induces  . Now the faithful action of D on N  forces an inner automorphism on N  x  ∈ N , so x ∈ N . Since D ≤ N x, we conclude that D ≤ N , and the lemma is proved.  We next prove Lemma 5.14. x is weakly closed in D with respect to G.

5. THEOREM 3: 3/2-BALANCE

53

Proof. Suppose false, so that there is h ∈ N such that y = xh ∈ D− x. Let Ky be the pumpup of K in CG (y), so that K covers Ky /O3 (Ky ) and x ∈ O3 3 (Ky ). But also L = K h is a 3-component of CG (y) and y ∈ O3 3 (L). Since y = x, it follows that L = Ky and that m3 (C(y, L)) ≥ m3 (Ky x) = 3. Conjugating by h−1 , we get m3 (C(x, K)) ≥ 3. But this  contradicts the fact that m3 (Q) = 2. ∼ Lemma 5.15. We have N = HJ and ∗ ∼ (K/O3 (K)) ∗ (M/O3 (M )) ∗ Z3m F (CG (x)/O3 (CG (x))) = for some m ≥ 1, where M is a 3-component and M/O3 (M ) ∼ = 3A6 . ∼ Proof. By Lemma 5.14 and [V17 , 3.10], the only alternative to N = HJ  ∼ is N = G2 (q) , q ≡ ±1 (mod 3).    = L3 (C  (d)) | d ∈ I3 (D) . If q > 2, then by [IA , 7.3.3, 4.7.3A], N N  , x] = 1, we deduce that [Nd /O3 (Nd ), x] = 1 for some d ∈ I3 (D), Since [N where Nd = L3 (CN (d)). But as d ∈ C(x, K), x ∈ O3 3 (CG (d)) by the 3∼ terminality of (x, K), a contradiction. Thus, q = 2 and N = U3 (3). = G2 (2) ∼  3 ∼  )) = U3 (3) ∈ Chev(3), N Now N was a pumpup of H. Since O (Aut(N is a trivial pumpup of H, which was in turn a 3-component of CJ (f ). Thus U3 (3) 3 = mp (C(x, K)). However, this again contradicts the fact that (x, K) ∈  TJop (G). The proof is complete. Lemma 6.6. Proposition 6.1b holds. Proof. Suppose false and continue the above argument. Since I = I1 ≤ K, I is a 3-component of CK (a) and hence I is a component of CK (a). Suppose I = K, in which case a ∈ E = A ∩ Q. Then

58

10. THEOREMS C6 AND C∗6

the pumpup of K in CG (a) is the diagonal pumpup J ∗ , contradicting the Ipo terminality of (x, K). Thus I < K and now [V17 , 13.19] yields that p = 3, I∼ = L3 (q1 ), K ∼ = L3 (q13 ) with a inducing a nontrivial field automorphism on K. In particular, a ∈ / R. Suppose now that mp (A) = 5, whence AQ := A ∩ Q ∼ = E32 . Since AQ permutes the 3-components J1 , J2 , . . . , Jm of J ∗ and m = p = 3, some  e ∈ A# Q leaves each Ji invariant. But e centralizes the diagonal I as e centralizes I ≤ K, so e centralizes J∗ . As (x, K) is 3-terminal, the pumpup of K in CG (e) is trivial, however, and hence centralized by x modulo core. Since [J∗ , x] = J∗ this is a contradiction. Thus mp (A) = mp (T ) = 4. This completes the proof.  Now assume that Proposition 6.1a fails. Let B ∈ B(T ) and let (a, J) again be a 5/2-obstruction with a ∈ B, J/O3 (J) not locally 2-balanced with respect to NB (J)/ a, and B = NB (J) × a  for some a ∈ B # . Let J = J1 , J2 , J3 again be the a -conjugates of J, and set Ca = CG (a), a = Ca /O3 (Ca ), and J ∗ = J1 J2 J3 . C Lemma 6.7. d0 = 2 or 3. Proof. In case d0 = 4 or 5, every element of B(T ) lies in Ep∗ (T ), so Lemma 6.6 implies the proposition in these cases. The same holds if d0 = 1j. Hence we need only rule out d0 = 1g. By Lemma 6.6, we may assume that mB = 4. Now by definition of B(T ), B = (B ∩ P ) × (B ∩ Q) with both factors of ∼ order 32 . Let L = L3 (CK (a)), so that by [IA , 4.7.3A], L = L2 (8) or SU3 (8). 3 # As B ∈ S (KQ), there is b ∈ L ∩ B . Since B/ a acts faithfully on J∗ , b acts nontrivially on J∗ , so the pumpup of L in Ca , being a -invariant, is diagonal. Hence J ∼ = L2 (8) or SU3 (8). But both these groups are locally 2-balanced for p = 3 by [V17 , 4.17], contradiction. The lemma follows.  It follows that J ∼ = L3 (q), q ≡  (mod 3),  = ±1. Lemma 6.8. x normalizes J. Proof. Suppose false, let I1 = L3 (CJ ∗ (x)), and let I be the subnormal closure of I1 in CG (x). Then I1 /O3 (I1 ) ∼ = J. As m3 (Q) = 2 and x ∈ Q, we must have I1 ≤ K, so I = K. But then the pumpup of K in Ca is J ∗ ,  contradicting the I3o -terminality of K. Now we can complete the proof of Proposition 6.1. As J is simple, x centralizes a 3-central subgroup D ∼ = E34 of J ∗ a, so that m3 (CJ ∗ a (D)) = 7. Since m3 (Aut(K)) ≤ 3, there exists 1 = d ∈ CD (K). As m3 (CG (d)) ≥ 7, it follows that m3 (C(d, Kd )) ≥ 4. Using Konvisser’s theorem [IG , 10.17] we conclude that x centralizes an E33 -subgroup E of C(d, Kd ). But Kd is a pumpup of K, so E ≤ C(x, K). This contradicts m3 (Q) = 2, however. The proof of Proposition 6.1 is complete.

7. NONQUASISIMPLE PUMPUPS OF K

59

7. Nonquasisimple Pumpups of K We preserve the above notation: see (5A), (5B), (5C), (5E), and the definitions of B(T ) and B0 (T ) at the beginning of Section 6. We show in this section and the next that the B-signalizer functors and B-functors that are available, thanks to Propositions 5.2 and 6.1, are nontrivial except in specific cases where Theorem 3 can be verified easily. One of those cases is that K is terminal in G. We quickly obtain: Proposition 7.1. If G is a counterexample to Theorem 3, then K is not terminal in G (for the prime p). Proof. Suppose for a contradiction that K is terminal in G. Let [K] be the product, as in [IG , 18.2, 18.3], of K and all its conjugates with which it commutes elementwise. Set M = NG ([K]). By [IG , 18.4], M is a Kpreuniqueness subgroup of G, and obviously the pumpup of K in M is K. Moreover, since the components of [K] form an equivalence class under a   G-invariant equivalence relation on K G [IG , 18.2], K M = [K]. We argue that M is a maximal subgroup of G. Let N be a maximal subgroup of G containing M . Then for all x ∈ Ip (Q), K is a component of CN (x) and in particular lies in Lp (N ), by Lp -balance. Moreover, if we put W = Op (N ), then for all x ∈ Ip (Q), K  K, CW (x), so [K, CW (x)] ≤ Op (K) ≤ Z(K) and then [K, CW (x)] = 1 by the Three Subgroups lemma. As mp (Q) ≥ 2, [K, W ] = 1. Therefore K ≤ E(N ). Let L be the subnormal closure of K in N . We see with [IG , 6.19] that L is a single component. Again as mp (Q) > 1, and K ∈ Tp is a component of CL (x) for all  x ∈ INp(Q), we M conclude from[V17 , 13.16] that K = L   N . Then K ≤ K ≤ [K], so [K] = K N  N and M = N , as desired. Hence M satisfies the conclusions of Theorem 3, contrary to assumption. The proof is complete.  ∼ U  T1 = NT (K) Let B ∈ B(T ), so that by definition, there is Ep2 = with x ∈ U ≤ B ∩ Q. Since mp (T1 ) = mp (T ) ≥ 4, U # ⊆ Ipo (G) by [V9 , 5.11]. The next result is the main result of this section. It dovetails with Proposition 7.1 to apply to counterexamples to Theorem 3. However, it will also be useful in proving Theorem 4, so we avoid mentioning Theorem 3 in its hypothesis. Proposition 7.2. Suppose that K is not terminal in G (with respect to the prime p). Then, for some choice of B and U and some u ∈ U # , Ku is not quasisimple. Remark 7.3. Except in the case p = 3, and Q ∼ = 31+2 , the proposition will be proved for any given B and U . Moreover, the proof is valid in both the normal and the exceptional situations. We argue by contradiction in a sequence of lemmas. Set Q0 = CQ (U ),

10. THEOREMS C6 AND C∗6

60

so that |Q : Q0 | ≤ p. We first prove Lemma 7.4. The following conditions hold: (a) For all y ∈ Ip (Q) with mp (CQ0 (y)) ≥ 2, we have Ky = K; and (b) For some t ∈ Ip (Q), we have mp (CQ0 (t)) = 1 and Kt is not quasisimple, but Kt is a trivial pumpup of K. Proof. Suppose first that y ∈ Ip (Q) and y ∈ Ipo (G). As K is Ipo terminal, it follows for such y that Ky = K. In particular, this holds for y ∈ U # . In particular, as Ku is quasisimple for all u ∈ U # , we have that K is quasisimple. It then follows for all y ∈ Ip (Q0 ) that Ky = K is quasisimple. If y ∈ Ip (Q − Q0 ) with either mp (CQ0 (y)) ≥ 2 or mp (K) ≥ 2, then again we have y ∈ Ipo (G) and so Ky = K. This proves (a). Moreover, it implies, since K is not terminal, that mp (K) = 1 and there exists t ∈ Ip (Q) such that Kt > K and mp (CQ0 (t)) = 1. As x = [U, t], we see, using mp (K) = 1 and [V17 , 3.1], that x = Ω1 (Z(T )) ∩ [Ω1 (T ), Ω1 (T )], whence x  NG (T ). As T ∈ Sylp (CG (x)), we have T ∈ Sylp (G). Recall that P = T ∩ K ∈ Sylp (K). Then Ω1 (Z(T )) =

x, z := Z1 , where z = Ω1 (P ). By Burnside’s lemma, x is weakly closed in x, z. In particular, Kt is not a diagonal pumpup of K, for if it were, x would be Kt -conjugate to xz. Similarly, if Kt is a vertical pumpup with mp (Kt ) = 1, then by [V17 , 3.1], x induces a nontrivial field automorphism on Kt /Op (Kt ) and again we reach the contradiction xz ∈ xKt . We claim that Kt is a trivial pumpup of K. Since Kt > K this will complete the proof of (b). To prove the claim, it suffices in view of the previous paragraph to rule out the case that (7A)

Kt is a vertical pumpup of K, and mp (Kt ) > 1.

Suppose then that (7A) holds. Note that if t ∈ Ipo (G), then since d0 = 5, K ∈ Tp5 (G) ∩ TJp , and K ↑p (Kt /Op (Kt )), [V17 , 8.3] implies that Kt /Op (Kt ) ∈ Tp5 (G), i.e., mp (Kt ) = 1, contradicting (7A). So t ∈ Ipo (G) and mp (CG (t)) ≤ 3. Now let St ∈ Sylp (CG (t)) and St ≤ S ∈ Sylp (G). As mp (S) ≥ 4, t ∈ Z(S). But Ω1 (Z(S)) ∼ = Ep2 , and as mp (St ) ≤ 3 it follows that Zt := Ω1 (St ) = Ω1 (Z(St )) = t Ω1 (Z(S)) ∼ = Ep3 . By [V17 , 20.21] we deduce that t ∈ Kt , and so Zt = t × Yt with Yt = Ω1 (St ∩ Kt ) ∼ = Ep2 . Then by [V17 , 20.22], AutKt (Zt ) is irreducible on Yt , and of course centralizes t. We will eventually see that Yt = Ω1 (Z(S)), and a contradiction wil quickly follow. Let A = AutG (Zt ) and AS = AutS (Zt ). As mp (G) > 3, AS = 1, and then as Z(S) is noncyclic, A# S consists of transvections. Then A acts reducibly on Zt , since otherwise by McLaughlin’s theorem [McL2] A would be transitive on Zt# , contradicting t ∈ xG . Since x is not A-invariant, A must normalize Yt , the only other AutKt (Zt )-invariant proper subgroup of Zt . Then [t, A] = Yt , so tA ⊇ tYt . It follows that Ω1 (Z(S)) ≤ Yt , and hence Ω1 (Z(S)) = Yt . But x is weakly closed in Z(T ), so some element

7. NONQUASISIMPLE PUMPUPS OF K

61

of E1 (Yt ) is weakly closed in Yt , contradicting the irreducibility of A on Yt . This completes the proof of (b) and the lemma.  / Q0 , We fix an element t as given in (b). Then mp (CQ0 (t)) = 1, so t ∈ whence (7B)

Q = Q0 t and mp (CQ (t)) = 2.

We set Y = Op (CG (t)). Lemma 7.5. There is b ∈ P − Q such that (a) b, x# ⊆ Ipo (G); (b) Either b ∈ Z(T1 ) or d = 1a, in which case b ∈ V  T1 for some V ∼ = Ep2 ; and (c) b ∈ Kt − Op p (Kt ). In particular, b = x. Proof. By (5D), mp (T1 ) = mp (T ) ≥ 4. Of course x ∈ Z(T1 ). If p does not divide |Z(K)|, we can take b ∈ Z(T1 ) ∩ P . Otherwise, mp (P ) > 1 = mp (Z(K)) and by [V9 , 5.12], there is V ≤ P with V ∼ = Ep2 and V  T1 . We choose b ∈ V − Z(K) in this case. As mp (CT1 ( b, x)) ≥ mp (CT1 (V )) = mp (T1 ) ≥ 4 by [V9 , 5.11], (a) and (b) follow. Then since Q centralizes P , (c) holds as well.  We fix b as in Lemma 7.5, and immediately obtain Lemma 7.6. b centralizes CY (x). Proof. Since b centralizes x and t, b normalizes CY (x). Let X = [CY (x), b] ≤ CY (x) ≤ Op (CG ( x, t)). But K is a component of CG ( x, t) and not a p -group, so [X, K] = 1. As b ∈ K, X = [X, b] = 1, proving the lemma.  Now we can prove Lemma 7.7. For some y ∈ b, x − x , CG (y) contains a t-invariant p-component J such that if we put J = J/Op (J), then (a) J is not weakly locally balanced with respect to t and x;  and (b) b does not centralize J;  (c) Ω1 (Q0 ) normalizes J and acts faithfully on J. Proof. Consider the action of b, x (∼ = Ep2 ) on Kt . We have CKt (Y ) ≤ CKt (Op (Kt )) ≤ Op p (Kt ) by [V9 , 6.1b] as Kt is not quasisimple. Since b∈ / Op p (Kt ), b therefore acts nontrivially on Y . Hence by [V9 , 9.2], there is y ∈ b, x such that X := [CY (y), b] = 1. By the preceding lemma, we have y = x. Of course y = b, so b and x have equivalent actions on CY (y). Therefore, X = [CY (y), x] = [X, x]. Furthermore, X ≤ Op (CG (t)) ∩ CG (y) and y centralizes t (as b, x centralizes Q). Hence by [IG , 20.6], either X ≤ Op (CG (y)) or CG (y) contains

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an X t-invariant p-component J such that X does not centralize J/Op (J), and J/Op (J) is not locally balanced with respect to t. In the latter case, since X = [X, b], b does not centralize J/Op (J). Hence (b) holds in that case, and so will (a), since X = [X, x], once we prove (c) (which implies that x normalizes J). Suppose, on the other hand, that X ≤ Y1 := Op (CG (y)). Again as

b, x centralizes Q, Q ≤ CG (y) and, in particular, U ≤ CG (y). Since b does not centralize X, b does not centralize Y1 . Hence again by [V9 , 9.2], there is u ∈ U # such that X1 := [CY1 (u), b] = 1. Again X1 = [X1 , b], and as y ∈ b, x − x − b and y centralizes X1 , X1 = [X1 , x]. By L∗p -balance [V9 , 6.2], X1 leaves Ku = K invariant. But x centralizes Ku and therefore so does X1 . Since b ∈ Ku , it follows that b centralizes X1 , contrary to X1 = [X1 , b] = 1. Hence this possibility is excluded. By the previous paragraph, it remains only to prove (c). Set y = Cy /Op (Cy ). Cy = CG (y) and C  Setting I = Lp (CJ (v)), Suppose first that some v ∈ Ip (Q0 ) centralizes J.   we have I = J and hence I ≤ [I, b]. As usual, since y ∈ b, x − x − b  so I ≤ [I, x]. and y centralizes I, b and x have the same action on I, But Kv = K by Lemma 7.4, so x centralizes Kv . It follows therefore by Lp balance that I ≤ [I, x] centralizes Kv . Since b ∈ Kv , b therefore centralizes I, contrary to I ≤ [I, b]. Since Suppose finally that some w ∈ Ip (Q0 ) does not leave J invariant.   mp (Q0 ) ≥ 2, we may choose w so that w = x. Set J ∗ = J w,x ≤ x ], where this time we set Lp (Cy ). It follows from [V9 , 9.4] that I = [I, I = Lp (CJ ∗ (w)). Then x leaves I invariant, and it follows from [V9 , 6.1a] that I = [I, x]. But then as above, I = [I, b], and we reach a contradiction as in the preceding paragraph, with w in place of v. Thus Ω1 (Q0 ) leaves J invariant. Then Ω1 (Q0 ) acts faithfully on J by the preceding paragraph. Hence (c) holds, and the lemma follows.   Now we can pin down J. Lemma 7.8. We have p = 3 and J ∼ = L2 (27), L3 (4) (with U inducing  inner automorphisms on J), A10 (with U acting on 10 letters with a regular orbit), or M12 . Proof. Suppose J is not of one of the specified isomorphism types (with p = 3). We have U t ∼ = p1+2 with U t acting faithfully on J and J not weakly locally balanced with respect to t and  x. Hence if we set Xu = Op (CJ (u))Lp (CJ (u)) for u ∈ U # , and X = Xu | u ∈ U # , we

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conclude from [V17 , 6.15] that one of the following holds: (7C)

 = J;  or (1) X  induces inner automor(2) p = 3, J ∼ = SL3 (4), and U  phisms on J.

Suppose that (7C1) holds. Each Xu leaves Ku = K invariant by Lp balance and L∗p -balance. Since x centralizes K, it follows that X0 := [X, x] centralizes K. Hence b ∈ K centralizes X0 . But  x x  0 = [X, ] = [J, ] = J. X Hence b centralizes J, contrary to Lemma 7.7b. Therefore, (7C2) must hold.  = 1, d0 = 1 and By Lemma 7.5, y ∈ I3o (G). Since J ∈ T31 and Z(J) 1 consequently K is also a T3 -group. Thus m3 (K) ≥ 2 and if 3 divides |Z(K)|, then by our choice of (x, K), x ∈ K.  We claim that for any u ∈ U # , either Ku (= K)  O3 (CG (u)) or else Q is connected of rank ≥ 3. Suppose, indeed, that Ku has at least 3 conjugates K = Ku , K1 , K2 in CG (u). If 3 does not divide |Z(K)|, then as m3 (K) > 1, m3 (CK1 K2 U (K x)) ≥ 4, whether u = x or not; thus m3 (Q) ≥ 4 and Q is connected by Lemma 3.1. If 3 does divide |Z(K)| (so that x ∈ K) and

u = x, then using elements of order 3 in Ki − Z(Ki ), i = 1, 2, we see that m3 (CK1 K2 U (K)) ≥ 4 again, as claimed. Finally if u = x, we quote Lemma 3.3b to obtain our claim. However, Q cannot be connected of rank at least 3. For if it were, then m3 (CQ (t)) ≥ 3, contradicting (7B). By the previous paragraph, Ku = K is  invariant under O3 (CG (u)) for each u ∈ U # . Now let W be a Sylow 3-subgroup of J invariant under Ω1 (Q0 ) and set  Since U induces inner

w = Z(W ), so that W ∼ = 31+2 and w centralizes J. automorphisms on J, w ∈ [CW (u), U ] for some (indeed any) u ∈ U # . But we know that CW (u) leaves Ku invariant, and as U centralizes Ku /O3 (Ku ), we conclude that [CW (u), U ] and hence w does as well. But w centralizes Ω1 (Q0 ) and mp (Q0 ) = mp (Q) (as Q0 = CQ (U )), so w ∈ Q0 , contrary to the  fact that Q0 acts faithfully on J, and the lemma is proved. To analyze these cases, we first prove  Lemma 7.9. Q leaves J invariant and acts faithfully on J.  has Z3 Z3 Sylow 3-subgroups if J ∼ Proof. Now Aut(J) = L2 (27) or  has 31+2 Sylow 3-subgroups if J ∼ A10 , while m3 (J) = 2 and Aut(J) = L3 (4) or M12 . In particular, m3 (J) ≤ 3 and as x = [U t , U t], x induces an inner automorphism on J. Furthermore, as Ω1 (Q0 ) acts faithfully on  m3 (Ω1 (Q0 )) ≤ 3. But mp (Q) = mp (Q0 ), so mp (Q) ≤ 3. We conclude J, therefore from Lemma 3.5 that mp (CG (x)) = mp (T ) ≤ 6. This immediately implies that Q leaves J invariant. Indeed, if false, then J Q contains three distinct 3-components J = J1 , J2 , J3 . Since x ∈ Z(Q)

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 x therefore induces an inner and x induces an inner automorphism on J,  automorphism on each Ji , 1 ≤ i ≤ 3. It follows that x centralizes an E36 subgroup A of J1 J2 J3 . But x also centralizes y, and as J is simple, y ∈ / A, ∼ so A y = E37 , whence mp (CG (x)) ≥ 7, contradiction.  = 1. Since Q1  Q (as Q leaves J invariFinally, suppose Q1 = CQ (J) ant), there is thus z ∈ I3 (Q1 ) with z ∈ Z(Q). Then z ∈ Q0 , contrary to the  Thus Q1 = 1 and the lemma is proved.  fact that Q0 acts faithfully on J. We treat the exceptional cases of Lemma 7.8 separately. In Lemmas 7.10–7.12, we assume (7D)

J ∼ = L2 (27).

We let A be a Q-invariant Sylow 3-subgroup of J, so that A ∼ = E27 . We shall make a more specific choice of A below. As J is not locally balanced with respect to t, t induces an outer automorphism on J and A t ∼ = Z3 Z3 . We set (7E)

F =A∩QQ

and first prove Lemma 7.10. A can be chosen so that F = U or Q0 . In particular, x ∈ A. Proof. As noted above, x induces an inner automorphism on J since

x = [U t , U t]. Thus A centralizes x and so A ≤ CG (x). As U t ≤ Q  either Q ∼ and Q is faithful on J, = 31+2 or Q ∼ = Z3 Z3 . Hence by Lemma 3.3b, A leaves K invariant. Thus A leaves Q invariant, so [A, Q] ≤ A ∩ Q = F . But [A, Q] ≥ [A, t] ∼ = Z3 Z3 , so F = E32 and A t ∼ is elementary abelian of order 9 or 27. Also F  Q, so if Q ∼ = Z3 Z3 , it is immediate that F = U or Q0 (= J(Q)).  Hence we can assume that Q = U t ∼ = 31+2 , whence K  O3 (CG (x)) and so T1 = T . Let T0 = CT (y). We argue that (7F)

T0 ∈ Syl3 (CG ( x, y)).

Since x, y = x, b, this is clear if b ∈ Z(T ), so by Lemma 7.5b we may assume that d0 = 1a, in which case |T : T0 | = 3. But in that case by [V17 , 12.1], CT (P ) ≤ Q, so Z(T ) = Z(Q) = x. Hence |CG ( x, y)|3 < |T |, and (7F) follows. Expand T0 to S ∈ Syl3 (Cy ), so that T0 = CS (x) by (7F). Since Q ≤ T0 , we may assume that we chose A = S ∩ J. Thus A ≤ CS (x) = T0 ≤ T . By choice, U  T1 = T . Hence [A, U ] ≤ x. Since F ≤ A and A is abelian, it follows that [A, F U ] ≤ x. But then if F = U (whence Q = F U ), we have [A, Q] ≤ x, contrary to t ∈ Q and [A, t] ∼ = E32 . We conclude that F = U and the lemma is proved. 

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By [V17 , 11.2.1g], NJ (A) contains a cyclic 13-subgroup D such that D transitively permutes the subgroups of A of order 3 and D/CD (A) ∼ = Z13 . In particular, CA (D) = 1 and D acts irreducibly on A. We next prove Lemma 7.11. D leaves K invariant. Proof. Again set F = A ∩ Q, so that F = U or Q0 . In view of the action of D on A, there are elements d1 = 1, d2 , . . . , dh , h = 4 or 13, in D such that if we set ei = xdi , 1 ≤ i ≤ h, then the ei are generators of the h distinct subgroups of order 3 in F . Moreover, as 13 is a prime, D = di  for −1 all i > 1. We set Hi = K di and Li = K di , 1 ≤ i ≤ h, so that K = H1 = L1 and Hi is a component of CG (ei ), 1 ≤ i ≤ h. On the other hand, as F ≤ Q0 , K = Kei for each i, 1 ≤ i ≤ h. If K = Hi for any i > 1, then di leaves K invariant and as D = di , the lemma holds. Hence we can assume that K = Hi for any i > 1, whence K = H1 centralizes Hi for all i > 1. Conjugating by d−1 i , Li centralizes K for all i > 1. As K = Kei is a component of CG (ei ), Li is a component of CG (x).   −1 If Li = Lj for any i = j, then K di dj = K, and D = d−1 i dj normalizes K, as desired. So we may assume that L1 , . . . , Lh are distinct components of CG (x). In particular, 3 ≥ mp (Q) = mp (C(x, K)) ≥ mp (L2 · · · Lh x) ≥ h ≥ 4, contradiction. The proof is complete.  Now we can prove Lemma 7.12. We have J ∼ = L2 (27). Proof. Suppose J ∼ = L2 (27) and continue the above analysis. Since D leaves both K and A invariant, D leaves CA (K) = F invariant. But D acts irreducibly on A, so F = A. [In particular, Q ∼ = Z3 Z3 and F = Q0 .] But D ≤ J centralizes y and hence centralizes the projection y0 of y on Q. Since y ∈ b, x, y0 ∈ x ≤ A. But CA (D) = 1, forcing y0 = 1 and y = b,  contrary to the fact that b acts nontrivially on J.  We next eliminate the possibility J ∼ = A10 , thus proving Lemma 7.13. We have J ∼ = L3 (4) or M12 . ∼ A10 and the projection W

of U on Proof. Suppose false, so that J = ∼  J, in its action on 10 letters, has a regular orbit. Let AW = Z3 Z3 be

and A = a Q-invariant Sylow 3-subgroup of J with W mapping onto W ∼   J(AW ) = E33 . Since x ∈ [U t , U t], x acts on J as an element of A. Hence x ∈ CG (A). Also since J is not locally balanced with respect to t,  by [V17 , 11.6.1a]. t also acts on J as an element of A,

has a regular orbit, u acts nontrivially Let u ∈ U − x. Then since W on A. We now repeat the second paragraph of the proof of Lemma 7.10, with only one alteration: we change t to u. We conclude that if Q ∼ = Z3 Z3 , then A∩Q = U or Q0 . Since u acts nontrivially on the abelian group A, this

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66

is a contradiction. Thus, Q ∼ = 31+2 . Next, we repeat the third paragraph of the proof of Lemma 7.10. We again expand T0 = CT (y) to S ∈ Syl3 (Cy ), so that T0 = CS (x) by (7F). We may assume that we chose AW = S ∩ J. In particular, A ≤ CS (x) ≤ T . But then [A, u] ≤ x, contradicting the fact  that |[A, u]| = 32 . The proof is complete. The proof in these remaining cases is also somewhat easier. First, as  has 31+2 Sylow 3-subgroups and Q acts faithfully on J,  we have Aut(J) 1+2 ∼ Q = U t = 3 . Hence again K is T -invariant, so Q  T . If t, x  T , it follows from the definitions of B(T ) in the various cases that there is B ∗ ∈ B(T ) with U ∗ = x, t ≤ B ∗ . But then the proposition holds with B ∗ , U ∗ in place of B, U , contrary to assumption. Thus (7G)

t, x  T .

Lemma 7.14. For some a ∈ CT (y), we have x ∈ [Q, a, Q]. Proof. If b ∈ Z(T ), then obviously CT ( b, x) covers T /R. In the contrary case, b, x  T by Lemma 7.5, and CT ( b, x) covers T /R since b, x ≤ Z(P ) by [V17 , 12.1] and P ≤ R. Since y ∈ b, x, it follows that CT (y) covers T /R. But t ∈ Q centralizes P and t, x  Q. Since t, x is not normal in T , we conclude that there is a ∈ CT (y) such that a does not centralize Q/ x. But then the action of a on Q ∼  = 31+2 implies that x ∈ [Q, a, Q], and the lemma is proved. Using the element a, we quickly derive a final contradiction. We have Q a ≤ Cy . If a leaves J invariant, then Q a leaves J invariant. Again  has 31+2 Sylow 3-subgroups and x ∈ [Q, a, Q], x must therefore as Aut(J)  contrary to the faithful action of Q on J. Hence a does not centralize J, leave J invariant. But x ∈ [Q, Q] induces an inner automorphism on J and hence on each component of Ja . Since mp (J) = 2 with J simple and x centralizing y, it follows that mp (CCy (x)) ≥ 7, whence mp (CG (x)) ≥ 7, contrary to Lemma 3.5. This completes the proof of Proposition 7.2. 8. Nontrivial Signalizer Functors In this section, like the last, we prove results that not only apply to a counterexample to Theorem 3, but also will be useful in the proof of Theorem 4. We continue the notation of the previous sections. We continue to analyze a pair B ∈ B(T ) and U ≤ B, chosen in accordance with Proposition 7.2. In particular, mB (= mp (B)) ≥ 3 if d0 = 1a and mB ≥ 4 if d0 = 1a. We set r = 2 if d0 = 1a. On the other hand, if d0 = 1a, we set r = 3 either if mB = 4 or if d0 = 4 and mB = mp (T ) = 5, and otherwise we set r = 4. Then in all cases, (8A)

mB ≥ r + 1.

By Propositions 5.2 and 6.1, G is 3/2- or 5/2-balanced with respect to B according as r = 2 or 3, respectively, while if r = 4, then G is 3-balanced

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with respect to B by Proposition 4.3. Correspondingly we let ΘB be the associated 3/2-balanced or 5/2-balanced B-functor Θ3/2 or Θ5/2 on G or the 3-balanced B-signalizer functor Θ3 on G, and we let (8B)

ΘB (G; B) = the closure of ΘB .

See [V9 , 3.1] for the definitions of these terms. We fix the notation r, ΘB , and ΘB (G; B) for the rest of this chapter. We shall prove Proposition 8.1. Suppose that K is not terminal (for the prime p). Then ΘB (G; B) = 1. We argue by contradiction in a sequence of lemmas. We know that K is a Tpd0 -group., Also by Proposition 7.2, we have (8C)

Ku is not quasisimple for some u ∈ U # ,

where Ku is the pumpup of K in CG (u). Now B ≤ CG (u) leaves K invariant and as K covers Ku /Op (Ku ), likewise B leaves Ku invariant. Using [V17 , 8.5] and (6B1), we quickly eliminate the cases d0 = 1a, 1g, 1j, 2, and 5. Thus we prove Lemma 8.2. We have d0 = 3 or 4. Proof. Suppose false. If d0 = 1a, 1g, 1j, or 2, then B induces inner automorphisms on K by definition of B(T ). Moreover, by [IA , 5.3g], (5F), and (6B), there exists a hyperplane D of B such that L3 (CK (D)) = 1, so I := L3 (CKu (D)) = 1. Note that in each case, B = (B ∩K)D. On the other 5 hand, if d0 = 5, then as K ∼ L2 (8) with p = 3 or 2B2 (2 2 ) with p = 5 as K is = a Tp -group, it follows from the definition of B(T ) and [V17 , 8.5] that there is a hyperplane D of B such that Lp (CK (D)) = 1, so I := Lp (CKu (D)) = 1 in this case as well. Again B = (B ∩ K)D, since any element of B acting nontrivially on K acts by a field automorphism. Since Ku is not quasisimple, we conclude therefore from [V9 , 6.3] that D contains a hyperplane F such that (8D)

[ΔF , I] = 1.

But now if r = 4, then mp (F ) ≥ 3, so ΔF ≤ Θ3 (G; B) = ΘB (G; B), whence ΘB (G; B) = 1, contrary to our assumption that the proposition fails. Thus r = 2 or 3. Observe that B leaves I invariant and also that B1 = B ∩ K ∈ E∗ (P ). It follows now from [IG , 16.11] that there is b ∈  B1 with [I/Op (I), b] = I/Op (I). But I/Op (I) is quasisimple, so I ≤ bI . Since I does not centralize ΔF , neither therefore does b. Thus [ΔF , b] = 1. Since mp (F ) ≥ r − 1, it follows from the definitions of Θ3/2 and Θ5/2 that [ΔF , b] ≤ Θr− 1 (G; B), so ΘB (G; B) = 1, giving the same contradiction. 2 We note that the proof is valid in both the normal and the exceptional situation. 

10. THEOREMS C6 AND C∗6

68

Now that d0 = 3 or 4, the maximality result Proposition 2.6 can come into play. We next treat the case d0 = 3, so that we are in the normal situation. We have p = 3, and in Lemmas 8.5–8.9, we assume ∼ A7 , M12 , M22 , or L (q), q ≡ 4 K is a T33 -group, whence K = 3 (8E) or 7 (mod 9),  = ±1, q > 4. Also by definition, B ∈ E∗ (R) in this case, so B = B1 E, where B1 = B∩K ∼ = E32 and E = B ∩ Q. In particular, mp (E) ≥ 2. The argument is similar to that of [IV9 , Proposition 7.8]. We therefore conform as closely as possible to the notation of that proposition. However, the possibility of quadratic action of X1 on X0 makes it necessary for us to distinguish large from small elements of T33 , as follows. Definition 8.3. T33+ = {L ∈ T33 | |L| > |L3 (7)|} T33− = T33 − T33+ . We then have Lemma 8.4. T33+ = {L3 (q) | q ≡  (mod 3),  = ±1, q ≥ 8}. Moreover, if p = d = 3, then T33+ ∩Lo3 (G) is closed under the pumping up process within all elements of Lo3 (G). Proof. The first statement is easily checked from the definitions of T33 and T33+ . For the second statement, note as in the proof of [V7 , 3.1] that as p = d = 3, all pumpups of elements of T33+ in Lo3 (G) are in either TG3 or T33 , and in the latter case, they are actually in T33+ by the defining condition for T33+ . In particular as A7 ∈ T33+ , [V8 , 16.4, 16.5] imply that the lemma holds.  We now prove Lemma 8.5. There exist subgroups W < V < B with |B : W | = 9, and B-invariant 3 -subgroups X1 , X0 of O3 (CG (V )) and CCG (u) (W ), respectively, with the following properties: (a) (1) X1 is a p1 -group for some prime p1 = 3; (2) X1 /(X1 ∩ O3 (Ku )) ∼ = Zp1 , p1+2 1 , or Ep21 ; and (3) X1 = [X1 , b] for some b ∈ B1 acting irreducibly on the Frattini quotient of X1 /(X1 ∩ O3 (Ku )); and (b) (1) X0 ≤ O3 (CG (u)) and X0 is a p0 -group for some prime p0 = 3; (2) X1 normalizes, but does not centralize, X0 ; and (3) If K ∈ T33+ , then [X0 , X1 , X1 ] = 1. u = Cu /O3 (Cu ). By [V17 , 4.23] and the Proof. Set Cu = CG (u) and C 1 does not centralize definition of B(T ) in the case K ∼ = M12 , it follows that B 1+2 ∼   a minimal B1 -invariant subgroup X1 = Zp1 , p1 , or Ep21 of O3 (CK u (b1 )), for

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1 centralizes V := b1  E some b1 ∈ B1# and for some prime p1 = 3. Then X and if we choose X1 to be a minimal B-invariant subgroup of O3 (CKu (b1 ))∩ 1 , we see that X1 satisfies the conditions of (a) (for CKu (V ) that covers X # suitable b ∈ B1 ) and that X1 ≤ O3 (CKu (V )) ≤ O3 (CG (V )) (as u ∈ E ≤ V ). Setting F = F (O3 (CG (u))), suppose first that CKu (F ) ≤ O3 (K). Then X1 does not centralize Op0 (F ) = Op0 (CG (u)) or F0 := Op0 (F )/Φ(Op0 (F )) for some prime p0 = 3. Moreover, by [V17 , 4.24], if K ∈ T33+ , X1 does not even act quadratically on F0 . Hence X1 does not centralize X0 = COp0 (CG (u)) (W ) for some hyperplane W of V , and if K ∈ T33+ , [X0 , X1 , X1 ] = 1. As |B : V | = 3, |B : W | = 9, so (b) holds for X0 in this case. We therefore may assume that CK (F ) ≤ O3 (K), so that F ≤ Z(K). As K is not quasisimple, F ∗ (K) = F ∗ (O3 3 (K)), and O3 (K) ≤ Z(K), so CK (L1 ) ≤ O3 3 (K), where L1 = E(O3 (K)). Let S ∈ Syl2 (L1 ) be B invariant, and N = O3 (NK (S)). Then K = O3 3 (K)N and B ≤ N . In particular O3 3 (K)CK (S) = O3 3 (K) or K. In the latter case, since CAut(L1 ) (S) is a 2-group by [V17 , 10.7.1], O3 3 (K)CK (L1 ) = K, so CK (L1 ) = K, which is absurd as L1 is not abelian. So CK (S) ≤ O3 3 (K). But then as B ≤ N and N covers K/O3 (K), we could have chosen X1 to lie in N , and we consider the action of X1 on S. As CK (S) ≤ O3 3 (K), [S, X1 ] = 1. Moreover,  := N/CN (S/Φ(S)) in the role if K ∈ T33+ , then applying [V17 , 4.24] with N   of L there, and X1 , B, and S/Φ(S) in the roles of R, B1 , and M there, 1 ] = 1. Whether or not K ∈ T 3+ , we then 1 , X we conclude that [S/Φ(S), X 3 complete the proof as in the previous case by taking X0 = CS (W ) for a suitable hyperplane W of V .  We now define (8F)

X01

 [X0 , X1 , X1 ] = [X0 , X1 ]

if K ∈ T33+ if K ∈ T33− .

Thus X01 = 1. The argument of [IV9 , Lemma 7.10] applies without essential change to yield: Lemma 8.6. One of the following holds: (a) X01 ≤ O3 (CG (w)) for some w ∈ W # ; or (b) V centralizes X01 , and X01 ≤ O3 (CG (v)) for some v ∈ V # . Proof. Indeed, otherwise we check that either X01 ≤ ΘB (G; B) or that 1 = [X01 , B] ≤ ΘB (G; B), contrary to our assumption that the proposition fails.  According as (a) or (b) of Lemma 8.6 holds, put y = w or y = v. Since X01 ≤ X0 ≤ O3 (CG (u)) ∩ CG (y) and X1 ≤ O3 (CG (V )) with y ∈ V , we conclude now by [IG , 20.6] and L∗3 -balance [V9 , 6.2]:

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Lemma 8.7. CG (y) contains a 3-component L with the following properties: (a) L is invariant under X1 X0 u; (b) X01 and u do not centralize L/O3 (L); and (c) L/O3 (L) is not locally balanced with respect to u. The next step is to pin down the structure of L/O3 (L) and the action y = Cy /O3 (Cy ). of B on L. Set Cy = CG (y) and C Lemma 8.8. The following conditions hold:        ∼ (a) L = L3 (q  ), L− n (q ), n = 4 or 5, or P Sp4 (q ), q ≡  (mod 3),   = ±1; (b) E leaves L invariant and induces faithful inner-diagonal automor and phisms on L; (c) B normalizes L.

 = N/CN (L),  so that Proof. Set X = X1 X0 , N = NCy (L), and N  ≤ Aut(L)  and X u ≤ N with X  nonabelian (as [X1 , X0 ] does not N    u) with X a 3 -group. But by the hypothesis centralize L). Also X ≤ CN (  is either a C3 -, a T3 -, or a flat TG3 -group. We conclude of Theorem C∗6 , L  is isomorphic to one of the following groups therefore from [V17 , 4.21] that L 



 −  (here q  ≡  (mod 3) and  = ±1): L2 (q  ), L± 3 (q ), Ln (q ), n = 4 or 5,  or P Sp4 (q  ). Note that SL3 (q  ) is ruled out here since d0 = 3.  For if it did, then so would [X1 , b] = Observe that b does not centralize L.  as any X1 and X01 , contrary to assumption. Indeed b does not act on L e ∈ E normalizing L. For if it did, then as X1 ≤ CG (V ) and E ≤ V , E,  Then X1 = [X1 , b] and hence b, would centralize the image of X1 in Aut(L).  whence X01 would as well, contradiction. would act trivially on L, Next, suppose (b) fails, so that by [IA , 2.5.12, 4.9.1], some e ∈ E # either  moves L, or induces a nontrivial field automorphism on L.  In centralizes L,  with u acting nontrivially the first case set I = L3 (CL (e)), so that I = L  on I. But as e ∈ Q, K covers Ke /O3 (Ke ), so u centralizes Ke /O3 (Ke ). However, by L3 -balance, either I ≤ Ke or I centralizes Ke /O3 (Ke ). The  and the second is first possibility is excluded as u does not centralize I, likewise excluded as b ∈ Ke does not centralize I/O3 (I). Thus e does not  centralize L. The argument in the remaining cases is similar. If e leaves L invariant  and we set I = (and hence induces a nontrivial field automorphism on L)  u does L3 (CL (e)), then as u induces an inner-diagonal automorphism on L, not centralize I by [IA , 7.1.4c]. As in the preceding paragraph, b is forced to  whence it normalizes L  and its action on L  is the same as that centralize I,

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of an element of e by [IA , 7.1.4c]. But we showed above that this is not the case. We conclude that e does not leave L invariant. Let L∗ = Le  b] = 1. and I = L3 (CL∗ (e)). Again I = [I, u], so in CG (e) we see that [I, ∗  like an element of e. It follows, as [X1 , e] = 1, that Hence b acts on L    ∗ , so X01 centralizes L,  contradiction. Thus  X1 = [X1 , b] acts trivially on L (b) holds. Since mp (E) ≥ 2, and E induces inner-diagonal automorphisms  L  ∼ on L, L2 (q  ) or L− = 3 (q) by [IA , 4.10.3]. Thus (a) holds as well, and  = 2. But mp (E) = mp (Q) by definition of B(T ), so it follows that mp (L) mp (CG (x)) = mp (T ) ≤ 6 by Lemma 3.5.  Now, Outdiag(L)  is cyclic Finally suppose that B does not normalize L. #  and by [IA , 2.5.12], so some e ∈ E induces an inner automorphism on L  b   is simple of 3-rank 2 and b does not leave hence on L0 := L . But as L L invariant, it follows that mp (CL∗ y (e)) = 7. Thus with [V17 , 19.1, 8.4b], m3 (C(e, Ke )) ≥ 5. However, as mp (CG (x)) ≤ 6 and so m3 (C(x, K)) ≤ 4, this contradicts Proposition 2.6. The proof is complete.  We conclude: Lemma 8.9. We have d0 = 4. Proof. Otherwise, by Lemma 8.8c, B normalizes L. Given the possi in Lemma 8.8a, we immediately deduce from [V17 , 4.22], and bilities for L  ∈ T 3+ , then X01 centralizes L,  contrathe definition of X01 (8F), that if L 3 3+ 3+  diction. Hence, assume that L ∈ T3 . We shall argue that K ∈ T3 as well. This will suffice, for then X01 = [X0 , X1 , X1 ] by (8F); and again by [V17 ,  4.22], we will reach the contradiction that X01 centralizes L. 3+ o o  ∈ L (G) ∩ T . By Lemma 8.4, for Now, (y, L) ∈ IL3 (G) and L 3 3 ∗ ∗ o any (y, L) mp (CG (x)). It follows that mp (C(e, Ke )) > mp (Q), again contradicting Proposition 2.6. The proof is complete. 

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74

Next we prove ∼ Lemma 8.15. L = A10 . ∼ Proof. Suppose that L = A10 , and let  E = {e ∈ E # | e acts as a 3-cycle on L}. Suppose first that E = ∅ and let e ∈ E. Set Le = L3 (CL (e)), so that Le /O3 (Le ) ∼ = A7 . The pumpup of Le in Ce then centralizes Ke /O3 (Ke ), since it is not true that A7 ↑3 L3 (4) [IA , 4.9.6]. As b ∈ Ke , it follows that Le,b := L3 (CLe (b)) covers Le /O3 (Le ). Furthermore, as Le e ≤ C(e, Ke ), m3 (C(e, Ke )) ≥ 3, whence m3 (E) = m3 (Q) ≥ 3, as usual by Proposition  it follows that there is e ∈ E such that 2.6. As E acts faithfully on  L,     e, L  e = L.  e ,b = L,  Hence L  e,b , L  so

e = e , and so such that L b ∈ C(y, L). But this is impossible by (8G) and Lemma 8.10b: X = [X, b]  This contradiction shows that E = ∅. acts nontrivially on L.  it follows that m3 (E) < 3, so m3 (Q) = Since E acts faithfully on L, m3 (E) = 2. As E = ∅, [V17 , 11.6.1b] implies that some e ∈ E # centralizes a Sylow 3-subgroup S of L. Since m3 (C(e, K)) ≤ m3 (Q) = 2 as usual, S ≤ CG (e) normalizes Ke . Notice that if m3 (C(y, L)) > 1, then m3 (CG (e, y)) ≥ 5, so m3 (C(e, Ke )) ≥ 3 and m3 (Q) ≥ 3 as usual, a contradiction. Thus, m3 (C(y, L)) = 1 so a Sylow 3-subgroup Sy of Cy containing S satisfies Ω1 (Sy ) = S × y ∼ = (Z3 Z3 ) × Z3 . In particular J(Sy ) ∼ = E34 . But y ∈ B, so e centralizes S y, whence B and J(Sy ) are conjugate in CG (e, y) and in particular in CG (e). Therefore AutCG (e) (B) contains an element f of order 3 with a free submodule on B. But Ke  CG (e) as m(C(e, Ke )) = 2. Hence B = B1 ×E with B1 = B ∩Ke  NCe (B) and E = B ∩C(e, Ke )  NCG (e) (B). So f acts quadratically on B, a contradiction. The proof is complete.  ∼ Next, assume that L = L3 (4). Then u induces an outer diagonal auto morphism on L by [V17 , 11.4.1g]. Again m3 (Q) = m3 (E) = 2 by Lemma 8.11b. Then B leaves L invariant, otherwise as usual m3 (CG (e)) ≥ 7 >  and we m3 (CG (x)) for some e ∈ E # inducing an inner automorphism on L obtain a contradiction to Proposition 2.6. Let A be a B-invariant Sylow 3-subgroup of L, so that A ∼ = Z3 . But B cen= E32 and A1 := [A, u] ∼ tralizes A1 and B ∈ E∗ (CG (x)), so A1 ≤ B, which in turn implies that A  normalizes B. Again some e ∈ E # induces an inner automorphism on L, so e ∈ B centralizes A. As m3 (C(e, Ke )) ≤ m3 (Q) = 2 by Proposition 2.6, again A leaves Ke invariant. Since u centralizes Ke /O3 (Ke ), so therefore does A1 . But A1 ≤ B and CB (Ke /O3 (Ke )) = E as K is I3o -terminal, so A1 = e1  ≤ E. ∼ On the other hand, as L = L3 (4), NL (A) contains a 2-element t which inverts A and normalizes A u (but not u). Also by [V17 , 11.4.1g],  and by a Frattini argument we can choose t to centralize B = e1 , u CB (L)  We set B ∗ = ABB t . Then B ∗ is t-invariant and B ∗ = A u CB (L),  CB (L).

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so [B ∗ , t] = A and [B ∗ , B ∗ ] = e1  with B ∗ ≤ CG (e1 ). We fix this notation and next prove:  ∼ Lemma 8.16. Suppose that L = L3 (4). Then the following conditions hold: (a) Ke1 is the unique 3-component of Ce1 of its isomorphism type; (b) A centralizes Ke1 /O3 (Ke1 ); and (c) y ∈ B1 . Proof. As already noted, m3 (Q) = m3 (E) = 2 by Lemma 8.11b; it follows using Proposition 2.6 as usual that m3 (C(e1 , Ke1 )) ≤ 2, and so (a) holds and Ke1  NG ( e1 ). Now t normalizes e1 , so t acts on Ke1 and C(e1 , Ke1 ). We have B1 ≤ B ∩Ke1 ≤ B ∗ ∩Ke1 . But e1  = [B ∗ , B ∗ ] and so A0 := AutB ∗ (Ke1 /O3 (Ke1 )) is abelian. It follows that A0 ∈ Syl3 (Inn(Ke1 /O3 (Ke1 ))) and B1 maps onto A0 with B1 ≤ Ke1 . Now B ∗ = B1 × (B ∗ ∩ C(e1 , Ke1 )). Both direct factors are t-invariant, as B1 ≤ Ke1 . Now (b) follows quickly. Indeed, as e1  = [B ∗ , B ∗ ] ≤ A, either A ≤ C(e1 , Ke1 ), as desired, or A ∩ B1 = 1, so assume the latter. Since u ∈ B ∗ and [A, u] = e1 , it follows that [A ∩ B1 , u] = e1 . But B1 ≤ Ke1  CG (e1 ), so e1 ∈ Ke1 , a contradiction. Thus (b) holds.  and In particular, it follows that B1 centralizes A u. But B1 acts on L   L u ∼ = P GL3 (4) with A u ∈ Syl3 (L u). Now [V17 , 11.4.1g] yields that  = 1. Thus there is b1 ∈ B # centralizing L.  Then b1 ∈ V and by CB1 (L) 1 [V9 , 4.4], if L1 is a 3-component of the pumpup of L in CG (b1 ), then L1 has the same properties (see Lemma 8.10) as L, so by our choice of y, we have y ∈ B1 , proving (c).  By a similar analysis we can now eliminate the L3 (4) case, thus proving ∼ Lemma 8.17. L = A7 and the exceptional situation occurs. Proof. Suppose false and continue the above argument. NKe1 (B1 ) contains a 2-element t1 inverting B1 by [V17 , 11.4.1a], and we can choose t1 to centralize A u, as A u centralizes Ke1 /O3 (Ke1 ). Then t1 normalizes

y ≤ B1 and so acts on CG (y). Moreover, t1 centralizes e1 ∈ A and so as e1 ∈ L, t1 leaves L invariant. But as t1 centralizes A u ∈ Syl3 (L u)     t1 therefore centralizes a Sylow 3with L u ∼ = O3 (Aut(L)), = P GL3 (4) ∼  subgroup of Aut(L). However, B1 leaves L invariant and t1 inverts B1 .  that Since B1 u is a 3-group, it follows at once, if we pass to AutCG (y) (L),  But b ∈ B1 acts nontrivially on L  by Lemma 8.10b, B1 must centralize L. contradiction.  In this final case, set N = NG (B) and N ∗ = N/CG (B). We use the “star” convention, writing Y ∗ for the image of Y ≤ N in N ∗ . Then as K∼ = L3 (4), there is H ≤ NK (B) such that H ∗ ∼ = Q8 with [B, H ∗ ] = B1 and

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76

CB (H ∗ ) = E. Also there is t ∈ NL (B) with t∗ of order 4, [B, t∗ ] = B ∩ L,  =: F . Note that and CB (t∗ ) = CB (L) (8I)

F ∩ E = 1.

 ∼ For if 1 = e ∈ F ∩ E, then CG (e) involves e × L = Z3 × A7 . However, m3 (C(e, Ke )) ≤ m3 (Q) = 2 by Proposition 2.6, and by [V17 , 10.6.6], CG (e)/C(e, Ke ) ≤ Aut(L3 (4)) does not involve A7 , so CG (e) cannot involve Z3 × A7 , contradiction. Thus (8I) holds. We first prove Lemma 8.18. We have Q = E ∼ = E32 .  ∼ to Proof. As E acts faithfully on L = E32 . Itsuffices = A7 , E = U ∼  Q = L, show that E ≤ Z(Q). For then, [Q, y] = 1, so Q acts on L; if L   then E would normalize every 3-component of LQ and then m3 (CG (E)) ≥ m3 (CCy (E)) ≥ 5, which contradicts Proposition 2.6. As Ω1 (Z(Q)) ≤ E and  = 1, also CQ (L)  = 1, whence Q embeds in Aut(A7 ) and Q ∼ CE (L) = E32 , as desired. Suppose that E ≤ Z(Q) and choose v ∈ E−Z(Q). Then v ∗ is a transvection with axis B1 x and center x. Any transvection in N ∗ must fix some I3o -terminality of element of E # and so must normalize E and B1 , by the ∗ ∗ ∗ t t (x, K). In particular, the center of (v ) , which is x , lies in E or B1 . If it lies in E, then t∗ normalizes E, so as E ∩F = 1, E = [B, t∗ ] = B ∩L. Then

v ∗ , t∗  induces SL(E) on E. In particular, some transvection t1 ∈ NN (E)  centralizes y. Then E t1  ∼ = 31+2 must normalize L and act faithfully on L, 2   since Z(E t1 ) ≤ E and CE (L) = 1. But | Aut(L)|3 = 3 , a contradiction. ∗ Therefore the center of (v ∗ )t lies in B1 . It follows, by conjugation by H, that every subgroup of B1 of order 3 is the center of some conjugate of v ∗ ; and then by conjugation by t∗ again, the same must hold for E. Hence every subgroup of E of order 3 is the center of some transvection in N ∗ . Since N ∗ conjugates transvections to transvections, N ∗ preserves the decomposition B = B1 × E. Hence NN (E) induces SL(E) on E and we reach the same contradiction as above. Therefore, E ≤ Z(Q) and the proof is complete.  Now as (x, K) is I3o -terminal, Q ∈ Syl3 (C(e, Ke )) for all e ∈ Q# . Lemma 8.19. Let d ∈ B − B1 . Then dN ∩ Q = ∅. Proof. Let g ∈ N with Q ∩ Qg = 1, say y ∈ Q and y g = z ∈ Q. Then Qg = (B ∩ C(y, Ky ))g = B ∩ C(z, Kyg ) = B ∩ C(z, Kz ) = Q. Here Kyg = Kz is the only 3-component of CG (z) isomorphic mod core to K (because Q ∈ Syl3 (C(z, Kz ))). Thus, Q is a TI-set in N . Let Q = {Qg | g ∈ G, Qg ≤ B}; N acts on Q. Note that |B # | = 10|Q# |. We have H ∗ ∼ = Q8 acting faithfully on B1 and trivially on Q. If Q0 ∈ Q is invariant under the involution s∗ of H ∗ , then either Q0 = CB (s∗ ) = Q or CQ0 (s∗ ) = 1, in which case Q0 = [B, s∗ ] = B1 . Hence the orbits for H ∗ on Q − {Q} are a fixed point {B1 } and a size 8 orbit.

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77

Let S ∈ Syl3 (N ) and S0 ∈ Syl3 (G) with T ≤ S ≤ S0 , and set T0 = NS0 (Q). Since Q ∈ Syl3 (C(e, Ke )) for all e ∈ Q# , T0 /Q acts faithfully on K. Hence |T0 : B| ≤ 3 and T0 /B, if nontrivial, acts quadratically on Q. But G contains Z3 Z3 , a Sylow subgroup of A10 , as we are in the exceptional situation. Hence T0 < S0 . If T0 ∼ = E33 . Hence for any g ∈ NS0 (T0 ), = 31+2 × E32 , then Q ≤ Z(T0 ) ∼ g Q ∩ Q = 1. A similar argument to that above, using CT0 (K) = Q, shows that Q = Qg , so g ∈ NS0 (Q) = T0 . Thus, S0 = T0 , a contradiction. Therefore, T0 = B or T0 /B ∼ = Z3 with |[B, T0 ]| = 32 . In any event, B = J(T0 ) char T0 so NS0 (T0 ) ≤ NS0 (B) = S. Therefore |S : T0 | ≥ 3. Given the H-orbits on Q, we deduce that |Q| = 9 or 10, which implies the lemma.  Lemma 8.20. The element y, but no element of E # , is 3-central in G. Proof. Since B ∈ S3 (G), B contains a 3-central element of order 3. Let e ∈ E # and expand B ≤ Se ∈ Syl3 (CG (e)). Then E = Q ∈ Syl3 (C(e, Ke )) by Proposition 2.6, so |Se : B| ≤ 3 and Se /B, if nontrivial, acts quadratically on B. Hence Se has class at most 2. As G contains A10 and thus Z3 Z3 , Se ∈ Syl3 (G). Also CG (e)/O3 (CG (e)) has no A7 component, since m3 (C(e, Ke )) = 2 and Ke /O3 (Ke ) ∼ = A7 . As ∪g∈N Qg covers B − B1 , and K fuses B1# completely, B1# must contain y and consist of 3-central elements, while B − B1 has none.  Lemma 8.21. y is not 3-central in G. y has the Proof. Suppose false and let B ≤ Sy ∈ Syl3 (Cy ). Since C  ∼ component L = A7 , m3 (Z(Sy )) ≥ 3, even if Sy is not normal in Cy . As 3 B ∈ S (G), Ω1 (Z(Sy )) ≤ B. But then Ω1 (Z(Sy )) ∩ E = 1, contradicting Lemma 8.20.  The conflict between Lemmas 8.20 and 8.21 finally completes the proof of Proposition 8.1. 9. The Subgroup M We let B, r, and ΘB (G; B) be as in the preceding section, so that ΘB = Θ3/2 , Θ5/2 , or Θ3 , the corresponding associated B-functor or Bsignalizer functor on G, and if K is not terminal in G with respect to p, then ΘB (G; B) = 1 by Proposition 8.1. By Propositions 4.3, 5.2, 6.1, and by [V9 , 3.2a1,b1], ΘB (G; B) is a p -group. (9A)

Whenever ΘB (G; B) = 1, we define M to be any maximal subgroup of G containing NG (ΘB (G; B)).

Such a subgroup M will almost always be shown to be a p-component preuniqueness subgroup of G. However, in a special configuration to arise in the next section when d0 = 4 and mB = 4, we shall have to construct an alternative subgroup M ∗ to satisfy the conclusion of Theorem 3.

78

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Thus, what our proof of Theorem 3 will show, when complete, is actually the following. It is important to have this straightened out before going on to Theorem 4. Theorem 9.1. Under the assumptions of Theorem C∗6 , and with (x, K), T , B(T ), B, mB , and d0 defined as above, one of the following holds: (a) ΘB (G; B) = 1, K is terminal in G (for the prime p), and the conclusion of Theorem 3 holds for the maximal subgroup NG ([K]), where [K] is generated by a maximal set, containing K, of mutually elementwise commuting G-conjugates of K. (See Propositions 8.1 and 7.1.) (b) ΘB (G; B) = 1, d0 = 4 and mB = 4, and the conclusion of Theorem 3 holds for some maximal subgroup M ∗ of G containing K, ΓA,3 (G) for all A ∈ B(T ). (See Proposition 10.1b.) (c) ΘB (G; B) = 1, and the conclusion of Theorem 3 holds for any maximal subgroup M of G containing NG (ΘB (G; B)). (See Propositions 7.2, 8.1, 9.2 − 9.6, 10.1a, and the development in Sections 11 − 16). Here we list some general consequences of signalizer functor theory as they apply in the present context, when ΘB (G; B) = 1. First, by [V9 , 3.2a3,b3], we obtain Proposition 9.2. We have ΓB,r (G) ≤ M . By Propositions 5.2 and 6.1, according to the values of r = 2, 3, or 4, we know that G is k-balanced with respect to B, where k = 3/2, 5/2, or 3, respectively. Using [V9 , 3.4], we also have Proposition 9.3. Let A ∈ Er (G) and assume that A is r-connected to B by a chain B = B1 , B2 , . . . , Bn of elements of Er+1 (G) with A ≤ Bn . If G is k-balanced with respect to Bi for each 1 ≤ i ≤ n, then we have (a) ΘB (G; B) = Θk (G; B) = Θk (G; A) = ΘA (G; A); and (b) ΓA,r (G) ≤ M . In particular, the balance hypotheses of the proposition hold if r = 4 by Proposition 4.3. Using Propositions 5.2 and 6.1, we have the following particular case of the proposition. Proposition 9.4. If A ∈ B(T ) and mp (A ∩ B) ≥ r, then ΘA (G; A) = ΘB (G; B) and ΓA,r (G) ≤ M . We also have the following variation of this last result. Proposition 9.5. Let X be a p-subgroup of G of rank ≥ r. If each element of Er (X) is r-connected to B by a chain of elements of Er+1 (G) with respect to each of whose members G is k-balanced, then NG (X) ≤ M .

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Proof. For any A ∈ Epr (X), Θk (G; A) = Θk (G; B) by Proposition 9.4. If g ∈ NG (X), then Ag ∈ Epr (X), so Θk (G; A) = Θk (G; B) = Θk (G; Ag ) = Θk (G; A)g . Hence g ∈ NG (Θk (G; B)) = M , as claimed.  In particular, using [V9 , 3.4], we have the following result in the case d0 = 1a with K ∗ = K1 K2 · · · Km having the usual meanings (whence if m ≥ 2, then B ≤ K ∗ ). See the definitions preceding Lemma 5.1. ∗  Proposition 9.6. Assume d0 = 1a and m ≥ ∗2. Let J be∗ a product i∈I Ki of at least 2 of the 3-components Ki of K , with m3 (J ) ≥ 4. Let X be a B-invariant Sylow 3-subgroup of J ∗ . Suppose that N is a subgroup of NG (X) permuting the set {X ∩ Ki | i ∈ I}. Then N ≤ M .

Proof. Without loss, I = {1, 2, . . . , k}, k ≥ 2. Let g ∈ N and for each i = 1, . . . , k, let Ai be the unique element of {Bjg | 1 ≤ j ≤ k} lying in Ki . g Let B+ = B1 · · · Bk and A+ = B+ = A1 · · · Ak ≤ X. Then m3 (B+ ) ≥ 4 by hypothesis. Moreover, B+ is 2-connected to A by the chain B+ = D0 , D1 , . . . , Dk = A+ , where D = A1 · · · A B+1 · · · Bk , 0 ≤ ≤ k. For any , 0 ≤ ≤ k, the group D Bk+1 · · · Bn is J ∗ -conjugate to an element of B0 (T ) (one needs only to conjugate Ai into T ∩ Ki for i = 1, . . . ,

). Hence by Proposition 5.2, G is 3/2-balanced with respect to D . As G is 3/2-balanced with respect to B by the same proposition, we have g ) = Θ3/2 (G; A+ ) Θ3/2 (G; B)g = Θ3/2 (G; B+ )g = Θ3/2 (G; B+ = Θ3/2 (G; Dk ) = Θ3/2 (G; Dk−1 ) = · · · = Θ3/2 (G; D0 ) = Θ3/2 (G; B+ ) = Θ3/2 (G; B) so g ∈ NG (Θ3/2 (G; B)) ≤ M . The proof is complete.



For the next proposition, see the definitions preceding (6H). Proposition 9.7. Suppose that d0 = 1g, 1j, 2, 3, or 4. Let B0 ∈ B0 (T ). According as r = 3 or 4, let ΘB0 (G; B0 ) = Θ5/2 (G; B0 ) or Θ3 (G; B0 ). Then ΘB0 (G; B0 ) = ΘB (G; B). Proof. We have B = B1 × E and B0 = B1∗ × E ∗ , with E, E ∗ ∈ E∗ (Q) and B1 , B1∗ allowable subgroups of P . By Propositions 6.1a and 4.3 it suffices to r-connect B0 and B by a chain of elements of B0 (T ). Now E ≥ U  Q with U ∼ = Ep2 . We use the chain B, B1∗ E, B1∗ U CE ∗ (U ), B0 .  10. K Lies in M We let B, ΘB , M , and r be as in the preceding section, and assume that ΘB (G; B) = 1. Thus M is a maximal subgroup of G containing NG (ΘB (G; B)).

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We shall prove Proposition 10.1. One of the following holds: (a) K ≤ M ; (b) There exists a maximal subgroup M ∗ of G such that K, ΓB,3 (G) ≤ M ∗ for all B ∈ B(T ). Moreover, M ∗ is an L-preuniqueness subgroup of G, where L is the subnormal closure of K in M . Furthermore, d0 = 4 and mB = 4. As usual, we argue in a sequence of lemmas. As before, we set B1 = B ∩ K and E = B ∩ Q. We have r = 2, 3, or 4 and mB = mp (B) ≥ r + 1. Assume that Proposition 10.1a fails. Since B leaves K invariant and ΓB,r (G) ≤ M , it follows from [V9 , 9.2] that Op (K) ≤ M . Then as K ≤ M by assumption, M does not cover K, whence (10A)

ΓB,r (K) < K.

We immediately obtain Lemma 10.2. The following conditions hold: (a) d0 = 4; and (b) r = 3 and mp (B) = 4. Proof. Indeed, if d0 = 4, it follows at once from the definition of B(T ) and (5F), (6B), (6C), and [V17 , 3.1f] that ΓB,r (K) = K, contrary to (10A), so d0 = 4. We reach the same contradiction by [V17 , 11.4.1e] if mp (B) > r + 1, so mp (B) = r + 1. Hence either (b) holds or r = 4 and mp (B) = 5. However, the latter possibility is excluded by definition of r, so (b) also holds.  In particular, we have (10B)

Either p = 3 and K ∼ = L3 (4) or p = 5 and K ∼ = F i22 .

These cases are considerably more difficult, requiring delicate permutability arguments. The analysis is modeled closely on the proof of [GL1, Part II, Chapter 2, Proposition 7.1]. The aim will be to construct directly a p-component preuniqueness subgroup M ∗ satisfying the conditions of Theorem 3, thus verifying alternative (b) of Proposition 10.1. Note that in the present cases, B ∈ E∗ (R) and mp (K) = 2. Since mp (B) = 4, it follows that (10C)

mp (E) = mp (Q) = 2.

Using the definition of TJop (G), we then have Lemma 10.3. Let (a, H) ∈ ILop (G) with H/Op (H) ∼ = K. Then (a, H) o is Ip -rigid, mp (C(a, H)) ≤ 2, and in case of equality, |C(a, H)|p ≤ |Q|. Proof. Since d0 = 4, L3 (4) and F i22 (and A7 in the exceptional situation) are the only elements of Lop (G) that are Tp -groups of p-rank 2. Furthermore, every element of Lop (G) is in Cp ∪ Tp ∪ TGp , and if p = 3 any TG3 -group

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in Lo3 (G) must be flat. Hence by [V17 , 11.4.1h], L3 (4) and F i22 have no vertical pumpups within the set Lop (G). Consequently, (a, H) is Ipo -rigid. Hence by [V9 , 6.4], there exists an Ipo -terminal and Ipo -rigid pair (x∗ , K ∗ ) such that K ∗ /Op (K ∗ ) ∼ = H/Op (H) ∼ = K and mp (C(x∗ , K ∗ )) ≥ mp (C(a, H)). o But (x, K) ∈ TJp (G), and so mp (C(x∗ , K ∗ )) ≤ mp (C(x, K)) = mp (Q) = 2. Thus, mp (C(a, H)) ≤ 2. In case of equality, we have mp (C(x∗ , K ∗ )) = 2, so  |C(a, H)|p ≤ |C(x∗ , K ∗ )|p ≤ |C(x, K)|p = |Q|, as claimed. For the proof of Proposition 10.1, we consider for each A ∈ B(T ) the set J (A) of all pairs (y, J) such that (10D)

(1) y ∈ A# and J is a p-component of CG (y); (2) J/Op (J) ∼ = K; and (3) J ≤ ΓA,3 (G).

In particular, (x, K) ∈ J (B). We first establish some properties of elements of J (A). Again as d0 = 4, each A ∈ B(T ) lies in R, so A = B1 × CA (K). Also x ∈ A as x ∈ Z(Q) and mp (A) = 4 as mp (Q) = 2. Lemma 10.4. If A ∈ B(T ) and (y, J) ∈ J (A), then the following conditions hold: (a) (b) (c) (d)

A induces inner automorphisms on J/Op (J); mp (C(y, J)) = mp (CA (J/Op (J))) = 2; J  CG (y); and If a ∈ CA (J/Op (J))# , then the subnormal closure L of J in CG (a) is a trivial pumpup of J, and (a, L) ∈ J (A).

Proof. By Lemma 10.3, mp (C(y, J)) ≤ 2. In particular, it follows that J  CG (y). But now if A does not induce inner automorphisms on J/Op (J), then p = 3 and as A ≤ CG (y) leaves J invariant, it follows from [V17 , 11.4.1e] and [V17 , 5.8], applied to J/Op (J), that J ≤ ΓA,3 (G), contrary to the definition of J (A). Since mp (J) = 2, and mp (C(y, J)) ≤ 2, this forces mp (CA (J/Op (J))) = 2, and we see that (a), (b), and (c) hold. Finally, let a and L be as in (d), and set M0 = ΓA,3 (G). By Lemma 10.3, (y, J) is Ipo -rigid. Moreover, L cannot be a diagonal pumpup of J; if it were, again with [V9 , 6.4], (x∗ , K ∗ ) would exist with K ∗ /Op (K ∗ ) ∼ =K and mp (C(x∗ , K ∗ )) ≥ 3, contradicting (x, K) ∈ TJop (G). Thus L is a trivial pumpup of J. In particular, it follows that L covers J/Op (J). But then if L ≤ M0 , M0 would cover J/Op (J) and as Op (J) ≤ ΓA,3 (G) ≤ M0 , this  would give J ≤ M0 , which is not the case. Hence (d) also holds. We introduce the following notation. If A ∈ B(T ) and (y, J) ∈ J (A), and we put F = CA (J/Op (J)), we denote by Jf the pumpup of J in CG (f ) for each f ∈ F # . This notation applies in particular for y = x and J = K.

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We next prove Lemma 10.5. Let A ∈ B(T ), let (y, J) ∈ J (A), and let H be a proper subgroup of G containing A and covering J/Op (J). If L denotes the subnormal closure of J0 = Lp (J ∩ H) in H, then L/Op (L) ∼ = J0 /Op (J0 ) ∼ = K.  = H/Op (H). Proof. By assumption, J/Op (J) ∼ = J0 /Op (J0 ). Set H By the preceding lemma, if we set F = CA (J/Op (J)), then y ∈ F ∼ = Ep2 and # ∼ (f, Jf ) ∈ J (A) for each f ∈ F , so Jf /Op (Jf ) = J/Op (J). It follows that  is quasisimple, J0 is a component of CL (f) for each f ∈ F # . But now if L ∼ L3 (4) or F i22 with L  a K-group, [III8 , 1.16] then as mp (F ) = 2 and J0 = ∼  is not quasisimple,   implies that L = J0 (= K), as asserted. However, if L then L is the product of p p-components cycled by y by Lp -balance. But  and as J0 is a component of CL (f) for each f ∈ F # , A acts faithfully on L  in the roles of w and X we get a contradiction from [V9 , 9.5], with y and L there. The proof is complete.  Lemma 10.6. Assume that there exists a maximal subgroup M ∗ of G containing K, ΓB,3 (G) | B ∈ B(T ). Then Proposition 10.1(b) holds. Proof. Let M ∗ be a maximal subgroup of G as in the lemma. By the preceding lemma, if L denotes the pumpup of K in M ∗ , then L/Op (L) ∼ = K. We shall argue that M ∗ is a p-component preuniqueness subgroup (with respect to L), in which case Proposition 10.1b will hold. Now, Q ≤ NG (B1 x) ≤ ΓB,3 (G) ≤ M ∗ . So Q ∈ Sylp (CM ∗ (L/Op (L))), / Sylp (C(z, Kz )), otherwise it would follow that for some z ∈ Ip (Z(Q)), Q ∈ contrary to the fact that (x, K) is Ipo -terminal. Thus by definition of an L-preuniqueness subgroup, it suffices now to prove that ΓQ,1 (G) ≤ M ∗ . Let 1 = V ≤ Q and set V1 = Ω1 (Z(V )). Since V1 char V , NG (V ) ≤ NG (V1 ), so it suffices to prove that NG (V1 ) ≤ M ∗ . Hence without loss we can assume that V = V1 , whence V is elementary abelian. Since mp (Q) = 2, we have V ∼ = Zp or Ep2 . If V = x, let A be any element of B(T ), and in the contrary case, set A = B1 , x, V , in which case A ∈ B(T ). Again by the preceding lemma, if K1 denotes the pumpup of K0 = Lp (CK (V )) in NG (V ), then K1 /Op (K1 ) ∼ = K0 /Op (K0 ) ∼ = K. But B1 ≤ K1 as B1 ≤ K0 (since B1 centralizes V ≤ Q). As in the proof of Lemma 10.4, mp (CNG (V ) (K1 /Op (K1 ))) = 2, otherwise there would exist an Ipo rigid pair (x∗ , K ∗ ) with mp (C(x∗ , K ∗ )) ≥ 3, leading to a contradiction with (x, K) ∈ TJop (G) as usual. In particular, it follows that K1  NG (V ). Hence by a Frattini argument, NG (B1 V ) covers NG (V )/K1 V (as B1 ∈ Sylp (K)). But B1 V ≤ A with mp (B1 V ) ≥ 3, so NG (B1 V ) ≤ ΓA,3 (G) ≤ M ∗ and consequently M ∗ covers NG (V )/K1 V and hence NG (V )/K1 . Finally, as V ≤ A, A ≤ NG (V ) so A acts on Op (K1 ), whence Op (K1 ) ≤ ΓA,3 (G) ≤ M ∗ by [V9 , 9.2]. But K0 covers K1 /Op (K1 ) and K0 ≤ K ≤ M ∗ . Hence K1 ≤ M ∗ and therefore NG (V ) ≤ M ∗ . We conclude that ΓQ,1 (G) ≤  M ∗ , as required, and the lemma is proved.

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Thus for the rest of the section we assume that G = K, ΓA,3 (G) | A ∈ B(T ) .

(10E)

We will get a contradiction at the end of the section, thus proving Proposition 10.1. Lemma 10.7. For some A ∈ B(T ), K is not permutable with ΓA,3 (G). Proof. If false, then K is permutable with M1 = ΓA,3 (G) | A ∈ B(T ). We shall argue that KM1 is a proper subgroup of G, which will contradict (10E). Observe first that B1 x ≤ A for each A ∈ B(T ) and B1 x ∼ = Ep3 . Moreover, r = 3. It follows therefore from Proposition 9.4 that ΓA,3 (G) ≤ M for each A ∈ B(T ), whence M1 ≤ M . Finally, x ≤ M1 and K centralizes x, so KM1 < G by [V9 , 13.2]. The proof is complete.  As we have noted in the preceding proof, for every A ∈ B(T ), we have A∩B ∼ = Ep3 , so ΘB (G; B) = ΘA (G; A), again by Proposition 9.4. Hence without loss we can assume the preceding lemma holds for B. Thus for the balance of the proof, we assume K is not permutable with ΓB,3 (G).

(10F) We set

M0 = ΓB,3 (G) and N = ΓB,3 (G) = CG (D) | D ∈ E3 (B), so that N ≤ M0 . We fix this notation as well. Our aim will be to prove that B(T ) = {B} and that K, M0  is contained in a proper subgroup of G, which together will contradict our assumption completing the proof. We first prove Lemma 10.8. If (y, J) ∈ J (B), then JN is a proper subgroup of G. Proof. Set F = CB (J/Op (J)), so that F ∼ = Ep2 and for each f ∈ F # , (f, Jf ) ∈ J (B) with Jf /Op (Jf ) ∼ = J/Op (J) ∼ = K and Jf  CG (f ) by Lemma 10.4. Set I = Lp (CJ (F )), so that I covers each Jf /Op (Jf ), whence Jf = IOp (Jf ).   Next set Nf = CN (f ) for f ∈ F # and N0 = Nf | f ∈ F # . By [V9 , 9.2], Op (CG (f )) ≤ CG (D) | f ∈ D ∈ E3 (B) ≤ CN (f ) = Nf . But Jf  CG (f ) and Nf ≤ CG (f ), with Jf = IOp (Jf ), so I is permutable with Nf for each f ∈ F # . Thus I is permutable with N0 . Since y ∈ F # with J = IOp (J) and Op (J) ≤ N0 , we conclude that J is permutable with N0 . We claim next that N0 = N . Indeed, clearly N0 ≤ N . On the other hand, N = CG (D) | D ∈ E3 (B). For any such D, we have D ∩ F = 1 as mp (F ) = 2 and mp (B) = 4, so if we choose f ∈ D ∩ F # , it follows that CG (D) ≤ N ∩ CG (f ) = Nf ≤ N0 , so N ≤ N0 . Thus N = N0 , proving our claim. Hence J is permutable with N , so JN is a group.

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But now as y ∈ N with y centralizing J, we conclude from [V9 , 13.2] as in Lemma 10.7 that JN < G, and the lemma is proved.  We next prove Lemma 10.9. For some g ∈ M0 and e ∈ E # , if we set y = eg and J = (Ke )g , then y ∈ B, (y, J) ∈ J (B), and CB (J/Op (J)) = E. Proof. Suppose false. We argue that K is permutable with M0 , contrary to (10F). To prove this, it will suffice to prove that KNG (D) ≤ M0 K for every D ∈ E3 (B), inasmuch as M0 is generated by the set of such NG (D). Thus for any such D, we need only show that Kg ≤ M0 K for every g ∈ NG (D). Since NG (D) ≤ M0 , this reduces to showing that (10G)

K g = g −1 Kg ≤ M0 K.

Since mp (D) ≥ 3, D ∩ E = 1. Let e ∈ D ∩ E # and consider Ke . Since (x, K) ∈ J (B), (e, Ke ) ∈ J (B) by Lemma 10.4d. In particular, Ke ≤ M0 . Since g ∈ M0 , it follows that J = (Ke )g ≤ M0 . But y = eg ∈ D ≤ B, J is p-component of CG (y), and J/Op (J) ∼ = K/Op (K) = K, so likewise (y, J) ∈ J (B). Since the lemma is assumed false, E = CB (J/Op (J)), so y ∈ E. Thus Ky is defined. Since mp (C(y, Ky )) = 2 by Lemma 10.4b (as (y, Ky ) ∈ J (B)), this forces Ky = J. Finally, Ky ≤ KOp (Ky ) ≤ KN and KN = N K by Lemma 10.8, so J = (Ke )g ≤ N K ≤ M0 K (as J = Ky and N ≤ M0 ). But as g ∈ M0 and Op (K) ≤ N , Op (K g ) ≤ M0 . Furthermore, K g ≤ Op (K g )(Ke )g , so K g ≤  M0 K, and (10G) is proved. This completes the proof of the lemma. We immediately obtain Lemma 10.10. If e, g, y, and J are as in Lemma 10.9 and we set F = CB (J/Op (J)), then the following conditions hold: (a) B = E × F ; and (b) J ≤ M. Proof. Indeed, suppose (a) is false. Since E ∼ =F ∼ = Ep2 and mp (B) = # 4, it follows that there is v ∈ (E ∩ F ) . Then Kv and Jv are each pcomponents of CG (v). Also, K covers Kv /Op (Kv ) and J covers Jv /Op (Jv ), so E = CB (Kv /Op (Kv )) and F = CB (Jv /Op (Jv )). Since E = F , it follows that Kv = Jv , so mp (C(v, Kv )) ≥ 3. However, as usual, this contradicts Lemma 10.3. Thus (a) holds. −1 Furthermore, if J ≤ M , then as g ∈ M0 ≤ M , likewise Ke = J g ≤ M . But Ke covers K and Op (K) ≤ M0 ≤ M , so K ≤ M , contrary to our assumption that the proposition fails. Thus (b) also holds.  Now we can prove Lemma 10.11. We have Q = E. In particular, B(T ) = {B}.

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Proof. If Q = E, then R = B1 E = B, whence {B} = B(T ), so the first assertion implies the second. Suppose then that Q > E and let e, g, y, J, and F be as in the preceding lemma. We have B = B1 × E, where B1 = B ∩ K, and B = F × E. Hence some f ∈ F # projects into x, with respect to the decomposition B1 × E. Since B1 centralizes Q and x ∈ Z(Q), it follows that Q ≤ CG (f ). Thus Q acts on Jf  CG (f ). Set Cf = CG (f ) f = Cf /Op (Cf ). If Q is not faithful on Jf , then some z ∈ Ip (Z(Q)) and C centralizes Jf . But z ∈ E as mp (E) = mp (Q) = 2, so also z ∈ F = CB (Jf ), contrary to the preceding lemma. Hence Q acts faithfully on Jf . Since Q > E, we conclude now from [V17 , 11.4.1e] that p = 3, K ∼ = Jf ∼ = L3 (4) 1+2 ∼ and Q = 3 . But now if we let a ∈ Q − E, then as a centralizes x and B1 , A =

B1 , x, a ∼ = E81 and A ∩ B ∼ = E27 . Furthermore, as B induces inner auto morphisms on Jf , a induces a noninner automorphism on Jf as |Jf |3 = 9 [V17 , 11.4.1]. It follows therefore from [V17 , 5.8] that Jf ≤ ΓA,3 (G). ∼ E27 , and NG (ΘB (G; B)) ≤ M , again Proposition However, as A ∩ B = 9.4 implies that ΓA,3 (G) ≤ M , so Jf ≤ M . As usual, Jf covers J/O3 (J)  and O3 (J) ≤ M , so J ≤ M , contrary to the preceding lemma. This immediately yields: Lemma 10.12. If (y, J) ∈ J (B) and F = CB (J/Op (J)), then Jf is not quasisimple for some f ∈ F # . Proof. It follows once again from Lemma 10.3 that F ∈ Sylp (C(y, J)), so (y, J) ∈ TJop (G). But then if each Jf were quasisimple, J would be terminal in G by definition of this term. Hence by our original choice of (x, K) (1F), K would be terminal in G, contrary to Proposition 7.1, applied to (y, J), and the lemma is proved.  Note that for each f ∈ F # , (f, Jf ) ∈ J (B) and F = CB (Jf /Op (Jf )). Hence for triples (y, J, F ) with (y, J) ∈ J (B) and F = CB (J/Op (J)), one can always assume without loss that J itself is not quasisimple. Likewise, by Proposition 7.1, we may assume that K is not quasisimple. We next prove Lemma 10.13. If (y, J) ∈ J (B) with J not quasisimple, then there exists a proper subgroup H of G that covers both K/Op (K) and J/Op (J). Proof. By Lemma 10.8, KN and JN are each proper subgroups of G and K, J are contained in p-components K ∗ , J ∗ of KN, JN , respectively, with K ∗ /Op (K ∗ ) ∼ = K/Op (K) and J ∗ /Op (J ∗ ) ∼ = J/Op (J). In particular, this forces Op (K) ≤ Op (K ∗ ) and Op (J) ≤ Op (J ∗ ). Hence by assump tion, neither K ∗ nor J ∗ is quasisimple. Furthermore, Op (K ∗ ), Op (J ∗ ) ≤ ΓB,3 (G) ≤ N .

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Consider first the case that Op (K ∗ ) or Op (J ∗ ) is not a 2-group. Since B leaves Sylow q-subgroups of Op (K ∗ ) and Op (J ∗ ) for each prime  invariant ∗ q and Op (K ), Op (J ∗ ) ≤ N , there thus exists an odd prime q = p such that a maximal B-invariant q-subgroup X of N is nontrivial. We argue that H = NG (X) has the required properties. We claim first that X leaves both K ∗ and J ∗ invariant. Since the proofs are identical, we treat only K ∗ . We have B1 = B ∩ K ∼ = Ep2 with K ≤ K ∗ . Hence [B1 , X] ≤ Lp (KN ) and consequently [B1 , X] leaves K ∗ invariant. Since K ∗ /Op (K ∗ ) ∼ = K is simple, likewise CX (B1 ) leaves K ∗ invariant, so by [V9 , 9.1] X = [B1 , X]CX (B1 ) leaves K ∗ invariant, and our claim is proved.  ∗ = K ∗ /Op (K ∗ ) ∼ Again as K = K and as q is odd, it follows from [V17 ,  ∗ , so K ∗ normalizes Y = Op (K ∗ )X. But 11.4.1f] that X centralizes K Y ≤ N and Y is B-invariant. Since X is a maximal B-invariant q-subgroup of N , it follows from [V9 , 8.4] that X ∈ Sylq (Y ). But now we conclude by a Frattini argument that NK ∗ X (X) covers K ∗ /Op (K ∗ ) and hence covers K. An entirely similar argument yields that NJ ∗ X (X) also covers J/Op (J), so the lemma holds with H = NG (X), as asserted. We can therefore assume that Op (K ∗ ) and Op (J ∗ ) are 2-groups. This time we let X be a maximal B-invariant 2-subgroup of N . It follows by the same argument as above that X leaves K ∗ and J ∗ invariant.  both ∗ But now the maximality of X implies that Op (K ), Op (J ∗ ) ≤ X. Since ∗ ∼ B1 ∈ Sylp (K ∗ ), and K = L3 (4) or F i22 , it follows from [V17 , 11.4.1f] ∗  that |X/CX (K )| ≤ 2, and if equality holds, then X/CX (K ∗ ) induces an  ∗ ; and the corresponding assertion holds for J∗ = outer automorphism on K ∗ ∗ J /Op (J ). But F ∗ (K ∗ ) = O2 (K ∗ ) = Op (K ∗ ) and F ∗ (J ∗ ) = O2 (J ∗ ) = Op (J ∗ ), as ∗ K and J ∗ are not quasisimple. Hence if we set X1 = J(X), X2 = Ω1 (Z(X)), and X3 = Ω1 (Z(J1 (X))), where J1 (X) denotes the subgroup of X generated by the elementary abelian subgroups of X of rank ≥ mp (X)−1, we conclude  ∗ for at least two values of i, 1 ≤ from [V17 , 11.4.1i] that NG (Xi ) covers K ∗  i ≤ 3, and that NG (Xj ) covers J for at least two values of j, 1 ≤ j ≤ 3. Hence for some k, 1 ≤ k ≤ 3, NG (Xk ) covers both K ∗ and J ∗ and hence covers both K/Op (K) and J/Op (J). Thus the lemma holds in this case  with H = NG (Xk ). Now we can prove Lemma 10.14. If (y, J) ∈ J (B), and F = CB (J/Op (J)), then either F = E or F = B1 = B ∩ K. Proof. Let H be a proper subgroup of G that covers both K/Op (K)  = H/Op (H) and K0 = Lp (K ∩ H), J0 = Lp (J ∩ H). and J/Op (J). Set H

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By Lemma 10.6, K0 ≤ K ∗ and J0 ≤ J ∗ , where K ∗ , J ∗ are p-components 0 ∼ ∗ = K of H with K = K/Op (K) and J∗ = J0 ∼ = J/Op (J). In particular, ∗ ∗   CB (K ) = E and CB (J ) = F .  ∗ = J∗ , then  ∗ = J∗ , then E = F . On the other hand, if K But now if K  But then J∗ centralizes  ∗ centralizes J∗ as each is a component of H. K ∗ ∼ ∼ B1 ≤ K0 ≤ K . Since B1 = F = Ep2 , this forces F = B1 . Thus either F = E or F = B1 , as asserted.  By Lemma 10.9 and the preceding lemma, there is (y, J) ∈ J (B) such that CB (J/Op (J)) = B1 . For such a choice of (y, J), we prove Lemma 10.15. JKN is a proper subgroup of G.  be as in the preceding lemma. Given the Proof. Let H, K0 , J0 , and H  0 centralizes J0 . Hence choice of (y, J), we have shown there that K (10H)

J0 K0 Op (H) = K0 J0 Op (H).

Furthermore, we have K = Op (K)K0 = K0 Op (K) and J = Op (J)J0 = J0 Op (J).   Also by Lemma 10.8, KN = N K and JN = N J. As Op (K), Op (J) ≤ N , it follows that

(10I)

(10J)

KN = K0 N = N K0 and JN = J0 N = N J0 .

But also Op (H) ≤ ΓB,3 (G) ≤ N , so by (10J) and (10H), we have (10K)

J0 K0 N = K0 J0 N = K0 N J0 .

Now we conclude from (10I) and (10K) that (10L)

JKN = KN J.

Thus J is permutable with the group KN and hence JKN is a group. But y centralizes J and y ∈ KN , which is a proper subgroup of G (by Lemma 10.8). Again we conclude from [V9 , 13.2] that JKN is a proper subgroup of G.  We also immediately obtain Lemma 10.16. If (z, L) ∈ J (B), then L ≤ N K or N J. Proof. If B0 = CB (L/Op (L)), we know from Lemma 10.14 that B0 = E or B1 . Correspondingly we argue that L ≤ N K or N J. Since the argument is the same in both cases, we treat only the case B0 = E. Then z ∈ E and L and Kz are each p-components of CG (z). But as mp (Q) = 2, it follows, as usual, that L = Kz . Thus K covers L/Op (L), so L ≤ Op (L)K.  But Op (L) ≤ N , so L ≤ N K, and the lemma is proved. Finally we prove Lemma 10.17. JKM0 is a proper subgroup of G.

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Proof. The proof follows the line of argument of Lemma 10.9. Thus let D ∈ E3 (B) and let g ∈ NG (D). As in that lemma, there is e ∈ D ∩ E # such that if we set z = eg and L = (Ke )g , then (z, L) ∈ J (B). Hence by the preceding lemma, L ≤ N K or N J. But L covers K g /Op (K g ), so K g ≤ Op (K)g L. Since Op (K) ≤ N ≤ M0 with g ∈ M0 , it follows that K g ≤ M0 L, whence K g ≤ M0 K or M0 J. Consequently Kg ⊆ M0 K ∪ M0 J, and similarly, Jg ⊆ M0 K ∪ M0 J.

(10M)

But the set of all such g generates M0 , so KM0 ⊆ M0 K ∪ M0 J and JM0 ⊆ M0 K ∪ M0 J.

(10N) Hence (10O)

JKM0 ⊆ JM0 K ∪ JM0 J ⊆ (M0 K ∪ M0 JK) ∪ (M0 KJ ∪ M0 J).

On the other hand, by Lemma 10.15, N KJ = N JK, whence (10P)

M0 KJ = M0 N KJ = M0 N JK = M0 JK.

But now combining (10O) and (10P), we conclude that (10Q)

JKM0 ⊆ M0 JK.

Since JKN is a group with N ≤ M0 , it follows now that M0 is permutable with JKN and hence that JKN M0 = JKM0 is a group. Finally set H = JKN (a proper subgroup of G by Lemma 10.15). As usual, Lemma 10.5 implies that Op (K) ≤ Op (H), so X = Op (H) = 1. But X ≤ N ≤ M0 . Thus by [V9 , 13.2], the subgroup JKM0 = HM0 is proper in G. The lemma is proved.  We have therefore shown that B(T ) = {B} (Lemma 10.11) and that

K, M0  = K, ΓB,3 (G) is a proper subgroup of G (Lemma 10.17). However, this is in conflict with (10E), and the proposition is finally proved. 11. The Pumpup of K in M We now assume that Theorem 3 fails. As a result, Theorem 9.1 fails. This assumption will be in force through Section 16, in which the proofs of those theorems are completed. By Proposition 7.1, K is not terminal in G for the prime p. Moreover, alternative (b) of Proposition 10.1 does not hold. Thus Propositions 7.2 and 8.1 apply, together with the results of Section 9 and conclusion (a) of Proposition 10.1. It is this main strand of the argument that we pursue, without further branches, through Section 16. We preserve the notation (x, K), M , B, d0 , r, T , T1 , P , Q, and R of the preceding sections. Thus by Propositions 9.2 and 10.1a, we have (11A)

ΓB,r (G), K ≤ M.

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In this section we prove Proposition 11.1. If L denotes the subnormal closure of K in M , then L/Op (L) ∼ = K and Q ∈ Sylp (CM (L/Op (L))). We set

= M/Op (M ) M and proceed in a sequence of lemmas, first checking that  = K.  Lemma 11.2. L/Op (L) ∼ = K, i.e., L Proof. By definition of B(T ), CB (K) ∼ = Epm with m ≥ 2. # Choose any y ∈ CB (K) . As mp (B) ≥ 4, y ∈ Ipo (G). Now (x, K) is Ipo -terminal, so the subnormal closure Ly of Lp (CK (y)) in CG (y) satisfies Ly /Op (Ly ) ∼ = K. Therefore the subnormal closure of Lp (CK (y)) in  is a component of CM (y) is also isomorphic modulo p -core to K. Hence K Lp (CL (y)). As y was arbitrary in CB (K), it follows from [III11 , 1.16], [V17 ,  = K,  as desired, or p = 3 and K ∼ 5.16], and [V9 , 9.5] that either L = A6 or ∼ ∼    m 3A6 , or L = Akp +δ with K = Aδ and CB (K) acting on L as a subgroup of  with δ fixed points and k regular orbits (of size pm ). L The A6 case is impossible as K ∈ T3 but A6 ∈ T3 , and the 3A6 case   ∼  ∈ O3 3 (L), happens only with L = 3O  N , which is impossible as then x contradicting the fact that K   CG (x). In the Akpm +δ case, since K ∼ = Aδ d 0 and K ∈ Tp , we must have d0 = 3 with δ = 7, or d0 = 5, i.e., p ≤ δ < 2p. But in either case, mp (B) = mp (CG (x)) by definition of B(T ). This implies  + mp (CL (x)) = mp (C  (L)) + that mp (B) = mp (CBL (x)) ≥ mp (CB (L)) B  acts on L with k kpm−1 + c, where c = mp (Aδ ). However, since CB (K)  + km + c, so pm−1 ≤ m, a contradiction regular orbits, mp (B) = mp (CB (L)) as m ≥ 2. The lemma follows.  We next observe  Lemma 11.3. If Q ≤ M , then Q ∈ Sylp (CM (K)).  with Q ≤ Q1 and |Q1 : Proof. Suppose false and choose Q1 ≤ CM (K) Q| = p. Since Q ∈ Sylp (C(x, K)), CQ1 (x) = Q. Let z ∈ Ip (Z(Q1 )), so that z ∈ Z(Q). Then as (x, K) is Ipo -terminal, the definition of that term [IG , 6.9] yields Q ∈ Sylp (C(z, Kz )), a contradiction as Q1 ≤ C(z, Kz ). The lemma follows.  To complete the proof of Proposition 11.1, we therefore assume that Q ≤ M and argue to a contradiction. Next, using M , we can improve our balance estimates when d0 = 5. Lemma 11.4. If p = 3 and d0 = 5, then m3 (Q) ≤ 6 and m3 (CG (x)) ≤ 8.

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Proof. Recall that by the definition of B(T ), BQ = z × Q∗ , where

z = Ω1 (P ) and Q∗ = Q b, with b3 = 1 and b inducing a field automorphism on K. We have m3 (B) = m3 (T1 ) ≥ 1 + m3 (Q). Suppose that m3 (Q) ≥ 7. Then m3 (B ∩ Q∗ ) ≥ m3 (B) − 1 ≥ m3 (Q) ≥ 7. By [IG , 10.17], there exists A  Q∗ such that A ∼ = E34 . Set D = CB∩Q∗ (A). As m3 (Aut(A)) = m3 (GL4 (3)) = m3 (L4 (3)) = 4 [IA , 3.3.3], m3 (D) ≥ 3. Hence B is 4-connected to A by the chain B, z D, z A, A. By Proposition 9.3, NG (A) ≤ M . As Q ≤ NG (A) but Q ≤ M , this is a contradiction, so m3 (Q) ≤ 6. As m3 (Aut(K)) ≤ 2 by [V17 , 3.1], m3 (CG (x)) = m3 (T ) = m3 (T1 ) ≤ 8. The proof is complete.  Now we can prove Lemma 11.5. If d0 = 5 and r = 4, then G is 5/2-balanced with respect to B. Proof. Suppose false. Then by Lemmas 4.2 and 4.1a, for some y ∈ B # and p-component J of CG (y), J/Op (J) is not locally 2-balanced for p, and y = Cy /Op (Cy ). Then Lemma 4.1b not B-invariant. Set Cy = CG (y) and C  and the action of B on it. But applies to give the isomorphism type of J,  J)  is not isomorphic to L (q),  = ±1, q ≡  (mod 3) since d0 = 5, J/Z( 3 (when p = 3) or to F i22 (when p = 5). Therefore p = 3 and J ∼ = A11 . If x normalizes J, then it normalizes each B-conjugate of J, and so m3 (CJ B  (x)) ≥ 9, contradicting Lemma 11.4. Hence x does not normalize J, so any 3-component Jx of L3 (CJ B  (x)) satisfies Jx ∼ = J ∼ = A11 . Let I be the subnormal closure of Jx in CG (x). Then I is not a diagonal pumpup of Jx as m3 (Q) ≤ 6, so as A11 ∈ TG3 , I must be a flat TG3 -group. Hence I∼ = A11 . Again as m3 (Q) ≤ 6, I  CG (x). Thus Q = (Q ∩ I) × CQ (I). Since E = B∩Q ∈ E∗ (Q) and Q∩I ∼ = Z3 Z3 , B contains D := J(Q∩I) ∼ = E33 . Now D x  Q so Q ≤ NG (D x) ≤ ΓB,4 (G) ≤ M , a contradiction completing the proof.  Finally we prove Lemma 11.6. We have Q ≤ M . Proof. If d0 = 1a, then B contains some U ≤ K with U ∼ = Ep2 , and r = 2, so Q ≤ NG (U ) ≤ ΓB,2 (G) ≤ M . If d0 = 1g, 1j, 2, 3, or 4, then B ≥ B1 × U with B1 ≤ K and Ep2 ∼ = U  Q, so Q ≤ NG (B1 U ) ≤ ΓB,4 (G) ≤ ΓB,r (G) ≤ M , as r ≤ 4. Finally if d0 = 5, then B ≥ B0 := (B ∩ K) × U with Ep2 ∼ = U  Q. Then if r = 3, Q ≤ NG (B0 ) ≤ ΓB,r (G) ≤ M . So suppose that r = 4. Now G is 5/2-balanced with respect to B by Lemma 11.5, so using [V9 , 3.2b2], if we let Θ5/2 be the 5/2-balanced functor on B, the completions Θ5/2 (G; B) and ΘB (G; B) (where ΘB is the 3-balanced signalizer functor) coincide. Hence Q ≤ NG (B0 ) ≤ NG (Θ5/2 (G; B)) = NG (ΘB (G; B)) = M , completing the proof. 

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Lemma 11.6 contradicts our assumption that Q ≤ M and completes the proof of Proposition 11.1. 12. The L-Preuniqueness of M : the Nonsimple Case We are now in a position to prove that M is an L-preuniqueness subgroup of G–that is, a p-component preuniqueness subgroup of G with respect to L. We treat the three cases K nonsimple, K simple of p-rank 2, and K (simple) of p-rank 1 separately. Here we prove Proposition 12.1. If d0 = 1a, then M is an L-preuniqueness subgroup of G. Under this assumption, we have p = 3 and K ∼ = SL3 (q), q ≡  (mod 3),  = ±1, 3A6 , 3A7 , (12A) or 3M22 . By Lemma 11.2 and our choice of (x, K), we have x ∈ O3 3 (K) ≤ O3 3 (L) in these cases. Also, either B ≤ K ∗ or B ≤ KU , where K = K1 , K2 , . . . , Km , K ∗ = K1 K2 · · · Km , and U are defined in Section 5, and the two possibilities for B correspond to the cases m > 1 and m = 1. Moreover, U ≤ B in either case. Furthermore, as defined in Section 8, r = 2, so by Proposition 9.2, ΓB,2 (G) ≤ M . We set B1 = B ∩ K, so that B1 ∼ = E32 and NG (B1 ) ≤ M .  where M

= M/O3 (M ). Hence By Proposition 11.1, Q ∈ Sylp (CM (L)), by definition of the term “p-component preuniqueness subgroup” [II3 , 1.1], to establish the proposition it remains to prove that ΓQ,1 (G) ≤ M . We first prove Lemma 12.2. We have NG ( x) ≤ M . Proof. Set N = NG ( x) and N = N/O3 (N ). Then M contains ∗ O3 (N ) ≤ ΓB,2 (G), K, and NG (B1 ), hence N0 := K ∗ CN (K ) ≤ M . Let ∗ ∗ ∗ T ∈ Syl3 (K ), so that by a Frattini argument, N = K NN (T ∗ ). But if m3 (T ∗ ) ≥ 4, then by Proposition 9.6, NG (T ∗ ) ≤ M so N ≤ M , as desired. Hence we may assume that m3 (K ∗ ) ≤ 3, so K ∗ = K1 or K1 K2 . In particular T normalizes every 3-component of K ∗ , and N/N0 has a subgroup of index at most 2 embeddable in Out(K1 ) or Out(K1 ) ×Out(K2 ). By [V17 , 11.3.2a], Out(Ki ) is 3-closed for each i, so N/N0 is 3-closed. Therefore N = N0 T NN (T ) = N0 NN (T ), so it suffices to show that NN (T ) ≤ M . Let g ∈ NN (T ). Now B = B1 U where B1 ∼ =U ∼ = E32 , B1  T , U  T , B1 ≤ K1 ∩ T , and U ≤ K2 ∩ T or CT (K1 ), depending on whether K ∗ = K1 or K ∗ = K1 . Then B g has a factorization satisying the same conditions as the factorization B = B1 U ; if g interchanges K1 and K2 , then B g = U g B1g , with U g and B1g in the roles of B1 and U , respectively, and otherwise B g = B1g U g , with B1g

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and U g in the roles of B1 and U , respectively. Moreover, B g is 2-connected to B by a chain of elements of B(T ). For instance if K1g = K2 , a chain is B g = U g B1g , U g U, B1 U = B. It follows therefore from Proposition 9.3 that ΘB (G; B)g = Θ3/2 (G; B)g = Θ3/2 (G; B g ) = Θ3/2 (G; B) = ΘB (G; B), so g ∈ NG (ΘB (G; B)) ≤ M . This completes the proof.



We next prove Lemma 12.3. Let 1 = V ∈ E3 (Q). If O3 (CG (V )) ≤ M , then NG (V ) ≤ M. Proof. Since K is I3o -terminal in G, K covers J/O3 (J), where J denotes the pumpup of K in CG (V ). Also as Q centralizes B1 , we have B1 ≤ J. Since O3 (J) ≤ O3 (CG (V )) ≤ M by hypothesis, it follows that J ≤ M . But B1 centralizes L3 (CG (V ))O3 (CG (V ))/JO3 (CG (V )), so as NG (B1 ) ≤ M , this in turn implies that L3 (CG (V )) ≤ M . Now CG (V )  NG (V ), so NG (V ) induces a group of permutations of the 3-components of CG (V ). We let J = J1 , J2 , . . . , Jn be all the NG (V )conjugates of J and put J ∗ = J1 J2 · · · Jn . Then J ∗ ≤ M by the preceding paragraph and J ∗  NG (V ). Furthermore, as x ∈ B1 , x ∈ O3 3 (J) and hence x centralizes J ∗ /O3 (J ∗ ). We let Hi be the pumpup of Ji in CG (x), 1 ≤ i ≤ n, so that H1 = K. We claim that each Hi is a 3-component of K ∗ . Indeed, for a given i, we have Ji = J g for some g ∈ NG (V ). Then xg ∈ O3 3 (Ji ) and we can choose g so that x centralizes xg . Since K is the pumpup of J in CG (x), likewise K g is the pumpup of Ji in CG (xg ). Moreover, as K is I3o -terminal in G, so also is K g . But x ∈ CG (xg ) centralizes K g /O3 (K g ) as x centralizes Ji /O3 (Ji ). Hence if Ni denotes the pumpup of K g in CG (x), then Ni /O3 (Ni ) ∼ = K g /O3 (K g ) ∼ = K as K g is I3o -terminal, so Ni is a 3∗ component of K by definition of K ∗ . On the other hand, if we set Ii = L3 (CJi ( x, xg )), then Ii covers Ji /O3 (Ji ), whence Ii ≤Hi ∩K g . Therefore

Ii ≤ Ni and Ii covers Ni /O3 (Ni ). But also Hi = IiHi and as Ni is subnormal in CG (x), it follows that Hi ≤ Ni , forcing Hi = Ni . Thus Hi is a 3-component of K ∗ , proving our claim. If mp (J ∗ ) ≤ 3, then clearly x is a Sylow 3-subgroup of O3 3 (J ∗ ). But then NG ( x) covers NG (V )/O3 (J ∗ ) by a Frattini argument. Since NG ( x) ≤ M by Lemma 12.2 and O3 (J ∗ ) ≤ M , it follows that NG (V ) ≤ M , as asserted. On the other hand, if mp (J ∗ ) ≥ 4, then a suitable product K1∗ of Ki ’s shares a Sylow 3-subgroup X with J ∗ by the preceding paragraph. Moreover, X is then invariant under a suitable K ∗ -conjugate of B. But then NNG (V ) (X) permutes the set {X ∩ J1 , . . . , X ∩ Jn } so lies in M by Proposition 9.6 as mp (K1∗ ) = mp (J ∗ ) ≥ 4. However, NNG (V ) (X) covers NG (V )/J ∗ by a Frattini argument and J ∗ ≤ M , so NG (V ) ≤ M in this case as well, and the lemma is proved. 

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Lemma 12.4. Set Q0 = CQ (U ). Then we have (a) CG (t) ≤ M for every t ∈ I3 (Q0 ); and (b) If V ∈ E3 (Q) and for some 1 = V0 ≤ V , mp (CQ0 (V0 )) ≥ 2, then NG (V ) ≤ M . Proof. We first prove that CG (u) ≤ M for every u ∈ U # . We have B ≤ CG (u) with mp (B) ≥ 3 and ΓB,2 (G) ≤ M , so O3 (CG (u)) ≤ M by [V9 , 9.2]. But then CG (u) ≤ NG ( u) ≤ M by the preceding lemma. Now let t ∈ I3 (Q0 ). Then U ≤ CG (t) acts on O3 (CG (t)), whence again by [V9 , 9.2] and the preceding paragraph, O3 (CG (t)) ≤ ΓU,1 (G) ≤ M . Thus CG (t) ≤ NG ( t) ≤ M by the preceding lemma, proving (a). Similarly let V, V0 be as in (b) and let U0 be an E32 -subgroup of CQ0 (V0 ). By (a), we have ΓU0 ,1 (G) ≤ M as U0# ⊆ I3 (Q0 ). But U0 ≤ CG (V0 ) acts on O3 (CG (V0 )), so again by [V9 , 9.2] and the preceding lemma, NG (V0 ) ≤ M . In particular, O3 (CG (V )) ≤ CG (V0 ) ≤ M as V0 ≤ V . Once again the  preceding lemma implies that NG (V ) ≤ M , so (b) also holds. Now we assume the proposition fails. Thus there is 1 = V ≤ Q such that NG (V ) ≤ M . Set V1 = Ω1 (Z(V )). Since V1 char V , NG (V ) ≤ NG (V1 ), so NG (V1 ) ≤ M . Hence without loss, we can assume that V = V1 , whence V ∈ E3 (Q). Using the preceding lemma, we obtain Lemma 12.5. The following conditions hold: (a) V ∼ = Z3 : (b) Ω1 (CQ (V )) = x, V  ∼ = E32 ; and (c) [O3 (CG (V )), x] ≤ M . Proof. If m3 (V ) ≥ 2, then as U  Q, V0 = CV (U ) = 1. Thus m3 (CQ0 (V0 )) ≥ 2 (as U ≤ Q0 ), so NG (V ) ≤ M by the preceding lemma, contrary to the choice of V . Thus V ∼ = Z3 , proving (a). The same argument proves that m3 (CQ0 (V )) ≤ 1, whence x = Ω1 (CQ0 (V )). In particular, V ≤ Q0 , so Q = Q0 V (as |Q : Q0 | ≤ 3) and consequently

x, V  = Ω1 (CQ (V )) ∼ = E32 , proving (b). Finally, set X = O3 (CG (V )). By [V9 , 9.1], X = [X, x]CX (x). But CX (x) ≤ CG (x) ≤ M by Lemma 12.2. Hence if (c) fails, then X ≤ M , whence NG (V ) ≤ M by Lemma 12.3, contrary to the choice of V . Thus (c) also holds.  As a consequence, we obtain Lemma 12.6. For some y ∈ B1 − x , CG (y) contains a V -invariant 3component J such that J ≤ M and J/O3 (J) is not locally balanced with respect to V . Proof. Again set X = O3 (CG (V )). By Lemma 12.5c, [X, x] ≤ M . By [V9 , 9.1], [X, x] = [CX (y), x] | y ∈ B1 − x, so for some such y, if we set Xy = [CX (y), x], we have Xy ≤ M . But B ≤ CG (y) and O3 (CG (y)) ≤

94

10. THEOREMS C6 AND C∗6

ΓB,2 (G) ≤ M , so Xy ≤ O3 (CG (y)). On the other hand, Xy ≤ O3 (CG (V ))∩ CG (y) and V ∼ = Z3 by Lemma 12.5a. Hence by [IG , 20.6], CG (y) contains an Xy V -invariant 3-component J such that Xy normalizes but does not centralize J/O3 (J). In particular, J/O3 (J) is not locally balanced with respect to V . Hence to complete the proof, it remains to prove that J ≤ M . Suppose false. By L3 -balance, J ≤ L3 (M ). But x ∈ O3 3 (K) ≤ O3 3 (L), so x centralizes J/O3 (J). However, Xy = [Xy , x], so Xy centralizes J/O3 (J), which is not the case. Thus J ≤ M , and the lemma is proved.  Note that as y ∈ B1 and B1 centralizes Q, we have Q ≤ CG (y). Since J ≤ M , this in turn implies: Lemma 12.7. The following conditions hold: (a) Ω1 (Q0 ) leaves J invariant and acts faithfully on J/O3 (J); (b) J/O3 (J) ∼ = L2 (27), L3 (4) (with U inducing inner automorphisms on J/O3 (J)), A10 (with U acting on 10 letters with a regular orbit), or M12 ; and (c) K ∗ = K. Proof. By Lemma 12.3a, Y = ΓQ0 ,1 (G) ≤ M , so J ≤ Y . Since mp (Q0 ) ≥ 2, (a) now follows from [V9 , 9.3]. In particular, x normalizes J. But Xy = [Xy , x] ≤ O3 (CG (V )) ∩ CG (y) acts nontrivially on J/O3 (J). Thus, J/O3 (J) is not even weakly locally balanced with respect to V and x. Moreover, U V ∼ = 31+2 as V ∼ = Z3 normalizes, but does not centralize U . Since J/O3 (J) is either a C3 -, a T3 -, or a TG3 -group, with J ≤ Y , we conclude from [V17 , 6.15] as in Lemma 7.8 that J/O3 (J) ∼ = L2 (27), (S)L3 (4), A10 , or M12 , with the action of U as claimed. In the SL3 (4) case as well, which it remains to rule out in order to prove (b), U induces inner automorphisms on J/O3 (J). Before completing the proof of (b) we prove (c). Suppose K ∗ > K, in which case K2 exists. Since y ∈ B1 ≤ K, y centralizes K 2 . Let J2 be the pumpup of I2 = L3 (CK2 (y)) in CG (y), so that I2 is a 3-component of CJ2 (x) and I 2 = K 2 is nonsimple. Since x leaves J invariant, either J2 ≤ J or J2 centralizes J/O3 (J) by L3 -balance. But clearly CJ (x) does not have a 3component of the shape of I2 , so J2 must centralize J/O3 (J). On the other hand, if we set Q2 = Q ∩ K2 , then Q2 ∈ Syl3 (K2 ) and Q2 is nonabelian, so [Q2 , Q2 ] = 1. But clearly [Q2 , Q2 ] ≤ Q0 . Since Q2 ≤ I2 centralizes J/O3 (J), we see that (a) is contradicted. Thus (c) also holds. Returning to (b), we claim that for any u ∈ U # , Ku  CG (u). Namely, suppose g ∈ CG (u) and Kug = Ku . We may assume g chosen so that [x, xg ] = 1. The pumpup of Kug in CG (xg ) is then K g , and (xg , K g ) is I3o (G)-terminal. Now x centralizes K g /O3 (K g ), so the pumpup L of K g in CG (x) is a trivial pumpup. Hence L ≤ K ∗ = K by (c), so L = K. Now I := L3 (CL ( x, xg , u)) covers both Ku /O3 (Ku ) and Kug /O3 (Kug ), so I has

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(at least) two composition factors isomorphic to K. However, as I ≤ L, I has only one such composition factor, contradiction. This proves our claim. Now if J/O3 (J) ∼ = SL3 (4), we can repeat the last paragraph of the proof of Lemma 7.8 to reach a contradiction, so (b) also holds.  y = Cy /O3 (Cy ), we see that If we set V = t, Cy = CG (y), and C we have reached the same situation encountered in Lemmas 7.7 and 7.8 (with the additional conclusion K ∗ = K here). Moreover, the arguments of Lemmas 7.9–7.14 (and the subsequent argument) will apply with only minor changes to yield a final contradiction. Indeed, first of all, the proof of Lemma 7.9 applies verbatim to yield:  Lemma 12.8. Q leaves J invariant and acts faithfully on J. We follow the line of argument of Lemmas 7.10–7.12 to prove Lemma 12.9. We have J ∼ L2 (27). = Proof. Assume false and again let A be a Q-invariant Sylow 3-subgroup of J, so that A ∼ = E27 . The proof of Lemma 7.10 applies without change to yield that A can be chosen so that (12B)

A ∩ Q = U or Q0 .

Again we let D be a cyclic 13-subgroup of NJ (A) with D/CD (A) ∼ = Z13 . ∗ In the present situation, the fact that K = K allows for a quicker finish to the proof. Namely, choose e ∈ (A ∩ Q) − x. Then there is d ∈ D # such  d that e = x . Then H := K d is an I3o -terminal 3-component of CG (e). But Ke is also a 3-component of CG (e), and x ∈ O3 3 (Ke ), so x centralizes H/O3 (H). By the 3-terminality of H and the definition of K ∗ , the pumpup of H in CG (x) is a 3-component of K ∗ . As K ∗ = K, also H = K. But

e ∈ Syl3 (O3 3 (H)), and [e, x] = 1, so e = x, contradicting our choice of e. The proof is complete.  Next, the proof of Lemma 7.13 applies without change to yield that ∼  J= A10 , whence Lemma 12.10. We have J ∼ = L3 (4) or M12 . Finally, we eliminate these last two cases, essentially as in the corresponding cases of Proposition 7.2. This time Q = U t ∼ = 31+2 , while ∗ K = K is T -invariant and Q  T . We first establish the analogue of (7D). Lemma 12.11. x, t is not normal in T . Proof. If false, then B ∗ := B1 x, t ∈ B(T ) and mp (B ∩ B ∗ ) ≥ mp (B1 ) = 2. Hence by Proposition 9.4, ΓB ∗ ,2 (G) ≤ M . On the other hand, B ∗ ≤ CG (y) as y ∈ B1 centralizes x, t. Note that if J ∼ = L3 (4), then from Lemma 12.7 we deduce that t induces an outer automorphism on J. In any case, since J is not locally balanced with respect to t and t ∈ B ∗ , we conclude therefore from [V17 , 11.4.2] and [V17 , 10.11.4b] that J ≤ ΓB ∗ ,2 (G), whence J ≤ M , contradiction. 

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Lemma 12.12. We have J ∼ = L3 (4) or M12 . Proof. Suppose false. As K is T -invariant, we can assume without loss that B1  T . We write B1 = b, x and repeat the second and third paragraphs of the proof of Lemma 7.14 to conclude once again that x ∈ [Q, a, Q] for some a ∈ CT (y).

(12C)

 has We have Q a ≤ CG (y). Since Q acts faithfully on J and Aut(J) Sylow 3-subgroups, a cannot leave J invariant, because of (12C). But again x induces an inner automorphism on J as x ∈ [U t , U t], and it follows that mp (CG (x)) ≥ 7, which is not the case. The proof is complete.  31+2

The contradiction from Lemmas 12.10 and 12.12 establishes Proposition 12.1. 13. The L-Preuniqueness of M : the Simple Rank 2 Case In this section we continue the notation B, r, and M of the previous section and we prove Proposition 13.1. Suppose that d0 = 1g, 1j, 2, 3, or 4. Then M is an L-preuniqueness subgroup of G. In these cases we have either (1) p = 3 and K ∼ = G2 (8), HJ, L3 (q), q > 2, q ≡  (mod 3),  = ±1, A7 , M12 , or M22 ; or (2) p = 5 and K ∼ = F i22 .

(13A) In particular, (13B)

K is simple and mp (K) = 2.

Indeed by definition of d0 (and even in the exceptional situation), (13C)

If d0 = 1g or 1j, then every Tp -group in Lop (G) is simple.

Again we must prove ΓQ,1 (G) ≤ M, and we set

= M/Op (M ). M

We need some preliminary results. Note that if we write B = B1 E ≥ B1 U as in the definition of B(T ), then B1 U  T1 and so T1 ≤ ΓB,r (G) ≤ M , as r ≤ 4. Expand T1 to S ∈ Sylp (M ). Thus R ≤ S. First, the following lemma allows us to replace (x, K) by (x∗ , Kx∗ ) whenever x∗ ∈ Ip (Z(Q)), without changing B or M . Lemma 13.2. Suppose that x∗ ∈ Ip (Z(Q)). Then (x∗ , Kx∗ ) ∈ TJop (G) and B0 (T ) = B0 (T ∗ ) for any T ∗ ∈ Sylp (CG (x∗ )) such that R ≤ T ∗ .

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Proof. As (x, K) is Ipo -terminal, Q ∈ Sylp (C(x∗ , Kx∗ )) by definition of Ipo -terminality. Likewise, if (x, K) is Ipo -rigid, then so is (x∗ , K ∗ ). Then as (x, K) ∈ TJop (G), it is clear from the definition of this set (Definition 1.9) that (x∗ , Kx∗ ) ∈ TJop (G). Now in the definition of B0 (T ) (following (6G)), or B0 (T ∗ ), allowability is only defined in terms of the component K, or the component K ∗ , respectively. So for B1 ∈ E2 (P ), the two notions of allowability coincide. Both B0 (T ) and B0 (T ∗ ) consist of all products B1 E with B1 ≤ P allowable and  E ∈ E∗ (Q), so B0 (T ) = B0 (T ∗ ). Since L/Op (L) ∼ = K, [V17 , 15.2] yields the following connnection result: Lemma 13.3. Let r = 3 or 4, and let A ∈ Ep∗ (S), so that mp (A) ≥ r + 1. Then there exists a chain A = A0 , A1 , . . . , An−1 , An = B such that Ai ∈

Ep∗ (S)

∪ B0 (T ) and mp (Ai ∩ Ai+1 ) ≥ r for all 0 ≤ i < n.

∼ K, A normalizes  = Proof. By [V17 , 19.1b] and the simplicity of L ∗       L, Q = CS(L), and P = S ∩ L. Choose x ∈ Ip (CZ(Q) (A)) (perhaps x∗ = x). By Lemma 13.2, we may replace x by x∗ if necessary to assume that B ∈ B0 (T ) and AP Q ≤ T . Note that if the membership B ∈ B0 (T ) required one of the nontrivial conditions (6B) for B ∩ P to be allowable, then in fact B ∩ P  T1 , by [V17 , 5.10]. We now apply [V17 , 15.2] to the  normal P Q,  with B  in the role of B, U  (any Ep2 subgroup of Q group X = A  in the role of U , and B  ∩ P in the role of V if B  ∩ P  T (otherwise, in X)   any normal Ep2 in P will do). The desired conclusion follows directly. Using Proposition 6.1, we easily prove Lemma 13.4. We have S ∈ Sylp (G). Proof. The argument is simple if either (13D)

(1) r = 4; or (2) r = 3 and G is 5/2-balanced with respect to every element of Ep∗ (S).

For then we can r-connect any A ∈ Ep∗ (S) to B as in Lemma 13.3, and conclude from Proposition 4.3 and Proposition 6.1a that Θ(G; A) = ΘB (G; B), where Θ = Θ3 or Θ5/2 according as r = 4 or 3. As NG (S) permutes Ep∗ (S), it normalizes ΘB (G; B) and hence lies in M . But S ∈ Sylp (M ), so S ∈ Sylp (G). For any A ∈ Ep∗ (S) for which (13D) fails, A normalizes L by [V17 , 19.1], and so replacing x by an element of CQ (A), thanks to Lemma 13.2, we may assume that A ≤ T . Then by Proposition 6.1b, p = 3, mB = 4, and K∼ = L3 (q), q ≡  (mod 3),  = ±1, or G2 (8).

10. THEOREMS C6 AND C∗6

98

 Thus mp (T ) = 4. Then S leaves L, P = S ∩ L, and Q = CS (L) invariant, so T = T1 ≤ M . Replacing x by an element of I3 (Z(S)) ∩ Q, which is again permitted by Lemma 13.2, we may assume that S = T . Then by the definition of B(T ) and [V17 , 5.10], B(T ) consists of those elements B ∈ E3∗ (P Q) such that B  T , when K ∼ = L3 (q); when K ∼ = G2 (8), the condition B  T is relaxed to just B ∩ Q  T and B ∩ K  P . Now choose V ≤ P and U ≤ Q with V ∼ = U ∼ = E32 , V  T , and ∗ ∗ U  T , and set B := V U  T . Then B ∈ B(T ). By Proposition 9.7, ΘB ∗ (G; B ∗ ) = ΘB (G; B). Expand S to W ∈ Syl3 (G) and let w ∈ NW (S). Set A = (B ∗ )w , so that A ∈ E∗ (T ) and A  T . By Proposition 6.1, G is 5/2-balanced with respect to B ∗ and hence to A. If A ≤ P Q, then the chain A, V CA (V ), V U = B ∗ 3-connects B ∗ to A by normal subgroups of T which are therefore in B(T ). Hence G is 5/2-balanced with respect to all subgroups in the chain, so ΘB ∗ (G; B ∗ ) = Θ5/2 (G; A) = ΘB ∗ (G; B ∗ )w . Thus w ∈ NG (ΘB ∗ (G; B ∗ )) ≤ M in this case. If A ≤ P Q, then by [V17 , 11.3.2b], A0 := A ∩ P Q ∼ = E33 . Let V0 and U0 be the projections of A0 on P and Q, respectively. Then A0 , V0 , and U0 are normal in T . If |V0 U0 | = 34 , let A1 = V0 U0  T , so that A1 ∈ B(T ). Then arguing as in the last paragraph but with the chain (13E)

A, A1 , V CA1 V, B ∗

we deduce that w ∈ M . Finally if |V0 U0 | = 33 , we expand V0 U0 to an arbitrary A1 ∈ E∗ (P Q) such that A1  T , and again use the chain (13E) to conclude that w ∈ M . Thus in any case, NG (S) ≤ M , so S ∈ Syl3 (G). The lemma is proved.  Lemma 13.5. One of the following holds: (a) G is 5/2-balanced with respect to every element of Ep4 (G); or (b) Q is connected of rank ≥ 4. Proof. Suppose (a) fails for some A ∈ Ep4 (G), and let (a, J) be a 5/2a = Ca /Op (Ca ). Then a ∈ A# , obstruction for A. Set Ca = CG (a) and C and by [V17 , 4.9], p ∈ {3, 5} and one of the following holds: (13F)

∼ L (q), q ≡   J)  = (1) p = 3 and J/Z( 3 ±1; or (2) (p, J) ∼ = (3, A11 ) or (5, F i22 ).

(mod 3),  =

Also J is weakly locally 2-balanced by [V17 , 4.9], so some a ∈ A does not leave J invariant. Furthermore, replacing a by a suitable G-conjugate, we can assume without   loss that S0 = CS (a) ∈ Syl3 (CG (a)). ∗ Set J = J a  , so that J ∗ is the product of p p-components J = J1 , J2 , . . . , Jp cycled by a .   ≥ 4, and We claim that mp (Q) ≥ 4. If L  Op (M ), then mp (CM (L))  p  so mp (Q) ≥ 4 as Q ∈ Sylp (CM (L)). So assume that L  O (M ). Then Q  S so there is 1 = z ∈ Q ∩ Z(S) ≤ Z(Q). By Lemma 13.2, we may

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99

replace x by z and assume that x ∈ Z(S). On the one hand, consider the case p = 3, J ∼ = SL3 (q). As x ∈ Z(S0 ), x centralizes J ∗ /O3 (J ∗ ) by [V17 , ∗ 12.1]. Let I be the subnormal closure of I0 = L3 (CJ ∗ (x)) in CG (x). Now I0 = I1 I2 I3 , where Ii = L3 (CJi (x)). As m3 (K) = 2, at least two of I1 , I2 , I3 , say the first two must centralize K, and indeed unless K ∼ = G2 (8), all three centralize K. If mp (Q) < 4, then K ∼ = G2 (8), p = 3, and x ∈ I1 ∩ I2 . But then [x, a ] = 1 and so [K, I 0 ] = 1, whence m3 (Q) ≥ m3 (I0 ) ≥ 4. On the other hand, in every other case J∗ is the product of p components which a ) ≥ 7, so mp (S) ≥ 7. are simple of p-rank at least 2. It follows that mp (C By Lemma 3.5, this forces mp (Q) ≥ 4, completing the proof of our claim. In particular, mp (B) ≥ 5. Another consequence, by [IA , 10.22(iii)], is that if p = 3, then Q is connected. So we may now assume that p = 5. But now if we let B ∗ ∈ E5∗ (J ∗ ) with B ∗ ≤ S0 , then B ∗ is 4-connected to B by Lemma 13.3 and consequently ΓB ∗ ,4 (G) ≤ M by Propositions 4.3 and 9.3. However, clearly mp (CB ∗ (Ji )) ≥ 4, 1 ≤ i ≤ 5, so J ∗ ≤ M . By  for at least 4 values of i, L5 -balance, J ∗ ≤ L5 (M ). Hence Ji centralizes L  1 ≤ i ≤ 5. Thus, by [V17 , 13.20], CM (L) possesses at least four components H1 , H2 , . . . isomorphic to F i22 . Let y ∈ I5 (Q). If y does not normalize some Hi , then CM (y) contains a copy of F i22 , so m5 (CQ (y)) ≥ 3. The same inequality clearly holds if y normalizes H1 and H2 . Hence Q is connected. This completes the proof of the lemma.  Suppose now that the proposition fails, in which case again there is an elementary abelian subgroup V ≤ Q such that (13G)

NG (V ) ≤ M.

Expand V to W ∈ Sp (Q) and set A = B1 W , where again B1 = B ∩ K. We have x ∈ W and mp (W ) ≥ 2 (the latter as mp (Q) ≥ 2), so mp (A) ≥ 4. Furthermore, if Q is connected of rank ≥ 4, then mp (W ) ≥ 3 and mp (A) ≥ 5. Now with Lemma 13.5, we have

(13H)

(1) If mp (A) ≥ 5, then mp (B) ≥ 5 and A and B are 4-connected in S; (2) If mp (A) = 4, then A and B are 3-connected in S, and G is 5/2-balanced with respect to every element of E4 (G).

Indeed, if mp (A) = 4 then mp (A ∩ B) ≥ mp (B1 Z(Q)) ≥ 3, and (13H2) holds by Lemma 13.5. If mp (A) ≥ 5, on the other hand, then mp (B ∩ Q) = mp (Q) ≥ mp (A ∩ Q) ≥ 3, so B ∩ Q and A ∩ Q are 2-connected in Q. As B = B1 × (B ∩ Q) and A = B1 × (A ∩ Q), (13H1) follows. Correspondingly set h = 4 or 3,

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100

so that by Propositions 9.3 and 4.3, we have (13I)

ΓA,h (G) ≤ M.

Using our general signalizer functor theory results, we reduce to the following situation. Lemma 13.6. The following conditions hold: (a) G is not 5/2-balanced with respect to A; and (b) CG (B1 V ) ≤ M .  = N/Op (N ). Also let H be the Proof. Set N = NG (V ) and N  ∼ pumpup of K in CG (V ), so that H is a p-component of N and H = K o as K is Ip -terminal. Furthermore, B1 ≤ H as B1 centralizes V ≤ Q. Set    ∗ ), so that H ∗ and Y are each normal in H ∗ = H N and Y = CN (V H N . Since A is abelian with V ≤ A, A ≤ N , and we expand B1 V to an A-invariant Sylow p-subgroup X of Y H ∗ . Finally, put A1 = B1 V = B1 × V , so that mp (A1 ) ≥ 3. Suppose now that (b) fails, so that CG (A1 ) ≤ M . Since Op (N ) ≤  ∗ = HC    ∗ (A1 ), and M covers H  ΓA,h (G), Op (N ) ≤ M by (13I). But Y H YH  Since CG (A1 ) ≤ M , it follows that M covers Y H  ∗, as K ≤ M covers H. ∗ ∗ and as Op (N ) ≤ M , we conclude that Y H ≤ M . But N = Y H NN (X) by a Frattini argument, so NG (X) ≤ M by (13G). On the other hand by [V9 , 5.14], if mp (A) ≥ 5, then all elements of Ep5 (X), B among them, are 4-connected in X, while if mp (A) = 4, then Q is not connected, and all elements of Ep4 (X), B among them, are 3-connected. We conclude that NG (X) ≤ NG (ΘB (G; B)) ≤ M , using Proposition 4.3 and Lemma 13.5. This contradiction completes the proof of (b). But then by (13I), as mp (A1 ) ≥ 3, we have h = 4 and mp (A1 ) = 3. In particular mp (A) ≥ 5. Suppose G is 5/2-balanced with respect to A. As A is 4-connected to B by (13H), and mp (A1 ) ≥ 3, it follows from Proposition 9.3 and [V9 , 3.2b2] that Θ5/2 (G; A) = Θ3 (G; A) = Θ3 (G; B) = ΘB (G; B). Thus NG (A1 ) ≤ ΓA,3 (G) ≤ NG (ΘB (G; B)) ≤ M , so (a) also holds and the lemma is proved.  We shall derive a contradiction by arguing that CG (B1 V ) ≤ M . Clearly we need only prove that CG (B1 ) ≤ M . Since G is not 5/2-balanced with respect to A, h = 4 and mp (A) ≥ 5, mp (B) ≥ 5. A = CA /Op (CA ). Lemma 13.7. Let 1 = A0 ≤ A, CA0 = CG (A0 ), and C 0 0  0 ∗ Suppose that CA0 has a p-component J such that J := J A > J and J is  not locally 2-balanced with respect to NA (J). Then J ∗ ≤ CM (L). Proof. Suppose false and choose a counterexample maximizing |CA0 |3 .  Fix a ∈ A such that J a = J.   J)  ∼ We have J/Z( = L3 (q  ), q  ≡  (mod 3),  = ±1, or A11 , with p = 3, or F i22 with p = 5.

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It suffices to show that J ∗ ≤ M , whence J ∗ ≤ Lp (M ) by Lp -balance. For then, as A leaves L invariant by definition of A, either the lemma holds or else some p-component J1 of J ∗ lies in L. However, as a leaves L invariant  and J1 = J1a , while mp (L) = 2, the latter possibility is excluded. By Lemma 13.5, Q is connected of rank at least 4. As A ≤ B1 × Q, it follows that mp (A) ≥ 5 and A is 4-connected to B. Hence it will suffice to find A∗ ≤ Ep5 (CA0 )) such that A∗ is 4-connected to A and J ∗ ≤ ΓA∗ ,4 (G).  Suppose that J is simple. Then mp (J ∗ A0 ) ≥ 7. Now mp (Aut(J)) ≤ mp (J)+ 1 (see [V17 , 8.4]). Hence by [V9 , 5.17], A is 4-connected to some A∗ ∈ E∗ (J ∗ A0 ). As mp (A∗ ) > pmp (J) ≥ 2p, J ∗ ≤ ΓA∗ ,4 (G) by [V17 , 11.4.1c], as desired.   = 1, whence p = 3 and J ∼ We may therefore assume that Z(J) = SL3 (q  ).

, in which Q  is connected of rank at least Applying [V9 , 5.17], this time to M 4, we conclude that m3 (S) ≥ 6 and all elements of E5 (S) are 4-connected. Thus ΓS,5 (G) ≤ M . If A ∈ S3 (G), we can take A < A∗ ∈ E36 (CG (a)); then J ∗ ∈ ΓA∗ ,4 (G) by [V17 , 11.4.1c], as desired. So assume A ∈ S3 (G). Let SA ∈ Syl3 (G) with CSA (A0 ) ∈ Syl3 (CG (A0 )). By the previous paragraph, SA ≤ M , so we may assume SA ≤ S. Let z ∈ I3 (Z(S)). Then z ∈ A and by [V17 , 12.1], [J∗ , z] = 1. Then by [V9 , 4.4] and our choice of A0 , A0 is 3-central in M , so SA = S. We can then take any A∗ ∈ E3∗ (S). By the previous paragraph, m3 (A∗ ) ≥ 6 and A∗ is 4-connected to B, so J ∗ ≤ ΓA∗ ,4 (G) ≤ M , again using  [V17 , 11.4.1c]. The proof is complete. Now let (a, J) be a 5/2-obstruction for A. Then there is D ∈ E2 (A) leaving J invariant such that XD := [ΔD ∩ CG (a),  A]  leaves J invariant, ∗ A but does not centralize J/Op (J). We set J = J . By Lemma 13.7,  J ∗ ≤ CM (L). In particular, as B1 ≤ K ≤ L, B1 centralizes J/Op (J). Since XD = [XD , B1 ]CXD (B1 ), X1 := CXD (B1 ) does not centralize J/Op (J). But X1 ≤ ΔD with D leaving J invariant. Hence if we let I be a 3-component of the pumpup of J in CG (B1 ), it follows from [V9 , 4.4] that I/O3 (I) is not locally 2-balanced with respect to D. We use the existence of I to attain our objective, thus proving Lemma 13.8. We have CG (B1 ) ≤ M .

  Proof. Set C = CG (B1 ), so that I, A ≤ C. Also let I ∗ = I C .  Let Since a moves J, a moves I. By Lemma 13.7 again, I ∗ ≤ CM (L). ∗ X be an A-invariant Sylow p-subgroup of I B1 . Then as Q is connected, every element of Ep∗ (X) is 4-connected to B in G, whence NG (X) ≤ M by Proposition 9.5. Since C = I ∗ NC (X) by a Frattini argument, we conclude that C ≤ M , as asserted.  Since CG (B1 ) ≤ M , likewise CG (B1 V ) ≤ M , contrary to Lemma 13.6b. This completes the proof of Proposition 13.1.

10. THEOREMS C6 AND C∗6

102

14. The Rank 1 Case: 3/2-Balance In view of Propositions 12.1 and 13.1, to complete the proof of Theorem

= 3, it remains to treat the case d0 = 5. We continue the notation M , M M/Op (M ), and B from the previous sections. In this case we have the following important restriction on elements of Lop (G) by definition of d0 : (14A)

Every element of Lop (G) that is a Tp -group has p-rank 1, or (in the exceptional case only) is isomorphic to A7 with p = 3.

In particular, mp (K) = 1. Furthermore, as usual, by (14A) and the hypotheses of Theorem C∗6 , (14B)

Every element of Lop (G) is a Cp -group, a Tp -group, or a TGp group (flat if p = 3, and of p-rank 2 if in Chev).

Thus, if J ∈ Lop (G) and J is a TGp -group with mp (J) > 2, then J ∈ Alt ∪ Spor. In this section we prove a series of results improving the level of balance with respect to elementary abelian p-subgroups of G. The principal result is Proposition 14.8 below, which is a sharpened version of Lemma 13.5. Recall that R = P Q with P ∈ Sylp (K) and Q ∈ Sylp (C(x, K)). Lemma 14.1. Suppose that p = 3. Then m3 (CG (x)) ≥ m3 (G) − 2. Proof. We have (y, Ky ) ∈ TJo3 (G) for all y ∈ I3 (Z(Q)) ∩ I3o (G), and we fix such an element y to maximize |CG (y)|3 . Without loss we may replace x by y and assume |CG (x)|3 ≥ |CG (z)|3 for all z ∈ I3 (Z(Q)) ∩ I3o (G). Expand T to S ∗ ∈ Syl3 (G). Obviously we are done if T = S ∗ , so assume T < S ∗ . Fix U  S ∗ and E  S ∗ such that U ∼ = E32 and E ∼ = E33 [IG , 10.17]. If U ∩ Q = 1, then as U ∩ Q  Q, there is z ∈ I3 (U ∩ Z(Q)). As z ∈ U , z ∈ I3o (G). Then (u, Ku ) ∈ TJo3 (G) so m3 (Q) = m3 (C(u, Ku )) ≥ m3 (NCG (u) (Ku )) − m3 (Aut(Ku /O3 (Ku ))). But then by [V17 , 19.1, 3.1a], m3 (Q) ≥ m3 (CG (u)) − 2 = m3 (G) − 2. Thus we are done in this case, so we may assume that U ∩ Q = 1. Notice that [P, U ] is then a nontrivial subgroup of P . Since |[S ∗ , U ]| ≤ p, it follows that [S ∗ , U ] = [P, U ] ≤ P , so [Q, U ] ≤ P ∩ Q = 1. In particular, whether Ω1 (Z(Q)) = x or m3 (Z(Q)) > 1, I3 (Z(Q)) ⊆ I3o (G). As [U, Z(Q)] = 1, U ≤ CG (x). Hence Ω1 (QU ) = U × Ω1 (Q) and J(T1 ) = J(QU ) = U × J(Q). Now, J(T ) ≤ T1 by [V17 , 19.1], so [J(T ), J(T )] = [J(Q), J(Q)]. If this commutator group is nontrivial, it contains some w ∈ I3 (Z(NS ∗ (T ))). Then w ∈ Z(Q) and the maximality of |CG (x)|3 is contradicted. Therefore J(Q) is abelian. Next, CQ (E)  Q. Suppose that CQ (E) = 1 and choose 1 = v ∈ CZ(Q) (E). Then in CG (v) we see first that E normalizes Kv , since [P, E, E] = 1; and then that E ∩ Q = 1, as |Ω1 (T1 /Q)| ≤ 32 . Since E  S ∗ , E ∩ Z(Q) = 1, so we may assume that we chose v ∈ E. Now m3 (S ∗ /CS ∗ (E)) ≤ m3 (GL3 (3)) = 2, so m3 (CG (v)) ≥ m3 (G) − 2, as desired. So we may assume

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that CQ (E) = 1, whence Q embeds in GL3 (3) and hence in 31+2 . As J(Q) is abelian and m3 (CG (v) ≥ 4, Q ∼ = E32 . But now Q stabilizes the chain E > U > 1, while [U, P ] = 1. Therefore P Q/Φ(P ) embeds in Aut(E) ∼ = GL3 (3). As P Q/Φ(P ) ∼ = E33 , this is absurd. This contradiction completes the proof.  We expand BR ≤ SL ≤ S ≤ S ∗ , where SL ∈ Sylp (NM (L)), S ∈  Q  SL . Sylp (M ), and S ∗ ∈ Sylp (G). Then as Q ∈ Sylp (CM (L)), Now we can prove Lemma 14.2. The following conditions hold: (a) Any two elements of Ep4 (S) are 3-connected in S; (b) G is 5/2-balanced with respect to any element of Ep4 (G); (c) ΓS,4 (G) ≤ M , and for any A ∈ E4 (S), we have ΓA,3 (G) ≤ M . Moreover, if S0 ≤ S, mp (S0 ) = 3, and every element of Ep3 (S0 ) lies in an Ep4 -subgroup of M , then NG (S0 ) ≤ M ; and (d) S ∈ Sylp (G). Proof. Choose U  SL with Ep2 ∼ = U ≤ Q. Let A ∈ E4 (S). If A does not normalize L, then let A0 be a hyperplane of A containing A ∩ SL = NA (L). Set A1 = A0 D, where D = Ω1 (CR∩K A  (A0 )). Since A0 has p orbits on K A , D ≤ A, and so A1 > A0 , whence A1 ∈ Ep4 (S). As A is 3-connected to A1 , we may assume by descending induction that A normalizes L. Replacing A by AΩ1 (P ) we may assume that A = Ω1 (P )F for some hyperplane F of A. Then F is 2-connected to F0 := U CF (U ) so A is 3-connected to AF0 and hence to Ω1 (P )U , which implies (a). It remains to prove (14C)

G is 5/2-balanced with respect to any element of Ep4 (S).

For if (14C) holds, then by (a) and [V9 , 3.4, 3.2b2], Θ5/2 (G; A0 ) = Θ5/2 (G; A) = Θ5/2 (G; B) = ΘB (G; B) for any A ∈ E4 (S) and any A0 ∈ Ep3 (A). This immediately implies (d), and (c) follows as well. Namely, NG (S) permutes the set Ep4 (S) and thus normalizes ΘB (G; B), so NG (S) ≤ M . Hence S ∈ Sylp (G), and so (b) holds by Sylow’s theorem. Suppose then that (14C) fails, and let (a, J) be a 5/2-obstruction with a ∈ A ∈ Ep4 (S). By Proposition 4.5d, p = 3 and J/O3 (J) ∼ = A11 . Moreover, by [V17 , 4.9a], A does not normalize J, so m3 (G) ≥ 3m3 (A11 ) + 1 = 10. By Lemma 14.1 and the definition of B(T ), m3 (B) ≥ 8, and so m3 (S) ≥ 8. But by [V9 , 5.16], all elements of E38 (S) are 4-connected. As G is 3balanced for p = 3 by Proposition 4.3, (14D)

Θ3 (G; E) = Θ3 (G; B) and so ΓE,4 (G) ≤ M for all E ∈ E38 (S).

Also as NG (S) permutes E38 (S), NG (S) ≤ M and so S = S ∗ ∈ Syl3 (G). Returning to our 5/2-obstruction (a, J), we may assume that S contains a Sylow 3-subgroup of CG (a). Then as m3 (J ∗ a) ≥ 10 we easily see that

104

10. THEOREMS C6 AND C∗6

), and as J/Z(J)  is obviously not J ∗ a ≤ M . By L3 -balance, J∗ ≤ E(M  a  ∗ ∗   and b ∈ I3 (CL∗ ∩S (a)), we involved in K, [J , L] = 1. Letting L = L

. Since A11 has have that b centralizes the subnormal closure of J∗ in M ∗  where we define no proper pumpups satisfying the criterion (14B), J ≤ H, H to be the product of all 3-components of L3 (M ) isomorphic modulo  b] = 1 and Hb := L3 (CH (b)) is a product of 33 -core to A11 . Indeed, [H, components of CG (b), by L3 -balance and the fact that M contains a Sylow 3-subgroup of CG (b), as m3 (CM (b)) ≥ m3 (J ∗ ) ≥ 8. It follows easily that all elements of E5 (S) are 4-connected in S [V9 , 5.15], so in view of Proposition 4.3 and [V9 , 3.4], (14E)

ΓS,5 (G) ≤ M , and ΓF,4 (G) ≤ M for all F ∈ E35 (M ).

We will now use the structure of M to show that M is an L-preuniqueness subgroup of G. This will contradict our assumption that Theorems 3 and 9.1 fail, and complete the proof of the lemma.  and show We let b1 ∈ I3 (S) be such that b1 is the projection of b on L, that (14F)

NG ( b1 ) ≤ M.

Indeed, set N = NG ( b1 ) and N1 = N ∩ M . As m3 (N1 ) ≥ m3 (HL∗ ) ≥ 10, certainly O3 (N ) ≤ N1 and |N1 |3 = |N |3 , by (14E). Let H1 be the product of all 3-components of N isomorphic modulo 3 -core to A11 . Again the criterion (14B) and L3 -balance imply that L3 (CH (b1 )) ≤ H1 and then by a Frattini argument, H1 ≤ HNN (R1 ) ≤ M , for R1 ∈ Syl3 (H). And then N = O3 (N )H1 NN (R2 ) ≤ M for R2 ∈ Syl3 (H1 ), proving (14F).  and set C = CG (d) and C1 = C ∩ M . We need Now let d ∈ I3 (CM (L)) only prove that C = C1 . Replacing d by a conjugate we may assume that b1 ∈ C1 . Using the fact that m3 (A11 ) = 3, we see that m3 (C1 ) ≥ 5, so O3 (C) ≤ C1 and C1 contains a Sylow 3-subgroup of C. Since (x, K) is I3o terminal, Ld = L3 (CL (d)) is a component of C1 . Then with (14F), L3 (C) ≤ C1 . Let RC ∈ Syl3 (L3 (C)O3 3 (C)); it suffices by a Frattini argument to show that NC (RC ) ≤ M , so m3 (RC ) ≥ 5 will suffice. But whether d acts  m3 (L3 (CH (d))O3 3 (CH (d))) ≥ 3, with strict nontrivially or trivially on H, inequality if d normalizes a 3-component of H. As RC contains b1 and d as well, we get the desired inequality m3 (RC ) ≥ 5. The lemma is finally proved.  Lemma 14.3. Suppose that I ∈ TGp (G) = TGp ∩ Lop (G). Then mp (I) = 2 and the Sylow p-subgroups of I are abelian. Proof. Suppose false. Keep in mind that by the hypotheses of Theorem C∗6 , I is flat if p = 3, and mp (I) ≤ 2 if I ∈ Chev. Using [V17 , 2.2], we see

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that I is isomorphic to one of the following groups: (14G)

(1) An , 10 ≤ n ≤ 12, with p = 3; (2) An , 3p + 1 ≤ n ≤ 4p + 1, with p ≥ 5; or (3) F1 , with p = 13.

o We call the set of all such groups TG− p . Notice that if I ↑p I1 ∈ Lp (G), then I1 ∈ Gp (as I ∈ Gp ), so I ∈ TGp (G); since Sylow p-subgroups are nonabelian or of p-rank at least 3, the same holds for I1 , and so I1 ∈ TG− p o (G). by [V17 , 2.2] again. That is, TG− is closed under pumping up in L p p The case (14G3) is a bit different from the others, so we postpone its consideration till the end of the proof. In the other cases, we have the following properties. Let SI ∈ Sylp (I), D = J(SI ), and mD = mp (I).

(14H)

(1) (2) (3) (4) (5)

D is elementary abelian; Every element of Ip (I) has an I-conjugate in D; Out(I) is a p -group and Z(I) = 1; mD ≥ 3; For every d ∈ D # , one of the following holds: (a) D ≤ E(CI (d))Op (CI (d)); (b) p = 3 and O3 (CI (d)) ∼ = 32 is contained in an E33 -subgroup of I; and (6) I = ΓD,2 (I).

Now since the lemma fails, we may assume by Lemma 14.2d that there exists 1 = a ∈ A ∈ Ep4 (S) and a p-component J of CG (a) such that J := J/Op (J) ∈ TG− p − {F1 } and Sa := CS (a) ∈ Sylp (CG (a)). We choose such a counterexample with mp (A) as large as possible, so that by (14H1,3), A ∩ J = J(Sa ∩ J).  ∈ Let H be the product of all p-components H1 of M such that H − TGp − {F1 }. We argue that H = 1 by showing that J ≤ H. Since M contains ΓA,3 (G) by Lemma 14.2c, M contains Op (J) and (since a ∈ J) M covers ΓA∩J,2 (J/Op (J)); thus, J ≤ M by (14H6). By Lp -balance, J lies in a

. Since mp (J)  > 1, every minimal a-invariant product J∗ of components of M     M , J∗ = 1. Now component of J∗ has p-rank at least 2. In particular, L   there is a0 ∈ Ip ( LM ) centralizing A, and therefore a0 ∈ C(a, J). (Possibly a0 = a.) Also, by the maximality of A, a0 ∈ A and so a0 ∈ Ipo (G). As [ a0 , J∗ ] = 1, the subnormal closure J0 of Lp (CJ (a0 )) in CG (a0 ) covers J∗ . o Since TG− p − {F1 } is closed under pumping up in Lp (G), the components of J0 /Op (J0 ) lie in TG− p − {F1 }. By [V17 , 5.15], J0 = ΓD,2 (J0 ) ≤ M . Hence  as desired; we have proved that H = 1.  J0 is a product of components of H, Let B1 = B ∩ K = B ∩ L. We claim that (14I)

CG (B1 ) ≤ M.

106

10. THEOREMS C6 AND C∗6

We have B1 ≤ C(a0 , J0 ) and mp (CM (B1 )) ≥ 4, so in a similar way we can argue that the subnormal closure J1 of Lp (CJ0 (B1 )) in CG (B1 ) lies in M , so that the pumpup is trivial and J1 = J0 . Let S1 be a Sylow p-subgroup of CG (B1 ) containing a Sylow p-subgroup S1,M of CM (B1 ). Then mp (S1,M ) ≥ mp (B) ≥ 4, so by Lemma 14.2c, S1 ≤ M . Let D1 ∈ Ep4 (S1,M ). Then as B1 J1 ≤ M and M ≥ ΓD1 ,3 (CG (B1 )), M contains Op p (CG (B1 ))Lp (CG (B1 )). As a Sylow subgroup S0 of this last group has rank at least 4, NG (S0 ) ≤ M by Lemma 14.2c, and a Frattini argument implies (14I). In a similar way we can argue that M is an L-preuniqueness subgroup of G, contradicting our assumption that Theorem 3 fails and completing the proof of the lemma unless perhaps J/O13 (J) ∼ = F1 . Namely, let  d ∈ Ip (CM (L)). Without loss we may assume that [B1 , d] = 1. Then d acts on H, and by properties (14H2,3,4), mp (CH (d)) ≥ 3. As B1 ≤ H, CM (d) contains some Ad ∈ Ep4 (M ). Then Op (CG (d)) ≤ ΓAd ,3 (G) ≤ M , d ∈ Ipo (G), and CM (d) contains a Sylow p-subgroup Sd of CG (d). As  and (x, K) is I o -terminal, CG (d) has a p-component Q ∈ Sylp (CM (L)) p  d = L.  Let X = Lp (CG (d))Op p (CG (d)) and RX ∈ Sylp (X). Ld ≤ M with L Then X ≤ Op (CG (d))Ld CG (B1 ) ≤ M . Moreover, by (14H5), and the existence of B1 , mp (RX ) ≥ 3, and in the event of equality, every Ep3 subgroup of RX lies in an Ep4 -subgroup of M . In either case NG (RX ) ≤ M by Lemma 14.2c. By a Frattini argument CG (d) ≤ M , as required. It remains to consider the case p = 13, J/O13 (J) ∼ = F1 . Note that by [V17 , 13.20], ↑13 (F1 ) = ∅. We repeat the previous four paragraphs with appropriate changes to make the proof. Throughout, TG− p − {F1 } should be 1+2 ∼ and A ∩ Sa ∼ replaced by {F1 }. In the first paragraph, Sa = 13 = E132 .  In the second, J ≤ ΓA,3 (G) ≤ M by [IA , 7.5.5]. In the third, it may be ∼ K × F1 , so every

) = L  × J0 = that m13 (S0 ) = 3, but if so, then F ∗ (M E133 -subgroup of S0 lies in an E134 -subgroup of M , and again NG (S0 ) ≤ M by Lemma 14.2c. In the fourth, a similar remark handles the possibility that  m13 (X) = 3. This completes the proof of the lemma. Since the exceptional situation entails the existence of an element of Lo3 (G) isomorphic to A10 , an immediate consequence is: Lemma 14.4. The exceptional situation does not occur. Another quick consequence is: Lemma 14.5. G is 2-balanced with respect to any element of Ep4 (G). Proof. Indeed, by (14B), Lemma 14.3, and [V17 , 4.13], every element  of Lop (G) is locally 2-balanced, so the lemma follows. Lemma 14.6. Let F ∈ E3 (B) and suppose that G is 3/2-balanced with respect to F . Then Θ3/2 (G; F ) = ΘB (G; B). Moreover, ΓF,2 (G) ≤ M . Proof. We set WF = Θ3/2 (G; F ). Note that by [V9 , 3.2], WF is a p group that is invariant under Γ := ΓF,2 (G). Hence the first assertion implies

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the second. We claim that (14J)

CWF (b) ≤ Op (CG (b)) for all b ∈ B # .

To prove this, fix b ∈ B # and set Cb = CG (b), Wb = CWF (b), and b = Cb /Op (Cb ). Expand B to B o ∈ Sp (G). Then as F ≤ B, Wb is CG (B)C invariant, hence B o -invariant. By the Bender-Thompson lemma [IG , 23.3] b )W

b , [Op (C b ), W

b ] = 1. Also, Γb := ΓF,2 (Cb ) ≤ Γ ∩  o Op (C applied to B

b is invariant under Γ  b := Γ  (C b ). Hence by [IA , 7.7.17], any Cb , so W F ,2

b must be isomorphic to L3 (4), b ) not centralized by W component of E(C ± D4 (4), F4 (4), E6 (2), E6 (4), A9k+2 , or A9k+4 , all with p = 3, or F i22 , with p = 5. Since d0 = 5, L3 (4) and F i22 are not possible; the rest all have 3-rank at least 3 [IA , 4.10.3, 5.2.10] so are not possible by Lemma 14.3. Therefore b ), W

F ] = 1, whence [F ∗ (C b ), W

F ] = 1. Hence W

F ≤ F (C b ) = Op (C b ); [E(C  but WF is a p -group, so (14J) follows. Now a routine argument establishes WF = Θ5/2 (G; B) when mp (B) = 4. For any 1 = D ≤ B, CWF (D) ≤ ΔD by (14J). Hence taking any D ∈ Ep3 (B), WF = [CWF (D1 ), D]CWF (D) | |D : D1 | = p ≤ [ΔD1 , D]ΔD | |D : D1 | = p ≤ Θ5/2 (G; B). On the other hand, as G is 5/2-balanced with respect to B, Θ5/2 (G; B) = [ΔF1 , F ] | F1 ∈ E2 (F ) ΔF , all terms of which are easily seen   to lie in Θ3/2 (G; F1 ) | |F : F1 | = p = Θ3/2 (G; F ), the last by 3/2-balance with respect to F . The proof that WF = Θ3 (G; B) when mp (B) ≥ 5 is similar. Namely, taking any D ∈ Ep4 (B), we use (14J) to get WF = CWF (D1 ) | |D : D1 | = p ≤

ΔD1 | |D : D1 | = p ≤ Θ3 (G; B). Conversely, taking anyE ∈ E2 (F ), we have, for any D ∈ Ep3 (B), ΔD = [ΔD ∩ CG (e), E] | e ∈ E # (ΔD ∩ CG (E)),   which, by 3-balance, lies in [Op (CG (e)), E] | e ∈ E # ΔE ≤ Θ3/2 (G; F ).  Now we set up the situation for Proposition 14.8. We let B ∗ be the subgroup of B inducing inner automorphisms on K, so that by definition of B, B ∈ E∗ (T ) and B ∗ ∈ E∗ (R). Also |B : B ∗ | ≤ p. We also put B1 = B ∩ K. Since P ∈ Sylp (K) is cyclic in the present case, B1 ∼ = Zp . We also put E = B ∩ Q = B ∗ ∩ Q, so that B ∗ = B1 × E. Since mp (B) ≥ 4, we have mp (B ∗ ) ≥ 3. Next, let L = L1 , L2 , . . . , Lm be the p-components of M such that  i is a TGp -group of p-rank 2 with abelian Sylow peither mp (Li ) = 1 or L subgroups, 1 ≤ i ≤ m. Also set L∗ = L1 L2 · · · Lm . Since B ∈ E∗ (T ) and  with Q ≤ T , B leaves each Li invariant, 1 ≤ i ≤ m, by Q ∈ Sylp (CM (L)) [V17 , 19.1]. Furthermore, by [V17 , 3.6a], B ∩ Li = Ω1 (S ∩ Li ), 1 ≤ i ≤ m, so B ∩ L∗ = B ∗ ∩ L∗ = Ω1 (S ∩ L∗ ). Finally, we define an important subgroup A∗ of B ∗ . B∗.

Definition 14.7. If G is 3/2-balanced with respect to B ∗ , we set A∗ = In the contrary case, we put Z = Ω1 (Z(Q)) and set A∗ = (B ∗ ∩ L∗ )Z.

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108

Since B ∗ ∈ S(R), Z ≤ B ∗ , so in either case of the definition, A∗ ≤ B ∗ and B1 x ≤ A∗ . Furthermore, by [V17 , 3.6b], Z induces inner automorphisms on each Li and hence on L∗ , so ∗. A∗ induces inner automorphisms on L We fix this notation. We shall prove Proposition 14.8. The following conditions hold: (a) mp (A∗ ) ≥ 3; (b) G is 3/2-balanced with respect to A∗ ; and (c) Either B ∗ = A∗ or Q is connected of p-rank ≥ 3. We prove Proposition 14.8 in a sequence of lemmas. We first eliminate a minimal configuration by a simple fusion argument.

) = L  ∗ , then L∗ has at least 3 p-components. In Lemma 14.9. If F ∗ (M  ∗ ) ≥ 3. particular, mp (L  so we must have m = 2.  ∗ = L, Proof. Suppose false. Since Q = 1, L Let M1 be the subgroup of M leaving L = L1 and L2 invariant, so that |M : M1 | ≤ 2. In particular, S ≤ M1 . By the F ∗ -theorem [IG , 3.6],

1 ≤ Aut(L  1 ) × Aut(L  2 ). Therefore by [V17 , 3.6c], Ω1 (S) is elementary M abelian. As NG (Ω1 (S)) ≤ ΓS,4 (G) ≤ M , it follows by [IG , 16.20] that M controls fusion, and hence transfer in G. As G is simple, M = Op (M ).  i ) is p-nilpotent with We will contradict this last equality. Namely, Out(L

1 /M

2 is an abelian cyclic Sylow p-subgroups, i = 1, 2, by [V17 , 3.6d], so M p-group, where we have defined M2  M by the conditions M2 ≥ Op (M )L∗

2 /L  ∗ = O p ( M

1 /L  ∗ ). Since mp (L  ∗ ) ≤ 3 and mp (S) ≥ mp (B) ≥ and M 4, M1 /M2 is a nontrivial abelian p-group. Hence, if M = M1 , we have O p (M ) < M , contradiction. Therefore |M : M1 | = 2, whence L1 and L2 are isomorphic and thus mp (L∗ ) = 2. So mp (M1 /M2 ) = 2, and any element of M − M1 interchanges two generators of Ω1 (M1 /M2 ). Hence, M again has a normal subgroup of index p, and the proof is complete.  We now assume the proposition is false. This assumption immediately yields: Lemma 14.10. G is not 3/2-balanced with respect to B ∗ . Proof. Indeed, suppose false. Then B ∗ = A∗ in this case by definition, and as mp (B ∗ ) ≥ 3, all parts of the proposition hold, contrary to assumption.  Let (b, J) be a 3/2-obstruction for B ∗ , chosen if possible to be a 3/2obstruction for A∗ . Given the possible isomorphism types of elements of

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Lop (G), and in view of Lemma 14.3, [IA , 7.7.1] yields that one of the following holds: (1) mp (J) = 1; (2) J/Op (J) is a TGp -group of p-rank 2 with abelian (14K) Sylow p-subgroups; or (3) p = 3 and J/Op (J) ∼ = L2 (27). Furthermore, if we set B0∗ = CB ∗ (J/Op (J)), [IG , 20.6, 20.7], [V17 , 4.13, 4.10c, 3.6e] also imply that (14L)

(1) mp (B ∗ /B0∗ ) ≥ 2; and (2) If J/Op (J) is a TGp -group, then some element b∗ of B ∗ either moves J or induces a nontrivial field automorphism on J/Op (J).

In (14L2), we choose b∗ ∈ A∗ , if possible. We use the existence of J to produce a p-component I of M distinct from L, of the same general form as J in (14K1) or (14K2), and satisfying the analogues of the conditions of (14L). We first prove  Lemma 14.11. We have J ≤ M with J/Op (J) centralizing L.   Proof. Since B ∗ ≤ B, B ≤ CG (b). Set J ∗ = J B . As ΓB,3 (G) ≤ M by Lemma 14.2c, Op (J ∗ ) ≤ M , and indeed J ∗ ≤ M by [V17 , 6.13] and [V9 , 9.3]. Thus J   CM (b), so by Lp -balance, J ≤ Lp (M ), and the set of p on which J projects nontrivially is permuted transitively components of M  In particular b by b. But b ∈ B ∗ = B1 × E, where E ≤ Q centralizes L.  L]  = 1, as desired, or J ≤ L.  In the latter normalizes L. Hence either [J, ∗ ∗ ∗ case, however, E ≤ B0 , whence |B : B0 | ≤ p, contrary to (14L). Hence this possibility is excluded and the lemma is proved.   b1 centralizes Now set B1 = b1 . Since b1 ∈ L and J ≤ M centralizes L, ∗ J/Op (J). But also b1 ∈ B , so by [V9 , 4.4], if J1 is a p-component of the pumpup of J in CG (b1 ), then likewise (b1 , J1 ) is a 3/2-obstruction for B ∗ . Hence without loss of generality, we can assume to begin with that b = b1 .  and b ∈ L. Thus J is a p-component of CG (b) with J ≤ M centralizing L  so that I = J.  Hence there exists a p-component I of M that maps on J, Our argument has therefore proved:  Lemma 14.12. J is contained in a p-component I of M with J = I = L.  We can immediately eliminate the L2 (27) possibility for I. Lemma 14.13. If p = 3, then J ∼ = L2 (27). Proof. Suppose false. Since B ∗ = B ∩R ∈ E∗ (R), B ∗ leaves I invariant by [V17 , 19.1]. As (b, J) is a 3/2-obstruction for B ∗ , there exists b0 ∈ B ∗ inducing a nontrivial field automophism on I = J, by [IA , 7.7.1]. Hence,

10. THEOREMS C6 AND C∗6

110

 = 3. Hence, b0 is not mp (CI (b0 ) b0 ) = 2. But mp (I) = mp (Aut(I))  contained in an element of E∗ (R), contradiction. We have therefore shown that either mp (I) = 1 or I is a TGp -group of p-rank 2 with abelian Sylow p-subgroups. Now as I = L, I = Li for some i, 2 ≤ i ≤ m. For definiteness, assume i = 2. In particular, L∗ > L. Now using Lemma 14.9, we obtain Lemma 14.14. We have mp (A∗ ) ≥ 3. Proof. If mp (L∗ ) ≥ 3, then as mp (A∗ ∩ L∗ ) = mp (L∗ ), the lemma  ∗ ) = CQ ( L  2 ). By holds, so we can assume that mp (L∗ ) = 2. Set Q0 = CQ (L ∗ ∗

 Lemma 14.9, L = F (M ), so Q0 = 1. But Q leaves L2 invariant, so Q0  Q and consequently Z0 = Q0 ∩ Z = 1. Since A∗ ∩ L∗ has rank 2 and Z0 ≤ A∗ with Z0 ∩ L∗ = 1, it follows that mp (A∗ ) ≥ 3 in this case as well.  We next prove Lemma 14.15. G is 3/2-balanced with respect to A∗ . Proof. Suppose false. Then by our choices before (14K) and in (14L), (b, J) is a 3/2-obstruction for A∗ and b∗ ∈ A∗ . But then (14L2) is con ∗ . The proof is tradicted because A∗ induces inner automorphisms on L complete.  Finally we prove Lemma 14.16. Q is connected of rank ≥ 3.  = 1. Let t ∈ Ip (Q). Proof. Note that by Lemma 14.9, Q0 := CQ (I) By [V9 , 5.13], we need only show that mp (CQ (t)) ≥ 3. Suppose first that  2 = I is a TGp -group of p-rank 2. If t moves I, it is immediate that L mp (CQ (t)) ≥ 3. On the other hand, if t leaves I invariant, then t normalizes Q0 = 1, so CQ0 (t) = 1. As mp (CI (t)) ≥ 2 in ths case by [V17 , 3.6a], mp (CQ (t)) ≥ 3, as required. Thus we can assume that mp (I) = 1. If t leaves I invariant, we check that mp (CQ (t)) ≥ 3 except possibly in the case that mp (Q0 ) = 1 and t  so consider the latter case. Then Q induces an inner automorphism on I, leaves I and Q0 invariant, whence t ∈ Z(Q). Thus we are done again unless mp (Q) = 2. However, in that case, Ω1 (Q) ≤ IQ0 , whence |B ∗ /B0∗ | ≤ p, contrary to (14L).   Hence we can assume that t moves I. Set I ∗ = I t and put Q∗ = CQ (I∗ ). It is immediate that mp (CQ (t)) ≥ 3 if Q∗ = 1, so we can assume that Q∗ = 1. Thus L∗ = LI ∗ , mp (L∗ ) = p + 1 (as I ∗ is the product of p

). Finally, let Q1 be  ∗ = F ∗ (M p-components of p-rank 1 cycled by t), and L the subgroup of Q leaving I invariant, whence Q1 leaves each p-component of I ∗ invariant, and hence Q = Q1 t. If Q1 > Q ∩ I ∗ , then by [V17 , 3.3] and [V17 , 3.1e], Q ∩ I ∗ has an abelian complement in Q1 and J(Q1 ) ∼ = Eps

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111

for some s ≥ p + 1. But then as t acts on J(Q1 ), mp (CJ(Q1 ) (t)) ≥ 2 and hence mp (CQ (t)) ≥ 3, so we can also assume that Q1 = Q ∩ I ∗ . But then as B ∗ ≤ Q1 B1 we see that, in fact, B ∗ = Ω1 (Q1 )B1 whence |B ∗ /CB ∗ (I)| = |B ∗ /B0∗ | = p, once again contradicting (14L). The lemma is proved.  Together Lemmas 14.14, 14.15, and 14.16 show that all parts of the proposition hold (when G is not 3/2-balanced with respect to B ∗ ), thus completing its proof. The proposition has the following key consequence. Proposition 14.17. If A ∈ E3 (S) with B1 x ≤ A and G is 3/2balanced with respect to A, then ΓA,2 (G) ≤ M . In particular, ΓA∗ ,2 (G) ≤ M . Proof. Indeed, we have A∗ ≤ B with G 3/2-balanced with respect to A∗ and mp (A∗ ) ≥ 3, so Θ5/2 (G; B) = Θ3/2 (G; A∗ ) by Lemma 14.6. But also A ∩ A∗ ≥ B1 x ∼ = Ep2 and G is also 3/2-balanced with respect to A, so Θ3/2 (G; A∗ ) = Θ3/2 (G; A) by Proposition 9.3. Since ΓA,2 (G) ≤ NG (Θ3/2 (G; A)), we have ΓA,2 (G) ≤ NG (Θ5/2 (G; B)), whence ΓA,2 (G) ≤ M , as asserted.  15. The Rank 1 Case: Controlling Cores

, B ∗ , B1 , E, and A∗ of the preceding section. We preserve the notation M In particular, B ∗ = B1 × E with E = B ∩ Q = B ∗ ∩ Q and ΓA∗ ,2 (G) ≤ M . Expand B1 to A ∈ Sp (R). Then x ∈ A and as mp (Q) ≥ 2, mp (A) ≥ 3. Also set A1 = B1 x, so that A1 ≤ A ∩ A∗ . We fix this notation. We shall need the following result, which will enable us to handle a minimal configuration. Proposition 15.1. If mp (A) = 3, then Op (CG ( a, a1 )) ≤ M for every a ∈ A# and a1 ∈ A# 1 . To begin the proof of the proposition, we quickly obtain Lemma 15.2. If mp (A) = 3, then the following conditions hold: (a) B ∗ = A∗ ; (b) If p = 3, then mp (S) = mp (G) ≤ 6; and (c) For all e ∈ E # , CG (e) ≤ M . Proof. In this case, mp (A ∩ Q) = 2, so certainly Q is not connected of rank ≥ 3. But then (a) follows from Proposition 14.8c. Furthermore, if p = 3, then mp (Q) ≤ 3 by [IG , 10.22], so mp (S) = mp (G) ≤ 6 by Lemma 3.5. Thus (b) also holds. Finally, if e ∈ E # , then as usual, K covers Ke /Op (Ke ). Also B1 ≤ Ke and NG (B1 e) ≤ ΓB ∗ ,2 (G) ≤ M (as B ∗ = A∗ ), so M covers NCG (e) (Ke )/Ke by a Frattini argument. But as B ∗ ≤ CG (e), also Op (Ke ) ≤ ΓB ∗ ,2 (G) ≤ M , and as K covers Ke /Op (Ke ) with K ≤ M , likewise Ke ≤ M . Thus NCG (e) (Ke ) ≤ M , so we may assume that Ke  CG (e), and in any event,

10. THEOREMS C6 AND C∗6

112

  C (e) H := Ke G

e ≤ M . Now mp (H) ≥ 3, and in case of equality, Ke has just two conjugates in CG (e), so B centralizes Ω1 (S ∩ H). Hence in any case, NG (S ∩ H) ≤ M by Lemma 14.2c, and as H  CG (e), (c) follows by a Frattini argument.  Now we assume the proposition is false. Thus (15A)

Op (CG ( a, a1 )) ≤ M for some a ∈ A# and

a1 ∈ A# 1 .

a = Ca /Op (Ca ), and fix this notation. We set Ca1 = CG (a1 ) and C 1 1 1 a containing Our first result reduces the analysis to the case p = 3 with C 1 a component isomorphic to L2 (27). Lemma 15.3. The following conditions hold: (a) p = 3; (b) Ca1 contains a 3-component J with J ∼ = L2 (27); (c) J ≤ M ; (d) J is the only 3-component of its isomorphism type in Ca1 ;  and (e) a induces a nontrivial field automorphism on J;  (f) B ≤ Ca and B induces inner automorphisms on J. 1

Proof. Set X = Op (CG ( a, a1 )), so that X ≤ M . We have B ≤ Ca1 as A1 ≤ B. Also as mp (A) = 3, B ∗ = A∗ by Lemma 15.2, so ΓB ∗ ,2 (G) ≤ M . In particular, it follows that Op (Ca1 ) ≤ M , so X ≤ Op (Ca1 ). Thus, a ) by L∗ -balance. By [IG , 20.6], X does  ≤ Op (C  ( a)) ≤ L∗p (C 1 = X 1 p Ca1  not centralize J for some a-invariant p-component J of Ca1 , and J is not locally balanced with respect to a. As usual, J is either a Cp - or a Tp -group or a TGp -group of p-rank 2, by Lemma 14.3. Furthermore, if J is a Tp -group, then mp (J) = 1, by (14A) and Lemma 14.4. We conclude therefore once again from (14C) and [IA , 7.7.1] that (14K) holds for J. In particular, J is simple. We claim next that J ≤ M . Suppose false. Since A1 centralizes P and Q, likewise R = P Q ≤ Ca1 . Also B leaves R invariant and by [V17 , 3.1e], Ω1 (Z(RB)) ≤ B ∗ . Set J ∗ = J RB , so that J ∗ ≤ M . Then RB centralizes some z ∈ Ip (J ∗ ). But x ∈ B and B ∈ Ep∗ (CG (x)), so z ∈ B. Since z also centralizes R, it follows that z ∈ Z(RB) and hence that z ∈ B ∗ . But

z = a1  as J is simple, so z, a1  ∼ = Ep2 and z, a1  ≤ B ∗ . However, z, a1    ∗ centralizes CL  (Ca ) (J ) and ΓB ∗ ,2 (G) ≤ M . Since Op (Ca1 ), J ∗ ≤ M , we p

1

conclude that Op (Ca1 )Lp (Ca1 ) ≤ M . However, X ≤ L∗p (Ca1 ), and by definition, L∗p (Ca1 ) = Lp (Ca1 )F , where F is a B-invariant p -subgroup of L∗p (Ca1 ) [V9 , (6A)]. Then F ≤ ΓB ∗ ,2 (G) ≤ M , so L∗p (Ca1 ) ≤ M and consequently X ≤ M , contrary to assumption. This proves our claim. In particular, J ∗ ≤ ΓB,3 (G), so by (14K) and [V17 , 6.12], the only possibility is p = 3 and J ∼ = L2 (27) with every element of B either moving

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113

 But J is the only 3-component J or inducing an inner automorphism on J. of its isomorphism type in Ca1 , otherwise clearly m3 (Ca1 ) ≥ 7, contrary to Lemma 15.2b. In particular, B induces inner automorphisms on J. Hence all parts of the lemma hold.  Remark 15.4. In the remainder of the argument, we shall derive a contradiction from the properties in Lemma 15.3 of the 3-component J of Ca1 . The only property of M we use is that B1  NG (B), which follows directly from the fact that NG (B) ≤ M , and in the present case that L  M .  and first prove We set Y = CM (L) Lemma 15.5. With the assumptions and notation of Lemma 15.3, the following conditions hold: (a) Q acts faithfully on J and Q ∼ = Z3 Z3 ; (b) B1 = a1 ; (c) Let Y0 = NAut(L2 (27)) (V ), where V is a Borel subgroup of L2 (27). Then Y ∼ = Y0 or O2 (Y0 ); and (d) S = P × Q (whence B = B ∗ ).  = 1, Proof. Since Q ≤ Ca1 and J  Ca1 , Q leaves J invariant. If CQ (J)  then some y ∈ I3 (Z(Q)) centralizes J. But then y ∈ E as Ω1 (Z(Q)) ≤ B ∗ , so CG (y) ≤ M by Lemma 15.2c. Thus M covers J and as Op (Ca1 ) ≤ M ,  = 1, so Q ≤ Aut(J),  which J ≤ M , which is not the case. Therefore CQ (J) has Z3 Z3 Sylow 3-subgroups by [V17 , 11.2.1e]. Observe next that as B induces inner automorphisms on J and B ∈ 3 E∗ (CG (x)), it follows that F := B ∩ J ∈ Syl3 (J), so NG (F ) ≤ ΓB,3 (G) ≤ M . But, as usual, NJ (F ) contains a cyclic 13-subgroup H such that H/CH (F ) ∼ = Z13 . Thus H ≤ M and as m3 (M ) ≤ 6, clearly H must leave L invariant.  ≤ 2, F = [F, H] centralizes L,  so F ≤ Q. On But then as m3 (Aut(L))  the other hand, a ∈ A ≤ R. As J is not locally balanced with respect to

a, a induces a noninner automorphism on J and F a ∼ = Z3 Z3 , by [V17 , ∼ ∼ 11.2.1e]. We conclude that F = E = E33 and Q = Z3 Z3 . Thus (a) holds. Also B ∗ = B1 E ∼ = E34 . Furthermore, H normalizes F a1  = E a1  ∼ = E81 with H centralizing a1 , whence B ∗ = E a1  and H normalizes B ∗ . But H centralizes B1 = B ∗ ∩ L as H is a 13-group leaving L invariant, so we must have a1  = B1 ,  proving (b). This in turn implies that Ca1 covers Y = CM (L). Setting Y1 = CY (a1 ), then Y1 ≤ Ca1 leaves J invariant, whence by [V17 , 11.2.1f],  Since Q acts faithfully on J,  it O3 (Y1 ), which is F -invariant, centralizes J.  But F H   a is contained (with follows that Y1 /O3 (Y1 ) acts faithfully on J.  This implies (c). index 2) in a Borel normalizer in Aut(J)). Now Q  S and Q maps isomorphically onto a Sylow 3-subgroup of  and containing Aut(J), so S = S1 × Q for some S1 acting faithfully on L a1 . Let S0 = Ω1 (S1 ). If S0 ∼ = E34 . But then, as = E32 , then A0 := S0 A ∼

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usual, ΓA0 ,3 (G) ≤ M . However, A0 ≤ Ca1 acts on J and as a ∈ A ≤ A0 , it follows from [V17 , 11.2.1a] that J ≤ ΓA0 ,3 (G), so J ≤ M , contradiction.  it follows from [V17 , 3.1e] that Hence S0 ∼ = E32 . As S1 acts faithfully on L, ∼ S1 = P , whence S = P × Q. Thus (d) holds and the proof is complete.  We next prove Lemma 15.6. We have ag1 ∈ A − A1 for some g ∈ NG (A). Proof. We have B1 ≤ Z(S) and B1 = a1  by Lemma 15.5. If B1 is weakly closed in S with respect to G, then by Gr¨ un’s theorem [V9 , 1.3] and the simplicity of G, NG (B1 ) has no normal subgroup of index 3. However, J  NG (B1 ) as J char CG (B1 ) by Lemma 15.3d. Since a induces a nontrivial field automorphism on J, it follows from [IA , 2.5.12] that a ∈ / [NG (B1 ), NG (B1 )], so NG (B1 ) has a normal subgroup of index 3, contradiction. We conclude that B1 is not weakly closed in S. But by Lemma 15.5cd, we have Ω1 (S) ∼ = Z3 × (Z3 Z3 ), and it is immediate that every element of S3 (S) other than B is S-conjugate to A. Hence by [V9 , 1.7] and the preceding paragraph, there is g ∈ NG (B) or NG (A) such that B1g = B1 . As NG (B) ≤ M and L  M , g ∈ NG (B), and the lemma follows.  This immediately yields: Lemma 15.7. We have P = B1 . Proof. By the structure of S, CS (A) = P A and CS (A) ∈ Syl3 (CG (A)) (as CG (A) ≤ CG (A1 ) ≤ ΓB,2 (G) ≤ M ). But by a Frattini argument, we can choose g in the preceding lemma to leave P A invariant. Now P A ∼ = P × E32 and P ∈ Syl3 (L) is cyclic, so if P > B1 , then B1 = Ω1 (1 (P A)) char P A,  whence B1g = B1 , contradiction. Thus P = B1 , as asserted. Now set SA := NS (A). Again by the structure of S and M , we have |SA : A| = 3 with A1 = Z(SA ) and SA ∈ Syl3 (NM (A)). But NG (A1 ) ≤ ΓB,2 (G) ≤ M and A1 char SA , so likewise our conditions imply that SA ∈ ∗ ∼ Z and Syl3 (NG (A)). Set N = NG (A) and N ∗ = N/CG (A), so that SA = 3 ∗ ∗ SA ∈ Syl3 (N ). Also g ∈ N . We prove  Lemma 15.8. We have O3 (N ∗ ) ∼ = SL2 (3).

∗ Proof. Suppose false. We have that N ∗ ≤ Aut(A) ∼ = GL3 (3) with SA ∗ ∗ centralizing A1 of index 3 in A and Z3 ∼ = SA ∈ Syl3 (N ). Since the lemma ∗ is assumed false we conclude that SA = O3 (N ∗ ) by [V17 , 16.4]. But then CG (A)SA  N . Since A1 ≤ A and A1 = Z(SA ), it follows by a Frattini argument that A1  N . But then as B1 ≤ A1 , B1g ≤ A1 , again contradicting the choice of g. 

Now at last we can derive a final contradiction. let z be a 2-element  ∗ centralizes z ∗ and so by of N such that z ∗  = Z(O3 (N ∗ )). Then SA

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a Frattini argument, we can choose z to normalize SA , in which case z leaves A1 = Z(SA ) invariant. But then z ∈ NG (A1 ) ≤ M , so z leaves B1 = A1 ∩ L invariant. Thus z ∈ NG (B1 ) acts on J, whence [z, A] induces  But B induces inner automorphisms on J and inner automorphisms on J. A = A1 a with a inducing a field automorphism on J and A1 ≤ B, so [z, A] ≤ A1 .  On the other hand, clearly O3 (N ∗ ) has one trivial and one faithful irreducible constituent in its action on A, so |[z, A]| = 9. We conclude therefore from the preceding paragraph that [z, A] = A1 . But z ∗ ∈ Z(N ∗ ), so N ∗ leaves A1 invariant. Hence A1  N , whence B1g ≤ A1 , again contradicting the choice of g. This completes the proof of Proposition 15.1. 16. The L-Preuniqueness of M : the Rank 1 Case Now we are in a position to complete the proof of Theorem 3 by establishing the following result. Proposition 16.1. If d0 = 5, then M is an L-preuniqueness subgroup of G. Again we must prove that ΓQ,1 (G) ≤ M . We assume false, so that again there is 1 = V ∈ Ep (Q) with NG (V ) ≤ M .

, B ∗ , B1 , E, and A∗ of the preceding two secWe preserve the notation M  so that Q ∈ Sylp (Y ). Replacing V by a suittions. We again set Y = CM (L), able Y -conjugate, we can assume without loss that CQ (V ) ∈ Sylp (CY (V )). We expand V to W ∈ Sp (CQ (V )), so that W ∈ Sp (Y ). Also x ∈ W and mp (W ) ≥ 2. We put A = B1 W = B1 × W , so that A ∈ Sp (R) and mp (A) ≥ 3. Finally, again set A1 = B1 x, so that A1 ≤ A ∩ A∗ . We fix all this notation. Lemma 16.2. Let v ∈ A# and suppose that mp (CG (v)) < 4. Let X ∈ Sylp (CG (v)). Then the following conditions hold: (a) Ω1 (X) = Ω1 (Z(X)) ∼ = Ep3 ; and (b) For any p-component J of CG (v), mp (J) = 1, and no element of Ip (CG (v)) induces a nontrivial field automorphism on J/Op (J). Proof. As mp (CG (v)) < 4 ≤ mp (G), v is not p-central in G, Thus  (a)  S = is clear if Z(S) is noncyclic, so suppose that Z(S) is cyclic. Then L

) has 3m components, m ≥ 1. If v has more than 3 orbits on F ∗ (M

), then mp (CM (v)) ≥ 4, contradiction. Since v the components of F ∗ (M

) = L 1 × L 2 × L  3 where each L i ∼  is v-invariant. But normalizes L, F ∗ (M =L ∗ then as mp (CG (v)) ≤ 3, v ∈ F (M ). By [V17 , 19.1], J(S) normalizes each Li and hence centralizes v. So mp (CG (v)) ≥ mp (J(S)) ≥ 4, a contradiction. This proves (a).

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In (b), if mp (J) = 1 and some f ∈ Ip (CG (v)) induces a nontrivial field automorphism on J/Op (J), then f acts nontrivially on a Sylow psubgroup of J by [V17 , 3.1], and (a) is contradicted. So we may assume that mp (J) > 1, and argue to a contradiction. By (a) and [V17 , 3.9], J/Op (J) is simple. Hence mp (J) = 2, and by [V17 , 3.9], J/Op (J) has abelian Sylow p-subgroups. Let X0 = Ω1 (X) and XJ = X0 ∩ J. Thus X0 = v × XJ . Also, as A1 (= B1 x) centralizes v, we may assume that X was chosen to contain A1 . We argue next that (16A)

A1 = XJ .

Recall that Q contains a subgroup U ∼ = Ep2 such that U  T1 . As mp (T1 ) ≥ 4 and P ∩ T1 ≤ Z(T1 ) with P ∩ U = 1, it follows that Ip (CT1 (U )) ⊆ Ipo (G). Hence, [v, U ] = 1, so the group HU := AutU (X0 ) is generated by a transvection moving v. Let HJ = AutJ (X0 ), which is irreducible on XJ by [V17 , 3.9], and centralizes v. We have HU , HJ  ≤ AutG (X0 ). If NG (X0 ) is irreducible on X0 , then by McLaughlin’s theorem [McL2], NG (X0 ) is transitive on X0# , so mp (CG (v)) = mp (CG (x)) ≥ 4, contradiction. Thus, NG (X0 ) is reducible on X0 , and the only possible nontrivial H-invariant subgroup of X0 is XJ . Then by [V9 , 1.8], NG (X0 ) is transitive on E1 (X0 ) − E1 (XJ ). Consequently mp (CG (x )) = 3 for all x ∈ X0 − XJ . As mp (CG (A1 )) ≥ 4, (16A) follows. Therefore NJ (X0 ) ≤ ΓA∗ ,2 (G) ≤ M , by Proposition 14.17. Again as NJ (XJ ) is irreducible on XJ , B1 has a CM (v)-conjugate B2 = B1g ≤ A1 with [v, g] = 1 and B2 = B1 . Thus Lg = L and Lg is v-invariant. Now

) = L ×L  g × I,  with I = 1 by Lemma 14.9, and v normalizing I.  But F ∗ (M ∗

 then as mp (CG (v)) < 4, v ∈ F (M ) with mp (I) = 1. As J(S) normalizes L by [V17 , 19.1], v ∈ Z(J(S)), whence mp (CG (v)) = mp (S) ≥ 4, a final contradiction. The proof is complete.  Our first application of Lemma 16.2 is Lemma 16.3. The pumpup of K in CG (V ) is trivial. Proof. As (x, K) is Ipo (G)-terminal, the pumpup of K in CG (y) is trivial for all y ∈ Ip (Q) such that mp (CG (y)) ≥ 4.  splits over Inn(L),  we may write T1 Since a Sylow p-subgroup of Aut(L) ∗ ∗ as a semidirect product P Q where Q ≤ Q . Then mp (Q∗ ) = mp (T1 )−1 ≥ 3 and we choose U  Q∗ with Q ≥ U ∼ = Ep2 . Set Q∗0 = CQ∗ (U ). Then for all ∗ y ∈ Ip (Q0 ), mp (CG (y)) ≥ 4 so the pumpup of K in CG (y) is trivial. Hence if V ∩ Q∗0 = 1, the assertion of the lemma holds. So we may assume that V ∩ Q∗0 = 1, whence |V | = p. Write V = v, so that Lemma 16.2 applies. Let J be the pumpup of K in CG (V ). By Lemma 16.2b, mp (J) = 1 and x does not induce a nontrivial field automorphism on J/Op (J). Hence by  [V17 , 3.1d], x centralizes J/Op (J), which proves the lemma.

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We next prove Lemma 16.4. The following conditions hold: (a) G is not 3/2-balanced with respect to A; (b) If mp (A) ≥ 4, then V ∼ = Zp and V ≤ A∗ ; and (c) Op (CG (V ))NG (B1 ) ≤ M .  = N/Op (N ), let H Proof. Assume false and set N = NG (V ) and N    ∼ be the pumpup of K in N , and put H ∗ = H N . By Lemma 16.3, H = K. We argue first that (16B)

H ∗ ≤ M.

If G is 3/2-balanced with respect to A, then ΓA,2 (G) ≤ M by Proposition 14.17 since A1 ≤ A and mp (A) ≥ 3. On the other hand, if mp (A) ≥ 4, then by Lemma 14.2c we have ΓA,3 (G) ≤ M . Correspondingly put h = 2 or 3, so that ΓA,h (G) ≤ M . In particular, Op (N ) ≤ ΓA,h (G) ≤ M . Now A acts on  ∗ are isomorphic to K. Moreover, mp (K) = 1  ∗ , and the components of H H 5 L2 (8) (p = 3) or 2B2 (2 2 ) (p = 5) as K is a Tp -group. Hence by and K ∼ =  ∗ , so (16B) holds if (a) or (b) fails. [IA , 7.3.8], ΓA,h (G) covers H Suppose then that Op (CG (V ))NG (B1 ) ≤ M . Since H ∗  N , Op (H ∗ ) ≤  and CG (B1 ) covers Op (NG (V )) = Op (CG (V )) ≤ M . But K covers H ∗ H /H. Since K ≤ M and NG (B1 ) ≤ M , we conclude finally that (16B) holds. Suppose H ∗ = H, in which case H and HV are normal in N . Since H has cyclic Sylow p-subgroups, NN (B1 ) and hence NN (B1 V ) covers N/H. Since N ≤ M , it follows that NG (B1 ) ≤ M (whence (c) holds) and likewise NG (B1 V ) ≤ M . But NG (B1 V ) ≤ ΓA,2 (G), so G is not 3/2-balanced with respect to A by Proposition 14.17, proving (a). Hence it is (b) that fails. Furthermore, V ≤ A∗ , otherwise as B1 ≤ A∗ , NG (B1 V ) ≤ ΓA∗ ,2 (G) ≤ M . Thus we have mp (A) ≥ 4 and mp (V ) ≥ 2 (as (b) fails). But then mp (B1 V ) ≥ 3. However, as noted above, ΓA,3 (G) ≤ M in this case, whence NG (B1 V ) ≤ M , which is not the case. We conclude that H < H ∗. Let J ∗ be the product of the p-components of H ∗ other than H. Since  J ∗ ≤ Y = CM (L).  But W leaves H and hence H ∗ ≤ M and K covers H, ∗ p J invariant. Since W ∈ S (Y ), this yields W ∩ J ∗ = 1. Thus if we set A0 = A ∩ V H ∗ , then as B1 V ≤ W , we see that mp (A0 ) ≥ 3. Expand A0 to X ∈ Sylp (V H ∗ ), so that X is abelian of the same p-rank as V H ∗ (as the p-components of H ∗ have cyclic Sylow p-subgroups). If mp (X) ≥ 4, then as X ≤ M , it follows from Lemma 14.2c that NG (X1 ) ≤ M , where X1 = Ω1 (X). Hence as usual N ≤ M by a Frattini argument, contradiction. Thus mp (X) = 3, whence A0 = Ω1 (X). But then NG (A0 ) ≤ M , otherwise we reach the same contradiction as A0 char X.

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This forces A = A0 , otherwise NG (A0 ) ≤ ΓA,3 (G) ≤ M . In particular, mp (H ∗ ) = 2, so J ∗ consists of a single p-component (of p-rank 1). To derive a contradiction in this remaining case, we use the definition of A∗ (Definition 14.7) to force A∗ ∩ H ∗ to be noncyclic. First, we know that B1 centralizes J∗ , and we let J be the pumpup of J ∗ in CG (B1 ). Since the components of J/Op (J) are either Cp - or Tp -groups or TGp -groups of L2 (8) (p = 3) or p-rank 2 with abelian Sylow p-subgroups, and since J∗ ∼ = 2B (2 25 ) (p = 5), and since B ≤ C (B ), it follows from [V , 6.12] that 2 1 17 G J ≤ ΓB,3 (G), so J ≤ M . Hence J is a product of p-components of CM (B1 ).  so J ≤ Y and consequently Y contains a In particular, J centralizes L,  product J1 of p-components such that J1 = J. ∗ ∗ But now by definition of A , we have A ∩ J1 = Ω1 (Q ∩ J1 ) with Q ∩ J1 ∈ Sylp (J1 ). Since B1 centralizes Q, this in turn implies that A∗ ∩ J = Ω1 (Q ∩ J) with Q ∩ J ∈ Sylp (J). But J1∗ = Lp (CJ ∗ (B1 )) is a p-component of CJ (V ), and J1∗ is either a diagonal of J or else J/Op (J) is quasisimple by Lp -balance. It follows therefore in either case that A∗ ∩ J1∗ = 1, whence A∗ ∩ J ∗ = 1. Since B1 ≤ A∗ , we conclude that A∗1 = A∗ ∩ H ∗ ∼ = Ep2 with ∗ ∗ ∗ A1 characteristic in a Sylow p-subgroup of H . Thus N = H NN (A∗1 ) by a Frattini argument and NG (A∗1 ) ≤ ΓA∗ ,2 (G) ≤ M , so N ≤ M , giving a final contradiction.  As an immediate corollary we obtain Lemma 16.5. If mp (A) = 3, then ΓE,1 (G) ≤ M . Proof. Indeed, A∗ = B ∗ in this case by Lemma 15.2a, so E ≤ A∗ . Let 1 = E ∗ ≤ E. Then Lemma 16.4b fails with E ∗ , B in the roles of V, A, respectively (as E ∗ ≤ A∗ and mp (B) ≥ 4). Since Lemma 16.4 was proved under the assumption that NG (V ) ≤ M , it follows that NG (E ∗ ) ≤ M . Since E ∗ was arbitrary, we conclude that ΓE,1 (G) ≤ M , as asserted.  Furthermore, using Proposition 15.1, we can prove Lemma 16.6. We have Op (CG (V )) ≤ M . Proof. If mp (A) ≥ 4, then as above ΓA,3 (G) ≤ M , and as A ≤ CG (V ) acts on X = Op (CG (V )), it follows that X ≤ M . Hence we can assume that mp (A) = 3. But then if mp (V ) > 1, we have V = W , whence x ∈ V . But CG (x) ≤ M by the preceding lemma as x ∈ E. Since X ≤ CG (x), X ≤ M in this case as well. Therefore we can also assume that mp (V ) = 1, whence V = a for some a ∈ A# .   Thus CG (V ) = CG (a). Moreover, we have X = CX (a1 ) | a1 ∈ A# 1 and as each CG ( a, a1 ) ≤ CG (a), each CX (a1 ) ≤ Op (CG ( a, a1 )). Since mp (A) = 3 and A ∈ Sp (R), it follows now from Proposition 15.1 that each  CX (a1 ) ≤ M , whence X ≤ M in this case as well.

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Now Lemmas 16.4c and 16.6 yield: Lemma 16.7. We have NG (B1 ) ≤ M . We complete the proof of the proposition by showing, to the contrary, that NG (B1 ) ≤ M . To accomplish this, we use the fact that G is not 3/2balanced with respect to A. Hence there exists a 3/2-obstruction (a, J) for A in G. As usual, for some a0 ∈ A# leaving J invariant, J/Op (J) is not locally balanced with respect to a0 , and mp (A/CA (J/Op (J))) ≥ 2. By Lemma 16.2b, this last condition implies that mp (CG (a)) ≥ 4. Hence again J/Op (J) is given by (14K). Our goal will be to prove that J ≤ M . Thus in Lemmas 16.8–16.12, we assume J ≤ M.

(16C) We immediately obtain

Lemma 16.8. The following conditions hold: (a) mp (A) = 3; and (b) If p = 3, then mp (S) ≤ 6. Proof. If mp (A) ≥ 4, then as J is not locally balanced with respect to a0  and a0 ∈ A, it follows once again from [V17 , 6.12, 6.13], given the possibilities for J/Op (J), that J ≤ ΓA,3 (G), so J ≤ M , contrary to (16C). Thus (a) holds, and (b) follows by Lemma 15.2b.  We set

  J∗ = JA .

L2 (27). Lemma 16.9. If p = 3, then J/O3 (J) ∼ = Proof. Suppose p = 3 and J/O3 (J) ∼ = L2 (27). By [V17 , 11.2.1d],   (16D) J ∗ = O3 (CJ ∗ (a1 ))L3 (CJ ∗ (a1 )) a1 ∈ A# 1 . But J ∗   CG (a), so Op (CJ ∗ (a1 )) ≤ Op (CG ( a, a1 )) ≤ M for any a1 ∈ A# 1 , again by Proposition 15.1. We thus conclude (16E)

L3 (CJ ∗ (a1 )) ≤ M for some a1 ∈ A# 1 .

We fix a p-component I of CJ ∗ (a1 ) with I ≤ M and let H be a pcomponent of the pumpup H ∗ of I in CG (a1 ). If H ≤ M , then as H ∗ =  a  and I ≤ H ∗ , it follows that I ≤ M , contradiction. Thus also H ≤ M . H On the other hand, a1 ∈ B, so B ≤ CG (a1 ). Since ΓB,3 (G) ≤ M , we conclude that (16F)

ΓB,3 (G) does not cover H/Op (H).

a = Ca /Op (Ca ). We set Ca1 = CG (a1 ) and C 1 1 1  By [V17 , 11.2.1b], I/O3 (I) ∼ = L2 (27). Also by Lemma 16.8, mp (Ca1 ) ≤  ∼ mp (S) ≤ 6 (as S ∈ Syl3 (G)), so by [V17 , 11.2.1c], H = L2 (27) and H is

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the only 3-component of its isomorphism type in Ca1 . Thus a0 induces a  Likewise by (16F) and [V17 , 11.2.1a], nontrivial field automorphism on H.  Thus H, a0 satisfy the same B must induce inner automorphisms on H. conditions as J, a in Lemma 15.3. Hence by Remark 15.4, the remainder of the proof of Proposition 15.1 applies without change to yield the same contradiction. The proof is complete.  Lemma 16.10. The following conditions hold: (a) a ∈ A − A1 ; (b) mp (CM (a)) = 3; (c) If J < J ∗ , then mp (J) = 1; and (d) A ∈ Sp (G). Proof. If mp (CM (a)) ≥ 4, choose D ∈ Ep4 (M ) with a ∈ D. By Lemma 14.2c, ΓD,3 (G) ≤ M . But as J/Op (J) is given by (14K) and by Lemma 16.9, mp (J) = 1 or J/Op (J) is a TGp -group of p-rank 2 with abelian Sylow 2-subgroups. As J/Op (J) is not locally balanced for p, [V17 , 6.12] implies as usual that J ≤ ΓD,3 (G) ≤ M , contrary to assumption. Therefore (b) holds. As A1 ≤ B ∈ Ep4 (M ), (a) holds as well. Suppose that J < J ∗ . Then J ∗ /Op (J ∗ ) is the direct product of p ≥ 3 components, centralized by a. It follows easily that there exists D0 ∈ Ep4 (J ∗ a) such that J ∗ A ≤ ΓD0 ,3 (G). But as S ∈ Sylp (G), D0 ≤ M g for some g ∈ G, so by Lemma 14.2a, ΓD0 ,3 (G) ≤ M g and J ∗ A ≤ M g . By Lp balance, J ∗ ≤ Lp (M g ). Note, however, that by (b), mp (CM g (ag )) = 3. If mp (J) = 2, then [Lg , J ∗ ] ≤ Op (M g ), and, using [V17 , 3.6c], mp (CM g (ag )) ≥ 4, a contradiction. So mp (J) = 1, proving (c). Finally, since x ∈ E ∩ A,  CG (A) ≤ M by Lemma 16.5. As a ∈ A, (d) follows from (b). Now by fusion analysis we can prove Lemma 16.11. We have mp (CG (a)) = 3. Proof. Suppose false and let D ∈ Ep4 (CG (a)). As in the previous lemma, mp (J) = 1 or J/Op (J) is a TGp -group of p-rank 2 with abelian Sylow p-subgroups. Suppose that the latter holds. By Lemma 16.10c, J = J ∗ is A-invariant. By [V17 , 3.6a], A centralizes an Ep2 -subgroup V0 of J as well as a. Hence A = V0 a by Lemma 16.10d. Let Pa ∈ Sylp (CG (a)) with A ≤ P . By [V17 , 3.6a] again, any element of Ip (Pa ) normalizing J centralizes A. As we are assuming that mp (Pa ) ≥ 4, it follows that mp (CPa (A)) ≥ 4. But x ∈ A so mp (CM (A)) ≥ 4 by Lemma 16.5, contradicting Lemma 16.10b. Therefore, mp (J) = 1.   If J ∗ > J, then J ∗ =



J a 

for some a ∈ A, and we set J0 =

Lp (CJ ∗ (a )). If J ∗ = J, we set J0 = J and choose any a ∈ A − a (A ∩ J0 ). In both cases A = a, a  (A ∩ J0 ). Write A ∩ J0 = z. Thus, A = a, a , z =

a, x, b1  by Lemma 16.10a, where b1  = B1 . Whether J = J ∗ or not, there is a p-element y ∈ Ip (J ∗ ) such that [A, y p ] = 1, [ a, z , y] = 1, and [A, y] = z (by [V17 , 3.1dg] and the structure of Zpm Zp ).

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We let NA = NG (A) and N A = NA /CG (A) ∼ = AutG (A), and study the structure of N A . Thus y is a transvection in N A with center z and axis

a, z. We show that (16G)

x ∈ a, z (whence [ x , y] = z).

For if x ∈ a, z and J < J ∗ , then mp (CM (a)) ≥ mp (CG ( a, x)) ≥ 4 by Lemma 16.5, contradiction. If x ∈ a, z and J = J ∗ , then a, z = a, x, so b1 ∈ A − a, z. Then b1 is sheared to z by y ≤ CG (x) ≤ M (Lemma  b1 is not

, b1 lies in the p-rank 1 component L, 16.5). But because in M sheared to any subgroup of its centralizer of order p, contradiction. This proves (16G). In addition to y, there is a second transvection u ∈ N A with center x and axis A1 = b1 , x. It arises from the splitting T1 = P Q0 with Q ≤ Q0 and P ∩ Q0 = 1, which exists by [V17 , 3.1e]. As mp (M ) ≥ 4, mp (Q0 ) ≥ 3. We take U  Q0 with Ep2 ∼ = U ≤ Q, and take any u ∈ U − Z(Q0 ). Because mp (CM (A)) = 3, [U, A] = 1, and so Z(Q) ∩ U = Z(Q) ∩ A = x = [A, u] with [b1 , u] = 1, as claimed. We will show that (16H)

N A acts irreducibly on A.

This will imply by McLaughlin’s theorem [McL2] that NA induces at least SL(A) on A. Hence there will be g ∈ NA such that bg1 = b1 x and xg = x. By Lemma 16.5, g ∈ M . However, we saw above that no element of M can shear b1 to x, contradiction. Thus (16H) will complete the proof of the lemma. Suppose that (16H) fails, and let A0 be a proper NA -invariant subgroup | = p.Then A0 ≤ CA (y) ∩ CA (u) = a, z ∩ A1 . of A. Suppose first that |A0 by a Frattini argument, and so NL (A) However, NL (A) covers NL b1

contains a p -element h ∈ CG (A ∩ Q) acting nontrivially on b1 . Since A0 is an h-invariant subgroup of A1 , it equals A0 = b1  or x. The latter is impossible by (16G), so A0 = b1 . Likewise, there is a p -element h ∈ J0 acting nontrivially on z and centralizing a. Hence A0 = z or a, with the latter impossible as a ∈ A1 . Thus, b1  = A0 = z. But x is sheared by

y to z; it cannot, however, be sheared by y to b1  because that would force y ∈ NG (A1 ) ≤ M and then xb1 would centralize fewer components of

than x. Thus b1  = z, contradiction. M We conclude that |A0 | = p2 . The transvections y and u act trivially on A/A0 , so their centers lie in A0 , whence A0 = z, x. Now h ∈ NA acts nontrivially on b1 and trivially on A ∩ Q, so either A0 = b1 , x = A1 or A0 = A ∩ Q. In the first case, xy ∈ b1 , x − x and y ∈ NG (A0 ) ≤ M again,

, contradiction. so x and xy centralize different numbers of components of M Thus A0 = A ∩ Q. Now xy ∈ x by (16G), but [A0 , u] = x. It follows that ∼ SL2 (p).

y|A , u|A  = 0

0

122

10. THEOREMS C6 AND C∗6

 ∼ SL2 (p) or Ep2 SL2 (p). In the latter case, Op (N ) is the Hence Op (N ) = full stabilizer of the chain A > A ∩ Q > 1 in Aut(A). As such, it contains a transvection t shearing b1 ( ∈ A ∩ Q) to x ∈ A ∩ Q, and centralizing A ∩ Q. A preimage t of t in N then lies in NG (A1 ) ≤ M . But as shown five paragraphs above, b1 is not sheared in M to any element of Ip (CM (b1 )).  This contradiction shows that Op (N ) ∼ = SL2 (p). Let s ∈ N map onto the  (central) involution of Op (N ). Then s ∈ NG ( x) ≤ CG (x)ΓS,4 (G) ≤ M by Lemma 16.5 and a Frattini argument. Suppose that s normalizes K. Then s normalizes b1  = A ∩ K, and so b1  = CA (s). Hence b1  is N -invariant. We get a contradiction in this case by playing off the element h above against a p -element h ∈ CJ ∗ (x) normalizing but not centralizing z. (Note that x induces a field automorphism on J0 /Op (J0 ) if J = J ∗ , and we can take a = x if J < J ∗ .) Since A = x, z, a, h ∈ N . Then CA (h ) = x, a and [A, h ] = z are h -invariant, as are b1  and A ∩ Q. As a ∈ A1 = b1 , x, b1 ∈ CA (h ), so

b1  = z ≤ J0 and A ∩ Q = x, a. In particular, a ∈ Q. But then Ka is a p-component of CG (a). Whether Ka = J or not, |A/A ∩ C(a, J)| ≤ p, a contradiction. Hence, finally, K s = K, so L = Ls . Then as Q ∩ A = [Q ∩ A, s], L  s ). But this implies that mp (CM ( b1  (Q ∩ A))) ≥ 4. As Q ∩ A ≤ CM ( L a ∈ A = b1  (Q ∩ A), mp (CM (a)) ≥ 4. This contradicts Lemma 16.10b and establishes (16H). The lemma is at last proved. 

We immediately deduce Lemma 16.12. We have J ≤ M . Proof. Otherwise, mp (CG (a)) = 3, so Lemma 16.2b applies and J   O p (CG (a)). Then by [V17 , 3.1d], (a, J) cannot be a 3/2-obstruction, contrary to our choice.   Lemma 16.13. J centralizes L. Proof. Indeed, if false, then as J ≤ M , it follows by Lp -balance that  Since mp (L) = 1, it J ≤ L. But A ≤ R induces inner automorphisms on L.  ≤ mp (A/CA (L))  ≤ 1, contrary to the fact that follows that mp (A/CA (J))  mp (A/CA (J/Op (J))) ≥ 2 as (a, J) is a 3/2-obstruction for A. Now b1 centralizes J and a. As in the discussion preceding Lemma 14.13, using [V9 , 4.4] we can assume without loss that b1 = a. Since J is a  it follows that J is contained p-component of CG (b1 ) and J ≤ Y = CM (L),  in a p-component I of Y with I = J. Now we can attain our objective: Lemma 16.14. We have NG (B1 ) ≤ M .  = N/Op (N ). Then Proof. Suppose false and set N = NG (B1 ) and N  N J is a p-component of N . We set H = J . We know that J = I is not

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123

5 locally balanced with respect to a0  (whence J ∼ L2 (8) (p = 3) or 2B2 (2 2 ) = (p = 5)). Hence again unless J ∼ = L2 (27) (with p = 3), H ≤ ΓB,3 (G) ≤ M by [V17 , 6.12]. In the remaining case, mp (J) = 3, and as B ≤ N , it follows from [V9 , 5.18] that B is 3-connected to some D ∈ E4 (J b1 ). However, G is 2-balanced with respect to every element of E4 (G) by Lemma 14.5, so ΓD,3 (G) ≤ M by Proposition 9.3. Thus M covers H/J. Since J ≤ M , H ≤ M in this case as well. If mp (HB1 ) ≥ 4, then again by Proposition 9.5 and a Frattini argument, M covers N/HB1 , whence N ≤ M , contradiction. Thus mp (HB1 ) ≤ 3, so L2 (27) with p = 3. But I is a p-component mp (H) ≤ 2. In particular, J ∼ =  Hence by (14K), J  L  ∗ . It follows therefore from the of Y with J = I. definition (14.7) of A∗ that A∗1 := A∗ ∩JB1 ∈ Ep∗ (JB1 ) with A∗1 characteristic in a Sylow p-subgroup of JB1 . Since NG (A∗1 ) ≤ ΓA∗ ,2 (G) ≤ M , we reach the same contradiction N ≤ M by a Frattini argument if H = J. Thus H > J, forcing mp (J) = 1 with H consisting of exactly two p-components. In particular, B leaves each p-component of H invariant. Since B ∈ Ep∗ (S), it follows that B ∩ H ∼ = Ep3 with B0 = Ep2 , whence B0 := B ∩ HB1 ∼ characteristic in a Sylow p-subgroup of HB1 . But then NG (B0 ) ≤ ΓB,3 (G) ≤ M , so again a Frattini argument yields that N ≤ M , giving a final contradiction. 

Since Lemmas 16.14 and 16.7 are in conflict, Proposition 16.1 is proved. This at last completes the proof of Theorem 3, and also the proof of Theorem 9.1. 17. Theorem 4: Almost Strong p-Embedding, the General Case For the proof of Theorem 4, we shall refer to the preuniqueness subgroup arising from Theorem 3 as M , regardless of which outcome of The = orem 9.1 gave rise to it. The rest of our notation is unchanged: M o M/Op (M ), (x, K) ∈ TJp (G), L the subnormal closure of K in M , Q ∈  Q ≤ S ∈ Sylp (M ), P = S∩L, R = P Q ≤ T1 ∈ Sylp (C(x, K))∩Sylp (CM (L)), Sylp (NCG (x) (K)). As M is an L-preuniqueness subgroup of G, ΓQ,1 (G) ≤ M . We fix this notation for the proof of Theorem 4. We must prove that M is a strong p-uniqueness subgroup of G. According to the definition of this term [I2 , 8.7], this means proving the following three statements: (1) If mp (L) > 1, then M is strongly p-embedded in G; (2) If mp (L) = 1, then M is strongly p-embedded in G or almost strongly p-embedded in G of wreathed or (17A) almost p-constrained type; and (3) Op (M ) = 1. In this section and the next, we prove (17A1,2), by contradiction. If either one fails, then it follows from [II3 , Theorem PU3 , Proposition 16.1]

10. THEOREMS C6 AND C∗6

124

that (17B)

(1) L  M ; and (2) If (L x)g ≤ M for g ∈ G, then g ∈ M .

Furthermore, by [II3 , Theorem PU2 ], V := Qg ∩ M has rank ≥ 2 for some g ∈ G − M . As usual, we choose g so that (17C)

(1) mp (V ) is maximal; and (2) Subject to (1), |V | is maximal.

Replacing g by gy for some y ∈ M , we can assume without loss that  invariant. In particular, V centralizes V ≤ S, whence V leaves Q = CS (L) x ∈ Z(S). In this short section, we reduce the analysis to specific residual config∼ urations with L = L3 (4) (with p = 3) or F i22 (with p = 5). We shall =K∼ prove Proposition 17.1. Suppose that either (17A1) or (17A2) fails. Then the following conditions hold: ∼ ∼ Either p = 3 and L = F i22 ; = L3 (4), or p = 5 and L ∼ ∼ Q = V = Ep2 ; V ≤ R and V ∩ Q = 1; If we set A = V Q then A = R ∼ = Ep4 and A is weakly closed in S with respect to G; (e) We can choose g ∈ NG (A); (f) L = K is quasisimple; and (g) Θ5/2 (G; B) = 1.

(a) (b) (c) (d)

We first prove Lemma 17.2. The following conditions hold: ∼ ∼ (a) Either p = 3 and L = F i22 ; = (S)L3 (4) or M12 , or p = 5 and L (b) V ∼ E ; and 2 = p (c) V ≤ R and V ∩ Q = 1. Proof. We have x ∈ CG (V ) ≤ ΓV,1 (G) ≤ M g (as V ≤ Qg and −1 ≤ M, ΓQ,1 (G) ≤ M ). But then if L ≤ M g , it follows that (L x)g −1 whence g ∈ M by (17B2), contrary to g ∈ G − M . Therefore L ≤ M g . But L  M by (17B1), so V acts on L. Then Op (L) ≤ ΓV,1 (G) ≤ M g ,  In particular, V acts faithfully on L,  whence so ΓV,1 (G) does not cover L.   is a Tp -group, we conclude now from [V17 , 5.9] that L V ∩ Q = 1. Since L ∼  is as in (a), and that V = Ep2 with V inducing inner automorphisms on L. This last condition implies that V ≤ R, so all parts of the lemma hold. 

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125

 = M/CM (L)  ∼  and quickly obtain We now set M = AutM (L), Lemma 17.3. The following conditions hold: (a) Q ∼ = Ep2 ; (b) If we set A = V Q, then A = R ∼ = Ep4 ; ∼  (c) L = SL3 (4); and ∼ (d) L M12 . =  is simple or L  ∼ Proof. By the preceding lemma, either L = SL3 (4). 1+2 ∼ Moreover, in the latter case, P = 3 and hence P is of exponent 3; and  is outer well-generated for p = 3 by [V17 , 5.8]. Since V induces inner L  it follows now from our maximal choice of V and automorphisms on L, [II3 , 16.13] in all cases that V = Qg . As V ∼ = Ep2 , likewise Q ∼ = Qg ∼ = Ep2 , so (a) holds. ∼ We next prove (e). Suppose that L = M12 . Then P ∼ = 31+2 . By Lemma  is locally balanced with respect to V . Hence by [V17 , 10.11.4], there 4.7, L exists an involution in NL (V ) ∩ NL (P ) inverting Z(P) ≤ V and centralizing V /Z(P ). Set A = V Q. Pulling back to M and using a Frattini argument, we see that there is a 2-element t ∈ NL (A) such that [Q, t] = 1, t inverts Z(P ), and t centralizes A/Z(P ). Let W = CV (t). Then W = 1 so t ∈ CG (W ) ≤ M g . Thus t normalizes V and so Z(P ) = [A, t] = [V, t] ≤ V . −1 −1 −1 Hence P ≤ ΓV,1 (G) ≤ M g . So P g ≤ M with Z(P g ) ≤ V g = Q. But as Out(M12 ) is a 3 -group, R ∈ Syl3 (M ). As L  M , it follows that Z3 ∼ = [R, R] ≤ Q. But [R, R] = [P, P ] and P ∩ Q = 1, a contradiction.  ∼ Hence L = (S)L3 (4) (with p = 3) or F i22 (with p = 5). Now, as Q centralizes P , Q ≤ Z(R). But A = V Q is abelian as V ∼ = Ep2 with V ∩ Q = 1. By Lemma 17.2c, A ≤ R, so A = R by orders. It follows that ∼ A∼ = L3 (4) if p = 3. This proves (b) and (c). The proof is = Ep4 . Hence L complete.  By the preceding two lemmas, (a), (b), and (c) of Proposition 17.1 hold, so it remains to verify parts (d), (e), (f), and (g). Lemma 17.4. A is weakly closed in S with respect to G. Proof. By the preceding lemma, A = R. Suppose now that Ah ≤ S for  =Q some h ∈ G. If h ∈ M , then h leaves L invariant, whence Qh ≤ CS (L) h h and P ≤ S ∩ L = P . Thus, R = R, as required. Suppose then that h ∈ M . Then W := Qh = Qh ∩ M satisfies the conditions (17C) in place of V , with h in place of g. Hence Lemma 17.2 applies to W as well as to V . We conclude that W induces faithful inner  In particular, E := W Q ∼ automorphisms on L. = Ep4 , so E = R = A. By 1+2 ∼    ∼ if p = 3 and S = E52 if p = 5. Since W [V17 , 11.4.1], S ≤ 3 = Ep2 ,    it follows that CS(W ) = W = P , whence CS (W ) = W Q = A. Then  Ah ≤ CS (Qh ) = CS (W ) = A, so Ah = A, completing the proof.

10. THEOREMS C6 AND C∗6

126

Now we prove Lemma 17.5. We can choose g ∈ NG (A). Proof. Since CG (V ) ≤ M g and V centralizes Q, A = V Q ≤ M g , so −1 ≤ M . But then Ag y ≤ S for some y ∈ M . The preceding lemma A −1 forces Ag y = A. Hence if we set g  = g −1 y, we have g  ∈ NG (A) and, in  particular, Qg ≤ S. Furthermore, as g ∈ / M and y ∈ M , g  ∈ / M . Thus we  can replace g by g , if necessary, and hence can assume without loss that g ∈ NG (A).  g −1

We use this to prove Lemma 17.6. The following conditions hold: (a) Θ5/2 (G; B) = 1; and (b) K is terminal in G. In particular, K is quasisimple. Proof. By Lemma 17.3b, A = R. But also B = (B ∩ K) × CB (K) ≤ S, so B = P × Q = A. Now by Theorem 9.1, if (a) fails, then M is a maximal subgroup of G containing ΓB,3 (G), and in particular containing NG (A). Hence by Lemma 17.5, g ∈ M , a contradiction. Finally, (a) implies (b), again by Theorem 9.1. The lemma is proved.  By the preceding three lemmas, parts (d), (e), and (g) of Proposition 17.1 hold. Thus it remains only to prove (f)–namely, that K = L. Since K  this will be the case if L is quasisimple, so assume not. Since K covers L, is terminal in G, we have K = Kt for each t ∈ Q# . But K ≤ L acts on X = Op (L). Since L is not quasisimple, [X, K] = 1 by [V9 , 6.1b], whence Y = [CX (t), K] = 1 for some t ∈ Q# . But K = Kt is a component of CG (t) and Y = [Y, K] by [IG , 4.3(i)], whence Y ≤ K. However, K leaves Y invariant, so Y ≤ Op (K) and as K is quasisimple, it follows that K centralizes Y , contrary to Y = [Y, K] = 1. Thus L is quasisimple, and the proposition is proved. 18. The Residual Cases We now eliminate the residual configurations of the preceding section, thus proving Proposition 18.1. The conditions (17A1,2) hold. We assume false, so that M satisfies the conditions of Proposition 17.1. In particular, K = L is quasisimple with K ∼ = L3 (4) (p = 3) or F i22 (p = 5), ∼ and with Q = Ep2 and Θ5/2 (G; B) = 1. Also A = R = P × Q ∼ = Ep4 and / M. there is g ∈ NG (A) − M with g ∈ We let D = NK (P ) = NK (A) and set D1 = CD (P ) = CD (A). Then by [V17 , 11.4.1a], according as K ∼ = L3 (4) or F i22 , we have ∼ Q8 or (Z4 ∗ SL2 (3))#2. (18A) D/D1 =

18. THE RESIDUAL CASES

127

In particular D acts absolutely irreducibly on P . Using [V17 , 16.3], we immediately obtain Lemma 18.2. Some 2-element h ∈ NG (A) interchanges P and Q. ∼ Ep4 , N  = N/CG (A). Since A =  ≤ Proof. Set N = NG (A) and N ∼  GL4 (p), p = 3 or 5. Furthermore, D ≤ N and D = D/D1 . Also as D   C  (t) for centralizes Q and ΓQ,1 (G) ≤ M with K  M , it follows that D N  is not normal in N  , otherwise g every t ∈ Q# . In addition, g ∈ N , so D  invariant, whence g ∈ M , which is not the case. would leave Q = CA (D)  Now [V17 , 16.3] yields the existence of the desired element h. We set J = K h, so that Q ≤ J and P = Qh centralizes J. Our goal will be to force J, h ≤ M , which will contradict the fact that K  M . First of all, as CG (Q) ≤ ΓQ,1 (G) ≤ M , K is the unique component of CG (Q). Hence J is the unique component of CG (P ) and consequently (18B)

h interchanges K and J.

We next prove Lemma 18.3. Each of P and Q centralizes any p -subgroup of G that it normalizes. Proof. Since P and Q are interchanged by h, it suffices to prove the lemma for Q. Suppose false and let X be a Q-invariant p -subgroup of G chosen minimal subject to being not centralized by Q. Since Q ∼ = Ep2 , Z and X = [X, Q]. Since C (Q it follows that Q1 = CQ (X) ∼ = p 1) ≤ G  ΓQ,1 (G) ≤ M , X ≤ M . But Q centralizes K = L  M , so X centralizes K, whence X ≤ Y = CM (K). In particular, P centralizes X. Therefore, X ≤ ΓP,1 (G) ≤ M h . As Q ≤ J  M h , X = [X, Q] ∈ NJ (Q; p ) = {1}, the last by [V17 , 11.4.1f] as J ∼ = K. This contradiction establishes the lemma.  Next, set E = D h , and E1 = CE (Q), so that h interchanges D and E and E/E1 ∼ = D/D1 with E/E1 acting faithfully on Q and centralizing P . Hence by [V17 , 11.4.1d], [E, E] centralizes K. By the structure of E/E1 , [E, E] contains a 2-element y that inverts Q (and centralizes P ). Setting −1 z = y h , it follows that z inverts P and centralizes Q. We fix y and z. Our argument yields: (18C)

[y, K] = [z, J] = 1.

Now we prove Lemma 18.4. We have K  CG (y) and J  CG (z).

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Proof. As y = z h and K = J h , we need only prove that J  CG (z). Set Cz = CG (z) and X = F (Cz )Op (E(Cz )). As G is of even type, F (Cz ) = z ), so by the preceding lemma, Q centralizes X. But Q ≤ J, so J = O2 (C QJ centralizes X. We prove a preliminary result. Let W be a Q-invariant p-subgroup of Cz with Q ∩ W = 1. We claim that |W | ≤ p. Indeed, if not, then W contains a Q-invariant subgroup U of order p2 , in which case some t ∈ Q# centralizes U . But then U ≤ CG (t) ≤ ΓQ,1 (G) ≤ M , so QU ≤ CM (z) and |QU | = p4 . However, z centralizes Q ∼ = Ep2 with Q ∈ Sylp (CM (K)) and z inverts P . Since K  M , it follows therefore from [V17 , 11.4.1b], applied to M/CM (K), that a Sylow p-subgroup of CM (z) has order ≤ p3 , contradiction. This proves our claim.  Now set I = Op (E(Cz )) and suppose J ≤ I, in which case Q ∩ I = 1. But then we can take W ∈ Sylp (I) and conclude that |W | ≤ p. In particular, either I = 1 or I is quasisimple with |I|p = p. Since Q ≤ J with Q ∼ = Ep2 , it follows in either case from [V17 , 3.1dg] that J centralizes I, whence J centralizes XI = F ∗ (Cz ). But then J ≤ F (Cz ) by the F ∗ -theorem [IG , 3.6], which is absurd. Therefore, J ≤ I. Let I1 be a component of I with J not centralizing I1 and let Q1 be a Qinvariant Sylow p-subgroup of I1 . If Q1 ∩ Q = 1, then we can take W = Q1 and conclude that |Q1 | ≤ p, which is impossible as J projects nontrivially on I1 and Ep2 ∼ = Q ≤ J. Thus Q0 := Q ∩ I1 = 1. But J ∩ I1  J, so J ≤ I1 . Let I ∗ be the product of all components of I distinct from I1 (if any), with I ∗ = 1 if I = I1 . Then we can take W so that W ∈ Sylp (I ∗ ) and again conclude that |W | ≤ p. Hence either I ∗ = 1 or I ∗ is a single p-component of p-rank 1. However, as Q ≤ J ≤ I1 , mp (I1 ) ≥ 2, so I1 char I and hence I1  Cz . Thus to complete the proof, it remains only to show that J = I1 . Expand Q to V ∈ Sylp (I1 ) and now let U be a normal Ep2 -subgroup of V . As usual, we must have U ∩ Q = 1, otherwise again we could take W = U and contradict |W | ≤ p. This time put Q0 = U ∩ Q and set V0 = CV (Q0 ). Then V0 ≤ ΓQ,1 (G) ≤ M , so V0 ≤ CM (z) and therefore |V0 | ≤ p3 , again by [V17 , 11.4.1]. Since |V : CV (U )| ≤ p, it follows that |V | ≤ p4 . Suppose next that I1 = ΓQ,1 (I1 ), in which case I1 ≤ M . But then as   I1 = QI1 , it follows that I1 centralizes K. In particular, I1 centralizes P . On the other hand, J ≤ I1 , and J is a component of CG (P ), as noted above, so J = I1  Cz in this case, as required. Thus we can assume that ΓQ,1 (I1 ) < I1 . Note that as G is of even type, I1 ∈ C2 . Since J ≤ I1 and |I1 |p ≤ p4 , we conclude now from [V17 , 11.4.3]  that J = I1 , so J  Cz in this case as well, and the lemma is proved. Now we can prove Lemma 18.5. J is a component of M .

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129

Proof. By the preceding lemma, K  CG (y). Since K  M , and M is a maximal subgroup of G, M = NG (K), so CG (y) ≤ M . But also NG (Q) ≤ M . However, by [V17 , 11.4.1b], J = NJ (Q), CJ (y), so J ≤ M . Since  J = QJ , it follows that J ≤ Y = CM (K). On the other hand, z ∈ K centralizes Y , so Y ≤ CG (z). But again by the preceding lemma, J  CG (z), so J  Y . Thus J is a component of Y and as Y  M , we conclude that J is a component of M , as asserted.  Now we quickly derive a final contradiction. Indeed, as Q ≤ J and Q ∈ Sylp (CM (K)), clearly KJ = Lp (M ). But again as M is a maximal subgroup of G, M = NG (KJ). On the other hand, h interchanges K and J, so h ∈ NG (KJ), whence h ∈ M . However, this is impossible as K  M , but K h = K. This completes the proof of Proposition 18.1. 19. The Case Op (M ) = 1 In this final section, we shall prove Proposition 19.1. Op (M ) = 1. Together with Proposition 18.1, this will complete the verification of the conditions(17A), thereby proving Theorem 4. Hence it will complete the proofs of Theorems C∗6 , C6 , and C∗∗ 6 , as discussed in Section 1. It will also complete the proofs of Theorems C∗7 and C7 by verifying the hypothesis of the nonexistence of p-thin configurations. (See [V7 , Section 2] for a wrap-up discussion of Theorems C7 and C∗7 .) Throughout this section we assume that Proposition 19.1 fails, i.e., Op (M ) = 1. Thus K is quasisimple and is a component of M . We first handle the case mp (K) > 1. Lemma 19.2. If mp (K) > 1, then M is not strongly p-embedded in G. We assume that Lemma 19.2 fails, through Lemma 19.13 and the subsequent argument, so that M is strongly p-embedded in G. We let S ∈ Sylp (M ) ⊆ Sylp (G), R = S ∩ K, and we assume, as we may, that S was chosen so that Q = CS (K) ∈ Sylp (CM (K)). Note that p ≤ 5, as K ∈ Tp with mp (K) > 1. Lemma 19.3. Let z ∈ I2 (CG (K)). Then CG (z) ≤ M . In particular, if z exists, then (z, K) is terminal (for the prime 2) and O2 (K) = 1. Proof. Let Rz ∈ Sylp (CM (z)), so that Rz contains some Sylow psubgroup of K, and mp (Rz ) ≥ mp (K) > 1 by hypothesis. If CG (z) ≤ M , then CM (z) is strongly p-embedded in CG (z). As K   CM (z), [IA , 7.6.2] implies that K/Op (K) ∼ = Ap or (with p = 5) K/O5 (K) ∼ = D4 (2). This contradicts K ∈ Tp , however, so CG (z) ≤ M . As K   M , O2 (K) ≤  O2 (CG (z)) = 1, G being of even type. The lemma is proved.

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Lemma 19.4. Let V be a nontrivial p -subgroup of K such that NK (V ) contains some p-element y ∈ Z(K). Then NG (V ) ≤ M . Proof. Let N = NG (V ) and N0 = N ∩ M . If N ≤ M , then N0 is strongly p-embedded in N , and y Q ≤ N0 with mp ( y Q) ≥ 3. Hence by [IA , 7.6.2], N has a p-component I with I/Op (I) ∈ Chev(p) of twisted rank 1 and mp (I) ≥ 3. In particular, Op p (N0 ) has no normal subgroup with a Sylow subgroup of order p, so Z(K) is a p -group. But then N0 /Op (N0 ) has at least two minimal normal subgroups, one in K and one in CM (K). This  contradicts the structure from [IA , 7.6.2], so the lemma holds. Lemma 19.5. CG (K) has even order and K is simple. Proof. Since Op (K) ≤ Op (M ) = 1, it suffices by Lemma 19.3 to show that CG (K) has even order. Suppose CG (K) has odd order, whence [K] = K, K is standard, M = NG (K), and CG (K) = CM (K) = O2 (M ). Then ΓQ,1 (K) ≤ M as K is standard. Choose z ∈ I2 (K) and set R0 = CR (z); indeed, choose z if possible so that R0 ≤ Z(K). Set Cz = CG (z). If Cz ≤ M , then CM (K) = CG (K)  Cz . Hence CM (K) ≤ O2 (Cz ) = 1, a contradiction as G is of even type. Therefore, Cz ≤ M. In particular, by Lemma 19.4, R0 ≤ Z(K), so R0 ≤ Q. By our choice of z and R0 , it follows that K/Z(K) has no element of order 2p. As a result, by [V17 , 17.4], p = 3, and either K ∼ = 3A6 or K/Z(K) ∼ = L (q), q = 2n , (q − )3 = 3,  = ±1. 3

and the only outer automorphisms of K of order 3 are Thus |R/Z(K)| = diagonal automorphisms of K/Z(K) ∼ = L3 (q). Also, all involutions of K are K-conjugate. Let C0 = Cz ∩ M , so that C0 is a strongly p-embedded subgroup of Cz containing Q ∈ Sylp (CM (K)). We assume, as we may, that S contains a Sylow p-subgroup Sz of C0 . Now Op (Cz ) ≤ Op (C0 ) and the structure of C0 /Op (Cz ) is given in [IA , 7.6.2]. We have [O2 (M ), K] = 1, so O2 (M ) ≤ Cz . Hence if F ∗ (Cz ) ≤ M , then ∗ [F (Cz ), O2 (M )] ≤ O2 (Cz ) = 1, a contradiction. Thus F ∗ (Cz ) ≤ M . As O2 (Cz ) ≤ ΓQ,1 (G) ≤ M there is a component J of Cz such that J ≤ C0 . Since J ∩M is strongly 3-embedded in J and J ∈ C2 , we have J/O3 (J) ∼ = A6 , U3 (3), L2 (8), L3 (4), or M11 , by [IA , 2.2.10, 7.6.1] and [V11 , 3.1]. Moreover, J = L3 (Cz ) and of course CQ (J) = 1. Next, we claim that 32

(19A)

A Sylow 2-subgroup T of NJ (Sz ) embeds in Out(K).

Indeed T ≤ C0 , and in every case CT (Q) = 1. If (19A) fails, then as CM (K) has odd order, some h ∈ T # induces a nontrivial inner automorphism on K. Let h0 ∈ K induce that automorphism of K by conjugation. Then

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131

[h, h0 ] = 1 and hh−1 0 is a 2-element acting nontrivially on Q and centralizing K. But CM (K) has odd order, contradiction. This proves (19A). As a consequence, noting that Sylow 2-subgroups of Out(K) are abelian, T is abelian as well. This rules out J/O3 (J) ∼ = L3 (4) or M11 , so (19B) J/O3 (J) ∼ = A6 , U3 (3), or L2 (8). Next, we restrict the possible isomorphism types of K even further. Suppose that K contains a subgroup A ∼ = A4 . Let V = O2 (A), so that NG (V ) ≤ M by Lemma 19.4. In particular Q ≤ O2 (CM (K)) ≤ O2 (CG (V )). Choose any v ∈ V # , write V = v, v  , and set Cv = CG (v). Then Q ≤ O2 (CCv (v  )). As G is of even type, O2 (Cv ) = 1, so by [IG , 20.6] there exists a locally unbalancing (for the prime 2) component H of Cv such that QV normalizes H and AutQ (H) is a nontrivial 3-subgroup of O2 (CAutQV H (H) (v  )). Since K has one class of involutions, v ∼K z. It follows from (19B) that H/O2 (H) ∼ = A6 , n 2 U3 (3), L2 (8), or B2 (2 2 ) for some n ≥ 3. The only locally unbalanced group among these (for the prime 2) is A6 , and the only instances of local unbalancing are from involutory automorphisms w such that O2 (CAut(A6 ) (w)) ∼ = Z5 [IA , 7.7.1ac]. Hence the nontrivial 3-group AutQ (H) does not exist, contradiction. We conclude that K contains no subgroup isomorphic to A4 . Consequently, K/Z(K) ∼ = U3 (2n ), n ≡ ±1 (mod 6). As n is odd, | Out(K)|2 = 2, so by (19A), |NJ (Sz )|2 ≤ 2. Given the choices for J in (19B), this forces J/O3 (J) ∼ = L2 (8). Now K has one class of involutions, so C0 covers OutM (K). It follows that 



O 3 (M ) = KO3 (C0 )

(19C) 



and |O3 (C0 ) : O3 (C0 ) ∩ CM (K)| = 1 or 3; in the latter case a 3-element of  O 3 (C0 ) − CM (K) induces an outer diagonal automorphism on K, and so Sz > Q. Suppose that Z(K) = 1, so that R ∼ = 31+2 . If Q = Sz , then by  3 the previous paragraph O (C0 ) ≤ CM (K), whence S = RQ. As Q is extraspecial or elementary abelian, with m3 (Q) = 2 in either case, and [R, Q] = 1, m3 (S) = 3. This is a contradiction since S ∈ Syl3 (G) (by strong 3-embedding) and m3 (G) ≥ 4. Thus Q < Sz . But in this case E32 ∼ = Q  Sz , and again by (19C), S = RSz . Thus Q  S, and as m3 (G) ≥ 4, m3 (CS (Q)) ≥ 4 by [V9 , 5.11]. However, CS (Q) = QR, and again m3 (QR) = 3, contradiction. Therefore Z(K) = 1. By (19C), |S : QR| ≤ 3, and this time QR = Q×R ∼ = Q × E32 . If Q = Sz , then S = QR with Q either elementary abelian of order 32 or extraspecial of order 33 . If Q < Sz , then QR ∼ = E34 with Q  S and R  S. In any case, S does not map onto Z3 Z3 . Hence by Yoshida’s theorem [IG , 15.19], S ≤ [NG (S), NG (S)] ≤ [M, M ]. However, M/K has extraspecial Sylow 3-subgroups of order 33 and exponent 9, so by

132

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Huppert’s theorem [IG , 15.21], S ≤ [M, M ], a contradiction. This completes the proof of the lemma.  This narrows the choices for K somewhat. ∼ L (q), q = 2n ≡  (mod 3), Lemma 19.6. Either p = 3 with K = 3  = ±1, or HJ, G2 (8), M12 , or M22 ; or p = 5 and K ∼ = F i22 . Proof. Let z ∈ I2 (CM (K)), by Lemma 19.5. Then Lemma 19.4 implies that CG (z) ≤ M , so K is a component of CG (z). As G is of even type, K ∈  C2 . But K/O2 (K) ∈ Tp with mp (K) > 1. The result follows directly. We can improve Lemma 19.5. Lemma 19.7. We have m2 (CG (K)) > 1. Proof. Suppose false and set C = CG (K). Then m2 (C) = 1, whence K does not embed in C and so, with Lemma 19.3, K is standard. We let t ∈ I2 (C). Also let z ∈ I2 (K) be 2-central in K and set Cz = CG (z) and Ct = CG (t). By the Z ∗ -theorem [IG , 15.3], C = O2 (C)CC (t). As Ct ≤ M by Lemma 19.3, and G is of even type, t inverts O2 (C) elementwise. If CC (t) is nonsolvable then it has a component isomorphic to SL2 (q) for some odd q, or 2A7 . None of these, however, is a C2 -group, and as CC (t)  CM (t) = CG (t) and G is of even type, CC (t) and C are solvable. As CC (t) is corefree, |CC (t)|2 ≤ 3, and so O2 (C) = 1. But O2 (Cz ) = 1. It follows that F ∗ (Cz ) ≤ M , and as O2 (Cz ) ≤ ΓQ,1 (G) ≤ M , Cz has a component J ≤ M . We have Q ∈ Sylp (C) and let Q ≤ Pz ∈ Sylp (Cz ). Then Pz ≤ M by strong p-embedding. Then ΓPz ,1 (Cz ) ≥ J, whence J ∗ := J, Pz  has a strongly p-embedded subgroup, namely J ∗ ∩ M . Therefore (19D) J/O2 (J) ∼ = A6 , U3 (3), L2 (8), L3 (4), or M11 , if p = 3, 1 5 and J/O2 (J) ∼ = 2F4 (2 2 ) , F i22 , or 2B2 (2 2 ), if p = 5. This is by the definition [V , 3.1]of C2 -groups and [IA , 7.6.1]. In every case as CQ (J) = 1, Ω1 (Pz ) ≤  11 N J Q (Pz ) ≤ C. Let T ∈ Syl2 (K) with z ∈ Z(T ). Then T maps into CAut(J) (Ω1 (Pz )), in which Sylow 2-subgroups are trivial unless J/O2 (J) ∼ = L3 (4) or F i22 , in which cases they are of order 2, by [V17 , 12.2]. Hence T0 := CT (J) satisfies |T : T0 | ≤ 2, with equality possible only if J/O2 (J) ∼ = L3 (4) or F i22 . We claim that K has no element y of order 2p. For if it did, we could take y so that u := y p ∈ T0 , by the Thompson transfer lemma, and then CG (u) ≤ M . But then by Lemma 19.4, CG (u) ≤ M , contradiction. Given the possibilities for K in Lemma 19.6, we see by [V17 , 17.4] that p = 3 and K ∼ = L3 (2n ), with n ≡ ±2 (mod 6) if  = +1 and n ≡ ±1 (mod 6) if  = −1.  Let E ∈ E2∗ (T ). Then E # ⊆ z K . For all e ∈ E # let Je = O3 (E(CG (e))), so that Je /O2 (Je ) ∼ = J/O2 (J) is as in (19D). Moreover, E maps into CAut(Je /O2 (Je )) (AutQ (Je /O2 (Je ))), whose Sylow 2-subgroup is of order at most 2 by [V17 , 12.2]. Thus Je /O2 (Je ) is centralized by a hyperplane Ee

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of E so Je = Je for all e ∈ Ee# . As m2 (E) ≥ 4, Je = J for all e ∈ E # , so J ≤ CG (E). If  = +1, then NK (E) contains an element v of order 3 and then J ≤ CG (E) ≤ M by Lemma 19.4, contradiction. So  = −1, i.e., K∼ = U3 (2n ), n ≡ ±1 (mod 6). Let H ≤ NK (E) with H cyclic of order (q+1)/3, a {2, 3} -number. Then  [H, E] = 1 so H normalizes J = O3 (CG (e)) while centralizing a Sylow 3subgroup Pz of J. Hence by [V17 , 12.2], [J, H] = 1. Now CK (H) contains a 3-element v, so by Lemma 19.4, NG (H) ≤ M . Therefore J ≤ M , a final contradiction. The proof is complete.  Lemma 19.8. We have p = 3 with K/O2 (K) ∼ = L3 (q), q ≡  (mod 3), n  = ±1, q = 2 , or G2 (8); or p = 5 with K/O2 (K) ∼ = F i22 . Moreover, K is standard in G. Proof. By Lemma 19.3, K is terminal in G, relative to the prime 2; and Lemma 19.6 gives the possible isomorphism types of K. In particular m2 (K) > 1, so by [II3 , Theorem PU∗4 ], K is standard in G, relative to the prime 2. Let Q2 ∈ Syl2 (CG (K)). Then as M is not strongly embedded in G, there is g ∈ G−M such that m2 (Qg2 ∩M ) ≥ 2, by [II3 , Theorem PU∗2 ]. If the lemma fails, then by Lemma 19.6 and as K ∈ C2 , p = 3 with K/O2 (K) ∼ = HJ, M12 , or M22 . As | Out(K)| = 2, there is u ∈ I2 (Q2 ) such that ug induces an inner automorphism on K, and hence by [IA , 5.3bcg], CM (ug ) contains an element w of order 3. But CG (u) ≤ M by Lemma 19.3 so w ∈ CG (ug ) ≤ M g . Thus w ∈ M ∩ M g , so by strong 3-embedding, g ∈ M , contradiction. The lemma follows.  Again let Q2 ∈ Syl2 (CM (K)). Lemma 19.9. The following conditions hold: (a) K/O2 (K) ∼ = L3 (2n ), n ≡ ±2 (mod 6) if  = +1 and n ≡ ±1 (mod 6) if  = −1, or G2 (8), with p = 3; or K/O2 (K) ∼ = F i22 with p = 5; (b) There exists g ∈ G − M such that Qg2 ≤ M , and for any such g, Ω1 (Qg2 ) induces faithful inner automorphisms on K; and (c) For every involution v ∈ CG (K), CCG (K) (v) is a p -group. Proof. As in [II3 , Section 16] we choose g ∈ G − M to maximize m := m2 (Qg2 ∩ M ) and then |Qg2 ∩ M |. By [II3 , Theorem PU∗2 ], m ≥ 2. Then Ω1 (Qg2 ∩ M ) induces inner automorphisms on K. If p = 3, this is because K, as given by the previous lemma, is outer well-generated for the prime 2 [V17 , 5.8]. If p = 5, it is because of strong 5-embedding and the fact that every outer involutory automorphism of K/O2 (K) ∼ = F i22 centralizes an element g of order 5 [IA , 5.3t]. Since Ω1 (Q2 ∩ M ) induces inner automorphisms on K and K is simple, we may apply [II3 , 16.11]. We conclude that g ∈ G − M may be chosen so that Qg2 ≤ M and Q2 ≤ M g . In particular, (b) holds. Now if there is u ∈ I2 (Q2 ) with CM (ug ) containing a nontrivial p-element

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134

y, then y ∈ M ∩ M g , contradicting the strong p-embedding of M in G; so for every ug ∈ I2 (Qg2 ), CM (ug ) is a p -group. When p = 3, this condition forces the congruences on n modulo 6 in (a). It also implies in any case that Qg2 ∩ K = 1. It remains to prove (c). We now have that the projection of Ω1 (Qg2 ) on Ω1 (Q2 ) is injective, hence surjective. Thus, every involution v of Q2 is the projection of some involution ug on Q2 . Then CCM (K) (v) = CCM (K) (ug ) is  a p -group, as asserted. The proof is complete. Lemma 19.10. F ∗ (CG (K)) is a simple group of order divisible by p. However, CG (K) has no elements of order 2p. Proof. The last sentence is a restatement of Lemma 19.9c. As K is standard, K  M , so Op (CG (K)) = Op (CM (K)) ≤ Op (M ) = 1. Hence F (CG (K)) = Op (CG (K)). But m2 (CG (K)) > 1, so as CG (K) has no elements of order 2p, F (CG (K)) = 1. Therefore, F ∗ (CG (K)) is the direct product of simple groups, all of which have order divisible by p. But then the lack of elements of order 2p implies that F ∗ (CG (K)) is simple. The proof is complete.  Set I = F ∗ (CG (K)). Lemma 19.11. We have mp (I) > 1. Proof. Otherwise, since mp (CG (K)) ≥ 2 and CG (K)  M , Op (M ) < M by [V17 , 3.1]. But M is strongly p-embedded in G, so Op (G) < G by  [V9 , 11.1], contradicting the simplicity of G. Lemma 19.12. M is a 2-component I-preuniqueness subgroup. Proof. Suppose that y ∈ I2 (CG (I)) and set Cy = CG (y) and Cy0 = Cy ∩ M . As p divides |I|, y ∈ M . Suppose that Cy0 < Cy . Then Cy0 is strongly p-embedded in Cy and has the simple component I. By [IA , 7.6.2], this forces p = 5 and I ∼ = D4 (2) (we are using the fact that A10 ∈ C2 ). But D4 (2) has elements of order 10, against Lemma 19.10. Therefore Cy0 =  Cy ≤ M , and the lemma follows. Lemma 19.13. We have p = 3. Moreover, I ∼ = A6 or L3 (2n ),  = +1 with n ≡ ±2 (mod 6), or  = −1 with n ≡ ±1 (mod 6). Proof. We have I ∈ C2 as G is of even type; I has no element of order 2p by Lemma 19.10; and mp (I) > 1 by Lemma 19.11. Now [V17 , 17.2] implies the result.  We now complete the proof of Lemma 19.2. Recall the restrictions on K and I from Lemmas 19.9a and 19.13. We have I  M . Let Q1 ∈ Syl2 (CG (I)) with Q1 , Q2  a 2-group. By [V17 , 5.8] and Lemma 19.13, I is outer wellgenerated (for the prime 2). Hence by [II3 , Theorem PU∗2 ] and [II3 , 16.13], there is h ∈ G − M such that Qh1 ≤ M and Ω1 (Qh1 ) embeds in Aut(I). Moreover, ΓQh ,1 (I) < I. 1

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135

If K ∼ = G2 (8), then Q1 = Ω1 (Q1 ) embeds in Inn(I) as I is outer wellgenerated (for the prime 2). Hence Q1 must be of class 2, contradiction. Therefore K ∼ G2 (8). Likewise, if I ∼ = = A6 , then Ω1 (Q1 ) embeds in Q2 ∩ I ∼ = D8 , which is not the case. Let R1 = Q1 ∩ K and R2 = Q2 ∩ I. By outer wellgeneration, Ω1 (R1 ) and Ω1 (R2 ) embed in each other, so Ω1 (R1 ) ∼ = Ω1 (R2 ). ∼ This forces K = I, given the restrictions on the isomorphism types of K and I. Let Zi = Z(Ri ), i = 1, 2. Then M fuses each of Z1# , Z2# , and Z1 Z2 − Z1 − Z2 completely. As Z1h ≤ Z1 Z2 with Z1h ∩Q1 = 1, and Z2g ≤ Z1 Z2 with Z2g ∩Q2 = 1, there must exist k ∈ G such that Z1k ∩ Z2 = 1. But both Z1 and Z2 centralize elements of M of order 3, so 3 divides |M ∩ M k |, whence k ∈ M . But Z1k ≤ K  M and Z2 ≤ I, and K ∩ I = 1, so we have a contradiction. This completes the proof of Lemma 19.2. Now we turn to the case mp (K) = 1. Let S ∈ Sylp (M ), chosen so that Q = CS (K) ∈ Sylp (CM (K)), and set P = S ∩ K. We first prove Lemma 19.14. M is not almost p-constrained. Proof. Suppose M is almost p-constrained [I2 , 8.2]. As Op (M ) = 1, this means that S = P ×Q, F ∗ (M ) = K×Q, and Q ∈ Sylp (J), where J is a pcomponent of CG (P ) and J/Op (J) ∈ Chev(p) has twisted Lie rank 1 and prank at least 3. Moreover, ΓQ,1 (G), ΓS,2 (G) ≤ M . In particular, Op (J) ≤ ΓQ,1 (G) ≤ M , so [Q, Op (J)] ≤ Q ∩ Op (J) = 1 and J is quasisimple. Let z ∈ I2 (K) and Cz = CG (z). Then Q ≤ CM (z) so Op (Cz ) ≤ ΓQ,1 (G) ≤ M , whence [Op (Cz ), Q] ≤ Op (Cz ) ∩ Q = 1. In particular, [Op (F ∗ (Cz )), Q] = 1. As G is of even type, F ∗ (Cz ) = O2 (Cz )E(Cz ), so if  we set X = Op (E(Cz )), then CQ (X) = CQ (F ∗ (Cz )) ≤ Op (Cz ) = 1. Let I then be a component of X such that [I, Q] = 1, so that I ≤ [I, Q]. If I ≤ M , then I ≤ [I, Q] ≤ Q, which is absurd. Hence I ≤ M , so I ≤ ΓQ,1 (X). As mp (Q) ≥ 3, this implies that I ∈ Chev(2), by [IA , 7.3.7] and [V9 , 9.3]. Let Q∗ ∈ Sylp (Cz ) with Q ≤ Q∗ . Since ΓS,2 (G) ≤ M and Q is not cyclic, Q∗ ≤ M , whence Q∗ = Q × T with T cyclic. Now Q normalizes I and CQ (I) = 1, for otherwise I ≤ ΓQ,1 (X) ≤ M , contradiction. Moreover, Op (I) ≤ O2 (Cz ) = 1 and by [V17 , 5.7], p does not divide | OutG (I)|, so mp (I) ≥ mp (Q) ≥ 3. Let QI and Q∗I be the projectionsof Q and Q∗ on  I, so that Q∗ ∈ Syl (I) and Q∗ /QI ∼ = P/T is cyclic. Then Γ (I), ΓQ∗ ,2 ≤ I

p

I

QI ,1

I

NI (QI ) and mp (QI ) ≥ 3. But by [V17 , 9.11], there is no such group I ∈  C2 − Chev(2). This contradiction completes the proof. We next prove Lemma 19.15. M is not wreathed.

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Proof. Suppose that M is wreathed [I As Op (M ) = 1, this  2 , M8.1].  ∗ ∗ = K1 × · · · × Kp , with means that p ≥ 5 and K := F (M ) = K Ki for all 1 ≤ i ≤ p. Moreover, if S ∈ Sylp (M ) and we set K1 = K ∼ =   R = S ∩ K M , then |S : R| = p and ΓR,1 (G), ΓS,2 (G) ≤ M . We write Ω1 (R) = y1 , . . . , yp  with yi ∈ Ip (Ki ), 1 ≤ i ≤ p. We shall reach a contradiction by showing that K1 is standard in G. (Since K1 and K2 are M -conjugate, K1 is not standard in G, however.) For every u ∈ K ∗ , let n(u) be the number of non-identity ui ’s when we write (19E)

u = u1 · · · up with ui ∈ Ki for all i.

First, suppose that u ∈ I2 (K ∗ ) with n(u) ≤ p − 3. We claim that for any i such that ui = 1 in (19E), Ki   CG (u). In proving this it suffices to assume that u1 = u2 = u3 = 1 and show that K1   CG (u). Now for any y ∈

y2 , y3 # , CG (y) ≤ ΓR,1 (G) ≤ M , so K1 ≤ CCG (u) (y) ≤ CM (u). Therefore K1 is a component of E(CCG (u) (y)). Set Cu = CG (u) and C u = Cu /Op (Cu ). Then by Lp -balance and [III8 , 1.16], and as p > 3, the subnormal closure J of K1 in Cu satisfies either J = K 1 or J ∼ = An ∈ C2 for some n > p2 ≥ 25. Moreover, as K1   M , K1 centralizes every p -subgroup of M that it normalizes. This includes Op (Cu ) ≤ Γy2 ,y3 ,1 (G) ≤ ΓR,1 (G) ≤ M , so K1 is a component of Cu . In particular K1 ∈ C2 as G is of even type, so K1 = J   Cu , as claimed. Next, let z ∈ I2 (K ∗ ) with n(z) < p. We again argue that if z projects trivially on some Ki , then Ki   CG (z) =: Cz . For specificity assume that i = 1. Then since Op (K ∗ ) ≤ Op (M ) = 1, m2 (Kj ) ≥ 2 for all j, and we choose four-groups Vj , 2 ≤ j ≤ p, such that [Vj , z] = 1 for all j ≥ 2. Let V = Vp−1 Vp . As p ≥ 5, the previous paragraph implies that K1   CG (v) for all v ∈ V # . Then K1   CCz (v) for all such v. Again as all components of Cz are C2 -groups with O2 (Cz ) = 1, and since m2 (V ) ≥ 4, we deduce from L2 -balance and [III8 , 1.16] that K1   CG (z). Finally, pick any involution z ∈ CG (K1 ). Then z ∈ CG (y1 ) ≤ ΓR,1 (G) ≤ M , so z permutes K2 , . . . , Kp in orbits of length 1 or 2. As p − 1 ≥ 4, and O2 (M ) ≤ Op (M ) = 1, it follows that m2 (CK2 ···Kp (z)) ≥ 4. This time let E24 ∼ = V ≤ CK2 ···Kp (z). By the previous paragraph, K1 is a component of CG (v) for all v ∈ V # , and we deduce as before that K1 is a component of CG (z). Thus, K1 is terminal in G (with respect to the prime 2). As M is not strongly embedded in G by the Bender-Suzuki theorem, K1 is standard in G by [II3 , Theorem PU4 ]. But this is obviously false, as noted above, so the proof is complete.  Finally we deal with the strongly p-embedded case. Lemma 19.16. If mp (K) = 1, then M is not strongly p-embedded in G.

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137

We proceed in a sequence of lemmas. Let C = CG (K). Then by strong Since mp (K) = 1 and Op (M ) = 1, K ∩ C = 1. p-embedding, C =CM (K).  ∗ M We also set K = K . Lemma 19.17. Op (C) = 1 and K is simple.    Proof. Since K   M , K ∗ = K M = KOp (E(K ∗ ∩ C)). But  [O p (E(K ∗ ∩ C)), Op (C)] = 1 so Op (C) ≤ CM (K ∗ )  M . Therefore Op (C) ≤ Op (CM (K ∗ )) ≤ Op (M ) = 1, as claimed. Since mp (K) = 1, Z(K) ≤ Op (C) = 1, so K is simple.  Lemma 19.18. Let z ∈ I2 (C). Suppose CC (z) contains an element e of order p, or that CM (z) contains an element e of order p inducing a field automorphism on K. Then CG (z) ≤ M . Proof. Under either hypothesis set C ∗ = C e, so that C ∗ ∩ K = 1. Let Cz = CG (z), C0 = Cz ∩ M , and Pz ∈ Sylp (CM (z)). Then PK := Pz ∩ K and PC := Pz ∩ C ∗ are nontrivial, so mp (C0 ) ≥ 2. Moreover, if the lemma fails, then C0 is strongly p-embedded in Cz . As K is a component of C0 , we see from [IA , 7.6.1, 7.6.2] that K ∼ = Ap ∈ Cp or (for p = 5) D4 (2) ∈ G5 . But  K ∈ Tp , a contradiction in either case. Lemma 19.19. There exist involutions t ∈ K such that mp (CM (t)) ≥ 3. For any such t, CG (t) ≤ M . Proof. By [V17 , 3.1], there exists t ∈ I2 (K) such that CAut(K) (t) covers a Sylow p-subgroup of Out(K). Fix such an element t, so that CM (t) ≤ NM (K), and let Pt ∈ Sylp (CM (t)). By [V17 , 19.1a], mp (NM (K)) = mp (M ) = mp (G) ≥ 4. As mp (Aut(K)) ≤ 2 by [V17 , 3.1], mp (C) ≥ 2. If p divides |CK (t)|, then mp (CKC (t)) = mp (CK (t)) + mp (CC (t)) ≥ 3, as claimed. If CK (t) is a p -group, then Pt is a complement to a Sylow p-subgroup of K in a Sylow p-subgroup of NM (K), so again mp (Pt ) ≥ mp (NM (K)) − 1 ≥ 3. Thus the first statement holds. For the second statement, let Ct = CG (t), C t = Ct /Op (Ct ), and C0 = CM (t), and suppose that C0 < Ct . Then C0 is strongly p-embedded in Ct . Then by [IA , 7.6.1], Ct has a p-component I such that I ∈ Chev(p) has twisted Lie rank 1. Moreover, by strong p-embedding, I ≥ Pt ∈ Sylp (C0 ) ⊆ Sylp (Ct ). Now NI (Pt ) acts irreducibly on Pt /Φ(Pt ), by [V17 , 12.3], whereas Pt /Pt ∩ C is metacyclic. Hence Pt ≤ C ≤ Ct , whence Pt ∈ Sylp (C). Then Op (Ct ) ≤ ΓPt ,1 (Ct ) ≤ C0 , so as C  C0 , [Op (Ct ), Pt ] ≤ Op (C) = 1 by Lemma 19.17. Hence [Op (Ct ), I] = 1 so I  E(Ct ). Since G is of even type, I ∈ C2 . But this is impossible, given the possible isomorphism types of I.  Therefore C0 = Ct and the lemma is proved.

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We fix any t ∈ K satisfying the conditions of Lemma 19.19 and use it to prove: Lemma 19.20. The following conditions hold: (a) F (C) = 1; and (b) Every component of E(C) is a simple C2 -group. (c) Every component of E(C) either is a Cp -group or a TGp -group, or has p-rank 1. Proof. By Lemma 19.17, F (C) = Op (C). Just as in Lemma 19.17, K ∗ = K(E(K ∗ ) ∩ C) ≤ KE(C) so [K ∗ , F (C)] = 1. Consequently, F (C) ≤ F (CM (K ∗ )) ≤ F (M ), so F (C) ≤ Op (M ). With t as in Lemma 19.19, we have F (C) ≤ CG (t) ≤ M , so F (C) ≤ Op (CG (t)) ≤ O2 (CG (t)) = 1, as G is of even type. Thus (a) holds. In (b) and (c), every component I of C is a component of CM (Ω1 (S ∩ K)) = CG (Ω1 (S ∩ K)). As mp (NM (K)) ≥ 4, I ∈ Lop (G). Now d0 = 5 as mp (K) = 1, so (c) holds by the hypothesis of Theorem C∗6 , unless the exceptional case holds and I ∼ = A7 . As A7 ∈ C2 , it suffices to prove (b) to complete the proof of the lemma. Let K0∗ = E(K ∗ ∩ C), a (possibly trivial) product of M -conjugates of K. Let I be a component of E(C) which is not a component of K0∗ . Then [I, K0∗ ] = 1 = [I, K ∗ ] so I is a component of E(CM (K ∗ )). Hence I is a component of E(M ), as K ∗  M . But CG (t) ≤ M and [t, I] = 1, so I is a component of CG (t). Thus I ∈ C2 , again as G is of even type. It remains to show that the components of K0∗ , all isomorphic to K, are C2 -groups. We may assume that K0∗ = 1; let K2 be a component of K0∗ . It is enough to show that there exists u ∈ Ip (C) and z ∈ I2 (C) such that [u, z] = 1. For then, CG (z) ≤ M by Lemma 19.18, whence K  CG (z) and K ∈ C2 . But if u and z do not exist, then K2 = E(C) and F ∗ (M ) = KK2 . As mp (M ) ≥ 4 and mp (Aut(K)) ≤ 2, some u ∈ Ip (C) must induce an outer automorphism on K2 . Then CK2 (u) has even order and we can take z ∈ I2 (CK2 (u)), contrary to our assumption. This contradiction completes the proof of (b) and the lemma.  Lemma 19.21. K is terminal with respect to the prime 2, i.e., K is a component of CG (z) for every z ∈ I2 (CG (K)). Proof. Recall that C = CG (K) = CM (K). If there is f0 ∈ Ip (NM (K)) inducing a nontrivial field automorphism on K, set f = f0 . Otherwise set f = 1, and in any case set C ∗ = C f . Since mp (NM (K)) ≥ 4, mp (C ∗ ) ≥ 3. Let Y = {z ∈ I2 (C ∗ ) | K   CG (z)}, so that we must prove Y = I2 (C ∗ ). We first assume that F ∗ (C ∗ ) = F ∗ (C). Set Y0 = {z ∈ I2 (C ∗ ) | mp (CC ∗ (z)) > 0}.

19. THE CASE Op (M ) = 1

139

By Lemma 19.18, Y0 ⊆ Y . We wish to quote [V17 , 15.1] with C ∗ in the role of X there. Thus we set   Yi+1 = Yi ∪ z ∈ I2 (X) A# ⊆ Yi for some A ∈ E22 (CX (z)) , i = 0, 1, 2 . . . , and by [V17 , 15.1] it suffices to prove that if Yi ⊆ Y , then Yi+1 ⊆ Y . That is, we need to show that if a four-group A ≤ C ∗ satisfies a ∈ Y for all a ∈ A# , then y ∈ Y for any y ∈ I2 (CC ∗ (A)). Fix such a subgroup A and an involution y ∈ CC ∗ (A). Let I be the subnormal closure of K in CG (y). Then by [V9 , 6.6], I is not a diagonal pumpup of K, and furthermore, I is quasisimple, so we need only rule out the possibility that I is a vertical pumpup of K. But in that case, [III8 , 1.16] applies, yielding (19F)

(K, I) ∼ = (An , An+4k ), k > 0, (A5 , M12 ), (A5 , HJ), 3 (G2 (4), Co1 ), (L3 (4), Suz), (22 L3 (4), He), or (2B2 (2 2 ), Ru).

The An+4k and He cases are impossible as I ∈ C2 as G is of even type. In the remaining cases, note that the prime p, for which mp (K) = 1, does not divide | Out(K)|. Hence mp (KC) ≥ 4 and mp (C) ≥ 3. Thus, by Lemma 19.19, CG (z) ≤ M for every z ∈ I2 (K). Indeed as Z(K) = 1, CG (z) ≤ NG (K) for all such z. In particular, ΓS,1 (I) normalizes K, where S ∈ Syl2 (K). But this is impossible by [V17 , 5.14]. Thus the conclusion of the lemma holds if F ∗ (C ∗ ) = F ∗ (C). Finally assume that F ∗ (C ∗ ) > F ∗ (C). As C ∗ = C f  with f p = 1, F ∗ (C ∗ ) = F ∗ (C) × e for some e of order p, e ∈ C. But C  NM (K), so C  C ∗ and thus [C, e] ≤ C ∩ e = 1. That is, e ∈ Z(C ∗ ). As a result,  Lemma 19.19 yields Y0 = I2 (C ∗ ) and the lemma is proved. This has several quick consequences. Lemma 19.22. The following conditions hold: (a) (b) (c) (d)

K is standard in G and normal in M ;   O p (M ) = K × Op (C); mp (C) ≥ 3; For T ∈ Syl2 (M ), ΓT ∩K,1 (G) ≤ M and ΓT ∩C,1 (G) ≤ M .

Proof. Part (a) follows directly from Lemma 19.21 and [II3 , Theorem PU4 ]. Since M is strongly p-embedded in G, it controls p-transfer in G by [V9 , 11.1]. As G is simple, M = Op (M ), which implies by [V17 , 3.1b]  that Op (AutM (K)) = Inn(K), as mp (K) = 1. This implies (b), which immediately implies (c) since mp (M ) = mp (G) ≥ 4. Then Lemma 19.19 and (c) imply that ΓS∩K,1 (G) ≤ M , while ΓS∩C,1 (G) ≤ M holds as K is standard. The proof is complete. 

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We number the components of M as K = K1 , K2 , . . . , Ks , s ≥ 2. Now we can prove Lemma 19.23. Each Ki is standard in G (with respect to the prime 2). In particular, Ki  M = NG (Ki ). Proof. As usual it suffices to prove that Ki is terminal in G with respect to the prime 2, and we may assume by the previous lemma that i > 1. As Op (M ) = 1 and M is strongly p-embedded in G, CG (Ki ) ≤ M . Hence we take any z ∈ I2 (CM (Ki )) and aim to prove the pumpup I of Ki in CG (z) equals Ki . This certainly holds if CG (z) ≤ M , so by Lemma 19.22d it holds for all z ∈ K. Note that I is semisimple since G is of even type. As in the proof of Lemma 19.20, we claim that if A is a four-subgroup of CM (Ki z) such that Ki is a component of CG (a) for all a ∈ A# , then Ki is a component of CG (z). Indeed by L2 -balance and [IG , 6.19], the pumpup I of Ki in CG (z) is trivial or vertical, with trivial core as G is of even type. But in the vertical case again by [III8 , 1.16], (Ki , I) must be one of the pairs in (19F). This time Lemma 19.22d implies that ΓS∩Ki ,1 (I) normalizes Ki , which violates [V17 , 5.14]. Thus our claim holds. If z ∈ KC, then z centralizes a four-subgroup A of K, so by our claim Ki is a component of CG (z). Now let z ∈ I2 (CM (Ki )) be arbitrary. If there are at least three Kj ’s, then any z ∈ I2 (CM (Ki )) normalizes K and some product I of Kj ’s with 1 = j = i, so CKI (z) contains a four-group A ≤ KC. Thus our claim again implies that Ki is a component of CG (z). This finishes the proof unless F ∗ (M ) = KK2 and i = 2. But in that case as mp (C) ≥ 3 by Lemma 19.22c, mp (Aut(K2 )) ≥ 3 and so mp (K2 ) ≥ 2 by [V17 , 3.1]a. If I > K2 , then I ≤ M , so I ∩ M is a strongly p-embedded subgroup of I, with component K2 . Then by [IA , 7.6.2], the only possibilities are K2 ∼ = Ap or D4 (2), the last with p = 5. But in both cases mp (Aut(K2 )) < 3, a  contradiction. Hence I = K2 and the proof is complete. Now let Qi ∈ Syl2 (CG (Ki )) for each i = 1, . . . , s; note that CG (Ki ) ≤ NG (Ki ) = M . Thus  m2 (Kj ). (19G) m2 (Qi ) ≥ j =i

Since Ki is standard, (19H)

ΓQi ,1 (G) ≤ NG (Ki ) = M ,

so M is a Ki -preuniqueness subgroup of G. Now M is not strongly embedded in G, by the Bender-Suzuki theorem [II2 , Theorem SE] and the structure of M . Therefore by [II3 , Theorem PU∗2 ], for each i there is gi ∈ G − M such that m2 (Qgi i ∩ M ) ≥ 2. We set Vi = Qgi i ∩ M. Replacing gi by a suitable element of gi M , we may and shall assume that

Vi , Qi  is a 2-group.

19. THE CASE Op (M ) = 1

141

By the simplicity of Ki and [II3 , 16.11], if gi is chosen to maximize m2 (Vi ), and then subject to this, to maximize |Vi |, then the following hold: (1) m2 (Vi ) = m2 (Qi ); and (2) For any A ∈ E2∗ (Vi ), m2 (A0 ) ≥ m2 (Vi )−1, where A0 (19I) is the subgroup of A inducing inner automorphisms on Ki . Also by [II3 , 16.9a] and the strong p-embedding of M in G, (19J)

ΓVi ,1 (G) ≤ M gi , Vi acts faithfully on Ki , and ΓVi ,1 (Ki ) is a p -group.

Indeed if p divides |ΓVi ,1 (Ki )|, then it divides |M gi ∩ M |, so gi ∈ M , a contradiction. Now we can prove Lemma 19.24. We have F ∗ (M ) = K1 K2 , with K1 = K, K2 ∈ Chev ∩ C2 ∩ Cp , and mp (AutM (K2 )) ≥ 3. Proof. First suppose that s ≥ 3. Among K1 , . . . , Ks , fix an index i so that m2 (Ki ) ≤ m2 (Kj ) for all j = i. Then by (19G), and as all Kj are simple, m2 (Qi ) ≥ m2 (Kj ) + 2 ≥ m2 (Ki ) + 2 (for any j = i). But then using (19I), m2 (Ki ) = m2 (Inn(Ki )) ≥ m2 (Vi ) − 1 = m2 (Qi ) − 1 ≥ m2 (Ki ) + 1, a contradiction. Hence as s > 1, s = 2, and we may assume that K = K1 . We have Ki  M , i = 1, 2.  Using Lemma 19.22b, mp (AutM (K2 )) = mp (Op (C)) = mp (M ) − 1 ≥ 3. Hence mp (K2 ) > 1 by [V17 , 3.1a]. Suppose that K2 is a TGp -group in Chev. Then as mp (K) = 1, hypothesis (d) of Theorem C∗6 implies that mp (K2 ) = 2. Hence by [V17 , 3.7], Op (M ) < M . As usual, since M is strongly p-embedded in G, this leads to the contradiction Op (G) < G, by [V9 , 11.1]. So K2 ∈ TGp ∩ Chev. Also if K2 ∈ Spor, then | Out(K2 )| ≤ 2 so mp (K2 ) ≥ 3. One easily checks from [IA , 5.6.1] and [IA , 5.3] that for any involution z ∈ Aut(K2 ), p divides |CK2 (z)|. Hence ΓV2 ,1 (K2 ) is not a p -group, contradicting (19J). So K2 ∈ Spor. Likewise if K2 ∼ = An for some n, then mp (K2 ) ≥ 3, so n ≥ 3p. Hence any involution of Aut(K2 ) ∼ = Σn centralizes an element of order p in K2 , and we reach the same contradiction as in the sporadic case. We have shown that K2 ∈ Chev − TGp . It follows  that K2 ∈ Chev ∩ Cp ∩ C2 , proving the lemma. Lemma 19.25. We have p = 3 and K2 ∼ = U4 (2), U5 (2), U6 (2), Sp6 (2), D4 (2), or F4 (2). Moreover, if S2 ∈ Syl2 (K2 ), then Ω1 (S2 ) has class at least 3 and sectional 2-rank at least 4. Proof. We use Lemma 19.24 and the definitions of Cp and C2 [V11 , 3.1]. If K2 ∈ Chev(p) − Chev(2), then p = 3 and K2 ∼ = L3 (3), L± 4 (3), or  G2 (3). Hence ΓV2 ,1 (K2 ) = K2 by [IA , 7.3.3], contradicting (19J). Thus K2 ∈

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Chev(2) ∩ Cp . As in the previous lemma, mp (AutM (K2 )) ≥ 3 but AutM (K2 ) contains no field automorphism of order p or graph automorphism of K2 ∼ = 3D (q) of order p = 3. The lemma then follows by [V , 7.15].  4 17 n Lemma 19.26. K1 ∼ = L2 (2n ) for some n ≥ 2, L− 3 (q) for some q = 2 ≡  (mod 3),  = ±1, or L2 (q), q ∈ F M. In particular K1 is outer wellgenerated for the prime 2.

Proof. As p = 3, m3 (K1 ) = 1. Then since K1 ∈ C2 , the lemma follows from [V17 , 3.4] and [V17 , 5.8].  Now we quickly complete the proof of Lemma 19.16 by examining the action of Qg11 ∩ M on K1 . As K1 is outer well-generated, Qg11 ≤ M by [II3 , 16.13], so V1 ∼ = Q1 . Indeed Ω1 (V1 ) induces inner automorphisms on K1 . Hence by Lemma 19.26 and [IA , 3.2.2], Ω1 (V1 ) has class at most 2 or is dihedral. But V1 ∼ = Q1 contains a copy of S2 ∈ Syl2 (K2 ), so Ω1 (V1 ) has class at least 3 and has sectional 2-rank at least 4, by Lemma 19.25. This is a contradiction, so Lemma 19.16 is proved. Together, Lemmas 19.2, 19.14, 19.15, and 19.16 show that whether mp (K) > 1 or not, one of (17A1,2) fails, thus contradicting Proposition 18.1. This completes the proof of Theorem 4. As discussed at the beginning of this section, the proofs of Theorems C∗6 , ∗ C6 , C∗∗ 6 , C7 , and C7 are now complete.

CHAPTER 11

Theorems C4 and C∗4 : Introduction 1. Theorem C∗4 and Its Cases The partitions Kp = Cp ∪ Tp ∪ Gp , one for each prime p [I2 , Definitions 12.1, 13.1] have led to three theorems dealing with K-proper simple groups G of restricted even type: Theorems C5 [V1 ], C∗∗ 6 [this volume, Chapter 10], and C∗7 [III1 , p. 5]. Together they imply that (1A)

If p is an odd prime such that m2,p (G) ≥ 4, then G ∈ K or G has a strong p-uniqueness subgroup.

∗ Indeed, in the situations of Theorems C∗∗ 6 and C7 , the strong p-uniqueness subgroup in question is of component type. In any case, if G is of restricted even type and e(G) ≥ 4, it remains to consider the case in which for every prime p > 2 such that m2,p (G) ≥ 4, G has a strong p-uniqueness subgroup. Recall that e(G) = max{m2,p (G) | p an odd prime}.

In this chapter we begin consideration of the last remaining theorem of the seven that form the “Classification Grid” [I2 , Section 3], namely Theorem C4 , the treatment of the “quasithin type” case for groups of special even type. For a variety of reasons – stronger results in the literature, better proofs, overlooked cases – as our project has evolved, the context and content of several theorems have changed, and most of these changes have reshaped the precise theorem now necessary to close out the “quasithin type” case in the Classification Grid. We call this new result Theorem C∗4 , formulate it below, and prove it in this volume and the next. For example, we are fortunate that the “classical” quasithin case, defined by the condition e(G) ≤ 2, has been completely settled by Aschbacher and Smith [ASm1], [ASm2] for groups of restricted even type. Having added this reference to our Background Results, and in view of (1A), we can now immediately assume that e(G) = 3 in the proof of Theorem C∗4 . In Theorems C∗7 and C∗∗ 6 , moreover, we have proved more than Theorems C7 and C6 . Namely, in the restricted even type case, Theorems C∗7 and C∗∗ 6 apply to any odd prime p such that Lop (G) ∩ Gp = ∅, and Lop (G) ∩ Tp = ∅ = Lop (G) ∩ Gp , respectively. As a result, the assumption m2,p (G) ≥ 4 is effectively weakened in these stronger theorems to mp (G) ≥ 4 (for which the notion of strong p-uniqueness subgroup is still well-defined [I2 , Section 8]). 143

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Finally, although we originally conceived of a proof of Theorem C4 by 2-local analysis, cf. [I2 , Section 22], we now shall give a treatment of Theorem C∗4 using odd local analysis as well, following the fundamental groundbreaking papers of Aschbacher [A13], [A24]. Now, Theorem C∗4 has a purpose analogous to (1A), but for the case e(G) = 3. The strength of Theorems C∗7 and Theorems C∗∗ 6 makes it possible to divide the proof of Theorem C∗4 into three cases, in one of which e(G) ≤ 2 and in the other two of which e(G) = 3. Define λ(G) = {p > 2, p prime | m2,p (G) = 3} λu (G) = {p ∈ λ(G) | G has a strong p-uniqueness subgroup} We remark that the term “strong p-uniqueness subgroup” has only been defined so far for primes p > 2 such that mp (G) ≥ 4 [I2 , Section 8]. This will not be an issue in the current volume, where we shall always have mp (G) > 3, and so we shall postpone defining it in the case mp (G) = 3 until the next volume. Of course this leaves the full meaning of λu (G) and Theorem C∗4 unspecified for the moment. Now we state our Theorem C∗4 . Theorem C∗4 . Let G be a K-proper simple group of restricted even type with e(G) ≤ 3. Then either e(G) = 3 and λ(G) = λu (G), or one of the following holds: (a) G ∼ = A12 ; or ∼ Co2 , Co3 , Suz, F3 , or F5 ; or (b) G = ∼ (c) G = U5 (2), U6 (2), D4 (2), Sp8 (2), or F4 (2); or (d) G ∼ = L4 (2n ) (n > 1), Sp6 (2n ) (n > 1), 2 D4 (2n ) (n ≥ 1), L6 (2), or L7 (2); or (e) G ∼ = L4 (3), U4 (3), or G2 (3); or ∼ (f) G = M22 , M23 , M24 , HJ, J3 , J4 , HS, He, or Ru; (g) G ∼ = L2 (2n ) (n > 2), L3 (2n ) (n > 1), U3 (2n ) (n > 2), U4 (2n ) (n ≥ 2n+1 2n+1 1), 2B2 (2 2 ) (n ≥ 1), Sp4 (2n ) (n > 1), G2 (2n ) (n > 1), 2F4 (2 2 ) (n ≥ 0), or 3 D4 (2n ) (n ≥ 1); (h) G ∼ = L4 (2), L5 (2), Sp6 (2), or U5 (4); or (i) G ∼ = U5 (2n ) (n ≥ 3). In [I2 ], it was planned that the entire case e(G) ≤ 3 would be treated by 2-local analysis. Since we are instead using p-local analysis with p odd when e(G) = 3, we must retain the case λ(G) = λu (G) pending the treatment of the odd uniqueness case. Thus the discussion of Theorem C4 which appears in [I2 , Sections 15 and 22], must be replaced by the discussion in this section. As noted above, the major subcase of Theorem C∗4 in which e(G) ≤ 2 has been completed in the Quasithin Theorem of Aschbacher and Smith. The version relevant to Theorem C∗4 is the following theorem, which is one of our Background Results.

1. THEOREM C∗4 AND ITS CASES

145

Theorem 1.1. [ASm1], [ASm2] Let G be a K-proper simple group of restricted even type with e(G) ≤ 2. Then G ∼ = G∗ , where G∗ is one of the groups listed in conclusions (e), (f ), (g), or (h) of Theorem C∗4 . We remark that our list given here is somewhat shorter than Aschbacher and Smith’s list because for us simple groups G with m2 (G) = 2 and also M12 are groups of small odd type, while J1 is of 2-uniqueness type. In any case, thanks to Theorem 1.1, it remains for us to assume that e(G) = 3, i.e., λ(G) = ∅, and to identify G as one of the groups listed in conclusions (a) - (d) or (i) of Theorem C∗4 . We now define κ(G) = {p ∈ λ(G) | mp (G) > 3} C∗4

∗ Because of Theorems C∗∗ 6 and C7 , we may assume in proving Theorem that κ(G) is covered by the following two subsets:

κc (G) = {p ∈ κ(G) | Lop (G) ⊆ Cp }

(1B)

κu (G) = κ(G) ∩ λu (G)

Indeed we have Theorem 1.2. Let G be a K-proper simple group of restricted even type such that e(G) = 3. Then either G ∼ = U5 (2n ), n ≥ 3, or κ(G) = κc (G) ∪ κu (G). Proof. Let p ∈ κ(G). If Lop (G) ∩ Gp = ∅, then we may assume that G has a strong p-uniqueness subgroup, for otherwise by Theorem C∗7 , G ∈ K, and by [V17 , 13.35], G ∼ = U5 (2n ), n ≥ 3. So p ∈ κu (G) in this case. Similarly, o if Lp (G)∩Gp = ∅ but Lop (G)∩Tp = ∅, then by Theorem C∗∗ 6 , G has a strong puniqueness subgroup, so p ∈ κu (G) again in this case. Thus we may assume  that Lop (G) ∩ (Tp ∪ Gp ) = ∅, i.e., Lop (G) ⊆ Cp , so p ∈ κc (G). As λ(G) = λu (G) is one of the conclusions of Theorem C∗4 , we may subdivide the remainder of the proof of Theorem C∗4 into two cases, focusing on an odd prime p ∈ λ(G): (Case A): p ∈ κc (G); (Case B): p ∈ λ(G) − λu (G) and κc (G) = ∅. Our method is to analyze the {2, p}-local structure of G, eventually obtaining enough detail either to determine the isomorphism type of G as a known simple group, or to show that G possesses a strong p-uniqueness subgroup (in a sense, as noted above, to be defined in the next volume). In Case A, in this volume, we shall actually avoid the p-uniqueness case and prove that G ∈ K when κc (G) = ∅. Thus we shall prove the following result identifying the groups in conclusions (a), (b), and (c) of Theorem C∗4 .

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Theorem C∗4 (Case A). Let G be a K-proper simple group of restricted even type such that e(G) = 3, and assume that κc (G) = ∅. Let p ∈ κc (G) (so that Lop (G) ⊆ Cp ). Then p = 3 and one of the following holds: (a) G ∼ = A12 ; ∼ Co2 , Co3 , Suz, F3 , or F5 ; or (b) G = ∼ (c) G = U5 (2), U6 (2), D4 (2), Sp8 (2), or F4 (2). In Case B, in a subsequent volume, we aim to prove the following theorem. Theorem C∗4 (Case B). Let G be a K-proper simple group of restricted even type such that e(G) = 3 and κc (G) = ∅. Then either λ(G) = λu (G) or one of the following holds: (a) G ∼ = L4 (2n ) (n > 1), Sp6 (2n ) (n > 1), or 2 D4 (2n ) (n ≥ 1); or (b) G ∼ = L6 (2) or L7 (2). Thus we shall obtain the groups listed in conclusion (d) of Theorem C∗4 . We remark that Case B is the only part of the proof of Theorem C∗4 where we shall require the p-uniqueness alternative. 2. The Stages of Theorem C∗4 (Case A) In the remaining chapters of this volume, we shall prove Theorem C∗4 (Case A). As usual, we divide the proof into stages. In Stage A1, we establish some important information about p-signalizers. Theorem C∗4 : Stage A1. Let G and p be as in Theorem C∗4 (Case A). Let P ∈ Sylp (G) and U  P with U ∼ = Ep2 , and set P0 = CP (U ). If Z(P ) is noncyclic, assume that U ≤ Z(P ). Then the following hold: (a) Op (CG (A)) has odd order for all elementary abelian p-subgroups A ≤ G such that mp (CG (A)) ≥ 3; (b) For any B ≤ G with B ∼ = Ep3 and NG (B; 2) = {1}, B # ⊆ Ipo (G); and (c) If Op (CG (x)) = 1 for some x ∈ Ip (G), then ΓoP,2 (G) ≤ M for some subgroup M < G such that Op (M ) = 1, assuming any one of the following additional hypotheses: (1) p = 3; (2) mp (CG (x)) ≥ 3 and Z(P ) is noncyclic; or (3) x ∈ P0 . The main conclusions of Stage A1 are (a) and (b), which are used throughout the subsequent analysis. Conclusion (c) is a technicality needed in only one place in the proof of Stage A2. The next definitions will permit the division of the problem into two major subcases, the first of which is treated in Stage A2.

2. THE STAGES OF THEOREM C∗4 (CASE A)

147

Definition 2.1. H = {H ≤ G | H is a 2-local subgroup of G and mp (H) = 3}. Hinv = {H ∈ H | H = CG (z) for some z ∈ I2 (G)} Hcons = {H ∈ H | F ∗ (H) = O2 (H)} Theorem C∗4 : Stage A2. Let G and p be as in Theorem C∗4 (Case A), and suppose that Hinv ⊆ Hcons . Then p = 3 and G ∼ = Co3 . We have now reached the situation where Hinv ⊆ Hcons . This is a generalization of a problem treated by K. Klinger and G. Mason [KMa1]. Definition 2.2. A KM -singularity in a 2-local subgroup H ∈ Hcons is an elementary abelian p-subgroup 1 = B0 ≤ H such that mp (CH (B0 )) = 3 and Lp (CG (B0 )) = 1. The rank of the singularity is mp (B0 ), and the type of the singularity is the isomorphism type of E(CG (B0 )/Op (CG (B0 ))). Theorem C∗4 : Stage A3. Let G and p be as in Theorem C∗4 (Case A). Assume that Hinv ⊆ Hcons . Then either (a) or (b) holds: (a) (1) p = 3; and (2) There exists H ∈ Hcons and a KM -singularity B0 in H of rank 2 and type A6 ; or (b) (1) Hinv = ∅; and (2) For every H ∈ Hcons and KM -singularity B0 in H, mp (B0 ) = 1. We are now ready to achieve in Stage A4 the final subdivision of the problem into two subcases as defined below. Definition 2.3. We say that G has an A9 -Sp6 (2) setup if and only if (1) There is H ∈ Hcons and B0 ≤ H such that B0 ∼ = E32 , m3 (CH (B0 )) = 3, and B0 is a KM -singularity of type A6 with respect to p = 3; that is, CG (B0 ) has a 3-component J such that J/O3 (J) ∼ = A6 ; # (2) For any H, B0 , and J as in (1), and any x ∈ B0 , J pumps up vertically or trivially in CG (x) to a 3-component Jx ; (3) For some x as in (2), Jx ∼ = A9 or Sp6 (2); and (4) For any x as in (2) with Jx a vertical pumpup of J, m3 (C(x, Jx )) = 1 and Jx /O3 (Jx ) ∼ = A9 , Sp6 (2), J3 , Sp4 (8), or L2 (36 ). In the L2 (36 ) case, or for any x ∈ B0# such that Jx is a trivial pumpup of J, we have [CO2 (H) (x), B0 ] = 1. Definition 2.4. We say that G has an extraspecial setup if and only if (1) For any z ∈ I2 (G) such that A ≤ CG (z) for some A ∼ = E33 , z is 2-central in G, and Qz := F ∗ (CG (z)) = O2 (CG (z)) is the central product of z = 3 or 4 copies of Q8 ; and

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(2) For any z and A as in (1), there exists x ∈ A# such that |CQz (x)| ≥ 25 , and for any such x, the 3-layer L := L3 (CG (x)) satisfies m3 (C(x, L)) = 1 and L/O3 (L) ∼ = U4 (2), G2 (3), or 32 U4 (3). Here, 32 U4 (3) refers to one of the two isomorphism classes of 3-fold covering groups of U4 (3); it is specified in [V17 , 10.10.12]. Theorem C∗4 : Stage A4. Let G and p be as in Theorem C∗4 (Case A). Assume that Hinv ⊆ Hcons . Then p = 3, and either (a) or (b) holds: (a) G has an A9 -Sp6 (2) setup; or (b) G has an extraspecial setup. Finally, in Stage A5, we identify the groups arising in both cases of Stage A4. Theorem C∗4 : Stage A5. Let G and p be as in Theorem C∗4 (Case A), with Hinv ⊆ Hcons . (a) If G has an extraspecial setup, then G ∼ = F3 , Suz, Co2 , U5 (2), U6 (2) or D4 (2); and (b) If G has an A9 -Sp6 (2) setup, then G ∼ = A12 , F5 , Sp8 (2), or F4 (2). Clearly, Theorem C∗4 (Case A) is a direct consequence of Stages A2, A4, and A5 of Theorem C∗4 . 3. Some Definitions For the convenience of the reader, we give the definitions of Cp for arbitrary primes p [I2 , 12.1], and relevant parts of the definition of restricted even type. Definition 3.1. Let p be a prime (possibly p = 2). The set Cp is comprised of all isomorphism types of Cp -groups. A simple K-group K is a Cp -group if and only if K ∈ Chev(p), or K ∼ = Ap , A2p , or A3p , or one of the following holds: (1) p = 2 and either (a) K ∼ = L2 (q), q ∈ F M9 (i.e., q a Fermat or Mersenne prime or 9); (b) K ∼ = L3 (3), L4 (3), U4 (3), or G2 (3); or (c) K ∈ Spor, but K ∼ J1 , Co3 , M c, Ly, He, O N , or F5 . = (2) p = 3 and either 1 (a) K ∼ = U5 (2), U6 (2), Sp6 (2), D4 (2), 3D4 (2), F4 (2), 2F4 (2 2 ) , or P Sp4 (8); or (b) K ∼ = M11 , J3 , Co1 , Co2 , Co3 , M c, Suz, Ly, O N , F i22 , F i23 ,  F i24 , F5 , F3 , F2 , or F1 . (3) p = 5 and either 1 5 5 (a) K ∼ = 2F4 (2 2 ) , 2F4 (2 2 ), or 2B2 (2 2 ); or (b) K ∼ = J2 , Co1 , Co2 , Co3 , HS, M c, Ly, Ru, He, F5 , F3 , F2 , or F1 .

3. SOME DEFINITIONS

149

∼ Co1 , He, O N , F i , F3 , or F1 . (4) p = 7 and K = 24 (5) p = 11 and K ∼ = J4 . Furthermore, a quasisimple K-group K ∈ Kp is a Cp -group if and only if K/Op (K) is a Cp -group, with the following exceptions: For p = 2: The groups SL2 (q), q ∈ F M9, 2A8 , SL4 (3), SU4 (3), Sp4 (3), and [X]L3 (4), X = 4, 4 × 2, or 4 × 4, are not included among the C2 -groups. For p = 3: The groups 3A6 and 3O N are not included among the C3 -groups. Thus, for example, C2 contains U3 (3) ∼ = G2 (2) , P Sp4 (3) ∼ = U4 (2), and 1 2 G (3 2 ) ∼ L (8). = 2 2 Definition 3.2. A Co2 -group is a C2 -group that is not isomorphic to L2 (q) for any odd q > 17. Moreover, Co2 is the set of all isomorphism types of Co2 -groups. Finally, parts of the definition [I2 , 8.8] of restricted even type immediately give the following lemma, which captures the relevance of that definition for Theorem C∗4 : Lemma 3.3. Suppose that G is of restricted even type. Then (a) G is of even type; (b) Every component of the centralizer of every involution of G is a Co2 -group; and (c) If x ∈ I2 (G) and K ∼ = L2 (9) is a component of CG (x) that is terminal in G, then m2 (C(x, K)) = 1.

CHAPTER 12

Theorem C∗4 : Stage A1. First Steps 1. Introduction Throughout this chapter we make the following assumptions:

(1A)

(1) G is a K-proper simple group of restricted even type such that e(G) = 3; (2) p is an odd prime such that m2,p (G) = 3 and mp (G) ≥ 4; and (3) Lop (G) ⊆ Cp .

We use the following notation: (1B)

(1) P ∈ Sylp (G); (2) U  P with U ∼ = Ep2 , chosen if possible so that U ≤ Z(P ); and (3) P0 = CP (U ).

Our goal in this chapter is to prove the following theorem. Theorem C∗4 : Stage A1. Let G, p, Lop (G), P , U , and P0 be as in (1A) and (1B). Then the following conditions hold: (a) Op (CG (A)) has odd order for all elementary abelian p-subgroups A of G such that mp (CG (A)) ≥ 3; (b) Suppose B ∈ Ep3 (G) and NG (B; 2) = {1}. Then B # ⊆ Ipo (G); and (c) If p = 3 and O3 (CG (x)) = 1 for some x ∈ I3 (G), then ΓoP,2 (G) ≤ M < G for some subgroup M such that O3 (M ) = 1. In proving (a), we will first prove it for |A| = p, thus proving (1C)

|Op (CG (x))| is odd for all x ∈ Ip (G) such that mp (CG (x)) ≥ 3.

Most of the work on (1C) and (c) will be devoted to proving the following proposition: Proposition 1.1. Assume (1A) and (1B). Then G is balanced with respect to any element of Ep3 (G). In the remainder of this section we show that Proposition 1.1 implies (1C) and (c) of Theorem C∗4 : Stage A1. Since G is balanced with respect to every element A ∈ Ep3 (G), ΓoP,2 (G) normalizes the completion W of the Op functor, and the p -group W equals 151

152

12. THEOREM C∗4 : STAGE A1. FIRST STEPS

  Op (CG (a)) | a ∈ A# for every A ∈ Ep3 (P ). In particular, P normalizes W . As mp (P ) > 3 = m2,p (G) and W is a p -group, W has odd order, which, together with Sylow’s theorem, implies (1C). Furthermore, the conclusion of (c) holds with M = NG (W ), as long as W = 1. Since m3 (G) ≥ 4, P has a normal E34 -subgroup by Konvisser’s theorem [IG , 10.17]. Hence taking x ∈ P with O3 (CG (x)) = 1, as we may by hypothesis and by Sylow’s theorem, there exists A ∈ Ep3 (P ) containing x, and so W = 1, as desired. 2. Initial Remarks Recall from [V9 , 1.6] that Sp (G) is the set of all elementary abelian psubgroups of G which are maximal with respect to inclusion. The following important lemma gives us a head start on the proof of Proposition 1.1: Lemma 2.1. Let B ∈ Ep4 (G). Then G is balanced with respect to B. If B ∈ Sp (G), then G is strongly balanced with respect to B. Proof. The development in [V3 , Section 3] actually proves the second statement, and the first statement follows immediately. Indeed, the proofs of Proposition 3.3 and Lemmas 3.4–3.6 in [V3 , Section 3] may be copied verbatim; neither of the extra hypotheses that hold there, viz., that p ∈ σ0 (G) and that m2,p (G) ≥ 4, is used in those proofs. The result is to prove that if Lemma 2.1 fails, then p = 3 and for some d ∈ D ∈ E34 (G), O3 (CG (d)) has even order. Hence m2,p (G) ≥ 4. This contradicts our current assumption that e(G) = 3, and hence Lemma 2.1 holds.  Before beginning the proof of Theorem C∗4 : Stage A1, we introduce the set [A13] of possible isomorphism types of proper simple sections of G, given that e(G) = 3. We call this set A, and abuse notation by writing K ∈ A to mean that the isomorphism class of K is in A. Definition 2.2. A is the set of the following isomorphism classes of simple groups: (a) L2 (q), q odd; e (b) L± 3 (p ), p an odd prime and e ≤ 3; 3 (c) 2 G2 (3 2 ); (d) G2 (p), P Sp4 (p), or L± 4 (p), p an odd prime; (e) An , n ≤ 12; (f) A sporadic group but not F i22 , F i23 , F i24 , Co1 , F2 , or F1 ; (g) A group in Chev(2) of twisted Lie rank at most 2; (h) L4 (q), Sp6 (q), or Ω− 8 (q), q even; or (2), L (2), Sp8 (2), D4 (2), or F4 (2). (i) L5 (2), L± 7 6 Theorem 2.3 (Aschbacher [A13]). Suppose G is K-proper, e(G) ≤ 3, and K is a proper simple section of G. Then K ∈ A. Proof. This is an immediate consequence of [V17 , Theorem 1.3c].



3. AN EXCEPTIONAL CONFIGURATION INVOLVING L3 (4)

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3. An Exceptional Configuration Involving L3 (4) We begin the proof of Proposition 1.1 by analyzing two exceptional configurations. In this section we consider the first of these, always assuming (1A):

(3A)

(1) p = 3 and (b, L) ∈ IL3 (G); (2) Cb = CG (b) and C b = Cb /O3 (Cb ); (3) t ∈ Cb is a 2-element and t induces a unitary-type automorphism on L ∼ = L3 (4) (i.e., CL (t) ∼ = U3 (2)).

Proposition 3.1. The configuration (3A) does not occur. Proof. Suppose that it does occur. Clearly m3 (Cb ) ≥ 1 + m3 (L) = 3, and as L ∈ C3 , b ∈ I3o (G). Thus, m3 (Cb ) = 3, so C(b, L) is a cyclic 3-group. Let T ∈ Syl3 (Cb ); we may assume that T < P (m3 (P ) ≥ 4). Set B = T ∩ L and A = B b ∼ = E33 . Then T = (B × Q) d, where Q ∼ = Q = C(b, L), and either d = 1 or d3 ∈ Q with d inducing an outer diagonal automorphism on L. As T < P , b is notcharacteristic in T . It follows that Q = b and d3 = 1. For, if Q > b or d3 = b, then b = Ω1 (1 (T )) char T , contradiction. As Q = b, T = A d. In particular, CG (A) = A × O3 (CG (A)).   Write NLT (B) = BH d × b with H ∼ = Q8 and, if d = 1, then H d ∼ = SL2 (3). Also, choose E ∼ = E34 with E  P , using [IG , 10.17], and set E0 = CE (b). Then m3 (E0 ) = 2.  = N/O3 (CG (A)), and N  = N/CG (A) ∼ Set N = NG (A), N = AutG (A) ≤ ∼ Aut(A) = GL3 (3). Since A  T , we may assume that R := P ∩N ∈ Syl3 (N ). Suppose first that

b  N.

(3B)

 of N  has orbits { b}, E1 (B), and E1 (A) − E1 (B) − { b} on The subgroup H E1 (A), of respective lengths 1, 4, and 8. Since N is not transitive on E1 (A), the only possibility is that B  N , bN = bR = E1 (A) − E1 (B), and   ∼ R = R1 d with  A ≤ R1 = CR (B), R1 ∩ d = 1, and O3 (N ) = R1 = E32 . 1 is the stabilizer of the chain  dˆ acts faithfully on R 1 and R Moreover, H A > B > 1. In particular, B ≤ Z(R1 ). Commutation with b induces an isomorphism (3C)

 as C  (b)-modules. 1 ∼ R =B N

  1 dˆ ∼ Hence, R = 31+2 or E32 .

# . As E  P Next, we show that there exists w ∈ E ∩ N such that w ∈R 1 # . If w 1 , 1 ∈ R 1 ∈ R and E ≤ A = CP (A), there is w1 ∈ E ∩ N − A, so that w  1 , 1 ∈ R1 , then d = 1 and w 1 acts quadratically on R we take w =w 1 . If w 1 , w 1 ], proving our assertion. and we can take w  in [R

154

12. THEOREM C∗4 : STAGE A1. FIRST STEPS

Now, [R1 , w] ≤ E ∩ A. But as bR = E1 (A) − E1 (B), all elements of A − B have centralizer of 3-rank 3. Since m3 (E) = 4, E ∩ A ≤ B and so  (see (3C)), and [R1 , w, R1 ] = 1. Thus, w ∈ Z2 (R1 ). By the action of H since A ≤ Z2 (R1 ), R1 = Z2 (R1 ). In particular, R1 /B is abelian. Then by 0 := [R  is an abelian 1 , Z(H)]  on R 1 , using (3C) again, R the action of Z(H)  0 in T . Again by 1 . Let R0 be the inverse image of R complement to ˜b in R 0  R 0 ∼ the action of H, = E34 or Z9 × Z9 . Since t ∈ CN (B b), t centralizes R 0 ∼ 1 ) = A.  by (3C). But e(G) = 3 by assumption, so R = Z9 × Z9 and Ω1 (R However, w ∈ I3 (R1 ) − A, a contradiction. Hence, (3B) must fail, i.e., (3D)

b  N.

If d = 1, then A ∈ Syl3 (Cb ), and (3D) implies that A ∈ Syl3 (G), which is absurd. Thus, d = 1 and T ∼ = 31+2 × Z3 . Choose v ∈ NE (T ) − T . By v (3D), A = A. As E0 ≥ [A, v], we may assume that d was chosen in E0 , so that E0 = d, z, where z = [T, T ] = Z(T v) = Z(P ). As E ≤ CG (d), d ∈ I3o (G) and so every component of CG (d)/O3 (CG (d)) is a C3 -group. By L3 -balance and [IA , 7.1.10], so is every component of CC b (d). As A5 ∈ C3 , CL (d) ∼ = A5 . Write B = z, h. By [V17 , 10.6.5],  

z, d ∼L z, dh or z, d ∼L z, dh−1 . But as b, z, d = Z(T )(E ∩ T )  NP (T ), we have   A = b, z, h ∼NP (T ) b, z, dh ∼NP (T ) b, z, dh−1 . Hence in each of these E33 -groups, the G-conjugates of b# fill the complement of a totally G-fused hyperplane. If d ∼L dh, then b is not G-conjugate to either  −1  z or dh, so if

d ∼ , then arguing

z ∼G dh ∼G d. On the other hand L dh  −1  −1 we get z ∼G dh ∼G d. Hence in all cases, similarly in b, z, dh  

z ∼G d ∼G zd ∼G z −1 d . Thus, every e ∈ z, d# is 3-central in G. However, we saw above that |Z(P )| = 3, so CG (e)/O3 (CG (e)) has no simple component. In particular, it has no L2 (27)-component. As all its components are C3 -groups, we conclude in particular by [V17 , 4.26] that (3E)

O3 (CG (b, e)) ≤ O3 (CG (e)) for all e ∈ E0# = z, d# .

Now, O3 is an E-signalizer functor, by Lemma 2.1, with completion Θ1 (G; E). Let V = O3 (CL (d)), so that V ∼ = Z7 . Then V ≤ O3 (CG (b, d)) ≤ O3 (CG (d)) ≤ Θ1 (G; E), by (3E). In particular, Θ1 (G; E) = 1. Set M = NG (Θ1 (G; E)), so that T ≤ P ≤ NG (E) ≤ M . Therefore T,V  ≤T O3 (M ) so T, V  has a normal 3-complement. But by [V17 , 10.6.3], T , V contains  the simple group L, a final contradiction. The proposition is proved.

4. AN EXCEPTIONAL CONFIGURATION INVOLVING L2 (pp )

155

4. An Exceptional Configuration Involving L2 (pp ) In this section we consider the following configuration, assuming (1A):

(4A)

(1) (2) (3) (4)

p is an odd prime and (b, L) ∈ ILp (G); H = NG ( b) and H = H/Op (H); L∼ = L2 (pp ); and a ∈ Ip (H) and a induces a nontrivial field automorphism on L.

Rather than rule this configuration out unconditionally, we give some consequences of it that will enable us to rule it out in the contexts in which we encounter it. We shall prove Proposition 4.1. Suppose that the configuration (4A) occurs, and p ≤ 7. Then the following conditions hold: (a) mp (CG (a)) > 3; (b) If p = 3, then O2 (CL (a)) is not covered by O3 (CG (a)) ∩ CG (b); (c) |G : CG (b)|p ≥ pp ; (d) b is neither p-central nor half p-central in G. The assumption p ≤ 7 is unfortunately needed at a unique point in the proof (Lemma 4.12). Note that part (d) of the proposition is immediate from part (c), and part (b) is immediate from part (a) since if O3 (CG (a)) has even order, then 3 = e(G) ≥ m2,3 (G) ≥ m3 (CG (a)). Our analysis will yield first part (c), and eventually part (a). Supposing that (4A) holds, let T ∈ Sylp (H). We may assume that a ∈ T ≤ P . Set E = T ∩ L ∈ Sylp (L), D = E b, and d = CE (a) ∼ = Zp . For any x ∈ Ip (C(b, L)), let Lx be the pumpup of L in CG (x). We proceed in a sequence of lemmas. Set (4B)

pp − 1 1 . m = (pp − 1) ≥ 13 and m0 = 2 p−1

2 Lemma 4.2. Let x ∈ Ip (C(b, L)). Then Lx ∼ = L2 (pp ) or L2 (pp ). In particular, L ∼ = L2 (pp ). 2 Proof. Let X = Op (Lx ), so that Lx /X ∼ = L2 (pp ) or L2 (pp ) by [V17 , 13.14] (a diagonal pumpup would violate m2,p (G) = 3). Let S ∈ Syl2 (X). Then by a Frattini argument, mp (NCG (x) (S)) = mp (CG (x)) ≥ 4. As e(G) = 3, S = 1 and X has odd order. Suppose that X = 1. Since Lx /X has Schur multiplier of order 2 [IA , 6.1.4], there is an odd prime r = p such that if we put R = Or (X) and let R0 be a critical subgroup of R and R1 = Ω1 (R0 ), then CLx (R1 ) ≤ X. Applying Clifford’s theorem to the action of a Sylow p-normalizer of Lx on R1 , we have mr (R1 /Φ(R1 )) ≥ m ≥ 13. Hence by [V9 , 5.9], mr (CR1 (s)) ≥ 4 for some involution s ∈ Lx , contradicting e(G) = 3. Therefore X = 1, completing the proof. 

12. THEOREM C∗4 : STAGE A1. FIRST STEPS

156

Set M = NG (D), so that M ∩ L = EY ∼ = Fpp .m with Y ∼ = Zm , ∼ and Y may be chosen to be a-invariant. Thus Y a = Fm0 .p × Z(p−1)/2 . Lemma 4.3. CT (L) = b and b  M . Proof. First suppose that CT (L) is noncyclic, and choose Ep2 ∼ =F ≤ CT (L). Fix an involution s ∈ L and let C = CG (s). Then CL (s) ∼ = Dpp − and indeed for any x ∈ F # , CL (s) ≤ CLx (s) ∼ = Dpp − or Dpp2 − ,  = ±1, p ≡  (mod 4). Let be an odd prime divisor of pp − , and put V = O (CL (s)). Then   1 = V  CCG (x) (s) x ∈ F # = ΓF,1 (CG (s)). In particular, O2 (CG (s)) normalizes and then centralizes V . As G is of even type, there is a component I ∈ C2 of CG (s) with [I, V ] = 1, so I ≤ ΓF,1 (CG (s)) and I does not have the weak p-signalizer property with depth 2 [IA , (7.7.4)ff.]. It follows from [IA , 7.7.17] and as I ∈ A ∩ C2 that p = 3 and I ∼ = E32 and V maps = L3 (4). But then by [V17 , 10.6.4], I0 := ΓF,1 (I) ∼ faithfully into CAut(I) (I0 ). As this last group is a {2, 3}-group, we have a contradiction. Hence, CT (L) is cyclic. It follows that Ja (T ) = E × CT (L). Now suppose that b  NG (Ja (T )). Then NG (T ) ≤ NG (Ja (T )) ≤ H, so T ∈ Sylp (G). But Ja (T ) is abelian and weakly closed in T . As a ∈ [H, H], the Hall-Wielandt-Yoshida theorem [IG , 15.27] implies that a ∈ [G, G], contradicting the simplicity of G. Hence,

b  NG (Ja (T )). Finally, if CT (L) > b, then b = Ω1 (Φ(Ja (T ))) char Ja (T ) so b  NG (Ja (T )), contradiction. Therefore CT (L) = b, Ja (T ) = D, and the lemma is proved.  Lemma 4.4. The following conditions hold: (a) T = D a ∼ = Zp × (Zp Zp ); and M (b) b = E1 (D) − E1 (E), dM = E1 (E), and E  M . Proof. Lemma 4.3 immediately implies (a). Since d ∈ [T, T ] but b ∈ [T, T ], d ∈ bG . By [V9 , 1.4], M controls G-fusion of b in D. The NL (D)-orbits on E1 (D) are { b}, E1 (E), O1 , and O2 , where |O1 | = |O2 | = m (4B). As | bM | divides |GLp+1 (p)|, but 1 + m = 12 (pp + 1) does not (by  Zsigmondy’s theorem [IG , 1.1]), the only possibility is that (b) holds. Lemma 4.5. F ∗ (H) = L × b ∼ = L2 (pp ) × Zp . Moreover, CG (D) = D. Proof. The second statement is an immediate consequence of the first. As mp (H) = p + 1 > 3, Op (H) has odd order. Then since CT (L) = b and L is simple, CCG (b) (L) = b × X where X = O{2,p} (H). It suffices to show that X = 1, so suppose that X = 1 and let N = NG (X) < G. Then L = Lp (CN (b)) has a subnormal closure L∗ in N , and L∗ has the usual structure by Lp -balance. Since m2,p (G) = 3, L∗ is a trivial or vertical pumpup of L. By Lemma 4.4b, bd ∈ bG ∩ CG (X). Therefore L1 := E(CG (bd)) ∼ = L and

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157

X ≤ CG (bd) with [D, X] = 1, so [L1 , X] = 1 and L1 ≤ N . Moreover, L1 ∩ L = 1, so L1 ≤ L∗ . Hence L∗ is a vertical pumpup of L, whence 2 L∗ /Op (L∗ ) ∼ = L2 (pp ) by [V17 , 13.14]. Nontrivial field automorphisms are then induced on L∗ /Op (L∗ ) by b and also every x ∈ a, b − b (since x induces a field automorphism on L). Thus a, b embeds in the cyclic group  Out(L∗ /Op (L∗ )), a contradiction. The proof is complete. Set M = M/D ∼ = AutG (D) and let RD = Op (M ). Lemma 4.6. The following conditions hold: (a) (b) (c) (d)

d = Z(P ); RD ∼ = Epp acts regularly on E1 (D) − E1 (E); M = RD (M ∩ H); and RD ∼ = E as CM (b)-modules, via x → [x, b].

Proof. By Lemma 4.4b, |M : M ∩ H| = |M : M ∩ H| = pp , so b is not p-central in G. Then (a) holds. By Lemma 4.4b, M , regarded as a subgroup of GL(D), lies in the parabolic subgroup PE which is the stabilizer of E. We have PE = QE HE where QE = Op (PE ) ∼ = Epp is acted on irreducibly by Y , and HE = GL(E) × GL( b). We argue that M ∩ QE = 1. If p > 3, then Z := Z(HE ) ∩ Y is nontrivial (of order 12 (p − 1)) and CD (Z) = b; if M ∩QE = 1 in this case, then Z ≤ Z(M ) and so M normalizes CD (Z) = b, contradiction. If p = 3, then as |M ∩ H|p = | a | = p, |M |p = pp+1 so again M ∩QE = 1. As Y acts irreducibly on QE , (b), (c), and (d) follow easily.  Set V = [RD , Y ], N = NG (V ), and C = CG (V ). Lemma 4.7. The following conditions hold: (a) (b) (c) (d)

RD = V b with RD /E = (V /E) × ( b E/E) ∼ = Epp+1 ; V /E ∼ E as ( b × Y

a)-modules; = V is homocyclic abelian of rank p or 2p; and If D  P , then P = V a, b.

∼ E as ( b × Y a)-modules. By Proof. By Lemma 4.6d, RD /D = Lemma 4.4b, E  M , and it follows that RD /E is elementary abelian. Thus, (a) and (b) hold. If D  P , then P ∈ Sylp (M ), so P = RD (P ∩ H) by Lemma 4.4b. This implies (d). Suppose that V is not abelian; then by the action of Y , [V, V ] = E. Write Y = y. Then with (b), the eigenvalues of y both on V /E and on E are the elements of (4C)

E := {αp | 0 ≤ i < p} for some α ∈ F× pp of order m0 i

(see (4B)). By a Lie ring argument, the eigenvalues of y on [V, V ] are among the elements of the set {ββ  | β, β  ∈ E}, which is disjoint from E, contradiction. Hence, V is abelian, and the action of Y on V implies (c). 

12. THEOREM C∗4 : STAGE A1. FIRST STEPS

158

At this point Proposition 4.1c is evident from Lemma 4.7b. It therefore remains to prove Proposition 4.1a. We assume by way of contradiction that it fails, so that (4D)

mp (CG (a)) = 3.

Lemma 4.8. Suppose that X = X/Op (X) is a Y a, b-invariant psection of G such that E ≤ X and Y acts fixed-point-freely on X. Then X is abelian. Proof. Let 1 = X 0 < X 1 < · · · < X n = X be part of a chief series for XY a, b. Since Y is fixed-point-free on X, every chief factor X j /X j−1 is isomorphic to Epp , and of course is centralized by b ∈ Z(Y a, b). In particular, X 1 = E. For every j = 2, . . . , n, by [IG , 9.16], |CX j /X j−2 (b)| ≤ |E| so CX j /X j−2 (b) = X j−1 /X j−2 . Therefore the mapping x → [x, b] induces a Y a-isomorphism from X j /X j−1 to X j−1 /X j−2 . Write Y = y. It follows that the set of eigenvalues of y on any X j /X j−1 (1 ≤ j ≤ n) is as in (4C). Hence, discounting multiplicities, the set of eigenvalues of y on any Y -invariant elementary abelian section of X is just E. Again {ββ  | β, β  ∈ E} ∩ E = ∅, and hence by a Lie ring argument, X must be abelian. (There can be no nontrivial Y -invariant map from X/Φ(X) ⊗ X/Φ(X) to [X, X]/Φ([X, X]).) The proof is complete.   = N/C ∼ Set N = N/Op (N ) and N = AutG (V ). Also let W = Op (C)   in N .  and R = Op (N ). Let R be the full preimage of R Lemma 4.9. W = C is abelian and Ω1 (W ) = Ω1 (V ). Proof. By Lemma 4.8 with W in the role of X, W is abelian. Next we claim that if |Ω1 (W )| > pp , then V is elementary abelian. Indeed define Z by Z/E = CΩ1 (W )/E (b). Then [Z, D] = E so M covers Z. Hence V = [RD , Y ] covers Z, whence V is elementary abelian, as claimed. Notice that |Ω1 (W )| ≤ p2p . This is because mp (CG (a)) ≤ 3 (4D), whereas a, which acts fixed-point-freely on the subgroup of Y of order m, acts freely on Ω1 (W ). By the previous paragraph, Ω1 (W ) = Ω1 (V ). Suppose for a moment that p = 3. We argue that V is elementary abelian. Suppose false, so that V is homocyclic of exponent 9 and rank 3. By [IG , 10.17], there is F  P with F ∼ = E34 . Then NF (D) is a normal elementary abelian subgroup of NP (D) = V a, b, so NF D (D) = DNF (D) ≤ DΩ1 (V b) = D. Hence D = F  P , whence P = V a, b. Suppose that V0 ≤ P with V0 abelian and |V0 | = |V |. Then |V0 ∩ V | ≥ 34 so |V0 ∩ E| ≥ 32 . Since every element of a, b − b acts freely on E, V0 ≤ V b. As |CV (b)| = 33 , V0 = V . We have proved that V = Ja (P ). Since RD = CP (V /Φ(V )),

b C = RD C  N , and so N = C(N ∩ H). Hence a ∈ [N, N ]. By the HallWielandt theorem, a ∈ [G, G], contradicting the simplicity of G. Therefore, V ∼ = E36 if p = 3.

4. AN EXCEPTIONAL CONFIGURATION INVOLVING L2 (pp )

159

In any event, mp (V ) > 3, and so as e(G) = 3, C has odd order. Hence, W = F ∗ (C) and as Ω1 (W ) = Ω1 (V ), W = C.  Let H1 = NN ( b) = NH (V ). By the construction of V , H1 = NH (E) =  = N/C ∼ EH2 where H2 = NH1 (Y ) and Op (H2 ) = b. We have N = N /W , ∗   and we let Hi be the image of Hi in N . Then F (CH2 (E)) = b so CH2 (V ) = 1. Hence, 1 = H 2 ∼ (4E) H = H2 . Then in view of [IG , 9.16], (4F) |CN (b)| ≤ |E||CH2 (b)| = pp |CH 2 (b)|.  b = Op (N 0 ). 0 = C  (b) and R Set N N    ≤ b = R b . Lemma 4.10. R  Since b ∈ Sylp (CG (x)) for all x ∈ Y # , Proof. Suppose that b ∈ R.  But then R is abelian by Lemma CR (x) = 1 and Y is fixed-point-free on R.  = 1, as desired. 4.8, whence R ≤ C and R    = b × [R,  Y ] and again  By [V9 , 8.2], R Assume next that b ∈ R.  Lemma 4.8 implies that the  preimage of [R, Y ] in N is abelian, forcing   Y ] = 1. Therefore R  = b . The same argument, with R b in place of R, [R,   b = b , and the proof is complete.  shows that R Let Y0 be the subgroup of Y of order m0 . Then Y = Y0 × YZ , where YZ ∼ = Z(p−1)/2 induces scalar transformations on E and V /E, hence on V . In particular NN (Y ) = NN (Y0 ). Lemma 4.11. The following conditions hold: 0 ; and (a) Y0 is a TI-set in N 0 : N  (Y )| ≡ 1 (mod m0 ). (b) |N N0  Proof. First note that   the stabilizer in N0 of the chain V > E > 1   lies in Rb , which equals b by Lemma 4.10. Let x  ∈ Y0# and w  ∈ CN0 ( x), 0 ∈ Y0 such that w 0 with w  having order dividing m0 . Then there exists w and w  have the same action on E. As w 0 and w  centralize b, they also have the same action on V /E. Hence w  = w 0b0 , where b0 stabilizes the chain =w 0 ∈ Y0 . It V > E > 1 and hence is a power of b. As w  is a p -element, w  x) of order m0 . This implies follows that Y0 is the unique subgroup of C  ( N0

(a). Now NN0 (Y ) = NN0 (Y0 ) so as |Y0 | = m0 , (b) follows as well. From this point on in this section we assume that (4G)

p ≤ 7.



12. THEOREM C∗4 : STAGE A1. FIRST STEPS

160

This enables us to prove 0 : N  (Y )| = 1 or pp . In the latter case, N  (Y ) = Lemma 4.12. |N N0 N0 CH 2 (b). 0 : N  (Y )|. By Lemma 4.11, n ≡ 1 (mod m0 ). But Proof. Let n = |N N0  2 , C  (b) ≤ N  (Y ), so n ≤ pp by (4F). As pp = 1+(p−1)m0 , also as H2 ∼ =H H2

N0

n = 1 + km0 , 0 ≤ k ≤ p − 1, and we must show k = 0 or p − 1. But if 0 are determined 0 < k < p−1, then n ≡ 0 (mod p), and as p -subgroups of N by their faithful action on E, n must divide |GLp (p)|p . It is routine to check that this divisibility is false if p ≤ 7, whence the lemma.  0 . Lemma 4.13. Y  N Proof. We may assume, by Lemma 4.12, that NN0 (Y ) = CH 2 (b) and 2 ∼ 0 : C  (b)| = pp . Recall from (4E) that H |N = H2 . Then CH 2 (b) = H2   b × Y  0 is 0 does as well. Thus N a has a cyclic Sylow 2-subgroup, so N   0 ) = b ≤ Z(N 0 ). As  solvable. By Lemma 4.10, Op (N a acts nontrivially 0 ) is not nilpotent, hence there is an odd prime s = p such that on Y , O 2 (N  0 ). Let x 0 ) − Z(N 0 ). Then s divides m0 , and as we  ∈ Os (N Os (N0 ) ≤ Z(N saw above in the proof of Lemma 4.11, Y0 is the unique subgroup of C  ( x) N0

0 , so Y = Y0 (Y ∩Z(N 0 ))  N 0 . This completes of order m0 . Therefore Y0  N the proof.    Y   induces scalar mappings 0 = b × (Z a), where Z Lemma 4.14. N on V . Proof. If V is of exponent p2 , then Aut(V )/Op (Aut(V )) ∼ = GLp(p)  0 / b by restriction to either Ω1 (V ) or V /Φ(V ). Hence by Lemma 4.13, N maps injectively into NGL(Ω1 (V )) (YS ) where YS ∼ = Z(pp −1)/2 embeds in turn in a Zpp −1 subgroup of GL(Ω1 (V )). This implies the desired result. Suppose then that V ∼ = GL2p (p) and let PE = = Ep2p . Let GV = GL(V ) ∼ NGV (E), a parabolic subgroup of shape Epp2 (GLp (p) × GLp (p)). Because [Y, b] = 1, Y embeds diagonally in PE /Op (PE ) and so CGV (Y ) ∼ = GL2 (pp )   [IA , 4.8.2], on which   a induces a field automorphism of order p. As b ∈ Z(N0 ) 0 ) = b , this implies the result. and Op (N   . Moreover, a ∈ [N, N ]. Lemma 4.15. We have Y  N

  ) = R  ≤ b . Proof. By Lemmas 4.10 and 4.14, Y centralizes Op (N  )), Y0 ] = 1. Thus, [F (N  ), Y0 ] = 1. Also by [V9 , 13.3], [Op (F (N    the chain V > E > 1. But If Ω1 (V ) = E, then    b∈ Op (N ) as b stabilizes 0 ) so b  N  and N  =N 0 is solvable. In then by Lemma 4.14, b = Op (N

5. BALANCE WITH RESPECT TO P0

161

 ) = 1 in this case. If Ω1 (V ) = V , on the other hand, then particular, E(N  ), Y0 ] = 1, so by the   [E(N ), Y ] = 1 by [V17 , 20.7]. Hence in any case, [F ∗ (N  ). F ∗ -theorem, Y0 ≤ F (N  ) where π is Since b acts quadratically on V , it must act trivially on Oπ (N  the set of prime divisors of m0 . Therefore Oπ (F (N )) ≤ N0 and Y = Y0 YZ =  . As Aut(Y ) is abelian and [Y ,  a] = 1, a ∈ [N, N ].  Oπ (F (N ))YZ  N    = b in N , and let R be an a-invariant Let R be the full preimage of R Sylow p-subgroup of the preimage of R in N , with b ∈ R. Lemma 4.16. R a = W a, b ∈ Sylp (N ).   Proof. By Lemma 4.14, b ∈ Sylp (CN (Y )). Then the lemma follows directly from Lemmas 4.15 and 4.14.  Lemma 4.17. W is the unique abelian subgroup of maximum order in R a. Proof. Let W1 be an abelian subgroup of R a with |W | = |W1 |. Then |W : W ∩ W1 | ≤ |W a, b : W1 | = p2 . If W1 ∩ W b ≤ W , then as [V, b] = E, |W : W ∩ W1 | ≥ pp , contradiction. Therefore W W1 = W a  for some a ∈ a, b − b, and |W : W ∩ W1 | = p. But a acts freely on E, so |W : W ∩ W1 | ≥ pp−1 , again a contradiction. Therefore W = W1 , as claimed.  Now we reach a final contradiction. By Lemma 4.17, W is a weakly closed abelian subgroup of R a ∈ Sylp (N ). According as V is elementary abelian or not, V = Ω1 (W ) or Ω2 (W ), so V char W and NG (W ) ≤ N . It follows that R a ∈ Sylp (G), and the Hall-Wielandt theorem and Lemma 4.15 imply that a ∈ [G, G]. This contradicts the simplicity of G and finishes the proof of Proposition 4.1. 5. Balance with Respect to P0 Recall the terminology (1B): P ∈ Sylp (G), Ep2 ∼ = U  P , and P0 = CP (U ). We begin our analysis of balance with the following proposition. Proposition 5.1. Op (CG (a)) ∩ CG (b) ≤ Op (CG (b)) for any a, b ∈ Ip (P0 ). We proceed in a sequence of lemmas. Note that as mp (G) ≥ 4, and by [IG , 10.20], (5A)

U # ⊆ Ipo (G) and mp (P0 ) = mp (P ) ≥ 4.

Lemma 5.2. Let U ≤ A ∈ Ep3 (G) ∩ Sp (G). Let u ∈ U # and a ∈ A# . Then (a) Op (CG (a)) ∩ C  G (u) ≤ Op (CG (u)); and (b) NCG (u) (A; p ) ≤ Op (CG (u)).

12. THEOREM C∗4 : STAGE A1. FIRST STEPS

162

Proof. Since a, u ≤ A, (b) implies (a). Suppose that (b) fails. Then CG (u) has a p-component L such that L/Op (L) is not strongly locally balanced with respect to p [IA , 7.7.10]. As u ∈ U # ⊆ Ipo (G), L/Op (L) ∈ Cp by (1A3), so by [IA , 7.7.12], p = 3, L/O3 (L) ∼ = L2 (27) and some a ∈ A induces a nontrivial field automorphism on L/O3 (L). This violates Proposition 4.1d, however. Thus (b) holds and the proof is complete.    Lemma 5.3. Θ1 (G; U ) = Op (CG (u)) | u ∈ U # is a p -group. Moreover, CΘ1 (G;U ) (u) = Op (CG (u)) for all u ∈ U # . p (G). By Lemma Proof. As mp (G) ≥ 4, there exists U ≤ B ∈ Ep4 (G)∩S   2.1, G is balanced with respect to B, so Θ1 (G; B) = Op (CG (b)) | b ∈ B # is a p -group and CΘ1 (G;B) (b) = Op (CG (b)) for all b ∈ B # , by the Signalizer Functor theorem [IG , 21.4]. Then     Θ1 (G; B) = CΘ1 (G;B) (u) | u ∈ U # = Op (CG (u)) | u ∈ U # = Θ1 (G; U ),

and the lemma is proved.



Recall that Γoo P,2 (G) = NG (D) | D ≤ P, mp (D) ≥ 2, mp (DCP (D)) ≥ 4 . Lemma 5.4. Γoo P,2 (G) ≤ NG (Θ1 (G; U )). Proof. This is immediate from Lemma 2.1 and [V2 , 1.3d].



The focus of our analysis is a subgroup A such that (5B)

U ≤ A ∈ Ep3 (G) ∩ Sp (G) and G is not balanced with respect to A.

Now we can prove Lemma 5.5. Assume (5B). Then the following hold: (a) There exist a ∈ A − U and u0 ∈ U # such that Op (CG (u0 )) ∩ CG (a) ≤ Op (CG (a)); (b) A = U × a;   (c) For any u ∈ U # , Vu := NCG (u,a) (A; p ) ≤ Op (CG (u)) ∩ CG (a), and CVu (U ) ≤ ΔCG (a) (U ); and   (d) NCG (a) (U ; p ) is a p -group. Proof. Choose a, b ∈ A# such that W := Op (CG (b)) ∩ CG (a) ≤ Op (CG (a)). By Lemma 2.1, mp (A) = 3. Since W is U -invariant, CW (u0 ) ≤ Op (CG (a)) for some u0 ∈ U # . But CW (u0 ) ≤ Op (CG (u0 )) by Lemma 5.2b. Then (a) and (b) hold, with the help of Lemma 2.1. Likewise Lemma 5.2 implies (c), which in turn implies (d) by Lemma 5.3.  Fix u0 ∈ U # and a ∈ A# satisfying Lemma 5.5a, so that V := Op (CG (u0 )) ∩ CG (a) ≤ Op (CG (a)).

5. BALANCE WITH RESPECT TO P0

163

Set Ca = CG (a) and C a = Ca /Op (Ca ). By [IG , 20.6], C a has a u0 invariant component L such that [L, V ] = 1 and V maps nontrivially into Op (CAut (L) (u0 )). In particular, LV

(5C)

L is locally unbalanced with respect to U and A.

Lemma 5.6. mp (Ca ) = 3. Proof. Otherwise a ∈ Ipo (G), so L ∈ Cp . Hence by [IA , 7.7.9], p = 3, L ∼ = L2 (27), and |V | = 4. Hence O3 (CG (u0 )) has even order, so with a Frattini argument and [IG , 10.20], m2,3 (G) ≥ m3 (CG (u0 )) = m3 (G) ≥ 4, contradiction.  Lemma 5.7. CU (L) = 1. Proof. Suppose false and let u1 ∈ CU (L)# . Then [AutV (L), u1 ] = 1, so CV (u1 ) acts nontrivially on L. But CV (u1 ) ≤ Op (CG (u0 )) ∩ CG (u1 ) ≤ Op (CG (u1 )) by Lemma 5.2. As CL (u1 ) covers L, CV (u1 ) centralizes L, a contradiction. The lemma follows.  Lemma 5.8. Let u ∈ U # and suppose that CL (u) has a p-component Lu . Let K be the subnormal closure of Lu in CG (u). Then the following conditions hold: (a) Lu /Op (Lu ), and all components of CG (u)/Op (CG (u)), lie in Cp ; (b) Suppose E ∈ E4 (CP (u)) and K ≤ ΓE,2 (G). Then Lu normalizes CΘ1 (G;U ) (a) = Θ1 (Ca ; U ). Proof. We have U # ⊆ Ipo (G) so any component I of K lies in Cp . Now either Lu /Op (Lu ) ↑p I/Op (I) or Lu /Op (Lu ) is a central quotient (possibly an isomorphic copy) of I/Op (I). Hence (a) follows by [IA , 7.1.10]. In (b), ΓE,2 (G) ≤ Γoo P,2 (G) ≤ NG (Θ1 (G; U )) by Lemma 5.4. Finally, CΘ1 (G;U ) (a) is generated by the subgroups CΘ1 (G;U ) (u, a) as u ranges over U # . As CΘ1 (G;U ) (u, a) = COp (CG (u)) (a) ≤ Op (CG (u, a)) = Op (CCa (u)), we have CΘ1 (G;U ) (a) ≤ Θ1 (Ca ; U ). The reverse inclusion follows from Lemma 5.5c. This completes the proof.  Lemma 5.9. L ∈ Chev(p). Proof. If L ∈ Chev(p), then as L is locally unbalanced for p, p = 3 and L∼ = L2 (27) by [IA , 7.7.1], a contradiction as mp (Ca ) = 3 by Lemma 5.6.  Lemma 5.10. Z(L) = 1, mp (L) ≤ 2, and mp (AutCG (a) (L)) ≥ 2. Proof. Let A ≤ S ∈ Sylp (Ca ). Since m3 (S) = 3, a is not characteristic in S, so a < Ω1 (Z(S)) =: Z. As Z ≤ A = U × a, Z ∩ U = 1. Hence by Lemma 5.7, [L, Z] = 1. Suppose that Z(L) = 1. If L ∈ Spor ∪ Alt, then as mp (L) ≤ 3 by Lemma 5.6, p = 3 and L ∼ = 3A6 , 3A7 , 3M22 or 3O N , by [IA , 6.1.4, 6.5.1]. If L ∈ Chev(r) for some r, then r = p by Lemma 5.9, and so again as mp (L) ≤

164

12. THEOREM C∗4 : STAGE A1. FIRST STEPS

3, L ∼ = SL3 (rn ), rn ≡  = ±1 (mod 3). In all cases, by [V17 , 20.10], [Z, L] = 1, contradiction. Hence, Z(L) = 1, and so mp (L) < mp (Ca ) = 3.  As CU (L) = 1, mp (AutCG (a) (L)) ≥ 2, and the proof is complete. Lemma 5.11. L ∈ Chev(r) − Alt for some r = p. Proof. It suffices by Lemma 5.9 to assume that L ∈ Spor ∪ Alt and derive a contradiction. Suppose first that L ∼ = A2p+k with 2 ≤ k < p, or J4 , HS, or He with p = 3, or Suz with p = 5. Recall that V = Op (CG (u0 )) ∩ CG (a), and set VU = CV (U ). Then V U = V = 1. (In the alternating case, V induces a root Ak or Σk subgroup on L; for the sporadic cases, see [IA , 7.7.1].) Then VU ≤ ΔCa (U ) by Lemma 5.5c. Hence L is not locally 2-balanced with respect to p, a contradiction by [IA , 7.7.7c]. Again using [IA , 7.7.1] and Theorem 2.3, we see that as L is locally ∼ unbalanced for p, p = 3 and  L = A7 , M12 , M22 , or HJ. As CU (L) = 1,  # covers L by [V17 , 6.1], but W is a 3 -group by W := O3 (CL (u)) | u ∈ U Lemmas 5.5c and 5.3, contradiction. The proof is complete.  Lemma 5.12. L ∈ Chev(r) for any odd r. Proof. Otherwise, by Lemma 5.11, L ∈ Chev(r), r ∈ {2, p}. As e(G) = 3, L ∈ A (Definition 2.2), so in view of Lemmas 5.9 and 5.10, one of the following holds, with  = ±1: ∼ L2 (rpn ), n ≥ 1, and some uf ∈ U # induces a (1) L = field automorphism on L; 3 (2) L ∼ = L− 3 (r ), p = 3, r ≡  (mod 3), and some # uf ∈ U induces a field automorphism on L; (5D) (3) L ∼ = L3 (rn ), n ≤ 3, rn ≡  (mod p), and AutU (L) ≤ Inndiag(L); or (4) L ∼ = G2 (r), P Sp4 (r), or L4 (r), U ≤ a L, and p divides r2 − 1, r2 − 1, or r + , respectively. Since r and p are distinct and odd, and p divides |L|, rn > 3 in (5D1,3), and r > 3 in (5D4). Suppose that (5D1) or (5D2) holds. By Lemma 5.8a, Lf = E(CL (uf )) ∼ = (r) lies in C . This is impossible as r and p are odd [V , 3.1], L2 (rn ) or L− p 11 3 [IA , 2.2.10]. Suppose that for some u ∈ U # , CL (u) has a p-component Lu ∼ = SL2 (rn ) n with n ≤ 3. Hence again by Lemma 5.8a, L2 (r ) ∈ Cp , which is again impossible. This rules out (5D4), and shows that if (5D3) holds, then p = of 3 and AutU (L) is the image in P GL3 (rn ) of a nonabelian 3-subgroup  GL3 (rn ). Hence by [V17 , 11.3.5], as r is odd, O3 (CL (u)) | u ∈ U # is not  a 3 -group. This contradicts Lemma 5.5d and completes the proof.

5. BALANCE WITH RESPECT TO P0

By Lemmas 5.11 and 5.12, L ∈ Chev(2) −



165

Chev(r) − Alt.

r>2

Lemma 5.13. No element of U # induces a field automorphism on L. Proof. Suppose the contrary: some uf ∈ U # induces a field automorphism on L. As L ∈ A, either L has twisted Lie rank 1 or 2, or L ∼ = L4 (q), p − n Sp6 (q), or Ω8 (q), q = q0 , q0 = 2 for some n. In the last three cases, by Lemma 5.8a, E(CL (uf )) ∼ = L4 (q0 ), Sp6 (q0 ), or Ω− 8 (q0 ), must lie in Cp . The only possibility [V11 , 3.1] is that p = 3 and L ∼ = Sp6 (8). But then m3 (Ca ) > 3, contradicting Lemma 5.6. So L has twisted Lie rank at most 2. Note that since L is locally unbalanced for p and lies in Chev(2), L ∈ Cp [IA , 7.7.9]. Again we apply Lemma 5.8a and conclude from [V17 , 14.5], keeping in mind that mp (Ca ) = 3, that (L, p) is one of the following pairs: (L2 (83 ), 3), 52

(L2 (45 ), 5), (2B2 (2 2 ), 5), or (L3 (27 ), 7). Let K be the subnormal closure of Lf := CL (uf ) in CG (uf ). Suppose first that K is a trivial or diagonal pumpup of Lf . Choose any E ∈ E4 (P0 ). Then K ≤ ΓE,2 (G) by [IA , 7.3.6]. Hence by Lemma 5.8b, Lf normalizes W := Θ1 (Ca ; U ). But W is a p -group, so it follows from [V17 , 4.35] that InnW (L) = 1, and then from [IA , 7.1.4c] that W centralizes L, contradiction. Therefore K is a vertical pumpup of Lf . By Lemma 5.8a, K/Op (K) ∈ Cp . Now by [V17 , 13.24], and using that e(G) ≤ 3 to dispose of the case p = 3, we get K/Op (K) ∼ = L2 (55 ) with p = 5 and L ∼ = L2 (45 ), or K/Op (K) ∼ = L2 (77 ) with p = 7 and L ∼ = L3 (27 ), or K/Op (K) ∈ Chev(2), or mp (Aut(K/Op (K))) = 2. Choose any E ∈ E∗ (P0 ). Unless p = 5 or 7 with K/Op (K) ∼ = L2 (pp ), and mp (C(uf , K)) = 1, we have K ≤ ΓE,2 (CG (uf )) by [IA , 7.3.7], so Lemma 5.8b applies, Lf normalizes W = Θ1 (Ca ; U ), and we again reach a contradiction with [V17 , 4.35] and [IA , 7.1.4c]. Hence, p = 5 or 7, L ∼ = L2 (45 ) or L3 (27 ), respectively, K/Op (K) ∼ = p L2 (p ), and a induces a nontrivial field automorphism on K/Op (K). By Proposition 4.1, mp (CG (a)) > 3 and uf is neither p-central or half p-central in G. Either one of these conclusions is a contradiction, and the lemma follows.  The next lemma rules out several exceptional possibilities for L from A (Definition 2.2). Lemma 5.14. L has twisted Lie rank 1 or 2, and either L ∼ = 3D4 (q) or AutA (L) ≤ Inndiag(L). 3D4 (q) for any q. Also Proof. Suppose false. In particular, L ∼ = L∼ = D4 (q) with p = 3, since m3 (Ca ) = 3. With Lemma 5.13 and [IA , 2.5.12], we conclude that AutA (L) ≤ Inndiag(L). Likewise if p divides | Outdiag(L)|, then since mp (Ca ) = 3, L/Z(L) ∼ = L± 3 (q), of twisted rank at most 2, contrary

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to assumption. Hence as CU (L) = 1, U faithfully induces inner automorphisms on L. By Lemma 5.10, L is simple. Thus, mp (L) = 2. Suppose that L is as in Definition 2.2i, so that (L, p) is one of the following: (L6 (2) or L7 (2), 7), (Sp8 (2), 5), (D4 (2), 5), (F4 (2), 5 or 7). In all cases |L|p = p2 and by [V17 , 4.29], L is balanced for the prime p, contradiction. Thus as L ∈ A, L ∼ = L4 (q), Sp6 (q), or Ω− 8 (q), q even. As mp (L) = 2, ∼ we see using [IA , 4.10.3] that L = L4 (q) with p dividing q + 1. But then for some u1 ∈ U # , E(CL (u1 )) ∼ = L2 (q 2 ), so L2 (q 2 ) ∈ Cp by Lemma 5.8a. Hence  q = 2, so L ∈ Alt, contradiction. The proof is complete. Lemma 5.15. L has twisted Lie rank 2. Proof. Otherwise, since mp (AutA (L)) = 2 and AutA (L) ≤ Inndiag(L) by Lemma 5.14, L ∼ = U3 (q), q > 2 with p dividing q + 1. Indeed q > 4 as U3 (4) is locally balanced with respect to the prime 5 [V17 , 4.29]. By [V17 , 6.6] applied to X = CG (a)/Op (CG (a)), one of the following holds: (1) q > 8 and for some u ∈ U # , E(CL (u)) ∼ = L2 (q);  (5E) (2) p = 3 and Θ1 (CG (a); U ) is not a 3 -group; or (3) p = 3 and Θ1 (CG (a); U ) ≤ O3 (CG (a)). If (5E1) holds, then L2 (q) ∈ Cp by Lemma 5.8a, a contradiction as q > 8. If (5E2) holds, then Lemma 5.5d is contradicted. And if (5E3) holds, then L is locally balanced with respect to U , and hence with respect to A = U a, contradiction. The lemma follows.  Lemma 5.16. L ∼ = L3 (q), q ≥ 4. Proof. By Lemma 5.15, L has twisted rank 2. Also mp (Ca ) = 3 and A = a × U ∈ Ep∗ (Ca ) with CU (L) = 1. And L is locally unbalanced with respect to A. If the lemma fails, then these conditions force L ∼ = Sp4 (q) or G2 (q), q > 2, p dividing q ± 1, but not Sp4 (8) ∈ C3 if p = 3; L ∼ = U4 (q) or n 3 6 2 U5 (q), q > 2, p dividing q − 1; D4 (q), p dividing q − 1; or F4 (2 2 ), n > 1, p dividing 24n − 1. In most cases we choose a particular u1 ∈ U # and set L1 = E(CL (u1 )), using [IA , 4.8.2, 4.7.3A] and [III11 , Table 13.1]. If L ∼ = Sp4 (q) or G2 (q), q > 2, then for suitable u1 , L1 = E(CL (u1 )) ∼ = L2 (q), but if p = 3 and L∼ = G2 (q), then we can arrange L1 ∼ = SL3 (q), ±1 =  ≡ q (mod 3). By Lemma 5.8a, L1 ∈ Cp , whence q = 4 and p = 5. But then by [V17 , 4.29], L is balanced for the prime p, contradiction. Likewise if L ∼ = U4 (q) or U5 (q), q > 2, with p dividing q − 1, we can choose u1 so that L1 ∼ = L2 (q 2 ) ∈ Cp , contradiction. Suppose that L ∼ = 3D4 (q), q > 2. Then if p = 3, A induces inner automorphisms on L since m3 (Ca ) = 3, and we can take u1 with L1 ∼ = SL3 (q), q ≡  (mod 3). If p > 3 we can arrange either L1 ∼ = L2 (q 3 ) or (when ordp q = 3 or 6) L3 (q) or U3 (q), respectively. Again L1 ∈ Cp , contradicting

6. THE SUBGROUP M , AND THE SECOND CASE OF BALANCE

167

∼ 3D4 (2), then p = 3 or 7, and L is Lemma 5.8a. On the other hand if L = locally balanced for p as L ∈ C3 and by [V17 , 4.31]. n Finally, if L ∼ = 2F4 (2 2 ), n > 1, then L is locally balanced for p = 3 (see [IA , 4.7.3A]). Assuming p > 3 we have p dividing 2n − 1 with L1 ∼ = L2 (2n ) n or 2B2 (2 2 ); or p dividing 2n + 1 with L1 ∼ = L2 (2n ); or p dividing 22n + 1 n 2 with L1 ∼ = B2 (2 2 ). As L1 ∈ Cp , the only possibility is p = n = 5 with L1 ∼ = 2B (2 25 ), and then L is locally balanced for p = 5 as L ∈ C , contradiction. 2 5 The proof is complete.  Lemma 5.17. q = 4, and some 2-element of Ca induces a unitary-type involutory outer automorphism of L. Proof. Again as mp (AutA (L)) = 2 and by Lemma 5.13, p divides q −1; also, either p = 3 or A induces inner automorphisms on L. Then L2 (q) ∈ Cp , so E(CL (u1 )) = 1 for all u1 ∈ U # . By [IA , 7.3.3], p = 3. Now [V17 , 6.7] implies that either the conclusion of the lemma holds, or (5E2) or (5E3) holds. The last two possibilities lead to the same contradictions as in the proof of Lemma 5.15, so the proof is complete.  As the configuration stipulated in Lemma 5.17 cannot exist, by Proposition 3.1, the proof of Proposition 5.1 is complete. By Sylow’s theorem, any commuting p-elements have (simultaneous) conjugates in P . Therefore the following corollary is immediate from Proposition 5.1: Corollary 5.18. Assume (1A) and (1B), and suppose that Z(P ) is noncyclic. Then G is balanced for the prime p. Proposition 5.1 and Lemma 2.1 also immediately yield: Corollary 5.19. Assume (1A) and (1B). Set V = Θ1 (G; U ) and M = NG (V ). Then V is a p -group, and one of the following holds: (a) Op (CG (a)) = 1 for all a ∈ Ip (P0 ); (b) For any D ∈ E2 (P0 ), V = Θ1 (G; D). Moreover, M contains ΓP0 ,2 (G) as well as Γoo P,2 (G). 6. The Subgroup M , and the Second Case of Balance Throughout this section, we continue the notation P ∈ Sylp (G), Ep2 ∼ = U  P , P0 = CP (U ), V = Θ1 (G; U ), (6A)

M = NG (V ), and M = M/Op (M ). For this section, we assume (6B)

V = 1, so that 1 = V  M < G.

By Corollary 5.19, (6C)



In particular, P ≤ M .

 ΓP0 ,2 (G), Γoo P,2 (G) ≤ M.

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Lemma 6.1. Assume (6B). Let J be a p-component of M . Then J ∈ Cp , or else p = 3 and J ∼ = (S)U3 (8), A12 , Sp8 (2), Sp6 (8), or U4 (8). Proof. Using [V9 , 6.7] we can find E ∈ Ep∗ (P ) such that U ≤ E ≤ NM (J). Suppose that the lemma fails. Let D ∈ E2 (E). Then NG (D) ≤ ΓE,2 (G) ≤ M . We argue that (6D)

For any p-component JD of CJ (D), JD /Op (JD ) ∈ Cp .

Namely, JD is a p-component of CM (D) = CG (D). By Lp -balance, for any d ∈ D # , the subnormal closure Jd of JD in CG (d) is a product of pcomponents of CG (d), and JD is a p-component of CJd (D). As d ∈ E ∈ Ep4 (G), d ∈ Ipo (G), so the components of Jd /Op (Jd ) lie in Cp . Hence so does JD /Op (JD ) by [III11 , 11.1c], proving (6D).  Now [V17 , 7.14] yields the lemma. Lemma 6.2. Suppose that p = 3, d ∈ I3 (M ), and m3 (CM (d)) ≥ 4. Let  J be a 3-component of CG (d). If O3 (CG (d)) does not normalize J, then CG (d) ≤ M . Proof. Since m3 (CM (d)) ≥ 4 and Γoo P,2 (G) ≤ M , M contains a Sylow  3-subgroup S of CG (d). Let C0 = L3 (CG (d)) d. As some 3-component of CG (d) has at least 3 conjugates by assumption, S∩C0 contains some E ∼ = E34 such that m3 (CE (L/O3 (L))) = 3 for each 3-component L of CG (d). Then CG (d) ≤ C0 NG (S ∩ C0 ) ≤ Γoo  S,2 (G) ≤ M . The proof is complete. Let us call an ordered pair (a, b) of commuting elements of Ip (G) balancing if and only if (6E)

Op (CG (a)) ∩ CG (b) ≤ Op (CG (b))

and non-isolated in the subgroup X of G if and only if a, b ≤ E ≤ X for some E ∈ Ep3 (X). A goal that we shall not quite be able to reach is to show that every ordered pair of commuting elements of order p that is non-isolated in G is balancing. (It would be enough to show this for pairs that are non-isolated in a fixed Sylow p -subgroup P , by Sylow’s theorem.) This goal, together with the condition that Op (CG (a)) = 1 for some a ∈ Ip (G) such that mp (CG (a)) ≥ 3, would imply that ΓoP,2 (G) ≤ M < G where M = NG (Θ1 (G; a, b)) for any particular non-isolated pair (a, b). Instead, however, we establish a balancing result with a similar consequence. In Proposition 5.1 we have verified the balancing condition (6E) for ordered pairs (a, b) with a, b ∈ Ip (P0 ). We now consider the three further cases: (1) a ∈ P − P0 , b ∈ P0 ; (2) a ∈ P0 , b ∈ P − P0 ; (6F) (3) a, b ∈ P − P0 .

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The main result of this section is as follows. Proposition 6.3. Assume (6A) and (6B). Then all pairs (a, b) satisfying (6F1) are balancing. We assume the proposition fails and fix a counterexample (a, b), so that a ∈ Ip (P ) − P0 , b ∈ Ip (P0 ), mp (CG ( a, b)) ≥ 3, and (6G)

W := Op (CG (a)) ∩ CG (b) ≤ Op (CG (b)).

Let Cb = CG (b) and C b = Cb /Op (Cb ). By [IG , 20.6], there is a pcomponent K of Cb such that W a normalizes K and the image of W in X := AutKW a (K) is nontrivial and lies in Op (CX (a)). We proceed in a sequence of lemmas. Lemma 6.4. (a, b) is non-isolated in P . Proof. If Ω1 (Z(P )) ≤ a, b, the assertion is obvious. So we may assume that Ω1 (Z(P )) ∩ P0 ≤ b, whence b ∈ Z(P ). In particular b ∈ Ipo (G), so by [V17 , 4.26], p = 3 and K ∼ = L2 (27). But then Z(P ) is noncyclic, so by our choice of U (6B2), U ≤ Z(P ) and P = P0 , a contradiction since  a ∈ P − P0 . Expand a, b to A ∈ Sp (G). By Lemma 2.1, mp (CG ( a, b)) < 4.

(6H) Hence, using Lemma 6.4,

mp (A) = 3 and we may assume that A was chosen to lie in P . Since U ≤ Z(P0 ) and mp (P ) ≥ 4, there is E ∈ Ep3 (P0 ) such that b ∈ E and ΓE,2 (G) ≤ M. The last condition holds by (6C). We choose E of rank mp (CP0 (b)), as we may. In particular, ΓE,2 (Cb ) ≤ CM (b). We can reduce fairly quickly to a single case. Lemma 6.5. K ∼ = M12 , p = 3, and m3 (Cb ) = 3. Proof. Note first that if K ∼ = L2 (27) and p = 3, then m3 (CG (a)) ≥ 4 by Proposition 4.1a. As m2,3 (G) ≤ 3, O3 (CG (a)) has odd order, whence W does as well. But the image of W in K is a four-group, contradiction. Thus (6I)

(K, p) = (L2 (27), 3).

Now suppose (K, p) = (M12 , 3). If K ∈ Chev, then by [V17 , 6.2] applied to H = AutC b (K), K ≤ ΓE/b,1 (Cb ) ≤ ΓE,2 (Cb ) ≤ CM (b), so (6J)

K is a component of CM (b).

We argue that the same conclusion holds if K ∈ Chev. Indeed if K ∈ Spor then by [IA , 7.7.1b] and Theorem 2.3, we have p = 3 with K/Z(K) ∼ = M12 , M22 , HJ, J4 , HS, or He, or p = 5 with K ∼ = Suz. As we are assuming

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K ∼ = M12 with p = 3, it follows with [IA , 7.5.5] that K = ΓE,2 (K) ≤ M , so (6J) holds. On the other hand, if K ∈ Alt, then K/Z(K) ∼ = Akp+r , 1 ≤ r < p, by failure of local balance. As mp (CCb (a)) ≤ 3, k ≤ 2. Then K = ΓE,2 (K) ≤ M by [IA , 7.5.2b], and again (6J) holds. If K ∈ Cp , then by [V17 , 4.26], p = 3 and K ∼ = L2 (27), contradicting (6I). Hence p = 3 and K is one of the groups listed in Lemma 6.1 or a subcomponent thereof. As K is not locally balanced for p = 3, K ∼ = SU3 (8) or U3 (8), by [V17 , 4.30]. Let L be the subnormal closure of K in M . As m2,3 (G) = 3, L is not a diagonal pumpup of K, and indeed all 3-components of M are P -invariant. =K  or L ∼ By L3 -balance, L = Sp6 (8) or U4 (8). Moreover,  ≤ 1. m3 (CM (L))  with E32 ∼ Otherwise, there exists F ≤ CP (L) = F  P . Then m3 (CP (F )) ≥ 4 oo  and CG (F ) ≤ ΓP,2 (G) ≤ M , so L is a subcomponent of a group in C3 and then itself is in C3 , contradiction.  ∼ Suppose L = Sp6 (8) or U4 (8). If CM (L) = 1, then m3 (CG ( a, b)) ≥ 4, ∗   contradicting (6H). Hence CM (L) = 1 and F (M ) = L. As m3 (L) = 3 <  Now every m3 (M ), some f ∈ I3o (M ) induces a field automorphism on L. E32 -subgroup E ≤ M satisfies m3 (CM (E)) = 4, so again m3 (CG ( a, b)) ≥ 4, contradiction.  = K.  As m3 (CM (L))  = 1, L M

.   ∼ Therefore L/Z( L) = U3 (8), whence L o But then b  P , so b ∈ I3 (G) and K ∈ C3 , a contradiction. Thus (K, p) =  (M12 , 3) and as M12 ∈ C3 , m3 (Cb ) = 3. The proof is complete. Lemma 6.6. CP (b) ∈ Syl3 (Cb ), U ≤ K, and AutK (U ) ∼ = GL2 (3). Proof. Since m3 (Cb ) = 3 and m3 (K) = 2, m3 (C(b, K)) = 1, and then  C(b, K) = b since b is not 3-central in G. Thus the Sylow 3-subgroups of Cb are isomorphic to Z3 × 31+2 . We have AU ≤ CP (b) with [a, U ] = 1, so AU = CP (b) ∈ Syl3 (Cb ). The Cb -conjugacy classes of subgroups of Cb of order 3 are then represented by b, z = [CP (b), CP (b)] = Ω1 (Z(P )), some y ≤ P ∩ K with y ∼K z, bz, and by. Moreover, a ∼Cb y or by by (6G) [IA , 5.3b]. Write U = z, u. If u ∼Cb y or by, then a, b is Cb -conjugate to a subgroup of CP (U ) = P0 , and so a, b is balancing by Proposition 5.1, contradiction. Also as m3 (Cb ) = 3, u ∼G b. Moreover, b ∼ bz in NP (CP (b)), so u ∼Cb bz. Therefore u ∼Cb z, which implies the  lemma by the structure of M12 [IA , 5.3b]. From [IA , 5.3b], we see that besides U , there is exactly one other E32 subgroup U1 of P ∩ K whose automizer in K is isomorphic to GL2 (3). Moreover, from the itemization of maximal subgroups of K [IA , 5.3b], (6K)

NK (U ), NK (U1 ) = K.

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171

Lemma 6.7. G is balanced with respect to U b, and with respect to U1 b. Proof. As U b is abelian, Proposition 5.1 implies the lemma for U b. Write Ω1 (Z(P )) = z. We next show that (6L)

z G ∩ CP (b) = U # ∪ U1# .

Let z1 = z g ∈ CP (b). As z is 3-central in G, z1 ∈ b, so z1 has a nontrivial projection z2 on P ∩ K. Again since z1  is conjugate to Ω1 (Z(P )), no component of CG (z1 ) is isomorphic modulo core to L2 (27), and so X := O3 (CG (z1 , b)) ≤ O3 (CG (z1 )), which has odd order as m2,3 (G) = 3 < m3 (CG (z)). Hence O3 (CK (z1 )) has odd order, forcing z2 ∈ z K ∩ P = U # ∪ U1# . Thus for a suitable h ∈ NK (U ) ∪ NK (U1 ), z2h = z. As b z is completely fused in G, b z ∩ z G = ∅, so z1 = z2 . This proves (6L). It follows that NP (CP (b)) permutes {U, U1 } and therefore normalizes U1 . Let F1 = U1 b. Then NP (F1 ) shears b to U1 . As NK (F1 ) = NK (U1 ) permutes U1# transitively, NP (F1 ) is transitive on bU1 . Since K is balanced with respect to F1 , any unbalancing pair in F1 × F1 must be conjugate to (f, z) for some f ∈ F1 . If the lemma fails, then such a pair must exist and so CG (z) must have a component I that is locally unbalancing with respect to f . As Z(P ) is cyclic and z ∈ Z(P ), Z(I/O3 (I)) = 1. On the other hand, z ∈ I3o (G) so I/O3 (I) ∈ C3 . By [IA , 7.7.1], I/O3 (I) ∼ = L2 (27), which is centerless, a contradiction. This completes the proof.  Now we can quickly reach a final contradiction to complete the proof of Proposition 6.3. Since U and U1 are connected via Z(CP (b)), M contains both ΓU b,2 (G) and ΓU1 b,2 (G), by Lemma 6.7. Therefore with (6K), K ≤

NG (U ), NG (U1 ) ≤ M . Let J be a 3-component of the subnormal closure  by L3 -balance. of K in M . Then J is a covering group of M12 , or M12 ↑3 J,  However, by Lemma 6.1, J ∈ C3 ∪ Chev(2) ∪ Alt. Hence by [IA , 7.1.10], M12 ∈ C3 ∪Chev(2)∪Alt, which is not the case. This contradiction completes the proof of Proposition 6.3. 7. Balance in the Remaining Cases In this section we complete the proof of Proposition 1.1, assuming that (6B) holds. The subsequent section deals with the case that (6B) fails. We choose any non-isolated pair (a, b) ∈ Ip (G) × Ip (G), assume that it is not balanced, and argue to a contradiction. Choose A ∈ Ep3 (G) with

a, b ≤ A and expand A to R ∈ Sylp (Cb ) and R to P g ∈ Sylp (G). (We again write Cb = CG (b) and C b = Cb /Op (Cb )). Replacing our choices by their conjugates under g −1 , we may assume that (7A)

A ≤ R ≤ P,

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so that b is extremal in P . We set A0 = A ∩ P0 = CA (U ). By Propositions 5.1 and 6.3, b ∈ P − P0 , since (a, b) is not balancing. Thus (a, b) satisfies either (6F2) or (6F3). As usual, (7B)

W := Op (CG (a)) ∩ CG (b) ≤ Op (CG (b)).

We assume our counterexample to be chosen so that if possible, after these reductions, a ∈ P0 . By Lemma 2.1, mp (A) < 4. Hence mp (A) = 3. Again by [IG , 20.6], there is a p-component K of Cb such that W a normalizes K and the image of W in X := AutKW a (K) is nontrivial and lies in Op (CX (a)). By [IA , 7.7.1] and Proposition 4.1, (7C)

K ∈ Cp ,

whence by our hypothesis, (7D)

mp (Cb ) ≤ 3.

Note also that by factoring Op (CG (a)) under A0 and using Proposition 6.3, we have (7E)   = Θ1 (G; A0 ) = Θ1 (G; U ) ≤ Op (M ). Op (CG (a)) ≤ Op (CG (x)) | x ∈ A# 0 We first prove Lemma 7.1. a ∈ P0 . Proof. If a ∈ P0 , then by our choice of (a, b), Op (CG (x)) ∩ CG (y) ≤ Op (CG (y)) for all x ∈ P0 and y ∈ P − P0 such that y is extremal in P and (x, y) is not isolated.   and for each For such x, let Wx = CW (x). Then W = Wx x ∈ A# 0 x ∈ A# 0 , Wx = (Op (CG (a)) ∩ CG (x)) ∩ CG (b) ≤ Op (CG (x)) ∩ CG (b) ≤ Op (CG (b)), the first inclusion by Proposition 6.3, and the second by the previous paragraph with y = b. Hence W ≤ Op (CG (b)), contradicting (7B) and completing the proof.  We need a little lemma. SL3 (q), q ≡  (mod 3),  = ±1, 3A6 , Lemma 7.2. If p = 3, then K ∼ = 3A7 , or 3M22 .

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173

Proof. Suppose false. We have R = CP (b) ∈ Syl3 (Cb ). Choose z ∈ I3 (Z(P )). Then z ∈ Z(R), so [K, z] = 1. Let J be a 3-component of the pumpup of K in CG (z). J/O3 (J) is then either a covering group of K or a trivial or vertical pumpup of K. Since K ∈ T3 , J/O3 (J) ∈ C3 by [IA , 7.1.10].  But z is 3-central so z ∈ I3o (G), whence J/O3 (J) ∈ C3 , contradiction.   Lemma 7.3. Let D ∈ E2 (CP0 (b)) and set KD = ΓD,1 ( K D ). Then K D ≤ CM (b). In particular, K ≤ CM (b). Proof. First, since a ∈ P0 , W ≤ Op (CG (a)) ≤ Θ1 (G; U ) ≤ Op (M ), so W ≤ Op (CM (b)). On the other hand, [K, W ] is nontrivial, so it contains K. If K D ≤ CM (b), then K D normalizes a nontrivial p -subgroup W 1 containing W , so p = 3 and K ∼ = L3 (4) or SL3 (4), with W containing a 2-element inducing a unitary-type automorphism on K, by [IA , 7.7.17] and the facts that K/Z(K) ∈ A (Theorem 2.3) and m3 (A11 ) = 3. This contradicts Proposition 3.1 and Lemma 7.2.  We lay out a “standard plan” for reaching a contradiction, which we shall follow for many possibilities for p and K. First, choose D ∈ E2 (CP0 (b)). Note that since U ≤ Z(P0 ) and mp (P0 ) > 2, there is D ∗ ∈ E3 (P0 ) such that D ≤ D ∗ . Indeed if p = 3 we can do better: for a suitable choice of D, there is D ∗ ∈ E4 (P0 ) such that D ≤ D ∗ . This is because there exists D ∗ ∈ E4 (P0 ) such that D ∗  P [IG , 10.17], and we can take D ≤ CD∗ (b). Notice that for all d ∈ D ∗ , Op (CG (d)) ≤ ΓD∗ ,2 (G) ≤ ΓP0 ,2 (G) ≤ M , by (6C). # Next, choose for each   d ∈ D a subgroup Id ≤ CK (d), in such a way # that K = I d | d ∈ D . Then show, using ΓD∗ ,2 (G) ≤ ΓP0 ,2 (G) ≤ M , that M covers each I d , implying that Id ≤ M , so M covers K and Lemma 7.3 is contradicted. Typically I d will be a subgroup of E(CK (d)) or Op (CK (d)), and the location of Id in CK (d) is restricted by Lp -balance or L∗p -balance. The following configuration will arise in carrying out the standard plan in both Lemma 7.5 and Lemma 7.6. (1) p ≥ 5, d ∈ D # , mp (K) = 1, and d induces a nontrivial field automorphism on K; (7F) (2) Id = Lp (CK (d)) and the pumpup J of Id in CG (d) has a Sylow p-subgroup isomorphic to p1+2 ; and (3) J/Op (J) ∈ Spor. Lemma 7.4. The configuration (7F) does not occur. Proof. Suppose it occurs and let T = CP (b) ∈ Sylp (Cb ). Since mp (T ) = 3 and b ∈ P − P0 , CT (K) = b and b is fused to all of bZ in G, where Z = Ω1 (T ∩K) ∼ = Zp . This implies that |NG (T ) : T |p = p and b D = Ω1 (T ). Also Z = Z(P ). Similarly let Ω1 (T ) ≤ T1 ∈ Sylp (CG (d)), so that by (7F2,3), T1 = b D ∗ = d × R1 with R1 = T1 ∩ J ∼ = p1+2 .

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Set Z1 = Z(R1 ). As mp (T1 ) = 3, d is fused to all of dZ1 , and a Sylow psubgroup Q1 of NG (T1 ) satisfies |Q1 : T1 | = p. If T1 = Ω1 (Q1 ) char Q1 , then Q1 ∈ Sylp (G), but mp (Q1 ) = mp (T1 ) = 3, contradiction. Therefore Q1 = Ω1 (Q1 ), and as |Q1 | = p5 and p ≥ 5, Q1 is regular and hence of exponent p. Also Z(T1 ) = d Z1  Q1 and as T1 ∈ Sylp (CG (d)), Q1 ∈ Sylp (NG (Q1 )). Write T ∩ K = x, so x has order ≥ p2 by (7F1), and hence x does not normalize Z(T1 ). Of course Z(T1 ) centralizes b and so is conjugate into Ω1 (T ) = b, d Z. Since Z(T1 ) is not normalized by a conjugate of x, it does not contain a conjugate of Z. But Z(T1 ) meets a conjugate of Z(T ) nontrivially, and hence contains a conjugate of b. As T1 is not embeddable in T , this is a contradiction. The proof is complete.  Lemma 7.5. We have p ≤ 5. Proof. Suppose that p > 5. By [IA , 7.7.1] and Theorem 2.3, if K ∈ Spor ∪Alt, then p = 7 and K ∼ = A7+r , 2 ≤ r ≤ 5. Then there is d0 ∈ CD (K), and we take Id0 = L7 (CK (d0 )) and Id = 1 for all d ∈ D − {d0 }. By Theorem 2.3, and as m2,7 (G) ≤ 3, the pumpup J of K in CG (d0 ) is trivial and normalized by a Sylow 7-subgroup in CG (d0 ). (J is the subnormal closure of Id0 in CG (d0 ).) Hence mp (CD∗ (J/Op (J))) ≥ 2, so J/Op (J) and Id0 /Op (Id0 ) are covered by ΓD∗ ,2 (G) and hence by M . Therefore M covers K, contradicting Lemma 7.3. Hence, K ∈ Chev(r) − Cp for some r = p. Let K1 , . . . , Kn be all the  d varies over D # . By [V17 , 5.18], K = p-components  of Lp (CK (d)) as # K i |1 ≤ i ≤ n . For each d ∈ D and p-component Ki of Lp (CK (d)), let Ji be the pumpup of Ki in CG (d). Provided we have Ji = ΓD∗ ,2 (Ji ) ≤ M for each such i (and d), M covers K, contradiction. Therefore there is d ∈ D # and a p-component Ki of CK (d) such that ΓD∗ ,2 (Ji ) < Ji . We fix this index i. By [IA , 7.3.4] and as p > 5, the pcomponents I of Ji do not satisfy I/Op (I) ∈ Chev − Chev(p). In particular, Ji is a vertical pumpup of Ki , and Ji /Op (Ji ) ∈ Chev(p) ∪ Alt ∪ Spor. d = Cd /Op (Cd ). Then since ΓD∗ ,2 (Ji ) < Ji , we Let Cd = CG (d) and C have mp (Aut(Ji )) ≥ 2. This condition, together with the conditions Ji ∈ A i ∼ and Lp (CJi (b)) = 1, implies that if Ji ∈ Spor, then Ji ∼ = He, K = L3 (2), ∼ and p = 7. (We have used [IA , 7.5.5] to rule out Ji = F3 .) By [V17 , 4.31], K ∼ = L3 (27 ), and so d induces a field automorphism on K. Now (7F) is satisfied with Ji in the role of J, and p = 7. Lemma 7.4 then provides a contradiction. Therefore, Ji ∈ Spor. If Ji ∈ Alt, say Ji ∼ = An , then n ≤ 12, and so as p ≥ 7, | AutD∗ (Ji )| ≤ p, whence ΓD∗ ,2 (Ji ) = Ji by [IA , 7.2.3a], giving the usual contradiction. Hence Ji ∈ Chev(p), so p = 7 and Ji ∼ = L2 (77 ), with b inducing a field automorphism on Ji . As mp (Cb ) = 3, this contradicts Proposition 4.1. The proof is complete. 

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Likewise we can prove Lemma 7.6. We have p = 3. Proof. We use the standard plan outlined before Lemma 7.5. We use Theorem 2.3 without comment. If p > 3, then p = 5 by Lemma 7.5. If K ∈ Spor, then by [IA , 7.7.1], K ∼ = Suz. Set Id = L5 (CK (d)) if L5 (CK (d)) ∼ = A6 , and Id = 1 otherwise. Given d with Id = 1, the pumpup J of Id in CG (d) then satisfies J/O5 (J) ∼ = A6 , A11 , L2 (310 ), U3 (9), Suz, Sp4 (25 ), Sp8 (2), ∗ or F4 (2)  by [V17#, 13.25], so J = ΓD ,2 (J) by [IA , 7.5.2, 7.5.5, 7.3.4]. As K = Id | d ∈ D by [V17 , 10.14.1], we get the usual contradiction. If ∼ K = An , then n = 7, 8, 9, 12, and we take Id = L5 (CK (d)), so that I d ∼ = An or An−5 or 1. Then the pumpup J of Id in CG (d) (modulo core) is An or A7 , or a pumpup of A8 in Chev(2), by [V17 , 13.26]; and J = ΓD∗ ,2 (J) by [IA , 7.3.4, 7.5.2], giving the usual contradiction. Hence K ∈ Chev(r) − Alt for some r. As K is not locally balanced, K ∈ C5 by [IA , 7.7.9], so r = 5   5 1 2B2 (2 2 ) or 2F4 (2 2 ) . By [V17 , 6.4], K = I d | d ∈ D # with I d and K ∼ = 1 unambiguously in Lie(r) ∩ Chev(r), or I d ∼ = A5 , A6 , A8 , U4 (2), or 2F4 (2 2 ) . Let J be the pumpup of any Id in CG (d). Assume J < ΓD∗ ,2 (J); otherwise we get the usual contradiction. 1 Then by [V17 , 5.19], J/O5 (J) ∼ = A10 , HS, Ru, L2 (55 ), or 2F4 (2 2 ) , 1 5 whence I d ∼ = 2F4 (2 2 ), which is bal= A5 or 2F4 (2 2 ) . In the latter case K ∼ anced for p = 5 by [IA , 7.7.8], contradiction. Also if J/O5 (J) ∼ = L2 (55 ), then Proposition 4.1 is contradicted as mp (CG (b)) = 3. Suppose that J/O5 (J) ∼ = A10 . Then F := D ∗ , b ≤ P is a 5-subgroup of CG (d) normalizing J, so AutF (J/O5 (J)) is abelian, whence D ∗ normalizes the image of Id in J/O3 (J). Hence I d is covered by ΓD∗ ,2 (J), and so M covers K, contradiction. Finally suppose that J/O5 (J) ∼ = HS or Ru, still with I d ∼ = A5 . Then 1+2 ∼ and as m5 (Cb ) = 3, Z(P ) = Z(P ∩ J). Hence P ∩J =5 (7G)

Z(P ) ∈ Syl5 (Id ).

Now, if L5 (CG (Z(P ))) = 1, then as Z(P ) is cyclic, some 5-component LZ of CG (Z(P )) would have to satisfy Z(LZ /O5 (LZ )) = 1. But LZ /O5 (LZ ) ∈ C5 , so Z(LZ /O5 (LZ )) = 1 by [V11 , 3.1] and [IA , 6.1.4], contradiction. Therefore L5 (CG (Z(P ))) = 1. By L5 -balance, L5 (CK (Z(P ))) = 1. Since K is not locally balanced for p = 5, [V17 , 6.5], with Cb in the role of X there, implies that K ∼ = L2 (45 ), with d inducing a field automorphism on K. But then (7F) is satisfied and Lemma 7.4 gives a contradiction. The proof is complete.  Now that p = 3, recall that D, for which we had free choice before, is now chosen so that m3 (D ∗ ) = 4; we have no a priori knowledge of the embedding of D in Cb . Lemma 7.7. K ∈ Chev(r) − Chev(3) for some r = 3.

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Proof. Suppose that K ∈ Spor ∪ Alt − Chev. As m3 (Cb ) = 3 (7D), and in view of Lemma 7.2 and [IA , 7.7.1] (unbalancing), one of the following holds: (1) K ∼ = A7 , M22 , or HJ; (7H) (2) K ∼ = M12 ; or (3) K ∼ = J4 , HS, or He. Suppose that (7H1) holds. We follow the standard plan. We take o = O3 (CK (d)), so that I d = O2 (CK (d)) ∼ = E22 or 1. Let Sd be a D bo and set I = [S , D]. Then I = I o , so invariant Sylow 2-subgroup of I d d d d d   Id | d ∈ D # = K by [V17 , 6.1]. Let d ∈ D # with I d = 1, and set Cd = d = Cd /O3 (Cd ). Then Id = [Id , D] ≤ [O3 (CC (b)), D]. Hence CG (d) and C d ∗ by L3 -balance, Lemma 6.2, and the fact that m3 (CCd (b)) ≤ m3 (Cb ) = 3, we conclude with [V17 , 4.33] that either Id ≤ M , or Id is contained in a product   ∼  of C d with L/Z( L) of components L = A7 , M12 , M22 , HJ, or L3 (q), q odd,  q ≡  (mod 3),  = ±1. In the L3 (q) cases by Lemma 7.2 and [IA , 6.1.4],  lies in ΓD∗ ,2 (G) ≤ M ,  = 1. Then by [V17 , 6.8], the preimage L of L Z(L) giving the usual contradiction. Suppose next that (7H3) holds. This time we take Id = L3 (CK (d)) for all d ∈ D # . According to the isomorphism types in (7H3), I d ∼ = 3M22 , , or one of 3A , L (2), respectively [I , 5.3imp]. By [V , 5.20], K = A 7 3 17 A  5 I d | d ∈ D # . Let d ∈ D # and let J be the pumpup of Id in Cd . By [V17 , 5.21], J = ΓD∗ ,2 (J), giving the usual contradiction. Thus, we may assume that (7H2) holds. In that case we do not follow the standard plan. Recall that b is extremal in P , so CP ∩K (b) ∈ Syl3 (K). Then from [IA , 5.3b] we see that the images of CP ∩K (b) and W in Aut(K) generate Inn(K). As P ≤ M , M covers K, contradiction. Therefore if the lemma fails, then K ∈ Chev(3). By [IA , 7.7.1], K ∼ = L2 (27), so m3 (Cb ) ≥ 4. This contradicts (7D) and completes the proof.  Ido

By [V17 , 6.3] and Lemmas 7.7 and 7.2, one of the following holds:   (1) K = L3 (CK (d)) | d ∈ D # ; or (2) K/O3 (K) ∼ = L3 (q),  = ±1, q ≡  (mod 3), and D (7I) maps into Inndiag(K), where its image is the same as the image of a nonabelian 3-subgroup of GL3 (q). Lemma 7.8. Condition (7I1) does not hold. d ∈ Proof. Follow the standard plan with Id = L3 (C  K (d)) for#each  and suppose by way of contradiction that K = I d | d ∈ D . Fix d and let J be the pumpup of a 3-component L of Id in Cd . As m3 (D ∗ ) = 4, it follows by [IA , 7.3.7] that J ≤ ΓD∗ /d,1 (Cd ) ≤ ΓD∗ ,2 (Cd ) ≤ M – leading to a contradiction as usual – unless possibly J/O3 (J) ∈ Chev(3) ∪ Spor ∪ Alt − Chev(s) for all s = 3. Hence the latter condition holds, whence J is a D# ,

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vertical pumpup of L. For the same reason, m3 (Aut(J)) > 2. Note also that 3A6 as as m3 (Cb ) = 3 and [L, d, b] = 1, m3 (L) − m3 (Z(L)) ≤ 1. Also L ∼ = K ∈ Chev [V17 , 13.27]. The only possibility, by [V17 , 13.31], is L ∼ = A5 with 3 J/O3 (J) ∼ = A11 . (Note that if J/O3 (J) ∼ = Co3 or 2 G2 (3 2 ) with L ∼ = L2 (8), then some 2-local in J d contains I × d where I/O3 (I) ∼ = Sp6 (2) or L2 (27), so m2,3 (G) > 3, contradiction.) But then as m3 (D ∗ ) = 4, J ≤  ΓD∗ ,2 (G) ≤ M , giving the usual contradiction. The proof is complete. ∼ L (q), and D acts on K as the image of a By Lemma 7.8 and (7I), K = 3 nonabelian 3-subgroup of GL3 (q). Lemma 7.9. q > 4 and q = 8. Proof. As [K, W ] = K, q > 4, by Proposition 3.1. Suppose that q = 8. Then K is locally balanced with respect to any element of I3 (Inn(K)). Hence a induces an outer automorphism on K with O3 (CK (a)) ∼ = Z19 . But now since b is extremal in P , M ∩ K contains a Sylow 3-subgroup as well as a Sylow 19-subgroup of K. Hence using [IA ,  6.5.3], we see that M covers K, a contradiction completing the proof. Now for any d ∈ D # , let Ido = O3 (CK (d)) and Id = [Ido , Rd ] where Rd ∈ Syl3 (CK (d)). By [V17 , 11.3.5] and Lemma 7.9,   (7J) K = I d | d ∈ D# . Lemma 7.10. K ∈ Chev(r). Proof. Suppose false and continue the above argument. Then for some  d ∈ D # , Id ≤ M . By Lemma 6.2, O3 (Cd ) normalizes every 3-component of Cd . Hence [Id , Rd ] = Id does as well. Then by [IA , 7.7.6] and Theorem 1 · · · L  m , where each L  i ∈ Chev(s), s = d := Cd /O3 (Cd ), Id ≤ L 2.3, in C  i /Z(L i) ∼ 3, or L = L2 (27), M12 , M22 , HJ, A7 , or A10 . As Id ≤ M , and  i , so O3 (Cd ) ≤ ΓD∗ ,2 (G) ≤ M , it follows that M does not cover some L d ). Using [IA , 7.3.7], we see that this implies that L  i /Z(L  i ) ∈  i ≤ ΓD∗ ,2 (C L  i ) ∈ Alt∪Spor, then as m3 (CD∗ (L  i )) = 1, m3 (Aut(L  i )) ≥  i /Z(L Chev(s). If L ∗ ∼   m3 (AutD∗ (Li )) ≥ 3, whence Li = A10 with D acting as the product of three  i = ΓD∗ ,2 (L  i ) by [IA , 7.5.2b, groups generated by 3-cycles. But then again L i ∼ 7.2.3a], contradiction. Therefore L = L2 (27), with b inducing a nontrivial  field automorphism on Li as Id ≤ O3 (CG ( b, d)). By Proposition 4.1, m3 (Cb ) > 3. But m3 (Cb ) = 3 by (7D), contradiction. This completes the proof.  Since Lemma 7.10 and Lemma 7.7 are in conflict, we have completed the proof of Proposition 1.1 under the assumption (6B).

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8. The Residual Case In this section we complete the proof of Proposition 1.1. By the previous section, we may assume that (6B) fails, that is, (8A)

Op (CG (d)) = 1 for all d ∈ Ip (P0 ).

Lemma 8.1. Assume that (8A) is true. Then one of the following holds: (a) Op (CG (d)) = 1 for all d ∈ Ip (G) with mp (CG (d)) ≥ 3; or (b) p ≥ 5, Sylow p-centers in G are cyclic, and Θ3/2 (G; D) = 1 for all D ∈ Ep2 (G) with mp (CG (D)) ≥ 3. Proof. Note first that if a ∈ Ip (P ) and b ∈ Ip (P0 ) ∩ Ipo (G) with [a, b] = 1, then W := Op (CG (a)) ∩ CG (b) = 1, and indeed if p > 3, then W ∗ := Op (CG ( a, b)) = 1. For, Op (CG (b)) = 1 by (8A). Hence W or W ∗ , if nontrivial, acts nontrivially on some locally unbalanced component K of CG (b). But as b ∈ Ipo (G), K ∈ Cp , whence p = 3 and K ∼ = L2 (27), and Proposition 4.1b is contradicted. This proves our assertion. Now if p = 3, then since m3 (P ) ≥ 4, there exists E  P with E ∼ = Ep4 and E ≤ P0 , by [V9 , 5.6]. Then for any a ∈ I3 (P ), CE (a) is noncyclic. But by the previous paragraph, for any b ∈ CE (a)# , O3 (CG (a)) ∩ CG (b) = 1. It follows that O3 (CG (a)) = 1, and conclusion (a) holds. We may therefore assume that p ≥ 5, and in view of (a) and (8A), that P0 < P . Thus, Z(P ) is cyclic. We show that (b) holds. Let A ≤ P with A∼ = Ep3 , and suppose that G is not 3/2-balanced with respect to A. Notice that as |P : P0 | = p, D ∩ P0 = 1 for any D ∈ E2 (P ), so ΔG (D) = 1 by (8A). Hence the failure of 3/2-balance is due to some elements a, b, x ∈ A# such that V := [Op (CG (a)) ∩ CG (b), x] ≤ Op (CG (b)), and in particular V = 1. As A ∩ P0 is noncyclic, there is d ∈ A# ∩ P0 such that Vd := [CV (d), x] = 1. But V ≤ Op (CG (a)), so Vd ≤ [Op (CG ( a, d)), x]. Let Cd = CG (d). Then by [IG , 20.6], Cd has an a Vd -invariant component J such that Vd maps into Op (CAutC (J) (a)). In particular, J is not d balanced for p, whence J ∈ Cp and then mp (Cd ) = 3. Hence x normalizes J, so J is not weakly locally balanced with respect to A. As p ≥ 5, it follows by [V17 , 4.34] that A = d, a, x with x inducing a nontrivial field automorphism on J ∈ Chev, and mp (J) = 1 with a inducing an inner automorphism on J. Therefore for A ≤ Q ∈ Sylp (Cd ), Ω1 (Z(Q)) = a, d. We conclude that a, d = z, d for some z ∈ Ip (Z(Q)) that is p-central in G. By (8A), Op (CG (z)) = 1, so by the first paragraph of the proof, Op (CG ( z, d)) = 1. Therefore Op (CG ( a, d)) = 1, a contradiction as 1 = Vd ≤ Op (CG ( a, d)). It follows that G is 3/2-balanced with respect to any A ∈ Ep3 (G). Now if D ≤ A ≤ P with A ∼ = Ep2 , then there exists D1 ≤ = Ep3 and D ∼ ∼ A ∩ P0 such that D1 = Ep2 . By 3/2-balance and [IG , 21.10b], Θ3/2 (G; D) =  Θ3/2 (G; D1 ) = 1, the last because of (8A). The lemma is proved. Of course, if Lemma 8.1a holds, then G is balanced with respect to every element of Ep3 (G), as Proposition 1.1 asserts. So to complete the proof of

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179

Proposition 1.1, it suffices to show that if Op (CG (d)) = 1 for some d ∈ Ip (G) with mp (CG (d)) ≥ 3, and (8A) and Lemma 8.1b hold, then a contradiction ensues. We may assume that CP (d) ∈ Sylp (CG (d)), and choose A ∈ Ep3 (P ) with d ∈ A and A ∈ Sp (G). Choose a ∈ A ∩ Z(P )# and set D = a, d. Then [Op (CG (d)), a] ≤ [Op (CG (d)), D] ≤ Θ3/2 (G; D) = 1. Thus Op (CG (d)) ≤ Op (CCG (a) (d)) ∈ NCG (a) (A; p ). But as a ∈ Z(P ), a ∈ Ipo (G), and as p ≥ 5, it follows from [IA , 7.7.12] that all components of CG (a)/Op (CG (a)) are strongly locally balanced for p. Hence Op (CG (d)) = Op (CCG (a) (d)) = 1. This is the desired contradiction. The proofs of Proposition 1.1, (1C) and part (c) of Theorem C∗4 : Stage A1 are thus complete as shown in Section 1. 9. B-Signalizers and Ipo (G): The Centralizing Case We define H = {H ≤ G | H is a 2-local in G and mp (H) = 3} In the remaining sections of this chapter we prove parts (a) and (b) of Theorem C∗4 : Stage A1. Thus we prove the following two propositions: Proposition 9.1. Assume (1A). Let Ep3 ∼ = B ≤ H ∈ H. Then B # ⊆ # i.e., mp (CG (b)) ≥ 4 for all b ∈ B .

Ipo (G),

Proposition 9.2. Assume (1A). Then Op (CG (A)) has odd order for every elementary abelian p-subgroup A of G such that mp (CG (A)) ≥ 3. For any group X of even order, we define (9A)

mIp (X) = max mp (CX (z)). z∈I2 (X)

Thus mIp (G) ≤ m2,p (G) = 3. We begin the proof of Proposition 9.1 with the following case. Proposition 9.3. Let z ∈ I2 (G). If Ep3 ∼ = B ≤ CG (z), then B # ⊆ Ipo (G). Assume Proposition 9.3 is false, and fix a counterexample (B, z). Lemma 9.4. We have Lp (CG (B)) = 1 and z ∈ Op (CG (B)) = 1. Proof. Let C = CG (B) and C = C/Op (C). Assume that L := E(C) = 1. Whether Z(L) = 1 or not, there exists x ∈ Ip (L) − Op p (L) [IG , 16.11], whence x B ∼ = Ep4 and so B # ⊆ Ipo (G), contrary to hypothesis. Hence F ∗ (C) = Op (C). If z acts nontrivially on Op (C), then the Baer-Suzuki theorem implies that mp (CG (B)) ≥ 4, again contrary to hypothesis. This contradiction finishes the proof.  Lemma 9.5. There exists b ∈ B such that b ∈ Ipo (G).

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Proof. Let P ∈ Sylp (G) with B ≤ P . If Z(P ) ∩ B = 1, then we have mp (BZ(P )) ≥ 4, and as BZ(P ) is abelian, B # ⊆ Ipo (G), contrary to assumption. Hence, there exists 1 = b ∈ B ∩ Z(P ). Since mp (P ) ≥ 4,  b ∈ Ipo (G). Fix b as in Lemma 9.5. and reach a contradiction Cb /Op (Cb ). As mp (Cb ) ≥ Cp by (1A3), and |Op (Cb )|

We will now study the structure of Cb := CG (b) by showing that mp (CCb (B)) ≥ 4. Let C b = 4 > m2,p (G), every component of E(C b ) lies in is odd by (1A2). Hence z is an involution of C b .

Lemma 9.6. Let L = E(C b ) have components L1 , . . . , Lk . Then the following conditions hold: (a) [z, Op (C b )] = 1; (b) L = 1 and z acts nontrivially on L; and (c) B normalizes, but does not centralize, each Li , 1 ≤ i ≤ k. Proof. By Lemma 9.4 and L∗p -balance [IA , 5.18], z ∈ Op (CCb (B)) ≤ L∗p (CCb (B)) ≤ L∗p (Cb ).

(9B)

But [L∗p (C b ), Op (C b )] = 1, so (a) holds. Then the F ∗ -theorem [IG , 3.6] implies (b). Assume that some Li is not normalized by some c ∈ B # . Since by Lemma 9.4 and L∗p -balance [IG , 5.18], mp (B) = 3 ≥ m2,p (C b ), Proposition 8.7(i) of [IG ], combined with the fact that Li ∈ Cp , yields that p = 3 and Li ∼ = L2 (8) and m3 (L3 (Cb ) b) ≥ 4. Indeed as m2,3 (C b ) ≤ 3, E(C b ) = L1 × L2 × L3 with Li ∼ = L2 (8) for i = 1, 2, 3. Thus B = a, b, c where a ∈ L3 (Cb ) projects nontrivially on each component of E(C b ). On the other hand, by (9B) and the definition of L∗p (Cb ), z ∈ L3 (Cb )O3 (Cb )O3 (CCb (P )) where P ∈ Syl3 (L3 (Cb )). Since m3 (P b) ≥ 4 > m2,3 (G), and X := O3 (Cb )O3 (CCb (P )) is P b-invariant, X has odd order. Hence 1 = z ∈  O 2 (L∗3 (C b )) = E(C b ). As [B, z] = 1, z ∈ CE(C b ) (c). But by the structure of E(C b ), CE(C b ) (c) is a 3-group, contradiction. Thus B normalizes every Li . Finally, if [B, Li ] = 1 for some i, then by L3 -balance, L3 (CG (B)) has a 3-component K such that K/O3 (K) ∼ = Li . This contradicts Lemma 9.4 and completes the proof of the lemma.  We now complete the proof of Proposition 9.3. By Lemma 9.6, z centralizes Op (C b ) and acts nontrivially on L = L1 ∗ · · · ∗ Lk , with each Li ∈ Cp . Suppose that for some i0 ∈ {1, . . . , k}, z normalizes and acts nontrivially 0 = C 0 /C (Li ). Then z = 1 = B  and on Li0 . Let C 0 = NC b (Li0 ) and C 0 Cb  = 1, i ∼ L = Li /Z(Li ). Now [V17 , 7.11] implies that either Op (C  (B)) 0

0

0

C0

3  i ∼ But the latter must hold, as or p = 3 and L 0 = L2 (3 ) and |B| = 3.

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 by [IG , 5.19]. It z ∈ Op (CG (B)) by Lemma 9.4, and so z ∈ Op (CC0 (B)) follows that there exists E32 ∼ = B0 ≤ B with b ∈ B0 and [B 0 , Li0 ] = 1. Then 3 ∼ BLi0 = E32 × P ΣL2 (3 ). It follows immediately that m3 (CG (B)) ≥ 4, a contradiction. z Thus without loss of generality we may assume that L1 = L2 . If CL1 (B) contains a p-element x ∈ Z(L1 ), then [x, z] = x−1 xz is a nontrivial p-element inverted by z, a contradiction since z ∈ Op (CC b (B)) by Lemma 9.4. Thus  O p (CL1 (B)) ≤ Z(L1 ). In particular, Z(L1 ) = 1 and no element of B can induce a nontrivial inner automorphism on L1 . However, as L1 ∈ Cp , [V17 , 7.12] implies that CL1 (B)−Z(L1 ) contains a p-element, a final contradiction. This completes the proof of Proposition 9.3. A direct corollary is: Corollary 9.7. Let y ∈ Ip (G) − Ipo (G). Then mIp (CG (y)) ≤ 2. For if mIp (CG (y)) > 2, then CG (y) contains an involution t and Ep3 subgroup B such that [B, t] = 1, and we may assume that y ∈ B. Then y ∈ Ipo (G) by Proposition 9.3, contrary to assumption. 10. The Non-Centralizing Case: A Dichotomy We continue the proof of Proposition 9.1, by contradiction. Thus, (10A)

Ep3 ∼ = B ≤ H ∈ H, but B # ⊆ Ipo (G).

Throughout this section, B is fixed. We have Ip (CG (B)) ⊆ B,

(10B) and by Proposition 9.3, (10C)

CG (B) has odd order.

Let P ∈ Sylp (G) and Ep2 ∼ = U  P be as in (1B); we may assume that B ≤ P . By [IG , 10.20], (10D)

U # ⊆ Ipo (G).

Let T(B) be the set of minimal B-invariant 2-subgroups of G, and for any T ∈ T(B), let BT = CB (T ). Using [IG , 11.12] together with (10C), we obtain that for any T ∈ T(B) and bT ∈ B − BT , (10E)

BT ∼ = Ep2 , B = BT × bT  , and bT  acts irreducibly on T .

In particular, T bT  ∼ = A4 if p = 3, and m2 (T ) ≥ 3 if p > 3. There may be many choices of T , and we shall further specify our choice below.

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We have the following dichotomy. Lemma 10.1. Let T ∈ T(B). Then one of the following holds: (a) BT ∩ Ipo (G) = ∅, and there exists y ∈ B # with mp (CG (y)) = 3; or (b) BT ∩ Ipo (G) = ∅, and there exist y, c ∈ BT# , bT ∈ Z(P ) ∩ B − BT ⊆ Ipo (G), and u ∈ U # such that [c, u] = bT and [y, u] = 1. In particular, c, u ∼ = Zp × p1+2 . = p1+2 and B u ∼ Proof. By (10A), there exists y ∈ B # with m3 (CG (y)) = 3. Hence Ω1 (Z(P )) ≤ B. If U ∩ BT = 1, then (a) holds by (10D). So assume that U ∩BT = 1. Therefore U ≤ B and Z(P ) is cyclic, whence U ∩B = Ω1 (Z(P )). Write U ∩ BT = bT . If [U, B] = 1, then 4 = mp (U B) ≤ mp (CG (B)), a contradiction. Hence, B acts nontrivially on U . Thus there exist y, c ∈ BT and u ∈ U − bT  such that [c, u] = bT and [y, u] = 1. It follows that

c, bT , u ∼  = p1+2 and B u ∼ = Zp × p1+2 , so (b) holds, as required. Note that in either case of the lemma, y ∈ Ipo (G). We make some elementary observations. Lemma 10.2. Let T ∈ T(B) and let bT be as in (10E). Let w ∈ BT# w = Cw /Op (Cw ). Then the following conditions and set Cw = CG (w) and C hold:   (a) T bT ∼ = T bT ; w )] = 1; (C (b) [T , Op w ); and (c) T bT normalizes every component of E(C   w ). (d) T bT acts faithfully on some component of E(C ∼ Proof. Since |Op (Cw )| is odd by (1C), T ∼ = T . Of course B = B, so (a) holds. By (10B) we may apply the Thompson-Bender lemma [IG , 23.3(ii)] w )TB  and conclude that T ≤ Op (X), whence (b) to the group X = Op (C holds. If (c) fails, then as T = [B, T ], B does not normalize some component w has at least p ≥ 3 components, each of which contains  of Cw . Hence C  into account, a non-central element of order p by [IG , 16.11]. Taking w o this implies mp (Cw ) ≥ p + 1 ≥ 4, so w ∈ Ip (G). Hence by the hypothesis w are Cp -groups. But as T ≤ Cw , we can apply (1A3), all components of C w . That result implies that p = 3, and gives the possible [IG , 8.7(ii)] to C w , all of which are T3 -groups. This isomorphism types for components of C contradiction proves (c). Together with the F ∗ -theorem [IG , 3.6], (b) and (c) imply (d). The proof is complete. 

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183

In the remainder of this section, we shall prove that conclusion (a) of Lemma 10.1 is impossible, that is: Proposition 10.3. Let T ∈ T(B). Then conclusion (b) of Lemma 10.1 holds. Assume then that for some T ∈ T(B), conclusion (a) holds, and let y ∈ B be as there. We assume that T and y have been chosen so that, if possible (subject to conclusion (a) holding), (10F)

[T, y] = 1, i.e., y ∈ BT .

Lemma 10.4. We have p = 3. Moreover, choose any x ∈ BT ∩ Ipo (G) and set Cx = CG (x) and C x = Cx /Op (Cx ). Then the following conditions hold: (a) L := E(C x ) ∼ = L2 (33 ), and m3 (O3 (C x )) = m3 (C(x, L)) ≤ 2; (b) y induces a nontrivial field automorphism on L, y ∈ BT , and CG (BT ) is solvable; and G (c) CL∩B (y) =: b0  ∼ = Z3 and b0  ∈ x .   Proof. By Lemma 10.2 applied wih w = x, T bT ∼ = T bT  acts faithfully on some component L of C x . Since T ∈ NCx (B; 2) and mp (CG (B)) = 3, L is not strongly locally 1-balanced with respect to p. But L ∈ Cp as x ∈ Ipo (G). Hence by [IA , 7.7.12], p = 3 and L ∼ = L2 (33 ). Then since m2,3 (G) = 3 and m2,3 (L) = 1, m3 (C(x, L)) ≤ 2 and C(x, L) has odd order, so L = E(C x ), proving (a). Now m3 (CCx (y)) ≤ m3 (CG (y)) = 3, while if y induces an inner automorphism on L, then m3 (CCx (y)) ≥ 4. Hence y induces a nontrivial field automorphism on L. Let T0 = O2 (CL (y)), and R ∈ Syl3 (Cx ) with B ≤ R. Then Ω1 (Z(R)) ∼ = E32 and so B = y Ω1 (Z(R)) = y, x, b0  with b0 ∈ NL∩B (T0 ). In particular, x, b0  is a hyperplane of B with m3 (CCx ( x, b0 )) ≥ 4, so that x, b0 # ⊆ I3o (G). Now T0 ∈ T(B). Moreover, BT0 ∩ x, b0  = 1, so by the previous paragraph, had we chosen T0 , we would still be in case (a) of Lemma 10.1, and y ∈ BT0 . Therefore by our choice of T , we have y ∈ BT , whence BT = x, y. As |C(x, L)| is odd, (b) follows. Hence to complete the proof it remains to assume that b0  ∈ xG and derive a contradiction. But if b0  ∈ xG , then R ∈ Syl3 (CG (b0 )), so b0  and x are NG (R)-conjugate. Now clearly J(R) ≤ LC(x, L), so J(R) = RL × J(RC ), where RL = R ∩ L ∼ = E33 and RC ∈ Syl3 (C(x, L)). As b0 ∈ [J(R), J(R)], also x ∈ [J(R), J(R)]. Hence [J(R), J(R)] ∩ Z(R) = 1, so J(R) is abelian. Thus J(RC ) ∼ = E3a , a = m3 (C(x, L)) ≤ 2. But then x ∈ [J(R), R, R]. As b0 ∈ [J(R), R, R], b0  and x are not NG (R)-conjugate, contradiction. The proof is complete.  y = Cy /O3 (Cy ), and K = L3 (Cy ). By Corollary 9.7, Set Cy = CG (y), C (10G)

mI3 (Cy ) ≤ 2.

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e  ∼ Lemma 10.5. K = A7 , M12 , M22 , HJ, or L± 3 (r ), where r = 3 is  = 2, and B induces inner or inner-diagonal an odd prime, e ≤ 3, m3 (K)  automorphisms on K.

Proof. By Lemma 10.4, p = 3 and T bT ∼ = A4 . As y ∈ BT , T bT  ≤   Cy . By (1C), O3 (Cy ) has odd order, so T bT ∼ = A4 . We may assume that CP (y) ∈ Syl3 (Cy ). Using the Bender-Thompson again [IG , 23.3(ii)], we see that   lemma  and hence (as y ), T] = 1. Therefore T bT acts faithfully on K, [O3 (C  0 of K.  m3 (Cy ) ≤ 3) on some component K  0 )) ≥ 2, then since T acts faithfully on K  0 , CB ( K  0 ) = BT . If m3 (CB (K But by Lemma 10.4, CG (BT ) is solvable, a contradiction. Therefore (10H)

 0 ) =  y . CB (K

Moreover, from the structure of Cx , T ≤ O3 (CG (BT )) = O3 (CCy (x)). We thus have    0 )) ≥ 2, as  x × T bT ∼ (1) m2,3 (Aut(K = Z3 × A4 acts (10I)  0 ; and faithfully on K T .  0 is not weakly 1-balanced with respect to B (2) K Also by (10G) and as m3 (Cy ) = 3, (10J)  0 ) ≤ 3, and if K  0 is simple, then m3 (K  0 ) ≤ 2 and mI (K  0 ) ≤ 1. m3 (K 3 0 ∼  0 ∈ C3 , then by (10I2) and [IA , 7.7.1], K Now if K = L2 (33 ). But then  0 ) = 3, contradicting (10J). Thus, K  0 ∈ C3 . However, by Theorem m3 (K  0 ) ∈ A.  0 /Z(K 2.3, K  0 ) = 1. Using [IA , 6.1.7], the definition of A Assume next that Z(K (Definition 2.2), and (10J) (together with [IA , 5.6.1] for 3-ranks), we find 0 ∼ K = 3A6 , 3A7 , 3M22 , 3O N , or SL3 (q), q ≡  (mod 3),  = ±1. But recall  0 ] = 1. By [V17 , 20.10], [bT , K  0 ] = 1, which that bT ∈ Z(P ), so [bT , P ∩ K  0 is simple. contradicts (10H). Hence, K  If K0 ∈ Spor ∪ Alt, then by failure of weak local 1-balance and [IA , 0 ∼ 7.7.6b], K = M12 , M22 , HJ, as desired, or A3n+4 , n ≥ 1. In the latter case,  0 ) ≤ 2, so K 0 ∼ n = 1 as m3 (K = A7 , as desired.  0 is simple and not in C3 , r =  Assume then that K0 ∈ Chev(r); as K ∼ 3. Since T = [T, x ] = E22 , it follows from [IA , 2.5.12] that T injects  x)). However, using [IA , into Inndiag(K0 ) and hence into O3 (CInndiag(K 0 ) (  0 , we see that if x  does not map into 4.9.1, 4.7.3A] and the simplicity of K   x)) = 1, so T = 1, contradiction. Thus, Inndiag(K), then O3 (C  ( Inndiag(K0 )

 0 . Likewise b, which is x  induces an inner-diagonal automorphism on K

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185

y , induces an inner-diagonal automorphism on K  0 by [V17 , 3-central in C  0 )|. Hence 20.8]. It now follows from [IA , 7.7.6c] that 3 divides | Outdiag(K e  n 0 ∼  0 ∈ A, K as K = U6 (2), L± 3 (r ) as in the statement of the lemma, or L3 (2 ), n I 2 ≡  (mod 3),  = ±1, n > 1. But m3 (U6 (2)) = 3 by [V17 , 18.3]. And the  x)) ∈ Lie(2) by [IA , 4.2.2], whence last case is impossible as then O2 (CK 0 ( x)) = 1, a contradiction. O2 (CK 0 (  0 ) = 2.  0 is as asserted in the lemma, and in particular, m3 (K Thus K    If E(Cy ) had any other component K1 , then K1 would contain an element y ) ≥ 4, contradiction. Therefore  1 ), and so m3 (C of order 3 outside Z(K y ) = K  0 and the proof is complete. E(C   ∼ Lemma 10.6. K = M22 or L3 (re ), r > 3. Proof. Suppose false. As in the previous lemma we may assume that CP (y) ∈ Syl3 (CG (y)). Set A = B ∩ K.  has  ∼ Suppose that K = L3 (re ). If re ≡  (mod 9), then by [IA , 6.3.5], K e I  a subgroup isomorphic to Zre − × Z re − , and as r −  is even, m3 (K) = 2, 3  3 = 32 . Since contradicting (10J). Thus 9 does not divide re − , so |K| ∼   x)) = E22 , x  induces an inner automorphism on K, and so does b, O2 (CK ( y . Hence as it is 3-central in C (10K)

B = A × y and Q := AutK (B) ∼ = Q8 .

 ∼ If K = M22 , then (10K) holds by [IA , 5.3c]. In either case, A ∈ Syl3 (K), and the Q-orbits on B # are {y}, {y −1 }, A# , yA# , and y −1 A# . Since y is not 3-central in G, y is G-conjugate to some element of B − y. If y is conjugate to an element of yA# or y −1 A# , then the same holds for y −1 . Since m3 (CG (y)) = 3 while m3 (CG (x)) ≥ 4 ≤ m3 (CG (b)), we have x, b ∈ A. In particular, x ∈ bG , which contradicts Lemma 10.4c. Thus y is conjugate to an element of A# . Let B ≤ S ∈  ∼ Syl3 (Cy ). If B < Ω1 (S), then K = L3 (re ) and Z := Ω1 (Z(S)) = z, y with z ∈ A# . But then y is conjugate to z in NG (S), which is impossible as [Ω1 (S), Ω1 (S)] contains z but not y. Thus B = Ω1 (S) and so NG (S) normalizes y G ∩ B = {y, y −1 } ∪ A# and hence normalizes y (and A). Therefore S ∈ Syl3 (G), a contradiction as m3 (S) = 3. The lemma is proved.  Again set A = B ∩ K. As m3 (Cy ) = 3, we have BT = y, a where

a = BT ∩ K.  ∼ Lemma 10.7. K = A7 .  ∼ Proof. Otherwise K = M12 or HJ, and we use [IA , Tables 5.3bg].   of type 3B, and N  (  a is an element of K a) =  a,  t × Since [T, BT ] = 1,  K   t and invert a. Set Ca = CG (a) T bT ∼ = Σ3 × A4 . Let t ∈ I2 (Cy ) map onto  a = Ca /O3 (Ca ). and C

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Suppose that a ∈ I3o (G). Then Lemma 10.4 applies with a in place of x. Let Ka = L3 (Ca ). Thus, y induces a nontrivial field automorphism on a ∼ K = L2 (33 ). As CK (t) contains a copy of A4 , it follows that t centralizes  a , with O2 (La ) =  a . As O3 (Ca ) has odd order, La := L2 (CK (t)) covers K K a O{2,3} (La ). Now for any prime s dividing |O2 (La )|, ms (CG (t)) ≤ 3 and so the sectional s-rank of a critical subgroup of Os (La ) is at most 4. But L2 (33 ) contains a Frobenius group of order 33 .13. Using Clifford’s theorem we conclude that Os (La ) ≤ Z(La ). Hence F (La ) ≤ Z(La ), so La is quasisimple. Thus La y contains Frobenius groups F33 .13 and F13.3 of the indicated orders, with y ∈ F13.3 . Still assuming a ∈ I3o (G), let V = O2 (CG (t))/Φ(O2 (CG (t))) and suppose that [V, La ] = 1. Because of F33 .13 , a Sylow 13-subgroup S13 of La has a free submodule on V of order 226 (as it contains all 13 2-dimensional modules for the Frobenius kernel). Then because of F13.3 , m2 (CV (y)) ≥ 8. However, C(y, K) has odd order (otherwise |CG (B)| is even, against (10C)). So the  which is less than 8 by sectional 2-rank of Cy is at most that of Aut(K), [V17 , 18.2], contradiction. Therefore [V, La ] = 1. As G is of even type, [E(CG (t)), La ] = 1, and then since m3 (CG (t)) ≤ 3 = m3 (La ), La ≤ Kt for some component Kt ∈ C2 of E(CG (t)). But m3 (Kt ) = 3 and Kt contains a Frobenius group F33 .13 , so by [V17 , 9.9], Kt /Z(Kt ) ∈ A, a contradiction. We have proved that m3 (CG (a)) = 3. a ) ∼ Hence, Lemmas 10.5 and 10.6 apply to a in place of y. Thus E(C =   ∼     a, t E(Ca ) = Σ3 × E(Ca ) or Σ3 Y Aut(E(Ca )). M12 , HJ, or A7 , and so  In either case, there exists an involution t0 ∈ NG ( a) inverting BT . Hence t0 ∈ NG ( x) inverts y, contradicting the structure of Cx . The proof is complete.  Now we can finish the proof of Proposition 10.3. As y ∈ BT , BT = y, a  ∼ where  a is a 3-cycle in K = A7 . If m3 (CG (a)) = 3, then we may replace y by a; by Lemma 10.7 applied to a, E(CG (a)/O3 (CG (a))) ∼ = A7 . Hence, there exists an involution t inverting y and acting as an outer automorphism on  It follows that there exists an involution inverting BT , which contradicts K. the structure of Cx . Therefore a ∈ I3o (G) and we may apply Lemma 10.4 with x = a. Let v ∈ NK (A) be of order 4, so v 2 ∈ NG ( x) ∩ Cy . It follows that A = [B, v 2 ] = Ω1 (Z(R)) for some R ∈ Syl3 (Cx ). Observe that R ∈ Syl3 (CG (A)). Hence by a Frattini argument, NG (R) covers NG (A)/CG (A), so there exists v  ∈ NG (R) acting the same way as v – and hence irreducibly – on A. On the other hand, B ∩ L = [Z(J(R)), R, R] is a characteristic subgroup of R properly contained in A, so v  normalizes B ∩ L. This is a contradiction, and the proof of Proposition 10.3 is complete. 11. The Case BT ∩ Ipo (G) = ∅ By Proposition 10.3, we have BT ∩ Ipo (G) = ∅. Recall that B ≤ P ∈ Sylp (G) and Ep2 ∼ = U  P . Since BT ∩ Ω1 (Z(P ))# ⊆ BT ∩ Ipo (G) = ∅,

11. THE CASE BT ∩ Ipo (G) = ∅

187

Ω1 (Z(P )) = bT  is cyclic. Now Proposition 10.3 implies that there exist y, c ∈ BT# and u ∈ U # such that [c, u] = bT and [y, u] = 1. We abstract this last condition, discarding the condition on U , as follows: Definition 11.1. BT∗ is the set of all y ∈ BT# such that for some c ∈ BT# and some u ∈ Ipo (G), [c, u] = bT and [y, u] = 1. Thus for such y, c, and u, c, u ∼ = Zp × p1+2 . = p1+2 and y, c, u ∼ Lemma 11.2. We have BT∗ = ∅. Moreover, if p = 3, then BT∗ = BT# . Proof. By Proposition 10.3, the first statement holds, so suppose that p = 3. By Konvisser’s theorem [IG , 10.17], there is E  P with E ∼ = E34 . As BT ∩ I3o (G) = ∅, BT ∩ E = 1 and CE (BT ) = B ∩ E = bT . Let y0 ∈ BT# and write BT = y0 , c0  and E1 = CE (y0 ). Then CE1 (c0 ) = CE1 (BT ) = bT . But as p = 3, m3 (E1 ) > 1, so there is u0 ∈ E1 such that [c0 , u0 ] = bT . Then  c0 and u0 demonstrate that y0 ∈ BT∗ . This completes the proof. For the rest of this section we let y ∈ BT∗ be arbitrary but fixed, and let c and u be as in Definition 11.1. We continue the notation (11A)

y = Cy /Op (Cy ), and K  = E(C y ). T ∈ T(B), Cy = CG (y), C

Again by Corollary 9.7, (11B)

mIp (Cy ) ≤ 2.

Also let R ∈ Sylp (Cy ) with B u ≤ R. Then Ω1 (Z(R)) ≤ CB (u) =

y, bT . Since y ∈ Ipo (G), y is not p-central in G. Thus, Ω1 (Z(R)) > y so Ω1 (Z(R)) = y, bT .  will evenThe constraints on the structure of Cy , and especially on K, tually lead to a contradiction. Two knotty cases, both with p = 3, will be handled in the next two sections. We shall reduce to those cases in this section, thus proving:  ∼ Proposition 11.3. Let y ∈ BT∗ . Then p = 3 and either K = L3 (4a ),  ∼ with |K|3 = 32 , or K = M12 . As usual, we proceed in a sequence of lemmas, terminating with Lemma 11.8. y must satisfy. We begin with several conditions that C Lemma 11.4. The following conditions hold:  ∈ A is simple; (a) K  = 2 and mI (K)  ≤ 1; (b) mp (K) p  contains subgroups isomorphic to p1+2 and c × T bT ; (c) AutCy (K)

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 ≥ 2; (d) m2,p (AutCy (K))  ∈ Cp ; and (e) K (f) If KbT is a component of CK (bT ), then KbT /Op (KbT ) ∈ Cp . Proof. By Lemma 10.2d, we may fix a p-component K0 of Cy such  0 . By Theorem 2.3, K  0 /Z(K  0 ) ∈ A. that T bT  acts faithfully on K  Using [IG , 16.11] and the fact that mp (Cy ) = 3, we conclude that Op (Cy ) normalizes K0 . Then as Z( c, u) = bT  with c, u ∼ = p1+2 , c, u acts    faithfully on K0 , and (c) follows for K0 instead of K. This immediately 0. implies (d) for K  0 ) = 1. By [IA , 6.1.7] and the definition of A Suppose now that Z(K 0 ∼ (Definition 2.2), either p = 3 with K = 3A6 , 3A7 , 3M22 , 3O N , or SL3 (ra ),  0 /Z(K 0) ∼ ra ≡  (mod 3),  = ±1; or K = G2 (3), U4 (3), Ω7 (3), J3 , M c, or   Suz, all with p = 3; or K0 /Z(K0 ) ∼ = U5 (2a ) with p = 5 dividing 2a + 1.  0 , bT ] = 1  0 is in the first collection of groups, then as bT ∈ Z(R), [K If K  0 /Z(K  0 ) is one of by [V17 , 20.10], contradicting the first paragraph. If K  the remaining listed groups, then mp (K0 ) ≥ 4 by [V17 , 18.5], contradicting  0 is simple, proving (a) for K0 . mp (CG (y)) = 3. Therefore K  0 ) ≤ 1. Then as mp (CG (y)) = 3 and by (11B), mp (K0 ) ≤ 2 and mIp (K  0 )), Ω1 (S) Suppose that mp (K0 ) = 1. Then by [V8 , 1.1], if S ∈ Sylp (Aut(K is abelian, which is impossible in view of (c). Therefore mp (K0 ) = 2. In view  0 , it follows that K  =K  0 , completing of [IG , 16.11] and the simplicity of K the proof of (a)–(d).  ∈ Cp , then K  is strongly locally 1-balanced with respect to Next, if K p by [IA , 7.7.12] (note that m3 (L2 (33 )) = 3). Hence by definition of strong  T] = 1, a contradiction. Thus (e) holds. local 1-balance [IA , 7.7.10], [K, Finally, let KbT be as in (f), and let I be the subnormal closure of KbT in CG (bT ). Since bT is p-central in G, bT ∈ Ipo (G) and so every p-component of I lies in Cp . By Lp -balance, KbT is a p-component of CI (y), and so by [IA , 7.1.10], KbT /Op (KbT ) ∈ Cp . Hence, (f) holds and the lemma is proved.   ∈ Spor ∪ Alt, then p = 3 and K  ∼ Lemma 11.5. If K = M12 .  ∼ Proof. If K = An , then parts (b) and (c) of Lemma 11.4 are in conflict:  = 2, and p is odd, Sylow p-subgroups of Aut(K)  are isomorphic as mp (K)  to Ep2 , and Lemma 11.4c is contradicted. Thus K ∈ Alt.  ∈ Spor. We use the tables [IA , 5.3, 5.6.1, 5.6.2] freely Suppose that K  is a Mathieu  ∼ and show that p = 3 with K = M12 , M24 , HJ, or Ru. If K  is a Janko group Mn , then by Lemma 11.4c, p = 3 and n = 12 or 24. If K group Jn , then for p > 3, we contradict Lemma 11.4d, and for p = 3, either  ∼  ∼ K = HJ or K = J4 , in which mI3 (J4 ) = 2, against Lemma 11.4b. Otherwise,  = 2 and mI (K)  ≤ 1 force K  ∼ if p = 3, the conditions m3 (K) = Ru. Then 3

11. THE CASE BT ∩ Ipo (G) = ∅

189

assuming p ≥ 5, note that |Suz|5 = 52 , against Lemma 11.4c, and all the  = 2 (and K  ∈ A) satisfy K  ∈ Cp , against Lemma other cases in which mp (K)  ∼ 11.4e. Finally we must rule out the groups K = M24 , HJ, and Ru when p = 3. But in those cases, since bT is 3-central in Cy , L3 (CK (b)) ∼ = 3A6 ∈ C3 , contradicting Lemma 11.4f. The proof is complete.   ∈ Chev, first proving We next consider the possibility K  ∈ Chev(r), then r = 2. Lemma 11.6. If K  is  ∈ Chev(r), r odd. Since K  ∈ Cp and K Proof. Suppose that K 3  = 2, K  ∼  ∈ A. As mp (K) L2 (ra ) or 2 G2 (3 2 ) simple, r = p. Again, K =  ∼  [IA , 6.5.5]. If K = G2 (r), P Sp4 (r), or L± 4 (r), then since mp (K) = 2, p 2  divides r − 1 (see [IA , 4.10.3]). But K contains SL2 (r) ∗ SL2 (r) [IA , 4.5.1],  ≥ 2, contradicting Lemma 11.4b. From Definition 2.2 we therefore so mIp (K)  ∼  = 2, p divides re − . Then, unless have K = L3 (re ),  = ±1, and as mp (K) 2  contains  3 = 3 , a maximal diagonalizable subgroup of K p = 3 and |K| I   Zp × Zp × Z2 , so mp (K) ≥ 2, contradiction. So p = 3 and |K|3 = 32 . In  admits no field automorphism of order 3, and m3 (C(y, K)) = particular, K  is nonabelian where R ∈ Syl3 (Cy ). Hence, there exists 1. But Ω1 (AutR (K))   v ∼ v ∈ I3 (Cy ) such that K = P GL3 (re ), which in turn contains Z3 ×Z3 ×Z2 , so again mI3 (Cy ) = 3. This contradicts (11B) and completes the proof.   ∈ Chev(2), then K  has twisted rank 2 and is not Lemma 11.7. If K a a U4 (2 ) or U5 (2 ).  is simple. If K  ∼ Proof. Suppose false. Recall that K = Sp6 (2a ) or  = 2, contradicting Lemma then using [IA , 4.10.3] we see that mp (K)  a ∼  = 2 forces p  11.4b. Likewise if K = L4 (2 ),  = ±1, or U5 (2a ), then mp (K)  has a subgroup (M1 × to divide 2a +  or 2a − 1, respectively. Now Aut(K)  a ∼ ∼ M2 ) g of p -index, where M1 = M2 = L2 (2 ) and g is a p-element. Hence  of C y has Ω1 (R)  abelian, contradicting by [V8 , 1.1], the Sylow p-subgroup R Lemma 11.4c.  ∼ Suppose that K = L5 (2), L± 6 (2), L7 (2), Sp8 (2), D4 (2), or F4 (2). If   = 2 and p3 divides |K|,  we find that (p, | OutCy (K)|) = 1, then as mp (K)    ∼ y ) > 3, conp = 3 and K L5 (2), whence in every case m3 (Cy ) ≥ m3 (K =  whence again p = 3 (see [IA , 2.5.12]), tradiction. So p divides | OutCy (K)|,  ≥ 4 gives a contradiction. and m3 (K)  ∈ A, we may now assume that K  has twisted rank 1. As Since K  = 2, the only possibility is that K  ∼ mp (K) = U3 (2a ) with p dividing 2a + 1  of Aut(K)  is an (see [IA , 6.5.3, 6.5.4]). If p > 3, then a Sylow p-subgroup R extension of a diagonalizable group by a group of field automorphisms, and  is abelian, contradiction. Hence, p = 3, and so bT , being 3-central so Ω1 (R)  acts on K  like a diagonalizable element of I3 (SU3 (2a )) with distinct in R, a Ω− 8 (2 ),

12. THEOREM C∗4 : STAGE A1. FIRST STEPS

190

  eigenvalues. But then NK ( bT ; 2) = {1} by [V17 , 4.28], contradicting Lemma 11.4c. The proof is complete.   ∼ Lemma 11.8. K = M12 or L3 (4a ), both with p = 3.  ∼ Proof. By Lemmas 11.5, 11.6, and 11.7, we may assume that K = a a a  a  3 a 2    L3 (2 ), Sp4 (2 ) , G2 (2 ) , D4 (2 ), or F4 (2 2 ) . Now R ∈ Sylp (Cy ) with  nonabelian. If some element f ∈ Ip (R)  induces a field or Ω1 (AutR (K))  then by [IA , 4.9.1, 4.7.3A, 4.10.3], and using graph automorphism on K,  = 2, forcing mp (Cy ) ≥ [V8 , 1.1], it follows easily that mp (CK (f)) = mp (K)  maps into Inndiag(K).  In particular, unless 4, a contradiction. Hence Ω1 (R) a ∼ ∩K  is  possibly p = 3 and K = L3 (4 ) (which is an allowed conclusion), R  ∼ Sp4 (2a ). nonabelian. With [IA , 4.10.3], this implies that p = 3 and K = a 3 a 2 ∼  = G2 (2 ), D4 (2 ), and F4 (2 a2 ), with We are now reduced to considering K p = 3 and a > 1, the last by Lemma 11.4e. But then as bT is 3-central in Cy , L3 (CK (bT )) ∼ = SL3 (2a ), 2a ≡  (mod 3),  = ±1. This contradicts Lemma   11.4f as SL3 (2a ) ∈ T3 . The proof is complete. Lemma 11.9. E(CG (y  )/O3 (CG (y  ))) ∼ = E(CG (y)/O3 (CG (y))) for all #  y ∈ BT . Proof. Let y  ∈ BT# . Then E(CG (y  )/O3 (CG (y  ))) ∼ = M12 or L3 (4a ), and these occur according as E(CG (BT )/O3 (CG (BT ))) ∼ = 1 or L2 (4a ), respectively, by Lemmas 11.2 and 11.8. This implies the result.   ∼ Lemma 11.10. If K = L3 (4a ), then |K|3 = 32 . Proof. Suppose false. Since y ∈ BT , we have T ∈ NCy (B; 2), and we know T = [T, B]. Now no x ∈ I3 (Cy ) can induce a nontrivial field  since if it did, m3 (CC (x)) ≥ m3 ( x, y × CK (x)) = automorphism on K, y 2 + 2 = 4, contradiction. One consequence is that [Ω1 (R), Ω1 (R)], which is nontrivial as |K|3 > 32 , lies in K; hence bT ∈ K. Another is that B  whence T does as well. As O3 (Cy ) has odd order, maps into Inndiag(K),  has rank 2 and lies in a parabolic subgroup of the image of B in Aut(K)  whence it lies in a Cartan subgroup, and then in Inn(K)  since Inndiag(K), 2 ∼  y = E32 is a diagonalizable |K|3 > 3 . Hence, B = y × Dy where D   subgroup of K containing bT . Notice that the three subgroups of Dy of order 3 other than bT  are K-conjugate to one another, but to no other subgroups of R. Thus, Dy  R. By Lemma 11.9, the same analysis applies to any y ∈ BT# and its centralizer. We now select such a y1 ∈ BT# so that |CG (y1 )|3 is maximal. We choose R1 ∈ Syl3 (CG (y1 )), P1 ∈ Syl3 (G), and U1  P1 so that B ≤ R1 ≤ P1 and U1 ∼ = E32 . If possible, we make these choices of R1 , P1 , and U1 so that (11C)

U1 ≤ R1 .

12. THE M12 CASE

191

Set b1  = Z(P1 ), so that b1 ∈ L3 (CG (y1 )) =: K1 . Set K 1 = K1 /O3 (K1 ). By the preceding paragraph, B = y1 × D  y1 where Dy1  R1 is a diagonalizable subgroup of K 1 containing bT . In particular, B  R1 and |R1 : CR1 (B)| = 3. We claim that in fact, (11C) holds. Suppose not. Then y1 ∈ CBT (U1 ). However, |BT : CBT (U1 )| ≤ 3, so [U1 , y2 ] = 1 for some y2 ∈ BT − y1 . As y2 ∈ B, we have |R1 : CR1 (y2 )| = 3, so R2 := CR1 (y2 )U1 centralizes y2 and |R2 | ≥ |R1 |. Hence by our choice, R2 ∈ Syl3 (CG (y2 )) and |R2 | = |R1 |. And still, R2 ≤ P1 . So we select P2 = P1 and U2 = U1 . Then as U2 ≤ R2 , our choice of R1 , U1 , and P1 is violated. This proves our claim. Thus, replacing y by y1 , we may assume without loss that U ≤ R ≤ P ∈ Syl3 (G) with R = CP (y) ∈ Syl3 (Cy ). As U  R and |RK | > 32 , it follows by [V17 , 10.6.9] that U y = B. But now U ∩ BT = 1. As U # ⊆ I3o (G), this contradicts Proposition 10.3 and completes the proof of the lemma.  Lemmas 11.8 and 11.10 prove Proposition 11.3. 12. The M12 Case We continue the notation of the previous section, and next rule out the  = E(C y ) ∼ M12 case of Proposition 11.3. Thus, we assume p = 3, K = M12 , and R ∈ Syl3 (Cy ). We set RK = R ∩ K ∈ Syl3 (K). By [IA , 5.3b], E2 (RK ) = {A1 , A2 , B1 , B2 }, where for both i = 1 and 2, A# i consists of ∼ i = C  3-central elements of K, A ( A ), Aut (A ) GL = i i 2 (3), and for K  Aut(K)  x)) ∼ every x ∈ Bi − Z(RK ), O3 (CK ( = Z3 × A4 . Set A∗i = Ai × y and ∗ Bi = Bi × y, i = 1, 2. Finally, let R ≤ P0 ∈ Syl3 (G). Thanks to Lemmas 11.2 and 11.9, the next two lemmas hold for all  R, RK , Ai , Bi , A∗ , B ∗ , and P0 as given in the previous y ∈ BT# (with K, i i paragraph).  ∼ Lemma 12.1. Suppose that K = M12 . With the above notation, the following conditions hold for any y ∈ BT# : (a) [RK , RK ] = Z(P0 ); (b) R = RK B ∼ = Z3 × 31+2 ; (c) For any x ∈ Bi∗ − Z(P0 ), x is not 3-central in G; (d) No A∗i is G-conjugate to any Bj∗ ; and (e) Any 3-element of NG (R) normalizes each A∗i and Bi∗ . Proof. Since m3 (G) ≥ 4 > m3 (Cy ), y is not 3-central in G, so y is  is cyclic. As Sylow 3-subgroups of K not characteristic in R. Also CCy (K) 1 are isomorphic to 31+2 , (b) holds, forotherwise  y = Ω1 ( (R)) char R, contradiction. Since R < P0 , y < y NP0 (R) ≤ Z(R) = [R, R], y. As [R, R] ≤ Z(NP0 (R)), it follows that [R, R] = Z(NP0 (R)), which implies (a). It also implies that y NP0 (R) = y[R, R]. If (c) fails, there is x ∈ Bi∗ −  [R, R] y for some i = 1, 2 such that x is 3-central. Now O3 (CK (x)) ∼ =

192

12. THEOREM C∗4 : STAGE A1. FIRST STEPS

Z3 × A4 , and in particular, O3 (CG (x, y)) has even order. As x is 3-central, m3 (CG (x)) = m3 (G) ≥ 4 > m2,3 (G). Hence O3 (CG (x)) has odd order, and all components of E(CG (x)/O3 (CG (x))) are C3 -groups. But then by [IG , 20.6], some component of CG (x)/O3 (CG (x)) is locally unbalanced and therefore, by [IA , 7.7.9], isomorphic to the simple group L2 (33 ). As x is 3central, it follows that a Sylow 3-center of G is noncyclic, contradicting (a). Thus, (c) holds. Finally, in Bi∗ any complement to [R, R] is a hyperplane consisting of non-3-central elements, while in A∗i , Ai is a hyperplane consisting of 3-central elements. This implies (d), which in turn immediately implies (e). The proof is complete.  Lemma 12.2. Write T = t, t , and let Ct = CG (t), y ∈ BT# , Jy = L3 (CCt (y)), and Et = E(Ct ). Then the following conditions hold:  (a) Jy ∼ = A5 and O3 (CCt (y)) = Jy × y; (b) Jy T / t ∼ = Σ5 ; (c) |Ct : Et T | = 2n , 0 ≤ n ≤ 2, and one of the following holds: (1) Et ∼ = A6 ; or = Sp4 (4) and E(CEt (t )) ∼ (2) Et /Z(Et ) ∼ = HS and E(CEt (t )) ∼ = A6 or A8 ;  (d) O 3 (NG (T )) = bT  T × E(CEt (t )) ∼ = A4 × Am , m = 6 or 8; and (e) O3 (Cy ) = 1.

Proof. Let Ct,y = CCt (y) = CCy (t). Then by [V17 , 10.11.6], and as O3 (Cy ) has odd order, O3 (Ct,y ) = O{2,3} (Ct,y ) t, ty  where [t, ty ] = 1 and t2y = 1; and E(Ct,y /O{2,3} (Ct,y )) ∼ = A5 . It follows that CO3 (Ct ) (y) has a normal 2-complement and a Sylow 2-subgroup t or t, ty , and [CO3 (Ct ) (y), BT ]

has odd order. As this holds for all y ∈ BT# , [O3 (Ct ), BT ] has odd order. Thus [O3 (Ct ), BT ] ≤ O2 (O3 (Ct )) ≤ O2 (Ct ) = 1 as G is of even type. But now O3 (Ct ) has a normal 2-complement, so O3 (Ct ) = O2 (Ct ) = t or

t, ty . Now by L3 -balance, L3 (Ct,y ) is a component of CEt (y). Also BT ∈ Syl3 (Ct,y ). As these statements hold for all y ∈ BT# , BT ∈ Syl3 (Ct ), in particular. Now all components of Et lie in C2 , so we see by [V17 , 13.29] that Et ∼ = Sp4 (4), HS, 2HS, or M23 , and y ∈ Et . Using [IA , 4.8.2, 4.2.3, 7.7.1], we see that |O3 (CAut(Et ) (y))| ≤ 2. Hence O{2,3} (Ct,y ) centralizes E(Ct )O2 (Ct ) = F ∗ (Ct ), so O{2,3} (Ct,y ) = 1 and then CO3 (Cy ) (t) = 1. Letting t range for a moment over T # , we get O3 (Cy ) = 1. Thus, F ∗ (Cy ) = K × y and (a) and (e) follow. Furthermore, as T ∈ NK (BT ; 2), (b) holds. Suppose that Et ∼ = Sp4 (4). Then since t induces a field automorphism ∼ on E(CK (t)) = L2 (4), it induces a field automorphism on Et (see [IA , 4.2.2, 2.5.12]). As | Out(Et )| = 4 and |O2 (Ct )| ≤ 4, (c1) holds in this case. On the other hand, if Et /Z(Et ) ∼ = HS, then again since t induces a field au∼ tomorphism on E(CK (t)) = L2 (4), by [V17 , 13.28], E(CEt (t )) ∼ = A6 or A8 . M23 , as no involutory automorphism of M23 centralizes an Finally, Et ∼ = E32 -subgroup, by [IA , 5.3d]. Thus, (c) holds.

12. THE M12 CASE

193

Let N = NG (T ) and N0 = CG (T ). Then E(N ) = E(N0 ) = E(CEt (t )) by L2 -balance. In particular, | Out(E(N ))| is not divisible by 3. But B ≤ N  with B ≤ N0 , so O3 (N ) contains T b1 ×E(N0 ) for some b1 ∈ B # , But bT  is weakly closed in B by Lemma 12.1c. As B ∩E(N0 ) is inverted elementwise in E(N0 ), b1  = bT . Hence (d) holds.  Now we can prove Lemma 12.3. There exists an involution z ∈ G inverting y and centralizing K. Proof. By Lemma 12.2d, there exists an involution z1 centralizing bT and inverting B ∩ E(Ct ) = CB (T ) = BT elementwise. Hence z1 normalizes

y, K, and RK = O3 (CK (bT )) ∼ = 31+2 . Since bT  = Z(RK ), z1 either centralizes or inverts RK / bT . In either case there is z0 ∈ NK (RK ) acting as z1 does on RK , centralizing z1 , and with z02 = 1. Then z := z1 z0 is an  involution inverting y and mapping into O2 (CAut(K) (RK )) = 1, as required.  Lemma 12.4. For some y0 ∈ BT∗ , at least one of A1 and A2 is normal in some Sylow 3-subgroup of G. Proof. We have U  P and Z(P ) = bT  ≤ U . Hence U normalizes both R and, by Lemma 12.1e, bT  BT = B. We have |BT : CBT (U )| ≤ 3 and choose y0 ∈ CBT (U )# . Without loss, y0 = y. Since m3 (CG (w)) ≥ 4 for all w ∈ U # , U g ∩ BT = 1 for all g ∈ G. But BT ≤ B1∗ or B2∗ , and B1∗ and B2∗ are conjugate. Hence U ≤ Bi∗ for either i, and so U y = Ai for some i.  Then U ⊆ {w ∈ A∗i | m3 (CG (w)) ≥ 4} = Ai . The lemma follows. We may now fix y ∈ BT# again such that for some P ∈ Syl3 (G), CP (y) = R ∈ Syl3 (Cy ) and A2  P . Let N = NG (A2 ), C = CG (A2 ), and PC = P ∩ C ∈ Syl3 (C). Thus P = PC A1 . We may write N ∩ K = A2 L where L∼ = Aut(A2 ) acting naturally on A2 , and A1 = bT  (A1 ∩ L). = GL(A2 ) ∼ Let Q = O2 (L) ∼ = Q8 . We also put N ∗ = NG (A∗2 ). Notice that as O3 (Cy ) = 1 by Lemma 12.2e, C ∗ := CG (A∗2 ) = A∗2 , by [V17 , 10.11.5]. Let W = O3 (N ∗ ), W0 = J(W ), and W ≤ S ∈ Syl3 (N ∗ ). Lemma 12.5. The following conditions hold: (a) [W0 , y] = CW0 (y) = A2 ; (b) W = W0 y with W0 ∼ = E34 the direct sum of two irreducible Qmodules; (c) W0 = J(S); and  (d) W  O 3 (NG (S)). ∗ ∗ Proof. Let x ∈ NP (R) − R. Then y x = yb±1 T and x ∈ N . But Q ≤ N ∗ and Q permutes transitively the 8 elements of E1 (A2 ) − E1 (A2 ) − { y}. As y G ∩ A2 = ∅, |N ∗ : NN ∗ ( y)| = 9. It follows that N ∗ /A∗2 ∼ = AutG (A∗2 )

12. THEOREM C∗4 : STAGE A1. FIRST STEPS

194

∗

stabilizes A2 and contains O2 (NSL(A∗2 ) (A2 )). With Lemma 12.3, N := N ∗ /A2 contains, with index 2, an extension of y by the holomorph of E32 . Since CN ∗ (y) contains an involution acting on A2 with determinant −1, W is abelian, indeed W = W 0 × y with W 0 inverted by Z(Q), and [W0 , y] = CW0 (y) = A2 , which is (a). Now W0 ∼ = E34 or Z9 × Z9 , and to prove (b), it remains to prove that Z9 × Z9 is impossible. But in that case NP (R) contains an element v of order 9 centralizing A2 and shearing y to

bT . Note that v normalizes A1 , so [A1 , v] ≤ bT , and hence v induces an inner automorphism on RK . Therefore NP (R) = RK ∗ V , where V is nonabelian of order 33 . As v has order 9, Ω1 (NP (R)) < NP (R) and so R = Ω1 (NP (R)) char NP (R). But this forces NP (R) = P , a contradiction as m3 (P ) ≥ 4. Thus (b) holds. Clearly, W0 = J(W0 y)  S. Now [L, y] = 1 so W0 /[W0 , y] and [W0 , y] are isomorphic as L-modules. It follows that |CW0 (v)| = 32 for all v ∈ (S/W0 )# , and |Z(S)| = 3. These imply (c). Finally, by [V17 , 10.10.20], S has at most two maximal subgroups containing W0 and isomorphic to W = W0 y. As W0  NG (S) by (c), part (d) follows.  Since z inverts y and centralizes K, z ∈ N ∗ . Then as z centralizes NK (A2 ) and normalizes W0 , and m2,3 (G) ≤ 3, and in view of Lemma 12.5b, we must have CW0 (z) = A2 . Therefore [W0 , z] ∼ = E32 is a natural L-module. We set

= M/A2 . M = NG (W ) and M Note that A2 = [W, W ]. Lemma 12.6. One of the following holds: (a) |G|3 = 36 ; or (b) There exists W ∗ ≤ CG (z) such that W ∗ ∼ = Z9 × Z9 , Ω1 (W ∗ ) = A2 , ∗ ∗ and W is L-invariant with CL (W ) = 1. Proof. Suppose that (a) fails, so that S ∈ Syl3 (G). By Lemma 12.5d, ≤ M , so |M |3 > |S|.

) ≤ CM ( 

) ≤ y ) ≤ NM (A∗2 ). Since [W0 , y] = A2 , CM (W Now CM (W ∗ ∼

NM (A2 ) ∩ CM (A2 ) = W . Thus M /W embeds in Aut(W ) = GL3 (3). Moreover, since W0 ∼ = E34 and m3 (Cy ) = 3, and the coset A2 y is completely fused,  M

−W

0 . Also L ≤ M and L acts naturally on W

0 , as [W0 , y] = A2 . y ⊆ W 



0 

0 ) =

y  /W0 , where (W1 /W0 ) × (W y  /W It follows that M /W0 = LW1 

/W

0 ) ∼

1 /W

0 = [O3 (M

/W

0 ), Z(L)] is a natural L-module, O3 (M = E33 , and W ∼ again thanks to the fact that L = GL2 (3) centralizes y. Let W1 be the full

1 in M . Then CW (Z(L)) = 1 so W1 is abelian. Since L is preimage of W 1 irreducible on W1 /W0 , either W1 ∼ = E36 or W1 ∼ = E32 × (Z9 × Z9 ). Now [L, z] = 1 and z inverts y. Also commutation with y induces an

0 to W

0 , and also from W

0 to A2 . As z centralizes

1 /W isomorphism from W

0 and centralizes W

1 /W

0 . Thus W ∗ := CW (z) is L-invariant A2 , z inverts W 1  O 3 (NG (S))

13. THE L3 (4a ) CASE

195

and of order 34 . As m2,3 (G) = 3, W ∗ ∼ = Z9 × Z9 , and (b) holds. The proof is complete.  Lemma 12.7. We have |G|3 = 36 . Proof. Otherwise Lemma 12.6b holds and we reach a contradiction by examining Cz := CG (z). We have K ≤ Cz with K ∼ = M12 , and LW ∗ ≤ Cz ∗ ∗ ∗ ∗ ∼ with F (LW ) = W = Z9 × Z9 and K ∩ LW = A2 L. Set V = O2 (Cz ) and V = V /Φ(V ), and suppose first that [V, A2 ] = 1, so that CK (V ) = CLW ∗ (V ) = 1. Since L is transitive on A# 2 , AutW ∗ (CV (x)) # contains Z9 for each x ∈ A2 , so d := m2 (CV (x)) ≥ 6. Factoring V under A# 2 thus yields m2 (V ) ≥ d + 18. On the other hand, as CK (V ) = 1, factoring V under B1 yields m2 (V ) ≤ d + 3m, where m is the sectional 2-rank of CG (x0 ) for x0 ∈ B1 − bT . All such x0 are in y G , so we get 18 ≤ 3m = 3r2 (Cy ) ≤ 3r2 (Aut(M12 )). However, by [V17 , 18.2], r2 (Aut(M12 )) < 6, a contradiction. Therefore [V, A2 ] = 1, whence [E(Cz ), A2 ] = 1 as G is of even type. In particular K and LW ∗ both act faithfully  on E(Cz ). Further, m3 (E(Cz )) ≤ E(Cz ) m2,3 (G) ≤ 3 and e(G) ≤ 3, so A2 contains a component I ∈ C2 such that 2 ≤ m3 (I) ≤ 3 and CK (I) = CLW ∗ (I) = 1. By [V17 , 9.10], however, there is no such group I, a final contradiction completing the proof of the lemma.  Now finally we can prove:  ∼ Proposition 12.8. K = L3 (4a ) for some a. Proof. Suppose false and continue the above argument, so that K ∼ = M12 and |G|3 = 36 , by Lemmas 11.8, 12.2e, and 12.7. Moreover, by ∼  Lemma 12.2d, CbT := CG (bT ) contains   a subgroup J = A6 . Set CbT = b )/ bT |3 ≤ 36 /33 = 33 , so J centralizes Cb /O3 (Cb ). Then |O3 (C T

T

T

b ). As |C b |3 = 36 , J normalizes every component of E(C b ), and O3 (C T T T  From Lemma 12.1a, a Sylow acts nontrivially on some such component I.  = 1. Since m3 (Cb ) = 3-center of G, and hence of CbT , is cyclic, so Z(I) T m3 (G) ≥ 4, I ∈ C3 . But then by [V17 , 7.16], I ∼ = 3J3 . Using [IA , 5.3h], however, we see that this implies that m3 (CI(y)) = 4, a contradiction as  m3 (Cy ) = 3. The proof is complete. 13. The L3 (4a ) Case In this section we continue the argument of the preceding section, and complete the proof of Proposition 9.1 by proving  ∼ Proposition 13.1. It is not the case that K = L3 (4a ) with |K|3 = 32 . Together with Propositions 11.3 and 12.8, this will provide a contradiction from which Proposition 9.1 will follow.

12. THEOREM C∗4 : STAGE A1. FIRST STEPS

196

∼ L2 (4a ) and with We have T ≤ L := L3 (CG (BT )) with L/O3 (L) = O3 (L) of odd order. Expand T to Z ∈ Syl2 (L), so that [Z, BT ] = 1 and Z = NZ /O3 (CZ ), and fix Z∼ = E22a . Set CZ = CG (Z), NZ = NG (Z), and N all this notation. We shall analyze CZ to reach a contradiction. By Lemma 11.9, for each y ∈ E1 (BT ), Ly /O3 (Ly ) ∼ = L3 (4a ), where Ly is the pumpup of L in CG (y); and we fix a B-invariant Ty ∈ Syl2 (Ly ) with center Z. Note that Ty is special of complexion 22a+4a . Also by the structure of Cy , Ty ∈ Syl2 (O3 (CCy (Z))). As CO3 (CZ ) (y) ≤ O3 (CCy (Z)), we may assume, conjugating within O3 (CCy (Z)) ∩ CG (B), that Ty was chosen to contain a Sylow 2-subgroup of CO3 (CZ ) (y). That is, (13A)

Sy := Ty ∩ O3 (CZ ) ∈ Syl2 (CO3 (CZ ) (y)).

We shall return to Sy presently, and show eventually that Sy = Ty . Note that for any y ∈ BT# , if we let Ry ∈ Syl3 (Cy ), then as in the M12 case, Ω1 (1 (Ry )) lies in C(y, Ly ), and hence is trivial as Ry ∈ Syl3 (G). Thus

y ∈ Syl3 (C(y, Ly )). In particular, BT ∈ Syl3 (CCy (Z)) = Syl3 (CCZ (y)). Since y was arbitrary, (13B)

BT ∈ Syl3 (CZ ).

Lemma 13.2. We have O2 (CZ ) = 1. Proof. Suppose false. As G is of even type, O2 (CG (z)) = 1 for all z ∈ Z # . Choose a subgroup Z0 ≤ Z maximal with respect to O2 (CG (Z0 )) = 1; thus, 1 < Z0 < Z. Set CZ0 = CG (Z0 ), so that F ∗ (CZ0 ) = O2 (CZ0 )E(CZ0 ). Write Z = Z0 × Z1 . For any z ∈ Z1# , Xz := O2 (CG (Z0 z)) = 1. Thus by [IG , 20.6], there exists an Xz z-invariant component Jz of CZ0 such that [Jz , Xz ] = 1 and Xz maps into O2 (CAutJz Xz z (Jz ) (z)). In particular, Jz is not locally 1-balanced with respect to z. On the other hand, G is of even type, so by L2 -balance and [IA , 7.1.10], Jz ∈ C2 . Thus by [IA , 7.7.9], either Jz ∼ = P GL2 (q), or Jz /Z(Jz ) ∼ = L3 (4), with z = L2 (q), q ∈ F M9, with Jz z ∼ inducing a graph-field automorphism on Jz /Z(Jz ). In either case, by [V17 , 18.1], (13C)

r2 (CAut(Jz ) (z)) ≤ 3,

where r2 denotes sectional 2-rank. Recall that BT ≤ CZ ≤ CZ0 . Also O3 (Cy ) has odd order for any y ∈ BT# , y , CC ( y) = CC (Z0 ) is solvable. In particular, so by the structure of C y Z0

y normalizes Jz for any z ∈ Z1# . Since y was arbitrary, BT acts faithfully on any such Jz . Thus m3 (Aut(Jz )) ≥ 2, and we conclude that Jz ∼ = L2 (9) or L3 (4). As CG (Z0 BT ) is solvable, Jz = CG (Z0 )(∞) , independent of z, and we write Jz = J. The possible images of z in Out(J) generate a subgroup of Out(J) of order only 2 or 6, respectively, and so Z1 = z has order 2. We claim that CTy (J) ≤ Z. For, Ty centralizes Z and hence normalizes J. If there exists t ∈ CTy (J) − Z, then as Ty is of type L3 (4a ), Z = [Ty , t] ≤ CTy (J), a contradiction as [J, z] = J. Thus, the claim holds. Since Ty maps

13. THE L3 (4a ) CASE

197

into CAut(J) (z), 4a = m2 (Ty /Z) ≤ r2 (CAut J (z)). As a ≥ 1, this contradicts (13C) and so the lemma is proved.  Lemma 13.3. The following conditions hold: (a) F ∗ (CZ ) = O2 (CZ ); and (b) For each y ∈ BT# , y is sheared in G to bT , whence every element of B − bT  is G-conjugate to an element of BT . Proof. For (b), let B ≤ R ∈ Syl3 (Cy ). Since m3 (Cy ) = 3 < m3 (G), R∈ / Syl3 (G), so there exists a 3-element v ∈ NG (R) − R. As Z(R) = bT , y and Z(R) ∩ [R, R] = bT , v shears y to bT . Since B = bT  × BT , (b) follows. Note that CCZ (y) = CCy (Z) is solvable for every y ∈ BT# . In particular, n CCZ (y) involves no 2B2 (2 2 ), n ≥ 3. Considering a chief series of O3 (CZ ), and that BT is noncyclic, we deduce that O3 (CZ ) is solvable. Next, suppose that I is a component of CZ . By the previous paragraph, 3 divides |I|, so with (13B), there is y ∈ BT# ∩ I. Let y  ∈ BT − y. Then Ty = [Ty , y] so Ty ≤ I. In particular, Z ≤ Z(I). But as G is of even type, and with [IA , 7.1.10] and L2 -balance, I ∈ C2 . Since m3 (I) ≤ 2, it follows by [V17 , 9.12] that I ∼ = [2 × 2]L3 (4). Now B ≤ NZ so I is B-invariant; and CB (Z) = CB (T ) = BT . Therefore CB (I) = CBT (I) = 1, the last because CG (y  ) does not involve I for any y  ∈ BT# . Hence m3 (AutB (I)) = 3, a contradiction as m3 (Aut(I)) = 2.  Therefore F ∗ (CZ ) = O2 (CZ ), so (a) holds as well. Now we choose a B-invariant Sylow 2-subgroup S of O3 (CZ ). Recall from (13A) that Sy = Ty ∩ O3 (CZ ) ∈ Syl2 (CO3 (CZ ) (y)) for all y ∈ BT# . Replacing Ty by a suitable CO3 (CZ ) (B)-conjugate, we may assume that Sy ≤ S for all y ∈ BT# . Hence (13D)

Sy = CS (y) and |Ty ||Sy | = |Ty | = 43a for all y ∈ BT# .

# Note that since bT ∈ L = E(CLy (BT )) ∼ = L2 (4a ), CTy (bT ) = 1 for all y ∈ BT , by [V17 , 10.6.8]. Hence CSy (bT ) = 1 and so

(13E)

CS (bT ) = 1.

Now we can prove Lemma 13.4. For each y ∈ BT# , Ty = Sy , i.e., Ty ≤ O3 (CZ ). Proof. Since BT ∈ Syl3 (CZ ) and Ty is B-invariant, the conclusion is a consequence of the Bender-Thompson lemma if CZ is 3-constrained. So we may assume that CZ has a 3-component I. If m3 (I) = 1, then since BT ∈ Syl3 (CZ ) with BT ∼ = E32 , it follows by  = 1. However, for each y ∈ B # , CC (y) = CC (Z) [V8 , 1.1] that CBT (I) y Z T is 3-solvable, a contradiction. Hence, BT ∈ Syl3 (I) and I = L3 (CZ ). For any y ∈ BT# , Ty = [Ty , BT ], and therefore Ty ≤ I. But Ty ≤ O3 (CCy (Z)) =

12. THEOREM C∗4 : STAGE A1. FIRST STEPS

198

y := [O3 (C ( T ]. We can therefore O3 (CCZ (y)) and therefore Ty ≤ U y )), B I apply [IA , 7.7.6bc] if Ty = 1. In that case, if I ∈ Chev(r), r = 3, then since b ∼ BT ∈ Syl3 (I), we conclude that I ∼ = L± 3 (r ) and r = 2. Otherwise, I = A7 or y ) ≤ 2, and since Ty is elementary abelian, |Ty | ≤ 4. M22 . In any event, m2 (U But as CNZ (BT )-module, Ty has at most two composition factors, both of order 4a . Hence a = 1. ∼  Unless I = A7 , a case we shall consider below, NI (BT ) is transitive on BT# , and therefore either the conclusion of the lemma holds or |Ty | = 4 for all y ∈ BT# . In the latter case |CO3 (CZ )/Z (y)|2 ≤ |Sy Z/Z| = 22 , and in particular |O2 (CZ )/Z| ≤ 28 . But I contains a nonabelian r-group, r > 3, or a group of order 11, so CI (O2 (CZ )) must cover I, a contradiction as F ∗ (CZ ) = O2 (CZ ) (Lemma 13.3). Hence, the lemma holds unless possibly I ∼ = A7 , which we now assume. If the lemma fails, then Ty1 has a nontrivial image in O3 (CAut(I)  (y1 )) # for some y1 ∈ B . By [IA , 7.7.1], y1 ∈ I is a 3-cycle and Ty is a four-group. T

1

On the other hand, if y2 ∈ BT# and y2 is the product of two disjoint 3-cycles, 2 4  then O3 (CAut(I)  (y2 )) = 1, so Ty2 = 1. Hence |Sy1 /Z| = 2 and |Sy2 /Z| = 2 . As BT has two subgroups conjugate to y1  and two conjugate to y2 , |S/Z| = 212 .  and hence CB (I)  = 1. But We have I = L3 (CZ )  NZ so B acts on I,  A7 is not involved in L3 (4) by [V17 , 10.6.6], so CB (I) contains no conjugate of any element of BT# . In view of Lemma 13.3b,  bT ] = 1. [I,  =N Z and N0 := CN (bT ) Let N = NNZ (S). Then by a Frattini argument, N   covers B I. Moreover, by (13E), we can consider S/Z as a 6-dimensional F4 module for N0 , in which dimF4 (CS/Z (yi )) = 1 or 2 for i = 1 or 2, respectively. But then as CS (bT ) = 1, dimF4 (CS/Z (bT y1 )) ≥ 3 for some choice of  = ±1. Consequently r2 (CG (bT y1 )) ≥ 6. However, by Lemma 13.3b, bT y1  ∈

yG for some y ∈ BT# , and r2 (Cy ) = 4. This contradiction completes the proof.  Lemma 13.5. Write E1 (BT ) = { yi }4i=1 . Then S = Ty1 ∗ Ty2 ∗ Ty3 ∗ Ty4 and Z(S) = Z. Proof. First, CS (bT ) = 1 by (13E), so by [V9 , 13.4], S has class at most 2. Also since CS (y) = Sy = Ty for all y ∈ BT# ,   Z(S) = CZ(S) (yi ) | 1 ≤ i ≤ 4 ≤ Z(Tyi ) | 1 ≤ i ≤ 4 ≤ Z, so Z(S) = Z. Then for 1 ≤ i < j ≤ 4, Tyi normalizes Tyj and so Tyi =  [Tyi , yj ] centralizes Tyj . The lemma follows.

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199

∼ E24a be one We can now finish the proof of Proposition 13.1. Let A1 = of the two maximal elementary abelian subgroups of Ty1 . Then as Z ≤ A1 , CS (A1 ) = A1 ∗ Ty2 ∗ Ty3 ∗ Ty4 ≤ O3 (CG (A1 )). Moreover, A1 is B-invariant, so CS (A1 ) ≤ S1 for some B-invariant S1 ∈ Syl2 (O3 (CG (A1 ))). For i > 1, Tyi ∈ N∗CG (yi Z) (B; 2) by [V17 , 10.6.7], as O3 (CG (yi )) has odd order. Also by [V17 , 10.6.7], A1 maps onto a Sylow 2-subgroup of its centralizer in Aut(Lyi /O3 (Lyi )). Therefore S1 = CS1 (yi ) | 1 ≤ i ≤ 4 = CS (A1 ). By a Frattini argument, NG (A1 ) ∩ NG (S1 ) covers AutG (A1 ). Now, for some w ∈ NLy1 (A1 ), Z ∩ Z w = 1. Hence for some w ∈ NG (A1 ) ∩ NG (S1 ),   Z ∩ Z w = 1. But then S1 /Z is abelian and S1 /Z w is not, a contradiction. The proof of Proposition 13.1 is complete. As noted at the beginning of this section, the proof of Proposition 9.1 is now complete.

14. Completion of the Proof In this short section we prove Proposition 9.2. As we have just proved Proposition 9.1, this will complete the proof of Theorem C∗4 : Stage A1. First, we require a lemma which will also be needed in Chapter 14. It depends only on the hypotheses of Theorem C∗4 . Here we use the notation C(A, K) := CCG (A) (K/Op (K)) for an elementary abelian p-subgroup A of G and a product K of p-components of CG (A). Lemma 14.1. Suppose that p = 3 and A is an elementary abelian 3subgroup of G such that L3 (CG (A)) has a 3-component K with K/O3 (K) ∼ = L2 (3n ), n > 2. Then m3 (A) = m3 (C(A, K)) = 1. Proof. Suppose A is noncyclic, and expand A to A∗ ∈ E3∗ (C(A, K)). Then L3 (CK (A∗ )) covers K/O3 (K) and is a 3-component of CG (A∗ ). Hence there is no loss in assuming that A = A∗ and m3 (A) ≥ 2. Now m3 (CG (A)) > 4, so as e(G) = 3, O3 (CG (A)) has odd order, whence O3 (K) does as well. Since K/O3 (K) contains U b ∼ = A4 (with U = O2 (U b)), we have m2,3 (G) ≥ m2 (A) + 1. As e(G) = 3, we may assume that m3 (A) = 2. Note also that for every a ∈ A# , the pumpup Ka of K in CG (a) is not diagonal (that would imply m2,3 (G) > 3, since n ≥ 3 and the Schur multiplier of L2 (3n ) is a 3 -group [IA , 6.1.4]). Hence either Ka is a trivial pumpup of K, or Ka /O3 (Ka ) ∼ = L2 (33n ), by [V17 , 7.2]. Write U = u, y and define Y = [O2 (CK (y)), u] = [Y, u]. We have CK/O3 (K) (y) ∼ = D3n ±1 , so Y = O{2,3} (Y ), and as n > 2, Y = 1. Put Cy = CG (y), so that Y ≤ O{2,3} (CG (A y)) = O{2,3} (CCy (A)). Then for any a ∈ A# , whether Ka /O3 (Ka ) ∼ = L2 (3n ) or L2 (33n ), we  2 have Y y ≤ O (CK (a)) ≤ Ka and CKa (y) has a normal {2, 3}-complement.

200

12. THEOREM C∗4 : STAGE A1. FIRST STEPS

Thus Y ≤ O{2,3} (CCG (a) (y)) = O{2,3} (CCy (a)). Hence  O{2,3} (CCy (a)) ≤ ΔCy (A) (14A) Y ≤ a∈A#

However, O2 (Cy ) = 1 as G has even type. By [V9 , 4.5], [O2 (Cy ), Y ] = 1 and Y ∩ O3 (Cy ) ≤ O{2,3} (Cy ) = 1. Thus as Y = 1, it follows by [IG , 20.6] that some AY -invariant 3-component J of Cy is not locally 2-balanced with respect to A, and the image of Y in JY A := AutJY A (J) is a nontrivial subgroup of ΔJY A (A). As [O2 (Cy ), Y ] = 1 and [J, Y ] = J, we have [O2 (Cy ), J] = 1. Therefore as e(G) = 3, J ≤ E(Cy ), and so J is a component of Cy . As G is of even type, J ∈ C2 . With [IA , 7.7.4], we conclude that J/Z(J) ∼ = L3 (2r ), with  = (−1)r , and Y is a nontrivial group of field automorphisms of J . In particular Y ∩ [Aut(J), Aut(J)] = 1. Since Y = [Y, u], it follows that u ∈ NCy (J). However, m3 (J) = 2 and O2 (J) ≤ O2 (Cy ) = 1. Hence m3 (JJ u ) = 4. As JJ u ≤ Cy , this contradicts e(G) = 3 and completes the proof of the lemma.  Now we prove Proposition 9.2. We assume that 1 < A ≤ B ∈ Ep3 (G)  and must prove that X := O 2 (Op (CG (A))) = 1. We suppose false and assume that A is a minimal counterexample. Then NG (B; 2) contains a Sylow 2-subgroup of X, which is nontrivial. Hence, A# ⊆ Ipo (G) by Proposition 9.1. By (1C), |A| > p. Let A0 be any hyperplane of A, so that Op (CG (A0 )) has odd order but X = 1. Let C0 = CG (A0 ) and C 0 = C0 /Op (C0 ), so that X = 1. By [IG , 20.6] applied in C 0 , X acts nontrivially on some component K of E(C 0 ) that is unbalanced with respect to A/A0 . For any a ∈ A# 0 , any pcomponent L of the subnormal closure of K in CG (a) satisfies L/Op (L) ∈ Cp by (1A3), so by Lp -balance and [IA , 7.1.10], K ∈ Cp . Hence, by [IA , 7.7.9], p = 3 and K ∼ = L2 (33 ). Thus A/A0 acts by field automorphisms on K and the image of X in Aut(K) is that of O2 (CK (A)) ∼ = E22 . Since K is simple with m3 (K) = 3 = e(G), C(A0 , K) = CC0 (K) has odd order. Consequently, Sylow 2-subgroups of C0 embed in Aut(L2 (33 )) and so are dihedral of order at most 8. In particular, the only 2-subgroups of C0 possessing an automorphism of order 3 are four-groups, and m2 (C0 ) = 2. Therefore for any involution t ∈ CG (A), A centralizes every A-invariant 2-subgroup of CC0 (t). Fix E ∈ E22 (CG (A)), and consider C := CG (t) for some t ∈ I2 (E). Then A centralizes every element of NCC (A0 ) (A; 2). As A0 is an arbitrary hyperplane of A, A centralizes NC (A; 2). In particular A centralizes O2 (C) and  O3 (E(C)), by [V17 , 10.7.2] and as G has even type. So CA (O3 (E(C))) = 1. Let P ∈ Syl3 (CC (A0 )). Then P = A1 A2 , where A1 ∼ = Z3 is a complement to A0 in A, and A2 ∈ Syl3 (C(A0 , K)). By Lemma 14.1, A2 is cyclic. Then A = Ω1 (P ). Thus if we expand A to A∗ ∈ Syl2 (C), then Ω1 (CA∗ (A0 )) = A. As A0 < A is an arbitrary hyperplane, A = Ω1 (R) for some R ∈ Syl3 (C). Then by [IG , 16.20], NC (A) controls C-fusion in R.

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201

Hence there is some 2-element y in NC (A) − CC (A). For a suitable y and hyperplane A0 of A, y normalizes A0 and inverts A/A0 . But this is absurd as Out(L2 (33 )) is abelian. This contradiction completes the proof of Proposition 9.2 and with it the proof of Theorem C∗4 : Stage A1.

CHAPTER 13

Theorem C∗4 : Stage A2. Nonconstrained p-Rank 3 Centralizers 1. Introduction Let G and p be as in Theorem C∗4 (Case A). That is, G is of restricted even type, m2,p (G) = e(G) = 3, mp (G) > 3, and Lop (G) ⊆ Cp . Recall that H = {H ≤ G | H is a 2-local subgroup of G and mp (H) ≥ 3} (1A)

Hcons = {H ∈ H | F ∗ (H) = O2 (H)} Hinv = {H ∈ H | H = CG (z) for some z ∈ I2 (G)}.

The analysis in Theorem C∗4 (Case A) naturally breaks up into the two cases Hinv ⊆ Hcons and Hinv ⊆ Hcons . In this chapter we treat the second of these cases. Theorem C∗4 : Stage A2. Let G and p be as in Theorem C∗4 (Case A), and suppose that Hinv ⊆ Hcons . Then p = 3 and G ∼ = Co3 . By assumption, therefore, and the fact that G is of restricted even type, there is an involution z ∈ I2 (G) such that mp (CG (z)) = 3 and F ∗ (CG (z)) = E(CG (z))O2 (CG (z)) with E(CG (z)) = 1; moreover, every component of E(CG (z)) is a Co2 -group. The proof of Theorem C∗4 : Stage A2 will then be accomplished in the following two major steps. Theorem 1. There exist z ∈ I2 (G), B ∈ Ep3 (CG (z)), and a component K of CG (z) such that B acts faithfully on K. Theorem 2. Let z, B, and K be as in Theorem 1. Then p = 3, K ∼ = Co . 2Sp6 (2), and G ∼ = 3 Of particular utility is Lemma 2.4, which is a form of cross-characteristic L-balance. It hypothesizes the existence of z ∈ I2 (G), K, a component of CG (z), A, an elementary p-subgroup of CG (z) for some odd prime p, and KA , a p-component of CK (A). Under suitable additional hypotheses, Lemma 2.4 guarantees the existence of a p-component J of CG (A) such that either J = J z and KA is a component of CJ (z), or J = J z and KA is a diagonal of JJ z . Typically, KA is a C2 -group, while J/Op (J) is a Cp -group, putting severe limitations on the possibilities for K and J/Op (J). However, numerous “small” possibilities remain which must be treated case by case. An interesting case which arises in the proof of Theorem 2 is the case when p = 3 and K = E(CG (z)) ∼ = Sp6 (2). This would arise were G ∼ = O8+ (2) 203

204 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

with z acting as an outer automorphism on E(G). We eliminate this case by using the Thompson transfer lemma to prove that z ∈ [G, G], but to do this we must first construct a subgroup G0 of G with G0 ∼ = O8+ (2) starting from x ∈ I3 (G) with NG ( x) =: Nx = x, z1  × J t where z1 is an involution inverting x, J t ∼ = W (E7 ). = W (E6 ), and CG (z1 ) =: Cz1 = z1  × K1 ∼ In Lemma 16.9, we then construct G0 = Nx , Cz1  as a central quotient of W (E8 ) using the Coxeter presentation for W (E8 ). 2. Components and p-Components In proving both Theorem 1 and Theorem 2, our general hypotheses are: (2A)

(1) G is of restricted even type and e(G) = 3; (2) p is an odd prime, m2,p (G) = 3, and mp (G) > 3; and (3) Lop (G) ⊆ Cp .

Notice that p is fixed. Also since Hinv ⊆ Hcons , (2B)

There exists (z, K) ∈ IL2 (G) with mp (CG (z)) ≥ 3.

Definition 2.1. K(G) = {(z, K) ∈ IL2 (G) | mp (CG (z)) ≥ 3}. For any element (z, K) ∈ IL2 (G), we use the following auxiliary notation: (1) S ∈ Syl2 (CG (z)) with Q1 := CS (K) ∈ Syl2 (C(z, K)); (2) Q ∈ Syl2 (CG (K)) with Q1 ≤ Q, so that Q1 = (2C) CQ (z); (3) Q0 := O2 (CG (z)); and (4) SK := S ∩ K. We assume (2A) and (2B) throughout this chapter. We first observe: Lemma 2.2. Let (z, K) ∈ K(G) and B ∈ Ep3 (CG (z)). Then B normalizes K. Proof. Note that by Theorem C∗4 : Stage A1, a ∈ Ipo (G) and Op (CG (a)) has odd order for all a ∈ B # . Assume the lemma fails. Since e(G) = 3, p = 3, NB (K) ∼ = E32 and m3 (K) ≤ 1. Let b ∈ B − NB (K). m If (3, |K|) = 1, K/Z(K) ∼ = 2B 2 (2 2 ) for some odd m > 1. Then i m3 (Aut(K)) ≤ 1, and so there exists a ∈ NB (K)# such that [a, K b ] = 1, 2 and CG (a) ≥ a×KK b K b . Since by [IA , 3.3.3], m2 (K z) ≥ 4, by Thompi son’s dihedral lemma at least two of the K b ’s, i ∈ {0, 1, 2}, must centralize O3 (CG (a)) and hence act nontrivially on E(CG (a)), where CG (a) = i CG (a)/O3 (CG (a)). Using [V9 , 2.8], we obtain that each such K b is a i component of CJi Jiz (z) for some 3-component Ji of CG (a). Since K b is a 3 -group, z normalizes Ji but does not centralize Ji /O3 (Ji ). Applying [V17 ,

2. COMPONENTS AND p-COMPONENTS

205

i 14.1] to Ji /O3 (Ji ), with K b in the role of Y there, yields Ji ∼ = Ru and so m2,3 (CG (a)) ≥ m2,3 (Z3 × Ru × Ru) ≥ 4, a contradiction. Thus 3 divides |K|. Let x ∈ CB (K) (perhaps x = 1). If x = 1 or m3 (K) > 1, then CG (z) ≥ 2

x × KK b K b and m3 (CG (z)) ≥ 4 > e(G), a contradiction. Hence, NB (K) acts faithfully on K and m3 (K) = 1, so there exists c ∈ NB (K) inducing an outer automorphism on K. But then there is a ∈ I3 (CK (c)), whence   2 b b ∼ E34 , contrary to m3 (CG (z)) = 3, completing the proof.  c, a, a , a =

Lemma 2.3. Assume (z, K) ∈ K(G). Then p divides |K|. Proof. Assume the contrary. By [IA , 7.1.2], we have mp (Out(K)) ≤ 1. Hence, if B0 := CB (K), then |B : B0 | ≤ p. Let C = CG (B0 ) and C := C/Op (C). By Theorem C∗4 : Stage A1, Op (C) has odd order, so K z ∼ = K z. By [V17 , 9.2], m2 (K z) ≥ 3. If K z acts faithfully on Op (C), Thompson’s dihedral lemma implies that m2,p (G) ≥ mp (B0 ) + m2 (K z) − 1 ≥ 4 > e(G), a contradiction. Hence, L := E(C) = 1. Then mp (Op (C)) ≤ 3 and again by the dihedral lemma, [Op (C), K] = 1. So K acts faithfully on E(C). If [E(C), z] = 1, then K is a component of E(C). But every component of E(C) has order divisible by p. So this is not the case. Hence, K z acts faithfully on E(C) and so by [V9 , 2.8], K is a component of CJ (z) for some p-component J of CG (B0 ). As K is a p -group, we see from [V17 , 14.1] – with K in the role of Y there – that (p, K, J) is one of the following: 3 (3, 2B2 (2 2 ), Sp4 (8)), (5, L2 (7), HJ), (7, A5 , A7 ), or (11, A5 , A11 ). In every case, CJ (z) is a p -group. Thus if B0 < B, then some b ∈ B induces a nontrivial outer automorphism on J . Hence, p = 3 and J ∼ = Sp4 (8) and, in this case, b induces a field automorphism on J . But then m2,3 (CG (B0 )) ≥ 4 > e(G), a contradiction. Hence, [B, J] = 1. Likewise if J ∼ = Sp4 (8) or HJ, then m2,p (J) > 0 so again m2,p (BJ) ≥ 4, contradiction. Hence, J ∼ = Ap with p = 7 or 11. Let J0 = Lp (CJ (B)), and for each x ∈ B # , let Jx be the subnormal closure of J0 in CG (x). Since e(G) = 3 it follows from [V17 , 13.12] that Jx = Op (Jx )J0 . Now suppose that for some such x, CC(z,K) (x) contains an involution tx . Then tx acts on Jx , and we also suppose that E(CJx /Op (Jx ) (tx )) ∼ = A7 . Let Kx be the subnormal closure of K in CG (tx ). Then CKx (x) has a component isomorphic to A7 . But Kx ↓2 A5 via z, and Kx ∈ Co2 , so we have a contradiction from [V17 , 13.13]. Thus, no such x and tx exist. If p = 11, then z acts on J0 /Op (J0 ), and hence on Jx /Op (Jx ), as a product of three disjoint transpositions. Hence z centralizes an involution tx ∈ C(z, K) mapping on a root involution of Jx /Op (Jx ), whence E(CJx /Op (Jx ) (tx )) ∼ = A7 , contradicting the previous paragraph. Therefore p = 7, and we similarly deduce that for any x ∈ B # , C(x, Jx ) has odd

206 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

order. Now any nontrivial 2-subgroup Ux of CC(z,K) (x) must act faithfully on Jx /O7 (Jx ) ∼ = A7 and centralize the isomorphic image of K ∼ = A5 , whence Ux = z. This holds for any x ∈ B # , so O2 (CG (z)) = z and B acts faithfully on E(C(z, K)), with CE(C(z,K))z/z (x) of odd order for all x ∈ B # . Let I be a component of E(C(z, K)). Then B normalizes I and CB (I) = 1. If I ∈ Co2 − Chev(2), then it follows easily from [IA , 4.10.3, 5.6.1] that m7 (Aut(I)) < 3, contradicting CB (I) = 1. Hence I ∈ Chev(2) and  Γ2B,1 (I) = 1. But then by [IA , 7.3.3], mp (B) = 2, a final contradiction.  The next lemma, a form of cross-characteristic L-balance peculiar to the environment of Theorem C∗4 (Case A), will be used very frequently. Lemma 2.4. Let z ∈ I2 (G), K a component of CG (z), and A ≤ CG (z) with A ∼ = Epm , m > 0. Suppose that A# ⊆ Ipo (G). Let KA be a p∗ ≤ C (A z) component of E(CK (A)) such that KA ∈ Co2 . Also let KA G ∗ ∗ satisfy F (KA ) = KA . (a) If ∗

z) ≥ 5 − m, m2 (KA then KA is a component of CLp (CG (A)) (z); and (b) If KA is a component of CLp (CG (A)) (z), then one of the following holds for some p-component J of CG (A): (1) J z = J and KA is a component of CJ (z); or (2) J z = J and J/Op p (J) ∼ = KA . Moreover, either p = 3 and 5 ∼ KA = L2 (8), or p = 5 and KA ∼ = 2B2 (2 2 ), or p ∈ {5, 7, 17} and KA ∼ = L2 (p). Remark 2.5. Note that the assumption A# ⊆ Ipo (G) is satisfied, by Theorem C∗4 : Stage A1, as long as mp (CG (A z)) ≥ 3. Also, if [K, A] = 1, then KA = K. Proof. Let C = CG (A) and C := C/Op (C). By Theorem C∗4 : Stage A1, Op (C) has odd order. Then F ∗ (C) = P E(C), where P = Op (C) ≥ A. Assume that K A acts nontrivially on P . If actually CK A (P ) = 1, then ∗ z) − 1 ≥ CK ∗ (P ) = 1, and in (a) and by [V9 , 2.3], m2,p (G) ≥ m + m2 (KA A 4 > e(G), contradiction. So Z := CK A (P ) is a nontrivial subgroup of the 2group Z(K A ), and mp (P ) ≤ m2,p (G) ≤ 3. Since A = 1, the Thompson dihedral lemma now implies that m2 (K A /Z) ≤ 2. But as Z = 1, there is no such Co2 -group K A , by [V17 , 9.2]. Therefore [K A , Op (C)] = 1 and KA acts faithfully on E(C). As A# ⊆ Ipo (G), all components of E(CG (a)/Op (CG (a))) lie in Cp for all a ∈ A# , and hence by Lp -balance and [IA , 7.1.10], so do all components of E(C). By [V9 , 2.8], K A ≤ E(C). By assumption  KA = Op (KA ) = [KA , KA ], so KA ≤ Lp (C). Moreover, by assumption, KA is a component of CG ( z A) = CC (z), so KA is a component of CLp (C) (z), proving (a).

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207

In (b), since Op (C) has odd order, K A is a component of CE(C) (z), and z

so by L2 -balance, either K A ≤ J or K A is a diagonal subgroup of JJ for some component J of C. Let J be the p-component of C mapping onto J .  Then since KA = Op (KA ), K ≤ J or K ≤ JJ z , in the respective cases. As KA is a component of CC (z), KA is a component of CJ (z) or CJJ z (z), respectively. Now O2 (KA ) = 1 as KA ∈ Co2 , so in the diagonalizing case, KA ∼ = J/Op (J). = J /O2 (J) ∼ In subcase (b2), therefore, KA ∼ = K A is simple. We saw above that o J ∈ Cp , so K A ∈ Cp ∩C2 . Moreover, as shown in [V9 , Remark 2.7], mp (KA )+ m2,p (KA )+mp (A) ≤ 3. Hence by [V17 , 14.3], the final assertion of the lemma holds, completing the proof.  Finally, we note a quick consequence of the Thompson dihedral lemma. Lemma 2.6. Let x ∈ Ipo (G) and let J be a p-component of CG (x). If m2 (J) ≥ 5, then J is quasisimple. Proof. Since x ∈ Ipo (G), Op (J) has odd order, by Theorem C∗4 : Stage A1. If [Op (J), J] = 1, then [F (Op (J)), J] = 1 and there exists an odd prime r = p such that CJ (Or (J)) ≤ Op (J). Since m2 (J) ≥ 5, there exists E2m ∼ = E ≤ J with m ≥ 5. By Thompson’s dihedral lemma, m2,r (J) ≥ m2,r (Or (J)E) ≥ m − 1 ≥ 4 > e(G) which is not possible.  3. Theorem 1: Introduction Through Section 10, we suppose that Theorem 1 fails. In view of that and Lemma 2.2, we are going to work under the following additional hypothesis: (3A)

For every (z, K) ∈ IL2 (G) and B ∈ Ep3 (CG (z)), CB (K) = 1.

Furthermore, for a fixed element K0 ∈ Co2 , we define the following sets: K1 (K0 ) = {(z, K) ∈ K(G) | K ∼ = K0 }

D2n or SD2n } K2 (K0 ) = {(z, K) ∈ K1 (K0 ) | for Q1 ∈ Syl2 (C(z, K)), Q1 ∼ = K3 (K0 ) = {(z, K) ∈ K2 (K0 ) | E(CG (z)) > K} The sets K2 (K0 ) and K3 (K0 ) will be used only in Sections 6 and 9. Observe that K3 (K0 ) ⊆ K2 (K0 ) ⊆ K1 (K0 ) ⊆ K(G). Also, K(G) = ∪K0 ∈Co2 K1 (K0 ), and K(G) = ∅ by (2B). Our strategy is to show, by the analysis of several cases, that K1 (K0 ) = ∅ for every possible K0 ∈ Co2 . Hence, our ultimate contradiction will be that K(G) = ∅. We shall frequently use the theorem that if K is a component of CG (z) for some involution z ∈ G, and m2 (K) ≥ 2, and K is terminal in G, then K is standard in G [II3 , Corollary PU4 ]. Recall our notation (2C).

208 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

Lemma 3.1. If m2 (Q) = 1, then p = 3 and there exists b ∈ I3o (G) with [b, K] = 1. Moreover, F ∗ (CG (z)) = Q0 K, Q0 b ∼ = SL2 (3), and Q is quaternion. Proof. By (3A) and Theorem C∗4 : Stage A1, there exists b ∈ Ipo (G) ∩ C(z, K). As SL2 (q) ∈ C2 for all odd q, F ∗ (CG (z)) = Q0 K. Now b acts nontrivially on Q0 , and as Q0 ≤ Q, Q0 ∼ = Q8 , Q is quaternion, p = 3 and ∼  Q0 b = SL2 (3). Lemma 3.2. Let (z, K) ∈ K(G) and suppose that K ∼ = M11 , L± 3 (3) or 1 2F (2 2 ) . If K ∼ U (3), assume that p = 3. Then K is terminal in G. = 3 4 Proof. If not, then as m3 (K) = 2 and e(G) = 3, there exists an involution v ∈ CG (K) such that K pumps up vertically to L in CG (v) and 1 L ∈ Co2 . If K ∼ = M11 , L3 (3), or 2F4 (2 2 ) , this is not possible by [V17 , 13.6]. Suppose that K ∼ = = U3 (3), so that p = 3. By [V8 , Lemma 3.10], L/Z(L) ∼ U4 (3) or G2 (4), and as e(G) = 3 ≥ m3 (CG (v)) ≥ m3 (L), L/Z(L) ∼ = G2 (4). Then there exists a ∈ I3 (K) ≤ L with I := E(CL (a)) ∼ = SL3 (4) or 2SL3 (4) o [IA , 4.7.3A, 6.1.4]. Since a ∈ K, a ∈ I3 (G). As m2 (I) − m2 (Z(I)) = 4 by [IA , 3.3.3, 6.4.4], I is a component of CJ (v) for some 3-component J of CG (a), by Lemma 2.4 with a in the role of A there. But that is not  possible by [V17 , 14.1, 14.2], with I in the role of Y there. Lemma 3.3. Let (z, K) ∈ K(G). Then K ∼ U3 (4). = Proof. Assume the contrary. If p = 5, then mp (K) = mp (Aut(K)) = 1, and so there exists Ep2 ∼ = A := b, a ≤ CG (K z). If p = 5, then there is A = b, a ∼ = E52 such that b ∈ I5 (C(z, K)) and a ∈ I5 (K) with KA := E(CK (A)) ∼ = A5 . Also in this case, set Cb = CG (b) and C b = Cb /O5 (Cb ). If p = 5, on the other hand, let KA = K. Then, in either case, A# ⊆ I5o (G). Moreover, since m2 (KA z) = 3, Lemma 2.4 yields that KA is a component of CJ (z) for some 5-component J of CG (A). By [V17 , 14.1,14.2],  with KA K ≤ C (b), in the role of Y there, p = 5 and J = K A ∼ . But K = K A = 5 G A so K is a component of CJb (z), where Jb is the pumpup of J in CG (b). As  Jb /O5 (Jb ) ∈ C5 , this contradicts [V17 , 14.1], completing the proof. Lemma 3.4. Let (z, K) ∈ K(G). If m2 (K z) ≤ 3, then K is isomorphic to one of the following groups: L2 (q), q ∈ {5, 7, 9, 17}, L± 3 (3), 2HJ, M11 . Proof. The result follows immediately from [V17 , 9.2] and Lemma 3.3.  We denote sectional 2-rank by r2 , and then have: Lemma 3.5. Let (z, K) ∈ K(G). If r2 (K/Z(K)) ≤ 2, then K ∼ = L2 (q), (3), or M . In particular, if 5 divides |K|, then K∼ q ∈ {5, 7, 9, 17}, L± = A5 , 11 3 A6 , or M11 . Proof. This follows easily from Lemma 3.4.



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Lemma 3.6. If K is not terminal in G, then there are z  , u ∈ I2 (CG (K)) such that K is a component of CG (z  ) but not of CG (u), and z, u is a 2group centralized by z  . Remark 3.7. Warning. The condition that (z, K) ∈ K(G) does not guarantee that (z  , K) ∈ K(G). An Ep3 -subgroup B of CG (z) may not centralize z  . However, B ∩ K ≤ CG (z  ). Proof. Let y ∈ I2 (Z(Q)) ⊆ I2 (Q1 ). If K is not a component of CG (y), take z  = z and u = y. If K is a component of CG (y), then since it is not terminal, there exists u ∈ I2 (Q) such that K is not a component of CG (u). We take z  = y in this case.  ∼ Proposition 3.8. Let (z, K) ∈ K(G). Then K = M11 . Moreover, if ± ∼ p = 3, then K = A6 or L3 (3). We will divide the proof in a series of lemmas. But first, let us set some additional notation, which we use throughout this section. (1) (z, K) ∈ K1 (G) with K ∼ = A6 , M11 or L± 3 (3); ∼ (3B) (2) p = 3 or K = M11 , with p dividing |K|; and (3) t = Ω1 (Z(SK )). Observe that as m3 (K) = 2 and e(G) = 3, K  CG (z). Recall that Q ∈ Syl2 (CG (K)), Q0 = O2 (CG (z)) and Q1 = CQ (z) ∈ Syl2 (C(z, K)). Lemma 3.9. If K is terminal in G, then the following conditions hold: (a) p = 3, m2 (Q) = 1, and Q is quaternion with z ∈ Q; (b) F ∗ (CG (z)) = Q0 K with Q0 ≤ Q, Q0 ∼ = Q8 ; (c) There exists E33 ∼ = B ≤ CG (z) such that B = b, B0 , [b, K] = 1 and E32 ∼ = B0 ≤ K; and 2 (d) O (CG (z, K)) = Q0 b ∼ = SL2 (3). Proof. Since G is of restricted even type, if K ∼ = A6 , m2 (Q) = 1 by ± ∼ [I2 , Definition 8.8]. Assume now that K = M11 or L3 (3). If m2 (Q) ≥ 2, by [II3 , Corollary P U2 ], there exists E22 ∼ = U ≤ Q and g ∈ G−NG (K) such that U g ≤ NG (K). By [V9 , 7.3], U g acts faithfully on K and ΓU g ,1 (K) < K. However, if K ∼ = M11 , by 5.3a of [IA ], K has a unique conjugacy class of involutions and an involution centralizer is a maximal subgroup of K,  implying ΓU g ,1 (K) = K. And if K ∼ = L± 3 (3), K = ΓU g ,1 (K) by Theorem 7.3.4 of [IA ]. Thus m2 (Q) = 1 in all cases. The rest of the lemma follows immediately from Lemma 3.1.  Lemma 3.10. We have K ∼ M11 . = Proof. Assume K ∼ = M11 . By Lemma 3.2, K is a terminal component and the conclusions of Lemma 3.9 hold. It follows that Q ≤ S and since | Out(K)| = 1, CG (z) = Q b ×K and S = Q ×SK . By the Z ∗ -Theorem and Burnside’s Lemma, z NG (S) ∩ {z, t, zt} = {z}. Then, by the Krull-Schmidt

210 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS g Theorem, there exists g ∈ NG (S) with tg = z. But then SK ≤ S and g g ∼ SK ∩ SK = 1, whence SK embeds isomorphically into S/SK = Q, which is impossible, completing the proof. 

Lemma 3.11. Suppose that K is terminal in G. If either K ∼ = A6 or z is 2-central in G, then z G ∩ CS (Q0 ) = {z}. ∼ Q0 b  CG (z). Proof. By Lemma 3.9, Q is quaternion and SL2 (3) = g := Suppose for a contradiction that s z ∈ CS (Q0 ) for some g ∈ G with s = z. Then s normalizes and indeed centralizes Q0 b. As s = z and m2 (Q) = 1, Q0 b embeds into Aut(K g ). But Aut(A6 ) has no subgroup isomorphic to SL2 (3). Therefore K ∼ = L± 3 (3), z is 2-central by hypothesis, g and z acts on K as a 2-central involution. Hence, by symmetry, we may assume that s = t or tz. By Burnside’s lemma we may also assume that g ∈ NG (S) ≤ NG (Ω1 (Z(S))) = NG ( t, z) and that g has odd order. Thus g acts on O 2 (CG ( t, z)) = O2 (CK (t)) × Q0 b ∼ = SL2 (3) × SL2 (3). By the Krull-Schmidt Theorem, g interchanges z and t. But g has odd order, a final contradiction.  Lemma 3.12. Assume that K is terminal in G. Then z is not 2-central in G. Proof. Assume that z is a 2-central involution of G. By the Z ∗ Theorem, z G ∩ S = {z}. Let y ∈ z G ∩ S with y = z g = z. By Lemma 3.11, [Q0 , y] = 1, so yz ∈ y Q0 . If y induces an inner automorphism on Q, then z is a square in −1 −1 CG (y). Hence z g is a square in CG (z), so z g ∈ K × Q, as CG (z)/KQ −1 is elementary abelian. Thus, z g ∈ CG (Q0 ), contradicting Lemma 3.11. Therefore, CQ (y) = z. Moreover, replacing y by a CG (z)-conjugate, we may assume that CS (y) ∈ Syl2 (CG ( y, z)). ± 2 ∼ Suppose that K ∼ = L± 3 (3). Hence O (CG (y)) = SL2 (3) × L3 (3). By K ∼ [IA , 4.5.1], yt ∈ y and CK (y) = Σ4 . It follows that t ∈ O2 (CK (y)) ≤ K g . But then yt ∈ Qg ×K g , and so y G ∩(Qg ×K g ) = {y}, whereas z G ∩(Q×K) = {z}, a contradiction. Hence, K∼ = A6 . ∼ Suppose that CK (y) = D10 . Then again yt ∈ y K , and so y t, z ⊆ y CG (z) . As  O 5 (CG (y)) ≤ K g , CK g (z) ∼ = D10 . Also, y, z, t ∈ Syl2 (CG ( y, z)). Hence, if we put U = y, z, t ∩ (Qg × K g ), then z is fused to every involution in zU . But then z, t ∩ U = 1, which is impossible. Therefore CK (y) ∼ = Σ4 and CQ (y) = z. Write O2 (CK (y)) = V b1 , 2 V ∼ = SL2 (3) × A6 , whence V ≤ = E22 . Then O (CK (y)) ≤ O2 (CG (y)) ∼ G g CK (z). By symmetry, y ∩ y, V  = {y}. As t ∈ V ∩ Z(S), we see that CS (V ) = Q y × V ∈ Syl2 (CG (V )). As Q0 b and Qg0 bg  are in CG (V ), we see that CG (V )/V has a semidihedral Sylow 2-subgroup and one class of involutions. Hence CG (V )/V ∼ = L3 (3) or M11 , in view of the structure

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of CG (z). As CG (V )/V has trivial Schur multiplier, CG (V ) = L × V with L ∼ = L3 (3) or M11 . Now NKz (V ) contains a subgroup N × z, where CN (V ) = 1 and N ∼ = Σ3 . Then N z normalizes L and centralizes Q0 b = O 2 (CL (z)). Hence the image of N z in Aut(L) lies in the image of z, and so by judicious choice of N we may assume that LN = L × N . We now use [V17 , 13.6] and the fact that e(G) = 3 to deduce that the pumpups of L in the centralizers of involutions in V , and then in the centralizers of involutions in V N , are all trivial pumpups. Again as e(G) = 3, L  CG (s) for every involution s ∈ V N . Therefore L is normalized by ΓV N,1 (K) = K. As CK (L) = 1, [L, K] = 1 so K ≤ CG (y), which is not the case. This contradiction completes the proof.  Lemma 3.13. K ∼ = A6 . Proof. Suppose not. Then K ∼ = L± 3 (3). By Lemma 3.2, K is a terminal component. Then the conclusions of Lemma 3.9 hold. By Lemma 3.12, there exists u ∈ NG (S) such that u2 ∈ S and Q = Qu  S. Moreover, z = z u ∈ Z(S). As a result, CQu (K) = 1. Let V = SK ∩ Qu  S. If V = 1, then [SK , Qu ] = 1, so Qu ∼ = ∼ u AutQ (K) ≤ CAut(K) (SK ) = Z(SK ) by [V17 , 10.14.5], a contradiction as Qu is not abelian. Therefore V = 1 and so V ∩ Z(S) = 1. As V ≤ Qu , z u ∈ V ∩ Z(S) ≤ Z(SK ). But u2 ∈ S, so Ω1 (Z(SK ))u = z. Hence, u = 1 = [S , S u ]. Now Aut u (K) ∼ SK ∩ Ω1 (Z(SK ))u = 1, so SK ∩ SK = K SK K u ≤ Z(S ) × Q. Hence, Ω (S ) ∼ Ω (S u ) is abelian, Z(SK ), and so SK 1 K = 1 K K which is not the case, completing the proof.  Lemma 3.14. The following conditions hold: (a) K is not terminal; (b) There exist v ∈ I2 (Q) and z1 ∈ I2 (Q1 ) such that K is a component of CG (z1 ), [z1 , v] = 1, and the pumpup L of K in CG (v) is isomorphic to one of A8 , HS, 2HS, L5 (2) or Sp4 (4); and (c) m3 (CG (v)) = 2. Proof. Suppose that K is terminal in G. Then z is not 2-central by Lemma 3.12. But then if g ∈ NG (S) − S with g 2 ∈ S, we have z = z g ∈ Z(S) ≤ CS (Q0 ), contrary to Lemma 3.11. Hence, (a) holds. As K is not terminal, by Lemma 3.6 there exist z1 ∈ I2 (CQ1 (z)) and v ∈ I2 (CQ (z1 )) such that K is a component of CG (z1 ) and K pumps up nontrivially in CG (v). Since m3 (K) = 2 while e(G) = 3, the pumpup L is vertical and L  CG (v). By [V17 , 13.2], L ∼ = A8 , HS, 2HS, U4 (2), L5 (2) or U4 (2), proving (b). Sp4 (4). Since (3A) holds, m3 (L) ≤ 2, and so L ∼ = Finally, suppose that m3 (CG (v)) ≥ 3, so that (v, L) ∈ K(G). By [V17 , 9.2] and (3A), m2 (L) ≥ 4 and m3 (Aut(L)) = 2. Hence there exists x ∈ I3o (G)∩C(v, L), and by Lemma 2.4, with A = x, L is a component of CJ (v) for some 3-component J of CG (x). But given the possible isomorphism types for L, this is not possible, by [V17 , 14.1, 14.2], with L in the role of Y there.  Therefore m3 (CG (v)) = 2, completing the proof.

212 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

∼ E33 . Then B ∩ K ≤ L and |L|3 = 32 so Now let B ≤ CG (z) with B = B ∩ K ∈ Syl3 (L). Clearly, [B ∩ K, z, z1 , v] = 1. Choose 1 = a ∈ B ∩ K such that La := E(CL (a)) ∼ = A5 or L3 (2). Set Ca = CG (a) and C a = Ca /O3 (Ca ). By [V17 , 10.14.4], F ∗ (La z 1 ) = La . We next prove Lemma 3.15. La z 1  acts faithfully on O3 (C a ). Proof. Suppose not, so that La ≤ E(C a ) by [V9 , 2.8], and as a ∈ Lemma 2.4b applies with A = a. Thus, since La ∼ L2 (8), there = exists a 3-component J of CG (a) with La a component of E(CJ (v)). We apply [V17 , 14.1, 14.2] with La in the role of Y there. If L ∼ = L5 (2), then we (2), but no such Y is possible. Hence, L could have chosen a so that La ∼ = 3 ∼ L = A8 , HS, 2HS or Sp4 (4), and La z1  < L z1  with La z1  ∼ = Σ5 . Thus, using [V17 , 14.1, 14.2], we obtain that J ∼ = A9 and v acts as a product of two disjoint transpositions on J. Since e(G) = 3, m3 (O3 (C a )) = 1, J = E(C a ) and, in particular, v ∈ J. (By Theorem C∗4 : Stage A1, |O3 (CG (a))| is odd.) We claim that m3 (CCG (a) (z1 )) = 3. We may assume that z1 induces an inner automorphism on J, since otherwise the claim is obvious. Therefore CJa (z1 ) contains a Z3 ×A4 subgroup. On the other hand, since K  CG (z1 ), we may assume that B ∩ K ∈ Syl3 (CG (z1 )), and then since K ∼ = A6 , CG (z1 ) does not involve Z3 × A4 , contradiction. This proves the claim. It follows from the claim that m3 (CJ (z 1 )) = 2. If z 1 acts as a transposition on J, then CJ (z 1 ) ∼ = A7 , and as m3 (CJ (z1 ) ∩ B) > 1, it follows that CJ (z1 ) ∩ K = 1, which is impossible as K  CG (z1 ). Therefore z 1 acts on J as a product of three transpositions. Thus C a ≥ a × J z 1  ∼ = Z3 × Σ9 and I3o (G),

CC a (z 1 ) ∼ = Z3 × (E23 : Σ3 ) × Σ3 . On the other hand there exists an involution u ∈ L z1  inverting a. Hence, NG ( a)/O3 (NG ( a)) contains a subgroup isomorphic to Σ3 × Σ9 . Suppose that O3 (J) = 1 and choose a prime r > 3 such that CJ (Or (J)) ≤ O3 (J). As [z1 , J] = 1, COr (J) (z1 ) = 1. Hence, Or (CG ( z1 , a)) = 1. However, since a∈K ∼ = A6 , Or (CCG (z1 ) (a)) = Or (CG (z1 )) = 1. Thus O3 (J) = 1. Since m3 (O3 (C a )) = 1, we obtain that F ∗ (CG (a)) = O3 (CG (a)) × O3 (CG (a)) × J. In fact, by a similar argument, we obtain that O3 (CG (a)) is inverted by z1 , and so is abelian. Moreover, as a ∈ K ≤ CG (z1 ), we obtain that O3 (CG (a)) = a. Let u ∈ I2 (NG ( a)) be an involution inverting a and centralizing J z1 . Consider CG (u). Since m3 (A9 ) = 3, m3 (CG (u)) = 3. By hypothesis (3A), J acts faithfully on T := O2 (CG (u)). Now z1 ∈ CG (u). Since u inverts a and centralizes B/ a, K u ∼ = Z2 × Σ4 , and = Σ6 . Hence, CKu (u) ∼ CCG (z1 ) (u) = CC(z1 ,K) (u)CKu (u). Now, CC(z1 ,K) (u) ≤ C(z1 , K) ≤ J z1 . Hence, CC(z1 ,K) (u)∩T = 1. Therefore CT (z1 ) is isomorphic to a subgroup of

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213

O2 (CKu (u)) ∼ = E23 . Let V = T /Φ(T ). Then J acts faithfully on V , Lemma 9.17 of [IG ] implies that m2 (CV (z1 )) ≤ 3, and so m2 (V ) ≤ 6. On the other hand, A9 ≤ GL6 (2), as A9 contains the Frobenius group F9·8 ∼ = U3 (2). This is a contradiction completing the proof.  Note that the argument above when K ∼ = A6 and J ∼ = A9 is difficult because we are in the vicinity of the group Σ12 . But in that group hypothesis (3A) is violated. Finally, we can eliminate the case K ∼ = A6 . Lemma 3.16. K ∼ A6 . = Proof. We have shown above that m3 (CG (v)) = 2 and indeed L contains a Sylow 3-subgroup of CG (v), whence a × La contains a Sylow 3subgroup of CG (v), so v inverts O3 (C a )/ a. As a ∈ K and O2 (CG (z1 )) = 1, we have that CO3 ,3 (CG (a)) (z1 ) ≤ CK (a) = a, b1  ∼ = Z3 ×Z3 . Let W be a crita. Then [W, W ] ≤ a. ical subgroup of O3 (C a ) of exponent 3 and containing   As v, z1  acts faithfully on W , CW (z1 ) = a, b1 . Suppose first that Z(W ) is a faithful La z1 -module. As |La | does not divide |GL(3, 3)|, it follows that m3 (Z(W )) ≥ 5. But then m3 (CG (vz1 )) ≥ 4, contradicting e(G) = 3. Hence, La acts trivially on Z(W ) < W . Thus, [W, W ] = a and La z 1  acts faithfully on an extraspecial quotient E of W with dim(CE/a (z 1 )) = 1. But this dimension must be even as z 1 maps into  Sp(E/ a), yielding a final contradiction. Now Lemmas 3.10, 3.13 and 3.16 complete the proof of Proposition 3.8.

4. Some Nonsimple Cases Proposition 4.1. Let (z, K) ∈ IL2 (G). Assume that m2 (C(z, K)) > 1 or that (z, K) ∈ K(G). If p > 3, then K ∼ 2M12 , 2HJ, or 2Ru. = Proof. Suppose false. Let Q1 ∈ Syl2 (C(z, K)). Then in any case, m2 (Q1 ) > 1, since if (z, K) ∈ K(G), then |C(z, K)| is divisible by p > 3 by (3A), and O2 (C(z, K)) = 1 as G is of even type. As m3 (K) > 1 and e(G) ≤ 3, (z, K) has no diagonal pumpups in G. Then with [V17 , 13.6a], K is terminal in G. Let Q ∈ Syl2 (CG (K)). By [II3 , Corollary PU2 ] and [V9 , 7.3], there exists E22 ∼ = U ≤ Q and g ∈ G − NG (K) such that U g ≤ NG (K)  and ΓU g ,1 (K) < K. Then by [V17 , 5.5bce], we may write U g = u, v with ∗ u inducing an inner automorphism on K corresponding  ∗ 2 to some u ∈ K ∗ of order 4, and v inverting u . Therefore Z(K) = (u ) ≤ [CK (u), v] ≤ [NG (K g ), CG (K g )] ≤ CG (K g ). Hence for some S0 ∈ Syl2 (CG (K)), 1 =  Z(K) ≤ S0 ∩ S0g , violating [V9 , 7.3]. The proof is complete.

214 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

√ √ 5. 2F4 ( 2), and Preliminaries for 2F4 ( 32), Sp4 (8), and 3D4 (2) The main result of this section is as follows. Proposition 5.1. Let (z, K) ∈ K(G). Then K ∼ 2F4 (2 2 ) . Moreover, = 5 if either p = 3 and K ∼ = 2F4 (2 2 ), then = Sp4 (8) or 3D4 (2), or p = 5 and K ∼ m2 (Q) ≥ 2 and the following conditions hold: 1

5

(5A)

(1) (p, K) is (3, 3D4 (2)), (3, Sp4 (8)) or (5, 2F 4 (2 2 )); (2) K is a terminal component in G; and (3) There is b ∈ Ipo (G) ∩ C(z, K) such that K is a component of CG (b).

We assume that (z, K) ∈ K(G) but that the conclusions of Proposition 5.1 fail. 1 Lemma 5.2. If K ∼ = 2F4 (2 2 ) , then the following conditions hold:

(5B)

(1) (2) (3) (4)

p = 3 or 5; 1 K∼ = 2F4 (2 2 ) is a terminal component in G; Z(SK ) = t; and There is b ∈ Ipo (G) ∩ C(z, K) such that K is a component of CG (b).

1 Proof. Assume K ∼ = 2F4 (2 2 ) . Since m3 (K) = 2 and e(G) = 3, K has no diagonal pumpups. In view of [V17 , 13.6a], K is terminal in G. As | Out(K)| = 2, if mp (K) = 1, then mp (Aut(K)) = 1 and there exists Ep2 ∼ = A ≤ C(z, K). Since m2 (K z) ≥ 4 by Lemma 3.4, Lemma 2.4 yields that K is a component of CJ (z) for some p-component J of CG (A). Using [V17 , 14.1, 14.2], with K in the role of Y there, we obtain that p = 3 or 5. But m3 (K) = m5 (K) = 2 [IA , 4.10.3]. Hence, mp (K) = 2 and p = 3 or 5. By (3A), there exists b ∈ Ipo (G) ∩ C(z, K). Now Lemma 2.4 (with A =

b), together with [V17 , 14.1, 14.2], implies that K ≤ J where J/Op (J) ∼ = K and J is a p-component of CG (b). Since m2 (K) = 5 by [IA , 3.3.3], and K has trivial Schur multiplier by [IA , 6.1.4], J = K by Lemma 2.6. Then [V17 , 10.5.1] gives Z(SK ) = t, completing the proof. 

Lemma 5.3. If (p, K) is one of the pairs (3, 3D4 (2)), (3, Sp4 (8)) or 5 4 (2 2 )), then (5A) holds.

(5, 2F

Proof. Assume K is not terminal in G. Then there exists an involution u ∈ CG (K) such that K has a nontrivial pumpup L in CG (u). Since mp (K) = 2 and e(G) = 3, the pumpup must be vertical, and so by [V17 , 13.2], p = 3 and L ∼ = U4 (8), then = Sp4 (26 ), L4 (8), or 3D4 (22 ). (If L ∼ m3 (L) = 3 and (3A) is violated.) In every case there exists 1 = a ∈ B ∩ K with E(CK (a)) ∼ = L2 (8) [IA , 4.8.2, 4.7.3A] and La := E(CL (a)) ∼ = L2 (26 ). Since m2 (La ) = 6, by Lemma 2.4, La is a component of CJ (u) for some 3-component J of CG (a),

√ √ 5. 2F4 ( 2), AND PRELIMINARIES FOR 2F4 ( 32), Sp4 (8), AND 3D4 (2)

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which is not possible by [V17 , 14.1, 14.2], with La in the role of Y there. Hence, K is terminal in G. Because of (3A), there exists b ∈ I3o (G) ∩ CG (z) that centralizes K. Since m2 (K) ≥ 5 by [V17 , 9.2], Lemma 2.4 (with A = b) yields that K is a component of CJ (z) for some p-component of CG (b). Now [V17 , 14.1, 14.2], with K in the role of Y there, imply that J/Op (J) ∼ = K. Finally, by Lemma 2.6 and [IA , 6.1.4], J = K, completing the proof of the lemma.  Lemma 5.4. Assume (5A) or (5B). Then m2 (Q) ≥ 2. Proof. Assume that m2 (Q) = 1. By Lemma 3.1, p = 3, Q is quaternion with z ∈ Z(Q), F ∗ (CG (z)) = Q0 K, Q8 ∼ = = Q0 ≤ Q and Q0 b ∼ SL2 (3). Hence, CG (K) = O2 (CG (K)) Q, b is solvable and CG (K × b) ≤ O2 (CG (K)) z, b. Let Cb := CG (b) and C b := Cb /O3 (Cb ). By Theorem C∗4 : Stage A1, O3 (Cb ) has odd order. Since CG (K) is solvable, K is the only 3-component of CG (b), and as m2,3 (K) = 1, m3 (O3 (C b )) ≤ 2 and m3 (CG (K)) ≤ 2. Then F ∗ (C b ) = O3 (C b )K and, since z = 1, z inverts a nontrivial 3-element of O3 (C b ). It follows that O2 (CG (K)) = 1. As CO2 (CG (K)) (z) ≤ O2 (CG (z)) = 1, F ∗ (CG (K)) = O2 (CG (K)), an abelian group inverted by z. Let X := Ω1 (O3 (CG (K))). Then Q0 acts faithfully on X, implying X ∼ = E32 . In particular, CG (K) ≥ XQ0 b ∼ = E32 SL2 (3). Let 1 = x ∈ CX (b). By Lp balance and Lemma 2.6, and as m2,3 (G) ≤ 3, CG (x) has a component Jx with K ≤ Jx . Since z inverts x, z ∈ NG ( x) normalizing Jx with K a component of CJx (z). Using [V17 , 14.1, 14.2], with Jx in the role of Y there, we obtain that Jx = K = E(CG (x)). By the action of Q0 on X, K = E(CG (y)) for every y ∈ X # , and so ΓX,1 (G) ≤ NG (K). Also, F ∗ (CG (y)) ≤ O3 (CG (y))X × K for all y ∈ X # . We shall use this to contradict the fact that CG (K) is solvable. Let t ∈ I2 (K) be an involution such that 3 divides |CK (t)|. Set Ct = CG (t) and St = O2 (Ct ). Then St ≤ ΓX,1 (St ) ≤ NG (K). Since Q ∈ Syl2 (CG (K)) and CQ (X) = 1, and CG (K) is solvable, we have CSt (K) = 1. Then |St : St ∩ K| ≤ | Out(K)|2 ≤ 2. As X normalizes St and centralizes K, [St , X] = 1. Therefore E(Ct ) contains a component K1 which is not centralized by X. Let H = XQ0 b ∼ = P GU3 (2). Then CX (K1 ) = 1 and XQ0 normalizes K1 , whence m3 (K1 ) > 1, H normalizes K1 , and CH (K1 ) = 1. Now CG (t) ≥ X × CK (t), so by our choice of t, m3 (CG (t)) ≥ 3. Hence by (3A), there exists c ∈ I3o (G) ∩ CG (t) centralizing K1 . Assume first that K1 is a component of CLc (t), where Lc = L3 (CG (c)). 1 Then by [V9 , 2.8], [V17 , 14.4], and Proposition 3.8, K1 ∼ = 2F4 (2 2 ) or 3D4 (2). Hence by [IA , 7.3.3], Γ := O2 (ΓX,1 (K1 )) ∼ = L3 (3) or K1 . Therefore some nonabelian composition factor Γ0 of Γ contains a 31+2 -subgroup. On the

216 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

other hand, Γ ≤ NG (K) ∩ CG (t). Thus Γ/CΓ (K) embeds in CAutG (K) (t). As CG (K) is solvable, Γ0 is involved in some composition factor of CAutG (K) (t). However, t ∈ K, and using the Borel-Tits theorem, we see that every composition factor of CAut(K) (t) has 3-rank at most 1, a contradiction. Hence, K1 acts faithfully on O3 (CG (c)/O3 (CG (c))). By Lemma 2.4 (with A = c), m2 (K1 ( t)) ≤ 3, and by Lemma 3.4, we have that K1 ∼ = (3), 2HJ, or M . Since m (K ) = 2 and because of L2 (q), q ∈ F M9, L± 11 3 1 3 Proposition 3.8, we obtain that K1 ∼ = 2HJ. But Aut(HJ) has a weakly closed subgroup of order 3 [IA , 5.3g] so does not contain a copy of P GU3 (2). This contradiction completes the proof.  We now complete the proof of Proposition 5.1. Suppose that K ∼ = 2F (2 21 ) . By [II , Corollary P U ], there exists E ∼ 4 3 2 22 = U ≤ Q and g ∈ G − NG (K) with U g ≤ NG (K). By [V9 , 7.3], ΓU g ,1 (K) < K. However, by [V17 , 10.5.6], ΓU g ,1 (K) = K, contradiction. So Proposition 5.1 is proved. √ 6. The L2 (8) and 2B2 ( 32) Cases In this section we continue the proof of Theorem 1. We work under the following hypothesis: (6A)

(1) (z, K) ∈ K(G), and (p, K) is (3, L2 (8)) or 5 (5, 2B2 (2 2 )); and (2) If p = 5, then K1 (A5 ) = ∅.

Thus | Out(K)| = p [IA , 2.5.12], and by (3A) (or since mp (Aut(K)) = 2), there exists (6B)

b ∈ Ipo (G) ∩ C(z, K).

Set R0 = Ω1 (SK ) ∼ = E2p . Then R0 ≤ Z(SK ) [IA , 2.4]. Moreover, as | Out(K)| = p, (6C)

S = Q × SK .

Lemma 6.1. We have m2 (Q) ≥ 2. Proof. Assume m2 (Q) = 1. By Lemma 3.1, p = 3, F ∗ (CG (z)) = Q0 K with Q0 = O2 (CG (z)) ∼ = Q8 , Q0 b ∼ = SL2 (3), and Q is a quaternion group with z = Z(Q). With (6C), z = Ω1 ([S, S]) char S, so S ∈ Syl2 (G). By the Z ∗ -theorem, z is conjugate to an involution of Ω1 (S) − z = Ω1 (Z(S)) − z. By Burnside’s lemma (cf. [IG , 16.2]), this fusion happens  in NG (S). But z char S, so we have a contradiction. The argument depends on whether K is terminal in G. Lemma 6.2. If K is terminal in G, then the following conditions hold: (a) K = E(CG (z)) and Q = Q1 = O2 (CG (z)); (b) Either Q ∼ = Ω1 (SK ); = SK or Q ∼

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217

(c) There exist a, c ∈ Ip (CG (z)) such that B = a, b, c ∼ = Ep3 and (1) a ∈ K; and (2) c induces a nontrivial field automorphism on K and [Q, c] = 1; and (d) S = Q × SK , and |CG (z) : (Q b × K c)| divides p − 2. Proof. By terminality, and as m2 (Q) ≥ 2, and | Out(K)| is odd, we may use [II3 , Corollary PU2 ] and [V9 , 7.6, 7.3] to obtain that there is g ∈ G − NG (K) such that Qg ≤ NG (K) and Qg acts faithfully on K. In particular, Q is isomorphic to a subgroup of SK , so Ω1 (Q) ≤ Z(Q). Hence, as z ∈ Q, z ∈ Z(Q) and Q1 = Q. By (6C), Ω1 (S) ≤ Z(S). Hence Ω1 (Q)g ≤ Z(S). Then S, S g  ≤ CG (Ω1 (Q)g ) and so by Sylow’s theorem S = S gv for some v ∈ CG (Ω1 (Q)g ). Hence, Ω1 (Q)gv = Ω1 (Q)g and gv ∈ NG (S). Now if (6D)

Φ(Q) = 1,

then p = 5 and either 1 = Φ(Z(S)) ≤ Q or [Q, Q] = 1. Since Ω1 (Q)g ∩ Ω1 (Q) = 1, the Krull-Schmidt theorem implies that Ω1 (Q) ≤ [Q, Q] and gv interchanges Ω1 (Q) and Ω1 (SK ). It follows in that case that Q ∼ = SK , so (b) holds. Moreover, K gv contains Ω1 (Q) and is a component of CG (t) for any t ∈ Ω1 (SK ). We claim that this implies that E(CG (z)) = K. For if K1 = K is a component of CG (z), then K1 is a component of CG (t, z) centralizing z ∈ K gv . As E(CK gv (z)) = 1, [K1 , K gv ] = 1 and so [K1 , Ω1 (Q)Ω1 (SK )] = 1. Thus every involution of K1 lies in Z(K1 ), which is not possible as K1 ∈ C2 [V17 , 9.4]. This proves our claim. We continue to assume (6D), and we establish (a) and (c) in that case. By the claim, Q∗ := F ∗ (C(z, K)) = O2 (C(z, K)) ≤ Q. Then Q stabilizes the chain Q∗ > Ω1 (Q∗ ) > 1, which is normal in C(z, K), so Q∗ = Q, and (a) holds. As the squaring map from Q/Ω1 (Q) to Ω1 (Q) is injective, any 5-subgroup of C(z, K) must act faithfully on Ω1 (Q0 ) and so |C(z, K)|5 ≤ 5. As m5 (Aut(K)) ≤ 2, we may choose a and c as described in (c). Now suppose (6D) fails, so that Q is elementary abelian. We argue again that K = E(CG (z)). Otherwise a component K1 of E(CG (z)), lying in C2 with elementary abelian Sylow 2-subgroups, must be K1 ∼ = L2 (2a ), 2 ≤ a < m2 (Q) ≤ p. Then B normalizes K1 , so A := CB (K1 ) is noncyclic as mp (Aut(K1 )) ≤ 1. As m2 (K1 z) ≥ 3, Lemma 2.4 implies that K1   CJJ z (z) for some p-component J of CG (A). The only possibility, by [V17 , 14.1, 14.2], with K1 in the role of Y there, is p = 5 and K1 ∼ = A5 ; but this is ruled out by our hypothesis (6A2). Therefore, K = E(CG (z)), and (a) holds. Now B acts faithfully on F ∗ (CG (z))/ z ∼ = K × (Q/ z), so m2 (Q/ z) ≥ p − 1 and Q ∼ = Ω1 (SK ), proving (b). Now (c) follows easily. It remains to prove the second assertion of (d). Now F ∗ (CG (K z)) = Q, so CG (K z)/Q is a group of odd order containing b, and acting faithfully on Q/Φ(Q), with a fixed point. It follows easily that CG (K z)/Q has order p or p(p − 2), as desired. The proof is complete. 

218 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

Lemma 6.3. Suppose that K is terminal in G and set A = a, c and  J = E(CG (A)). Then z ∈ J = O2 (CG (A)). Moreover, J ∼ = K. Proof. Let Ca = CG (a) and C a = Ca /Op (Ca ). Then NCa (Q) ≤ CNG (K) (a), so Q ∈ Syl2 (NCa (Q)) and hence Q ∈ Syl2 (Ca ). If CQ (Op (C a )) = 1, then the action of Q on COp (C a ) (A) (which is also faithful by the A × Blemma) yields m2,p (G) > p by the Thompson dihedral lemma, contradicting e(G) = 3. So CQ (Op (C a )) = 1, and thus E(C a ) = 1. Now b acts nontrivially on Ω1 (Q) = Ω1 (O2 (CG (z))). It then follows by [V17 , 7.25] that I := Lp (Ca ) is a single simple p-component containing Q, and I ∼ = K. In particular z ∈ I  and I = O2 (C a ). As [Q, A] = 1, furthermore, [I, A] = 1. Now Op (Ca ) has odd order by Theorem C∗4 : Stage A1, so to complete the proof it remains only to show that I is quasisimple (and quote [IA , 6.1.4] to conclude that I ∼ = I). If p = 5, this follows Lemma 2.6. If p = 3, then by Lemma 6.2d, CG ( z, a) is a {2, 3}-group, so CO3 (I) (z) = 1. This implies that z inverts O3 (I), so I = [I, z] centralizes O3 (I), completing the proof.  Lemma 6.4. Suppose that K is terminal in G and let J be as in Lemma 6.3. Let t ∈ I2 (SK ). Then J is a component of CG (t), S ∈ Syl2 (CG (t)), and [J, K] = 1. Proof. Since all involutions of K are conjugate, and CK (c) is a Frobenius group of order p(p − 1), we may assume that [t, c] = 1, whence t inverts a. As t normalizes a, c = A and centralizes Q ∈ Syl2 (J), t normalizes J. Let s ∈ NJ (Q) be of order 2p − 1. Then s normalizes E(CG (Q)) = K and centralizes a. But CAut(K) (a) is a p-group, so [K, s] = 1. In particular [t, s] = 1. As CAut(J) (Q s) = 1, [J, t] = 1. It suffices to show that J is a component of CG (t). For then, as z ∈ J, E(CG (t)) ≤ JCG (t, z) so z is 2-central in E(CG (t)) t. Consequently |E(CG (t))) t |2 ≤ |NG (K)|2 = |Q|2 = |K|22 = |J|22 . Hence J is the unique component of CG (t) of its isomorphism type, so J  CG (t). Then by a Frattini argument, CG (t) = JNCG (t) (Q) = JCNG (K) (t). As S contains Q ∈ Syl2 (J), S ∈ Syl2 (CG (t)). Moreover, as t ∈ Z(SK ), SK normalizes J and maps into CAut(J) (Q s) = 1. Thus we have [SK , J] = 1 and [a, J] = 1. But K = SK , a, so [K, J] = 1, as desired. To show that J is a component of CG (t) we first let J ∗ be the subnormal closure of J in CG (c). Since z ∈ J ≤ J ∗ and m2 (CG (z)) = 2p, and since Op (CG (c)) has odd order, J ∗ is a single p-component and J ∗ /Op (J ∗ ) ∈ Cp , by Theorem C∗4 : Stage A1. By [V17 , 7.2], J ∗ /Op (J ∗ ) ∼ = L2 (8), Sp4 (8), or 3D (2) if p = 3, and 2B (2 52 ) or 2F (2 25 ) if p = 5. Since a  C (c), a ≤ 4 2 4 K O2 (CG (c, z)). But if J ∗ = JOp (J ∗ ), then a induces an inner automorphism on J ∗ /Op (J ∗ ) and hence a ∈ O2 (CJ ∗ (z)). However, J ∗ /Op (J ∗ ) ∈ Chev(2), so O2 (CJ ∗ (z)) ≤ Op (J ∗ ) by the Borel-Tits theorem. Hence a ∈ Op (J ∗ ); but a has order p, contradiction. It follows that J ∗ = JOp (J ∗ ). Now Op (CG (c)) has odd order and O{2,p} (CG (c, z)) = O{2,p} (CNG (K) (c, z)) = 1.

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As NCG (c) (Q) permutes Ω1 (Q)# transitively, we get Op (CG (c)) = 1 and J  CG (c). Therefore CG (c) = JCCG (c) (z) = (J × c) a, τ , where a, τ  = CK (c) is a Frobenius group of order p(p − 1) and t ∈ τ  = CS (c) ∼ = Zp−1 . Then CCG (t) (c) = CCG (c) (t) = J c, τ  ≥ J × c, t. Thus if we set V = O2 (CG (t)), then t ∈ CV (c) = τ . Suppose that [V, J] = 1. Then J acts faithfully, and c fixed-point-freely, on V ∗ = [V t /Φ(V ), c]. As [J, c] = 1, but the simple group J has elements of order p2 and so cannot 2 embed in SLp (2p−1 ), we must have |V ∗ | ≥ 2p −1 . Now J ∩ V = 1. Thus 2 as z ∈ V , |CV ∗ z (z)| > 2(p −1)/2 . On the other hand, using [IG , 9.16], |CV ∗ z (z)| ≤ |CV z (z)| = 2|CV (z)| ≤ 2|CG (z)|2 /|CJ (z)|2 . As |CG (z)|2 = 2 |Q|2 and |CJ (z)| = |Q|, we conclude that 2|Q| > 2(p −1)/2 . Hence according as p = 3 or 5, 24 > 24 or 211 > 212 , contradiction. This proves that [V, J] = 1. Since e(G) = 3 and G is of even type, it follows easily that J ≤ X := E(CG (t)). Hence, J is a component of C Lc  (c) for some component L of X.     In particular, z ∈ J ≤ Lc . Thus Lc = L, for otherwise Lc ∼ = J p and 2 m2 (CG (z)) ≥ p , contradicting m2 (CG (z)) = 2p. If [L, c] = 1, then J b× c acts faithfully on L and has p-rank 3, contradicting (3A). Hence, [L, c] = 1,  whence J = L   CG (t). This completes the proof of the lemma. Lemma 6.5. Assume (6A). Then K is not terminal in G. Proof. Suppose that K is terminal in G. We have S = Q × SK , z ∈ Z(S) = Z(Q) × Z(SK ) ∼ = E22p , and CG (Z(S)) = CNG (K) (Z(S)) = S. Set V = Z(S) and X = NG (S)/S, so that X acts faithfully on V . For any g ∈ NG (S), let g be its image in X. Put V0 = Z(Q) and V1 = Z(SK ). Take y ∈ NK (SK ) of order 2p − 1. Then [Q, y] = 1 and so y ∈ NG (S) and y ∈ X. Observe that Y := y acts transitively on V1# and trivially on V0 , and Y  NX (V0 ). Indeed for any g ∈ NG (S) such that Z(Q)∩Z(Q)g = 1, g ∈ NNG (K) (S) ≤ NNG (K) (Z(Q)). That is, if we let Ω be the set of X-conjugates of V0 in V , then distinct elements of Ω intersect trivially. Hence by definition [IG , 14.1], (X, V ) is a Goldschmidt-O’Nan pair. Then by O’Nan’s theorem [IG , 14.2], one of the following holds: (6E)

(1) V0 is X-invariant; (2) V0X = {V0 , V1 }; and (3) V1 is X-invariant, V − V1 = ∪g∈X V0g , and |Ω| = 2p .

Assume first that S ∈ Syl2 (G). Then |X| is odd. Since |X| is odd and X is transitive on Ω, |Ω| is odd, so (6E1) holds. But recall that as K is terminal, there is u ∈ G − NG (K) such that Qu ≤ S and Qu acts faithfully on K. In particular, Ω1 (Q) = Ω1 (Q)u . By Burnside’s lemma (cf. [IG , 16.2]),

220 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

there exists g ∈ NG (S) such that Qu = Qg . In particular, V0g = V0 , and (6E1) is contradicted. Therefore there is g ∈ NG (S) − S such that g 2 ∈ S. If Q ∩ Qg = 1, then K = K g and so g ∈ NG (K) which is not possible as S ∈ Syl2 (NG (K)). In particular, (6E1) does not hold. Assume that (6E3) holds. Then V1g = V1 , and so (because of the action of Y on V1# ), we may assume that g ∈ CG (t), a contradiction as S ∈ Syl2 (CG (t)) by Lemma 6.4. Hence (6E2) holds, and there exists x ∈ X such that V0x = V1 , and thus g ∈ NG (S) with Qg = SK . Then K g = E(CG (Q))g = E(CG (Qg )) = E(CG (SK )) = J. But this is a contradiction as K is standard and [K, J] = 1. The proof is complete.  Lemma 6.6. There exists t ∈ I2 (Q1 ) such that K has a nontrivial pumpup in CG (t). Moreover, if Q1 is neither dihedral nor semidihedral, such an involution t exists in Z(Q1 ). 5 Proof. Since K ∼ = L2 (8) or 2B2 (2 2 ), K is semirigid in G by [V17 , 13.8]. If Q1 is neither dihedral nor semidihedral, then the result holds by Lemma 6.5 and [IG , 7.4]. So assume that Q1 is dihedral or semidihedral. If |Q1 | = 4, then as z ∈ Z(C(z, K)), C(z, K) has a normal 2-complement; but O2 (CG (z)) = 1, so |C(z, K)| = 4, a contradiction as B ∩ C(z, K) = 1. Thus |Q1 | > 4 so Z(Q1 ) = z. As Q1 = CQ (z), Q1 = Q. As K has a nontrivial pumpup in CG (t) for some t ∈ I2 (Q), the lemma follows. 

Lemma 6.7. If L is a nontrivial pumpup of K, then LLz is isomorphic to one of the following groups: L2 (8) × L2 (8), L2 (26 ), L3 (8), or U3 (8), if 5 5 p = 3, and 2B2 (2 2 ) × 2B2 (2 2 ) or Sp4 (25 ), if p = 5. Proof. Suppose that L is a vertical pumpup of K. Then the desired conclusion holds by [V8 , 3.86] and [V17 , 13.5], unless p = 3 and L ∼ = G2 (3).  But that case is impossible as m3 (G2 (3)) = 4 > 3 = e(G) [IA , 3.3.3]. Lemma 6.8. If L is a vertical pumpup of K in CG (t) for some t ∈ Q1 , then there is no x ∈ Ipo (G) ∩ CG (t) such that [L, x] = 1. Moreover, mp (CG (t)) ≤ 2. Proof. Assume that such t, L, and x exist with [L, x] = 1. Then m2 (L t) ≥ 4 by Lemma 6.7 and [IA , 3.3.3]. Hence Lemma 2.4 (with A = x) implies that there is a p-component J of CG (x) such that L is a component of CJ (t). Now [V17 , 14.1 and 14.2], with L in the role of Y there, give us a contradiction. If mp (CG (t)) ≥ 3, then there is Ep3 ∼ = B0 ≤ CG (t). Because of (3A), # there exists x ∈ B0 such that [L, x] = 1. By Theorem C∗4 : Stage A1,  x ∈ Ipo (G), contrary to the previous paragraph. The lemma is proved. 5

For the rest of this section, we set K0 = L2 (8) if p = 3, and K0 = 2B2 (2 2 ) if p = 5. Lemma 6.9. If K1 (K0 ) = ∅, then K2 (K0 ) = ∅.

√ 6. THE L2 (8) AND 2B2 ( 32) CASES

221

Proof. Assume the contrary. Let (z, K) ∈ K1 (K0 ). Then by definition of K2 (K0 ), Q1 is dihedral or semidihedral. Since G is of even type, K = E(CG (z)) and Q0 = F ∗ (C(z, K)). Then b acts nontrivially on Q0 , and so Q0 ∼ = L2 (8), and Q1 ∼ = Q8 . Hence, p = 3, K ∼ = SD16 . As | Out(K)| = 3 while m3 (CG (z)) = 3, there exists an element c ∈ CG (z) of order 3 such that CG (z) = Q1 b × K c with K c ∼ = Aut(L2 (8)). Hence, a nontrivial pumpup of K occurs in CG (u) for some u ∈ I2 (Q1 − Z(Q1 )). Let L be this pumpup. If L = L1 × L2 with Li ∼ = L2 (8), then (u, L1 ) ∈ K2 (L2 (8)), contrary to assumption. Hence, by Lemma 6.7, L ∼ = U (8), then m (L c) = 3, which is not L2 (26 ), L3 (8) or U3 (8). If L ∼ = 3 3 possible by (3A). Hence, we are in one of the first two cases, and CL (c) z ∼ = Σ5 or Aut(L3 (2)) correspondingly. Let C = CG (c), C = C/O3 (C) and X = O3 (C). Since, by Lemma 6.8, m3 (CG (u)) = 2 = m3 ( c × CL (c)), we have that CX (u) is cyclic. As O 2 (C(z, c)) = Q0 b × c, a, with a ∈ CL (c), we have CX (z) = c. As z normalizes CX (u), we have CX (u) = CX (z) = c. Hence, [X, uz] = 1 and so as CL (c) = [CL (c), uz], we have [CL (c), X] = 1. Therefore F ∗ (C) = XE(C) with E(C) = 1, and CL (c) is a component of CJ (u) for some 3-component J of CG (c). By [V17 , 14.1, 14.2], with CL (c) in the role of Y there, J ∼ = A9 . Since e(G) = 3, J = E(C) and m3 (X) = 1. As O3 (CG (c)) has odd order by Theorem C∗4 : Stage A1, we conclude that O2 (CJ (z)) = Q0 b × a, which is not the case for any  involutory automorphism of A9 , a final contradiction. Lemma 6.10. Let (z, K) ∈ K2 (K0 ) and suppose that x ∈ Ipo (G) ∩ C(z, K). Suppose that (z, K) < (t, L) ∈ IL2 (G) is a nontrivial pumpup and [x, t] = 1. Then LLz is a diagonal pumpup of K in CG (t), (t, L) ∈ K3 (K0 ), and [LLz , x] = 1. Proof. Suppose to the contrary that L is a vertical pumpup of K in CG (t). Then L ∼ = L2 (26 ), U3 (8) or L3 (8), if p = 3, and Sp4 (25 ) if p = 5. By Lemma 6.8, [L, x] = 1. But then [K, x] = 1, a contradiction.  Lemma 6.11. Let (z, K) ∈ K2 (K0 ) and suppose that x ∈ Ipo (G) ∩ CG (z) induces a nontrivial field automorphism on K. Suppose that (z, K) < (t, L) ∈ IL2 (G) is a nontrivial pumpup and [x, t] = 1. Then LLz is a diagonal pumpup of K in CG (t), (t, L) ∈ K3 (K0 ), and x induces a field automorphism on L. Proof. Suppose not. Then, by Lemma 6.7, L ∼ = L2 (26 ), L3 (8) or U3 (8), 5 if p = 3, or Sp4 (2 ), if p = 5. Since x ∈ CG (t) normalizes K, x normalizes L, and its action on K implies that it must act as a field automorphism on L. If L ∼ = U3 (8), 3 = m3 (L x) ≤ m3 (CG (t)), which contradicts Lemma 6.8. Hence, L ∼ = L2 (26 ) or L3 (8), or Sp4 (25 ). Let Lx := E(CL (x)). Then ∼ Lx = A5 or L3 (2), or A6 , and Lx z ∼ = Σ5 or Aut(L3 (2)), or P GL2 (9), correspondingly. In particular, D2p ∼ = CLx (z) =: Lxz ≤ CL (z) = K. Thus, Lxz has index 1 or 2 in CK (x), and Lxz  CG ( x, z).

222 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

Let Cx = CG (x) and C x := Cx /Op (Cx ). Then F ∗ (C x ) = Op (C x )E(C x ). Assume first that [Lx , Op (C x )] = 1. By [V9 , 2.8], Lx ≤ E(C x ). We conclude from [V17 , 14.1, 14.2], with Lx in the role of Y there, that there exists a component J of C x such that (p, L, Lx , J) = (3, L2 (26 ), A5 , A9 ) or (5, Sp4 (25 ), A6 , A10 or HS). In all cases Lx is a component of CJ (t). Now, D2p ∼ = Lxz  CJ (z). Hence, we may assume that p = 3 and z acts as (12)(34)(56) in J z ∼ = Σ9 . If y is an involution in Lxz , then y ∈ Lx and so y is a product of two transpositions in J . On the other hand, as Lxz  CG ( x, z), we must have y ∈ z × Σ{7,8,9} ≤ J, whence y is a transposition or a product of four transpositions, a contradiction.  Hence t ×Lx z acts faithfully on Op (C x ). Let X be a critical subgroup of Op (C x ) of exponent p with x ∈ X. By Lemma 6.8, mp (CG (t)) ≤ 2. Hence, t inverts X/ x. Also, mp (CC(z,K) (x)) = 1 and CK (x) ≤ CL (x), whence mp (CX (z)) ≤ 2 and x ∈ [CX (z), CX (z)]. Assume first that Lx acts nontrivially on Z(X). Then mp (Z(X)) ≥ 5, whence mp (CX (tz)) ≥ 4 > e(G), a contradiction. Therefore [Lx , Z(X)] = 1. Choose X1 to be an Lx zinvariant complement to [X, X] in Z(X). Then X0 := X/X1 is extraspecial with xX1 ∈ Z(X0 ) and Lx acts faithfully on X0 . As mp (CX (z)) ≤ 2 and x ∈ [CX (z), CX (z)], dim(CX0 /xX1  (z) ≤ 1. As this dimension is even, z inverts X0 / xX1 , asdoes  t, whence tz centralizes X0 / xX1 , contrary to  the faithful action of t × Lx z, completing the proof. Lemma 6.12. The following conditions hold: (a) Let (z, K) ∈ K2 (K0 ). Then there is a diagonal pumpup (z, K) < (t, L) ∈ IL2 (G) for some (t, L) ∈ K3 (K0 ); (b) In (a), if (z, K) ∈ K3 (K0 ), then |C(t, L)|2 > |C(z, K)|2 ; and (c) If K1 (K0 ) = ∅, then K3 (K0 ) = ∅. Proof. By Lemma 6.9, (c) follows from (a). Hence it suffices to prove (a) and (b). Let (z, K) ∈ K2 (K0 ). Again by Propositions 7.3 and 7.4 of [IG ], (6F)

Q1 = z × P1

for some P1 such that K has a nontrivial pumpup in CG (y) for every y ∈ I2 (P1 ). One consequence of this is that if H is a perfect subgroup of C(z, K), then K has a nontrivial pumpup in CG (y) for every y ∈ I2 (H). Indeed, to check this we may assume that Q1 ∩H ∈ Syl2 (H). Then |Q1 ∩H : P1 ∩H| ≤ 2 and by the Thompson transfer lemma, y is H-conjugate into P1 ∩ H. Let a ∈ Ip (K). Then a ∈ Ipo (G). Assume first that E(CG (z)) = L1 K and mp (L1 ) = 2. Then L1  CG (z), as e(G) = 3. If m2 (L1 z) ≤ 3, then by Lemma 3.4 and Proposition 3.8, L1 ∼ = 2HJ and z ∈ Z(L1 ) ≤ [Q1 , Q1 ], contrary to (6F). Thus m2 (L1 z) ≥

√ 6. THE L2 (8) AND 2B2 ( 32) CASES

223

4. By Lemma 2.4 (with A = a), L1 is a component of CJ (z) for some p-component J of CG (a). Then, as usual, L1 is given (as Y ) in [V17 , 14.1, 14.2]. Using the ranks given in [V17 , 9.2], and keeping Propositions 3.8 and 5.1 in mind, we see that if p = 3, then L1 ∼ = 3D4 (2) or Sp4 (8), while if p = 5, 5 then L1 ∼ = 2F4 (2 2 ), HJ, HS, 2HS, or Ru. By [V17 , 17.1] and the tables [IA , 5.3gmr], there exists a 2-central involution t ∈ L1 with mp (CL1 (u)) > 0. Let x1 ∈ Ip (CL1 (u)). Since mp (L1 ) = 2 and mp (K) > 0 (Lemma 2.3), x1 ∈ Ipo (G) by Theorem C∗4 : Stage A1. As t ∈ L1 = [L1 , L1 ], K has a nontrivial pumpup in CG (t), which is K1 ×K1z by Lemma 6.10, with K1 ∼ = K. In particular mp (CG (t)) ≥ mp (K1 K1z CL1 (t)) ≥ 3. Let Q2 ∈ Syl2 (C(t, K1 )). Since t is 2-central in L1  C(z, K) and | Out(L1 )|2 ≤ 2, 1 1 |Q2 | ≥ |K1z |2 · |CC(z,K) (t)|2 ≥ 8 · |Q1 | = 2|Q1 |. 2 4 Thus all the conclusions of Lemma 6.12 hold in this case. Suppose next that E(CG (z)) = L1 L2 K. Then mp (L1 ) = mp (L2 ) = 1. If Li  C(z, K) for i = 1, 2, let ai ∈ Ip (Li ) ⊆ Ipo (G) and let ti ∈ Q1 ∩ Li be a 2-central involution of C(z, K). Let t = t1 . Then [t, a2 ] = 1 and t ∈ Z(Q1 ). Again as t ∈ L1 = [L1 , L1 ], K has a nontrivial pumpup in CG (t), which, thanks to a2 and Lemma 6.10, is K1 × K1z with K1 ∼ = K. As t is 2-central in C(z, K), |C(t, K1 )|2 ≥ |K1z |2 · 12 |C(z, K)|2 > |C(z, K)|2 , and so the lemma holds in this case. If, however, L1  C(z, K), then L1 ∼ = L2 and there exists u ∈ Q1 such that u2 normalizes both L1 and L2 , Lu1 = L2 , and Q1 = NQ1 (L1 ) u. This time take t ∈ I2 (Q1 ∩ L1 ) to be 2-central in NQ1 (L1 ). Using a2 ∈ Ip (L2 ) and Lemma 6.10 as before, we again get a diagonal pumpup K1 × K1z of K in CG (t), and this time |C(t, K1 )|2 ≥ |K1z |2 · 12 |NQ1 (L1 )|2 ≥ 8 · 14 |Q1 | > |Q1 |, as desired. Next, suppose that E(CG (z)) = L1 K and mp (L1 ) = 1. Let a1 ∈ Ip (L1 ). Since mp (CG (z)) = 3, there exists c ∈ Ip (CG (z)) such that B := a, a1 , c ∼ = Ep3 , and either c induces a nontrivial field automorphism on K or [c, K] = 1 [V17 , 3.1d]. Replacing c by a C(z, K)-conjugate if necessary, we may choose t ∈ I2 (CL1 (c) ∩ Z(Q1 )), which is possible by [V17 , 9.8]. Since t ∈ L1 = [L1 , L1 ] and by Lemma 6.10 or Lemma 6.11, the pumpup of K in CG (t) is K1 × K1z , with mp (K1 K1z c) ≥ 3. Thus, (t, K1 ) ∈ K3 (K0 ). As t ∈ Z(Q1 ), |C(t, K1 )|2 ≥ |K1z |2 |Q1 |/2 > |Q1 |, as required. Finally, suppose that E(CG (z)) = K. Then (z, K) ∈ K3 (K0 ) so we need only verify (a). Since mp (CG (z)) = 3, we can find E = a, b, c ∼ = Ep3 with E ≤ CG ( a, z), b ∈ CG (K) and either [c, K] = 1 or c inducing a nontrivial field automorphism on K. Let A := b, c. Let P0 := Q0 ∩ P1 , so that Q0 = z × P0 . Choose x0 ∈ A# so that |CQ0 (x0 )| is as large as possible. Then as Q0 = CQ0 (x) | 1 = x ∈ A > z, there exists t ∈ I2 (P0 ) such that [t, x0 ] = 1 for some p-element 1 = x0 ∈ A and K pumps up nontrivially to L in CG (t). By Lemmas 6.10 and 6.11, the pumpup of K in CG (t) is K1 × K1z , K1 ∼ = K. Also mp (CG (t)) ≥ mp (K1 K1z x0 ) = 3, so (t, K1 ) ∈ K3 (K0 ). The proof is complete. 

224 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

Now we can finish this section. Lemma 6.13. Conditions (6A) do not occur. Proof. Suppose that they do occur and continue the above argument. Then K1 (K0 ) = ∅, so by Lemma 6.12c, K3 (K0 ) = ∅. Choose (z, K) ∈ K3 (K0 ) with |C(z, K)|2 maximal. By Lemma 6.12ab, there is (t, L) ∈  K3 (K0 ) such that |C(t, L)|2 > |C(z, K)|2 , contradiction. Thus we have proved: Proposition 6.14. If p = 3, then K1 (L2 (8)) = ∅. If p = 5 and 5 K1 (2B2 (2 2 )) = ∅, then K1 (A5 ) = ∅. 7. Some Cases of p-Rank 2 In this section we continue the proof of Theorem 1. We work under the following hypothesis: (7A)

(z, K) ∈ K(G), and (p, K) is (3, 3D4 (2)), (3, Sp4 (8)), or 5 (5, 2F4 (2 2 )).

Under this assumption, by Proposition 5.1, m2 (Q) ≥ 2, K is terminal, and there exists b ∈ Ip (C(z, K)) such that K is a component of CG (b). Now set C = CG (K) and M = NG (K). We study the structure of C. We have Q ∈ Syl2 (C) and ΓQ,1 (G) ≤ M. Lemma 7.1. The following conditions hold: (a) K is a component of CG (x) for every x ∈ Ip (C); (b) O2 (C) = 1 = O2 (C); (c) F ∗ (C) = L1 × · · · × Ln , n ≥ 1, where each Li is nonabelian simple, 1 ≤ i ≤ n; and (d) mp (C) = 2. Proof. Let R0 ∈ Sylp (C). If Ω1 (Z(R0 )) = b, then for every x ∈ Ip (R0 − b), K is a component of CG ( x, b). The pumpup Jx of K in CG (x) is not diagonal since e(G) = 3, and so it is a trivial pumpup by [V17 , 7.22b], i.e., Jx /Op (Jx ) ∼ = K. By Lemma 2.6 (and [V17 , 9.2]), Jx = K, as desired. On the other hand if Ω1 (Z(R0 )) = b, take x ∈ Ip (Z(R0 )− b). By the same argument, K is a component of CG (x). Now take any y ∈ Ip (R0 − Z(R0 )), and consider x, y. By the same argument, K is a component of CG (y). Hence, (a) holds. As e(G) = 3, mp (K) = 2 and m2,p (K) > 0, we have K  CG (b). Now  Op (C(b, K)) ≤ Op (CG (b)), which has odd order by Theorem C∗4 : Stage A1. As z ∈ C(b, K), either z inverts some element of Ip (Op p (C(b, K))) or C(b, K) has a p-component. In both cases there exists Ep2 ∼ = A0 ≤ C. As m2,p (K) > 0, mp (C) = 2 and (d) holds.

7. SOME CASES OF p-RANK 2

225

  Since m2 (Q) ≥ 2, O2 (C) = CO2 (C) (y) | y ∈ I2 (Q) . But for any y ∈ I2 (Q), K  CG (y) by terminality of K, so CO2 (C) (y) ≤ O2 (C(y, K)) ≤ O2 (CG (y)) = 1, asG is of even type. Hence,  O2 (C) = 1. Now O2 (C) = CO2 (C) (x) | x ∈ Ip (A0 ) , where again A0 ∈ Ep2 (C). But for any x ∈ Ip (A0 ), CO2 (C) (x) ≤ O2 (C(x, K)) ≤ O2 (CG (x)) = 1 by Theorem C∗4 : Stage A1. Hence (b) holds, and (c) is an immediate consequence.  Now fix A ≤ K with A ∼ = Ep2 and such that there is x0 ∈ A# such that 5 E(CK (x0 )) ∼ = L2 (8) or 2B2 (2 2 ), according as p = 3 or 5. (See [IA , 4.8.2, 4.7.3A] and [III11 , Table 13.1].) We set L = F ∗ (C) and CA = CG (A). Lemma 7.2. The following conditions hold: (a) Q ∈ Syl2 (CA ); (b) O p (CA ) = Op (C); and  (c) L = E(C) = O p (E(C)) = E(CA ). Proof. By [V17 , 7.22a], CAut(K) (A) is a p-group. Let u be a 2-element of NCA (Q). Then u ∈ Op (NG (K) ∩ CA ) ≤ C, by the previous paragraph. As Q ∈ Syl2 (C), u ∈ Q, and (a) holds. Next, let Γ = ΓQ,1 (CA ), C. Then as K is standard, Γ ≤ NCA (K). Since CAut(K) (A) is a p-group, Op (Γ) ≤ C. As C ≤ Γ, it follows that O p (Γ) = Op (C). Thus to prove (b) it suffices to show that Γ = CA . To show this we may assume that ΓQ,1 (CA ) < CA . But Q ∈ Syl2 (CA ) by (a), and m2 (Q) > 1. Hence, by the Bender-Suzuki theorem [II2 , Theorem  SE], O2 (CA /O2 (CA )) is a simple Bender group and ΓQ,1 (CA ) is a solvable maximal subgroup of CA . But F ∗ (C) is nonsolvable by Lemma 7.1c, so Γ = CA , as desired. Hence, (b) holds. It follows that E(C) = E(CA ) and Op (C) = Op (CA ). The latter group is of odd order by Theorem C∗4 : Stage  A1, so E(C) = O p (E(C)), implying (c). The lemma is proved.  Lemma 7.3. One of the following holds: ∼ 2B2 (2 52 ) × A5 ; (a) p = 5 and L = 1 5 (b) p = 5 and L ∼ = L2 (52 ), 2F4 (2 2 ) , or 2B2 (2 2 ); or ∼ L2 (8), A6 , M11 , or L± (3). (c) p = 3 and L = 3 Proof. By Lemmas 7.1c and 7.2c, 

L = L1 × · · · × Ln = Op (L), and so mp (L) ≥ n. Hence by Lemma 7.1d, n ≤ 2 ≤ mp (Aut(L)). Moreover, each Li is a component of E(C) = E(CA ) and so lies in Cp . Suppose that n = 2. Then mp (L1 ) = mp (L2 ) = 1. If p = 3, then by [V17 , 3.5], L1 ∼ = L2 ∼ = L2 (8). But then as m7 (K) = 2, m2,7 (KL1 L2 ) = 4 > e(G),

226 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS 5 contradiction. Therefore p = 5, whence Li ∼ = 2B2 (2 2 ) or A5 by [V17 , 3.5] 5 (and K ∼ = L2 , then accordingly m2,31 (KL1 L2 ) = 4 or = 2F4 (2 2 )). If L1 ∼ L2 and (a) m2,3 (KL1 L2 ) = 4, again giving contradictions. Hence L1 ∼ = holds in this case. Hence, we may assume that n = 1. Of course as [K, L] = 1, mr (L) ≤ 5 e(G) = 3 for all odd r. Also, if p = 5, then since m3 (K) = m3 (2F4 (2 2 )) = 2, we have m2,3 (L) ≤ e(G) − 2 = 1. Thus, if p = 5, then by [V17 , 7.21], 5 5 2 ∼2 2 2 either (b) holds or L ∼ = L± 3 (5) or F4 (2 ). But if L = F4 (2 ), then again m2,31 (KL) = 4, contradiction. If L ∼ = L± 3 (5), then for any involution u ∈ L, ∼ Lu := E(CL (u)) = SL2 (5). But K is standard in G, so K  CG (u) and then CL (u)  CG (u). Hence Lu is a component of CG (u), so Lu ∈ C2 , a contradiction. Therefore, if p = 5 (and n = 1), (b) holds. We may therefore assume that p = 3 and (c) fails. Again [V17 , 7.21] 1 applies and yields that L ∼ = 2F4 (2 2 ) , 3D4 (2), or Sp4 (8). In particular |L|2 ≥ 211 . Now by [V17 , 5.8], K is outer well-generated for p = 2. Since m2 (Q) > 1, it follows from [II3 , Theorem PU2 ] and [V9 , 7.6, 7.3] that there exists g ∈ G−M such that Qg ≤ M and ΓQg ,1 (K) < K. In particular CQg (K) = 1; but then ΓQg ,1 (K) = K by [V17 , 5.22], contradiction. This completes the proof.  Recall that A ∈ Ep (K) was chosen to contain x0 such that E(CK (x0 )) ∼ = 5

2

L2 (8) or 2B2 (2 2 ), according as p = 3 or 5. Fix t ∈ I2 (E(CK (x0 ))) and set Ct = CG (t), so that C ≤ Ct ∩ C0 , where we have put C0 = CG (x0 ). Finally, fix PC ∈ Sylp (C), so that mp (PC ) = 2 by Lemma 7.1d. Note also that 

O2 (CAut(K) (t)) ≤ Inn(K). Indeed, Out(K) has odd order unless K ∼ = Sp4 (8), in which case | Out(K)|2 = 2, and by choice, t is a short root involution or a long root involution. But any involution in Out(K) interchanges the K-conjugacy classes of short and long root involutions, so (7B) holds. (7B)

Remark 7.4. For later use we remark also that by [V17 , 10.5.7], if p = 5, then t is 2-central in K. Furthermore, in every case CG (t) ≥ x0  C so mp (CG (t)) ≥ 3. Lemma 7.5. We have O2 (Ct ) ≤ K. Proof. Set W = O2 (Ct ) and again let A0 ∈ Ep2 (C). By Lemma 7.1a and the standardness of K, K  CG (a) for all a ∈ A# 0 . Hence CW (a) maps  2 into O (CAut(K) (t)) and hence CW (a) ≤ K × C by (7B). As CW (a) ≤ O2 (CG ( t, a)) and O2 (CC (a)) = 1 by [V17 , 4.26] and Lemma 7.3, CW (a) ≤ K. Therefore W ≤ K, as claimed. 

7. SOME CASES OF p-RANK 2

227

Lemma 7.6. Suppose that L is a single component. Then the following conditions hold: (a) There exist a component Kt of Ct and a component J of C0 such that L ≤ Kt ≤ J; and (b) ΓPC ,1 (G) ≤ M , so L  L, ΓPC ,1 (G). Proof. Choose A0 ∈ Ep2 (PC ) such that mp (A0 ∩ L) = mp (L). Then CA0 L (L) = 1 and [A, A0 ] = 1. By Lemma 7.2c, L is a p-component of CA . By Lp -balance, and the facts that e(G) = 3 and CA0 L (L) = 1, the subnormal closure J of L in C0 is a single p-component. By Lemma 7.5, [O2 (Ct ), A0 L] = 1. Thus CA0 L (E(Ct )) = 1, and since e(G) = 3 we conclude that L ≤ E(Ct ). Indeed mp (Ct ) ≤ e(G) = 3, and B0 := x0  A0 ∈ Ep3 (Ct ). Since [A0 L, x0 ] = 1 and mp (Ct ) ≤ 3, L lies in a single component Kt of Ct . Now by our hypothesis (3A), CB0(Kt ) = 1. As CB0 (L) = x0 , we conclude that Kt ≤ C0 . Then Kt = LKt ≤ J. Hence to prove (a) it remains to show that J is quasisimple. In (b), for any 1 = P0 ≤ PC , K is a component of CG (P0 ) by Lemma 7.1a, so as K is standard, K  NG (P0 ). Therefore NG (P0 ) ≤ NG (K) = M . As L = E(C)  M , (b) holds. As a result, since PC is noncyclic, Op (C0 ) ≤ Op (C0 ). As Op (L) = 1, ΓPC ,1 (G) ≤ M . But [L, x0 ] = 1 soL normalizes  J L centralizes Op (C0 ), whence J = L does as well, and J is quasisimple. The proof is complete.  Now we can prove Lemma 7.7. We have p = 5. Proof. Suppose false, so that p = 3, J/O3 (J) ∈ C3 , and L is given by Lemma 7.3c. If Kt = L, then in view of Remark 7.4, either Proposition 3.8 or Proposition 6.14 is contradicted. Hence L < Kt and by Lemma 7.6a, L < J. Now [V17 , 7.23] applies. As we know that L  L, ΓPC ,1 (J), that lemma implies that e(G) ≥ e(J x0 ) > 3, a contradiction. This completes the proof.  Similarly we obtain 5 Lemma 7.8. If L is simple, then L = Kt = J ∼ = 2B2 (2 2 ).

Proof. Now, the isomorphism type of L is given by Lemma 7.3b, and again J/O5 (J) ∈ C5 . Suppose first that L < J. By [V17 , 7.23], either L  L, ΓPC ,1 (J) or e(G) ≥ e(J x0 ) > 3, contradicting our general hypothesis or Lemma 7.6b. Therefore L = Kt = J. L2 (52 ). By Remark 7.4, m5 (CG (t)) ≥ 3, so In particular L ∈ Co2 , so L ∼ = 1 5 2F4 (2 2 ) . Therefore Kt ∼ (t, Kt ) ∈ K(G). By Proposition 5.1, Kt ∼ = = 2B2 (2 2 ), as claimed. 

228 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS 5 Lemma 7.9. If L is not simple, then 2B2 (2 2 ) × A5 ∼ = L  Ct . 5 Proof. By Lemma 7.3, L = L1 × L2 with L1 ∼ = 2B2 (2 2 ) and L2 ∼ = A5 . Write A0 = a1  × a2  with ai ∈ I5 (Li ), i = 1, 2. Since CG (ai ) ≤ M by Lemma 7.6b, L3−i is a component of CCt (ai ), i = 1, 2. By Lemma 7.5, [O2 (Ct ), L] = 1. As e(G) = 3 it follows that L ≤ E(Ct ), and thus by L5 -balance, L3−i lies in some component H3−i of Ct . If H1 = H2 , then as ai ∈ Hi , Hi = Li , i = 1, 2, and the lemma holds. So we may assume 5 that H1 = H2 . Then A5 ↑5 H1 and 2B2 (2 2 ) ↑5 H1 . The latter implies that 2 ∼ 2B2 (2 52 ) or 2F4 (2 52 ), whence by [III11 , Table 13.1], it is false that H1 =  A5 ↑5 H1 . This contradiction completes the proof of the lemma.

Summarizing, Lemmas 7.7, 7.8, and 7.9, together with Remark 7.4, establish the following proposition. Proposition 7.10. Let (z, K) ∈ K(G). If p = 3, then K ∼ 3D4 (2) or Sp4 (8). = 5 If p = 5 and K ∼ = 2F4 (2 2 ), then there is a 2-central involution t ∈ K 5 and a subgroup Kt  E(CG (t)) such that Kt = E(CG (K)) ∼ = 2B2 (2 2 ) or 2B (2 25 ) × A . Moreover, (t, L) ∈ K(G) for every component L of K . 2 5 t 8. Configurations Involving M12 and HJ Our next goal is to rule out the possibility that for some (z, K) ∈ K(G), K∼ = L2 (q), q ∈ {5, 7, 17}. When q = 5 and p = 5, this possibility leads to configurations involving M12 and HJ that we shall rule out in the present section. First of all, we rule out (z, K) ∈ K(G) such that K ∼ = HJ. The argument is similar for K ∼ = Ru or K/Z(K) ∼ = HS, so we do all three in one lemma. Proposition 8.1. Let (z, K) ∈ K(G). If p = 5, then K ∼ HJ or Ru, = and K/Z(K) ∼ HS. = Proof. Assume the contrary. As C(z, K) contains some x of order p, m2 (Q) ≥ m2 (Q1 ) ≥ 2. As usual, K has no diagonal pumpups, as m3 (K) > 1; and K is then terminal in G, since ↑2 (K) ∩ C2 ∩ L2 (G) = ∅ by [V17 , 13.6a]. As K is terminal, by [II3 , Corollary P U2 ] and [V9 , 7.3] there exists E22 ∼ = U ≤ Q and g ∈ G − NG (K) such that U g ≤ NG (K) =: M and  ΓU g ,1 (K) < K. Suppose that K ∼ = HJ or Ru. By [V17 , 5.8], K is outer well-generated for the prime 2. Hence, applying [V9 , 7.6, 7.3], we may assume that Qg ≤ M , and from [V17 , 5.5] we obtain that Q ∼ = E22 . If CG (z) has a component K1 = K, then m2 (Q) ≥ m2 (K1 z) ≥ 3, a contradiction. Therefore, K = E(CG (z)) and x acts nontrivially on Q0 ≤ Q ∼ = E22 , which is obviously not possible. Thus, K/Z(K) ∼ = HS. But then by [V17 , 5.3], ΓU g ,1 (K) = K, a final contradiction. 

8. CONFIGURATIONS INVOLVING M12 AND HJ

229

Now the main result of this section concerns the following configuration.

(8A)

(1) (2) (3) (4)

p = 5; (z, K) ∈ K(G) with K ∼ = A5 ; Q ∈ Syl2 (CG (K)) with z ∈ Q; u ∈ I2 (Q) and L is the subnormal closure of K in CG (u) with L/Z(L) ∼ = M12 or HJ; (5) L is terminal in G and M = NG (L); and (6) u is 2-central in CG (L).

We shall prove Proposition 8.2. No configuration (8A) occurs. We suppose that (8A) holds and argue to a contradiction in a sequence of lemmas, as usual. Lemma 8.3. We have L  CG (u) and O2 (CCG (K) (u)) = 1. Proof. Since L is terminal in G, it is standard in G by [II3 , Corollary PU4 ]. Hence L  CG (u). Set X := O2 (CCG (K) (u)). As K < L  CG (u) and O2 (CG (u)) = 1 and O2 (CG (u)) ≤ LCCG (u) (L), we have that O2 (CL (K)) =  1 = X by [V17 , 4.7]. Since u is 2-central in CG (L), we may choose U ∈ Syl2 (CG (L)) and Qu ∈ Syl2 (CG ( u K)) with U ≤ Qu . We may assume that Qu = CQ (u). Lemma 8.4. Suppose Z(L) = 1. Then either U = u or m2 (U ) ≥ 2. Proof. Suppose on the contrary that m2 (U ) = 1 but U = u. If  ∼ L = HJ, then Qu := U × E with E = O2 (CL (K)) ∼ = E22 . If L ∼ = M12 , then Qu := (U × v) s with v ∈ I2 (L) and with [v, s] = 1, s2 ∈ U , and either s = 1 or s inducing an outer automorphism on L. In either case, u ∈ 1 (Qu ) ≤ U and so u = Ω1 (1 (Qu )). Then u  NQ (Qu ), whence Q = Qu . As m2 (Q/U ) ≤ 2, m2 (Q) ≤ 3. Now m5 (CG (K)) ≥ 2 and either Q is abelian or u = Ω1 ([Q, Q] ∩ Z(Q)). Also CG ( u K) is a {2, 3}-group, since its image in Aut(L) is a {2, 3}-group, as is CG ( u L). Therefore [V17 , 20.6], applied to H = CG (K), yields O2 (CC(z,K) (u)) = 1, contradicting Lemma 8.3.  Lemma 8.5. Suppose Z(L) = 1. Then m2 (U ) ≥ 2. Proof. Suppose not. Then u ∈ U ∼ = 2HJ, then = Z2n or Q2n . If L ∼ ∼ Qu = U ∗ Q0 with Q0 := CL (K) = Q8 . It follows that u = Ω1 ([Qu , Qu ] ∩ Z(Qu )), and so Qu = Q. Again, as in the preceding lemma, CG ( u K) is a {2, 3}-group. But then again, [V17 , 20.6] applied to H = CG (K) contradicts Lemma 8.3. Hence, L ∼ = Z4 , u = = 2M12 . Then Qu = (U ∗ Z0 ) s with L ≥ Z0 ∼ U ∩ Z0 , and s2 ∈ U , with possibly s = 1. If U s is nonabelian, then as |Qu /U | ≤ 4, u ∈ [Qu , Qu ] ≤ U and again u = Ω1 (Z(Qu )) and Qu = Q.

230 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

If U s is abelian but Qu is nonabelian, then u = [Qu , Qu ], while if Qu is abelian, then u ∈ 1 (Qu ) ≤ U and so u = Ω1 (1 (Qu )). Thus in all cases, we have Qu = Q. Again we check that m2 (Q) ≤ 3, that either Q is abelian or u = Ω1 ([Q, Q] ∩ Z(Q)), and that CG ( u K) is a {2, 3}-group. Again  [V17 , 20.6] contradicts Lemma 8.3, completing the proof. Lemma 8.6. We have Z(L) = 1. Proof. This is immediate from Lemma 8.5 and Proposition 4.1.



Lemma 8.7. We have U ∼ = Z2 or E22 . Proof. Suppose that U ∼ Z2 or E22 . Then by Lemma 8.4, we have = m2 (U ) ≥ 2 and |Q| ≥ |U | > 4. We use (8A5) to apply [II3 , Corollary PU2 ] and [V9 , 7.3]. Using [V9 , 7.7] if L ∼ = M12 , and [V17 , 5.8] and [V9 , 7.6] otherwise, we find g ∈ G − M with Qg ≤ M and ΓQg ,1 (L) < L. This  contradicts [V17 , 5.5]. The following argument in the M12 case is due to Aschbacher and Seitz. Lemma 8.8. F ∗ (CG (u)) = u × L. Proof. Assume false, so that m2 (U ) = 2. Then for some g ∈ G − M , E(CL (U g )) ∼ = A5 , by [V17 , 5.5]. ∼ If L = M12 , U g = u1 , u2  where u1 acts as an inner automorphism on L and u2 acts as an outer automorphism on L. We may assume g chosen so that U, U g  is a 2-group, so u1 = v1 v with v1 ∈ L and v ∈ U . Now U ≤ M and so U g ≤ M g . Since [u1 , U ] = 1, U ≤ M g . Thus [U, U g ] ≤ U g ∩ U = 1, and so NG (U ) = U × L u2  or U y × L u2  for some y ∈ I3 (G). In both  cases U ≤ Z(O 2 (NG (U ))). On the other hand, v1 ∈ [CL (u1 ), u2 ] ≤ U g , so u1 = v1 ∈ L. But then there exists s ∈ NL (U g ) with s inducing an automorphism of order 2 on U g , which is a contradiction. Hence, the lemma holds in this case. Now suppose L ∼ = = HJ. Let x ∈ I5 (K). Then x ∈ L and CAut(L) (x) ∼ CL (x) = x × K1 where K1 = E(CL (x)) ∼ = A5 , by [V17 , 10.11.7]. Note that as U ∈ Syl2 (CG (L)), for any v ∈ U # it follows that (8B)

CCG (v) ( x × K1 ) = U × x .

Now m2 (CCG (u) (x)) ≥ m2 (K1 U ) = 4. As e(G) = 3, it follows, with the Thompson dihedral lemma, that L5 (CG (x)) = 1. Let Cx = CG (x) and C x := Cx /O5 (Cx ). Assume first that K 1 acts nontrivially on O5 (C x ). By Thompson’s dihedral lemma again, m5 (O5 (C x )) ≥ 3, and so, as E(C x ) = 1, m5 (O5 (C x )) = 3. Hence, [U , O5 (C x )] = 1, and so m5 (CG (u)) = 3 and (u, L) ∈ K(G), contradicting Proposition 8.1. Thus [K 1 , O5 (C x )] = 1 and so, as usual, K 1 ≤ E(C x ) as e(G) = 3. In fact, K 1 is a component of CE(C x ) (u) for every u ∈ U # . This implies that there exists a 5-component J of CG (x) such that K1 is a component of CJ (u) for every u ∈ U # . Now as J/O5 (J) ∈ C5 , it follows from [V17 , 14.1, 14.2], with K1 in the role of

8. CONFIGURATIONS INVOLVING M12 AND HJ

231

∼ HJ or J = ∼ A5 . Note that since Y there, and [IA , 7.3.3] that either J = o ∗ x ∈ K ≤ CG (z), x ∈ I5 (G) by Theorem C4 : Stage A1. Suppose first that J ∼ = HJ. Then as m5 (Cx ) ≥ 4, m5 (CC x (J)) ≥ 2. Now U acts faithfully on J . If there exists an involution v ∈ U # that centralizes CC x (J), then m5 (CG (v)) = 3, so (v, L) ∈ K(G), a contradiction as above. Thus U acts faithfully on CC x (J). It follows that m2 (CG ( x×U ×K1 )) ≥ 4, which contradicts (8B). Hence, J = K 1 ∼ = A5 . Let C = CCx (J). Then (8B) implies that U ∈  Syl2 (C) and ΓU ,1 (C) = x × U . The Bender-Suzuki theorem [II2 , Theorem SE] applied to C/ x yields that C/ x ∼ = U , A4 , or A5 . In any case  m5 (Cx ) ≤ 3, contradicting x ∈ I5o (G) and completing the proof. We now introduce a case division according to the pattern of pumpups Ky of K in the centralizers of involutions y ∈ Q. We choose an involution y satisfying the following conditions: (8C)

(1) y ∈ Z(Q); (2) Subject to (1), Ky ∼ = M12 or HJ, if possible; and (3) Subject to (1) and (2), Ky = K, if possible.

If (1) and (2) are not possible, any y ∈ I2 (Z(Q)) will do. The analysis revolves about z, y, and u. We introduce involutions z0 and u0 : (1) If Ky ∼ = M12 or HJ, set z0 = z, u0 = y, and L = Ky ; (2) If Ky = K, set z0 = y and u0 = u; and (8D) (3) Otherwise, set z0 = z and u0 = u. Thus in every case, Ku0 = L, Kz0 = K, and Q1 ≤ C(z0 , K). Now we can prove Lemma 8.9. Either (8D1) or (8D2) holds. In particular, [z0 , u0 ] = 1. Proof. Suppose on the contrary that (8D3) holds. By [V17 , 13.2a], (8E)

Ky /Z(Ky ) ∼ = L2 (24 ) or L± 3 (4).

In every case, K is semirigid in Ky , indeed |CAut(Ky ) (K)|2 = 2, by [V17 , 13.8]. As Q centralizes K y, it normalizes Ky and so Q = Qy zwhere we have putQy = CQ (Ky ). Also, [Ky , u] = 1, for otherwise Ky = K Ky lies in L = K L , contradicting [V17 , 10.11.8]. Therefore u ∈ Q − Qy so Q = Qy u as well. As CQ (L) = u (Lemma 8.8) and |CAut(L) (K)|2 = 4, |CQ (u)| = 8 so |CQy (u)| = 4. Now by [V9 , 5.5], (8F)

r2 (Q1 ) ≤ r2 (Q) ≤ 5.

We have E := E(C(z, K)) = 1, K1 , or K1 K2 where the Ki are components, by Lemma 2.3 and the fact that e(G) = 3. Now let A ∈ Syl5 (C(z, K)),

232 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

so that m5 (A) = 2. Then A normalizes each Ki and centralizes Z(Ki ) [IA , 6.3.1]. Letting Z = Z(E), and T = O2 (C(z, K)), we have (8G)   r2 (Ki /Z(Ki )) + 4m5 (A/CA (T )) ≤ r2 (Ki /Z(Ki )) + r2 (T /Z) ≤ 5. i

i

If [T, A] = 1, this implies m2 (A/CA (T )) = 1, whence CA (T ) = 1 and so K1 exists. But then r2 (K1 /Z(K1 )) ≤ 1, which is absurd by Burnside’s transfer theorem. Therefore, [T, A] = 1, so CA (E) = 1. Suppose next that E = K1 , whence r2 (K1 ) ≤ 5 and m5 (Aut(K1 )) ≥ 2. But also (z, K1 ) ∈ K(G). If m2 (K1 z) ≥ 4, then by Lemma 2.4, K1 is given (as Y ) in [V17 , 14.1, 14.2], and so as m5 (Aut(K1 )) ≥ 2, K1 ∼ = HJ, Ru, HS, or 2HS. But these are impossible by Proposition 8.1. If m2 (K1 z) ≤ 3, then by Lemma 3.4 and as m5 (Aut(K1 )) ≥ 2, K1 ∼ = 2HJ, contradicting Proposition 4.1. Therefore E = K1 K2 , and so m5 (K1 ) = m5 (K2 ) = 1. By (8G), we may assume that r2 (K1 /Z(K1 )) ≤ 2, so by Lemma 3.5 and Proposition 3.8, K1 ∼ = A5 or A6 . Note that as m2,3 (CG (z)) ≤ e(G) = 3 and m2,3 (K) = 1, we have m3 (K1 K2 ) ≤ 2. Now if K1 ∼ = A6 , it follows n 2 ∼ 2 that K2 /Z(K2 ) = B2 (2 ) for some n ≥ 3, and hence r2 (C(z, K)) ≥ 2 + r2 (K2 z) = n + 3 ≥ 6 by [V17 , 9.2], contradiction. Therefore, K1 ∼ = A5 , so m2 (K2 z) ≤ r2 (K2 z) ≤ 3 by (8F), and m3 (K2 ) ≤ 1. As m5 (K2 ) = 1, K2 ∼ = A5 × A5 . = A5 by Lemma 3.4. Thus E ∼ ∼ Now F := Q1 ∩ K1 K2 z = E25 so as CAut(Ky ) (K) ∼ = Z2 , m2 (CF (Ky y)) ≥ 4. Therefore Fi := CF (Ky )∩CF (Ki ) = 1, i = 1, 2. Let v ∈ Fi# . As Ky is terminal, Ky   CG (v), and by L2 -balance, Ki ≤ CE(CG (v)) (K) so [Ky , Ki ] = 1. It follows that Ky × K1 × K2 ≤ G. In particular, m2,3 (G) ≥ m3 (Ky ) + 2, so m3 (Ky ) ≤ 1. Let z1 ∈ I2 (K1 ) and let I be the pumpup of K in CG (z1 ). Then Ky = K Ky ≤ I, so I is a vertical pumpup of K. If Ky ∼ = U3 (4), then it follows from [V17 , 13.2a] and Lagrange’s theorem that I = Ky . In that case m5 (CG (z1 )) ≥ m5 (Ky K2 ) ≥ 3, so (z1 , Ky ) ∈ K(G), and Lemma 3.3 is U3 (4), so m2 (I) ≥ m2 (Ky ) ≥ 4. contradicted. Hence, Ky ∼ = Finally, let x2 ∈ I5 (K2 ). Then applying Lemma 2.4 to the component I of CG (z1 ) and the subgroup x2  of CG (z1 ), we see that I is a component of JJ z1 for some 5-component J of CG (x2 ). Thus I is given (as Y ) by [V17 , 14.1, 14.2]. Given (8E) and that Ky is a component of CI (y), Ky /Z(Ky ) ∼ = L3 (4). Therefore m3 (Ky ) > 1, contradicting what we saw above. The proof is complete.  Lemma 8.10. Let t0 ∈ I2 (K) and g ∈ t0 , z0 # . Let K1 be a component of CG (g) isomorphic to A5 . Then g = t0 , and if g = z0 , then K1 = K. Moreover, if g = z0 t0 , then K, K1  = L. Proof. By Lemma 8.8, we have that CG (u0 ) = ( u0  × L) v, with ∼ M12 or HJ. If L ∼ L = = M12 or v = 1, then v 2 ∈ u0  with v inducing

8. CONFIGURATIONS INVOLVING M12 AND HJ

233

an outer automorphism on L. Moreover, in any case, we may assume that CG ( u0 , z0 , K) = u0 , z0 , w, where w = v if L ∼ = M12 , and if L ∼ = HJ, the ∼ projection of z0 , w on L is O2 (CL (K)) = E22 . Suppose that K1 is a component of CG (z0 ) with K1 ∼ = K but K1 = K. Let Qz0 be a u0 -invariant Sylow 2-subgroup of CCG (z0 ) (K1 × K). As CK1 (u0 ) ≤ u0 , z0 , w, u0 induces an inner automorphism on K1 and also CQz0 (u0 ) = z0 . Then Qz0 = z0 . Now Q1 is contained a Sylow 2-subgroup T of C(z0 , K), and T / z0  is embeddable in Aut(K1 ) and hence in D8 . But Q1 ∈ Syl2 (C(z, K)), with F ∗ (C(z, K)) = O2 (C(z, K))E(C(z, K)) and every component of C(z, K) a Co2 -group. Hence m5 (C(z, K)) ≤ 1 by [V17 , 9.6], so m5 (CG (z)) ≤ 2, a contradiction. Thus, the lemma holds for g = z0 . As t0 ∈ I2 (K), z0 t0 ∈ z0L [IA , 5.3bg], and E(CL (z0 t0 )) =: Kz0 t0 ∼ = K. But Kz0 t0 = K. By [IA , 5.3b], NL (K) is a maximal subgroup of L and NL (K)(∞) = K. Hence, L = K, Kz0 t0 , so the lemma holds for g = z0 t0 . It remains to show that CG (t0 ) has no component K0 with K0 ∼ = A5 . Suppose the contrary. Suppose first that K ∗ := K0 × K1 × K2  E(CG (t0 )) with {K0 , K1 , K2 } transitively permuted by a 3-element of CL (t0 ). As E(CG ( u0 , t0 )) = 1, u0 normalizes each Ki . We have that m2 (CG (u0 )) ≤ 1 + m2 (Aut(L)) ≤ 5, so u0 must induce an outer automorphism on some Ki . But then O3 (CG ( u0 , t0 ))) = 1, which is not the case as F ∗ (CG (t0 , u0 )) is a 2-group. Hence, Z := O2 (CG ( u0 , t0 )) normalizes K0 , as does u0 . It follows that Z centralizes K0 . But then CK0 (u0 ) ≤ CG ( u0  × Z) = u0 , t0 , whence |CK0 (u0 )| ≤ 2, which is impossible. Hence, CG (t0 ) has no component  isomorphic to A5 . Now we can reach a contradiction completing the proof of Proposition 8.2, using an argument of Finkelstein and Solomon. Set W := { z0 , t0  | t0 ∈ I2 (K)}. For any involution w of G, let K(w) denote the product of all components of CG (w) which are isomorphic to A5 . Finally for any W ∈ W, set K(W ) = K(w) | w ∈ W # . It follows from Lemma 8.10 that K(W ) = L for all W ∈ W. Hence, NG (W ) ≤ NG (L) for all W ∈ W. As CG (z0 ) ≤ NG (K), CG (z0 ) permutes the members of W, and so CG (z0 ) ≤ NG (L). But then CG (z0 ) maps injectively into NG (L)/O2 (NG (L)) ∼ = CG (u0 ). Hence C(z0 , K) embeds in C(u0 , K) ∼ = E23 . As Q1 embeds in C(z0 , K), Q1 embeds in E23 . Using [V17 , 9.6] again, we see that m5 (CG (z)) ≤ 2, contradicting (z, K) ∈ K(G). The proof of Proposition 8.2 is complete.

234 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

9. L2 (q), q ∈ {5, 7, 17} In this section, we continue the proof of Theorem 1, proving the following proposition via a sequence of lemmas. L2 (q), Proposition 9.1. Suppose that (z, K) ∈ K1 (K0 ). Then K0 ∼ = q ∈ {5, 7, 17}. Assume the contrary, and choose (z, K) with K ∼ = L2 (q) for some q ∈ (z). Since p divides |K| by Lemma 2.3 and B ≤ C {5, 7, 17}. Let Ep3 ∼ = G B normalizes every component of CG (z) by Lemma 2.2, while p does not divide | Out(K)|, we may write ∼ Ep2 and a = B ∩ K = 1. B = A × a with A = CB (K) = Lemma 9.2. We have q = p ≥ 5 and K ∼ = L2 (p). Moreover, there exists a p-component J of CG (A) such that one of the following holds: (a) J = J z with J/Op (J) ∼ = L2 (p), L2 (p2 ), L3 (p), or U3 (p) z (b) J = J , p = 5 and J/O5 (J) ∼ = HJ; or z ∼ (c) J = J and J/Op (J) = K. Proof. Since by [V17 , 9.2], m2 (K z) ≥ 3, Lemma 2.4 yields that either K is a component of CJ (z) for some p-component J of CG (A), or CG (A) has p-components J1 and J2 (possibly equal) with J1z = J2 and J1 /Op (J1 ) ∼ = K. In the latter case q = p, since L2 (q) ∈ Cp for q = p. We use the fact that p divides |K| (Lemma 2.3) and the lemmas [V17 , 14.1, 14.2], with K in the role of Y there, to determine the possibilities for K and J. If p = 3, then (K, J/O3 (J)) = (A5 , A9 ) or (L2 (17), J3 or 3J3 ) and m2,3 (AJ) = 4 > e(G) which is not possible. Thus p ≥ 5. Then q = p, K ∼ = L2 (p), L2 (p2 ), L± = L2 (p), and either J/Op (J) ∼ 3 (p), P Sp4 (p) ± z z ∼ ∼ or L4 (p), or p = 5 and J/O5 (J) = HJ, or JJ /Op (JJ ) = L2 (p) × L2 (p). Since m2,p (P Sp4 (p)) ≥ 2 ≤ m2,p (L± 4 (p)) [IA , 4.5.1], while mp (A) = 2 and ∼ P Sp4 (p) or L±  e(G) = 3, J/Op (J) = 4 (p). This completes the proof. Observe that the following conditions and notations hold: (1) (z, K) ∈ K(G) and K ∼ = L2 (p); (2) O2 (CG (z)) = Q0 ≤ Q1 ≤ Q with Q1 ∈ Syl2 (C(z, K)) and Q ∈ Syl2 (CG (K)); and (3) Ep2 ∼ = A ≤ CG (K z) and one of the following holds: (9A) (a) A acts faithfully on Q0 ; (b) A acts faithfully on E(C(z, K)); or (c) There exist b, c ∈ A such that A = b, c, b acts faithfully on O2 (CG (z)), and c ∈ K1 for a component K1 of CG (z) with K1 = K. Lemma 9.3. The following conditions hold: (a) m2 (Q1 ) ≥ 3; and (b) K is not terminal in G.

9. L2 (q), q ∈ {5, 7, 17}

235

Proof. If (9A3a) holds, [V9 , 8.1] implies that m2 (Q0 ) ≥ 3. If (9A3b) or (9A3c) holds, there is a component K1 = K of CG (z) and m2 (K1 z) ≥ 3 by [V17 , 9.2]. Hence, in all cases, m2 (Q1 ) ≥ 3. Suppose that K is terminal in G. Then [II3 , Corollary PU2 ] together with [V9 , 7.6, 7.3] implies that Qg ≤ NG (K) and CQg (K) = 1, for some g ∈ G − NG (K). But m2 (Qg ) ≥ 3 > 2 = m2 (Aut(K)), a contradiction completing the proof.  Now, through Lemma 9.7 below, we assume (see [IG , Definition 7.2]) that (9B)

For all (z ∗ , K ∗ ) ∈ K1 (K), (z ∗ , K ∗ ) is semirigid in G.

M12 or HJ Remark 9.4. In particular, if K ∼ = = L2 (5), then L/Z(L) ∼ for any pumpup L of K in CG (u), u ∈ I2 (C(z, K)). As m2 (Q1 ) ≥ 3, Q1 is neither dihedral nor semidihedral, so we may apply [IG , 7.4, 7.3] to obtain the following: (9C)

(1) K1 (K) = K2 (K); and (2) For any (z ∗ , K ∗ ) ∈ K1 (K), if Q∗1 ∈ Syl2 (C(z ∗ , K ∗ )), then Q∗1 = z ∗  × P1∗ for some P1∗ such that K ∗ has a nontrivial pumpup in CG (u) for all u ∈ I2 (P1∗ ).

In particular, Q1 = z × P1 where K has a nontrivial pumpup in CG (u) for all u ∈ I2 (P1 ). Lemma 9.5. Assume (9B). Then there exists t ∈ I2 (P1 ) and 1 = x ∈ such that [t, x] = 1. Moreover, if (9A3b) or (9A3c) holds, then one of the following holds: (a) E(CG (z)) = K1 K with A acting nontrivially on K1 ; or (b) E(CG (z)) = K1 K2 K, mp (K1 ) = mp (K2 ) = 1 and we may choose A = a1 , a2  ∼ = Ep2 with ai ∈ A ∩ Ki , i = 1, 2. A corresponding conclusion holds for all (z ∗ , K ∗ ) ∈ K1 (K).

Ipo (G) ∩ CG (z, K)

Proof. First consider the case that (9A3a) holds or, more generally, that [Q0 , A] = 1. We claim that there exists t ∈ I2 (Q0 ) and x ∈ A# with z = t ∈ CQ0 (x). Since [Q0 , A] = 1 there exist x, a ∈ A# such that T := [CQ0 (x), a] = 1. Then T = [T, a] so |T | > 2. Any t ∈ I2 (T ∩ P1 ) then satisfies the claim (note that z ∈ P1 ). Finally, as x ∈ A, x ∈ Ipo (G). Thus the first assertion of the lemma holds if (9A3a) or (9A3c) holds. We may assume now that E(CG (z)) > K and (9A3b) holds. Since e(G) = 3, O2 (CG (z)) = 1, and p divides the order of every component of CG (z) by Lemma 2.3, the number of components of CG (z) is at most 3. Consider the case in which E(CG (z)) = K1 K, A acting faithfully on the component K1 . As Q1 = z × P1 , z ∈ Z(K1 ). Thus, if Z(K1 ) > 1, as A is noncyclic, we may choose an involution t ∈ Cz,Z(K1 ) (x) for some x ∈ A# ,

236 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

and the conclusions of the lemma hold. So assume that Z(K1 ) = 1. If m2 (K1 z) ≤ 3, then by Lemmas 3.4 and 3.3, K1 ∼ = L2 (q), q ∈ F M9, L± 3 (3), or M11 . Since p ≥ 5 and CA (K1 ) = 1, we have a contradiction. Hence, m2 (K1 z) ≥ 4, and by Lemma 2.4 (with a in the role of A there), K1 is a component of CLp (CG (a)) (z) (recall that a = B ∩ K). By [V17 , 14.1, 14.2], with K1 in the role of Y there, and as Z(K1 ) = 1, we have K1 ∼ = = J/O5 (J) ∼ 5 2B (2 2 ) (with some element of A# inducing a field automorphism on K ), 2 1 2F (2 21 ) , 2F (2 52 ), HJ, HS or Ru. By the Thompson transfer lemma applied 4 4 to P1 ∩ K1 ≤ Q1 ∩ K1 ∈ Syl2 (K1 ), every involution in K1 is K1 -conjugate into P1 . It therefore suffices to show that for some x ∈ A# , CK1 (x) has even order. This follows directly from [IA , 4.9.1, 4.8.7, 5.3gms]. The only remaining case is that E(CG (z)) = K1 K2 K for some components K1 and K2 . Again, we may assume that Z(K1 ) = Z(K2 ) = 1. Since e(G) = 3 and p divides the order of every component, mp (K1 ) = mp (K2 ) = 1 and A = a1 , a2  ∼ = Ep2 with ai ∈ Ip (Ki ), i = 1, 2. Now we may choose t ∈ I2 (K1 ) ∩ P1 and x = a2 . The result follows immediately.  Lemma 9.6. Let t ∈ I2 (Q1 ) be such that the pumpup L of K in CG (t) is nontrivial, and there exists x ∈ Ipo (G) ∩ CG (z, K) with [x, t] = 1. Then L is a diagonal pumpup of K, and mp (CG (t)) = 3. Proof. Suppose on the contrary that L is a vertical pumpup of K. The possible isomorphism types of L are given in [V17 , 13.2]. If p = 5, then L∼ = L3 (4), with Z(L) elementary abelian (see = L2 (24 ) or U3 (4), or L/Z(L) ∼ Remark 9.4). Moreover, K ≤ L and z acts as a field automorphism on L if L ∼ = L± = L2 (24 ), and as a graph automorphism if L/Z(L) ∼ 3 (4), If p = 7, ∼ ∼ then L/Z(L) = L3 (4) or HJ, and if p = 17, L = J3 . In those cases z acts as an outer automorphism on L. As x ∈ CG (t) and [x, K] = 1, x normalizes L and therefore centralizes it U3 (4) by Lemma 3.3, as m5 (CG (t)) ≥ by [V17 , 13.7]. Then for p = 5, L ∼ = m5 (L × x) = 3. Therefore in all cases, m2 (L z) ≥ 4 by [V17 , 9.2]. Since x ∈ Ipo (G), Lemma 2.4 (with A = x) implies that L is a component of CJJ t (t) for some p-component J of CG (x). Moreover z normalizes L and [x, z] = 1, so z normalizes JJ t and K is a component of CJJ t (z). If J = J t , then L is a diagonal of JJ t and as K ≤ L, we must have J z = J. But then K/Z(K) ∼ = L/Z(L), a contradiction. Therefore J = J t . Using [V17 , 14.1, 14.2], with L in the role of Y there, and using Remark 9.4, we obtain that p ∈ {5, 7} with L/Z(L) ∼ = L3 (4) and J/Op (J) ∼ = He. But then ∼ K = L2 (p) is a component of CJ (z), contradicting the structure of He from [IA , 5.3p]. Thus the pumpup L of K in CG (t) is L = K1 × K1z with K1 ∼ = K. Since x ∈ CG (t) and [x, K] = 1, [x, K1 K1z ] = 1, whence mp (CG (t)) = 3, completing the proof.  Still assuming (9B), we now consider the general case of a pair (t1 , L1 ) ∈ K2 (K) and let S1 ∈ Syl2 (CG (t1 , L1 )). We suppose further that CG (t1 )

9. L2 (q), q ∈ {5, 7, 17}

237

 1 = L1 with L1 and L  1 interchanged by an involution has a component L ∼  t0 ∈ CG (t1 ). In particular, L1 = L1 . We may apply (9C) with (t1 , L1 , S1 ) in place of (z, K, Q1 ) to conclude that S1 = t1  × T1 for some T1 such that L1 pumps up nontrivially in CG (u) for every u ∈ I2 (T1 ). Lemma 9.7. There exists an involution t2 ∈ S1 such that L1 pumps up  2 in CG (t2 ). Furthermore, (t2 , L2 ) ∈ K2 (K) and if diagonally to L2 × L S2 ∈ Syl2 (CG (t2 , L2 )), then |S2 | ≥ 2|S1 |.  1 ). As L  1 is perfect, t2 is L  1 -conjugate into Proof. Take t2 ∈ I2 (L  1 ∩ T1 , by the Thompson transfer lemma. Hence L1 has a nontrivial L  1 ) = mp (Aut(L1 × L  1 )) = 2 pumpup in CG (t2 ). Also, since mp (L1 × L  1 ). In particuwhile mp (CG (t1 )) = 3, there exists x ∈ Ipo (G) ∩ C(t1 , L1 × L  2 with lar, [x, t2 ] = 1. By Lemma 9.6, the pumpup of L1 in CG (t2 ) is L2 × L t 1 ∼  = L1 and mp (CG (t2 )) = 3. Then (t2 , L2 ) ∈ K2 (K) by (9C1). L2 = L 2  1 has 1 or 2 S1 -conjugates, and accordingly, t2 is either As e(G) = 3, L central or half-central in S1 . Since |L2 (p)|2 = 2ap with a5 = 2, a7 = 3 and  2 |2 = 2ap −1 · |S1 |, or |S2 | = 1 |S1 | · |L  2 |2 = a17 = 4, either |S2 | ≥ 12 |S1 | · |L 4 a −2 p · |S1 |. Thus the result holds unless perhaps p = 5 and there exists 2  v = L  1 . We show that the existence of v leads v ∈ CG (t1 , L1 ) such that L 1  v is the only L2 (p)-component of to a contradiction. As m2,p (G) = 3, L 1  1 . Hence it is invariant under t0 and we can choose CG (t1 ) centralizing L1 L  1 ), we get that L1 tv ∈ I2 (CLv (t0 )). Just as we argued with t2 ∈ I2 (L 1  1 also diagonalizes in CG (tv ). Since diagonalizes in CG (tv ). As [t0 , tv ] = 1, L  [L1 , L1 ] = 1 it follows that CG (tv ) has at least four L2 (p)-components. This  contradicts m2,p (G) = 3 and completes the proof of the lemma.

We now complete the proof of Proposition 9.1 in the case when (9B) holds. By Lemma 9.3, K is not terminal in G. By Lemmas 9.5 and 9.6, there exists an involution t1 and a p-element x ∈ CG (z, K) such that x ∈  1 in CG (t1 ) and Ipo (G), [x, t1 ] = 1 and K pumps up diagonally to K1 × K (t1 , K1 ) ∈ K2 (K). We are now in the situation of Lemma 9.7, and so there 2 exists t2 ∈ I2 (CG (t1 , K1 )) such that K1 pumps up diagonally to K2 × K in CG (t2 ), (t2 , K2 ) ∈ K2 (K), and if Si ∈ Syl2 (CG (ti , Ki )), i = 1, 2, then |S2 | ≥ 2|S1 |. Hence, (t1 , K1 ) < (t2 , K2 ). Since (t2 , K2 ) ∈ K2 (K), again the conditions of Lemma 9.7 hold, and we may repeat the argument to obtain (t3 , K3 ) ∈ K2 (K) satisfying the conditions of Lemma 9.7, with (t2 , K2 ) < (t3 , K3 ) and with |S3 | ≥ 2|S2 |. We continue the process, which does not stop; this is obviously not possible as sooner or later we will have a subgroup of order 2M > |G|. This completes the proof of the case of Proposition 9.1 when (9B) holds.

238 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

It remains to treat the cases when some (z, K) ∈ K1 (K) is not semirigid in G. This is where we need the previous section. By [V17 , 9.7], we have ∼ A5 , and there exists an involution Lemma 9.8. We have p = 5, K = u ∈ CG (K) such that the subnormal closure L of K in CG (u) is isomorphic to M12 , 2M12 , HJ, or 2HJ. Since m3 (L) = 2 and e(G) = 3, L does not have a diagonal pumpup in the centralizer of any involution. With [V17 , 13.6a], we then deduce: Lemma 9.9. L is terminal in G. Now we check: Lemma 9.10. The conditions (8A) hold, for suitable u, Q, and z. Proof. Choose u and L as in Lemma 9.8. Then u ∈ CG (L) ≤ CG (K) and we choose a 2-centralinvolution u of CG (L). The pumpup J of K in   L CG (u ) then contains L = K , and by [V17 , 13.6a], J = L. Replacing u by u we see that (8A1,5,6) are satisfied. Let R ∈ Syl2 (CCG (L) (u)) and expand R to Q ∈ Syl2 (CG (K)). Then (8A4) holds. Our original (z, K) ∈ K(G) has a CG (K)-conjugate (z  , K) ∈ K(G) such that z  ∈ Q. Replacing z by z  , we have the lemma.  Lemma 9.10 and Proposition 8.2 contradict each other, thereby establishing Proposition 9.1. 5

5

Corollary 9.11. If p = 5, then K1 (2B2 (2 2 )) = K1 (2F4 (2 2 )) = ∅. Proof. This is immediate from Propositions 6.14, 7.10, and 9.1.



10. The Remaining Cases of Theorem 1 Lemma 10.1. Let (z, K) ∈ K(G). If p > 3, then m2 (Q1 ) > 1. Proof. If E(CG (z)) = K, then by (3A), there is b ∈ Ip (C(z, K)) and b acts nontrivially on O2 (CG (z)). As p > 3, O2 (CG (z)) is neither cyclic nor quaternion, so m2 (O2 (CG (z))) > 1. Since O2 (CG (z)) ≤ Q1 , the lemma holds in this case. On the other hand, if CG (z) has a component K1 = K,  then m2 (K1 ) > 1 as K1 ∈ C2 , and so m2 (Q1 ) > 1, as claimed. Our next main result is: Proposition 10.2. Let (z, K) ∈ K(G). Then m2 (K z) ≥ 4. To prove the proposition, assume that m2 (K z) ≤ 3. By Lemma 3.4, K∼ = L2 (q), q ∈ F M, A6 , L± 3 (3), U3 (4), M11 or 2HJ. Now Proposition 9.1, Lemma 3.3, and Proposition 3.8 imply that ∼ (10A) K ∼ = A6 or L± 3 (3), all with p ≥ 5, or z ∈ K = 2HJ, with p = 3. We have also used Proposition 4.1 to restrict p. Note that in every case, Out(K) is a 2-group and m2 (K z) = 3. Recall that p divides |K|, by Lemma 2.3. We proceed in a sequence of lemmas.

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Lemma 10.3. If mp (K) = 1, then the following conditions hold: (a) p = 5 and K ∼ = A6 ; (b) m2 (Q1 ) ≥ 3; and (c) Q1 = z×P1 for some P1 such that K pumps up vertically in CG (t) for every t ∈ I2 (P1 ), and the vertical pumpup L is isomorphic to one of the following groups: A8 , Sp4 (4), HS, 2HS, L5 (2) or U4 (2). In all cases m2 (L) ≥ 4. Proof. The condition mp (K) = 1 implies that p > 3, and, since Out(K) is a 2-group, that mp (Aut(K)) = 1 and there exists Ep2 ∼ = A ≤ C(z, K). Hence, either A acts faithfully on O2 (CG (z)), or E(CG (z)) > K. By [V9 , 8.1] and [V17 , 9.2], m2 (Q1 ) ≥ 3, i.e., (b) holds. Since m2 (K z) = 3, by Lemma 2.4, K is a component of CJ (z) for some p-component J of CG (A). By [V17 , 14.1, 14.2], with K in the role of Y there, we get p = 5, K ∼ = A10 or HS. In particular, = A6 , and J/O5 (J) ∼ (a) holds. Since m2 (Q1 ) ≥ 3, by definition of restricted even type [I2 , 8.8], K is not terminal in G. As (z, K) is semirigid in G by [V17 , 9.7], it follows from [IG , 7.4, 7.3] that Q1 = z × P1 for some P1 such that K pumps up nontrivially in CG (t) for every involution t ∈ P1 . As m3 (K) = 2 while e(G) = 3, that pumpup L is vertical, and so, by [V17 , 13.2], L ∼ = A8 , Sp4 (4), HS, 2HS, L5 (2) or U4 (2), as claimed. Finally, m2 (L) ≥ 4 by [V17 , 9.2].  Lemma 10.4. Assume mp (K) = 1. Then there exist t ∈ I2 (P1 ) and x ∈ I5o (G)∩C(z, K) such that m5 (CCG (z) (x)) = 3, [x, t] = 1 and x centralizes the (vertical) pumpup L of K in CG (t). Proof. Suppose that there exist t ∈ I2 (P1 ) and x ∈ I5o (G) ∩ K such that m5 (CCG (z) (x)) = 3 and [t, x] = 1. Since x ∈ CG (t) centralizes K, it normalizes L, and indeed centralizes L by [V17 , 14.6]. Thus it is enough to show the existence of commuting elements t ∈ I2 (P1 ) and x ∈ I5o (G) ∩ C(z, K) with m5 (CCG (z) (x)) = 3. Indeed it is enough to know, for t, just that t is conjugate in C(z, K) into P1 . By Sylow’s theorem and the Thompson transfer lemma, it is also enough to show the existence of t ∈ I2 ([C(z, K), C(z, K)]) and x ∈ I5o (G) ∩ C(z, K) such that [t, x] = 1 and m5 (CCG (z) (x)) = 3. Suppose first that E(C(z, K)) has two components, K1 and K2 . Then 5 divides |K1 | and |K2 | and so we choose x ∈ I5 (K1 ) and t ∈ I2 (K2 ), thereby satisfying our conditions. Next, suppose that E(C(z, K)) = K1 is quasisimple. If m2 (K1 z) ≤ 3, then Lemma 10.3 applies to K1 and implies that K1 ∼ = A6 . Hence CA (K1 ) = 1. Just assuming that CA (K1 ) = 1, we can then take x ∈ CA (K1 )# and any t ∈ I2 (K1 ). So assume that m2 (K1 z) ≥ 4 and CA (K1 ) = 1. Also as e(G) = 3, m3 (K1 ) ≤ 1. Applying Lemma 2.4 to K1 and a subgroup of K of order 5, we deduce from [V17 , 14.1, 14.2], with K1 in the role of 5 Y there, that K1 ∼ = 2B2 (2 2 ). Hence some x ∈ I5 (CG (z)) induces a field

240 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

automorphism on K1 , and we take t ∈ I2 (CK1 (x)), thereby satisfying our conditions. Finally, suppose that F ∗ (C(z, K)) = O2 (C(z, K)), which then equals Q0 . Then CA (Q0 ) = 1, so there is x ∈ A# such that [CQ0 (x), A] = 1. We take t ∈ I2 ([CQ0 (x), A]) ⊆ I2 ([C(z, K), C(z, K)]), and the proof is complete.  Now we can prove Lemma 10.5. We have mp (K) = 2. Proof. Let x, t, and L be as in Lemmas 10.4 and 10.3. Since [x, K] ≤ [x, L] = 1, without loss we may alter A and assume that x ∈ A. By Lemma 10.3, L ∼ = A8 , Sp4 (4), HS, 2HS, L5 (2), or U4 (2), and m2 (L) ≥ 4. Applying Lemma 2.4 to L and x, we see that L is a component of CJx (t) for HS or 2HS, for otherwise some 5-component Jx of CG (x). Note that L ∼ = m5 (CG (t)) ≥ m5 (L x) = 3, and Proposition 8.1 is contradicted. Therefore by [V17 , 14.1, 14.2], with L in the role of Y there, we have L ∼ = A8 , and ∼ Jx /O5 (Jx ) = A10 or HS, on which t induces an outer automorphism in either case. Let Cx = CG (x) and C x = Cx /O5 (Cx ). We have A × K ≤ Cx . Since Aut(J x ) does not contain Z5 × A6 [V17 , 14.6], [J x , A] = 1. Therefore CG (A) has a 5-component JA ≤ Jx with K ≤ L ≤ JA and J A = J x . Set A = CA /O5 (CA ). CA = CG (A) and C  = Ω1 (O5 (C A )), and so z centralizes Since m2,5 (JA ) = 1 and e(G) = 3, A A ). O5 (C  Suppose that JA ∼ = A10 . Since K   CJA (z),  z ∈ JA is a root involution. ∼ Therefore z ∈ J x is a root involution. Since J t = Σ10 , CJx t (z) ≥ D × K where D ∼ = D8 and z ∈ Z(D). As Q1 = z × P1 , z is C(z, K)-conjugate into [Q1 , Q1 ] = [P1 , P1 ], so K pumps up nontrivially in CG (z), which is absurd. Hence, JA ∼ = Jx ∼ = HS, and z ∈ JA ≤ Jx . Let a ∈ A# ; we claim that C(a, Ja ) has odd order. Let

a = Ca /O5 (Ca ). Ca = CG (a) and C By L5 -balance and [V17 , 13.20], and as e(G) = 3, the subnormal closure Ja of JA in CG (a) satisfies Ja = JA ∼ = HS. If the claim is false, then ta , Ja ] = 1. Now Op (Ca ) has there is an involution ta ∈ Ca such that [ ∗ odd order by Theorem C4 : Stage A1, and m2 (Ja ) ta  ≥ 5. As e(G) = 3, it follows, with the Thompson dihedral lemma, that Ja is quasisimple, so that JA = Ja ∼ = HS. As K ≤ JA and K ≤ E(CG (ta )) by L2 -balance, JA = K Ja ≤ E(CG (ta )). Since e(G) = 3, JA lies in a single component La

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∼ HS. This contradicts Proposition of CG (ta ), and by [V17 , 13.6a], La = JA = 8.1, so the claim holds. Now consider CG (z). Recall that A ≤ C(z, K). If there exists a component K1 = K of CG (z) with m5 (K1 ) = 1, then there is a0 ∈ A# such that either [a0 , K1 ] = 1 or a0 induces an outer automorphism on K1 . In  1 , Ja ] = 1 as C  (K)  is solvable. Thus the former case the former case, [K 0 Ja0 is impossible by the previous paragraph. In the latter case, as K1 admits an outer automorphism of order 5, m2 (K1 z) ≥ 4. Hence, by Lemma 2.4, K1 is a component of CY Y z (z) for some 5-component Y of CG (b), where b ∈ I5 (K). Since b ∈ I5o (G), it follows from [V17 , 14.1, 14.2], with K1 in the role of Y there, that a0 induces a field automorphism on K1 and 5 K1 ∼ = 2B2 (2 2 ). Hence, there exists an involution w0 ∈ CK1 (a0 ). Since z ∈ K1 , m2 (CCG (a0 ) (K)) ≥ 2. By the previous paragraph, w0 acts nontrivially on Ja0 . But w0 ∈ NK1 (A), and so there exists v0 ∈ NK1 (A) of order 4 with v02 = w0 . Therefore Z4 ∼ v0  injects into CAut(JA ) (K) ∼ = E22 [V17 , =  13.28]. This is absurd. Suppose now that there exists a component K1 = K of CG (z) with m5 (K1 ) ≥ 2. Since e(G) = 3 and m5 (K) = 1, m5 (K1 ) = 2; and as m3 (K) = 2, m3 (K1 ) ≤ 1. The latter condition together with [V17 , 9.2] gives us that m K1 ∼ = L2 (q), q ∈ F M, L2 (2a ), L3 (22a+1 ), U3 (22a ), or 2B2 (2 2 ) for some a and m. Now the former condition, and Lemma 3.3, imply that K1 ∼ = U3 (24a+2 ), a ≥ 1. Then m2 (K1 z) ≥ 4, so by Lemma 2.4, K2 is a component of CM (z) for some 5-component M of CG (b), where b ∈ I5 (K). As b ∈ I5o (G), this is not possible by [V17 , 14.1, 14.2]. Hence, K = E(CG (z)), and so A acts faithfully on Q0 . By [IG , 11.12], there exist a, b ∈ A# such that Q0 A ≥ a × X b, with X = [X, b] special and m2 (X/Φ(X)) ≥ 4. In a ), particular, m2 (X) ≥ 2. Consider CG (a). If X acts nontrivially on O5 (C   then by the Thompson dihedral lemma, m5 (O5 (Ca )) ≥ 3, and so m2,5 (Ca ) ≥ a )) + m2,5 (Ja ) ≥ 3 + 1 > e(G), which is obviously not possible. m5 (O5 (C  b , K]  = 1, and C  ∼ a ). As [X (K) Hence, X b acts faithfully on E(C = Aut(Ja )   E22 [V17 , 13.28], [X, Ja ] = 1. But this is not possible as C(a, Ja ) has odd order. This completes the proof of the lemma.  We can now complete the proof of Proposition 10.2. Since mp (K) = 2, the following conditions hold by (10A): (1) p = 3 and K ∼ = 2HJ with z ∈ Z(K); and (10B) (2) There exists x ∈ Ipo (G) ∩ CG (z, K). Let Cx = CG (x) and C x = Cx /O3 (Cx ). By [V17 , 14.1], K acts trivially on E(C x ) and hence faithfully on O3 (C x ). Take a ∈ I3 (K) with CK (a) ∼ = 3A6 ×Z2 . Then m2 (CG ( a, x)) ≥ 3, and so by Lemma 2.4, Ka := E(CK (a))

242 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

is a component of CG (z) ∩ L3 (CG ( a, x)). As m3 (Ka ) > 1 and e(G) = 3, Ka ≤ J where J is a 3-component of CG (x), and so m2,3 (CG ( a, x)) ≥ 4, an obvious contradiction. This completes the proof of Proposition 10.2. Collecting our prior results, we are now reduced to the following cases for elements of K(G). Lemma 10.6. Let (z, K) ∈ K(G). Then p > 3 and one of the following holds: 3 (a) p = 5 and K ∼ = A8 , 2B2 (2 2 ), [2 × 2]L3 (4), 2Sp6 (2), or M12 ; (b) p = 7 and K ∼ = [2 × 2]L3 (4); or (c) p = 11 and K ∼ = J4 . Proof. By Proposition 10.2 and Lemma 2.4, K is a component of CJ (z) for some Epm ∼ = A ≤ C(z, K), m ∈ {1, 2}, and some p-component J of 5 CG (A). (The L2 (8), 2B2 (2 2 ), and L2 (p) exceptions in Lemma 2.4 are ruled out by Propositions 6.14 and 9.1 and Corollary 9.11.) The possibilities for p and K are then given by [V17 , 14.1, 14.2], with K here playing the role of Y there. These are further restricted by Propositions 3.8, 5.1, 7.10, and 8.1. Moreover, 0 < mp (K) < 3 by Lemma 2.3 and (3A). Furthermore, Proposition 10.2 rules out some small cases. The result is the claimed list of possibilities for p and K. In particular, p > 3.  We will now deal with these cases one by one. Notice that by Lemmas 10.6 and 10.1, (10C)

If (z, K) ∈ K(G), then m2 (Q1 ) ≥ 2.

Proposition 10.7. Let (z, K) ∈ K(G). If p = 5, then K ∼ A8 . = Proof. Assume the contrary. As m5 (Aut(K)) = 1 while m5 (CG (z)) = 3, there exists A ∼ = E52 with A# ⊆ I5o (G) ∩ CG (z, K). As in the proof of Lemma 10.6, K is a component of CJA (z) for some 5-component JA of CA := CG (A). Invoking [V17 , 14.1, 14.2], with K playing the role of Y there, we obtain that JA /O5 (JA ) ∼ = A10 or HS. Likewise for each x ∈ A# , K is a component of CJx (z), where Jx is a 5-component of CG (x), also with Jx /O5 (Jx ) ∼ = A10 or HS. As neither A10 nor HS involves the other, by order considerations, it follows that Jx = JA O5 (Jx ). Let C A := CA /O5 (CA ). Take t ∈ I2 (JA ) such that E(CJ A (t)) ∼ = A6 . ∼ Then t ∈ I2 (Jx ) and E(CJx /O5 (Jx ) (t)) = A6 . Let Jt,A = L5 (CJA (t)) and Jt,x = L5 (CJx (t)). It follows that Jt,x = Jt,A O5 (Jt,x ). Now O5 (CG (x)) has odd order by Theorem C∗4 : Stage A1. Furthermore, Jt,x /O5 (Jt,x ) is simple, so O2 (Jt,x ) = 1. On the other hand Jt,x = O 2 (Jt,x )   CG ( t, x). Let W = [O2 (CG ( t, x)), Jt,x ]. By [IG , 4.3], W = # [W, Jt,x ] ≤ Jt,x ∩ O  2 (CG ( t, x)) ≤ O2 (J#t,x  ) = 1. As x ∈ A was arbitrary, and O2 (CG (t)) = CO2 (CG (t)) (x) | x ∈ A , we have [O2 (CG (t)), Jt,A ] = 1. Since e(G) = 3 and m3 (Jt,A ) > 1, Jt,A therefore lies in a component L of E(CG (t)) and is a 5-component of CL (A). Since Jt,x is a 5-component

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of CCG (t) (x), it also lies in L, and is a 5-component of CL (x), for each x ∈ A# . By the (B5 )-property [IA , 7.1.3], Jt,x and Jt,A are quasisimple, and so Jt,x = Jt,A . As L ∈ Co2 , it follows by [V17 , 5.2] that L = Jt,A ∼ = A6 . Note that m5 (CG (t)) ≥ m5 (Jt,A A) = 3, so (t, L) ∈ K(G), and by Proposition 10.2, m2 (L t) ≥ 4, a contradiction as m2 (A6 ) = 2. The proof is complete.  Proposition 10.8. Let (z, K) ∈ K(G). If p = 5, then K ∼ M12 . = Proof. Assume the contrary. Since m3 (K) = 2 and e(G) = 3, K has no diagonal pumpups. By [V17 , 13.6a], K is terminal in G, and as m5 (K) = m5 (Aut(K)) = 1, there exists E52 ∼ = A ≤ C(z, K). Whether A acts faithfully on O2 (CG (z)) or E(C(z, K)) = 1, we conclude by [V9 , 8.1] and [V17 , 9.2] that m2 (Q1 ) ≥ 3. We may now use [II3 , Corollary PU2 ] and [V9 , 7.3] to obtain E22 ∼ = U ≤ Q and g ∈ G − NG (K) such that U g ≤ NG (K), U g acts faithfully on K and ΓU g ,1 (K) < K. Then by [V17 , 5.5d], U g can be written as u, v where [CK (u), v] has even order. By [V9 , 7.7], we may then assume that −1 Qg ≤ NG (K) acting faithfully on K. Since m2 (Q) ≥ m2 (Q1 ) ≥ 3, [V17 , 5.5] now implies that Γ g−1 (K) = K, which contradicts [V9 , 7.3]. The Q ,1 proposition follows.  3 Proposition 10.9. Let (z, K) ∈ K(G). If p = 5, then K ∼ 2B2 (2 2 ). =

Proof. Suppose the contrary. As m5 (Aut(K)) = 1, there exists A ∈ E52 (C(z, K)). We have m2 (K z) = 4. Thus by Lemma 2.4, K is a component of CJA (z), where JA is a 5-component of CG (A); and by [V17 , 14.1, 14.2], with K playing the role of Y there, and by Lemma 2.6, JA ∼ = Ru. Moreover, as e(G) = 3, |CG (A × JA )| is odd as m5 (A × JA ) = 4, and so z ∈ J. By [V17 , 13.20e] and Lemma 2.6 again, and as e(G) = 3, JA   CG (a)  for all a ∈ A# , indeed JA  O5 (CG (a)). We argue that (10D)

For every a ∈ A# , CCG (a) (JA ) has odd order.

Suppose, in fact, that there exists an involution t ∈ CCG (a) (JA ). Then m5 (CG (t)) = 3. As z ∈ JA , t ∈ C(z, K). By L2 -balance, K ≤ E(CG (t)), and so JA ≤ E(CG (t)). Moreover, JA is a component of CE(CG (t)) (a) and since ↑5 (Ru) = ∅ by [V17 , 13.20], JA is a component of CG (t). But then (t, JA ) ∈ K(G), which is impossible by Proposition 8.1. Thus (10D) holds. Now consider CG (z). If K = E(CG (z)), then A acts faithfully on Q0 . Hence there exist a0 , b0 ∈ A# such that R0 := [CQ0 (a0 ), b0 ] = 1. In particular, |R0 | ≥ 24 and [R0 , K] = 1. But in CG (a0 ), we see that R0 maps into CAut(JA ) (K) ∼ = E22 . Hence CR0 (JA ) = 1, contrary to (10D). Thus, K = E(CG (z)). Let K1 = K be a component of CG (z). If [K1 , b] = 1 for some b ∈ A# , then CCG (z) (b) ≥ b×K1 ×K. Since CAut(JA ) (K) is a 2-group [IA , 5.3r], we obtain that CG (b) ≥ b × K1 × JA , contrary to (10D).

244 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

Therefore CA (K1 ) = 1. But (z, K1 ) ∈ K(G), so by Lemma 10.6 and Propositions 10.7, 10.8, and 8.1, K1 /Z(K1 ) ∼ = L3 (4) or Sp6 (2). Hence m5 (Aut(K1 )) = 1 [IA , 4.10.3], so CA (K1 ) = 1, a final contradiction.  ∼ Proposition 10.10. Let (z, K) ∈ K(G). If p = 5, then K = 2Sp6 (2). Proof. Assume the contrary. Since m5 (K) = 1, there exists E52 ∼ = A ≤ C(z, K). Also m2 (K) = 4 by [V17 , 9.2]. By Lemma 2.4, as usual, and [V17 , 14.1, 14.2], K is a component of CJA (z) for some 5-component JA of CG (A) with JA /O5 (JA ) ∼ = Co3 . Let t ∈ I2 (JA ) with Jt,A := L5 (CJA (t)) and Jt,A /O5 (Jt,A ) ∼ = M12 . For any x ∈ A# , K is similarly a component of CJx (z) for some 5-component Jx of CG (x) with Jx /O5 (Jx ) ∼ = Co3 , and we set Jt,x = L5 (CJx (t)). Since K ≤  JA , JA ≤ Jx = JA O5 (Jx ) and Jt,x = Jt,A O5 (Jt,x ). The third paragraph of the proof of Proposition 10.7 may be repeated verbatim, and we conclude that [O2 (CG (t)), Jt,A ] = 1. Hence, as there, Jt,A ≤ L for some component L of CG (t), whence Jt,x ≤ L for all x ∈ A# , and by the (Bp )-property, Jt,A and all Jt,x ’s are quasisimple and isomorphic to M12 . Then by [V17 , 13.6a], L∼ = M12 . But m5 (CG (t)) ≥ m5 (LA) = 3, so (t, L) ∈ K(G), contradicting Proposition 10.8. This completes the proof.  Proposition 10.11. Let (z, K) ∈ K(G). If p = 5 or 7, then K ∼ = E22 L3 (4). Proof. Assume the contrary. Since mp (K) = 1, there exists Ep2 ∼ =A≤ C(z, K). Since m2 (K z) ≥ 4, applying Lemma 2.4 and [V17 , 14.1, 14.2], with K in the role of Y there, we obtain that CG (A) has a p-component J with J/Op (J) ∼ = He and K = E(CJ (z)). Moreover, as e(G) = 3 and by [V17 , 13.6a], K is a terminal component. Likewise by Lemma 2.6 and [V17 , 13.9], J is a component of CG (a) for all a ∈ A# . Consider CG (z). By now, given that p = 5 or 7, the only possible components of CG (z) are isomorphic to E22 L3 (4). But e(G) = 3 while m3 (K) = 2, and so we have that K = E(CG (z)). Similarly, J  CG (a). Now, A acts faithfully on Q0 . Choose a, b ∈ A# such that Tb := [CQ0 (a), b] = 1. Since J  CG (a) and [J, b] = 1, [Tb , J] = 1. Hence, NCG (a) (Tb ) ≥ J × A and  so m2,p (CG (a)) ≥ 4 > e(G), which is not possible. Proposition 10.12. Let (z, K) ∈ K(G). If p = 11, then K ∼ J4 . = Proof. Assume the contrary. As m3 (K) = 2, K has no diagonal pumpups. Using [V17 , 13.10], we see that K is terminal in G. For any component K1 of C(z, K), (z, K1 ) ∈ K(G), so K1 ∼ = K and e(G) > 3, contradiction. Therefore K = E(CG (z)), and since m11 (Aut(K)) = 2, m2 (Q1 ) ≥ 2. Then since | Out(K)| = 1, it follows from [II3 , Corollary PU2 ] and [V9 , 7.6, 7.3] that for some g ∈ G − NG (K), Qg ≤ NG (K) and ΓQg ,1 (K) < K. Note also that Q0 admits some b ∈ I11 (C(z, K)) faithfully, so Qg has a subgroup T such that Φ(T ) ≤ Ω1 (Z(T )) and |T /Φ(T )| = 210 .

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Suppose first that some u ∈ I2 (Qg ) acts on K like a 2-central involution of K. Then by [V17 , 5.4], ΓQg ,1 (K) = CK (u). Let v ∈ I2 (Qg ) − {u}. Then CK (v) ≤ CK (u), so u, v ≤ Z(CK (v)). However, it is clear from [IA , 5.3i] that Z(CK (v)) = v, a contradiction. Therefore every involution in Qg acts on K as an element of class 2B of K [IA , 5.3i]. Let v, w be a four-subgroup of Qg with v ∈ Z(Qg ). Since F ∗ (CK (v)) = O2 (CK (v)), it follows that either w acts on K like an element of E := O2 (CK (v)), or [O2 (CK (v)), w] = 1. If the former holds for all w ∈ I2 (Qg ), then the image R of Qg in Aut(K) lies in O2 (CAut(K) (v)) and hence is elementary abelian. Because of b and as v ∈ Qg , R = O2 (CAut(K) (v)). But R then contains some 2-central involution, contrary to what we saw above. Therefore, for some w ∈ I2 (Qg ) − {v}, [O2 (CK (v)), w] = 1 whence by [V9 , 7.7], we may assume that C(z, K)g ≤ NG (K). Hence, (Q0 b)g acts faithfully on K, centralizing z g , which acts on K like an element of class 2B. As elements of I11 (Aut(M22 )) normalize no nontrivial 2-subgroup of Aut(M22 ), it follows that the image of Qg0 in Aut(K) is O2 (CAut(K) (z g )). Again this image contains some 2-central involution of Aut(K), giving a final contradiction.  Now Lemma 10.6, together with Propositions 10.7, 10.8, 10.9, 10.10, 10.11, and 10.12, imply that K(G) = ∅. This completes the proof of Theorem 1. 11. Theorem 2: Introduction By Theorem 1, the following set is nonempty: (11A) Kf (G) = {(z, K, B) | (z, K) ∈ IL2 (G), B ∈ Ep3 (CG (z)), CB (K) = 1}. The main step in proving Theorem 2 is the following proposition, whose proof will occupy us through Section 15. Then in Section 16, we shall rule out conclusion (b), thereby establishing Theorem 2 and completing the proof of Theorem C∗4 : Stage A2. Proposition 11.1. Let (z, K, B) ∈ Kf (G). Then p = 3 and one of the following holds: (a) K ∼ = Co3 ; or = 2Sp6 (2) and G ∼ ∼ (b) K = Sp6 (2) and m2 (C(z, K)) = 1. At this point the reader may review the first two sections of this chapter, particularly (1A), (2A), (2B), (2C), and the lemmas of Section 2. For any (z, K, B) ∈ Kf (G), we have the auxiliary notation (2C) and we introduce the following further notation. (11B)

(1) For any x ∈ B # , Kx := E(CK (x)); and (2) P ∈ Syl2 (G) with S ≤ P .

246 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

Lemma 11.2. Assume (z, K, B) ∈ Kf (G). Then K = E(CG (z)), and any Sylow p-subgroup of CG (z) acts faithfully on K. Proof. Let X ∈ Sylp (C(z, K)), and assume that X = 1. By Lemma 2.2 and our assumption, B acts faithfully on K. Hence, we may choose X to be B-invariant, and B ∩ X = 1. Then CX (B) = 1, and so mp (CG (z)) ≥ 4 > e(G), which is not possible. Since every component of CG (z) has order divisible by p by Lemma 2.3, the result now follows.  In particular, (11C) If (z, K, B) ∈ Kf (G) and B0 ∈ Ep3 (CG (z)), then (z, K, B0 ) ∈ Kf (G). Here are the components K with which we must deal in the remainder of this chapter. Proposition 11.3. Let (z, K, B) ∈ Kf (G). If p = 5, then K ∼ = 2F4 (2 2 ). 3 ∼ If p = 5, then p = 3 and K = J3 , U3 (8), Sp4 (8), D4 (2), U4 (2), Sp6 (2), or 2Sp6 (2). 5

Proof. Since K is a component of CG (z), K ∈ Co2 . As mp (Aut(K)) ≥ 3 and mr (K) ≤ 3 for all odd r, it follows from [V17 , 9.2] that we may assume that K ∈ Co2 ∩ Chev(2) (otherwise, p = 3 and K ∼ = J3 ). Since mp (Aut(K)) = 3, the untwisted Lie rank of K is at least 2 (see [IA , 4.10.3]). Because we may adjust our choice of B ∼ = Ep3 , by (11C), we shall subdivide our proof of Proposition 11.3 in the following way, noting that mp (Aut(K)) ≥ mp (AutB (K)) = 3: (1) If mp (Inndiag(K)) ≤ 2, we will show that mp (K) = 2 and there exists b ∈ B # inducing a field or graph automorphism on K. (a) If b induces a field automorphism on K, we will show that either p = 3 and K ∼ = U3 (8) or Sp4 (8), or p = 5 and K ∼ = 2F (2 52 ). 4 (b) If b induces a graph automorphism on K, we will show that p = 3 and K ∼ = 3D4 (2). (2) If mp (Inndiag(K)) > 2, we will show that p = 3 and K ∼ = U4 (2), Sp6 (2) or 2Sp6 (2). We now proceed with the proof. (1) If mp (Inn(K)) = 1, then mp (Aut(K)) ≤ 2 by [V17 , 3.1], a contradiction. So mp (Inn(K)) ≥ 2. As we are assuming that mp (Inndiag(K)) ≤ 2, mp (K) = mp (Inndiag(K)) = 2 and there exists b ∈ B # that induces a nontrivial field or graph automorphism on K. (1a) By [IA , 4.9.1],  O2 (CK (b)) ∼ = d Σ(2 p ), a

where K ∼ = d Σ(2a ) and p divides a.

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∼ A± (2a ), B2 (2a ), G2 (2a ) or 2F4 (2 a2 ), then by [III11 , Table 13.1], If K = 2 a there exists c ∈ K ∩ B # with Kc ∼ = A1 (2a ), or possibly Kc ∼ = 2B2 (2 2 ) in the last case. (See (11B1).) By [IA , 6.1.4], K ∈ Lie(2). Since p ≥ 3, a ≥ 3 and so m2 (CKz (c)) ≥ 4. By Lemma 2.4 (with A = c), either Kc is a component of CJ (z) or J/Op (J) ∼ = Kc for some p-component J of CG (c). Now J/Op (J) ∈ Cp while Kc ∈ Chev(2). We apply Lemmas [V17 , 14.1] and [V17 , 14.2], with Kc in the role of Y there. we conclude that either 5 p = a = 3 and K ∼ = 2F4 (2 2 ), as = U3 (8) or P Sp4 (8), or p = a = 5 and K ∼ desired, or p = 3 and K ∼ = G2 (8). In the last case, however, we could have # chosen c ∈ K ∩ B such that Kc ∼ = SU3 (8), and then Lemmas [V17 , 14.1] and [V17 , 14.2] would lead to a contradiction. a 2 a If K ∼ = A± 3 (2 ) or D4 (2 ), then  2 p p Kb = O2 (CK (b)) ∼ = A± 3 (2 ) or D4 (2 ), a

a

respectively. By [IA , Theorem 3.3.3], m2 (Kb ) ≥ 4; and Lemma 2.4 (with A = b) implies that Kb is a component of CJ (z) for some p-component J of CG (b). Using Lemmas [V17 , 14.1] and [V17 , 14.2], with Kb in the role of Y there, we obtain that either p = a = 3 and K ∼ = U4 (8) or p = a = 5 and ∼ K = L4 (32). In the former case mp (Inndiag(K)) = 3 and in the latter case mp (Inndiag(K)) = 1, both contrary to our assumptions. If K ∼ = B3 (2a ), by [IA , Theorem 4.10.3], then either mr (Inndiag(K)) ≤ 1 or mr (Inndiag(K)) = 3 for all odd primes r, a contradiction. Finally, since K ∈ A by [V17 , 1.3], the only remaining case is K∼ = 3D4 (2a ),  with p dividing a and Kb = O2 (CK (b)) ∼ = 3D4 (2 p ). By [IA , Theorem 3.3.3], m2 (Kb ) ≥ 5 and by Lemma 2.4 (with A = b), Kb is isomorphic to a component of CJ (z) for some p-component J of CG (b). Now Lemmas [V17 , 14.1] and [V17 , 14.2], with Kb in the role of Y there, imply that p = 3. Hence, K has no field automorphism of order p, a contradiction. (1b) In this case p = 3 and K∼ = 3D4 (2a ) ∈ Lie(2), a



and we assume that a ≥ 2. By [IA , 4.7.3A], Kb = O2 (CK (b)) ∈ Lie(2), with Kb ∼ = G2 (2a ) or A2 (2a ) and 2a ≡  (mod 3),  = ±1. As a ≥ 2, m2 (Kb z) ≥ 4 and by Lemma 2.4 (with A = b), Kb is a component of CJ (z) for some 3-component J of CG (b), which is not possible by Lemmas [V17 , 14.1] and [V17 , 14.2], again with Kb in the role of Y there. Hence, K ∼ = 3D4 (2), as desired. (2) Again by [IA , 4.10.3], since mp (Inndiag(K)) ≥ 3, the untwisted Lie rank of K is at least 3. Moreover, if p divides | OutdiagCG (z) (K)|, then mp (CG (z)) > 3 = e(G), contradiction. Therefore mp (Inn(K)) = mp (Inndiag(K)) = 3, as Op (CG (z)) = 1, and so we may take B ≤ K. Suppose that K ∼ = A3 (2a ),  = ±1. Then 2a ≡  (mod p) and there exists b ∈ B with Kb ∼ = A2 (2a ). If, further, a ≥ 3, or a = 2 and  = +1,

248 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

then m2 (Kb z) ≥ 4, and by Lemma 2.4 (with A = b), Kb is a component of CJ (z) for some p-component J of CG (b), contrary to [V17 , 14.1, 14.2]. If a = 2 and  = −1, then (p, K) = (5, U4 (4)), in which case Kb ∼ = U3 (4), and there exists B0 ∈ E2 (B) such that KB0 := E(CK (B0 )) ∼ = A5 . Then m2 (KB0 z) = 3 and so by Lemma 2.4 (with A = B0 ), KB0 ≤ J0 for some 5-component J0 of CG (B0 ). By L5 -balance, KB0 ≤ L5 (CG (b)), and as KB0 ≤ Kb , we have Kb ≤ L5 (CG (b)). Hence, Kb is a component of CL5 (CG (b)) (z), which together with [V17 , 14.1] and [V17 , 14.2], again with Kb in the role of Y there, gives us a contradiction. Therefore a = 1, and as mp (K) = 3, p = 3 and K ∼ = U4 (2), as desired. a ∼ If K = U5 (2 ), then as mp (K) = 3, a = 2 and p = 5. Again Lb ∼ = U3 (4) for some b ∈ B, and we reach a contradiction as in the case K ∼ = U4 (4). Next, suppose that K ∼ = C3 (2a ). If a ≥ 2, then there exists b ∈ B with a Kb ∼ = C2 (2 ). By [IA , 3.3.3], m2 (Kb z) ≥ 4. By Lemma 2.4 with A = b, Kb is a component of CLp (CG (b)) (z). Now [V17 , 14.1] and [V17 , 14.2], again with Kb in the role of Y there, give us a contradiction. Therefore a = 1, so p = 3 and K ∼ = (2)Sp6 (2) [IA , 6.1.4, 4.10.3], as desired. Suppose that K ∼ = 2D 4 (2a ). Then as mp (K) = 3, there exists b ∈ B  a ∼ with Kb = L4 (2 ), with 2a ≡ − (mod p),  = ±1 [IA , 4.8.2, 4.10.3]. Again as m2 (Kb z) ≥ 4, a combination of Lemma 2.4 (with A = b), and [V17 , 14.1,14.2], yields a contradiction. Finally, as mp (K) = 3, and K ∈ A, we are reduced to the case in which p = 3 and K ∼ = L6 (2) or L7 (2). Then there exists b ∈ K ∩ B with Kb ∼ = Ln−2 (2), n ∈ {6, 7}. Again a combination of Lemma 2.4 and [V17 , 14.1] gives a contradiction.  12. Components with Outer p-Automorphisms For the rest of this chapter, we assume that (12A)

(z, K, B) ∈ Kf (G), and we set Cz = CG (z).

In this section we shall be eliminating the following configurations:

(12B)

5 (1) Either p = 5 and K ∼ = = 2F4 (2 2 ), or p = 3 and K ∼ U3 (8), Sp4 (8), or 3D 4 (2); (2) c ∈ B # induces a graph automorphism on K if (p, K) = (3, 3D4 (2)), and a field automorphism otherwise; (3) b ∈ B # ∩ K with Kb ∼ = L2 (8) if p = 3, and Kb ∼ = 2B (2 25 ) if p = 5; and 2 (4) Cz := CG (z), Cb := CG (b) and C b := Cb /O3 (Cb ).

Lemma 12.1. Assume (12A) and (12B). Then O{2,p} (CG ( b, z)) = 1 and z inverts Op (Cb ).

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Proof. Let W = O{2,p} (CG ( b, z)). Since Op (Cb ) has odd order, COp (Cb ) (z) = CO{2,p} (Cb ) (z) ≤ W , so it suffices to show that W = 1. In Cz , K is locally balanced with respect to p [IA , 7.7.8], and so W ≤ Op (CCz (b)) ≤ C(z, K) ≤ CG ( b, z), the last as b ∈ K. Therefore W ≤ O{2,p} (C(z, K)) ≤ O2 (CCz (K)) ≤  O2 (Cz ) = 1 as K  Cz and G is of even type. The lemma is proved. Our first main result is: Proposition 12.2. Assume (12A) and (12B). Then Lp (Cb ) = 1, and if Kb ≤ J for some p-component J of Cb , then Kb ∼ = J/Op (J). We will prove Proposition 12.2 in a sequence of eight lemmas. First note that by Lemma 2.4, Lp (CG (b)) = 1, and there exists a p-component J of Cb such that either Kb ≤ JJ z > J or Kb ≤ J = J z . We may assume that J/Op (J). J = J z , and we assume for a contradiction that Kb ∼ = Lemma 12.3. The following conditions hold: (a) J/Z(J) ∼ = G2 (3); (b) K ∼ = U3 (8); and (c) F ∗ (Cz ) = z × K. Proof. By [V17 , 14.1], p = 3, J /Z(J) ∼ = G2 (3) and z induces an outer automorphism on J. By [IA , 7.1.4c], (12C)

|CAut(J) (K b )| = 2.

Suppose that K ∼ = P Sp4 (8) or 3D4 (2), or there exists a 2-element t ∈ Cz inducing an outer automorphism on K ∼ = U3 (8). Then t, z is an abelian 2-group, and there exists a 2-element u ∈ NCz ( b) that inverts b and centralizes Kb . (This follows from [IA , 4.8.2, 4.7.3A] for K ∼ = P Sp4 (8) or 3D4 (2). For U3 (8), u induces a graph automorphism of order 2 on K, and so inverts a torus conjugate to b × a with a of order 9.) In particular, as [z, b] = 1, | u, z |2 > 2. Hence by (12C), Cu,z (J) has even order. But m3 (J) ≥ 4 by [V17 , 10.10.3], and so m2,3 (G) ≥ 4 > e(G), a contradiction. Therefore K ∼ = U3 (8) and | OutCz (K)| is odd. By (12C) again, and as C(b, J) has odd order, |C(z, K)|2 ≤ |CCb (K b )|2 = |CAut(J) (K b )|2 = 2. This implies (c) since G is of even type. The proof is complete.  Since b ∈ K, [b, O2 (CG (z))] = 1 and [b, L] = 1 for every component L of CG (z) with L = K. As e(G) = m3 (CG ( z, b), looking at CG (b), we

250 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

conclude that the following hold:

(12D)

(1) |Cz : z × K| is odd; (2) There exists c ∈ B such that CK (c) ∼ = P GU3 (2); (3) F ∗ (C b ) = O3 (C b )J, and (J, m3 (O3 (C b ))) = (G2 (3), 1) or (3G2 (3), 2); (4) z induces an outer automorphism on J and b ∈ J; and (5) There exists t ∈ I2 (CK ( b, c)) with T := O2 (CK (t)) ∼ = 23+8 and Z := Z(T ) = CT (b) ∈ Syl2 (Kb ).

In fact we can make things a bit more precise. Lemma 12.4. The following conditions hold: (a) Cz = F ∗ (Cz ) c; (b) O3 (CG ( b, z)) = z; (c) J ∼ = J; # (d) If J ∼ = 3G2 (3) and b0 ∈ I3 (Z(J)), then for every x ∈ b, b0  , the pumpup of J in CG (x) is trivial;   and (e) Let C = C(b, J). Then C = b × C 0 , where C 0 = [C, z] is a cyclic 3-group. If C 0 = 1, then 1 = Z(J) ≤ C 0 . Proof. To prove (b), O3 (CG ( z, b)) = z × O3 (CK (b)) = z by [IA , 7.7.8]. Next, z inverts O3 (J) by (b), and [z, J] = J, so O3 (J) ≤ Z(J). Thus (c) holds, with the help of [IA , 6.1.4]. # Suppose that Z(J) = b0  ∼ = Z3 , and let x ∈ b, b0  . By [V17 , 13.20], ↑3 (J) = ∅. By L3 -balance, and as e(G) = 3, the pumpup Jx of J in CG (x) is trivial, proving (d). If (a) fails, then by (12D1), there exists w ∈ Cz inducing an outer diagonal automorphism on K. We may assume that w3 = b = [w, c]. If c ∈ b J, then [w, c] ∈ J, contradicting (12D4). Thus c ∈ b J. If Z(J ) = 1, then as O2 (C b ) = JCC b (J), (12D3) implies that c ∈ b J, contradiction.   Thus Z(J ) = b0 ∼ J). Then m3 (C) = 2 and = Z3 for some b0 . Let C = C(b,  

b, b0  is central in C. Hence, b, b0  contains I3 (C) . Therefore c = c1 c2 with c1 ∈ C, c2 ∈ J, c31 = b0 and c32 = b−1 0 . But now as [z, c] = 1 and z normalizes both C and J, z must normalize c1  and c2 . As bz0 = b−1 0 , z = c−1 , contrary to [z, c] = 1. So c ∈ b, J, for i = 1, 2. But then c czi = c−1 i contradiction. This proves (a). Let C = C(b, J), as in (e). As e(G) = 3 < m3 (J), |C| is odd, and m3 (C) ≤ 2. Hence by Frobenius’s theorem, C has a normal 3-complement, i.e., C is a 3-group. We have CC (z) ≤ CO2 (Cz )∩Cb (Kb ) = b, by (a) and as   K ∼ = U3 (8). Thus z inverts C/ b , which is abelian. Suppose that b ≤ of z, C 0 ≤ Ω1 (C). But this C0 ≤ C with C 0 / b ∼ = E32 . Then by the action   is impossible since m3 (C) ≤ 2 and Ω1 (C) = b, b0 ≤ Z(C). Therefore, C0

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    cannot exist, whence C/ b is cyclic and C is abelian. As z inverts C/ b ,     C∼  = b × (C/ b ), and the lemma is proved. Lemma 12.5. Suppose x ∈ B # induces a field automorphism on K, and x ) = O3 (C x ). x := CG (x)/O3 (CG (x)). Then F ∗ (C set C  is a component Proof. By [IA , 4.9.1], CK (x) ∼ = P GU3 (2) or U3 (2). If L  and x , then from looking at CG ( z, x) we see that z normalizes L of C ∗ F (CL ( z )) embeds in z× x×O3 (CK (x)). This implies, in particular, that  ∈ C3 ∩ Chev(2), then L/Z(   ∼  ∈ Chev(3) ∪ Spor, by [IA , 4.5.1, 5.3]. If L L) L = A5 or L3 (4), and so O3 (CL (y)) = 1 for all the involutory automorphisms y of  ∈ C3 −Chev(2)−Chev(3)−Spor.  [IA , 7.7.1ca], a contradiction. Therefore L L ∼  = A9 and z acts on L  as a product of three transpositions. But Hence, L now m2 (CG ( z, x)) ≥ 4 contrary to the structure of CCG (z) (x). The proof is complete.  Recall (12D5) that t ∈ I2 (CK ( b, c)), T = O2 (CK (t)), and t ∈ Z(T ) = Z ∈ Syl2 (Kb ). Lemma 12.6. We have that F ∗ (CG (t)) = O2 (CG (t)), and t is a 2-central involution of J. Proof. Since Z ≤ Kb ≤ J, and G2 (3) has a unique conjugacy class of involutions [IA , 4.5.1], t is a 2-central involution of J. In particular, m3 (CG (t)) = 3, and so by Lemma 2.3, 

E(CG (t)) = O3 (E(CG (t))).

(12E)

By (12D1,5) and Lemma 12.4a, Cz = z × K c ∼ = Z2 × P ΣU3 (8). Therefore CG ( z, t) = CCz (t) = z × T b, c. We may assume that S =

z × T , so Ω1 (S) ∼ = E24 . Also, T is indecomposable, for if T = T1 × T2 with Ω1 (T1 ) = [T1 , T1 ] ∼ = Z2 , then a 7-element of NK (T ) cannot act transitively on Z # , by the Krull-Schmidt theorem. Set Xb = O3 (Cb ). By Lemma 12.4be, Xb is inverted by z and either (1) Cb = b × (Xb × J) z, with J ∼ = G2 (3); or (2) Cb = b × (Xb × (Z0 ∗ J)) z, with J ∼ = 3G2 (3), and (12F) Z0 a cyclic 3-group inverted by z with Ω1 (Z0 ) = Z(J) =: b0 . Hence, (12G)

CG ( b, t)/O2 (CG ( b, t)) ∼ = (SL2 (3) ∗ SL2 (3))E22 ,

with no normal Q8 -subgroup, and Sylow 3-subgroups of O2 (CG ( b, t)) are abelian of rank at (12H) most 2. In particular, (12I)

CG ( b, t) has no chief factor of order 4.

252 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

Since both CG ( b, t) and CG ( z, t) are solvable, every component of CG (t) is normalized by b, z as well as by T = [T, b]. Now if the lemma fails, then CG (t) has a component L. Assume first that L is simple. As b normalizes every component of CG (t), T = [T, b] normalizes L, and T ≤ L as t ∈ L. Suppose that T ≤ L∗ := L × CCG (t) (L). Then T or T z is a Sylow 2-subgroup of CL∗ (z) = CL (z)×CCG (t,z) (L). As T is indecomposable, it follows that T ∩ L = 1 and z ∈ Syl2 (CL (z)), which is absurd. Therefore, T ≤ L∗ . Hence OutT b (L) has a section isomorphic to A4 , forcing L ∼ = D4 (q), q odd [V17 , 20.9]. But then L ∈ Co2 , a contradiction as G is of even type. We conclude that Z(L) = 1. In view of (12E), we conclude from [V17 , 9.2] that L is isomorphic to one of the following, with X ∼ = Z2 or E22 : (12J)

[X]L3 (4), 2G2 (4), 2Sp6 (2), 2M12 , 2M22 , 4M22 , 2HJ, 2HS, 2Ru.

In every case m3 (L) ≥ 2, so since e(G) = 3, L  CG (t). Set H = CL (b)   be the image of H modulo O2 (CG ( b, t)). By (12G) CG ( b, t) and let H and (12H), O3 (H) is abelian, m3 (O3 (H)) ≤ 2, and H is solvable. By [V17 , 9.5], it follows that CG ( b, t) has a chief factor of order 4, contrary to (12I), or finally L/Z(L) ∼ = L3 (4). In this final case, as CCG (t) (z) = CCz (t) is 2closed, so is CL (z). But then z induces an inner automorphism on L, whence m2 (CL (z)) ≥ 5 [IA , 6.4.4a]. However, m2 (CCG (t) (z)) = m2 (CCz (t)) = 1 + m2 (K) = 4, contradiction. The proof is complete.  Now set (12K)

R = F ∗ (CG (t)) = O2 (CG (t)) and Rz = CR (z),

and fix (12L)

E33 ∼ = A ≤ CCb (t) with b, c ≤ A.

Corollary 12.7. We have J ∼ = G2 (3), and R is of symplectic type. Proof. Suppose that J ∼ = 3G2 (3). Take b0 ∈ I3 (Z(J)), and consider CG (x) for any x ∈ b, b0 # . By Lemma 12.4d, the pumpup Jx of J in CG (x) is trivial. As e(G) = 3, Jx  CG (x) and |C(x, Jx )| is odd. Hence, any 2-subgroup of CG (x) acts faithfully on Jx /O3 (Jx ). As b, b0  ≤ CG (t) and b centralizes Jx /O3 (Jx ), [CR (x), b] = 1. Thus by [IG , 11.25], [R, b] = [CR (x), b] | x ∈ b, b0 #  = 1. Hence b ∈ CCG (t) (R) ≤ R, which is impossible. We have proved that J ∼ = G2 (3). Finally, take any hyperplane A0 of A. Then A0 ∩ b, c = 1. From the structure of CG (b) and CG (x) for every x ∈ b, c − b (Lemma 12.5 and (12F)), it follows that F ∗ (CG (A0 )/O3 (CG (A0 ))) is a 3-group [IG , 5.12]. As m2,3 (G) = 3, and by Thompson’s dihedral lemma, m2 (CR (A0 )) ≤ m2 (CG (A0 )) ≤ 2. Similarly, m2 (CR (A)) = 1. Then [IG , Lemma 24.4] implies that R is of symplectic type. The lemma is proved. 

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Observe the following: (12M)

If d ∈ b, c − b, then CK (d) ∼ = CK (c), by [IA , 4.9.1d].

Lemma 12.8. We have b ∼G b−1 . Proof. Assume b−1 ∼ b. Then by the structure of CG (b) ((12F) and Corollary 12.7), there exists an involution s ∈ CG (J) with bs = b−1 . But  then m3 (CG (s)) ≥ 4, a contradiction, so the lemma holds. Lemma 12.9. The following conclusions hold: (a) If CR (b) > t, then CR (b) ∼ = Q8 ∗ Q8 with Z = CCR (b) (z); (b) R is extraspecial of width at least 3; (c) CT (d) := Qd ∼ = Q8 and CZ (d) = t for all d ∈ b, c − b; (d) Z = t × [Z, c] and Z ∩ Rz = t or Z; and (e) For each d ∈ b, c − b, either CRz (d) = t or CRz (d) = Qd . b = J  b := CG (b)/O2 (CG (b)). Then C z ∼ Proof. Let C = Aut(G2 (3)),  t)) ∼ and CR (b) is isomorphic to an A, z-invariant subgroup of O2 (CCb ( = Q8 ∗Q8 . Let R1 be an A, z-invariant 2-subgroup of CG ( b, t) isomorphic to t)). Then R1 = Q0 ∗ Qz0 with Z = CR1 (z) Q8 ∗ Q8 and mapping onto O2 (CCb ( and Q0 ∼ = Q8 . As A ∩ J acts faithfully on R1 and CR (b) is A z-invariant, it follows that either CR (b) = t or CR (b) = R1 , proving (a). In either case, as [R, b] is a central product of quaternion groups, and m3 (CG (t)) = 3, it follows that R is extraspecial of width at least 3, proving (b). Now as O 3 (CK (d)) ∼ = U3 (2) with t ∈ CK (d), we have CT (d) =: Qd ∼ = Q8 with CZ (d) = t, proving (c). Also, Z = t × [Z, c]. As Z ∩ R = Z ∩ Rz is c-invariant, it follows that Z ∩ Rz = t or Z, proving (d). Finally, as  CRz (d) is a b-invariant subgroup of CT (d), (e) follows as well. and ( z × T )R/R is a subgroup of Lemma 12.10. We have R ∼ = 21+8 + CG (t)/R isomorphic to E25 . Proof. T z ∈ Syl2 (Cz ) is b, c-invariant with [T z , b, c] = T and CT z ( b, c) = t, z. As Rz is b, c-invariant and t ∈ Rz , Rz = [Rz , b, c](Rz ∩ z, t) = (R ∩ T )(R ∩ z). Now Lemma 12.9b implies that |Rz | > 8, so |T ∩ R| > 4. Suppose that Z ≤ R. As Z ∩ R is c-invariant, Z ∩ R = t, whence Q8 ∗ D8 , it follows that z ∈ R and Rz = T ∩ R ∼ T ∩R∼ = = Q8 . = Q8 . As R ∼ But then R = Rz ∗ R0 where R0 = CR (Rz ), so CR0 z (z) = z, t. Hence R0 is cyclic or of maximal class, contradicting Lemma 12.9b. Hence Z ≤ R, and Z ≤ CR (b) ∼ = Q8 ∗ Q8 by Lemma 12.9a. Let Rb = [R, b], the central product of m ≥ 1 copies of Q8 , and set R0 = CRb (z). If R0 = t, then Rb ∼ = Q8 , and z × T /Z maps injectively into CG (t)/R and hence into O6− (2). But m2 (T /Z) = 6 and m2 (O6− (2)) = 4, so such

254 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

  a map cannot exist. Therefore R0 > t. As R0 = CR0 (a) | a ∈ b, c# and CR0 (b) ≤ CRb (b) = t, we may assume without loss of generality that Qc ≤ R0 , by Lemma 12.9e. We argue that R0 = Qc . Otherwise, there exists d ∈ b, c − b − c such that CR0 (d) = t. In T we see that Qc ∩ Qd ≤ t, and in R, that [Qc , Qd ] ≤ t. Hence Qc Qd ∼ = Q8 ∗ Q8 . But then Qc Qd = Ω1 (Qc Qd ) ≤ Ω1 (T ) = Z ∼ = E23 , which is absurd. Thus, R0 = Qc . Let V = R/ t, a nondegenerate symplectic space, and set Vb := Rb / t. Then Vc := Qc / t is a nonsingular subspace of Vb , and so Vc⊥ is z-invariant. Let Rc⊥ be the preimage of Vc⊥ in Rb ; then CRc⊥ (z) = t, and so Rc⊥ = t or Rc⊥ ∼ = Q8 . So m = 1 or 2 and R ∩ T z = Qc Z. Therefore z × (T /R ∩ T) ∼ = E25 embeds in CG (t)/R. As m2 (O6− (2)) = 4, we must have m = 2, completing the proof.  Our final lemma is: Lemma 12.11. We have b ∼G b−1 . Set H = CG (t)/R. We may regard H as a subgroup of O8+ (2) with O2 (H) = 1 and m2 (H) ≥ 5. Moreover, as A ∼ = AR/R and e(G) = 3, we have m3 (H) = 3. Now by [V17 , 10.1.6], it follows that H ∼ = Sp6 (2). But then b ∼ b−1 in CG (t). The conflict between Lemmas 12.8a and 12.11 completes the proof of Proposition 12.2. Proposition 12.12. Assume (12A) and (12B). If J is a p-component of CG (b) on which Kb acts nontrivially, then z normalizes J. As usual we break the proof into a sequence of lemmas (six lemmas this time). We assume that J z = J. Recall that by Lemma 11.2, K = E(CG (z)). We have J/Op p (J) ∼ = L2 (8) 5 2 or B2 (2 2 ) according as p = 3 or 5, by Lemma 2.4; thus by [IA , 6.1.4], J/Op (J) has trivial Schur multiplier, so Op (J) = [Op (J), J]. By Lemma 12.1, z inverts Op (JJ z ), so JJ z = [J, z] centralizes Op (JJ z ). Thus, J is simple. Also, as c induces a field automorphism on Kb , c ∈ Cb and c induces a field automorphism on both J and J z . Let us summarize this and set some notation.  z z (1) J, J = J × J ∼ = Kb ; = J × J z and J ∼ (2) c induces a field automorphism on both J and J z ; (12N) c := Cc /Op (Cc ); and (3) Cc := CG (c) and C  := C/Op (C). (4) C := CG ( b, c), and C Lemma 12.13. The following conditions hold:   z (a) C b = Op (C b ) × (J × J ) z c; (b) Op (C b ) is cyclic of order p or p2 ;

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(c) F ∗ (Cz ) = K × z; and (d) |CC b (c)| = 2p p4 and Op (CC b (c)) = CJ (c) z ∼ = Fp(p−1) Z2 . z

Proof. If E(C b ) > JJ , then there exists yet another component Y of C b and as e(G) = 3, mp (Op (C b Y c ≤ 2. It follows that b ∈ Z(Y ). But z then by [V17 , 7.9h], mp (Y ) > 2, contradiction. Therefore E(C b ) = J J , and mp (Op (C b ) c) ≤ 2. Suppose that mp (Op (C b )) ≥ 2. Then COp (C b ) (c) is cyclic. Also a Sylow p-subgroup of CG (Kb b, z) acts faithfully on K, so embeds in CAut(K) (Kb b) ∼ = Zp or Zp2   by [V17 , 7.22c]. Thus, mp (COp (C b ) (z)) = 1 as well, and COp (C b ) (c) = b .

Let Y be a critical subgroup of Op (C  b) of exponent p with b ∈ Y . Then

z, c acts on Y with z inverting Y / b . Thus we may choose d ∈ Y with     [d, c] = b and with d inverted by z. Let X = b, d = b × [X, z]. Then c   normalizes and hence centralizes [X, z] = d , a contradiction. Therefore Op (C b ) is cyclic and (b) holds. If (a) fails, then C b /Op (C b ) ∼ = z  Aut(J × J ). Then as [Op (C b ), z] = 1, there is c ∈ Ip (Cb ) such that z inverts c , whence [c, c ] = 1. For an involution y ∈ J, CCb (y) ≥ b, c, c , J z , contradicting m2,p (G) = 3. Thus (a) holds, and (d) follows. Also C(z, K) ≤ CCb ( z Kb ) ∼ = CC b (K b ) so by (a), z ∈ Syl2 (C(z, K)). This implies (c) as  O2 (Cz ) = 1, and the lemma is proved. Lemma 12.14. We have p = 3 and K ∼ = U3 (8).

5 Proof. If not, then K ∼ = 2F4 (2 2 ) = 3D4 (2) or Sp4 (8) with p = 3, or K ∼ with p = 5. Assume first that K ∼ = G2 (2), = 3D4 (2). As we chose c so that CK (c) ∼ ∗ = C (c). By ∼ Kc = U3 (3). Now Lemma 2.4 applies with A = c and KA K that lemma and and [V17 , 14.1], there exists a 3-component L of CG (c) such ∼ ∼   that Kc ≤ L and L = U3 (3), L± 3 (9) or U4 (3). Assume L = Kc . Then L  Cc 3 as e(G) = 3. It follows that O (CK c (b))  CJ (c) z, which is impossible as ± ∼ O 3 (CK c (b)) = CJ×J z (c, z). Hence, L = L± 3 (9) or L4 (3) and [z, L] = 1. Since b ∈ L while |CG ( b, c)|3 = 34 , |CL (b)| ≤ 33 , and we have a contradiction by [V17 , 10.10.2]. 5 1 Similarly, if p = 5 and K ∼ = 2F4 (2 2 ) and Kc = = 2F4 (2 2 ), then CK (c) ∼  c = L.  As [CK (c), CK (c)]; and Lemma 2.4 and [V17 , 14.1] imply that K  C c . As in the previous case, this leads to the impossibility e(G) = 3, L z CJ×J (c, z)  CJ (c) z. Therefore p = 3 and K ∼ = Sp4 (8). Let b1 ∈ I3 (CJ (c)) and b2 ∈ I3 (CJ z (c)). We may choose  notation so that b1 b2  ∈ Syl3 (Kb ), so that b ∼K −1

b1 b2  ∼Cb b1 b2 . Now some Sylow 3-subgroup R of Cb has the form R = ( h× h1 × h2 ) c, where h3 = b, h3i = bi , i = 1, 2, A1 := h, h1 , h2  ∼ =

256 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

Z9 × Z9 × Z9 , and ac = a4 for all a ∈ A# 1 . Thus A := b, b1 , b2  = Ω1 (Ja (R)). G Let NA = NG (A). Then b ∩ E1 (A) = bNA . Now the irreducible constituents of the 2-group W0 := AutNG (b) (A) on A are b and b1 , b2 ; since

b ∼ b1 b2 , NA is irreducible on A. Since WA := AutG (A) has order divisible by 24 , WA ∼ = Z  2 × Σ4 is the stabilizer in Aut(A) of the frame }, by [V9 , 16.5]. Hence for any x ∈ E1 (A) − F , F := { b , b1 b2  , b1 b−1 2 ∼ AutNG (x) (A) = E23 (for example for x = bb1 b2 ) or Z2 × Σ3 (for example for x = bb1 ). Let Wbb1 = NWA ( bb1 ) ∼ = Z2 × Σ3 and let T ∈ Syl3 (NNA ( bb1 ) and T ≤ ∗ T ∈ Syl3 (NG ( bb1 )). Since WA contains A4 and |WA |3 = 3, T /CT (A) ∼ = Z3 ∗ acts freely on A. As A = Ω1 (Ja (R)), bb1  = CA (T ) = Ω1 (Z(T )). In particular, Z(T ∗ ) is cyclic. On the other hand, J z ≤ L3 (NG ( bb1 )). It follows that O3 3 (L3 (NG ( bb1 ))) = 1, so some 3-component M of the pumpup of J z in NG ( bb1 ) satisfies Z(M/O3 (M )) = 1. Clearly bb1 ∈ I3o (G), so M/O3 (M ) ∈ C3 , M/O3 3 (M ) ∈ A as e(G) = 3, and J z ∼ = L2 (8) ↑3 M/O3 (M ). But by [V17 , 13.34], there is no such group M . This contradiction completes the proof of the lemma.  Therefore we are in the following situation: (1) p = 3 and K ∼ = U3 (8); (2) CK (c) ∼ = E32 : SL2 (3) and b ∈ CK (c) − O3 (CK (c)); (12O) and  := E(C c ). (3) Let FK := O3 (CK (c)) and L Observe that CG ( z, c) is solvable, and so z normalizes and acts nonc . trivially on every component of C  = 1. Lemma 12.15. We have L  FK ] = 1. Because Proof. Assume the contrary. Suppose first that [L, ∗  of the structure of CK (c), CFK (L) = 1. By Theorem C4 : Stage A1, O3 (Cb ) has odd order, and hence by Lemma 12.13, CG ( b, c) is solvable. Hence, b    0 be a c , as does bCK (c) = CK (c). Let L normalizes every component of C  component of L on which CK (c) acts faithfully. Then z acts nontrivially on  0 is  0 and FK ≤ O3 (C  ( z )). In particular, m3 (O3 (CL0 ( z ))) ≥ 2. Since L L L0 0 ∼  0 ), a a C3 -group, [V17 , 4.27] implies that L = [3 × 3]U4 (3) and FK ≤ Z(L  = 1. contradiction. Thus [FK , L]   is quasisimple If Z(L) = 1, as e(G) = 3, [V17 , 7.9h] implies that L  ∼ and Z(L) = X ≤ CK (c) fusing all nontrivial = Z3 . Since there exists Q8 ∼  = 1. Again using [V17 , 7.9h] we elements of FK , we obtain that FK ∩ L observe that  ≥ 4 > e(G), m2,3 (CG (c)) ≥ m3 (FK ) + m2,3 (L)  is a product of simple components and a contradiction. Thus L  3 ≥ m2,3 (CG (c)) ≥ m3 ( c, FK ) + m2,3 (L).

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 = 0 and again by [V17 , 7.9], L  ∼ Hence m2,3 (L) = L2 (8). As CCG (z) (c)  via an inner automorphism. It follows that is solvable, z must act on L m2 (O2 2 (CG ( z, c)) ≥ m2 (O2 2 (CL (z)) = 3. However, CG ( z, c) = CCz (c), and with Lemma 12.13c, m2 (O2 2 (CCz (c))) = 2. This contradiction completes the proof.  c ). Looking at CK (c), we observe that b ∈ F, and using Let F := O3 (C Lemma 12.13d, we have |CF (b)| ≤ 33 . If z inverts F /  c, then the image  of z in Cc /O3 3 (Cc ) lies in the center of this quotient. On the other hand, from Lemma 12.13a, we see that there exists D8 ∼ = D ≤ CG ( b, c) ≤ Cc with z ∈ D − Z(D). Therefore, CF ( z ) >  c, and FK ∩ CF ( z ) = 1. Now the z ). Write Z(D) = t. Then we structure of CK (c) implies that FK ≤ CF ( have c ) and C  ( z ) =  c × FK ; (1) F = O3 (C F ∼ (2) D8 = D ≤ CG ( b, c) ≤ CG (c) with z ∈ D − Z(D); (12P) and (3) t ∈ I2 (Z(D)) and tz ∈ z D . Since t ∈ CJ×J z (z) = Kb , t ∈ K is 2-central in K and t inverts FK .  Now (12P1) implies that CC (z) (zt)  =  z) ∼ c, Since tz ∈ z D , CF ( = CF (tz).  F  z ) ∩ CF (tz) =  c. and so CF ( There exist d1 , d2 such that (12Q)

d1 ∈ I3 (CJ (c)), d2 ∈ I3 (CJ z (c)), and 1 = d1 d2 ∈ FK .

    We fix d1 and d2 . Then d 1 d2 ∈ F , and as [D, d1 d2 ] ≤ F , it follows that E32 ∼ = d1 , d2  ≤ F. Lemma 12.16. The following conditions hold: (a) di ∈ I30 (G), i = 1, 2; and (b) If I1 is the pumpup of J z in CG (d1 ), then I1 ≤ L3 (CG (d1 )) and one of the following holds: (1) I1 is a trivial pumpup; or (2) I1 is a vertical pumpup, I1 /O3 (I1 ) ∼ = Sp4 (8) or 3D4 (2), and c induces a non-trivial outer automorphism on I1 /O3 (I1 ). Proof. Since d1 , d2  ≤ b, c, d1 , d2  ∼ = E34 , (a) holds. 1 := C1 /O3 (C1 ). Since J z is a 3-component Let C1 = CG (d1 ) and C of CG ( b, d1 ), J z ≤ L3 (CG (d1 )) and we let I1 be the subnormal closure c induces a field of J z in CG (d1 ). If I1 is a diagonal pumpup of J z , then  1 ) ≥ 4, a contradiction. automorphism on every component of I1 and m2,3 (C Suppose that I1 is a vertical pumpup of J z . By [V17 , 7.2], I1 ∼ = Sp4 (8) or 3D (2), and because of (12N2), (b2) holds. The proof is complete.  4 Lemma 12.17. We have m3 (CG ( c, d1 )) ≥ 5.

258 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

Proof. Let F2 be a critical subgroup of F of exponent 3 containing  c.   =  z ) =  c, then CF2 (tz) c and so F2 = CF2 ( t), a contradiction as If CF2 (

 t must act faithfully on F2 by [IG , 11.11]. z )| ≥ 32 , and so F2 ∩ FK = 1, and the structure of CK (c) Hence, |C  ( F2

z ) ≤ F2 . Since F2 is characteristic in F and tz ∈ z D , implies that CF (    ≤ F2 . Finally, as d D ≤ CG (c), we have that CF (tz) 1 d2 ∈ FK ≤ F2 , the  on d1 , d2  implies that d1 , d2  ≤ F2 . irreducible action of D Suppose F2 is elementary abelian. We have that CK (c) = FK W b FK with W ∼ z ) =  c × FK . As = Q8 and t ∈ Z(W ) inverting FK . Also, CF2 (  = 33 . As tz inverts FK , m3 (F2 ) ≥ 5. In particular, tz ∈ z D , |CF2 (tz)| m3 (CG (c, d1 )) ≥ 5. Assume next that F2 is non-abelian. Then c ≤ Z(F2 ) F0 := [F2 , F2 ]  c . Consider F0z := F0 ∩ C  ( z ) = 1. is a non-trivial normal subgroup of C F2 If |F0z | ≥ 32 , F0z ∩ FK = 1, and because of the action of CK (c) on FK , we   ≤ F0 . Hence, t CF2 (tz) obtain that F0z = CF2 (z). Since F0 is D-invariant, centralizes F2 /Φ(F2 ), an obvious contradiction. c, and so F0 /  c is inverted by both z and tz, whence Thus F0z =    [F0 , t] = 1. As t inverts d1 , d2 , we have d1 , d2  ∩ F0 = 1, and as e(G) = 3, m3 (F0 ) ≤ 3. If m3 (F0 ) = 3, then F2 ≥ d1 , d2  × F0 ∼ = E35 and so   c and F2 is an m3 (CG ( c, d1 )) ≥ 5, as desired. If m3 (F0 ) = 1, then F0 =  extraspecial group. But now z acts on an extraspecial group centralizing the z ) abelian, which is impossible. Therefore it remains to center and with CF ( consider the case when m3 (F0 ) =  2.    ∼ E35 , contradicting c contains b × d1 , d2 × F0 = If [b, F0 ] = 1, C  := F0 , b is a non-abelian m3 (CG (b)) = 4. Hence, [b, F0 ] = 1, and so R    R].  Observe that [ F0 ,  c ∈ Z(R), c ∈ [R, c, d1 ] = 1, group of order 33 . Since  implying that c is contained in the commutator subgroup of a subgroup R∼ = 31+2 of CG (d1 ). 1 := CG (d1 )/O3 (CG (d1 )), We shall apply Lemma 12.16. As there, let C and let I1 be the (vertical or trivial) pumpup of J z in CG (d1 ). Whether I1 is vertical or trivial, the image of c in Aut(I1 ) is outside [Aut(I1 ), Aut(I1 )]. Thus if R normalizes I1 , then c ∈ [R, R], a contradiction.   So I1 has at least 3 R-conjugates, all c-invariant. But then m3 ( I1R , d1 , c ) ≥ 5. The lemma is proved. 

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1 = C1 /O3 (C1 ), we can apply Keeping the notation C1 = CG (d1 ) and C Lemma 12.16 again and prove: Lemma 12.18. m2,3 (C1 ) ≥ 4. Proof. Consider again the pumpup I1 of J z in C1 . If I1 is a vertical 1 as pumpup, then by [V17 , 7.2], I1 ∼ = Sp4 (8) or 3D4 (2). Hence, I1  C   e(G) = 3 and m3 (I1 ) = 2. Since m3 (Aut(I1 )) = 3 and m3 (CG (d1 )) ≥ 5, c induces m3 (CC1 (I1 )) ≥ 2. Moreover, m3 (CG ( c, d1 )) ≥ 5, and so, as   c)) ≥ 2. Hence, m2,3 (C1 ) ≥ 4, an outer automorphism on I1 , m3 (C  ( CC (I) 1

1 ) ≥ as claimed. So assume that the pumpup I1 is trivial. Since m3 (C m3 (CG ( c, d1 )) ≥ 5 and  c induces a field automorphism on I1 , there exists   I1  1 ) ≥ 4, as c] = 1. Again m2,3 (C X ≤ CG (d1 ) such that X ∼ = E33 and [X, claimed.  Since e(G) = 3, the conclusion of Lemma 12.18 is of course impossible, and Proposition 12.12 is proved. Corollary 12.19. Assume (12A) and (12B). Then Kb is a component of CG (b). Proof. Recall that C b = CG (b)/Op (CG (b)). By Propositions 12.2 and 12.12, K b is a component of C b . It remains to show that if L is the pcomponent of CG (b) with L = K b , then Op (L) ≤ Z(L). If p = 5, then m2 (Lb ) = 5 [IA , 3.3.3], and the result follows from Lemma 2.6. Suppose then that p = 3, and by (12B3), L ∼ = L2 (8). By Lemma 12.1, z inverts O3 (L). As Kb contains a Frobenius subgroup of order 8.7, we conclude from [V9 , 2.10] that L is quasisimple, completing the proof.  We will now prove the main result of this section. Proposition 12.20. Assume (12A). Then p = 3 and K ∼ = J3 , U4 (2), Sp6 (2), or 2Sp6 (2). We assume that the proposition fails, and argue to a contradiction in a series of lemmas, in the rest of this section. By Proposition 11.3, 12B holds and by Corollary 12.19, Kb is a component of CG (b). We extend our notation as follows: (12R)

(1) (2) (3)

Cb = CG (b) and C b := Cb /Op (Cb ); B0 := B ∩ K and C0 := CG (B0 ); and 0 := C0 /Op (C0 ). C

0 ) = Op (C 0 )E(C 0 ) and every component of C 0 is an As usual, F ∗ (C element of Cp .

260 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

Lemma 12.21. If Lp (C0 ) = 1, then the following conditions hold: 5  0 where (p, L 0 ) ∼ 0 ) = L (a) E(C = (3, L2 (8)), (5, A5 ) or (5, 2B2 (2 2 ));  0 has a simple preimage L0 in E(C0 ), and Lp (CG (b)) = E(Cb ) = (b) L Kb × L0 ; and (c) [z, L0 ] = L0 and [L0 , K] = 1. Proof. Let L0 be a p-component of Lp (C0 ). By Lp -balance, L0 ≤ Lp (Cb ), and so L0 ≤ E(C b ). Let L be the subnormal closure of L0 in Cb . Suppose that L is a diagonal pumpup of L0 . Then as e(G) = 3, p = 3 and m3 (L0 ) = 1. In particular, L0 ∼ = L2 (8) by [V17 , 7.9]. Now if [K b , L] = 1, then as c induces a field automorphism on K b , some element of B # induces a field automorphism on each component of L, and so m2,3 (Cb ) ≥ 4. And if [K b , L] = 1, then again m2,3 (Cb ) ≥ 4. Thus the pumpup is either vertical or trivial. If the pumpup is vertical, then CB0 (L) = b ≥ B0 ∩Kb , and so [L, K b ] = 1. But L and K b are components of C b , so L = K b . But then L0 is a component of E(CL (B0 )) = E(CK b (B0 )) = 1, which is absurd. Therefore the pumpup is trivial, [B 0 , L] = 1 and L = L0 . In particular, [K b , L] = 1, and as e(G) = 3, mp (L b) ≤ 3. This in turn implies, using [V17 , 7.9h], that L is simple and so mp (L) ≤ 2. We argue that (12S)

5 (p, L) ∼ = (3, L2 (8)), (5, A5 ), or (5, 2B2 (2 2 )).

Suppose that c induces an inner automorphism on L, and write c = c1 c2 where cp1 = cp2 = 1, [L, c2 ] = 1, c1 ∈ L, and c2 induces a nontrivial field automorphism on Kb . Then B 0 K b c2  ≤ CC b (L), and so mp (CC b (L)) ≥ 3 and m2,p (CC b (L)) ≥ 2. As e(G) = 3, mp (L) = 1, and (12S) follows from [V17 , 7.9] in this case. Otherwise, L admits an outer automorphism of order p, induced by c. Then by [V17 , 7.9] again, either (12S) holds or L ∼ = Sp4 (8), 3D4 (2), or 5 2F (2 2 ), with p = 3, 3, or 5, respectively. In every case m (C (c)) > 0 and 4 2,p L as [B0 , L] = 1, m2,p (C b ) ≥ mp (B) + m2,p (CL (B)) > 3 = e(G), an obvious contradiction. This establishes (12S). Because CG ( z, b, Kb ) = CCz ( b, Kb ) is solvable, z normalizes and acts nontrivially on L. By Lemma 12.1, however, z inverts Op (Cb ), whence L is quasisimple and L = L0 . As Op (Cb ) has odd order, L is simple [IA , 6.1.4]. 0 ) = L  01 ∗ . . . ∗ L  0m , a product of m components for some Let E(C m ≥ 1. From the argument above, E(C b ) = K b L1 . . . Lm M 1 . . . M k with  0i , (p, Li ) = (3, L2 (8)), (5, A5 ) or (5, 2B2 (2 25 )), 1 ≤ i ≤ m, and M j Li ∼ =L other components of C b , 1 ≤ j ≤ k. Since e(G) ≥ m2,p (C b ) ≥ m2,p (K b B) + m + k = 2 + m + k, m = 1 and k = 0. In particular, (a) and (b) hold. Finally, consider K0 := Kb | b ∈ B0# . By [V17 , 5.6], K0 = K, and so  [K0 , L0 ] = [K, L0 ] = 1. This completes the proof.

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Lemma 12.22. Suppose that Lp (C0 ) = 1. Choose Y ∈ Sylp (Op p (CG (b))) normalized by z, and let Y0 := [Y, z]. Then Y0 = 1 and [Y0 , K] = 1. Moreover, Y0 is abelian and is inverted by z. Proof. Consider C b . If K b is not normal in C b , then as B0 = b, B0 ∩ Kb , Lp (C0 ) = 1, a contradiction. Thus Kb  CG (b). Let Y and Y0 be as in 0 ) = Op (C 0 ), Y0 = 1. As Y normalizes the lemma. Since z ∈ C0 and F ∗ (C Kb and z centralizes Kb , Kb is centralized by [Y, z] = Y0 . Now by [V17 , 5.6], K = Kb | b ∈ B0# , and so Y0 centralizes K. Finally, Lemma 11.2 implies  that z inverts Y0 and so Y0 is abelian. This completes the proof. Lemma 12.23. There exists z  ∈ I2 (CG (B, z)) and d ∈ Ip (CG (B)) such that z  inverts d. Moreover, (z  , K, B) ∈ Kf (G). 0 ) = 1, this follows from the A × B-lemma applied to the Proof. If E(C 0 ) (see Lemma 12.22). Suppose that E(C 0 ) =: action of z × c on Op (C 5 2 ∼ L0 = 1, so that L0 = L2 (8), A5 , or B2 (2 2 ), with p = 3, 5 or 5, respectively. and [L0 , z] = L0 by Lemma 12.21. If z acts on L0 like an involution z1 ∈ L0 , then [z1 , c] = 1 and z1 lies in a D2p -subgroup of CL0 (c) ∼ = L2 (2), A5 , or 2B (2 21 ), as desired. Otherwise | Out(L )| is even, so p = 5, L ∼ A , 2 0 0 = 5 and L0 z ∼ = Σ5 with [L0 , B] = 1. But then there exists a D10 -subgroup

d, z   ≤ L0 with [B, f ] = [ B, z , z  ] = 1. Since z  ∈ L0 , [K, z  ] = 1, and the pumpup of K in CG (z  ) is either trivial or isomorphic to F4 (25 ), by [V17 , 13.11]. As F4 (25 ) ∈ A, the latter is impossible, so (z  , K, B) ∈ Kf (G) and the lemma follows.  Since we are aiming for a contradiction in our proof of Proposition 12.20, there is no loss in replacing z by z  and fixing d as in Lemma 12.23. Thus the following hold: (12T)

(1) d ∈ Ipo (G) ∩ CG (K), [d, c] = 1, and dz = d−1 ; and d := Cd /Op (Cd ). (2) Cd = CG (d), and C

We will now study CG (d). Lemma 12.24. The following conditions hold: (a) K = Lp (CG (d)); and (b) If p = 3, then K ∼ = = Sp4 (8) or 3D4 (2), and if p = 5, then K ∼ 5 2F (2 2 ). 4 Proof. Assume first that K ∼ = U3 (8). Lemma 11.2 implies that z in   acts faithfully on O3 (C d ). verts O3 (Cd ), and so O3 (Cd ) is abelian. Suppose K Since K contains a Frobenius group of order 8.7, we may argue as at the end of the proof of Corollary 12.19 to a contradiction, using that e(G) = 3. Hence, K acts faithfully on L3 (CG (d)). Since e(G) = 3, there exists a 3component L of CG (d) such that K is a component of CL (z) Since this is not possible by [V17 , 14.1, 14.2], K ∼ = U3 (8) and (b) holds.

262 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

Then by [V17 , 9.2], m2 (K) ≥ 5, and by Lemma 2.4, with d in the role of A there, there is a unique p-component L of CG (d) such that K ≤ L, L is normalized by z and K is a component of CL (z). Since d ∈ Ipo (G), L/Op (L) ∈ Cp and we may apply [V17 , 14.1] and [V17 , 14.2] to conclude that L/Op (L) ∼ = K. Since m2 (K) ≥ 5, Op (L) = 1 by Lemma 2.6 and L = K. Finally, assume there exists a p-component H = K of CG (d). Then  ≤ 2. By [V17 ,  d)  K]  = 1 and as e(G) = 3 while m2,p (K) = 1, mp (H [H,   ∼  7.9], H is simple and, as now mp (H) = 1, either p = 3 and H = L2 (8), or 5 2 ∼  p = 5 and H = A5 or B2 (2 2 ). Then d ) ≥ mp (C   (  c) ≥ 2 + 2 > e(G), c)) + m2,p (K  m2,p (C Op (Cd )H 

a contradiction. This completes the proof.

Lemma 12.25. There is a c-invariant Sylow p-subgroup Rd of C(d, K) such that (a) Rd ≤ Op p (CG (d)) and Rd is inverted by z; (b) Rd is abelian and mp (Rd ) ≤ 2; and (c) If mp (Rd ) = 2, then Ω1 (Z(Rd c)) = d. Moreover, there exists R ∈ Sylp (CG (d)) such that R = (Rd × RK ) c with b extremal in RK = R ∩ K. d ), K]  = 1, Lemma 11.2 implies that z inverts Proof. Since [Op (C d ). Hence, Op (C d ) is abelian and z acts as an element of Z(C d /Op (C d )). Op (C d /Op (C d ). It follows that In particular, z centralizes a Sylow p-subgroup of C ∗ ∗   

 cF (Cd ) is a Sylow p-subgroup of Cd /F (Cd ). It follows immediately that there exists a c-invariant Sylow p-subgroup Rd of C(d, K) satisfying (a). Since in addition m2,p (K) = 1, (b) holds. If mp (Rd ) = 2, [c, Rd ] = 1 for otherwise m2,p (CG (d)) ≥ 4 > e(G). Hence,  Ω1 (Z(Rd c)) = d and the rest of the lemma follows. Our next goal is to prove that R ∈ Sylp (G). 5

Lemma 12.26. If (p, K) = (3, Sp4 (8)) or (5, 2F4 (2 2 )), then R ∈ Sylp (G) and NG (R) ≤ NG (K). Proof. It suffices to show that NG (R) ≤ NG ( d). Using the previous lemma, we observe that Z := Ω1 (Z(R)) = d, B0 . Hence it suffices to show that NG (Z) ≤ NG ( d). Assume that NG (Z) ≤ NG ( d). 5 # By [V17 , 13.33], Kx ∼ = L2 (8) or 2B2 (2 2 ) for all x ∈ B0 , and Kx is a component of CG (x). Hence dG ∩ E1 (B0 ) = ∅. On the other hand, AutK (Z) has two orbits of size 4 on B0# (if p = 3) and is transitive on B0# if p = 5 [IA , 4.8.7]. Since 5 does not divide | Aut(Z)| when p = 3, the only possibility, in either case, is that dNG (Z) = E1 (Z) − E1 (B0 ),

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263

of cardinality p2 . It follows that AutG (Z) contains the full stabilizer of the chain Z > B0 > 1 (on which AutK (Z) acts irreducibly). Let b0 ∈ B0# and let x ∈ B0# ∩ Kb0 . By Corollary 12.19, Kb0 is a component of CG (b0 ). By the previous paragraph, there is y ∈ NG (Z) such that dy = dx and [B0 , y] = 1. Since y centralizes both b0 and x, y normalizes Kb0 . Finally, [d, Kb0 ] = 1 as Kb0 ≤ K. Therefore 1 = [d, Kb0 ]y = [dy , Kby0 ] =  [dx, Kb0 ] = [x, Kb0 ] = 1, a contradiction. This completes the proof. Lemma 12.27. Let p = 3 and K ∼ = 3D4 (2). Then NG (R) ≤ NG (K). In particular, R ∈ Syl3 (G). Let R ≤ W ∈ Syl3 (NG (K)). Then W = CW (K)(W ∩ K) c. Then Rd = CCW (K) (d) contains Z := Ω1 (Z(CW (K))). As Ω1 (CRd (c)) = d by Lemma 12.25c, it follows that CZ (c) = d, and so d ∈ Z(W )and W = R. It remains to show that NG (R) ≤ NG (K). We assume not and break the proof into a series of five lemmas. Observe that d = Ω1 (Z(R)) ∩ Rd , and let a ∈ I3 (Z(RK )). Let x ∈ NG (R) with K = K x . If e ∈ I3 (Rd ∩ Rdx ), then both K and K x are components of CG (e) and so m2,3 (CG (e)) ≥ 1 + m2,3 (L) + m3 (L) = 4, a contradiction. Hence, Rd ∩ Rdx = 1 and Rdx maps isomorphically into a normal subgroup of R/Rd . As Aut(3D 4 (2)) has 3-exponent 9 by [V17 , 12.4], it follows that exp(Rd ) ≤ 9. Lemma 12.28. We have Rdx ≤ Rd × RK . Proof. If not, let c1 ∈ Rdx − (Rd × RK ). Then [Rd , c1 ] ≤ Rd ∩ Rdx = 1. It follows by Lemma 12.25c that Rd is cyclic and Rd < Z(R). But then Rdx < Z(R), contrary to c1 ∈ Rdx .  Lemma 12.29. The group Rd is cyclic, and d ∼G a. Proof. If d ∼G a, then d ∼NG (R) a by Burnside’s argument, and so a is not characteristic in R. However,  (12U)

a = Z(R) ∩ [M, M ], M

where the intersection is over all maximal subgroups M of R. (Note that M = Rd RK shows that the right side of (12U) lies in a, while every maximal subgroup of a Sylow 3-subgroup of Aut(3D4 (2)) is nonabelian.) Therefore a ∼G d. Suppose now that Rd is noncyclic. If CRd (c) = h1  with h31 = d, then h1 ∈ Z(R) and hx1 ∈ Z(R). Let h2 be the projection of hx1 into RK . Then h2 ∈ Z(RK ) = a, and so dx = (hx1 )3 ∈ Rd , which is not the case. Hence, by Lemma 12.25c, CRd (c) = d and CR (c) has exponent 3. Let R0 = Ω1 (Rd ). Then R0x ≤ Rd × RK by the previous lemma, and the projection R1 of R0x into RK is a normal E32 -subgroup of R. So dx  = [R0 , R]x ≥ [R1 , R] = Z(RK ) = a, contradiction. The proof is complete. 

264 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

Lemma 12.30. We have Rd = d. In particular, d ∈ [R, R] and   = { d , ad , a−1 d }. Also, Z(R) = a, d and so, if R < R∗

d for some R∗ ∈ Syl3 (G), then Z(R∗ ) = a. NG (R)

Proof. Suppose that Rd > d. Then by Lemma 12.29, there exists h1 ∈ Rd with h31 = d. As Rdx ≤ Rd × RK by Lemma 12.28, we may assume that hx1 = h0 h2 with h0 ∈ Rd and h2 ∈ RK with h32 = a. Since Rd is cyclic, h1  R and so h2   RK . But this contradicts [V17 , 20.17]. Hence Rd = d, and the rest follows immediately.  Lemma 12.31. CG (a) has no 3-component whose simple quotient is isomorphic to L2 (8). Proof. By [IA , 6.1.4] the Schur multiplier of L2 (8) is trivial. If the lemma is false, let E be the product of all 3-components of CG (a) whose simple quotient is isomorphic to L2 (8). Then E  CG (a) and so Z(R∗ )∩E =  1, for any R∗ ∈ Syl3 (CG (a)). But then a ∈ E, a contradiction. Lemma 12.32. Let A = CR (b) = d, b, h  R with h3 = a. Let x ∈ NG (R) − NG (K) with dx = da. Then E(CK (bx )) = 1. Proof. Suppose not. Then K0 := E(CK x−1 (b)) ∼ = L2 (8) or U3 (3). By Lemma 12.21, every component of E(Cb ) is isomorphic to L2 (8). But since U3 (3) does not embed in L2 (8), and e(G) = 3, L3 -balance implies that −1 −1 K0 ∼ = L2 (8) is a component of E(Cb ). As dx = da−1 , [Kb , dx ] = 1. Hence, Kb ≤ K0 . As Kb  E(CG (b)), we must have [K0 , Kb ] = 1. Thus, a ∈ K0 and so a = ax ∈ K0x = E(CK (bx )). On the other hand, CR (b) ∈ Syl3 ( d×CK (b)), so as x ∈ NG (R), CR (bx ) ∼ = CR (b) ∼ = Z9 ×Z3 ×Z3 , whence x CR (b ) contains a Sylow 3-subgroup of CK (bx ). Then ax  = Φ(CR (bx )) ∈  E(CK (bx )), a contradiction. The lemma is proved. Now we can quickly reach a final contradiction proving Lemma 12.27. If bx induces an outer automorphism on K, then CR (bx ) = d, bx  × CRK (bx ) has exponent 3 by [IA , Table 4.7.3A] , contrary to the fact that h ∈ CR (b) ∼ = x x x K CR (b ). Suppose that b ∈ RK . Then b ∈ a by Lemma 12.32 and [IA , 4.7.3A]. But Kbx is a component of CG (bx ), contrary to Lemma 12.31. Hence, we may assume bx = dw for some w ∈ aK . But then bx ∈ (da)K ⊆ dG , again contrary to CG (bx ) having a component isomorphic to L2 (8). This completes the proof of Lemma 12.27. Therefore R ∈ Sylp (G) and NG (R) ≤ NG (K). In particular, as Out(K) is abelian [IA , 2.5.12], c ∈ [NG (R), NG (R)]. By [V17 , 20.1], R has no homomorphic image isomorphic to W := Zp Zp . Now Yoshida’s theorem [IG , 15.19] asserts that NG (R) controls p-transfer in G. Hence c ∈ [G, G], contradicting the simplicity of G and completing the proof of Proposition 12.20.

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13. K Is Not J3 We begin by proving several facts relevant to the components K appearing in the conclusion of Proposition 12.20, before focussing on J3 . Lemma 13.1. Suppose that (z, K, B) ∈ Kf (G). Then (a) There exists an involution z  that is 2-central in CG (K) and such that (z  , K, B) ∈ Kf (G); and (b) K is standard in G. Proof. By Proposition 12.20, p = 3 and K ∼ = J3 , U4 (2), Sp6 (2), or 2Sp6 (2). Thus by [IA , 5.6.1, 4.10.3] and [IA , 5.3, 2.5.12], m3 (K) = 3 and  Out(K) is a 3 -group. Hence as e(G) = 3, O3 (CG (z)) = K, so B ≤ K. Choose any involution z  ∈ CG (K z) and let L be the subnormal closure of K in CG (z  ). By L2 -balance, and as m3 (CG (z  )) ≤ e(G) = m3 (K), L is a single component, so (z  , L, B) ∈ Kf (G). Thus Proposition 12.20 applies to L, and B ≤ K ≤ L. So [V17 , 20.20a] implies that L = K. Taking z  to be 2-central in CG (K), as we may, we have (z  , K, B) ∈ Kf (G), proving (a). Repeating the previous argument shows that K is terminal in G. As e(G) = 3 and m2,3 (K) > 0, we must have [K, K g ] = 1 for all g ∈ G. Hence  (b) holds by definition of “standard” [IG , 18.5], completing the proof. We consider the following situation (recall the notation (2C)): (1) (z, K, B) ∈ Kf (G) with Z(K) = 1; and (2) z is 2-central in CG (K), so that Q = Q1 . Thus by Proposition 12.20, p = 3 and K ∼ = J3 , Sp6 (2), or U4 (2).

(13A)

Lemma 13.2. Assume (13A). Suppose that m2 (Q) > 1. If A ∈ E33 (K) and L := Lp (CG (A)), then Q ≤ L, L/O3 (L) ∼ = L2 (8) and Q ∼ = E23 . Proof. By (13A2), Q = Q1 and z ∈ Z(Q). Since A# ⊆ I3o (G), |O3 (CG (A))| is odd, by Theorem C∗4 : Stage A1. Let CA = CG (A) and A ) = O3 (C A )L.  Since m2 (Q) > 1 and A := CA /O3 (CA ). Then F ∗ (C C  O3 (C A )] = 1, L  = 1 e(G) = 3, Thompson’s dihedral lemma implies that [Q,  = 0. Hence, L  is  acts faithfully on L.  Since e(G) = 3, m2,3 (L) and Q ∼ quasisimple and, by [V17 , 7.9], L = L2 (8). As | Out(L2 (8))| = 3 is odd, z ∈ Q ≤ L. Let QL ∈ Syl2 (L) with Q ≤ QL . As the Sylow 2-subgroups of L are abelian, QL ≤ CG (z). In particular, QL normalizes K. Since CAut(K) (A) = A (see [IA , 5.3h, 4.8.1]), we have [QL , K] = 1, and so QL = Q. In particular, Q ∼ = E23 . The proof is complete.  Lemma 13.3. Assume (13A). If m2 (Q) > 1, then z is 2-central in G and Q = Z(CG (z)). Proof. Assume m2 (Q) > 1. Then Lemma 13.2 applies; let L be as in that lemma. Moreover, K is standard in G, by Lemma 13.1.

266 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

Let M = NG (K) and C = CG (K), and expand Q to T ∈ Syl2 (M ). Then Q = T ∩ C  T . We claim that (13B)

Q ≤ Z(T ).

Namely, NL (Q) ≤ NG (Q) ≤ M . Hence 7 divides | AutM (Q)|. If (13B) fails, then AutT (Q) = 1, so as T ≤ M , AutM (Q) = Aut(Q). Then AutM (∞) (Q) = Aut(Q). But Out(K) is solvable, so M (∞) ≤ K × C. Therefore Aut(Q) ≤ AutKC (Q), which is solvable as Q ∈ Syl2 (C). This contradiction establishes (13B). Next, T /Q embeds in Aut(K), containing a Sylow 2-subgroup of Inn(K). Hence by Proposition 12.20 and [V17 , 20.20b], m2 (Z(T /Q)) ≤ 2. Thus, m2 (Z(T )) ≤ 5, and so Q ∩ Qg = 1 for all g ∈ NG (T ). As K is standard, NG (T ) ≤ NG (K) = M , so T ∈ Syl2 (G). By (13B), z is 2-central in G. It remains to show that Q = Z(CG (z)). Since Q ∈ Syl2 (CG (K)) and | Out(K)| ≤ 2, and by (13B), it suffices to show that Q ≤ Z(CG (K z)). As e(G) = 3 = m3 (K), any 2-local subgroup of CG (K z)/ z is a 3 group. But E22 ∼ = Q/ z ∈ Syl2 (CG (K z)/ z), so by Burnside’s theorem, CG (K z) = QO2 (CG (K z)). As G is of even type, O2 (CG (K z)) ≤  O2 (CG (z)) = 1. Thus, CG (K z) = Q, and the proof is complete. Lemma 13.4. Assume (13A). Then m2 (Q) = 1. Proof. Suppose not. By [II3 , Theorem PU∗2 ], there exists g ∈ G − M −1 −1 −1 such that z g ∈ M and [z g , z] = 1. Then by Lemma 13.3, [Q, z g ] = 1 so Qg ≤ CG (z) ≤ M . Since m2 (Z(T )) ≤ 5 and Qg ∩ Q = 1, we have Qg ≤ Z(T ). In particular, g ∈ NG (Z(T )), so g ∈ NG (T ). By Burnside’s −1 −1 theorem, z g ∈ Z(T ). As z is arbitrary in Q, Qg ∩ Z(T ) = 1, for any −1 g ∈ G − M . Let ZK = Z(T ∩ K). It follows that Qg contains no K−1 conjugate of any element of Z(T ). Consequently the image E of Qg in M/C ∼ = AutM (K) contains no 2-central involution of AutM (K). If K ∼ = Sp6 (2), then Out(K) = 1, so E ≤ Z(CInn(K) (e)) for all e ∈ E # , by Lemma 13.3. But by [V17 , 10.2.4e], m2 (Z(CInn(K) (e))) ≤ 2 < m2 (E), a contradiction. If K ∼ = J3 , then by [IA , 5.3h], all involutions of E ∩ Inn(K) = 1 are 2-central in Aut(K), contradiction. And if K ∼ = U4 (2), then by [IA , 7.3.3], K = ΓE,1 (K), contradicting [IA , 18.8]. The lemma is proved.  Corollary 13.5. Assume (13A). If A ∈ E33 (K), then L3 (CG (A)) = 1. Proof. Set L = L3 (CG (A)) and suppose that L = 1. Let CA = CG (A) A ) = O3 (C A )L.  Since A  ≤ O3 (C A ) and A := CA /O3 (CA ). Then F ∗ (C and C    ∼ e(G) = 3, m2,3 (L) = 0. Hence, L is quasisimple and, by [V17 , 7.9], L =  z , O3 (CA )] = 1 L2 (8). Moreover, Thompson’s dihedral lemma implies that [ and so z ∈ L. Let QL ∈ Syl2 (L) with z ∈ QL . Since QL is abelian, QL ≤ CG (z). In particular, QL normalizes K. As in Lemma 13.2, CAut(K) (A) = A, whence

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[QL , K] = 1, and so QL ≤ Q. In particular, m2 (Q) > 1, contrary to Lemma 13.4.  Corollary 13.6. Assume (13A). Let p = 3 and K ∼ = J3 , Sp6 (2) or ∗ ∗ U4 (2). Then F (CG (z)) = Q × K and |CG (z) : F (CG (z))| ≤ 2. Proof. By Lemma 11.2, K = E(CG (z)) and a Sylow 3-subgroup of CG (z) acts faithfully on K. Hence CG (K z) is a 3 -group, so equals Q. In particular, F ∗ (CG (z)) = Q×K, and since | Out(K)| = 2, |CG (z) : (Q×K)| ≤ 2.  The main result of this section is as follows. Proposition 13.7. If (z, K, B) ∈ Kf (G), then K ∼ J3 . = As usual we will break the proof into a series of lemmas (this time nine lemmas). Assume Proposition 13.7 fails. After replacing z by a 2-central z  ∈ I2 (CG (K)), if necessary, we can work under the following assumptions and notation: (1) p = 3 and K = E(CG (z)) ∼ = J3 with z 2-central in CG (K); (2) B ≤ K; (13C) (3) X = {x ∈ B # | Kx := E(CK (x)) ∼ = A6 }; and x := Cx /O3 (Cx ). (4) For any x ∈ X, Cx := CG (x) and C Our proof will analyze the embedding of Kx in Cx for any x ∈ X. Note that B ∈ Syl3 (CK (x)) for any x ∈ X, whence (13D)

NK (B) permutes X transitively by conjugation.

Lemma 13.8. Assume (13C). Let x ∈ X. Assume that Kx acts nontrivx ). ially on O3 (C x ) such that the following  of O3 (C Then there exists a normal subgroup X conditions hold:  and x  and (a) z, Kx  acts faithfully on X,  ∈ X; (b) One of the following holds: ∼ (1) X = E3m with m ∈ {4, 5}; or 2 with |X 1 | ≤ 3, X 2 extraspecial, and |X/Z(   ∈  =X 1 × X X)| (2) X 6 8 {3 , 3 }. x )) ≥ 4. In particular, m3 (O3 (C x ) of exponent 3  to be a critical subgroup of O3 (C Proof. Choose X  Then X  is z-invariant. By Theorem C∗ : Stage A1, O3 (Cx ) with x  ∈ X. 4 has odd order. Hence, if Xz is a z-invariant Sylow 3-subgroup of the preimage in O3 3 (Cx ) of CX (z), then X ≤ O3 (CO2 (CG (z)) (x)) = O3 (CK (x)) = x z =   x. In particular, (a) x, and so z inverts X/  [IA , 5.3h]. Hence, X holds.

268 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

 ∼ Let E22 ∼ = E3m . If t ∈ I2 (E) and = E ≤ Kx . Assume first that X m3 (CX (t)) = k, then m3 (CX (tz)) = m − k. As e(G) = 3, k ≤ 3 and m − k ≤ 3. In fact, as t ∈ [Kx , Kx ] − Z(Kx ), m − k is even and m − k > 0.   x, Hence, m − k = 2 and so m ≤ 5. Finally, as Kx acts nontrivially on X/ m ≥ 4 as 5 divides |Kx |. Thus (b1) holds in this case.  is nonabelian. Then [X,  X]  =   =X 1 × Now assume that X x. Hence, X     X2 with X1 = [Z(X), z] elementary abelian and X2 extraspecial of exponent   then Kx = [Kx , t] centralizes X), 3. Let t ∈ I2 (E). If t inverts X/Z(   a contradiction as t ∈ Kx . So m3 (C  (t)) > m3 (C  (t), whence X/Z( X), Z(X) X   Kx ] = 1. m3 (C  (t)) ≤ e(G) − 1 = 2. Thus m3 (Z(X)) ≤ 3, so [Z(X), Z(X)

1 | ≤ 3, as desired. Also as [ x, t] = 1, Since e(G) = 3 and m2,3 (Kx ) = 1, |X (t) and C (tz) are extraspecial, of width at most 2 since e(G) = 3. CX/  X 1  X 1 X/  X 1 | = 36 or 38 , so (b2) holds and the lemma follows.  As C  (z) =  x, |X/ X

Lemma 13.9. Assume (13C). For any x ∈ X, Kx ≤ L3 (Cx ) = 1.  x acts trivially on E(C x ) and hence Proof. If false, then as e(G) = 3, K     faithfully on O3 (Cx ). Let X, X1 , and X2 then be as in Lemma 13.8. By [IA , 5.3h], there is s ∈ I2 (K ∩ NG ( x)) such that xs = x−1 and Kx s ∼ = P GL2 (9). As s centralizes an element of Kx of order 5, Kx s has  no faithful four-dimensional representation in characteristic 3. Hence, X must be non-abelian with Kx s acting faithfully on the extraspecial group  X 1 ) ≥ 6. Replacing s by sz if necessary, we may assume  X 1 with dim(X/ X/ , and so CX (s) is abelian. As that m3 (CX/  X 1 (s)) ≥ 3. But s inverts x   = 36 with |C   (s)| = 33 . But again s e(G) = 3, we have |X/Z( X)| X/Z(X)

 centralizes an element v ∈ Kx of order 5. As v normalizes CX (s) and Z(X),  a final contradiction. The lemma is proved. v acts trivially on X,  Lemma 13.10. Let x ∈ X. Then there exists a 3-component Jx of Cx such that Kx ≤ Jx , and either Jx /O3 (Jx ) ∼ = A6 or L4 (3), or Jx /O3 3 (Jx ) ∼ = U4 (3). The isomorphism type of Jx is independent of x ∈ X.

Proof. By Lemma 13.9, Kx is a component of CL3 (Cx ) (z). Hence by Lemma 2.4b, Kx is a component of CJx (z) for some z-invariant 3-component Jx of Cx . By [V17 , 14.1, 14.2], Jx /O3 (Jx ) ∼ = A6 , L2 (34 ), U4 (2), L± 3 (9), or L4 (3), or Jx /O3 3 (Jx ) ∼ = U4 (3). By (13D), the isomorphism type of Jx is independent of x. Again as K ∼ = J3 , there exists s ∈ I2 (NK ( x)) such that xs = x−1 and Kx s ∼ = P GL2 (9). Then Aut(Jx ) contains a subgroup isomorphic to P GL2 (9). In particular, Jx ∼ = U4 (2) by [V17 , 10.9.5]. x := Nx /O3 (Nx ). Assume that Jx ∼ Let Nx := NG ( x) and N = L2 (34 ) ± or L3 (9). Then z induces a field or a graph automorphism (respectively) on z) ∼ Jx , and CJx ( = P GL2 (9). Hence, there exists an involution w ∈ Jx with

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 x is identical to that of s. Now,  x w  = CJx ( z ), and the action of w  on K K x contains a subgroup isomorphic to  x ] = 1. Hence, N [ x, w]  = 1 and [w s, K Σ3 × Jx , violating e(G) = 3, a contradiction. This finishes the proof of the lemma.  Lemma 13.11. The following conditions hold: (a) z is not a 2-central involution in G; (b) z ∈ [Cz , Cz ], and Q = z; and (c) Let x ∈ X. If Jx /O3 (Jx ) ∼ = A6 , then there exists d ∈ I3 (CG (K)) such that dz = d−1 . The proof of this lemma will occupy the next three lemmas. Fix x ∈ X and let Jx be as in Lemma 13.10. Thus Kx ≤ Jx   Cx and Jx /O3 (Jx ) ∼ = A6 or L± (3). 4 Let t ∈ I2 (Kx ) and Ct = CG (t), so that t is a 2-central involution of K. Observe that t ∈ z G .

(13E)

(∞) Indeed, CK (t) ∼ = 21+4 A5 , so t ∈ Ct , while F ∗ (Cz ) = Q × K and m2 (Q) = (∞) 1, whence z ∈ K = Cz . We now continue the proof of Lemma 13.11 by considering the three possible isomorphism types of Jx /Z(Jx ).

Lemma 13.12. Lemma 13.11 holds if Jx /Z(Jx ) ∼ = U4 (3). Proof. By [IA , 4.5.1], Jx has a unique conjugacy class of involutions, and their centralizers are 2-constrained in Jx . Hence, z induces an outer automorphism on Jx , and as t ∈ Kx ≤ Jx , t is a 2-central involution of Jx . Observe that as Q centralizes K ≥ x × Kx , Q ≤ Cx , and |Q| = 2 by [V17 , 10.10.4]. In particular, z ∈ [Cz , Cz ] and 28 ≤ |S| ≤ 29 . Assume that z is a 2-central involution of G. Since z must centralize a subgroup isomorphic to Σ3 Y P GL2 (9) in Nx , |Nx |2 ≥ |U4 (3)|2 · 22 = 29 . It follows that Nx contains a Sylow 2-subgroup S0 of G. But Z(S0 ) is cyclic by [V17 , 10.10.5], while we can see in Cz that Z(S) is not. This is a contradiction that completes the proof.  Lemma 13.13. Lemma 13.11 holds if Jx ∼ = L4 (3). Proof. Take w ∈ I2 (NK ( x) that inverts x. Note that C(x, Jx ) has odd order (as e(G) = 3 while m3 (Jx ) = 4). Therefore w induces a nontrivial outer automorphism on Jx . z )) ∼ Assume first that E(CJx ( = A6 and z induces an inner automorphism z ). Let Sz ∈ Syl2 (C  ( z )) with  t ∈ Z(Sz ) and SJ ∈ on Jx . Consider C  ( Jx

Jx

Syl2 (Jx ) with Sz ≤ SJ . Observe that by [V17 , 10.10.6], Sz ∼ = D8 × D8 with z and t corresponding to the central involutions in the two D8 direct factors. Since z is not 2-central in Jx , there exists g ∈ NSJ (Sz ) − Sz . By the

270 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

Krull-Schmidt theorem, g interchanges z and  t, contradicting (13E). Thus  z induces an outer automorphism on Jx . As Q ≤ Cx , it follows that |Q| = 2 and so |S| ≤ 29 . In particular, z ∈ [Cz , Cz ]. Finally, assume that z is 2-central in G. As |Jx z|2 = 28 and w ∈ Nx − Cx , |Nx |2 ≥ 29 . As |S| ≤ 29 , equality holds. Let S0 ∈ Syl2 (Nx ). Then by [V17 , 10.10.7], m3 (CJx S0 (Z(S0 )) = 2. But Z(S) = z, t and  m3 (CG ( z, t)) = 1, a contradiction completing the proof. Lemma 13.14. Lemma 13.11 holds if Jx ∼ = A6 . Proof. Suppose that Jx ∼ = A6 . By Lemma 12.1, z inverts O3 (Cx ). As Nx contains P GL2 (9), which in turn contains a Frobenius group of order 9.8, it follows by [V9 , 2.10] that Jx is quasisimple, so that Jx = Kx . By Corollary 13.5, Lp (CG (B)) = 1. Hence, z acts nontrivially on O3 3 (CG (B))/O3 (CG (B)), and using the Baer-Suzuki theorem, we obtain that there exists d ∈ I3 (CG (B)) inverted by z. Now d ∈ CG (B) normalizes the component Kx of Cx , and [Kx , z] = 1, so [Kx , d] = 1. As x ∈ X is arbitrary, and K = Kx | x ∈ X by [V17 , 10.13.3], we have [K, d] = 1. By Lemma 13.4 and Corollary 13.6, C(z, K) = Q and m2 (Q) = 1. Thus, CG (K) = XQ with X = O2 (CG (K)) and d ∈ X. As m2,3 (K) = 2, a Sylow 3-subgroup R of X is cyclic, whence Q = z and z ∈ [Cz , Cz ]. It remains to prove that z is not 2-central in G. Assume it is 2-central and consider Ct = CG (t). If F ∗ (Ct ) = O2 (Ct ), then z ∈ O2 (Ct ). But then X = [X, z] ≤ O2 (Ct ), an impossibility. Hence, E(Ct ) =: L = 1. If CK (t)∩L ≤ t, then as CK (t) is perfect, Ct contains a subgroup isomorphic to CK (t) × L or CK (t) ∗ L with t ∈ L in the latter case. In either case, |L/Z(L)|2 ≤ 2, which is impossible. Hence, O2 (CK (t)) ≤ L and there is a component L1 of L with O2 (CL1 (z)) = CK (t). As CCt ( z, CK (t)) = z, t, it follows that F ∗ (Ct ) = L1 = L and O2 (CL (z)) = CK (t) with t = Z(L)  = L/Z(L). As L  ∈ C2 , [V17 , 20.18] implies that and 27 ≤ |L|2 ≤ 29 . Let L L cannot exist, a contradiction completing the proof.  The proof of Lemma 13.11 is complete. Since t is 2-central in K, we may assume see [IA , 5.3h] that

t = Z(SK ). Recall that Ct = CG (t) and S ≤ P ∈ Syl2 (G). Without loss, P ∈ Syl2 (Ct ). Lemma 13.15. F ∗ (Ct ) = O2 (Ct ). Proof. Suppose false, so that L := E(Ct ) = 1. Since F ∗ (CG ( z, t)) = O2 (CG ( z, t)), z normalizes every component of L. Also, as z is not 2central in G and t, z = Z(S), we have Z(P ) = t. Since Z(P ) is cyclic, every component L1 of L satisfies Z(L1 ) = 1, and of course L1 ∈ Co2 . Let C0 = CG ( z, t)∞ = O2 (CG ( z, t)) = CK (t) ∼ = 21+4 − A5 . Since e(G) ≤ 3 and C0 is perfect, C0 normalizes all components of L. Then as

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C0  CG ( z, t) and with the Schreier property, C0 ≤ L1 for some component  1 = L1 /Z(L1 ); then L1 of L. In particular t ∈ L1 . Set L X  O 2 (CL1 (z)) = E ∼ ∼ where X = A5 and E = E24 is the core of the natural permutation module  over F2 . This restricts the isomorphism type of L1 ∈ Co . If L1 ∈ Spor, for X 2 then by [IA , 5.3], L1 ∼ = 2HS; but then L1  Ct as e(G) ≤ 3, and then a Sylow 2-center of G is noncyclic, by [IA , 6.4.1, 6.4.2], a contradiction. If  X  must be a central product of Lie L1 ∈ Chev(3), then O3 (CL1 (z)) = E X  is not such a components in characteristic 3, by [IA , 4.2.2, 4.9.2]. But E product, contradiction. As Z(L1 ) = 1, and by definition of Co2 , we therefore must have L1 ∈ Chev(2). Since t = Z(P ), no 2-central involution of L1 / t splits and is stable relative to L1 . Using [IA , 6.1.4, 6.4.1, 6.4.2] and [IA , 4.10.3], and the fact that mr (L1 ) ≤ 3 for all odd primes r, we see that L1 ∼ = 2Sp6 (2). Hence Aut(L1 ) = Inn(L1 ). But then since [z, v] = 1 for some v ∈ I5 (L1 ), it follows easily that z acts on L1 as a transvection, i.e., a long root element. Hence CL1 (z) involves A6 . But CG ( z, t) does not involve  A6 , a contradiction completing the proof. Lemma 13.16. We have z ∈ O2 (Ct ), and either O2 (Ct ) is of symplectic type, or z, t  Ct . Proof. We have S ∈ Syl2 (CG ( z, t)) ⊆ Syl2 (Cz ). Then Z(S) =

z, Z(E) = z, t. By Lemma 13.11b, z ∈ [S, S], and so t = Z(S) ∩ [S, S]. We have P ∈ Syl2 (G) with S ≤ P . Now F ∗ (CG ( z, t)) = z × E where E is an extraspecial 2-group with [E, E] = t, so t = Z(P ). By Lemma 13.11a, z ∈ tG and S < P , so z and tz are P -conjugate. Therefore t is weakly closed in Z(S). As z ∈ G = [G, G], it follows by [V9 , 1.2] that (13F)

z ∈ [Ct , Ct ].

Set Qt := O2 (Ct ) = F ∗ (Ct ). The action of CG (t, z) on E implies that CQt (z) = E or E z. Suppose that Qt is of symplectic type. If z ∈ Qt , then one of the allowed conclusions holds. So we may assume that CQt (z) = E. If Qt is extraspecial, then z also acts on a complement C := CQt (E) centralizing only t. Therefore C is of maximal class, and z induces a transvection on C/ t and hence, also on Qt / t since Qt = EC. But then (by the Dickson invariant) z ∈ [Ct , Ct ], a contradiction. Therefore Qt is not extraspecial. Therefore there is a characteristic subgroup Y ≤ Qt such that Y is cyclic of order at least 4, |Qt : CQt (Y )| ≤ 2, and if equality holds, then Qt is the central product of an extraspecial group with a group of maximal class. Moreover, if |Y | = 4, then Y = Z(Qt ). Now since Y  Ct and Aut(Y ) is abelian, [Y, z] = 1, by (13F). On the other hand, Z(CQt (z)) = Z(E) = t. Thus if [Qt , Y ] = 1, then CY (z) = t, a contradiction. Therefore |Qt : CQt (Y )| = 2 and |Y | ≥ 8. Again, as CQt (z) = E is of exponent 4, z acts nontrivially on Y , contradicting (13F). We have proved that Qt is not of symplectic type.

272 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

Therefore by Philip Hall’s theorem [IG , 10.3], there exists a characteristic but noncyclic elementary abelian subgroup U ≤ Qt . Then CU (z) lies in a normal abelian subgroup of CG ( z, t) and contains t. Therefore CU (z) = t or z, t. In the first case U must be a four-group on which z acts nontrivially, and so z ∈ [Ct , Ct ], a contradiction. Therefore z ∈ U = z, t and we are done.  We now complete the proof of Proposition 13.7. Thanks to Lemma 13.11, CG (z) = ( z × K) w where w ∈ S and 2 w ∈ z. Thus, S = ( z × SK ) w. As before, let t = Z(SK ). Then Z(S) = z, t. Since z is not 2-central in G while Z(S) is characteristic in S, there exists u ∈ NP (S) − S such that z = z u ∈ Z(S). Since t ∈ [Ct , Ct ] while z ∈ [Cz , Cz ], z u = t, and so Z(S)# = {z, z u = zt, t}. Since Z(P ) ≤ Z(S), Z(P ) = t. By Lemma 13.16, z ∈ O2 (CG (t)) and either O2 (CG (t)) is of symplectic type, or z, t  CG (t). Let x ∈ X, so that Kx ∼ = A6 . Without loss of generality we may assume that [t, x] = 1, and so t ∈ Kx . By Lemma 13.10 there exists a 3-component Jx of Cx with Kx ≤ Jx , and Jx ∼ = U4 (3). = A6 or L4 (3), or Jx /Z(Jx ) ∼ , then by Lemma 13.11, there exists a 3-element d ∈ CG (K) A If Jx ∼ = 6 inverted by z. Then d ∈ Ct , and as z inverts d, z ∈ O2 (Ct ), a contradiction. Hence, Jx /Z(Jx ) ∼ = L± 4 (3). Then m3 (CCx (t)) = 3. If z, t  CG (t), as G z ∈ t , m3 (CG ( z, t)) ≥ 3. Since m3 (CCz (t)) = 1, this is not possible. Hence, Qt is of symplectic type. As CQt (z) ∼ = O2 (Cz×K (t)), and m3 (Ct ) = 3, we must have Qt ∼ = Q8 ∗ Q8 ∗ Q8 . Also, Ct /Qt contains a subgroup isomorphic to A5 (as can be seen in K). By [V17 , 10.9.8], Ct /Qt contains ∼ U4 (2) ∼ = Ω− 6 (2) = [Out(Qt ), Out(Qt )]. Therefore z is Ct -conjugate to all the involutions in CK (t) ∩ Qt − t. But all involutions in K are K-conjugate to t. Therefore z ∈ tG , a contradiction. This completes the proof of Proposition 13.7. 14. From U4 (2) to 2Sp6 (2) In this section we will prove the following statement. Proposition 14.1. Assume (z, K, B) ∈ Kf (G), with p = 3 and K ∼ = U4 (2). Then there exists u ∈ I2 (G) with E(CG (u)) ∼ = 2Sp6 (2). Assume the contrary. As usual we break the proof into a series of lemmas. Because m3 (K) = 3, and by Lemma 13.4, we are now working under the following conditions. (1) p = 3, K ∼ = U4 (2) and B ≤ K; and (14A) (2) m2 (Q) = 1. Lemma 14.2. For all x ∈ B # , L3 (CG (x)) = 1.

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Proof. By Lemma 13.6, F ∗ (CG (z)) = Q × K and |CG (z) : F ∗ (CG (z))| ≤ 2. x = Cx /O3 (Cx ) Let x ∈ B # with L3 (CG (x)) = 1. Set Cx = CG (x) and C   and take Lx to be a component of Cx . As CG ( z, x) is a {2, 3}-group, z  x and C  ( z ) is a {2, 3}-group. Moreover, as K ∈ acts nontrivially on L Lx

Chev(3), O3 (CLx ( z )) = 1 by the Borel-Tits theorem, and Ω1 (O2 (CLx ( z ))) ≤  x ∈ C3 , we have m2 (O2 (C  ( z ))) ≥ 2, a

 z . By [V17 , 4.27], however, as L Lx contradiction that completes the proof. 

By the Z ∗ -theorem, z G ∩ S − {z} = ∅. The next two lemmas are due to K. Gomi. Lemma 14.3. For any u ∈ z G ∩ S with u = z, CQ (u) = z. Proof. Assume not, and let Q0 ≤ CQ (u) with |Q0 | = 4. By [V17 , 10.9.3a], if u ∈ SK × z, then CSK (u) contains a subgroup X ∼ = E24 , while if u ∈ S − QSK , then CSK u (u) contains X ∼ = E24 . Thus in any case, CS (u) ≥ Q0 × X with X ∼ = E24 . Since u = z g for some g ∈ G, Q0 × X ≤ S g with Q0 ∩ Qg = 1. Hence, (Q0 × X)/Qg embeds in AutG (K g )  with (Q0 × X)/Qg ∼ = Z4 × E23 , contrary to [V17 , 10.9.3b]. Lemma 14.4. We have Q = z. Proof. Assume Q = z. Let u = z g ∈ S − z. As CQ (u) = z by Lemma 14.3, we have that u ∈ Q × SK and u is Q-conjugate to zu. Now, z ∈ Qg K g . Hence, by [V17 , 10.9.3d], there exists y ∈ I2 (CK g (z)) g with zy ∈ z K . Then y = z(zy) ∈ z G . But z G = uG and uG ∩ K g = ∅, a contradiction.  We now choose a 2-central involution t ∈ K so that we have the following: (1) t ∈ I2 (Z(SK )) with A := CB (t) ∼ = E32 ; (2) O 2 (CG ( z, t)) = (Q1 ∗ Q2 )A with Q1 ∼ = Q2 ∼ = Q8 and (Q1 ∗ Q2 )A ≤ CK (t); (14B) (3) |CG ( z, t) : O2 (CG ( z, t))| = 4 or 8; (4) R0 := Q1 ∗ Q2 ; and (5) Ct := CG (t). Lemma 14.5. We have z ∈ [Ct , Ct ]. Proof. Since z, t = Z(S) and z ∈ [Cz , Cz ], t is 2-central in G and = {t}. The result follows by the simplicity of G and [V9 , 1.2]. 

tG ∩ Z(S)

Lemma 14.6. Let N  Cz ∩ Ct . Then either N ≤ z, t or R0 ≤ N . Proof. Assume that N ≤ z, t. We have Q0 := O2 (Cz ∩Ct ) = z×R0 . Now Q0 / z, t is a sum of two non-isomorphic irreducible A-modules interchanged in CK (t)/ t. So, Q0 / z, t is the unique minimal normal subgroup of (Cz ∩ Ct )/ z, t. Hence, N z, t contains Q0 . It follows that  N ≥ [N, A] ≥ [Q0 , A] = R0 .

274 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

Lemma 14.7. We have E(Ct ) = 1. Proof. Assume now that L = E(Ct ) = 1. If 1 = N  Ct with CN (z) ≤

z, t, then as CN (z)  CCt (z), we have R0 ≤ N by Lemma 14.6. Let L0 be the product of all simple components of Ct . Then L0  Ct and t ∈ L0 , whence CL0 (z) ≤ t1  by Lemma 14.6, where t1 ∈ {z, zt}. If L0 = 1, then t1  ∈ Syl2 (L0 ) which is absurd. Hence, L0 = 1. Thus for all components L1 of Ct , we have Z(L1 ) = 1 and by [V17 , 9.3], we have 3

L1 /Z(L1 ) ∈ {M12 , M22 , 2B2 (2 2 ), L3 (4), Sp6 (2), G2 (4)}. ∼ 2Sp6 (2). As we are assuming that Proposition 14.1 is false, L1 = 3 2 ∼ If L1 /Z(L1 ) = B2 (2 2 ), A normalizes L1 , for otherwise z normalizes every A-conjugate of L1 (as CG ( t, z) is a {2, 3}-group) and as L1 admits only inner 2-automorphisms, m2 (CG ( t, z) > 5 = m2 (Cz ), contradiction. But now there exists a ∈ A# centralizing L1 (as m3 (Aut(L1 )) = 1), and as m2 (L1 ) = 4 or 5 by [V17 , 9.2], L1 is a component of CJ (t) for some 3-component J of CG (a), by Lemma 2.4. Since this is not possible by [V17 , 3 2B2 (2 2 ). 14.1], L1 /Z(L1 ) ∼ = Therefore the 3-rank is 2 for each component of Ct , and so, as m3 (Ct ) ≤ 3, L is quasisimple with L/Z(L) ∈ {M12 , M22 , L3 (4), G2 (4)}. If CL (z) ≤ z, t, then a Sylow 2-subgroup of L, z is dihedral or semidihedral, which is clearly not the case. Hence, R0 ≤ CL (z) and A acts faithfully 2M12 , 2M22 , or 2L3 (4). Thus L ∼ on CL (z), whence L ∼ = = 2G2 (4). ∼

 Let Ct := Ct /C(t, L). We claim that A = A. For otherwise, choose a ∈ CA (L)# . As m3 (Ct ) ≥ m3 ( a L) = 3, and m2 (2L3 (4)) > 4 [IA , 6.4.4], Lemma 2.4 implies that L  E(CJ (t)) for some 3-component J of CG (a),  ≤ C  ( contradicting [V17 , 14.1]. Thus the claim holds and E32 ∼ z ). By =A L [V17 , 20.19], z induces a field automorphism on L, so CL (z) is nonsolvable, contradicting the structure of CG ( t, z) given in (14B2). This completes the proof.  We now let T := O2 (Ct ) = F ∗ (Ct ) and C t = Ct /T . Lemma 14.8. Either |Ct : CCt (z)| ≤ 2, or T is extraspecial with R0 ≤ T . Proof. If CT (z) ≤ z, t, then T z is dihedral or semidihedral, by [IG , 10.24]. But A acts faithfully on T z, a contradiction. Hence, R0 ≤ CT (z) by Lemma 14.6. Suppose that T is not of symplectic type. By [IG , 10.3], there exists a characteristic elementary abelian subgroup U of T with |U | ≥ 4. If z ∈ U , U = z, t by the structure of CCz (t). If z ∈ U , then CU (z) = t and so |U | = 4. But then z ∈ [Ct , Ct ] contrary to Lemma 14.5. Hence, z ∈ U and |Ct : CCt (z)| ≤ 2 as claimed.

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If T is of symplectic type but not extraspecial, then by [V9 , 5.1], T has a cyclic characteristic subgroup C0 with |C0 | ≥ 4. As z ∈ [Ct , Ct ], by Lemma 14.5, we have C0  CCz (t) which is impossible by Lemma 14.6, completing the proof.  Lemma 14.9. Either |Ct : CCt (z)| ≤ 2, or T = Q1 ∗Q2 ∗D0 with D0 ∼ = D8 or Q8 . Proof. Suppose |Ct : CCt (z)| > 2. By Lemma 14.8, T is extraspecial and CT (z) = R0 or R0 × z. If z ∈ T , then T = R0 ∗ D0 with D0 ∼ = D8 , as claimed. If z ∈ T , then CT (z) = R0 and D0 := CT (R0 ) is extraspecial. Then z normalizes D0 , and CD0 (z) = D0 ∩ R0 = t, so with [IG , 10.24], D0 ∼ = D8 or Q8 .  Lemma 14.10. We have |Ct : CCt (z)| = 2. Proof. Assume not. If Ct < Cz , then Ct ∩ K  Ct of index 2 or 4, whence z ∈ [Ct , Ct ] contrary to Lemma 14.5. Thus by Lemma 14.9, T = Q1 ∗ Q2 ∗ D0 with D0 ∼ = D8 or Q8 . Now Ct is isomorphic to a subgroup ± of O6 (2), and CCt (z) contains A w as a normal subgroup of index 1, 2, or 4, where w ∈ I2 (K) interchanges Q1 and Q2 . Let V := T / t. Suppose first that z ∈ T . As z ∈ [Ct , Ct ] by Lemma 14.5, z ∈ Ω± 6 (2). Hence, by the Dickson invariant, z does not induce a transvection on V . But Q1 ∗ Q2 = CT (z), and z normalizes D0 = CT (A), so z induces a transvection on V , contradiction. Hence, z ∈ T with D0 ∼ = D8 and C is isomorphic to a subgroup of O6+ (2). Now the hypotheses of Lemma [V17 , 20.2] are satisfied with H = Ct . As z ∈ [Ct , Ct ], we conclude that z, t  Ct , whence |Ct : CCt (z)| = 2 as claimed.  Lemma 14.11. Let z, t ≤ E ∼ = E25 with E ≤ z×K. Let N := NG (E). A or Σ6 . Then |z N | = 6 and N/E ∼ = 6 Proof. We have S  P = S w ∈ Syl2 (G) with w2 ∈ S and t ∈ Z(P ). t = Ct /O2 (Ct ) = Ct /O2 (Cz,t ) by Lemma 14.10. Let Cz,t = CG (z, t) and C Then t = C z,t w, z,t ; (1) C  w 2 ∈ C  =  z,t =  z , e, u , E e, u , e ∈ E ∩ K # ; (2) C (14C) (3) Cz = z × K u, u2 = 1; and z,t ∼  = 1 or u  = 1. (4) C = E23 or E22 according as u t ]| ≤ 2 and as z ∈ [Ct , Ct ], [C t , C t ] =  z,t , w] t , C z . So [C  =  z . So |[C  Suppose that w does not normalize E. Note that E = J( z SK ). So w   z  =  z , e. But [  z  SK , w]  ≤  z , contradicdoes not normalize  z  SK =  tion. Thus, w ∈ N , so zt ∈ z N .

276 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

As |tN ∩K | = 5, |z N | ≡ 1 (mod 5), so |z N | = 6, 11, or 16. But 11 does not divide | Aut(E)|, and |G : CN (z)| is not divisible by 4. Hence |z N | = 6.  As CN (z)/E ∼ = A5 or Σ5 , finally, N/E ∼ = A6 or Σ6 . Now we complete the proof of Proposition 14.1 by a transfer argument. Lemma 14.12. z ∈ [G, G]. Proof. Let P ∈ Syl2 (N ) with t ∈ Z(P ). Now |Ct : Cz,t | = 2 and we may assume that |P : CP (z)| = 2. We have Cz = ( z × K) v with K ∼ = U4 (2), v ∈ P , v 2 ∈ z, and either v = 1 or Cz / z ∼ = Aut(U4 (2)). Thus, Cz,t = ( z × CK (t)) v and Z(CK (t) v) = t, so z, t = Z(Cz,t )  Ct . Now, E is the quotient of a standard permutation module for N/E by the trivial submodule. As |z N | = 6, z ∈ E0 , where E0 is the irreducible N/E-submodule of E of codimension 1. We consider N := N/E0 . As N contains a subgroup EJ with J ∼ = A5 , we see that N contains a subgroup ∼ H × z = A6 × Z2 of index 1 or 2. Let H be the preimage of H in N , so that P1 := P ∩ H  P with |P/P1 | = 2 or 4 and with z ∈ P1 . Hence, P has a subgroup P0 of index 2 or 4 with z ∈ P0 and P0 cyclic. Therefore if w ∈ [G, G], the Thompson transfer lemma [IG , 15.15] implies that there exists x ∈ G such that w := z x ∈ P0 with |P : CP (w)| ≤ 2 and CP (z)x ≤ CP (w). As z is not 2-central in G, equality holds, and we have w, t = Z(CP (w)) and z, tx = w, t. In particular tx = t. But

z, t  CG (t), whence w = z x ∈ z, t. However, w ∈ P0 and z, t∩P0 = t, a contradiction. The lemma is proved.  Of course, since G is simple, Lemma 14.12 cannot hold. This contradiction completes the proof of Proposition 14.1. 15. K ∼ = Co3 = 2Sp6 (2) Implies G ∼ Proposition 15.1. Let (x, K, B) ∈ Kf (G). If p = 3 and K ∼ = 2Sp6 (2), ∼ then G = Co3 . In addition to our usual notation (2C), let us set (15A)

H := CG (K),

and Cz = CG (z), and write Z(K) = z1 . By Lemma 13.1, K is standard in G. In particular, (z1 , K, B) ∈ Kf (G). Replacing z by z1 , we may assume that

z = Z(K), whence z ∈ Z(Q). Also Out(K) = 1 [IA , 2.5.12], so Cz = KH. Lemma 15.2. The following conditions hold:  (a) B ≤ K = O 3 (Cz ) and |H|3 = 1; (b) m2 (Q) = 1; (c) H ∩ H g has odd order for all g ∈ G − NG (K); and

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277

(d) z G ∩ H = {z}. Proof. Since e(G) = 3 = m3 (CG (z)), B ≤ K and |H|3 = 1; then (a) follows as Cz = KH. As K is standard by Lemma 13.1, (c) holds. Therefore z G ∩ H = z NG (K) = {z} as Z(K) = z. Assume that m2 (Q) > 1. By Lemma 13.2, J := L3 (CG (B)) satisfies J/O3 (J) ∼ = L2 (8) and E23 ∼ = Q ∈ Syl2 (J). As Q ≤ H, (c) implies that NJ (Q) ≤ NG (K) ≤ NG ( z), contradicting the fact that NJ (Q) permutes  Q# transitively. Thus m2 (Q) = 1 and the proof is complete. Let Cz := Cz /H ∼ =K∼ = Sp6 (2). Let ti ∈ K, 0 ≤ i ≤ 3, be such that the images ti represent the four classes of involutions in K, as in [V17 , 10.2.4a]. Thus t0 is not 2-central in K. By [V17 , 10.2.4d], ti ∼K ti z for each i = 0, 1, 2, 3.

(15B)

Lemma 15.3. Let y ∈ z G ∩ S − {z}. Then (a) AutG ( y, z) ∼ = Σ3 ; and C

(b) y ∈ t0 z . Proof. By Lemma 15.2d, y = 1. Hence, by (15B), y ∼Cz yz. Likewise in CG (y), z ∼CG (y) zy. This implies (a). Suppose that (b) fails. We may then assume that y = t0 . Let X =   3 O (CG ( y, z))  NG ( y, z). Then as O3 (Cz ) = K, and by [V17 , 10.2.4e],  X = O 3 (CK (t0 )) = O2 (X) u with z ∈ O2 (X) ∼ = Q8 ∗ Q8 and u3 = 1. But then z  NG ( y, z), contradicting (a).  Lemma 15.4. We have Q = z and Cz = K. Proof. Suppose false. Since G is of even type and Cz = KH, we have Q > z. By the Z ∗ -Theorem, there exists y = z g ∈ S − {z}. Write y = y1 y2 with y1 ∈ Q and y2 ∈ SK − z and y12 = y22 ∈ {1, z}. In either case, let Qy = CQ (y). As z ∈ Qy , we have y ∈ R0 := CQg (z). In either case, |R0 | > 2, by assumption. By Lemma 15.2b, R0 ∩ H = 1. Then R0 ∼ = R0 < CCz (y).   g 3 3 As |H |3 = 1, [R0 , O (CG (y))] = 1, so [R0 , O (CK (y))] = 1. Now we use Lemma [V17 , 10.2.4]. We may assume that y = ti for some i, 0 ≤ i ≤ 3. By Lemma 15.3b, y = t0 . If y = ti with i ∈ {1, 2}, then by  [V17 , 10.2.4fg], CCK (y) (O3 (CK (y))) = y and so |R0 | = 2, a contradiction. Hence, y = t3 . By [V17 , 10.2.4b], t3 has order 4 in K. Hence y = t3 yQ with yQ ∈ Q of order 4. Let b1 ∈ I3 (CK (t3 )) be as in [V17 , 10.2.4h]. Then b1  ∈ Syl3 (CG ( y, z)) so NG ( b1 ) covers NG ( y, z)/CG ( y, z). Set Y = O2 (CCG (y,z) (b1 )). Thus, NG (Y ) covers NG ( y, z)/CG ( y, z). By Lemma [V17 , 10.2.4h], Y embeds in Y 1 := O2 (CL (t3 )) ∼ = D8 × Z2 , and t3 is a square in Y . Hence t3 is a square in Y , so z is a fourth power in Y . Since NG (Y ) covers AutG ( y, z), y is a fourth power in Y . Hence y is a fourth  power in Y . But Y ≤ Y 1 , which has exponent 4, contradiction.

278 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

By [V17 , 10.2.4c], Z(S) = z, so S ∈ Syl2 (G). Now let t = t0 and u = t1 . By Lemma 15.4 and [V17 , 10.2.4ab], z, t, and u represent the Cz conjugacy classes of involutions in S. We may replace t and u by extremal conjugates lying in S. By [V17 , 10.2.4f], u is 2-central in K, so we may assume that z, u  S. Then by the Thompson transfer lemma, we may replace t by an extremal conjugate and assume that [t, u] = 1. Consider Ct := CG (t). Lemma 15.5. The following conditions hold: (a) t ∈ z G ; (b) u ∈ z G ; (c) CG ( z, t) = t × C0 , with Z(C0 ) = z and C0 ∼ = CM12 (z0 ), where z0 is a 2-central involution in M12 ; (d) I2 (G) = z G ∪ tG ; (e) z G ∩ Ct = z Ct ; and (f) u ∈ z Ct . Proof. The Z ∗ -Theorem and Lemma 15.3b imply (a), (b), and (d), and also that tG ∩ Cz = tCz . Lemma 15.4 and [V17 , 10.2.4e] imply (c). Since tG ∩ Cz = tCz , G is transitive on the set of all commuting pairs  (z · , t· ) ∈ z G × tG , which implies (e). As u ∈ Ct , (e) implies (f). We now complete the proof of Proposition 15.1. Let T0 ∈ Syl2 (CCt (z)) t = Ct / t. Then T0 = T0 / t ∈ and set z0  = Z(T0 ) ∩ [T0 , T0 ]. Also let C t ) with  z  = Z(T0 ) and CCt ( z) ∼ Syl2 (C = CM12 (z0 ) for z0 a 2-central involution in M12 .  t has the involution fusion pattern of Also, u  ∈ zCt by Lemma 15.5, and C  2 M12 . As O2 (Ct ) = 1 and O (CCt (z)) = CCt (z), it follows by [V17 , 10.11.1] t is a simple C2 -group. Hence, by Lemma [V17 , 10.11.2], C˜t ∼ that C = M12 . ∼ Then, as T0 splits over t, we conclude that Ct = Z2 × M12 . Moreover, the fusion of involutions in Cz has been shown to be uniquely determined. It follows that G has the centralizer of involution pattern [I1 , Section 16, footnote] of Co3 , i.e., G ≈ Co3 . Hence by our Background Result [I1 , (16.1)], G∼ = Co3 , so Proposition 15.1 is proved. 16. The Z2 × Sp6 (2) Case Our goal here is to rule out this case, by constructing a U6 (2) subgroup – which has a graph automorphism whose fixed subgroup is Sp6 (2) – and using the Thompson transfer lemma to obtain a contradiction. That is, we prove: Sp6 (2). Proposition 16.1. Let (z, K, B) ∈ Kf (G). Then K ∼ = Throughout this section, we assume (z, K, B) ∈ Kf (G) and K ∼ = Sp6 (2), and we argue to a contradiction. By Lemmas 13.1 and 13.4, we may assume that z is 2-central in CG (K), and that m2 (Q) = 1. As Out(K) = 1 [IA ,

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2.5.12], and m3 (K) = 3 = e(G), CCz (K) is a 3 -group, hence a 2-group since G is of even type. Thus, Cz = K × Q. The main work is to show that G is of U6 (2)-type, see [V16 , 7.4]. Proposition 16.2. Let z ∈ I2 (G) with Cz = Q×K where Q is a 2-group of 2-rank 1 and K ∼ = Sp6 (2). Then G is of U6 (2)-type. That is, there exist t ∈ I2 (G), D ∈ E34 (G), x ∈ D # , and subgroups L, o Nx , and N o of G such that the following conditions hold: and F ∗ (CG (t)/F ∗ (CG (t))) ∼ (a) F ∗ (CG (t)) ∼ = 21+8 = U4 (2); +  o o o ∼  := N /D = Σ6 , and O3 (NG (D)) (b) CG (D) = D ≤ N ≤ NG (D), N ≤ N o; (c) L ∼ = U4 (2), F ∗ (CG (x)) = x × L, and L  Nxo ≤ NG ( x) with o Nx ∼ = Σ3 × L; and  o on a 6-element set such that (d) There is a transitive action of N o ∩ N o stabilizes a 2-element subset and D is isomorphic as F N  oN 3 x module to the core of the corresponding permutation module V . As in (2C), we let S ∈ Syl2 (Cz ), SK = S ∩ K, and P ∈ Syl2 (G) with S ≤ P . We proceed in a sequence of lemmas. Lemma 16.3. Q = z. Proof. By the Z ∗ -Theorem, there is g ∈ G such that t1 := z g ∈ Cz −{z} with t1 = z m t2 , t2 ∈ I2 (K), and m ∈ {0, 1}. In any case, Q ≤ CG (t1 ) −1  1  C  ( with Q ∩ Qg = 1. So Q1 := Qg ≤ Cz with Q1 ∼ = Q K t1 ), where ∼    for some   Cz = Cz /Q. If |Q| > 2, then there exists F = Z4 with t1 ∈ F  R  ∈ Syl2 (C  ( t )). But this is impossible by [V17 , 10.2.7], so Q = z.  R K 1 Lemma 16.4. We have z G ∩ Z(S) = {z}. Proof. Suppose not. Let S ≤ P ∈ Syl2 (G). If S < P , there exists g ∈ NP (Z(S)) with z = z g ∈ Z(S), contrary to assumption. Hence S ∈ Syl2 (G). Then by the Thompson transfer lemma, z has an extremal conjugate z g ∈ SK . As CS (z g ) = CS (z) = S, we have z g ∈ Z(S) − z, contrary to assumption.  We also note Lemma 16.5. A Sylow 2-subgroup of G has no normal Z4 -subgroup. Proof. This is immediate from [V17 , 10.2.7] and [V9 , 1.5]. Now let x ∈ I3 (K) and t ∈ I2 (K) be such that NK ( x) = x, t × H1 with H1 ∼ = Sp2 (2) ∼ = Σ3 . Thus = Sp4 (2) and x, t ∼ (16A)

t is a transvection (long root element) in K.



280 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

Set x = Nx /O3 (Nx ), H = [H1 , H1 ] ∼ Nx = NG ( x), N = A6 , and Ct = CG (t). Lemma 16.6. We have Nx = O3 (Nx ) x, t J z for some component J  Nx , CJ (z) ∼ = H1 ∼ = Sp4 (2), and one of the following holds: (a) zt centralizes J ∼ = U4 (2); (b) t centralizes J ∼ = U4 (2); or (c) CJ/Z(J) (t) ∼ = CJ/Z(J) (zt) ∼ = U4 (2) and J/O3 (J) ∼ = U4 (3). Proof. As m2 (CG (H1 z)) ≥ 4, it follows from Lemma 2.4 (with A =

x) that H is a component of CJ (z) for some 3-component J of CG (x).  ∼ Since H = A6 , U4 (2), L2 (34 ), = A6 , we see by [V17 , 14.1, 14.2] that J ∼ ±  J)  ∼ L3 (9) or L4 (3), or J/Z( = U4 (3). As e(G) = 3 and m2,3 (J) > 0, J  Nx .   4 ∼  there must exist  ,  t ≤ CNx (H), Suppose that J = L2 (3 ) or L± 3 (9). As z  = 1, [IA , 7.1.4c, 4.5.1]. This contradicts m2,3 (G) = 3. t1 , J] t1 ∈ {t, zt} with [ ∼  ∼ For the same reason, if J/Z(J) = L± 4 (3), then z, t = E22 acts faithfully  As m2 (J z) = 5 (recall that O3 (Nx ) has odd order on J centralizing H. ∗ by Theorem C4 : Stage A1), J is quasisimple by [V9 , 2.4]. Then by [V17 , 13.3b], either (c) holds or J ∼ = L4 (3), notice that = L4 (3). To rule out J ∼ in that case, CCx (z) = CCz (x) ∼ = Z6 × Σ6 . Hence H = E(CJ (z)) ∼ = A6 is centralized by t, so z induces an inner-diagonal automorphism on J by [V17 , 13.3a]. Thus by [IA , 4.5.1], CJ (H z) ∼ = Z4 , contrary to the structure of  J)  ∼ CCx (z). Therefore, if J/Z( (3), then (c) holds. = L± 4 ∼  Suppose next that J = A6 . By [V17 , 10.2.8b], every 2-central involution of K has a K-conjugate in H1 . Hence by Lemma 16.4, z g =: s ∈ z H1 −{z} −1 for some g ∈ G. Then [x, s] = 1, so if we put y = xg , then y ∈ I3 (CG (z)). Using [IA , 4.8.2], we see that CG (s, x) ∼ = CG (z, y) = z × X where X = CK (y) ∼ = Z3 × Sp4 (2), Z3 × Σ3 × Σ3 , or GU3 (2). Thus O3 (CG (s, x)) ∼ = Z2 . C x  = J = HO3 (J) as J  Nx . Since O3 (Cx ) On the other hand, H has odd order by Theorem C∗4 : Stage A1, CO3 (Cx ) (s) = 1 and CG (s, x) ∼ = CCx /O3 (Cx ) (s). Now s induces an inner automorphism on H1 so CH (s) ∼ = D8 or Σ4 . It follows that |O3 (CG (s, x))| ≥ 4, a contradiction. The only remaining possibility for J is J ∼ = U4 (2),   x = x z  with ,  t1 × J  and then for a suitable choice of t1 ∈ {t, zt}, N ∼   ( z ) H . Again J is quasisimple by [V , 2.4]; and J  z ∼ Aut( J) and C = 1 = 9 J ∼ again |O3 (Nx )| is odd. Hence J = U4 (2) and either (a) or (b) holds. The proof is complete.  Lemma 16.7. We have O3 (Nx ) = 1. Proof. Suppose false. Then by part (c) of Theorem C∗4 : Stage A1, there is a subgroup M < G such that Γ := ΓoP0 ,2 (G) ≤ M , where P0 is a Sylow 3-subgroup of G; we choose P0 , as we may, so that P0 ∩ K ∈ Syl3 (K) and

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281

x ∈ J(P0 ∩ K). Write J(P0 ∩ K) = x1  × x2  × x3 , where all three xi ’s lie in xK . For each i = 1, 2, 3, expand  J(P0 ∩ K) to Pi ∈ Syl3 (NG ( xi )). As m3 (J(P0 ∩ K)) = 3, Pi ≤ ΓM ≤ M , and then ΓoPi ,2 (G) ≤ M . In particular, since Pi = xi  (Pi ∩ Ji ), where Ji := E(NG ( xi )) ∼ = U4 (2) or L4 (3) or Ji /Z(Ji ) ∼ = U4 (3), Ji ≤ ΓoPi ,2 (G) ≤ M by [V17 , 5.1]. Then since Ji  NG ( xi ) by the previous lemma,  NG ( xi ) ≤ M by a Frattini  argument. By [IA , 7.3.2], Cz = K z = NKz ( x1 ), NKz ( x2 ) ≤ M . Now K ≤ L2 (M ) by L2 -balance, and then as m2 (K) = 6 > 4 and e(G) = 3, K ≤ E(M ) by the Thompson dihedral lemma. Indeed as m2,3 (G) ≤ 3, M has a component M0 such that K = E(CM0 (z)). As NK ( xi ) ≤ M0 , it follows that Ji ≤ M0 for each i = 1, 2, 3. Hence J1 ↑3 M0 /O3 (Z(M0 )), and K ↑2 M0 /O2 (Z(M0 )). Since e(G) = 3, we conclude that E(M ) = M0 ∼ = (2), by [V , 13.4]. Moreover, z induces a graph autoU6 (2), SU6 (2), or Ω+ 17 8 morphism on M0 [IA , 4.9.1, 4.9.2], and M = (W × M0 ) g, z, where (16B)

W = O{2,3} (M ) is inverted by z, and if g = 1, then g induces a triality on M0 ∼ = D4 (2).

Note that O{2,3} (Nx ) ≤ O{2,3} (M ) = W since M0 is strongly locally 1-balanced with respect to the prime 3 [IA , 7.7.11d]. Hence W = 1. Let u be a Sylow 2-center in M0 (and M ) and set Cu = CG (u). Then m3 (CM (u)) = m3 (CM0 (u)) = 3 and since ΓoP0 ,2 (G) ≤ M , M contains some R ∈ Syl3 (Cu ). Let E ∈ E33 (R). Then O2 (Cu ) = ΓE,2 (O2 (Cu )) ≤ M , so O2 (Cu ) ≤ O2 (CM (u)) ≤ M0 . Hence [O2 (Cu ), W ] = 1. As G is of even type and |W | is odd, there therefore exists a component I of Cu such that [I, W ] = 1. However, W is invariant under CM (u) and hence under ΓE,2 (Cu ). We are now in the general setup of the weak psignalizer property [IA , (7.7.4)], with p = 3, and with Cu , I, W here in the roles of X, K, and H there. We conclude immediately that I does not have the weak 3-signalizer property with depth m3 (E) = 3. As I ∈ Co2 and m3 (I) ≤ m3 (Cu ) ≤ 3, it follows from [IA , 7.7.17] that I/Z(I) ∼ = L3 (4). Moreover, AutE (I) is a subgroup of Inn(I) of order 9, since otherwise I ≤ ΓE,2 (I) ≤ CM (u) and so [I, W ] = 1, contradiction. Given that action of E on I, we conclude from [V17 , 4.5] that AutW (I) ∈ NAut(I) (AutE (I); {2, 3} ) = 1. That is, [I, W ] = 1, a contradiction completing the proof of the lemma.  By Lemmas 16.6 and 16.7, (16C)

Nx = x, t, z J.

We now begin to consider the different cases of Lemma 16.6. Through Lemma 16.12, we assume that Lemma 16.6a holds, that is, (16D) Nx = x, t1  × J z ∼ = Σ3 × Aut(U4 (2)), and t1 = zt. Set Ct1 = CG (t1 ). We first prove

282 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

Lemma 16.8. Assume (16D). Then Ct1 = t1  × K1 with K1 ∼ = Sp6 (2). Moreover, t is a long root involution in K1 . Proof. Let Rt1 = O2 (Ct1 ). We have CCt1 (z) = CCz (t) = t1  × V H1 with E25 ∼ = V  V H1 and with t the unique minimal normal subgroup of V H1 . Also, Ct1 ≥ L z. In particular, z ∈ Rt1 , whence also t ∈ Rt1 , and so (V H1 ) ∩ Rt1 = 1. Thus, CRt1 (z) = t1  and so Rt1 is cyclic or of maximal class. In any case, J acts faithfully on E(Ct1 ), whence by Proposition 11.1, E(Ct1 ) =: K1 ∼ = Sp6 (2) and m2 (C(t1 , K1 )) = 1. Then, as in Lemma 16.3, Ct1 = t1  × K1 , as claimed. The final statement holds by (16A) and as CCt1 (t) = CG (z, t) = CCG (z) (t).  Now we can prove Lemma 16.9. Assume (16D). Then there exists G0 ≤ G such that G0 ∼ = and Ct1 , Nx  ≤ G0 .

O8+ (2)

Proof. We shall find involutions z2 , z3 , . . . , z9 satisfying the Coxeter relations for the Weyl group of E8 . Let z1 = t1 and Cz1 = CG (z1 ). Let z3 , z4 , . . . , z9 be Coxeter generators for z1  K1 ∼ = W (E7 ), in such a way that z4 , . . . , z9 are Coxeter generators for z J (with z3 a long end node adjoining z4 in the E7 diagram). In order to do this we need to know that z ∈ K1 .

(16E)

To prove (16E) observe that by (16A), t is 2-central in Cz . Therefore as Cz 1 ∼ = Cz , |CCG (z1 ) (z)|2 = |CCz (t)|2 = |Cz |2 = |Cz1 |2 , So if z ∈ K1 , then z is 2-central in K1 , whence z is in [CK1 (z), CK1 (z)] ≤ [Cz , Cz ] by [V17 , 10.2.8a]. But obviously z ∈ [Cz , Cz ], a contradiction. Thus (16E) holds and we have z3 , . . . , z9 as claimed, by [V17 , 10.2.9], according to which z = t1 t is a Coxeter involution in Ct1 . (Note that z is also a Coxeter involution in J z ∼ = W (E6 ).) Fix z2 ∈ I2 (Sol(Nx )) − {z1 }. Then (z1 z2 )3 = 1 and [ z1 , z2  , z4 , . . . , z9 ] = 1 = [ z1  , z3 , . . . , z9 ]. Set M = z5 , . . . , z9  ∼ = W (D5 ) ∼ = 24 Σ5 . We have

z2 , z3  ≤ CG (M ) =: CM . Notice that x ∈ CM and Also z1 ∈ CM

CCM (x) = x × CJz (M ) = x . and CCM (z1 ) = CCz1 (M ) ∼ = E22 . Indeed

CCM (z1 ) = z1 , v where v is a long root involution in K1 ∼ = C3 (2), and so a Sylow 2-subgroup SM of CM is of maximal class. As z1 ∼G v, v lies in a cyclic maximal o of index 2 such that subgroup of SM , and then CM has a subgroup CM

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283

o , by Thompson transfer. Since z z ∈ O 2 (C ) ≤ C o , z ∈ z C o . z1 ∈ CM 1 2 2 1 M M M   o = xC M by a Frattini argument, and CM = As NCM ( x) ∼ = Σ3 , CM O3 (CM ) x, z1  with O3 (CM ) nilpotent and admitting x fixed-point-freely. o (z1 ) ∼ Since CCM = Z2 , O3 (CM ) = O2 (CM ), and indeed the only possibility o o . We ∼ is CM = A4 , CM ∼ = Σ4 . Now v is weakly closed in z1 , v, so v ∈ CM know [z1 , z3 ] = 1 so by inspection of Σ4 , either z2 , z3  ∼ = Σ3 or z3 ∈ C o . M

Then z3  = C (z1 ) = v so z3 = v. Then Suppose z3 ∈  

z1  K1 = z3 , . . . , z9  ≤ v z1 K1 ≤ K1 , a contradiction. Therefore z2 , z3  ∼ = Σ3 , whence z2 , z3 , . . . , z9  is a quotient of W (E8 ), containing Sp6 (2) × Z2 . As m2,3 (G) = 3, G0 := z2 , z3 , . . . , z9  ∼ = O+ (2). o . CM

o CM

8

By construction z1 ∈ Ct1 = z1  K1 = z3 , . . . , z9  ≤ G0 , and then x ∈

z1 z2  ≤ G0 . Also J z = z4 , . . . , z9  ≤ G0 and Nx ≤ G0 by (16D). The proof is complete.  Now K1 = E(CG0 (t1 )). Choose a 2-central involution y of E(CK1 (x)) = E(CG (x)) ∼ = U4 (2). Then y is also 2-central in G0 . Thus y is a short root y = Cy / y, involution in K1 ∼ = C3 (2). Set Cy = CG (y), Ry = O2 (Cy ), C and C y = Cy /Ry . Lemma 16.10. Assume (16D). Then Ry = O2 (CG0 (y)). Proof. Let R0 = O2 (CG0 (y)) ∼ = 21+8 + . Suppose first that Ry ≤ R0 . G By [V17 , 10.1.2], there exists w ∈ x 0 ∩ CK1 (y). In particular, G0 contains CG (w) as well as CG (t1 ). Hence, neither w nor t1 has any fixed points on Ry R0 − R0 , but w and t1 normalize Ry and R0 and commute. Let C0 = CRy R0 (R0 ), so that CC0 (t1 ) ≤ C0 ∩ R0 = y. Hence by the A × B-lemma, [C0 , w] = 1, and so C0 = y. Now let N0 = NRy R0 (R0 ) > R0 , N = NG (R0 ), 0 embeds as a normal  = N/R0 . Then N ≤ Cy so N0  N . Hence N and N + ∼   in subgroup in the image of N in Out(R0 ) = O8 (2), and this image of N + turn embeds in a parabolic subgroup P0 of O8 (2), by the Borel-Tits theorem.  contains X ∼ But N ≥ NG0 (R0 ) so N = Σ3 × (Σ3 Z2 ) acting faithfully on R0 .  0 embeds in Hence P0 is of type A1 × A1 × A1 , X covers P0 /O2 (P0 ), and N  2-subgroup. By [V17 , 10.1.3], Z(O2 (P0 )) ∼ P0 as an X-invariant = Z2 is then  X-invariant, hence w-invariant. Therefore CN0 (w) = 1. But we saw above that CN0 /R0 (w) = 1, a contradiction. Therefore Ry ≤ R0 , and so, if the lemma fails, then Ry = y. Now CG0 (y)/ y has a unique minimal normal subgroup R0 / y, so CG0 (y)/ y acts faithfully on E(Cy ). We have CCy (x) = CCG (x) (y) = QEF where Q = ∼ ∼ [Q, E] = O2 (QEF ) ∼ = 21+4 + , E = E33 , and F = E22 . Thus Q = O2 (CCy (x)),  = O2 (C  ( x)). Let I be a component of Cy such that so by [IG , 5.19], Q Cy  Q]  = 1. Since 5 does not divide |CC (x)|, I is not a Suzuki group, so 3 [I, y

284 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

 Likewise as CC (x) is solvable, x normalizes I. If I = I g for some divides |I|. y  2 g ∈ E # , then as m3 (Cy ) ≤ e(G) = 3, m3 (I) = 1 and O3 (E(Cy )) = II g I g . But then NE (I) ∼ = E32 and m3 (Cy ) ≥ m3 (NE (I)E(Cy )) > 3 = e(G), a contradiction. Therefore, E normalizes I. By [V17 , 20.9], Out(I) is 2nilpotent. As Q = [Q, E], Q induces inner automorphisms on I. For the same reason, AutQ (I) must have Q8 or Q8 ∗Q8 preimage in I, and AutQ (I) = [AutQ (I), AutE (I)]. In particular, y ∈ I, so Z(I) = 1. Now as I ∈ Co2 , [V17 , 4.4] yields I/Z(I) ∼ = 21+4 = M12 , M22 , or HJ and | AutQ (I)| = 4. As Q ∼ + , there must be a second component I1 of Cy with | AutQ (I1 )| = 4, and then m3 (Cy ) ≥ m3 (II1 ) = 4 > e(G), again a contradiction. The proof is complete.  Now let G1 = [G0 , G0 ] ∼ = D4 (2), and let M = NG (G1 ). As x ∈ G1 and NG ( x) = NG0 ( x), we see that CG (G1 ) = 1, whence M = G0 or M ∼ = Aut(D4 (2)). Lemma 16.11. Assume (16D). Then Cy = CM (y). Proof. Let Ey = E(Cy ). As CG (t1 ) ≤ G0 , CEy (t1 ) ≤ CG0 (Ry ) = y. Then, as components of Cy are Co2 -groups, it follows that Ey = 1 and Ry = F ∗ (Cy ). Thus C y embeds in O8+ (2), with m3 (C y ) = 3 and O2 (C y ) = 1. The possible structures of C y are then given by [V17 , 10.1.5]. We consider the various cases of that lemma. Set Y = CCy (t1 ), the full preimage in Cy of Y = CC y (t1 ). Also set Y0 = CCy (t1 ), so that Y0 ≤ Y and |Y : Y0 | = |tY1 |. As CG (t1 ) ≤ G0 , we have Y0 = CCG0 (y) (t1 ) = CCG0 (t1 ) (y), so Y 0 ∼ = Z2 × Σ3 × Σ3 by [V17 , 10.1.1ab]. Furthermore, Y0 is an extension of CRy (t1 ) = (Q1 ∗ Q2 ) × E by Y 0 , where Q1 ∼ = Q2 ∼ = Q8 and t1 interchanges the Q8 -subgroups Q3 and Q4 of CRy (Q1 ∗Q2 ) = Q3 ∗ Q4 ∼ = Q8 ∗ Q8 . Also Y is an extension of Ry by Y . So |Y : Y0 | = 4|Y : Y 0 | = |tY1 | ≤ |I2 (t1 Ry )|. Since |Y 0 | = 72, tY1 is a union of Y -orbits and |tY1 | = |Y |/18. Every involution in t1 Ry is in the set Xt1 = {t1 ce | c ∈ Q1 ∗ Q2 , c2 = 1, and e ∈ E}. The elements t1 ce with c ∈ y are permuted by Y0 in four Y0 -orbits, each of length 18. These orbits are fused by Y , which contains Ry . Also, the four elements t1 ye are fused by Y , as are the four elements t1 e. Hence the Y -orbits on Xt1 have lengths 72, 4, and 4. In particular (16F)

|Y | = 18|tY1 | = 72.18, 76.18, 80.18, 72, or 144.

16. THE Z2 × Sp6 (2) CASE

285

∼ A9 , then these condiBut Y = CC y (t1 ) and 9 divides |Y |. If F ∗ (C y ) = 4 tions imply that |Y | = |CC y (t1 )| = 2.7! or 2 .3.3!. contradicting (16F). If F ∗ (C y ) ∼ = Sp6 (2), then similarly, and with [V17 , 10.2.6], |Y | = 29 .32 .5 or 29 .32 , contradiction. If F ∗ (C y ) ∼ = U4 (2) ∼ = P Sp4 (3), then with [IA , 4.5.1], 6 2 7 2 |Y | = 2 .3 , 2 .3 , or 2.6!. So by (16F), CC y (t1 ) ∼ = Z2 × Σ6 . But then by [V17 , 10.1.4], t1 acts freely on Ry / y, which it evidently does not. (Note that some element of I3 (Cy ) is fixed-point-free on Ry / y by [V17 , 10.1.1d].) Suppose next that Z3 × A6 ≤ C y ≤ Σ3 × Σ6 . Again since |CC y (t1 )| is divisible by 9, |Y | = 2.6!, 6!, 2.4!.3! = 288, or 144. Hence, with (16F), C y ∼ = Σ3 × Σ6 with t1 ∈ Sol(C y ), or C y ∼ = Z3 × Σ6 . In both cases t1 ∈ [C y , C y ]. The latter case is impossible as C y contains W , where W = CG1 (y). Here W = W 1 × W 2 × W 3 with W i ∼ = Σ3 and CRy (O3 (W i )) = 1 for all i = 1, 2, 3. Moreover, for the former case, note that by [V17 , 10.1.1c], t1 centralizes every W i that it normalizes. But Sol(C y ) = W i for some i by the KrullSchmidt theorem, and t1 inverts O3 (C y ), a contradiction. The last two paragraphs, and [V17 , 10.1.5], imply F ∗ (C y ) = O3 (C y ). Let W be as in the previous paragraph. Then O3 (W ) contains no element w with |CRy (w)| = 26 . As a Sylow 3-subgroup Ay of Out(Ry ) ∼ = O8+ (2) is isomorphic to Z3 ×(Z3 Z3 ), it follows that O3 (W ) is conjugate to a subgroup of J(Ay ), and if Ay is chosen so that Ay ∩ W ∈ Syl3 (W ), then O3 (W ) = J(Ay ∩ C y ) and Ay ∩ C y ∼ = E33 or Z3 Z3 . As m2 (C y ) ≥ 3, m3 (O3 (C y )) ≥ 3 by the Thompson dihedral lemma, so O3 (W ) = J(O3 (C y ))  C y . Let W0 be a Sylow 3-subgroup of the preimage of W in Cy . Then Cy = Ry NCy (W0 ) so it suffices to show that NCy (W0 ) ≤ M . Let Wi , 1 ≤ i ≤ 3, be the preimage of W i in W0 . Then NCy (W0 ) permutes {W1 , W2 , W3 } since W1 , W2 , W3 are the only subgroups of W0 of order 3 without fixed points y . Thus NC (W0 )/ y W0 is a subgroup of Z2 Σ3 ∼ on R = Z2 × Σ4 , and y ∼ NCy ∩G0 (W0 )/ y W0 = Z2 × D8 . Hence, we are done unless NCy (W0 ) ∼ = Z2 Σ3 permutes {W1 , W2 , W3 } transitively. In that case, NCy (W0 ) permutes W := {w ∈ W0 | CRy (w) ∼ = Q8 ∗ Q8 } transitively. By our original choice of y, m2 (CRy (x)) = 3 so x has an Ry conjugate in W. By the transitivity on W, E(CG (w)) ∼ = E(Cx ) ∼ = U4 (2) for all w ∈ W. Hence E(CG (w)) ≤ G0 for all w ∈ W. Using [IA , 7.3.2] we conclude that NCy (W0 ) normalizes

E(CG (w)) | w ∈ W = E(CG0 (w)) | w ∈ W = G0 , so NCy (W0 ) ≤ M . The proof is complete.



Lemma 16.12. Assume (16D). Then t1 ∈ [G, G]. Proof. There is S1 ∈ Syl2 (CG (t1 )) with Z(S1 ) ∩ K1 = y, u, for some long root involution u of K1 ∼ = C3 (2). Thus, Z(S1 ) = t1 , y, u and CK1 (u) is

286 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

an extension of F ∗ (CK1 (u)) ∼ = E25 by Σ6 . As CG (y) is solvable, y G ∩Z(S1 ) ⊆ Z(S1 ) − t1 , u. Suppose that y g ∈ Z(S1 ) − {y}. Then y g ∈ {t1 y, yu, t1 yu}. Let R1 = O2 (CG (y g )). Now, S1 ≥ Q1 ∗ Q2 with Qi ∼ = Q8 and Z(Q1 ) = Z(Q2 ) = y. Also, CG (y g )/R1 ∼ = C y does not contain a Q8 subgroup. So, y ∈ R1 , whence yy g ∈ y R1 . So, y has a conjugate in t1 , u, contrary to the above. Hence y G ∩ Z(S1 ) = {y}. Then as t1 ∈ [CG (y), CG (y)], we have  t1 ∈ [G, G] by [V9 , 1.2], completing the proof. As G = [G, G], we conclude: Lemma 16.13. Case (b) or (c) of Lemma 16.6 occurs. We set L = E(CJ (t)). Thus, in both case (b) and case (c), L = CJ (t)(∞) ∼ = U4 (2); moreover, as H = [H, H] ≤ J, also H = E(CL (z)). Furthermore, L z ∼ = Aut(L). In case (b), of course, L = J. The analyses of these two cases are similar, up to a point where we can rule out case (c). We continue to let Ct = CG (t), and we set Lt = E(Ct ) and Rt = O2 (Ct ). Lemma 16.14. The following conditions hold: (a) F ∗ (Ct ) = Rt ∼ = 21+8 + ; (b) Ct = Rt L z with L ∼ = U4 (2) and L z ∼ = Aut(L); and 3 (c) For any A ∈ E3 (L), CRt (A) = t. Proof. By Lemmas 16.6 and 16.7, Ct ≥ L z ∼ = Aut(L), L ∼ = U4 (2). Suppose that [Lt , L] = 1. Since m3 (Ct ) ≤ e(G) = 3, L ≤ I for some component I of Lt . As m3 (L) = 3, Proposition 11.1 implies that I = Lt ∼ =  Sp6 (2). As m3 (Ct ) = 3, it follows that O3 (Ct ) = Lt and so H = [H1 , H1 ] ≤ L ≤ Lt . But V = [V, H], where t ∈ V = O2 (CK (t)), and so t ∈ Lt , which is impossible. Hence, [Lt , L] = 1. Now, there is x1 ∈ xG ∩ H ⊆ xG ∩ L, so Lt ≤ CG (x1 )(∞) ∼ = J. In particular, if Lt = 1, then by order considerations, Lt is not a product of Suzuki groups, so 3 divides |Lt |. But then m3 (Ct ) ≥ m3 (Lt L) ≥ 4 > e(G), contradiction. Hence, Lt = 1 and (a) holds. As L = [L, z], z ∈ Rt . But CRt (z)  CG (t, z) = CCz (t) and so either CRt (z) = t or (16G)

CRt (z) = V.

Since L acts faithfully on Rt , Rt z cannot be of maximal class, so (16G) holds. We next claim that (16H)

For any A ∈ E33 (L), CRt (A) is cyclic.

We may assume that A contains an element x1 ∈ xG ∩ H1 , say x1 = xg , g ∈ G. To prove (16H), it is enough to show that CO2 (CC (x ) (t)) (A)) is cyclic. But A0 := Ag to show that (16I)

−1

G

1

contains x and so, conjugating by g −1 , it is enough CO2 (CG (x,y)) (A0 ) is cyclic,

16. THE Z2 × Sp6 (2) CASE −1

287

−1

where y = tg . We have [A0 , y] = [A, t]g = 1 and y ∈ CG (x) = x×J z, with J ∼ = U4 (3). Now (16I) holds by [V17 , 4.3], so = U4 (2) or J/O3 (Z(J)) ∼ (16H) is established. Using (16H) and Lemma 16.5, we show that Rt is extraspecial. Suppose that t ∈ U , a characteristic elementary abelian 2-subgroup of Rt with |U | > 2. By (16H), CU (L) = t. As L has a 31+2 -subgroup, |U | ≥ 27 . It follows that V = CU (z). As |U/V | > 2, z shears U/V onto the indecomposable H1 -module V . But then by [V17 , 10.2.8c], |CU ( x1 )| ≥ |CV (x1 )|2 = 26 , whereas m2 (CG (x1 )) = m2 (CG (x)) = m2 (J z) < 6, a contradiction. Hence, every characteristic abelian subgroup of Rt is cyclic, so Rt is of symplectic type. Let A ∈ E33 (L). Since CRt (A) is cyclic, either Rt = [Rt , A] is extraspecial or Rt has a characteristic Z4 -subgroup. But the latter is impossible by Lemma 16.5. Thus, Rt is extraspecial and CRt /t (A) = 1. As V = CRt (z) ≤ Rt , Rt has width at least 4 and m2 (CRt /t (z) ≤ 5, so m2 (Rt / t) ≤ 10. On the other hand, as CRt (A) = t, |Rt / t | = 210 by [V17 , 10.9.1], and so Rt ∼ = Q48 , as claimed. Now (c) follows. Then Ct /Rt embeds in O8+ (2), containing L z, and having 3-rank 3, with O2 (Ct /Rt ) = 1. Then by [V17 , 10.1.5], either Ct /Rt ∼ = Aut(U4 (2)) or Ct /Rt ∼ = Sp6 (2). (U4 (2) does not embed in A9 ; it would have index 7.) Now CG ( z, t) acts irreducibly on Rt /V ∼ = E24 , so z acts freely on Rt / t. Let  z ) in Ct . The free action of Ct = Ct / t and let C0 be the preimage of CCt ( z on Rt / t implies that CCt /Rt (zRt ) ≤ C0 Rt /Rt . Hence CCt /Rt (zRt ) has a Σ6 -subgroup of index at most 2. By the Borel-Tits theorem, this rules out Ct /Rt ∼ = Sp6 (2). Thus Ct /Rt ∼ = Aut(U4 (2)), whence (b) holds. The proof is complete.  Lemma 16.15. Case (b) of Lemma 16.6 holds. Proof. By Lemma 16.13, we must rule out case (c) of Lemma 16.6. So assume that J/O3 (Z(J)) ∼ = U4 (3). We let x1 ∈ xG ∩ H, let Jx1 = ∼ E(CG (x1 )) = J, and claim that C := CG ( x1 , t) satisfies |C|2 ≥ 28 . If t induces an inner automorphism on Jx1 , then t is 2-central in Jx1 z, and our claim holds. Otherwise, by the action of t on Jx1 [IA , 4.5.1], either E(C) = 1 or C has a normal E24 -subgroup. But C = CCt (x1 ) has neither of these structures, being an extension of Z2 or an extraspecial group of 2-rank at most 3 by a group X such that F ∗ (X) = O3 (X). This contradiction proves our claim. On the other hand, |CRt (x1 )| ≤ 25 . For otherwise, |CRt (x1 )| = 27 . Take A ∈ E33 (Ct ) containing x1 . Then A is generated by three conjugates of x1 in Ct , so CRt (A) contains Q8 , contradicting Lemma 16.14c. Now 28 ≤ |C|2 ≤ 2|CRt (x1 )||C[Ct ,Ct ]/Rt (x1 )|2 =: n. If x1 is 3-central in L, then x1 lies in the center of a 31+2 -subgroup of L, whence n ≤ 2.23 .22 < 28 , contradiction. Thus x1 has 0- or 2-dimensional fixed point space on the

288 13. THEOREM C∗4 : STAGE A2. NONCONSTRAINED p-RANK 3 CENTRALIZERS

natural F4 U4 (2)-module Vnat . Accordingly we get n ≤ 2.|CRt (x1 )|.22 ≤ 28 or n ≤ 2|CRt (x1 )|.2 ≤ 27 . The only possibility is that CVnat (x1 ) = 0 and CRt (x1 ) ∼ = Q8 ∗ Q8 . But this is impossible, by [V17 , 16.2]. The proof is complete.  Now that Cx ∼ = E34 = Z3 × Aut(U4 (2)), we choose D ≤ Cx such that D ∼ z and D = D. We can take D = O3 (CCx (D0 )), where D0 = x × (D0 ∩ H) ∈  = NG (D)/D. Syl3 (CK (x)). Set N = NG (D) and N ∼ Lemma 16.16. We have that CG (D) = D and N = Z2 × Σ6 . Proof. We have D with D ∩ H ∈ Syl3 (H) and D ∩ L = CL (D ∩ H). Then CG (D) = D. Also, NNx (D)/D ∼ = Z2 × Z2 × Σ4 . Moreover, x and D ∩ L are the unique proper NNx (D)-invariant subgroups of D. Also,  acts irreducibly on D.

x × H ≤ K and xK ∩ (D ∩ H) = ∅. Hence, N In the notation of [V9 , Section 10], t is a reflection on D with center

x ∈ [1], while z is a reflection on D with center u ∈ [3]. If [1] ∼ [3] (again in the notation of that section), then x is N -conjugate to u. But then CN (x⊥ ) = D t is N -conjugate to CN (u⊥ ) = D z, and so t ∈ z G , which  acts irreducibly on D, it follows is not the case. Hence, [1] ∼ [3]. Since N  ≤ Z2 × Σ5 or A6 ≤ N  ≤ GO− (D) ∼ by [V9 , 10.1] that either N = Z2 × Σ6 . Z × Z × Σ , we see that the latter case holds with As NNx (D)/D ∼ = 2 2 4 ∼  N = Z2 × Σ6 , completing the proof.  We now complete the proof of Proposition 16.2. Let N o = N (∞) t,  ∼  ) acting as −1 on D, it o ∼ so that N = Σ6 × Z2 with Z(N = Σ6 . Since N o follows from [V17 , 10.8.2] that for some permutation representation of N on a set Ω of cardinality 6, D is isomorphic to the core of the corresponding  o . Also let N o be the subgroup of Nx inducing permutation module for F3 N x   o ). Then, o ∼ t = Z(N Z inner automorphisms on L. Thus N = 2 × Σ4 with  x x with Lemmas 16.14 and 16.16, we see that (a), (b), and (c) of Proposition o ∩ N  o stabilizes a 2-element subset 16.2 hold. Moreover, by [V17 , 10.8.3], N x of Ω. Hence part (d) of Proposition 16.2 holds, so the proof of the proposition is complete. Corollary 16.17. There is no (z, K) ∈ IL2 (G) such that K ∼ = Sp6 (2) and m2 (C(z, K)) = 1. Proof. If such a (z, K) existed, then since Out(K) = 1, Proposition 16.2 would imply G is of U6 (2)-type. Hence by Proposition 7.5 of Chap that o ter 16,1 Go := LN ∼ = U6 (2). Then Ct = CGo (t) z. Since z centralizes x, it normalizes CGo (t), x = Go (note that CGo (t) is a maximal parabolic subgroup of Go ). Thus M := Go z ≤ G and Ct ≤ M . Finally, let R ∈ Syl2 (Ct ) with z ∈ R, so that R ∈ Syl2 (M ). As t = Z(R), R ∈ Syl2 (G). By the Thompson transfer lemma, there exists u ∈ 1 The proof of Proposition 7.5 in Chapter 16 may be inserted at this point. There is no circularity of reasoning.

16. THE Z2 × Sp6 (2) CASE

289

z G ∩ Go . But then by [V17 , 10.3.1], either m2 (CG (u)) ≥ 9 or |CG (u)|2 ≥ 211 . As m2 (CG (z)) = 7 [IA , 3.3.3] and |CG (z)|2 = 2|K|2 = 210 , we have a contradiction, and the corollary is proved.  Corollary 16.17 and Proposition 11.1 complete the proof of Theorem 2, and with it, the proof of Theorem C∗4 : Stage A2.

CHAPTER 14

Theorem C∗4 : Stage A3. KM -Singularities 1. Introduction We have now reached the situation where all involution centralizers of p-rank 3 are 2-constrained. We now subdivide the problem into two cases determined by the p-local structure of G, or more precisely by the maximum rank of a KM -singularity in some H ∈ Hcons . The initials KM are for K. Klinger and G. Mason [KMa1], who first investigated groups simultaneously of characteristic 2-type and characteristic p-type for some odd prime p such that m2,p (G) ≥ 3. Their ideas guarantee that for some elementary abelian p-subgroup B0 = 1 of some H ∈ Hcons , Lp (CG (B0 )) = 1. The dichotomy established in this chapter will be (a) For some such B0 , mp (B0 ) = 2; or (b) For all such B0 , mp (B0 ) = 1. More precisely, we shall prove the following theorem. Theorem C∗4 : Stage A3. Let G and p be as in Theorem C∗4 (Case A). Assume that Hinv ⊆ Hcons . Then either (a) or (b) holds: (a) (1) p = 3; and (2) There exists H ∈ Hcons and a KM -singularity B0 in H of rank 2 and type A6 ; or (b) (1) Hinv = ∅; and (2) For every H ∈ Hcons and KM -singularity B0 in H, mp (B0 ) = 1. We remind the reader that a KM -singularity in H ∈ Hcons [V11 , Definition 2.2] is an elementary abelian p-subgroup B0 = 1 of H such that mp (CH (B0 )) = 3 and Lp (CG (B0 )) = 1. The rank of the singularity is mp (B0 ), and the type of the singularity is the isomorphism type of the layer E(CG (B0 )/Op (CG (B0 ))). Throughout this chapter we assume the hypotheses of the theorem, and in particular (1A) For every t ∈ I2 (G) such that mp (CG (t)) = 3, F ∗ (CG (t)) = O2 (CG (t)). 2. Centralizers of Noncyclic p-Groups Definition 2.1. For any A ∈ Ip (G) or A ∈ Ep (G), LA = Lp (CG (A)). 291

292

14. THEOREM C∗4 : STAGE A3. KM -SINGULARITIES

We begin by analyzing LA for elementary abelian p-subgroups of G, particularly noncyclic ones lying in 2-local subgroups. Note that the possible structures for such subgroups LA /Op (LA ) can be found among the groups in [V17 , 7.1, 7.10]. Likewise the possible vertical pumpups in G of components of LA /Op (LA ) can be found in [V17 , 7.2]. In this section our main result is the following: Proposition 2.2. Let A ∈ Ep (G) have p-rank r ≥ 2 and act nontrivially on a 2-subgroup S of G. Suppose that A# ⊆ Ipo (G) and LA = 1. Then there exist a hyperplane A1 of A and a 2-group R ≤ CS (A1 ) such that R = [R, A] = 1. For any such A1 and R, let J be a p-component of LA and K the pumpup of J in LA1 , i.e., the subnormal closure of J in CG (A1 ). Then the following conclusions hold: (a) r = 2 and p ≤ 7; (b) R normalizes K; (c) J = LA and K = LA1 with (J/Op (J), K/Op (K)) as in [V17 , 7.2]; and (d) If J/Op (J) ∼ = L2 (pm ) and K/Op (K) ∼ = L2 (ppm ), then p = 3 and m = 2. Before proving the proposition, we mention several corollaries. Corollary 2.3. If A ∈ Ep3 (G), then LA = 1 and m2 (CG (A)) ≤ 1. Proof. If LA = 1, then for any involution z ∈ LA , A acts faithfully on O2 (CG (z)) by (1A). In particular, A# ⊆ Ipo (G) by Theorem C∗4 : Stage A1. But then mp (A) = 2 by Proposition 2.2a, contradiction. Therefore LA = 1.  Then m2 (CG (A)) ≤ 1 by [V9 , 2.2]. Corollary 2.4. Let A ∈ Ep (G) have p-rank 2. If CG (A) has a pcomponent J, then C(A, J) := CCG (A) (J/Op (J)) has odd order. Moreover, Ip (C(A, J)) = A# . Proof. Expand A to a maximal elementary abelian subgroup A∗ of C(A, J). Then LA∗ = Lp (CLA (A∗ )) covers J/Op (J), so A = A∗ by Corollary 2.3. Consequently Ip (CC(A,J) (A)) ⊆ A ≤ Z(C(A, J)). As p is odd, C(A, J) is p-nilpotent by [IG , 16.11]. Thus J = LA  CG (A) and C(A, J) ≤ Op p (CG (A)). By [IA , 16.11], there is an element of order p in J − Op p (J), so mp (CG (A)) ≥ 3. Hence Op (CG (A)) has odd order by Theorem C∗4 : Stage A1, completing the proof.  

Corollary 2.5. Op (CG (A)) normalizes each p-component of CG (A), for any elementary abelian p-subgroup A of G. Proof. Otherwise, CG (A) has at least p ≥ 3 p-components L1 , L2 , . . . , and there is E ∼ = Ep3 such that E ≤ AL2 L3 and LE ≥ Lp (CCG (A) (E)) = 1, contradicting Corollary 2.3.  For the final corollary we need a lemma.

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293

Lemma 2.6. Let (x, K) ∈ ILop (G), i.e., let x ∈ Ipo (G) and let K be a p-component of Lp (CG (x)). Then the following hold: (a) mp (K) − mp (Z(K)) ≥ 1, and in case of equality, mp (K) = 1 and Z(K) = 1; and (b) If b ∈ Ip (Inn(K)) and Lp (CK (b)) = 1, then b acts on K/Op (K) like an element of Ip (K). Proof. Without loss Op (K) = 1, and also Z(K) = 1, since otherwise there is nothing to prove. By [V17 , 7.1], p = |Z(K)| = 3 and K/Z(K) ∼ = U4 (3), G2 (3), J3 , or M c. In the first two cases, K/Z(K) has a U3 (3) subsystem subgroup [IA , 2.6], whose Schur multiplier is trivial [IA , 6.1.4]; thus, m3 (K) − m3 (Z(K)) ≥ m3 (U3 (3)) > 1, so (a) holds. Also in these cases the hypothesis of (b) does not hold, by the Borel-Tits theorem. On the other hand, if K/Z(K) ∼ = J3 or M c, every element of I3 (K/Z(K)) splits over Z(K) [IA , 5.3hn], so (b) holds, and (a) holds by  [IA , 5.6.1]. Now we can prove Corollary 2.7. Let K be a p-component of CG (a) for some a ∈ Ipo (G). Suppose that b ∈ Ip (CG (a)) and Lp (CK (b)) = 1. If K/Op (K) ∼ = L2 (ppm ), then assume that p = 3. Then mp (C(a, K)) = 1, C(a, K) has odd order, and C(a, K)  CG (a). Proof. Let J be a p-component of Lp (CK (b)). Again by [IG , 16.11], there are elements of order p in J − Op p (J), and so mp (CG ( a, b)) ≥ 3. If b induces an inner automorphism on K/Op (K), then we may assume that b ∈ K by Lemma 2.6b. Let E ∈ Ep∗ (C(a, K)) be b-invariant and set R = b × E. Then 1 = Lp (CK (R)) ≤ LR , so mp (R) = 2 by Corollary 2.3. Then mp (C(a, K)) = mp (E) = 1, whence C(a, K) is p-nilpotent, and of odd order by Theorem C∗4 : Stage A1, so the lemma holds in this case. Now assume that b induces an outer automorphism on K/Op (K). Since a ∈ Ipo (G), K/Op (K) ∈ Cp , and so J/Op (J) ∈ Cp by [IA , 7.1.10]. Then as p divides | Out(K/Op (K))|, and by [V17 , 7.2], K/Op (K) ∼ = L2 (ppm ), or 5 3D (2) or Sp (8) with p = 3, or 2F (2 2 ) with p = 5. In all but the first case, 4 4 4 there is b ∈ Ip (K) such that Lp (CK (b )) = 1. Replacing b by b , we are done by the previous case. Finally, if K/Op (K) ∼ = L2 (ppm ), then p = 3 by assumption. Hence m3 (C(a, K)) = 1 by [V12 , 14.1], and C(a, K) has odd  order as m3 (K a) = 3m + 1 > 3 = e(G), completing the proof. Proof of Proposition 2.2. As mp (A) ≥ 2 and LA = 1, mp (CG (A)) ≥ 3 by [IG , 16.11]. Also if S is any A-invariant 2-subgroup of G such that [S, A] = 1, then R := [CS (A1 ), A] = 1 for some hyperplane A1 of A, and A1 and R satisfy the first assertion of the proposition. Suppose that the proposition fails, and choose (A, A1 , R, J, K) to be a counterexample, with r > 2 if possible. Note that if there is B ≤ G with B∼ = Epr+1 and LB = 1, and t is an involution in CG (B), then by (1A), B

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acts faithfully on O2 (CG (t)) and so B is a counterexample to (a). Hence by our choice, r > 2, whence m2,p (G) ≥ r + 1 > 3, a contradiction. Therefore (2A)

For every B ∈ Epr+1 (G), CG (B) is p-constrained, and m2 (CG (B)) ≤ 1.

The last inequality follows from [V9 , 2.2], with B here in the role of A there. Now suppose that CG (A1 ) has at least two p-components L1 , L2 . Let Li , i = 1, 2, be the product of all p-components of CG (A1 ) other than Li . Then mp (Li A1 ) ≤ mp (C(A1 , Li )) ≤ r by (2A), for i = 1, 2. As mp (A1 ) = r − 1, we have mp (Li ) − mp (Op p (Li )) ≤ 1, for i = 1, 2. Hence by Lemma 2.6a, mp (Li ) = 1 and Op p (Li ) = Op (Li ), i = 1, 2, and LA1 = L1 L2 . We may choose notation so that K = L1 . Then as J ≤ Lp (CK (A)), we also have mp (L2 A) ≤ r by (2A). Hence A contains every element of order p in CL2 (A). As mp (L2 ) = 1, A ∩ L2 = 1 and then A = A1 × (A ∩ L2 ). Since mp (CG (A1 )) ≥ 3, Op (CG (A1 )) has odd order by Theorem C∗4 : Stage A1, and R = [R, A] = [R, A ∩ L2 ] ≤ L2 as L2   CG (A1 ). Thus (2B)

m2,p (L2 /Op (L2 )) = mp (L2 ) = 1.

# ⊆ I o (G) by But then with [IA , 7.7.12], L2 /Op (L2 ) ∈ Cp . As A# p 1 ⊆ A hypothesis, however, L2 /Op (L2 ) ∈ Cp , with the help of [IA , 7.1.10], contradiction. We have proved that K = LA1 is a single p-component, which implies (b). Hence K is either a trivial or a vertical pumpup of J. If K is a trivial pumpup of J, then R = [R, A] centralizes K/Op (K), so R acts faithfully on Q = Op (CG (A1 )), where CG (A1 ) = CG (A1 )/Op (CG (A1 )). As A1 centralizes Q and A,

mp (A1 ) + m2 (R) ≤ mp (Q) ≤ r, by the Thompson dihedral lemma applied to RQ and (2A), respectively. Therefore m2 (R) = 1 and mp (Q) = r. In particular, p = 3, R ∼ = Q8 , and [A, Z(R)] = 1. As R centralizes K/O3 (K) and e(G) = 3, m3 (AK) ≤ 3, so r = 2. By the critical subgroup lemma [IG , 11.11] there is Q1 char Q with A1 ≤ Q1 , Q1 of class 2 and exponent 3, and CR (Q1 ) = 1. As m3 (Q) = 2, it follows that Q1 ∼ = 31+2 . But now m3 (Q1 A) = m3 (Q1 RA) = 3 by [V9 , 5.7]. As [J, QA] = 1 and r = 2, (2A) is again contradicted. Therefore K is a vertical pumpup of J = Lp (CK (A)) normalized by RA. Since A# ⊆ I3o (G), K/O3 (K) and J/O3 (J) are as in [V17 , 7.2], so (c) holds. Suppose next that K/Op (K) ∼ = L2 (ppm ). Then J/Op (J) ∼ = L2 (pm ), so pm > 3. Consequently mp (K) > 3, and |C(A1 , K)| is odd as e(G) = 3. Therefore R acts faithfully on K/O3 (K); indeed, since R = [R, A] and Out(K/O3 (K)) ∼ = Z2 × Zpm is abelian [IA , 2.5.12], R ≤ K, so R embeds in a dihedral group. Then R = [R, A] forces p = 3. As m3 (A) > 1, m = 2 by [V12 , 14.1], so (d) holds. By (c) and (d) and [V17 , 7.2], p ≤ 7. It remains to show that r = 2, so suppose that r > 2. Let A2 be any hyperplane of A1 , and K ∗ the pumpup of

3. KM -SINGULARITIES OF RANK 2 AND TYPE A6 : THE CASE Hinv = ∅

295

K in CG (A2 ). By [V17 , 7.7], applied to J and K, mp (K1 ) − mp (Z(K1 )) > 1 for any covering group K1 of K/Op (K). Therefore as e(G) = 3, K ∗ is not a diagonal pumpup of K. By (d) and [V17 , 7.2] again, if K ∗ is a vertical pumpup of K, then J/O3 (J) ∼ = L2 (3m ) and K ∗ /O3 (K ∗ ) ∼ = L2 (39m ). As A/A2 is noncyclic, L3 (CK ∗ (A)) = 1, a contradiction as J ≤ L3 (CK ∗ (A)). Hence K ∗ is a trivial pumpup of K, so A1 ≤ C(A2 , K ∗ ). Choose an involution t ∈ J. As A ≤ CG (t) and r = 3, A acts faithfully on V := O2 (CG (t)), by (1A). In particular, we may choose A2 so that V1 := [CV (A2 ), A1 ] = 1. As noted in the previous paragraph, m  p (K) > 1, so    p as e(G) = 3, O (CG (A2 )) normalizes every p-component of (K ∗ )CG (A2 ) . Thus V1 = [V1 , A1 ] normalizes K ∗ and then lies in C(A2 , K ∗ ). Hence NCG (A2 ) (V1 ) covers K ∗ A/Op (K ∗ ) and in particular covers JA/Op (J). Consequently e(G) ≥ mp (JA) > mp (A) ≥ 3, a contradiction completing the proof.  3. KM -Singularities of Rank 2 and Type A6 : The Case Hinv = ∅ We begin with a refinement of Proposition 2.2. Then we give a proof of Theorem C∗4 : Stage A3 under the extra assumption Hinv = ∅. Recall that for any p-element or elementary abelian p-subgroup A of G, LA = Lp (CG (A)). Proposition 3.1. Let A ∈ Ep2 (G) be such that A acts faithfully on some 2-subgroup R of G and A# ⊆ Ipo (G). Suppose that J := LA = 1. Then the following conditions hold: (a) Suppose in addition that there is an elementary abelian p-subgroup B of G such that B > A and B normalizes some nontrivial 2subgroup U of CG (A). Then p = 3 and one of the following holds: (1) J/O3 (J) ∼ = A6 . For any a ∈ A# , La /O3 (La ) ∼ = A6 , L2 (36 ), A9 , J3 , 3J3 , Sp6 (2), or Sp4 (8); or (2) J/O3 (J) ∼ = E32 , B o ∼ = E33 = L2 (8). Moreover, there exist Ao ∼ o o o such that A ≤ B ≤ G, CG (B ) has even order, and J o := LAo satisfies J o /O3 (J o ) ∼ = A6 ; and (b) If p = 3, then either (a1) or (a2) holds, or J/O3 (J) ∼ = L2 (8) and every element of I3 (CG (A)) induces an inner automorphism on J/O3 (J). Proof. We have J/Op (J) ∈ Cp , as A# ⊆ Ipo (G) by assumption, and by Lp -balance and [IA , 7.1.10]. It suffices to prove (a). For then, in proving (b), we may assume that the hypothesis of (a) fails. Thus for any b ∈ I3 (CG (A)) − A, NJ ( b ; 2) = {1}, since otherwise, B = A b satisfies the hypothesis of (a). In particular, m2,3 (J/O3 (J)) − m3 (O3 3 (J)) = 0. Hence with [V17 , 3.12], the conclusion of (b) holds.

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By Corollaries 2.3 and 2.4, J is a single p-component, and C(A, J) has odd order and p-rank 2. Therefore the image of BU in Aut(J/Op (J)) shows that m2,p (Aut(J/Op (J))) ≥ 1, and in particular, J/Op (J) ∼ L2 (p). = Since CA (R) = 1 by assumption, we may write A = a1 , a2  with Ri := [CR (ai ), A] = 1 for i = 1 and 2. Let Ji be the pumpup of J in CG (ai ). By Lp -balance and Corollary 2.5, Ji is a single p-component of CG (ai ), and J is a p-component of CJi (a3−i ), i = 1, 2. In particular, J and Ji satisfy the conclusions of Proposition 2.2 for J and K. Thus if J/Op (J) ∼ = L2 (pm ), pm ∼ then as m > 1, Ji /Op (Ji ) = L2 (p ), by [V17 , 7.2]; by Proposition 2.2, this implies that p = 3. Hence, Corollary 2.7 implies that in any case, C(ai , Ji )  CG (ai ) and C(ai , Ji ) has p-rank 1 and odd order, i = 1, 2. Therefore Ji = Lai is a vertical pumpup of J (otherwise A ≤ C(ai , Ji )). As ai ∈ Ipo (G) by hypothesis, p and (J/Op (J), Ji /Op (Ji )) are as in [V17 , 7.2]. As J/Op (J) ∼ = ∼  A5 . L2 (p), that lemma yields p ≤ 5 and J/Op (J) = We consider various possibilities for (p, J/Op (J), J1 /Op (J1 )), in three batches: 1

(3A)

5

(1) (3, U3 (3), 3D4 (2)), (5, 2F4 (2 2 ) , 2F4 (2 2 )); 5 5 (2) (3,L2 (8),Sp4 (8)), (3,L2 (8),3D4 (2)), (5,2B2 (2 2 ),2F4 (2 2 )); (3) (5, U3 (5), F5 ).

Suppose (p, J/Op (J), J1 /Op (J1 )) is as in (3A1). By [V17 , 7.2] again, J1 /Op (J1 ) ∼ = J2 /Op (J2 ). Choose c ∈ Ip (J), and if p = 3 choose c to be non-2-central in J. As mp (J) ≥ 2 = mp (A), c ∈ Ipo (G). Set (3B)

Yi = Lp (CJi (c)), i = 1, 2,

5 so that Yi /Op (Yi ) ∼ = L2 (8) or 2B2 (2 2 ), respectively [IA , 4.7.3A, 4.8.7], and  Yi = O2 (CJi (c)) [IA , 4.9.5]. Likewise, we let 

Y = O2 (CJ (c)) ≤ Y1 ∩ Y2 ∩ J, 1 so that Y /Op (Y ) ∼ = L2 (2) or a subgroup of index at most 2 in 2B2 (2 2 ). (We have used the non-3-central choice when p = 3.) In any case, 2p divides |Y |, so 2p divides |Y1 ∩ Y2 |. This implies in turn that Y1 and Y2 have the same subnormal closure H in CG (c). Since A acts nontrivially on each Yi /Op (Yi ), H is a vertical pumpup of Yi , i = 1, 2. Then (Yi /Op (Yi ), H/Op (H)) occurs in [V17 , 7.2]. It follows that (p, H/Op (H)) is (3, 3D4 (2)), (3, Sp4 (8)), or 5 (5, 2F4 (2 2 )), whence a1 and a2 induce inner automorphisms on H/Op (H). In every case, CH/Op (H) (a2 ) is the direct product of a cyclic p-group and the image of Y2 [IA , 4.7.3A, 4.8.2, 4.8.7]. Therefore a1 induces an inner automorphism on Y2 /Op (Y2 ). However, a1 induces a nontrivial field or graph automorphism on J2 /Op (J2 ) and hence induces a nontrivial field automorphism on Y2 /Op (Y2 ) [IA , 4.7.3A, 4.2.3]. This is a contradiction, so the case (3A1) does not occur.

3. KM -SINGULARITIES OF RANK 2 AND TYPE A6 : THE CASE Hinv = ∅

297

Next consider the case (3A2) for p, J, and J1 . As B normalizes the 2-group U , there is c ∈ B # inducing a field automorphism on J/Op (J), and hence inducing a field or graph automorphism on J1 /Op (J1 ). Moreover, if J1 /O3 (J1 ) ∼ = 3D4 (2), we may ensure that CJ1 /O3 (J1 ) (c) ∼ = G2 (2) (perhaps sacrificing the condition c ∈ B) by replacing c by a suitable element of CG (A) of order 3 inducing a field automorphism on J/O3 (J) [IA , 4.7.3A]. With this possible change of c, in case (3A2) we have CJi /Op (Ji ) (A c) ∼ = 1

Σ3 × Z3 or 2B2 (2 2 ) × Z5 . Let zi be an involution in CJi (A c), and Si = O2 (CJi (zi )), i = 1, 2. The hypotheses of the proposition and of part (a) are satisfied by Ao = a1 , c, B o = A c, J o = LAo , U o = CS1 (c) and Ro = S2 (we use (1A)), in place of A, B, J, U , and R, respectively. In the (3, L2 (8), Sp4 (8)) case of (3A2), J o /O3 (J o ) ∼ = A6 , so (a2) holds. In the other two cases of (3A2), we are reduced to the two cases of (3A1), which led to a contradiction. Now consider case (3A3), which we shall rule out, using [IA , 5.3w] without comment. The argument is similar to that for (3A1), except that we choose c ∈ J to have order 3, not p. We again have J2 /O5 (J2 ) ∼ = J1 /O5 (J1 ). Also as m2 (F5 ) > 4, Ji ∼ = F5 , i = 1, 2, by [V9 , 6.1c] and [IA , 6.1.4], whence J is also quasisimple. We have Yi := E(CJi (c)) ∼ = A9 , i = 1, 2, ∼  and Y1 ∩ Y2 ∩ J = A4 . Set Cc = CG (c) and Cc = Cc /O3 (Cc ). If Yi acts c ), then by [V9 , 2.2] (with A = c) and as e(G) = 3, faithfully on O3 (C c ), Yi ] = 1. Now by [V9 , m2 (Yi ) < 4, which is not the case. Thus, [O3 (C  2.5], Yi ≤ L3 (Cc ). Let Hi be the subnormal closure of Yi in Cc , a product of 3-components such that (3C)

Yi is a component of CHi (ai ).

Since m2,3 (G) ≤ 3, Hi is a single 3-component, and H1 = H2 since Y1 ∩ Y2 ∩ J ∼ = A4 contains involutions. As [Y2 , a1 ] = 1, a1 acts nontrivially on H1 /O3 (H1 ). Now (3C) has the following consequences: H1 /O3 (H1 ) ∈ Spor ∪ Chev by [IA , 5.3, 4.9.6], and then H1 /O3 (H1 ) ∼ = A14+5k , k ≥ 0. Hence, H1 contains E33 × A4 , so m2,3 (Cc ) ≥ 4, contradicting e(G) = 3. We have shown that for the cases (3A), the conclusion of the proposition is valid. But if none of the cases (3A) holds, then by [V17 , 7.2] and Proposition 2.2d, p = 3 and conclusion (a1) holds. This completes the proof of the proposition.  Definition 3.2. H∗ (G) is the set of all quadruples (H, B, A, J) with (a) H ∈ Hcons ; (b) A ≤ B ≤ H with A ∼ = Ep3 ; and = Ep2 and B ∼ (c) J = LA = 1. B#

Note that if (H, B, A, J) ∈ H∗ (G), then as B normalizes O2 (H), A# ⊆ ⊆ Ipo (G) by Theorem C∗4 : Stage A1.

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298

Proposition 3.3. If H∗ (G) = ∅, then p = 3 and there is (H, B, A, J) ∈ H∗ (G) such that J/O3 (J) ∼ = A6 . Proof. Begin with any (H, B, A, J) ∈ H∗ (G), chosen if possible so that CG (B) has even order. If NCG (A) (B; 2) = {1}, then Proposition 3.1a applies with R = O2 (H). Thus, p = 3, and according as (a1) or (a2) of that proposition holds, J/O3 (J) ∼ = A6 , as desired, or (CG (z), B o , Ao , J o ) ∈ H∗ (G) for any involution z ∈ CG (B o ), in view of (1A). As J o /O3 (J o ) ∼ = A6 , the desired conclusion holds. We may therefore assume that NCG (A) (B; 2) = {1}. In particular, CG (B) has odd order. Let U ∈ NG (B; 2) be minimal nontrivial. Thus U ≤ CG (A). Set C := CB (U ), so that CA = B. By minimality, U = [U, B] is elementary abelian and mp (C) = 2. If I := LC = 1, then (H, B, C, I) ∈ H∗ (G), and we are done as in the first paragraph. So we may assume that LC = 1. By [V9 , 2.2], m2 (CG (C)) ≤ 4 − mp (C) = 2. Therefore U ∼ = E22 and p = 3, and CG (C) contains a subgroup U E ∼ = Σ3 × Σ3 , by [V9 , 2.1]. By Proposition 3.1b, we are again done unless possibly J/O3 (J) ∼ = L2 (8) and I3 (CG (A)) ⊆ AJ. In this last case, choose 1 = x ∈ C ∩ A, and let Jx be the pumpup of J in CG (x). With Corollary 2.5 and [V17 , 7.2], we have that Jx /O3 (Jx ) ∼ = L2 (8), 3D (2), or Sp (8) with A inducing inner automorphisms on J /O  (J ). 4 4 x x 3 Then m3 (Jx C(x, Jx )) ≤ 3, by Corollary 2.3 in the first case and Corollary 2.7 in the other two cases. However, m3 (CG (x)) ≥ m3 (CE) = 4, so there is y ∈ I3 (CG (x)) − x such that m3 (CG ( x, y)) ≥ 4 and J0 := CJx (y) satisfies J0 /O3 (J0 ) ∼ = L2 (8), G2 (2), or Σ6 , respectively. Hence, some subgroup F ≤ CG (x) with y ∈ F ∼ = E33 centralizes an involution z ∈ J0 . Now (CG (z), F, x, y , L3 (J0 )) ∈ H∗ (G) and CG (F ) has even order. As CG (B) has odd order, this contradicts our original choice of (H, B, A, J). The proof is complete.  Corollary 3.4. If Hinv = ∅, then the conclusion of Theorem C∗4 : Stage A3 holds. Proof. If condition (b2) of Theorem C∗4 : Stage A3 holds, then (b) of the theorem holds. So assume that (b2) fails. By definition of KM -singularity [V11 , 2.2], there then exists (H, B, A, J) ∈ H∗ (G), and by Proposition 3.3, p = 3 and we may assume that J/O3 (J) ∼ = A6 . Hence (a) of the theorem holds, as claimed.  4. The Case Hinv = ∅: {2, 3}-Local Subgroups For the rest of the proof of Theorem C∗4 : Stage A3, we assume, as we may by Corollary 3.4, that (4A)

Hinv = ∅, i.e., mIp (G) ≤ 2.

We begin with a few simple consequences of (4A). The first is immediate. Lemma 4.1. If A ∈ Ep3 (G), then CG (A) has odd order.

5. p = 3; L2 (3n ) 3-COMPONENTS

299

Lemma 4.2. Let N be a p-local or 2-local subgroup of G. Then N has at most two p-components. Moreover, O2 (N ) normalizes every p-component of N. Proof. Let P ∈ Sylp (N ). Suppose L1 , . . . , Ln are the p-components of N . For each i = 1, . . . , n, choose xi ∈ Ip (P ∩ Li ) − Op p (Li ) [IG , 16.11], and set B = x2 , . . . , xn . Suppose that n > 1. Choose any x ∈ Op (N )O2 (N ) of prime order. If x has order p, then x2 , . . . , xn , x ∼ = Epn centralizes an  1 . If x2 = 1, then x1 , . . . , xn  ∼ n E centralizes x. In either involution of L = p I  case, n ≤ mp (G) ≤ 2 by (4A), and the result follows. Lemma 4.3. Let A ≤ B ≤ G with B ∼ = Ep3 , |B : A| = p, and A# ∩ = ∅. Suppose that CG (A) contains a noncyclic elementary abelian 2-subgroup E. Then LA /Op (LA ) is simple, E acts nontrivially on it, and mIp (LA ) = 0. Ipo (G)

Proof. Set CA = CG (A) and C A = CA /Op (CA ). By Theorem C∗4 : Stage A1, Op (CA ) has odd order, so E ∼ = E. If CE (Op (C A )) = 1, then C contains Ep2 × D2p × D2p by the Thompson dihedral lemma, so mIp (CA ) ≥ 3, contrary to (4A). Therefore CE (Op (C A )) = 1, so [LA , E] = 1. As A contains an element of Ipo (G), any component L of LA lies in Cp by hypothesis and [IA , 7.1.10]. Thus L is listed in [V17 , 7.1]. Since mp (A) = 2 = mIp (C A ), mIp (LA ) = mp (Z(LA )). Thus LA = L, since otherwise an involution in CLA (L) centralizes an element of Ip (L) − Z(L). Now if Z(L) = 1, the only possibilities are L ∼ = 3J3 , 3M c, 3U4 (3), and 3G2 (3), all with p = 3. But then for some 2-central involution t ∈ L, CL (t) contains Σ3 , so mI3 (L) > m3 (Z(L)), contradiction [IA , 5.3hn, 4.5.1]. Therefore, L is simple.  As A ≤ Z(C A ) and e(G) = 3, the result follows. Lemma 4.4. Let A ∈ Ep3 (G). Then any A-invariant p -subgroup of G is solvable. Proof. Otherwise, let X be a minimal nonsolvable A-invariant p -subgroup of G. By minimality, X/ Sol(X) is the direct product of simple groups permuted transitively by A. If a ∈ A# permutes them nontrivially, then CX (a) is nonsolvable and contradicts the minimality of X. So X/ Sol(X) is simple. By [IA , 7.1.2], X/ Sol(X) ∈ Chev and AutA (X/ Sol(X)) is generated by a field automorphism, whence CX (A) has even order. This contradicts  mIp (G) ≤ 2 and completes the proof. 5. p = 3; L2 (3n ) 3-Components Continuing the case Hinv = ∅, we prove the following two results together. Proposition 5.1. We have p = 3.

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300

Lemma 5.2. Let H ∈ H. Choose any B ∈ E33 (H), let V be any minimal B-invariant subgroup of O2 (H), and set B0 = CB (V ). Then there is b0 ∈ B such that the following conditions hold: (a) V ∼ = A4 × E32 ; and = E22 and V B = V b0  × B0 ∼ (b) V ≤ LB0 ; (c) LB0 /O3 (LB0 ) ∼ = L2 (3n ), n ≥ 2. (d) m2 (CG (B0 )) = 2, and Sylow 2-subgroups of O2 (CG (B0 )) are dihedral. Proofs of Proposition 5.1 and Lemma 5.2. Let H ∈ H. Choose any B ∈ Ep3 (H), let V be any minimal B-invariant subgroup of O2 (H), and set B0 = CB (V ). By (4A), CV (B) = 1. Therefore B0 ∼ = Ep2 , there exists b0 ∈ B acting irreducibly on V , V is elementary abelian, and V B = V b0  × B0 . 0 = C0 /Op (C0 ), and L0 = LB . By Theorem C∗ : Set C0 = CG (B0 ), C 0 4 # o Stage A1, B ⊆ Ip (G). Since m2 (C0 ) ≥ m2 (V ) ≥ 2, Lemma 4.3 implies  0 is simple with mI (L  0 ) = 0; moreover, AutV (L  0 ) is nontrivial and that L p  0 ). Also as mI (G) ≤ 2, acted on nontrivially by the image of b0  in Aut(L p CL0 (B) = CL0 (b0 )

has odd order. By [V17 , 7.8], p = 3 and LB0 /O3 (LB0 ) ∼ = L2 (3n ), n ≥ 2. It remains to prove (b) and (d) of Lemma 5.2. Let t ∈ C0 be any in 0 , then volution. If t induces a (possibly trivial) field automorphism on L m3 (CG (t)) ≥ m3 (CL0 (t)) + m3 (B0 ) > 2, contradiction. Therefore t in 0 . As t was arbitrary, duces a nontrivial inner-diagonal automorphism on L n C(B0 , L0 ) has odd order and m2 (C0 ) ≤ m2 (P GL2 (3 )) = 2. We use the fact that Out(L2 (3n )) is abelian [IA , 2.5.12]. For one thing, Sylow 2-subgroups of O 2 (C0 ) embed in O2 (Aut(P GL2 (3n ))) and hence in L2 (3n ), so are dihedral. 0 lies in L  0 , so Hence (d) holds. For another, the image of V = [V, b0 ] in C  2 V ≤ O (L0 C(B0 , L0 )) = L0 , which proves (b).  Proposition 5.3. If Hcons = ∅, then Theorem C∗4 : Stage A3 holds. Proof. By Proposition 5.1, take any H ∈ Hcons and apply Lemma 5.2 to it. As F ∗ (H) = O2 (H), Proposition 3.1a applies with B0 in the role of A there. We conclude that n = 2, so alternative (a) of the theorem holds.  6. 2-Subgroups of G Normalized by E33 -Subgroups By Proposition 5.3, Theorem C∗4 : Stage A3 will be completely proved once we show that F ∗ (H) = O2 (H) for some H ∈ H. We prove this by contradiction in the next three sections, thus making the following assumption. (6A)

For all H ∈ H, E(H)O2 (H) = 1.

In this section we make our only use of (6A). We prove the following result, from which strong restrictions on the {2, 3}-local structure of G will be deduced in the next section. Note that p = 3 by Proposition 5.1.

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Proposition 6.1. Let B ≤ G with B ∼ = E33 , and suppose that T is a nontrivial B-invariant 2-subgroup of G. Then T ∼ = E22 . We prove the proposition by contradiction in a sequence of lemmas. Assuming that T ∼ E22 , we first prove = Lemma 6.2. There exist B-invariant four-subgroups U1 , U2 of G and elements a1 , a2 , b ∈ B generating B such that N := U1 U2 B = U1 a1  × U2 a2  × b ∼ = A4 × A4 × Z3 . Proof. By (4A), CT (B) = 1. For each hyperplane B0 ≤ B set TB0 = CT (B0 ), so that TB0 = [TB0 , B] ≤ O2 (CG (B0 )). Thus if TB0 = 1, then TB0 B contains a copy of A4 centralizing B0 . Hence Lemma 5.2 applies with H = NG (TB0 ), giving that TB0 is dihedral and hence TB0 ∼ = E22 by the action of B. As T = 1, there is a hyperplane B1 of B such that 1 = TB1 ≤ Z(T ). Since T ∼ E22 , T > TB1 and so there is a hyperplane B2 = B1 of B such = that TB2 = 1. The lemma follows with Ui = TBi , i = 1, 2, and with any  choices of b ∈ B1 ∩ B2# , a1 ∈ B2 − B1 and a2 ∈ B1 − B2 . We set H = NG (U1 U2 ) ∈ H, C = CG (U1 U2 ), L = L3 (CG (b)), and  L = L/O3 (L). For i = 1, 2, set Ii = L3 (CG ( b, ai )). Lemma 6.3. The following conditions hold: ∼ (a) L = A9 , J3 or 3J3 ; and (b) U1 U2 ∈ Syl2 (NG ( b) ∩ C). Proof. Let i = 1 or 2, and {i, j} = {1, 2}. Apply Lemma 5.2 with NG (Uj ), B, and Uj in the roles of H, B and V . We conclude that Ii ∼ = L2 (3n ), n ≥ 2, and Uj ≤ Ii . By L3 -balance and Lemma 4.2, the subnormal closure Li of Ii in CG (b) is a 3-component of L. Moreover m3 ( b Li ) ≥ m3 ( b Ii ) = n + 1 ≥ 3, so C(b, Li ) has odd order by Lemma 4.1. Therefore L1 = L = L2 . In particular Ii is a 3-component of L3 (CL (ai )) for i = 1, 2. Moreover U1 U2 ≤ I1 , I2  ≤ L. Now [V17 , 7.6a] implies (a). In particular  and conclude that C0 has m3 (L) ≥ 3 [IA , 5.6.1]. We set C0 = CNG (b) (L) I odd order, since m3 (G) < 3. Suppose that (b) fails, so that NG ( b) ∩ CG (U1 U2 ) contains a 2-element   t ∈ U1 U2 . Then t normalizes L. But by [V17 , 7.6b], CAut(L)  (U1 U2 ) is the  of U 1 U 2 . As t ∈ U1 U2 , t U1 U2 ∩ C0 = 1. This contradicts image in Aut(L)  the fact that C0 has odd order, so the proof is complete. Lemma 6.4. We have E(H) = 1. Proof. Clearly C ∩ B = b. Suppose first that m3 (C) > 1. Let P ∈ Syl3 (H), so that m3 (P ) = m3 (B) = 3. By [IG , 10.11] there is A ≤ C such that A ∼ = E32 and A  P , and then there exists 1 = a ∈ Ca1 ,a2  (A). Then A a ∼ = E33 and for some i = 1, 2, Ui = [A a , Ui ] and A = CAa (Ui ). Now

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Lemma 5.2 applies with NG (Ui ), Ui , A a and A in the roles of H, V , B and B0 . By part (d) of that lemma, m2 (CG (A)) = 2. But A centralizes U1 U2 ∼ = E24 , a contradiction. Therefore m3 (C) ≤ 1, and equality holds as b ∈ C. By Lemma 6.3b, NC ( b)/U1 U2 has odd order, so b ∈ Z(NC ( b)). Thus C has a normal 3-complement by Burnside’s normal complement theorem [IG , 16.5]. But by Lemma 4.4, O3 (C) is solvable, so C is solvable. Hence  E(H) ≤ C (∞) = 1, as required. Now we use (6A). Set W = O2 (H) = O2 (C). By Lemma 6.4 and (6A), W = 1. Also set Ni = NG (Ui ), Ci = CG (Ui ), and N i = Ni /O2 (Ni ), i = 1, 2. Then W = O2 (CCi (Uj )) where, as before, {i, j} = {1, 2}. Obviously E(N i ) = E(C i ). By (4A), m3 (Ci ) = 2. Lemma 6.5. W ≤ O2 (Ci ) for i = 1 and i = 2. Proof. If false, then as C = CCi (Uj ), the theory of balance [IG , 20.6] provides a W U1 U2 -invariant 2-component Ii of L2 (Ni ) such that [W , I i ] = 1,

≤ and such that if we let Y = W Ii U1 U2 and Y = AutY (I i ), then W 1 U j )). By Lemma 4.4, no components of E(N i ) 2 )) = O2 (C  (U O2 (CY (U Y   are 3 -groups. Thus E(N i ) = O3 (E(N i )) and so O2 (Ci ) normalizes I i by Lemma 4.2. Let Y ∗ = Y O2 (Ci ) and write Y ∗ = AutY ∗ (I i ). Thus j  aj  ∼ U = A4 . As G is of restricted even type, I i ∈ Co2 , and I i ∈ A by [V17 , Theorem 1.3]. If b centralizes I i , then I i is involved in A9 or J3 by

= 1 or Lemmas 6.3 and 4.4. Hence in any case by [V17 , 4.2], either W j ) = U j × Σ with 1 = b ∈ Σ  ∼ CIi (U = Σ3 and I i ∼ = M12 . In the first case [W , I i ] = 1, contradiction. Hence the second case holds. As m3 (Ci ) = 2,  has a Ii   L3 (Ci ) and so CCi (I i ) is a 3 -group. Hence an involution of Σ preimage t ∈ CIi (U1 U2 ) ∩ NG ( b) − U1 U2 such that t is a 2-element. This contradicts Lemma 6.3b and proves the lemma.  Now we complete the proof of Proposition 6.1. Write U1 = u, v and set Cu = CG (u). Notice that U2 a2  ≤ Cu , and as W ≤ O2 (C2 ) and [W, U1 ] = 1, we have W ≤ O2 (CCu (U2 )), with W being U2 a2 -invariant. Using [V17 , 4.2] again, we conclude that [W, J] = 1 for every component J of Cu , unless possibly J/Z(J) ∼ = M12 . But by Lemma 6.5, W ≤ O2 (CG (U1 )) = O2 (CCu (v)), and M12 is locally 1-balanced with respect to the prime 2 by [IA , 7.7.1]. Hence, in any case [W, J] = 1, by [IG , 20.6]. We have therefore shown that [W, E(Cu )] = 1. However, [W, O2 (Cu )] = 1 by L∗2 -balance [IG , 5.18]. As G has (restricted) even type, O2 (Cu ) = 1. Therefore, [W, F ∗ (Cu )] = 1, which contradicts the F ∗ -Theorem [IG , 3.6]. This completes the proof of Proposition 6.1.

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7. Subgroups H ∈ H and Involution Centralizers We choose any H ∈ H, set VH = O2 (H) and CH = CG (VH ), and choose any involution z ∈ VH . Set Ez = E(CG (z)). Also let B ≤ H with  0 = L0 /O3 (L0 ), and B∼ = E33 , and set B0 = CB (VH ), L0 = L3 (CG (B0 )), L Lb = L3 (CG (b)) and Lb = Lb /O3 (Lb ) for every b ∈ B0# . Using Proposition 6.1 we shall now prove the following structure theorem for our arbitrary H ∈ H. Proposition 7.1. VH ∼ = E22 , B0 ∼ = E32 , and either (a) or (b) holds: ∗ (a) F (H) = VH × E(H) with E(H) ∼ = A6 , M11 or L2 (8), and CH ∼ = 2 ) or L (34 ); 0 ∼ E22 ×Aut(L2 (8)) in the last case. Moreover, L (3 L = 2 2 or (b) F ∗ (H) = VH ×O3 (H), O2 (CH ) = O3 (H) ∈ Syl3 (Ez ), Ez /O2 (Ez ) ∼ = 2 ∼ ∼  L3 (4), CEz /O2 (Ez ) (VH ) = U3 (2), and L0 = L2 (3 ). By Proposition 6.1, VH ∼ = E22 and m2,3 (H/VH ) < 3. By (4A) and the fact that B/B0 ≤ Aut(VH ), B0 ∼ = E32 . Notice that several choices may be possible for B0 . In particular any E32 -subgroup B ∗ ≤ CH that is normal in some Sylow 3-subgroup of H is a possible choice for B0 , because m3 (CH (B ∗ )) = 3 by [IG , 10.20(ii)]. We proceed in a sequence of lemmas, the first of which describes normalizers of subgroups of B0 . By Lemma 5.2c, 0 ∼ L = L2 (3n ), n ≥ 2. Lemma 7.2. Let b ∈ B0# . Then (a) Lb ∼ = L2 (3nb ) with nb = n or 3n, and C(b, Lb ) has odd order; (b) If there is 1 = S ≤ CG (b) such that S is a 2-group, [S, VH ] = 1 and S ∩ VH = 1, then S ∼ = Z2 , n = nb = 2 and [S, B0 ] = 1; (c) CCH /VH (b) has a normal 2-complement and cyclic Sylow 2-subgroups of order dividing n; (d) AutG (B0 ) does not contain SL(B0 ) ∼ = SL2 (3); and 0 ∼ (e) If AutG (B0 ) contains a Q8 -subgroup, then L = A6 . Proof. Choose any b0 ∈ B −B0 . Then VH B = B0 ×VH b0  ∼ = E32 ×A4 . By Lemma 5.2, VH ≤ L0 . Write B0 = b, b . By L3 -balance [IG , 5.17] and [V17 , 7.3a], the pumpup L of L0 in CG (b) is a single 3-component, and, 0 ∼  = L/O3 (L), L3 (C  (b )) is a covering group of L setting L = L2 (3n ). FurL  thermore, CL (b ) = CL (B0 ) does not contain an isomorphic copy of Σ6 , for if it did, then mI3 (G) ≥ m3 (B0 ) + mI3 (Σ6 ) = 2 + 1, contradicting (4A). And by  b  does not contain any subgroup isomorphic to A4 ×A4 . Proposition 6.1, L ∼ Therefore by [V17 , 7.3b], L = L2 (3n ) or L2 (33n ), with b inducing a nontriv in the latter case. In particular, m3 (L b) ≥ 3 ial field automorphism on L in any case, so by (4A), (7A)

C(b, L) has odd order.

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This implies that L = Lb , so (a) holds. Also, (7A) implies that any 2-subgroup R of CG (b) acts faithfully on Lb . If S is as in (b), we take R = S × VH , and [V17 , 11.1.1a] implies that any involution s ∈ S induces a field automorphism on Lb . Thus CbLb /O3 (bLb ) (s) contains Z3 × L2 (3nb /2 ). As m3 (CG (s)) ≤ mI3 (G) < 3, we conclude that nb = 2, and it follows that S = s, and n = 2 by (a). Similarly, as m3 (CLb (s)) = 1, the fact that m3 (CG (s)) < 3 implies that [s, B0 ] = 1, and (b) is completely proved. In (c), we take R ∈ Syl2 (CCH (b)), so that R = VH F where F is a group of field automorphisms of Lb , by [V17 , 11.1.1b]. In particular R/VH is cyclic. Hence CCH /VH (b) has a normal 2-complement [IG , 16.7] and (c) holds.  0 ) is abelian, N := [NG (B0 ), NG (B0 )] induces inner auSince Out(L  0 . If AutG (B0 ) contains Q8 , then the image of N in tomorphisms on L Aut(B0 ) contains an involution, so C := CNG (B0 ) (L0 /O3 (L0 )) has even or0 ∼ der. As mI3 (G) ≤ 2, L = A6 in this case, proving (e). Continuing, we have 2    0 ). Therefore O (Aut(L)) = Inn(L) and so O2 (NG (B0 )) maps into Inn(L if AutG (B0 ) contains SL2 (B0 ), then AutC (B0 ) contains SL2 (B0 ), and in particular C contains commuting elements z, y of orders 2 and 3, respec 0 , so m3 (CG (z)) ≥ 3, a tively. But then CG (z) contains y and covers L contradiction. Hence (d) holds, and the proof is complete.  Lemma 7.3. Suppose that O2 (H) = 1. Then Proposition 7.1b holds. Proof. Set W = O2 (H) and write VH = z, z  . Then W = O2 (CCG (z) (z  )). Since G has restricted even type, O2 (CG (z)) = 1, and hence by [IG , 20.6], W acts faithfully on the product J ∗ of all z   W -invariant components of Ez = E(CG (z)) that are locally unbalanced with respect to z  . Let J be a component of J ∗ . Again as G has restricted even type, J ∈ C2 , and so by [V17 , 4.6], either J ∼ = L2 (q), q a Fermat or Mersenne prime or 9, with CJ (z  ) ∼ = Dq±1 , respectively, or J/O2 (J) ∼ = L3 (4) with CJ/O2 (J) (z  ) ∼ = U3 (2).  In all cases z induces a noninner automorphism on J. Denote by S a VH invariant Sylow 2-subgroup of J. L3 (4). Now B0 ≤ CH ≤ CG (z). If CB0 (J) Assume first that J/O2 (J) ∼ = contains a nonidentity element b, then since VH ≤ Lb , we have J = [J, z  ] ≤ Lb and then J ≤ E(CLb (z)) = 1, a contradiction. Therefore CB0 (J) = 1, so m3 (CAut(J) (z  )) ≥ 2. Given the possible isomorphism types of J, however, this is impossible. Therefore J/O2 (J) ∼ = L3 (4). Then CCG (z) (J) is a 3 -group as mI3 (G) ≤ 2. Since every component of  J ∗ has order divisible by 3, J = J ∗ = O3 (Ez ). If J = Ez , then Ez has a n component I such that I/Z(I) ∼ = 2B2 (2 2 ), n ≥ 3. But then m2 (CVH I (b)) ≥ 5, contradicting Lemma 7.2b. Therefore Ez = J. Finally, O2 (CH )E(CH ) ≤ L∗2 (CG (z)) by L∗2 -balance [IG , 5.18]. But Sylow 2-subgroups of L3 (4) are self-centralizing in Aut(L3 (4)) [IA , 3.1.4], and so L∗2 (CG (z)) = Ez . Thus O2 (CH ) = O2 (CEz (z  )) ∼ = E32 . By the

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305

∼ remark before Lemma 7.2, we may take B0 = O2 (CH ). Then AutEz (B0 ) = Q8 , but AutG (B0 ) does not contain SL(B0 ), by Lemma 7.2d. Therefore 0 ∼ AutG (B0 ) is a 2-group, whence B0 = O2 (CH ). By Lemma 7.2e, L = L2 (32 ) and the lemma is proved.  For the rest of the proof of Proposition 7.1, we may assume that O2 (H) = 1, whence F ∗ (H) = VH E(H) and E(H) = 1, because m3 (H) = 3. Lemma 7.4. Either E(H) is quasisimple with m3 (E(H)) = 2, or Proposition 7.1a holds with CH ∼ = Aut(L2 (8)). Proof. Let H1 , . . . , Hm be the components of E(H). By Lemma 4.4, 3 divides |Hi | for each i = 1, . . . , m. But m3 (E(H)) ≤ m3 (CH ) = m3 (B0 ) = 2. Thus if the first alternative of the lemma fails, m3 (Hi ) = 1 for each i, and m ≤ 2. If m = 2, take a non-trivial 3-element b ∈ H1 and S ∈ Syl2 (H2 ). Then S ≤ CG (b), [S, VH ] = 1 and S ∩VH = 1. (By [V17 , 9.1a] and [IA , 6.1.4], Z(Hi ) is cyclic for each i; then as VH = [VH , B], VH ∩ Hi = 1 for each i.) As S is non-cyclic, we have a contradiction with Lemma 7.2b. Thus m = 1. Because of Lemma 7.2c, for all b ∈ B0# , CH1 (b) is solvable with cyclic Sylow 2-subgroups. In particular, CB0 (H1 ) = 1 and so 3 divides | OutB0 (H1 )|. By [V17 , 9.1a], H1 ∼ = L2 (2n ), n ≥ 2, or L3 (2n ), n ≥ 1, with  = ±1 and 2n ≡ − (mod 3). Hence CH1 (b) ∼ = L2 (2n/3 ) or L2 (2n/3 ). As this must have cyclic Sylow 2-subgroups, the only possibility is H1 ∼ = L2 (8). Then H1 b ∼ = Aut(H1 ), and since CH1 (b) has a Sylow 2-subgroup S of order 0 ∼ 2, Lemma 7.2b implies that L = A6 . Thus Proposition 7.1a holds, as asserted.  Lemma 7.5. If E(H) is quasisimple with m3 (E(H)) = 2, then E(H) ∼ = A and A6 , with n = 2 or 4, or M11 , with n = 2. Moreover, if E(H) ∼ = 6 n = 4, then a Sylow 2-subgroup T of NE(H) (B0 ) is isomorphic to Z4 , and CL0 (T ) is 3-solvable. Proof. Set X = E(H), a quasisimple group by assumption. By [V17 , 9.1b], X ∈ Chev(2), or X/Z(X) ∼ = L3 (3), a Mathieu group, HJ, J4 , HS, or Ru. Most of these will be ruled out using Lemma 7.2c, reducing us to the following cases: 1 (7B) X ∼ = A6 , L3 (3), U3 (3), 2F4 (2 2 ) , or M11 , or X/O2 (X) ∼ = L3 (2m ),

where  = ±1, 2m ≡  (mod 3), and m > 1. Indeed all the sporadic groups mentioned above except X ∼ = M11 violate Lemma 7.2c; see [IA , Table 5.3]. Suppose then that X ∈ Chev(2). If some b ∈ B0# acts as an element of Aut(X) − Inndiag(X), then either b induces a graph automorphism on X, in which case m2 (CX (b)) > 1 by [IA , 4.7.3A], or b induces a field or graph-field automorphism, in which case the facts that CX (b) has cyclic Sylow 2-subgroups and |X|3 = 1 imply with [IA , 4.9.1] that

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X∼ = L2 (23 ). This is a contradiction as m3 (L2 (23 )) = 1. Hence B0 induces inner-diagonal automorphisms on X. Then X is as in (7B) by [V17 , 20.4]. So we are indeed reduced to the cases (7B). 1 If X ∼ = U3 (3) or 2F4 (2 2 ) , then some CX (b) contains a Z4 -subgroup, contradicting Lemma 7.2b. If X ∼ = A4 × X. More= L3 (3), then Out(X) is a 2-group, so O2 (H) ∼ over, a maximal parabolic subgroup Y ≤ X satisfies F ∗ (Y ) ∼ = E32 and ∗ ∗ ∗ ∼ Y /F (Y ) = GL(F (Y )) ≥ SL(F (Y )). By the remark before Lemma 7.2, we may assume that B0 = O3 (Y ), and then Lemma 7.2d is contradicted. Suppose that X ∼ = L3 (2m ). If 2m ≡  (mod 9), then there is B1 ≤ ∼ X such that B1 = E32 , B1 is normal in a Sylow 3-subgroup of X, and for some b ∈ B1# , CX (b) contains L2 (2m ); namely, we can take B1 to be the image of a diagonalizable subgroup of SL3 (2m ). By the remark before Lemma 7.2, we may take B0 = B1 . However, as m > 1, this contradicts Lemma 7.2c. Therefore 2m ≡  (mod 9), so |X|3 = 9 and we may take B0 ∈ Syl3 (X), whence AutX (B0 ) ∼ = Q8 . But in this case m ≡ 0 (mod 3), so  O 3 (Aut(X)) ∼ = P GL3 (2m ). Hence CB (X) = 1. Choose b ∈ CB (X). If X acted faithfully on O3 3 (CG (b))/O3 (CG (b)), then by the Thompson dihedral lemma, CG (b) would contain the direct product of b and m2 (X) copies of Σ3 , so mI3 (G) ≥ m2 (X). This is absurd as m2 (X) > 2 but mI3 (G) < 3. Thus X centralizes O3 3 (CG (b))/O3 (CG (b)), whence X ≤ Lb := L3 (CG (b)), which contradicts [V17 , 7.3c]. Consequently X ∼ = M11 , then for any = A6 or M11 as claimed. If X ∼ # 0 ∼ b ∈ B0 , CX (b) contains an involution and so L = A6 by Lemma 7.2b. If X ∼ = Z4 acting faithfully = A6 , then NX (B0 ) contains a subgroup T = t ∼ 0 ∼ on B0 . Then t normalizes L0 and t ∩ VH = 1. Suppose that L = L2 (3n ),  0 ) has odd order as mI (G) < 3 ≤ m3 (L0 ). Hence by [V17 , n > 2. Then CG (L 3  0 . Thus 11.1.1a], the involution t0 ∈ t induces a field automorphism on L n/2 I ∼ CL0 (t0 ) = L2 (3 ), so n/2 ≤ m3 (G) ≤ 2, whence n = 2 or 4. Moreover, if n = 4, then CL0 (t) is solvable. Thus the final assertion holds and the lemma is proved.  Lemmas 7.3, 7.4 and 7.5 prove Proposition 7.1. 8. The Residual Cases Fix H ∈ H. First suppose that E(H) ∼ = L2 (8), M11 , or A6 as in Propo0 ∼ sition 7.1a. We keep this assumption in Lemmas 8.1–8.3. Since L = L2 (3n ) with n even, L0 ∩ H = NL0 (VH ) contains a subgroup Σ ∼ = Σ4 , which we fix. Since [Σ, B0 ] = 1, it is immediate by [V17 , 4.1] that [Σ, E(H)] = 1, as Σ has no nontrivial quotient of odd order. Lemma 8.1. For any involution y ∈ Σ, we have E(H) = E(CG (y)). Proof. First suppose V is any four-subgroup of Σ such that E(H) = E(CG (V )). We claim that E(H) = E(CG (y)) for all y ∈ V # . Fix y and

8. THE RESIDUAL CASES

307

write V = y, y  . Let Ey be the subnormal closure of E(H) in CG (y). As G is of even type and by L2 -balance, Ey is a component of E(CG (y)) or the product of two such components interchanged by y  , and E(H) is a component of CEy (y  ). Since B0 acts faithfully on E(H), it acts faithfully on each component of Ey . But m3 (B0 ) = 2 = m3 (CG (y)), and so Ey is quasisimple. By Lemma 7.2a, CEy (b) is solvable for all b ∈ B0# . Given that E(H)B0 ∼ = Aut(L2 (8)), A6 or M11 , and that Ey ∈ Co2 , we conclude by [V17 , 20.3] that E(H) = Ey , proving our claim. For any involution y ∈ Σ, choose an involution z ∈ VH such that Y := z, y ∼ = E22 . Then E(CG (z)) = E(H) by our claim, and it follows immediately using L2 -balance again that E(CG (Y )) = E(CG (z)) = E(H).  Then E(CG (y)) = E(H) by our claim again, as desired. Lemma 8.2. L0 = E(CG (E(H))) ∼ = E(H) ∼ = A6 .   Proof. Let L∗0 = CL0 (y) y is an involution of Σ . Then L∗0 = L0 . Indeed, O3 (L0 ) has odd order by Theorem C∗4 : Stage A1, and L∗0 covers  0 . (If it did not, then by [IA , 7.3.4], L 0 ∼ L = A6 , and then NL0 (L∗0 ) would be strongly embedded in L0 , contradicting the Bender-Suzuki theorem [II2 , Theorem SE].) By Lemma 8.1, L0 = L∗0 normalizes E(H), and of course L0 centralizes B0 . As CAut(E(H)) (B0 ) is a 3-group [V17 , 7.24] and L0 = O3 (L0 ), we have [E(H)B0 , L0 ] = 1. If E(H) ∼ = L2 (8) or M11 , then E(H)B0 contains Z3 × Σ3 . But m3 (L0 ) ≥ 2, so mI3 (E(H)B0 L0 ) ≥ 3, contradicting (4A). Therefore, E(H) ∼ = A6 . Let W be an A4 -subgroup of E(H). Then H ∗ := NG (O2 (W )) contains L0 W , so H ∗ ∈ H. But L0 ≤ CG (E(H)) ≤ H ∗ ∼ A6 . Applying Proposition 7.1 and the prior argument in this 0 = with L lemma to H ∗ , we deduce that E(H ∗ ) ∼ = A6 is the unique nonsolvable composition factor of H ∗ . Therefore L0 = E(H ∗ ) = E(CG (E(H))), completing the proof.  ∼ A6 for all b ∈ B # , and E(H) = E(CG (y)) for all Lemma 8.3. Lb = 0 involutions y ∈ CG (E(H)). 0 ∼ Proof. Let T ∈ Syl2 (NE(H) (B0 )), so that T ∼ = Z4 . If L = L2 (34 ), then by Lemma 7.5, CL0 (T ) is 3-solvable. However, this is absurd because 0 ∼ E(CG (E(H))) ∼ = A6 lies in L0 and centralizes T . Therefore L = A6 . # Likewise for any b ∈ B0 , the involution t ∈ T inverts b and centralizes  0 in Lb would be of order 3, so  0 . If Lb ∼ L = L2 (36 ), then the centralizer of L t would centralize Lb , contradicting (4A). Therefore Lb ∼ = A6 , proving the first assertion of the lemma. Let y ∈ CG (E(H)) be any involution. Then [y, y  ] = 1 for some involution y  ∈ E(CG (E(H))) = L0 , and we know by Lemma 8.1 that E(H) =

14. THEOREM C∗4 : STAGE A3. KM -SINGULARITIES

308

∼ A6 , CG ( y, b) is E(CG (y  )). Hence we may assume that y = y  . As Lb = # solvable for all b ∈ B0 , by Lemma 7.2a. Then as in Lemma 8.1, using [V17 ,  20.3], we may again argue that E(H) = E(CG ( y, y  )) = E(CG (y)). Lemma 8.4. Proposition 7.1b holds. Proof. Assume false and continue the above analysis. By Lemma 8.3, E(H) is terminal in G and E(H) ∼ = A6 . Hence by definition of restricted even type [I2 , 8.8(3c)], m2 (CG (E(H))) = 1. But E(CG (E(H))) ∼ = A6 by Lemma 8.3, a contradiction.  Now fix z ∈ VH# and set C = CG (z), Q = O2 (C) and K = E(C). By Proposition 7.1b and as K ∈ Co2 , O2 (K) is elementary abelian and 0 ∼ K/O2 (K) ∼ = L3 (4). Also L = A6 . Expand VH Q to R ∈ Syl2 (CG (B0 )). Lemma 8.5. The following conditions hold: (a) Ω1 (Q ∩ L0 ) = z = Ω1 (Z(Q)); (b) z is 2-central in G; (c) For any involution y ∈ K − Z(K), no two involutions in y, z are G-conjugate; and (d) For any P ∈ Syl2 (CH ), there is y ∈ I2 (P ∩ K − Z(K)) such that

y char P .  0 ), and since mI (G) ≤ 2, Proof. By Lemma 7.2a, R embeds in Aut(L 3 0) CL0 (u) is a 3 -group for all u ∈ R# . Therefore the image of RL0 in Aut(L  0 , M10 or P GL2 (9), so R is dihedral or semidihedral of is isomorphic to L order at most 16. In any case NR (VH ) ∈ Syl2 (L0 ). As L0 has only one class of involutions, we could have chosen R so that z ∈ Z(R). In particular, we may expand R to S ∈ Syl2 (C). By [V17 , 10.6.1b], CAut(K) (B0 ) ∼ = B0 × Z2 , and so R ≤ VH CR (K). Set R0 = CR (K) ∩ NR (VH ), so that |R0 | = 4. If R0 ∼ = E22 , then as R0 ≤ L0 , R0 lies in an A4 -subgroup of L0 . Hence NG (R0 ) ∈ H, so Proposition 7.1 applies to NG (R0 ) in place of H. This yields a contradiction since CG (R0 ) involves K/O2 (K) ∼ = L3 (4). Therefore R0 ∼ = Z4 . Thus [R0 , z K] = 1 so Q > z. Hence z < NQ (VH ) ≤ R0 , so R0 ≤ Q. Since R is of maximal class, Q is cyclic or of maximal class, and there is Q0  S such that Z4 ∼ = Q0 ≤ Q. Thus (a) holds. Moreover (8A)

for any Q1  S such that Q1 ∼ = Z4 , z ∈ Q1 .

For otherwise, in C = C/CC (K) ≤ Aut(K), we would have Q1  S with Q1 ∼ = Z4 , contradicting [V17 , 10.6.2]. Now (8A) implies that z char S, so S ∈ Syl2 (G), proving (b). Let y ∈ K − Z(K) be an involution. By [V17 , 10.6.1ca], y has a Kconjugate in Z(S). Also, no two involutions of y, z are NG (K)-conjugate

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(we use [V17 , 10.6.1e] if Z(K) = 1). By Burnside’s Lemma [IG , 16.2], Gfusion in y, z is controlled in NG (S), and hence in C, since z char S. As C ≤ NG (K), (c) follows. Finally, let P ∈ Syl2 (CH ) and P ∗ ∈ Syl2 (CG (z)) with P ≤ P ∗ . Let ∗ P0 = CP ∗ (K), so that Q ≤ P0∗ and P0∗ embeds in R. Thus P0∗ is cyclic, dihedral of order at least 8, or semidihedral. Let P1∗ be the largest cyclic ∗ subgroup of P0∗ , so that Q ≤ P1∗  P ∗ , and set P = P ∗ /P1∗ (P ∗ ∩ K). Then ∗ ∗ ∗ ∗ P is an extension of P 0 , of order at most 2, by P /P 0 , which embeds in a Sylow 2-subgroup of Out(K), which is a four-group. As V H ≤ Z(P ), ∗ P is abelian. Moreover, if P 0 = 1, then AutP0∗ (P1∗ ) ≤ Φ(Aut(P1∗ )), so ∗ ∗ ∗ P 0 ∩ Φ(P ) = 1. It follows that Φ(P ) = 1. Therefore Φ(P ) ≤ P1∗ (P ∗ ∩ K). In particular, VH ∩ Φ(P ) < VH , so since NG (VH ) ∩ NG (Φ(P )) is transitive on VH# by a Frattini argument, VH ∩ Φ(P ) = 1. In particular, z ∈ Φ(P ). Now CP ∗ ∩K (VH ) is of exponent 4 and embeddable (mod Z(K)) in Q8 . As

z = Ω1 (P1∗ ), we conclude that

y := Ω1 (Φ(P )) ≤ K. Thus (d) holds and the proof is complete.



Lemma 8.6. Proposition 7.1b does not hold. Proof. Suppose that it does hold and continue the preceding argument. Write VH = z, z  , so that CK/O2 (K) (z  ) ∼ = U3 (2). By Lemma 8.5d, there is an involution y ∈ CK (z  ) − Z(K) such that y is characteristic in a Sylow 2-subgroup of CH . Since AutG (VH ) ≥ AutL0 (VH ) = Aut(VH ), a Frattini argument shows that AutCG (y) (VH ) = Aut(VH ). On the other hand, by [V17 , 10.6.1c], z  and z  y are K-conjugate, modulo Z(K). By Lemma 8.5a, Z(K) ≤ z. Therefore z and zy are G-conjugate, modulo z  . As z ∼G zz  , either z ∼G zy or zz  ∼G z ∼G zz  y. In either case, z ∼G zy, as z ∼CG (y) zz  . This contradicts Lemma 8.5c, so the lemma is proved.  Now Lemmas 8.4 and 8.6 are in conflict. Therefore, as noted at the beginning of Section 6, Theorem C∗4 : Stage A3 is proved.

CHAPTER 15

Theorem C∗4 : Stage A4. Setups for Recognizing G 1. Introduction In this chapter we develop separately the two strands of the conclusion of Theorem C∗4 : Stage A3. We pursue them to points from which the final identification of the target groups takes off, in Stage A5. Theorem C∗4 : Stage A4. Let G and p be as in Theorem C∗4 (Case A). Assume that Hinv ⊆ Hcons . Then p = 3, and either (a) or (b) holds: (a) G has an A9 -Sp6 (2) setup; or (b) G has an extraspecial setup. We recall the detailed definitions of setups referred to in Theorem C4 : Stage A4 [V11 , 2.3, 2.4]. G has an A9 –Sp6 (2) setup if and only if (1) There is H ∈ Hcons and B0 ≤ H such that B0 ∼ = E32 , m3 (CH (B0 )) = 3, and B0 is a KM -singularity of type A6 ; that is, CG (B0 ) has a 3-component J such that J/O3 (J) ∼ = A6 ; (2) For any H, B0 , and J as in (1), and any x ∈ B0# , J pumps up vertically or trivially in CG (x) to a 3component Jx ; (1A) (3) For some x as in (2), Jx ∼ = A9 or Sp6 (2); and (4) For any x as in (2) such that Jx is a vertical pumpup of J, m3 (C(x, Jx )) = 1 and Jx /O3 (Jx ) ∼ = A9 , Sp6 (2), J3 , Sp4 (8), or L2 (36 ). In the L2 (36 ) case, or for any x ∈ B0# such that Jx is a trivial pumpup of J, we have [CO2 (H) (x), B0 ] = 1. Likewise we say that G has an extraspecial setup if and only if (1) For any z ∈ I2 (G) such that A ≤ CG (z) for some A ∼ = E33 , z is 2-central in G, and Qz := F ∗ (CG (z)) = O2 (CG (z)) is the central product of

z = 3 or 4 copies of Q8 ; and (1B) (2) For any z and A as in (1), there exists x ∈ A# such that |CQz (x)| ≥ 25 , and for any such x, the 3-layer L := L3 (CG (x)) satisfies m3 (C(x, L)) = 1 and L/O3 (L) ∼ = U4 (2), 32 U4 (3), or G2 (3). 311

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An A9 -Sp6 (2) setup arises from conclusion (a) of Theorem C∗4 : Stage A3, and gives rise to the target simple groups A12 , F5 , Sp8 (2), and F4 (2). An extraspecial setup arises from conclusion (b) of Theorem C∗4 : Stage A3, and gives rise to the target simple groups U5 (2), U6 (2), D4 (2), Co2 , Suz, and F3 . We recall also that H is the set of all 2-local subgroups H of G such that mp (H) ≥ 3; Hcons = {H ∈ H | F ∗ (H) = O2 (H)}; and Hinv = {H ∈ H | H = CG (z) for some z ∈ I2 (G)}. Remark 1.1. The assumptions of Stages A3 and A4 of Theorem C∗4 are identical. Therefore we may use any result from Chapter 14 in the current chapter, provided of course that any additional hypotheses are satisfied. In particular, we shall soon use some preliminary results from [V14 , Section 2]. 2. Pumpups of a Rank 2 KM -Singularity of Type A6 We begin by assuming that conclusion (a) of Theorem C∗4 : Stage A3 holds. Thus p = 3, and we consider an arbitrary H ∈ Hcons containing some B ∼ = E32 such = E33 , which in turn contains a KM -singularity B0 ∼ # that J := L3 (CG (B0 )) satisfies J/O3 (J) ∼ = A6 . For each x ∈ B0 , we consider the pumpup Jx of J in CG (x). We again write Lx = L3 (CG (x)) and LA = L3 (CG (A)) for any A ∈ E3 (G). Lemma 2.1. If x ∈ B0# , then Jx is a trivial or vertical pumpup of J. Proof. This is immediate from [V14 , Corollary 2.5]. (See Remark 1.1.)  Our first main result is: Proposition 2.2. Let x ∈ B0# . Then the following conditions hold: (a) m2 (CG (B0 )) ≤ 3 and C(B0 , J) has odd order; (b) Any B0 -invariant 2-subgroup of CG (x) acts faithfully on Jx /O3 (Jx ); (c) If Jx is a vertical pumpup of J, then m3 (C(x, Jx )) = 1 and one of the following holds: (1) CG (B) has odd order, and either Jx ∼ = J3 or 3J3 , or Jx /O3 (Jx ) ∼ = L2 (36 ); (2) CG (B) has even order, and Jx ∼ = Sp6 (2) or Sp4 (8); or ∼ (3) Jx = A9 ; (d) If Jx /O3 (Jx ) ∼ = A6 or L2 (36 ), then [CO2 (H) (x), B0 ] = 1; and (e) Jx = Lx . Note that except for the two cases in (d), it is being asserted that O (J) = 1. 3

Proof. By [V14 , Corollary 2.4], C(B0 , J) has odd order. Therefore m2 (CG (B0 )) ≤ m2 (Aut(A6 )) = 3, proving (a). Note that if Lx has a 3component K besides Jx , and A ∈ E3∗ (C(x, K)), then m3 (A) ≥ m3 (J x) =

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3 and LA covers K/O3 (K) by L3 -balance. But LA = 1 by [V14 , Corollary 2.3], contradiction. Therefore (e) holds. Since m3 (B0 Jx ) ≥ m3 (B0 J) = 4 > e(G), no 2-local subgroup in CG (x) contains B0 and covers Jx /O3 (Jx ) (or even J/O3 (J)). This implies (b), which in turn yields (d) in the A6 case. If Jx is a vertical pumpup of J, then by [V17 , 7.2], the isomorphism type of Jx /O3 (Jx ) is as claimed in (c). By [V14 , Corollary 2.7], m3 (C(x, Jx )) = 1. L2 (36 ). Then m2 (Jx ) ≥ 4, and if equalNow suppose that Jx /O3 (Jx ) ∼ = ∼ ity holds, then Jx /O3 (Jx ) = J3 , 3J3 , or A9 , whence Jx /O3 (Jx ) contains a Frobenius group of order 9.8. Hence by [V9 , 2.4, 2.9], Jx is quasisimple, and indeed O3 (Jx ) = 1 as the Schur multipliers of these groups are {2, 3}-groups and O3 (CG (x)) has odd order by Theorem C∗4 : Stage A1. Furthermore, B = B0 × b for some b ∈ I3 (J), as m3 (C(B0 , J)) = 2. As C(B0 , J) has odd order, CG (B) has even order if and only if CG (B0 )/CG (B0 , J) contains Σ6 . By [V17 , 13.32], |CG (B)| is therefore odd if Jx /O3 (Jx ) ∼ = J3 , but if Jx ∼ = Σ6 , so |CG (B)| is = Sp6 (2) or Sp4 (8), then CJx (B0 ) contains Sp4 (2) ∼ even in these two cases. This completes the proof of the proposition unless Jx /O3 (Jx ) ∼ = L2 (36 ), which we assume for the remainder of the proof. We need to prove (c) and (d). Again if |CG (B)| is even, CG (B0 )/CG (B0 , J) must contain Σ6 . Therefore some involution t ∈ CG (B0 ) induces a nontrivial field automorphism on J/O3 (J), and hence also on Jx /O3 (Jx ). But then m3 (CG (t)) ≥ m3 (CJx (t) x) = 1 + m3 (L2 (33 )) = 4, contradicting e(G) ≤ 3. So (c) holds. It remains to prove (d). Set CG (x) = CG (x)/O3 (CG (x)) and V = [CO2 (H) (x), B0 ] and assume for a contradiction that V = 1. As Out(J x ) is abelian and B0 / x induces nontrivial field automorphisms on J x , we have (2A)

[B0 , NG ( x)] ∩ B0 ] ≤ x . 

For the same reason V = [V, B0 ] ≤ O2 (Jx C(x, Jx )) = Jx , and as Jx has dihedral Sylow 2-subgroups, V is a four-group. Thus CInn(J x ) (V ) ∼ = V . Set B1 = CB (V ), so that every element of B1 induces a possibly trivial field automorphism on J x . Therefore E(CJ x (B1 )) ∼ = A6 , and the hypotheses of  := our proposition hold with B1 and J1 L3 (CJx (B1 )) in place of B0 and J. Thus, the previous part of our proof is valid for (H, B, B1 , J1 ). Set W = CO2 (H) (V ). As F ∗ (H) = O2 (H), and by the A × B-lemma, CB1 (W ) = 1. Since V ≤ W , CB (W ) = 1 and B1 = CB (CW (B1 )). Thus as CG (B) has odd order, it follows by [V9 , 8.5] that there exists a subgroup I = I1 × I2 × I3 ≤ W B such that Ii ∼ = A4 for each i = 1, 2, 3, and B1 ∈ Syl3 (I2 I3 ). But x ∈ B1 = B0 , so O2 (I1 ) = [O2 (I1 ), B0 ] ≤ [CO2 (H) (x), B0 ] = V . Thus equality holds throughout. In particular, by (b), x = b2 b3 for some bi ∈ I3 (Ii ), i = 2, 3. Likewise for i = 2, 3, [B1 , CO2 (I) (bi )] = 1, so (c) applies (recall that CG (B) has odd order) to give that each Lbi ∼ = A9 , J3 , or 3J3 ; the possibility

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15. THEOREM C∗4 : STAGE A4. SETUPS FOR RECOGNIZING G

∼ L2 (36 ) does not occur as m2 (CG (bi )) ≥ m2 (CO (I) (bi )) = 4. Lbi /O3 (Lbi ) = 2 In particular, bi  ∈ xG . Moreover, B1 = bi , ci  with ci ∈ Lbi − O3 (Lbi ), and so there is a 2-element ti ∈ NLbi (B1 ) inverting ci . We claim that

ci  = x. Indeed, otherwise the ti -orbits on E1 (B1 ) are bi , ci , and two conjugates of x, whence ci  = b3−i  because b3−i  ∈ xG . Thus t3−i inverts bi , and t2 t3 inverts B1 elementwise, contradicting the B1 -version of (2A) as x ∈ B1 ∩ B0 . This proves the claim. Now ti inverts x for i = 2, 3, and we set X = t2 , t3  ≤ NG ( x) and  X = X/CX (B1 ). As the actions of t2 and t3 on B1 are distinct, we must  ∼ have X = Σ3 , so there is a 3-element y ∈ X such that [B1 , y] = x and [x, y] = 1. But B1 maps onto the Sylow 3-subgroup of Out(Jx /O3 (Jx )). Hence we may modify y by an element of B1 and take y ∈ Jx C(x, Jx ), and then y ∈ C(x, Jx ). On the other hand, as m3 (C(x, Jx )) = 1, x ∈ y and y has order 3n , n > 1. Let R ∈ Syl3 (CG (x)), so that R0 := R ∩ Jx C(x, Jx ) is clearly the largest abelian subgroup of R, and then x = Ω1 (Φ(R0 )) char R. Therefore R ∈ Syl3 (G). Moreover, |R/R0 | ≤ | Out(Jx /O3 (Jx ))|3 = 3. Hence any nontrivial 3-subgroup of G has an abelian maximal subgroup. But Sylow 3-subgroups of J3 do not have this property [V17 , 10.13.1] and so Lbi ∼ = A9 , i = 2, 3. Thus t2 ∈ Lb2 may be chosen to be a root involution with J t2  ∼ = Σ6 , i.e. t2 induces a field automorphism on J ∼ = L2 (9). So t2 b2  ∼ = Z6 acts faithfully on Jx /O3 (Jx ) by field automorphisms. Let C = CG (t2 ), so that C involves P ΣL2 (33 ), an extension of L2 (33 ) by a field automorphism of order 3. Hence C ∈ Hinv , so C ∈ Hcons by the hypothesis of Theorem C∗4 : Stage A4. Let Q = O2 (CG (t2 )) = F ∗ (CG (t2 )), and let U ∈ Syl3 (CJx (t2 )), so that NJx ∩C (U ) has a b2 -invariant subgroup Y such that Y /O3 (Jx ) ∼ = Z13 acts faithfully on U and is not centralized by b2 . By Clifford’s theorem, Q/Φ(Q) has 13 distinct homogeneous U -submodules permuted regularly by Y . Then b2 permutes them in four disjoint 3-cycles and a fixed point, and so |CQ (b2 )| ≥ 28 . However, |CG (b2 )|2 ≤ | Aut(A9 )|2 = 27 , a contradiction. The lemma is at last proved.  Lemma 2.3. Suppose that x ∈ B0# and Jx ∼ = Sp4 (8). Then for any = Sp4 (8). y ∈ B0 − x, Jy ∼ Proof. Suppose false, so that we may write B0 = x1 , x2  with Jxi ∼ = Sp4 (8), i = 1, 2. By L3 -balance, J = E(CJxi (x3−i )), i = 1, 2, with x3−i inducing a field automorphism on Jxi . We have B = B0 y for some y ∈  I3 (J). Set Ji = E(CJxi (y)) = O2 (CJxi (y)) ∼ = L2 (8), i = 1, 2. Thus any element of B − xi , y induces a nontrivial field automorphism on Ji . So x3−i induces a nontrivial field automorphism on Ji . In particular, J1 =  := CG (B0 )/O3 (CG (B0 )) has a unique subgroup Y ∼ J2 . Moreover, C = Σ6 ,

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and O2 (CC (B)) = O2 (CY (y)) ∼ = Σ3 . Thus O2 (CG (B)) has a unique Z3 subgroup W , and W ≤ J1 ∩ J2 . The subnormal closures of J1 and J2 in CG (y) therefore coincide; we call them I. Since J1 = J2 , and by [V14 , 2.5], I is a vertical pumpup of both J1 and J2 . By [V17 , 7.2], I/O3 (I) ∼ = Sp4 (8) or 3D4 (2). Every element of B0 − xi  induces a nontrivial field automorphism on Ji , and hence induces a field or graph automorphism on I/O3 (I), according to the isomorphism type of I/O3 (I). Applying this for i = 1 and 2 we find that B0 faithfully induces a group of field or graph automorphisms on I/O3 (I), respectively, forcing B0 to be cyclic [IA , 2.5.12]. This contradiction completes the proof.  





Now we can prove: Proposition 2.4. For some x ∈ B0# , Jx ∼ = A9 or Sp6 (2). Proof. Suppose not. Since CB0 (O2 (H)) = 1, there exist generators b1 , b2 of B0 such that [CO2 (H) (bi ), B0 ] = 1 for i = 1 and 2. By Proposition 2.2dc, we may assume that for each i = 1, 2, Jbi /Z(Jbi ) ∼ = J3 or Jbi ∼ = Sp4 (8), according as CG (B) has odd or even order. By Lemma 2.3, Jbi /Z(Jbi ) ∼ = J3 for both i = 1 and 2, and CG (B) has odd order. Let z ∈ I2 (J). We study Cz := CG (z). For any b ∈ B0# , set Cb = CCz (b), so that Cb = CCG (b) (z) with z ∈ J ≤ Jb . ∼ J3 (such as b1 or b2 , for example). Take any b ∈ B0# for which Jb /Z(Jb ) = (∞) ∼ Set Db = Cb = Qb Eb , where Qb = [Qb , Eb ] ∼ and Eb ∼ = 21+4 = Ω− − 4 (2) = A5 [IA , 5.3h]. In particular, Db is a 3-component of CCz (b), so by L3 -balance and the B3 -property [IA , 7.1.3], Qb ≤ O3 (Cz ), while the image of Db in z := Cz /O3 (Cz ) is a component of C  (b) isomorphic to A5 . Next we C Cz show that (2B)

O3 (Cz ) is solvable.

For otherwise, Cz has a chief factor F = M/N within O3 (Cz ) which is the direct product of Suzuki groups, and then CF (B0 ) is also such a 1 nontrivial direct product, possibly including some 2B2 (2 2 ) direct factors [IG , 1 3.27(iii)], [IA , 4.9.1a]. Let S0 ∈ Syl2 (CM (B0 )), so that |S0 | ≥ |2B2 (2 2 )|2 > 2.  We have S0 ≤ O2 (O3 (Cb1 )). But by Proposition 2.2b, S0 acts faithfully on Jb1 /Z(Jb1 ) ∼ = J3 , and CAut(Jb1 ) (z) is an extension of O2 (CJb1 (z)) ∼ = Qb1 

by Σ5 [IA , 5.3h]. Thus O2 (O3 (Cb1 )) = Qb1 , so S0 ≤ Qb1 . As |S0 | > 2,  Eb S0 1 = Qb1 . Therefore Qb1 ≤ M , and Qb1 projects faithfully into F since the projection of S0 is not of exponent 2. But this is absurd; Ω1 (Qb1 ) is not abelian, but Ω1 (S1 ) is abelian for any 2-subgroup S1 ≤ F . Thus (2B) holds, and consequently F ∗ (O3 (Cz )) = O2 (Cz ) as G is of even type. Moreover, (2C) Qb ≤ O2 (Cz ) for any b ∈ B # such that Jb /Z(Jb ) ∼ = J3 . 0

Indeed, in view of the irreducible action of Db on Qb / z, the only alternative to (2C) would be that Qb ∩ O2 (Cz ) = z. As Qb ≤ O3 (Cz ) and

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Qb / z is the unique minimal normal subgroup of Db / z, Db / z would act faithfully on Or (Cz /O2 (Cz )) for some prime r > 3, and them mr (Cz ) ≥ 4 by the Thompson dihedral lemma, contradicting e(G) = 3. We now have F ∗ (Cz ) = O2 (Cz ) =: V. Otherwise, as B0 ∈ Syl3 (Cbi ), i = 1, 2, and [B0 ∩ Dbi , E(Cz )] ≤ [Dbi , E(Cz )] = 1, we would have B0 ∩ E(Cz ) = CB0 (Dbi ) = bi  for both i = 1 and 2, which is impossible. Using Proposition 2.2b and [IG , 5.19(i)], we see that for each b ∈ B0# , CV (b) injects into O3 (CAut(Jb )) ( z ), where Jb = Jb O3 (CG (b))/O3 (CG (b)). By the first paragraph and Proposition 2.2, Jb /Z(Jb ) ∼ = J3 or Jb ∼ = A6 6 ∼ or L2 (3 ). For any b such that Jb /Z(Jb ) = J3 , we have CV (b) = Qb by (2C). Similarly if Jb ∼ = L2 (36 ), then CV (b) acts faithfully on Jb and contains CV (B0 ) ∈ Syl2 (J) ⊆ Syl2 (Jb ); thus as Out(Jb ) is abelian, [CV (b), B0 ] ≤ CV (B0 ), whence [CV (b), B0 ] = 1 and CV (b) = CV (B0 ). Therefore, V is generated by all the subgroups Qb such that b ∈ B0# and Jb /Z(Jb ) ∼ = J3 , and for all such b, Qb = Rb ∗ CV (B0 ) with Rb = [Qb , B0 ] ∼ = Q8 . In particular, V = CV (B0 )W , where W = CV (CV (B0 )) and CW (B0 ) =

z. Moreover, for any b ∈ B0# , CW (b) = [CW (b), B0 ] ∼ = Q8 or CW (b) =

z ≤ CW (B0 ) according as Jb /Z(Jb ) ∼ = J3 or not. It follows that W B0 contains no subgroup isomorphic to B0 × E22 or Z3 × A4 , so every characteristic abelian subgroup of W is cyclic. Thus W is of symplectic type by Philip Hall’s theorem [IG , 10.3], and from the structure of the subgroups CW (b), W is the central product of k copies of Q8 , where k ≤ |E1 (B0 )| = 4. For i = 1 and 2, W = CW (bi )[W, bi ] and so [V, bi ] = [W, bi ] is the central product of k − 1 ≤ 3 copies of Q8 . Hence CAut([W,bi ]) (bi ) embeds in GU3 (2), so it is solvable. As Ebi ∼ = A5 , [V, bi , Ebi ] = 1. Therefore B0 ∩ Ebi centralizes [V, bi ]CV (B0 ), so 2.4k = |[V, bi ]CV (B0 )| ≤ |CV (B0 ∩ Ebi )| ≤ 25 . Hence, k ≤ 2. Since CB0 (V ) = 1, we must have k = 2, V ∼ = 21+6 + , and Cz /V + ∗ ∼ ∼ ∼ embeds in Out(V ) = O6 (2) = Σ8 . As F (CCz /V (bi )) = Z3 × A5 for i = 1 and 2, Cz /V ∼ = A8 or Σ8 , and consequently AutCz (B0 ) ∼ = D8 . Finally, since CV (B0 ) ∈ Syl2 (J) with J  NG (B0 ), and | Out(J)| = 4 < | AutCz (B0 )|, CNG (B0 ) (J) contains an involution t inverting B0 elementwise. But CAut(Jb1 ) (J) has odd order by [V17 , 10.13.2c], so [Jb1 , t] = 1. This is  impossible as t inverts B0 ∩ Jb1 = 1, so the proof is complete. = 3J3 . More generally, for any Corollary 2.5. For any x ∈ B0# , Jx ∼ y ∈ I3 (G), it is impossible that m3 (C(y, Ly )) = 1 and Ly /O3 (Ly ) ∼ = 3J3 , where Ly = L3 (CG (y)). Proof. By Proposition 2.2c, the second assertion implies the first. Suppose that m3 (C(y, Ly )) = 1 and Ly /O3 (Ly ) ∼ = 3J3 . Let Q ∈ Syl3 (CG (y)).

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Then Ω1 (Z(Q)) = y and Ω1 (Q) is abelian by [V17 , 10.13.2ab]. Therefore Q ∈ Syl3 (G) and every exponent-3 subgroup of G is abelian. But this is absurd by Proposition 2.4, since A9 and Sp6 (2) both contain Z3 Z3 .  Combining our results, we have shown Proposition 2.6. If conclusion (a) of Theorem C∗4 : Stage A3 holds, then G has an A9 -Sp6 (2) setup. Proof. Condition (1A1) is trivial, conditions (1A2,3) hold by Lemma 2.1 and Proposition 2.4, and condition (1A4) holds by Proposition 2.2cd and Corollary 2.5.  3. Centralizers of 2-Central Involutions In view of Proposition 2.6, we assume for the remainder of this chapter that conclusion (b) of Theorem C∗4 : Stage A3 holds. Thus, (3A)

(1) ∅ = Hinv ⊆ Hcons ; and (2) For every H ∈ Hcons , every B ∈ Ep3 (H), and every B0 ∈ Ep2 (B), CG (B0 ) is p-constrained.

We shall complete the proof of Theorem C∗4 : Stage A4 by proving Proposition 3.1. Assume (3A). Then G has an extraspecial setup. Property (3A2) has the following quick consequence. Lemma 3.2. Let H ∈ Hcons , B ∈ Ep3 (H), and B0 ∈ Ep2 (B). Then m2 (CG (B0 )) ≤ 4 − m3 (B0 ). Proof. Let C = CG (B0 ), in which Op (C) has odd order by Theorem C∗4 : Stage A1. We apply the Thompson dihedral lemma in the p-constrained group C/Op (C). If our lemma fails, we get mp (CC (t)) ≥ 4 for some t ∈  I2 (C), contradicting e(G) = 3. We rather quickly shall find that p = 3 and that centralizers of 2-central involutions are 2-constrained with Fitting subgroups of symplectic type. We shall use the following notation.

(3B)

(1) If T is an extraspecial group, then wd T is its width, i.e., wd T = 12 dimF2 (T /Z(T )); (2) Let z ∈ I2 (G) and B ∈ Ep3 (CG (z)). Let Qz = F ∗ (CG (z)) and assume that Qz is of symplectic type. Then # (a) B0 (z) = {(b, B)| B ∈ Ep3 (CG (z)), b ∈ B };  (b) β(z) = max wd[CQ (b), B] (b, B) ∈ B0 (z) ; and   (c) B(z) = (b, B) ∈ B0 (z) wd[CQ (b), B] = β(z) .

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318

We first prove: Proposition 3.3. Let H ∈ Hcons and set Q = F ∗ (H) = O2 (H). Let B ∈ Ep3 (H). Then the following conditions hold: (a) (b) (c) (d)

p = 3; Either CQ (B) = 1 or Q is of symplectic type; There is a unique class z G of involutions such that m3 (CG (z)) = 3; For z as in (c), Qz = F ∗ (CG (z)) is of symplectic type. Moreover, for any (b, B) ∈ B(z), CQz (b) ∼ = Q8 ∗ Q8 ∗ S, where m2 (S) ≤ 1. (e) If z is as in (c) and (b, B) ∈ B(z) with Lb := L3 (CG (b)) = 1, then CQz (b) acts faithfully on Lb /O3 (Lb ), β(z) = 2, and Lb /O3 3 (Lb ) is isomorphic to U4 (2), L± 4 (3), or G2 (3).

When the group B is clear from context, we may write b ∈ B(z) instead of (b, B) ∈ B(z). Proof. If BQ does not normalize any noncyclic elementary abelian subgroup of Q, then by Philip Hall’s theorem [IG , 10.3], Q is of symplectic type. Now assume that Q is not of symplectic type, whence BQ normalizes some P0 ∈ E22 (Q). Note that m2 (CQ (B)) ≤ 1 by Lemma 3.2. Set QB = Ω1 (CQ (B)) and P = QB Ω1 (Z(QB P0 )), which is again B-invariant and noncyclic elementary abelian. Then [P, B] = 1, B acts completely reducibly on P , and |CP (B)| = |QB | ≤ 2. Moreover, for every hyperplane B0 of B, |CP (B0 )| ≤ 4 by Lemma 3.2. It follows that p = 3 and CQ (B) = 1. In particular, (b) holds. Now (b) implies in turn that for every H1 = CG (z) ∈ Hinv (and such H1 ’s exist by (3A1)), F ∗ (H1 ) = O2 (H1 ) = Qz is of symplectic type. Then for T1 ∈ Syl2 (H1 ), z = Ω1 (Z(T1 )), implying that T1 ∈ Syl2 (G) and z G is the unique class of 2-central involutions in G. This proves (c) and (d), except for the assertion in (d) concerning CQz (b). For the rest of the proof we may assume that H = CG (z) and Qz is of symplectic type. By [IG , 24.4(ii,iv)], which applies because of Lemma 3.2, p ≤ 5, and Qb := CQz (b) is of symplectic type for every b ∈ B. We may assume that B was chosen so thatthere is (b, B) ∈ B(z). For any such b, CB (Qb ) = b, for otherwise Q = CQ (b1 ) | b1 ∈ CB (Qb )# = Qb , contrary to CB (Q) = 1. Thus B/ b acts faithfully on Qb . If p = 5, this implies that 1+4 ∼ 1+8 Qb contains 21+4 = 2+ . If p = 3, on the other hand, it implies that − ∗ 2− 1+4 ∼ Qb = 2+ ∗ S for some S. With (b, B) ∈ B(z) as in the preceding paragraph, let Cb = CG (b) and C b = Cb /Op (Cb ). By Theorem C∗4 : Stage A1, Op (Cb ) has odd order, so Qb ∼ = Qb . Set Lb = Lp (CG (b)) and suppose first that Lb = 1. If [Lb , z] = 1, then [Lb , Qb ] ≤ O2 (Lb ) = 1. As z is the unique minimal normal subgroup of Qb , Qb acts faithfully on Op (C b ). Using the Thompson dihedral lemma and

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[IG , 16.11], and as b ∈ Op (C b ), we reach the contradiction m2,p (Qb Op (C b )Lb ) ≥ m2 (Qb ) + mp (Lb ) − mp (Z(Lb )) > m2 (Qb ) + 1 ≥ 4. Thus if Lb = 1, then Qb acts faithfully on Lb , which is the first part of 0 )) = 1 b = Cb /C(b, Lb ). By (3A2) and Lp -balance, Lp (C  (B (e). Set C Lb ∼ for every hyperplane B0 of B. In particular, B = Ep2 and Lp (C  (b0 )) = 1 Lb

 # . Moreover, Q b ∼  b . subgroup of C for all b0 ∈ B = Qb is a B-invariant ∼   Now by [V17 , 7.5], Lb ∈ Chev(p) or (p, Lb ) = (3, M c), (5, F5 ), or (5, Ly). b) ∼ Suppose that (p, L = (5, F5 ). As m2 (F5 ) > 4 [IA , 5.6.1], it follows with the Thompson dihedral lemma that Lb is quasisimple, indeed Lb ∼ = F5 by [IA , 6.1.4]. Then z acts on Lb as a 2-central involution, so L5 (CCG (z) (b)) contains two A5 5-components J, J  interchanged by a 2-element [IA , 5.3w]. On the other hand there is t ∈ I2 (Lb ) such that E(CLb (t)) ∼ = 2HS. In particular, m5 (CG (t)) ≥ 1 + m5 (HS) = 3, so t ∈ z G by (c). Hence there is b ∈ I5 (CG (z)) such that E(CCG (z) (b )) has an HS component I. Using L5 -balance and [IA , 5.3], and the fact that m5 (CG (z)) = 3, we see that I lies in a 5-component I ∗ of CG (z) such that I ∗ /O5 (I ∗ ) ∼ = I or F2 , and I ∗  CG (z). In either case CI ∗ (b) does not contain 5-components J and J  , and so [I ∗ , JJ  ] ≤ O5 (CG (z)), forcing m5 (CG (z)) ≥ 4, contradiction. Thus, b) ∼ (p, L = (3, M c) or (5, Ly). As mp (CG (z)) = 3, Lb  Cb . Since Op (CG (b)) has odd order, Qb ≤ O2 (CC b (z)). But O2 (CLb (z)) ∼ = Z2 , so [Lb , Qb ] = 1, b ∼ contradiction. Therefore by [V17 , 7.5], p = 3 and L = U4 (2), L± 4 (3), or  b ) is a 2-group, so Q8 . For each of these groups, Out(L G2 (3), and in (d), S ∼ =  b . Then [Q  b , B]  ≤ O2 (C  ( z )) ∼ [Qb , B] induces inner automorphisms on L = Lb Q8 ∗Q8 in each case, so β(z) = 2. Furthermore, in the group Qb ∼ ∗Q ∗S Q = 8 8 of symplectic type, m2 (S) = m2 (CQb (B)) ≤ m2 (CG (B)) ≤ 1 by Lemma 3.2. This completes the proof if Lb = 1. Finally, assume that Lb = 1, so that Qb acts faithfully on Op (C b ). If ∗ 21+4 p = 5, then then Qb contains 21+4 − − , so that m2 (Qb ) ≥ 4. Also if p = 3 but the assertion in (d) fails, then m2 (Qb ) ≥ 4. But then using the Thompson dihedral lemma, m2,p (Qb Op p (Cb )) ≥ 4, contradicting e(G) = 3. Therefore (a) and (d) hold, and the proof is complete.  We refine our knowledge of the structure of Qz . First, the configuration in Proposition 3.3d is knotty when S ∼ = Q8 . We escape it by using an argument basically due to Klinger and Mason [KMa1]. Lemma 3.4. In Proposition 3.3d, S ∼ Q8 . Moreover, β(z) = 2. = Proof. By Proposition 3.3 it suffices to prove the first assertion. Suppose that S ∼ = Q8 . Let z, b, and B be as in Proposition 3.3cd. Then Qb := CQz (b) ∼ = Q8 ∗ Q8 ∗ Q8 . By Proposition 3.3e, Lb = 1, so Qb acts faithfully on O3 (C b ), where again Cb = CG (b) and C b = Cb /O3 (Cb ). Since O3 (Cb ) has odd order, Qb normalizes some R ∈ Syl3 (O3 3 (Cb )). Let R0 be a characteristic subgroup of R minimal with respect to [R0 , z] = 1. Then R0

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is of class at most 2 and exponent 3. If R0 is abelian, then by the representation theory of Qb , R0 has a Qb -absolutely-irreducible subgroup of order 38 inverted by z. But then for some t ∈ I2 (Qb ) − {z}, m3 (CR0 (t)) ≥ 4, contradicting e(G) = 3. So R0 is not abelian, whence z centralizes Z(R0 ). Then in CG (z) we see that [Qb , Z(R0 )] ≤ Qb ∩ Z(R0 ) = 1, so Qb centralizes Z(R0 ). Again let t ∈ I2 (Qb ) − {z}. Then m3 (CR0 /Z(R0 ) (t)) ≥ 4. Since m3 (CR0 (t)) ≤ 3, it follows from [V9 , 5.10] that |Z(R0 )| = 3, whence R0 ∼ = 31+8 and CR0 (t) ∼ = 31+4 with Z(R0 ) = b. With t as in the preceding paragraph, m3 (CG (t)) ≥ m3 (31+4 ) = 3 so t ∈ z G by Proposition 3.3c. Therefore CG (z) contains a conjugate R1 of CR0 (t). Now F ∗ (CG (z)) = Qz is of symplectic type and by Proposition 3.3d, m2 (CQz /z (y)) ≤ 6 for all y ∈ R1 − Z(R1 ). But the minimal dimension of a faithful F2 R0 -module is 18. It follows that wd Qz = 9 and CQz (y) ∼ = Qb for all y ∈ R1 − Z(R1 ). Write Z(R1 ) = v and let P ∈ Syl3 (CG (z)) with R1 ≤ P . Then v ∈ Z(P ), so by the action of P on Qz / z, P embeds in GL9 (4). Hence by [V9 , 13.1], R1 = P ∈ Syl3 (CG (z)). Since Qb is absolutely irreducible on R0 / b, R centralizes R0 / b, and so R = R0 ∗T for some T ≥ b. But CR0 (t) ∼ = R1 ∈ Syl3 (CG (z)) and t ∈ z G , so t inverts T / b elementwise. Again as Q contains a four-group consisting of such t’s, this forces T = b and R0 = R ∈ Syl3 (O3 3 (Cb )). Choose x ∈ R1 − Z(R1 ) and let s ∈ I2 (CQz (x)) − {z}. Thus, CQz (xs) = and Z(D) = z. Let E ∈ Syl3 (CO3 3 (x) (s)) ∼ D × s with D ∼ = 31+4 = 21+4 − be CQz (x)-invariant. Since z inverts E/ x, CDs (E) = s. Set Qs = O2 (CG (s)). Then CG (s) contains Qs (ED) and Qs ∩ (ED) = 1. Set L = CG (s) and N = NL (E). Choose y noncentral in E with |CN (y)|2 maximal. Let I2 ∈ Syl2 (CN (y)) and expand I2 to I ∈ Syl2 (N ). Then |I| ≥ |D s | = 26 . However, |E − Z(E)| = 240 = 16.15, so some N -orbit O on E − Z(E) has |O|2 ≤ 16. Therefore by choice of y, |y N |2 ≤ 16. Letting |I| = 2α and |I2 | = 2β , we conclude that 2β ≥ 2α−4 . Let H2 = CO2 (L) (y) ∼ = Q8 ∗Q8 ∗Q8 . Write CE (y) = y×E0 with E0 ∼ = 31+2 . As Z(E0 ) = Z(E) acts fixed-point-freely on H2 / s, E0 acts faithfully on H2 . Now I2 × y acts on E and, by the A × B lemma, CI2 (E0 ) = CI2 (E) =

s. So CE0 I2 (H2 ) = s. Note that H2 / s has 27 nonzero singular vectors, and E0 I2 permutes them. Hence I2 must fix a vector, i.e., there is t2 ∈ I2 (H2 ) − s such that |CH2 I2 (t2 )| = |CH2 (t2 )||I2 / s | ≥ 25+β ≥ 2α+1 . Finally, CO3 3 (CG (y)) (t2 ) has a Sylow 3-subgroup E2 ∼ = E. But then (t2 , E2 ) is G-conjugate to (s, E) and so 2α = |I| = |CG (s) ∩ NG (E)|2 = |CG (t2 ) ∩ NG (E2 )|2 ≥ |CH2 I2 (t2 )| ≥ 2α+1 , a final contradiction. The proof is complete.



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We next set M = CQz (B), choose b ∈ B and let Q be a 21+4 + -subgroup of CQz (b). Let B0 be a complement to b in B. As b ∈ B, B0 is faithful on Q. Set Cb = CG (b), Lb = L3 (Cb ), C b = Cb /O3 (Cb ), and Qb = CQz (b). Lemma 3.5. The following conditions hold: (a) M is cyclic; (b) wd Qz ≤ 5; for all b ∈ B(z). (c) If Lb = 1, then M = z and CQz (b ) ∼ = 21+4 + Proof. Set Ez = [Qz , B]. By [V9 , 9.2], we may choose a hyperplane B1 of B with CEz (B1 ) ≥ Q1 ∼ = Q8 . By Lemma 3.2, m2 (CQz (B1 )) ≤ 2, so CEz (B1 ) = Q1 and M is cyclic, proving (a). Now β(z) = 2 by Lemma 3.4. Hence for any b1 ∈ B1# , either CEz (b1 ) = Q1 or wd CEz (b1 ) = 2. This implies (b). For (c), suppose by way of contradiction that z < Z1 ≤ M , so that CQz (b) ≥ Q ∗ Z1 . Since CQz (b) acts faithfully on Lb by Proposition 3.3e, b = and QB0 ≤ Cb with Q = [Q, B0 ], QZ1 B0 acts faithfully on Lb . Let C ∼    Cb /C(b, Lb ), so that Cb embeds in Aut(Lb ). By Proposition 3.3e, Lb = U4 (2),  L± 4 (3), or G2 (3). As QZ1 B0 embeds in Cb , it follows from [V17 , 7.26] that b ∼ L = Z3 ×U3 (3). = U4 (3) and Cb contains an involution u such that C bLb (u) ∼ Since m3 (CG (u)) = 3, we have u ∈ z G by Proposition 3.3c. As wd Qz ≤ 5, CG (z) has a subgroup a × J ∼ = Z3 × J with J/O3 (J) ∼ = U3 (3), acting faithfully on a symplectic F2 -space V of dimension at most 10. Moreover, if [CQz (a), J] = 1, then since 7 divides |J|, it follows that wd[CQz (a), B] ≥ 3, so β(a) ≥ 3, contradicting β(z) = 2. Thus [Qz , a, J] = 1 so J embeds in U5 (2). As 7 does not divide |U5 (2)|, this is impossible. The proof is complete. 

We now achieve our main goal about the structure of Qz = O2 (CG (z)), and the action of b ∈ B(z) on it. Proposition 3.6. The following conditions hold: (a) Qz is the central product of = 3 or 4 copies of Q8 ; (b) CQz (b) ∼ = 21+4 + ; (c) CQz (B) = z; (d) If B1 is a hyperplane of B with CQz (B1 ) > z, then CQz (B1 ) ∼ = Q8 ; ∗ a characteristic subgroup (e) Either E(C b ) = 1 or F (C b ) contains  Rb ∼ = E35 such that CRb (z) = b . Proof. We first claim that if (b) and (d) hold, then (a) holds. Namely, suppose that (a) fails. As wd Qz ≤ 5 by Lemma 3.5b, (d) implies that there exist five hyperplanes A of B such that CQz (A) ∼ = Q8 . Any three of these hyperplanes have trivial intersection by (b). Dualizing, we must have five nonzero vectors e1 , . . . , e5 ∈ B ∗ such that any three of them span B ∗ . But then, writing coordinates relative to the ordered basis {e1 , e2 , e3 } over the

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field F3 , we may assume that e4 = (1, 1, 1) and e5 = (1, 1, −1). But then e5 = e3 + e4 , a contradiction. This proves our claim. Next, note that if Lb = 1, then the lemma follows from Lemmas 3.5 and 3.2. So we may assume that Lb = 1. Let Rb be a characteristic subgroup of O3 (C b ), minimal subject to the conditions b ∈ Rb and [Rb , z] = 1. Then by [IG , 11.11], Rb has class at most 2 and exponent 3. Suppose first that Rb is elementary abelian and let E = [Rb , z]. Then z inverts E, and so Qb = CQz (b) acts faithfully on E. Let t ∈ I2 (Qb ) − {z}. Then tz ∈ tQ and so E = CE (t) × CE (tz) with CE (t) ∼ = CE (tz). Since m2,3 (G) = 3, m3 (E) ≤ 4. Hence m3 (E) = 4 and Qb acts absolutely irreducibly on E, whence z = CGL(E) (Qb ) and so M = z, implying (b), (c), and (d). Hence (a) holds by the first paragraph of the proof. Let E0 = CRb (z). Then m3 (E0 ) ≤ 3 and b ∈ E0 . As Qb B acts on E0 it follows that Qb acts trivially on E0 . But then as Qb acts completely reducibly on Rb and m3 (CRb (t)) = 3, we conclude that E0 = b and Rb ∼ = E35 , as claimed in (e). Thus, we are done if Rb is abelian. We may now assume that Rb has class 2 and [Z(Rb ), z] = 1. As Qb ≤ O2 (CG (z)), [Qb, Z(R  b )] = 1. Then since m2 (Qb ) = 3, Lemma 3.2 implies that Z(Rb ) = b and so Rb is extraspecial. But Sp4 (3) has no subgroup isomorphic to Qb , so |Rb | > 35 . Thus if we again let t ∈ Qb be any involution distinct from z, then CG (t) contains a subgroup isomorphic to 31+4 . In particular m3 (CG (t)) ≥ 3, so t ∈ z G . But then CG (z) contains 31+4 , which cannot act faithfully on Qz as wd Qz ≤ 5 by Lemma 3.5b. This contradiction completes the proof of the proposition.  Lemma 3.7. Let B ≤ T ∈ Syl3 (CG (z)). Then one of the following holds: (a) B = T ; or (b) T ∼ = Z3 Z3 , B = J(T ), and a generator of Z(T ) is not in B(z). Proof. Suppose that B = T . If Qz ∼ = 21+6 − , then the result follows from the structure of a Sylow 3-subgroup T0 of O6− (2) and the fact that CV (Z(T0 )) = 0, where V is the natural F2 O6− (2)-module. So assume that Qz ∼ = 21+8 + . Then T leaves invariant an orthogonal decomposition Qz / z = U × W with W an asingular plane. By Proposition 3.6b, T acts faithfully on U and so again the isomorphism type of T is as claimed and CQz /z (Z(T )) ≤ W , completing the proof.  Lemma 3.8. We have Lb = 1. b = Cb /O3 3 (Cb ). Proof. Suppose not. Let W = O3 (C b ) = F ∗ (C b ) and C Let B ≤ T ∈ Syl3 (CG (z)). Since b ∈ B(z), Lemma 3.7 implies that B ∈ Syl3 (CCG (z) (b)) = Syl3 (CCb (z)). Since B/ b is faithful on Qb , B ∩ W = b   and so z inverts W / b . Now by Proposition 3.6e, there is Rb char W such that b ∈ Rb ∼ = E35 . Let t be any involution of Qb −{z}. Then CRb (t) ∼ = E33 . Thus |CW /Rb (t)| ≤ 3

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∼ 21+4 acts faithfully and so |W /Rb | ≤ 9. But if |W /Rb | = 1, then CQz (b) = + on W /Rb , which is impossible. Therefore W = Rb .   b ). Then since Qb  CC (b), Q b  C b . As z inverts W / b , z ∈ Z(C z    b ) =   b acts absolutely irreducibly on W / b , C  (Q z . In particular, As Q Cb    ∈ Syl3 (C b ) and W B ∈ Syl3 (C b ). Now H := BQb / b ∼ B = SL2 (3) ∗ SL2(3)   and H acts faithfully on W / b . It follows that as F3 H-module, W / b is the tensor product of two faithful SL2 (3)-modules, and we infer that |CW / b (B)| = 3. Therefore m3 (CG (B)) ≤ 4. However, A := CW (t) ∼ = E33 and so, as t ∈ z G and B = J(T ), A and B are G-conjugate. But m3 (CG (A)) ≥ m3 (W ) = 5, a contradiction proving the lemma.  Lemma 3.9. Lb ∼ 3G2 (3). = Proof. Suppose that Lb ∼ = 3G2 (3). Then as B ≤ NCb (Lb ) and Out(Lb ) is a 3 -group, CT (b) contains Z3 × R with R ≤ Lb and R ∼ = 31+2 [IA , p. 319]. But this contradicts Lemma 3.7.  Propositions 3.3 and 3.6, and Lemmas 3.8 and 3.9, almost establish that G has an extraspecial setup. It remains only to prove that for b ∈ B(z), m3 (C(b, Lb )) = 1 and Lb ∼ = L4 (3), U4 (3), or 31 U4 (3). Lemma 3.10. We have m3 (C(b, Lb )) = 1. Moreover, B = b × (B ∩ Lb ) with Qb ≤ Lb and Qb (B ∩ Lb ) ∼ = SL2 (3) ∗ SL2 (3). ∼ Proof. We have Lb ∼ = U4 (2), L± 4 (3), or G2 (3), or Lb /Z(Lb ) = U4 (3), by Proposition 3.3, [IA , 6.1.4], and Lemma 3.9. Then, with [V17 , 18.6], 3 = e(G) ≥ m2,3 (Cb ) ≥ (m2,3 (Lb ) − m3 (Z(Lb ))) + m3 (C(b, Lb )) ≥ 2 + m3 (C(b, Lb )). Hence m3 (C(b, Lb )) = 1. It follows that C(b, Lb ) has a normal 3-complement, so C(b, Lb ) has odd order. As Out(Lb ) is a 3 -group, B ≤ Lb × C(b, Lb ). Then Qb = [Qb , B] ≤ Lb and the lemma follows.  L4 (3). Lemma 3.11. Lb ∼ = Proof. Suppose that Lb ∼ = L4 (3) and choose t ∈ I2 (Qb ) − {z}. Set t = Ct /Qt . By [V17 , 10.10.8], Ct = CG (t), Qt = O2 (CG (t)), and C ∼ D8 Y Σ6 , CL (t) = C (t) = ( v × J) y = b

Lb

where = t, J ∼ = A6 , and y 2 = 1. In particular, [J, z] = J . Then H := CCb (t) ≥ b CLb (t) so m3 (Ct ) ≥ 3. Thus t ∈ z G so Qt = F ∗ (Ct ) ∼ = 21+6 − 1+8 ∼       or 2+ . Hence H = CCt (b) has a 3-component J with J/O3 (J) = A6 . and so as Since 3-elements of O6− (2) have solvable centralizers, Qt ∼ = 21+8 + 1+6 1+6  z] = J, S ∼ t ∈ Qb ≤ Qz , S := CQz (t) ∼ = Z2 × 2+ . As [J, = 2+ .  t Of course O2 (Ct ) = 1, and using [V17 , 10.1.5] we see that either J  C t ) = E(C t ) ∼ or F ∗ (C = A9 or Sp6 (2). In any case, [V17 , 10.1.5] implies that t embeds in Σ9 or Sp6 (2), in both of which a Sylow 2-subgroup contains C v2

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a normal elementary abelian subgroup with a D8 quotient. Hence S has a normal elementary abelian subgroup with quotient isomorphic to a subgroup of D8 . But as S ∼ = 21+6 + , no such elementary abelian subgroup exists, contradiction. This completes the proof.  To complete the proof of Proposition 3.1, we must prove Proposition 3.12. If Lb /Z(Lb ) ∼ = U4 (3), then Lb ∼ = 32 U4 (3). If the proposition fails, then as m3 (C(b, Lb )) = 1, Lb by [V17 , 10.10.11]. Then by [V17 , 10.10.16],     1 2 1 (3C) In C b , b, Lb contains b × J × J with J

∼ = U4 (3) or 31 U4 (3), 2 ∼ =J ∼ = A4 .

Instead of proving Proposition 3.12, we shall prove the stronger statement: Proposition 3.13. There do not exist z1 ∈ z G , (b1 , B1 ) ∈ B(z1 ) and a 3-component Lb1 of CG (b1 ) such that (a) Lb1 /O3 (Lb1 ) ∼ = U4 (3) or 3U4 (3); (b) m3 (C(b1 , Lb1 )) = 1; (c) Lb1 has a subgroup isomorphic to A4 × A4 ; and (d) B1 = b1  × (B1 ∩ Lb1 ); (e) z1 ∈ Lb1 and O2 (CG (z1 )) ∩ Lb1 ∼ = 21+4 + . Suppose that Proposition 3.13 is false. For simplicity, we drop the 1subscript and continue in the proof with the notation Lb , Cb , and C b , and with the assumption (3C), which we show leads to a contradiction. Notice that conditions (b), (d), and (e) in Proposition 3.13 are satisfied for our b, by Lemma 3.10. 1 Since Lb has one class of involutions, we may assume that z ∈ J . As O3 (Cb ) has odd order by Theorem C∗4 : Stage A1, we may find U1 × U2 ≤ Lb with U1 ∼ = E33 , such = U2 ∼ = E22 , and a1 , a2  ≤ A ≤ b, Lb  with A ∼ that z ∈ U1 , U1 U2 A = b × J 1 × J 2 and J i = Ui ai  ∼ = A4 . Again as U1 × U2 ≤ Lb , and Lb has one class of involutions, (U1 × U2 )# ⊆ z G . Hence F ∗ (CG (u)) = O2 (CG (u)) for all u ∈ (U1 × U2 )# . It follows that D := F ∗ (NG (U1 )) is a 2-group, so that NG (U1 ) ∈ Hcons . Lemma 3.14. CD (A) = 1. Proof. Otherwise, choose t ∈ I2 (CD (A)) ⊆ z G . Set A1 = b, a1 . Then  CG (A1 ) ≥ U2 × t ∼ = E23 , contrary to Lemma 3.2. By Lemma 3.14, NG (U1 ) ≥ J 1 × J 2 × J 3 ≥ A with J 3 = U3 b ∼ = A4 . For uniformity we sometimes write a3 = b, so that J i = Ui ai , i = 1, 2, 3. Now set (3D)

E = U1 × U2 and QE = O2 (CG (E)).

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As U1 ≤ E, F ∗ (CG (E)) = QE . Then E ≤ Lb and by [V17 , 10.10.17a],   (3E) N b,Lb  (E) = E( b × J) with J ∼ = A6 , J acting irreducibly on E and E ≤ Z(QE ). Since [QE , b] = 1, E < QE .

(3F) We now prove Lemma 3.15. CQE (b) = E.

Proof. Let F = CQE (b) and suppose for a contradiction that F > E. b = Cb /C(b, Lb ), F ∼ In C = F . Hence by [V17 , 10.10.17b], F J ∼ = E25 A6 , and then there is f ∈ F − E such that CLb (f ) ∼ U (2). But then CC b (f ) = 4  contains Z3 × P Sp4 (3), so m3 (CG (f )) ≥ 4, contrary to e(G) = 3. Recall that b ∈ B(z), so O2 (CLb (z)) = (B ∩ Lb )(Qz ∩ Lb ) ∼ = SL2 (3) ∗ SL2 (3). We have E = [E, A] ≤ CLb (z). Let E0 be the hyperplane of E such that E ∩ Qz ∩ Lb = E 0 . Then E = E0 × e where e interchanges the two SL2 (3) central factors of O2 (CLb (z)). It follows that QE ∩ Qz ≤ CQz (E0 ) = E0 ∗ Q0 with Q0 ∼ = Q8 or Q8 ∗ Q8 . Thus (3G)

|QE ∩ Qz | ≤ 27 , and if QE is abelian, then |QE ∩ Qz | ≤ 25 .

z = Cz /Qz . By Lemma 3.15, b acts fixed-pointWe set Cz = CG (z) and C  = 2 and |Q  E : [Q  E , b]| ≤ 2. freely on QE /E. In particular, |CQE (b)| = |E| Before proceeding we note the following argument of F. Smith [SmF]. z ∼ = Sp6 (2). Moreover, |G|2 ≤ 216 . Lemma 3.16. C z |2 ≤ z ∼ Proof. Assuming C = Sp6 (2), we see from [V17 , 10.1.5] that |C 16  which implies that |G|2 = |Cz |2 = 2 . z ∼  ∼ Suppose then that C = A6 . = Sp6 (2). Let w ∈ I3 (Cz ) with E(CCz (w)) 5 Since CQz (B) = z and β(z) = 2, |CQz (w)| ≤ 2 , and hence by inspection of 3-centralizers in Out(Qz ) [IA , 4.8.2], CQz (w) = z. It follows easily that if z ) with dim CV (w z ∼ 3 ∈ I3 (C i ) = 2i, where V is the natural C w 2 , w = Sp6 (2)4 2 module, then |CQz /z (w 2 )| = 2 and |CQz /z (w 3 )| = 2 . Hence by [V16 , z . Let t ∈ E0 − z ⊆ E # ⊆ z G , Qt = 10.2.1], Qz / z is the spin module for C F ∗ (CG (t)) = O2 (CG (t)), Cz,t = CG ( z, t), Qz,t = O2 (Cz,t ), R = CQz (t) and R0 = CQt (z). (Note that the argument to come does not disprove the existence of Co2 . In Co2 , the involution corresponding to t is not in z G .) Then by [V16 , 10.2.2], Cz,t /R is isomorphic to a parabolic subgroup of Sp6 (2) of shape E26 L3 (2), and Qz,t /R has a unique minimal Cz,t -invariant z is transitive subgroup, which is of order 8. Moreover, by [V16 , 10.2.2], C on I2 (Qz ) − {z}, whence z, t is normal in a Sylow 2-subgroup of Cz . Hence by [IG , 16.21], AutG ( z, t) ∼ = Σ3 . In particular, z ∈ Qt , whence Cz,t /R0 ∼ = Cz,t /R. Thus, Qz,t /R0 ∼ = E26 , so z, t ≤ Φ(Qz,t ) ≤ R ∩ R0 . = Qz,t /R ∼ 27 ,

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We claim that Φ(Qz,t ) > z, t. Let y ∈ Qz,t −R with CCz ( y ) a parabolic z ) acts z of shape E25 Σ6 . By [V17 , 10.2.1], an element of I5 (C subgroup of C 4 fixed-point-freely on Qz / z, and so |Qz / z : CQz /z (y)| = 2 . Therefore |R : CR (y)| ≥ 8. But if Qz,t / z, t were abelian, then |R : CR (y)| ≤ 4 by [IG , 9.16], contradiction. This proves the claim. As Φ(Qz,t ) ≤ R∩R0 and R∩R0 is elementary abelian with Φ(Qz,t )/ z, t a faithful Cz,t /Qz,t -module, we have Φ(Qz,t ) = R ∩ R0 ∼ = E25 and CR (Φ(Qz,t )) = Φ(Qz,t ). If CQz,t (Φ(Qz,t )) = Φ(Qz,t ), then Cz,t /Φ(Qz,t ) is isomorphic to a subgroup of GL5 (2). But |Cz,t /Φ(Qz,t )|2 = 212 > |GL5 (2)|2 . Therefore CQz,t (Φ(Qz,t )) ≤ R. Let T1 /R be the unique irreducible Cz,t -submodule of Cz,t /R. Also let T2 = CQz,t (Φ(Qz,t )). Then R is contained properly in RT2 and RT2 is Cz,t invariant, whence RT2 ≥ T1 . Also RR0 is Cz,t -invariant with |RR0 /R| = 8, so RR0 = T1 ≤ RT2 . Finally, we may pick x ∈ R ∩ R0 and x0 ∈ R0 − R with [x, x0 ] = t. But x0 ∈ R0 ≤ RT2 and so x0 = uv with u ∈ T2 and v ∈ R. Then t = [x, x0 ] = [x, uv] = [x, v] ∈ z, a contradiction. This completes the proof.  Now we can prove Lemma 3.17. We have |QE | ≤ 212 , and QE is nonabelian in case of equality. Proof. Assume that either |QE | ≥ 214 , or QE is abelian with |QE | = z , |Q  E | ≥ 27 and |[Q  E , b]| ≥ 26 . Hence Then (3G) implies that in C 0 := z ∼ Sp6 (2) by Lemma 3.16, it follows from [V17 , 10.1.5] that C as C = ∗ z ) ∼  E , b] ≤ C 0 , so [Q  E , b] is isomorphic to a F (C = U4 (2) or A9 . Then [Q   Sylow 2-subgroup of A8 . But then Z([QE , b]) has order 2, so equals C  (b). 212 .

QE

 E , b]| = 2m with m odd, a contradiction. Hence |[Q Since b acts fixed-point-freely on QE /E, |QE | ≤ 212 and QE is nonabelian if equality holds. The proof is complete.  Recall (3E) that J ∼ = A6 is a complement to E ∼ = E24 in NLb (E). We now take a minimal preimage J of J in NL (E). Thus J/O{2,3} (J) ∼ = A6 b

and [E, O{2,3} (J)] = 1. Lemma 3.18. J ∼ = A6 . Proof. Since z ∈ E, F ∗ (NG (E)) = O2 (NG (E)). Since J acts irreducibly on E, O2 (NG (E)) centralizes E and hence equals QE . Therefore CJ (QE ) = 1. In particular, if we put X = O{2,3} (J), then as [E, X] = 1, X acts faithfully on QE /E. By Lemma 3.17, X embeds in GL8 (2). In particular, for any prime r dividing |X|, r ≥ 5, so mr (X) ≤ 2. Let Xr be a critical subgroup of exponent r in Or (X). Then mr (Xr /Φ(Xr )) ≤ 2, so since J contains a Frobenius group of order 9.4, Xr ≤ Z(J). It follows that

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F ∗ (X) = F (X) ≤ Z(J), so X = 1 as A6 has no nontrivial {2, 3} -covering. The lemma is proved.  Let R ∈ Syl3 (J). Thus R ∼ = E32 . As R b ∼ = E33 , O3 (CG (R)) has odd ∗ order by Theorem C4 : Stage A1. Lemma 3.19. Let L = L3 (CG (R)) and suppose that L = 1. Then m3 (C(R, L)) = 2, and L/O3 (L) ∼ = L2 (3n ), n ≥ 2. Moreover, Sylow 2subgroups of CG (R) are dihedral or semidihedral. Proof. It follows from [V17 , 18.4] that there exists x ∈ R# such that m3 (CG (x)) ≥ m3 (CLb b (x)) ≥ 4. Let I be a 3-component of CG (R). Then I has a pumpup I ∗ in CG (x), and since x ∈ I3o (G), any component of I ∗ /O3 (I ∗ ) is in C3 , by Theorem C∗4 : Stage A1. Therefore by [IA , 7.1.10], I/O3 (I) ∈ C3 . On the other hand, if some t ∈ I2 (CG (R)) centralizes an element v ∈ I3 (I) − O3 3 (I), or if m3 (C(R, I)) ≥ 3, v ∈ I3 (C(R, I)) − R, and we take any t ∈ I2 (CI (v)), then B := v R ≤ CG (t) and m3 (B) = 3. Hence by (3A2), L3 (CG (R)) = 1, contradiction. So m3 (C(R, I)) = 2 and (3H)

For every t ∈ I2 (CG (R)), I3 (CI (t)) ⊆ O3 3 (I).

Therefore by [V17 , 7.17], I/O3 (I) ∼ = L2 (8) or L2 (3n ), n ≥ 2. Also (3H) and [IG , 16.11] imply that I = L. Suppose that I/O3 (I) ∼ = L2 (8). If m3 (CG (R)) ≥ 4, then some element v ∈ I3 (CG (R)) induces a nontrivial field automorphism on I/O3 (I), so CI (R v) has even order, giving a contradiction as above. By [V17 , 18.4], we may then assume that m3 (CG (x)) ≥ 5 for some x ∈ R# . Let Ix be the subnormal closure of I in CG (x). If Ix is a diagonal pumpup, then as m3 (CG (x)) ≥ 5, CCG (x) (B) is not 3-constrained for some B ∈ E33 (CG (x)), contradicting (3A2). Otherwise Ix /O3 (Ix ) ∈ C3 , so by [V17 , 7.2], Ix /O3 (Ix ) ∼ = 3D4 (2) or Sp4 (8). As m3 (CG (x)) ≥ 5, m3 (C(x, Ix )) ≥ 2, and some B ∈ E33 (CG (x)) is easily found [IA , 4.7.3A, 4.8.2] such that L3 (CG (B x)) = 1, again contradicting (3A2). Thus I/O3 (I) ∼ = L2 (3n ), and we assume for a contradiction that a Sylow 2-subgroup S of CG (R) is not dihedral or semidihedral. As m3 (C(R, I)) = 2, S centralizes O3 3 (CG (R))/O3 (CG (R)), so S acts faithfully on I/O3 (I). But then some t ∈ I2 (S) induces a field automorphism on I/O3 (I), so t centralizes some v ∈ I3 (I). Then again CG (R v) has even order so (3A2) is contradicted. The lemma follows.  Lemma 3.20. Let P = CQE (R). Then P = 1 or P ∼ = E22 . Proof. Suppose that P = 1. Since E = [E, R], P ∩ E = 1 and so by Lemma 3.15, CP (b) = 1. In particular CΩ1 (Z(P )) (b) = 1, so Ω1 (Z(P )) contains a four-group V . If L3 (CG (R)) = 1, then P embeds in a dihedral or semidihedral group by Lemma 3.19, and since V ≤ Z(P ), P = V , as desired.

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So assume that L3 (CG (R)) = 1 and let R1 be a P -invariant Sylow 3subgroup of O3 3 (CG (R)). Thus CP (R1 ) = 1. In particular V R1 lies in a Σ3 × Σ3 subgroup V R2 of P R1 by the Thompson dihedral lemma. Thus for some involution w ∈ V (and hence for all w ∈ V # , by the action of b), m3 (CR1 (w)) ≥ 3. By Lemma 3.7, and since R ≤ Z(R1 ), |CR1 (w)| ≤ 33 . Hence |CR1 /R (w)| ≤ 3 and so |R1 /R| ≤ 33 . Thus, P b embeds in SL3 (3), whence P = V . The proof is complete.  Now we can prove Lemma 3.21. QE = E × P ∼ = E26 . Proof. If [QE /E, J] = 1, then QE = EP . Since E ∩ P = 1, and by (3F) and Lemma 3.20, the lemma holds in this case. Otherwise, as CQE /E (b) = 1, QE /E has a faithful b × J-chief factor V . Since J contains a Frobenius group RH0 of order 9.4, |V | ≥ |[V, R]| ≥ 28 . But using Lemma 3.17, we have |QE /E| ≤ 28 , so this implies that P = 1, |QE | = 212 , and J × b is irreducible on QE /E. In particular, by Lemma 3.17, QE is not abelian. We show that this leads to a contradiction. In characteristic 2, besides the trivial representation, A6 has two self-dual and quasiequivalent irreducible 4-dimensional modules V0 and V1 , writable in F2 , and two 8-dimensional irreducible modules. Moreover, V0 and V1 extend to F2 Σ6 modules whose tensor product is the Steinberg module of Σ6 ∼ = Sp4 (2); and as J-module, V0 ⊗ V1 is the sum of the two 8-dimensional irreducibles. Since CV (b) is trivial, V ∼ =F2 J Vi ⊕ Vi for some i = 0, 1, and also E ∼ = Vj for some j = 0, 1. We argue that i = j. Otherwise, with all Hom’s over F2 J, 0 = Hom(Vi , Vi ⊗ Vj ) ∼ = Hom(Vi ⊗ V ∗ , Vj ) ∼ = Hom(Vi ⊗ Vi , Vj ). i

But the commutator map from QE /E × QE /E to E yields a nontrivial element of Hom(Vi ⊗ Vi , Vj ), contradiction. Thus i = j. Hence there is h ∈ I3 (J) such that CQE (h) = 1. Choose J0 ≤ J with h ∈ J0 ∼ = A5 . (At this point a theorem of Graham Higman about 2-groups on which A5 acts, an element of order 3 being fixed-point-free, implies that QE is abelian.) In detail, let W be the restriction of Vi to J0 . Since QE is not abelian, (3I)

Hom(W ⊗ W, W ) = 0. F2 J0

However, (3I) is false. Namely, W ⊗F2 F4 ∼ = W0 ⊕ W 0 , where W0 is the natural F4 L2 (4)-module and W 0 its nontrivial Galois twist. But neither W0 ⊗ W0 nor W0 ⊗ W 0 has a homomorphic image isomorphic to W0 or W 0 ; the second is the Steinberg module, and if the first had such an image, the kernel would be the direct sum of two trivial modules, contradicting absolute  irreducibility of W0 . This contradiction completes the proof. Set N = NG (QE ) and N = N/QE ∼ = AutG (QE ).   Lemma 3.22. F ∗ (N ) = b × J, and QE = U1 U2 U3 .

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Proof. Now N embeds in AutG (QE ) and hence in L6 (2), and J is a component of CN (b). Suppose first that O2 (N) = 1, and let W = CQE (O2 (N )), so that QE > W > 1. Then W is b × J-invariant, whence W = [QE , b] or [QE , J]. In either case N is irreducible   on QE /W and W , so O2 (N ) stabilizes the chain QE > W > 1. But b × J acts irreducibly on the full stabilizer of this chain in Aut(QE ), which has order 28 . Hence |N |2 = |QE ||O2 (N )|2 |J|2 = 217 , contradicting Lemma 3.16. ) = 1. It follows by [V17 , 10.4.1] that that either F ∗ (N ) =   Thus, O2 (N b × J or F ∗ (N ) ∼ = Sp6 (2). But in the latter case NN ( b) ≥ b, t × EJ, where t is an involution. Then t normalizes Lb and centralizes EJ, whence But then m3 (CG (t)) ≥ 4, contradicting [Lb , t] = 1 by [V17 , 10.10.17b].   e(G) = 3. Hence F ∗ (N ) = b × J . Since U1 U2 U3 ≤ CG (E) and NN (B; 2) = {1}, U1 U2 U3 ≤ QE and then equality holds by orders. The lemma is proved.  It follows that (3J)

Ui  NG (U1 U2 U3 ) ⇐⇒ i = 3.

Our eventual contradiction will come by proving U1  NG (U1 U2 U3 ).

(3K)

Recall AU1 U2 U3 = a1  U1 × a2  U2 × a3  U3 ∼ = A4 × A4 × A4 , and b = a3 . The key to (3K) is to set La1 = L3 (CG (a1 )) and show the following:

(3L)

(1) CG (a1 ) has a subgroup H ∼ = A4 × A4 with H ∩ C(a1 , L1 ) = 1; (2) m3 (C(a1 , La1 )) = 1; (3) La1 /O3 (La1 ) ∼ = U4 (3) or 3U4 (3); (4) (a1 , B1 ) ∈ B(z1 ) for some involution z1 of La1 and some E33 ∼ = B1 ≤ a1  La1 ; and (5) O2 (CG (z1 )) ∩ La1 ∼ = 21+4 + .

Once these are established, we can observe that (3M)

H can be taken to be a subgroup of La1 .

This observation allows us simply to repeat the proof of Proposition 3.13, through (3J), replacing b by a1 , and B by B1 , and interchanging subscripts 1 and 3, to obtain (3K). (Lemma 3.16 does not need to be reproved.) By analogy we set Ca1 = CG (a1 ) and C a1 = Ca1 /O3 (Ca1 ).  To prove (3M), we have H = O3 (H) so H ≤ La1 C(a1 , La1 ), and C(a1 , La1 ) is a cyclic 3-group, which we may assume contains Z = Z(La1 ) ∼ = Z3 , since otherwise the assertion holds trivially. Thus, HZ/Z projects onto

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a subgroup H 1 /Z ≤ La1 Z/Z. Notice that elements of order 3 in the A4 direct factors of H 1 /Z are images of elements of I3 (La1 ), since they centralize involutions in La1 [V17 , 10.10.9]. Let P ∈ Syl3 (HZ) with Z ≤ P . Then P is abelian, so P Z ∩ H1 is abelian, where H1 is the preimage of H 1 in La1 . Therefore H1 ∼ = H, as desired. Now we set about to verify the conditions (3L). Set A1 = a2 , a3  = a2 , b . Lemma 3.23. The following conditions hold: (a) m3 (Ca1 ) ≥ 5; in particular, a1 ∈ I3o (G); (b) La1 ∈ C3 and CA1 U2 U3 (La1 ) = 1; (c) m3 (C(a1 , La1 )) = 1; and (d) m2,3 (La1 ) − m3 (Z(La1 )) ≤ 2. Proof. We have a1 ∈ Lb and CLb (a1 ) contains U2 so has even order. As Lb /O3 3 (Lb ) ∼ = U4 (3), m3 (CLb b (a1 )) ≥ 5 by [V17 , 10.10.9], proving (a). By hypothesis, all components of La1 then lie in C3 . Since m2,3 (G) = 3, every 3-component of La1 is normalized by A = a1  × A1 , by [IG , 8.7(ii)]. Let P1 ∈ Syl3 (O3 3 (Ca1 )). If La1 = 1, then U2 U3 acts faithfully on P 1 , and by the Thompson dihedral lemma, m2,3 (Ca1 ) ≥ 4, a contradiction. Thus La1 = 1, and we let L be a component of La1 . Now CA1 U2 U3 (La1 )  A1 U2 U3 so if CA1 U2 U3 (La1 ) = 1, then without loss, [La1 , U2 ] = 1. But then by the Thompson dihedral lemma applied to U 2 P 1 , m3 (P 1 ) ≥ 3. As [P 1 , La1 ] = 1 and L3 (CG (P1 )) = 1 by L3 -balance, (3A2) is contradicted. Therefore CA1 U2 U3 (La1 ) = 1.  3 (L1 ) ≥ 2, or Similarly, if La1 has a second 3-component L1 , and m if La1 has at least two 3-components besides L, then m3 (C(a1 , L)) ≥ 3 and (3A2) is again contradicted. Thus, if (b) fails, then La1 = LL1 with  3 (L1 ) = 1. In particular, as L ∈ C3 , L ∼ m  3 (L) = m = L2 (8) by [V17 , 3.5]. Now Ca1 contains some F ∼ = E34 , and some hyperplane F0 ≤ F centralizes an involution of L with |CF0 (L)| ≥ 32 , again contradicting (3A2). Hence (b) holds. If (c) fails, then by the same argument, we reach a contradiction if some involution of La1 centralizes an element of I3 (La1 ) − O3 3 (La1 ). By [V17 , 7.17], we deduce that La1 ∼ = L2 (8) or L2 (3n ), n ≥ 2. In particular m2 (La1 ) ≤ 3. But as Out(L) is abelian, U2 U3 = [U2 U3 , A1 ] embeds in Inn(La1 ) and has 2-rank 4, a contradiction. Finally, if (d) fails, then obviously m2,3 ( a1  La1 ) ≥ 4, contrary to e(G) = 3. The lemma follows.  Lemma 3.24. The following conditions hold: (a) U2 U3 ≤ La1 ; and (b) La1 ∼ = L± 4 (3) or 3U4 (3). Proof. Lemma 3.23bd yields that CA1 U2 U3 (La1 ) = 1, La1 ∈ C3 , and m2,3 (La1 ) − m3 (Z(La1 )) ≤ 2. It follows from [V17 , 3.13] then that Out(La1 )

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1

is 3-closed. Hence U 2 U 3 = [U 2 U 3 , A ] ≤ O3 (La1 C(a1 , La1 )) = La1 and then (a) follows as O3 (Ca1 ) has odd order. Thus, with Lemma 3.23 and (3A2), we have (1) m2 (La1 ) ≥ 4 and CA1 U2 U3 (La1 ) = 1; (2) m2,3 (C a1 ) ≤ 3; (3) m3 (C a1 ) ≥ 5; and (3N) (4) For any D ∈ E33 (X) with CLa (D) of even order, 1 L3 (CLa1 (D0 )) = 1 for all D0 ∈ E32 (D). Therefore by [V17 , 20.11], either (b) holds or La1 ∼ = M c or 3M c. In these two exceptional cases, for an involution t ∈ La1 , CLa1 a1  (t) contains z is a copy of Z3 × 2A8 . In particular m3 (CG (t)) ≥ 3 so CG (t) ∼ = Cz . But C as in [V17 , 10.1.5], and so does not involve Z3 × A8 , a contradiction. Hence the lemma is proved.  Now we can prove Lemma 3.25. The conditions (3L) hold. Proof. Lemma 3.23bc implies (3L1,2). In view of Lemma 3.24b and [V17 , 10.10.10], La1 contains a subgroup L ∼ = SL2 (3) ∗ SL2 (3) (disjoint from a1 ) with L a1   CG (a1 , z1 ). Also there is e ∈ I2 (NLa1 (L)) interchanging the two solvable components of L. Let B1 ∈ Syl3 (L a1 ); then (3L4) holds. Write O2 (L) = Qa1 , Z(Qa1 ) = z1 , Cz1 = CG (z1 ), and Qz1 = O2 (Cz1 ). By Proposition 3.3c, z1 ∈ z G , so Qz1 ∼ = 21+8 + . Now because of e, either Qa1 ≤ Qz1 or Qa1 ∩ Qz1 = z1 . Suppose that the latter holds and let z satisfies   C  ( z = Cz /Qz , so that A4 × A4 ∼ a ). However, C C = L 1 1 1 1 Cz1 1 z ) the conclusion of [V17 , 10.1.5], and in no case does some element of I3 (C 1 have such a normal subgroup in its centralizer. Therefore Qa1 ≤ Qz1 , which is (3L5). It remains to rule out the possibility La1 ∼ = L4 (3). We repeat the proof of Lemma 3.11, substituting a1 for b, z1 for z (even in subscripts), and reach a contradiction as there. This completes the proof of the lemma.  As noted following (3L) and (3M), we now obtain (3K), which contradicts (3J). This completes the proof of Proposition 3.13, which in turn implies Proposition 3.12, the last step in the proof of Proposition 3.1. Then Propositions 2.6 and 3.1 yield Theorem C∗4 : Stage A4.

CHAPTER 16

Theorem C∗4 : Stage A5. Recognition 1. Introduction In this chapter, we show that the conclusions of Stages A2 and A4 of Theorem C∗4 imply the conclusion of Theorem C∗4 (Case A). Namely, we must show that the existence of an extraspecial setup or an A9 -Sp6 (2) setup in G implies that G is isomorphic to one of the simple K-groups in the conclusion of Theorem C∗4 (Case A). In Section 2, we will outline the strategy for the extraspecial setup, which will be completed in Sections 3–9. Then in Section 10, we will outline the strategy for the A9 –Sp6 (2) setup, which will be completed in Sections 11–18. 2. The Extraspecial Setup In the next eight sections we finish the case in which p = 3 and F ∗ (CG (z)) is extraspecial for some z ∈ I2 (G) such that m3 (CG (z)) = 3. In view of Theorem C∗4 : Stages A1, A2, and A4, assuming the extraspecial setup, we set (2A)

L = {U4 (2), G2 (3), 32 U4 (3)},

and work with the following hypotheses and notation. Recall that by Theorem C∗4 : Stage A1, O3 (CG (x)) = O{2,3} (CG (x)) for all x ∈ I3o (G).

(2B)

(1) G is of restricted even type with e(G) = m2,3 (G) = 3 and Lo3 (G) ⊆ C3 ; (2) For any B ∈ E33 (G) such that NG (B; 2) = {1}, B # ⊆ I3o (G); (3) For any involution z of G such that CG (z) contains some A ∼ = E33 , z is 2-central in G, and Qz := F ∗ (CG (z)) = O2 (CG (z)) is the central product of = 3 or 4 copies of Q8 (note that z is then a Sylow 2-center in G, hence determined up to Gconjugacy); and (4) With z and A as in (3), there exists x ∈ A# such that |CQz (x)| ≥ 25 , and for any such x the 3-layer L := L3 (CG (x)) satisfies L/O{2,3} (L) ∈ L and m3 (C(x, L)) = 1. 333

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We let Lactual be the subset of L consisting of those L/O{2,3} (L) actually occurring in (2B4) for some x, z, and A as in (2B3,4). It will turn out that Lactual is a singleton, i.e., L/O{2,3} (L) is unique up to isomorphism in (2B4). However, the various target groups G cover all three possibilities for L/O{2,3} (L). We shall prove: Proposition 2.1. Under the conditions (2B), one of the following holds: (a) Lactual D4 (2), (b) Lactual (c) Lactual

= {U4 (2)}. Moreover, G is isomorphic to U5 (2), U6 (2), or Co2 ; = {32 U4 (3)} and G ∼ = Suz; or = {G2 (3)} and G ∼ = F3 .

We remind the reader that our assumed Background Results include the assertion that each of the 26 sporadic groups is determined up to isomorphism by its centralizer of involution pattern. (2C)

We assume (2B) through Section 9.

At this point, cores present little difficulty. Lemma 2.2. Let x, A, z, and L be as in (2B3,4). Then z ∈ L. Furthermore, O3 (NG ( x)) = O{2,3} (NG ( x)) = 1. x = Nx /O{2,3} (Nx ). By Proof. Set Nx = NG ( x), Cx = CG (x), and N ∗ Theorem C4 : Stage A1, O{2,3} (Nx ) = O3 (Nx ) = O3 (CG (x)). By (2B4),   ∈ L, and O3 (L),which contains x , is a cyclic 3-group centralized by z. L By assumption CQz (x) is the central product of at least two copies of Q8  ∈ L, | Out(L)| ≤ 8 < |CQ (x)|. Hence z whose centers equal z. But as L z  By the F ∗ -theorem [IG , 3.6], z ∈ L.  induces an inner automorphism on L. As |O3 (Cx )| is odd, z ∈ L, as claimed. Next, let V = O{2,3} (CG ( x, z)); we claim that V = 1. As [ x, z , A] = 1, V ∈ NCG (z) (A; {2, 3} ). Given the structure of F ∗ (CG (z)), V A embeds in O8+ (2) with V  V A. Suppose the claim is false. As |O8+ (2)|{2,3} = 52 7, some hyperplane A0 of A centralizes an element v ∈ I5 (V ) ∪ I7 (V ). With [IA , 4.8.2], however, we find m3 (A0 ) ≤ m3 (CO+ (2) (v)) ≤ 1, a contradiction. 8 This proves the claim. Now let W = O3 (NG ( x)) = O3 (CG (x)) = O{2,3} (CG (x)) = O{2,3} (NG ( x)) (we have used (a) of Theorem C∗4 : Stage A1). Then CW (z) ≤ V = 1, so z inverts W . But then as z ∈ L = [L, z], z centralizes W . Since W has odd order, W = 1. 

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3. The 32 U4 (3) and G2 (3) Cases: G ∼ = Suz or F3 In this section we reduce to the case Lactual = {U4 (2)} by proving: Proposition 3.1. Assume the setup (2B). If Lactual ∩ {32 U4 (3), G2 (3)} = ∅, then G ∼ = Suz or F3 (and accordingly, Lactual = {32 U4 (3)} or {G2 (3)}). We fix z ∈ I2 (G), A ∈ E33 (CG (z)), and x ∈ A# satisfying (2B) and such that L ∼ = 32 U4 (3) or G2 (3) (see Lemma 2.2). Thus, Qz = F ∗ (CG (z)) = O2 (CG (z)) ∼ = Q8 , = 3 or 4. We set Cx = CG (x), Nx = NG ( x), Cz = CG (z), and C z = Cz /Qz . We first prove Lemma 3.2. The following conditions hold: (a) m3 (CG (A)) = 5; (b) O3 (Nx ) = 1; (c) F ∗ (Nx ) = x E(Nx ) with E(Nx ) ∼ = 32 U4 (3) or G2 (3). Moreover, CNx (E(Nx )) = x; (d) CQz (x) = O2 (CE(Nx ) (z)) ∼ = 21+4 + ;  # (e) For any x ∈ A such that |CQz (x )| ≥ 25 , conclusions (b), (c), and (d) hold with x in place of x. Proof. By Lemma 2.2, z ∈ L and (b) holds. By our choice of x, and (2B4), F ∗ (Nx /O3 (Nx )) ∼ = Z3r ∗ 32 U4 (3) or Z3r × G2 (3), for some r ≥ 1. Indeed the same holds for any x as in the lemma, except that it is also  possible that O3 (CG (x )) ∼ = Z3r × U4 (2), and r may depend on x . By (b), [IA , 4.5.1] and [V17 , 10.10.9], (3A)

O 2 (CCx (z)) ∼ = Z3r × (SL2 (3) ∗ SL2 (3)).

But then m3 (CCx (A)) = 5 (see [IA , 3.3.3, 6.4.4]), and (a) holds. In particular, m3 (CG (x )) ≥ 5, which rules out the possibility E(CG (x )) ∼ = U4 (2), as e(G) = 3. Thus (c) holds, for x and x . Next, |O2 (CCz (x))| = |O2 (CCx (z))| ≤ 25 , and in case of equality, O2 (CC (z)) = O2 (CL (z)) ∼ = Q8 ∗ Q8 , x

with z 2-central in L, by [V17 , 10.10.15]. But |O2 (CCz (x))| ≥ |CQz (x)| ≥ 25 , and the same holds for x , by our assumptions. Therefore O2 (CL (z)) = CQz (x) ∼ = Q8 ∗ Q8 , and similarly for x . Note that by [IA , 4.5.1], L has just one class of involutions. Hence (d) holds, and it remains to prove that r = 1. But if r > 1, or 21+6 then Cz contains Z9 × E32 . As Qz = F ∗ (Cz ) ∼ = 21+8 + − , Sylow 3+ subgroups of Out(Qz ) embed in O8 (2) and hence in (Z3 Z3 ) × Z3 , so they have no Z9 × E32 -subgroup, contradiction. Thus, r = 1, and the proof is complete. 

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Lemma 3.3. There exists t0 ∈ z L ∩ CL (z) such that [CL (z), t0 ] = Q1 x1  ∗ Q2 x2  ∼ = SL2 (3) ∗ SL2 (3), with x1 , x2  = [A, t0 ] inverted by t0 , and A = x, x1 , x2 . Proof. By [V17 , 10.10.14c], AO2 (CL (z))/ x ∼ = SL2 (3) ∗ SL2 (3) with  some involution of CL (z) inverting A/ x. Now (xx1 )t0 = xx−1 1 with t0 ∈ Cz . We factor Qz under x, x1 . Using that = 3 or 4 (2B3), and that |CQz (x )| ≤ 25 for all x ∈ x, x1 # (Lemma 3.2de), we see that one of the following holds: (1) = 3, CQ (x1 ) ∼ = Q8 ∗ Q8 , and z

(3B)

CQz (xx1 ) = CQz (xx−1 1 ) = CQz ( x, x1 ) = Q2 ; or (2) = 4, CQz (x1 ) = Q2 = CQz ( x, x1 ), and ∼ CQz (xx1 ) ∼ = CQz (xx−1 1 ) = Q8 ∗ Q8 .

These will lead to G ∼ = Suz and G ∼ = F3 , respectively. First we determine L and C z in the two cases. Lemma 3.4. Suppose that = 3. Then L ∼ = Ω− = 32 U4 (3) and C z ∼ 6 (2) or − ∼ C z = O6 (2). Proof. By (3B1), x and x1  are the only subgroups A0 of x, x1  of order 3 with CQz (A0 ) ∼ = Q8 ∗Q8 . By Lemma 3.2, we have symmetry between x and x1 . Hence, setting Cx1 = CG (x1 ) and L1 = E(Cx1 ), we see by Lemma 3.3 that there is t1 ∈ z G ∩ I2 (Cx1 ) such that xt1 = x−1 . Then Lemma 3.3c implies that t1 induces an outer automorphism on L. If L ∼ = G2 (3), then 1 1 2 2 ∼ by [IA , 2.5.12, 4.9.1], CL (t1 ) = G2 (3 2 ). Hence G2 (3 2 ) embeds in Cz , C z , and O6− (2). But this is impossible as 7 does not divide |O6− (2)|. Therefore L∼ = 32 U4 (3). By [V17 , 10.1.10], CL (t1 ) is nonsolvable. As t1 ∈ z G , C z is a nonsolvable subgroup of O6− (2) with m3 (C z ) = 3. Finally, [V17 , 10.9.8] −  yields C z ∼ = Ω− 6 (2) or O6 (2), as claimed. The next argument is based on [SmF, Lemma 2.5ff]. Lemma 3.5. Suppose that = 4 and C z ∼ = Sp6 (2). Write NC z (A) = AΓ, 2 = 1 and [A, γ ] ∼ Z for each Γ =

γ , γ , γ 

ρ, τ , with γ where Z2 Σ3 ∼ = 1 2 3 i = 3 i i, ρ3 = 1, τ 2 = 1, ρ, τ  permuting {γ 1 , γ 2 , γ 3 } and γ τ3 = γ 3 . Then τ has an involutory preimage in NCz (A). Proof. Since CQz (A) = z, the preimage Γ of Γ in NCz (A) satisfies Γ ∩ Qz = z. Let γi and τ be preimages of γ i and τ in Γ. Let A0 = [A, γ3 ], so that X := O 3 (CC z (A0 )) ∼ = Σ6 , with τ a root involution in X. By [V17 , 10.1.11f], CQz (A0 ) = z. Hence the preimage X of X in CCz (A0 ) satisfies X ∩ Qz = z. We have τ ∈ X. In particular, X = [X, X] τ  and γ1 ∈ X − [X, X], with γ 1 and τ representing the two X-classes of involutions in X − [X, X].

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We have two cases: (1) z ∈ [X, X] ∼ = 2A6 ; or (3C) ∼ (2) [X, X] = A6 . In case (3C1), the only involution in [X, X] is z. Hence, as γ12 = γ22 (thanks to ρ), [γ1 , γ2 ] = (γ1 γ2 )2 = z. Now by the action of Γ, [γi , γj ] = z for all 1 ≤ i < j ≤ 3. Thus [γ1 γ2 , γ3 ] = 1 and so (γ1 γ2 γ3 )2 = γ32 z. Consider the involutions γ 1 γ 2 γ 3 and γ 3 τ . Their commutators on the natural C z -module are both 3-dimensional, so they are conjugate by [V17 , 10.2.5a], whence (γ3 τ )2 = (γ1 γ2 γ3 )2 = γ32 z. It follows that [γ3 , τ ] = τ 2 z. Thus τ γ3 = τ −1 z. If τ 2 = z, then γ3 centralizes [X, X] τ  = X. But [γ3 , γ1 ] = z and γ1 ∈ X, contradiction. Hence τ 2 = 1 in this case. In case (3C2), we argue similarly. We have z ∈ [X, X] and |X/[X, X]| = 4, and if X/[X, X] ∼ = E22 the desired conclusion is clear, so assume that X/[X, X] ∼ = Z4 . This time, [γ1 , γ2 ] ∈ [X, X] ∩ z = 1 so [γi , γj ] = 1 for all 1 ≤ i < j ≤ 3. Hence, CX (γ3 ) ≥ [X, X] γ1  = X. But also (γ3 τ )2 = (γ1 γ2 γ3 )2 = γ32 and thus τ γ3 = τ −1 . Since [X, γ3 ] = 1, τ = τ −1 . The proof is complete.  Lemma 3.6. Suppose that = 4. Then L ∼ = A9 . = G2 (3) and C z ∼ Moreover, if a ∈ C z is a 3-cycle, then CQz /z (a) = 1. Proof. By (3B2), x1  is weakly closed in x, x1  with respect to Cz . Let L2 = E(CG (xx1 )). Then x1 ∈ L2 and x1 is inverted in L2 . Hence in Cz , −1 x = (xx1 )x−1 1 ∼ (xx1 )x1 = xx1 ∼ xx1 , the last conjugacy in Cx . Likewise −1 x ∼Cz xx2 ∼Cz xx2 . Also, x is not Cz -conjugate to x1 or x2 . −1 Let δ = x±1 1 x2 . Then xδ ∼Cz xδ , with CQz ( x, δ) = CQz ∩Cx (δ) = z.  ∼ Hence if CQz (x δ) = Q8 ∗ Q8 , then m2 (Qz / z) ≥ 12, contradicting = 4.   Therefore, xCz does not contain xδ or xδ −1 for either choice of δ. So 5 ≤ | xCz ∩ A| ≤ 7. As A ∈ Syl3 (CCz (x)), xCz ∩ A = xNCz (A) . But 5 and 7 do not divide |GL3 (3)|, so | xNCz (A) | = 6 and we may choose notation so that x1 x2 ∈ xCz but x−1 x2 ∈ xCz . In particular, x, like x1 x2 , is inverted by an involution u ∈ z G ∩ Cz and we may choose u ∈ NCz (A). Again Lemma 3.3c implies that u induces an outer automorphism on L. Suppose now that L∼ = 32 U4 (3). By [V17 , 10.10.18a], CCG (u) (x) = CCx (u) then has a 3-component isomorphic mod 3 -core to A6 or U3 (3). Also | AutL (A)| = 4. Since u inverts x, AutNx ∩Cz (A) contains a subgroup of order 8, whence | AutCz (A)| is divisible by 48. In particular, 34 divides |C z |. As u ∈ z G , we may apply [V17 , 10.1.11abe] and conclude that C z ∼ = Sp6 (2) or A9 . Actually C z ∼ = Sp6 (2) since in the A9 case, | AutCz (A)| would be 24, contradicting what we just saw. We saw in the previous two paragraphs that xG ∩E1 ( x1 , x2 ) consists

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of the single subgroup x1 x2 . As x1 , x2  = A ∩ L, AutNx ∩Cz (A) normalizes x1 , x2 , x1 x2 , and x. As it has order 8, it contains an involution τ centralizing x1 , x2  and inverting x. Let τ be a 2-element of NNx ∩Cz (A) mapping on τ. Then τ must centralize O2 (CL (z)) ∼ = Q8 ∗ Q8 . Hence by [V17 , 10.10.18b], τ 2 (= z) = 1. On the other hand, by [V17 , 10.2.10], since | xNCz (A) | = 6 and x = [A, τ ], Autτ  (A) is a subgroup of the complement in the wreath product AutCz (A) ∼ = Σ3 Σ3 . Thus Lemma 3.5 applies 2 and yields τ = 1, a contradiction. 1 Therefore, L ∼ = 2 G2 (3 2 ). By [V17 , 10.1.11d], = G2 (3), whence CL (u) ∼ Cz ∼ = A9 , and a 3-cycle in C z is fixed-point-free on Qz / z, or C z ∼ = Sp6 (2). This time | AutNx ∩Cz (A)| ≤ 4 since | Out(L)| = 2 and |CL (z)/O2 (CL (z))| = 2 · 32 . Hence | AutCz (A)| ≤ 24. But if C z ∼ = Sp6 (2), then | AutCz (A)| = 48, . The lemma is proved.  A contradiction; so C z ∼ = 9 Now, for = 3, set G∗ = Suz or Aut(Suz) according as C z ∼ = Ω− 6 (2) − ∗ ∗ ∗ or O6 (2). For = 4, set G = F3 . In any case let z ∈ G be a 2-central involution, and set C ∗ = CG∗ (z ∗ ) and Q∗ = O2 (C ∗ ). Lemma 3.7. Cz ∼ = C ∗. Proof. We have an isomorphism β : Qz → Q∗ , which induces an isomorphism β∗ : Aut(Qz ) → Aut(Q∗ ) such that (3D)

β(y δ ) = (β(y))β∗ (δ) for all y ∈ Qz and δ ∈ Aut(Qz ).

Identifying Cz / z and C ∗ / z ∗  with their images in Aut(Qz ) and Aut(Q∗ ), we see that if = 3, then β∗ (Cz / z) = C ∗ / z ∗ , since Cz / z and C ∗ / z ∗  are either both the full automorphism groups or both their commutator subgroups. If = 4, then by Lemma 3.6 and [V17 , 10.1.10], β∗ (Cz / z) = (C ∗ / z ∗ )α for some α ∈ Aut(Q∗ ). Replacing β by α−1 ◦ β, we may assume in all cases that β was chosen so that (3E)

β∗ (Cz / z) = C ∗ / z ∗  .

We can then (in many ways) define a bijection γ : Cz → C ∗ , one coset of Qz at a time, so that γ(wy) = γ(w)β(y) for all y ∈ Qz and w ∈ Cz , γ(1) = 1, and γ induces β∗ modulo z and z ∗ . Thus γ|Qz = β. We wish to choose γ to be an isomorphism. Let T be a transversal to Qz in Cz . Each such γ is then determined by its values on T . For every g ∈ C z , let tg be the corresponding element of T . We choose t1 = 1. For any u, u ∈ Cz , we write u = tg y and u = tg y  for unique y, y  ∈ Qz and g, g  ∈ C z , and compare γ(uu ) with γ(u)γ(u ). We have uu = tg tg y tg y  . Also as β∗ is an isomorphism, γ(tg tg ) = γ(tg )γ(tg )f (g, g  )

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for some f (g, g  ) ∈ z ∗ . Thus using (3D) and the multiplicativity of β, we have γ(u)γ(u ) = γ(tg )β(y)γ(tg )β(y  ) = γ(tg )γ(tg )β(y)β∗ (tg z) β(y  ) γ(uu ) = γ(tg tg )β(y tg y  ) = γ(u)γ(u )f (g, g  ). A standard calculation shows that f ∈ Z 2 (C z , z ∗ ), and that all cohomologous cocycles arise from merely changing the choice of transversal. Therefore to achieve our goal that γ can be chosen to be an isomorphism, it suffices to show that f is cohomologically trivial. This we do by identifying H 2 ([C z , C z ], z ∗ ) with the exponent 2 subgroup of the Schur multiplier of C z [IG , 31.9]. That multiplier in both cases is of order 2 [IA , 6.1.4]. Then we restrict to M = CG (x) or NG ( x) and further to a Sylow 2-subgroup R of M ∩ NG (A) ∩ Cz , and compare results to the covering group 2C z , to achieve our goal. Let R∗ be the 2-subgroup of G∗ such that β∗ (R/ z) = R∗ / z ∗ . Thus R∗ is Sylow in NC ∗ (A∗ ) ∩ NC ∗ ( x∗ ) where A∗ and x∗  are Sylow in the preimages in C ∗ of β∗ (A z / z) and β∗ ( x, z / z), respectively. Suppose that f is cohomologically nontrivial. Then f is a factor set for the covering group 2C z . Hence if s ∈ I2 (C z ) pulls back to an element of order 4 in 2C z , then f (s, s) = z ∗ . So γ(t2s ) = (γ(ts ))2 z ∗ . Thus if (ts )2 = 1, then γ(ts )2 = z ∗ , and if t2s = z, then γ(ts )2 = 1. Now consider the case = 4. We take M = NG ( x) ∼ = Σ3 Y Aut(G2 (3)) by Lemma 3.3ab. Recall R ∈ Syl2 (NM (A) ∩ CM (z)). By the structure of CM (z) [IA , 4.5.1], R ∼ = E23 ∼ = R∗ . However, R ∈ Syl2 (NCz ( x) ∩ NCz (A)), so as x ∈ C z ∼ = A9 is the product of two disjoint 3-cycles, R ≤ C z ∼ = A9 contains a root involution s of A9 . By [IA , 5.2.4e], s pulls back to an element of order 4 in 2A9 , so the previous paragraph shows that f is cohomologically trivial, as desired. Similarly, if = 3 and G∗ = Suz, let M = CG (x), so that L = F ∗ (M ) ∼ = ∼ 32 U4 (3). Now R ∈ Syl2 (CCz (x) ∩ NCz (A)). As C z ∼ (2), R E = = Ω− 22 . 6 Viewing R in NM (A) ∩ CM (z), we see |R| = 8|M : L|, so L = M and R is elementary abelian. Likewise R∗ is elementary abelian. But 2C z ∼ = Sp4 (3) is of 2-rank 2, so some s ∈ R pulls back to an element of order 4 in Sp4 (3), and as in the previous case, f is again cohomologically trivial. Finally, suppose = 3 and G∗ = Aut(Suz). By an argument like the previous paragraph’s, we can argue that when restricted to [C z , C z ], f is cohomologically trivial. Replacing f by a cohomologous cocycle, we may assume that γ|[Cz , Cz ] is an isomorphism. Let M = Nx . Again L = F ∗ (M ) ∼ = 32 U4 (3) and now |M : L| = 4, so M/ x ∼ = Aut(L). Set R0 = CCz (A/ x) and R0∗ = CC ∗ (A∗ / x∗ ). Then β∗ (R0 / z) = R0∗ / z ∗ . But R0 ≤ M and as gcd(|M : L|, |Z(L)|) = 1, M is determined up to isomorphism. This forces R0 ∼ = R0∗ . In fact, by [V17 ,

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10.10.19a], R0 = A u where u2 = z and u inverts x. Similarly R0∗ = A∗ u∗  with (u∗ )2 = z ∗ . As A is self-centralizing in Aut(C z ), and similarly for A∗ in Aut(C ∗ /Q∗ ), conjugation of [Cz , Cz ] by u corresponds under β∗ to conjugation of [C ∗ , C ∗ ] by u∗ . Then γ|[Cz , Cz ] extends to an isomorphism  γ  : Cz → C ∗ by setting γ  (u) = u∗ . This completes the proof. Lemma 3.8. If = 4, then G ∼ = F3 . Proof. By our Background Results [I1 , Secs. 15–18], it is enough to show that G ≈ F3 , i.e., G and F3 have the same centralizer of involution pattern. In view of Lemma 3.7, it suffices to show that G (like F3 , [IA , 5.3x]) has just one class of involutions. Now 3-cycles of C z ∼ = A9 have no fixed points on Qz / z, while x fixes Q8 ∗ Q8 , and 3-central elements fix Q8 on Qz . Thus, every subgroup of C z of order 9 contains an element fixing no involution of Qz − z. Also, a 5-cycle of C z centralizes a 3-cycle and hence has no fixed points on Qz / z. It follows that Cz is transitive on the 270 involutions of Qz − z. Since L has one class of involutions, I2 (Qz ) ⊆ z G . Next, let t ∈ I2 (Cz ) − Qz . Whether t ∈ Cz ∼ = A9 is a product of two or four disjoint transpositions, it inverts a conjugate of x, so with the Baer-Suzuki theorem, t inverts a conjugate of x. As NG ( x)/ x ∼ = Aut(L), all involutions inverting a conjugate of x are conjugate to one another (see [IA , 4.9.1]). Finally, CL (z) contains some  involutions outside Qz , and as I2 (L) ⊆ z G , the lemma follows. Lemma 3.9. Suppose = 3 and t, u ∈ I2 ([Cz , Cz ]) − {z}. Then exactly one of the following holds: (a) t ∈ Qz − {z}; (b) t ∈ [Cz , Cz ] − Qz and t is 2-central in [C, C]; (c) t ∈ [Cz , Cz ] − Qz and t is not 2-central in [C, C]; moreover, CG (t) contains a copy of A6 . Moreover, if u satisfies the same of these conditions as t, then t ∼G u. Proof. The first assertion is trivial. If t and u both satisfy (a), then t ∼Cz t by Witt’s lemma applied to the action of C z on Qz . In cases (b) and (c), the fact that C z ∼ = P Sp4 (3) has unique classes of 2-central and non-2-central involutions allows us to assume that t = u. We must show in (b) that I2 (Qz t) ⊆ tG . By the structure of [C z , C z ] ∼ = − Ω6 (2) and the Baer-Suzuki Theorem, we may replace t by a conjugate and assume that t normalizes A and inverts x0 = xx1 , with [t, A] = x0  and x ∼Cz x1 . Then all involutions in Qz t are fused into CQz (x0 )t under [Qz , x0 ] and A acts nontrivially on CQz (x0 ) ∼ = Q8 . If t acts nontrivially on CCz (x0 ), then t CCz (x0 ) is quasidihedral or Q8 ∗ Z4 , and the desired fusion is clear. If t centralizes CCz (x0 ), then we need only show that t and tz are conjugate. Now t normalizes ACCz (x0 ) and ACCz (x0 ) t ∼ = Z3 × Σ3 × SL2 (3). A conjugate of t then centralizes a conjugate of x which acts nontrivially on CCz (x0 ), so again replacing t by a conjugate we may assume that t ∈ CG (x)∩

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CG (z). Thus either t ∈ L or L t = CG (x). In either case, by [V17 , 10.10.19c], all involutions in the coset Lt are conjugate. Thus t and tz are conjugate, as asserted. Similarly, by the structure of [C z , C z ] ∼ = Ω− 6 (2) and the Baer-Suzuki Theorem, we may replace any involution t of type (c) by a conjugate inverting x, normalizing each of the subgroups x1  and

x2 , and centralizing exactly one of them. Then t ∈ NG ( x), inverting

x = Z(L), centralizing z and acting on O2,3 (CL (z)) = L1 ∗ L2 ∼ = SL2 (3) ∗ SL2 (3), normalizing both L1 and A with |CA (t)| = 3. By [V17 , 10.10.19b], this uniquely determines the coset Lt and this coset has a unique class of involutions; moreover, each such involution centralizes a copy of A6 in L. Thus t is determined up to G-conjugacy, and the lemma is proved.  Lemma 3.10. Suppose that = 3. Then the following conditions hold: (a) If t ∈ I2 ([Cz , Cz ]) − {z} and t satisfies Lemma 3.9a or 3.9b, then t ∈ zG; (b) For all t ∈ I2 ([Cz , Cz ])−Qz satisfying Lemma 3.9c, |C[Cz ,Cz ] (t)|2 ≤ 29 . Moreover, for some such t, equality holds, and a Sylow 2subgroup P of C[Cz ,Cz ] (t) is isomorphic to one of CSuz (2B) = (V × Lt ) v, where V ∼ = E22 , Lt ∼ = D8 , = L3 (4), v 2 = 1, V v ∼ and v induces a field automorphism on Lt ; (c) Suppose that C z ∼ = O6− (2) and t is as in (b), with |C[Cz ,Cz ] (t)| = 9 2 . Then a Sylow 2-subgroup of CCz (t) is isomorphic to a Sylow 2-subgroup of CAut(Suz) (2B) = (V × Lt ) v, w where V , Lt , and v are as in (b), w2 = 1, [w, V v] = 1, and w induces a graph automorphism on Lt ; and (d) If C z ∼ = O6− (2), then for some y ∈ I2 (Cz ) − [Cz , Cz ], C[Cz ,Cz ] (y) is by Σ5 , namely the centralizer of a 2-central an extension of 21+4 − involution in Aut(HJ). Proof. We have z ∈ L, and CL (z) contains involutions of types (a) and (b) from Lemma 3.9, with one of type (b) acting nontrivially on A. Since L has one class of involutions, (a) follows. Our analysis in Lemma 3.9 applies to the groups Suz and Aut(Suz) as well as to G. In Suz there is an involution t ∈ z Suz , necessarily of type (c) from Lemma 3.9, with CSuz (t) = (V × Lt ) v as described in (b). Replacing t by a suitable conjugate, we obtain (b), as a Sylow 2-subgroup of [Cz , Cz ] is a Sylow 2-subgroup of Suz. Similarly (c) holds, by [V17 , 10.14.2]. Finally, (d) follows from similar reasoning and the fact that there is  y ∈ I2 (Aut(Suz)) − Suz such that CSuz (y) ∼ = Aut(HJ) [IA , 5.3o]. Now we can prove ∗ Lemma 3.11. Suppose that = 3. Then C z ∼ = Ω− 6 (2), i.e., G = Suz. Proof. Suppose for a contradiction that C z ∼ = O6− (2). Let y ∈ I2 (Cz )− [Cz , Cz ] be as in Lemma 3.10d. By the Thompson transfer lemma, there is g ∈ G such that y g ∈ [Cz , Cz ] and y g is extremal in a Sylow 2-subgroup

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of Cz . Now CG (y, z) = y × C[Cz ,Cz ] (y), so z is characteristic in a Sylow 2-subgroup of CG (y, z). Hence z is 2-central in CG (y), so y and z are not G-conjugate (otherwise y would be the only involution to be 2-central in CG (y)). Hence y g must be of type (c) in Lemma 3.9. Since y g is extremal in a Sylow 2-subgroup of Cz , we may set t = y g and apply Lemma 3.10c to t. Let P2 ∈ Syl2 (CCz (y g )) ⊆ Syl2 (CG (y g )), and Q ∈ Syl2 (CCz (y)). Without loss, Qg ≤ P2 . By Lemma 3.10c we have P2 = (V × (P2 ∩ Lyg )) v, w with V  P2 , v, w ∼ = E22 , and Lyg v, w of index 3 in Aut(L3 (4)). By [V17 , 10.14.3], [P2 , P2 , P2 , P2 , P2 ] = Z(P2 ) ∩ Lyg = z. On the other hand from the structure of C[Cz ,Cz ] (y), z ∈ [Q, Q, Q, Q, Q]. Hence z g = z. Thus g ∈ Cz , which is absurd. The lemma is proved.  Now fix t ∈ I2 (G) satisfying Lemma 3.9c. Recall that we have an isomorphism γ : Cz → C ∗ . Lemma 3.12. If u, v ∈ I2 (Cz ), then u ∼G v if and only if γ(u) ∼G∗ γ(v). Proof. As Suz has two classes of involutions [IA , 5.3o], it suffices in view of the characterizations in Lemma 3.9 to show that there exists more than one class of involutions in G. But by [IA , 4.5.1] there exists an involution u ∈ Inndiag(L), and hence in NG (L), such that CL/Z(L) (u) ∼ = U3 (3). G As 7 divides |U3 (3)| but not |Cz |, u ∈ z , as required.  Finally we prove Lemma 3.13. Let t∗ ∈ I2 (Suz) be a non-2-central involution. Then CG (t) ∼ = CSuz (t∗ ). Proof. We have already seen that a Sylow 2-subgroup P = (V ×PL ) v of Ct := CG (t) is of the correct isomorphism type. As is visible in G∗ , the subgroup PL , isomorphic to a Sylow 2-subgroup of L3 (4), can be taken to have all its involutions conjugate to z, in view of the previous lemma. However, F ∗ (Cz ) = O2 (Cz ), so F ∗ (CCt (v)) is a 2-group for every involution v ∈ PL , by [IG , 5.12]. In particular PL normalizes and acts faithfully on every component of E(Ct ). As m2 (PL ) = 4, Ct has no A6 component, and as E(CL (t)) ∼ = A6 and m3 (Ct ) ≤ 3, some component Kt of Ct properly contains a copy of A6 . We have Kt ∈ C2 since G is of even type. Every PL -invariant subgroup of P disjoint from PL has order at most 4, and so Kt = E(Ct ) with |CCt (Kt )|2 ≤ 4 and hence | Aut(Kt )|2 ≥ 27 . Let P0 = P ∩ Kt . Then |P0 /Z(Kt )| ≤ 28 , while Kt contains A6 and m3 (Kt ) ≤ 3. Also, we may compute in C ∗ that CCt (z) = CCz (t) = P . Hence by [V17 , 9.15], Kt /Z(Kt ) ∼ = L3 (4) or Sp4 (4). Now P is isomorphic to a Sylow 2-subgroup of CSuz (t∗ ). Thus J(P ) is maximal in P , Z(J(P )) ∼ = E24 , and Z(J(P )) ≤ Φ(J(P )). The last condition rules out Sp4 (4), so Kt /Z(Kt ) ∼ = L3 (4). In that case Z(J(P )) induces inner automorphisms on Kt via 2-central involutions and V := CZ(J(P )) (Kt ) ∼ = E22 . As Kt ∈ C2 , Z(Kt ) is elementary abelian, so P ∩ Kt ≤ J(P ). But

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∼ E22 , so Kt ∼ [J(P ), J(P )] = = L3 (4), whence J(P ) = V × (P ∩ Kt ). Finally 2 P = J(P ) v where v = 1 and CJ(P ) (v) ∼ = Z2 × D8 . Hence V v ∼ = D8 and v induces a field automorphism on Kt , proving that Ct ∼ = CSuz (t∗ ), as claimed.  Summing up: Lemma 3.14. If = 3, then G ∼ = Suz. Proof. By Lemmas 3.7, 3.12, and 3.13, G ≈ Suz. Hence G ∼ = Suz by our Background Results [I1 , Secs. 15–18].  Proposition 3.1 follows from Lemmas 3.8 and 3.14, and (2B3). 4. The U4 (2) Setup In the next six sections we finish the case in which p = 3 and Lo3 (G) contains U4 (2). Thus there exists z ∈ I2 (G) such that m3 (CG (z)) ≥ 3. As we still assume (2B) we work with the following hypotheses and notation:

(4A)

(1) G is of restricted even type with e(G) = m2,3 (G) = 3 and Lo3 (G) ⊆ C3 ; (2) For any B ∈ E33 (G) such that NG (B; 2) = {1}, B # ⊆ I3o (G); (3) For any involution z of G such that CG (z) contains some A ∼ = E33 , z is 2-central in G, and ∗ Qz := F (CG (z)) = O2 (CG (z)) is the central product of = 3 or 4 copies of Q8 (note that z is then a Sylow 2-center in G, hence determined up to Gconjugacy); and (4) With z and A as in (3), there exists x ∈ A# such that |CQz (x)| ≥ 25 , and for any such x we have F ∗ (CG (x)) ∼ = Z3r × U4 (2), r > 0.

We assume (4A) through Section 9. Note in (d) that by Lemma 2.2, O3 (NG ( x)) = 1. We shall prove: Proposition 4.1. Under the conditions (4A), G is isomorphic to U5 (2), U6 (2), D4 (2), or Co2 . Our setup for the proof is as follows. Note that any elementary abelian 3-subgroup of U4 (2) is diagonalizable with respect to an orthonormal basis of the natural module, and so lies in an element of E33 (U4 (2)). (4B)

We fix z, A, and x as in (4A3,4), choose D ≤ CG (x) such that A ≤ D ∼ = E34 , and set N = NG (D) and N = N/CG (D). Thus, A = CD (z).

The case division resulting in the various target groups G0 in Proposition 4.1 is according to the structure of N and NG ( x). In each case we then

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determine important 2-local information about G. In the case G0 = Co2 , we obtain the entire involution fusion pattern in G. (This had been done by Fredrick Smith [SmF].) In the cases G0 = U5 (2), U6 (2), or D4 (2), we identify a subgroup of G as isomorphic to G0 , by constructing a Phan-type amalgam and using theorems of Bennett and Shpectorov [BeSh1, Theorem 1.3] and Gramlich, Hoffman, Nickel, and Shpectorov [GHNSh, Theorem B]. If this subgroup is proper in G, we show that its normalizer isstrongly 3-embedded in G, and reach a contradiction using Holt’s theorem [II2 , Theorem SF]. 5. 3-Local Structure Set Nx = NG ( x), Cx = CG (x), Cz = CG (z), and Qz = O2 (Cz ). By Lemma 2.2, O3 (Nx ) = O3 (Cx ) = 1. As Aut(U4 (2)) = U4 (2) γ where γ is a graph automorphism of order 2 [IA , 2.5.12], there is an elementary abelian 2-subgroup Sx ≤ Nx such that the following conditions hold:

(5A)

(1) Nx = Rx LSx with x ∈ Rx ∼ = Z3r and [Rx , L] = 1; (2) Sx embeds in Ω1 (O2 (Out(Rx × L))) ∼ = Z2 × Z2 , so that one of the following holds: (a) Sx = 1; (b) Sx ∼ = Z2 with Sx ≤ Cx ; (c) Sx ∼ = Z2 inverts x, and [Sx , L] = 1; (d) Sx ∼ = Z2 inverts x, and AutNx (L) = Aut(L); or (e) Sx ∼ = E22 .

In all subcases except (5A2d), Nx = Rx∗ × L∗ with F ∗ (Rx∗ ) = Rx and F ∗ (L∗ ) = L. Lemma 5.1. We have CG (D) = D, so N = N/D. Proof. D = x × DL where DL = D ∩ L ∼ = E33 is a full diagonalizable subgroup of L ∼ = U4 (2). It follows easily that CAut(L) (DL ) = DL . Then as  F ∗ (Cx ) = x L, CG (D) ≤ DL CG ( x L) = DL x = D, as claimed. Lemma 5.2. Suppose that B and B ∗ are G-conjugate subsets of D and B ∩ xG = ∅. Then B and B ∗ are N -conjugate. Proof. By [V17 , 10.9.4], D ∩ L is weakly closed in a Sylow 3-subgroup of L, so D is weakly closed in a Sylow 3-subgroup of Nx or of CG (F ) for any  F ≤ D containing x. Therefore the result follows from [V9 , 1.4].

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We next prove: Lemma 5.3. The following conditions hold: (a) Let R ∈ Syl3 (Cz ). Then J(R) ∼ = E33 ; and (b) G permutes transitively the set of all pairs (z, B) such that z ∈ I2 (G) and B ∈ E33 (Cz ). Proof. Recall that Qz = O2 (Cz ) = F ∗ (Cz ) is the central product of  ∼ O(−1) (2) and so

= 3 or 4 copies of Q8 . Hence Cz /Qz embeds in Out(Qz ) = 2 its Sylow 3-subgroups, of 3-rank 3, embed in Z3 Z3 or (Z3 Z3 ) × Z3 . Hence if (a) fails, then = 4 and there is A1 ∈ E33 (Cz ) such that AA1 ∼ = Z3 × 31+2 . But then Z(AA1 ) must map to a Sylow 3-center of Out(Qz ), whence for some y ∈ Z(AA1 ), CQz (y) ∼ = Q8 ∗ Q8 ∗ Q8 . However, this contradicts (4A3), and (a) follows. Now by (4A3), G is transitive on the set of involutions z in (b). By (a) and Sylow’s theorem, for a given z, Cz is transitive on the set of possible B’s, so (b) is proved.  Lemma 5.4. Set DL = D ∩ L. Then the following conditions hold: (a) AutL (D) ∼ = AutL (DL ) ∼ = Σ4 , and DL is isomorphic to the core of the natural permutation module for AutL (D); (b) N L := N ∩ L contains reflections; and (c) All reflections in N are N -conjugate. Proof. By [V17 , 10.9.4], DL ∼ = Σ4 preserves a non= E33 , AutL (DL ) ∼ singular quadratic form on DL , and NL (DL ) contains an involution y acting as a reflection on DL . Then y induces a reflection on D as well, since D = x × DL ≤ x × L. This proves (b). Furthermore y lies in a Σ3 subgroup y, u of AutL (DL ), where u3 = 1. As y is a reflection, y centralizes [DL , u, u]. But then if we let x ∈ x [DL , u, u] − x − L, x has exactly 4 N -conjugates, and they generate D. This implies (a). Let y1 and y2 be involutions inducing reflections on D. By Lemma 5.3b, CD (y1 ) and CD (y2 ) are G-conjugate. Since they contain conjugates of x, Lemma 5.2 implies that they are then N -conjugate. So to prove that y1 and y2 are N -conjugate (modulo D) we may assume that CD (y1 ) = CD (y2 ). This hyperplane contains a conjugate of x, which we may assume is x itself. As the action of each yi on DL lies in AutNx (DL ) ≤ O(DL ) with respect to some fixed quadratic form, the action of each yi on DL , and then on D, is determined by its fixed point set. Hence y 1 = y 2 , and (c) follows as well.  We come to the main result of this section. Define X = xN and Z = Z(Aut(D)) ∼ = Z2 . Lemma 5.5. One of the following holds: (a) N ∼ = Σ5 × Z2 or Σ5 , and |X | = 5; (b) N ∼ = Σ6 or GO− (D), and |X | = 15 or 30, respectively; (c) N = W (D4 ) and |X | = 4; or (d) N ∼ = W (D4 ).Z3 or GO+ (D), and |X | = 12 or 24, respectively.

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Proof. We show that N is irreducible on D and then quote [V9 , 10.1, 10.3]. If x  N , then obviously x is the unique subgroup of D of order 3 such that E(Cx ) has a U4 (2)-type 3-component; hence x is the unique subgroup of D of order 3 such that CO2 (Cz ) (x) contains Q8 ∗ Q8 . But every pair of D-invariant Q8 -subgroups of O2 (Cz ) is centralized by some nonidentity element of D, and as there are at least 3 such pairs, we have a contradiction. Therefore x  N . If DL = D ∩ L  N , then by [V9 , 10.1b], there is E33 ∼ = E  N such that ∼ E = DL as modules for N L . Let E be a Sylow 3-subgroup of the preimage of E in N . By a Frattini argument, NE := NN (E) covers at least N L . Then N E acts irreducibly on E/D ∼ = E and trivially on D/DL . Hence E/DL is elementary abelian and E0 := [E, O2 (N L )] satisfies DL ≤ E0 and E0 /DL is an NL -invariant complement to D/DL in E/DL . Since N L is irreducible on E0 /DL and DL , E0 is special or E0 ∼ = Z9 × Z9 × Z9 or E36 . Let y ∈ N L be a reflection (on D). Then y acts as a reflection on both E/DL and DL . If E0 is abelian, then CE0 (y) ∼ = E34 or Z9 ×Z9 . But e(G) = 3 and CG (y)/O2 (CG (y)) embeds in O8+ (2), which does not contain Z9 × Z9 , contradiction. Hence E0 is special, whence DL ∼ = ∧2 (E0 /DL ), forcing y to have determinant 1 on DL . This is impossible as y is a reflection on DL . Consequently, N is irreducible on D.  Now the lemma follows directly from [V9 , 10.1, 10.3]. 6. G ∼ = U5 (2) In this section we assume (6A) N∼ = Σ5 or Σ5 × Z2 , and prove Proposition 6.1. If (6A) holds, then G ∼ = U5 (2). By Lemma 5.2 and [V9 , 10.1], xG ∩ D = xN = x ∪ [41 ], where [41 ] is an N ∩ Nx -orbit on E1 (D) consisting of elements of the form xyi , 0 ≤ i ≤ 3, such that yi ranges over an NL (D)-orbit (of length 4) of 3-central elements of L lying in D. Thus in particular, no element of xG lies in L. Choose s ∈ xN − x and set E = x, s ∼ = E32 . Without loss we may assume that E = x, y0  . Lemma 6.2. We have Cx = x × L, and Nx = Cx v where v 2 = 1 and Cv ( x) = Cv (L) = 1. That is, either (5A2a) or (5A2d) holds, according as v = 1 or v = 1. Proof. By (6A), AutNx (D), which is the stabilizer of x in N , is isomorphic to Σ4 or Σ4 × Z2 , and in the latter case contains −1. Since AutL (D) ∼ = Σ4 and the image of D in Cx is self-centralizing, the result follows. 

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347

Lemma 6.3. The following hold: (a) There exists Σ ≤ N such that Σ ∼ = Σ5 and D is the core of the natural permutation Σ-module, in which x has exactly 5 Σ-conjugates; (b) There exists e ∈ E # such that NG ( e)/O{2,3} (NG ( e)) ∼ = Σ3 × GU3 (2) or Σ3 × ΓU3 (2); and

(c) Moreover, E = Z(P ) for some P ∈ Syl3 (G), P ∼ = Z3 × (Z3 Z3 ). Proof. By Lemma 5.4a, any element of AutL (D) ∼ = Σ4 outside the commutator subgroup has determinant −1 on D. Consequently any element w ∈ N ∩ L acting as a transposition on X = xN lies outside SL(D). Hence if we let Σ be the Σ5 -subgroup of N containing w, then D is the core of the natural permutation module for Σ. Therefore, E = CD (v) for some 3-cycle v ∈ Σ fixing x, and a transposition t ∈ Σ centralizing v interchanges x with an element of E − x. Therefore t inverts some e ∈ E # with e ∈ xG . We have E = x × (E ∩ L), where E ∩ L = CDL (v) lies in the class [4], in the notation of [V9 , 10.1]. In particular E ∩ L is 3-central in L, whence NL ( e)/O3 (NL ( e)) ∼ = GU3 (2). As t inverts e and centralizes E ∩ L, e ∈ E ∩ L and we have t e NL ( e)/O3 (NL ( e)) ∼ = Σ3 × GU3 (2). Let P ∈ Syl3 (CG (E)) with D ≤ P . Then P ∼ = Z3 × (Z3 Z3 ) and |P | = 35 = |N |3 = |Cx |3 . As D = J(P ), P ∈ Syl3 (G), so P ∈ Syl3 (CG (e)) with P = e × (P ∩ L) = x × (P ∩ L), and P ∩ L ∼ = Z3 Z3 . Hence it remains only to show that CG (e)/ e is 3-constrained, in view of the fact that Out(31+2 ) ∼ = GL2 (3). Suppose on the contrary that L3 (CG (e)) has a 3-component J. Then e ∈ J, for otherwise e ∈ O3 3 (J) ∩ P ≤ [P, P ] by [IG , 15.12(i)], a contradiction. But CG (E) = CCx ( e) is 3-solvable, so Z(P ∩L) acts faithfully on J/O3 (J). Hence P ∩ L acts faithfully on J/O3 (J) and covers a Sylow 3-subgroup of CG (e)/ e. Moreover J/O3 (J) ∈ C3 by (4A1,2). As CG (e) ≥ CL (E) with CL (E) ∼ = GU3 (2), it follows from [V17 , 7.19] that AutJ (D) ∼ = Σ4 or Z2 × Σ4 . But Σ5 has only one class of Σ4 subgroups. There is a Σ4 -subgroup of AutJ (D) fixing only e, and a similar subgroup of AutL (D) fixing only

x. Hence e ∈ xG , a final contradiction completing the proof of the lemma.  Lemma 6.4. The following conditions hold: (a) = 3; (b) Cz /Qz ∼ = GU3 (2) or ΓU3 (2); and (c) O{2,3} (CG (e)) = 1, with e as in Lemma 6.3. Proof. Suppose first that = 4. Recall that A = CD (z) ∼ = E33 . Using (4A4), we see that [Qz , A] = Qz , and then with [V17 , 10.1.7], there are 12 elements a ∈ A# such that CQz (a) ∼ = Q8 ∗ Q8 . However, for any such a, by  3 ∼ (4A) and Lemma 2.2, O (CG (a)) = Z3r × U4 (2), and so as D ∈ E34 (CG (a)), there exists N a ≤ N such that N a ∼ = A4 and a = CD (N a ). By (6A), O 2 (N ) ∼ = A5 , which has only 5 subgroups isomorphic to A4 . Therefore there

16. THEOREM C∗4 : STAGE A5. RECOGNITION

348

∼ Q8 ∗Q8 , a contradiction. are at most 10 elements a ∈ A# such that CQz (a) = Hence = 3, by (4A3), and so Cz /Qz embeds in Out(Qz ) ∼ = O6− (2). Let e be as in Lemma 6.3, and let w ∈ I2 (NG ( e)) map into the Σ3 direct factor of NG ( e)/O3 (NG ( e)). Then m3 (CG (w)) ≥ 3, so w ∈ z G by (4A3). Consequently Cz involves GU3 (2). Now, any GU3 (2)-subgroup of O6− (2) is a maximal parabolic subgroup of Ω− 6 (2) considered in the guise − of P Sp4 (3). Therefore if (b) fails, then Cz /Qz = Ω− 6 (2) or O6 (2). As + CQz (x) ∼ = Q8 ∗ Q8 , NCz ( x) involves a section O4 (2) ∼ = Σ3 Z2 . But then Nx ∩ Cz / x must involve a semidirect product of E32 by D8 , with faithful action of the D8 . Consequently CAut(L) (z) must involve such a semidirect product, which contradicts [V17 , 10.9.2]. Therefore (b) holds. Since O{2,3} (Cx ) = 1 by Lemma 2.2, it follows from [V17 , 10.9.5d] that NCx (D; {2, 3} ) = {1}. Now let H be a Hall {2, 3}-subgroup of NG ( e). Thus H = H1 ×H2 with e ∈ H1 ∼ = GU3 (2), and O3,2 (H2 ) ≤ L. = Σ3 , O2 (H2 ) ∼ 2 Let t1 , t2  be a four-subgroup of H1 × O (H2 ) such that t1 ∈ H1 and t2 ∈ L. Then m3 (CG (ti )) = 3 for both i = 1 and i = 2 (note that t2 lies in a Q8 subgroup of L). Hence ti ∈ z G , so CO{2,3} (CG (e)) (ti ) ≤ O{2,3} (CG (ti )) = 1 by (b). Thus [O{2,3} (CG (e)), t1 t2 ] = 1. But [O3 (H2 ), t1 t2 ] = O3 (H2 ), so [O{2,3} (CG (e)), O3 (H2 ) e] = 1. However, e O3 (H2 ) contains x by construction, and e ∈ D, so O{2,3} (CG (e)) ∈ NCx (D; {2, 3} ) = {1}. Thus (c) holds, and the proof is complete.  Now set I = {0, 1, 2, 3, 4}. Fix a 5-cycle g ∈ N ∼ = Σ5 and set xi = xg , i ∈ I. For any i, j ∈ I with i = j, set yij = xi xj and eij = x−1 i xj . By the construction in Lemma 6.3, we may assume that e01 = e and y01 is 3-central in L. Conjugating in Σ, eij ∈ eG and yij is 3-central in E(CG (xi )) and E(CG (xj )), for all i, j ∈ I. Set Li = E(CG (xi )) and write i

O3,2 (NG ( eij )) = Kij × Jij

(6B)

with Kij ∼ = Σ3 and Jij ∼ = SU3 (2). Note that Jij is the unique subgroup of CG ( eij ) of its isomorphism type. Since CLi (eij ) ≥ CLi ( xi , xj ) ∼ = GU3 (2), it follows that Jij = O3 (CLi ( xi , xj )) ≤ Li , and by symmetry Jij ≤ Lj . Lemma 6.5. Let i, j, k, be distinct elements of I. Then (a) Kij ≤ Jk ; and (b) Jij , Jik  = Li . Proof. In (a), let tij ∈ I2 (N ) interchange xi and xj , and fix xk and x . Then tij ∈ N acts as a reflection on D with center eij . Given the structure of NG ( eij ), it follows that tij ∈ Kij . As such involutions generate Kij , it suffices to show that tij ∈ Jk . Suppose false. Clearly  tij ∈ O2 (CG (ek )) − Jk . It follows that CG (ek ) ∼ = Z3 × ΓU3 (2) and tk 2 induces a graph automorphism on X := O (ΓU3 (2)) ∼ = GU3 (2). But then |CX (tk )|3 = 3 by [V17 , 11.3.6], so |CD (tk )| ≤ 32 . As tk , like tij , induces a reflection on D, this is a contradiction, so (a) holds.

6. G ∼ = U5 (2)

349

In (b), as noted just before this lemma, Jij = CLi ( xi , xj ) ≤ Li . In particular Jij , Jik  ≤ Li . Moreover, Jij = Jik , for otherwise, [Jij , xi , xj , xk ] = 1, whence every Sylow 2-subgroup of Jij is generated by a reflection and so has order 2, which is absurd. Now Jij and Jik are D-invariant and so Jij (D ∩ Li ) and Jik (D ∩ Li ) are maximal parabolic subgroups of Li ∼ = P Sp4 (3) with no common Sylow 3-subgroup. (A Sylow 3-subgroup Z3 Z3 of Li has a unique normal 31+2 -subgroup.) Therefore Jij , Jik  (D ∩ Li ) =

Jij (D ∩ Li ), Jik (D ∩ Li ) = Li , and (b) follows by the simplicity of Li .  Now set Gi = Ki,i+1 , 0 ≤ i ≤ 3. Also set G01 = J34 , G12 = J04 , G23 = J01 , G012 = L4 , and G123 = L0 . Also put G∗ = G012 , G123  and M = NG (G∗ ). Lemma 6.6. G = G∗ ∼ = U5 (2). Proof. By Lemma 6.5 and (6B), Gi , Gi+1  ≤ Gi,i+1 ∼ = SU3 (2), 0 ≤ i ≤ 2; [G0 , G23 ] = [G01 , G3 ] = 1; and G01 , G12  = G012 ∼ = G12 , G23  = = U4 (2) ∼ G123 . Therefore G∗ ∼ (2), by a theorem of Bennett and Shpectorov U = 5 [BeSh1, Theorem 1.3]. As E(CG∗ (y)) | y ∈ X  = G012 , G123  = G∗ , and N ∩ G∗ is transitive on X , N ≤ NG ( E(CG (y)) | y ∈ X ) = NG ( E(CG∗ (y)) | y ∈ X ) = M. We apply Holt’s theorem [II2 , Theorem SF] to M to show that G = G∗ . Obviously G is not isomorphic to A9 or a simple Bender group, so it is enough to show that M controls G-fusion in T ∈ Syl2 (Cz ), and that Cz ≤ M . As CG∗ (z)/Qz ∼ = GU3 (2), Cz ≤ M unless possibly Cz /Qz ∼ = ΓU3 (2) (Lemma 6.4b). In that case, by a Frattini argument, Cz ≤ CG∗ (z)NG (A), where A = J(CP (z)) = CD (z) ∼ = E33 . As x ∈ A, CG (A) ≤ Cx ≤ G∗ and then NG (A) ≤ CG (A)NG (D) ≤ M by another Frattini argument, as needed. Now T ∈ Syl2 (Cz ) ⊆ Syl2 (G) as z = Z(T ). We show that G and M both have two classes of involutions, from which the desired control of fusion follows trivially. In the following paragraphs, H can be either G, G∗ , or M . Note that CH (x) = Cx , and we have just seen that CM (z) = Cz . Let u ∈ I2 (Qz ) − {z} with u, z  T . We claim that z H ∩ Qz = {z}. Suppose false. Since Qz ∼ = Q8 ∗ Q8 ∗ Q8 and CH (z)/Qz contains GU3 (2), all involutions of Qz − {z} are CH (z)-conjugate, so u ∈ z H . By [IG , 16.21], there is a 3-element g ∈ NH ( u, z) such that ug = z. By a Frattini argument we may take g to normalize CT ( u, z) = CT (u). Write CQz (u) = u × Qu , so that Qu ∼ = Q8 ∗ D8 and z = Z(Qu ). Then R := Qgu ≤ T , uz = Z(R), and z ∈ R. In particular, R ∩ Qz is elementary abelian and m2 (R) ≤ m2 (Qz ) − 1 = 2. Hence Rg /Rg ∩ Qz contains E23 . However, Rg /Rg ∩ Qz ∼ = Rg Qz /Qz embeds in T /Qz and thus in SD16 , g g so m2 (R /R ∩ Qz ) ≤ 2. This contradiction proves our claim. It follows immediately that I2 (Qz ) − {z} ⊆ uH .

350

16. THEOREM C∗4 : STAGE A5. RECOGNITION

Any 2-central involution t of L centralizes an E33 -subgroup of Cx and so is in z G . Moreover by [V17 , 10.9.2], CL (t) is an extension of SL2 (3) ∗ SL2 (3) by a 2-central involution t of L interchanging the two SL2 (3) factors, but with tt being L-conjugate into O2 (CL (t)) − t and hence H-conjugate to u. Since t ∈ O2 (CL (t)), t ∈ O2 (CH (t)). Moreover by [V17 , 10.9.2], F :=  O 3 (CCH (x) ( t, t )) ∼ = Z3 ×L2 (3), while m3 (CCH (z) (v)) = 1 for all involutions v ∈ CH (z) − O22 2 (CH (z)) by [V17 , 11.3.6]. Hence t ∈ O22 2 (CH (t)) − O2 (CH (t)). Moreover, O2 (CL (t)) = CO2 (CH (t)) (x) is the product of two of the Q8 central factors of O2 (CH (t)) interchanged by t , so t normalizes Q := CO2 (CH (t)) (O2 (CL (t))) ∼ = Q8 . Since [t , x] = 1 but [Q, x] = Q, we have   [t , Q] = 1. As t acts freely on O2 (CH (t))/Q, all involutions in t O2 (CH (t)) are O2 (CH (t))-conjugate into t Q and hence into t, t , hence to t or tt . As t ∈ z H and tt ∈ uH , we conclude that I2 (O22 2 (CH (z))) ⊆ z H ∪ uH . If M = G∗ , then T ≤ O22 2 (CH (z)) and our assertion that M and G each have two classes of involutions holds, so the lemma holds. To complete the proof, we assume for a contradiction that M > G∗ , so that M = G∗ γ where γ 2 = 1 and CG∗ (γ) ∼ = Sp4 (2) ∼ = Σ6 . By the Thompson transfer lemma [IG , 15.16], γ has a G-conjugate u ∈ O22 2 (Cz ), and hence in z G ∪uG . As Cz is solvable but CG (γ) is not, u ∈ uG . Hence CG (u), like Σ6 , contains a subgroup B = b1 , b2  ∼ = E32 in which no subgroup of order 3 is weakly closed. On the other hand, u is G-conjugate to a non-2-central involution u∗ ∈ L, which, by [V17 , 10.9.6], can be chosen so that v := CDL (u∗ ) lies in the N -orbit [3] on E1 (D). Then x, v ∈ Syl3 (CCG (u∗ ) (x)), containing two subgroups in the N -orbit [61 ] besides x and v. But X = [1] ∪ [41 ], so

x is weakly closed in x, v with respect to G. In particular, x, v ∈ Syl3 (CG (u∗ )). Therefore x, v and B are conjugate. But no element of E1 (B) is weakly closed in B with respect to G, a final contradiction.  This completes the proof of Proposition 6.1. 7. G ∼ = U6 (2) In this section our main goal is to prove the following:   Proposition 7.1. If N ∼ = Σ6 and |X | = 15, then G = LN ∼ = U6 (2).   We set G0 = LN . Furthermore, we shall also abstract part of the identification of G0 for use elsewhere.

7. G ∼ = U6 (2)

351

For the proof of Proposition 7.1, we continue the notation Cz = CG (z) and Qz = O2 (Cz ) and first prove: Lemma 7.2. The following conditions hold: (a) For some involution t1 , Nx = x, t1  × L ∼ = Σ3 × U4 (2); and 1+8 ∼ ; and C /Q (2). 2 U (b) = 4, i.e., Qz ∼ = + z z = 4 Proof. First, by [V9 , 10.1c2], X = [1] ∪ [8] ∪ [6], and since elements of [6] are inverted by a reflection in NL (DL ), x is inverted by a reflection of N . Therefore Nx contains Σ3 × U4 (2). Since NN ( x) has order |N |/|X |, it is isomorphic to Z2 × Σ4 , and so Nx ∼ = Σ3 × U4 (2), proving (a). Next, let t1 ∈ I2 (Nx ) with [t1 , L] = 1. Then m3 (CG (t1 )) ≥ m3 (L) = 3, so by (4A3), t1 ∈ z G . Let x1 ∈ [6]. If = 3, then as L ∼ = Ω− 6 (2), of index 2 in ∼ Out(Qz ), we must have z ∈ CQz (x1 ) = Q8 . But x1 ∈ X so z is 2-central in L1 := E(CG (x1 )), and thus x1 = xg for some g ∈ G which can be adjusted so that g ∈ Cz . Hence Q8 ∼ = CQz (x) ∼ = Q8 ∗ Q8 , a contradiction. = CQz (x1 ) ∼ Therefore = 4. Now CG ( x, z) = CCx (z) is an extension of Q8 ∗ Q8 by a group of order 2.33 . Hence |CCz /Qz (x)|2 ≤ 2, and so Cz /Qz ∼ = U4 (2) by  [V17 , 10.1.11cb]. Thus (b) holds, and the proof is complete. Lemma 7.3. There is a transitive action of N on a 6-element set such that N ∩ Nx stabilizes a 2-element subset and D is isomorphic to the core of the corresponding permutation module V over F3 . Proof. Now N ∩ Nx contains Z2 ×Σ4 , whose center inverts x. There fore [V17 , 10.8.3] implies (b). We abstract the conclusions of the preceding two lemmas as follows. Definition 7.4. We say that G is of U6 (2)-type with respect to D, z, Nxo , L, and N o if and only if the following conditions hold: ∗ ∗ ∼ (a) z ∈ I2 (G), F ∗ (Cz ) ∼ = 21+8 + , and F (Cz /F (Cz )) = U4 (2); o (b) D = CG (D) ∈ E34 (G), D ≤ N o ≤ NG (D), N := N o /D ∼ = Σ6 , and  O 3 (NG (D)) ≤ N o ; (c) x ∈ D # and L  Nxo ≤ Nx with Nxo ∼ = U4 (2), and = Σ3 × L, L ∼ ∗

x × L = F (Cx ); o (d) There is a transitive action of N on a 6-element set such that o N o ∩ Nxo stabilizes a 2-element subset and D is isomorphic as F3 N module to the core of the corresponding permutation module V . Thus by the preceding two lemmas, G is of U6 (2)-type with respect to D, z, Nx , L, and N . We shall prove: Nxo ,

Proposition 7.5. Suppose that G is of U6 (2)-type with respect to D, z, o o N ∼ L, and N . Then L = U6 (2).

o We assume that an isomorphism N ∼ = Σ6 has been chosen and fixed so o that V has basis B := {v1 , . . . , v6 } permuted naturally by N , and N o ∩ Nxo stabilizes {v1 , v2 }.

352

16. THEOREM C∗4 : STAGE A5. RECOGNITION

Let T ∈ Syl3 (N o ), so that D ≤ T ∈ Syl3 (NG (D)). We may assume that T permutes B as (v1 , v2 , v3 ), (v4 , v5 , v6 ). Then D = J(T ), so T ∈ Syl3 (G). o Moreover, we may identify D with the irreducible F3 N -submodule of V /V0 , where V0 = F3 v0 and v0 = v1 + v2 + · · · + v6 . In particular, x is identified with v1 − v2  + V0 /V0 . Finally, we let β ∈ D be the image of v1 + v2 + v3 modulo V0 , so that Z(T ) = β . Note that by declaring B to be an orthonormal basis in V , we obtain an inherited nondegenerate bilinear form f on D which is preserved by N o . Lemma 7.6. CG (β) has a normal subgroup H1 ∗ H2 with H1 ∼ = H2 ∼ = SU3 (2) and Z(H1 ) = Z(H2 ) = β. Proof. By the structure of L, there is H2 ≤ L with H2  CG ( x, β), Z(H2 ) = β, H2 ∼ = SU3 (2), and T ∩ H2 ∈ Syl3 (H2 ). Let b45 be the commutator of (v4 , v5 , v6 ) ∈ CT ( x, β) with v3 − v4 + V0 ∈ D. Then v4 − v5 + V0 = b45 ∈ R2 := O3 (H2 ) and b45 = xg , where g ∈ N o and −1 g = (v1 , v4 )(v2 , v5 )(v3 , v6 ) ∈ CN o (β) ∩ NN o (T ). Therefore H1 := H2g  g −1 CG ( b45 , β) with H1 ∼ = H2 , Z(H1 ) = β, and R1 = R2 = O3 (H1 ). Thus o R1 = x, ρ and R2 = b45 , σ, where the images of ρ and σ in T ≤ N ∼ = Σ6 are commuting elements of order 3. Let Y1 = O3 (CG (β)); we argue that Y1 = 1. Suppose that Y1 = 1. As m3 (CG (β)) = m3 (D) = 4 > e(G), |Y1 | is odd. Let Q1 ∈ Syl2 (H1 ) and t0 ∈ I2 (Q1 ), so that t0 inverts R1 / β. As x ∈ R1 and CG ( x, β) is a {2, 3}-group, we have CH1 (Y1 ) = β. But Q1 ≤ H1 ≤ CG (b45 ) = Cxg , whence t0 ∈ z G , and so CG ( t0 , β) is also a {2, 3}-group. So t0 inverts Y1 and R1 = [R1 , t0 ] ≤ CH1 (Y1 ), a contradiction. Hence, Y1 = 1, as claimed. By solvable L3 -balance [IG , 13.8], for each i = 1, 2, Ri ≤ O3 (CG (β)) or Ri ≤ Li for some component Li of E(CG (β)). Suppose that R2 ≤ L2 . Since x ∈ R1 and CG ( x, β) is solvable, [L2 , R1 ] = 1. It follows that R1 ≤ L1 = L2 . Now H2 is a solvable 3-component of CCG (β) (x), and Z(H2 ) ≤ Z(L2 ). By [III17 , 10.23], L2 ∼ = SU3 (8). Now β ∈ I3o (G) but SU3 (8) ∈ C3 , contradicting the hypothesis of Part A of Theorem C∗4 . Therefore R2 ≤ O3 (CG (β)), and similarly R1 ≤ O3 (CG (β)). Since |T | = 36 , O3 (CG (β)) = R1 ∗ R2 . Set C β = CG (β)/R1 R2 . We have |C β |3 = 3 and CL (β) ∼ = GU3 (2). It follows that F ∗ (CG (β)) = R1 R2 . Now H1 centralizes b45 ∈ H2 and Z(H 1 ) inverts R1 / β, whence Z(H 1 ) centralizes CR1 R2 (R1 ) = R2 . Similarly H2 centralizes x and Z(H 2 ) inverts R2 / β and centralizes R1 . Indeed since CL (β) ∼ = GU3 (2), O2 (H 2 ) centralizes R1 , and similarly O2 (H 1 ) centralizes R2 . Therefore [O2 (H 2 ), O2 (H 1 )] = 1 and so  [H1 , H2 ] = 1, and the lemma holds.

7. G ∼ = U6 (2)

353

o ∼ Lemma 7.7. Let w ∈ N o and let σ be the image of w in N = Σ6 . Let ∼ o U1 = CNx (L) = Σ3 and let H1 be as in Lemma 7.6. Then the following hold:

(a) If σ({1, 2}) = {1, 2}, then U1w = U1 ; and (b) If σ({1, 2, 3}) = {1, 2, 3}, then H1w = H1 . Proof. Recall that Nxo = x, t1  × L ∼ = Σ3 × L. Thus t1 acts on D as a reflection inverting x = v1 − v2 + V0 and preserving the form f . Thus the o image of t1 in N ∼ = Σ6 is the transposition (1, 2). Now if σ({1, 2}) = {1, 2}, then σ commutes with this image, so w normalizes   D t1  as well as its abelian normal subgroup D. Hence w normalizes tD 1 = U1 , proving (a). As a result, for w ∈ N o , the conjugate U1w is determined uniquely by o the 2-element set w({1, 2}), where as usual we identify N with Σ6 . If w({1, 2}) = {i, j}, we write {1,2}→{i,j}

U1

= U1w ,

notation that is well-defined by (a). As L = E(CG (U1 )), the notation L{1,2}→{i,j} = Lw is also well-defined. {1,2}→{i,i+1} for i = 2, 3, 4,5, and define ti to be an We put Ui = U1 arbitrarily chosen involution of Ui . Thus, Ui = tD i , i = 2, 3, 4, 5, and ti is the transposition (i, i + 1). As x = v1 − v2 + V0 , we have ti ∈ O3 (Cx ) = L for i = 3, 4, 5. Thus since Ui = t D i , U3 , U4 , U5  ≤ L = CG (U1 ). In the same way we obtain (7A)

[Ui , Uj ] = 1 whenever 1 ≤ i < j − 1 ≤ 4.

Now t4 and t5 centralize v1 + v2 + v3 + V0 = β ∈ D. Hence U4 , U5  ≤ g −1 L (β)) = H2 . In the proof of Lemma 7.6, we defined H1 = H2 , where −1 g −1 ∈ CN o (β) and g −1 = (1, 4)(2, 5)(3, 6). Hence H1 contains U4 , U5 g =

U1 , U2 . But U1 , U2  ∩ D contains v1 − v2 , v2 − v3  + V0 /V0 and also covo ers the 3-cycle t1 t2 of N . Hence | U1 , U2  |3 ≥ 33 , whence U1 , U2  =

I2 (H1 ) ≥ O3 (H1 ). Now in (b), w centralizes v1 + v2 + v3 + V0 = β ∈ D. By Lemma 7.6, the only solvable 3-components of CG (β) are H1 and H2 , so H1w = H1 or H2 . On the other hand, since σ({1, 2, 3}) = {1,2, 3},w normalizes the Σ3 o subgroup of N supported on this set, which is t1 , t2 . Hence w normalizes  O 3 (C

t1 , t2 D = U1 , U2 . By the previous paragraph, w normalizes O3 (H1 )  and hence H1w = H1 . The proof is complete.

w , w ∈ N o, We now define U12 = H1 . By Lemma 7.7b, a conjugate U12 o depends only on w({1, 2, 3}) = {i, j, k} (again identifying N and Σ6 ); and {1,2,3}→{i,j,k} w . We set = U12 we write U12 {1,2,3}→{i,i+1,i+2}

Ui,i+1 = U12

Thus Ui , Ui+1  = I2 (Ui,i+1 ), i = 2, 3, 4.

, i = 2, 3, 4.

16. THEOREM C∗4 : STAGE A5. RECOGNITION

354

Now we can use the Phan presentation of SU6 (2) due to Bennett and Shpectorov [BeSh1] to prove:  o Lemma 7.8. Let G0 = LN . Then N o , Nxo  ≤ G0 ∼ = U6 (2). Proof. Let w ∈ N such that w({1, 2, 3}) = {4, 5, 6}, for example w = w , so U w (1, 4)(2, 5)(3, 6). Then [β, w] = 1 and U4 , U5  ≤ U12 45 = U12 = H2 . Hence (7B)

[U12 , U45 ] = 1.

In particular, [U2 , U45 ] = 1. Conjugating by g0 := (6, 5, 4, 3, 2, 1), [U1 , U34 ] = 1. Similarly [U23 , U5 ] = 1. These relations, together with (7A) and (7B), will give us a Phan-type presentation for U6 (2) once we verify the last required relations, namely ∼ U23 , U34  ∼ (7C)

U12 , U23  = = U34 , U45  ∼ = U4 (2). Let X = U34 , U45 . We have just seen that [U1 , U34 ] = 1, and as U1 ≤ U12 , [U1 , U45 ] = 1 by (7B). Thus X ≤ CG (U1 ) = L. Now Z(U45 ) = β, and conjugating by g0 again, Z(U34 ) = β g0  = v3 + v4 + v5 +V0 ∈ CL (β)− β. Hence by [IA , 7.3.2], U34 , U45  = L. Conjugating again by powers of g0 yields (7C).   Set G1 := U12 , U23 , U34 , U45  ≤ Lg0  ≤ G0 . Clearly U1 ≤ U12 ≤ G1 . By the theorem of Bennett and Shpectorov [BeSh1, Theorem 1.3], G1 is a central quotient of SU6 (2). As |G|3 = 36 , G1 ∼ = U6 (2). Clearly D ≤ G1 , and o o ∼ ∼ AutG1 (D) = Σ6 = N , so N ≤ G1 . Hence G0 ≤ G1 , so G0 = G1 , and the lemma is proved.  Lemma 7.8 completes the proof of Proposition 7.5. We return to the proof of Proposition 7.1. The key step in proving G0 = G is to show that G0 is strongly 3-embedded in G, should G0 < G. We first prove the following lemma. Note that U(G; G0 ; 3) is the set of all y ∈ I3 (G0 ) such that whenever g ∈ G and y g ∈ G0 , we have g ∈ G0 [V9 , (11A)]. Lemma 7.9. The following conditions hold: (a) NG (D) ≤ G0 ; (b) Nx ≤ G0 ; (c) β ∈ U(G; G0 ; 3); and (d) Let B = x, β. Then ΓB,1 (G) ≤ G0 . Proof. Parts (a) and (b) hold by Lemma 7.8. Since F ∗ (CG ( β)) = O3 (CG ( β)) and the Hi are solvable components of CG ( β), CG (β)/H1 H2 is 3-closed, so NG ( β) ≤ H1 H2 NNG (β) (T ) ≤ H1 H2 N ≤ G0 , as D = J(T ). Moreover, every 3-element of G0 is G0 -conjugate into D, by [IA , 4.8.2, 4.8.4]. As N ≤ G0 and N controls G-fusion in D, it follows that G0 controls Gfusion of elements of I3 (T ). In particular, (c) holds. Finally, as Z(T ) = β and B  T , we have xβ i ∈ xT , i = 1, 2, so (d) follows from (b) and (c). 

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Now with [IA , 4.8.2, 4.8.4], I3 (G0 ) = xG0 ∪ β G0 ∪ y G0 , where y ∈ G0 corresponds to diag(ω, ω, ω −1, ω −1 , 1, 1). Thus D  NG0 ( y), whence some Sylow 3-subgroup P of NG ( y) lies in NG (D) ≤ G0 , and replacing y by an N -conjugate, we may assume that β ∈ Z(P ). Moreover, by inspection of G0 , P = D w where w ∈ I3 (P ) and w has a free summand on D. Lemma 7.10. NG ( y) ≤ G0 . Proof. Let X = NG ( y). Then B ≤ D ≤ X so O3 (X) ≤ ΓB,1 (G) ≤ G0 by Lemma 7.9. But O3 (CG0 (y)) = 1, so O3 (X) = 1. If X is 3constrained, then O3 (X) contains β ∈ U(G; G0 ; 3) by Lemma 7.9, whence X ≤ NG (O3 (X)) ≤ G0 by [III8 , 6.1a], as required. Therefore we may assume that E(X) = 1, and derive a contradiction. Let J be a component of E(X). By (4A1), J ∈ C3 . As NG0 ( y) is a {2, 3}-group, so is ΓB,1 (X), by Lemma 7.9. In particular J ≤ ΓB,1 (X). This implies that J is B-invariant. Moreover, X0 := NX (Q) | Z(P ) ≤ Q ≤ P  is a {2, 3}-group, since again [III8 , 6.1a] forces X0 ≤ G0 . By the preceding paragraph and [IA , 7.3.4], if J ∈ Chev, then J ∈ Chev(3). But in that case, J must have twisted rank 1 (with a Borel subgroup which is a {2, 3}-group), as otherwise J is generated by parabolic subgroups containing P ∩ J, so J ≤ X0 , which is impossible. The only cases, as m3 (J) ≤ 3, are J ∼ = L2 (9) and U3 (3). Hence m3 (J) = 2, and it is impossible that w have a free summand on D. The upshot is that J ∈ Chev. If J ∈ Spor, then by [V11 , 3.1] and [IA , 7.5.5], J ∼ = M11 or J3 . But in the latter case, P ∩ J has no maximal subgroup which is elementary abelian [IA , 5.3h], contradicting the structure of P . And if J ∼ = M11 , again w has no free summand on D. Thus finally, J ∈ Alt, whence J ∼ = A9 as J ∈ C3 . Now [V9 , (10A)] applies with J and y in the roles of L and x0  there. We conclude from [V9 , 10.1] that | yN | = 15. Let N0 = N ∩ NJ (D). Now computing in J × y, we N  have that | y i (1, 2, 3) 0 | = 3 for i = 0, 1, and 2. These nine subgroups are in yG . For, they have nonsolvable centralizers in CG (y) (containing A6 ), while CG (β, y  ) and CG (x, y  ) are solvable for all y  ∈ I3 (CG (β)) and I3 (Cx ) − x. Moreover, since P < T and Z(P ) = y, β, subgroups of the N  form y i (1, 2, 3)(4, 5, 6)(7, 8, 9) 0 , i = 1, 2, account for 8 further elements  of yN . As 9 + 8 > 15, this is impossible, and the proof is complete. Now as G0 controls G-fusion in I3 (T ) and NG ( w) ≤ G0 for w ∈ {β, x, y}, we have: Lemma 7.11. Either G = G0 or G0 is strongly 3-embedded in G. Now with Lemma 7.2b, |CG0 (z)| = |Cz |, so Cz ≤ G0 . In view of Lemma 7.11 and [V17 , 17.6], we can apply [V9 , 11.2] with G0 in the role of M to conclude that G0 = G. This completes the proof of Proposition 7.1.

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8. G ∼ = D4 (2) Recall that N = NG (D) and N = N/D ∼ = AutG (D). The main result for this section is: Proposition 8.1. Suppose that O3 (N ) ∼ = W (D4 ). Then G ∼ = D4 (2). By Lemma 5.5, d := |N |/|W (D4 )| = 1 or 3, and |X | = 4d. Observe that (5A2d) holds, i.e., Nx ∼ = Σ3 Y Aut(U4 (2)). Thus, (8B) Sx ∼ = Σ3 , and Nx / x ∼ = Aut(L). = Z2 , Nx /L ∼

(8A)

Indeed, since N contains W (D4 ), some involution of N inverts D elementwise. Hence Nx /Cx ∼ = Aut(L), by [V17 , 10.9.4]. = Z2 and Nx /CNx (L) ∼ On the other hand, |N |2 = |W (D4 )|2 = 26 by assumption, with |X |2 = 4 by (8A). Therefore | AutNx (D)|2 = |N |2 /|X |2 = 24 = 2| AutxL (D)|2 , so |Sx | = | AutNx (D) : AutxL (D)|2 = 2. This implies (8B). We proceed as usual in a sequence of lemmas. Set (8C)

NA = NCz (A).

Lemma 8.2. NA permutes X ∩ E1 (A) transitively. Moreover, for any Y ∈ X ∩ E1 (A), we have CQz (Y ) ∼ = CQz ( x) ∼ = Q8 ∗ Q8 .   Proof. Let Y ∈ X ∩ E1 (A), so that for some h ∈ N , Y = xh ∈ E1 (A). Then [Y, z] ≤ [A, z] = 1 so z × A ≤ CG (Y ). As Y is conjugate to x,  it follows from (8B) that z ∈ O 2 (CG (Y )) = LY := E(CG (Y )). Hence m3 (CLY (z)) ≥ m3 (A ∩ LY ) = 2, so z is 2-central in LY . As U4 (2) has a unique class of 2-central involutions, it follows that Y is Cz -conjugate to

x. Now A is weakly closed in a Sylow 3-subgroup of Cz by Lemma 5.3, so Y and x are NCz (A) = NA -conjugate, by [V9 , 1.4]. It remains to show that NA permutes X ∩ E1 (A). But this follows immediately from Lemma 5.2.  Lemma 8.3. The following conditions hold: (a) = 4; (b) Cz /Qz has a normal subgroup of index d isomorphic to Σ3 ×Σ3 ×Σ3 , whose representation on Qz / z is the tensor product of natural 2-dimensional representations of the direct factors Σ3 ∼ = GL2 (2). Moreover, if d = 3, then Cz /Qz ∼ = Σ3 Z3 ; and (c) The 12 subgroups X ≤ D of order 3 in [1] ∪ [3] ∪ [41 ] ∪ [42 ] satisfy NG (X) ∼ = Nx , and are permuted by N in 1 or 3 orbits according as d = 3 or 1; and (d) D is weakly closed in a Sylow 3-subgroup of G.

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Proof. We use the notation [1], [3], [4], [41], [42 ], [6], [61 ], [12] of [V9 , Section 10] for the NCx (D)-orbits on E1 (D). By Lemma 5.5, either X = [1] ∪ [3] or |X | = 12, and in the latter case there is an element of N − Nx of order 3 normalizing a Sylow 3-subgroup Px of Nx , and hence permuting the three elements of E1 (D ∩ Z(Px )) − E1 (D ∩ Z(Px ) ∩ [Px , Px ]). Therefore, one of the following holds: (8D)

(1) d = 1 and X = [1] ∪ [3]; or (2) d = 3 and X = [1] ∪ [3] ∪ [41 ] ∪ [42 ].

We have A = CD (z) = x × CDL (z), with CDL (z) containing one element of [3], one of [6], and two of [4], by [V17 , 10.9.7a]. The elements of [3] and [6] are fixed-point-free on the Frattini quotient of O2 (CL (z)) ∼ = Q8 ∗ Q8 (see [V17 , 10.9.2c]). We set

y = E1 (A ∩ DL ) ∩ [3]. One consequence, as [3] ⊆ X , is that |CQz /z (y)| = 24 ; but C(L∩Qz )/z (y) = 1, so = 4, proving (a). As L ∩ Qz = CQz (x), CQz (xy) = CQz (xy −1 ) = 1. A further consequence is that E1 (A) = E1 ( x × CDL (z)) consists of [1], one element from each of [3] and [6], and two elements from each of [61 ], [12], [4], [41 ] and [42 ]. Notice that the orbits mentioned exhaust E1 (D). In (8D), we conclude that |X ∩ E1 (A)| = 2d. Recall that NA = NCz (A); set NA,x = NA ∩ Nx . Suppose that (8E)

d = 1.

Then X ∩E1 (A) = { x , y}, whence |NA : NA,x | = 2 in view of Lemma 8.2. But as NA,x ≤ Nx , we see from (8B) and [V17 , 10.9.7b] NA,x = CCz (A) i, t where i, t ∼ = E22 , i inverts A, and t acts on A as a reflection with [t, A] ∈ [6]. Then t is the only element of i, t acting as a reflection on A, so NA = CCz (A)U , where U is a 2-group, i, t ≤ U and U/ z is an abelian group of order 8. (Note that z ∈ Syl2 (CG (A))). Since N = W (D4 ), the N -orbits on E1 (D), besides X , include the centers of the twelve reflections in N (one of these lies in [6] by inspection of L, so this orbit is [6] ∪ [61 ]); two frames for D other than [1] ∪ [3] (these are [41 ] of E1 (D), constituting [4] ∪ [12]. and [42 ]); and the remaining 16 elements  ±1 ∈ [61 ], so there is a reflection ρ ∈ N As y ∈ [3], we have xy inverting xy. Since xy is on the diagonal of x × L, xρ = x. On the other hand as xy ≤ x, y ≤ A, ρ normalizes x, y and A. Now

z ∈ Syl2 (CG (A)), so by a Frattini argument ρ is the image of some 2element r ∈ NA with [r2 , A] = 1. It follows that U/ z is elementary abelian and is generated by t, r, and itr, acting as commuting reflections on A. Therefore, NA / z ∼ = Σ3 × Σ3 × Σ3 . The centers of these three

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  reflections are [t, A], xy, and xy −1 . We saw above that xy and xy −1 are fixed-point-free on Qz / z, and shall prove the same for [t, A]. As the two order 3 diagonal subgroups in [t, A] × xy are NA -conjugate, their fixed-point subgroups on Qz are isomorphic. Since CQz /z (xy) = 1, it follows easily that CQz ([t, A]) ∼ = Q8 . Hence if CQz ([t, A]) = z, then (4A4) and Lemma 2.2 imply that CG ([t, A]) has a normal subgroup M with M ∼ = U4 (2), and so there is symmetry between x and [t, A]. In particular z is 2central in M and so CG (x, z) ∼ = CG ([t, A], z). But CAutG (Qz ) ([t, A]) contains a copy of Σ3 ×Σ3 , whereas CAutG (Qz ) (x) does not. This contradiction proves that [t, A] is fixed-point-free on Qz / z. It is then immediate that as a module for NA /CNA (Qz ) ∼ = Σ3 × Σ3 × Σ3 , Qz / z is isomorphic to the tensor product of natural modules for the three direct factors. Moreover, by [V17 , 10.1.8], Cz /Qz ∼ = Σ3 × Σ3 × Σ3 and so (b) holds in this case. Now assume that d = 3. With (8D) it follows that |NA : NA,x | = 6. Again |NA |2 = 23 , and again we find a reflection ρ as above, and NA /CNA (Qz ) contains Σ3 × Σ3 × Σ3 . Then [V17 , 10.1.8] yields (b). Next we prove (c), whether d = 1 or 3. The group N is isomorphic to NG∗ (D ∗ ) for the almost simple group G∗ := D4 (2) γ ∗  , where D ∗ ≤ G∗ , D ∗ ∼ = E34 and γ ∗ ∈ ΓD4 (2) is a graph automorphism of order d [IA , 2.5.10]. Since the orbits [1] ∪ [3], [41 ], [42 ] are fused in the d = 3 case, the twelve subgroups in these orbits have isomorphic normalizers in N . Therefore like Nx , each of these normalizers, call it NG ( x ), contains an involution z  inducing a reflection on D. In particular m3 (CG (z  )) = 3. Hence by (4A3,4) and Lemma 2.2, z  ∼G z and NG ( x ) ∼ = Nx . Thus (c) holds. Furthermore, D ∗ is determined in G∗ up to conjugacy [IA , 4.10.3], and NG∗ (D ∗ ) contains a Sylow 3-subgroup S ∗ of G∗ . Using [IA , 4.9.2] we see that m3 (C[G∗ ,G∗ ] (τ )) = 2 for all τ ∈ I3 (S ∗ ) − [G∗ , G∗ ], so J(S ∗ ) ≤ [G∗ , G∗ ]. Likewise with [IA , 4.8.2], S ∗ ∩ [G∗ , G∗ ] ∼ = Z3 × (Z3 Z3 ), which implies that J(S ∗ ) = D ∗ . Hence D = J(S) for S ∈ Syl3 (N ), whence S ∈ Syl3 (G) and (d) follows, completing the proof of the lemma.  In view of Lemma 8.3c, we define (8F)

X ∗ := [1] ∪ [3] ∪ [41 ] ∪ [42 ].

Using Lemma 8.3d, we see that N controls G-fusion in D [IG , 16.9], and hence this fusion can be read off G∗ = D4 (2) γ ∗ . We conclude: Lemma 8.4. [6] ∪ [61 ] and [4] ∪ [12] are the N -orbits on E1 (D) − X∗ . Moreover, if d = 1, then [1] ∪ [3] is an N -orbit, while if d = 3, then X∗ is an N -orbit. The next lemma gives a different calculus for these N -orbits.

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Lemma 8.5. Let T ∈ Syl2 (Cz ). Then the following conditions hold: (a) T = Qz E, where E ∼ = E24 , E ∈ Syl2 (CN (z)), and Qz ∩ E = z; (b) Set F = ED. Then F = F1 × F2 × F3 × F4 , where Fi ∼ = Σ3 for i = 1, 2, 3, 4, the Fi are all N -conjugate, and NG (F )-conjugate as well; and (c) Write O3 (Fi ) = fi  for each i. Then   (1) X ∗ consists of the twelve subgroups fi fj of D such that 1 ≤ i < j ≤ 4 and  = ±1; ] consists of the four subgroups fi  and the eight sub(2) [6] ∪ [61   groups fi fj fk f , {i, j, k, } = {1, 2, 3, 4}, ,  ,  = ±1; and    (3) [4] ∪ [12] consists of the 16 subgroups fi fj fk , 1 ≤ i < j < k ≤ 4, ,  = ±1.

Proof. Let CA = CG (A). Then as already observed, z ∈ Syl2 (CA ). Since A z is self-centralizing in Cz by Lemma 8.3b, O2 (CA /A) is an abelian group of odd order inverted by z. But D ∈ Syl3 (CA ), as is visible in Cx . It follows that NG (A) ∩ NG (D) contains a 2-group E such that z ∈ Z(E) and E/ z covers AutCz (A) ∼ = E23 . Let f  = [D, z]; then E normalizes f , whence z is not a square in E. Hence E ∼ = E24 and ED has the structure 3 claimed in (b). Since |Cz : Qz |2 = 2 , (a) follows. Writing O3 (Fi ) = fi , E ∩ Fi = zi , and Ai = CD (zi ), for all i, we see by Lemma 5.3 that the pairs (zi , Ai ) are all conjugate in G. Indeed since D = J(Ri ) for some Ri ∈ Syl3 (CG (Ai )), a Frattini-type argument shows that those pairs are all N -conjugate. Moreover, since NG (A) ∩ Cz is the unique subgroup of Cz of its isomorphism type containing A, those pairs are all NG (F )-conjugate. This completes the proof of (b). Then (c1) follows with the help of Lemma 8.3bc. The 16 subgroups in (c3) are visibly NG (F )-conjugate, so they must constitute [4] ∪ [12], as asserted. Then (c2) follows easily, completing the proof of the lemma.  Following Lemma 8.5b we set (8G)

F 4 = F1 F2 F3 , and E 4 = F 4 ∩ E = z1 , z2 , z3  .

We may choose notation so that O3 (F 4 ) = A. It then follows that z4 = z, and Cz = Qz F 4 γ with Qz ∩ F 4 γ = 1. Here γ ∈ Cz is an element of order d in a Sylow 2-normalizer (cf. Lemma 8.3b). Set z0 = z1 z2 z3 z4 . By Lemma 8.5b, z1 ∼G z2 ∼G z3 ∼G z4 and z0 z1 ∼G z0 z2 ∼G z0 z3 ∼G z0 z4 . We next prove Lemma 8.6. Let {i, j, k, } = {1, 2, 3, 4}. Then the following conditions hold: (a) zi zj is not conjugate to zi , but zi zj is conjugate to a non-2-central involution of L3 (CG (x0 )) ∼ = U4 (2) for some x0 ∈ X ∗ ; and

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  (b) fk , f  ∈ Syl3 (CG (zi zj )). Also, f  ∈ fk G , fk f−1 ∈ fk f G , and fk G = fk f G . Proof. We use the transitivity of NG (F ) on {F1 , F2 , F3 , F4 }. Let x0 = fk f . Conjugating F4 to Fi in NG (F ), we see that x0 ∈ X ∗ , x0 is inverted in CF ( zi , zj ), and zi is a 2-central involution in E(CG (x0 )). Similarly, zj is a 2-central involution in E(CG (x0 )). Then by [V17 , 10.9.2d], zi zj is a non-2-central involution as claimed. If zi zj is conjugate to zi , then C1 := CG ( zi zj , x0 ) ∼ = CG ( z, x) =: C2 for some x ∈ I3 (Cz ). But O 2 (C1 /O{2,3} (C1 )) ∼ = Z3 × A4 , whereas O2 (C2 /O{2,3} (C2 )) is the central product of quaternion groups (or of order 2), contradiction. Thus (a) holds. We claim that fk f  and fk  are not conjugate. As  O 2 (CG (fk f )/O2 (CG (fk f ))) ∼ = U4 (2) ∼ = P Sp4 (3),

every involution u ∈ CG (fk f ) satisfies |CG (fk f , u)|2 ≥ 25 . However, z ∈ I2 (CG (fk )) and |CG (fk , z)|2 = |CCz (fk )|2 = 23 < |CG (fk f , u)|2 , proving the claim. Let Fk = fk , f . Clearly [Fk , zi , zj ] = 1. Note that Fk ∩ fk G = { fk  , f }, and the other two elements of E1 (Fk ) lie in Fk ∩ fk f G . By  the previous claim, O3 (NG (Fk )) ≤ CG (Fk ). As shown in the paragraph before last, Fk ∈ Syl3 (CG (zi zj , fk f )), and the first part of (b) follows. Clearly fk f−1 ∈ (fk f )N and f  ∈ fk N , so the lemma follows.  Let Ψ = { fi  | 1 ≤ i ≤ 4} and let NΨ be the (setwise) stabilizer of Ψ in N . We know that Ψ  is an orbit  of NΨ on E1 (D). Using the structure of j k Cz , we see that Ψ1 := { fi fj fk | 1 ≤ i < j < k ≤ 4; j = ±1, k = ±1} is also an NΨ orbit on E1 (D), of cardinality 24 . Let Ψ2 consist of the 8 subgroups in E1 (D) generated by elements f1 f2±1 f3±1 f4±1 . Then E1 (D) is the disjoint union of X ∗ , Ψ, Ψ1 , and Ψ2 , and we deduce with Lemma 8.5c that (8H)

[4] ∪ [12] = Ψ1 and f1 N = [6] ∪ [61 ] = Ψ ∪ Ψ2 .

For any i ∈ {1, 2, 3} set (8I)

xi = fi f4−1 ∈ X ∗ , f0 = f1 f2 f3 f4 , and ui = f0 fi−1 .

Lemma 8.7. Let {i, j, k} = {1, 2, 3}. Then CG (xi ) = xi  × Li , with Li ∼ = U4 (2). Proof. Using Lemma 8.3c, we conclude that CG (xi ) = Z ×Li with Li ∼ = U4 (2) and Z a cyclic 3-group containing xi . Since NG (D) acts irreducibly on D, |Z| = 3. The proof is complete.  Lemma 8.8. Let {i, j, k} = {1, 2, 3}. Then the following conditions hold: (a) f0  ∈ f1 G ; (b) fj , fk , fi f4  = fj , fk , f0  = D ∩ Li ;

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(c) With respect to an appropriate orthogonal basis B of the natural Li ∼ = U4 (2)-module, fj , fk , and f0 are represented by the following diagonal matrices, where ω 3 = 1 = ω: (ω, ω −1 , 1, 1), (1, 1, ω, ω −1 ), and (1, ω, ω −1 , 1); and (d) There exists y ∈ I2 (G), independent of i, such that with respect to the basis B, zj , zk , and some involution of y, f0  ∼ = L2 (2) are represented by permutation matrices corresponding to the following permutations of {1, 2, 3, 4}, respectively: (12), (34), and (23). Moreover, [y, f0 fi−1 ] = [y, f0 fj−1 ] = [y, f0 fk−1 ] = 1. Proof. Part (a) holds by (8H), since f0  ∈ Ψ2 . Next, zi z4 inverts

fi , f4  and centralizes fj , fk . In particular zi z4 ∈ NG ( xi ), so by Lemma 8.7, fj , fk  ≤ CO3 (NG (xi )) (zi z4 ) ≤ Li . Also if f ∈ Li for f = fi or f = f4 , then f ∈ [6] by (8H), so f is real in Li by [V17 , 10.9.4]. Therefore f fi f4−1 ∼G f −1 fi f4−1 . This says that some element of X ∗ is conjugate to fi  or f4 , contradiction. Hence neither fi nor f4 lies in Li , whence

fi , f4  ∩ Li = fi f4 . As f0 = (fi f4 )fj fk , (b) follows. In (c), fi f4  ∈ X ∗ ∩ E1 (D ∩ L) = [3], so diag(ω −1 , ω −1 , ω, ω) represents fi f4 with respect to a suitable basis. Then fj  and fk  are in f G ∩E1 (D ∩ L) = [6] and commute with fi f4 , so without loss fj and fk are represented by diag(ω, ω −1 , 1, 1) and diag(1, 1, ω, ω −1), respectively. Hence f0 = fj fk fi f4 is represented by diag(1, ω, ω −1 , 1), proving (c). Then zj is 2-central in Li , inverts fj , and centralizes fk , so without loss zj , and similarly zk , are represented by the permutation matrices (12) and (34). Let y ∈ I2 (Li ) be represented by the permutation matrix (23). Then [y, xj ] = [y, fj f4−1 ] = [y, fj x−1 i fi f4 ] = [y, fj fi f4 ] = 1 and similarly [y, xk ] = [y, fk fi f4 ] = 1. Now y centralizes xi , xj , xk , the trace-0 hyperplane of D with respect to the basis  {fi }4i=1 of D, and y inverts f0 . This proves (d). Now for {i, j, k} = {1, 2, 3}, define (8J)

Ki = fi , zi  and K0 = f0 , y,

where y is as in Lemma 8.8d. Thus K0 , K1 , K2 , K3 are all isomorphic to L2 (2). Set  Uik = O2 (CLi (uj )) (see (8I)). Lemma 8.9. Let {i, j, k} = {1, 2, 3}. Then the following conditions hold: (a) [Ki , Kj ] = 1; (b) Uik = Uki ∼ = SU3 (2); (c) K0 , Kj  ≤ Uik ; and (d) Uij , Uik  = Li . Proof. Part (a) is trivial; see Lemma 8.5. In the notation of the previous proof, uj = fi f4 fk ∈ Li is represented as diag(ω −1 , ω −1 , ω −1 , 1), so = fi f4 fk fi−1 f4 = xk Uik ∼ = SU3 (2). Moreover, Uik is centralized by uj x−1 i

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so Uik ≤ O2 (CG (xk )) = Lk . Then Uik ≤ O3 (CLk (uj )) = Uki , so (b) holds by symmetry. By Lemma 8.8d, K0 = y, f0  ≤ Uik , and as [uj , zj ] = 1 and  fj ∈ Li , Kj ≤ Uik , which is (c). Finally, Uij = O2 (CLi (uk )) with uk =  fi f4 fj ∈ Li represented as (1, ω, ω, ω), so (d) follows from [IA , 7.3.2]. Set G0 = U12 , U13 , U23  ≥ K0 , K1 , K2 , K3  . By Lemma 8.9, [GHNSh, Theorem B] applies and yields (8K)

G0 ∼ = D4 (2).

Now set M = NG (G0 ). To complete the proof of Proposition 8.1 we show Lemma 8.10. The following conditions hold: (a) M = G0 ; and (b) NG ( v) ≤ M for all v ∈ I3 (G0 ). Proof. For every x ∈ X ∗ , Nx ∼ = NM ( x) so (b) holds for v ∈ X ∗ ; and CG (M ) = Z(M ) = 1. As Out(G0 ) ∼ = Σ3 and T ∈ Syl2 (G) ∩ Syl2 (M ), M = G0 g where g = 1 or g induces a graph automorphism of G0 of order 3. Also N = NG (D) permutes X ∗ , so N normalizes E(CG (v)) | v ∈ X ∗  =

E(CM (v)) | v ∈ X ∗  = U12 , U13 , U23  = G0 . Thus N ≤ M . Let P ∈ Syl3 (M ) with D ≤ P . Using [IA , 4.9.2, 4.10.3] we see that D = J(P ). Hence NG (P ) ≤ N ≤ M and P ∈ Syl3 (G). We claim next that O3 (CG (v)) = 1 for all v ∈ D # . Indeed, G is balanced with respect to D. This is by [IG , 20.6] and because Lo3 (G) ⊆ C3 , and so by [IA , 7.7.9] the only possible locally unbalancing component of CG (v)/O3 (CG (v)) would be L2 (33 ), which is impossible as D = J(P ) and N has no element of order 13. Therefore by the Signalizer Functor theorem, N normalizes the 3 -group Θ1 (G; D), and CΘ1 (G;D) (v) = O3 (CG (v)) for all v ∈ D # . Taking v ∈ X ∗ , we conclude that v is fixed-point-free on Θ1 (G; D). Now D has a subgroup V ∼ = E32 (generated by x and an element of [4]) in which there is a unique element v0  of E1 (V ) − X ∗ . It follows that [Θ1 (G; D), v0 ] = 1. But N acts irreducibly on D, so Θ1 (G; D) ≤ O3 (CG (D)) = 1, proving the claim. Set C1 = CG (u1 ) and C1z = CC1 (z1 ). Then z1 ∈ C1 and T1 := O 3 (C1z ) = O3 (CCG (z1 ) (u1 )) ∼ = Q8 so T1 ∈ Syl2 (C1 ) and C1 = O2 (C1 )C1z . As O3 (C1 ) = 1, F ∗ (C1 ) = O3 (C1 ). By the way T1 acts on O3 (CG0 (u1 )), every chief factor of C1 within F ∗ (C1 ) has order at most 32 . Therefore C1 is a {2, 3}-group. As P ∈ Syl3 (G) and u1 is 3-central in M , C1 ≤ M and then NG ( u1 ) ≤ C1 N ≤ M . Now N has two orbits on E1 (D) − X ∗ , namely f4 N and u1 N . Thus, to complete the proof of (b), it remains to show that NG ( f4 ) ≤ M . Again as N ≤ M it suffices to show that C4 := CG (f4 ) ≤ M . We have S :=

z1 , z2 , z3  ≤ C4 with S0 := z1 z2 , z2 z3  ∼ = E22 disjoint from z G ∩ S, which

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in turn contains z1 , z2 , z3 , and possibly z1 z2 z3 (Lemma 8.6). We argue that S ∈ Syl2 (C4 ).

(8L)

Otherwise, there is u ∈ NC4 (S) − S with u2 ∈ S. Then there exist 1 ≤ i < j ≤ 3 such that ziu = zi or zj . If ziu = zj , then as u2 ∈ S, u normalizes CG ( zi , f4 )CG ( zj , f4 ) = F . Hence zku = zk where {i, j, k} = {1, 2, 3}. So without loss we may assume that ziu = zi , whence u ∈ CG ( zi , f4 ) ≤ F .  Hence u ∈ O 2 (NF (S) ∩ CG (f4 )) = S, contradiction. This proves (8L). Given the fusion information about elements of S # it follows from Burnside’s transfer theorem [IG , 15.12(i)] and [IA , 5.6.4] that in C 4 := C4 /O2 (C4 ), O 2 (C 4 ) is trivial or isomorphic to an odd order extension of L2 (q) for some q ≡ ±5 (mod 8) (possibly q = 3). If E(C4 ) = 1, then since Lo3 (G) ⊆ C3 and m3 (G) = 4, E(C4 ) ∼ = L2 (33 ); but this is impossible as 13 does not divide |N |. So Q4 := F ∗ (C4 ) = O3 (C4 ). Note that D ∈ Syl3 (C4 ∩ G0 ). If D ≤ Q4 , then as D = J(P ), C4 ≤ N ≤ M , so suppose that D ≤ Q4 . Then for D ≤ P4 ∈ Syl3 (C4 ), P4 = D g ∼ = (Z3 Z3 )×Z3 with g 3 = 1 and g ∈ M −G0 . Then P4 = Q4 D and D acts quadratically on Q4 /Φ(Q4 ), with |Q4 /Φ(Q4 )| ≤ |CP4 (g)| = 33 . Hence the image of C4 /Q4 in Aut(Q4 /Φ(Q4 )) ≤ GL3 (3) is a {2, 3}-group. Then C4 = SP4 ≤ M , and (b) is proved. Finally we prove (a) by a transfer argument. Suppose false; then g ∈ I3 (M − G0 ) can be chosen so that CM (g) = g × Mg , with Mg ∼ = G2 (2) and u1 3-central in Mg . Let R ∈ Syl3 (CCM (g) (u1 )); then u1  = Z(R ∩ Mg ) and R ∈ Syl3 (CM (g)). Expand R to R∗ ∈ Syl3 (CG (g)) and let R0 = NR∗ (R). Then u1 ∈ Z(R0 ) so R0 ≤ CM (g) by (b). Thus R = R0 ∈ Syl3 (CG (g)). Since |P | > |R|, NP (R) shears g to Z(R) ∩ [R, R] = u1 . But g ∈ [R, R] so u1  is weakly closed in Z(R). Hence by [III2 , 6.1b], g ∈ [NG ( u1 ), NG ( u1 )] ≤  [M, M ] = G0 , a final contradiction. Since NG (P ) ≤ NG (D) ≤ M , it follows from (8K) and Lemma 8.10 that (8M)

M∼ = D4 (2) is strongly 3-embedded in G.

Lemma 8.11. Cz ≤ M . Proof. Let C = Cz and C0 = CM (z). Then O2 (C) = Qz = O2 (C0 ) and we put C = C/Qz . We have C 0 ∼ = Σ3 × Σ3 × Σ3 . Let P ∈ Syl3 (C0 ). By  Lemma 8.10b, ΓP,1 (C) ≤ C0 , and in particular P ∈ Syl3 (C). Lemma 8.10 also rules out all possibilities for the structure of C from [V17 , 10.1.5], except that F ∗ (C) = P and |C : C 0 | divides 4. Since any 2-element acting on P normalizes a subgroup of order 3, C = C 0 by Lemma 8.10b, completing the proof.  Lemma 8.11, [V17 , 17.6], and (8M) allow us to apply [V9 , 11.2] to M and conclude that G = M ∼ = D4 (2). This completes the proof of Proposition 8.1.

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9. G ∼ = Co2 In this section we consider the cases of Lemma 5.5 left untreated by Propositions 6.1, 8.1, and 7.1, namely N ∼ = GO (D),  = ±1. We wish to prove that G ∼ = Co2 . For the most part we follow Fredrick Smith’s treatment [SmF]. Our plan for recognition problems of several sporadic target groups G∗ has been to establish that G ≈ G∗ , i.e., that G has the same centralizer of involution pattern as G∗ . This means, by definition, that there is an isomorphism φ : T → T ∗ between Sylow 2-subgroups T and T ∗ of G and Co2 , respectively; for all involutions z1 , z2 ∈ T , z1 ∼G z2 if and only if φ(z1 ) ∼Co2 φ(z2 ); moreover, CG (z1 ) ∼ = CCo2 (φ(z1 )). One of our assumed Background Results asserts that if G ≈ G∗ ∈ Spor, then G ∼ = G∗ . However, proving that G ≈ Co2 runs into difficult details. We shall establish a closely related condition which we call G ≈o Co2 , which is weaker in one way (the structure of CG (u)) but stronger in other ways; in particular, it includes the condition that |G| = |Co2 |. We add to our Background Results the assertion that (9A)

G ≈o Co2 =⇒ G ∼ = Co2 .

(Portions of the paper [SmF] establish this implication.) Specifically, Co2 has three classes of involutions; let z ∗ , u∗ , y ∗ ∈ I2 (Co2 ) represent these classes, with z ∗ 2-central in Co2 , and u∗ half 2-central in Co2 and CCo2 (z ∗ ). Then we define: Definition 9.1. G ≈o Co2 if and only if G has 3 classes of involutions, represented by z, u, and y, such that (a) There is an isomorphism φz : CG (z) → CCo2 (z ∗ ); (b) There is an isomorphism φy : CG (y) → CCo2 (y ∗ ); (c) Both φz and φy take all elements of z G , uG , and y G in their domains into (z ∗ )Co2 , (u∗ )Co2 , and (y ∗ )Co2 , respectively; (d) Qu := O2 (CG (u)) = Z0 × Q1 with E24 ∼ = Z0  CG (u), u ∈ Q1 ∼ = 1+6 ∼ 2+ , and CG (u)/Qu = A8 . Moreover, Z0 and Qu /Z0 u are, respectively, natural modules for CG (u)/Qu ∼ = L4 (2) and CG (u)/Qu + ∼ (2), respectively. Finally u is half 2-central in CG (z) and Ω = 6 ∗ φz (Qu ) = O2 (CCo2 (u )); and (e) |G| = |Co2 |. The extension problem for Qu by CG (u)/Qu is troublesome, but we are able to side-step it to do the calculations necessary to apply the Thompson order formula to obtain |G|. Our proof also requires us to add two properties of Co2 to our Background Results. These are well-known properties of Co2 that should have been included in [IA , 5.3k], with a reference to [CCPNW1]. Hence our

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additional Background Results are the following: (1) If G ≈o Co2 , then G ∼ = Co2 ; ∗ (2) Let T ∈ Syl2 (Co2 ). Then J(T ∗ ) ∼ = E210 and NCo2 (J(T ∗ )) splits over J(T ∗ ); and (9B) (3) With T ∗ as in (2), CCo2 (Z(T ∗ )) splits over O2 (CCo2 (Z(T ∗ ))). With these added Background Results in hand, we shall prove: Proposition 9.2. If N ∼ = GO  (D),  = ±1, then G ∼ = Co2 and  = −1. For the proof we assume that N = GO (D),  = ±1.

(9C)

and, in view of our Background Results, we must prove that  = −1 and G ≈o Co2 . According as  = +1 or −1, |X | = 24 or 30 (see Lemma 5.5). In either case |NN ( x)| = |N |/|X | = 25 · 3, so NN ( x) ∼ = E22 × Σ4 . This implies immediately that (9D) Nx ∼ = Σ3 × Aut(U4 (2)). Let z0 ∈ Nx be an involution such that z0 lies in the Σ3 direct factor in (9D). Set Cz0 = CG (z0 ), Qz0 := O2 (Cz0 ), and C z0 = Cz0 /Qz0 . Then Cz0 covers L and so, by (4A3), z0 is 2-central in G and C z0 involves U4 (2). On the other hand there is z  ∈ I2 (Cx ) such that CN (z  ) ∼ = Σ3 × Z2 × Σ6 . x

z

By (4A3) again, ∈ Therefore C z0 involves Z3 × A6 , and so is not embeddable in O6− (2). Hence Qz0 ∼ = 21+8 + , and C z0 is isomorphic to a subgroup of O8 (2) involving U4 (2) and with an element of order 3 whose centralizer has a 3-component of type A6 . Also m3 (C z0 ) = 3. The only possibility, by [V17 , 10.1.11b], is C z0 ∼ = Sp6 (2).  Moreover, z , x is G-conjugate to a group z0 , x1  with x1 ∈ Cz0 mapping onto an element of a long root L2 (2) subgroup of Sp6 (2), and z0G .

CQz0 (x1 ) = z0  , by the structure of CCz0 (x1 ) ∼ = CCx (z  ). Let x2 , x3 ∈ I3 (Cz0 ) be such that E := x1 , x2 , x3  ∼ = E33 and for the natural 6-dimensional F2 C z0 -module Vnat , dim[Vnat , xi ] = 2i, i = 1, 2, 3. Factoring Qz0 under various hyperplanes of E we find that CQ (x2 ) ∼ = Q8 ∗ Q8 and CQ (x3 ) ∼ = Q8 . z0

z0

Consequently Qz0 / z0  is uniquely determined as a module (namely, the spin module) for C z0 [V17 , 10.2.1]. Notice that L contains a Sylow 3subgroup of Cz0 , and so 3-central elements of Cz0 are 3-central in G. On the other hand both z  and a 2-central involution of L have centralizers of

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3-rank 3, so they are conjugate to z0 , and so Cz0 has at least two conjugacy classes of subgroups of order 3 in xG . Hence (9E)

All non-3-central subgroups of Cz0 of order 3 lie in xG .

Now fix a 2-central involution z of L. This is the involution on which we shall focus. We continue to let Cz := CG (z), Qz := O2 (Cz ), and C z := Cz /Qz . Note that O3,2,3 (CG (z, x)) ∼ = Z3 × (SL2 (3) ∗ SL2 (3)). In particular z and z0 are conjugate by (4A3). Hence from the preceding paragraphs we deduce Lemma 9.3. Under the assumption (9C), |G|2 = 218 , C z ∼ = Sp6 (2), Qz ∼ = 1+8 2+ , and Qz / z is the spin module for C z . All non-3-central elements of I3 (Cz ) are in xG . Finally, x ∈ C z has 2-dimensional fixed points on the natural 6-dimensional F2 C z -module. Our next major result is: Proposition 9.4. There exist involutions z, u, y ∈ G and involutions z ∗ , u∗ , y ∗ ∈ G∗ = Co2 such that (a) I2 (G) = z G ∪ uG ∪ y G , with z 2-central in G and u half 2-central in G and CG (z); (b) There is an isomorphism φz : CG (z) ∼ = CG∗ (z ∗ ) carrying z, u, ∗ z G ∩CG (z), uG ∩CG (z), and y G ∩CG (z) to z ∗ , u∗ , (z ∗ )G ∩CG∗ (z ∗ ), ∗ ∗ (u∗ )G ∩ CG∗ (z ∗ ), and (y ∗ )G ∩ CG∗ (z ∗ ), respectively; (we call the isomorphism “involution-fusion-preserving”; (c) CG (u) is an extension of Qu = Z0 × Q1 (where E24 ∼ = Z0  CG (u) ∼ ∼ ) by L (2) Aut(Z ) Ω(Z Q /Z and u ∈ Q1 ∼ = = = 21+6 4 0 0 1 0 u); more+ over, φz (Qu ) = O2 (CCo2 (u∗ )); and (d) There exists an involution-fusion-preserving isomorphism CG (y) ∼ = CG∗ (y ∗ ). Let z ∗ ∈ Co2 be any 2-central involution. Using the facts that CG (z  x)(∞) ∼ = A6 and that the Schur multiplier of A6 has a Sylow 2-subgroup of order 2, we may apply [IG , 31.12] to deduce the following. Lemma 9.5. Cz ∼ = CCo2 (z ∗ ). In particular we fix T ∈ Syl2 (Cz ) and T ∗ ∈ Syl2 (CCo2 (z ∗ )) and conclude that ∼ T ∗. (9F) T = We write Z2 (T ) = z, u and Z2 (T ∗ ) = z ∗ , u∗ , where u and u∗ correspond under the isomorphism (9F). In view of Lemma 2.2, O{2,3} (Cx ) = 1, and so ∼ Σ3 × Aut(U4 (2)). (9G) Nx =

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Furthermore, x1 and x2 represent the Cz0 -conjugacy classes in xG ∩ Cz0 ,  since x3 ∈ [CCz0 (x3 ), CCz0 (x3 )] cannot be conjugate to x. As O2 (Cx ) ∼ = Aut(U4 (2)), this quickly implies the next lemma, which introduces the involutions u, y ∈ Cx . Lemma 9.6. z G ∩ Cx is the union of two Cx -classes, represented by z and z  . There are two other Cx -classes in I2 (Cx ). These are represented by u ∈ O2 (CL (z)) − z and y ∈ L, with CN (u) ∼ = Σ3 × D8 × Σ4 and x

CNx (y) ∼ = Σ3 × E22 × Σ4 .

Moreover, u ∈ Qz . By [V17 , 10.2.2], we obtain further: Lemma 9.7. The following conditions hold: (a) I2 (Qz ) − {z} = uCz ; and (b) CG ( u, z)/CQz (u) ∼ = E26 L3 (2), a split 2-constrained extension with 2-chief factors of order 8, and CQz (u) ∼ = Z2 × 21+6 + ; 17 (c) |Cu |2 = 2 ; and (d) F ∗ (CG ( u, z)) = O2 (CG ( u, z)). Next, we consider the structure of Cu := CG (u). We may assume that T and u were chosen so that u, z  T ≤ Cz . Then Tu := CT (u) ∈ Syl2 (Cu ). We let   Qu = O2 (Cu ) and Z = z Cu . We next prove Lemma 9.8. The following conditions hold: (a) Z = Ω1 (Z(Qu )) ∼ = E25 ; (b) Z = Z0 × u for some Z0  Cu such that Z0# ⊆ uG and Z0 u−{u} ⊆ zG; (c) Qu = Z0 × Q1 for some Q1 ∼ = 21+6 + ; and + ∼ (d) Cu /Qu ∼ = L4 (2) = Ω6 (2), and the F2 Cu /Qu -modules Z0 and Qu /Z are, respectively, a natural L4 (2)-module and the natural Ω+ 6 -module. Proof. We have CNx (u) ∼ = Σ3 × D8 × Σ4 and in particular z ∈ Z(Cu ). However, using Lemma 9.7b, we compute that |CG (u, z, x)|2 ≥ 26 , so z is 2-central not only in Cu but also in CCx (u), whence z ∈ O2 (CCx (u)). Indeed since L ∼ = P Sp4 (3) has no four-subgroup all of whose involutions are 2-central, z is on the D8 -Σ4 diagonal, so [z, g] = 1 for some g ∈ I3 (Cu ∩ Cx ). Before proceeding we let R ∈ Syl3 (Cu ∩ Cx ), so that R ∼ = E32 , and note that (9H)

R ∈ Syl3 (Cu ).

Indeed, by [V17 , 10.9.6], E1 (R) contains [1], [3], and two elements of [61 ], in the notation of [V9 , Section 10]. Hence by [V9 , Corollary 10.4], according as

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 = + or −, all four elements of E1 (R) are in xG , or two are in xG and the other two are conjugate to each other but not to x. Now if R ∈ Syl3 (Cu ), then in either case R would contain some x ∈ xG lying in the commutator subgroup of some Sylow 3-subgroup of CG (x ). But this is impossible by the structure of Cx . Hence (9H) holds. Now, the structure of M := CG ( u, z) is given in Lemma 9.7b. Suppose that E(Cu ) = 1 and let E be any component of Cu . As E(CG ( u, z)) = 1, [E, z] = E. Likewise as z ∈ O2 (CG ( u, x), it follows from [V17 , 9.13] that E/Z(E) ∼ = M12 , M22 , HJ, or J4 . In every case the structure of M = CG ( u, z) is contradicted, having no subnormal subgroup isomorphic to CE (z) (see [IA , 5.3bcgi]). Thus as G is of even type, F ∗ (Cu ) = Qu = O2 (Cu ) and E23 ∼ = z, z g , u ≤ Z ≤ Ω1 (Z(Qu )),     the last since z, z g , u ≤ z g ≤ z Cu = Z. An element s ∈ CG (u, z) of order 7 satisfies CQz (s) ∼ = D8 . Also, since z is 2-central in Cu , Qu ≤ O2 (M ). These facts and Lemma 9.7b imply that CQu (s) = u, z. As Z > u, z, |Z| ≥ 25 and |Qu | = 2m , m ≡ 2 mod 3, m ≤ 14. We will eventually show that (9I)

Cu /Qu ∼ = L4 (2) and m = 11.

As 25 |L5 (2)|2 < |Cu |2 and Qu = F ∗ (Cu ), we have m = 5. Moreover, as Z ≤ Z(Qu ), Qu ≤ CO2 (M ) (Z). Whether Z ≤ Qz or not, [CQz (u), Z] = 1 as no element of M induces a transvection on Qz / z (by the perfectness of M Qz /Qz ). Therefore z, u ≤ CQz (Z) < Qz , so m = 14. u = E(C u ). Thus M ≤C u , |C u : M | is odd, and u = Cu /Qu and E Set C   M is a perfect extension of O2 (M ) by L3 (2), with all (1 or 2) chief factors ) of order 8. In particular M  contains a Frobenius group of order in O2 (M 11 8.7. On the other hand, as |Qu | ≤ 2 , mr (Aut(Qu )) < 7 for all odd primes  centralizes F (C u ) and normalizes every component of r. It follows that M u . Then M  induces inner automorphisms on every component of Eu , so E u is quasisimple. Since O2 (C u ) = 1 and C  (x) ∼   M ≤ Eu and E = Z3 ×Σ3 ×H Cu  )),  with H a 2-group and x  3-central, Eu is simple. As M = N  (O2 (M Cu

u = F ∗ (C u ). By [V17 , 20.12], C u = E u and (9I) holds. E u on Qu / u has By the action of R, every noncyclic chief factor of E 4 order at least 2 . Since Z ≥ u, z = CQu (s) for some s ∈ I7 (Cu ), and Z  Cu , we have Z ∼ = E2a , a = 5, 8, or 11. By (9F) and (9B1), J(T ) ∼ = E210 , ∼ so a = 11. Suppose that a = 8. Then Qu /Z = E23 so [Qu , Cu ] ≤ Z. But then |CQu (s)| ≥ 25 , a contradiction. So a = 5. The same analysis applied to Ω1 (Z(Qu )) shows that Z = Ω1 (Z(Qu )). u ∼ Since |Z/ u | = 24 , Z/ u is a natural C = L4 (2)-module. Now x)) = 1, so CQu (x) = O2 (CG ( x, u)) ∼ O2 (CCu ( = D8 × E22 and CCu (x) u ∼ is a {2, 3}-group. Thus x is a product of two 3-cycles in C = A8 . By [V17 , 20.16], Z = u × Z0 for some Z0  Cu . Then [CZ (x), R] = CZ0 (x), so

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u is CZ0 (x) = O2 (O2 (Cx ∩ Cu )) ≤ L. It follows that CZ0 (x)# ⊆ uG . But C # # # G G transitive on Z0 , so Z0 ⊆ u and uZ0 ⊆ z . We have proved (a) and (b). u (which has no nonAs CQu (s) = u, z ≤ Z, Qu /Z is irreducible for C trivial representation of dimension less than 4). Also u = [CQu (x), CQu (x)] so since u ∈ Z0 , Qu /Z0 is nonabelian. Hence it is extraspecial. Then Qu /Z u ∼ is a natural module for C = Ω+ 6 (2).  Write R = v, v  with v, v  ∈ xCu . Then v and v are disjoint 3-cycles u ∼ in C = CQu /Z (v) ∼ = E24 , and similarly for v  , = A8 . We have CQu /u (v) ∼  and CQu (v)CQu (v )Z = Qu . So ZCQu (x)/ u ∼ = E28 ∼ = ZCQu (x )/ u and Z/ u < ZCQu (R)/ u ≤ Ω1 (Z(Qu / u)). By the irreducibility of Qu /Z, Qu / u is elementary abelian. Hence (c) and (d) hold and the proof is complete.  We turn briefly to the 3-local structure of G. Now write R = v, x u ∼ with x ∼Cu v, so that in C = A8 , v is a 3-cycle, with CZ0 (v) = 1. Set Cv = CG (v). We next prove: Lemma 9.9. The following conditions hold: (a) CG (v) = Pv Hv , where F ∗ (CG (v)) = Pv ∼ = 31+4 , Hv = O2 (Hv )Av , ∼ with F ∗ (Hv ) = O2 (Hv ) ∼ = 21+4 − , Av = SL2 (5), and I2 (Hv ) ⊆ O2 (Hv ); and (b)  = −, i.e., N ∼ = GO − (D). Proof. We have CCv (u) = CCu (v) = v × Hv , where F ∗ (Hv ) = ∼ O2 (Hv ) = CQu (v) ∼ = CQu /Z0 (v) ∼ = 21+4 − , and Hv /O2 (Hv ) = A5 acts faithfully on O2 (Hv ). Thus O2 (Hv )/ u is the core of the natural A5 -permutation module, hence is a projective module, and even free for a Sylow 2-subgroup. Hence, Hv = O2 (Hv )Av where Av ∼ = A5 or SL2 (5), and all involutions of Hv − O2 (Hv ) are Hv -conjugate modulo u. As |Hv |2 = 27 , U4 (2) contains no copy of Hv , so v ∈ xG . By [V9 , Corollary 10.4], N has just two orbits on E1 (D), and x, v is Cx -conjugate into D, so v is 3-central in G. By [V17 , 20.13], either u ∈ Z ∗ (Cv ) or Cv /O2 (Cv ) ∼ = HJ or J3 . But ∼ CCv (x) = CCx (v) = Z3 × GU3 (2). This is inconsistent with the 3-structure of HJ and J3 [IA , 5.3gh], so u ∈ Z ∗ (Cv ). Let Wv = O2 (Cv ), Pv ∈ Syl3 (Wv ), and Wx = O2 (CG ( v, x)). By the previous paragraph, |CG ( v, x) : Wx |3 = 3. Also |P : Pv | = 3 for some P ∈ Syl3 (Cv ). Hence Pv contains a Sylow 3-subgroup of O2 (CG ( v, x)). In particular, x ∈ Pv . Now CWv (u) = v, so Wv / v is abelian. But with (9G), Wx is a 3-group, so Wv = Pv . By [IG , 9.16], |Pv / v | ≤ |CPv (x)| ≤ 34 . But CAv (Pv ) = 1, so |Pv | = 35 ; indeed CPv (x) contains 31+2 , so Pv ∼ = 1+4 3 . Hence Hv is isomorphic to a maximal proper 2-local subgroup of [Out(Pv ), Out(Pv )] ∼ = Sp4 (3). As m2 (Sp4 (3)) = 2, Av ∼ = SL2 (5) and no involution of Hv /O2 (Hv ) pulls back to an involution of Hv . Clearly Pv = F ∗ (Cv ), so (a) holds.

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It follows that P ∈ Syl3 (G) and that J(P ) ∼ = E34 . Hence J(P ) is conjugate to D. Since v is inverted in G (in Cu ) and |NCv (P )|2 = 24 we  must have |NN (P )|2 = 25 . This implies (b), so the lemma is proved. Lemma 9.10. The following conditions hold: (a) I2 (G) is the disjoint union of z G , uG , and y G ; (b) Cy := CG (y) has a unique class of elements of order 3, represented by x; and (c) If z0 ∈ I2 (CG (v)) and z1 is not 2-central in CG (v), then z0 ∈ z G . Proof. We have seen that I2 (Qz ) ⊆ {z} ∪ uG , so for (a), it remains to consider involutions t ∈ Cz − Qz . By [V17 , 10.2.3], and the Baer-Suauki theorem, t inverts some w ∈ I3 (Cz ) such that w ∈ C z ∼ = Sp6 (2) is not 3-central in C z . By (9E), w ∈ xG , so a conjugate t of t lies in Nx . In turn, t centralizes an element w of I3 (L), a conjugate of which lies in D and hence in xG ∪ v G , by [V9 , Corollary 10.4]. So t centralizes an element of xG ∪ v G . We count G-conjugacy classes of cyclic subgroups of order 6 containing an element of xG ∪ v G . In Cx there are four classes of involutions [IA , 4.5.1]: two inner, represented by z and u, and two outer, represented by z1 and y, say, where z1 ∈ z G , CL (z1 ) ∼ = Σ6 , and CL (y) ∼ = Z2 × Σ4 . In Cv there are two classes of involutions by Lemma 9.9, for a total of 6 classes of Z6 ’s. On the other hand in Cz there are three classes of subgroups of order 3, two conjugate to x by (9E) and the other central in a GU3 (2) subgroup of Cz and hence conjugate to v by the structure of Cv . Likewise I3 (Cu ) has two classes, represented by v and x. This gives only five Z6 classes, and the  missing one must be yxG . This implies all parts of the lemma. Continuing this argument, we let Ry ∈ Syl3 (Cy ) with x ∈ Ry , and Qy = O2 (Cy ), and next prove Lemma 9.11. The following conditions hold: (a) Ry ∼ = Ry × E22 , and AutCy (Ry ) ∼ = SD16 ; and = E32 , CCy (Ry ) ∼ ∗ ∼ (b) Qy = F (Cy ) = E29 or E210 . Proof. By Lemma 9.10b, x is 3-central in Cy . We have CG (x, y) =

xy × CL (y), with CL (y) ∼ = Z2 × Σ4 . Hence the first two assertions of (a) hold, and AutNx ∩Cy (Ry ) ∼ = E22 . By Lemma 9.10b again, | AutCy (Ry ) : AutCy ∩Nx (Ry )| = |E1 (Ry )| = 4, so | AutCy (Ry )| = 16 and (a) holds. We next show that E := E(Cy ) = 1. Suppose false. By the structure of CCy (x) and NCy (Ry ), E is quasisimple and Ry ≤ E. Then 1 = W := O 2 (C(x, y))) = Ry [W, Ry ] ≤ E and W ∼ = Z3 × A4 , so by [V17 , 9.14], E/Z(E) ∼ = M22 . But CL ( y Ry [W, Ry ]) contains an involution w. Then w ∈ I2 (L) ⊆ z G ∪ uG and by [IA , 5.3c], [E, w] = 1. But CG (w) ∼ = Cz

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371

or Cu , neither of which has order divisible by |E|. Therefore, E = 1, so F ∗ (Cy ) = Qy .  We have Qy = x ∈R# CQy (x ) with all factors conjugate under NCy (Ry ), y

and each CQy (x ) elementary abelian of order 23 or 24 . Moreover, each [CQy (x ), Ry ] is a four-group not containing y. Hence |Qy | = 29 or 210 . Moreover, whether Ry centralizes Z(Qy ) or not, it follows that Qy is in fact elementary abelian, as required.  Now we can prove

Lemma 9.12. The following conditions hold: (a) Cy /Qy ∼ = Aut(A6 ) and Qy ∼ = E210 ; and (b) Qy has the following [Cy /Qy , Cy /Qy ] ∼ = A6 -composition factors: two trivial modules, and two self-dual 4-dimensional modules which are quasi-isomorphic but not isomorphic. Proof. For each x ∈ Ry# , CCy (x ) ∼ = CCy (x) is a {2, 3}-group, so O3 (Cy ) = Qy . Let Γ = Qy NCy (Ry ). Then ΓRy ,1 (Cy ) ≤ Γ. Thus either Γ = Cy or F ∗ (Cy /Qy ) = J/Qy is a component such that Cy = JΓ and (J ∩ Γ)/Qy is strongly 3-embedded in J/Qy . Moreover, in the latter case, Qy ≤ Z(J), so J/Qy is involved in Aut(Qy / y) and hence in GL9 (2). M11 . Hence, with In particular, 11 does not divide |J/Qy |, so J/Qy ∼ = [IA , 7.6.1], Γ = Cy or F ∗ (Cy /Qy ) ∼ = L3 (4) or A6 . To rule out Γ = Cy and L3 (4) we will show that 5 divides |Cy | but 7 does not. Suppose that 5 does not divide |Cy |. Then since |Cz |5 = |Cu |5 = 5, and by Lemma 9.10, G has exactly two conjugacy classes of Z10 -subgroups. Let fx ∈ I5 (Cx ) and fv ∈ I5 (Cv ). From the structure of Cx and Cv ,

x ∈ Syl3 (CG (fx )) and v ∈ Syl3 (CG (fv )), so fx  ∼G fv . However, fx and fv both centralize involutions (z0 ∈ CNx (L) and u ∈ O3,2 (Cv ), respectively.) Therefore CG (fv ) and CG (fx ) each have a single conjugacy class of involutions. As z0 ∼ z, we see from the structure of Cz that CG (z0 fx ) ∼ = Z5 × D12 . As a group with a single class of involutions, and other subgroups of order 5 besides fx , CG (fx ) = fx  × I, I ∼ = L2 (q), q = 11 or 13, by [V17 , 20.14]. However, fx is rational in L, so |NG ( fx ) : CG (fx )| = 4. As Out(I) ∼ = Z2 , CCG (fx ) (I) has even order. So some involution centralizes an element of order q. This is impossible by Lemma 9.10 and the structures of Cz , Cu , and Cy . So 5 divides |Cy |, ruling out Cy = Γ. Next, let sz ∈ I7 (Cz ) and put Qsz = O2 (CG (zsz )). Then CG (zsz ) ∼ = D8 × Z7 with Qsz ≤ Qz . Hence Qsz ∈ Syl2 (CG (sz )), and z is weakly closed in Qsz so CG (sz ) has a normal 2-complement. The involutions in CG (sz ) are in z G ∪ uG , and centralizers of elements of order 7 in Cz and CG (u) are

16. THEOREM C∗4 : STAGE A5. RECOGNITION

372

of order 2a · 7, so O2 (CG (sz )) = sz  and in particular sz  ∈ Syl7 (G) with CG (sz ) = Qsz × sz . It follows that |CG (y)| is not divisible by 7. Thus, F ∗ (Cy /Qy ) ∼ = A6 , and since Cy has a single class of elements of order 3, Cy /Qy ∼ = Aut(A6 ). In particular, O2 (CCy /Qy (x)) = 1 so CQy (x) = O2 (CCy (x)) ∼ = E24 . Likewise CQy ( x, v) ∼ = E22 . By the Brauer-Wielandt formula, |Qy | = (24 )4 /(22 )3 = 210 . Using [V17 , 10.8.1], we see that (b) must hold, and the lemma is proved.  Let y0 ∈ y G ∩ T be extremal in T , so that CT (y0 ) ∈ Syl2 (CG (y0 )). The next lemma gives additional information about CG (y) and determines the isomorphism type of NG (J(T )). Lemma 9.13. The following conditions hold: (a) J(T ) = Qy0 ∼ = E210 ; ∼ (b) CG (y) = CCo2 (y ∗ ) for some y ∗ ∈ I2 (Co2 ); (c) NG (J(T )) ∼ = NCo2 (J(T ∗ )), a split and faithful extension of E210 by Aut(M22 ); (d) J(T ) ∩ Qz ∼ = E25 Σ6 ; and = E25 , and in C z , NCz (J(T )) ∼ (e) Let Qy ≤ Cyo ≤ Cy with Cyo /Qy ∼ = Σ6 . Then I2 (Cy − Cyo ) ⊆ y G . Proof. By (9F), T ∼ = E210 = T ∗ ∈ Syl2 (Co2 ). Then by (9B1), J(T ) ∼ and T splits over J(T ). Hence by Gasch¨ utz’s theorem, NJ := NG (J(T )) splits over J(T ). Also, Qy0 = J(T ) (whence (a) holds) and so Cy splits over Qy . To prove (b), it will therefore be enough to show that the action of Cy /Qy ∼ = Aut(A6 ) on Qy is uniquely determined. A similar analysis applies to Cu . Since u, z  T , m2 (CT (u)) = m2 (T ) and so J(T ) ≤ CT (u). Recall that we have a normal series (9J)

1 < Z0 < Z0 × u < Qu < Cu ,

u = Cu /Qu ∼ with Z0 ∼ = E24 a natural module for C = L4 (2), and Qu /Z0 ∼ = 1+6 + ∼ u = Ω (2) faithfully. If J(T ) ≥ Z0 , then J(T ) ≤ Qu ; 2+ admitting C 6 but m2 (Qu ) = 7 < m2 (J(T )), contradiction. So Z1 := J(T ) ∩ Z0 < Z0 . u ) = m2 (21+6 ) = 4, m2 (Z1 ) ≥ 2. If equality holds, then B := As m2 (C + (J(T ) ∩ Qu )/Z1 is a maximal elementary abelian subgroup of Qu /Z0 ∼ = 21+6 + ; but J(T )Qu /Qu ≤ CCu (B) ∼ = E23 , so |J(T )| ≤ 29 , contradiction. Thus,  , and PU := N  (U ) m2 (Z1 ) = 3, whence J(T )Qu /Qu ≤ C  (Z1 ) =: U Cu

Cu

u . Hence J(T ) meets the successive is a maximal parabolic subgroup of C 3 quotients (9J) in sections of order 2 , 2, 23 , and 23 . Moreover, the sections ∼ of order 23 are natural PU /U = L3 (2)-modules. It follows that (9K)

NCu (J(T ))/J(T ) is an extension of E24 by L3 (2) and has nontrivial center.

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373

We also claim that two of the above sections of order 23 are dual to  -modules. Indeed, as F2 C u -modules, Qu /Z0 u ∼ each other as PU /U = ∧2 Z0 . 2 ∼  Then taking fixed points of U , we get (J(T ) ∩ Qu )/Z1 u = ∧ Z1 ∼ = Z1∗ as ∼ modules for PU /U = L3 (2), proving our claim. Since J(T ) also has a trivial  -module, it follows that composition factor as PU /U (9L)

For any s ∈ I7 (NCu (J(T ))), s has a free summand on J(T ),

a fact that we shall need presently. Next, we have not yet determined NCz (J(T ))/J(T ); but by Tits’ lemma [IA , 2.6.5], and the fact that m2 (Qz ) < 10, it is a proper parabolic subgroup of C z ∼ = Sp6 (2). Hence its order is 29 b, where b = 1, 3, 9, 21, or 45. Now by [IG , 16.9], NJ controls G-fusion in J(T ). So by Lemma 9.10a,   1 1 1 # |J(T ) | = |NJ | + + |NJ ∩ Cz | |NJ ∩ Cu | |NJ ∩ Cy0 |     1 2 8 |NJ |2 315 + + = + 86 3.11.31 = |NJ |2 b 3.7 32 .5 315 b Since b divides 315, which divides |NJ ∩ Cu |, both factors on the right side are integers. The only solution is b = 45, giving |NJ /J(T )| = 28 .32 .5.7.11, NJ ∩ Cz ∼ = E25 Σ6 , and J(T ) ∩ Qz ∼ = E25 . So (d) holds. Also, (9M)

|z G ∩ J(T )| = 77, |uG ∩ J(T )| = 330, and |y G ∩ J(T )| = 616.

As Cy0 has one class of elements of order 3, so does NJ . Then since CO3 (NJ ) (x) is an extension of Sx := CJ(T ) (x) ∼ = E24 by an element of NNCx (Sx ) ( x, v ; 3 ), it follows that O3 (NJ ) is a 2-group. Because 11 divides |NJ |, NJ is irreducible on J(T ), so O3 (NJ /J(T )) = 1. Therefore I := F ∗ (NJ /J(T )) is simple. From Cx and Cu , I contains subgroups P1 := 24 Σ6 and 24 L3 (2). If I ∈ Chev(2), then P1 is a parabolic subgroup of I. So I has twisted rank at least 3 and a double bond in its Dynkin diagram. Clearly there is no such group satisfying the restriction that |I| divides 28 .32 .5.7.11. So I ∈ Chev(2). Again from the order of NJ /J(T ), it is clear that I ∈ Chev(r) for any odd r, I ∈ Alt, and if I ∈ Spor, then I ∼ = M22 . Again with Gasch¨ utz’s theorem, NJ is a split extension of J(T ) = F ∗ (NJ ) by Aut(M22 ). To prove (c), it suffices by [V17 , 10.12.1, 10.12.2] to show that the hypotheses of [V17 , 10.12.1] apply to J(T ) in the role of V there. For then, those lemmas imply that J(T ) ∼ = T10 or G10 . Since for any t ∈ J(T )# , CNJ (t) has no composition factor isomorphic to L3 (4), [V17 , 10.12.2b] implies that J(T ) ∼ = G10 , giving the uniqueness of the splitting extension NJ . Let fr ∈ Ir (NJ ) for r ∈ {3, 5, 7, 11}, and let dr = dim CJ(T ) (fr ). We have seen above that as J(T ) = Qy0 , d3 = 4. Also d11 = 0 since 2 is a primitive root modulo 11, and d5 = 2 by Lemma 9.12b. The discussion before (9K) and through (9L) establishes that d7 = 1 and subgroups of NJ of order 7 have free summands on J(T ). Thus we can apply [V17 , 10.12.1, 10.12.2], and (c) follows. Then (b) is immediate.

374

16. THEOREM C∗4 : STAGE A5. RECOGNITION

At last, consider (e). Using the structures of involution centralizers in G, we see that there are three G-conjugacy classes of elements of order 10, represented, say, by zgz , ugu , and ygy , where gz  ∈ Syl5 (CG (z)), and similarly for gu  and gy . In particular, any two Z10 -subgroups of G with conjugate Sylow 2-subgroups are G-conjugate. For t ∈ {z, u, y}, let Yt ∈ Syl2 (CG (tgt )); then Yz ∼ = E22 , Yu ∼ = Q8 , and Yy ∼ = D8 , visibly in CG (t). We have Cy ≤ NG (Qy ) with Qy ∼G J(T ) ∼ = E210 , and set Ay = CQy (gy ). Thus Ay ≤ Yy . Since m2 (Yy ) ≤ 2 so m2 (Ay ) ≤ 2, i.e., Ay ∼ = E22 . G Furthermore, by the action of gy , |Ay ∩z | ≡ 77 (mod 5) and |Ay ∩y G | ≡ 616 (mod 5), so G A# y = {y, z1 , z2 } where z1 , z2 ∈ z .

By a remark above, z1 gy and z2 gy are conjugate in G and indeed in CG (gy ). Since Qy is the unique E210 -subgroup of CG (zi ) for each i, z1h = z2 for some h ∈ NG (Qy ) ∩ CG (gy ) =: W ≤ NG (Ay ). However, by [IA , 5.3c], W/Ay ∼ = Z10 . Therefore we may take h to be an involution, and D8 ∼ = Ay h ∈ Syl2 (CG (ygy )). Now we can prove that h ∈ y G . Otherwise h, hy ∈ z G , and hgy , hygy , z1 gy , and z2 gy are all G-conjugate. So h, hy1 , z1 , z2 are conjugate in CG (gy ), in which Yy ∼ = D8 is a Sylow 2-subgroup with y = Z(Yy ) weakly closed in Yy . Hence h and z1 are conjugate in CG (y). But z1 ∈ Qy  CG (y) and h ∈ Qy , so we have a contradiction. Hence, h ∈ y G . Finally, to show I2 (Cy −Cyo ) ⊆ y G , it suffices to show that h acts freely on Qy , for then since Cy /Qy ∼ = Aut(A6 ), all involutions in Cy −Cyo are conjugate modulo Qy to h, and then genuinely conjugate to h. But h centralizes gy and therefore, using [IA , 5.3c], h inverts an element of I11 (Cy /Qy ), which is fixed-point-free on Qy ∼ = E210 . This implies the desired free action of h, whence (e) holds and the lemma is proved.  Next we consider the full fusion pattern of involutions in G. We already have an isomorphism Cz → CCo2 (z ∗ ). We will actually prove Lemma 9.14. The G-fusion pattern of involutions in Cz is uniquely determined. That is, there exists an isomorphism φz : Cz → CCo2 (z ∗ ) such that φz carries z to z ∗ , u to u∗ , T to T ∗ , and G-conjugates of z, u, and y, respectively, to Co2 -conjugates of z ∗ , u∗ , and y ∗ . First, by our Background Result (9B2) and Lemma 9.5, (9N) Cz = Qz H, with H ∼ = Sp6 (2). We fix such an H. By [V17 , 20.16], H z / z is determined up to Cz conjugacy, and hence so is [H z , H z] = H. By Lemma 9.7a, I2 (Qz ) ⊆ {z}∪uG . Next, by [V17 , 10.2.4], C z = Cz /Qz has four classes of involutions, three of which are 2-central. Let Vnat be the

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375

natural C z ∼ = Sp6 (2)-module. We first prove Lemma 9.15. Let t ∈ I2 (Cz ) be such that t is not 2-central in C z . Then t ∈ yG. Proof. By [V17 , 10.2.5a1], t acts freely on Qz / z. Hence I2 (Qz t) ⊆ ∪ (tz)Qz . By [V17 , 10.2.5a2], there are commuting elements w, x0 ∈ I3 (C x ), inverted and centralized, respectively, by t, and dim CVnat (x1 ) = 2 for all x1 ∈ x0 , w − w, and CVnat (w)) = 0. Thus commuting t-invariant preimages x0  and w of order 3 exist and lie in xG and vG , respectively. Moreover, Q8 ∼ = Q8 ∗ Q8 for all x1 ∈ x0 , w − w. = CQz (w) ≤ CQz (x1 ) ∼ Let L0 = E(CG (x0 )) and C = CG ( x0 , z), so that t, w ≤ C. As ∼ ∼ ∼ CQz (x0 ) ∼ = 21+4 + , C = CCx0 (z) = Z3 × (L1 ∗ L2 ) a, b, where L1 = L2 = a ∼ ∼ SL2 (3), a, b = E22 , L1 = L2 , a ∈ L0 , and Li b = GL2 (3), i = 1, 2. Then w ∈ Li for some i = 1, 2 as CQz ( w, x0 ) ∼ = Q8 . Since t inverts w, t ∼L0 b, ∼ whence t ∈ L0 and CL0 (t) = Z2 × Σ4 by [V17 , 10.9.2e]. By definition of y (Lemma 9.6), t ∈ y G . The lemma is proved.  t Qz

C

The three conjugacy classes of involutions in C z besides t z are all 2central. They are represented in Z(T ) by t , a long root involution, ts , a short root involution lying in O2 (CC z (t )), and a “hybrid” involution ts = t ts . We let t , ts , and ts be the preimages of these involutions in H. Let C1 = CH (t ), a split extension of E25 by Σ6 . By Lemma 9.13d, E25 ∼ = J(T ) = O2 (C 1 ). Hence O2 (C1 ) = J(T ) ∩ H, and t , ts  ≤ O2 (C1 ). Let Jz = J(T ) ∩ Qz ∼ = E25 . We shall determine the Gconjugacy class of each involution s of O2 (C1 ), and describe (9O) the distribution of I2 (Jz s), and then I2 (Qz s), among z G , uG , and y G .   The fifteen nonidentity cosets in J(T )/ t are permuted transitively by C 1 /J(T ) ∼ = Σ6 , and each such coset consists of a conjugate of ts and a conjugate of ts . One consequence is that we only need to consider the involutions s = t , ts , and ts in (9O). Another consequence is that (9P)

The number of conjugates of z, u, or y in J(T ) − Jz t  is a multiple of 15.

For any s ∈ O2 (C1 ), let γ(s) be the set of commutators γ(s) = {[s, r] | r ∈ Qz }. ⊆ and of course |sγ(s)| = |Qz : CQz (s)|. Clearly also Thus sγ(s) = Q G z zsγ(s) = (zs) ⊆ (zs) since [zs, r] = [s, r] for all r ∈ Qz . Moreover, if zsγ(s) ∩ sγ(s) = ∅, then for some r, r ∈ Qz , [s, r] = [s, r ]z. This would be a contradiction if |γ(s)| = |γ(s) z / z |. Hence sQz

(9Q)

sG

If |γ(s)| = |γ(s) z / z |, then zsγ(s) ∩ sγ(s) = ∅.

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376

By Lemma 9.7a, I2 (Qz ) − {z} ⊆ uG . Beyond that, let us begin with s = t and s = ts . Then s is a transvection, inverting some w ∈ I3 (H) with 2-dimensional support on the natural H-module. Hence CQz /z (w) = 1 and s acts freely on Qz / z. As s ∈ J(T ), [Jz , s] = 1. Thus |γ(s)| = 16 = |γ(s) z / z |, and so (9R)

sJz is the disjoint union sγ(s) ∪ zsγ(s), each subset being totally fused of order 16.

But by (9M) and Lemma 9.7, |z G ∩ J(T ) − Qz | ≡ |y G ∩ J(T ) − Qz | ≡ 1 + |uG ∩ J(T ) − Qz | ≡ 1 (mod 15), the only possibility when s = t is that (9S)

z Qz I2 (t Jz ) = I2 (t Qz ) = tQ  ∪ (zt ) , these last two sets of G G size 16 lying in z and y (respectively or vice-versa).

We will see below that t ∈ z G , whence zt ∈ y G . Now let s = ts . Again (9R) holds. Hence, again by (9M), (9T)

The fifteen conjugates of the coset ts Jz under NH (O2 (C1 )) together consist of 240 elements of y G , and either 240 further elements of y G or 240 elements of uG .

Which of these occurs, and which of these elements are in H, will be determined once we deal with the case (9U)

s = ts ,

which we do now. Clearly γ(s) ⊆ [Qz , O2 (C1 )] ≤ Jz . In the NH (O2 (C1 ))conjugates of the coset ts Jz , there are 60 elements of z G , 300 or 60 elements of uG , and 120 or 360 elements of y G , respectively, by (9M) and (9T). Now by [V17 , 10.2.5b], CH (s) contains ξ1 , ξ2  ∼ = E32 , where [Vnat , ξ1 ] and [Vnat , ξ2 ] are disjoint of respective dimensions 2 and 4, so that CQz (ξ1 ) = z, CQz (ξ2 ) ∼ = Q8 ∗ Q8 , and CQz (ξ1 ξ2±1 ) ∼ = Q8 are all s-invariant. Let Q12 = CQz (ξ1 ξ2 ), CQz (ξ1 ξ2−1 ) ∼ = Q8 ∗ Q8 and Q0 = CQz (Q12 ) ∼ = Q12 . Then as [Q12 , ξ1 ] = Q12 , we have [Q12 , s] = 1 and so γ(s) = {[s, r] | r ∈ Q0 } ⊆ Q0 ∩ Jz . Since [Q12 Jz , s] = 1, |γ(s)| = |γ(s) z / z | = 4. Thus γ(s) is a set of coset representatives for Q12 in CQz (s). It follows that we have a disjoint union  gsγ(s) I2 (sQz ) = (9V)

g∈Q12 , g 2 =1

with each term in the union a set of 4 conjugate involutions. As s, γ(s) ≤ J(T ), we also have  gsγ(s). I2 (sQz ∩ J(T )) = g∈Q12 ∩Jz , g 2 =1

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As Q12 contains 20 solutions of g 2 = 1, and Q12 ∩ Jz ∼ = E23 , (9W)

I2 (sQz ∩ J(T )) is the union of 8 sets of the form gsγ(s), and I2 (sQz ) is the union of 20 such sets.

Moreover, ξ1 , ξ2  Q12 has two trivial orbits on I2 (Q12 ), namely {1} and {z}, and is transitive on the rest of I2 (Q12 ), while centralizing s and stabilizing γ(s). We conclude for our case s = ts that (9X)

I2 (ts Qz ∩ J(T )) = ts γ(ts ) ∪ zts γ(ts ) ∪ IJ and I2 (ts Qz ) = ts γ(ts ) ∪ zts γ(ts ) ∪ I.

Here the unions are disjoint and each term consists of mutually conjugate elements. The cardinalities are 32 = 4 + 4 + 24 80 = 4 + 4 + 72 Comparing with (9V), we see that

(9Y)

z Qz = zt γ(t ) are contained (1) tQ s = ts γ(ts ) and (zts ) s s G G in z and u (respectively or vice-versa). The remaining involutions in ts Qz lie in y G ; (2) I2 (ts Jz ) = I2 (ts Qz ) consists of 16 elements each of z Qz (respectively uG and y G , lying in tQ s and (zts ) or vice-versa).

In view of Lemmas 9.7a and 9.15, (9S), and (9Y), to complete the proof of Lemma 9.14, it remains to decide whether (9Z)

(1) t ∈ z G or y G ; (2) ts ∈ uG or y G ; and (3) ts ∈ z G or uG .

The asymmetry that permits us to answer is that t , ts , and ts lie in the complement H to O2 (Cz ). Now t  z = {z1 , y1 } where z1 ∈ z G and y1 ∈ y G , by (9S). As y1 ∈ J(T ), CG (y1 ) = J(T )H1 , where H1 ∼ = Aut(A6 ). The module structure of J(T ) for [H1 , H1 ] is given in Lemma 9.12. For any f ∈ I5 (NH (J(T ))), CJ(T ) (f ) ∼ = E22 , containing z, t  = z, y1 . Let r ∈ CCG (y1 ) ( f ) be a 2-element outside the Σ6 -subgroup of H1 . Then r interchanges the two (nonisomorphic but quasiisomorphic) nontrivial A6 -composition factors in J(T ). We use r to establish the answers in (9Z). Namely, O2 (C1 ) is an indecomposable [H1 , H1 ]-module with CO2 (C1 ) ([H1 , H1 ]) = t , and Jz is an indecomposable [H1 , H1 ]-module with CO2 (C1 ) ([H1 , H1 ]) = z. Since r normalizes H1 and J(T ) and interchanges the two nontrivial A6 -composition factors in J(T ), it must interchange O2 (C1 ) and Jz . Since I2 (Qz )−{z} ⊆ uG and I2 (O2 (C1 ) − Z(C1 )) contains ts and ts , we have (9AA)

t = z r , trs ∈ uG , and trs ∈ uG .

378

16. THEOREM C∗4 : STAGE A5. RECOGNITION

As noted above, Lemmas 9.7a and 9.15, and (9S), (9Y), and (9AA), complete the proof of Lemma 9.14. There are two loose ends to tie up to finish the proof of Proposition 9.4. Lemma 9.16. Let g ∈ I5 (Cy ). Then The following conditions hold: (a) z G ∩ CQy (g) = ∅; (b) For any z1 ∈ z G ∩ CQy (g), [Cyo , z1 ] = 1, where Qy ≤ Cyo ≤ Cy with Cyo /Qy ∼ = Σ6 ; and (c) There exists an involution-fusion-preserving isomorphism from Cy to CG∗ (y ∗ ). Proof. Since Cy = Qy Ly with F ∗ (Cy ) = Qy and Ly ∼ = Aut(A6 ), it suffices for (a) and (b) to show that Cz contains some y1 ∈ y G such that CCz (y1 ) contains a copy of Σ6 . But by (9S), we can take y1 = zt . Then by (a) and (b) and Lemma 9.14, there exists an involution-fusion-preserving isomorphism from Cyo to the corresponding subgroup of CG∗ (y ∗ ) of index 2. This extends to an isomorphism of Cy onto CG∗ (y ∗ ). By Lemma 9.13e, the extension preserves fusion of involutions, proving (c).  Lemma 9.17. Let Su = CT (u) ∈ Syl2 (Cu ). Then Qu / u = J(Su / u). u ∼ u = Cu / u. By Lemma 9.8cd, Q Proof. Let C = E210 , with Z = u /Q u ∼  u /Z  a natural 0 ∼ Z = E24 a natural module for C = L4 (2) and V := Q + ∼   module for Cu /Qu = Ω6 (2).  u . Suppose that A  ∈ E10 (Su ) with A  = 1. Let u /Q u = Cu /Qu ∼ Let C =C  Q  u ∩ A)|,  so that m+n = m2 (Q  u /Q  u ∩A) ≤ :Z  ∩ A|  and n = |Q  u : Z( m = |Z +      m2 (A). As A = 1, m ≥ 1, and n ≥ 2, since Su ≤ Ω (Qu /Z) contains no  ≥ 3. transvections. Thus m2 (A)   must be the If m2 (A) = 3, then m = 1 and n = 2. In particular, A  of Z.  Then A  = O2 (N  (Z  ∩ A))  = u of the hyperplane Z ∩ A centralizer in C Cu

 is O2 (NCu (Vi )) for some totally singular 3-subspace Vi of V . But then A  = Vi . Hence n ≥ 3, the stabilizer of the chain V > Vi > 0 and so CV (A)  > 3. contradiction. Thus, m2 (A)  It follows that A = J(Su ) = O2 (P) ∼ = E24 , where P is a parabolic u equal to N  (Z 1 ) = C  (v) for some 2-subspace Z 1 of subgroup of C Cu

Cu

 contains v  and some nonzero singular vector v ∈ V . But then CV (A) Z properly (as n ≥ 2) and is invariant under the parabolic subgroup P , so  ≥ 5. This is a contradiction as A  contains no transvections. m2 (CV (A)) The proof is complete.  Note that since φz (T ) = T ∗ and φz (u) = u∗ , Lemma 9.17 implies that φz (Qu ) = O2 (CCo2 (u∗ )), as asserted by Proposition 9.4. Now Lemmas 9.14, together with Lemmas 9.5 and 9.13 and 9.16c, complete the proof of Proposition 9.4.

9. G ∼ = Co2

379

Next we use the Thompson order formula to prove: Proposition 9.18. |G| = |Co2 |. For any involution t ∈ G, let cG (t) = |CG (t)| and fG (t) = |FG (t)|, where FG (t) = {(z0 , u0 ) ∈ z G × uG | t ∈ z0 u0 }. By Lemma 9.10a, the Thompson order formula corresponding to the classes  z G and uG is |z G ||uG | = t∈I2 (G) fG (t), or equivalently   cG (u)cG (z)fG (y) . (9BB) |G| = cG (u)fG (z) + cG (z)fG (u) + cG (y) Our analysis applies to G and equally well to G∗ = Co2 . Hence it is enough to prove that the local structure shared by G and Co2 determines the right side of (9BB) uniquely. Lemmas 9.5, 9.8cd, 9.13b, and 9.14 establish the uniqueness of cG (z), cG (u), cG (y), and the G-fusion pattern of involutions in Cz = CG (z) . Lemma 9.19. fG (z) = fCo2 (z) and fG (y) = fCo2 (y). Proof. This is an immediate consequence of Lemmas 9.14 and 9.16c.  To compute fG (u) we need a lemma. Lemma 9.20. Let U1 and U2 be four-subgroups of G such that nz := |U1 ∩ z G | = |U2 ∩ z G | = 1 or 2, and nu := |U1 ∩ uG | = |U2 ∩ uG | = 3 − nz . Let ui ∈ Ui ∩ uG , i = 1, 2. Then there is g ∈ G such that U1g = U2 and ug1 = u2 . Proof. Replacing the Ui by conjugates, we may assume that z ∈ U1 ∩ U2 . Let vi = zui , i = 1, 2. By Lemma 9.15 and (9S), ui is 2-central in C z , but ui = t , i = 1, 2. Conjugating U1 and U2 independently in Cz , we may assume {u1 , u2 } ⊆ {1, ts , ts }. (See the remarks before (9O).) We have v i = ui , i = 1, 2. If ui = ts , then by (9Y2), either ui or vi is in y G , contrary to assumption. If u1 = 1, then U1 ≤ Qz . Then v1 ∈ uG by Lemma 9.7a. Hence nu = 2, so v2 ∈ uG . By Lemma 9.7a again, u2 = ug1 for some g ∈ Cz , as desired; note that u2 = 1. Finally, we may assume that u1 = u2 = ts . Then (9Y1) implies that u1 and u2 are Qz -conjugate. The proof is complete.  Lemma 9.21. fG (u) = fCo2 (u). Proof. Let G∗ = Co2 and let z ∗ be a 2-central involution of G∗ . By Lemma 9.14, there is a G-involution-fusion-preserving isomorphism Cz → CG∗ (z ∗ ) which we write as g → g ∗ and X → X ∗ for every g ∈ Cz and ∗ X ⊆ Cz . Thus (uG ∩ Cz )∗ = (u∗ )G ∩ CG∗ (z ∗ ), and so on for other classes. Let Su ∈ Syl2 (CG (u, z)); then Su∗ ∈ Syl2 (CG∗ (u∗ )) and u → u∗ . Lemma 9.17 implies that Q∗u = J(Su∗ mod u∗ ) = O2 (CG∗ (u∗ )).

380

16. THEOREM C∗4 : STAGE A5. RECOGNITION

Suppose (z0 , u0 ) ∈ FG (u). Then every involution in z0 , u0  is in z G ∪ uG and the four-group z0 , u ≤ z0 , u0  contains elements of z G and uG . On the other hand, we have four-groups z, u ≤ Qu , with uz ∼ u, and z1 , u ≤ with u = Z(CQ(u) (v)), so uz1 ∈ z G by Lemma 9.10c. CQu (v) ∼ = 21+4 − By Lemma 9.20, z0 , u is Cu -conjugate to z, u or z1 , u. In particular z 0 ∈ Qu . Since Cu /Qu acts faithfully as L4 (2) on Z(Qu ) = Z0 × u, Qu u0 ∈ CG (u)/Qu is either the identity, a 2-central involution, or a non-2-central involution. Accordingly, m2 (Z(Qu u0 )) = 5, 4, or 3. Choose involutions ui ∈ Su , i = 3, 4, 5, such that m2 (Z(Qu ui )) = i. For any v ∈ Cu with v 2 ∈ Qu , set hG (u, Qu v) := |FG (u) ∩ (z G × (uG ∩ Qu v))|, so that (9CC)

fG (u) =



hG (u, X).

X∈CG (u)/Qu

Also note that for every g ∈ CG (u) and Qu u0 ∈ I2 (CG (u)/Qu ), (9DD)

hG (u, Qu ug0 ) = hG (u, Qu u0 ),

since the two sides are the cardinalities of conjugate subsets. Set hi = hG (u, Qu ui ), i = 3, 4, 5, and let n3 and n4 be the number of non-2-central, respectively 2-central, involutions in CG (u)/Qu ∼ = A8 . Also ∼ set n5 = 1. Then by (9CC) and (9DD), and as CG (u)/Qu = A8 , (9EE)

fG (u) = n3 h3 + n4 h4 + n5 h5 = 210h3 + 105h4 + h5 .

Since Qu , u3 , u4 , u5  ≤ Su , the counts h3 , h4 , and h5 are determined by the isomorphism class of Su and its involution fusion pattern, which are the  same as in Su∗ . Thus (9EE) completes the proof of the lemma. Now Proposition 9.18 follows immediately from (9BB) and Lemmas 9.19 and 9.21. Together, Propositions 9.4 and 9.18 complete the proof that G ≈o Co2 . Hence by (9B1) and Lemma 9.9, G ∼ = Co2 and  = −1. Thus Proposition 9.2 is proved.

10. The A9 –Sp6 (2) Setup In the next nine sections we finish the case in which p = 3 and Lo3 (G) contains A9 or Sp6 (2). In view of Theorem C∗4 : Stages A1, A2, and A4, and assuming the A9 –Sp6 (2) setup, we work with the following hypotheses and

11. 3-LOCAL STRUCTURE

381

notation:

(10A)

(1) G is of restricted even type with e(G) = m2,3 (G) = 3 and Lo3 (G) ⊆ C3 ; (2) For any B ∈ E33 (G) such that NG (B; 2) = {1}, B # ⊆ I3o (G); and (3) For any z ∈ I2 (G) such that m3 (Cz ) = 3, we have F ∗ (Cz ) = O2 (Cz ).

and

(10B)

(1) There exists a 2-local subgroup H of G containing subgroups A ≤ B such that F ∗ (H) = O2 (H), A∼ = E33 , and a 3-component J of = E32 , and B ∼ CG (A) such that J ∼ = A6 ; (2) For any H, A, B, and J as in (1), there exists x ∈ A# such that J pumps up vertically in Cx to a 3-component Jx ; (3) For some x as in (2), Jx ∼ = A9 or Sp6 (2); and (4) For any x as in (2), m3 (C(x, Jx )) = 1 and Jx /O3 (Jx ) ∼ = A9 , Sp6 (2), J3 , L2 (36 ), or Sp4 (8). In 6 the L2 (3 ) case, or if J pumps up trivially in Cx , [A, CO2 (H) (x)] = 1.

Note that O3 (J) = 1 in (10B1) because of (10B3). We shall prove: Proposition 10.1. Under the conditions (10A) and (10B), G is isomorphic to A12 , F5 , Sp8 (2), or F4 (2). 11. 3-Local Structure ∼ A9 or Sp6 (2) and expand We fix an arbitrary x0 ∈ A such that Jx0 = 3 B to D ∈ E4 (G). As e(G) = 3, m3 (C(x0 , Jx0 )) = 1 and D = x0  × DJx0 , where DJx0 = D ∩ Jx0 . Then [V9 , (10A)] holds with Jx0 , ND := AutG (D), AutJx0 (D), and x0 ND in place of L, N , NL , and X , respectively. Hence [V9 , 10.1] applies. We put N0 := AutNG (x0 ) (D). We set x1  = A ∩ Jx0 and write x1 NJx0 (DJx0 ) = { x1  , x2  , x3 }. Also let Cx0 = CG (x0 ), C x0 = Cx0 /O3 (Cx0 ), D ≤ P ∈ Syl3 (Cx0 ). Thus P = R0 ×(P ∩Jx0 ), where R0 = CP (Jx0 ) is cyclic by (10B4), P ∩Jx0 ∼ = Z3 Z3 , D = J(P ) and R0 D = Ja (P ). As J   CJx0 (x1 ), x1 is a 3-cycle if J x0 ∼ = A9 , and dim[V, x1 ] = 2 if J x0 ∼ (2), where V is the natural F J -module. Note that for either Sp = 6 2 x0 isomorphism type of J x0 , all elements of I3 (Jx0 ) are real in Jx0 . Hence the N0 -orbits on E1 (D) are [1], [3], [4], [6], [61 ], [8], and [12], as in [V9 , (10A)ff.]. Lemma 11.1. Part (c) of [V9 , 10.1] applies.

382

16. THEOREM C∗4 : STAGE A5. RECOGNITION

Proof. Suppose first that x0   NG (D), so that P ∈ Syl3 (NG (D)). As D = J(P ), P ∈ Syl3 (G). Choose x ∈ A − x0  such that [CO2 (H) (x), A] = 1. Now P/Ω1 (P ) is cyclic and Ω1 (P ) ∼ = Z3 × (Z3 Z3 ), so the only possible pumpups Jx in (10B4) are A9 and Sp6 (2). But then NJx (D) is irreducible on D/ x, so x = x0 , contrary to our choice. Thus part (a) of [V9 , 10.1] does not hold. Suppose that part (b) of [V9 , 10.1] applies. Let R be as there, and let S be a Sylow 3-subgroup of a preimage of R in NG (D). Then by [V9 , 10.5], S has a NJx (D)-invariant subgroup S0 which is homocyclic abelian or special, and such that S = S0 x0 . We have D ∈ Syl3 (CG (D)), so S/D ∼ = R. Note that DJx0 is ND -invariant, hence S-invariant. Then [x0 , S] = DJx0 = Z(S). In particular, [x1 , S] = 1 and x1 ∈ [S, S]. If J pumps up trivially in CG (x1 ), then S normalizes CD (J) = A, whereas xS0 = D, a contradiction. Therefore J pumps up vertically in CG (x1 ) and (10B4) applies. Suppose that Jx1 := Jx1 /O3 (Jx1 ) ∼ = A9 , Sp6 (2), or J3 . By (10B4),  m3 (C(x1 , Jx1 )) = 1. As Out(Jx1 ) is a 2-group in each case, x1 ∈ [S, S], contradiction. Thus, Jx1 ∼ = L2 (36 ) or Sp4 (8), on which x0 induces a nontrivial field automorphism. Therefore S = SI x0 , where SI = InnS (Jx1 ). Because m3 (C(x1 , Jx1 )) = 1 and Jx1 has abelian Sylow 3-subgroups, SI is abelian. Hence SI ∩ S0 is abelian of index at most 3 in S0 , so S0 is abelian by [V9 , 10.5]. But [S, x0 ] = DJx0 , so SI = S0 is the unique abelian maximal subgroup of S. If Jx1 ∼ = L2 (36 ), then SI is elementary abelian, so S = Ω1 (S). As m3 (C(x1 , Jx1 )) = 1 = m3 (Out(Jx1 )), the image of S in CG (x1 )/Jx1 is abelian. Thus x1 ∈ [S, S], contrary to what we saw above. Finally we have Jx1 ∼ = Sp4 (8). Thus S0 contains Z9 ×Z9 , and so must be homocyclic abelian of rank 3, again by [V9 , 10.5]. By the action of NL and the fact that S/DJx0 is abelian, S has class at most 2. Hence Ω1 (S) = D. For some w ∈ I3 (Jx0 ), we have S w ∈ Syl3 (ND ) with w acting freely on DJx0 and then freely on S/D since [x0 , w] = 1. Hence D = J(S w) char S w and so S w ∈ Syl3 (G). Not every extremal conjugate of x0 in S w can lie in the coset S0 w x0 ; for if it did, the transfer VG→Sw/S0 w (x0 ) = 1 (see [IG , 15.15]), so x0 ∈ [G, G] = G, a contradiction. Now |CG (x0 )|3 = 35 , whereas for any x ∈ 6 4 S w − S0 x0 , x−1 0 , |CS (x)| ≥ 3 or |CS (x)| ≤ 3 according as x ∈ S0 or not. Therefore all extremal conjugates of x0 lie in Ω1 (S0 x0 ) = D, indeed in −1 D − DJx0 . Hence some element of xG 0 lies in DJx0 x0 , which is totally fused g by S0 . We can therefore choose g ∈ G with x0 = x−1 0 . Then g normalizes Jx0 . As J x0 ∼ = A9 or Sp6 (2), we may adjust g by an element of Jx0 and assume that xg1 = x1 . Hence Jxg1 = Jx1 . But g inverts x0 , which induces

11. 3-LOCAL STRUCTURE

383

a nontrivial field automorphism on Jx1 ∼ = Sp4 (8). This contradicts the fact   that Out(Jx1 ) is abelian, and completes the proof of the lemma.  Lemma 11.2. For any x as in (10B2), O3 (CG (x)) ∼ = Z3 × A9 or Z3 × Sp6 (2).

A9 or Sp6 (2), so that by Proof. Suppose first that Jx /O3 (Jx ) ∼ = (10B4), Jx /O3 (Jx ) ∼ = L2 (36 ), J3 , or Sp4 (8). Expand D to Qx ∈ Syl3 (CG (x)) and to Q ∈ Syl3 (CG (x0 )). As E(CJx /O3 (Jx ) (x0 )) ∼ = A6 , it follows by [IA , 4.9.1, 5.3h] that all subgroups d such that d ∈ x0 (D ∩ J) are fused in NQx (D), and |Qx : D| ≥ 9 > |Q : D|. But since D ∩ J = x2 , x3 , these subgroups d include some in the orbits [61 ] and [12], in addition to x0 . Also Q ∈ Syl3 (G), Z(Q) = x0 , x1 x2 x3 , and Z(Q) ∩ [Q, Q] = x1 x2 x3 . So

x0  has three NG (Q)-conjugates, and in particular, has conjugates in the orbit [8]. As D = J(Q), we have

x0 ND ⊇ [1] ∪ [61 ] ∪ [8] ∪ [12] = E1 (D) − E1 (DJx0 ). But then x1 x2 x3 ND ≤ DJx0 . By Lemma 11.1, ND acts irreducibly on D, a contradiction. Thus, Jx /O3 (Jx ) ∼ = A9 or Sp6 (2). Now O3 (Jx ) has odd order as m3 (CG (x)) > 3. Also A9 contains a Frobenius group of order 9.8 (indeed a semidirect product E32 SL2 (3)), m2 (A9 ) = m2 (A8 ) = 4, and m2 (Sp6 (2)) > 4 [IA , 3.3.3], so Jx is quasisimple by [V9 , 2.4, 2.9]. As the Schur multipliers of both groups are  2-groups [IA , 6.1.4], Jx ∼ = Z3n × A9 = A9 or Sp6 (2). Therefore, O3 (CG (x)) ∼ or Z3n × Sp6 (2), n ≥ 1. As x  ND by Lemma 11.1, n = 1. The proof is complete.  We now determine the fusion of x0  in D, first proving: Lemma 11.3. No element of I3 (A) is G-conjugate to x1 x2 x3 . Proof. Recall that D ≤ P ∈ Syl3 (Cx0 ), and P ∼ = Z3 × (Z3 Z3 ) with

x1 x2 x3  = Z(P ) ∩ [P, P ]. Therefore if the lemma fails, there is x ∈ A# and T ≤ CG (x) such that T ∼ = Z3 Z3 and x = Z(T ). Expand T to Px ∈ Syl3 (CG (x)). If Jx is a vertical pumpup of J, then with Lemma 11.2 and (10B4), x ∈ [Px , Px ], a contradiction. Thus the pumpup Jx of J in CG (x) is trivial, and Px = (Px ∩ Jx ) × (Px ∩ C(x, Jx )) with Px ∩ Jx ∼ = E32 . So T ∩ Jx = 1, and T embeds in C(x, Jx ), whence m2,3 (CG (x)) > 3, a contradiction. The proof is complete.  Since A = x0 , x1  and x1 x2 x3  ∈ [4], Lemma 11.3 implies that (11A)

[1] ∼ [4], [3] ∼ [4], and [61 ] ∼ [4].

This narrows the possibilities for x0 ND = X in [V9 , 10.1].

384

16. THEOREM C∗4 : STAGE A5. RECOGNITION

Lemma 11.4. (a) One of the following holds: (1) x0 ND = [1] ∪ [3], and ND is GL(D)-conjugate to one of W (D4 ), W (D4 )∗ , or W (C4 ); or (2) x0 ND = [1]∪[3]∪[8] or [1]∪[3]∪[8]∪[6]∪[61], and O2 (GO+ (D)) ≤ ND ≤ GO+ (D); (b) Let M ≤ GL(D) be the monomial group on [1] ∪ [3]. Then ND contains O2 ([M, M ]). There is an involution g ∈ NG (D) mapping into [M, M ], preserving [1] ∪ [3], and interchanging x0  and x1 . Moreover, [6] ∼ND [61 ] and [4] ∼ND [12]; and (c) For any g as in (b), g ∈ NG (J) and J g ∼ = Σ6 . Proof. Lemma 11.3 rules out cases (1) and (2) of [V9 , Proposition 10.1c]. Since [8] is an ND -orbit, that proposition implies all parts of (a), as well as the fusion claimed in (b). In particular T := O2 (GO+ (D)) ≤ ND ≤ GO+ (D). From the definition [V9 , (10B)] of our ND -invariant bilinear form, [1]∪[3] is an orthogonal frame in D all of whose 1-spaces are isometric. Thus M is conjugate in GL(D) to W (C4 ), and so O2 ([M, M ]) = T ≤ ND . As CG (D) has odd order, this implies the existence of g in (b), permuting [1] ∪ [3] without fixed points. Since g preserves [1] ∪ [3], J g has two conjugacy classes of subgroups of order 3, implying (c).  Lemma 11.5. Suppose that | x0 ND | = 12 or 24, and let d ∈ Z(P ) ∩ d = Cd /O3 (Cd ). Then the following condi[P, P ]# . Let Cd = CG (d) and C tions hold: d ) = O3 (C d ); (a) F ∗ (C d ) ∼ d (b) If Cd has a section isomorphic to A5 , then O3 (C = 31+4 and C contains an SL2 (5)-subgroup; (c) Cd contains no Q8 ∗ Q8 subgroup; and (d) For any x ∈ D # such that x ∈ [6], there is g ∈ G interchanging x and x0 . Moreover xx0 ∈ dG . Proof. Let Q ∈ Syl3 (NG (P )). Then Q normalizes d and Z(P ) =

x0 , d, so |Q : P | ≤ 3 as P = CQ (x0 ). As | x0 ND |3 = 3, |Q| = 36 and Q ∈ Syl3 (NG (D)). By [V9 , 10.1c], Q is a split extension of D by a Sylow 3-subgroup of O+ (D), acting naturally. Thus D = J(Q) by [V17 , 10.10.1b].  = O3 (C d ). Hence, Q ∈ Syl3 (G) ∩ Syl3 (Cd ). By [V17 , 7.18], (a) holds. Let R ∼ Suppose that Cd has a section isomorphic to A5 . Then R = 31+4 and     so Cd /R embeds in Sp4 (3). Let Y be the image of Cd /R in P Sp4 (3), and suppose that Y0 a subgroup of Y minimal subject to having a quotient Y0 /Y1 ∼ = A5 . To prove (b) it suffices to argue that Y1 = 1, for then as Sp4 (3) has 2-rank 2 and possesses a central involution, the preimage of Y0 in Sp4 (3) must be isomorphic to SL2 (5). Suppose then that Y1 = 1. As |P Sp4 (3)|{2,3} = 5, O2 (Y1 ) or O3 (Y1 ) must be nontrivial. By the Borel-Tits

12. THE MIXED CASE

385

∼ U4 (2), theorem and examination of the parabolic subgroups of P Sp4 (3) = the only possibility is that Y0 is a parabolic subgroup with a Levi factor isomorphic to A5 . This contradicts the minimality of Y0 and proves (b). ∼  Next,   suppose that Q8 ∗Q8 = W ≤ CG (d). As m2 (W ) = 3, RW contains S ∼ = d × Σ3 × Σ3 × Σ3 by the Thompson dihedral lemma. As D = J(Q),  ≤ R.  Then W normalizes the chain  = O3 (S) D   >D  > d > 1 R  This is impossible by so z must stabilize this chain and hence centralize R. (a), so (c) holds. Finally, [V9 , 10.1c5] holds, so d ∈ [4] ∼ [12]. But x0 x ∈ [12], so x0 x ∈ dG . Note that for every x  ∈ [61 ], x0 , x  contains two subgroups in [61 ] ∼ [6], one in [3] ∼ [1], and x0 . On the other hand if x ∈ [6], x0 , x contains two subgroups in [12] ∼ [4] ∼ [1], one in [6], and x0 . These fusion patterns are incompatible, so x0 , x  ∼G x0 , x. Hence the ND ∩ NG ( x0 )orbits [6] and [61 ] are not paired. Therefore [6] must be self-paired, proving the existence of g in (d). The proof is complete.  12. The Mixed Case We continue with the setup (10A) and (10B), fixing H, A, and B as in (10B1). We again fix x0 ∈ A such that Jx0 ∼ = A9 or Sp6 (2) and set

x1  = A ∩ Jx0 . We have A = x0 × x1  with x1 ∈ Jx0 a 3-element with minimal nontrivial support. Then x1 is real in Jx0 . Let y0 and y1 generate the two elements of E1 (A) besides x0  and x1 . It follows that y0  and y1  are NJx0 (A)conjugate. We can narrow the possibilities for the pair (Jx0 , Jy0 /O3 (Jy0 )) to those occurring in the four target groups. Lemma 12.1. Either Jy0 /O3 (Jy0 ) ∼ = A6 or Jy0 ∼ = Jx0 . Proof. Otherwise, by Lemma 11.2, and after possibly interchanging xi ’s and yi ’s, we may assume that Jx0 ∼ = A9 and Jy0 ∼ = Sp6 (2). Then 3 CCG (x0 ) (A) = CCG (y0 ) (A) contains O (CJy0 (A)) = J t ∼ = Σ6 for some involution t, whence Jx∗0 := Jx0 t ∼ = Σ9 . Some involution in Jx∗0 inverts x1 and centralizes J. By Lemma 11.4, x0  and x1  are interchanged in NG (A). Therefore there is an involution u inverting x0 and centralizing both Jx0 and Jx∗0 O{2,3} (CG (x0 ))/O{2,3} (CG (x0 )) ∼ = Σ9 . Set Cu = CG (u), Q = O2 (Cu ), and F = x1 , x2  ≤ D ∩ Jx0 ∼ = E33 . Since m3 (Cu ) ≥ m3 (Jx0 ) = 3, Q = F ∗ (Cu ) by (10A3), and CQ (f ) is D ∩ Jx0 -invariant for each f ∈ F # . Consider these groups CQ (f ). For i = 1 and 2, there is gi ∈ ND interchanging x0  and the 3-cycle xi  ≤ Jx0 , by Lemma 11.4b. Then ugi inverts xi and centralizes D/ xi , so ugi acts on Jx0 as a transposition. As CCu (xi ) is conjugate to CG ( x0 , ugi ), we have O2 (CCu (xi )) = u = CQ (xi ). On the other hand, for any f ∈ F − x1 − x2 ,

16. THEOREM C∗4 : STAGE A5. RECOGNITION

386

we have f  ∈ [6] and so f = y0h for some h ∈ G. Then CQ (f ) is conjugate −1 to a 2-subgroup of CCG (y0 ) (uh ) invariant under the E33 -subgroup (D ∩ −1 Jx0 )h . By the Borel-Tits theorem applied in Jy0 ∼ = Sp6 (2), CQ (c) embeds in the unipotent radical of a parabolic subgroup of 3-rank 2, i.e., of type B2 or A1 × A1 . In any case |CQ (f )| ≤ 27 , and in case of equality, CQ (f ) is neither abelian nor extraspecial. From the various values of |CQ/u (f )|, f ∈ F # , we get |Q/ u | ≤ 212 . But Cu contains CNG (x1 ) (u) = (CNG (x0 ) (ug1 ))g1 , so Cu contains W = Σ × F7.6 , with x1 ∈ Σ ∼ = Σ3 and F7.6 a Frobenius group of order 7.6. Now, CQ/u (x1 ) = 1 and CW (Q) = 1. The smallest faithful irreducible F2 F7.6 - and F2 Σ-modules have dimensions at least 6 and 2 respectively [IG , 9.12], so |Q/ u | = 212 with W acting irreducibly on Q/ u. This implies that Q is abelian or extraspecial, whence CQ (f ) is as well for all f ∈ F − x1  − x2 . However, since |Q/ u | = 212 , we have |CQ (f )| = 27 for each such f , contradicting the last paragraph and completing the proof of the lemma.  13. The A9 Case In the next three sections we shall prove Proposition 13.1. If Jx0 ∼ = A9 , then G ∼ = A12 or G ∼ = F5 . We are continuing the assumptions (10A) and (10B), and the notation of the last two sections. Thus, assuming that Jx0 ∼ = A9 , [1] ∪ [3] = { xi }3i=0 with xi (i > 0) a 3-cycle in Jx0 and [1] ∪ [3] = D ∼ = E34 . Also A = x0 , x1 , J = L3 (CG (A)) ∼ = A6 . By Lemma 11.4b there is an involution g ∈ ND interchanging x0  and x1  and preserving [1] ∪ [3]. We have J g ∼ = Σ6 . In particular g leaves invariant the J-conjugacy class of root A4 -subgroups of Jx0 lying in J. This class has odd cardinality, so g normalizes a root A4 subgroup V c ≤ J, ∼ which we fix, with V = E22 and c3 = 1. Then c is a 3-cycle in Jx0 , so # c ∈ xG 0 . By J-conjugation we may assume that c = x2 . Let v ∈ CV (g) be any involution and set 

Cv = CG (v), CV = CG (V ), I = O3 (E(Cv )), and I0 = E(CI (V )). For any y ∈ A# , we put Iy = E(CG ( v, y)). As A6 has one class of involutions, one of the following holds: (1) In Jy ∼ = A9 , v is a root involution. Moreover, (13A)

  O 3 (CCv (y)) = O3 (CCG (y) (v)) = y × Iy with Iy ∼ = A5 ; or

(2) Jy ≤ JO3 (CG (y)), Iy = 1, and v maps to a root involution of JO3 (CG (y))/O3 (CG (y)) ∼ = A6 .

In particular, Ix0 ∼ = Ix1 ∼ = A5 .

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Lemma 13.2. The following conditions hold:  (a) I0 = O3 (E(CG (V c))) = Ix0 , Ix1  ∼ = A8 ; (b) O2 (CV ) = 1; and (c) One of the following holds: (1) I = I0 ∼ = A8 ; or ∼ (2) I = HS or 2HS and Lemma 11.4a2 holds, with | x0 ND | = 24, and all elements of I3 (Cv ) are in xG 0. Proof. Let T (resp. TV ) be an A-invariant Sylow 2-subgroup of O3 (Cv ) (resp. O3 (CV )). Let y ∈ A# . If y satisfies (13A1), then any involution  t ∈ V centralizes A ∩ Iy , so CIy (t) ∼ = Iy or Σ3 . Therefore O3 (CCV (y)) =   A or y × Iy , and similarly for O3 (CCv (y)). Thus T ∩ O3 (CCv (y)) ≤   O3 (O3 (CCv (y))) = 1 and similarly TV ∩ O3 (CCV (y)) = 1. It follows that [CT (y), A] = [CTV (y), A] = 1. On the other hand, suppose that y satisfies (13A2). Then as m2,3 (G) < 4, NC(y,Jy ) (A; 2) = {1}, so again [CT (y), A] = [CTV (y), A] = 1. As y ∈ A# was arbitrary, [T, A] = [TV , A] = 1. By [V17 , 10.7.3], and since G has even type, [A, O3 (Cv )] ≤ O2 (Cv ) and [A, O3 (CV )] = 1. In particular, [A, Cv , O3 (Cv )] = 1. As O3 (Cv ) ≤ O2 (Cv ) = 1, we must have CA (I) = 1, whence I = 1 and so m3 (Cv ) = 2 by (10A3). Next, Ix0 ≤ L3 (CG ( v, x0 )) ≤ L3 (Cv ) by L3 -balance, and Ix0 = [Ix0 , A] centralizes O3 (Cv ), so Ix0 ≤ I. In particular x1 ∈ I, and then as x0 = xg1 , A Ix0 , Ix1  ≤ I. Since Ix0 ≤ CJx0 (v), V c and Ix0 are root A4 - and A5 -subgroups of Jx0 with complementary supports in the action of Jx0 on 9 letters. Therefore [V c , Ix0 ] = 1. By g-conjugation, (∞) ∼ A [V c , Ix1 ] = 1. As c ∈ xG = 9 0 ∩ V c we put Jc = E(CG (c)) = CG (c) and have

Ix0 , Ix1  ≤ CI (V c)(∞) ≤ CI0 (c)(∞) ≤ CG (c)(∞) = Jc . By construction [A, c] = 1, so A acts on Jc with A5 ∼ = Ixi ≤ CLc (xi ), i = 0, 1. Therefore xi acts on Jc as a 3-cycle, i = 0, 1. As m3 (A Ix0 , Ix1 ) ≤ m3 (Cv ) = 2 with A normalizing Ix0 and Ix1 , A lies in Jc ∼ = A9 as the subgroup generated by two 3-cycles x0 and x1 , and xi ∈ Ix1−i , i = 0, 1. In particular Ixi is a root A5 -subgroup of Jc , i = 0, 1. As m3 (Cv ) = 2,

Ix0 , Ix1  is then a root A8 -subgroup of Jc , CI (V c)(∞) = Ix0 , Ix1  ∼ = A8 and I is a single component. As I ∈ C2 , all the components of E(Cv ), as A8 or Suzuki groups, are locally balanced for the prime 2 [IA , 7.7.1]. Hence O2 (CV ) ≤ O2 (Cv ) = 1 by [IG , 5.18, 5.19], the latter with Cv , V , and CCv (E(Cv )) in the roles of X, P , and N . This proves (b). Now I0 = E(CI (V )) ≤ E(CV ) and as I ∈ C2 , also I0 ∈ C2 , by [IA , 7.1.10]. But I0 is c-invariant with E(CI0 (c)) ∼ = A8 , so I0 ∼ = A8 by [V17 ,  3 13.30]. As m3 (CV ) = 2, I0 = O (CV ). Thus all assertions in (a) hold. The conditions E(CI (V )) ∼ = A8 , m3 (I) = 2, and I ∈ C2 , imply that I = I0 ∼ = HS or 2HS, by [III11 , 1.6d] (note that m3 (L4 (4)) = 3). = A8 , or I ∼ If I/Z(I) ∼ = HS, then as A ∈ Syl3 (I), all elements of A# are I-conjugate

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[IA , 5.3h], so the orbits [1] and [61 ] are fused. Hence Lemma 11.4a2 holds  with | x0 ND | = 24, and the proof is complete. Lemma 13.3. The following conditions hold: (a) V ∈ Syl2 (CG (I0 )); (b) F ∗ (CV ) = V × I0 ;  (c) O 2 (CG (x0 )) = Jx0 ∼ = A9 ; and (d) NG (V ) ∼ Σ Y Σ . = 4 8 Proof. By Theorem C∗4 : Stage A1, O3 (CG (x0 )) has odd order. Thus with Lemma 11.2, if (c) fails, then there is an involution u ∈ CG (x0 ) acting on Jx0 as a transposition. Then m3 (CG (u)) ≥ m3 ( x0  Jx0 ) = 3, so by (10A3), F ∗ (CG (u)) = O2 (CG (u)). Replacing u by a Jx0 -conjugate, we may assume that in the natural action of Jx0 on 9 letters, u occurs in the cycle decomposition of v. Thus [u, v] = 1 = [u, Ix0 ], u normalizes V , and Ix0 ≤ CJx0 (u)(∞) ∼ = A7 . By Lemma 13.2a, Ix0 ≤ I0 ∼ = A8 . Hence (∞) = E(CI0 (u)) ≤ E(CG (u)), the last by L2 -balance and the Ix0 ≤ CI0 (u) fact that G has even type. But E(CG (u)) = 1, a contradiction. Therefore (c) holds. Let Q ∈ Syl2 (CG (I0 )) with V ≤ Q. As x0  Ix0 ≤ I0 , Q ≤ Jx0 by  (c), and then Q ≤ O2 (CJx0 (Ix0 )) = V , proving (a). By Lemma 13.2b, O2 (CV ) = 1, and (b) follows. Finally, NJ (V c) = V c, t ∼ = Σ4 , with ∼

c, t = Σ3 . Moreover t is a root involution in Jx0 and acts as a transposition on Ix0 . Then t normalizes I0 , and by (a), CVt (I0 ) = V , so I0 t ∼ = Σ8 and  V I0 t ∼ = Σ4 Y Σ8 . Then (a) and (b) imply (d). 14. G ∼ = A12 In this section we continue the notation of the previous section and prove: 

Proposition 14.1. If Lemma 13.2c1 holds, i.e., I = O3 (Cv ) is a component of CG (V ) isomorphic to A8 , then G ∼ = A12 . Recall from the previous section that A4 ∼ = V x2  ≤ CG (I). Then as 2 (C (x )) =: H ∼ A . Hence, I is , and by Lemma 13.3c, I ≤ O x2 ∈ xG = 9 G 2 0 the stabilizer of a point in the action of H on 9 letters. Clearly CH (V ) = I. By Lemma 13.3d, there is an involution t ∈ NG (V ) ∩ NG ( x2 ) acting nontrivially on V and x2 , and acting on I0 as a transposition. There then exists a sequence t4 = t, t5 , . . . , t10 of I0 -conjugates of t such that (14A) t2i = 1, 4 ≤ i ≤ 10; (ti ti+1 )3 = 1, 4 ≤ i < 10; (ti tj )2 = 1, 4 ≤ i < j−1 < 10. Let t1 = x2 and t2  = CV (t). Then NG (V ) = t1 , t2 , t4 , t5 , . . . , t10 , with t31 = 1, (t1 t2 )3 = 1, t22 = 1, (ti tj )2 = 1, i = 1, 2, 4 ≤ j ≤ 10. Moreover, as I0 is a root A8 -subgroup of Jx2 ∼ = A9 , we may write ∼ Σ3 Y Σ9 , where, in addition to relations NG ( x2 ) = t1 , t3 , t4 , . . . , t10  =

(14B)

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already noted, (14C)

(t1 t3 )2 = 1, (t3 t4 )3 = 1, (t3 ti )2 = 1, 5 ≤ i ≤ 10.

Let x = t9 t10 . Then x is a 3-cycle in Jx2 , so x ∈ xG 0 . We have CNG (V ) (x) = t1 , t2 , t4 , t5 , t6 , t7 , x ∼ = (Σ4 Y Σ5 ) × Z3 , and CNG (x2 ) (x) = 7

t1 , t3 , t4 , t5 , t6 , t7 , x ∼ = (Σ3 Y Σ6 ) × Z3 , with ti i=1 ≤ E(CG (x)) ∼ = A9 , in view of Lemma 13.3c. Therefore, by [V17 , 11.5.1], applied in E(CG (x)), (14D)

(t2 t3 )3 = 1.

Now set G0 = ti 10 i=1 . By [Hu1, I.19.8], the relations (14A)−(14D) present A12 . So we have proved: Lemma 14.2. G0 ∼ = A12 . We next prove Lemma 14.3. Cv ≤ G0 . 

Proof. We have I0 = I = O3 (Cv )  Cv . Let C = C(v, I). Then by Lemma 13.3 and [IG , 15.12(iii)], C = V O2 (C). But O2 (C) ≤ O2 (Cv ) = 1 as G is of even type. Hence C = V . As | Out(I)| = 2 and |CG0 (v) : I| = 8, the lemma follows.  Lemma 14.4. v G ∩ G0 = v G0 . Proof. Let z be a 2-central involution of I. Then I2 (G0 ) = v G0 ∪ z G0 ∪ (vz)G0 and all involutions of G0 are 2-central in G0 . Now z = Z(E) for some E ≤ I with E ∼ = Q8 ∗ Q8 . Let E ≤ T ∈ Syl2 (G0 ). We may assume that Z(T ) = v, z. As v has a nontrivial image in the quotient T I/I ∼ = D8 , v ∼G z. Thus it suffices to show that v ∼G vz. But Cvz := CG0 (vz) = AΣ where Σ ∼ = Σ6 and A  Cvz with A the trace-0 submodule of a natural (∞) permutation module for Σ. Hence zv = Z(Cvz ) ≤ Cvz ≤ CG (vz)(∞) . On (∞) the other hand, Cv = I and v ∈ I. Hence v ∼G (vz) and the lemma is proved.  Now we quickly complete the proof of Proposition 14.1. Suppose that G0 = G. Recall that T ∈ Syl2 (G0 ) with v ∈ Z(T ). By Lemmas 14.3 and 14.4, v ∈ U(G; G0 ; 2). In particular NG (T ) ≤ G0 , so T ∈ Syl2 (G). By Theorem SF and the remark following it [II2 , p. 24], G ∼ = A9 or G is a simple Bender group. As these conclusions are obviously false, G = G0 ∼ = A12 , completing the proof of the proposition. 15. G ∼ = F5 In this section we continue the notation of Section 13 and prove  Proposition 15.1. If Lemma 13.2c2 holds, i.e., I = O3 (Cv ) ∼ = HS ND G = [1] ∪ [3] ∪ [8] ∪ [6] ∪ [61 ], [4] ∼ [12], and or 2HS, I3 (I) ⊆ x0 , x0  ∼ , then G F . I3 (Cv ) ⊆ xG = 5 0

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We set

G∗ = F5 and Cv∗ = CG∗ (v ∗ ), where v ∗ is a fixed non-2-central involution of G. We first prove Lemma 15.2. I ∼ = Cv ∗ . = 2HS and Cv ∼ Proof. By Lemmas 13.2 and 13.3b, F ∗ (CV ) = V × I0 ∼ A8 . As [V, c] = V , we have CV = V × W with with I0 = E(CI (V c)) = ∼ W = CCV (c) = A8 or Σ8 . But by Lemma 13.3c, CG (c) does not contain a Σ8 -subgroup, so W ∼ = A8 . Now if I ∼ = HS, then CI (V ) ∼ = Σ8 [IA , 5.3m]. ∼ Therefore I = 2HS. We have O2 (Cv ) ≤ CG (I0 ) so O2 (Cv ) = CV (I) = v, with Lemma 13.3a. Therefore Cv = IV = I v   for any v  ∈ I2 (V − v), and Cv / v ∼ = Aut(I). But Cv∗ has a similar factorization in F5 ; Aut(HS) has a unique class of involutions centralizing an A8 -subgroup; and the Schur multiplier of HS has order 2 [IA , 5.3wm, 6.1.4]. Therefore Cv ∼  = Cv∗ , as claimed. ∼ A8 . Since x1−i ∈ Ix with Let z be a 2-central involution in I0 = i  3 ∼ Ixi = O (CI0 (xi )) = A5 for i = 0 and 1, inspection of A8 shows that a Sylow 3-subgroup of CI0 (z) is in yI0 , where y := x0 x1 . Replacing z by an I0 -conjugate, we may assume that y ∈ Syl3 (CI0 (z)). Set F = y, c ∈ Syl3 (CCG (c) (z)). Fix an isomorphism Cv → Cv∗ and let z ∗ be the image of z under it. Set Cz = CG (z) and Cz ∗ = CG∗ (z ∗ ). We begin the proof that Cz ∼ = Cz ∗ with the following facts about Cz : 

(15A)

(1) CCz (c) is solvable, and O3 (CCz (c)) ≤ c × CJc (z)  CCz (c), with CJc (z) = Wc y, sc , z ∈ Wc = [Wc , y] = W1 ∗ W2 , Wi ∼ = Q 8 , y ∈ I0 ≤ J c ,

y, sc  ∼ = Σ3 , and W1sc = W2 ; (2) CCz (y) ≥ y × Wy c, sy  with z ∈ Wy ∼ = Q8 ∗ Q8 , Wy = [Wy , c], c ∈ I3 (Jy ) and c, sy  ∼ = Σ3 ; (3) F = y, c, z ∈ Syl2 (CG (F )), and y g = c and cg = y for some g ∈ Cz ;

(15B)

(1) v ∈ Wy ; (2) O2 (CCz (v)) = F ∗ (CCz (v)) = v × Wv with Wv ∼ = D8 ∗ D8 ∗ D8 , and CCz (v)/O2 (CCz (v)) ∼ = Σ5 ; and (3) { y , c} ⊆ x0 G , but yc±1 are G-conjugate into Z(P ) ∩ [P, P ].

∼ Indeed as  holds  z is 2-central in I0 = CJc (V ) = A8 . By construc ±1 (15A1) ±1 = x0 x1 x2 ∈ [12], while y = x0 x1  ∈ [61 ] and c = x2  ∈ tion, yc [3], so (15B3) holds by the assumptions of Proposition 15.1. By Lemma 11.4b, there is h ∈ ND interchanging x0 with x2 , and x1 with x3 . Then

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Lemma 11.5d implies that some element of NG (F ) interchanges y = (x2 x3 )h and c = xh0 . Also Jc contains a Sylow 2-subgroup of CG (c) by Lemma 13.3c, and it is visible in A9 that |CJc (F )|2 = 2, whence z ∈ Syl2 (CG (F )). Therefore (15A3) holds by a Frattini argument, and (15A2) follows from (15A1) by g-conjugation. As y ∈ I0 ≤ Cv and c, v = V c ∼ = A4 , (15B1) follows from (15A2). Finally, (15B2) follows from the isomorphism Cv → Cv∗ and the structure of G∗ [IA , 5.3w]. Thus (15A) and (15B) hold. Now we can prove Lemma 15.3. F ∗ (Cz ) = O2 (Cz ). Proof. Otherwise, as G is of even type, Y := E(Cz ) is a nontrivial central product of C2 -groups, with m3 (Y ) ≤ m3 (Cz ) ≤ 2 by (10A3). Thus F = y, c contains every element of I3 (CCz (F )). By (15A3), CY z/z (F ) has odd order, so no component of Y is a Suzuki group. Hence Y ∩ F contains some f = 1. By (15A1,2), either Wc = [Wc , f ] or Wy = [Wy , f ]. Therefore z ∈ Y and Z(Y ) = 1. We use [IA , 6.1.4, 5.6.1, 4.10.3] to sift out all Y1 ∈ C2 such that Z(Y1 ) = 1 and m3 (Y1 ) ≤ 2, and find that Y1 /Z(Y1 ) ∼ = M12 , M22 , HJ, HS, Ru, L3 (4), or G2 (4), and in all cases m3 (Y1 ) = 2. Hence Y is of one of these isomorphism types, and CF (Y ) = 1. Unless Y /Z(Y ) ∼ = L3 (4), Out(Y ) is a 3 -group [IA , 2.5.12bc, 5.3], and in the L3 (4) case the structures of CCz (y) and CCz (c), together with [IA , 4.8.2, 4.8.4], imply that F induces inner automorphisms on Y . Therefore F ≤ Y . By (15A1), Wc = [Wc , F ] ≤ Y with Wc ∼ = Q8 ∗ Q8 . But CInn(Y ) (c) has no such F -invariant Q8 ∗ Q8 or E24 -subgroup Wc or Wc /Z(Y ), unless possibly Y ∼ = 2G2 (4) and Z(Y ) ≤ Wc (see [IA , 5.3] for the sporadic cases;  CInn(Y ) (c) is a 3-group if Y /Z(Y ) ∼ = L3 (4); and O2 (CInn(Y ) (c)) ∼ = L2 (4) or ∼ SL3 (4) if Y /Z(Y ) = G2 (4), neither of which contains Q8 ∗ Q8 ). Finally,  in the case O2 (CInn(Y ) (c)) ∼ = SL3 (4), the preimage of an E24 subgroup of Inn(Y ) in Y is not extraspecial but elementary abelian, by [IA , 6.4.4]. This completes the proof.   = R/Φ(R). Set R = F ∗ (Cz ) = O2 (Cz ) and R ∼ D8 ∗ D8 ∗ D8 ∗ D8 , and v ∈ R. Moreover, Lemma 15.4. R = Wy ∗ Wc = O2 (CG ( z, v)) = CR (v). Finally, z is a Sylow 2-center in G. Proof. Let R1 be any characteristic elementary abelian subgroup of R containing z. Then CR1 (F ) = CR (F ) = z by (15A3). By the structure of CCz (v) in (15B2), therefore, CR1 (v) = v, z = R1 or CR1 (v) =

z. (Note that if h ∈ I5 (CCz (v)), then [O2 (CCz (v)), h] ∼ = Q8 ∗ D8 and Ω1 (CO2 (CCz (v)) (h)) = v, z.) In either case, [R1 , F ] = 1, so R1 = CR (F ) =

z. As R1 was arbitrary, P. Hall’s theorem [IG , 10.3] implies that R is of symplectic type. But CR (F ) = z so R is extraspecial. Again as CR (F ) = z, CR (c) = [CR (c), y] z ≤ [O2 (CCG (c) (z)), y] z = Wc . Since R is extraspecial, CR (c) is as well, and the action of sc in (15A1)

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implies that CR (c) = 1 or Wc . As g ∈ Cz interchanges c and y, CR (y) = 1 or Wy , respectively. If CR (y) = 1, then R = CR (yc)CR (yc−1 ). But by (15B3) and Lemma 11.5c, CR (bc±1 ) does not contain Q8 ∗ Q8 . Hence, R = [R, F ] ∼ = Q8 or Q8 ∗ Q8 , so that Out(R) is 3-closed. Then Wy = [Wy , c] ≤ CR (y) = 1, a contradiction. Therefore Wy ≤ R, so Wc ≤ R and (15B1) implies that v ∈ R. Now CR (z) ≤ O2 (CCv (z)) ∼ = Z2 × D8 ∗ D8 ∗ D8 by (15B2). Therefore R = Wy ∗ Wc ∼ = D8 ∗ D8 ∗ D8 ∗ D8 and CR (v) = O2 (CCv (z)). Obviously z is a  Sylow 2-center in Cz , so it is a Sylow 2-center in G. Next, we analyze C z := Cz /R. We know the following:

(15C)

(1) F = c, y ∈ Syl3 (C z ); (2) S := CC z (v) ∼ = Σ5 ; (3) c and cy represent the C z -conjugacy classes in I3 (C z ); (4) NC z ( c)/O2 (NC z ( c)) ∼ = Σ3 × Σ3 ; (5) Elements of I3 (S) are in cC z ; and (6) CR (yc) = CR (yc−1 ) = 1.

Indeed by (15B3), E1 (F ) consists of two conjugates of x0  and two of Z(P ) ∩ [P, P ]. As F ∈ Syl3 (CCz (c)), (15C1) follows. Also g interchanges y and c, and y ∈ Jc is inverted in CJc (z), so (15C3) holds. Since z is 2-central in I0 , I, and Jc , with O2 (CG (v, z)) ≤ R by Lemma 15.4, (15C2,4) hold. As [V, x0 ] = 1 and Cv ∼ = Cv∗ has a unique conjugacy class of elements of order 3, (15C5) holds. Finally, (15C6) follows from the facts that R = Wy Wc , Wy = [Wy , c], and Wc = [Wc , y]. Lemma 15.5. C z ∼ = A5 Z2 . Moreover, for any u ∈ I3 (Cz ), u lies in a component of C z if and only if u ∈ x0 G . A5 Z2 . By [V17 , 10.1.12], E(C z ) ∼ Proof. Suppose C z ∼ = = A6 , A7 , or A8 . Now CC z (E(C z )) ≤ CC z (A) so by (15C4), CC z (E(C z )) ≤ O2 (C z ) = 1. Let H be an A6 -subgroup of E(C z ) containing F . Then a Sylow 5-subgroup f of H is a Sylow 5-subgroup of C z , so by (15C2), CR (f ) = 1. By [V17 , #

10.1.13], CR (x) = 1 for any x ∈ F , contradicting (15C6). Therefore, Cz ∼ = Σ5 , E(S) is a diagonal of E(C z ). The last assertion = A5 Z2 . As S ∼ of the lemma then follows from (15B3) and (15C3,5).  ∼ A5 , and We can be more specific. Write E(C z ) = X + × X − with X ± = (by Lemma 15.5) yc±1 ∈ X ± .

Lemma 15.6. The image of C z in Out(R) is determined up to Out(R)conjugacy. Proof. Let H = Out(R) ∼ = = O8+ (2). For any h ∈ I5 (H), E(CH (h)) ∼ A5 ; moreover, there are exactly two conjugacy classes of subgroups of H of

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order 5, distinguished by whether they have trivial or nontrivial fixed points  (see [IA , 4.8.2]). As C  (yc±1 ) = 1 by (15C6), X + and X − have no on R R  Therefore neither do F + ∈ Syl5 (X + ) and trivial composition factors on R. F − ∈ Syl5 (X − ). Hence the image of X + × X − in H is determined up to conjugacy as E(CH (F + ))E(CH (F − )), with F − ∈ Syl5 (E(CH (F + ))). Now,  CH (X + X − ) = 1, and there is an involution t ∈ [H, H] such that X +X − t ∼ = Σ5 Y Σ5 . As Out(X + X − ) ∼ = Out(A5 ) Z2 ∼ = D8 , there are X +X − exactly two subgroups of H  containing   and isomorphic to A5 Z2 . We write them as X + X − t1 and X + X − t2 , where the ti ∈ I2 (H), and we may assume that t2 = t1 t. Hence Sylow 5-subgroups of CX + X − (t1 ) and CX + X − (t2 ) are not X + X − -conjugate. This in turn implies that one of these  and the other does not. Sylow 5-subgroups has nontrivial fixed points on R Without loss we choose notation so that CX + X − (t1 ) has a Sylow 5-subgroup  without fixed points on R. But we know that the Σ5 -subgroup S of C z has a nontrivial fixed point  on R. Therefore a Sylow 5-subgroup ofa Z2 × A5 -subgroup of C z has no  whence C z = X + X − t1 and the proof is complete.  fixed points on R, Now we can identify Cz up to isomorphism using cohomology as in [IG , Section 31]. Lemma 15.7. Let t ∈ I2 (Cz − E(C z )) and x ∈ I2 (X + ). Then the restriction mappings from C z to t and x yield an injective homomorphism  ψ : H 2 (C z , Z2 ) → H 2 ( t , Z2 ) × H 2 ( x , Z2 ). Proof. It suffices to show that any extension 1 → Z2 → Y → C z → 1 splits if involutory preimages t and x of t and x exist in Y . As SL2 (5) is the universal covering group of A5 and x2 = 1, the preimage of X + is Z2 × Y1 with A5 ∼  = A5 Z2 , as desired. = Y1   Y . Then Y1 , t ∼ Lemma 15.8. We have Cz ∼ = Cz ∗ . Proof. By Lemma 15.6, there is an isomorphism φ : Cz / z → Cz ∗ / z ∗  which is obtained by identifying these groups with subgroups of Aut(R) and Aut(O2 (Cz ∗ )), respectively. Choose a bijection φ : Cz → Cz ∗  and which is an isomorphism if possible. Let T be a transversal which lifts φ, to R in Cz containing 1. Then the equation φ(t1 t2 ) = f (t1 , t2 )φ(t1 )φ(t2 ), , z ∗ ),

t1 , t2 ∈ T

and by our choice of φ, φ is defines a normalized cocycle f ∈ z an isomorphism if and only if f is cohomologically trivial (see [IG , Section 31]). Choose x and t as in Lemma 15.7; by Lemma 15.5 and (15A3), we may choose t to centralize c and invert y. We use Lemma 15.7 to show that f Z 2 (C

16. THEOREM C∗4 : STAGE A5. RECOGNITION

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is trivial. Let d ∈ F # map on a generator of F ∩ X − , so that by Lemma 15.5, d is 3-central in G. By Lemma 11.5b, the preimage X+ of X + in Cz is the product of O2 (Cz ) with CX+ (d) ∼ = SL2 (5), with centers identified. The same argument applies in G∗ , so X+ and its counterpart in Cz ∗ are isomorphic. As |H 2 (A5 , Z2 )| = 2, it follows from [IG , 31.11] that f ↓X + is cohomologically trivial, so f ↓x is as well. Likewise by a Frattini argument we may choose the preimage t of t to normalize c, y. Then t2 ∈ CO2 (Cz ) ( c, b) = z, and by our choice of t,  [t, c] = 1 and t inverts y. Thus, t ∈ O2 (CG (c)) = Jc ∼ = A9 by Lemma 13.3c, ∼ so t ∈ CJc (z). But as z is 2-central in Jc = A9 with [z, y] = 1, we have CJc (z) ∩ NJc ( y) ∼ = Z2 × Σ3 , whence t is an involution. Exactly the same reasoning in G∗ shows that φ(t) is an involution in Cz ∗ . Therefore f ↓ t is cohomologically trivial, and the proof is complete.  As the final step in proving G ≈ G∗ , we prove: Lemma 15.9. There is an isomorphism φ : Cz → Cz ∗ such that for all involutions t, t ∈ Cz , t ∼G t if and only if φ(t) ∼G∗ φ(t ). Proof. Let ψ : Cz → Cz ∗ be any isomorphism. It is enough to show that (15D)

(1) I2 (G) = v G ∪ z G ; and (2) For any t ∈ I2 (Cz ), t ∈ z G if and only if ψ(t) ∈ ∗ (z ∗ )G . (∞)

In order to prove (15D2) for t ∈ Cz − Cz , we may have to alter ψ. We first consider (15D) for involutions in R. Let d = yc ∈ I3 (Cz ). Then by (15C6) and Lemma 15.5, CR (d) = z and we may assume that d ∈ X + .  has 135 singular vectors as an orthogonal space, As R ∼ = D8 ∗D8 ∗D8 ∗D8 , R and they are permuted without fixed points by a Sylow 5-subgroup Y ∼ = E52 0 := C  (Y 0 ) = 1, R 0 is a nondegenerate of C z . For any Y 0 ∈ E1 (Y ) with R R 4-dimensional orthogonal space of − type, so Y transitively permutes its 5 singular vectors. There are exactly two such subgroups Y 0 , and they are interchanged in NC z (Y ) as NC z (Y )/Y ∼ = D8 . Therefore   (15E) Cz is transitive on J := u ∈ I2 (R − z) |CCz (u)|5 = 1 . The corresponding assertion holds in Cz ∗ , and so using the fact that (15C2) holds in G and G∗ , ψ(v) is a non-2-central involution of G∗ . As v ∈ J, J ⊆ v G . Note also that there are 62 choices for Y , each giving rise to 20 elements of J; but CC z (v) ∼ = Σ5 has 6 Sylow 5-subgroups, each contained in a diagonal of X + ×X − and hence contained in a unique Sylow 5-subgroup of C z . Hence, |J| = 62 · 20/6 = 120. Now O2 (CE(Cv ) (z))/ v is not of maximal class [IA , 5.3m], so it contains a four-group U/ v  T / v for some T ∈ Syl2 (CCz (v)) [IG , 10.11]. Then

15. G ∼ = F5

395

|CE(Cv )/v (U )|2 ≥ 12 |T / v |2 , whence the elements of (U/ v)# are E(Cv )conjugate to z v [IA , 5.3m]. Thus, z G ∩ U covers (U/ v)# , whence (R − {z} − J) ∩ z G = ∅. As CR (d) = z, every Cz -orbit on R − z − J has order divisible by 2 · 3 · 52 . As 120 + 150 = 2 · 135, there is a unique such orbit, it ∗ lies in z G , and likewise, its image under ψ lies in (z ∗ )G . Therefore (15E2) holds for all t ∈ I2 (R), and I2 (R) ⊆ z G ∪ v G . Now consider the case t ∈ Cz − O2 (Cz ). We show first that the image t ∈ C z = Cz /R does not lie in a component X of C z . For if it did, then t would invert a C z -conjugate of d, and therefore  Thus all involutions in the coset tR/ z would be conjugate, act freely on R. so all would be images of involutions of tR. However, by Lemma 11.5b, the preimage of X has an SL2 (5)-subgroup containing z; this subgroup in turn contains square roots of z, a contradiction. Therefore, either t ∈ E(C z ) with t ∈ X + ∪ X − , or t ∈ E(C z ). In either case, t inverts an element dt of order 3, which we fix, on the diagonal of X + × X − . Hence t inverts a preimage dt of dt of order 3. By Lemmas  15.5 and 11.2, O 3 (NG ( dt )) = dt  × Jdt ∼ = Z3 × A9 . If t induces an even permutation on Jdt , then for some u ∈ Jdt with [t, u] = u2 = 1, [Jdt , tu] = 1. G Choosing any 3-cycle x ∈ Jdt , we have x ∈ xG 0 as [3] ⊆ x0  , and CG (x) contains x × dt , tu × E(CG ( x, dt )) ∼ = Z3 × Σ3 × A6 . But by Lemma  13.3c, O 2 (CG (x)) ∼ = A9 , which has no Σ3 × A6 subgroup, contradiction. Therefore t acts on Jdt as a transposition or the product of three disjoint transpositions. In either case, t centralizes a 3-cycle, so tG ∩ Jx0 = ∅ by Lemma 13.3c again. But v ∈ Jx0 is a root involution and z ∈ Jx0 is 2-central, so I2 (Jx0 ) ⊆ v G ∪ z G . In particular we have established (15D1). If t acts as a transposition on Jdt , so that CJdt (t) contains A7 , unlike Cz , then t ∈ v G . If t acts as the product of three transpositions, then t centralizes a product f of three disjoint 3-cycles. Thus f  ∈ [4], so f  ∈ x0 G and then t ∈ z G by the assumptions of Proposition 15.1 and (15D1). (Note that ND controls fusion of x0 in D by [IG , 16.9].) Therefore to complete the proof it is enough to show that t acts on Jdt as a transposition if and only if ψ(t) acts on E(CG∗ (ψ(dt ))) as a transposition. Consider the case t ∈ E(C z ). Then t inverts a Sylow 3-subgroup F 0  so I2 (tR) ⊆ tR z. of C z . As before as CR (yc) = 1, t acts freely on R, Moreover, t and tz both invert an element w ∈ I3 (CJdt (z)) with w ∈ F 0 . Thus w ∈ Jdt ∼ = A9 is the product of two disjoint 3-cycles, so both t and tz act on Ldt as the product of three transpositions. But clearly ψ(t) maps into E(Cz ∗ /O2 (Cz ∗ )), and the same argument is valid in G∗ , so (15D2) is valid in this case.

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Finally, suppose that t ∈ E(C z ). Then t is uniquely determined up to conjugacy in C z ∼ = A5 Z2 [IG , 10.31]. Set Rdt = CR (dt ). In view of (15C6), Rd t ∼ = Q8 ∗ Q8 and Z(Rdt ) = z. As z G ∩ V = ∅, z is 2-central in Jdt , whence Q8 ∗ Q8 ∼ = O2 (CJd (z)) = Rdt . Note that t acts freely on R/Rdt , so all involutions in tR are R-conjugate into tRdt . Thus to check (15D2) on the entire coset tR, it suffices to check it on tRdt . Set I = I2 (tRdt ), Iz = I ∩ z G , and Iv = I ∩ v G . What we need is ∗

ψ(Iz ) = ψ(I) ∩ (z ∗ )G .

(15F)

We claim that Iz and Iv are Rdt -orbits and Iv = zIz . To see this, set Y = NG ( dt ), Y = Y /O2 (Y ) ∼ = Σ9 and X = CY (z). As t inverts dt ,  t inverts no element t ∈ O 2 (Y ); but dt = 1, and from the structure of C z ,       of X of order 3, so  t ∈ O2 (X). Thus Rdt  t ≤ O2 (X). By [V17 , 11.5.2],    Rd = O2 (C 2  (z)) and Rd permutes I in two orbits, one containing a t

O (Y )

t

transposition w  and the other containing the non-transposition w z . But we have seen that involutions inverting dt lie in z G if and only if they act on Jdt as transpositions. Hence the claim holds. Now ψ(Iz ) and ψ(Iv ) = ψ(Iz )z ∗ are the ψ(Rdt )-orbits in I2 (ψ(tRdt )). A similar argument in G∗ shows that exactly one of these orbits lies in ∗ ∗ (z ∗ )G . If ψ(Iz ) ⊆ (z ∗ )G , then (15F) holds. Otherwise, define ψ  : Cz → Cz ∗ by ψ  (x) = ψ(x)α(x), where α : Cz → z ∗  is the unique surjective homomorphism. Then ψ  ↓O2 (Cz ) = ψ ↓O2 (Cz ) , (15F) holds for ψ  , and the proof is complete.  Now the proof of Proposition 15.1 is complete. Indeed, Lemmas 15.2, 15.8, and 15.9 show that G ≈ F5 . By one of our Background Results [I1 , Secs. 15–18], this implies that G ∼ = F5 . Lemma 13.2 and Propositions 14.1 and 15.1 complete the proof of Proposition 13.1. 16. The Sp6 (2) Case

In the next three sections we return to the setup (10A) and (10B) and finish the last case, namely, that p = 3 and Lo3 (G) contains Sp6 (2). We fix H, A, and B as in (10B1), again fix x0 ∈ A with Jx0 ∼ = A9 or Sp6 (2), and set x1  = A ∩ Jx0 . We continue the notation introduced in the first paragraphs of Section 11. In view of Lemmas 11.2, 11.4, and 12.1, and Proposition 13.1, we assume that  ∼ O3 (CG (x1 )) = ∼ Z3 × Sp6 (2), (16A) O 3 (CG (x0 )) = and for all y, y  ∈ A − x0  − x1  = {x00 x11 | 0 , 1 = ±1}, we have Jy /O3 (Jy ) ∼ = Jy /O3 (Jy ) ∼ = A6 or (16B)    O 3 (CG (y)) ∼ = O3 (CG (y  )) ∼ = O3 (CG (x0 )). Thus we prove Proposition 16.1. If Jx0 ∼ = Sp8 (2) or F4 (2). = Sp6 (2), then G ∼

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397

Again x0 has an ND -conjugate in Jx0 by Lemma 11.1. Therefore x0 is inverted in G. Now Out(Jx0 ) = 1 [IA , 2.5.12], so ∼ Σ3 × Sp6 (2) and NG ( x0 )/O{2,3} (CG (x0 )) = (16C) CG (x0 ) = x0  × Jx0 × O{2,3} (CG (x0 )). We fix (16D)

z ∈ I2 (NG (D)) inverting x0 and centralizing Jx0 .

Since [D, z] = x0 , CND (z) = z × NJx0 ∼ = E22 × Σ4 , where NJx0 = AutL (D) ∼ = Z2 × Σ4 , the Weyl group of type C3 . Moreover, NJx0 contains two conjugacy classes of reflections, with centers conjugate to x1  ∈ [3] and

x1 x2  ∈ [6], respectively. Lemma 16.2. ND ∼ = W (C4 ), W (F4 ), or GO+ (D). Moreover, the stabilizer in ND of [1] ∪ [3] is the full monomial group 24 Σ4 ∼ = W (C4 ). Proof. We have |NND ( x0 )| = 25 .3, so |ND | = 25 .3| x0 ND |. But | x0 ND | > 2, so |ND | > 26 .3 = |W (D4 )| = |W (D4 )∗ |. Thus if Lemma 11.4a1 holds, then ND ∼ = W (C4 ), the monomial group on [1] ∪ [3]. So assume that Lemma 11.4a2 holds, whence O2 (GO+ (D)) ≤ ND ≤ GO+ (D) and | x0 ND | = 12c, c = 1 or 2. Therefore |ND | = 27 32 c and |GO+ (D) : ND | ≤ 2. But H := GO+ (D)/O2 (GO+ (D)) ∼ = D8 , and as O2 (GO+ (D)) ≤ + + [GO (D), GO (D)], a reflection in ND maps to a noncentral involution of H. Therefore there is a unique possibility for ND if ND < GO+ (D), and this possibility must be W (F4 ). Finally, in any case, the stabilizer of [1] ∪ [3] in ND contains NJx0 (the full monomial group on [3]), the reflection z, and the involution g from Lemma 11.4b, interchanging x0  and x1 . These generate the desired monomial group on [1] ∪ [3]. The lemma is proved.  17. G ∼ = Sp8 (2) To construct an Sp8 (2)-subgroup of G from ND and Jx0 we use a slight variant of the Gilman-Griess argument [IA , 2.9.9]. Let Σ be a root system of type C4 , W its Weyl group, and Σ1 a subsystem of Σ of type C3 . A variant is needed because there are pairs (β, γ) of roots in Σ for which no w ∈ W can be found such that {β w , γ w } ⊆ Σ1 . Indeed if we take Σ = {±2ai | 0 ≤ i ≤ 3} ∪ {±ai ± aj | 0 ≤ i < j ≤ 3}, and let Σ1 be the corresponding set in which the subscripts range only from 1 to 3, then up to W -conjugacy there is a unique such pair (β, γ), namely (17A)

(β, γ) = (a0 + a1 , a2 + a3 ).

Now in the group Σ(2) ∼ = Sp8 (2), we have [Xβ , Xγ ] = 1, and our variant will make an analogous assumption about “root groups” in G. We shall use the following theorem.

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Theorem 17.1 (cf. [IA , 2.9.9]). Let X be a group containing subgroups ∼ K1 = Sp6 (2) and R. We use standard notation [IA , 2.10] for K1 , with root system Σ1 . Let Σ be a root system of type C4 containing Σ1 , with Weyl group W ∼ = W (C4 ). Let W1 ≤ W be the Weyl group of Σ1 . Suppose that δ : R → W is an isomorphism, and let R act on Σ by αr = αδ(r) , r ∈ R. For each α ∈ Σ choose some r ∈ R and α ∈ Σ1 such that α = αr , and set −1 −1 Yα = (Xα )r (thus Yα ≤ X) and yα (1) = xα (1)r . Assume the following: (a) δ(R ∩ K1 ) = W1 . Moreover, R ∩ K1 permutes the set {Xα | α ∈ Σ1 } by conjugation in K1 the same way it acts on Σ1 via δ; (b) For each α ∈ Σ1 the stabilizer Rα of α in R normalizes Xα in the group X; and (c) [Ya0 +a1 , Ya2 +a3 ] = 1.   Set I = K1R ≤ X. Then I ∼ = Sp8 (2). Proof. The hypotheses differ from those of [IA , 2.9.9] only in the following two ways. First, we have here specialized to the case Σ(q) = C4 (2) and Σ1 (q) = C3 (2), with R now assumed isomorphic to W (C4 ). Second, we have added the assumption (c), in view of the discussion above. The proof of [IA , 2.9.9] may be copied to show that conditions (a) and (b) imply that:

(17B)

(1) The Yα are independent of the choices of α and r; (2) For any r ∈ R and α ∈ Σ, (Yα )r = Yαδ(r) ; and (3) For any linearly independent pair of roots (α, β) ∈ Σ not W -conjugate to (a0 + a1 , a2 + a3 ), the analogue of the Chevalley commutator formula for [Xα , Xβ ] in Sp8 (2) holds for [Yα , Yβ ] in X.

Now for any w ∈ W , if we let r = δ −1 (w), then [Y(a0 +a1 )w , Y(a2 +a3 )w ] = [Yar0 +a1 , Yar2 +a3 ] = [Ya0 +a1 , Ya2 +a3 ]r = 1. Together with (17B3), this implies that the Yα , α ∈ Σ, satisfy all instances of the Chevalley commutator formula. Since Σ has rank bigger than 2, I is a homomorphic image of the universal version C4 (2), by [IA , 2.9.6]. But that  universal version is the simple group Sp8 (2). The proof is complete. By [V17 , 10.2.12], D ∩ Jx0 is self-centralizing in Aut(Jx0 ). Hence by (16C), CG (D) = D × O{2,3} (CG (D)). Hence NG (D) is 3-solvable. Let N0 be a Hall {2, 3}-subgroup of NG (D), and let N1 be the subgroup of N0 mapping on the stabilizer of [1] ∪ [3], so that N1 /D ∼ = W (C4 ). Using N1 and Theorem 17.1, we shall prove:   Proposition 17.2. Let G1 = JxN01 . Then G1 ∼ = Sp8 (2). Moreover, CG (z) = Jx0 Q where Q = O2 (CG (z)). We have Z(Q) ∼ = E27 , and either Q = Z(Q) or Q/ z ∼ = E28 irreducible for Jx0 . = E214 with Q/Z(Q) ∼

17. G ∼ = Sp8 (2)

399

Using standard notation for elements of Jx0 , let R1 = nα (1) | α ∈ Σ1  ∼ = W (Σ1 ) = W1 , (17C) nα (1) → wα . Also, set Sα = Xα , X−α  for α ∈ Σ1 , and S2a0 = x0 , z, where z is as in (16D). Then R1 normalizes S := S2a0 × S2a1 × S2a2 × S2a3 , the direct product of four copies of Σ3 . We may assume that we originally chose D = O3 (S). Thus R1 is a complement to D ∩ Jx0 in NJx0 (D ∩ Jx0 ). Note that all such complements are D-conjugate, being normalizers of Sylow 2-subgroups of O3,2 (NJx0 (D ∩ Jx0 )). Lemma 17.3. R in N1 such that R1 = R ∩ Jx0 .  D has a complement  R Moreover, JxN01 = JxDR = J . x0 0 utz’s theorem implies that Proof. Since |N1 : D|3 = 3 = |R1 |, Gasch¨ N1 splits over D. Then R ∩ Jx0 = O3 (CR (x0 )) is a complement to D ∩ Jx0 in NJx0 (D ∩ Jx0 ), like R1 . Hence by the preceding paragraph R ∩ Jx0 is D-conjugate to R1 . Replacing R by a suitable D-conjugate, therefore, we have the lemma.  Lemma 17.4. Hypothesis (a) of Theorem 17.1 holds with Jx0 playing the role of K1 . Moreover, δ(R ∩ z, x0 ) = w2a0 . Proof. By Lemma 17.3, the isomorphism R1 → W1 taking nα (1) to wα for each α ∈ Σ1 (see (17C)) extends to an isomorphism δ : R → W . This implies the first assertion. Notice that w := δ(R ∩ z, x0 ) ≤ CW (δ(R1 )) = CW (W1 ). But CW (W1 ) = w2a0  × −1, with the diagonal element in W1 . As w ∈ W1 and w = −1, the lemma follows.  Lemma 17.5. Hypothesis (b) of Theorem 17.1 holds. Proof. Let α ∈ Σ1 . The stabilizer δ(Rα ) of α in δ(R) is generated by all the reflections wγ , γ ∈ Σ ∩ α⊥ . Thus it is certainly sufficient to show that [δ −1 (wγ ), Xα ] = 1 for each such γ. If γ ∈ Σ1 , then δ −1 (wγ ) = nγ (1) and the desired commutativity holds because γ ⊥ α implies [X±γ , Xα ] = 1 (as we are in characteristic 2). Likewise if γ = ±2a0 , then δ −1 (wγ ) ∈

x0 , z ≤ CG (Jx0 ) ≤ CG (Xα ), as required. So assume that γ ∈ Σ − Σ1 − {±2a0 }. Renumbering the indices 1, 2, 3 if necessary, we may assume that γ = ±a0 ± a1 and α ∈ Za2 + Za3 . For any σ ∈ Σ1 of the form σ = ±ai ± aj , we have [S2a0 , δ −1 (σ)] ≤ [S2a0 , R1 ] = 1, and conjugating this relation by various elements of R1 we get that δ −1 (γ) centralizes A∗ := O3 (S2a2 S2a3 ). Since δ −1 (γ) normalizes S, it normalizes O3 (S2a0 S2a1 ) = A. Moreover, Xα   is contained in O2 (CG (A)) = O2 (CJx0 (A)) ∼ = Σ6 , in which A∗ is a Sylow

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3-subgroup. As Sylow 3-subgroups of Aut(Σ6 ) are self-centralizing and γ is   an involution, [δ −1 (γ), Xα ] ≤ [δ −1 (γ), O2 (CG (A))] = 1. As remarked above, Lemma 17.5 implies that (17B) holds. To prove that G1 ∼ = Sp8 (2), it remains to check hypothesis (c) of Theorem 17.1, Choosing g ∈ R interchanging x0 and x1 , we see that g interchanges x0 , z and S2a1 , and hence δ(g) interchanges 2a0 and 2a1 . Conjugating I2 (S2a1 ) = {x2a1 (1), x−2a1 (1), δ −1 (w2a1 )} by g, we have I2 ( x0 , z) = {y2a0 (1), y−2a0 (1), δ −1 (w2a0 )}. There is then no loss in replacing z and assuming (17D)

z = y2a0 (1).

To check hypothesis (c) of Theorem 17.1, note that [Ya0 +a1 , Y2ai ] = [Ya2 +a3 , Y2ai ] = 1 for all 0 ≤ i ≤ 3, by (17B3); we set (17E)

E = Y2ai | 0 ≤ i ≤ 3, C = CG (E), and show that C is abelian.

Set z = Cz / z, Q = O2 (Cz ) and Z = Z(Q). Cz = CG (z), C By (17D), z ∈ E, so C ≤ Cz . As m3 (Cz ) ≥ m3 (Jx0 ) = 3, Q = F ∗ (Cz ) and CJx0 (Q) = 1, by (10A3). For distinct i, j = 1, 2, 3, set Ci = O2 (CCz (xi )), Cij ± = O2 (CCz (xi x±1 j )), and Ci,j = O2 (CCz ( xi , xj )). As Q  Cz , Qi := CQ (xi )  Ci , Qij ± := CQ (xi x±1 j )  Cij ± , and Qi,j := CQ ( xi , xj )  Ci,j . As R1 induces the full monomial group on [3], with [R1 , z] = 1, all Qi are isomorphic, as are all Qij ± , and all Qi,j . We next prove Lemma 17.6. The following conditions hold: (a) |Q| = |Q2 |2 |Q23+ |2 |Q2,3 |−3 ; (b) Qi = Ci ∼ = E25 and Qij ± = Cij ± for all distinct i, j = 1, 2, 3; and (c) |Q| = 27 or 215 , according as Jx0 x1 /O3 (Jx0 x1 ) ∼ = A6 or Sp6 (2). Proof. The isomorphisms just noted, and the Brauer-Wielandt formula applied to the action of x2 , x3  on Q, yield (a). Notice that |Q| ≥ 27 since Jx0 acts faithfully on Q. Choose g ∈ R interchanging x0 with x1 , and interchanging x2 with x3 . Then g interchanges Jx0 with Jx1 , S2a0 with S2a1 , and S2a2 with S2a3 . Hence CCz (x1 )g = CG ( z g , x0 ), with z g ∈ S2a1 . Thus z g is a long root involution (i.e., transvection) in Jx0 ∼ = C3 (2). = Sp6 (2) ∼  2 Therefore O (CCz (x1 )) = O2 (CCz (x1 ))P1 , where P1 ∼ = C2 (2) = Sp4 (2) ∼ contains S2a2 and S2a3 as long root A1 (2)-subgroups, and O2 (CCz (x1 )) = C1 ∼ = E25 is an indecomposable extension of z by a natural P1 -module. Hence, C1,2 ∼ = E23 , on which x3  acts faithfully. As Q1,2  CCz ( x1 , x2 ), |Q1,2 | = 8 or 2. Likewise Q1  CCz (x1 ), so Either Q1 ∼ = E25 with Q1,2 ∼ = E23 , or Q1 = Q1,2 ∼ = Z2 . (17F)

17. G ∼ = Sp8 (2)

401

Whichever of these alternatives holds is also valid with 1 and 2 replaced by any distinct i, j = 1, 2, 3, by the monomial action of R1 . If A centralizes Jx0 x1 /O3 (Jx0 x1 ), then by the monomial action of R on  D, Jx2 x3 /O3 (Jx2 x3 ) ∼ = A6 , and O2 (CG (x2 x3 )/O2 (CG (x2 x3 ))) ∼ = Z2 × Σ6 , with z ∈ S2a0 acting on Jx2 x3 /O3 (Jx2 x3 ) as a root involution. Moreover, as [z, x2 , x3 ] = 1, C23+ = C2,3 ∼ = E23 for some transposition = O2 (CΣ6 (z  )) ∼  z ∈ Σ6 . Therefore |Q2,3 | ≤ |Q23+ | = 8 or 2. Using (17F), we conclude that |Q2,3 | = |Q23+ | = 8, |Q2 | = 25 , and the lemma holds in this case with |Q| = 27 . We may therefore assume that Jx0 x1 ∼ = Sp6 (2). Now x0 was chosen in Section 11 to be an arbitrary element of A# with a centralizer having an Sp6 (2) (or A9 ) 3-component. So (16C) applies to x0 x1 , and then x2 x3 by the  action of R1 . Thus, O2 (CG (x2 x3 )) = Jx2 x3 . In particular x0  = [z, D] ≤ Jx2 x3 and z acts on D ∩ Jx2 x3 as a reflection. Therefore z lies in a root A1 (2) subgroup of Jx2 x3 ∼ = C3 (2), and as |CJx2 x3 (x0 )|2 ≤ |CCx0 (x2 x3 )|2 = 4, z is a short root involution in Jx2 x3 ∼ = C3 (2). Hence CJx2 x3 (z) is a parabolic subgroup of Jx2 x3 of type A1 × A1 , C23+ ∼ = (Q8 ∗ Q8 ) × E22 , C2,3 ∼ = E23 , and CCz (x2 x3 ) acts irreducibly on both Z(C23+ )/ z and C23+ /Z(C23+ ) (see [IA , 2.6]). In particular, if Q23+ = z, then Q2 = z by (17F), and then |Q| = 2 by (a). This contradicts |Q| ≥ 27 , so |Q23+ | > 2. Assume that Q23+ < C23+ . Then as Q23+  CCz (x2 x3 ), the only possibilities are Q23+ = Q2,3 = C2,3 ∼ = E23 or Q23+ ∼ = Q8 ∗Q8 . In the former case, as 7 7  ∼ |Q| ≥ 2 , |Q| = 2 by (a), and Cz /Q embeds in Aut(Q) = GL6 (2). In the latter case, x2 , x3  normalizes no E23 subgroup of Q23+ , so Q2 = z by (17F). and Cz /Q embeds in Out(Q) ∼ But then Q = Q23+ Q23− ∼ = O8+ (2). In = 21+8 + ∼ either case, as |C23+ : Q23+ | > 2, Sp6 (2) = Jx0 Q/Q < Cz /Q. We also have  m3 (Cz ) = 3, and O2 (CCz /Q (x1 )) ∼ = Σ6 , as seen above. [V17 , 10.2.13], applied with Cz /Q here in the role of M there, then yields the contradiction Jx0 Q/Q = Cz /Q. Therefore Q23+ = C23+ has order 27 , whence by (17F), Q2,3 ∼ = E23 , 15 ∼ Q2 = E25 and |Q| = 2 . So (b) and (c) hold and the proof is complete.  Now we can prove Lemma 17.7. The following conditions hold: (a) Cz = Jx0 Q; (b) C = CJx0 (E)CQ (E); (c) Z ∼ = E27 ;  (d) Q is elementary abelian; and (e) Q/Z(Q), if nontrivial, is irreducible under the action of Jx0 . Proof. By (17D) and definition of E (17E), E ≤ z × Jx0 , so E normalizes Jx0 and Q. Hence (a) implies (b).  Q is extraspecial or Suppose that |Q| = 27 . By the action of Jx0 on Q, elementary abelian. But m2 (Q) ≥ 5 by Lemma 17.6b, so Q ∼ = E27 , whence

402

16. THEOREM C∗4 : STAGE A5. RECOGNITION

 with Y2a ≤ (c). Moreover, R1 ∼ = Σ3 Σ3 so R1 acts irreducibly on Q, i 4  S2ai   R1 , |CQ (S2ai )| = 2 , and [Q, Y2ai , Y2aj ] = 1 for each i, j = 1, 2, 3. Since CCz (x1 ) ∼ = E25 (Z3 × Sp4 (2)), [V17 , 10.2.13] implies (a). As (d) and (e) are obvious, the lemma holds in this case. By Lemma 17.6c, we may assume that |Q| = 215 , and as we saw in the proof of that lemma, Q23+ ∼ = (Q8 ∗Q8 )×E22 , Q2,3 = Z(Q23+ ), and Q2 ∼ = E25 . # In particular, Q2,3 ≤ Z(CQ (y)) for all y ∈ x2 x3  , so Q2,3 ≤ Z. Therefore CZ (x2 x±1 3 ) = Q2,3 , while Q2,3 ≤ CZ (xi )  CCz (xi ) and thus CZ (xi ) = Qi for i = 2, 3. Then CQ/Z (xi ) = 1, i = 2, 3, and Z ∼ = E27 by the Brauer-Wielandt formula, proving (c). As any nontrivial F2 Jx0 -module has dimension at least  is elementary abelian by [V17 , 10.2.14], proving (d). 6, Q/Z ∼ = E28 . Then Q 23− are conjugate under Jx0 , and Jx0 does not embed in Since Q23+ and Q L4 (2), Q/Z(Q) is irreducible, proving (e). Let y ∈ x2 , x3 # be arbitrary. Since |Q| = 215 , Jy ∼ = Sp6 (2), so CG (y) = y × Jy × Wy , where Wy = O{2,3} (CG (y)). It follows that Wy = O{2,3} (CG (D)). As y is arbitrary,    O{2,3} (CG (D)) centralizes O2 (CG (y)) | y ∈ x2 , x3 # and hence O{2,3} (CG (D)) ≤ O2 (CG (Q)) = 1. Thus CG (y) = y × Jy . Let W = CG (Z) = CCz (Z). Now Jx0 ∼ = W Jx0 /W ≤ Cz /W and Cz /W acts faithfully on Z. Another application of [V17 , 10.2.13] yields Cz = W Jx0 . As Jx0 contains the Sylow 3-subgroup x1 , x2 , x3  of CCz (x1 ), and W ∩Jx0 1, W is a 3 -group. Then for any y ∈ x2 , x3 # , CW (y) ≤ O3 (CG (y)) ∩ Cz CJy (z), in which all 3 -composition factors are 2-groups. As W  Cz , W a 2-group, so W = Q. Thus (a) holds and the proof is complete.

= = is 

 is abelian. Lemma 17.8. C  = 26 and Y2a ≤ S2a for all i, it follows from [V17 , Proof. As |Z| i i  Y2a ] ∼  := 10.2.11, 10.2.15] that  ui  := [Z, Z2 for all i = 1, 2, 3, and R = i 3 ∼ 2 , u 3  = ∩i=1 CZ (Y2ai ) = E23 . Then [ ui , C] ≤ [ ui , CCz (Y2ai )] = 1 for all

 u1 , u  So C stabilizes the chain  C] ≤ [Z,  CC ( Y2a , Y2a , Y2a )] ≤ R. i, and [Z, z 1 2 3 >R  > 1. The Three Subgroups lemma implies that [C, C] ≤ CC (Z).  In Z z  = 1. As C = (C ∩ Jx )(C ∩ Q) and particular [C ∩ Jx0 , C ∩ Jx0 ] ≤ CJx0 (Z) 0  ∩ Q]  = 1.  Q is abelian, by Lemma 17.7bd, it suffices to show that [C ∩ Jx0 , C  shows that C  ∩ Z = R,  which is centralized The action of the Y2ai ’s on Q by C ∩ Jx0 . In particular, we are done if |Q| = 27 . ∗ = Suppose then that |Q| = 215 . Let E ∗ = S2a1 S2a2 S2a3 , so that E S2a1 × S2a2 × S2a3 , with CQ/Z (O3 (S2ai )) = CQ/Z (xi ) = 1 for each i = 1, 2, 3, by Lemma 17.6b. It follows that Q/Z ∼ = V1 ⊗ V2 ⊗ V3 , with Vi a natural S2ai ∼ = L2 (2)-module centralized by the other two direct factors. Therefore

17. G ∼ = Sp8 (2)

403

 a free module of rank 1. Hence Q =B  ⊕ Z,   as F2 E-modules, Q/Z ∼ = F2 [E], ∼       with B = F2 [E]. Now [Z, E, E] = [R, E] = 1, whereas CQ (E) = B0 R, with  E, E]. Therefore B 0 = C  (E) ∩ [Q,  E, E] is centralized 0 = C  (E) ≤ [B, B B

Q

0 R  is then centralized by CJ (E), and the by CJx0 (E). But CQ (E) = B x0 proof is complete. 

Now we quickly complete the proof of Proposition 17.2. By Lemma 17.7, it suffices to show that G1 ∼ = Sp8 (2). As noted after (17C), S is R1 -invariant. Since E ∈ Syl2 (S), NR1 S (E) covers R1 S/S by a Frattini argument. In particular NR1 S (E) permutes transitively the set {Y2ai }3i=0 . But NR1 S (E) normalizes C = CG (E). As C/ z = C/Y2a0 is abelian by Lemma 17.8, C is therefore abelian. As noted before (17E), this completes the verification of the hypotheses of Theorem 17.1. Hence G1 is isomorphic to a covering group of Sp8 (2). But the Schur multiplier of Sp8 (2) is trivial [IA , 6.1.4], so G1 ∼ = Sp8 (2), completing the proof of Proposition 17.2. We finally get rid of cores. Lemma 17.9. We have NG (D; 3 ) = 1. In particular, O3 (CG (D0 )) = 1 for all 1 = D0 ≤ D.   Proof. Since A ∈ E2 (D), it suffices to show that NCG (x) (D; 3 ) = 1 for each x ∈ A# . For any x ∈ A# , CG (x)/O3 (CG (x)) has no component I/O3 (CG (x)) ∼ = L2 (33 ), as I would then centralize the pumpup of J = E(CJx0 (A)) ∼ = A6 in CG (x) modulo O3 (CG (x)), and so m2,3 (CG (x)) > 3, contrary to assumption. Therefore by [IA , 7.7.12], every component of CG (x)/O3 (C  G (x)) is strongly  locally balanced with respect to the prime 3. Hence NCG (x) (D; 3 ) = O3 (CG (x)). This holds, similarly, for any x ∈ x2 , x3 # . Again let x ∈ A# and suppose that W := O3 (CG (x)) = 1. Since e(G) = 3, W is a {2, 3} -group. Now choose M ≤ J ∼ = A6 such that x3 ∈ M ∼ = A4 .  (2) or J = JO (J ). Also, C (W ) = 1 or M as Sp Then we have Jx ∼ = 6 x x M 3 Jx /O3 (Jx ) is simple. In either case, using [IG , 9.12(iii)], CW (x3 ) = 1. By the previous paragraph, O3 (CG (x3 )) = 1. As x3 ∈ xG 0 , we may assume that x = x0 . g Since G1 ∼ = Sp8 (2), S2a1 = S2a0 for some g ∈ G1 . Let z  = z g , so that [W, z  ] ≤ [W, Jx0 ] = 1 by (16B). Moreover, J ∼ = A6 centralizes z  and W . But ∼ by Proposition 17.2, Cz /O2 (Cz ) = Sp6 (2). Hence as z ∼ z  , Sp6 (2) has an r-local subgroup, r > 3, containing an A6 subgroup. As |Sp6 (2)|{2,3} = 5.7, Sp6 (2) must then contain an element of order 5.7. But even GL6 (2) contains no such element, a final contradiction.  Now define (17G)

  Sp (2) . G0 = E(CG (d)) | d ∈ D # , E(CG (d)) ∼ = 6

404

16. THEOREM C∗4 : STAGE A5. RECOGNITION

In view of Lemma 16.2, we set G∗ = Sp8 (2), F4 (2), or F4 (2), according as (17H) ND = W (C4 ), W (F4 ), or GO+ (D). Lemma 17.10. G0 = G. Proof. Suppose false, set M = NG (G0 ) < G, and let P ∈ Syl3 (NG (D)). We shall show that P ∈ Syl3 (G) and ΓP,1 (G) ≤ M . Indeed, NG (D) ≤ M by (17G), and D = J(P ) by [V17 , 10.10.1b], so P ∈ Syl3 (G). Let Ed = E(CG (d)) for any d ∈ D # . Thus by Lemma 17.9, for d ∈ A# , Ed ∼ = Sp6 (2) or A6 ∼ = Ed ≤ Jx0 (in the latter case, G∗ = Sp8 (2), so D ∈ Syl3 (CG (d)) and there are no additional components, which would have to have Sylow 3-subgroups of order 3 and be in C3 , an impossibility). In any case Ed ≤ M for any d ∈ A# . Also O3 (M ) = 1 by Lemma 17.9. Since NM (D) = NG (D) is irreducible on D = J(P ), F ∗ (M ) = E(M ) must be simple, with Ed = E(CE(M ) (d)) for all d ∈ A# by L3 -balance and the Bp -property. In particular, G0 ≤ E(M ). Now, Ex0 = Jx0 ∼ = Sp6 (2) ∈ Alt∪Chev(r) for any r > 2. We conclude from [IA , 5.2.8, 5.3, 3.1.3, 4.9.3], respectively, that E(M ) ∈ Alt, E(M ) ∈ Spor, E(M ) ∈ Chev(3), and finally, E(M ) ∈ Chev(2). Then using [IA , 4.7.3A, 4.8.2], E(M ) ∼ = F4 (2) or Sp8 (2) according as E(M ) is an exceptional or classical group of Lie type. Then | Out(E(M ))| = 2 or 1, respectively, by [IA , 2.5.12], so M ∼ = G∗ or Aut(F4 (2)). Moreover, with [IA , 7.3.4], G0 = E(M ). To show that ΓP,1 (G) ≤ M , suppose that g ∈ G and M ∩ M g contains an element x = y g of order 3, x, y ∈ M . We must show that g ∈ M . By [IA , 4.7.3A, 4.8.2], m3 (CG0 (u)) = 4 for all u ∈ I3 (G0 ). As D = J(P ), we may replace x and y by M -conjugates and assume that x, y ∈ D. Then as NG (D) ≤ M controls G-fusion in D, we may assume that x = y, and we are reduced to showing that CG (d) ≤ M for representatives d of the ND -orbits on E1 (D). We have shown this for d = x0 and d = x0 x1 . Since Nd ∼ = W (C4 ) or W (F4 ), remaining representatives are d = x1 x2 x3 ∈ Z(P ) and, if G∗ = Sp8 (2), d1 = x0 d. For d = x1 x2 x3 , set Cd = CG0 (d) and H1 = CJx0 (d) ∼ = GU3 (2). Then P ≤ Cd and NG (P ) ≤ NG (D) ≤ M , so by a Frattini argument it is enough  to show that O3 (CG (d)) ≤ M . If G0 = Sp8 (2), then on the natural G0 module V , dim(CV (x0 )) = 6 and V = CV (x0 ) ⊥ CV (d), whence Cd = H1 × x0 , z ∼ = GU3 (2) × Σ3 . If G0 ∼ = F4 (2), then the structure of Cd is given in [IA , 4.7.3A], and since CG0 ( d, x0 ) contains x0  × H1 , we have Cd = H1 H2 with H2 ∼ = H1 ∼ = GU3 (2), x0 ∈ O3 (H2 ), [O3 (H1 ), O3 (H2 )] = 1, and Z(H1 ) = Z(H2 ) = d. By Lemma 11.5a, F ∗ (CG (d)) = O3 (CG (d)). But a Sylow 2-subgroup of Cd acts faithfully on no proper subgroup of O3 (Cd ), so O3 (CG (d)) = O3 (Cd ). If G0 ∼ = Sp8 (2), then Z(O3 (Cd )) = x0 , d contains only one conjugate

17. G ∼ = Sp8 (2)

405

of x0 , so CG (d) ≤ NG ( x0 ) ≤ M in that case. On the other hand, if ND ≥ W (F4 ), then H1 H2 /O3 (H1 H2 ) contains Q8 × Q8 and so all elements of O3 (CG (d)) − H1 − H2 are Cd -conjugate. We saw above that  x0 ∈ O3 (CCG (d) (O3 (H1 ))) = O3 (H2 ). From Jx0 we see that x1 x2 ∈ O3 (H1 ) for some  = ±1. As x0 x1 x2  ∈ [12] ⊆ dG by [V9 , 10.1c5], it follows that xG 0 ∩ O3 (CG (d)) = O3 (H2 ) − d or (O3 (H1 ) ∪ O3 (H2 )) − d. There fore O 3 (CG (d)) normalizes O3 (H2 ) and O3 (H1 ) = CO3 (CG (d)) (O3 (H2 )). As  P ∈ Syl3 (CG (d)) it folows that O3 (CG (d)) = Cd , and then CG (d) ≤ M by a Frattini argument, as desired. The final case is to consider G0 ∼ = Sp8 (2) and d1 = dx0 . In this case |ND |3 = 3 and so x0 , d = Z(P ), whence P = d1  × (P ∩ Jx0 ). Again it is  enough to show that O3 (CG (d1 )) ≤ M . We have E(CG0 (d1 )) ∼ = U4 (2) by  [IA , 4.8.2], so clearly O3 (CG (d1 )) = d1  Ed1 , where Ed1 = E(CG (d1 )). As

d1  is a direct factor of P , d1  Ed1 = d1  × Ed1 by [IG , 15.12(i)]. Thus Ed1 has Z3 Z3 Sylow 3-subgroups, contains U4 (2), and is in C3 . By [V17 , 7.20], Ed1 ∼ = U4 (2) or Sp6 (2). Now, every subgroup of P/ d1  of order 3 G G is of the form d1 , u / d1  for some u ∈ xG 0 ∪ (x0 x1 ) ∪ (x0 x1 x2 ) . Hence by the results obtained already and the fact that N controls G-fusion in D, NCG (d1 )/d1  ( d1 , u / d1 ) ≤ CM (d1 )/ d1 . That is, ΓP/d1 ,1 (CG (d1 )/ d1 ) ≤ CM (d1 )/ d1  . But if Ed1 ∼ = Sp6 (2), then Ed1 d1  / d1  = ΓP/d1 ,1 (Ed1 d1  / d1 ), 

whence Ed1 ≤ M , a contradiction. Therefore O3 (CG (d1 )) = E(Cd1 ) d1  ≤ M , completing the proof of the strong 3-embedding of M . Let z ∈ I2 (M ) be a 2-central involution of M . We may choose z so  that O 3 (CM (z)/O2 (CM (z))) ∼ = Sp6 (2) or Sp4 (2), the last being necessary only if G0 ∼ = Aut(G0 ). In any case, if CM (z) < CG (z), = F4 (2) and M ∼ then CM (z) is strongly 3-embedded in the K-group CG (z), so its structure is given by [IA , 7.6.2], which is not the case. Therefore CG (z) ≤ M . Now with [V17 , 17.6], we may apply [V9 , 11.2] to M and conclude that M = G, a contradiction. The proof is complete.  Lemma 17.11. If G∗ = Sp8 (2), then G ∼ = Sp8 (2). Proof. If G∗ = Sp8 (2), then ND = W (C4 ) is a monomial group on the frame { xi }3i=0 in D. It follows that x0 ND is the unique ND -orbit on E1 (D) of length as small as 4. Now suppose that d ∈ D # and E(CG (d)) ∼ = Sp6 (2). N We claim that d ∈ x0  . Indeed the setup [V9 , (10A)] holds with d playing the role of both x and x0 and D ∩ E(CG (d)) playing the role of DL . Then as ND = W (C4 ), [V9 , 10.1] implies that | dND | = 4, and the claim follows. It follows that G0 = G1 , so G0 ∼ = G∗ by Proposition 17.2. By Lemma 17.10, the lemma follows. 

406

16. THEOREM C∗4 : STAGE A5. RECOGNITION

18. G ∼ = F4 (2) We complete the proof of Proposition 16.1 in this section. By Lemmas 16.2, 17.10, and 17.11, we may assume that (18A)

G0 = G

and ND = W (F4 ) or GO+ (D), i.e., G∗ = F4 (2). Note that by (16C), [8] is an ND ( x0 )-orbit on E1 (D), and by [V9 , 10.1c5], [6] ∼ND [61 ] [1] ∼ND [4] ∼ND [12], and | x0 ND | = 12 or 24. The only possibility, as |W (F4 ) : W (C4 )| = 3, is that the W (F4 )-orbits on E1 (D) are [1] ∪ [3] ∪ [8], [6] ∪ [61 ], and [4] ∪ [12], represented by x0 , x0 x1 , and x1 x2 x3 , respectively. Moreover, GO+ (D) fuses the first two of these. In any case, NND ( x0 x1 ) ∼ = NND ( x0 ), and so it follows from (16B) and Lemma 17.9 that (18B) NG ( x0 x1 ) = Sol(NG ( x0 x1 )) × Jx x ∼ = Σ3 × Sp6 (2). 0 1

Moreover, x2 ∈ J ≤ Jx0 x1 , but by the W (C4 ) action on [1] ∪ [3], x2 , x0 x1  ∼

x0 , x2 x3 , so O3 (CJx0 x1 (x2 )) = O3 (CG ( x0 x1 , x2 )) ∼ = O3 (CG ( x0 , x2 x3 )) =   ±1 ∼ lie in long root O 3 (CJx0 (x2 x3 )). Therefore CJx0 x1 (x2 x±1 3 ) = Σ6 and x2 x3 subgroups of Jx0 x1 ∼ = C3 (2). Now if x0 ∈ Jx0 x1 , then NG ( x0 x1 ) contains an involution inverting x0 and centralizing x0 x1 . But this is impossible by the structure  of NG ( x0 ). . So x0 ∈ Jx0 x1 , and similarly x1 ∈ Jx0 x1 . Hence A ∩ Jx0 x1 = x0 x−1 1 Now our previous development starting with x0 , A, J, Jx0 , and the played by x2 x−1 frame { xi }3i=0 in D is validif these 3 , x2 , x3 ,  roles are −1   E(CG ( x2 , x3 )) =: J , E(CG ( x2 x3 )) =: Jx x−1 , and the frame 2 3     −1  }. F := { x0 x1  , x0 x1 , x2 x3  , x2 x−1 3 In particular,

(18C)

(1) ND contains a copy W  of W (C4 ) acting monomially on D with respect to the frame F  ; (2) The root L2 (2)-subgroups S2ai , i = 2, 3 of J are long root subgroups of Jx0 ∼ = C3 (2), but short root ∼ subgroups of Jx0 x1 = C3 (2); (3) There is an involution z  ∈ NJx0 (D) acting on D as   and axis a reflection with center x2 x−1 3   , x x ; x0 x1 , x0 x−1 2 3 1 (4) CG (z  ) = Jx x−1 Q , where Q = O2 (CG (z  )), and 2 3 as in Lemmas 17.6 and 17.7, |Q | = 215 , Z(Q ) ∼ = E27 , and Q / z   is elementary abelian with Jx2 x−1 3 composition factors of dimensions 6 and 8.

18. G ∼ = F4 (2)

407

Following [FinFr2], we construct the final piece of F4 (2). We first strengthen Lemma 17.7d. Lemma 18.1. Q ∼ = 21+8 + × E26 . Proof. Let DJ = D ∩ Jx0 . Then Z(Q) = z × Z0 where Z0 := ±1 [Z(Q), DJ ] = [Z(Q), x1 x±1 2 x3 ] for any choice of signs. We also know that Q/Z(Q) = CQ/Z(Q) (x2 x3 ) × CQ/Z(Q) (x2 x−1 3 ), with CQ/Z(Q) (xi ) = 1 for i = 2, 3, and i = 1 as well. Factoring Q/Z(Q) under  then by conjugation for±1 ±1 ∼ (x x then yields C x1 , x2 x±1 1 Q/Z(Q) 3 2 x3 ) = E22 , so in fact |CQ (x1 x2 x3 )| = 8. Now x1 x2 x3  has exactly four NJx0 (DJ ) ∼ = Σ3 Σ3 -conjugates, whose fixed points on Q (all isomorphic) therefore generate a group R such that R ∩ Z(Q) ≤ z, RZ(Q) = Q, and NJx0 (DJ ) is irreducible on R. Then Z(R) = R ∩ Z(Q) = z so R ∼ and R × Z0 = Q, as desired.  = 21+8 + Lemma 18.2 ([FinFr2]). The following conditions hold:   ≤ I for some I ≤ Jx0 such that I ∼ (a) z  , x2 x−1 = L3 (2); and 3 (b) For any such I, x0 , z ≤ CG (I) ∼ = L3 (2).   is a short root L2 (2)-subgroup of Jx0 ∼ Proof. By (18C2,3), z  , x2 x−1 = 3 C3 (2), so (a) is obvious from the Dynkin diagram C3 . Let γ ∈ I7 (I) with x2 x−1 3 normalizing γ. Then CI (γ) = γ, so CJx0 (I) = 1 and CNG (x0 ) (I) =

x0 , z. By [V17 , 10.2.16], CZ(Q) (γ) = z. But then Q/[Z(Q), γ]  ∼ by Lemma 18.1, so Qγ := O2 (CCz (γ)) = CQ (γ) ∼ = D8 with z = = 21+8 + Z(Qγ ). Thus Qγ ∈ Syl2 (CG (γ)). Also, obviously Qγ is normalized, then −1 3  by x2 x−1 3 , and then Qγ ≤ O (CG (x2 x3 )) ≤ CG (z ). As I = centralized,  −1 z  , x2 x3 , γ , CG (I) contains Qγ , x0 , Qγ ∈ Syl2 (CG (I)), and z ∈ [Qγ , Qγ ]. Also x0  is self-centralizing in CG (I) and inverted by z. By [V17 , 20.15],  CG (I) ∼ = L3 (2). Now we construct A1 (2)-subgroups Ki , 0 ≤ i ≤ 4, corresponding to the nodes of the extended F4 -diagram. Set   . K0 = z, x0  , K2 = S1 = z1 , x1  , and K4 = z  , x2 x−1 3 Thus [K0 , K2 ] = [K0 , K4 ] = [K2 , K4 ] = 1. We may identify Jx0 with B3 (2) in such a way that K2 is a short root A1 (2)-subgroup and K4 is a long root A1 (2)-subgroup. Then using [V17 , 10.2.17], there is a long root A1 (2)-subgroup K3 of Jx0 ∼ = Sp4 (2) and K3 , K4  ∼ = A2 (2), and = B3 (2) such that K2 , K3  ∼ [K0 , K3 ] = 1. On the other hand, K0 and K2 lie in O3 (CG (K4 )) = Jx2 x−1 ∼ = C3 (2), and 3 they are short root A1 (2)-subgroups there. By Lemma 18.2, CG ( K3 , K4 ) ∼ = L3 (2), and so we may write CG ( K3 , K4 ) = K0 , K1  ,

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16. THEOREM C∗4 : STAGE A5. RECOGNITION

∼ C3 (2). with K0 and K1 (conjugate) short root A1 (2)-subgroups of Jx2 x−1 = 3 Now by [V17 , 10.2.19], all A2 (2)-subgroups of Jx2 x−1 containing a short root 3 A1 (2)-subgroup are conjugate. From regarding Jx2 x−1 for an instant as B3 (2) and looking at the ex3 tended B3 -diagram, we see that Jx2 x−1 ∼ = C3 (2) has an A3 (2)-subgroup 3 generated by three short root A1 (2)-subgroups forming an A3 -diagram. By the previous paragraph, one such subgroup is K0 , K1 , K2∗  for some short root A1 (2)-subgroup K2∗ commuting with K0 . Then by [V17 , 10.2.17], K2∗ = K2 . The result is that K1 , K2 , K3 , K4 form a weak Curtis-Tits system of type F4 (2) in G [III13 , Defn. 1.2]. By the Blok-Hoffman-Shpectorov theorem [III13 , 1.3] and the Curtis-Tits theorem [IA , 2.9.3], (18D) G := K1 , K2 , K3 , K4  ∼ = F4 (2). 0

∼ L3 (2), so CG ( K3 , K4 ) ≤ G , and in particular Note that CG0 ( K3 , K4 ) = 0 x0 ∈ G0 . Hence D ≤ x0  Jx0 ≤ G0 . Lemma 18.3. G0 ≤ G0 . Proof. Let x ∈ D # with E(CG (x)) ∼ = Sp6 (2). Then |CG (x)|3 = 35 .  Now x ∈ I3 (G0 ), so by [IA , 4.7.3A], either CG0 (x) ∼ = Z3 × Sp6 (2) or 6 |CG0 (x)|3 = 3 . As the latter is impossible, E(CG (x)) ≤ G0 . Hence by  definition of G0 , G0 ≤ G0 . Now by Lemma 18.3, (18A), and (18D), G ∼ = F4 (2). This completes the proof of Proposition 16.1 and with it the proof of Proposition 10.1. As discussed in the opening of this chapter, Propositions 2.1 and 10.1 complete the proof of Theorem C∗4 : Stage A5, and with it the proof of Theorem C∗4 (Case A).

CHAPTER 17

Properties of K-Groups As has been the case throughout this series, this chapter of preliminary lemmas logically precedes Chapters 9–16. We also note that occasionally in this chapter, the proof of a Lemma A will refer to a later Lemma B. This is merely due to the organization of the chapter, and does not affect the validity of the logic. For in every such instance, the proof of Lemma B will be seen to refer to no other results in this chapter.

1. Simple Sections of G, e(G) = 3 The following definition repeats [V12 , Definition 2.2]. Definition 1.1. A is the set of all isomorphism classes of the following simple groups: (a) (b) (c) (d) (e) (f) (g) (h) (i)

L2 (q), q odd; e L± 3 (p ), p an odd prime and e ≤ 3; 2 G (3 32 ); 2 G2 (p), P Sp4 (p), or L± 4 (p), p an odd prime; An , n ≤ 12; A sporadic group but not F i22 , F i23 , F i24 , Co1 , F2 , or F1 ; A group in Chev(2) of twisted Lie rank at most 2; L4 (q), Sp6 (q), or Ω− 8 (q), q even; or (2), L (2), Sp8 (2), D4 (2), or F4 (2). L5 (2), L± 7 6

Lemma 1.2. Let p be an odd prime and let K ∈ A. Let X be a p-covering group of K, i.e. a covering group of K with Z(X) a p-group for some odd prime p. Let H be a p-covering group of X. Then the following conditions hold: (a) m2,p (X) ≤ m2,p (H); (b) mp (X) ≤ mp (H), unless X ∼ = O N , H ∼ = 3O N , and p = 3; (c) e(H) ≤ 3 unless p = 3 and H ∼ = [3 × 3]U4 (3), SU6 (2), or 3Suz. Proof. Obviously (a) and (b) are trivial unless p divides the order of the Schur multiplier of K. Also, (c) follows trivially in any case when for all primes p, (1A)

mp (H) ≤ 3, or mp (CH (z)) ≤ 2 for all z ∈ I2 (H). 409

410

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∼ An , then 3-cycles of K split in 3An , and (a) and (b) follow. Since If K = K ∈ A, n ≤ 12, and one easily checks (1A). If K ∈ Spor, Tables [IA , 5.6.1, 5.6.2] contain the information needed to verify all parts of the lemma. Next, suppose that K ∈ Chev(p) − Alt. As K ∈ A, we may assume that p = 3 and H is a quotient of [3 × 3]U4 (3) or 3G2 (3) [IA , 6.1.4]. In the 3G2 (3) case, the structure of CK (z) for z ∈ I2 (K) shows that m3 (CK (z)) = 2 and m3 (CK (V )) = 0 for any four-group V ≤ K. Hence m2,3 (K) = 2. It is immediate that m3 (CH (z)) ≥ m3 (CX (z)) ≥ 2, so (a) and (c) hold. Also m3 (3G2 (3)) = m3 (G2 (3)) = 4 by [V8 , 1.5] and [IA , 3.3.3], so (b) holds. In the U4 (3) case, let P ∈ Syl3 (K). Then J(P ) ∼ = E34 and J(P ) splits over the elementary abelian 3-groups Z(H) and Z(K). Moreover, for any involution z ∈ K, a Sylow 3-subgroup of CK (z) lies in a conjugate of J(P ). Again we have m3 (CK (z)) = 2 and m3 (CK (V )) = 0 for any four-group V ≤ K, so m3 (X) = 4 + m3 (Z(X)) and m2,3 (X) = 2 + m3 (Z(X)). This implies all parts of the lemma in this case. Now suppose that K ∈ Chev − Chev(p) − Alt. Using [IA , 6.1.4], we see that having excluded A6 , we have that X and H lie in Lie(r) for some r = p, i.e., they are quotients of a universal version J of a group of Lie type. We may assume that |Z(J)| is not a power of 2. If K ∈ Chev(r), r odd, then p = 3 and G ∼ = L3 (rn ), rn ≡  (mod 3),  n ≤ 3, and we may assume that H = X = SL3 (rn ) and p = 3. It is then routine that m3 (CK (z)) = 1 or 2 for any z ∈ I2 (K), m3 (CH (z)) = 2, and m3 (K) = m3 (H) = 2. So (a) and (b) hold. Also m2,p (H) and m2p (K) are at most 2 for all primes p except possibly r = p, for which m2,r (H) = m2,r (K) = n, from the centralizer of an involution. Therefore we may assume K ∈ Lie(2), whence the universal version is J ∼ = SU6 (2) (p = 3), SL3 (2n ) (p = 3, n  2 ≡  (mod 3),  = ±1), or SL5 (2n ) (p = 5, 2n ≡  (mod 5)). In all cases |Z(J)| = p so we only have to compare J and K in (a) and (b). By the Borel-Tits theorem, m2,r (X) and m2,r (H) are the maximal r-ranks among all (proper) parabolic subgroups of X and H. If J ∼ = SU6 (2), then we have m3 (J) = 5 > 4 = m3 (K), m2,3 (J) = 4 > 3 = m2,3 (K), and m2,s (J) < 3 for all primes s > 3. Thus, e(H) ≤ 3. Similarly for J ∼ = SL3 (2n ), we have m3 (J) = 2 = m3 (K), m2,3 (J) = 2 ≥ m2,3 (K) if  = +, and m2,3 (J) = 1 ≥ m2,3 (K) if  = −. Also ms (J) ≤ 2 ≥ ms (K) for all odd primes s, so e(H) ≤ 3. Finally, suppose that p = 5 and K = U5 (2n ), n ≡ 2 (mod 4). Then mp (J) = 4 ≥ mp (K). Also m2,p (J) and m2,p (K) are achieved as the p-ranks of the normalizer of the center U of a long root subgroup: a Levi factor The other maximal parabolics are stabilizers of isotropic 2-subspaces of the natural module. We find the m2,p (J) = 3 ≥ m2,p (K). Also m2,s (J), for prime divisors s of 2n + 1, is achieved in the same parabolics as 3. As ms (J) ≤ 2 for non-divisors s of 2n + 1 [IA , 4.10.3],

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e(J) = 3. Sylow p-subgroups of NJ (U ) are abelian, so e(K) ≤ e(J) = 3. The lemma follows.  Theorem 1.3 (Aschbacher [A13]). Let X = H/K with H a K-group such that e(H) ≤ 3. Then the following conditions hold: (a) If X = E(X) and p is a prime, p > 3, then m2,p (X) ≤ m2,p (H). Moreover, if mp (H) ≤ 3, then mp (X) ≤ mp (H); (b) If X is quasisimple and p = 3, then m2,p (X) ≤ m2,p (H). Moreover, if m3 (H) ≤ 3, then m3 (X) ≤ m3 (H), unless X ∼ = O N and X is  covered by a section of H isomorphic to 3O N ; and (c) If X is simple, then X ∈ A. We write X = H/K and assume the theorem fails. Lemma 1.4. If Y is a quasisimple K-group with |Z(Y )| odd and e(Y ) ≤ 3, then Y /Z(Y ) ∈ A. In particular, if the theorem fails, then either (a) or (b) of the theorem fails. Proof. Supposing (a) and (b) hold, we deduce that X is a simple Kgroup such that e(X) ≤ 3. Hence, the second statement will follow from the first. If Y ∼ = An for some n, then n ≤ 12 since A13 contains A9 × A4 , so m2,3 (A13 ) = 4. If Y ∈ Spor, then the information in [IA , 5.6.2] shows that Y ∈ A. So we may assume that Y ∈ Chev(r) for some r. Suppose that r is odd. Then as m2,r (Y ) ≤ 3, Y cannot contain SL2 (rn ) or Z2 × P SL2 (rn ), n ≥ 4; SL2 (ra ) ∗ SL2 (rb ) or SL2 (ra ) × SL2 (rb ), a + b ≥ 4; SL2 (r)∗(P )Sp4 (r) or SL2 (r)×(P )Sp4 (r)or SL± 4 (r), all of which have r-rank at least 4. It is straightforward to find involution centralizers or subsystem subgroups of one of these types in Y unless Y ∈ A is of one of the types (a)–(d) in Def. 1.1. Therefore we may assume that Y ∈ Chev(2). We may assume that Y has twisted Lie rank at least 3, and is not L4 (2n ), Sp6 (2n ), or 2 D4 (2n ). If Y has level q(Y ) = 2n , n > 1, then for some prime p dividing 2n − 1, a parabolic subgroup M with mp (M ) ≥ 4 is available with M a Borel subgroup if Y has twisted rank at least 4, and M the normalizer of a long root subgroup if Y = U6 (2n ) or U7 (2n ). Finally suppose that q(Y ) = 2. Then m3 (M ) ≥ 4 for a suitable parabolic subgroup M of Y for the following  pairs (Y, O 2 (M/O2 (M ))): (A7 (2), A3 (2) × A3 (2)); (B5 (2), B4 (2)); (D5 (2), D4 (2)); (E6 (2), D4 (2)); (U7 (2), U5 (2)); (2E6 (2), 2 A5 (2)); (32E6 (2), SU6 (2)); (2 D5 (2), A1 (2) × U4 (2)). The same M ’s exist for larger groups of the various types. This completes the proof.  Now assume the theorem fails and choose a counterexample to (a) or (b) of the theorem, with H minimal. If (a) fails, let p be a prime for which it fails; and if (b) fails, let p = 3. Let H0 be defined by H0 /K = Z(H/K). We proceed in a sequence of lemmas, following [A13, Theorems 2, 3] closely.

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Lemma 1.5. The following conditions hold: (a) (b) (c) (d) (e)

No proper subgroup of H covers X; H0 and K are nilpotent; O2 (H) = 1; H/Op (H) is simple; and If m2 (H) > 4, then H is quasisimple.

Proof. Part (a) is immediate from the minimality of H, and (b) follows by a Frattini argument. Let q be an odd prime different from p. Then mp (H/Oq (H0 )) = mp (H) and m2,p (H/Oq (H0 )) = m2,p (H), so Oq (H0 ) = 1 by minimality of H. Now suppose O2 (H) = 1. Then m2,p (H) = mp (H) and m2,p (X) ≤ mp (X). Hence, as (a) or (b) of Theorem 1.3 is false, we must have mp (X) > mp (H). Now replace H by H1 := H/O2 (H) and X by X1 = H1 /(KO2 (H)/O2 (H)). Then mp (H1 ) = mp (H) < mp (X) = mp (X1 ). Hence, by the minimality of H, H1 ∼ = O N . But then H is not a counterexample to = 3O N and X ∼ Theorem 1.3, contrary to assumption. Hence, Or (H) = 1 for all r = p. In particular, (c) holds and H0 = Op (H). If (d) holds and m2 (H) > 4, then e(H) > 3 by the Thompson dihedral lemma, contradiction. Therefore (d) implies (e), so it remains to prove (d). Now, H := H/Op (H) = H 1 × · · · × H n , with each H i simple and the image of a p -component Hi  H. In particular, m2 (Hi ) ≥ 2 for all i. Let Vi be a four-subgroup of Hi . We must prove that n = 1. Suppose false, so that it is (a) that fails. Hence, p > 3. As Hi ≤ H, e(Hi ) ≤ 3. By Lemma 1.4, H i ∈ A for all i = 1, . . . , n.  1 , is not simple. As  = H/[K, H] and suppose that some H  i , say H Set H p > 3, it follows from the fact that H 1 ∈ A and from [IA , 6.1.4] that p = 5 1 ∼ and H = SU5 (2m ) for some m ≡ 2 (mod 4). Then m2 (H1 ) = 4m ≥ 8 so if ∗ F (H1 ) = O5 (H1 ), then by the Thompson dihedral lemma, m2,5 (H1 ) > 3, contradiction. Therefore H1 is quasisimple, whence [H1 , H2 ] = 1, and so m2,5 (H) ≥ m5 (H1 ) > 3, contradicting e(H) ≤ 3. We have shown that every  ∼  i is simple, so K = [K, H] and H H = H. If Op (H) = 1, then as K is nilpotent, H = X and the theorem holds trivially. Since the theorem fails, we may assume that F ∗ (H1 ) = Op (H1 ) ≤ Op (H). It follows that for some z ∈ I2 (Vn ), Rz := COp (H) (z) is normalized but not centralized by H1z := CH1 (z)(∞) , and H 1z = H 1 . Now mp (Rz ) ≤ m2,p (H) ≤ 3. Let Q be a critical subgroup of Rz of exponent p and class at  1z := H1z /CH (Q/Φ(Q)), a one-headed group with head most 2, and set H 1z H 1 . Then mp (Q) ≤ 3. Note that (1B)

 1z ) = Op (H  1z ). Z ∗ (H

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413

 1z embeds Suppose that mp (Q/Φ(Q)) > 3. By [V9 , 5.8], Q ∼ = p1+4 , so H in Sp4 (p). By (1B), its image does not contain −1. But m2 (Sp4 (p)) = 2, so  1z ) = 1. The Z ∗ -theorem then contradicts (1B). m2 (H Therefore mp (Q/Φ(Q)) ≤ 3, and (1B) implies that equality holds. Thus  1z embeds in SL3 (p), so by [V9 , 5.8], mp (Q) = 3 = m2,p (H). Also H mp (H 1 ) ≤ 2 and m2,p (H 1 ) ≤ 1.

(1C)

If Hi is not simple for some i with 1 < i < n, then the above argument shows  1z × H  iz embeds in SL3 (p), which is impossible.  iz has head H i and H that H Hence, E(H) = H2 × · · · × Hn . As [E(H), Q] = 1, it follows that n = 2 and m2,p (H2 ) = 0. As mp (H2 ) ≥ 1, we have mp (H) ≥ 4. Therefore, since the theorem fails, we may assume that m2,p (H) ≥ 4, with H = H1 × H2 . Let t ∈ I2 (H1 ). Then t ∈ Z ∗ (H1 ) and so COp (H) (t) = 1. It follows that mp (H2 ) ≤ mp (CH (t)) − 1 ≤ m2,p (H) − 1 = 2. As m2,p (H2 ) = 0, it follows that m2,p (H) ≤ 3, a final contradiction proving (d).  Lemma 1.6. The following conditions hold: (a) (b) (c) (d)

F ∗ (H) = Op (H); X is quasisimple with |Z(X)| odd and m2 (X) ≤ 4; mp (Q) ≥ 4 for Q a critical subgroup of Op (H); and m2,p (X) > m2,p (H) ≥ m2 (X) − 1.

Proof. Suppose F ∗ (H) = Op (H). As H/Op (H) is simple by Lemma 1.5, we have H = R ∗ L, with R = Op (H) and L quasisimple with Z(L) = Op (L). As X is quasisimple, X ∼ = L/Z for some Z ≤ Z(L). As mp (L) ≤ mp (H) and m2,p (L) ≤ m2,p (H), the pair (L, L/Z) affords a counterexample, whence H = L and 1 = K ≤ Z(H) = Op (H), by minimal choice of H. Also, H/Z(H) ∈ A by Lemma 1.2. But then e(H) > 3 by Lemma 1.4, contrary to assumption. Thus (a) holds. As H is not quasisimple, we have m2 (H) = m2 (X) ≤ 4 by Lemma 1.5e. Moreover as X = E(X) and H/Op (H) is simple, X is quasisimple, proving (b). Suppose that mp (Op (H))] ≤ 3. Let Q be a critical subgroup of Op (H) of exponent p and class at most 2. As mp (Q) ≤ 3, [V9 , 5.8] yields that Q∼ = p1+4 or mp (Q/Φ(Q)) ≤ 3. Accordingly H = H/Op (H) is a quotient by a p-subgroup of a subgroup of Sp4 (p) or SL3 (p). As Z(H) = 1, H must be isomorphic to a simple subgroup of L3 (p), whence H ∼ = L2 (p), L3 (p), or L3 (2). In any case, mp (X) ≤ mp (Q) ≤ mp (H), so the failure of Theorem 1.3 is that m2,p (H) < m2,p (X). But m2,p (X) ≤ 1 in all cases, yielding a contradiction proving (c). As mp (H) ≥ 4, it now follows that m2,p (X) > m2,p (H). Finally, by the Thompson dihedral lemma, m2,p (H) ≥  m2 (H) − 1 = m2 (X) − 1, proving (d). Lemma 1.7. H/Op (H) ∈ Chev(r) for some odd prime r.

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Proof. Suppose not. As m2 (X) ≤ 4 and m2,p (X) ≥ m2 (X) by Lemma 1.6, we may use Tables 3.3.1, 5.6.1 and 5.6.2 of GLS3 to reduce quickly to p = 3 and X ∼ = 3A6 , A7 , 3A7 or 3O N . In the first three cases, m2,3 (X) = 2 and so m2,3 (H) ≤ 1. Let Q be a critical subgroup of Op (H) of exponent p with mp (Q) ≥ 4. If Q is nonabelian, then CQ (t) = Z(Q) for t ∈ I2 (H) and so t inverts Q/Z(Q), which is impossible. Hence Q is abelian and |CQ (t)| = 3. Thus |Q| ≤ 33 . But 5 does not divide |GL(3, 3)|, a contradiction. Hence, X∼ = 3O  N and m2,3 (X) = 3, whence m2,3 (H) ≤ 2. As before, m2,3 (H) = 1. So, m2,3 (H) = 2 and either Q is extraspecial or Q is abelian. In the former case, |CQ/Z(Q) (t)| = 3 for all t ∈ I2 (H), and so |Q/Z(Q)| = 32 , which is absurd. So, we get Q ∼ = E3n , n ≤ 6. But X contains a Frobenius group of order 11 · 10, whence n ≥ 10, a final contradiction.  By minimality of H, and as O2 (H) = 1 (Lemma 1.5c), we have Lemma 1.8. Let M be a 2-local subgroup of H and L = E(L) a section of M . Let q be an odd prime, and if q = 3 assume that L is quasisimple. Then mq (L) ≤ m2,q (H) ≤ 3, unless possibly q = 3 and L ∼ = O N . Finally we prove Lemma 1.9. H ∈ Chev(r) for any odd r. Proof. First suppose that H ∈ A. Cleary, COp (H) (t) = 1 for t ∈ I2 (H), and so m2,p (X) > m2,p (H) ≥ 1. If X ∼ = L2 (re ), then m2,p (X) ≤ 1, a ∼ contradiction. If X = G2 (r), P Sp4 (r), L4 (r), or U4 (r), then m2,p (X) ≤ 2 and so m2,p (H) = 1. Let Q be a critical subgroup of Op (H) of exponent p. If m2,p (H) = 1 and 1 = Z(Q) = Q, then t inverts Q/Z(Q) for t ∈ I2 (H), again not the case. So Q is abelian and then Q ∼ = Epm with m ≤ 3. Hence, G2 (r), H/CH (Q) embeds in SL3 (p). In particular, m2 (H) = 2. Thus X ∼ = 3 2 ∼ P Sp4 (r), L4 (r) or U4 (r). Suppose X = G2 (3 2 ). Then p = 3 and m2,3 (X) = 3. So, m2,3 (H) = 2. If 1 = Z(Q) = Q, then m3 (Q/Z(Q)) = 2, which is certainly not the case. Hence Q is abelian and m3 (Q) ≤ 6. But X contains a Frobenius group of order 8 · 7, and hence X does not embed in GL(6, 3). ± e e ∼ Thus either X ∼ = L± 3 (r ) for some e ≤ 3, or p = 3 and X = SL3 (r ) for some e ≤ 3 and r > 3. Then either m2,p (X) ≤ 2 or p = r, e = 3, and m2,p (X) = 3. In the former case, m2,p (H) = 1 and H/CH (Q) embeds in SL3 (p). But then 3 X ∼ = (S)L± = SL3 (p) and m2,p (X) = 1 = m2,p (H). Finally, if X ∼ 3 (p ), we have m2,p (H) = 2. As before, we conclude that Q is abelian and mp (Q) ≤ 6, 9 So H/CH (Q) embeds in GL6 (p). But |H/CH (Q)| ia divisible by p 3±1 and by Zsigmondy’s Theorem, this is divisible by some prime s which does not divide pm − 1 for any m ≤ 6. Hence, |H/CH (Q)| does not divide |GL6 (p), yielding a contradiction which establishes that H ∈ A. We complete the proof by arguing that if H ∈ Chev(r) − A, r > 2, then m2,r (H) > 3, contrary to e(H) ≤ 3.

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Since 2 G2 (3 2 ) ≥ Z2 × L2 (3n ), H ∼ = 2 G2 (3 2 ) by Lemma 1.8 if n > 3, and by definition of A otherwise. As L2 (rn ) ∈ A, H has an involution z such that CH (z) has a (solvable) component L ∼ = SL2 (rn ), where rn = q(H). If r > 3, then by Lemma 1.8, mr (E(CH (t))) ≤ 3 for every involution t ∈ H. But then examination of the table [IA , 4.5.1] shows that H ∈ A, contrary to assumption. Therefore r = 3 and Lemma 1.8 applies only to single components L of centralizers CH (t), t ∈ I2 (H). It yields m3 (L) ≤ 3, so L/Z(L) ∼ = L2 (3e ), e ≤ ± 3, L3 (3), or P Sp4 (3). Using [IA , 4.5.1] and the definition of A again, we conclude that H ∼ = G2 (3a ), P Sp4 (3a ), or U4 (3a ), a = 2 or 3, L4 (33 ), P Sp6 (3), 3 or D4 (3), and that for the (classical) involution z, that m3 (CH (z)) ≥ 4. As e(H) ≤ 3, this last condition implies that p = 3. Let Cz = CH (z) and let Q be a critical subgroup of O3 (Cz ) of exponent 3 and class at most 2. Then as m3 (O3 (Cz )) ≤ 3, AutCz (Q/Φ(Q)) embeds  in SL3 (3) or Sp4 (3), by [V9 , 5.8]. But O3 (C z )/ z contains W / z = L2 (3a ) × L2 (3a ), a = 2 or 3, P Sp4 (3), or L2 (33 ). Let W be the full preimage of W in Cz , and set Y = W (∞) . Then Y  Cz . Moreover, either [Q, Y ] = 1 or AutY (Q/Φ(Q)) ∼ = L2 (9), with Q ∼ = 31+4 in the latter case (as SL3 (3) does not contain L2 (9)). Still in the latter case, CO3 (Cz ) (Q/Φ(Q)) = QCO3 (Cz ) (Q) with CO3 (Cz ) (Q) cyclic, as m3 (Cz ) ≤ 3. Hence CY (Q)(∞) =: R is quasisimple with R/Z(R) ∼ = L2 (9), so m2,3 (H) ≥ m3 (QR) > 3, contradiction. Therefore, [Q, Y ] = 1. As Y = O3 (Y ), [O3 (Cz ), Y ] = 1 and so Y = E(Y ). As m3 (Y ) ≤ 3, one of the following occurs: ∼ G2 (9), P Sp4 (9), or U4 (9); ∼ 3A6 ∗ 3A6 , H = (1) Y / z = 3 3 ∼ ∼ (1D) (2) Y = SL2 (3 ), H = D4 (3); or (3) Y ∼ = P Sp6 (3). = Sp4 (3), H ∼ n

n

In the last two cases, Y centralizes an SL2 (3)-subgroup of C z , whence m3 (Cz ) > 3, contradiction. Thus (1D1) holds. This time set P = O3 (H) = F ∗ (H) and let E be a critical subgroup of P of exponent 3 and class at most 2. As CP (z) ≤ O3 (Cz ) and Y  Cz , [Y, CP (z)] = 1. As m3 (Cz ) = 3, CP (z) is cyclic. If [E, E] = 1, then [E, E] = CE (z) and z inverts E/[E, E], contrary to z ∈ Z ∗ (H/CH (E)). Hence, E is abelian. Suppose that H ∼ = G2 (9) and let a ∈ I2 (Y )− z. Then a ∈ I2 (H) = z H . By the previous paragraph, CP (z) ≤ CP (a). Hence equality holds, so P =

CP (a) | a ∈ I2 (Y ) = CP (z). Thus, P is cyclic, contrary to P = F ∗ (H), a contradiction. Therefore H ∼ = P Sp4 (9) or U4 (9). Take g ∈ H such that z = z g ∈ CG (z) and set A = z, z g  and t = zz g . If Ω1 (CP (z)) ≤ CG (t), then CP (z) ≤ CG (t) and, conjugating by g, CP (z g ) ≤ CG (t). Hence P = ΓA,1 (P ) ≤ CG (t), so H is quasisimple, a contradiction. As CP (z) is cyclic, CP (z) ∩ CP (t) = 1, so z inverts CP (t).

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Suppose that H ∼ = U4 (9). Then regarding H as Ω+ 6 (9), we see that g − ∼ ∼ may be chosen so that L = E(CH (t)) = Ω4 (9) = L2 (81) and z induces a nontrivial field automorphism on L. Let L be the 2-component of CH (t) mapping onto L. Then as z inverts CP (t), L = [L, z] centralizes O3 (L), i.e.,  L is quasisimple. Let J = O3 (CH (A)) ∼ = L2 (9). Then J ≤ L ∩ Y ∩ Y g and # so [CP (a), J] = 1 for all a ∈ A . Hence [P, J] = 1, a contradiction. Therefore H ∼ = P Sp4 (9). This time, regarding H as Ω5 (9), we may choose g so that L = E(CH (t)) ∼ = Ω3 (9) ∼ = L2 (9), but [L, z] = 1. Again let L be the 2-component of CH (t) mapping on L. If L is quasisimple, then  letting J = O 3 (CH (A)) ∼ = A4 , we reach a contradiction just as in the last two sentences of the previous paragraph. So L is not quasisimple. Now, for any d ∈ I2 (L), we have d ∈ tH and td ∈ z H , so CP (td) is cyclic. Let D be a four-subgroup of L. Then     CP (t) ≤ CP ( t, d) | d ∈ D # ≤ CP (td) | d ∈ D # , forcing m3 (E) ≤ 3, a final contradiction.



As Lemmas 1.7 and 1.9 contradict each other, this completes the proof of Theorem 1.3. 2. Flat TGp -Groups Lemma 2.1. Let K ∈ TG3 be flat, i.e., such that whenever J ∈ K3 and J 4. Thus, assume that K is a classical group. Again, if q > 2 and q ≡  (mod 3), then ↑3 (SL3 (q)) contains Sp2n (q) for n ≥ 3 and Dn± (q) for n ≥ 4, as well as Ln (q) for n ≥ 4. Also for q ≡ − (mod 3), ↑3 (SL3 (q 2 )) contains SL− n (q), n ≥ 6. The only alternative for K, since the lemma fails, is that q(K) = 2,

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and specifically, if K ∼ = Ln (2), then n < 6. The conditions K ∈ G3 and m  3 (K) ≤ 4 then rule out all possibilities except those listed in the lemma, contradiction. The proof is complete.  Lemma 2.2. Suppose that p is an odd prime and K ∈ TGp . Suppose that if p = 3, then K is flat. Suppose also that if K ∈ Chev, then mp (K) ≤ 2. Then one of the following holds: (a) mp (K) = 2 and K has abelian Sylow p-subgroups; moreover, if K ∈ Chev, then p does not divide | Outdiag(L)|, and K is p-saturated; (b) p = 3 and K ∼ = An , 10 ≤ n ≤ 12; ∼ (c) p > 3 and K = An , 3p + 1 ≤ n ≤ 4p + 1; or (d) p = 13 and K ∼ = F1 . Proof. If K ∈ Spor, then from the lists of sporadic Gp -groups [IA , p. 103] and flat TG3 -groups (Lemma 2.1), and from the chart of sporadic p-ranks [IA , 5.6.1] and their group orders, we see that p = 13 and K ∼ = F1 , ∼ or else (a) holds. Likewise if K = An , then (a) holds if n ≤ 3p (recall that Akp ∈ Cp if k ≤ 3), while (b) or (c) holds otherwise, by definition of G6p [III3 , (5B)]. So assume that K ∈ Chev, whence mp (K) = 2 by assumption (if mp (K) = 1, then K ∈ Cp ∪ Tp ). Since K ∈ Gp , K ∈ Chev(p). If p = 3, then again from Lemma 2.1 and [IA , 4.10.3a], the condition m3 (K) = 2 − forces K ∼ = Sp4 (q), L− 4 (q), or L5 (q), and then using [IA , 2.5.12, 4.8.1] we see that (a) holds. So assume that p ≥ 5. If p divides | Outdiag(K)|, then K∼ = Lpk (q), q ≡  (mod p),  = ±1, and then mp (K) ≥ pk − 2 > 2, contradiction. Hence p does not divide | Outdiag(K)| and it remains to show that any P ∈ Sylp (K) is abelian.  be the universal version of K. Then |Z(K)|  p = | Outdiag(K)|p = Let K  N n i  1. Write |K| = q i Φi (q) as in [IA , (4.10.1)]. Here the Φi are cyclotomic polynomials. Let m0 = ordp q, the multiplicative order, i.e., the least integer such that p divides Φm0 (q). Then by [IA , 4.10.1, 4.10.2c], nm0 > 0, and it suffices to show that if nm0 ≤ 2, then ni = 0 for every i divisible by pm0 ; moreover, we only need check this for prime divisors p of |W |, where W is the Weyl group of the overlying algebraic group. This is straightforward to check if K is an exceptional type groups in Chev. If K is a classical group, one can alternatively use the language of [IA , Sec. 4.8]. Let V be the  natural K-module, of type τ = L, U , S, or O. Since K contains Ep2 , there is a decomposition V = V0 ⊥ V1 ⊥ · · · ⊥ Vn , where the isometry group of V0 is a p -group, and each Vi , 1 ≤ i ≤ n affords an orthogonally indecomposable representation of Zp of type τ , and with  contains commuting groups n ≥ 2. But then n = 2 as mp (K) = 2 and K acting individually on each Vi , i ≥ 1. Then P is a Sylow p-subgroup of the isometry group of V1 ⊥ V2 . As p is odd, P stabilizes both V1 and V2 and hence is the direct product of two cyclic p-groups. The proof is complete. 

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3. Groups of Low p-Rank, p Odd Lemma 3.1. Suppose F ∗ (X) = K is a simple K-group of p-rank 1, for some odd prime p. Let P ∈ Sylp (X). Then the following hold: (a) If mp (X) > 1, then Op (X) < X and mp (X) = 2;  (b) If X = O p (X), then Op (X) = K; (c) For some t ∈ I2 (K), CX (t) covers a Sylow p-subgroup of X/K; (d) Any Ep2 -subgroup of Aut(K) contains a nontrivial field automorphism;  (e) P splits over P ∩ K, Op (CX (P ∩ K)) = P ∩ K, P/P ∩ K is cyclic, and Ω1 (P ) ∼ = Zp or Ep2 ; 5 5 (f) If K ∈ Tp , so that K ∼ L2 (8) or 2B2 (2 2 ) for p = 3 and 5, re= spectively, and B ∼ = Epr acts on K, r ≥ 2, then K = ΓB,r−1 (K); and (g) If P contains a field automorphism f of order p, then [P ∩K, f ] = 1. Proof. If p does not divide | Out(K)|, then the assertions hold trivially. So assume that p divides | Out(K)|. Then K ∈ Alt ∪ Spor. If K ∈ Chev(p), 1 the only possibility is p = 3 and K ∼ = L2 (8). So we may = 2 G2 (3 2 ) ∼ assume that K ∈ Chev(r) for some r = p. Then by [IG , 15.12(i)], the Schur multiplier of K is a p -group, whence by [IA , 6.1.4], Outdiag(K) is also a p -group. Evidently D4 (q) and 3D4 (q) do not have cyclic Sylow 3-subgroups [IA , 4.10.3], so all outer automorphisms of K of p-power order are images of field automorphisms, and thus P/P ∩ K is cyclic and P splits over P ∩ K. As field automorphisms lie outside [Aut(K), Aut(K)] by [IA , 2.5.12], (a) and (b) hold. With [IA , 4.9.1], (d) then holds. As the centralizer of a field automorphism of any order is of even order, (c) also holds. We may assume that K is the universal version. Suppose f is a field automorphism of order p. Then as p is odd, Ω1 (P ) = Ω1 ((P ∩ K) f ) ∼ = Ep2 ,  N ni as asserted in (e). Let the order formula for K be |K| = q i Φi (q) p [IA ,(4.10.1)]. Then q = q0 for some prime power q0 , and |CK (f )| = q0N i Φi (q0 )ni . Let m0 = ordp (q) = ordp (q0 ), the multiplicative order (note q ≡ q0 (mod p)). Then by [IA , 4.10.2c], nm0 = 1 = mp (K). Hence |K|/|CK (f )| is divisible by integer Φm0 (q)/Φm0 (q0 ), which in turn is divisible by p. This proves (g), which implies the remaining part of (e). Finally, (f) follows from [IA , 7.7.8].  Lemma 3.2. Let p be an odd prime and K ∈ Kp ∩ Chev with mp (K) = 1. Let f be a field automorphism of K of order p and P an f -invariant Sylow p-subgroup of K. If K is not locally balanced for p, then Lp (CK (f )) = 1 and Ω1 (P ) ≤ Lp (CK (f )). Proof. Let L = Lp (CK (f )) and first suppose that L = 1. Then 1 CK (f ) ∼ = L2 (2), L2 (3), or 2B2 (2 2 ) (note that m3 (U3 (2)) > 1). As p divides 5 |CK (f )|, p = 3, 3, or 5, respectively, and K ∼ = L2 (8), L2 (27), or 2B2 (2 2 ), respectively. But m3 (L2 (27)) > 1, so that case is out. By [IA , 7.7.8cd],

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however, K is then locally balanced for p, contradiction. Thus L = 1. As CP (f ) = 1 and p divides |L|, while mp (CK (f )) = 1, Ω1 (P ) ≤ L, completing the proof.  Lemma 3.3. Let p be an odd prime and K = K1 × · · · × Kn , where each Ki ∈ K is simple of p-rank 1. Suppose that K  X and P ∈ Sylp (X), with P normalizing each Ki , 1 ≤ i ≤ n. Then P splits over P ∩ K. Proof. Without loss X = KP . Applying Lemma 3.1e to X/CX (K1 ) and pulling back a complement to a subgroup Q1 ≤ P , we have X = KQ1 with K1 ∩ Q1 = 1 and P = (P ∩ K1 )Q1 . Then by induction on n, X1 := (K2 · · · Kn )Q1 contains a complement Q in Q1 to Q1 ∩ (K2 · · · Kn ). Evidently, Q is a complement to P ∩ K in P .  Lemma 3.4. Let K ∈ C2 . If m3 (K) = 1, then K ∼ = L2 (2n ), n ≥ 2, n 2 ≡  (mod 3),  = ±1, or L2 (q), q ∈ F M.

n L− 3 (2 ),

Proof. If K ∈ Chev(2), this holds by the information in [IA , 4.10.3ab]. If K ∈ Spor∩C2 , then K ∼ J1 so m3 (K) > 1 by [IA , 5.6.1]. If K ∈ Chev(3)− = Chev(2), then as L2 (3) is solvable and L2 (8) ∈ Chev(2), m3 (K) > 1. These remarks imply the lemma.  Lemma 3.5. Suppose that K ∈ Cp and mp (K) = 1. If p = 3, then 5 ∼ K = L2 (8). If p = 5, then K ∼ = 2B2 (2 2 ) or A5 . Proof. See [V11 , 3.1] and [IA , 3.3.3, 4.10.3, 5.6.1].



Lemma 3.6. Let p be an odd prime and K ∈ TGp with abelian Sylow psubgroups of p-rank 2. Let P ∈ Sylp (Aut(K)). Then the following conditions hold: (a) Ω1 (P ∩ Inn(K)) ≤ Z(P ); (b) CP (P ∩ Inn(K)) ≤ P ∩ Inn(K); (c) Ω1 (P ) is elementary abelian; (d) Out(K) is p-nilpotent with cyclic Sylow p-subgroups consisting of images of field automorphisms; (e) If B ∈ Ep (Inn(K)), then K is weakly locally balanced with respect to B; and (f) If f is a field automorphism of order P in P , then Ω1 (P ∩Inn(K)) = [P ∩ Inn(K), f ]. Proof. If K ∈ Alt, then K ∼ = An , 2p < n < 3p, except that if p = 3, ∼ then K = A7 . Also Out(K) is a 2-group, and P = P ∩ K ∼ = Ep2 . Thus (a)–(d) hold. In (e), if K is not weakly locally balanced with respect to B, then B = P , and as B ≤ K, [Op (CK (b)), B] = 1 for some b ∈ B # . If b is a p-cycle, then CK (b) ∼ = Zp × An−p and since n = 7 has been excluded, Op (CK (b)) = 1, contradiction. Otherwise B has thhe same support on n letters as b, so [Op (CK (b)), B] = 1, again a contradiction. So the lemma holds if K ∈ Alt.

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∼ Ep2 , this time from [IA , 5.3], and again If K ∈ Spor, then P = P ∩ K = (a)–(d) follow. This time if (e) fails, then by [IA , 7.7.6b], K ∼ = M12 , M22 , or HJ, with p = 3, so K ∈ Tp , contradiction. Since m3 (L2 (27)) > 2, we may now assume, by [IA , 7.7.1c], that K ∈ Chev(r) for some r = p. Note that K cannot have a graph automorphism of order p = 3, since m3 (D4 (ra )) = 4 for r = 3 [IA , 4.10.3a], and 3D4 (ra ) has nonabelian Sylow 3-subgroups (see [IA , 4.7.3A]). Suppose that p divides | Outdiag(K)|. Then by [IA , 2.5.12], K/Z(K) ∼ = Lpk (q), or E6 (q) with p = 3, q ≡  (mod p),  = ±1. Since P is abelian, the only possibility is K ∼ = L3 (q) with p = 3; but then K ∈ T3 , contrary to assumption. Thus, (3A)

| Outdiag(K)|p = 1,

and so any outer automorphisms of p-power order are field automorphisms. With [IA , 2.5.12], this implies (d). Another consequence of (3A) is that the center of the universal version of K is a p -group. This implies (e), by [IA , 7.7.7d]. We next prove (a). Thus we assume f ∈ Ip (P ) is a field automorphism and it is enough to show that mp (CK (f )) = 2. There is no loss  in assuming that K is the universal version, and we write |K| = q N i Φ(q)ni as in [IA , (4.10.1)]. We also set m0 = ordp (q), the multiplicative order, so that nm0 by [IA , 4.10.2c]. We have q = q0p , and the order of 2 = mp (K) =  CK (f ) is q0N i Φ(q0 )ni . As q ≡ q0 (mod p), ordp (q0 ) = ordp (q) = nm0 . Hence mp (K0 ) = nm0 = 2 by [IA , 4.10.2c] again, and (a) is proved. Also Φm0 (q)/Φm0 (q0 ), like (q m−0 − 1)/(q0m0 − 1), is divisible by p to the first power exactly, so |K|/|CK (f )| is divisible by p2 and (b) holds. Moreover, P ∩ Inn(K) and C¶∩Inn(K) (f ) are each homocyclic, by [IA , 4.10.2c]. The mapping x → [x, f ] therefore has image isomorphic to Ep2 , proving (f). It remains to prove (c), and we may assume that f exists. Hence [IA , 7.3.3] applies and yields that there exist at least two subgroups P0 and P1  of P ∩ K of order p such that Ki := Or (CK (Pi )) = 1, and Ki ∈ Lie(r) by [IA , 4.2.2], i = 0, 1. By [IA , 4.2.3], f induces a field automorphism of order p on Ki . Also, since P ∩ K is abelian, Z(Ki ) is a p -group [IA , 15.12(i)]. Thus  by [IA , 4.2.2], Op (CK (Pi )) induces inner-diagonal automorphisms on Ki . It follows that CKf  (Ki ) = CK (Ki ) and mp (Ki ) = 1 = mp (CKf  (Pi Ki )). Now Ω1 (P ) = Ω1 (P ∩K f ), and it follows from 3.1 that the image of Ω1 (P ) modulo CKf  (Pi Ki ) is abelian. But Pi is the only subgroup of CKf  (Pi Ki ) of order p, and P0 ∩ P1 = 1. Hence Ω1 (P ) is elementary abelian, completing the proof.  Lemma 3.7. Let p be an odd prime and K ∈ Kp with mp (K) = 2. Suppose that K = F ∗ (X) and mp (X) = 3. Then Op (X) < X. Proof. Since p divides | Out(K)|, K ∈ Chev. We regard X ≤ Aut(K). If X contains a graph or graph-field automorphism of order p, then as p > 2 and m3 (D4 (q)) > 2 for any q [IA , 4.10.3a, 3.3.3], K ∼ = 3D4 (q). But then Out(K) is abelian [IA , 2.5.12], so the result holds in that case. If X contains

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a field automorphism f of order p, then f ∈ [X, X] by [IA , 2.5.12], and the  result holds. Hence we are reduced to the case Op (X) ≤ Inndiag(K) and p divides | Outdiag(K)|. As mp (K) = 2 and p > 2, we see by [IA , 4.10.3ab] that K ∼ = L3 (q), q ≡  (mod 3),  = ±1, and p = 3. But then by [IA , 4.10.3a], m3 (Inndiag(K)) = 2, contradicting m3 (X) = 3. The proof is complete.  Lemma 3.8. Let p be an odd prime and K ∈ TGp ∩ Chev with mp (K) = 2. If p = 3, assume that K is flat. Suppose that f is a nontrivial field automorphism of K of order p. Let P ∈ Sylp (K) be f -invariant. Then P is abelian, [Ω1 (P ), f ] = 1, and Ω1 (P ) = [P, f ]. Proof. By Lemma 2.2, P is abelian, and then Lemma 3.6 finishes the proof.  Lemma 3.9. Let p be an odd prime, K ∈ Kp , and P ∈ Sylp (K). If Ω1 (P ) is abelian and mp (K) ≤ 3, then K is simple. If in addition mp (K) = 2, then P is abelian and NG (P ) is irreducible on Ω1 (P ). Proof. Once it is established that P is abelian, the irreducibility of NG (P ) on Ω1 (P ) holds by [GL1, I: (12-1)]. Suppose that Z(K) = 1. We use [IA , 6.1.4] without comment. If K ∈ Alt, then p = 3 and K ∼ = 31+2 is not abelian, = 3A6 or 3A7 , so P = Ω1 (P ) ∼ contrary to assumption. If K ∈ Spor, then from [IA , 5.6.1] we get p = 3 and K ∼ = 3M22 or 3O N , as mp (K) ≤ 3. In both cases, from [IA , 5.3cs], P = Ω1 (P ) is not abelian, contradiction. [Note, however, that Ω1 (P ) ∼ = E34 ∼ is abelian in 3J3 .] If K ∈ Chev(p) − Alt, then p = 3 and K/Z(K) = U4 (3), Ω7 (3), or G2 (3). By [IA , 6.4.4], m3 (K) ≥ 5 in the first case, and this implies the same equality in the second case as 2U4 (3) = Ω− 6 (3) ≤ Ω7 (3). In the G2 (3) case, every root element is inverted by a Cartan involution, so P = Ω1 (P ), but even P/Z(K) is not abelian, contradiction. Still assuming Z(K) = 1, assume that K ∈ Chev(r) − Alt, r = p. If K ∼ = E6± (q) with p = 3, then K contains L ∼ = F4 (q) and m3 (L) = 4 by [IA , 4.10.3a], contradiction. So suppose that K/Z(K) ∼ = Lkp (q), q ≡  (mod p),  = ±1, k ≥ 1. If k = 1, then a monomial subgroup of K contains HN with H ∼ = Σp acting faithfully on H, so Ω1 (P ) is not = Epp−1 and N ∼ abelian. If k > 1, then K contains L ∼ = SLp (q) with L ∩ Z(K) = 1, so again Ω1 (P ) is not abelian, contradiction. Hence K is simple. Now suppose that mp (K) = 2 and Ω1 (P ) is abelian. A survey of the tables [IA , 5.3] shows that if K ∈ Spor, then P = Ω1 (P ) when mp (P ) = 2, so P is abelian in the present case. If K ∈ Alt, then K ∼ = An , 2p ≤ n < 3p, and |P | = p2 , hence P is abelian. If K ∈ Chev(p), 1+2 , yielding ∼ then from [IA , 3.3.3], K ∼ = L2 (p2 ) or L± 3 (p), and so P = Ep2 or p the desired result. Suppose then that K ∈ Chev(r), r = p. Then as TGp = G4p ∪ G5p ∪ G6p , K ∈ Cp ∪ Tp ∪ TGp .

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Hence by Lemma 2.2, we may assume that either K ∈ Cp ∪ Tp , or p = 3 and K is a non-flat TGp -group. If p > 3, then by definition of Cp and Tp ([V11 , 1 5 3.1], [V10 , 1.1]) and as mp (K) = 2, p = 5 and K ∼ = 2F4 (2 2 ) , 2F4 (2 2 ), or F i22 . In these cases, [IA , 4.10.2c, 5.3t] show that P is abelian, as required. Assume finally that p = 3. If K is of exceptional Lie type, we see from [IA , 4.7.3A] that K contains SL3 (q) for some q ≡  (mod 3) and  = ±1, so K contains 31+2 and Ω1 (P ) is not abelian. Thus K is a classical group. If K∼ = L3 (q) with q and  as usual, then P has a maximal subgroup P0 which is diagonal with respect to some frame F, and every element of P − P0 has order 3 (because itcycles the three lines in F and has determinant 1.) Thus, P = Ω1 (P ) is abelian in this case. In the other cases where m3 (K) = 2, we use [IA , 4.8.1, 4.8.2] to see that K is a flat TG3 -group. The proof is complete.  Lemma 3.10. Let K ∈ K3 and P ∈ Syl3 (K) be such that m3 (P ) > 1 and some subgroup of P outside Z(K) of order 3 is weakly closed in P with respect to K. Then K ∼ = G2 (q) , q ≡ 0 (mod 3), or HJ. Proof. We may pass to K/Z(K) and assume that Z(K) = 1 (note that the Schur multipliers of G2 (q) and HJ are 3 -groups [IA , 6.1.4]). Suppose that Z = z ∼ = Z3 is weakly closed in P with respect to K. Then Z commutes with none of its distinct K-conjugates. Also by [GL1, I: (121)] we may assume that P is not abelian, for then NK (P ) is irreducible  on Ω1 (P ). In many cases we proceed inductively, exhibiting O3 (CK (y)) ∼ = C × L for some non-3-central y ∈ I3 (K), where C is a cyclic 3-group and F ∗ (L) is simple with m3 (L) > 1 and L ∼ G2 (q) for any q ∼ 0 (mod 3). = = Then Z ≤ C and the projection of Z on L is weakly closed in a Sylow 3-subgroup of L, giving a contradiction by induction. Now if K ∼ = An , then n ≥ 9. Obviously Z commutes with some distinct Σn -conjugate Z ∗ , and CΣm (Z) contains an odd permutation, so Z ∼K Z ∗ , contradiction. Suppose K ∈ Spor. If K is a Mathieu group, then m3 (K) = 2, while K contains M11 or M22 , which have a completely fused E32 -subgroup. NK (P ) has no normal subgroup of order 3 if K ∼ = J3 . NK (J(P )) has no normal subgroup of order 3 if K ∼ = Co3 , M c, Suz, or Ly. There is one class of elements of order 3 in J4 , HS, Ru, and O N . We make the inductive argument above for the following pairs (K, L): (Co2 , U4 (2)), (Co1 or F5 , A9 ), (F i22 , U4 (3)), (F i23 , Ω7 (3)), (F i24 , D4 (3) · 3), (F3 , G2 (3)), (F i22 , F2 ), (F3 , F1 ). ∼ He. Thus, K ∼ And [IA , 5.3p, Note 4] rules out K = = HJ. If K ∈ Chev(3) with m3 (K) > 1, then K ∈ Lie(3), and for any parabolic subgroup M ≥ P , Z(P ) ≤ O3 (M ) so Z  M . Hence M ≤ NG (Z), so K has twisted rank 1. Since a Cartan subgroup normalizes Z, q(K) = 3. The only example is thus U3 (3) ∼ = G2 (2) .

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Suppose then that K ∈ Chev(r) − Chev(3), r = 3. If K is of exceptional Lie type, we again argue inductively with (K, L) as follows: (E8 (q), E7 (q)), ∼ 3 (E7 (q), D6 (q)), (E6± (q), L± 6 (q)), and (F4 (q), C3 (q)). If K = D4 (q), then  ∼ m3 (K) = 2 and K contains P GL3 (q), q =  (mod 3),  = ±1. Finally, suppose that K is a classical group. If K ∼ = Ln (q), q ≡  (mod 3),  = ±1, then z is the image of an element of GLn (q) of order 3 (otherwise [IA , 4.8.4] applies and |CK (z)|3 < |K|3 , contradiction). Hence we may assume that z is diagonal. Then a distinct diagonal conjugate of Z is easily found, unless n = 3, in which case a permutation matrix of order 3 commutes with z, completing the Ln (q) case. In the remaining cases by [IA , 4.8.2], there are n = m3 (K) elements z1 , . . . , zn of order 3 with linearly independent (mutually orthogonal) 2-dimensional supports on the natural module such that for some 1 ≤ k ≤ n, z = z1 · · · zk . If k = 1, then z = z1 is conjugate to z2 . If k > 1, then z is conjugate to z1 · · · zk−1 zk−1 . This completes the proof.  Lemma 3.11. Let K ∈ K3 and x ∈ I3 (Aut(K)). Suppose that CK (x) has a component L ∼ = 3A6 such that Z(L) ≤ Z(K). Then m3 (K) > 2. Proof. Suppose K ∈ Chev(r). If r = 3, then as Z(K) = 1, q(K) = 3 by [IA , 6.1.4]. By the Borel-Tits theorem, x is a graph automorphism, and then by [IA , 4.9.2], L ∼ = G2 (q), contradiction. So r = 3. As A6 ∈ Chev(r) by [IA , 4.9.6], r = 2. Since Z(K) is a 3-group, K ∈ Lie(2). But then if x ∈ Aut0 (K), L ∈ Lie(2) by [IA , 4.2.2], which is not the case. Hence x is a field or graph-field automorphism, so K/Z(K) ∼ = Sp4 (8) by [IA , 4.9.1], and hence Z(K) = 1 by [IA , 6.1.4], contradiction. Therefore K ∈ Alt ∪ Spor. If K ∈ Alt, then K/Z(K) ∼ = An , n ≥ 9, so Z(K) = 1 by [IA , 6.1.4], contradiction. Finally, if K ∈ Spor, then from [IA , 5.6.1], the only possibility with Z(K) = 1 and m3 (K) ≤ 2 is K ∼ = 3M22 , but in that case, CK (x) is  solvable by [IA , 5.3c], contradiction. The proof is complete. Lemma 3.12. Let K ∈ C3 and assume that m2,3 (K) = m3 (Z(K)). Then K∼ = L2 (8). Proof. If K ∈ Spor, then this follows from [IA , 5.6.2] (and the observation that M11 is 3-saturated and contains elements of order 6). If K ∈ Alt − Chev(3), then K ∼ = A9 , which shares these properties with M11 . If K ∈ Chev(2) − Chev(3), then (see [V11 , 3.1]) some parabolic subgroup of K contains Σ3 , which implies the desired conclusion. If K ∈ Lie(3), then K contains A4 or SL2 (3). Finally assume that K ∈ Chev(3) − Lie(3) but 1 K∼ 2 G2 (3 2 ) ∼ = L2 (8). Then Z(K) = 1 and K/Z(K) contains (S)U4 (3) or = G2 (3). So the result follows by [IA , 6.4.4] and the fact that G2 (3) contains  SL2 (3) ∗ SL2 (3). Lemma 3.13. Let K ∈ C3 and suppose that m2,3 (K) − m3 (Z(K)) ≤ 2. Then Out(K) is 3-closed. Proof. We use [IA , 2.5.12] and [V11 , 3.1]. If K ∈ Chev(3) and 3 divides Out(K), then K/Z(K) ∼ = 3D4 (2), Sp4 (8), or U6 (2), and in every case

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| Out(K)| divides 6. So we may assume that K ∈ Chev(3). The condition on m2,3 (K) rules out K ∼ = D2n (3n ), n ≥ 2, which contains the central product of four copies of SL2 (3n ). Hence Out(K)/ Outdiag(K) and Outdiag(K) are cyclic. If thes result is false, then Outdiag(K) has a subgroup of prime order n p ≡ 1 (mod 3), so p ≥ 7 and K ∼ = L± m (3 ), m ≥ 7. But then K contains three commuting copies of SL2 (3n ), so m2,3 (K) ≥ 3 and m3 (Z(K)) = 0, contrary to assumption. The lemma follows.  4. Balance and Signalizers Lemma 4.1. Let K = A6 , M11 or L2 (8). Let D be an elementary abelian 3-subgroup of K of maximal rank. Then CAut(K) (D) has odd order. Proof. This is immediate from [IA , 3.1.4] for K = A6 ∼ = L2 (9), from [IA , 5.3a] for K = M11 , and from [IA , 6.5.1] for K = L2 (8), in the last case using the fact that | Out(K)| is odd [IA , 2.5.12].  Lemma 4.2. Let L ∈ Co2 ∩ A. Let F and W be subgroups of Aut(L) such that F ∼ = A4 , F normalizes W , and W ≤ O2 (CInn(L)F W (O2 (F ))). Suppose also that either L/Z(L) is involved in A9 or J3 , or that CAut(L) (F ) contains an element b of order 3. If W = 1, then L/Z(L) ∼ = M12 , b exists, and O (F ) × Σ . CInn(L) (O2 (F )) ∼ = 2 3 Proof. If L is embeddable in A9 or J3 , set b = 1. Assume that W = 1. Note first that as L ∈ C2 , Lemma 20.9 below implies that Out(L) is 2nilpotent, and indeed a 2-group unless L ∈ Chev(2) [IA , 2.5.12]. In any case O2 (F ) induces inner automorphisms on L. Hence if L ∈ Chev(2), then W = 1 by the Borel-Tits theorem, a contradiction. Otherwise W, F × b maps into O2 (Aut(L)) = Inn(L). Replacing W and F × b by their images, we may assume that L is simple and work within L. Write O2 (F ) = u, v and set C = CL (u). Then W ≤ O2 (CC (O2 (F ))) ≤ L∗2 (C) by L∗2 -balance [IG , 5.18], so L∗2 (C) = 1. But the structure of C may be found in [IA , 4.5.1, 6.5.1] or [IA , 5.3] according as L ∈ Chev or L ∈ Spor. From these data, we see that either CC (O2 (F )) = O2 (F ) or L∗2 (C) = 1, unless L ∼ = M12 , HS, 3 ∼ J2 or Ru, with C = Z2 × Σ5 , Z2 × Aut(A6 ) E22 × A5 , or E22 × 2B2 (2 2 ), respectively. In particular the order of L does not divide |A9 | or |J3 |, so our hypothesis yields b = 1. 3 Now 1 = W ≤ O2 (CC (v)). If C ∼ = E22 × A5 , or E22 × 2B2 (2 2 ), then the Borel-Tits theorem is contradicted. Thus L ∼ = M12 or HS, whence CC (v) ∼ = Σ3 or D10 , respectively [IA , 6.5.1]. Since by assumption b ∈ CC (v)  has order 3, we have L ∼ = M12 and the proof is complete. Lemma 4.3. Let F ∗ (X) = K = P Sp4 (3) or U4 (3) and X = K t with t2 = 1 and E(CK (t)) ∼ = A6 . If K ∼ = U4 (3), assume in addition that m2 (CAut(K) (E(CK (t))) = 2. Let y ∈ I2 (X) and B0 ∈ E32 (CX (y)). Then CO2 (CX (y)) (B0 ) is cyclic.

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Proof. We have t ∈ Inndiag(K), by Lemma 13.3a if K = U4 (3), and as Aut(P Sp4 (3)) = Inndiag(P Sp4 (3)).  Let L = O 3 (CX (y)). From [IA , 4.5.1], CInndiag(K) ( y L) is a 2-group, and since m3 (CX (y)) ≥ 2, L ∼ = A6 or SL2 (3) ∗ SL2 (3), or K ∼ = U4 (3) and ∼ L = U3 (3). Moreover, CX (t) is solvable for any 2-element t ∈ K # . In the cases where CX (y) is not solvable, CAut(K) (B0 ) ≤ CAut(K) (L). Therefore if CX (y) is not solvable, then CCX (y) (B0 ) embeds in Outdiag(K), so is cyclic. Suppose then that L ∼ = SL2 (3) ∗ SL2 (3). Let Q = O2 (L) and R = CO2 (CX (y)) (B0 ). By the A × B-lemma applied to the action of R × B0 on Q, B0 acts faithfully on CQ (R), whence R centralizes QB0 = L. Then again  by [IA , 4.5.1], R is cyclic. The proof is complete. Lemma 4.4. Let K ∈ Co2 with Z(K) = 1 and mr (K) ≤ 3 for all odd primes r. Let E ∈ E3 (Aut(K)) with CZ(K) (E) = 1, and let Q = O2 (CK (x)) for some x ∈ E # . Suppose that Q = [Q, E]. Suppose also that 3 divides |K| and CK (x) is solvable. Then |Q/Z(K)| = 4 and K/Z(K) ∼ = M12 , M22 , or HJ. Proof. Since K ∈ C2 , Z(K) = O2 (K). Let K = K/Z(K). By [IA , 6.1.4], m2 (Z(K)) ≤ 2. Therefore as CZ(K) (E) = 1, we have [Z(K), E] = 1. Since Q = [Q, E], 1 = Q ≤ O2 (CK (x)). If K ∈ Chev(3), then it follows by [IA , 7.7.1] that K ∼ = L2 (33 ); but then Z(K) = 1, contradiction, by 3 [IA , 6.1.4], since SL2 (3 ) ∈ C2 . So K ∈ Chev(3). Note also that K ∼ [L, L] = 1 2 2 for L = B2 (2), G2 (2), or F4 (2 ), since Z(K) = 1 and K ∈ C2 . Hence if K ∈ Chev(2), then K ∈ Lie(2). In that case, if x induces an element of  Aut0 (K), then by [IA , 4.2.2], O2 (CK (x)) is a product of Lie components in  characteristic 2, so O2 (CK (x)) = O2 (O2 (CK (x))) = 1, contradicting Q = 1. Thus if K ∈ Chev(2), then x induces a field or graph-field automorphism on 3 K, so q(K) ≥ 8. Again by [IA , 6.1.4], this forces K ∼ = 2B2 (2 2 ), contradicting our assumption that 3 divides |K|. Thus, if K ∈ Chev(r) for some r, then r > 3, so K ∼ = L2 (q), q ∈ F M9, and again by definition of C2 , K ∈ C2 , contradiction. We have proved that K ∈ Chev. If K ∈ Alt, then K ∼ = 2An for some n ≥ 5, so K ∈ C2 , contradiction. Hence, finally, K ∈ Spor. By our hypothesis K is not weakly locally balanced with respect to the prime 3, so the proof is finished by [IA , 7.7.1].  Lemma 4.5. If K = L3 (4) and W ∈ Syl3 (K), then NAut(K) (W ; {2, 3} ) = {1}. Proof. Let X ∈ NAut(K) (W ; {2, 3} ). We have W ∼ = E32 . As Out(K) is a {2, 3}-group [IA , 2.5.12] and CK (x) = W for every x ∈ W # ,   X ≤ O 3 (CInn(K) (x)) | x ∈ W # = 1, as claimed.



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Lemma 4.6. Let K ∈ C2 be simple and suppose that Inn(K) ≤ H ≤ Aut(K), t ∈ I2 (H) and O2 (CH (t)) = 1. Then one of the following holds: (a) K ∼ = L3 (4), t ∈ Inn(K), O2 (CH (t)) = O2 (CK (t)) ∼ = E32 , and CK (t) ∼ = U3 (2); or (b) K ∼ = L2 (q), q = 2a +  ∈ F M9,  = ±1, O2 (CH (t)) = O2 (CK (t)) ∼ = Z2a−1 + , and t ∈ Inndiag(K) − Inn(K). Proof. This is a direct consequence of [IA , 7.7.1, 6.5.1], and the defi nition of C2 [V11 , 3.1]. Lemma 4.7. Let H = Aut(L) for L ∼ = M12 , HJ, or Ru. Let K = E(CL (s)) = 1 for s ∈ I2 (L). Then O2 (CH (K)) = 1. Proof. In all cases, CH (K) ≥ U ∼ = Z2 × Z2 . If X := O2 (CH (K)) = 1, then for some u ∈ U # , CX (u) := Y = 1. Then CL (u) ≥ u × Y × K. But  this is impossible by [IA , 5.3bgr]. Lemma 4.8. Let L ∈ Cp ∪ Tp ∪ TGp (G), p an odd prime. Suppose that k ≥ 1 is an integer or half-integer, A ∈ Ep[k]+2 (Aut(L)), and L is not locally k-balanced with respect to A. Then the following conditions hold: (a) k ≤ 2; (b) If k = 2, then one of the following holds: (1) p = 3, L/Z(L) ∼ = L3 (q),  = ±1, q ≡  (mod 3), A ≤ Inndiag(L) and A is the image of a nonabelian p-subgroup of GL3 (q); (2) p = 3, L ∼ = L has a regular orbit and = A11 , and A ≤ Inn(L) ∼ 2 fixed points on 11 letters; or (3) p = 5, L ∼ = F i22 , and A ∈ Syl5 (Aut(L)). (c) If k = 3/2 but L is locally 2-balanced with respect to A, then one of the following holds: (1) L ∈ Chev − Chev(p), and some a ∈ A is a nontrivial field, graph, or graph-field automorphism of L; (2) p = 3, L/Z(L) ∼ = Ln (q), n = 3 or 6,  = ±1, q ≡  (mod p), A ≤ Inndiag(L) ∼ = P GLn (q), and A is the image of some nonabelian 3-subgroup of GLn (q); or (3) p = 3 and L ∼ = L2 (27), A7 , 3A7 , M12 , M22 , 3M22 , HJ, or A10 . (d) If k = 1 and L ∈ Cp , then p = 3, L ∼ = L2 (27), and some element of # A is a nontrivial field automorphism. Proof. Part (a) holds by [IA , 7.7.7ab], and part (d) by [IA , 7.7.1, 7.7.8]. Part (b) holds by [IA , 7.7.4] and the observation that if L/Z(L) ∼ = Ap−1 (q), q ≡  (mod p),  = ±1, then m  p (L/Z(L)) = p − 2; so if p > 5, then L ∈ Gp − TGp = Kp − CTGp , contradiction. By (d) and Lemma 4.12, (c) holds if we assume that K ∈ Chev. Finally, if L ∈ Alt ∪ Spor, then (c)  follows from [IA , 7.7.1bc] and the definitions of Cp , Tp , and TGp .

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Lemma 4.9. Suppose that p is an odd prime and K ∈ CTGp . Then the following hold: (a) K is 3-balanced and weakly 2-balanced with respect to p; (b) If K is not 2-balanced with respect to p, then p ∈ {3, 5} and either (1) K/Z(K) ∼ = Lp (q), q ≡  (mod p),  = ±1, or ∼ (2) (p, K) = (3, A11 ) or (5, F i22 ); (c) Suppose in (b) that if K ∈ TGp ∩ Chev, then mp (K) = 2. Then L5 (q) should be removed from the conclusion (b1) for p = 5; and (d) Suppose, in addition to the assumption in (c), that if K ∈ Tp , then mp (K) = 1. Then p = 3 and K ∼ = A11 . Proof. By [IA , 7.7.7ab], (a) holds unless possibly K ∼ = An with either n > p3 , or n = 9k + 4, k ≥ 1, and p = 3. But then K ∈ G1p ∪ G3p [III3 , (5B)], so K ∈ Gp − TGp , contrary to hypothesis,. Hence (a) holds. Similarly, (b) follows from [IA , 7.7.7c], excep that we need to prove in (b1) that p ≤ 5. But if p ≥ 7, then m  p (SLp (q)) = mp (SLp (q)) − 1 = p − 2 ≥ 5, so K ∈ G2p and we again reach the contradiction K ∈ Gp − TGp . Here as usual,  = ±1 and q ≡  (mod p). Likewise in (c), if K/Z(K) ∼ = L5 (q) and p = 5, then m5 (K) ≥ 3, contrary to assumption. Finally in (d), the groups (S)L3 (q) for p = 3 and F i22 for p = 5 lie in Tp − Tp5 , and so they are assumed away, leaving only A11 and p = 3. The proof is complete.  Lemma 4.10. Let p ∈ {3, 5} and K ∈ Cp ∪ TGp . If K ∈ TGp , assume that K is flat if p = 3, while if p = 5 and K ∈ Chev, assume that mp (K) = 2. Then the following conditions hold: (a) If K is not locally balanced with respect to p, then one of the following holds: (1) p = 3 and K ∼ = L2 (27), P Sp4 (q), L− n (q), n = 4 or 5, q ≡  (mod 3),  = ±1, D4− (2), D5 (2), HS, A10 , or A11 ; or  is a TG5 -group of 5-rank 2, or isomorphic to An , (2) p = 5 and L 17 ≤ n ≤ 19; (b) If K is as in (a), then mp (Out(K)) ≤ 1; (c) Suppose that K is not weakly locally balanced with respect to p, and let Inn(K) ≤ H ≤ Aut(K) and [Op (CH (x)), y] = 1 for some commuting x, y ∈ Ip (H). Then K ∼ = L2 (27) or A10 , or y is a field automorphism. ∼ L2 (27) with p = 3, and Proof. If K ∈ Cp , then by [IA , 7.7.9], K = the assertions all hold. So assume that K ∈ TGp . If p = 3, note that Sp8 (2), M23 , and A12 are locally balanced by [IA , 7.7.8, 7.7.1b, 7.7.1a], so the flat TG3 -group K is as in (a1). If p = 5, then we may assume in (a) that K ∈ Alt ∪ Spor, i.e., K ∈ G65 [III3 , (5B)], and we again quote [IA , 7.7.1ab] to get the alternating groups in (a2), together with Suz, of 5-rank 2. Then (b) follows, see [IA , 2.5.12]. Suppose that K is not weakly locally balanced, as in (c). If p = 3 and K ∈ Alt ∪ Spor, or p = 5 and K ∼ = Suz, we quote [IA , 7.7.6b] to get

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∼ A10 as the only possibility. If p = 5 and K = ∼ An , 17 ≤ n ≤ 19, then x K= is the product of three disjoint 5-cycles, and y and O5 (CΣn (x)) have disjoint supports, so [O5 (CΣn (x)), y] = 1, contradiction. As L2 (27) is allowed when p = 3, we may assume that K ∈ Chev(r), r = p. If p = 3, then the listed groups in (a1) all are p-saturated by [IA , 6.4.1], and the same conclusion holds for the groups in (a2), by Lemma 2.2. Hence by [IA , 7.7.6c], either  x or y induces an automorphism of K outside Op (Inndiag(K)) = Inn(K)). By Lemma 3.6d, such automorphisms are field automorphisms. Hence if (c) fails, then y induces an inner automorphism on K, and x induces a field automorphism. But then Op (CK (x)) ≤ Z(CK (x)), and as y acts as an element of CK (x), [Op (CK (x)), y] = 1, contradiction. This completes the proof of (c). Finally, (b) is trivial unless K ∈ Chev, in which case it holds with the help of Lemma 3.6d. This completes the proof.  Lemma 4.11. Suppose that p is an odd prime, K ∈ Tp5 is not weakly locally balanced with respect to p, and let Inn(K) ≤ H ≤ Aut(K) and [Op (CH (x)), y] = 1 for some commuting x, y ∈ Ip (H). Then y is a field automorphism. Proof. Suppose y is not a field automorphism. Then by Lemma 3.1ad and [IA , 4.9.1], y is an inner automorphism and x is a field automorphism. If K ∈ Chev(p), then by [IA , 7.7.1c], p = 3 and K ∼ = L2 (33 ), against mp (K) = 1. So K ∈ Chev(r), r = p. Then by [IA , 7.7.1d], however, x ∈ Inndiag(K), a contradiction. The proof is complete.  Lemma 4.12. Let p be an odd prime and K ∈ Chev(r)∩CTGp −Cp , r = p. Suppose that E ∼ = Ep2 and E ≤ Inndiag(K), and Inn(K) ≤ H ≤ Aut(K) with H being E-invariant. Suppose that e ∈ E # and [Op (CH (e)), E] = 1. Then K/Z(K) ∼ = Ln (q), q ≡  (mod p), n = 3, 5, or 6, with p = 3, 5, or 3, respectively, and E is the image of a nonabelian subgroup of GLn (q).  p (Ln (q)) = n − 2, so Proof. Observe that if K/Z(K) ∼ = Ln (q), then m  3 (E6 (q)) = 5 so K ∼ E6 (q), if q ≡  n ≤ 6 as L/Z(L) ∈ CTGp . Likewise m = (mod 3) and p = 3. If e ∈ Inn(K), then by [IA , 7.7.6d], the Schur multiplier of K/Z(K) is divisible by p, so by the previous paragraph, K/Z(K) ∼ = Ln (q), n = 3, 5, or 6, and p = 3, 5, or 3, respectively. Again by [IA , 7.7.6d], E is the image of a nonabelian subgroup of GLn (q), so the lemma holds in this case. Suppose then that e ∈ Inn(K). Then p divides | Outdiag(K)| so again by the first paragraph, K and p are as claimed in the lemma. It remains to  of GL (q) and derive assume that E is the image of an abelian subgroup E n  := GL (q), then since a contradiction. If e has a preimage e of order p in H n  is abelian it preserves the eigenspaces of e on the natural H-module.  E Now  acting as scalars on each such eigenspace, Op (CH (e)) has a preimage in H so [Op (CH (e)), E] = 1, contrary to hypothesis. Hence for some preimage   ep =: f is a generator of a Sylow p-subgroup of Z(H). By e of e in H,

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 [IA , 4.8.4], CH (e) ∼ = GLn/p (q p ). Thus Op (CH (e)) ≤ Z(CH (e)). Since E is abelian, it centralizes Op (CH (e)), whence [Op (CH (e)), E] = 1, a final contradiction.  Lemma 4.13. Let p be an odd prime. Suppose K ∈ Cp ∪ Tp5 ∪ TGp . If K ∈ TGp , assume that K has Sylow p-subgroups of rank 2. Then K is locally 2-balanced for p. ∼ A11 ∈ TG3 , Proof. This follows from Lemma 4.9d. Notice that if K =  then m3 (K) = 3. Lemma 4.14. Suppose that K ∈ TG3 ∩ Chev(2) with q(K) ≤ 2 and K locally unbalanced for p = 3. Then K ∼ = D3 (2), D4− (2), or D5 (2), and K is locally 3/2-balanced for p = 3. ∼ A2 (2) or D  (2), m ≥ 3,  = (−1)m+1 . Proof. By [IA , 7.7.8b], K = m  (2)) = m (D  (2)) ≥ m − 1, so D  (2) ∈ TG if  3 (Dm But A2 (2) ∈ G3 and m 3 3 m m  m ≥ 6. Finally, K is locally 3/2-balanced for p = 3 by [IA , 7.7.7d]. Lemma 4.15. Suppose (K, p) = (L3 (q), 3) or (F i22 , 5), B ∈ E32 (Aut(K)), K is not locally 3/2-balanced with respect to B, and m3 (CAut(K) (B)) ≥ 3. Then (K, p) = (L3 (q), 3), q > 8, and some element of I3 (CAut(K) (B)) is a field automorphism. Proof. Since m5 (Aut(F i22 )) = m5 (F i22 ) = 2, we may assume that p = 3 and K ∼ = L3 (q). In particular, m3 (Inndiag(K)) = m3 (P GL3 (q)) = 2, so a nontrivial field automorphism f ∈ CAut(K) (B) exists as asserted. Hence it remains to rule out the possibility q = 8. But in that case the only failure of local 3/2-balance can be in B = i, d, where i and d are, respectively, inner and outer diagonal (and O3 (CK (d)) ∼ = Z19 ). Since CInndiag(K) (f ) ≤ Inn(K) by Lemma 11.3.1, [f, d] = 1, contradicting [f, B] = 1. The proof is complete.  Lemma 4.16. Let K ∈ K3 with U3 (3) 4, and in particular B1 ∈ Syl3 (K). For any b ∈ B1# , b fixes an (orthogonal) frame F pointwise, in the natural module V , and an element b ∈ B1 − b cycles the three lines forming the frame. Hence   CK (b) = A b with A  CK (b) and A abelian with invariant factors q −  and 13 (q − ). Moreover, CA (b ) = b. Given the congruences on q and as q > 4, q −  is not a power of 3. We take any prime divisor p1 of 13 (q1 − ) and let X be a  minimal b -invariant p1 -subgroup of A. The proof is complete.  := Lemma 4.24. Let K ∈ T33+ and let L be a 3-component with L # ∼ L/O3 (L) = K. Let B1 ∈ Syl3 (L) and b1 ∈ B1 , and let R be a B1 -invariant  = p0 -subgroup of O3 (CL (b1 )) for some prime p0 = 3, such that 1 = R  B 1 ] ∈ Sylp (O3 (C  (b1 )), and chosen if possible so that p0 = 2. Let M [R, 0 L be a faithful Fp1 L-module for some prime p1 = 3. Then [M, R, R] = 1. Proof. Since M is faithful and R = 1, [M, R] = 1. If p0 = p1 , then [M, R, R] = [M, R] and we are done, so assume that p0 = p1 . Suppose that p0 ≥ 5. Since K ∈ T33+ , some y ∈ R# normalizes but does not centralize a Sylow q-subgroup of L, where K ∼ = L3 (q a ). Then by Theorem B of Hall and Higman [G1, 11.1.1], [M, y, y] = 1, as required. Therefore we may assume  is a homocyclic abelian 2-group of rank that p0 < 5, whence p0 = 2. Now R

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2. But if [M, R, R] = 1, then by [V9 , 5.2], R is elementary abelian. Hence |R| = 4. Since R was chosen if possible so that p0 > 2, K ∼ = U3 (5) or L3 (7). But then K ∈ T33+ , contradiction.  Lemma 4.25. Let K ∈ TG3 ∩ (Alt ∪ Spor) and suppose that K is locally unbalanced with respect to some B ∈ E33 (Aut(K)). Then K ∼ = A10 or A11 . Proof. By assumption, and as Out(K) is a 2-group, m3 (K/Z(K)) = m3 (Aut(K)) ≥ 3. However, if K ∈ Spor, then by [IA , 7.7.1b] and [IA , 5.6.1], m3 (K/Z(K)) = 2. So K ∼ = An , with 10 ≤ n ≤ 12 as K ∈ TG3 . But A12 is  locally balanced for p = 3, by [IA , 7.7.1a]. Hence the lemma holds. Lemma 4.26. Let p be an odd prime and K ∈ Cp . Then K is locally balanced for p unless p = 3 and K ∼ = L2 (27). Proof. This holds by comparison of [V11 , 3.1] with [IA , 7.7.8].



Lemma 4.27. Let K be a C3 -group and t an involutory automorphism of K such that O3 (CK (t)) = 1. Then either K ∼ = A9 or O3 (CK (t)) ≤ Z(K). In particular, if m3 (O3 (CK (t))) ≥ 2, then K ∼ = [3 × 3]U4 (3). Finally, if K/Z(K) ∈ A and CK (t) is a {2, 3}-group, then m2 (O2 (CK (t))) > 1. Proof. We begin with the first of the three assertions of the lemma. As K ∈ C3 , Z(K) is a 3-group, which we may factor out. Thus, assume  that K is simple. If K ∈ Chev(3), then [IA , Sec. 4.9] O3 (CK (t)) is a product of Lie components in Lie(3). So O3 (CK (t)) = 1, as desired. If K ∈ Spor ∪ Chev(2) − Chev(3), and O3 (CK (t)) = 1, then K is locally unbalanced for p = 2. Then either K is one of the first three (unbalanced) exceptions in [IA , 7.7.7e], or K ∼ = He by [IA , 7.7.1b]; in either case, K ∈ C3 , contradiction. If K ∈ Alt − Chev(3), then as K ∈ C3 , K ∼ = A9 , as claimed. Then the second assertion of the lemma follows, with [IA , 6.1.4]. Suppose that K ∼ = A9 and O3 (CK (t)) = 1. Using [IA , 5.2.8a] we see that t acts on K like the product (in Σ9 ) of three disjoint transpositions. Then O2 (CK (t)) ∼ = E23 , so the last assertion of the lemma holds in this case. Finally, suppose that K ∈ C3 [V11 , 3.1], K/Z(K) ∈ A (Definition 1.1) and Z(K) = 1 [IA , 6.1.4], so that in particular, K/Z(K) ∼ A9 . We get K ∼ = =  3G2 (3), 3U4 (3), [3 × 3]U4 (3), 3J3 , 3M c, 3Suz, 3O N , or SU6 (2). Assume by way of contradiction that CK (t) is a {2, 3}-group and m2 (O2 (CK (t)) ≤ 1. Then m2 (O2 (CK/Z(K) (t))) ≤ 1, as |Z(K)| is odd. These conditions on CK (t), however, are impossible: for K ∈ Chev(3), by inspection of [IA , 4.5.1, 4.9.1]; for K ∈ Spor, by inspection of [IA , 5.3]; and for K ∼ = SU6 (2), because [Z(K), t] = 1 forces t ∈ Inn(K), so by the Borel-Tits theorem, CK (t) ≤ P for some parabolic subgroup P of K with t ∈ O2 (P ). Then m2 (CO2 (P ) (t)) = 1, whence t ∈ Z(O2 (P )) and m2 (O2 (P )) = 1; but m2 (P ) = 9 and P/O2 (P )  embeds in Out(O2 (P )), which is impossible. The proof is complete. Lemma 4.28. Let K = U3 (2a ), a odd, and let b ∈ K be the image of an element h ∈ I3 (SU3 (2a )) with distinct eigenvalues. Then NK ( b , 2) = {1}.

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Proof. If false, then as a Sylow 2-subgroup T of SU3 (2a ) is a TI-set, we could choose T to be h-invariant. Then since 3 divides 2a +1, Z(T ) ≤ CT (h). But CT (h) has odd order as h has distinct eigenvalues, a contradiction.  Lemma 4.29. Suppose that (K, p) = (L6 (2) or L7 (2), 7), (Sp8 (2), 5), (Sp4 (4), 5), (D4 (2), 5), (U3 (4), 5), (G2 (4), 5) or (F4 (2), 5 or 7). Then K is locally balanced for p. Proof. Let P ∈ Sylp (K). Note that in every case, P ∼ = Ep2 . We first claim that it suffices to prove that (4B)

Op (CK (x)) = 1 for all x ∈ Ip (P ).

This is obvious if K = Sp8 (2), as then Out(K) = 1. For the rest, suppose that H is a group with F ∗ (H) = K simple and W = Op (CH (x)) = 1. As Out(K) is a p -group in every case, x ∈ K. Then [W, CK (x)] ≤ W ∩ K = 1, the last by (4B). In particular W centralizes P . If K = D4 (2), note that by a Frattini argument, CAut(K) (P ) Inn(K)  Aut(K). As Out(K) ∼ = Σ3 , it suffices to show that CH (P ) does not contain a graph automorphism w of order 3; this in fact holds since CK (w) ∼ = G2 (2) or P GU3 (2) is a 5 -group, by [IA , 4.9.2]. In the cases K = L6 (2), L7 (2), and U3 (4), CH (P ) does not contain a graph automorphism w of order 2, since CK (w) embeds in Sp6 (2) in the first two cases, and CK (w) ∼ = L2 (4) in the third. Similarly if K = Sp4 (4) oor G2 (4), CH (P ) does not contain a field automorphism w of order 2. Finally a similar argument works for K = F4 (2) if p = 7. For p = 5, 1  CK (w) ∼ = 2F4 (2 2 ) does not contain x O2 (CK (x)) ∼ = Z5 × Σ6 [IA , 4.8.6]. This proves our claim.  Still with K = F4 (2), we set N0 := x O2 (CK (x))  CK (x) and use [IA , 4.8.6b]. By a Frattini argument, CK (x) = N0 NCK (x) (P ). But |NCK (x) (P )| = |NK (P )|/|xNK (P ) | = 25 3|P |/23 3 = 4|P | = |NN0 (P )| so CK (x) = N0 and O5 (CK (x)) = 1. Similar arguments show that for p = 7, O7 (CK (x)) = 1, and, using [IA , 4.8.8b], for p = 5 and K = G2 (4), O5 (CK (x)) = 1. Finally, for the classical cases for K, we use [IA , 4.8.1, 4.8.2] to compute CK (x) ∼ = L2 (8) or Z7 × Ln−3 (2) for K = Ln (2), n = 6 or 7; and CK (x) ∼ = GU2 (4) ∼ = Z5 × L2 (4) for K = D4 (2), Sp4 (4), or U3 (4). In all cases  Op (CK (x)) = 1, so the proof is complete.  Lemma 4.30. Suppose that K ∼ = A12 , Sp8 (2), = L or K ↑3 L, where L ∼ Sp6 (8), or U4 (8). Then K is locally balanced with respect to p = 3. Proof. This is immediate from [IA , 7.7.1a, 7.7.8bc].



Lemma 4.31. Let L3 (2) ↑7 K ∈ K7 and suppose that K is not locally balanced for p = 7. Suppose also K ∈ A. Then K ∼ = L3 (27 ). Proof. As L3 (2) ↑7 K, K ∈ Chev(2) ∪ Chev(7) ∪ Spor by [IA , 4.9.6, 5.2.9]. Using [IA , 7.7.1] and that K ∈ A we find that K ∈ Chev(2). Assuming that K ∼ L3 (27 ), we have by [IA , 4.2.2] that K has level q(K) = 2. As = K ∈ A, and | Aut(K)|7 ≥ 72 , K ∼ = L6 (2), 3D4 (2), or F4 (2). If K ∼ = L6 (2)

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∼ or F4 (2), we appeal to Lemma 4.29. Indeed with [IA , 4.8.6b], CF4 (2) (x) = Z7 ×L3 (2)) for all x ∈ I7 (F4 (2)), which implies that for x ∈ P ∈ Syl7 (F4 (2)), NCF (2) (x) (P ; 7 ) = {1} and then NF4 (2) (P ; 7 ) = {1}. Hence the lemma will 4 follow for K = 3D4 (2) if we show that K embeds in F4 (2). Let (L, σ) be a σ-setup for F4 (2). Let (B, T ) be a σ-invariant Boreltorus pair, and let K be the subgroup of L generated by all long T -root subgroups. Then K ∼ = D4 and K is invariant under N = NL (T ). Moreover, N /N ∩ K ∼ = Σ3 maps onto Out(K). Choose n ∈ N mapping to an element of I3 (Out(K)), and set τ = σn. Then (K, τ ) and (L, τ ) are σ-setups for 3D (2) and F (2), respectively. As K ≤ L, the desired embedding follows. 4 4 The proof is complete.  Lemma 4.32. Let H be a K-group such that O2 (H) = 1. Let p be an odd prime and D ∈ Ep2 (H). Set  V = O{2,p} (CH (y)). y∈D # 

If V = 1, then there exists a DV -invariant component J of Op (E(H)) such that in DV J := AutDV J (J), we have 1 = V ≤ ΔDV J (D). 

Proof. We have F ∗ (H) = O2 (H)E(H) = O2 (H)Op (E(H))E(Op (H)). As V has odd order and is nontrivial, and as Z(F ∗ (H)) ≤ O2 (H), V ∗ acts nontrivially on at least one of the three  above factors of F (H). #  # For any y ∈ D , V ≤ O2 (CH (y)), so O2 (CH (y))E(CH (y)) | y ∈ D ≤ CH (V ). Since CO2 (H) (y) ≤ O2 (CH (y)) and D is noncyclic, [O2 (H), V ] = 1. Next, we show that [E(Op (H)), V ] = 1 let I be any component of E(Op (H)). By [IA , 7.1.2], mp (Aut(I)) ≤ 1, so there is y ∈ D # such that either I ≤ E(CH (y)) or I y = I. In the first case [I, V ] = 1 by the preceding paragraph, so assume that I y = I and set Iy = E(CI y  (y)), a diagonal of I y . Then I ≤ Iy CE(H) (I). Moreover, [Iy , V ] = 1 by the previous   E(H) , which has paragraph, so [I, V ] ≤ [CE(H) (I), V ], and V normalizes Iy p components. But V is by construction a p -subgroup of CH (y), so V normalizes I. Then [I, V ] ≤ I ∩ CE(H) (I) ≤ Z(I) so [I, V ] = 1, as claimed.  Finally, there is a component J of Op (E(H)) with [J, V ] = 1. Obviously V ≤ ΔH (D), and the result follows from [IG , 20.6].  Lemma 4.33. Suppose that X is a K-group with O3 (X) = 1 = Z(X), and b ∈ I3 (X) with m3 (CX (b)) ≤ 3. Suppose that V ≤ X is a 2-group such that 1 = V = [V, d ] ≤ O3 (CX (b)) for some d ∈ I3 (CX (b)). Assume that  every component of E(X) lies in A and is normalized by O3 (X). Then V is contained in a product of components centrally isomorphic to A7 , M12 , M22 , HJ, or L3 (q),  = ±1, q ≡  (mod 6). Proof. By [IG , 20.6], V b, d  acts faithfully on some component L of ∼ E(X) which is not weakly locally balanced for p = 3. As L ∈ A, L =

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D4 (q) for any odd q, so Out(L) has a normal 2-complement by Lemma 20.9. Therefore V induces inner automorphisms on L. By [IA , 7.7.1, 7.7.4d], L has one of the asserted isomorphism types and the projection of V on L lies in O3 (CL (b)). (If L ∼ = L3 (q), q ≡  (mod 3), then q must be odd; otherwise  O 2 (CL (b)) is a product of Lie components by [IA , 4.2.2], so O3 (CL (b)) has odd order, contradiction.) The same holds fo any component L of E(X) on which V acts faithfully, for then CV b,d  (L) ≤ b; but V ≤ O3 (CX (b)) so CV b,d  (L) = 1. Hence V is contained in the product of components on which it acts faithfully. The proof is complete.  Lemma 4.34. Let p ≥ 5 be a prime. Suppose that K ∈ Kp , and K  X with mp (X) = 3, Op (X) = 1, and Z(X) = 1. Suppose that Ep2 ∼ = a, x ≤ X and [Op (CX (a)), x] does not centralize K. Then mp (K) = 1, K ∈ Chev, x induces a nontrivial field automorphism on K, and a induces an inner automorphism on K. Proof. By [IA , 7.7.6ab], and since p > 3, K ∈ Chev(r), r = p, with a mapping into Inndiag(K). Note that if Op (K) = 1 or p divides | Outdiag(K)|, then with [IA , 6.1.4] and as mp (X) = 3, we have p = 5 and K/Z(K) ∼ = L5 (q), q ≡  (mod 5); but then m5 (X) > 3, contradiction. Thus, a maps into Inn(K) and K is p-saturated. If also x maps into Inndiag(K), then it maps into Inn(K), and by [IA , 7.7.6], [Op (CX (a)), x] = 1, contrary to assumption. Therefore as p > 3, x induces a field automorphism on K. Finally if mp (K) > 1, then mp (CK (x)) > 1 (a consequence of [IA , 4.10.3]) and so mp (X) > 3, contrary to assumption. Therefore  mp (CK (x)) = 1 and the proof is complete. 52

Lemma 4.35. Let (K, p) = (L2 (83 ), 3), (L2 (45 ), 5), (2B2 (2 2 ), 5), or (L3 (27 ), 7). Let u ∈ Ip (Aut(K)) be a field automorphism and K0 = CK (u). Then K0 normalizes no nontrivial p -subgroup of K. Proof. This follows from [IA , 6.5.1, 6.5.3, 6.5.4].



5. Generation Lemma 5.1. Let P ∈ Syl3 (X) and J  X with O3 (X) = 1, J ∈ K3 , and o J/Z(J) ∼ = U4 (2) or L± 4 (3). If O3 (X) = 1, then J ≤ ΓP,2 (X). Proof. Unless Z(J) = 1 (whence J/Z(J) ∼ = U4 (3)), P J contains Z3 ×J, so ΓoP,2 (X) contains ΓP ∩J,1 (J), which equals J by [IA , 7.6.1]. If J/Z(J) ∼ = 4 U4 (3), let M ≤ J with M/Z(J) a parabolic subgroup of shape 3 A6 and M ∩ P ∈ Syl3 (M ). By [IA , 6.4.4], O3 (M ) is elementary abelian of rank at least 5, so M ≤ ΓoP,2 (X). Likewise the other maximal parabolic M1 /Z(J) with M1 ∩ P ∈ Syl3 (M1 ) satisfies m3 (O3 (M1 )) ≥ 3, with the help again of  [IA , 6.4.4], so J = M, M1  ≤ ΓoP,2 (X), as claimed. Lemma 5.2. Let K ∈ Co2 and let A ∼ = E52 act on K. Let JA be a component of E(CK (A)) such that JA ∼ = A6 . If JA is a component of E(CK (a)) for all a ∈ A# , then K = JA .

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Proof. If false, then m5 (Aut(K)) > 2. Then as K ∈ Co2 , it follows  that K ∈ Chev(2). But then JA  O2 (CK (a)) for all a ∈ A# . Hence by [IA , 7.3.3], JA  K, which is absurd.  Lemma 5.3. Let K = HS or 2HS, and let V ≤ Aut(K) be a four-group. Assume that K is K-proper. Then ΓV,1 (K) = K. Proof. We use [IA , 5.3m] without comment. Let Γ = ΓV,1 (K) and assume that Γ < K. Let z ∈ I2 (K) map on a 2-central involution of K/Z(K), so that z is 2-central in K. We argue that CK (z) is a maximal subgroup of K. For this we may assume that Z(K) = 1. Suppose that CK (z) < M < K. Let T ∈ Syl2 (K). The odd local structure of K shows that for any odd p-subgroup 1 = W  M , 29 = |T | = |M |2 > |CK (W )|2 | Aut(W )|2 , so O2 (M ) = 1. If F ∗ (M ) = O2 (M ), then z = Ω1 (Z(O2 (M ))) so M = CK (z), contradiction. Therefore E(M ) = 1. If Z(E(M )) = 1, then z ∈ Z(E(M )) and E(CM (z)) = 1, which is impossible. So E(M ) is a product of simple components, permuted transitively by T as Z(T ) = z ≤ E(M ). Let x ∈ I5 (CK (z)). Then Z4 ∼ = CCK (z) (x) and x normalizes all components of := E(M ), so J E(M ) is simple and M ≤ Aut(J). If J ∈ Spor ∪ Alt, then |J|2 = |M |2 or |M |2 /2 = 28 or 29 , and |J| divides |K|, giving a contradiction. If J ∈ Chev(r), r odd, then J ∼ = L2 (r), L2 (r2 ), or L± 3 (5) by r-structure, and then the sectional 2-rank r2 (M ) ≤ 4, whereas r2 (K) ≥ 5. So J ∈ Chev(2). or 21+4 ∗ Z4 . Hence q(J) = 2, but a Levi factor Then O2 (CJ (z)) ∼ = 21+4 − − of CJ (z) is A5 , so J is a twisted group. Again |J|2 = 28 or 29 , which is impossible. Therefore, CK (z) is maximal in K, so we may assume that no element of V acts on K as a 2-central involution. Therefore, for some u ∈ V # , L := E(CK (u)) ∼ = A6 for some u ∈ V # . 2 Then as |K|3 = 3 , L has a trivial or vertical pumpup L∗ in Γ, and L∗ /O2 (L∗ ) ∈ Chev(3); indeed L∗ /O2 (L∗ ) ∼ = A6 , A8 , L5 (2) or Sp4 (4), by [III11 , 1.6]. Since |L∗ | divides |K|, L∗ /O2 (L∗ ) ∼ = A6 or A8 . Moreover, since 15 divides |L|, and by the odd local structure of K, L centralizes every element of NK (L; 2 ). Therefore L∗ = L or L∗ ∼ = A8 . As m3 (K) = 2, L∗  Γ. Let A ∈ Syl3 (L). Also write K = K/Z(K). Then CK (A) = A × u1 , where u1  is the projection of u, Z(K) on K. . Hence if L = L∗  Γ, then Γ ≤ NK ( u1 ). But Γ ≥ CK (v) for any v ∈ V − u, so CK (v) ≤ NK ( u1 ). This forces v, Z(K) to project onto u1  also, so CV (K) = 1 and Γ = K, contradiction. Therefore, L∗ ∼ = A8 . Given the 3-local structure of Aut(K), and by a Frattini argument, we conclude that the image of Γ in Aut(K) lies in NAut(K) (L∗ ) ∼ = Z2 × Σ8 . This forces E(CK (v1 )) ∼ = A6 or A8 for any v1 ∈ V − u. But then V centralizes L, so E(CK (v1 )) = E(CK (v1 u)). This is impossible as u acts nontrivially on E(CK (v1 )), so the proof is complete.  Lemma 5.4. Let K = J4 and let z be a 2-central involution of K. Then CK (z) is a maximal K-subgroup of K. Proof. Suppose that CK (z) < M < K and M is a K-group. If O2 (M ) = 1, then as O2 (CK (z)) = 1, z inverts O2 (M ), and then as

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O2 (CK (z)) ∼ = 21+12 , mr (O2 (M )) ≥ 26 for some odd prime r. But mr (K) ≤ 2 [IA , 5.6.1]. Therefore O2 (M ) = 1. Note that O2 (CK (z)) is irreducible on Q/ z, where Q = O2 (CK (z)). For otherwise, by the action of an 11-element x ∈ CK (z) on Q, CK (z)/Q has a trivial composition factor on Q/ z, which is impossible as Q = [Q, O2,3 (CK (z))]. Therefore if O2 (M ) = 1, then z  M and M = CK (z), contradiction. Hence, O2 (M ) = 1 and so F ∗ (M ) = E(M ) = L1 × · · · × Ln is the direct product of simple groups. Since O2 (CK (z)) is perfect and mr (K) ≤ 2 for all odd primes, O2 (CK (z)) normalizes all the Li ’s. Then as z = Z(O2 (CK (z))), n = 1. As CL1 (z) has a sporadic composition factor, L1 ∈ Spor and then obviously L1 ∼ = K, a final contradiction.  Lemma 5.5. Let K ∈ K2 with K/Z(K) ∼ = M12 , HJ, or Ru. Assume that K is K-proper. Let U be a 2-subgroup of Aut(K) such that m2 (U ) ≥ 2 and Γ := ΓU,1 (K) < K. Then the following hold: 3 3 (a) |U | = 4 and E(Γ) ∼ = A5 ; A5 ; or 2B2 (2 2 ) or 22B2 (2 2 ), respectively; (b) If Z(K) = 1, then every involution in U ∩ Inn(K) has a preimage in K of order 4; and (c) If K/Z(K) ∼ = HJ or Ru, then U ≤ Inn(K); (d) If K ∼ = M12 , u ∈ U # ∩ Inn(K), and v ∈ U − Inn(K), then u ∈ [CK (u), v]; and (e) If K ∼ = 2M12 , and u and v are as in (d), then v inverts the preimage of u in K.

Proof. We use [IA , 5.3bgr] without comment. First suppose that there is u ∈ U # such that L := E(CK (u)) = 1. Thus L ∼ = A5 ; A5 or L2 (7); or 2B (2 23 ) or 22B (2 23 ), in the respective cases. By L  -balance, L ≤ L  (Γ). 2 2 2 2 Since mr (K) ≤ 2 for all odd primes r, [O2 (Γ), L] = 1, so L ≤ E(Γ). By order considerations, and by 3-local structure in the HJ cases, the pumpup of L in Γ is trivial or vertical. But in the M12 and HJ cases, no maximal subgroup of K involves a nontrivial pumpup of L. And in the Ru case, the only nontrivial pumpups of L are Sp4 (8), Ru, and 2Ru, by [IA , 7.1.10], the Borel-Tits theorem, and [IA , 5.3]. As |Sp4 (8)| does not divide |Ru|, we conclude in any case that L   Γ and then L  Γ.  In the HJ case, LCK (L) is maximal in K, with O2 (CAut(K) (L)) ∼ = E22 ,  2 ∼ and (a) follows. In the Ru case, O (NAut(K) (L)) = L × E22 , and again (a) follows. In the M12 case, similarly, |NK/Z(K) (L)|2 = 24 , so the image of U in Aut(K) contains only involutions v such that Lv := E(CK (v)) ∼ = L, and then Lv = L. This again implies (a), which is therefore true in all cases. Parts (b) and (c) follow directly from (a) and [IA , 5.3bgr]. In the K = M12 case, (d) holds because CK (u) ≤ CK (v). To prove (e), let K = 2M12 , K1 = K y where y ∈ I2 (Aut(K)), and K 1 = K1 /Z(K). There is x ∈ K of order 4 such that x and y play the roles of u and v in (d), in K 1 ∼ = Aut(K). Then xy ∈ I2 (K 1 − K), so xy and y are conjugate in K. Hence xy and y

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are both involutions. It follows that y inverts x, which is the assertion of (e). Therefore if the lemma fails, then every u ∈ I2 (U ) is a 2-central involution of Inn(K). Note that even if Z(K) = 1, CK (u) covers CK/Z(K) (u). Thus, we may assume that Z(K) = 1 and E22 ∼ = U ≤ K. If K ∼ = M12 or HJ, then CK (u) is maximal in K and Z(CK (u)) = u, implying Γ = K, contrary to assumption. Hence K ∼ = Ru. Let T ∈ Syl2 (CK (u)). Again Z(T ) ∼ = Z2 so u = Z(T ) and Z(CK (u)) =

u, and CK (u) < Γ < K. For every subgroup W ≤ K of odd prime power order, it is easily checked that |K|2 = 214 > |CK (W )|2 | Aut(W )|2 , so if W  Γ, then W ≤ O2 (CK (z)) = 1. Thus O2 (Γ) = 1. If E(Γ) = 1, then as E(CK (u)) = 1, E(Γ) is the product of simple components and u ∈ E(Γ). As u = Z(T ), F ∗ (Γ) = E(Γ) and T permutes transitively the set of components of E(Γ). The odd local structure of K implies that there are at most two such components, and if there are two, they are isomorphic to A5 . But in that case | Aut(F ∗ (Γ))|2 < 214 = |T |, a contradiction. Hence if E(Γ) = 1, then E(Γ) =: J and Γ ≤ Aut(J). By the odd local structure, J ∈ Alt. By the structure of CK (u), and as J < K, J ∈ Spor ∪ Chev(r) for any odd r. So J ∈ Chev(2). Let x ∈ I5 (CK (u)). Then x ∈ [Γ, Γ] so x ∈ J. Also P := CCK (u) (x) ∈ Syl2 (CK (x)). Thus, by the structure of    CK (x), O2 (CΓ (x)) ∼ = O2 (CK (x)) or Z5 × Q8 . But O2 (CInndiag(J) (x)) is a central product of groups in Lie(2), which is impossible. Therefore, F ∗ (Γ) = O2 (Γ). Let Q = O2 (Γ) and Z = Ω1 (Z(Q)). Then Z > u. On the other hand, CQ (x) embeds in Q8 so CZ (x) = u. As m2 (K) = 6 [IA , 5.6.1], |Z| = 25 . As a result, CG (Z) is a 2-group. Thus |Γ|5 = 5. Likewise for any v ∈ I3 (Γ), CQ (u) embeds in M10 so m2 (CZ (v)) ≤ 2. Thus CZ (v) = u and |Γ|3 = 3. Obviously NG (Z) contains no subgroup of order 13 or 29. Therefore |uΓ | = |Γ : CK (u)| = 7. But then x centralizes two elements of uΓ , contradicting CZ (x) = u. The proof is complete.  Lemma 5.6. Suppose that either p = 3 and K ∼ = U3 (8), Sp4 (8), or 3D (2), or p = 5 and K ∼ 2F (2 52 ). Assume that K is K-proper. Let = 4 4 Ep2 ∼ = L2 (8) (if = A ≤ K, and suppose that for some a0 ∈ A# , E(CK (a)) ∼ 5 2 # p = 3) or B2 (2 2 ) (if p = 5). Then K = E(CK (a)) | a ∈ A . 5 Proof. Suppose first that p = 5 and K ∼ = 2F4 (2 2 ), or p = 3 and 5  O2 (CK (a)) ∼ K∼ = 2B2(2 2 ) = Sp4 (8). Then by [IA , 4.8.7, 4.8.2], E(CK (a)) =  or L2 (8), respectively, for all a ∈ A# . Then K = E(CK (a)) | a ∈ A# by [IA , 7.3.3]. Similarly, if K ∼ = L2 (8) for = U3 (8), then by assumption E(CK (a0 )) ∼ # some a0 ∈ A , whence the full preimage of A in GU3 (8) is abelian. Again the desired generation follows from [IA , 7.3.3].

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∼ 3D4 (2). Then CK (a0 ) ∼ Finally, suppose that K = = Z3 × L2 (8) [IA , 4.7.3A], and a0 is not 3-central in K, whence A contains one 3-central subgroup Z of order 3 and three subgroups ai  ∈ a0 K , i = 0, 1, 2. Let L = E(CK (ai )) | 0 ≤ i ≤ 2 and L = L/O3 (L). By L3 -balance, L = E(L), and as m3 (K) = 2 [IA , 4.10.3], L is quasisimple. Since no 2-local subgroup of K involves Z3 × L2 (8) (Borel-Tits theorem), O3 (L) has odd order. Moreover, mr (K) ≤ 2 for all odd primes, which easily yields that L is quasisimple. Obviously [E(CK (a0 )), a1 ] = 1, so [L, a1 ] = 1 and so L2 (8) ↑ 3 L.

(5A)

Consequently, L ∈ Chev(2) ∪ Chev(3) ∪ Spor. Now from the Borel-Tits theorem we easily see that e(K) = 1. In particular, L/Z(L) ∈ A (Definition 3 1.1). In Spor ∪ Chev(3), (5A) forces L ∼ = Co3 or 2 G2 (3 2 ), contradicting e(L) = 1 [IA , 5.3, 4.5.1]. So L ∈ Chev(2). As mr (P ) ≤ 1 for every parabolic subgroup P of L and every odd prime r, the only possibilities from Definition 1.1 are groups L of twisted rank at most 2. If the twisted rank is 1, then L∼ = U3 (8), contradicting Lagrange’s theorem as L ≤ K. So L has twisted rank 2, and by the structure of parabolic subgroups of K and the BorelTits theorem, the Levi factors L1 , L2 of the two types P1 , P2 of maximal  parabolic subgroups of L satisfy O2 (Li ) ∼ = L2 (2) or L2 (8), i = 1, 2. If Li ∼ = L2 (8) for some i, then the only 2-subgroup of K on which Li acts faithfully is isomorphic to 21+8 , so |Li |2 = 212 = |K|2 ; by Tits’s lemma, L = K, as desired. Otherwise L1 ∼ = A2 (2), A6 , or = L2 ∼ = L2 (2), whence L ∼ U3 (3), contradicting L2 (8) ↑3 L. This contradiction completes the proof.  Lemma 5.7. Let K = F ∗ (X) ∈ Kp for some odd prime p. Suppose that mp (X) ≥ 3 and X has a strongly p-embedded subgroup M . Then p does not divide |X/K|. Proof. By [IA , 7.6.1, 7.6.2, 5.6.1, 4.10.3, 3.3.3], K ∼ = A1 (pa ), 2 A2 (pa ), n or 2 (2 2 ), n > 1, and with a Frattini argument, M ∩ K is a Borel subgroup of K, hence p-closed. By [IA , 2.5.12, 4.9.1], if p divides |X/K|, then M contains a field automorphism f of order p, whence CK (f ) ≤ M . But  CK (f ) is not p-closed, contradiction. Thus, p does not divide |X/K|. 2B

Lemma 5.8. The following quasisimple groups K ∈ Kp are outer wellgenerated for the designated primes p: (a) p = 2: K = L2 (2n ), n ≥ 2; (b) p = 2: K/Z(K) ∼ = L3 (4), Z(K) elementary abelian; K = L3 (2n );  K = L3 (q),  = ±1, q ≡  (mod 3); 1 (c) p = 2: K = L2 (q), q ∈ F M9; K = G2 (8); K = 2F4 (2 2 ) ; K = M11 ; (d) p = 2: K ∼ = HJ or Ru; (e) p = 3: K/Z(K) ∼ = L3 (4); (f) p = 5: K ∼ = F i22 ;

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(g) p > 2: K ∈ Alt ∪ Spor; and 1 (h) p = 2: K = 2F4 (2 2 ) , 3D4 (2), Sp4 (8). Proof. Let E ≤ Aut(K) with E ∼ = Ep2 and E ≤ Inn(K). We must show K = ΓE,1 (K) =: Γ. The cases in (c), (e), (f), (g), and (h) are either trivial (because p does not divide | Out(K)|) or covered by [V8 , 8.4]. Consider (d). It suffices to do the case K ∼ = SL3 (4). We use the classification of subgroups of K/Z(K) [IA , 6.5.3]. If there is e ∈ E # such that E(CG (e)) ∼ = L2 (4), then NK (E; 2) = {1} and so K = Γ by [IA , 7.3.1]. Assuming then that no such e exists, we see from [IA , 4.8.2, 4.8.4] that CK (e) ∼ = Z21 for all e ∈ E −Inn(K). As a Sylow 7-subgroup of K/Z(K) is self-centralizing in K/Z(K), Γ contains more than one subgroup of order 7, and so either equals K or is isomorphic to L3 (2). In the latter case, since Γ is E-invariant, CE (Γ) = 1; but no element of E # centralizes Γ by [IA , 4.8.2, 4.8.4], a contradiction. Thus the lemma holds in case (e). We now assume that p = 2. If some e ∈ E # induces a field or graph-field automorphism on K, then K = Γ by [IA , 7.3.8], so assume that no such e exists. With [IA , 2.5.12], we are done in case (a), and in case (b) we may assume that E = e, f  with e a graph automorphism and f ∈ Inn(K). In odd characteristic [IA , 7.3.3] completes the proof, so assume that K/Z(K) ∼ = n ). Then (using [I , 6.4.2b]) C (f ) contains a Sylow 2-subgroup T of (2 L± A K 3 K. Thus by Tits’s lemma [IA , 2.6.7], if Γ < K, then T < Γ ≤ M for some parabolic subgroup M of K. As CK (e) ∼ = L2 (2n ), E(M/O2 (M )) ∼ = L2 (2n ). This is impossible if K is a unitary group, and in the linear group case as well, since M must be e-invariant. Thus the lemma holds in case (b), completing the proof.  Lemma 5.9. Let E ∼ = Ep2 act on a simple Tp -group K, p odd. Suppose that ΓE,1 (K) < K. Then one of the following holds: (a) p = 3, K ∼ = M12 , and K is locally balanced with respect to E; (b) (p, K) = (3, L3 (4)) or (5, F i22 ), and E induces inner automorphisms on K. Proof. Since K ∈ Cp , K ∈ Chev(p). If K ∈ Chev − Chev(p), then by [IA , 7.3.4], either (b) holds with K ∼ = L3 (4), or K ∼ = Sp8 (2) with p = 3; but Sp8 (2) is a G3 -group, contrary to assumption. Next assume that K ∈ Spor ∪ Alt. Then | Out(K)| = 2 or 4, so as E acts faithfully on K, mp (K) ≥ 2. By definition of Tp [I2 , 13.1], either (b) holds with p = 5, or p = 3 with K ∼ = A7 , # ∼ HJ, M12 , or M22 . By [IA , 7.5.1, 7.5.5], K = M12 and every e ∈ E acts on K as a 3-central element. Thus O3 (CK (e)) = O3 (CAut(K) (e)) = 1 by [IA , 5.3b], so K is locally balanced with respect to E, and (a) holds. The proof is complete.  Lemma 5.10. Suppose K ∈ T32 . Then there is B1 ∈ E3∗ (K) such that  K = L3 (CK (b)) | b ∈ B1# . Moreover, for any such B1 and P ∈ Syl3 (K)

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such that B1 ≤ P , B1 is the weak closure in P of each of its non-3-central subgroups of order 3. In particular, B1  P .  = SL (q). Proof. We have K = L3 (q),  = ±1, q ≡  (mod 9). Let K 3 Since q ≡  (mod 9), every x ∈ I3 (K) is the image of a diagonalizable x  ∈ K (orthogonally diagonalizable if  = −1). It follows easily that there are just two K-conjugacy classes of subgroups x of order 3: C1 , for which the eigenvalues of x  are the three cube roots of unity, and C2 , for which two of the eigenvalues of x  are equal ninth roots of unity.  Let F be an (orthogonal) frame in the natural K-module V . Then the stabilizer of F in K contains P = P0 w ∈ Syl3 (K), where P0 is diagonal with respect to F, w3 = 1, and w cycles the three subspaces constituting F, as does every x ∈ P − P0 . Thus x ∈ C1 for all x ∈ P − P0 . On the other hand, P0 is abelian of rank 2, and P1 := Ω1 (P0 ) contains three subgroups with the fourth  subgroup Z(P ) ∼ in C2 , cycled by w, together = Z3 , with  # Z(P ) ∈ C1 . Then K = L3 (CK (x)) | x ∈ P1 . Moreover, the only E32 subgroup of P containing an element of C2 is P1  P . As L3 (CK (x)) = 1  for all x ∈ C1 , the lemma follows. Lemma 5.11. Suppose that K ∈ (T33 − {M12 }) ∪ {HJ}. Then for any  B1 ∈ E3∗ (K), K = CK (b) | b ∈ B1# . Proof. If K ∼ = L3 (q), q ≡  (mod 3),  = ±1, then q = 4 by definition 3 of T3 , so [IA , 7.3.4] yields the desired conclusion. Otherwise K ∼ = A7 , M22 ,  or HJ, and we appeal to [IA , 7.5.2b, 7.5.5] to complete the proof. ∼ Epm acts on K, Lemma 5.12. Suppose that K is a TGp -group and B =  m ≥ 2. If ΓB,1 (K) < K, then m = 2 and one of the following holds: (a) p = 3, K ∼ = An , 10 ≤ n ≤ 12, and E acts on K like an E32 -subgroup of K with a regular orbit on n letters; (b) K ∼ = A4p and all orbits of E on 4p letters have length p; (c) p = 3, K ∼ = Sp8 (2), and every element of E # acts on K like an element of K with 2-dimensional fixed-point space on the natural K-module; or (d) p = 3, K ∼ = 3A6 for every = 3ON , and E(CK (e)) = E(CK (e )) ∼  # e, e ∈ E . In particular if mp (B) ≥ 3, then K = ΓB,1 (K). Proof. If K ∈ Chev(p), then K ∈ Cp so K ∈ Gp . If K ∈ Chev−Chev(p), then as K ∈ Gp , we see from [IA , 7.3.4] and [V11 , 3.1] that p = 3 and (c) holds. If K ∈ Spor, then unless K ∼ = 3ON , K = ΓB,1 (K) by [IA , 7.5.5] and the list of generic sporadic groups [I2 , p. 103]. If K ∼ = 3ON , then p = 3, and the assertion follows by [IA , 5.3s]. If K ∈ Alt − Chev, then as K ∈ G6p , either p = 3 and K ∼ = An , 2p < n ≤ 4p + 1, = An , 10 ≤ n ≤ 12, or p > 3 and K ∼ n = 3p (as K ∈ Cp ). We may assume that B ≤ K. If B has a regular orbit on n letters, then p = 3 and (a) holds. Otherwise, all nontrivial orbits

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of E on n letters have length p. If n = 4p or if B has a fixed point on n letters, then K = ΓB,1 (K) by [IA , 7.5.1], contradiction. Hence (b) holds, completing the proof.  Lemma Z(X) = 1, CK (E) has q ≡ 0 (mod

5.13. Let K ∈ T3 C∗3 and K  X with m3 (K) ≥ 2, m3 (X) ≤ 4, and O3 (X) = 1. Suppose that E ≤ NX (K), E ∼ = E32 , and a component I such that I/O3 (I) ∼ = L2 (q) for some q > 2, 3). Then for some e ∈ E # , I is not a component of CK (e).

Proof. We may assume that CE (K) = 1. Suppose false. Then I    CK (e). If K ∈ Chev(r ), then I  ΓrE,1 (K) = K by [IA , 7.3.1], contradiction. If K ∈ Alt, then K ∼ = An , with n ≥ 5 + 6 = 11, and then n = 11 as Z(X) = 1 and m3 (X) ≤ 4. But then E(CK (e)) ∼ = A8 for some e ∈ E # , as required. Suppose finally that K ∈ Spor. As CE (K) = 1 and Z(X) = 1, m3 (K/Z(K)) ≥ 3. As m3 (X) ≤ 4, we see from [IA , 5.6.1] that K/Z(K) ∼ = J3  or O N . But then q = 9, contrary to assumption. The proof is complete.  Lemma 5.14. Suppose that K, L ∈ C2 with K ≤ L and K a component of CL (z) for some z ∈ I2 (Aut(L)). Let S ∈ Syl2 (K). If (K, L) ∼ = (A5 , M12 ), 3 2 (A5 , HJ), (G2 (4), Co1 ), (L3 (4), Suz), or ( B2 (2 2 ), Ru), then ΓS,1 (L) does not normalize K. Proof. Let t be a 2-central involution of CL (z) with t ∈ K, and let T ∈ Syl2 (CL (z)) with t ∈ Z(T ). If L ∼ = M12 or HJ, then by [IA , 5.3bg], ∼ NL (K) is maximal in K, with NL (K) = Z2 × Σ5 or A4 × A5 , respectively. Then CNL (K) (t) ∼ = Z2 × D8 or A4 × E22 , so CNL (K) (t) < CL (t) and hence CL (t) ≤ NL (K), as claimed. In the three other cases, V := CL ( z K) is a four group and V # ⊆ z L , so V ∈ Syl2 (CL (K)). Hence by a Frattini argument, CL (z) contains a Sylow 2-subgroup of NL (K). Hence if the lemma fails, then |CL (t)|2 ≤ |ΓS,1 (L)|2 ≤ |NL (K)|2 = |CL (z)|2 . In every case we see from [IA , 5.3lor] that this implies that t ∈ z L , say t = z g , g ∈ L. Moreover T ∈ Syl2 (CL (t)), so with Sylow’s theorem, we may assume g ∈ NL (T ). Thus t and z are Aut(T )-conjugate. However, this is impossible. To see this, let T0 = T ∩ K, so that z ∈ T0 Now if L ∼ = Co1 , then t ∈ [T, T, T ] (by the Chevalley commutator formula for G2 (4)) but z ∈ [T, T, T ] since T /T0 ∼ = Suz, then t lies in a Q8 -subgroup of T , indeed of T0 , = D8 . If L ∼ while again T /T0 ∼ = Ru, then z ∈ [T, T ] but = D8 so t does not. Finally if L ∼ t ∈ [T, T ]. The proof is complete.  Lemma 5.15. Let p be an odd prime. Let J ∼ = An1 and let I be a root An0 -subgroup of J for some n0 ≤ n1 . Suppose that for both i = 0 and 1, 3p + 1 ≤ ni ≤ 4p + 1. Let P ∈ Sylp (I) and D = J(P ). Then J = ΓD,2 (J). Proof. D is generated by disjoint p-cycles, of which we choose three, x1 , x2 , x3 . For each i = 1, 2, 3, let Di = xj | j = i ∼ = Ep2 and Ωi = the fixed point set of Di on a set Ω with |Ω| = n1 on which J acts naturally. As n0 ≥ 3p + 1, Ω1 ∩ Ω2 ∩ Ω3 = ∅, and clearly Ω1 ∪ Ω2 ∪ Ω3 = Ω. Hence  J = AΩi | 1 ≤ 1 ≤ 3 ≤ CJ (Di ) | 1 ≤ i ≤ 3 ≤ ΓD,2 (J), as required.

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∼ Epm Lemma 5.16. Let K = An , n ≥ 5, and let E ≤ Aut(K) with E = for some odd prime p and m ≥ 2. Suppose L ≤ K and L is a component of CK (e) for all e ∈ E # . Then L ∼ = Aδ for some δ ≥ 5, n = kpm + δ for some k ≥ 1, and E acts on K like a subgroup of K with δ fixed points and k regular orbits. Proof. Let Ω be a set of cardinality n on which K acts naturally. Let E1 ≤ E with E1 ∼ = Ep2 . By [III11 , 1.16], the fixed point set of E1 equals the fixed point set of e for every e ∈ E1# . Then for any e ∈ E # , choose e ∈ E1 − e  and note that the fixed point set of e is the same as those of 

e, e  and e. The result follows. Lemma 5.17. Let K = G2 (8) and B ∈ E32 (K). Then   K = L3 (CK (b)) | b ∈ B # . 

Proof. For any b ∈ B # ⊆ I3 (K), L3 (CK (b)) = O2 (CK (b)) (see [IA ,  4.7.3A]). Therefore the result holds by [IA , 7.3.3]. Lemma 5.18. Let p ≥ 7 be a prime and let K ∈ Chev(r) ∩ Kp be a K-group for some r = p. Suppose that K ∈ A. Let D ∼ = Ep2 act on K, and let K1 , . . . , Kn be all the p-components of E(CL (d)) as d varies over D # . Then K = K1 , . . . , Kn . Proof. Without loss, we assume that X = KD with F ∗ (X) = K simple. Let D0 = D ∩ K. Suppose first that some d ∈ D # induces a nontrivial field automorphism on K. As p ≥ 7 and |K| is divisible by p, |D : D0 | = p and for d ∈ D − D0 , CK (d) either is simple or has a nonabelian simple subgroup Kd of index 2. The latter can only hap1 pen if CK (d) ∼ = G2 (2) (p = 7) or 2F4 (2 2 ) (p = 13); and in those cases, mp (K) = 1 and so NK (D) covers CK (d)/K  d . Using [IA , 7.3.8], we deduce that K = E(CK (d)), NK (D) | d ∈ D # = E(CK (d)) | d ∈ D # , the latter group being normalized by NK (D). The result follows in this case. Since p ≥ 7 and K ∈ A, every noninner automorphism of K of order p is a field automorphism, so we may assume that D ≤ K. By [IA , 7.3.3], K is generated by all Lie components of all CG (d), d ∈ D # . Using [IA , 4.9.7b], we may restrict to just those Lie components L of order divisible by p. Any such Lie component is quasisimple, either because p ≥ 7 (ruling out Sp4 (2) and solvable Lie components) or because its Dynkin diagram embeds properly in the extended Dynkin diagram of K by [IA , 4.2.2] (ruling out G2 (2) and 2F (2 21 )). This implies the lemma.  4 Lemma 5.19. Suppose L ↑5 K or L ∼ = K, with L ∼ = A5 , A6 , A8 , U4 (2), 1 2  E act on K. Then unless possibly or F4 (2 2 ) , and K ∈ K5 ∩ A. Let D ∼ 2 = 5 1 5 2  ∼ K = A10 , HS, Ru, L2 (5 ), or F4 (2 2 ) , we have K = Γ (K). D,1

ΓD,1 (K).

Since 3 divides |L|, K is not a Suzuki ∼ 2F4 (2 12 ) by [IA , 7.3.4]. group. Hence if K ∈ Chev(r), r = 5, then K = Proof. Suppose K =

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Suppose that K ∈ Chev(5). Then L ∈ Chev(5) so L ∼ = A5 and L = ΓD,1 (L). If L ↑5 K, then by the Borel-Tits theorem and [IA , 2.5.12], L ↑5 K by a field automorphism, so K ∼ = L2 (55 ). Next, assume that K ∈ Spor. Using that L ↑5 K ∈ A, we conclude from M c. If m5 (K) = 1, then CD (K) = 1, [IA , 5.3] that m5 (K) ≤ 2 and K ∼ = while if m5 (K) ≤ 2, we can appeal to [IA , 7.5.5] to complete the proof.  Lemma 5.20. Let K = J4 , HS, or He. Let D ∈ E32 (Aut(K)) and  K0 = L3 (CK (d)) | d ∈ D # . Then K = K0 . Proof. As | Aut(K)| ≤ 2, we may regard D ≤ K. Let N = NK (D), Ld = L3 (CK (d)) and Kd = Ld (N ∩ CK (d)) for each d ∈ D # . If K = HS, or K = He with Ld ∼ = L3 (2), then D ∈ Syl3 (CK (d)) and so CK (d) = Kd by a Frattini argument. In the other two cases ((K, Ld ) = (J4 , 6M22 ) or (He, 3A7 )), D ≤ P ∈ Syl3 (K) ∩ Syl3 (CK (d)) with P ∼ = 31+2 and Ld controlling the CK (d)-fusion of E32 -subgroups of P . Hence in these cases as well, CK (d) =   Ld(N ∩ CK (d)) = Kd . Now N normalizes K0 , and N K0 ≥ Kd | d ∈ D # = CK (d) | d ∈ D # = K by [IA , 7.5.5]. Therefore K0  K,  so K0 = K. The proof is complete. Lemma 5.21. Let L ↑3 K ∈ K3 and K   X with L ∼ = 3M22 , A5 , 3A7 , or L3 (2), and K/Z(K) ∈ A. Let D ∈ E34 (X). Then K ≤ ΓD,2 (X). Proof. The desired conclusion holds if m3 (Aut(K/Z(K))) ≤ 2 (see [IA , 7.2.2d]), so we may assume that m3 (Aut(K/Z(K))) > 2. As ↑3 (3M22 ) = {J4 } and ↑3 (3A7 ) = {He} [IA , 5.3], these two candidates for L are therefore ruled out [IA , 5.6.1]. Indeed, when L ∼ = A5 or L3 (2) and K ∈ Spor, we see from [IA , 5.3] that m3 (Aut(K/Z(K))) = 2, contradiction. If K ∈ Chev(r), then L ∈ Chev(r) by [IA , 4.9.6], so r = 3, and then by [IA , 7.3.6], we are done unless possibly r = 2. But since a Sylow 3-subgroup of Aut(L) normalizes a nontrivial 2-subgroup of L, it follows from [IA , 7.3.1] that even K ≤ ΓD,3 (X), which is stronger than the desired generation. Hence, we may assume that K ∈ Alt. As K ∈ A, we may assume that K = A11 and, of course, that m3 (CD (K)) < 2. Then D acts on K as a subgroup of K generated by three mutually disjoint 3-cycles, and the  desired conclusion holds by [IA , 7.5.1]. 1

Lemma 5.22. Let K = 2F4 (2 2 ) , 3D4 (2) or Sp4 (8). Let Q ≤ Aut(K) with |Q| = 211 . Then K = ΓQ,1 (K). Proof. Let Γ = ΓQ,1 (K), and let Q ≤ P ∈ Syl2 (Aut(K)). If K = Sp4 (8), then |P : Q| = 4, so Q ∩ U = 1 = Q ∩ V , where U and V are, respectively, the high long and high short root subgroups of P . Then Γ ≥ CK (Q ∩ U ), CK (Q ∩ V ) ≥ P ∩ K, LU , LV , where LU is the derived group of a Levi factor of the maximal parabolic subgroup NK (U ), and similarly for LV . By Tits’s lemma [IA , 2.6.7], if H is a Cartan subgroup in NK (P ∩ K), then ΓH = K. As H is abelian and H ∩ Γ = 1, while K is simple, Γ = K by [V9 , 13.2].

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1 So assume that K ∼ = 2F4 (2 2 ) or 3D4 (2). Let M be the maximal parabolic 1 subgroup of K0 := O3 (Aut(K)) ∼ = 2F4 (2 2) or 3D4(2) containing P and such that M/O2 (M ) ∼ = Σ3 . Then U := Z(P )M ∼ = E22 and Z(P ) ≤ Φ(O2 (M )), for otherwise Z(P ) ∼ = Z2 would be a direct factor of O2 (M ) and so O2 (M ) would be elementary abelian, which is absurd. Therefore U ≤ Φ(O2 (M )) ≤ Φ(P ) ≤ Q, so Γ ≥ Γ0 := ΓU,1 (K). Finally, Γ0 is invariant under M and contains CK (Z(P )). But M ∩ K and CK (Z(P )) are the two maximal parabolic subgroups of K containing P ∩ K, so they generate K,  and hence Γ0  K. Hence, Γ ≥ Γ0 = K, completing the proof.

6. The Balance-Generation Dichotomy Lemma 6.1. Let K = A7 , M22 , HJ, or M12 , and let D ∈ E32 (Aut(K)). # If K ∼ =  M12 , assume also  that O3 (CK (d)) = 1 for some d ∈ D . Let # K0 = O2 (CK (d)) | d ∈ D . Then K = K0 . Proof. As | Out(K)| = 2, we may regard D as a subgroup of K. Let N = NK (D), so that K0  K1 := K0 N . Let Yd = O2 (CY (d)) for d ∈ D # , and set X = { d ≤ D | Yd = 1}. If K = A7 , then X consists of the two subgroups of D generated by 3-cycles, and these are interchanged by N . Likewise if K = M22 , HJ, or M12 , then from the information in [IA , 5.3cgb], |X| = 4, 3, or 3, respectively, and N is transitive on X. In all cases, for each

d ∈ X, DYd ∼ = Z3 × A4 . Suppose that Yd ∩ O3 (K1 ) = 1 for some d ∈ X. By the action of N , K0 is then a 3 -group. Let S ∈ Syl2 (K0 ). It follows that S = K0 has order 24 , 28 , 26 , or 26 , respectively. Comparing with |K| we conclude that K = HJ or M12 , in which cases N contains some P ∈ Syl3 (K) containing D. Now P ∼ = 31+2 and Z(P ) acts nontrivially on each O2 (CK (d)), d ∈ X. We may assume that S is P -invariant and hence S has sectional 2-rank at least 6. But |K|2 ≤ 27 , m2 (K) ≤ 4, and S is not extraspecial of order 27 , contradiction. Now pass to K 1 = K1 /O3 (K1 ). As m3 (K) = 2 in each case, all com ponents of K 1 are normalized by O3 (K1 ). Then by [IG , 20.6], for each

d ∈ X, DY d acts faithfully on a component J d of K 1 which fails to be weakly locally balanced for p = 3. By [IA , 7.7.1, 7.7.4d], and as m3 (J d ) ≤ 2, we have J d ∼ = A7 , M12 , M22 , HJ, or L3 (q), q ≡  (mod 6),  = ±1. (Note in the last case that if q were even, then Y d ≤ O2 (CJ d (d)); but O2 (CJ d (d)) = 1  since O 2 (CJ d (d)) is the commuting product of Lie components [IA , 4.2.2].) But K does not involve L3 (q), by order considerations. Among the remaining four possibilities for J d , which are also our possibilities for K, the only possible proper involvement, by orders and 3-fusion, is that of J d ∼ = A7 in K = M22 . But if that occurred, D ≤ Jd and so |X| = 2 (its value for A7 ), whereas |X| = 4, contradiction. Therefore J d ∼ = K, contradicting K1 < K and completing the proof. 

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Lemma 6.2. Let p be an odd prime and K = F ∗ (H) ∈ Chev with K simple. Let E ∈ Ep2 (H) and a ∈ Ip (H). Then one of the following holds: (a) Op (CH (a)) = 1; (b) K = ΓE,1 (K); (c) p = 3 and K ∼ = L2 (27), with a a field automorphism of order 3 and E ≤ K. Proof. Suppose that (a) and (b) fail. Then [IA , 7.7.1] and [IA , 7.3.4] apply, respectively. If K ∈ Chev(r) − Chev(p) for some r = p, then by [IA , 7.3.4] and 1 Proposition 3.1, K ∼ = = Sp6 (2), Sp8 (2), or 2F4 (2 2 ) , all with p = 3, or K ∼ 5 1 2B (2 2 ) or 2F (2 2 ) , both with p = 5. But these groups are all locally 2 4 balanced for p by [IA , 7.7.8], so (a) holds, contrary to assumption. Finally, if K ∈ Chev(p), then (c) holds by [IA , 7.7.1c]. The proof is complete.  Lemma 6.3. Let K ∈ Chev(r), r = 3, be a K-group, and suppose that K is not locally balanced for the prime 3. Suppose that D ∼ = E32 acts on K. Then one of the following holds:   (a) K = E(CK (d)) | d ∈ D # ; or (b) K/Z(K) ∼ = L3 (q),  = ±1, q ≡  (mod 3). The image of D in Aut(K) is the image in Inndiag(K) of a nonabelian 3-subgroup of GL3 (q). Proof. If r = 2, then this is immediate from [IA , 7.3.3]. So assume r = 2. If some d ∈ induces a nontrivial field or graph-field automorphism on K, or acts trivially on K, then (a) holds by [IA , 7.3.8], unless possibly K ∼ = 3 2 L2 (8), Sp4 (8), G2 (8), or F4 (2 2 ). But L2 (8) ∈ Chev(3) is locally balanced for the prime 3, and using [IA , 4.8.2, 4.7.3A] to compute centralizers of elements of D, we see that the other three groups are locally balanced as well, contrary to assumption. Next, if q := q(K) ≤ 2, then the unbalancing property forces K ∼ = L3 (2) (−1)n+1 (2), n ≥ 3, by [IA , 7.7.8]. In the first case, CD (K) = 1 so (a) or Dn holds trivially. In the other case, if n = 3, (a) follows since K ∼ = A8 is generated by two A5 -subgroups centralizing disjoint 3-cycles. In general, (a) then follows by an inductive argument using [IA , 7.3.2d]. We may therefore assume that q > 2, D injects into Aut0 (G), and K ∼ = L2 (8). Hence every Lie component L of CK (d), d ∈ D # , satisfies q(L) = 2m > 2 for some m, and so L = E(L). Then (a) holds by [IA , 7.3.3]. The proof is complete.  D#

Lemma 6.4. Let K ∈ Chev(r) − Alt for some r = 5, with K/Z(K) ∈ A. Suppose that K is not locally balanced for p = 5. Suppose that K   X, where X is a group such that O5 (X) = 1, Z(X) = 1, and m5 (X) = 3. Let

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D∼ there exist = E52 with D ≤ X. Then   subgroups Id ≤ CK (d), one for each # # and any Id either lies unambiguously d ∈ D , such that K = Id | d ∈ D 1 in Lie(r) ∩ Chev(r), or is isomorphic to A5 , A6 , A8 , U4 (2), 2F4 (2 2 ) , or 1. Proof. Since m5 (X) = 3, D clearly normalizes K, and we assume without loss that X = KDZ(X). If CD (K) = 1, we take Id = K for d ∈ CD (K)# and Id = 1 otherwise; the conclusion then follows from the list of ambiguous groups [IA , 2.2.10]. So assume that CD (K) = 1. We wish to apply [IA , 7.3.3]. Note that since m5 (X) = 3, we cannot have K/Z(K) ∼ = L5k (q), q ≡  (mod 5). In particular, with [IA , 6.1.4], K is simple and 5-saturated. Also, K ∈ C5 since K is assumed not to 5 1 be locally balanced for p = 5; in particular, K ∼ 2B2 (2 2 ), 2F4 (2 2 ) , or =   2F (2 25 ). Now [I , 7.3.3] implies that K = K | d ∈ D # , where for each 4 A d  d ∈ D # , Kd = Or (CK (d)), a central product of groups in Lie(r). Then by of all Lie components of CK (d) of [IA , 4.9.7b], if we let Jd be the product  order divisible by 5, we have K = Jd | d ∈ D # . If Jd has a Lie component 1 5 isomorphic to 2F4 (2 2 ) for some d, then we get the contradiction K ∼ = 2F4 (2 2 ), as follows from [IA , 4.2.2, 4.9.1, 4.9.2]. Hence every Jd is quasisimple except for those Jd isomorphic to Σ6 , when r = 2. Note also that since m5 (X) = 3, m5 (Jd ) = 1 for all d ∈ D # , for otherwise d ∈ CD (K) = 1, contradiction. Finally, set Id = [Jd , Jd ] for each d ∈ D # . Then Id = Jd or Id ∼ = A6 . Again since all ambiguous groups of order divisible by 5 are allowed as  possibilities for Id , it remains only to show that K = Id | d ∈ D # . We may therefore assume that r = 2 and Jd ∼ = Σ6 for some d ∈ D # . Hence by [IA , 4.2.2], the untwisted Dynkin diagram of K has a double bond, and either K ∼ = Sp8 (2) or F4 (2) (recall that = Sp4 (25 ) or q(K) ≤ 2, in which case K ∼ m5 (X) = 3). But F4 (2) is locally balanced for p = 5 (Lemma 4.29). It therefore suffices to show in H = Aut(K), K ∼ = Sp4 (25 ) or Sp8 (2),that if E ≤ H ∼ with E = E52 , then K = K0 where K0 = E(CK (e)) | e ∈ E # . But for each  e ∈ E # , a Frattini argument shows that O2 (CK (e)) ≤ E(CK (e))NK (E). Hence using [IA , 7.3.3],      K = O 2 (CK (e)) | e ∈ E # ≤ E(CK (e))NK (E) | e ∈ E # = K0 , NK (E) . As NK (E) normalizes K0 , K = K0 by simplicity of K. The proof is complete.  Lemma 6.5. Suppose K ∈ K5 ∩A and K ∈ Chev(r)−Alt for some r = 5. Assume that K  X, O5 (X) = 1 = Z(X), and m5 (X) = 3. Suppose that d ∈ I5 (X) − CX (K) and I   CK (d) with I/O5 (I) ∼ = A5 . Let R ∈ Syl5 (I). Then one of the following holds: (a) L5 (CK (R)) = 1; (b) K is locally balanced for p = 5; (c) K ∼ = L2 (45 ).

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Proof. By (c), we may assume that d induces an inner-diagonal automorphism on K, so K ∈ Chev(2) with level q(K) ≤ 2 or q(K) = 4, by [IA , 4.2.2]. Clearly then m5 (Inndiag(K)) ≥ 2. The only such element  5 (U5 (4)) = 3 = m5 (X), of A which is not 5-saturated is U5 (4); but as m K/Z(K) ∼ U5 (4). Hence K is simple and 5-saturated. For the same rea= son, K ∼ U4 (4), Sp6 (4), or Ω− = 8 (4). As for the remaining groups satisfying our conditions, including that m5 (Inndiag(K)) ≥ 2, conclusion (a) holds if K ∼ = Sp4 (4), G2 (4), 3D4 (4), L4 (4), Sp8 (2), or D4 (2), while conclusion (b) 1  holds if K ∼ = U3 (4), 2F4 (2 2 ) , or F4 (2). Lemma 6.6. Let K = U3 (q), q = 2n ≥ 8, and suppose K = E(X) where Op (X) = 1 and CX (K) is a nontrivial cyclic p-group, and p is an odd prime divisor of q + 1. Let U ≤ X with U ∼ = Ep2 and U mapping into Inndiag(K). Then one of the following holds: (a) For some u ∈ U # , CK (u) has a component isomorphic to L2 (q), q > 8;   (b) p = 3 and O3 (CX (u)) | u ∈ U # is not a 3 -group; or   (c) p = 3 and O3 (CX (u)) | u ∈ U # = 1. Proof. Suppose that U is the image of an abelian p-subgroup U ∗ of GU3 (q). Then U ∗ is diagonalizable and (a) holds, provided that q > 8. If q = 8, however, then p = 3. As q + 1 is a power of 3, it follows in this case that O3 (CInndiag(K) (u)) = 1 for all u ∈ U # . As Outdiag(K) ∼ = Z3 and Out(K)/ Outdiag(K) is generated by the image of a graph automorphism γ of order 2 [IA , 2.5.12], and as m3 (CK (γ)) = m3 (L2 (8)) = 1, O3 (CX (u)) = 1 for all u ∈ U # . (Note that F ∗ (X) ∼ = U3 (8)×Z3n for some n, and O3 (CX (u)) centralizes O3 (CX (u)).) Thus (c) holds in this case. Now we may assume that U is the image of a nonabelian p-subgroup of GU3 (q). It follows that p = 3. If q > 8, then (b) holds by Lemma 11.3.5. So assume that q = 8. By the previous paragraph we may assume that some u ∈ U # is the image of a non-diagonalizable element of GU3 (q). By [IA , 4.8.4], u maps to a nontrivial element of Outdiag(K), and CK (u) = QH is a Frobenius group with Q ∼ = Z3 . Thus U = u, h. = Z19 and H = h ∼ Hence there is v ∈ I3 (NK (U )) such that uv = uh. Then Qv = O3 (CK (uh)) and Q, Qv  = K, by an application of [IA , 6.5.3]. Thus (b) holds and the proof is complete.  Lemma 6.7. Suppose that K ∼ = L3 (q), q = 2n ≥ 4, p is a prime divisor ∗ of q − 1, and K = F (X). Let U ≤ X with U ∼ = Ep2 and U mapping into Inndiag(K). Assume also that E(CK (u)) = 1 for all u ∈ U # . Let  K0 = O3 (CX (u)) | u ∈ U # . Then p = 3 and one of the following holds: (a) q = 4 and some involution of X induces a unitary-type automorphism on K; (b) K0 is not a 3 -group; or (c) K0 = 1.

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Proof. Since K = F ∗ (X), we shall regard K ≤ X ≤ Aut(K). If U is the image of an abelian – hence diagonalizable – subgroup of GL3 (q), then E(CK (u)) ∼ = L2 (q) for some u ∈ U # , contrary to assumption. Hence U is the image of a nonabelian p-subgroup of GL3 (q), whence p = 3.  Then U embeds in Op (Inndiag(K)) = Inn(K) and its preimage in SL3 (q) is abelian, so U is diagonalizable and then is a maximal diagonalizable subgroup of exponent p. As q ≥ 4, E(CK (u)) ∼ = L2 (q) for some u ∈ U # , contrary to hypothesis. Hence p = 3, and q ≡ 1 (mod 3). If q > 4, then (b) holds by Lemma 11.3.5. So assume that q = 4. Let u ∈ U # . If u ∈ K then CAut(K) (u) = P × ϕ for some P ∈ Syl3 (K) and unitary-type automorphism ϕ of K. Hence if U ≤ K, then (a) or (c) holds. Otherwise any u ∈ U − K must satisfy O3 (CK (u)) ∼ = Z7 [IA , 4.8.4]. Moreover, K0 is NX (U )-invariant, hence invariant under a Sylow 3-subgroup of X. By Lemma 10.6.3, K0 NX (U ) ≥ K, whence K0 ≥ K and (b) holds. The proof is complete.  Lemma 6.8. Suppose K is a component of the group X and K ∼ = A7 , 3 (X). Then K ≤ (q), q ≡ 0 (mod 3). Let D ∈ E M12 , M22 , HJ, or L± 4 3 ΓD,2 (X). Proof. If m3 (Aut(K)) ≤ 2, the result is obvious (see [IA , 7.2.2d]). Thus, we may assume that K ∼ = L± 3 (q) [IA , 5.3]. It also suffices to prove the  result assuming that D normalizes K, by [IA , 7.2.3a]. But O3 (Aut0 (K)) ∼ = (q) has 3-rank at most 2, and Aut(K)/ Aut (K) is cyclic [I , 2.5.12], P GL± 0 A 3 so m3 (Aut(K)) ≤ 3 and then CD (K) = 1. The assertion then follows from  [IA , 7.3.7], applied to D/ d for any d ∈ CD (K). Lemma 6.9. Let K = F ∗ (X) with K ∈ TG3 a K-group. Let E ∈ E32 (X). Then one of the following holds:   (a) K = L3 (CK (e)) | e ∈ E # ; (b) K is locally balanced for the prime 3; (c) K ∼ = A11 and E has a regular orbit; or (d) K ∼ = A10 and E has no regular orbit.   Proof. Let H = L3 (CK (e)) | e ∈ E # E = L3 (H)E by L3 -balance. Suppose first that K ∈ Spor. If (b) fails, then K ∼ = J4 , HS, or He, by [IA , 7.7.1b]. In all three cases, from [IA , 5.3], | NK (L3 (CK (e))E, 2) | ≤ 2 for each e ∈ E # , so |CO3 (H) (e)| ≤ 2 and therefore |O3 (H)| ≤ 2. Then L3 -balance implies that H = E(H)E. In all three cases, m3 (K) = 2 and L3 (CK (e)) is (nontrivial and) quasisimple for all e ∈ E # . Hence E(H) is a single component, which is a nontrivial pumpup of 6M22 , A5 , or 3A7 , according to the isomorphism type of K. If K ∼ HS, then K is the only = such pumpup up to isomorphism, so H = K. If K ∼ = HS, then HNG (E) is not strongly 3-embedded in K by [IA , 7.6.1], so it equals K and (a) holds. Suppose next that K ∈ Chev(r) for some r. As K ∈ G3 , K ∈ Chev(3),  and by [IA , 7.3.3], either K = ΓrE,1 (K) or K ∼ = Sp8 (2). But Sp8 (2) is locally

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balanced for p = 3 by [IA , 7.7.8b], so we may assume that K = ΓrE,1 (K). Hence (a) holds unless possibly for some e ∈ E # , some Lie component of CK (e) is a 3 -group or not quasisimple. Note that as K ∈ G3 , K ∼ = 3 L2 (8), 2B2 (2 2 ), 2 A2 (8), Sp4 (8), or G2 (8). Hence if some e ∈ E # induces a nontrivial field or graph-field automorphism on K, then (a) holds unless 3 1 possibly K ∼ = 2F4 (2 2 ). But in that case H contains 2F4 (2 2 ) as well as CK (E ∩ K) ∼ = SU3 (8). Hence H has one class of subgroups of order 3. Since K has no strongly 3-embedded subgroup by [IA , 7.6.1], K = H and (a) holds. We may therefore assume that E maps into Aut0 (K). If q(K) > 2,  then L3 (CK (e)) = Or (CK (e)) for all e ∈ E # , so (a) holds. If q(K) ≤ 2, then by Lemma 4.14, K ∼ = D3 (2), D4− (2), or D5 (2). Thus we may assume that E ≤ K. Let V be the natural K-module; then dim CV (E) ≥ 2 and every irreducible constituent of E on V has dimension at most 2. For any e ∈ E # such that CV (e) > CV (E), CK ([V, e]) ∼ = Ω(CV (e)) is quasisimple and hence lies in H. Then it follows from [IA , 7.3.2] that H = K, so again (a) holds. Finally, if K ∈ Alt − Chev, then by definition of TG3 , K ∼ = An , n = 10, 11, 12, so by unbalance, n = 10 or 11, with no regular orbit for E in the first case. In the second case, let Ω be a set of cardinality n = 11 on which K acts naturally. If E has no regular orbit, then for any e ∈ E # with fixed point set Ωe on Ω satisfying |Ωe | > 2, AΩe ≤ H. Then as ∪e∈E # Ωe = Ω, K = H. This completes the proof.  Lemma 6.10. Let J ∈ TG3 (G)∩Chev−Chev(3). Let E ∈ E33 (Aut(J)) and D ∈ E2 (E). Suppose that B ∈ E3 (CAut(J) (E)) and J is locally unbalanced   with respect to B. Then J = L3 (CJ (d)) | d ∈ D # . Proof. Note that by our assumptions, (6A)

m3 (Aut(K)) ≥ 3.

Say J ∈ Chev(r), r = 3. Let q = q(J). We argue that we may assume that (6B)

q > 2.

 (2), Otherwise, by the failure of local balance, [IA , 7.7.8b], and (6A), J ∼ = Dm m+1 m ≥ 4,  = (−1) . In that case let V be the natural F2 K-module. Expand D to some D ∗ ∈ E∗ (J) and decompose V into CV (D ∗ ) and the sum of nontrivial irreducibles for D ∗ :

(6C)

V = V0 ⊥ V1 ⊥ · · · ⊥ Vm−1 ,

where dim Vi = 2 for all i, V0 is of + type and Vi is of − type for all i > 0. Let Wi = V0 ⊥ Vi for i > 0, and Ji = CJ (Wi⊥ ). Then Ji ∼ = = Ω(Wi ) ∼ − ⊥ ∼ ∼ Ω4 (2) = A5 . For any i > j > 0, let Wij = Wi + Wj and Jij = CJ (Wij ) = ∼ ∼ Ω(Wij ) ∼ = Ω+ 6 (2) = A8 . As Jij contains Ω(Wi ) × Ω(Vj ) = Z3 × A5 , ii follows easily that Ji , Jj  = Jij . Using [IA , 7.3.2d], we have J = J1j | 2 ≤ j < m = Ji | 1 ≤ i < m .

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For every i > 0, CD (Vi )  = 1 and we choose 1 = di ∈ CD (Vi ). Then Ji ≤ L3 (CJ (di )), so J = L3 (CJ (d)) | d ∈ D # , as desired. This justifies (6B).    Next, by [IA , 7.3.3], if we set Jd = Or (CJ (d)), then J = Jd | d ∈ D # . If D ≤ Aut0 (J), then for any Lie component I of any Jd , q(I) ≥ q > 2 by [IA , 4.2.2]. As r = 3, I is quasisimple. Moreover, |I|3 = 1. (Othern wise I = 2B2 (2 2 ) for some n, which never happens if D ≤ Aut0 (J); see [IA , 4.7.3A].) Hence Jd ≤ L3 (CJ (d)) for each d and the desired generation holds. We may therefore assume that some d ∈ D # induces a field  or graph-field automorphism on J, and Or (CJ (d)) is not quasisimple. The 3 only possibilities are J ∼ = L2 (8), (S)U3 (8), 2F4 (2 2 ), Sp4 (8), and G2 (8). But 3 L2 (8) ∈ C3 , while 2F4 (2 2 ), Sp4 (8), and G2 (8) are locally balanced for 3, by [IA , 7.7.8c]. So J/Z(J) ∼ = U3 (8). By our assumptions, J is locally unbalanced with respect to BE, and BE ∈ E33 (Aut(J)). We show that this is impossible for U3 (8). Namely, using [IA , 4.8.2, 4.8.4], we see that the only instance of local unbalancing is by an element x ∈ I3 (Inndiag(J)) of order 3 which satisfies CP GU3 (8) (x) ∼ = Z3 × F19.3 (F19.3 being a Frobenius group of order 19.3). A preimage x  of x in GU3 (8) has minimal polynomial ), so det x  is not a cube. Therefore x ∈ Inn(J). t3 − ω where ω ∈ Syl3 (F× 82 However, if x ∈ BE, then m3 (CAut(J) (x)) ≥ 3 so x centralizes a field automorphism f of order 3. Thus x ∈ CInndiag(J) (f ) ≤ Inn(J) by Lemma 11.3.1, a contradiction. The proof is complete.  Lemma 6.11. Let K  X with K ∈ C3 ∪ T3 ∪ {A10 } and m3 (K) > 1. Suppose that F ≤ B ∈ E3 (X) with K locally unbalanced with respect to B and F ∼ = E33 . Let F0 be the set of all f ∈ F # such that Kf := L3 (CK (f )) is a single 3-component and m3 (H) ≥ 2. Then K = Kf | f ∈ F0 . Proof. As CF (K)# ⊆ F0 we may assume that CF (K) = 1, whence m3 (AutF (K)) = 3. If K ∈ C3 , however, then by the assumed failure of balance, K ∼ = L2 (33 ) and m3 (AutB (K)) ≤ 2, a contradiction as F ≤ B. Thus K ∈ T3 , whence K/Z(K) ∼ = L3 (q 3 ) for some q. Likewise q > 2, for again, otherwise, m3 (AutB (K)) ≤ 2 by the assumed failure of balance, contradiction. Let F1 = InndiagF (K). Then m3 (F1 ) ≤ 2 and by [IA , 4.9.1], every f ∈ F −F1 induces a field automorphism on K. As q > 2, F −F1 ⊆ F0 . But by [IA , 7.3.8], K = Kf | f ∈ F − F1 , completing the proof.  Lemma 6.12. Let p be an odd prime and K ∈ Kp with either mp (K) = 1 or K a TGp -group with abelian Sylow p-subgroups of rank 2. Suppose that K   X and A ∈ Epn (X). If K is not locally balanced for p, then K ≤ 5 ΓA,n−1 (X). Also, if K is not a pumpup of L2 (8) (with p = 3) or 2B2 (2 2 ) (with p = 5), then K ≤ ΓA,3 (X). Proof. By [V9 , 9.3], if we set A0 = AutA (K) ∼ = NA (K)/CA (K) and m = mp (A0 ), it suffices to assume that m ≥ 2 and prove K = ΓA0 ,m−1 (K).

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∼ L2 (8) If mp (K) = 1, then by Lemma 3.1f, we may assume that K = 5 2 (with p = 3) or B2 (2 2 ) (with p = 5), so K is locally balanced for p, by [IA , 7.7.8cd]. So assume that K ∈ TGp has abelian Sylow p-subgroups of rank 2. If K ∈ Alt, then K ∼ = An , 2p < n < 3p and we may assume that A0 ∈ Sylp (Inn(K)), so that m = 2. Then K = ΓA0 ,1 (K) by [IA , 7.5.1a]. If K ∈ Spor, then the possibilities for K ∈ Gp are given in [I2 , p. 103], and K = ΓA0 ,1 (K) by [IA , 7.5.5]. As K ∈ Chev(p), we may assume that K ∈ Chev(r) − Chev(p), r = p. If m = 2, then K = ΓA0 ,1 (K) by [IA , 7.3.3] and the fact that K ∈ Gp . So assume that m > 2, whence by Lemma 3.6d, m = 3 and A0 = Ω1 (P ∩ K) f , where P ∈ Sylp (Aut(K)) and f is a field  automorphism of order p. For every x ∈ A0 ∩K, let Lx= Or (CK (x)). Then  again by [IA , 7.3.3] and the condition K ∈ Gp , K = Lx | x ∈ (A0 ∩ K)# . So it suffices to show that (6D)

Lx ≤ ΓA0 /x,1 (Lx )

for every x ∈ Ip (A0 ∩ K). Note that by [IA , 4.2.2, 4.2.3], Lx ∈ Lie(r) and f induces a nontrivial field automorphism on it. The desired conclusion (6D) then holds, by [IA , 7.3.8], unless for some x, Lx ∼ = L2 (8) with p = 3 or 2B (2 25 ) with p = 5. This proves the final assertion of the lemma. 2 To complete the proof, suppose that (6D) fails for some x. If p = 5, then 5 as x induces an inner automorphism on K, we must have K ∼ = 2F4 (2 2 ) ∈ C5 , a contradiction, because of the Dynkin diagram considerations in [IA , 4.2.2]. So p = 3 and L2 (8) ↑3 K, with K ∈ Chev(2) − Chev(3) having abelian Sylow 3-subgroups of rank 2. In particular K does not contain any element of B3 , so K ∈ B13 , cf. [V7 , (1G)]. Thus, K is a flat TG3 -group, so by Lemma 2.1, K∼ = Sp4 (8), or L4 (8), as L2 (8) ↑3 K. But Sp4 (8) ∈ C3 , so K ∼ = L4 (8). In this  := L3 (CK (y)) | y ∈ x, f  − x,. final case, apply [IA , 7.3.8] to get K = Ky Each such Ky ∼ = A8 , so Ky = ΓA0 ,2 (K) and thus K = ΓA0 ,2 (K). = L4 (2) ∼ The proof is complete.  Lemma 6.13. Let p be an odd prime. Suppose K ∈ Cp ∪ Tp5 ∪ TGp . If K ∈ TGp , assume that K has abelian Sylow p-subgroups of rank 2, and that if p = 3, then K is flat. Suppose that B ∼ = Epn acts on K. Then one of the following holds: (a) K = ΓB,n−1 (K); or (b) K is locally balanced with respect to B. Proof. We may assume that CB (K) = 1. If K ∈ Cp and (b) fails, then ∼ K = L2 (27) with some b ∈ B # inducing a field automorphism of order 3, and then (a) holds by Lemma 11.2.1a. If K ∈ Tp5 , then the lemma follows from Lemma 3.1f. In the remaining case, Lemma 6.12 supplies the proof. 

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Lemma 6.14. Let K = F ∗ (X) with K ∈ K3 and m3 (K) = 1. Let E ∈ E32 (X). Then one of the following holds:   (a) K = L3 (CK (e)) | e ∈ E # ; or (b) K is locally balanced for the prime 3. Proof. Since there exists x ∈ Aut(K) − Inn(K) of order 3, K ∈ Chev(r). As m3 (D4 (q)) > 1 < m3 (3D4 (q)) for any q, x is an inner-diagonal or field automorphism. In either case, r = 3 as mp (K) = 1. Indeed, if x ∈ Inndiag(K), then K involves L3 (q), q ≡  (mod 3),  = ±1, so m3 (K) > 1, contradiction. Thus, every element of I3 (Aut(K)) − Inn(K) is a field automorphism. If K ∼ = L2 (8), then (b) holds.   ∼ If K = L2 (8), then K = Or (CK (e)) | e ∈ E # , by [IA , 7.3.3]. For 

e ∈ E # ∩ K, since m3 (K) = 1, Or (CK (e)) = L × P where L is a 3 -group and P is a 3-group; and then P = 1. If L = 1, then the Lie component n m L must be 2B2 (2 2 ) for some n, so K ∼ = 2F4 (2 2 ) for some m, but this is  absurd by [IA , 4.7.3A]. Hence L = 1 and so K = O r (CK (e)) | e ∈ E − K . 

Therefore (a) holds unless Or (CK (e)) = L3 (CK (e)) for some e ∈ E − K, which we henceforth assume. As e induces a field automorphism on K, and 3 m3 (K) > 0, K ∈ Lieexc − Lie(3), so K ∼ = U3 (8), Sp4 (8), G2 (8), or 2F4 (2 2 ).  In every case m3 (K) > 1 by [IA , 4.10.3a], a final contradiction. Lemma 6.15. Let J  X with J ∈ Kp for some odd prime p, and J a K-group. Let p1+2 ∼ = P ≤ X be such that CP (J) = 1. Write P = U t with U ∈ E2 (P ). Set X = X/CX (J) and assume that [Op (CJ (t)), Z(P )] = 1. Assume also  p (J) ≥ 3, then J ∈ CTG  p.  that if m Set J0 = Op (CJ (u))Lp (CJ (u)) | u ∈ U # . Then one of the following holds: (a) J = J0 ; or (b) p = 3 and J/O3 (Z(J)) ∼ = L2 (27), L3 (4), A10 or M12 . Moreover if J/Z(J) ∼ = L3 (4), then U induces inner automorphisms on J, while if J/Z(J) ∼ = A10 , then U acts on it as a nines-subgroup with a regular orbit. Proof. Our hypotheses imply that J is not weakly locally balanced with respect to t Z(P ). Hence if J ∈ Cp , then p = 3 and J ∼ = L2 (27), by [IA , 7.7.9], so (b) holds. We assume then that J ∈ Cp . Suppose that J ∈ Alt, so that J ∼ = An for some n. Since P is nonabelian, n ≥ p2 . Then mp (J) ≥ 3 so J ∈ CTGp − Cp . This implies that p = 3 and n ≤ 12. The failure of weak local balance yields n = 10, so (b) holds if U acts on J with a regular orbit on 9 of the 10 letters[IA , 7.7.6, 7.7.1]. If U acts otherwise, then (a) holds by [IA , 7.5.2c]. Likewise if J ∈ Spor, we deduce from [IA , 7.7.6, 7.7.1] that p = 3 and J ∼ = M12 , M22 , or HJ. In the first case, (b) holds. In the second case, | Aut(J)|3 = |J|3 < 33 , contradiction. If J ∼ = HJ, let V ∈ E32 (J) act on J

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like U . But by [IA , 5.3g], CJ (v) = O3 (CJ (v))L3 (CJ (v))V for all v ∈ V # , so J0 V = ΓV,1 (J) = J by [IA , 7.5.5]. As J0  J0 V , (a) holds. As Chev(p) ∩ Kp − Alt ⊆ Cp , the remaining case to consider is J ∈ Chev(r) − Alt − Chev(p) for some prime r = p. If AutU (J) ∩ Inndiag(J) = 1, then by [IA , 2.5.12], the only possibility is J∼ = D4 (q 3 ) with p = 3. Then (a) holds by [IA , 7.3.9b]. So assume AutU (J) ∩ Inndiag(J) = 1. If some u ∈ U # induces a graph automorphism on J, then J ∼ = D4 (q) or 4 (q), and if q = 2, then the unbalancing hypothesis is contradicted by [IA , 7.7.11d]. Hence q > 2, and (a) follows from [IA , 4.7.3A] and [IA , 7.3.9b]. Suppose some u ∈ U # induces a field or graph-field automorphism on J, so that J has level q(J) = q p . If mp (J) = 1, then by Lemma 3.1e,  Aut(J) does not contain p1+2 , so mp (J) > 1. If J1 := Or (CJ (u)) ∈ Lieexc , then (a) follows easily from [IA , 4.9.1] and [IA , 7.3.9b]. So suppose J1 ∈ Lieexc . Using mp (J) > 1 we see that J/Z(J) ∼ = U3 (8), B2 (8), G2 (8), or 2F (2 23 ) with p = 3, or 2B (2 52 ) or 2F (2 52 ) with p = 5. Then [I , 7.7.11d] 4 2 4 A 3 ∼ and the unbalance hypothesis reduce us to J/Z(J) = U3 (8) or 2F4 (2 2 ); 3 indeed 2F4 (2 2 ) is locally balanced for p = 3 by inspection of [IA , 4.7.3A] 3 and the fact that Out(2F4 (2 2 )) ∼ = Z3 [IA , 2.5.12], so we need only rule out U3 (8) to conclude that (b) holds. But in the U3 (8) case, CAut(J) (t) = CInndiag(J) (t) ∼ = Z3 × F19.3 . This implies that the image V of t Z(P ) in Aut(J) is self-centralizing. Since | Inndiag(J)|3 > 32 , |NInndiag(J) (V )|3 = 33 = |NAut(J) (V )|3 . But V is normalized by a field automorphism from U , a contradiction. Hence, (b) holds. We are now reduced to the case 3D

AutU (J) ≤ Inndiag(J), and since Outdiag(J) has cyclic Sylow p-subgroups [IA , 2.5.12], in fact AutU (J) ∩ Inn(J) = 1.  3 (E6 (q)) ≥ It then follows from [IA , 7.7.6c] that p divides | Outdiag(J)|. As m  5 when q ≡  (mod 3) ( = ±1), E6 (q) ∈ CTG3 , so J/Z(J) ∼ = L (q), q ≡  (mod p), np

and again by [IA , 7.7.6c], t Z(P ) is the image of a nonabelian p-subgroup of GLnp (q). The condition J ∈ CTGp forces np = 3, 5, or 6. We have q > 2 as J ∈ Cp . If np = 6, [IA , 7.3.3] yields (a). So n = 1 and p = 3 or 5. Let J = P GLp (q) and let U be the image of AutU (J) in  the preimage of U in J.  If U  is abelian, it is J. Let J = GLp (q) and U  is not abelian. Let diagonalizable and (a) holds, so we may assume that U   0 H 0 = Op (CJ (u)) | u ∈ U and H = H 0 U . Note that H 0 ≤ Lp (q). Let H

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 =H 0U  . Since H 0 is H-invariant, be the preimage of H 0 in SLp (q), and H    contains SL (q). Let V be the natural J-module. it suffices to show that H p  =  is nonabelian, H  is (absolutely) irreducible on V . Thus Or (H) Since H 1 where q is a power of the prime r. Let s be a prime distinct from r and  Suppose that N  = 1. Then by Clifford’s theorem, N   = Os (H). p, and N is abelian, either preserving a unique ordered (orthogonal) frame F in V or  acts monomially on V and acting irreducibly on V . In the former case, H   ) =: C is cyclic. Let u ∈ U − C. Since u normalizes N  it preserves F CU ( N   and so centralizes N , contradiction. Hence N is irreducible on V , so cyclic ) of order dividing (q p − )/p(q − ). But then Sylow p-subgroups of NJ (N  = 1, are cyclic, contradicting the structure of U . We have proved that N ∗    whence F (H) = Op (H)E(H).  is not abelian, it follows that for each u ∈ U # , U / u acts Since U fixed-point-freely on the abelian group H u := Op (CLp (q) (u)). Suppose that Op (H) = 1.  Since H x is abelian, it Since U = CH (U ), there is 1 = x ∈ U ∩ Op (H).  follows that [E(H), H x ] = 1. And for any u ∈ U − x, H u = [H u , x]  = 1. But then   H  0 ] = 1, so E(H) centralizees E(H). Therefore [E(H), #

H u ≤ Op (CH 0 (u)) = 1 for all u ∈ U , so H 0 = 1. This implies that either q −  or (q p − )/(q − ) is a power of p. The latter is impossible by Zsigmondy’s theorem [IG , 1.1], so U ≤ Inn(J). The only possibilities are J∼ = L3 (4), U5 (4), and U3 (8). Suppose that J ∼ = U5 (4) or U3 (8). Then by Lemma 4.18, J is locally balanced with respect to any inner automorphism of order p. Hence t ∈  is nonabelian, U is selfInndiag(J) − Inn(J). On the other hand as U centralizing in Inndiag(J). Therefore as | Inn(J)|p > p2 , |NInn(J) (U )|p = p3 = |NInndiag(J) (U )|p . Hence t cannot normalize U , a contradiction. Thus, J∼ = L3 (4) and (b) holds. Finally suppose that Op (H) = 1, so that F ∗ (H) = E(H) is the direct product of simple groups I 1 , . . . , I m . If any such component I i is a p -group, then the fact that H u is abelian # for all u ∈ U implies that U normalizes and acts faithfully on I i ; but mp (Aut(I i )) ≤ 1 by Lemma 3.1, contradiction. For the same reason U normalizes each I i , and as U is self-centralizing, U ∩I i = 1. But the preimage of U is nonabelian, so m = 1 and I 1 has a covering group by Zp . Also mp (I 1 ) ≤ mp (J) ≤ p − 1.  If p = 5, then using [IA , 6.1.4, 4.10.3] we see that I 1 ∼ = Lp (q  ), where q  ≡  (mod p),  = ±1. If p = 3, we get the same conclusion with an additional restriction: q is a power of q  ; but there are also two additional

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∼ A6 or A7 , using [IA , 6.5.3]. In these exceptional cases, possibilities, I 1 = the definition of H implies that O3 (CH (u)) = 1 for some u ∈ U # . This condition must then hold for exactly two subgroups u of U , with I 1 ∼ = A7 and O3 (CH (u)) ∼ E . But then U must induce inner automorphisms on 2 = 2 # ∼ J, and so O3 (CH (u)) = E22 for all u ∈ U , a contradiction.  Therefore, I 1 ∼ = Lp (q  ). Write q  = (r )b and q = ra , where r and r are primes different from p. We claim that r = r . By the previous paragraph  we may assume that p = 5. There is E ∈ Er4 (I 1 ) such that CI 1 (E) is not abelian. (We can take E = Z(Or (P  )), where P  is a parabolic subgroup of I 1 of type A1 (q  ) × A1 (q  ) or A1 ((q  )2 ), according as  = 1 or −1. Then Or (P  ) is not abelian.) If r = r , on the other hand, then by [IA , 4.10.3a], mr (J) ≤ 4, and E is diagonalizable as a subgroup of J, as is CJ (E). Thus, CJ (E) is abelian, contradiction. Hence r = r , as claimed. If b ≥ a then b = a as |I 1 |r ≤ |J|r , whence  =  and I 1 = J, so # (a) holds. So we may assume that b < a. Now some u ∈ U induces an inner automorphism on J, so CJ (u) contains a Frobenius group with p−1 complement of order p and kernel A = [A, U ] ∼ = Zs , where s is the p -share of q − . Then as Out(I 1 ) is p-closed, A ≤ I 1 , so A ≤ A := Op (CI 1 (u)). p−1 As p − 1 > 1, we must have A ∼ = Zs , where s is the p -share of q  −  . But a similar argument shows that A ≤ A, so A = A and s = s . In particular, k := (q − )/(q  −  ) is a power of p (and p divides q  −  ). We use Zsigmondy’s theorem [IG , 1.1] to contradict this assertion. In particular if  = −1, then since J ∼ = U3 (8) has been ruled out above, k is not a power of p, contradiction. Thus,  = 1. Likewise if  = 1, then k can be a power of p only if r ∈ F M, a = 2, and b = 1. But then r = 2 and p = 3, which does not divide r − 1, contradiction. Therefore,  = −1. Suppose there exists a prime t dividing r2b − 1 but not rc − 1 for any 0 < c < 2b. Then t divides rb + 1 and hence ra − 1, so a = 2bn for some integer n. Then k = (q − )/(q  −  ) = (rb − 1)(r2bn − 1)/(r2b − 1). As p divides rb + 1, k is not a power of p, contradiction. Thus no such prime t exists. Note that if r = 2 and b = 3, then p = 3 and q  + 1 = 9, so q − 1 = 9k is a power of 3, which is impossible. Hence, by [IG , 1.1], b = 1 and r ∈ F M . As p divides r + 1, a is even; but p does not divide r − 1, so as k is a power of p, r = 2 and p = 3. But then I 1 ∼ = U3 (2), which is solvable, a final contradiction. The lemma is proved.  7. Cp -Groups, p odd Let p be an odd prime. Consider the following situation:

(7A)

(1) (2) (3) (4)

C is a group with Op (C) = 1 and mp (C) ≥ 3; C = C/Op (C), and Op (C) has odd order; K = Lp (C) and every component of K is in Cp ; and e(C) ≤ 3, so that every component of K is in A, by Theorem 1.3.

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Lemma 7.1. Assume (7A). Let L be a p-component of K. Then one of the following holds: (a) L ∈ Chev(p) and one of the following holds: (1) p > 3, L ∼ = L2 (p); (2) L ∼ = L2 (p2 ); (3) L ∼ = L2 (pn ), n ≥ 3; (4) L ∼ = L± 3 (p); ± 2 (5) L ∼ = L± 3 (p ), L4 (p), P Sp4 (p), or G2 (p); (6) p = 3, L ∼ = L2 (8); or (7) p = 3, L ∼ = 3U4 (3) or 3G2 (3); (b) L ∈ Chev − Chev(p) and one of the following holds: 1 (1) p = 3, L ∼ = 3D4 (2), 2F4 (2 2 ) , Sp6 (2), or Sp4 (8); or 5 1 5 (2) p = 5, L ∼ = 2B2 (2 2 ), 2F4 (2 2 ) , or 2F4 (2 2 ); (c) L ∈ Alt − Chev and one of the following holds: (1) p = 3, L ∼ = A9 ; (2) p = 5, L ∼ = A10 ; (3) p = 7, L ∼ = A7 ; or (4) p = 11, L ∼ = A11 ; or (d) L ∈ Spor and one of the following holds: (1) p = 3 and L ∼ = M11 , J3 , 3J3 , M c, or 3M c; (2) p = 5 and L ∼ = HJ, Co3 , Co2 , HS, M c, He, Ly, Ru, F5 , or F3 ; (3) p = 7 and L ∼ = He, ON , or F3 ; or (4) p = 11 and L ∼ = J4 . Proof. By assumption, Op (L) = 1 and L ∈ Cp . We let z ∈ I2 (L) and choose P ∈ Sylp (CL (z)). Suppose first that L ∈ Lie(p) ∩ Chev(p). Then Z(L) = 1 and m2,p (LOp (C)) ≤ e(C) ≤ 3. As Op (C) = 1, m2,p (L) ≤ 2. By (a1,2,3), we may assume that K = A1 (pr ). We use [IA , 4.5.1] to see the structure of CL (z). In particular, we see that z may be chosen so that mp (CL (z)) ≥ 3, contradicting m2,3 (L) ≤ 2, unless L is as in (a4,5). If L ∈ Chev(p) − Lie(p) and L is simple, then (a6) holds. Now assume that L ∈ Chev(p) with Z(L) = 1, but L is not as in (a7). Then p = 3 and L ∼ = 3Ω7 (3) or [3 × 3]U4 (3) (recall that 3A6 ∈ C3 , by definition). In the latter case P /Z(L) is isomorphic to E32 , and is inverted by an involution of CL (z), so |Φ(P )| ≤ 3. But CAut(L) (z) covers Out(L), which acts irreducibly on Z(L) [IA , 6.3.1]. Hence P ∼ = E34 , contradicting m2,3 (L) ≤ 3. Finally, if L ∼ = 3Ω7 (3), then z may be chosen so that CL/Z(L) (z) ∼ = 2U4 (3). Then m3 (E(CL (z))) ≥ 4 by [IA , 6.4.4], again a contradiction. So we may assume that L ∈ Chev(p).

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Next, assume that L ∈ Chev−Chev(p). By the definition of Cp ([V11 , 3.1] or [V3 , Sec. 1]) and [IA , 6.1.4], L is as in (b1,2) unless p = 3 and L ∼ = U5 (2), U6 (2), SU6 (2), D4 (2), or F4 (2). In every case a parabolic subgroup M such that m3 (M ) = 3 + m3 (Z(L)) is easily found, so m2,3 (LO3 (C)) ≥ 4, a contradiction. So we may assume that L ∈ Chev. If L ∈ Alt − Chev, then since C3 does not contain 3A6 , L ∼ = Ap (p > 5), ∼ A2p (p > 3), or A3p . Hence if L and p are not as in (c), then L = An , n ≥ 14. Let M be a minimal subgroup of L such that M = L. We get a contradiction by showing that m2,3 (M ) ≥ 4. By a Frattini argument, Op (M ) is nilpotent, so we may assume that Op (M ) is a {2, 3}-group. If O2 (M ) = 1 it is enough to show that m3 (M ) ≥ 4, so we may pass to M/O2 (M ) and assume that  M/O3 (M ) ∼ = L, and even that M = O3 (M ). If F ∗ (M ) = O3 (M ), then as m2 (M ) = 6, the Thompson dihedral lemma implies that m2,3 (M ) > 3, as desired. And otherwise, M ∼ = An contains A4 × A10 , so again m2,3 (M ) > 3. So we may assume that L ∈ Spor. We use the tables [IA , 5.6.1, 5.6.2]. Suppose that m2 (L) ≥ 5 and m2,3 (L) ≥ 4. We claim that m2,3 (L) ≥ 4, contrary to e(C) = 3. If p = 3, this is obvious, so assume that p > 3. Again let M be minimal in L subject to M = L. Then we may pass without loss to M/O{2,3} (M ) and even to M/O2 (M ) and assume that Op (M ) is a 3-group. If F ∗ (M ) = O3 (M ), then m2,3 (M ) ≥ 4 as m2 (L) ≥ 5, by the Thompson dihedral lemma; and otherwise L ∼ = L, so the claim holds. The claim completes the proof for p > 3; for p = 3 it is sufficient to use, in addition, the obvious condition  m2,p (L) − mp (Z(L)) ≤ 2. This completes the proof. Lemma 7.2. Suppose that p is an odd prime and L, J are in the conclusion of Lemma 7.1 for the prime p. Then J ↑p L if and only if (p, J, L) is one of the following: (a) (p, L2 (pn ), L2 (pnp )), np > 3; (b) (3, L2 (8), 3D4 (2) or Sp4 (8)); (c) (3, U3 (3), 3D4 (2)); (d) (3, A6 , Sp4 (8) or Sp6 (2) or A9 or J3 or 3J3 ); 1 5 5 (e) (5, 2F4 (2 2 ) or 2B2 (2 2 ), 2F4 (2 2 )); (f) (5, A5 , A10 or HJ or Co3 or Co2 or HS or He or Ru); (g) (5, U3 (5), F5 ); (h) (7, L2 (7), He or F3 ); Proof. It suffices to list, for each L in Lemma 7.1, the components of centralizers of elements of Ip (Aut(L)). This is readily done, using the  Borel-Tits theorem and [IA , 4.7.3A, 4.8.2, 4.9.1, 4.9.2, 5.2.6, 5.3]. Lemma 7.3. Assume (7A). Suppose also that p = 3 and mI3 (C) ≤ 2. Then the following hold: (a) One of the following holds: (1) K ∼ = L2 (8) or L2 (3n ) for some n ≥ 2, and mI3 (K) = 0;

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17. PROPERTIES OF K-GROUPS 1 3 2 2  (2) K ∼ = L± 3 (3), A9 , M11 , J3 , D4 (2), F4 (2 ) , Sp4 (8), or L2 (8) × I L2 (8), and m3 (K) = 1; or (3) K ∼ = 3G2 (3) or 3J3 , and mI3 (K) = 2. (b) Suppose that K  K1 = K c and c3 = 1, and J is a 3-component of CK (c). Suppose also that J/O3 3 (J) ∼ = L2 (3n ) for some n ≥ 2, CK (c) has no subgroup isomorphic to Σ6 , and K 1 has no subgroup isomorphic to A4 × A4 . Then K ∼ = L2 (3n ) or L2 (33n ), with c acting on K trivially or as a field automorphism, respectively; and (c) There is no I ≤ K such that I/Z(I) ∼ = L3 (2m ),  = ±1, 2m ≡  (mod 3), and |I/Z(I)|3 = 32 .

Proof. Without loss, O3 (C) = 1. We first prove (a). First suppose that mI3 (K) ≥ 2. Then mI3 (K) = mI3 (C) = 2, so as O3 (C) = 1, we have O3 (K) = 1. Let K1 be a component of K such that Z(K1 ) = 1. As K1 ∈ C3 ∩ A, we use [IA , 6.1.4] to deduce that K1 /Z(K1 ) ∼ = G2 (3), J3 , U4 (3), M c, or Suz. In the last three cases, m3 (CK1 (z)) ≥ 3 for any 2-central involution z of K1 , by Lemma 10.10.9 and [IA , 5.3no, 6.1.4], contradicting mI3 (C) = 2. So K1 ∼ = 3G2 (3) or 3J3 , mI3 (K1 ) = 2, and thus K = K1 . Hence, (a3) holds in this case. We may therefore assume that mI3 (K) ≤ 1. If K is not quasisimple, then it must have two components K1 , K2 , with m3 (Ki ) = 1 for i = 1 and 2. As Ki ∈ C3 , Ki ∼ = L2 (8), as desired. So we may assume that K is quasisimple and m3 (K) ≥ 2. If K ∈ Chev(3), then involution centralizers are given in [IA , 4.5.1], and so the condition mI3 (K) ≤ 1 forces (a1) or (a2) to hold. If K ∈ Chev(2) ∩ C3 − Chev(3), then m3 (CK (z)) ≥ 2 for a long root involution z ∈ K, unless K ∼ = 3D4 (2), 1 2F (2 2 ) , or Sp (8). If K ∈ Alt ∩ C − Chev(3), then K ∼ A . And if = 9 4 4 3 K ∈ Spor∩C3 , note that the condition mI3 (K) ≤ 1 implies that m2,3 (K) ≤ 2, so we can use [IA , 5.6.2]. Noting that K ∈ C3 and mI3 (M c) ≥ m3 (A8 ) = 2, we conclude that K ∼ = M11 or J3 , as desired. Thus, (a) holds. In (b), the possible pumpups of J are given in Lemma 7.2. Using this, we see that if (b) fails, then K ∼ = A6 . = A9 , J3 , 3J3 or Sp4 (8), with J ∼ But A9 contains A4 × A4 , as do J3 and 3J3 , inside a subgroup disjoint from the center and isomorphic to an extension of E24 by GL2 (4) [IA , 5.3h]. Thus K ∼ = Sp4 (8). But then c acts as a field automorphism, centralizing Sp4 (2) ∼ = Σ6 . This contradiction proves (b). In (c), since |I/Z(I)|3 = 32 , we have m = 3. Just the condition that |L3 (2m )| divides |K| reduces the possibilities in (a) to K ∼ = L2 (3n ), A9 and Sp4 (8). As m ≥ 2, a Sylow 2-subgroup S of I satisfies |S/Φ(S)| ≥ 24 [IA , 2.4.1, 3.3.1], so L2 (3n ), whose Sylow 2-subgroups are dihedral, is impossible. Since L3 (4) has an E24 -subgroup all of whose involutions are conjugate while |L3 (4)|2 = |A9 |2 , it does not embed in A9 . Neither do L3 (2m ) for m > 3, as m2 (L3 (2m )) > m2 (A9 ) [IA , 3.3.3]. Finally, if K ∼ = Sp4 (8), let P be a Sylow 3-subgroup of K. Then Ω1 (P ) ∼ = E32 and K has more than one conjugacy class of subgroups of order 3 [IA , 4.8.2]. On the

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other hand, Ω1 (S) ≤ I while I has a single such class [IA , 4.8.2]. The proof is complete.  The next lemma gives quite crude results, but they are good enough for our purposes. Lemma 7.4. Let K ∈ Chev(p) be simple, p an odd prime. Then the sectional 2-rank r2 (K) satisfies the following conditions: (a) r2 (L2 (pn )) = 2; n (b) r2 (L± 3 (p n)) ≤ 3; 2 (c) r2 ( G2 (3 2 ) ) = 3; and n n n (d) r2 (L± 4 (p )), r2 (B2 (p )) and r2 (G2 (p )) are all at most 7. Proof. We use the facts that if T is a 2-group and U  T , then r2 (T ) ≤ r2 (T /U ) + r2 (U ); if T involves U , then r2 (U ) ≤ r2 (T ); and if T is abelian, then r2 (T ) = m2 (T ). We use the structure of Sylow 2-subgroups T of K, as implied by [IA , n T has 4.5.1]. If K ∼ = L± = L2 (pn ), then T is dihedral. If K ∼ 3 (p ), then n 2  ∼ a quaternion subgroup U  T with T /U cyclic. If K = G2 (3 2 ) , then n n n T ∼ = E23 . If K ∼ = L± 4 (p ), B2 (p ), or G2 (p ), then T has a normal subgroup U = Q1 ∗ Q2 ∗ Z with Q1 and Q2 quaternion and Z cyclic (possibly trivial),  and T /U embeds in D8 . These imply the lemma. Lemma 7.5. Let K ∈ Kp and p be as in Lemma 7.1, with p ≤ 5. Suppose that Inn(K) ≤ X ≤ Aut(K) and X contains subgroups B ∼ = Ep2 and Q such that Q is a 2-group of symplectic type and B acts faithfully on Q. Suppose also that Lp (CK (b)) = 1 for all b ∈ B # . Then one of the following holds: (a) (p, K/Z(K)) ∼ = (3, M c), (5, F5 ), or (5, Ly); (b) p = 3 and K/Z(K) ∼ = U4 (2), L± 4 (3), or G2 (3). Moreover, in (b), Q does not contain a copy of Q8 ∗ Q8 ∗ Q8 . Proof. Since Z(K) is a p-group, we may assume that K is simple, i.e., K ≤ X. Moreover K ∼ D4 (q) for any odd q, so Out(K) is 2-nilpotent by = Lemma 20.9. Set Q0 = [Q, B], so that Q0 ≤ K. As CB (Q) = 1, and p = 3 or 5, Q0 contains the central product of p − 1 copies of Q8 , so m2 (Q0 ) ≥ p and r2 (Q0 ) ≥ 2(p − 1), where r2 denotes sectional 2-rank. If K ∈ Chev(p) in Lemma 7.1, then by Lemma 9.2 and Lemma 7.4, we have p = 3 and K∼ = L± 4 (3), P Sp4 (3), or G2 (3), as in (b). Next, suppose that K ∈ Chev − Chev(p), 5 1 whence by Lemma 7.1, K ∈ Chev(2). If p = 5 and K ∼ = 2B2 (2 2 ) or 2F4 (2 2 ) , then m2,5 (Aut(K)) = 1 by the Borel-Tits theorem and the structure of parabolic subgroups of K. So these cases cannot occur. Likewise if K ∼ = 2F (2 25 ), then by the Borel-Tits theorem, B ∩ K is cyclic. Hence B contains 4 a field automorphism b by [IA , 4.9.1]. But then L5 (CK (b)) = 1, contrary to hypothesis. So we may assume that p = 3. Similar arguments rule out

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1 K ∼ = 2F4 (2 2 ) and K ∼ = Sp4 (8); in the latter case, B must contain a field automorphism b, whence L3 (CK (b)) = 1. Suppose that K ∼ = Sp6 (2). Then B lies in a parabolic subgroup of 3-rank 2; there are two conjugacy classes of such parabolic subgroups P , represented by CK (z) and CK (y), where z and y are long and short root involutions, respectively. In either case B ∈ Syl3 (P ) and P contains a long root A1 (2)-subgroup L. Hence B contains a conjugate

b of O3 (L), and L3 (CK (b)) ∼ = Σ6 , contrary to assumption. The last case in this category is K ∼ = 3D4 (2). Again Aut(K) contains two conjugacy classes of parabolic subgroups P of 3-rank 2, one of which must contain B, and both of which contain a standard graph automorphism b, i.e., with L3 (CK (b)) ∼ = G2 (2) . For any R ∈ Syl3 (P ), Ω1 (P ) ∼ = E32 , so b has a K-conjugate in B and our hypothesis is again contradicted. Next, assume that K ∈ Alt − Chev, so that K ∼ = A9 or A10 with p = 3 or 5, respectively. Then Z(Q0 ) = z where z is of cycle shape (2)2 or (2)4 . It follows easily that mp (CAut(K) (Z(Q0 ))) = 1 in both cases, so they cannot satisfy our hypotheses. We are reduced to the cases K ∈ Spor, and use the tables [IA , 5.3]. We have B ≤ K so Q0 ≤ K. If p = 3, so that K ∼ = M11 , J3 , or M c, K has one class of involutions z, but in the first two cases m3 (CK (z)) = 1, so they are ruled out. Thus we are reduced to the case p = 5. Since m5 (CK (Z(Q0 ))) ≥ 2 and Z(Q0 ) ≤ K, m2,5 (K) ≥ 2, and this yields K ∼ = Ly or F5 , as claimed. As for the final assertion, suppose that

Q ≥ Q1 ∼ = Q8 ∗ Q8 ∗ Q8 . There is no loss in assuming Z(K) = 1. We ignore U4 (2) since Aut(U4 (2)) embeds in (the centralizer of an involutory automorphism in) L4 (3). Let  Z = Z(Q), H = CAut(K) (Z), and H0 = O3 (C). Recall that Q0 = [Q, B]. As  B is faithful on Q0 and Out(K) is a 2-group, H0 = BQ0 = O3 (CInn(K) (Z)) and Q0 ∼ = Q8 ∗ Q8 , with the help of [IA , 4.5.1]. Let C = CAut(K) (Q0 ) and C1 = C ∩ Q1 ∼ = Q8 . Then C1 ∩ Inn(K) ≤ C ∩ Inn(K) = Z, and [H0 , C] ≤ CBQ0 (Q0 ) = Z. In particular, C centralizes BQ0 /Q0 . Set Aut(K) = Out(K). Then C 1 = C1 /Z ∼ = E22 . In particular 4 divides (3). Moreover, Aut (B) contains D8 (coming from | Out(K)|, so K ∼ = L± H 4 ∼ (3)), but Aut (B) E . It follows that |C 1 | ≤ | Out(K)/2|. GL± = 22 H∩Inn(K) 4 ∼ Thus | Out(K)| ≥ 8, whence K = U4 (3) and |C| = 8, so C = C1 . Finally there exists a graph automorphism g ∈ I2 (Aut(K)) such that L := E(CK (g)) ∼ = P Sp4 (3) [IA , 4.5.1]. As K has one class of involutions, we may assume that Z is a Sylow 2-center in L, and then Q0 = O2 (CL (g)). Hence  g ∈ C − Z, so C cannot be isomorphic to Q8 . Lemma 7.6. Suppose that p = 3 and X = LB is a K-group such that L = E(X) is a quasisimple group in Lemma 7.1, B = b, a1 , a2  ∼ = E33 , b ∈ Z(X), [ai , L] = 1, and E(CL (ai )) has a component Li ∼ = L2 (3ni ), ni ≥ 2, for i = 1, 2. Suppose also that for both values of i, Ui is a four-subgroup of

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Li , and U1 U2 = U1 × U2 . Suppose finally that mI3 (X) ≤ 2. Then (a) L ∼ = A9 , J3 or 3J3 ; and (b) CAut(L) (U1 U2 ) is the image in Aut(L) of U1 U2 . Proof. Clearly L is a pumpup of Li for i = 1, 2. But the possible pumpups of Li ∼ = L2 (3ni ), ni ≥ 2, for i = 1, 2 are given in Lemma 7.2. Other than the desired isomorphism types of L, we must rule out L ∼ = L2 (33ni ), L∼ = Sp4 (8). The first is impossible since m2 (L2 (3m )) = 2 = Sp6 (2), and L ∼ for any m, while L ≥ U1 U2 ∼ = Sp4 (8) or Sp6 (2), then X = E24 . Finally if L ∼ contains E32 × Σ6 ≥ E32 × Σ3 × Σ3 , contradicting mI3 (X) ≤ 2. The proof is complete.  Lemma 7.7. Let p, J and L be as in Lemma 7.2. Let K ∈ Kp be a covering group of L. Then m  p (K) ≥ 2. Proof. By [IA , 6.1.4], the only case in which K is not simple is p = 3, L = 3J3 . The lemma then follows easily from the ranks given in [IA , 3.3.3, 4.10.3, 5.6.1].  Lemma 7.8. Let K ∈ Cp be one of the groups L in Lemma 7.1. Suppose that mIp (K) = 0 and there exists b0 ∈ Ip (Aut(K)) such that CK (b0 ) has odd order but b0 acts nontrivially on some V ∈ NAut(K) ( b0  ; 2). Then p = 3 and K ∼ = L2 (3n ) for some n ≥ 2. Proof. First, the condition mIp (K) = 0, i.e., centralizers of all elements of Ip (K) in K have odd order, narrows the list in Lemma 7.1 to the following 5 groups: L2 (pn ), n ≥ 1; L2 (8), p = 3; 2B2 (2 2 ), p = 5; and Ap , p ∈ {7, 11}. To check this we can use [IA , 4.5.1, 4.7.3A, 4.8.2, 5.2.6, 5.3]. In particular, K is simple. In all these remaining cases if b0 ∈ Inn(K), then b0 is a field automorphism and so CK (b0 ) has even order, contrary to assumption. So we may assume that b0 ∈ Inn(K). As [V, b0 ] = 1, NK ( b0  ; 2) = {1}, so m2,p (K) > 0.

(7B)

If K ∼ = L2 (pn ), then we may assume that p > 3. At this point, all remaining groups K are ruled out by the condition (7B). We can use the Borel-Tits theorem and [IA , 6.5.1, 7.7.11a] to verify this in the various cases.  Lemma 7.9. Sort the pairs (p; L) in Lemma 7.1 into classes as follows. (a) (b) (c) (d) (e) (f)

5

(p > 3; L2 (p)), (3; L2 (8)), (5; 2B2 (2 2 )); (p > 3; L2 (p2 )); (p > 3; L2 (pn ), n ≥ 3); (p = 5; A10 ); (p = 5; Co2 , Co3 , M c, F3 ), and (p = 7; O N, F3 ); 1 3 2 2  (p; L± 3 (p)), (p = 3; L2 (9), M11 , D4 (2), F4 (2 ) , Sp4 (8)), (p = 5; 1 5 HJ, HS, He, Ru, 2F4 (2 2 ) , 2F4 (2 2 )), (p = 7; He), and (p = 11; J4 );

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± 2 n (g) (p; L± 3 (p ), G2 (p), P Sp4 (p), L4 (p)), (p = 3; L2 (3 ), n ≥ 3, A9 , Sp6 (2), J3 , M c), and (p = 5; Ly, F5 ); (h) (p = 3; 3U4 (3), 3G2 (3), 3J3 , 3M c); (i) (p = 7; A7 ); and (j) (p = 11; A11 ). When this is done, the following rank conditions hold for the groups in the corresponding class: (a) mp (L) = 1, m2,p (L) = 0; (b) mp (L) = 2, m2,p (L) = 0; (c) mp (L) > 2, m2,p (L) = 0; (d) mp (L) = 2, m2,p (L) = 1, m2,3 (L) = 3; (e) mp (L) = 2, m2,p (L) > 0, m3 (L) > 3; (f) mp (L) = 2, m2,p (L) = 1, m3 (L) ≤ 2; (g) mp (L) > 2, m2,p (L) > 0; (h) m3 (L) > 3, m2,3 (L) > 1; (i) mp (L) = 1, m2,p (L) = 0, m2,3 (L) = 2; and (j) mp (L) = 1, m2,p (L) = 0, m2,3 (L) = 3.

Proof. All the ranks and 2-local ranks for L ∈ Spor come from [IA , 5.6.1, 5.6.2]. For L ∈ Alt, they come from [IA , 5.2.10a] and the normalizers of root four-groups. For L ∈ Chev(2), the p-ranks come from [IA , 4.10.3] (as does the 3-rank of L± 3 (p)), and the 2-local ranks are the maximum ranks among all parabolic subgroups, by the Borel-Tits theorem. Finally, let L/Z(L) ∈ Lie(p) − Chev(2). Then mp (L) is given in [IA , 3.3.3, 6.4.4] and [V8 , 1.5], and it remains to settle m2,p (L). In (a-c), the result follows from Dickson’s classification of subgroups of L2 (q) [IA , 6.5.1], as does the result for L2 (3n ) in (g). In the rest of (g), and in (h), we see from [IA , 4.5.1] that some involution centralizer in L/Z(L) has p-rank 2, which suffices. Finally, in L± 3 (p) in (f), we can use the classification by Hartley  and Mitchell of all subgroups of L [IA , 6.5.3]. The proof is complete. Lemma 7.10. Assume (7A) but that K is not quasisimple. Then either K is the direct product of two or three groups, from the following categories in Lemma 7.9: (a) × (a), (a) × (a) × (a), (a) × (b), (a) × (f ), (a) × (i), 5 (b) × (b), (b) × (i), (f ) × (i), (i) × (i), or K ∼ = 2B2 (2 2 ) × A10 for p = 5. Proof. Write K = L1 · · · Ln , n > 1. Without loss we may assume that C = KQ where 1 = Q ∈ Sylp (Op p (C)), so that Op (C) = Op (K). Trivially m2,p (C) ≥ m2,p (C), so as e(G) ≤ 3, (7C)

m2,p (KQ) ≤ 3.

Also (7D)

m2,3 (C) ≤ 3.

These are the only consequences of our hypothesis e(G) ≤ 3 that we shall use.

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Without loss we may replace C by any subgroup covering C, so by a Frattini argument Op (C) is nilpotent. Then as m2,r (C/O2 (C)) ≤ m2,r (C) ≤ 3 for r = 3 and p, we may pass to C/O2 (C) and assume that |Op (C)| is odd. Likewise we may pass to C/Os (C) for any s ∈ {3, p}, so we may assume that Op (C) = O3 (C). If mp (L2 ) ≥ 3 + mp (Z(L2 )), then mp (L2 Op (K)) ≥ 4, so as L1 has even order, m2,p (C) ≥ 4, contradicting (7C). Hence no Li is of class (c), (g), or (h). Suppose that no Li is of type (a), (i), or (j) in Lemma 7.9. Thus all Li are of type (b), (d), (e), or (f). If mp (L1 ) − mp (Z(L1 )) ≥ 2, then m2,p (L2 ) = 0, for otherwise the normalizer of some nontrivial 2-subgroup of L2 has p-rank 4, contradicting (7C). Hence by Lemma 7.9, L2 is in class (b). The same argument with L1 and L2 reversed shows that K is of class (b) × (b), an allowed conclusion. Therefore we may assume that mp (Li )−mp (Z(Li )) ≤ 1 for all i. But such components are only of type (a), (i), or (j). Therefore we may assume that L1 is of type (a), (i), or (j). Suppose it is of class (j). Then m3 (L1 ) = 3. As p = 11, L2 contains (S)L2 (p) (or J4 ) and thus contains SL2 (3) or A4 . Hence if L1 is quasisimple, then m2,3 (C) ≥ 4, contradiction. Therefore F ∗ (L1 ) = O3 (L1 ). As m2 (L1 ) = 4, there exists E ∈ E24 (L1 ) and z ∈ I2 (CL2 (E)) such that CEz (O3 (L1 )) ≤ z. The Thompson dihedral then implies that m2,3 (C) ≥ 4, whether [O3 (L1 ), z] = 1 or not. So L1 , and indeed any Li , is not of class (j). A similar argument shows that if L2 is of class (d), then L1 cannot 5 contain SL2 (3) or A4 , and so (as p = 5), L1 ∼ = 2B2 (2 2 ) (and there are no further components, since m2,5 (C) ≤ 3 and Q = 1). As (a) × (d) is an allowed conclusion, we may assume that no Li is of class (d). Thus classes (a), (b), (e), (f), and (i) remain. Suppose next that L1 is of class (e). First, if L1 ≤ E(K), then whether or not an involution of L2 centralizes O3 (L1 ), the Thompson dihedral lemma implies that m2,3 (L1 L2 ) ≥ m2 (L1 ). Hence, m2 (L1 ) ≤ 3, so with [IA , 5.6.1], L1 ∼ = O N . Let R char O3 (L1 ) be of exponent 3 and class at most 2, with CL1 (R) ≤ O3 (L1 ). Since R admits a 71+2 Sylow 7-subgroup of L1 , m3 (R/Φ(R)) ≥ 6 · 7 = 42. Factoring R under a four-group, we get |CR (t)| ≥ 314 for some t ∈ I2 (L1 ). But then m2,3 (L1 ) ≥ m3 (CR (t)) > 3 by [V9 , 5.8], a contradiction. Thus, L1 ≤ E(K). Now m3 (L1 ) ≤ m2,3 (L1 L2 ) ≤ 3. By [IA , 5.6.1], the only possibility is L1 ∼ = 3O N , whence p = 7. But as  L2 ∼ = L2 (7), L2 (72 ), L± 3 (7), O N , F3 , or A7 ,

L2 contains a four-group V with AutL2 (V ) ∼ = Σ3 . Since an element of I3 (NL2 (V )) is real in NL2 (V ), but Z(L1 ) ≤ Z(L1 L2 ), we conclude that m2,3 (K) ≥ m3 (NK (V )) ≥ 4, contradiction. We have ruled out components of class (e).

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The only case remaining to be considered is n ≥ 3. If n ≥ 4, then obviously m2,p (KQ) ≥ 4 (in view of [IG , 16.11]), contradiction. So suppose n = 3. We get a similar contradiction if mp (Li )−mp (Z(Li )) > 1 for any i, so with Lemma 7.9, and since class (j) has been ruled out, we may assume that p = 7 and L1 ∼ = L2 ∼ = A7 . If L1 is quasisimple, choose a four-group V ≤ L1 such that m3 (NL1 (V )) = 2 and let N = NK (V ). Then N contains real elements of L2 and L3 of order 3, so m2,3 (K) ≥ m3 (N ) ≥ 4, contradiction. Hence L1 , and similarly L2 , is not quasisimple. But m2 (L1 L2 ) ≥ 4 and there is an involution z ∈ L3 . Hence again, whether z centralizes O3 (L1 L2 ) or not, the Thompson dihedral lemma implies that m2,3 (L1 L2 z) ≥ 4, a final contradiction.  Lemma 7.11. Let K ∈ Cp and Inn(K) ≤ H ≤ Aut(K). If A ∼ = Epn acts on K, n ≥ 0, and we set C = CH (A), then one of the following holds: (a) F ∗ (C) = Op (C)E(C), and every component of E(C) lies in Cp ; or (b) p = 3, K ∼ = L2 (33 ), and the image of A in Aut(K) has order 3 and is a group of field automorphisms. Proof. By [IA , 7.1.10] and induction on |A|, every component of E(C) lies in Cp . Suppose that (a) fails, so that Op (C) = 1. Choose such an A with n minimal, and let B be a hyperplane of A and L = E(CH (B)). Let x ∈ A − B; then Op (CCH (B) (b)) = 1, but F ∗ (CH (B)) = Op (CH (B))L. Hence, using [IG , 20.6], there is an x-invariant component L0 of L which is not 1-balanced for p. As L0 ∈ Cp , p = 3 and L0 ∼ = L2 (33 ) [IA , 7.7.1]. But using a generating set of B, we get a sequence L0 2. Then m3 (Aut(J)) = 3, so CE (J) = 1, and E(CJ ( g CE (J))) ∼ = G2 (q) ∈ C3 or L± 3 (q) ∈ C3 , contradiction. So g cannot exist, whence mp (InndiagE (J)) ≥ 2. Next suppose that some f ∈ E # induces a nontrivial field automorphism on J, so that q ≥ 2p . If J/Z(J) ∼ = Lp (q), q ≡  (mod p),  = ±1, with InndiagE (J) the image of a nonabelian p-subgroup of GLp (q), then

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∼ U3 (8) with mp (InndiagE (J)) = 2 so CE (J) = 1. Then unless J/Z(J) = p = 3, an allowed conclusion, we have that E(CJ ( f  CE (J))) is a central 1 extension of Lp (q p ) ∈ Cp , contrary to assumption. So we may assume that [IA , 7.3.3a] does not hold, and by [IA , 7.3.3], there exists e ∈ InndiagE (J)# such that CJ (e) has a Lie component I. We choose such e and I so that |I| is maximal. Then CE (I)/CE (J) is cyclic. (Otherwise, taking a comple ment F to CE (J) in CE (I), we have I  O2 (CJ (h)) for all h ∈ F # , with  [III11 , 1.13b], so I  Γ2F,1 (J) = J by [IA , 7.3.1], contradiction.) If also CE (J) = 1, then any components of CJ (f ) = CJ ( f  CE (J)) and CJ (e) = CJ ( e CE (J)) must lie (modulo core) in Cp . As q(I) ≥ 2p , 5 5 it follows that I ∼ = 2B2 (2 2 ) or 2F4 (2 2 ) = L2 (8) or Sp4 (8) with p = 3, or I ∼ with p = 5. Moreover, q = 2p . If p = 5, then the only possibility for J 5 is J ∼ = Sp4 (8), = 2F4 (2 2 ) ∈ C5 by [IA , 4.2.2], as desired. If p = 3 and J ∼ Sp (8), an allowed then with [IA , 4.2.2, 4.8.2, 4.8.4, 4.7.3A], we get J ∼ = 6 ∼ ∼ conclusion. So assume that p = 3 and I = L2 (8). Therefore, J = U3 (8), Sp4 (8) ∈ C3 , L4 (8), or G2 (8). But the last two are out because in those cases, there was a large choice of I, viz., I ∼ = L2 (64) or SU3 (8), respectively. Thus the lemma holds if f exists and CE (J) = 1. Still assuming that f exists, assume now that CE (J) = 1, so that m3 (InndiagE (J)) ≥ 3.

(7E) As J ∈ A, (7F)

J ∼ = L4 (q) with p dividing q − 1, or Sp6 (q) or Ω− 8 (q) with p dividing q 2 − 1, or U4 (q) or (S)U5 (q) with p dividing q + 1.

Accordingly there exists D ∈ E2 (E) such that ID := E(CJ (D)) ∼ = L2 (q), L2 (q), L2 (q 2 ), L2 (q), or SU3 (q). As ID /Op (ID ) = JD ∈ Cp , we have ID ∼ = ∼ L2 (8), p = 3, and q = 8, so J = Sp6 (8) or U4 (8), both allowable conclusions. We have shown that we may assume that no nontrivial field automorphism f exists in E. Thus, E induces inner-diagonal automorphisms on J. In particular, (7E) holds as mp (CE (J)) ≤ 1. If q > 2, then (7F) still holds and the previous paragraph carries through, so we may assume that q = 2. If (7F) still holds, then we have p = 3 and either J ∈ C3 , as desired, or ∼ J ∼ = Ω− 8 (2) with JD = A5 ∈ C3 . Assume finally that (7F) does not hold. Then again p = 3 by (7E), with either J ∈ C3 , as desired, or J ∼ = L6 (2), L7 (2), or Sp8 (2). The last of these is an allowed conclusion, and in the other two cases, D ∈ E2 (E) exists such that JD ∼ = L4 (2) or L5 (2), so JD ∈ C3 , contradiction. The proof is at last complete.  Lemma 7.15. Let K ∈ Cp ∩ Chev(2) for some odd prime p, and assume that mp (K) ≥ 3. Then p = 3 and K ∼ = U4 (2), U5 (2), U6 (2), Sp6 (2), D4 (2), or F4 (2). Moreover, if S ∈ Syl2 (K), then Ω1 (S) has class at least 3 and sectional rank at least 4.

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Proof. The list follows by examination of the definition of Cp [V11 , 3.1]. All the groups in the list contain a copy of U4 (2), which in turn contains a split extension of E ∼ = E24 by A ∼ = A5 , with E the core of the permutation module. Hence a Sylow 2-subgroup V of A acts freely on E, whence [E, V, V ] = 1, and the lemma follows.  Lemma 7.16. Suppose that I ∈ C3 with Aut(I) containing A6 , Z(I) = 1, and |I|3 ≤ 36 . Then I ∼ = 3J3 . Proof. Since I contains A6 , |I|3 = 3n , 3 ≤ n ≤ 6. The lemma then  follows easily from [IA , 6.1.4]. Lemma 7.17. Let K ∈ C3 , but not K ∼ = L2 (8) or L2 (3n ) for any n ≥ 2. Then there is an involution t ∈ K such that I3 (CK (t)) ⊆ Z(K). Proof. First suppose that K ∈ Chev(3) − {3A6 }. If K ∼ = 2 G2 (3 2 ) for some n > 1, then K contains Z2 ×L2 (3n ) and the result is clear. If Z(K) = 1, so that K/Z(K) ∼ = U4 (3), Ω7 (3), or G2 (3), then K/Z(K) contains a copy of GL2 (3) or GU2 (3), which implies the result. If Z(K) = 1, then since K ∼ = L2 (3n ), n ≥ 2, K has untwisted Lie rank at least 2. Hence K contains SL2 (3) and the result is again clear. Therefore we may assume that K is one of the groups in Chev(2) ∪ Spor in [V11 , 3.1]. In the sporadic cases, according to the tables [IA , 5.3], the result holds. In the Chev(2) cases, using [IA , 4.8.2, 4.7.3A], an element x ∈ I3 (K)−Z(K) is easily found such that CK (x) has even order, completing the proof.  n

Lemma 7.18. Let X be a K-group and Q ∈ Syl3 (X) ∩ Syl3 (L4 (3)). Suppose that Z(Q) ≤ Z(X) and that every component of X/O3 (X) is in C3 . Then X is 3-constrained. Proof. Without loss, O3 (X) = 1. Suppose that the lemma is false and let K1 , . . . , Kn , n > 0, be the components of X. Let Qi = Q ∩ Ki for each 3 does i. If K1Q = {K1 }, then as m3 (Q) = 4 and |Q| = 36 , |Q1 | =3. Hence  not divide | Out(K1 )|, by Lemma 3.1g. In particular, Q ∩ K1Q is a direct

factor of a maximal subgroup of Q, so |Ja (Q)| = 35 , contradicting Lemma 10.10.1a. Thus as Z(Q) ∼ = Z3 , Z(Q) ≤ Ki for every i. Using [IA , 6.1.4] and the definition of C3 [V11 , 3.1], we see that the conditions |Q/Z(Q)| = 35 and K1 ∈ C3 force K1 ∼ = J3 and Q ≤ K1 . But then Ω1 (Q) < Q [IA , 5.3h], whereas Q ∈ Syl3 (L4 (3)) so Ω1 (Q) = Q [IA , 2.4]. This contradiction completes the proof.  Lemma 7.19. Let X be a K-group with O3 (X) = 1 and Z(X) = 1. Let J ∈ C3 be a normal component of X. Suppose that P ∈ Syl3 (X) with P ∼ = Z3 ×(Z3 Z3 ) and CX (Z(P )) containing a subgroup H such that H/O3 (H) ∼ = GU3 (2), CH (J) ≤ O3 (H), and P ∩ H ∈ Syl3 (H). Then AutJ (J(P )) ∼ = Σ4 or Σ4 × Z2 .

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Proof. Our conditions imply that P = Z(X) × (P ∩ H) and CP ∩H (J) = 1. If J ∈ Alt, then P ∩ H embeds in J/Z(J), so J ∼ = A9 as J ∈ C3 . But for ∼ E33 , so Aut(A9 ) ∼ a 3-central element y ∈ I3 (A9 ), O3 (CA9 (y)) = = Σ9 does 5 not involve GU3 (2). Likewise if J ∈ Spor, then |J|3 ≤ 3 and m3 (J) = 3 or 4, with Z(J) = 1 if m3 (J) = 4. Using [IA , 5.3, 5.6.1] we see that this forces J/Z(J) ∼ = J3 ; but then Ω1 (P ) is abelian, a contradiction. Suppose that J ∈ Chev(3). If some element of P induces a non-inner automorphism on J, then the only possibility, by [IA , 4.9.1, 4.9.2] and the fact that |P | ≤ 35 , is J ∼ = L2 (33 ). But that too is impossible since it does not involve Q8 , hence not O3 (H)/O3 (H). Hence P ∩ H induces inner automorphisms on J, whence |J|3 = 34 or 35 , and the only possibility is J ∼ = P Sp4 (3). Finally, suppose that J ∈ Chev − Chev(3). As J ∈ C3 , |J/Z(J)|3 ≤ 34 , and m3 (J) − m3 (Z(J)) ≤ 3, we can use [IA , 4.10.3] to determine 3-ranks, and deduce that 1 J∼ = Sp6 (2), 3D4 (2), 2F4 (2 2 ) , or Sp4 (8). If J ∼ = Sp4 (8) then J contains Z9 × Z9 , which is not embeddable in P . If 1 ∼ J = 2F4 (2 2 ) , then | Aut(J)|3 = 33 , impossible as P ∩ H acts faithfully on J. If J ∼ = 3D4 (2), then | Aut(J)|3 = 35 , m3 (Aut(J)) = 3, while |J|3 = 4 3 and m3 (J) = 2, so a Sylow 3-subgroup of X is not embeddable in P , contradiction. The only possibilities for J are thus U4 (2) and Sp6 (2). The proof is finished by Lemma 10.9.4 and the observation that if J ∼ = Sp6 (2), then  NJ (J(P )) ∼ = Sp2 (2) Σ3 , an extension of E33 by Z2 × Σ4 . Lemma 7.20. Let K ∈ C3 . Suppose that K ≥ L ≥ P where L ∼ = U4 (2) and P ∈ Syl3 (K). Then K = L or K ∼ = Sp6 (2). Proof. Assume that K = L. Since K ∈ C3 and L is simple and contains P , K is simple. If K ∈ Chev(3), then O3 (L) = 1 by Tits’s lemma [IA , 2.6.7], a contradiction. As m3 (K) = 3 and |K|3 = 34 , K ∈ Spor [IA , 5.6.1, 5.3h]. Of the remaining groups K ∈ C3 , the only ones with m3 (K) = 3 are Sp6 (2) A9 , and the lemma follows.  and A9 . But as |A9 |/|U4 (2)| = 7 < 9, K ∼ = Lemma 7.21. Let p ∈ {3, 5} and let L ∈ Cp be simple. Suppose that mp (L) ≤ 2 ≤ mp (Aut(L)). Assume also that mr (L) ≤ 3 for all odd primes r, and, if p = 5, that m2,3 (L) < 2. Then one of the following holds: 1 2 2  3 (a) p = 3 and L ∼ = L2 (8), A6 , L± 3 (3), M11 , F4 (2 ) , D4 (2), or Sp4 (8); or 5 1 5 2 2 2  2 (b) p = 5 and L ∼ = 2B2 (2 2 ), L2 (52 ), L± 3 (5), F4 (2 ) , or F4 (2 ). Proof. If L ∈ Chev(p), then the condition mp (L) ≤ 2 yields L ∼ = L2 (p), 2 G (3 12 ) ∼ L (8) (for p = 3 only), L (p2 ), or L± (p) [I , 3.3.3]. Then L (p) = 2 2 2 2 A 3 is eliminated as mp (Aut(L2 (p))) = 1. So the lemma holds in this case. If L ∼ = Akp , then k = 2 as mp (L) ≤ 2; but if p = 5, then m2,3 (A10 ) ≥ m2,3 (Z3 × A4 ) = 2, contradiction. So p = 3 and the lemma holds again.

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Suppose that L ∈ Spor. If p = 5, then the condition m2,3 (L) < 2 forces ∼ L = M11 or J1 [IA , 5.6.2]; but then m5 (L) = 1 [IA , 5.6.1], contradiction. So p = 3 and as | Out(L)| ≤ 2, m3 (L) = 2. Using the definition of C3 [V11 , 3.1] and the known ranks [IA , 5.6.1] yields just L ∼ = M11 , as asserted. Finally, suppose that L ∈ Chev(2) − Chev(p). Among the few groups of this type in Cp [V11 , 3.1], we note that m3 (U5 (2)) = m3 (U6 (2)) = m3 (D4 (2)) = m3 (F4 (2)) = 4 and m3 (Sp6 (2)) = 3, ruling these out for p = 3. The remaining groups are listed in this lemma, completing the proof.  Lemma 7.22. Suppose (p, K) = (3, 3D4 (2)), (3, Sp4 (8)), (3, U3 (8)), or 5 (5, 2F4 (2 2 )). Let A ∈ Ep2 (K), let a ∈ A# and let L be a component of CK (a). (Thus, if K ∼ = 3D4 (2) or U3 (8), assume that E(CK (a)) ∼ = L2 (8).) Then the following conditions hold: (a) CAut(K) (A) is a p-group; (b) ↑p (K) ∩ Cp = ∅; and (c) CAut(K) (L a) embeds in Zp2 . Proof. First consider (a), for which it suffices to show CInndiag(K) (A) to be a p-group. For then a Sylow r-subgroup R of CAut(K) (A) for a prime r = p embeds in O p (Out(K)/ Outdiag(K)). Hence R can only be nontrivial if K ∼ = Sp4 (8) or U3 (8), p = 3, r = 2, and R = g, where g is a graph-field 8 or graph automorphism, respectively. But then CK (g) ∼ = 2B2 (2 2 ) or L2 (8) so m3 (CK (g)) ≤ 1 and [A, g] = 1, a contradiction. According to our hypothesis, and [IA , 4.7.3A, 4.8.2, 4.8.7], and Lemma 10.5.8, there is a ∈ A# such that E(CK (a)) = 1, and for any such a, 5 CInndiag(K) (a) ∼ = Z3 × L2 (8), Z9 × L2 (8), Z9 × L2 (8), or Z52 × 2B2 (2 2 ), in the four respective cases. Then (using [IA , 6.5.4] in the case p = 5), CK (A) ∼ = Z3 × Z9 or Zp2 × Zp2 , proving (a). Moreover, (c) follows as well. For (b), suppose that K ↑p L ∈ Kp via x ∈ Ip (Aut(L)). As U3 (8) ∈ C3 , we may ignore the case K = U3 (8). By surveying [IA , 5.3], we see that L ∈ Spor. Now K ∈ Chev(2) unambiguously, so by [IA , 7.1.10], L ∈ Chev(2). If x is a field, graph-field, or graph automorphism, then from[IA , 4.9.1, 4.9.2] 52 we conclude that L ∼ = D4 (8), Sp4 (83 ), or 2F4 (2 2 ), none of which lie in Cp . So we may assume that x ∈ Inndiag(L). Suppose that L is of exceptional type. If p = 3, we see from [IA , 4.7.3A] that L ∼ = 2E6 (2) ∈ C3 . If p = 5, then by [IA , 4.2.2], the untwisted Dynkin diagram of L must contain the F4 -diagram as a proper subdiagram, which is impossible. So L is a classical group, whence K is as well, by [IA , 4.8.2, 4.8.4]. Thus p = 3 and K ∼ = Sp4 (8). Again by [IA , 4.2.2], the untwisted Dynkin diagram of L must have a double bond, so L ∼ = Sp2n (q) for some n and some even q. If L ∈ C3 , the only possibility is L ∼ = Sp6 (2). But by [IA , 4.8.2], it is false that Sp4 (8) ↑3 Sp6 (2). This completes the proof of (b). 

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Lemma 7.23. Let p ∈ {3, 5}. Let J ∈ Cp and x ∈ Ip (Aut(J)), and let L  be a component of Op (E(CJ (x))). Let S ∈ Sylp (L) and suppose that S ≤ S1 where S1 is a noncyclic p-group acting on J and L. If p = 3, suppose that 5 2 2 ∼ 2 L∼ = L2 (8), A6 , M11 , or L± 3 (3). If p = 5, suppose that L = L2 (5 ), B2 (2 ), 1 HJ, HS, He, Ru, or 2F4 (2 2 ) . Then one of the following holds: (a) L  L, ΓS1 ,1 (J); (b) m  2,p (J) ≥ 3; or (c) m2,r (J) > 3 for some odd prime r not dividing the order of the Schur multiplier of J. Proof. Suppose that J ∈ Chev(p). Then L ∈ Chev(p) so (L, J) ∼ = 3

(L2 (8), 2 G2 (3 2 )), (U3 (3), U3 (33 )), or (L2 (p2 ), L2 (p2p )). In the first two cases by [IA , 4.5.1], (b) holds. In the third case, ΓS,1 (J) contains a Sylow psubgroup of J, which does not normalize L, so (a) holds. Likewise if J ∈ Alt, then L ∈ Alt so (L, J) ∼ = (A6 , A9 ) with p = 3. But then J = ΓS,1 (J) so (a) holds. Next, suppose that J ∈ Spor. Using [IA , 5.3] we see that (J, L) ∼ = (Co3 , L2 (8)) or (Suz, A6 ) if p = 3; (Co1 , HJ) or (F2 , HS) if p = 5. Using  2,3 (J) = 3, so (b) holds; and in [IA , 5.6.2], we see that in the p = 3 cases m the p = 5 cases m2,3 (J) > 3, so (c) holds with r = 3 (see [IA , 6.1.4]). Hence, we may assume that J ∈ Chev − Chev(p). In particular L ∈ Chev(s) for some prime s = p. If J = ΓS1 ,1 (J), then (a) obviously holds, so assume that J > ΓS1 ,1 (J). Then by [IA , 7.3.4], J ∈ Chev(2), so L ∈ 5 1 Chev(2). If p = 5, then L ∼ = 2B2 (2 2 ) or 2F4 (2 2 ) . Hence, using [IA , 4.2.2], 5 52 J ∼ = 2F4 (2 2 ) or 2B2 (2 2 ), so J = ΓS1 ,1 (J) by [IA , 7.3.4], contradiction. So we may assume that p = 3. We may also assume that m  2,3 (J) ≤ 2. As J ∈ C3 ∩ Chev(2) − Chev(3), we compare the list of possible J’s in [V11 , 3.1] with the groups in [IA , 7.3.4]. We conclude that (J, L) = (Sp6 (2), A6 ). But then by [IA , 7.3.3], ΓS,1 (J) ∼ = O6− (2) is a maximal subgroup of J, so it cannot normalize L. Hence (a) holds and the proof is complete.  Lemma 7.24. Let K = A6 , M11 , or L2 (8) and B ∈ E32 (Aut(K)). Then B = CAut(K) (B). Proof. Let b ∈ B # , with b a field automorphism in the case K = L2 (8). Then in every case (see [IA , 4.9.1, 5.3a]), CAut(K) (b) embeds in Z3 × Σ3 , whence the lemma.  Lemma 7.25. Let X be a K-group and Q ∈ Syl2 (X), where Q is a Sylow 5 2-subgroup of L2 (8), L2 (32), or 2B2 (2 2 ). Accordingly set p = 3, 5, or 5, and assume that Op (X) = 1 and E(X) = 1. Suppose that NX (Q) − CX (Q) contains an element b of order p. Suppose also that every component of 5 E(X) lies in Cp . Then E(X) ∼ = L2 (8) or 2B2 (2 2 ) according as p = 3 or 5. Proof. Let K be a component of E(X). Since Ω1 (Q) = Z(Q) and 5 K ∈ Cp , K ∼ = L2 (8) if p = 3, and K ∼ = A5 or 2B2 (2 2 ) if p = 5. It suffices

8. Tp -GROUPS AND TGp -GROUPS

473

therefore to rule out A5 when p = 5. But in that case, since m2 (Q) = 5, b normalizes K and hence centralizes Q ∩ K ∼ = E22 . However, this forces b to centralize Ω1 (Q) and then Q, contrary to hypothesis.  Lemma 7.26. Let F ∗ (X) = K ∼ = U4 (2), L± 4 (3), or G2 (3). Suppose that ∼ ∼ Q ∗ Z1  QZ1 B0 ≤ X with Q = [QZ1 , B] ∼ = 21+4 + , Z1 = Z4 , B0 = E32 and QB0 ∼ = U4 (3) and there is u ∈ I2 (X) such that = SL2 (3) ∗ SL2 (3). Then K ∼ ∼ CK (u) = U3 (3). Proof. Write Z(Q) = z. Since QB0 ∼ = SL2 (3) ∗ SL2 (3), it is immediate from [IA , 4.5.1] that QB0 = O2 (CAut(K) (z)), and moreover in every case, CK (QB0 ) = z. Suppose K ∼ = G2 (3). Still using [IA , 4.5.1], we have that both solvable components of QB0 are normal in CK (z), being a long root SL2 (3) and a short root SL2 (3). As outer automorphisms of K interchange long and short roots, CAut(K) (z) = CK (z) t where t interchanges the two root SL2 (q)’s of QB0 . Therefore CAut(K) (QB0 ) = z, as claimed in this case. The argument is similar if K ∼ = Ω5 (3); here = U4 (2) ∼ + + ∗ ∼ ∼ CAut(K) (z) = Ω4 (3), and F (Ω4 (3)) = Q. Suppose next that K ∼ = L4 (3), in which case Aut(K) = P GL4 (3) γ, where γ is a graph automorphism. We may choose γ so that E(CK (γ)) ∼ = P Sp4 (3) [IA , 4.5.1], and indeed then so that [QB0 , γ] = 1. However, CP GL4 (3) (z)/ z ∼ = P GL2 (3) Z2 , so CAut(K) (QB0 ) = z, γ ∼ = E22 has no element of order 4. Therefore K ∼ = U4 (3). Then CK (z) = QB0 t, t  where t induces an outer automorphism on each solvable component of QB0 , and t interchanges those solvable components. So a generator a ∈ Z1 has an involutory image a ∈ Out(K). If a ∈ Outdiag(K), then from [IA , 4.5.1], we see that the preimage of a in Aut(K) contains an involution u such that E(CK (u)) ∼ = U3 (3), as desired. Otherwise, let a ∈ Q have order 4. Then t := aa is an involution in an Inn(K)-coset labelled g or g  in [IA , 4.5.1]. Let C = CK (t). Then z ∈ C and CC (z) has a single solvable component, and it contains z. In   particular, z ∈ O 3 (CC (z)). But O3 (C) ∼ = P Sp4 (3), SL2 (3) ∗ SL2 (3), or A6 , none of which contains such an involution z. This contradiction completes the proof.  8. Tp -Groups and TGp -Groups Recall that we are using the following notation. TGp (G) = Lop (G) ∩ TGp Tp (G) = Lop (G) ∩ Tp Tpi (G) = Lop (G) ∩ Tpi CTGp = Cp ∪ Tp ∪ TGp Lemma 8.1. If K ∈ Tp , p an odd prime, then mp (K) ≤ 2. Proof. This is immediate from the definition of Tp [I2 , 13.1] and [IA , 4.10.3, 5.6.1]. 

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Lemma 8.2. Let K ∈ Tp , p an odd prime, and let a ∈ Ip (Aut(K)). Then L (CK (a)) is quasisimple or trivial. p

Proof. By the (Bp )-property, all p-components of Lp (CK (a)) are quasisimple. Suppose the lemma is false. Then using [IG , 16.11] we see that mp (K) ≥ 2, and indeed either mp (K) ≥ 3, contradicting Lemma 8.1, or K is simple and a is non-inner. In that case K ∼ = L3 (q), q ≡  (mod 3),  = ±1, with p = 3. Hence a is a diagonal automorphism or a field automorphism, and CK (a), as given by [IA , 4.8.2, 4.8.4, 4.9.1], has no more than one pcomponent.  Lemma 8.3. Let p be an odd prime and let the sets Tpi , 1 ≤ i ≤ 5, be defined by [V12 , Definition 1.1]. Suppose that (x, K), (y, L) ∈ ILp (G) and (x, K) < (y, L). If L/Op (L) ∈ Tpi for some i, then K/Op (K) ∈ Cp or K/Op (K) ∈ Tpj for some j ≥ i. Proof. Set K = K/Op (K). We may assume that Op (L) = 1. We freely use the definitions of Cp and Tpi ([V11 , 3.1], [V10 , 1.1]). Notice that if i ≤ 4, then mp (L) > 1. If i = 5, then mp (L) = 1, whence mp (K) = 1. Hence K ∈ Cp ∪ Tp5 , as desired. If i = 4, then L ∼ = L3 (4) or F i22 ; thus mp (K) = 1 5 and again K ∈ Cp ∪ Tp . We may now assume that p = 3. If i = 3, then by [IA , 4.8.2, 4.8.4], K ∼ = L2 (q) for some q ≡ 0 (mod 3), so mp (K) = 1, yielding the desired result. If i = 2, the only additional possibility for K is L3 (q) for some q ≡  (mod 3),  = ±1, in which case K ∈ T32 ∪ T33 . Finally, similar reasoning applies if i = 1 unless (x, K) < (y, L) is a diagonal pumpup, in which case the lemma holds with j = 1, 2, 3, or 4.  Lemma 8.4. If K ∈ Tp and Z(K) = 1, then (a) mp (Aut(K)) ≤ 1 + mp (K); and (b) If K ∈ Tp3 ∪ Tp4 or K ∼ = HJ, then mp (Aut(K)) = mp (K). Proof. If K ∈ Tp5 , (a) holds by Lemma 3.1a. If K ∼ = L3 (q), q ≡  (mod 3),  = ±1, then by [IA , 4.10.3ab], m3 (K) = 2 = m3 (Inndiag(K)). As Out(K)/ Outdiag(K) is generated by images of ΦK and ΓK [IA , 2.5.12], m3 (Out(K)/ Outdiag(K)) ≤ 1, with equality if and only if q is a cube. Hence (a) and (b) hold in this case. If K ∈ Alt ∪ Spor, then Out(K) is a 2-group, so (a) and (b) hold. The only remaining case is K ∼ = G2 (8) with p = 3, in which case Out(K) is cyclic, consisting of images of field automorphisms. Hence (a) holds, and the proof is complete.  Lemma 8.5. Let p be an odd prime. Let K ∈ Tp with mp (K) = 1 or K ∼ = G2 (8) (with p = 3). Suppose that K  X, mp (CX (K)) ≥ 2, and mp (X) > mp (KCX (K)). Let T1 ∈ Sylp (X) and Q = CT1 (K). Let p U ∼ = Ep2 with U ≤ Q and U  T1 . Then there is B ∈ E∗ (T1 ) of the

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475

form B = B1 E b, where U ≤ E ∈ Ep∗ (Q), B1 ∈ Ep∗ (K), and b induces a nontrivial field automorphism on K. Moreover, Lp (CK (b)) = 1. Proof. Following [V9 , 5.11], choose B ∈ Ep∗ (T1 ) such that U ≤ B. Note that mp (Out(K)) ≤ 1, with any noninner automorphism of order p being a field automorphism, by Lemma 3.1ad and [IA , 2.5.12]. As Op (K) = 1, and by assumption, mp (B) = mp (X) ≥ mp (KCX (K)) + 1 ≥ mp (K) + 1 + mp (Q) ≥ mp (Aut(K)) + mp (Q) ≥ mp (X). Hence equality holds throughout. It follows that mp (B ∩ K) = mp (K), mp (B ∩ CX (K)) = mp (CX (K)), and some b ∈ B # induces a nontrivial outer automorphism, hence a field automorphism, on K. Finally, if 1 Lp (CK (b)) = 1, then CK (b) ∼ = L2 (2), L2 (3), or 2B2 (2 2 ) (with p = 3, 3, or 5 5, respectively), so K ∈ Tp by assumption. But L2 (8), L2 (33 ), and 2B2 (2 2 )  lie in Cp , contradiction. The proof is complete. Lemma 8.6. If K ∈ Tp , p an odd prime, then mp (Aut(K)) ≤ 3. Moreover, if equality holds, then p = 3 and every E33 -subgroup of Aut(K) contains a field automorphism of order 3 with K/Z(K) ∼ = L3 (q),  = ±1, q ≡  (mod 3). Proof. By Lemma 3.1a we may assume mp (K) ≥ 2. As Out(F i22 ) ∼ = Z2 by [IA , 5.3], with m5 (F i22 ) = 2, we may assume that p = 3 and K ∼ = L3 (q), q ≡  (mod 3),  = ±1. As m3 (P GL3 (q)) = 2, and K admits no graph automorphism of order 3, the lemma follows from [IA , 4.9.1].  Lemma 8.7. Let K ∈ T31 with Z(K) = 1. Let X be a group with O3 (X) = 1, K  X, m3 (X) ≥ 4, and m3 (CX (K)) = 2. Let T ∈ Syl3 (X) and B1 ≤ T ∩ K with B1 ∼ = E32 and B1  T . Then the following conditions hold: (a) K ∼ = SL3 (q),  = ±1, q ≡  (mod 3); and (b) If t ∈ I3 (T ) induces a field automorphism of order 3 on K, then there is t1 ∈ I3 (CT (B1 )) ∩ Kt. Proof. If (a) fails, then K ∼ = 3A6 , 3A7 , or 3M22 , and so Out(K) is Hence T = (T ∩ K)CT (K), with T ∩ K ∼ a = 31+2 . Let X = 3 X/CX (K), and let A ∈ E4 (X). As m3 (CX (K)) = 2, A = T ∩ K ∼ = E32 and m3 (CA (K)) = 2. Let a ∈ A with a = 1. Thus a = bc with b ∈ T ∩ K − Z(K) and c ∈ CT (K). Then 1 = [CA (K), a] = [CA (K), c]. But m3 (CT (K)) = 2, so c ∈ CA (K), whence b ∈ A. Letting a vary we find in this way that T ∩ K ≤ A, which is absurd as A is abelian. Hence (a) holds. In (b), q is a cube, so q ≡  (mod 9). Therefore there is a frame F preserved by T ∩ K = H w, where H is diagonal with respect to F and w cycles the three lines constituting F, with w3 = 1. Moreover, |CH (w)| = 3, so T ∩ K has maximal class and a unique normal E32 -subgroup, which is B1 . We choose an (orthogonal) basis of the natural module cycled by w and with 3 -group.

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respect to which H is diagonalized, and use it to coordinatize T ∩ K. Then NAut(K) (T ∩K) contains a field automorphism τ1 of order 3 of the base field, which centralizes B1 . Let τ be the image of t in Aut(K). Then by [IA , 4.9.1],

τ  and τ1  are Inndiag(K)-conjugate. But Sylow 3-subgroups of Out(K) are abelian, being extensions of the central subgroup Outdiag(K) by a cyclic group of field automorphisms. So, replacing τ by τ −1 if necessary, we may assume that τ ≡ τ1 (mod Inn(K)). Let N = NInn(K) (B1 ), so that τ ≡ τ1 (mod N ). Therefore there is S1 ∈ Syl3 (NKt (B1 )) containing an element s1 of order 3 mapping on τ1 . As S1 is NKt (B1 )-conjugate to (T ∩ K) t, (T ∩ K)t contains a conjugate t1 of s1 satisfying the desired conditions. This completes the proof.  Lemma 8.8. Suppose K ∈ CTG3 , Z(K) = 1, and m3 (K) − m3 (Z(K)) ≤ 1. Then K ∼ = SL3 (q), q ≡  (mod 3),  = ±1, or 3A6 , 3A7 , or 3M22 . Moreover, if T ∈ Syl3 (K), then CAut(K) (T ) is a 3 -group. Proof. We use freely the information about Schur multipliers in [IA , 6.4.1]. If K ∈ Alt ∪ Spor, then the first assertion holds, with the help of [IA , 5.6.1]. If K/Z(K) ∼ = U4 (3) or Ω7 (3), then m3 (K) − m3 (Z(K)) = m3 (K/Z(K)) > 1 by [IA , 6.4.4]. If K/Z(K) ∼ = G2 (3), then Z(K) ∼ = Z3 , and if P/Z(K) is the product of the two highest long root subgroups of K/Z(K), then AutK (P/Z(K)) ∼ = GL2 (3). Hence P splits over Z(K) and m3 (K) ≥ 3. Hence these cases lead to a contradiction, so we may assume for the first assertion that K ∈ Chev − Chev(3). Hence K is a quotient of SL3k (q) or E6 (q), q ≡  (mod 3),  = ±1, k ≥ 1. Since m3 (E6 (q)) ≥ m3 (SL6 (q)) = 5, the first assertion holds. The last assertion is proved in Lemma 12.1.  9. C2 -Groups Lemma 9.1. Let K ∈ Co2 ∩ A be simple. Then (a) m3 (K) = 1 if and only if K ∼ = L2 (2n ), n > 1; L3 (2n ), n > 1, n 2 ≡ − (mod 3); or L2 (17); and (b) If m3 (K) = 2 and K ∈ Chev(2), then K ∼ = L3 (3), a Mathieu group, HJ, J4 , HS, or Ru. Proof. The 3-ranks of the groups in Co2 are determined in [IA , 3.3.3, 4.10.3, 5.6.1, 6.5.1]. This yields the lemma.  Lemma 9.2. Let K ∈ C2 and suppose that mr (K) ≤ 3 for all odd primes r. Then the isomorphism type of K and the r-rank of K for various primes r are as listed in Table 9.1. Moreover, for every odd prime r, one has: mr (Aut(K)) ≤ mr (K) + 1. The notation a± ∈ N with a ≥ 1 stands for a+ :=

3+(−1)a 2

and a− :=

3−(−1)a . 2

Proof. If K ∈ Spor, the assertions of the lemma hold by the r-ranks given in [IA , 5.6.1] and the fact that Out(K) is a 2-group. If K ∈ Chev(r) − Chev(2) for some odd r, or K ∼ = L2 (9) or G2 (2) ∼ = U3 (3), then as = B2 (2) ∼

9. C2 -GROUPS

Table 9.1. The ranks of K K m2 (K) m3 (K) mr (K), r ≥ 5 L2 (q), q ∈ F M 2 1 ≤1 L2 (9) ∼ 2 2 ≤1 = P Sp4 (2) ∼ = A6 L3 (3) 2 2 ≤1 U3 (3) ∼ 2 2 ≤1 = G2 (2) L2 (2a+2 ) a+2 1 ≤1 L3 (2a ) 2a a+ ≤2 U3 (2a ) a a− ≤2 L4 (2a ) 4a 1 + a+ ≤3 a U4 (2 ) 4a 1 + a− ≤3 L5 (2) 6 2 ≤1 U5 (4) 8 2 ≤3 L6 (2) 9 3 ≤2 L7 (2) 12 3 ≤2 Sp4 (2a+1 ) 3(a + 1) 2 ≤2 Sp6 (2a ) 6a 3 ≤3 a Ω− 6a 3 ≤3 8 (2 ) 2B (2 2a+1 2 ) 2a + 1 0 ≤1 2 G2 (2a+1 ) 3(a + 1) 2 ≤2 2F (2) 5 2 ≤2 4 2F (2 2a+1 2 ) 5(2a + 1) 2 ≤2 4 3D (2a ) 5a 2 ≤2 4 2L3 (4) 5 2 ≤1 E22 L3 (4) 6 2 ≤1 2Sp6 (2) 4 3 ≤2 3 2 2 B 2 (2 2 ) 4 0 ≤1 3 2 E22 B 2 (2 2 ) 5 0 ≤1 2G2 (4) 5 2 ≤2 M11 2 2 ≤1 M12 /2M12 3/4 2 ≤1 M22 /2M22 4/5 2 ≤1 4M22 4 2 ≤1 M23 4 2 ≤1 M24 6 2 ≤1 HJ/2HJ 4/3 2 ≤2 J3 4 3 ≤1 J4 11 2 ≤2 HS/2HS 4/5 2 ≤2 Ru/2Ru 6/7 2 ≤2

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∼ L2 (q), q ∈ F M9, as listed, or K/O2 (Z(K)) ∼ K ∈ C2 , either K = = L± 3 (3), ± L4 (3), or G2 (3), the last two of which have 3-rank 4 and so are excluded. The ranks in these cases follow from [IA , 6.5.1, 6.5.3, 3.3.3], and again Out(K) is a 2-group [IA , 2.5.12]. As Alt ∩ C2 ⊆ Chev(2), we may then assume that K ∈ Chev(2). We use [IA , 4.10.3, 3.3.3] for r-ranks, for r odd and r = 2 respectively (and 1 [IA , 3.3.2d] for 2F4 (2 2 ) ). We may assume that K/Z(K) ∈ Lie(2). Clearly e(K) ≤ 3, so K/Z(K) ∈ A, an initial restriction. But K = D4 (2), B4 (2), and F4 (2) are excluded as m3 (K) = 4. Also K = U5 (2n ) is excluded for n = 2, since mr (K) = 4 for any prime r = 5 dividing 2n + 1. These remarks, and [IA , 6.1.4] for Schur multipliers, determine which K ∈ Chev(2) are listed. Finally, mr (Aut(K)) ≤ mr (K) + 1 for odd primes r unless possibly mr (Out(K)) > 1. But in that case, given the possible K’s, we must have r dividing | Out(K)| and q(K) a power of 2r . The only possibility is r = 3, K = L3 (2a ),  = ±1, 2a ≡  (mod 3). But then m3 (K) = 2 = m3 (Inndiag(K)), and as m3 (Out(K)/ Outdiag(K)) ≤ 1, again m3 (Aut(K)) ≤ m3 (K) + 1. The proof is complete.  Lemma 9.3. Let J ∈ C2 with Z(J) = 1 and mr (J) ≤ 3 for all odd primes r. Suppose that t ∈ I2 (Aut(J)) with CJ (t) a {2, 3}-group. Then J/Z(J) ∈ {M12 , M22 , Sz(8), L3 (4), Sp6 (2), G2 (4)}. Moreover, if J/Z(J) ∼ = L3 (4), then Z(J) has exponent 2. Proof. As J ∈ C2 with Z(J) = 1 and mr (J) ≤ 3 for all odd r, Lemma 9.2 reduces us, if the lemma fails, to J ∼ = 2HJ, 2HS, 2Ru, or an exponent 4 covering group of L3 (4). The last of these is not in C2 [V11 , 3.1], and the others do not have an involutory automorphism t such that CJ (t) is a  {2, 3}-group [IA , 5.3]. The lemma follows. Lemma 9.4. Let K ∈ C2 . Then I2 (K) ⊆ Z(K). Proof. Suppose that I2 (K) ⊆ Z(K). Thus Z(K) = 1. If Z(K) is cyclic then Sylow 2-subgroups of K are quaternion, whence K ∼ = SL2 (q) or 2A7 for some odd q. But then K ∈ C2 [V11 , 3.1]. So suppose that Z(K) is noncyclic. From [IA , 6.1.4] and [V11 , 3.1], we have Z(K) ∼ = E22 and 2B (2 23 ), D (2), U (2), or 2E (2). In every case there is K/Z(K) ∼ L (4), = 3 2 4 6 6 x ∈ I3 (Aut(K)) − Inn(K) acting transitively on Z(K)# : a diagonal, field, graph, diagonal, or diagonal automorphism, respectively. Moreover, x can be chosen so that CK/Z(K) (x) has even order, indeed CK/Z(K) (x) ∼ = L2 (4), 2B

1

G2 (2), U5 (2), or 2 D5 (2) [IA , 4.8.2, 4.9.1, 4.7.3A]. Then the preimage  in K of any involution of CK/Z(K) (x) is necessarily in I2 (K). 2 (2 2 ),



Lemma 9.5. Let K = O 3 (K) ∈ Co2 with Z(K) = 1 and e(K) ≤ 3. Suppose that b ∈ I3 (Aut(K)) and set H = CK (b). Suppose also that (1) O3 (H) is abelian; (2) m3 (O3 (H)) ≤ 2; and (3) H is solvable.

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Then one of the following holds: (a) H has a chief factor of order 4; or (b) K/Z(K) ∼ = Z2 or E22 . = L3 (4) and Z(K) ∼ Proof. The possible isomorphism types of K are those listed in Lemma 3 9.2 such that Z(K) = 1, excepting the 3 -case K/Z(K) ∼ = 2B2 (2 2 ). Now, the case K/Z(K) ∼ = Sp6 (2) is ruled out by condition (3), (2), or (1), according as dim[V, b] = 1, 2, or 3, where V is the natural F2 [K/Z(K)]-module [IA , 4.8.2]; K/Z(K) ∼ = G2 (4) is ruled out by condition (3) [IA , 4.7.3A]; K/Z(K) ∼ = L3 (4) satisfies conclusion (a); K/Z(K) ∼ = M12 satisfies conclusion (b) unless condition (1) fails; and the remaining groups, all sporadic, satisfy conclusion  (b) or fail conclusion (3) [IA , 5.3]. The lemma is proved. Lemma 9.6. Suppose that O2 (X) = 1, O2 (X) = 1, and every component of E(X) is a Co2 -group. Let Q ∈ Syl2 (X). If Q/ z embeds in D8 for some z ∈ I2 (Z(Q)), then m5 (X) ≤ 1. Proof. Suppose A ≤ X with A ∼ = E52 . Since Q has sectional rank at most 3, E(X) =: J is quasisimple and [O2 (X), A] = 1. Thus, CA (J) = 1. As O2 (X) = 1, (Q ∩ J)/Z(J) ∼ = E22 , E23 , or D8 . Hence by [IA , 4.10.5, 5.6.3,5.6.4, 2.4, 5.3a], J/Z(J) ∼ = L2 (q), q odd, L2 (8), or A7 . As J ∈ Co2 , J ∼ = L2 (q), q ∈ {5, 7, 9, 17}, or L2 (8). In every case, | Aut(J)|5 ≤ 5, so the lemma is proved.  Lemma 9.7. Suppose that K ↑2 J via x ∈ I2 (Aut(J)), with J ∈ C2 and mr (J) ≤ 3 for all odd primes r. Then one of the following holds: (a) K is semirigid in J; or 3 (b) (J/Z(J), K) ∼ = (M12 , A5 ), (HJ, A5 ), or (Ru, 2B2 (2 2 )). Proof. Suppose first that J ∈ Chev(2). If x is a field or graphfield automorphism, then x = CAut(J) (K) by [IA , 7.1.4c], so semirigidity holds by definition [IG , 7.1]. Otherwise, by the Borel-Tits theorem, x is a graph automorphism and J and K are given by [IA , 4.9.2]. By L2 -balance and the Borel-Tits theorem, CInn(J) (K) has odd order. Let R ∈ Syl2 (CAut(J) (K x)). If R = x, then x = R ∈ Syl2 (CAut(J) (K)) and semirigidity holds, so we may assume that R > x. But whether J is a twisted group or not, R/ x faithfully induces field automorphisms on K, so R = x and the lemma holds in this case. So we may assume that J ∈ Chev(2). If J ∼ = L2 (q), q ∈ F M9, then K ≤ L2 (CJ (x)) = 1, which is absurd. If J ∈ Chev(3), then as m3 (J) ≤ 3, J ∼ = L± 3 (3), and we reach the same contradiction. Finally, suppose that J ∈ Spor. We use [IA , 5.3, 5.6.1] without comment. Without loss, Z(J) = 1. The condition mr (J) ≤ 3 for all odd r, and the condition that E(CJ (x)) = 1, narrow the possibilities down to J = M12 , HJ, Ru, J3 , and HS. Note that in these cases, if x ∈ Inn(J), then either

x = CAut(J) (K), whence semirigidity holds, or (J, K) = (M12 , A5 ). So we

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∼ may assume that x ∈ Inn(J), whence J = J3 . If J is one of the groups in (b), ∼ then obviously (b) holds. If J = HS, then K ∼ = A6 and CInn(J) (K) = x; moreover, R := CAut(J) (K) = x, y where y ∈ I2 (Aut(J)) and CJ (y) ∼ = Σ8 , so again semirigidity holds (with Q = y in [IG , Def. 7.1]). This completes the proof.  Lemma 9.8. Let K ∈ Chev(2)∩C2 with mp (K) = 1 for some odd prime p. Suppose that K  X and x ∈ Ip (X) induces a nontrivial field automorphism on K. Then x centralizes some 2-central involution of X in K. Proof. If Z(K) = 1, then by [IA , Table 6.3.1], the assertion is obvious 3 since x centralizes Z(K) (note that K/Z(K) ∼ 2B2 (2 2 ), since mp (K) = = 1 by assumption). So assume that K is simple. Then we may assume that F ∗ (X) = K. By Lemma 3.1, Outdiag(K) is a {2, p} -group. But the image of x in Out(K)/ Outdiag(K) is central [IA , 2.5.12], so x normalizes a Sylow 2-subgroup T of X and [T, x] ≤ T ∩ K. It therefore suffices to show that CZ(T ∩K) (x) is nontrivial, since it is (T ∩ K)CT (x) = T -invariant. But Z(T ∩ K) contains a highest root group U , and CU (x) = 1, completing the proof.  Lemma 9.9. There is no K ∈ A satisfying the following conditions: (a) K ∈ C2 ; (b) m3 (K) = 3; and (c) K contains a Frobenius subgroup F of order 33 .13. Proof. Consider the various cases of Def. 1.1. Our hypothesis (b) rules out the groups in (a)–(d) of that definition (see [IA , 3.3.3, 4.10.3] for ranks) except for P Sp4 (3) and L4 (p), p ≡  (mod 3),  = ±1. We will consider P Sp4 (3) in case (g), and L4 (p) ∈ C2 . The alternating groups in case (e) violate our hypothesis (c), and the only sporadic group in case (f) which is in C2 and has 3-rank 3 is J3 , which violates our hypothesis (c) (see [IA , 5.6.1]). So we may assume that K ∈ Chev(2). Let m be the smallest degree of a faithful characteristic 2 representation of K. By our hypothesis (c), m ≥ 13. This rules out all the groups in cases (g), (h), and (i) of Def. 1.1, except 1 certain groups U5 (2n ), U6 (2), 2F4 (2 2 ) , and F4 (2). But these last groups (and most of the others in (g)–(i), for that matter) violate our hypothesis (b). This completes the proof.  Lemma 9.10. There exists no I ∈ K with the following properties: (a) I ∈ C2 ; (b) m3 (I) ≤ 3; (c) Aut(I) has a subgroup H with F ∗ (H) ∼ = Z9 × Z9 and H/F ∗ (H) ∼ = GL2 (3); and (d) I has a subgroup isomorphic to M12 . Proof. Suppose that I ∈ C2 − Chev(2). Conditions (b) and (c) prevent I∼ = L2 (q), q ∈ F M9, as well as I ∈ Chev(3). Thus I ∈ Spor, and conditions (b) and (c) again give a contradiction [IA , 5.6.1, 5.3].

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Thus, assume that I ∈ Chev(2). Without loss, I is simple. By Lemma 20.9, Out(I) has a normal 2-complement, so in (c), O3,2 (H) ≤ Inn(H). As O3,2 (H) is a Frobenius group with complement of order 8, I has no characteristic 2 representation of degree less than 8. Combined with condition (b) and [IA , 4.10.3ab], this reduces us to considering I ∼ = L3 (q), q ≡  (mod 3), n  = ±1, 3D4 (q), 2F4 (2 2 ) , and 2 D4 (q). By (d), I contains a subgroup isomorphic to Z2 × Σ5 . Hence by the Borel-Tits theorem, the Levi factor L0 of some (proper) parabolic subgroup of I contains a Σ5 -subgroup. In all but the 2 D4 (q) case, however, I has  twisted rank at most two, so O2 (L0 ) ∈ Lie has twisted rank 1 and hence does not contain Σ5 , contradiction. Thus, we may assume that I ∼ = 2 D4 (q) contains M12 and hence a Frobenius group F of order 11.5. Hence F embeds in I ∗ := Sp8 (q). On the natural I ∗ -module V , d := dim[V, O11 (F )] is a multiple of 5, but also is even as [V, O11 (F )] is nondegenerate. Hence d ≥ 10 > dim V , which is absurd. The lemma is proved.  Lemma 9.11. There is no group L ∈ C2 − Chev(2) satisfying all the following conditions for some odd prime p: (a) Q∗ = Q×P ∈ Sylp (L) with P cyclic and Q isomorphic to a Sylow psubgroup of a twisted rank 1 group K ∈ Chev(p) such that mp (Q) ≥ 3; and (b) ΓQ,1 (L), ΓQ∗ ,2 (L) ≤ NL (Q). Proof. Suppose for a contradiction that L exists. Since mp (Q) ≥ 3, ∼ L = L2 (q), q ∈ F M9. If L ∼ = L3 (3), G2 (3), or L± 4 (3), then as L has twisted Lie rank at least 2, it is generated by its parabolic subgroups containing Q∗ , so L = ΓQ∗ ,2 (L) ≤ NL (Q), contradiction. The only other possibility, by definition of C2 [V11 , 3.1], is that L ∈ Spor ∩ C2 . Using the tables [IA , 5.3], most cases are ruled out by the conditions that mp (Q∗ ) ≥ 3 and that for every x ∈ Ip ([Q∗ , Q∗ ]) (in particular for some x ∈ Z(Q∗ ) if Q∗ is nonabelian), mp (CL (x)/Op (CL (x))) ≤ 1. The remaining cases are (L, p) ∼ = (Co2 , 3), (Co1 , 5), (F3 , 3), (F2 , 5), and (F1 , 7). In these, there is a subgroup H = H1 × H2 ≤ L such that H1 ∼ = Zp , mp (H2 ) > 1, and H2 is simple without any strongly p-embedded subgroup. The groups H2 are, respectively, isomorphic to U4 (2), HJ, G2 (3), HS, and He. Therefore replacing H by a suitable Lconjugate, we have Q∗ ∩ H ∈ Sylp (H) and H2 ≤ ΓQ∗ ,2 (L). Thus, Q ∩ H2  H2 ; but mp (H2 /Q ∩ H2 ) ≤ 1, so the simplicity of H2 is contradicted. This completes the proof.  Lemma 9.12. Suppose K ∈ C2 , Z(K) is noncyclic, and m3 (K) = 1 or 2. Then K ∼ = [2 × 2]L3 (4). Proof. If K ∈ C2 − Chev(2), then Z(K) is cyclic (see [V11 , 3.1] and L3 (4), then by [IA , 6.1.4] and [IA , 6.1.4]). If K ∈ Chev(2) but K/Z(K) ∼ = [IA , 4.10.3a], m3 (K) = 0 or m3 (K) ≥ 3. As C2 excludes covering groups of  L3 (4) with centers of exponent 4, the result follows.

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Lemma 9.13. Let K  X = K z, x, where K ∈ C2 and z and x are commuting elements of order 2 and 3, respectively. Suppose that z is 2central in X, E(CK (z)) = 1, and z ∈ O2 (CX (x)). Then K/Z(K) ∼ = M12 , M22 , HJ, or J4 . Proof. Suppose that K ∈ Chev(2). Since z is 2-central, z induces an inner automorphism on K. Then using [IA , 4.2.2, 4.9.1, 4.9.2], we have z ∈ O2 (CKz (x)) ≤ CX (K), contradicting E(CK (z)) = 1. Therefore K ∈ Chev(2). If K ∈ C3 , then by [IA , 7.7.9], K ∼ = L2 (33 ) ∈ C2 , contradiction. If K ∼ = L2 (q), q ∈ F M, then CK (x) has odd order, so z cannot induce an inner automorphism on K. Hence K ∈ Spor ∩ C2 . By [IA , 7.7.1], either the lemma holds or K ∼ = HS with z an outer involution. But then z is not 2-central in X [IA , 5.3m], a final contradiction.  Lemma 9.14. Let K ∈ C2 and R ∈ Syl3 (K) with R ∼ = E32 . Suppose that ∼ A . Then K/Z(K) x ∈ R# and A  CK (x) with A ∼ = 4 = M22 . Proof. Since K ∈ C2 , Z(K) is a 2-group and hence the hypothesis carries over to K/Z(K). So we assume that K is simple. If K ∼ = An , then n = 5, 6, or 8 as K ∈ C2 , so A does not exist, contradiction. If K ∈ Chev(3), then as K is not locally balanced for p = 3, K ∼ = L2 (33 ) ∈ C2  by [IA , 7.7.1], contradiction. If K ∈ Lie(2), then by [IA , 4.2.2], O2 (CK (x)) is a central product of groups in Lie(2), contradicting the existence of A. 1 And K ∼ 2F4 (2 2 ) by [IA , 4.7.3A]. Since m3 (K) > 1 and K ∈ Chev(3), = K∼ L2 (q), q ∈ F M9. So K ∈ Spor ∩ C2 . Now by [IA , 7.7.1], K ∼ = = M22 as  |HJ|3 = |M12 |3 > 32 . The proof is complete. Lemma 9.15. Let K ∈ C2 . Suppose that K contains a subgroup L ∼ = A6 , |K/Z(K)|2 ≤ 28 , | Aut(K)|2 ≥ 27 , m3 (K) ≤ 3, and Aut(K) contains a copy of P ∈ Syl2 (L3 (4)). Suppose also that for some involution z of K centralizing a Sylow 2-subgroup of K, CK (z) is a 2-group. Then K/Z(K) ∼ = L3 (4) or Sp4 (4). Proof. If K ∈ Chev(2), we reach a contradiction, using z and [IA , 5.3] if K ∈ Spor, z and [IA , 4.5.1] if K ∈ Chev(3), and L if K ∼ = L2 (q), q ∈ F M. Hence, K ∈ Chev(2). The assumed order restrictions force K ∈ Lie(2). If z is a (long) root element, then since CK (z) is a 2-group, K/Z(K) ∼ = L3 (4) or L2 (2n ), and the latter groups are ruled out since they do not contain A6 . Thus, we may assume that z is not a long root element, whence the untwisted Dynkin diagram of K has a double bond. Our order restrictions  then force K ∼ = Sp4 (4), as claimed. 10. Some Familiar C2 -Groups and Their Subgroups 10.1. O8+ (2). Lemma 10.1.1. Let K = D4 (2) and let z be a graph automorphism of K such that CK (z) ∼ = Sp6 (2). Let y be a 2-central involution of K which is

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also 2-central in CK (z). Let Ry = O2 (CK (y)). Then the following conditions hold: (a) CRy (z) = (Q1 ∗ Q2 ) × E where Q1 ∼ = E22 , and = Q2 ∼ = Q8 , E ∼ z ∼ CRy (Q1 Q2 ) = Q3 ∗ Q3 , Q3 = Q8 ; (b) CCK (y) (z)/CRy (z) ∼ = Σ3 × Σ3 ; (c) CK (y)/Ry = W1 × W2 × W3 , where each Wi ∼ = Σ3 , z permutes {W1 , W2 , W3 }, and z centralizes every Wi that it normalizes; and (d) CRy (O3 (Wi )) = y for i = 1, 2, 3. Proof. By [IA , 4.9.2], z is determined up to K-conjugacy. In standard notation [IA , 2.10] we take the D4 root system Σ to consist of all ±ai ± aj , 1 ≤ i < j ≤ 4. We write ai + aj as aij and ai − aj as ai¯j , and also write xaij (1) as xij and xai¯j (1) as xi¯j . We take a fundamental system to consist of a1¯2 , a2¯3 , a3¯4 and a34 . We then take z to permute the xα (1), α ∈ Σ, by changing the sign of a4 . We choose y ∈ CK (z) ∼ = B3 (2) to be y = x12 , which is 2-central in both CK (z) and K. Then Ry = D1 ∗ D2 ∗ D3 ∗ D4 where D1 = x13 , x2¯3 , D2 = x1¯3 , x23 , D3 = x14 , x2¯4 , D4 = x1¯4 , x24 . Then Di ∼ = D8 , 1 ≤ i ≤ 4, and we write D1 ∗ D2 = Q1 ∗ Q2 and D3 ∗ D4 = Q3 ∗ Q4 where the Qi ’s are isomorphic to Q8 and uniquely determined up to order. Then Ry = Q1 ∗ Q2 ∗ Q3 ∗ Q4 , Qz3 = Q4 , and (a) holds with ∗ (c) E = x14x1¯4 , x24 x2¯4 . Moreover,  holds with W1 the image of W1∗ = ∗ x±a1¯2 (1) , W2 the image of W2 = x±a1¯4 (1) , and W3 the image of W3 =

x±a14 (1). Now, z centralizes W1∗ and interchanges W2∗ and W3∗ , which implies (b). Finally, to prove (d), since CRy /y (O3 (Wi∗ )) is Wi∗ -invariant, it suffices to show that CRy /y (Wi∗ ) = 1. To verify this, for i = 1, for example, this means that CRy /y (x1¯2 ) ∩ CRy /y (x−a1¯2 (1)) = 1, which is easily checked.  Lemma 10.1.2. Let K = O8+ (2), let y ∈ I2 (K) be a 2-central involution, and let z1 ∈ I2 (K) − [K, K] with K1 = CK (z1 ) ∼ = Sp6 (2). Let x ∈ I3 (K) and assume that the following conditions hold: (a) NK ( x) = x, z1  × L t, x, z1  ∼ = Σ3 , L ∼ = U4 (2) and L t ∼ = Aut(L); (b) y is 2-central in L. Then xK ∩ CK1 (y) = ∅. Proof. First, z1 stabilizes x[K,K] since x is real in [K, K]. Next, write C[K,K] (y) = Ry H with Ry ∼ = 21+8 and H = H1 × H2 × H3 . Here each Hi = xi , ti  ∼ = Σ3 with O3 (Hi ) = xi . We have CRy (xi ) = y for all i = 1, 2, 3. Moreover CAut(K) (y)/CInn(K) (y) ∼ = Σ3 permutes the Hi faithfully. ±1 ±1 This implies that the elements v = xi xj for i = j are the only elements of O3 (H) such that CRy (v) ∼ = Q8 ∗ Q8 . We assume without loss that H1z1 = H1 z1 and H2 = H3 . Since O2 (CL (y)) ∼ = Q8 ∗ Q8 and [O2 (CL (y)), S] = O2 (CL (y)) for some 3subgroup S ≤ CL (y), CRy (x) ∼ = Q8 ∗ Q8 , so we may assume that x = xi x±1 j

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for some i = j. As z1 inverts x, we may assume that x = x2 x−1 3 , with z1 z1 H x2 = x3 (and x1  = x1 ). Then let w = x2 x3 . Clearly w ∈ x 3 ⊆ xK ,  x ∈ CK (y), and x ∈ C[K,K] (z1 ) = K1 . The lemma is proved. Lemma 10.1.3. Let K = D4 (2), and let P be a parabolic subgroup with m3 (P ) = 3. Then P acts irreducibly on O2 (P )/Z(O2 (P )). Proof. Clearly P is uniquely determined up to conjugacy as having a Levi factor of type A1 × A1 × A1 . We adopt the notation of the proof of Lemma 10.1.1 and may assume that P = CK (y). Set Q = O2 (P ) and Q = # Q/ y ∼ = E28 . Let x ∈ Q be arbitrary. Taking an appropriate   Psequence of commutators of x with x1¯2 , x3¯4 , and x34 shows that x13 ∈ x . Then taking further commutators   with the opposites of these three fundamental  root elements yields xP = Q, as required. ∗ ∼ Lemma 10.1.4. Suppose Q = F ∗ (C) ∼ = 21+8 + and C/F (C) = Aut(U4 (2)). Suppose that CQ/Z(Q) (x) = 1 for some x ∈ I3 (C). Let z ∈ I2 (C) with CC/Q (z) ∼ = Z2 × Σ6 . Then z acts freely on Q/Z(Q).

 = C/Q, and identify [C,  C]  with the Proof. Let C = C/Z(Q) and C  is diagonalizable on the natural module V , matrix group U4 (2). Then x with eigenvalues on V that are cube roots of unity. On the other hand, z is uniquely determined up to conjugacy in Aut(U4 (2)), by [IA , 4.9.2], and  C]  as a so replacing z by a conjugate, we may assume that z acts on [C, generator of Gal(F4 /F2 ). Thus z inverts x. As CQ (x) = 1 by assumption,  z then acts freely on Q, as desired. Lemma 10.1.5. Let X be a K-subgroup of K = O8+ (2) and assume that O2 (X) = 1 and m3 (X) = 3. Then one of the following holds: (a) F ∗ (X) = O3 (X) and X is a {2, 3}-group. Moreover, |X|2 ≤ 25 and m2 (X) ≤ 3; (b) F ∗ (X) ∼ = Z3 × A6 and X embeds in Σ3 × Σ6 ; ∼ (c) F ∗ (X) ∼ = Ω− 6 (2) = U4 (2); ∗ (d) F (X) ∼ = A9 , and if X ∼ = Σ9 , then a 3-cycle in K centralizes a 6-dimensional subspace of the natural F2 K-module; or (e) X ∼ = Sp6 (2). Proof. Let E ∈ E33 (X). For any e ∈ E # , CK (e) is a {2, 3}-group or CK (e) = e × Ke , Ke ∼ = U4 (2) or Aut(U4 (2)) [IA , 4.8.2]. Hence if CX (e) is not a {2, 3}-group, then CK (e) ∼ = e × Le , where 32 .5 divides |Le |, ∼ m3 (Le ) = 2, and Le ≤ Ke . Thus, Le = A6 or Σ6 . It follows that for any E32 ∼ = E0 ≤ X, CX (E0 ) is a {2, 3}-group. Also, X has no element of order 3.7. Moreover, NX ( e) acts on Le so maps into Y = AutK (Ke ), in which AutY (Le ) ∼ = Σ6 . So AutNX (e) (Le ) ≤ Σ6 . In particular, NX (E; {5, 7}) = {1}. As O2 (X) = 1, O3 (X) = 1 and F (X) = O3 (X).

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Suppose F (X) = 1 and set Z = Ω1 (Z(F (X))). If |Z| = 3, then X ≤ NX (Z) and by the first paragraph, either the first statement of (a) holds or (b) holds. If m3 (Z) = 2, then by the first paragraph, CX (Z) is a {2, 3}group, so X is as well. If m3 (Z) = 3, then as Sylow 3-subgroups of K are isomorphic to (Z3 Z3 ) × Z3 , Z = F (X) = F ∗ (X), and as 13 does not divide |K| but Aut(Z) ∼ = GL3 (3), X is again a {2, 3}-group. When X is a {2, 3}-group, F ∗ (X) = O3 (X) =: R and a Sylow 2-subgroup T of X embeds in Aut(R/Φ(R)) ≤ GL3 (3), since R embeds in (Z3 Z3 ) × Z3 . Thus |T | ≤ |GL3 (3)|2 = 25 and m2 (T ) ≤ m2 (GL3 (3)) = 3, as claimed in (a).  Therefore we may assume that F ∗ (X) = O3 (E(X)) = K1 × · · · × Kk ,  where each Ki = O3 (Ki ) is simple. By [IG , 8.7], E normalizes each Ki , and so as m3 (X) = 3, E ∩ Ki = 1. If k > 1, then by the first paragraph, k = 2 and K2 ∼ = A6 ; but then m3 (E ∩ K2 ) = 2, so CX (E ∩ K2 ) is a solvable group by the first paragraph. As [K1 , K2 ] = 1, this is impossible. Therefore k = 1 and F ∗ (X) = L is simple. Since m3 (K) = 4 and |K|3 = 35 , |L|3 = 3b , b = 3 or 4. Then if 5.7 does not divide |L|, we have L ∼ = U4 (2) by [III17 , 9.9], and (c) holds. Therefore we have |L| = 2a 3b 5c 7, with a ≤ 13, b = 3 or 4, and c = 1 or 2. If L ∈ Spor, then order considerations imply that L ∼ = HJ, which contains a copy of 3A6 ; but by the first paragraph, K contains no such subgroup. So L ∈ Spor. If L ∈ Alt, then Out(L) is a 2-group, so E ≤ L and m3 (L) = 3. Thus if L ∼ = An , then n ≥ 9; but also n < 10 as X has no element of order 3.7, so L ∼ = Σ9 . = A9 . To prove (d) in this case, suppose in fact that X ∼ Let t ∈ X correspond to a transposition, and let v be a 7-cycle in CL (t). On the natural F2 K-module V , dim Vv = 6 where Vv = [V, v], and Sylow 7-subgroups of O(Vv ) ∼ = Σ8 are self-centralizing; so t must be a transvection on V . But a 3-cycle w ∈ J inverted by t then satisfies dim CV (w) = 6. Thus (d) holds if L ∈ Alt. We may therefore assume that L ∈ Chev(r) for some r. If r > 3, then L ∼ = L2 (rm ) for some m, and m3 (Aut(L)) ≤ 2, contradiction. If r = 3, then as m3 (L) = 3, L ∼ = L2 (33 ) or P Sp4 (3) [IA , 3.3.3]; but 13 does not divide |K|, so (c) holds. Therefore we may assume that L ∈ Chev(2) − Chev(3). By Lemma 3.1, m3 (L) > 1 as m3 (Aut(L)) ≥ 3. Suppose that m3 (L) ≤ 2. Then also m3 (Inndiag(L)) ≤ 2, so some e ∈ E # induces a nontrivial field, graph, or graph-field automorphism on L. By the first paragraph, either CL (e) is a {2, 3}-group or E(CL (e)) ∼ = A6 . We use [IA , 4.9.1, 4.9.2]. In the latter case L ∼ = Sp4 (8); but then |L e |3 > 34 , contradicting (10A). If CL (e) is a {2, 3}-group, then as m3 (L) = 2, L ∼ = 3D4 (2) or U3 (8), and we reach the same contradiction. Therefore, m3 (L) = 3. By [IA , 4.10.3], L has untwisted Lie rank at least 3. Using [IA , 4.10.3] to compute m3 (L), we see that L must be a classical (10A)

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group. If L is of type B, C, or D, then using |L|2 ≤ 213 in addition, we see that L ∼ = Sp6 (2) or 2 D4 (2). The latter violates (10A), so (e) holds. So we may assume that L is a linear or unitary group. If q(L) > 2, then |L|2 ≤ 213 and m3 (L) = 3 force L ∼ = U4 (4), which contradicts (10A). So q(L) = 2, whence by 3-ranks, L ∼ = Ln (2), n = 6 or 7, and (10A) is contradicted. This completes the proof.  Lemma 10.1.6. Let H ≤ O8+ (2) with O2 (H) = 1, m2 (H) ≥ 5, and m3 (H) = 3. Then H ∼ = Sp6 (2). Proof. Now, m2 (Aut(U4 (2))) = m2 (SO5 (3)) = 4 and m2 (Aut(A9 )) = m2 (D8 Z2 ) = 4. Hence the result is immediate from Lemma 10.1.5.  Lemma 10.1.7. Let A ∈ E3 (D4 (2)) be such that m3 (A) = 3, V = [V, A], and dim CV (a) ≤ 4 for all a ∈ A# , where V is a natural F2 D4 (2)-module. Then dim CV (a) = 4 for exactly 12 elements a ∈ A# . Proof. Since E34 (D4 (2)) is a single conjugacy class [IA , 4.10.3], there is a decomposition V =⊥4i=1 Vi such that each Vi is A-invariant and 2dimensional of − type. Let Ai = CA (Vi ) for each i. Since V = [V, A], we have |Ai | = 32 for all i. As dim CV (a) ≤ 4 for all a ∈ A# , the six intersections Aij = Ai ∩ Aj , i = j, are distinct subgroups of A of order 3. This implies the result.  Lemma 10.1.8. Let V ∼ = E28 have a nondegenerate quadratic form of + type. Let A ≤ N ≤ H ≤ O(V ) with H a K-group, O2 (H) = 1, m3 (H) = 3, A∼ = Σ3 ×Σ3 ×Σ3 , and d := |NH (A) : N | = 1 or 3. Suppose that as = E33 , N ∼ F2 N -module, V is the tensor product of the three natural modules for the Σ3 direct factors. Suppose also that for any x ∈ I3 (H) such that |CV (x)| ≥ 24 , we have |CV (x)| = 24 and CH (x) ∼ = E33 or E32 ×Σ3 ≤ NH (A). Then A  H, and H = N or H ∼ = Σ3 Z3 according as d = 1 or 3. Proof. Write N = N1 × N2 × N3 with Ni = ui , ti  ∼ = Σ3 and ui of order 3 for each i = 1, 2, 3. Then the ui  are the only subgroups in E1 (A) acting Frobeniusly on V . Hence NH (A)/A permutes the ui , so if d = 3, NH (A) ∼ = Σ3 Z3 (otherwise m3 (H) = 4, contrary to assumption). Let A ≤ P ∈ Syl3 (H). Since |O(V )|3 = 35 and m3 (O(V )) = 4, our hypotheses imply that |P | ≤ 34 , so P ≤ NH (A). It therefore suffices to show that A  H. We investigate the structure of CH (u) for various u ∈ A# . Note first ±1 4 that |CV (ui u±1 j )| = 2 for any 1 ≤ i < j ≤ 3, so |CH (ui uj ) : A| = 2 ±1 ±1 ±1 2 by hypothesis. Also, |CV (u1 u±1 2 u3 )| = 2 , so CH (u1 u2 u3 ) embeds in ±1 GU3 (2). But CNH (A) (u1 u±1 2 u3 ) is obviously a 3-group. It follows that ±1 ±1 CH (u1 u2 u3 ) is a 3-group, and hence contained in NH (A). Now factoring under u1 u2 , u2 u3  yields O3 (CH (ui )) = O3 (H) = 1 for all i. Finally consider CH (ui ), in which A is a Sylow 3-subgroup. Suppose CH (ui ) has a component J. Then as NCH (ui ) (A) is reducible on A/ ui , any component

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J of CH (ui ) must have 3-rank 1, and in view of our previous computations, J must be centralized by some u ∈ A# such that CH (u) is a {2, 3}group. This is impossible, so  CH (ui ) is 3-constrained. We have shown that ΓA,1 (H) = CH (u) | u ∈ A# ≤ NH (A). that If O3 (H) = 1 or E(H) has more than one component,   it follows F ∗ (H) ≤ NH (A), so H is 3-constrained. Then O3 (H) ≥ Z(P )N = A, so A = J(O3 (H))  H and we are done. So we may assume that J := F ∗ (H) is simple. If J ∼ = An , then n ≥ 9 as m3 (J) = 3, so J = ΓA,1 (J) ≤ NH (A), contradiction. If J ∈ Spor, then as m3 (J) = 3, J ∼ = J3 does not embed in O(V ) by order considerations, contradiction. If J ∈ Chev(r), r = 3, then again J = ΓA,1 (J) by [IA , 7.3.4], contradiction. So J ∈ Chev(3) − Chev(2)  ∼ and A ≤ O 3 ([NH (A), NH (A)]) ≤ J. Then m3 (J) = 3 so by [IA , 3.3.3], J = 3 ∼ P Sp4 (3) ∈ Chev(2), contradiction, or J = L2 (3 ), contradicting Lagrange’s theorem. The lemma is proved.  Lemma 10.1.9. Let K = Ω+ 8 (2). Then up to K-conjugacy, there is one subgroup L ≤ K such that L ∼ = A9 and such that any 3-cycle x ∈ L satisfies dim CV (x) = 6, where V is a fixed natural F2 K-module. Proof. Let Ω = {1, 2, . . . , 9}. Let W be the natural F2 Σ9 -module,    with permutation basis {wi }i∈Ω . For any S ⊆ Ω define q( i∈S wi ) = |S| 2 (mod 2). Then q is a Σ9 -invariant quadratic form on W with associated  bilinear form (wi , wj ) = 1 + δij . Let W0 = F2 9i=1 wi ∈ rad(W ). Then q lifts to a quadratic form on W = W/W0 which is nondegenerate since Σ9 is irreducible on W . One calculates that W has 135 nonzero singular vectors, whence q is of + type. Hence we have an embedding Σ9 ≤ O8+ (2) in which a transposition maps to a transvection. Intersecting with K gives an L ≤ K with L ∼ = A9 and (10B)

O8+ (2) = NO+ (2) (L)K. 8

Conversely, given any L with AΩ ∼ = L ≤ K, let L7 ∼ = A7 be the stabilizer 7 in L of 8 and 9. Since L is generated by three 3-cycles, dim CV (L7 ) ≥ 2. Let s ∈ I7 (L7 ). Then dim CV (s) = 2 so CV (L7 ) = CV (s) is nondegenerate and L7 , like s, acts irreducibly on [V, s], which is then necessarily of + type. The Brauer character ϕ7 of L7 on [V, s] is its Brauer character on V minus 2. Let xr ∈ Ir (L7 ), r = 3, 5, 7, with xr not a 3-cycle. Then x3 is the product of two commuting conjugates of x. Also elements of I5 (K) and I7 (K) are rational. So ϕ7 is uniquely determined: ϕ7 (1) = 6, ϕ7 (x) = 3, ϕ7 (x3 ) = 0, and ϕ7 (x5 ) = 1 = −ϕ(x7 ). By Witt’s lemma, [V, s] is unique up to K-conjugacy, and given [V, s], L7 , which is absolutely irreducible on [V, s], is unique up to O([V, s])-conjugacy (see [V9 , 12.1a]).   Let L6 be the stabilizer in L7 of 7. Then L = L7 , y for some y ∈ I3 (CL (L6 )). We argue that CK (L6 ) has a unique subgroup of order 3.  Namely, for f ∈ I5 (L6 ), O3 (CK (f )) ∼ = A5 . Also for disjoint 3-cycles x1 , x2 ∈ (2), which is 3-closed. Hence CK (L6 ) is 3-closed L6 , CK ( x1 , x2 ) ∼ = E32 ×Ω+ 4

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with a Sylow 3-subgroup of order 3. This proves the uniqueness of L up to  O8+ (2)-conjugacy, and then, by (10B), up to K-conjugacy, as claimed. Lemma 10.1.10. Let K = Ω+ 8 (2) and let V be a natural K-module. Up + to O8 (2)-conjugacy, there is a unique subgroup L ≤ K such that L ∼ = A9 and such that CV (x) = 0, where x ∈ L is a 3-cycle. Proof. There are two K-conjugacy classes C1 , C2 of elements x ∈ I3 (K) such that CV (x) = 0, interchanged in O8+ (2). For each i = 1, 2 there is a triality τi such that for xi ∈ Ci , dim CV (τi (xi )) = 6. Hence by Lemma 10.1.9, for each i = 1, 2, there is a unique A9 , up to K-conjugacy, in which a 3-cycle belongs to Ci . Since C1 and C2 are conjugate in O8+ (2), the result follows.  Lemma 10.1.11. Suppose H is a K-subgroup of O8+ (2) such that O2 (H) = 1 and m3 (H) = 3. Suppose that for every y ∈ I3 (H), dim CV (y) ≤ 4, where V is the natural O8+ (2)-module. Let L ≤ H be a 3-component (of itself ). Then the following conditions hold: (a) If |H|3 ≥ 34 , then H ∼ = A9 , U4 (2), Aut(U4 (2)), or Sp6 (2); A9 , and if L ∼ (b) In (a), if H involves U4 (2), then H ∼ = = A6 and ∼ L   CH (x) for some x ∈ I3 (H), then H = Sp6 (2) or A9 ; (c) In (a), if there is x ∈ I3 (H) such that |CH (x)|2 ≤ 2, then H ∼ = U4 (2) or A9 ; (d) If L/O3 (L) ∼ = L2 (8), then H ∼ = A9 or Sp6 (2); ∼  (e) If L/O3 (L) = U3 (3), then H ∼ = Sp6 (2); and (f) Suppose that H ∼ = Sp6 (2) = A9 and y ∈ H is a 3-cycle, or that H ∼ and y ∈ I3 (H) satisfies O3 (CH (y)) ∼ = Σ6 . Then CV (y) = 0. Proof. By our assumptions, H is not solvable, and H contains no element of order 3 fixing 6 dimensions on the natural F2 O8+ (2)-module. Moreover, the additional assumptions in (a), (d), (e), and (f) imply that a Sylow 3-subgroup of H is not elementary abelian. Therefore by Lemma 10.1.5, H∼ = A9 , U4 (2), Aut(U4 (2)), or Sp6 (2). Therefore (a) holds; moreover, order considerations imply (d) as well. U4 (2), and we need only Likewise in (e), since 7 divides |U3 (3)|, F ∗ (H) ∼ = rule out H ∼ A . But in this case, L contains a Z = 9 12 -subgroup containing a ∼ 3-central element of H; on the other hand, if H = A9 , then there is no such subgroup. So (e) holds. Two observations imply the restrictions in (b). First, |A9 |/|U4 (2)| = 7 but A9 has no subgroup of index 7; and second, U4 (2) ∼ = P Sp4 (3) ∈ Chev(3). − ∼ In (c), since Aut(U4 (2)) = O6 (2) ≤ Sp6 (2) with 3 -index, it suffices to consider the case H ∼ = Aut(U4 (2)). Then the [H, H]-fusion in I3 ([H, H]) differs from that in H only in that 3-central elements vc are real in H but not in [H, H]. But |C[H,H] (vc )|2 = 8 anyway. For any other v ∈ I3 ([H, H]), |CH (v)|2 = 2|C[H,H] (v)|2 ≥ 4 [IA , 4.8.2], proving (c). Finally, let H and y be as in (f). Then CH (y) contains a subgroup J∼ = A6 , which must act faithfully on [V, y] or CV (y). If these have respective

10. SOME FAMILIAR C2 -GROUPS AND THEIR SUBGROUPS

dimensions 2d1 and 2d2 , J must embed either in CO±

2d1 (2)

489

(y) ∼ = GUd1 (2)

± (2). But by hypothesis, d2 ≤ 2, so J must embed in [IA , 4.8.2] or in O2d 2  GUd1 (2). Therefore d1 = 4, so CV (y) = 0. The lemma is proved.

Lemma 10.1.12. Let X be a K-subgroup of O8+ (2) with O2 (X) = 1. Suppose that E32 ∼ = D8 . Suppose = A ∈ Syl3 (X), A = CX (A) and NX (A)/A ∼ that CX (A) has odd order, and for some a ∈ A# , CV (a) = 0, where V is a natural X-module. Suppose also that X has a subgroup S ∼ = Σ5 . Then X∼ = A6 , A7 , or A8 . = A5 Z2 , or E(X) ∼ Proof. We have |O8+ (2)| = 213 35 52 7. Since A5 cannot act faithfully on E32 , E52 , or Z7 , [F (X), S0 ] = 1, where we have put S0 = [S, S] ∼ = A5 . By orders, no component of E(X) is a Suzuki group, so  E(X) has at most 2 E(X) . components. As S0 is perfect, S0 ≤ E(X). Let I = S0 Suppose that I has more than one component. By orders and [III17 , 9.9], I = I1 I2 with I1 ∼ = I2 ∼ = A5 . Then S0 is a diagonal of I. Now S = S0 CS (A ∩ S0 ), so CS (A∩S0 ) ≤ I. Since CX (A) has odd order, NS (I1 )∩CS (A∩S0 ) = 1, so SI ∼ = A5 Z2 and AutX (Ii ) = Inn(Ii ), i = 1, 2. Hence X = SICX (I1 I2 ). But a Sylow 5-subgroup of O8+ (2) is self-centralizing, so X = SI, as claimed. Therefore I is quasisimple and S0 ≤ I. If A ≤ I, then |I|3 = 3 so Out(I) is a 3 -group [V8 , 1.1]. So I = I g for some g ∈ NX (A). Some t ∈ I2 (S − S0 ) centralizes A ∩ I, and a suitable involution in tI g centralizes A, contrary to hypothesis. Hence, A ≤ I so I  X. We have Or (I) = 1 for every prime r; by assumption for r = 2, and as Sylow r-subgroups of X are abelian for r > 2. If I ∈ Alt, then the desired conclusion holds by order considerations. If I ∈ Spor, then the conditions that A ∈ Syl3 (I) and |I| divides |O8+ (2)| are not satisfied [IA , 5.3]. If I ∈ Chev(r) − Alt, r odd, then I ∼ = A5 , L2 (52 ), or L2 (7), so A ≤ I, contradiction. Finally we may assume that I ∈ 12 2 Chev(2) − Alt. As I = [I, I], |I|2 ≤ |Ω+ 8 (2)|2 = 2 . Because |I|3 = 3 and ± n ∼ ∼ A = E32 , if I = L3 (2 ), then AutI (A) contains Q8 , contrary to assumption. Indeed, the constraints on 3-structure, and order considerations, rule out all possibilities for I, completing the proof.  ∼ Lemma 10.1.13. Suppose K ≤ L ∼ = Ω+ 8 (2) with K = A6 . Let V be a natural F2 L-module. Suppose that for some x ∈ I5 (K), CV (x) = 0. Then for every y ∈ I3 (K), CV (y) = 0. Proof. The F2 K irreducible modules include the trivial module and two quasi-equivalent modules of degree 4, namely, the cores M1 , M2 of the two permutation modules of degree 6. There are also two complex irreducible characters χ1 , χ2 of degree 8; being of defect 0, they remain distinct and irreducible modulo 2. This is a complete list as A6 has exactly 5 conjugacy classes of elements of odd order [V9 , 12.1b]. If V is irreducible, it follows that its Brauer character is the restriction of χ1 or χ2 to the set of elements of odd order. But χi (y) = −1 for i = 1, 2,

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so CV (y) = 0 in that case. Hence we may assume that V is reducible and has no trivial composition factors. Therefore every composition factor of V is M1 or M2 . But CMi (x) = 0 for i = 1 and 2, so CV (x) = 0, contrary to assumption. This contradiction completes the proof of the lemma.  10.2. Sp6 (2). Lemma 10.2.1. Let K = Sp6 (2) and let V be an 8-dimensional F2 Kmodule supporting a K-invariant quadratic form of + type. Let x1 , x2 , x3 ∈ I3 (K) with dim[Vnat , xi ] = 2i, where Vnat is the natural F2 K-module. Suppose that dim CV (x1 ) = 0, dim CV (x2 ) = 4, and dim CV (x3 ) = 2. Then V is determined uniquely (as the spin module). Moreover, if v ∈ I5 (CK (x1 )), then dim CV (v) = 0. Proof. Let v ∈ I5 (CK (x1 )). If CV (v) = 0, then V = CV (v) ⊥ [V, v] with both summands of dimension 4 and − type, and x1 -invariant. But x1 cannot act on a 4-dimensional space of − type without fixing a point, contradiction. Therefore CV (v) = 0. If V is not absolutely irreducible for K, then V ⊗F2 F2 has a K-invariant section of dimension at most 4, on which K must act trivially, contradicting dim CV (x1 ) = 0. So V is absolutely irreducible. We will calculate the Brauer character ϕ of K afforded by V and show that it is uniquely determined, whence V is determined up to isomorphism [V9 , 12.1a]. There are conjugacy classes of K of elements of order 3 (represented by x1 , x2 , x3 ), of order 5 (a single class v G , as Σ8 ∼ = O6+ (2) ≤ K), of G order 7 (a single class s for the same reason), of order 9, and of order 15 (a single class (x1 v)G , as Σ3 × Σ6 ≤ K). Clearly ϕ(1) = 8, ϕ(x1 ) = −4, ϕ(x2 ) = 2, and ϕ(x3 ) = −1. As CV (v) = 0 and v is rational in K, ϕ(v) = −2. Likewise ϕ(x1 v) = 1 and ϕ(s) = 1, using rationality in K in each case. This leaves elements s of order 9. Let P ∈ Syl3 (K) with J(P ) = x1 , x2 , x3 . Then Z(P ) is conjugate to x3 and V ↓ P = CV (Z(P )) ↓ P ⊕ [V, Z(P )] ↓ P . Let ϕ1 and ϕ2 be the Brauer characters of P on these two summands, so that ϕ ↓ P = ϕ1 + ϕ2 . Since P is nonabelian, ϕ2 is induced from J(P ), and so ϕ2 (s ) = 0. As dim CV (Z(P )) < 3, ϕ1 is a lift of a character ψ of P/[P, P ]. Since x1 ∈ 1+2 P − Z(P ), ϕ1 (x1 ) = ϕ1 (x−1 1 ) = −1. Likewise if Q is the unique 3 subgroup of P , then all subgroups w of Q of order 3 outside Z(P ) are conjugate in K (which contains SU3 (2)), and so conjugate to x2 . Hence ϕ1 (w) = ϕ1 (w−1 ) = 2. s is conjugate in P to an element of the form ±1 ±1   x±1 1 w . Therefore ϕ1 (s ) = ϕ1 (x1 ) = −1. Hence ϕ(s ) = −1 and we have shown that ϕ is uniquely determined, proving the lemma.  Lemma 10.2.2. Let V be an 8-dimensional F2 -space with a quadratic form of + type. Let K ≤ O(V ) with K ∼ = Sp6 (2). If the representation of K on V is the spin representation, then K is transitive on the set Vsing of singular vectors of V . Moreover, the stabilizer in K of an element of Vsing

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is a parabolic subgroup M of K which is a split extension of E26 by L3 (2), and such that Soc(M ) ∼ = E23 . Proof. 4 2 Since V is the orthogonal sum of four asingular planes, |Vsing | = + 2 3 = 135. Let O be a K-orbit of odd length on Vsing , and v ∈ O. By Tits’ lemma [IA , 2.6.5] and as a Cartan subgroup of K is trivial, Kv is a parabolic subgroup of K of index at most 135. Moreover, let x ∈ I3 (K) with dim CVnat (x) = 4, where Vnat is the natural F2 K-module. Then as V is the spin module, CV (x) = 0, so Kv ∩ xK = ∅. In particular Kv is not of type Sp4 (2). The only other possibility is Kv = E26 L3 (2), and then  |K : Kv | = 135, so K is transitive on Vsing . The lemma follows easily. 34

Lemma 10.2.3. Let K = Sp6 (2) and t ∈ I2 (K). Then there exist xc , xi ∈ I3 (K) such that xtc = xc and xti = x−1 i , and neither xc nor xi is 3-central in K. Proof. Let Vnat be the natural F2 K-module. It suffices to show that xc exists. For then as xc is not 3-central, CVnat (xc ) and [Vnat , xc ] are nontrivial and t-invariant, so t acts nontrivially on at least one of them, say W . Then by inspection of Sp(W ) ∼ = Sp2 (2) or Sp4 (2), t inverts an element xi of CVnat (W ⊥ ) ∼ = Sp(W ). Then xi is clearly not 3-central in K, so it satisfies our requirements. To show that xc exists, use the Baer-Suzuki theorem to find an element x ∈ K of odd prime order inverted by t. Let H = NK ( x). Let M ≤ K with M ∼ = O6+ (2) ∼ = Σ8 . If x has order 5 or 7, we may assume that x ∈ M , and it is easily checked that NM ( x) ∼ = NK ( x), so t ∈ NK ( x) ≤ M ∼ = Σ8 . But then CM (t) contains some xc ∈ I3 (M ). Since M preserves a +-type quadratic form on the natural F2 K-module Vnat , CVnat (xc ) = 0. Hence, as desired, xc is not 3-central in K. So, we may assume that x has order 3. If x is 3-central in K, then P := O3 (H) ∼ = 31+2 is nonabelian, and so CP (t) = 1. Let xc ∈ I3 (CP (t)). Then xc  = x since t inverts x. But E1 (P )−{ x} is permuted transitively by CK (x) ∼ = SU3 (2). Hence CVnat (xc ) = 0, for otherwise P would act Frobeniusly on Vnat , an impossibiity. Thus, we may assume that t inverts no 3-central element of I3 (K). But now [IA , 4.8.2], P = g1  × g2  × g3  ∼ = E33 , with X := { gi  | 1 ≤ i ≤ 3} = {E ∈ E1 (P ) | dim CVnat (E) = 4}. Thus t permutes X. If gi t = gi  for some i, then xc = gi git is the desired element of I3 (CP (t)). We may therefore assume that t normalizes, and then inverts, every gi . But then t  inverts the 3-central element g1 g2 g3 , a final contradiction.  ∼  Let Lemma 10.2.4. Let K ∼ = 2Sp6 (2) with z = Z(K). = Sp6 (2) and K Vnat be the natural F2 K-module. Then the following conclusions hold: (a) K has four classes of involutions, with representatives ti , 0 ≤ i ≤ 3; moreover, for any 1 ≤ i ≤ 3, ti is 2-central in K and does not act freely on Vnat ;

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492

 while   t0 and  t1 are involutions in K, t22 = z =  t23 ;  Z(T) = z for T ∈ Syl2 (K);   to  ti z for each 0 ≤ i ≤ 3; ti is K-conjugate ∼ CK ( t0 ) = Z2 × (Q8 ∗ Q8 ) : Σ3 and O2 (CK (t0 )) ∼ = E16 : Z3 with 7 |CK (t0 )| = 2 ·3. Also, if b ∈ CK (t0 ) has order 3, then CK ( b, t0 ) ∼ = Z3 × E22 ; (f) |CK (t1 )| = 29 ·32 , CK (t1 )/R1 ∼ = SL2 (2)×Sp2 (2), R1 := O2 (CK (t1 ))  3 = O2 (O (CK (t1 ))) = R × E with R ∼ = Q8 ∗ Q8 and E ∼ = E22 , and

t1  is the unique normal 2-subgroup of CK (t1 ) of 2-rank 1; t2 ) ∼ (g) CK (t2 ) ∼ = E25 : Σ6 and CK ( = (Z4 ∗ 21+4 )A6 ; and (h) |CK (t3 )| = 29 · 3 and if b1 ∈ CK (t3 ) has order 3, then R b1  × E  CK (t3 ), with R ∼ = Q8 ∗ Q8 , t1  = Z(R) = CR (b1 ), and CK ( b1 , t3 )) = b1  × D × t1 , with D ∼ = D8 , and t3 ∈ Z(D). Moreover, Z(CK (t3 )) = t1 , t3 .

(b) (c) (d) (e)

Proof. We use standard notation [IA , 2.10] for K = Sp6 (2). We use Π = {±2ai , ±aj ± ak | 1 ≤ i ≤ 3, 1 ≤ j < k ≤ 3} and Σ = {a1 − a2 , a2 − a3 , 2a3 }. Then U=



Xα where Xα = xα (1) ∼ = Z2 .

α∈Π∩Z+ Σ

∼ E22 , The commutator formula [IA , 2.4] easily yields Z(U ) = X2a1 Xa1 +a2 = and we put t1 = xa1 +a2 (1), t2 = x2a1 (1), and t3 = t1 t2 . As NK (U ) = U , Burnside’s lemma implies that t1 , t2 , and t3 are mutually nonconjugate. Moreover, t1 , t2 , and t3 centralize I3 := O3 ( X±2a3 ) as well as V3 := [Vnat , I3 ], which is a nondegenerate 2-subspace of Vnat . Therefore t1 , t2 , and t3 act nonfreely on Vnat , as asserted in (a). By Tits’s lemma, Pi := CK (ti ), for 1 ≤ i ≤ 3, is a parabolic subgroup of K containing U , and we easily find P1 = P{a1 −a2 ,2a3 } , P2 = P{a2 −a3 ,2a3 } , and P3 = P{2a3 } . Let Ui = O2 (Pi ), 1 ≤ i ≤ 3. ∼ 1+4 Then U1 is the direct product of R := X  a1 ±a3 , Xa2 ±a3 , Xa1 +a2  = 2+ and E := X2a1 X2a2 ∼ = = E22 . Moreover, O3 ( X±(a1 −a2 ) ) acts on U1 / t1  ∼ E26 without fixed points. Also xa1 −a3 (1)xa2 +a3 (1) and a2 −a3 (1)xa1 +a3 (1) are commuting elements of R of order 4, which are interchanged by an element of Xa1 −a2 , Xa2 −a1 . This implies that no Q8 subgroup of U1 is normal in P1 . Thus (f) holds. Similarly, P3 = CK (Z(U )) and U3 = U1 Xa1 −a2 . Since P3 ≤ P1 , U1  P3 . Let b1 be a generator of O3 ( X±2a3 ). Then U1 b1   P1 so U1 b1   P3 . Then b1 centralizes t1  × E. Calculating the action of x2a3 and x−2a3 on U1 / t1  E, we see that CU1 (b1 ) = t1  E, so CK ( b1 , t3  = b1  × Y where Y := Xa1 −a2 Xa1 +a2 X2a1 X2a2 ∼ = D8 ×Z2 is a Sylow 2-subgroup of B2 (2) with Z(Y ) = Xa1 +a2 X2a1 = t1 , t3  = Z(U ) and [Y, Y ] = xa1 +a2 (1)x2a1 (1) =

t1 t3 . As b1 centralizes Z(U ) = t1 , t3 , Z(P3 ) = Z(U ). Thus (h) holds.

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∼ Σ6 and U2 is ∼ B2 (2) = Next, P2 = P{a2 −a3 ,2a3 } = U2 K2 where K2 = elementary abelian, the direct product of the five positive root groups Xα such that α has a nonzero projection on a1 .  →K  ∗ := Spin7 (3) and the resulting Now consider the embedding σ : K ∗ embedding σ : K → K := Ω7 (3) [IA , 6.2.2], so that U ∗ := σ(U ) ∈ Syl2 (K ∗ ). Let W be the natural F3 K ∗ -module. Then U ∗ acts monomially on W ; indeed there is an orthogonal frame F in W composed of isometric 1-spaces such that U ∗ lies in the stabilizer P ∗ = E ∗ A∗ of F, with E ∗ ∼ = E26 stabilizing each 1-space in F (and acting with determinant 1) and A∗ ∼ = A7 permuting the seven 1-spaces in F naturally. Since a Sylow 2-subgroup of A7 has orbits of length 1, 2, and 4 on seven letters, Z(U ∗ )# = {e∗2 , e∗4 , e∗6 = e∗2 e∗4 }, where e∗i inverts exactly i of the seven 1-spaces in F. Note that the only element of Z(U ∗ )# lying in the center of a Q8 ∗ Q8 -subgroup of K ∗ is e∗4 , so σ(t1 ) = e∗4 . By [IA , 6.2.1d], e∗4 splits over Z(Spin7 (3)), but e∗2 and e∗6 do not.  but t2 and t3 do not. Likewise by [IA , 6.2.1e], Therefore t1 splits over Z(K), ∗ ∗ # each e ∈ Z(U ) is conjugate to e∗ z ∗ in Spin7 (3), necessarily in U ∗ (here  ∗ ); so t1 , t2 , and t3 are each sheared in K  to Z(K).  This implies

z ∗  = Z(K (c), and that the assertions in (b) and (d) not involving t0 . Also, P2 must t2 has order 4, all parts of act irreducibly on U2 since 5 divides |P2 |, and as  (g) hold as well. It remains to prove that there is a unique further class of involutions in K, represented by an element t0 satisfying (b), (d), and (e); for then (a) will K K hold as well. Let Γ = I2 (K) − tK 1 − t2 − t3 . Notice that if P = P1 or P2 and 3 D ∈ E2 (P ), then CO2 (P ) (D) ≤ Z(U ). Hence for any t ∈ Γ, m3 (CK (t)) = 1, in view of Lemma 10.2.3. Now, K has three classes of elements of order 3. We let B = b1 , b2 , b3  ≤ K with B ∼ = E33 , and with dim([V, bi ]) = 2 for each i, where V is the standard module for K. We also let β = b1 b2 b3 . Thus, b1 , b2 b3 and β are representatives of the three K-classes of elements of order 3. We have NK ( b1 ) = ∼

b1  X2a3 × K2 , with K2 ∼ = Σ6 , and x2a3 (1) ∈ tK 2 ; and CK (β) = GU3 (2); in both CK (b1 ) and CK (β), every involution centralizes an E32 -subgroup. Therefore by the previous paragraph, every involution in Γ centralizes an if necessary. we may assume element of (b2 b3 )K . Replacing b3 by b−1 3 that a Sylow 2-subgroup of CK (b2 b3 ) ∼ = L2 (2) × L2 (2) × Z3 [IA , 4.8.2] is S23 := X2a3 × t1 , and the three involutions of S23 are pairwise nonconjugate in CK (b2 b3 ). Setting t2 = x2a3 (1), it follows that t2 ∈ tK 2 and (10C)

t0 := t1 t2 = xa1 +a2 (1)x2a3 (1)

represents the fourth and last (by Lemma 10.2.3) conjugacy class in I2 (K). Since b2 b3 ∈ I3 (K2 ) with K2 ∼ = Σ6 , and t1 ∈ CK2 (b2 b3 ), it follows that  maps on t . t2 ), where  t2 ∈ K t1 ∈ [K2 , K2 ]. But by (g), [K2 , K2 ] = CK2 ( 2  is isomorphic to D8 , whence the preimage Thus the preimage of t1 , t2  in K of t0  is isomorphic to E22 , and (b) and (d) follow for t0 .

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With the Borel-Tits theorem, we have |CK (t0 )| = 2n .3, and 7 ≤ n < 9. This is because t0 is not 2-central, but t0 lies in the E26 subgroup E generated by all three Xaj +ak and all X2ai , and [t0 , Xa1 −a2 ] = 1. Also theorem, 2n−3 ≡ 1 (mod 3), NCK (t0 ) ( b2 b3 ) ∼ = E22 × Σ3 , so by Sylow’s   whence n = 7. Since CK (t0 ) contains E X±(a1 −a2 ) , they are equal. Finally,  of E because of the irreducible A7 action on E in Spin7 (3), the preimage E 1+6  is isomorphic to 2 . Then as |NC (t ) ( b2 b3 ) ∼ in K = E22 × Σ3 , all parts + K 0 of (e) follow. The proof is complete.  Lemma 10.2.5. Let K = Sp6 (2), and let Vnat and V be the natural and spin F2 K-modules, respectively. Then the following conditions hold: (a) If t ∈ I2 (K) is not 2-central in K, then (1) t acts freely on V and Vnat ; and (2) There exists w, x0  ∈ E32 (K) such that [x0 , t] = 1, t inverts w, dim CVnat (x) = 2 for all x ∈ w, x0  − w, and CVnat (w) = 0; and (b) If t is a short root involution of K, then t inverts some x ∈ I3 (K) such that dim CVnat (x) = 2, and [CVnat (x), t] = 0. Proof. Considering (10D)

Vnat =⊥3i=1 Wi ,

where the Wi are hyperbolic planes, we see that some elements of I2 (K) act freely on each Wi and hence freely on Vnat . Let ti , 0 ≤ i ≤ 3, be as in Lemma 10.2.4. By part (a) of that lemma, we conclude that the elements of ∼ tK 0 act freely on Vnat . Also, by (10C), t0 inverts O3 ( X±2a3 ) = Z3 , which is fixed-point-free on V , so t0 acts freely on V as well. Therefore (a1) holds. Let X ∼ = Sp2 (2) Σ3 be the stabilizer in K of the decomposition (10D). Some 2-central involution t∗ of X inverts a fixed-point-free element w of X on Vnat , so t∗ ∈ tK 0 . To prove (a2), it therefore suffices to prove it for t = t∗ . We may choose t∗ and w to be centralized by a wreathing element x0 ∈ I3 (X). Then every x ∈ w, x0  − w cycles the three Wi ’s and so acts freely on Vnat . Thus, dim CVnat (x) = 2, and (a2) holds. Finally, consider V = V1 ⊥ V2 where dim V1 = 4. Then CK (V2 ) ∼ = Sp4 (2) and a short root involution of CK (V2 ) is also a short root involution of K. We may choose an isomorphism CK (V2 ) ∼ = Σ6 such that long root A1 (2)subgroups map to root Σ3 -subgroups. Then short root A1 (2)-subgroups map to regular Σ3 -subgroups and (c) is proved.  Lemma 10.2.6. In K = Sp6 (2), there are two classes of involutions whose centralizers have order divisible by 9. Those centralizers have orders 29 32 5 and 29 32 . Proof. These orders are the orders of the parabolic subgroups of K which are the centralizers of a long root element and a short root element, respectively. Hence we need only show that for any E ∈ E32 (K), any z ∈ I2 (CK (E)) is a root element.

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Let V be the natural K-module and write E = e1 , e2 . If [V, e1 ] = V1 is 2-dimensional, then CK (e1 ) = e1  × K1 where K1 = Sp(e⊥ 1 ) and we may  assume that e2 ∈ K1 ∼ = Σ6 . Then O2 (CK (E)) = CK1 (e2 ), and the two choices for the conjugacy class of e2 both yield root involutions generating the centralizers. So we may assume that dim[V, e] ≥ 4 for all e ∈ E # . As a result we may assume that [V, e1 ] = V and the irreducible E-submodules of V are three orthogonal 2-dimensional subspaces Vi , 1 ≤ i ≤ 3. But then CK (E) normalizes each Vi and centralizes e1 , so CK (E) has odd order. The proof is complete.  Lemma 10.2.7. Let K = Sp6 (2) and z ∈ I2 (K). Let P ∈ Syl2 (CK (z)). Then there is no subgroup F  P such that z ∈ F ∼ = Z4 . Proof. Suppose that such z and F exist. If z acts freely on the natural (6-dimensional) F2 K-module, then z is not a square, by consideration of Jordan canonical form. So by Lemmas 10.2.4a and 10.2.5a1, we may assume that z is 2-central in K, i.e., P ∈ Syl2 (K). Let M be the parabolic subgroup of K containing P of type L3 (2). Then z ∈ O2 (M ) ∼ = E26 , so F ≤ O2 (M ) and as F  P , a generator of F acts as a transvection on O2 (M ). But M/O2 (M ) ∼ = L3 (2) has two nontrivial composition factors on O2 (M ), so no element of P − O2 (M ) can act as a transvection. This contradiction proves the lemma.  Lemma 10.2.8. Let K = Sp6 (2) ∼ = C3 (2) and T ∈ Syl2 (K). Then the following conditions hold: (a) Z(T ) ≤ [T, T ]; (b) Let x ∈ I3 (K) with O3 (CK (x)) =: H ∼ = Σ6 . Then every 2-central involution of K has a K-conjugate in H; and (c) Let X = x, z be a long root A1 (2)-subgroup of K, with O3 (X) =

x. Let H1 = CK (X) ∼ = Sp4 (2) and let x1 ∈ xK ∩ H1 . Let ∼ V = O2 (CK (z)) = E25 . Then CV (x1 ) ∼ = E23 . Proof. We may take K to be generated by involutions xα (1), α ∈ {2ai , aj ± ak | 1 ≤ i ≤ 3, 1 ≤ j < k ≤ 3} subject to the Chevalley commutator formula. It is easily checked from that formula that Z(T ) =

x2a1 (1), xa1 +a2 (1) ∼ = E22 . Now [T, T ] contains [xa1 −a3 (1), xa2 +a3 (1)] = xa1 +a2 (1) and [xa1 −a2 (1), x2a2 (1)] = xa1 +a  holds.  2 (1)x2a1 (1), whence (a) It follows also that Z(T ) ≤ H0 := x±2a2 (1), x±(a1 −a2 ) (1) ∼ = B2 (2) ∼ = Σ6 . But [H0 , L] = 1 where L = x±2a3 (1), which contains an element x0 of order 3. As there is a unique conjugacy class of subgroups of K of order 3 centralizing a Σ6 -subgroup, (b) follows. In (c), we  may assume that z = x2a1 (1) and X = x, z = z, x−2a1 (1). Then H1 = x±2a3 (1), x±(a2 −a3 ) (1) ; we may assume that x1 ∈ x±2a3 (1).    Then O 2 (CK (x1 )) = H0 , so O2 (CK (x1 , z)) = O2 (CH0 (z)) ∼ = Z2 × Σ4 .  However, CK (z) = O2 (CK (z))H1 with x1 ∈ H1 ∼ = = Σ6 , so O2 (CH1 (x1 )) ∼  Σ3 . It follows that CO2 (CK (z)) (x1 ) ∼ = E23 , as claimed in (c).

496

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∼ Z2 × Sp6 (2). Then the Coxeter Lemma 10.2.9. Let W = W (E7 ) = involutions of W are products tz where t = O2 (W ) and z is a long root involution of [W, W ] ∼ = C3 (2). Proof. Since the Coxeter involutions are all conjugate and generate W , they must project nontrivially on < t >, and hence have the form tz for various conjugate z ∈ I2 ([W, W ]). From the Dynkin diagram, the centralizer of one such involution contains W (D6 ) and hence has order divisible by 32 .5. The lemma then follows from Lemma 10.2.6.  Lemma 10.2.10. Let K = Sp6 (2), A ∈ E33 (K), and N = NK (A). Then the following conditions hold: (a) N = Nb Nc ∼ = Σ3 Σ3 . The base group Nb is the direct product of three orthogonal long root L2 (2) subgroups, permuted by Nc ∼ = Σ3 . (b) Suppose τ ∈ I2 (N ), let Aτ = [A, τ ], and let Oτ be the orbit of N on E1 (A) containing Aτ . Then |Aτ | = 3 if and only if one of the following holds: (1) |Oτ | = 3 and τ  Aτ ∼ = Σ3 is one of the Σ3 -direct factors of Nb ; or (2) |Oτ | = 6 and τ is N -conjugate to an involution of Nc . Proof. We use standard notation for K [IA , 2.10]. By [IA , 4.8.1], A is unique up to conjugacy. The three root L2 (2)-subgroups X±2ai  =: Li , i = 1, 2, 3, generate their direct product Nb , which we may assume contains A. The monomial subgroup of K contains a Σ3 -subgroup Nc =

na1 −a2 , na2 −a3  permuting the Li naturally (and [na1 −a2 , X±2a3 ] = 1). Now Li acts naturally on a hyperbolic plane Vi of the natural F2 K-module V and centralizes Vi⊥ . So NK (A) permutes the Vi , and as CK (A) = A, NK (A) = Nb Nc , proving (a). If τ = x2ai (1) or nai −aj , then it is straightforward to check that condition (b1) or (b2) holds, respectively, and conversely. Other involutions in Nb invert at least two of the O3 (Li )’s. Other involutions in N − Nb are in Nb nai −aj − Nc for some i = j, so are N -conjugate to σ := na1 −a2 x2a3 (1). But [A, σ] contains a diagonal of O3 (L1 ) × O3 (L2 ) as well as O3 (L3 ), so |[A, σ]| > 3. The lemma follows.  Lemma 10.2.11. Let K = Sp6 (2) and let V be a 6-dimensional faithful F2 K-module. Then V is isomorphic to the natural module. Proof. We consider K ≤ L = GL(V ) and show that the Brauer character χ of K on V is uniquely determined, and use [V9 , 12.1a]. Let Pr ∈ Sylr (K) for r = 3, 5, and 7. For x ∈ I3 (K), x is determined up to K-conjugacy by dx := dim[V, x] (see [IA , 4.8.2]), and χ(x) = (12 − 3dx )/2. If x ∈ K has order 9, then x is rational in K so χ(x) = 0. Similarly if x ∈ I5 (K), then x is rational in K and dim CV (x) = 2, so χ(x) = 1. There is one class of elements x of order 15, namely x = x3 x5 where xr has order r and [V, x3 ] ∩ [V, x5 ] = 0; this implies that χ(x) = −2. Finally, if x ∈ I7 (K),

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then x is also rational in K so χ(x) = −1. Indeed NK ( x) is a Frobenius group of order 7.6 so V is irreducible for K, and thus the isomorphism type of V is uniquely determined. The proof is complete.  Lemma 10.2.12. Let K = Sp6 (2), P ∈ Syl3 (K), and D = J(P ). Then CAut(K) (D) = D. Proof. By [IA , 2.5.12], Out(K) = 1. Now P ∼ = Z3 Z3 and D =

x1 , x2 , x3  where the natural module V is the direct product of the three  four-groups [V, xi ], i = 1, 2, 3. The result follows easily. ∼ L ≤ M ≤ K, with K = ∼ L6 (2) or O+ (2) Lemma 10.2.13. Let Sp6 (2) = 8 and M a K-group. Assume that O2 (M ) = 1 and m3 (M ) ≤ 3, and that there   is x ∈ I3 (L) such that I := O 2 (CM (x)) = O2 (CL (x)) ∼ = Sp4 (2). Then M = L. Proof. As |M | divides |K|, and m3 (M ) = m3 (L) = 3, |M |2 |/|L2 | divides 5.7.31. It follows easily that L centralizes F (M ), and normalizes every component of E(M ). Hence L ≤ E(M ), indeed L ≤ J for some component J  M . Then [I, I] ∼ = A6 ↑3 J via x. This implies in turn that if J ∈ Alt, then J ∼ = A9 , an impossibility as |L| does not divide |A9 |. As L contains a Sylow 3-subgroup Z3 Z3 of J, J ∈ Spor [IA , 5.3]. By considering 3-ranks, the only possible J ∈ Chev(r), r odd, is J ∼ = P Sp4 (3); but then again |L| does not divide |J|, a contradiction. Therefore J ∈ Chev(2). If x does not act on J as an element of Inndiag(J), then J ∼ = Sp4 (8) by [IA , 4.9.1, 4.9.2], again contradicting Lagrange’s theorem. So x acts as an element of Inndiag(J), whence by [IA , 4.2.2, 4.7.3A], J ∼ = Cn (2) for some n, or F4 (2). ∼ By 3-ranks, J = L. so L = J  M . As Out(L) = 1, M = L × CL (M ). Now L is not involved in any proper parabolic subgroup of K, so by the Borel-Tits theorem, |CL (M )| = |CL (M )|2 divides 5.7.31. But also K has no element of order 5.7 or 7.31 [IA , 4.8.2]. So CL (M ) = 1 and M = L, as claimed.  Lemma 10.2.14. Let X = Sp6 (2) ∼ = C3 (2). Let R ∼ = A1 (2) be a subgroup of X generated by a long root subgroup and its opposite, and let R0 = O3 (R). Let V and W be faithful F2 X-modules such that dim(V ) = 6, dim(W ) = 8, dim(CV (R0 )) = 4, and dim(CW (R0 )) = 0. Then HomF2 X (W ⊗F2 W, V ) = 0. Proof. In this proof, tensor products, Homs, dimensions, and isomorphisms are over F2 unless stated otherwise. Let Y = CX (R). On the natural module M , involutions of R are transvections, so [M, R] is nondegenerate and a natural R-module. Thus Y ∼ = Sp4 (2). Next set V  = CV (R0 ), of  dimension 4. We regard V as an F2 Y -module; it is faithful since an element of I5 (Y ) acts nontrivially on V . It then follows [IA , 2.8.10d] that if we choose any non-inner automorphism σ of Y and set W  = (V  )σ , then {V  , W  , V  ⊗ W  } is a complete set of pairwise nonisomorphic nontrivial irreducible F2 Y -modules. The natural module V  is self-dual, and so also is W .

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Let A and B be V  , W  , or a tensor product of these. As B is self-dual, HomF2 Y (A, B) ∼ = CA⊗B (Y ). This fact and Schur’s lemma give (10E) Hom(W  ⊗ W  , V  ) ∼ = CV  ⊗W  ⊗W  (Y ) ∼ = Hom(W  , V  ⊗ W  ) = 0. F2 Y

F2 Y

≤ Y . Since dim(CV (R0g )) = dim(CV (R0 )) = There is g ∈ X such that g gσ g 4, CV  (R0 ) = 0. Then CW  (R0g ) ∼ = CV  (R0 ) = 0, so CV  ⊗W  (R0 ) = 0. g As CW (R0 ) = 0, all composition factors of W ↓Y are isomorphic to W  , and then HomF2 Y (W ⊗ W, V /[V, R0 ]) = 0 by (10E). Thus for any φ ∈ HomF2 X (W ⊗W, V ), φ(W ) ⊆ [V, R0 ] < V , so φ = 0 as V is F2 X-irreducible. The lemma follows.  Rg

Lemma 10.2.15. Let K = Sp6 (2) have natural F2 K-module V . Let H ≤ K with H = (S1 × S2 × S3 )I ∼ = Σ3 Σ3 , Si ∼ =I∼ = Σ3 for i = 1, 2, 3. Let Yi ∈ Syl2 (Si ) and Vi = [V, Yi ], i = 1, 2, 3. Then |Vi | = 2 for each i, and 3 

CV (Yi ) = V1 × V2 × V3 ∼ = E23 .

i=1

Proof. Let P ∈ Syl3 (H), so that P ∼ = Z3 Z3 and J(P )  H. Then P ∈ Syl3 (K). Also we may write J(P ) = x1 , x2 , x3  and Wi := [V, xi ] ∼ = E22 , i = 1, 2, 3, so that V = ⊕3i=1 Wi . Thus NK (J(P )) ∼ = H so H = NK (J(P )) = (GL(V1 ) × GL(V2 ) × GL(V3 ))L ∼ = Σ3 Σ3 . By the Krull-Schmidt theorem, we may renumber the direct factors of O3,2 (H) if needed in order to achieve Si = GL(Wi ), i = 1, 2, 3. Then each Yi is generated by a transvection with center in Wi , and the lemma follows easily.  Lemma 10.2.16. Let K = Sp6 (2) act faithfully on V ∼ = E26 and let γ ∈ I7 (K). Then CK (γ) = 1. Proof. NK ( γ) is a Frobenius group of order 7.6, so all primitive 7th roots of unity are eigenvalues of γ on V . Therefore 1 is not an eigenvalue of γ on V , as required.  Lemma 10.2.17. Let K = Sp6 (2) ∼ = C3 (2). Let L be a short root A1 (2)subgroup of K. Then CK (L) contains a unique long root A1 (2)-subgroup of K, and a unique short root A1 (2)-subgroup of K. Proof. Without loss, L = xα (1), x−α (1), where α = a1 + a2 is the highest short root in a C3 root system {±2ai , ±ai ± aj | i, j = 1, 2, 3, i = j}. Let T ∈ Syl2 (CK (L)) be generated by the positive root subgroups. Then without loss, CG (L) = CG (xα (1)) ∩ CG (x−α (1)) is the Levi factor x±(a1 −a2 ) (1) × x±2a3 (1). In the natural representation of K, the three involutions in T ∩ CG (L) have different numbers of Jordan blocks, so they are pairwise non-conjugate. This implies the lemma.  Lemma 10.2.18. Let K = Sp6 (2) ∼ = C3 (2). Let X1 and X2 be short root A1 (2)-subgroups such that D1 := O3 (X1 ) = O3 (X2 ). Suppose that yi ∈ I2 (Xi ), i = 1, 2, and that y1 , y2  is a 2-group. Then X1 = X2 .

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Proof. By assumption X2 = X1g for some g ∈ K, so g ∈ NK (D1 ). Modifying g by an element of X2 , we may assume that y2 = y1g . Now O3 (NK (D1 )) =: D0 ∼ = E33 . We have NK (D0 ) = (S1 × S2 × S3 )I where Si = ti , zi  ∼ = Σ3 for each i = 1, 2, 3, with t3i = 1; and I = (123), (12) ∼ = Σ3 permutes {S1 , S2 , S3 } by permuting the subscripts on the S , t , and zi . i i  −1 Without loss, D1 = t1 t2 . Then NK (D1 ) = D0 z1 z2 , z3 , (12) has E23 Sylow 2-subgroups and so [NK (D1 ), NK (D1 )] ≤ D0 . Hence y1−1 y2 = [y1 , g] is a 3-element. But y1 , y2  is a 2-group, so y1−1 y2 = 1. Hence X1 = D1 , y1  =

D1 , y2  = X2 , as claimed.  Lemma 10.2.19. Let K = Sp6 (2) ∼ = C3 (2). Then any two A2 (2)subgroups of K containing short root A1 (2)-subgroups are K-conjugate. Proof. Let L1 and L2 be A2 (2)-subgroups of K containing short root subgroups. The irreducible F2 Li -modules are of dimensions 1, 3, 3, and 8. Moreover any element si ∈ I7 (Li ) is fixed-point-free on the natural F2 Kmodule V , so as F2 Li -module, V must have two 3-dimensional composition factors. In particular, Li acts faithfully on some 3-dimensional subspace Vi of V . The action of si forces Vi to be totally isotropic. Hence by Witt’s lemma, we may replace L2 by a K-conjugate and assume that V1 = V2 . Let M = NK (Vi ), so that L1 , L2  ≤ M . We have M = QL where Q = O2 (M ) ∼ = E26 and L ∼ = L3 (2). Conjugating L1 by an appropriate element we may assume that L1 ∩ L2 contains NL1 ( s1 , t), a Frobenius group of order 7.3, and a maximal subgroup of both L1 and L2 . Write NLi ( t) = t, yi , i = 1, 2, such that y1 ∈ Qy2 . Then t, yi  is a short root A1 (2)-subgroup of K for i = 1 and 2, and

y1 , y2  ≤ Q y1 , a 2-group. By Lemma 10.2.18, t, y1  = t, y2 . Hence  L1 = s1 , t, y1  = s1 , t, y2  = L2 , completing the proof. 10.3. U6 (2). Lemma 10.3.1. Let K = U6 (2) and t ∈ I2 (K). Then either m2 (CK (t)) ≥ 9 or |CK (t)|2 ≥ 211 . Proof. K has a parabolic subgroup P = QL with Q = F ∗ (P ) ∼ = E29 ,  2  ∼ ∼ O (L) = L3 (4), and as O (L)-module, Q = V ⊗ V , where V is the natural module and V  its Galois twist. One calculates that for every involution  u ∈ O 2 (L), m2 ([Q, u]) = 4 and so m2 (CQ (u)) = 5. Without loss, t ∈ P . Clearly we may assume that t ∈ Q, but t ∈ QS, where S ∈ Syl2 (L). Replacing t by an L-conjugate, we may assume that t ∈ QZ(S), so t = zu where u ∈ I2 (Z(S)) and z ∈ CQ (u). Now [Q, u]  QS, and we set QS = QS/[Q, u]. Also CQ (u)  QS, and as |CQ (u)| = 2, z ∈ Z(QS). Furthermore [Q, u] = 1 so u ∈ Z(QS). Therefore t ∈ Z(QS) and so  by [IG , 9.16], |CQS (t)| ≥ |QS| = 211 . The lemma is proved. 2

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10.4. L6 (2). Lemma 10.4.1. Let N be a K-subgroup of L6 (2) such that O2 (N ) = 1. Suppose b ∈ I3 (N ) and J ∼ = A6 is a component of CN (b). Then either Sp F ∗ (N ) = b × J or N ∼ = 6 (2). Proof. Sylow p-subgroups of O3 (N ) are cyclic for p > 3, except possibly for p = 7, where they are of order at most 72 . Since O2 (N ) = 1, it follows that [F (O3 (N )), J] = 1. Likewise a J-orbit on components of E(O3 (N )) can have length at most 2, so [E(O3 (N )), J] = 1 by the Schreier property. Therefore [O3 (N ), J] = 1. Then L3 -balance and the fact that m3 (L6 (2)) = 3 imply that L := J N is single component. Thus either L = J or J ↑3 L, where L = L/O3 (L). If L = J, then N normalizes CQE (J) = [QE , b] and [QE , J] = CQE (b), so F ∗ (N ) = b × J. So assume that J ↑3 L. Now m3 (L) ≤ 3 and |L|3 ≤ |L6 (2)|3 = 34 . So L ∈ Alt (A9 is the only candidate, but A9 contains a Frobenius group of order 9.8 so does not embed in L6 (2)). Likewise L ∈ Spor (see [IA , 5.3]; |Suz|3 > |J3 |3 > 34 and m3 (O N ) = 4). If L ∈ Chev(3), then by [IA , 4.9.1, 4.9.2], L ∼ = L2 (36 ) so m3 (L) > 4. Finally suppose that J ∈ Chev(2). Now ∼ A6 = [Sp4 (2), Sp4 (2)] is the only realization of A6 in Chev(2). Thus either |L|3 = |Sp4 (8)|3 > 34 or b maps into Inndiag(L) with q(L) = 2 and the untwisted Dynkin diagram of L has a double bond [IA , 4.2.2, 4.9.1, 4.9.2]. But Sp8 (2) ≤ F4 (2) and |Sp8 (2)|3 > 34 , so the only possibility is L ∼ = Sp6 (2). Then the Schur multiplier of L is a 2-group and Out(L) = 1 [IA , 6.1.4,  2.5.12], so N ∼ = L. The proof is complete. 10.5.

2F

4 (q)

.

∼ 2F 4 (2) and R ∈ Syl (K). Then K has two Lemma 10.5.1. Let K = 2 classes of maximal parabolic subgroups, represented by P1 and P2 , of respective orders 211 · 5 and 211 · 3, and with Z(P1 ) = Z(R) = z ∼ = Z2 and 1 2 ∼ ∼ ∼ 2 Z(O2 (P2 )) = E22 . Also P1 /O2 (P1 ) = B2 (2 ) = F20 and P2 /O2 (P2 ) ∼ = Σ3 . Finally Z2 (O2 (P2 )) ∼ = E23 and P2 permutes Z2 (O2 (P2 )) − Z(O2 (P2 )) transitively, all of whose involutions have centralizer of order 29 · 3. 1

Proof. Let K ∗ = 2F4 (2 2 ), which has two conjugacy classes of maximal parabolic subgroups, represented by P1∗ and P2∗ , with P1∗ /O2 (P1∗ ) ∼ = 2B (2 21 ) ∼ F ∗ ∗ ∗ ∼ = 20 and P2 /O2 (P2 ) = Σ3 . Set Pi = Pi ∩ K, i = 1, 2; then 2 |Pi | = |Pi∗ |/2 and by [IA , 3.3.2], Pi /Ui ∼ = Pi∗ /O2 (Pi∗ ), where we set Ui = O2 (Pi ). We adopt the notation of [IA , 2.4.5d], so that we may take K ∗ to be generated by involutions xi , i even, and yi , i odd, 0 ≤ i ≤ 15, as well as elements xi , i odd, such that x2i = yi ; P1 ∩ P2 =: R ∈ Syl2 (K) is generated by x2i and y2i−1 , 1 ≤ i ≤ 4, and xi xi+2 , i = 1, 3, 5; and the commutator relations among the xi and yi are as in [IA , 2.4.5d]. Moreover, using these commutator relations, we see that P1 = S, x9  with Z :=

y5  = Z(P1 ) = Z(S), and P2 = S, x0  with U := y3 , y5  = Z(O2 (P2 )) and V := U x4  = Z2 (O2 (P2 )). Using the Borel-Tits theorem in K ∗ , we

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see that P1 = CK (Z) and P2 = NK (U ) = NK (V ). Furthermore, for some w3 ∈ I3 (P2 ), CV (w3 ) = x4 . Set C4 = CK (x4 ). Then |C4 ∩ P2 | = 29 · 3, and xP4 2 = V − U . Using the Borel-Tits theorem in K ∗ again, we see that |C4 | divides 29 · 3, so C4 ≤ P2 . The remaining statements in the lemma hold by  [IA , 3.3.2e]. Lemma 10.5.2. Let K ∼ = 2F 4 (2) . Then K has 2 classes of involutions, # represented by z ∈ Z(P1 ) and t ∈ Z2 (O2 (P2 )) − Z(O2 (P2 )). Proof. Set z = y5 and t = x4 . We must show that any involution u ∈ K is conjugate to z or t. We first claim that |CK (u)| is divisible by either 3 or 5. Namely, by the Baer-Suzuki theorem and Lemma 10.5.4, u inverts some element x ∈ K of odd prime order p. If p = 3, then u acts on O3 (CK (x)) ∼ = 31+2 so 3 divides |CK (u)|. Likewise if p = 13, then as a Sylow 13-normalizer is a Frobenius group of order 13 · 6, 3 divides |CK (u)|. Suppose then that p = 5. Since NK (P5 ) is strongly 5-embedded in K, we may assume that u ∈ NK (P5 ). Then according as u does or does not invert all of P5 , 3 or 5 divides |CK (u)| by the structure of NK (P5 ), proving our claim. Accordingly, by the Borel-Tits theorem in K ∗ , we may replace u by a conjugate and assume that CK (u) ≤ P2 or CK (u) ≤ P1 . If CK (u) ≤ P1 , let w5 ∈ I5 (CK (u)). From [IA , 4.8.7], CK (w5 ) has cyclic Sylow 2-subgroups, so u is the unique involution in CP1 (w5 ). Hence u = z. Similarly, if CK (u) ≤ P2 , then we may assume that u ∈ CP2 (w3 ). By [IA , 4.7.3A], CK (w3 ) has Sylow 2-subgroups of 2-rank 1, so u =  CZ2 (O2 (P2 )) (w3 ) = t. The proofs are complete. Lemma 10.5.3. Let K ∼ = 2F 4 (2) . Then P1 and P2 are the unique maximal subgroups of K containing CK (z) and CK (t), respectively. Proof. We first claim that any subgroup H < K such that |H|2 ≥ 29 lies in a conjugate of P1 or P2 . To do this, it suffices by the Borel-Tits theorem (in K ∗ ) to show that O2 (H) = 1. Suppose then that O2 (H) = 1. Let S ∈ Syl2 (H). For any element xp ∈ Ip (K), CK (xp ) is a {2, p}-group, so F (H) is a p-group. Whether p = 3, 5, or 13, | Aut(F (H))|2 ≤ |GL2 (p)|2 < 29 , so CS (F (H)) = 1. Therefore E(H) = 1. By the Borel-Tits theorem, any 2-local subgroup of K, like P1 and P2 , is solvable; and the same holds for odd local subgroups, by our knowledge of p-local structure [IA , 4.7.3A, 4.8.7]. That is, H is an N -group. It follows that F ∗ (H) is a simple N 1 group with |H|2 ≥ 29 and |H| dividing |2F4 (2 2 ) |. Also m2 (H) ≤ 5, by [IA , 3.3.3]. If F ∗ (H) ∼ = An , then as 7 does not divide |H|, n ≤ 6 so |H|2 < 1 9 2 , contradiction. If F ∗ (H) ∈ Spor, then |H| does not divide |2F4 (2 2 ) |. If F ∗ (H) ∈ Chev(r) with r > 2, then F ∗ (H) ∼ = L2 (q) with q dividing 1 ± 2 2 ∗ ∼ 3 · 5 · 13 or F (H) = L3 (3), and again |H| does not divide |2F4 (2 2 ) | or |H|2 < 29 . So F ∗ (H) ∈ Chev(2) unambiguously. If F ∗ (H) has twisted rank 1 1, then the conditions that m2 (H) ≤ 5 and |H| divides |2F4 (2 2 ) | imply that F ∗ (H) ∼ = U3 (4), in which some 5-local subgroup involves A5 , contradiction.

502

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So F ∗ (H) has twisted rank at least 2. Since all parabolic subgroups are solvable, and H < K, q(F ∗ (H)) ≤ 2 and we easily reach the contradiction |H|2 < 29 . Hence our claim holds. In particular, P1 and P2 are maximal subgroups of K. Since P1 = CK (z), the lemma holds for z. Suppose then that P is a maximal subgroup of K containing CK (t), but with P = P2 . Then |P ∩ P2 | is divisible by 29 · 3, and in particular P = P2g for some g ∈ K − P2 . Let w3 ∈ I3 (CK (t)). Then t ∈ I2 (CP (w3 )), whence t ∈ Z2 (O2 (P )) − Z(O2 (P )). But also tg ∈ Z2 (O2 (P )) − Z(O2 (P )). Hence th = tg for some h ∈ P , by Lemma 10.5.1. −1 −1 Then gh−1 ∈ CG (t) ≤ P2 so P = P h = P2gh = P2 , contradiction. The proof is complete.  Lemma 10.5.4. Let K ∼ = 2F 4 (2) . Then K has one class of subgroups of order p, for p ∈ {3, 5, 13}, represented, say, by xp . NK ( x13 ) ∼ = F13·6 . Let xp ∈ Pp ∈ Sylp (K). Then CK (x3 ) is isomorphic to a subgroup of SU3 (2) of index 2; in particular we can take P3 = O3 (C(x3 )) ∼ = 31+2 . Finally NK (K5 ) = K5 .L5 is strongly 5-embedded in K, where K5 ∼ = E52 is selfcentralizing and L5 ∼ = SL2 (3) ∗ Z4 . Proof. For p = 3, every element of order 3 is 3-central with centralizer 1 SU3 (2) in 2F4 (2 2 ), which implies the desired statements [IA , 4.7.3A]. For 1 p = 5, the assertions follow easily from the corresponding facts for 2F4 (2 2 )  [IA , 4.8.7]. Then Sylow’s theorem implies the assertions for p = 13. Lemma 10.5.5. Let K ∼ = 2F 4 (2) and P2 the unique maximal subgroup of K containing CK (t). If s ∈ I2 (G) with CK (s) ≤ P2 , then s ∈ Z2 (O2 (P2 )) − Z(O2 (P2 )). Proof. By Lemma 10.5.2, t ∈ R0 := O2 (P2 ). By Theorem 2.1(iii) of [G1], [R0 , t]  R0 . Let x ∈ CK (t) of order 3. Then [R0 , t]  R0 x. Also, t ∈ Z(R0 ) by Lemma 10.5.2. So, [R0 , t] = 1 and then Z(R0 ) ∩ [R0 , t] = 1. Since x normalizes both Z(R0 ) and [R0 , t], x normalizes Z(R0 ) ∩ [R0 , t]. As x acts non-trivially on Z(R0 ) ∼ = E22 (for z ∈ Z(R0 ) and (3, |CK (z)|) = 1 0| = 4, and so by Lemma 10.5.1), Z(R0 ) ≤ [R0 , t]. Now |tR0 | = |C|R R0 (t)| |[R0 , t]| = 4. Thus Z(R0 ) = [R0 , t]. Suppose there exists an involution s ∈ R0 − Z(R0 ), t with CK (s) ≤ P2 . Then using the same arguments as above we obtain that Z(R0 ) = [R0 , s]. Thus Z(R0 )t, Z(R0 )s ≤ Z(R0 /Z(R0 )), and in fact, Z(R0 )t, Z(R0 )s ≤ Z(O 2 (P2 /Z(R0 ))). Consider W := Z(R0 ), t, s. Then W  O2 (P2 ) and [W, W ] ≤ Z(R0 ) of order at most 2. But x ≤ O2 (P2 ) acts irreducibly on Z(R0 ). Hence, [W, W ] = 1 and W is elementary abelian with [W, x] = Z(R0 ). Thus CK (x) contains a subgroup isomorphic to E22 which contradicts Lemma 10.5.4. The result now follows from Lemma 10.5.1g  1

Lemma 10.5.6. Let K = 2F4 (2 2 ) . Then for any four-subgroup U ≤ Aut(K), ΓU,1 (K) = K.

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Proof. If for at least one of the involutions u ∈ U , ug acts as a 2central element of K, then by Lemma 10.5.2, without loss of generality we may assume that CK (ug ) = P1 , a maximal parabolic of K with a center of order 2. Let v ∈ U − u. If v g also acts as a 2-central involution on K, CK (ug ) = CK (v g ), and so ΓU g ,1 (K) = K, a contradiction. Then by Lemmas 10.5.5 and 10.5.1, 3 divides |CK (v g )|. Since by Lemmas 10.5.1 and 10.5.2, |P1 | = 211 · 5, CK (ug ) ≤ P1 . Again, maximality of P1 implies that ΓU g ,1 (K) = K, a contradiction. Therefore for every involution u ∈ U , ug acts on K as a non-central involution. Hence, Lemma 10.5.3 and the fact that ΓU g ,1 (K) < K, implies that CK (wg ) ≤ P2 for every w ∈ I2 (U ). But this contradicts Lemma 10.5.5, completing the proof.  5

Lemma 10.5.7. Let K = 2F4 (2 2 ). Then any element of I5 (K) centralizes a 2-central involution of K. Proof. By [IA , 4.8.7], I5 (K) is a single conjugacy class. The lemma then follows directly from Lemma 17.1.  n

Lemma 10.5.8. Let K = 2F4 (2 2 ). Then I5 (K) is a single conjugacy n+1 n class. If x ∈ I5 (K), then CK (x) ∼ = Zm × 2B2 (2 2 ), where m = 2n + n 2 2 + 1 ≡ 0 (mod 5) and n = ±1. In particular, if n = 1 or 5, then m = 5 or 52 , respectively. Proof. By [IA , 4.8.7], I5 (K) is a single conjugacy class. Let Σ be the root system of F4 and Π a fundamental system in Σ as in [IA , 1.8.8], so that Π = {a2 − a3 , a3 − a4 , a4 , a+−−− }. Let Σ1 be the (B2 ) subsystem generated by a3 and a4 , and Σ2 the (B2 ) subsystem generated by a1 and a2 . The unique angle-preserving and length-changing map ρ : Σ → Σ carrying Π to Π then obviously preserves Σ1 , hence also its orthogonal complement Σ2 . Let K = F4 (F2 ) and let ψ ∈ Aut(K) be as in [IA , 1.15.4], i.e., ψ(xα (t)) = xαρ (t) or xαρ (t2 ) according as α is long or short. Then ψ 2 = φ2 , the Frobenius, and (K, ψ n ) is a σ-setup for K. Clearly ψ preserves the subsystem subgroups K 1 and K 2 of K corresponding to Σ1 and Σ2 , so 





K = O 2 (CK (ψ n )) ≥ O2 (CK 1 (ψ n )) × O2 (CK 2 (ψ n )) =: K1 × K2 , n with K1 ∼ = K2 ∼ = 2B2 (2 2 ). Let T1 ∈ Syl2 (K1 ) and let P1 be a parabolic subgroup of K (existing by the Borel-Tits theorem) containing NK (T1 ) and hence containing K2 × NK1 (T1 ). Then P1 contains K2 × Z, where Z ∼ = Z2n −1 acts faithfully on T1 . So K2 × Z must be isomorphic to P1 /O2 (P1 ) (the other class of maximal parabolic subgroups has Levi factor only L2 (2n ) × Z2n −1 , which does not contain a copy of K2 ). Now K2 ≤ CK (K1 ) ≤ CK (T1 ) ≤ P1 . But O2 (CK (K1 )) is K1 × K2 -invariant and hence trivial by the Borel-Tits theorem. As CZ (K1 ) ≤ CZ (T1 ) = 1,

CK (K1 ) = K2 .

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 ∼ K1 . Therefore Finally, let x ∈ I5 (K2 ). By [IA , 4.8.7], O2 (CK (x)) = O (CK (x)) = K1 . Also as NK (T1 ) ≤ P1 and NK1 K2 (T1 ) covers P1 /O2 (P1 ),  NK (K1 )/K1 K2 is a 2-group. As O2 (CK (x)) = K1 , it follows that CK (x) ≤ K1 K2 . Then CK (x) = K1 × CK2 (x), whose precise structure is given in  [IA , 6.5.4]. The lemma is proved.

2

10.6. L3 (4) and L3 (4n ). ∼ L3 (4), where Lemma 10.6.1. Let K be quasisimple with K := K/Z = ∼ Z = Z(K) = E22 . Let S ∈ Syl2 (K) and identify S with its image in Aut(K). Let u ∈ Aut(K) be a (noninner) involution such that u normalizes S and CK (u) ∼ = U3 (2). Then the following conditions hold: (a) Z(S) = Z(S); (b) If B ≤ K with B ∼ = E32 , then CAut(K) (B) = B × u , where  u ∈ Aut(K) is Aut(K)-conjugate to u; (c) All involutions in the coset uS are S-conjugate, and all involutions in S are K-conjugate; (d) u centralizes some involution y ∈ Z(S) − Z(K); and (e) In (d), if z ∈ Z(K) is an involution, then no two involutions in

y, z are Aut(K)-conjugate. Proof. Part (a) is a direct consequence of [IA , 6.4.2b], which also implies that Z(S) ∼ = E24 . Then as m2 (CS (u)) = 1, u acts freely on Z(S), which implies (d). Any conjugacy in (d) would have to occur in CK (y) = S. As y ∈ Z(S), (e) holds. In (b), B ∈ Syl3 (K) is self-centralizing in K since its preimage in SL3 (4) is absolutely irreducible on the natural module; the assertions of (b) and (c) follow from [IVK , 2.1ae] and the fact that u acts freely on S/Z(S) and Z(S).  Lemma 10.6.2. Suppoise K = F ∗ (X) ∼ = L3 (4) and let S ∈ Syl2 (X). Then S has no normal Z4 -subgroup. Proof. We have S = T φ, γ where T = S ∩ K, and φ and γ are, respectively, field and graph automorphisms of order 2. It is easily checked that (10F)

|S : CS (t)| ≥ |T : CT (t)| > 2

for all t ∈ S − T . Moreover, T = F1 F2 with F1 ∼ = F2 ∼ = E24 and F1 ∩ F2 = Z(T ) = [Fi , gi ] for any gi ∈ T − Fi , as is evident from the Chevalley commutator formula [IA , 2.4]. Hence (10F) holds for all t ∈ T − Z(T ) and in particular for any t ∈ S of order 4. The result follows.  Lemma 10.6.3. Let K = L3 (4), and let T ∈ Syl3 (K) and V ∈ Syl7 (K). The K = T, V . Proof. This follows easily from [IA , 6.5.3].



Lemma 10.6.4. Let K = L3 (4) and T ∈ Syl3 (K). Then T = CK (x) for all x ∈ T # , and |CAut(K) (T ) : T | = 2.

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Proof. It follows from [IA , 6.5.3] that the unique maximal subgroup of K containing T is NK (T ) ∼ = U3 (2), which is a Frobenius group with kernel T . This proves the first statement. Sylow 3-subgroups of P GL3 (4) are nonabelian of order 32 so CAut(K) (T )/T is a 2-group embedding in Out(K). As field automorphisms of K of order 2 do not centralize T , but unitary-type  automorphisms of order 2 do, the result follows (see [IA , 2.5.12]). Lemma 10.6.5. Let X = P GL3 (4), K = [X, X] = L3 (4), P ∈ Syl3 (X), A5 , and and B = P ∩ K ∼ = = E32 . Let d ∈ P − B with CK (d) ∼  let h ∈ B − Z(P ). Then Z(P ) d ∼K Z(P ) dh or Z(P ) d ∼K Z(P ) dh−1 .  and B  ∈ Syl3 (SL3 (4))  = GL3 (4), and let P ∈ Syl3 (X) Proof. Let X ∼  ∼ be the full inverse images of P and B. Then P = Z3 Z3 and B = 31+2 .  has order 3, then Using [IA , 4.8.1, 4.8.2, 4.8.4] we see that if x  ∈ P − B ∼ x) = A5 . Also, if x  has order 9 and maps on x ∈ P , then x is CK ( determined up to K-conjugacy (and CX (x) ∼ = Z21 ). As two of the four subgroups of P/Z(P ) of order 3 pull back to subgroups of P of exponent 9,  has exponent 3, the result follows. while B  Lemma 10.6.6. A7 is not involved in L3 (4). Proof. If it were, L3 (4) would have a proper subgroup of index at most A8 [IA , 2.2.10] and |L3 (4)| = |A8 |, 8 and so would embed in A8 . But L3 (4) ∼ =  contradiction. One could also quote [IA , 6.5.3]. Lemma 10.6.7. Let X = Aut(K) where K = L3 (4n ). Let B = U H be a Borel subgroup of Inndiag(X), with Sylow 2-subgroup U and Cartan subgroup H. Let A ≤ U with A ∼ = E24n . Then the following hold: ∗ (a) U ∈ NCX (Z(U )) (Ω1 (O3 (H); 2)); and (b) CX (A) = A. Proof. In (a), the group ΦK of field automorphisms acts faithfully on Z(U ), so CX (Z(U )) = U CH (Z(U )) g, where g is a graph automorphism. Here CH (Z(U )) ∼ = Z4n −1 is inverted by g, so (a) follows. We have NX (A) = Y ΦK , where Y is a maximal parabolic subgroup of X and A = O2 (Y ) is the product of two root subgroups, each admitting ΦK naturally. Thus  CX (A) = CY (A) = A, as required. Lemma 10.6.8. Let K = L3 (4n ) and let x ∈ I3 (Inndiag(K)) with L := E(CK (x)) ∼ = L2 (4n ). Let B = U H be a Borel subgroup of L, with |U | = 4n and |H| = 4n − 1. Let S = O2 (CK (U )). Then S ∈ Syl2 (K) and CS (h) = 1 for all h ∈ H # . Proof. Without loss, x = diag(1, ω, 1), where ω 3 = 1 = ω ∈ F4n . Then we may take U to consist of all matrices I + αe13 , α ∈ F4n (where e13 is a matrix unit), and H to consist of all diag(α, 1, α−1 ), α = 0. As a result, S is the unipotent upper triangular group and CK (h) has odd order for all h ∈ H, so the lemma holds. 

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∼ L3 (4a ). Let S ∈ Syl (X) and U  S Lemma 10.6.9. Let F ∗ (X) = K = 3 with U ∼ = E32 . Suppose that no element of I3 (S) induces a field automorphism on K. Then one of the following holds: (a) E(CK (u)) ∼ = L2 (4a ) for some u ∈ U # , and U is the unique diagonalizable E32 -subgroup of S ∩ K; or (b) |K|3 = 32 . Proof. We may assume that S = Ω1 (S)(S ∩ K), and hence, by [IA , 4.9.1], that X = K or Inndiag(K). We have S = T n where T = O3 (H), H is a Cartan subgroup, diagonal with respect to some basis (which we fix) in the natural SL3 (4a )-module, and n permutes that basis in a 3-cycle. Now T ∼ = Zh ×Zh or Zh ×Zh/3 , where h = (4a −1)3 . Hence |CS (u)| ≥ |S|/3 ≥ h2 /3 for every u ∈ U # . One computes that |CS (x)| = |CS (n)| = 32 for every x ∈ S − T , so if (b) fails, then U ≤ T . Hence U is diagonal and (a) follows easily.  n

10.7. 2B2 (2 2 ), n ≥ 3. n

Lemma 10.7.1. Let K = 2B2 (2 2 ), n ≥ 3, and let S ∈ Syl2 (K). Then CAut(K) (S) is the image of Z(S) in Inn(K). Proof. This is a special case of [IA , 2.6.5e], with S here in the role of  UJ there. Lemma 10.7.2. Let H be a K-group with O2 (H) = 1. Let T ∈ Syl2 (H). Then any 3-element a ∈ CH (T ) centralizes O3 (H). Proof. Obviously [O2 (H), a] = 1, and as each component of E(O3 (H)) is a Suzuki group and [T, a] = 1, [E(O3 (H)), a] = 1 [IA , 3.1.4]. Therefore  [O3 (H), a] = 1 by [IG , 3.17(ii)]. Lemma 10.7.3. Suppose that A is a 3-group acting on a K-group K = O3 (K). If [T, A] = 1 for some T ∈ Syl2 (K), then [K, A] ≤ O2 (K). Proof. We may assume O2 (K) = 1, so F ∗ (K) = O2 (K)E(K) with every component of E(K) a Suzuki group. Thus by [IA , 2.5.12], A induces field automorphisms on K. Then [T, A] = 1 implies [E(K), A] = 1 so  [F ∗ (K), A] = 1. By [IG , 3.17(ii)], [K, A] = 1, as required. 10.8. Sp4 (2) ∼ = Σ6 . Lemma 10.8.1. Let K = Σ6 . Up to isomorphism, there are four irreducible F2 K-modules. They are absolutely irreducible and self-dual. The degrees are 1, 4, 4, and 16, and the two modules of degree 4 are quasiisomorphic. On a given 4-dimensional module V , some elements v ∈ I3 (K) satisfy dim CV (v) = 2 and others satisfy dim CV (v) = 0. Upon restriction to A6 , the modules of degree 4 remain absolutely irreducible. Proof. One 4-dimensional module, V0 , is the core of a 6-dimensional F2 K-permutation module. The other is its twist under an outer automorphism of K, which interchanges the two conjugacy classes of Σ5 -subgroups

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of K. Both are absolutely irreducible, even after restriction to A6 , since A6 contains a Frobenius group of order 9.4. For any x ∈ I3 (K), dim CV0 (x) = 2 iff x is a 3-cycle; otherwise dim CV0 (x) = 0. These dimensions are reversed in the twisted module. Finally, Σ6 has a rational ordinary (absolutely) irreducible character of degree 16, which remains irreducible modulo 2 since it has defect 0. As Σ6 has four conjugacy classes of elements of odd order, there are just four absolutely irreducible modular characters modulo 2, by a basic theorem of Brauer [V9 , 12.1b], so the asserted list of modules is complete.  Lemma 10.8.2. Let K = Σ6 and let α ∈ Aut(K) − Inn(K). Fix a transitive permutation action of K of degree 6 and let V be the core of the corresponding permutation module over F3 . Let W be the nontrivial 1-dimensional F3 K-module. Then V ⊗ W ∼ = V α. Proof. Since V is irreducible, so is V α . As α interchanges the two faithful permutation representations of K of degree 6, V α is the core of the other permutation module of degree 6. By [V9 , 12.1], it is enough to show that β α = βσ, where β is the Brauer character of V and σ is the sign character on K (restricted to 3-regular elements). The nonidentity even 3-regular elements x ∈ K have cycle shapes (2)2 , (4)(2), or (5). Accordingly, β(x) = 0, −2, or −1, and xα has the same cycle shape, so the desired equality holds in this case. The odd 3-regular elements have cycle shapes (2), (2)3 , and (4), the first two interchanged by α and the third fixed (as a conjugacy class) by α. Accordingly, we have β(x) = 2, −2, and 0, so again the desired equality holds. The proof is complete.  Lemma 10.8.3. Let K = Σ6 , acting naturally on Ω = {1, 2, 3, 4, 5, 6}. Let V be the F3 K-module which is the core of the permutation module F3 Ω. Let H ≤ K with H ∼ = Σ4 × Z2 . Suppose that the involution in Z(H) induces a reflection on V . Then H is the stabilizer in K of some 2-element subset of Ω. Proof. Let α be a non-inner automorphism of K, and let W be the nontrivial 1-dimensional F3 K-module. By Lemma 10.8.2, V α ∼ = V ⊗ W. Now K has two conjugacy classes of Z2 × Σ4 subgroups, interchanged by α. Hence if H is not the stabilizer in K of some 2-element subset of Ω, then H α is the stabilizer in K of, say, {1, 2}. Then on V considered as H α -module, Z(H) = (12) induces a reflection ρ with center F3 (v1 − v2 ). But then on V considered as H-module, Z(H) induces −ρ, which is not a reflection. Since Z(H) ≤ A6 , as H-module, V is the sum of a projective 3dimensional module and a 1-dimensional module centralized by Z(H). This contradicts our hypothesis and completes the proof.  10.9. U4 (2) ∼ = P Sp4 (3). Lemma 10.9.1. Let K = U4 (2) and let V be an F2 K-module such that for some B ∈ E33 (K), CV (B) = 0. Then dim V = 10.

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Proof. Suppose that dim(V ) = 10. Let β be the Brauer character of V , restricted to B, where it is an ordinary character too. Then (β, 1B ) = 0 by hypothesis. This implies that   (10G) 10 + 6β(x1 ) + 12β(x2 ) + 4 β(x3 ) + β(x−1 3 ) = 0, where all xi ∈ B # and dim[W, xi ] = 2i, W being the 6-dimensional module arising from the isomorphism K ∼ = Ω− 6 (2). 1+2 ∼ Now x3  = Z(P ) for some 3 = P ≤ K. As V ↓P is faithful, the only possibility is β(x3 ) = 6ω + 4, and so β(x3 ) + β(x−1 3 ) = −6 + 8 = 2. Then (10G) yields (10H)

3 + β(x1 ) + 2β(x2 ) = 0.

K Also, xK 1 and x2 are the two conjugacy classes in I3 (K) represented in an Σ6 -subgroup H ≤ K. Now the (absolutely) irreducible F2 H-modules are 1, 4, 4, and 16, the modules themselves being X1 , X4 , X4α , and X16 , where α ∈ Aut(H) − Inn(H). Interchanging X4 and X4α if necessary, we may assume that the Brauer character βX4 of X4 satisfies

βX4 (x1 ) = βX4α (x2 ) = −2 and βX4 (x2 ) = βX4α (x1 ) = 1. The minimum value of the left side of (10H) occurs if the H-composition factors of V are two X1 ’s and two X4α ’s, in which case β(x1 ) = 2 and β(x2 ) = −2 and the left side of (10H) is positive. Hence it is positive in all cases. This contradiction proves the lemma.  Lemma 10.9.2. Let K = F ∗ (X) ∼ = U4 (2) and let z ∈ I2 (X) be 2-central. Then z ∈ K and the following hold: (a) CX (z)/O2 (CX (z)) ∼ = Σ3 × Σ3 or Σ3 × Z3 ;   3 3 (b) O (CX (z)) = O (CK (z)) = K1 ∗ K2 , where K1 ∼ = K2 ∼ = SL2 (3); 2 t (c) CK (z) = K1 K2 t, where t = 1, Ki = K3−i , and t ∈ z K ; (d) For t as in (c), tz ∈ z K , (tz)K ∩K1 K2 = ∅, and CK (tz) ∼ = D8 Y Σ4 ; and (e) If X ∼ = GL2 (3) = Aut(K) and b ∈ I2 (NCX (z) (K1 ))−K, then K1 b ∼ and CK (b) ∼ = Z2 × Σ4 . Proof. We have K ∼ = P Sp4 (3), with z the image of an involution in  K := Sp4 (3) with two eigenvalues equal to 1. Also, Aut(K) = Inndiag(K) [IA , 2.2.10, 2.5.12]. Part (a) (when X = Aut(K)) and part (b) follow from  contains Sp2 (3) Z2 , which implies the rest of (a) and (c) [IA , 4.5.1]. Also K  = 2, so non Now m2 (K) (t can be taken to pull back to an involution in K).  2-central involutions of K pull back to elements of K of order 4, and so there exists no four-group in K consisting of 2-central involutions. This implies all but the last statement of (d), and also that tz is half 2-central. Viewing K as Ω5 (3) and tz = diag((−1)2 13 ), the structure of CK (tz) is easily verified. Finally, by [IA , 4.5.1], to prove (e) it is enough to rule out the possibility CK (b) ∼ = Σ6 . But in that case, viewing X as SO5 (3), z and b both have four eigenvalues equal to −1 on the natural 5-dinensional F3 -module V5 .

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Hence on the support V4 of z on V5 , b has three eigenvalues equal to −1. This implies that b induces a graph automorphism on Ω(V4 ) ∼ = K1 ∗ K2 , so b interchanges K1 and K2 , contrary to assumption. The proof is complete.  Lemma 10.9.3. Let L = U4 (2), M = Aut(L), and u ∈ I2 (M ). Then the following conditions hold: (a) m2 (CM (u)) = 4; (b) There is no subgroup of M isomorphic to Z4 × E8 ; and (c) If u ∈ I2 (M − L), there exists x ∈ CL (u)# with ux ∈ uL . Proof. We identify M = SO(5, 3). Then every involution of M is diagonalizable with respect to a suitable basis for the natural module V of M . Hence, (a) holds. Moreover, if E ≤ M with E ∼ = E24 , then E is diagonalizable and, by consideration of eigenspaces, CM (E) is diagonal, whence CM (E) = E, proving (b). As CM (u) = u × CL (u) and uL ∩ L = ∅,  (c) follows by the Z ∗ -Theorem. Lemma 10.9.4. Let K = U4 (2) and P ∈ Syl3 (K). Then P ∼ = Z3 Z3 and J(P ) ∼ = E33 . Moreover, AutK (J(P )) ∼ = Σ4 preserves a nonsingular quadratic form on J(P ) and contains a reflection inverting a subgroup of J(P ) of order 3 with exactly six AutK (J(P ))-conjugates. Proof. With respect to a suitable orthonormal frame of the natural K-module V , P is contained in the monomial group, which quickly yields P ∼ = Z3 Z3 with J(P ) the diagonal subgroup. The frame consists of all the J(P )-invariant 1-subspaces. Then NK (J(P )) is the full monomial group within SU4 (2), proving A := AutK (J(P )) ∼ = Σ4 . A permutation matrix t corresponding to a transposition lies in NK (J(P )) (characteristic 2!) and its image t ∈ AutK (J(P )) is a reflection. Now J(P ) has exactly three subgroups P1 , P2 , P3 of order 3 such that CV (Pi ) = 0. These subgroups generate J(P ); A then acts monomially on J(P ) with respect to them. As t ∈ [A, A], t has exactly six conjugates, as claimed. Finally, the nonsingular quadratic form q on J(P ) such that q(x) = 1 for all x ∈ ∪i Pi# , and such that distinct Pi ’s are orthogonal, is clearly preserved by the monomial action of A. The proof is complete.  Lemma 10.9.5. Let K ∼ =B≤ = U4 (2), S ∈ Syl3 (K), b ∈ Z(S) and E33 ∼ S. Then the following conditions hold: (a) CK (b) is a semidirect product of 31+2 by SL2 (3); (b) NK (B), a maximal subgroup of K, is a semidirect product of E33 by Σ4 ; (c) Aut(K) does not contain a subgroup isomorphic to P GL2 (9); and (d) NK (B; 3 ) = {1}. Proof. We use the fact that K is isomorphic to the adjoint version of B2 (3), and from this point of view, Aut(K) = Inndiag(K). The two maximal parabolic subgroups P1 , P2 of containing have Levi decompositions Pi = Ui Li where U1 ∼ = E33 , L1 ∼ = Sigma4 , U2 ∼ = 31+2 , and L2 ∼ = SL2 (3). A

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Cartan subgroup of K is of order 2, while a Cartan subgroup H of Aut(K) is isomorphic to E22 . We may take H to normalize S. Choose h ∈ H −K; then Pi∗ = Pi h, i = 1, 2, are the two maximal subgroups of Aut(K) containing S. Clearly B is conjugate to J(P ) = O3 (P1 ) and b to Z(P ) = Z(O3 (P2 )). Thus NK (B) ∼ = P1 and NK ( b) ∼ = P2 , proving (a) and (b). Suppose that ∼ L = P GL2 (9) with L ≤ Aut(K). We may assume that S ∩ L ∈ Syl3 (L). Then NL (S ∩ L) ≤ Pig for some i = 1, 2 and some g ∈ G with E32 ∼ = S ∩L ≤ Ui , by the Borel-Tits theorem. But P1 /O3 (P1 ) has no element of order 8, and for any V ≤ U2 with V ∼ = E32 , |NP2 (V )|2 = 2, a contradiction. Thus  (c) holds as well. Finally, (d) follows from [IA , 7.7.11c]. Lemma 10.9.6. Let K = U4 (2) and let DK ∈ E33 (K). Suppose that  t ∈ I2 (K) and O3 (CK (t)) ∼ = A4 (equivalently, t is a non-2-central involution of K). If v ∈ I3 (DK ∩ CK (t)), then v has exactly three conjugates under the action of NK (DK ). Proof. On the natural F4 K-module V , dim CV (t) = 2 as t is not a transvection. This implies that [V, v] = V , which quickly implies the result.  Lemma 10.9.7. Let K = U4 (2), X = Aut(K), and C = CX (z) where z ∈ I2 (K) is 2-central in K. Let A0 ∈ Syl3 (C) and DK ∈ E33 (CK (A)). Let NX = NX (DK ) and NK = NK (DK ). Then the following conditions hold: (a) A0 ∼ = E32 . One element of E1 (A0 ) has 3 NK -conjugates; another, B, has 6 NK -conjugates; and the remaining two are NK -conjugate and have 4 NK -conjugates; and (b) NX (A0 ) = CX (A0 ) i, t where i, t ∼ = E22 , i inverts A0 , and t acts on A0 as a reflection with [A0 , t] = B. Proof. With respect to an orthonormal basis of the natural K = U4 (2)module, we may assume that z is the transvection I + e14 (where e14 is a matrix unit). Then A0 is the subgroup of the diagonal subgroup with equal first and fourth diagonal entries. Then (a) follows easily; and an element of B # is diag(1, ω, ω −1 , 1), where ω 3 = 1. In (b), X = K i where the graph automorphism i acts on the diagonal subgroup as Galois conjugation of F4 , so i inverts A0 . Also the permutation matrix t = (23) interchanges the two elements of E1 (A0 ) in the 4-orbit of NK . Any element of NX (A0 ) must preserve B and either preserve or interchange the subgroups interchanged  by t, so NK (A0 ) = CK (A0 ) i, t and (b) follows. ∼ Aut(P Sp4 (3)). Let C be a nonsolvable Lemma 10.9.8. Let K = O6− (2) = K-subgroup of K such that m3 (C) = 3. Then C = K or [K, K]. Proof. Suppose false. We may assume that C ≤ [K, K]. By [IA , 7.7.11c], [K, K] is strongly locally balanced for p = 3, so O3 (C) = 1. Also all proper parabolic subgroups of [K, K] are solvable, so all 3-local subgroups of [K, K] are solvable by the Borel-Tits theorem. Hence F ∗ (C) = E(C) is simple and so m3 (Aut(E(C))) = 3. But E(C) is a {2, 3, 5}-group so

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E(C) ∼ = A5 or A6 by [III17 , 9.9]. Thus m3 (Aut(E(C))) ≤ 2, contradiction. The lemma is proved.  10.10. L± 4 (3). Lemma 10.10.1. The following conditions hold: (a) Let P ∈ Syl3 (L4 (3)). Then Ja (P ) = J(P ) ∼ = E34 ; and (b) Let X be an extension of D ∼ = Ω+ = E34 by H ∼ 4 (3), acting naturally, and let P ∈ Syl3 (X). Then Ja (P ) = J(P ) = D. Proof. Since L4 (3) ∼ = Ω+ 6 (3) has a parabolic subgroup M such that 3 X = O (M ) satisfies the hypotheses in (b), the two parts of the lemma are equivalent. Let A be an abelian subgroup of P of order 34 . We must prove that A = D, so suppose not. It follows easily from [IA , 3.2.2a] that CD (H) = Z(P ) ∼ = Z3 Hence if AD = P , then A ∩ D ∼ = Z3 , so |A| < 4 3 , contradiction. Therefore |AD| = 35 and we may assume that A = (A ∩ D) a, with a ∈ H # centralizing the hyperplane A ∩ D of D. But H∼ = SL2 (3)∗SL2 (3), so two appropriate conjugates of a generate a subgroup H0 ≤ H such that either H0 ∼ = SL2 (3) or H0 ∼ = A4 , and in either case 2 |CD (O2 (H0 ))| ≥ 3 . If H0 ∼ = SL2 (3), this is impossible as Z(H0 ) inverts all of D; and if H0 ∼ = A4 , it is impossible as E22 ∼ = O2 (H0 ) ≤ [H0 , H0 ] must embed in SL(D/CD (O2 (H0 ))) ∼ = SL2 (3). These contradictions complete the proof.  ± Lemma 10.10.2. Let K ∼ = L± 3 (9) or L4 (3). Then for any x ∈ I3 (K), we have |CK (x)|3 ≥ 81.

Proof. Suppose that |CK (x)|3 ≤ 33 . Let P ∈ Syl3 (K) with CP (x) ∈ Syl3 (CK (x)), so that |P | = 36 . By [IG , 9.16], |[P, P ]| ≥ 33 . From the structure of root groups and the Chevalley commutator formula [IA , 2.4], 3 P = [P, P ]x. Replacing x by a K ∼ = L± 4 (3) and |[P, P ]| = 3 , whence x  suitable NAut(K) (P )-conjugate, we may assume that x = α∈Σ0 xα for some Σ0 ⊆ Σ, where Σ is the fundamental system of roots defining P , and 1 = xα lies in the root subgroup for α. If Σ0 = Σ, then it is easily seen that x is Kconjugate into J(P ) ∼ = E34 , so |CK (x)|3 ≥ 34 , contradiction; while if Σ0 = Σ, then, either using the commutator formula or the standard four-dimensional representation of P , we see that x3 = 1, a contradiction completing the proof.  Lemma 10.10.3. If K ∼ = L± 4 (3), G2 (3) or 3G2 (3), then m3 (K) ≥ 4. Proof. The result follows from [IA , 3.3.3] and [V8 , 1.5].



Lemma 10.10.4. Let L = U4 (3) and t ∈ I2 (Aut(L)), and suppose that K = E(CL (t))) ∼ = A6 . Then CAut(L) (K) is an elementary abelian 2-group. Proof. By [IA , 2.5.12], Aut(K) = Inndiag(K) γ, where γ is a graph automorphism. According to [IA , 4.5.1], we may assume that t = t2 or γ2 , in the notation of that table. If t = t2 , then CInndiag(L) (K) = t by

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that table. Moreover, regarding L as P Ω− 6 (3), CV (L) is a 2-dimensional nondegenerate subspace of + type of the natural 6-dimensional orthogonal module V . So CAut(L) (K) − Inndiag(L) contains a reflection, which implies that CAut(L) (K) ∼ = E22 , as needed. On the other hand, if t = γ2 , then CInndiag(L) (K) = 1 by [IA , 4.5.1], and so CAut(L) (K) ∼ = Z2 , completing the proof.  Lemma 10.10.5. Let S0 ∈ Syl2 (X), with F ∗ (X) ∼ = U4 (3) and |X|2 ≥ 29 . Then Z(S0 ) is cyclic. Proof. Let K = F ∗ (X), regard X ≤ Aut(K), and let z ∈ I2 (Z(S0 ) ∩ K). Then F ∗ (CK (z)) ∼ = 21+4 + , so z = Z(S0 ∩ K). Suppose that the lemma  fails, so that there exists y ∈ I2 (Z(S0 )) − K. Let Ky = O3 (CK (y)). If y ∈ Inndiag(K), then set K ∗= K y and SK = S0 ∩ K. Then by [IA , 4.5.1], Ky ∼ = U3 (3) or A6 . Then SK normalizes Ky , so since Z(SK ) is cyclic and SK ∈ Syl2 (K), CSK (Ky ) = 1. Hence | Aut(Ky )|2 ≥ |SK | = 27 , which is not the case for either isomorphism type of Ky . Therefore y ∈ Inndiag(K). From [IA , 4.5.1], Ky ∼ = P Sp4 (3), A4 × A4 , or A6 , and |CAut(K) (y)|2 ≤ 2| Aut(Ky )|2 = 27 , 27 , or 25 . But |CAut(K) (y)|2 ≥ |S0 | = 29 , a contradiction. The lemma follows.  Lemma 10.10.6. Let K = L4 (3) and z ∈ I2 (K) with L = E(CK (z)) ∼ = ∼ ∼ A6 . Let Sz ∈ Syl2 (CK (z)). Then Sz = D × E where z ∈ D = D8 = E and E ≤ L. Proof. We identify K with Ω(V ) where V is a 6-dimensional F3 orthogonal space of + type. Let W ⊆ V be a nondegenerate 2-dimensional ∼ subspace of − type; then W ⊥ is 4-dimensional of − type. Since Ω− 4 (3) = A6 , and by [IA , 4.5.1] the conjugacy class of z in K is uniquely determined, we may assume that z inverts W and CV (z) = W ⊥ . Then CO(V ) (W ⊥ ) =

e, t ∼ = D8 with e2 = z, t a reflection on W , and e of nontrivial spinorial norm. Likewise CO(V ) (W ) = L, u, v ∼ = Z2 × Σ6 with v inverting W ⊥ , u a ⊥ reflection on W , and L ∼ = A6 with L ≤ Ω(V ). Then CΩ(V ) (z) = ( z×L) < 2 2 ev, tu > with (ev) = e = z, [ev, L] = 1, (tu)2 = 1, and L tu ∼ = Σ6 . Thus  Sz ∩ L tu ∼ = Z2 × (Sz ∩ L) and ev, tu ∼ = D8 , and the result follows. Lemma 10.10.7. Let X = Aut(K), K = L4 (3), and T ∈ Syl2 (X). Suppose E22 ∼ = V ≤ CT (Inn(K) ∩ T ). Then m3 (CK (V )) = 2. Proof. Let z ∈ I2 (Z(T ) ∩ Inn(K)) and set F = O2 (CK (z)) ∼ = SL2 (3) ∗ SL2 (3). (Out(K) is a 2-group [IA , 2.5.12]; also see [IA , 4.5.1].) Calculating in P GL4 (3) shows that CCP GL (3) (z) (O2 (F )) = z. Then from [IA , 4.5.1] 4 for graph automorphisms we see that CCX (z) (O2 (F )) = z, g, where g 2 = 1 and CK (g) ∼ = Aut(P Sp4 (3)). In any case, F = O2 (CCK (g) (z)) so V = z, g  and m3 (CK (V )) = m3 (F ) = 2, as required.

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Lemma 10.10.8. Let K = L4 (3). Let z ∈ I2 (K) be such that CK (z) has a normal subgroup L ∼ = SL2 (3)∗SL2 (3) with z ∈ L. Let t ∈ I2 (O2 (L))−{z}. Then CK (t) = ( v × J) y with v ∼ = Z4 , v 2 = t, J ∼ = A6 , v, y ∼ = D8 , and ∼ J y = Σ6 . Proof. This can be seen by regarding L4 (3) as P SO6+ (3), and decomposing the underlying orthogonal space into eigenspaces of t as V = V2 ⊥ V4 , with dim Vi = i and both Vi ’s of − type. Then v ∼ = SO(V2 ), J ∼ = P SO(V4 ),  and y induces a reflection on V2 and V4 . Lemma 10.10.9. Let K ∈ K3 with K/Z(K) ∼ = U4 (3). Let Y /Z(K) be an elementary abelian 3-subgroup of K/Z(K) with CK/Z(K) (Y ) of even order. Then Y splits over Z(K) and m3 (CK (Y )) = m3 (K). Proof. We first calculate in L = SU4 (3). We may assume that L is defined with respect to the form e14 − e23 + e32 − e41 and that Y centralizes diag(1, −1, −1, 1). Then a Sylow 3-subgroup P of CL (Y ) consists of all I + ae23 + be14 , a, b ∈ F3 . We may assume that x ∈ P . Then Y lies in the E34 subgroup A ≤ L consisting of all unipotent upper triangular matrices in L with 0 entries in the (1, 2) and (3, 4) positions. But by [IA , 6.4.4], the preimage of A in the universal covering group of L has an E36 Sylow 3-subgroup. The result follows.  Lemma 10.10.10. Let K = L± 4 (3) or 3U4 (3). Then there is L ≤ K such ∼ that L = SL2 (3) ∗ SL2 (3) and LZ(K) is normalized by CAut(K) (Z(O2 (L))). Proof. Let z ∈ I2 (K) be 2-central. If Z(K) = 1 we simply take  L = O 3 (CK (z)). So assume that K ∼ = 3U4 (3) and set K = K/Z(K).  Our choice of L will satisfy L = O3 (CK (z)), so the normalization condition will be satisfied. By Lemma 10.10.9, a Sylow 3-subgroup of L splits over Z(K). The result follows.  ∼ Lemma 10.10.11. Out(U4 (3)) = D8 acts faithfully on the E32 Sylow 3-subgroup of the Schur multiplier of U4 (3). There are two isomorphism classes of 3-fold covering groups of U4 (3), labelled 31 U4 (3) and 32 U4 (3) (in Atlas notation). Proof. See [IA , 6.3.2, Table 6.3.1].



We give a definition now that distinguishes between the two isomorphism classes of quasisimple groups of the form 3U4 (3). The universal 3covering group of U4 (3) is K := [3 × 3]U4 (3) [IA , 6.1.4], and by [IA , Table 6.3.1], Out(K) ∼ = D8 acts faithfully on Z(K). It follows that Out(K) has two orbits on E1 (Z(K)), each stabilized by a different four-subgroup of Out(K). Let V1 and V2 be the two four-subgroups of Out(K), and choose any Z1 , Z2 ≤ Z(K) such that |Zi | = 3, i = 1, 2, and Vi leaves Zi invariant. According to [IA , 4.5.1], we can make the choice in such a way that the coset of Inn(K) in Aut(K) labelled “g” is contained in V1 , and the coset labelled “g  ” is contained in V2 . (Note that V1 ∩ V2 is the exponent 2 subgroup of Outdiag(K) ∼ = Z4 ).

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Definition 10.10.12. For i = 1 and 2, the quasisimple group 3i U4 (3) is defined as K/Zi , in the terms just discussed. The following lemma is then immediate. (An analogue holds of course for Out(31 U4 (3)).) Lemma 10.10.13. The image of Out(32 U4 (3)) in Out(U4 (3)) is the fourgroup generated by the involution in Outdiag(U4 (3)) and the coset g  in [IA , Table 4.5.1]. Lemma 10.10.14. Let K = 32 U4 (3) or G2 (3), and let z ∈ I2 (K). Then the following conditions hold: (a) O 2 (CAut(K) (z)) = O2 (CInn(K) (z)) ∼ = SL2 (3) ∗ SL2 (3); (b) Let A0 ∈ Syl3 (CK (z)). Then m3 (CK (A0 )) = 5 or 4, according as K = 32 U4 (3) or K = G2 (3); and (c) With A0 as in (b), some involution of CK (z) inverts A0 /A0 ∩Z(K). Proof. Note that Out(K) is a 2-group [IA , 2.5.12]. Let K = K/Z(K). We use standard notation for K [IA , 2.10]. Let Σ be the twisted root system for K ∼ = U4 (3), or the ordinary root system for K ∼ = G2 (3). Let α and β be orthogonal roots, with α long. Then we may assume that z ∈ X±α , whence O 2 (CK (z)) = X±α  ∗ X±β , proving (a). We may take A0 = Xα , Xβ , and then the involution hα+β (−1) inverts A0 , proving (c). Let A be the group generated by all Xγ such that γ is a positive linear combination of α and β. Then the Chevalley commutator formula implies that A ∼ = E34 . Note that by [IA , 3.3.3, 6.4.4], m3 (K) = 5 or 4 in the two respective cases, so it remains only to prove that if K ∼ = U4 (3), then the preimage of A in K is elementary abelian. Now AutK (A) contains A6 and is irreducible on A. Hence if A were not elementary abelian, then it would be extraspecial, so A6 would embed in Sp4 (3). This is impossible, however, as m2 (Sp4 (3)) = 2  and Z(Sp4 (3)) ∼ = Z2 . The lemma follows. Lemma 10.10.15. Let K = 32 U4 (3), G2 (3), or U4 (2) and let Inn(K) ≤ H ≤ CAut(K) (Z(K)). Also let z ∈ I2 (Inn(K)). Then |O2 (CH (z))| ≤ 25 . Also if O2 (CH (z)) ∼ = Q8 ∗ Q8 , then z is 2-central in Inn(K). Proof. For K = G2 (3) or U4 (2) ∼ = P Sp4 (3), this follows from the information in [IA , 4.5.1]; note that in the P Sp4 (3) case, m2 (O2 (CH (z)) = 4 if z is not 2-central in Inn(K). For K = 32 U4 (3), Inn(K) has one class of involutions z, and Qz := O2 (CInn(K) (z)) ∼ = Q8 ∗ Q8 , so it remains to show that O2 (CH (z)) = Qz . But the involution in Outdiag(K) inverts Z(K), so with Lemma 10.10.13, we may assume that H = Inn(K) γ with γ in the class labelled γ2 in [IA , 4.5.1]. We assume for a contradiction that O2 (CH (z)) = Qz y where y ∈ H − Inn(K), and of course y 2 ∈ Inn(K), so y 2 ∈ Qz . Let B ∈ Syl3 (CInn(K) (z)), so that Qz = [Qz , B] = [Qz y , B]. Replacing y, we may assume that [y, B] = 1, whence CQz y (B) = y, z and [Qz , y] = 1.

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∼ A6 . Hence BQz embeds If y 2 = 1, then by [IA , 4.5.1], F ∗ (CInn(K) (y)) = in Aut(A6 ), which does not even contain a copy of SL2 (3). Therefore, y 2 = z. Choose any y1 ∈ Qz of order 4; then there is b1 ∈ B # centralizing y1 . Similarly, F ∗ (CInn(K) (yy1 )) ∼ = A6 , and Aut(A6 ) must contain a copy ∼ of CQz (y1 ) b1  = Z4 ∗ SL2 (3). This is again impossible, and the proof is complete.  ∼ Lemma 10.10.16. Let K = 31 U4 (3). Then K has a subgroup H = A4 × A4 .  = [Z3 × Z3 ]U4 (3) ∈ K3 and consider K as P Ω− (3). Proof. Let K 6 Let V be the underlying orthogonal space. Then there are decompositions V = V1+ ⊥ V2+ and V = V1− ⊥ V2− such that dim Vi = 3 and Vi is of type  = ±, for i = 1 and 2. Moreover, there is a similarity s ∈ GO(V ) such that  and K. Let s interchanges Vi+ and Vi− , for each i = 1, 2. Then s acts on K      K be the stabilizer of V1 in K, and K its stabilizer in K, for each  = ±. Let γ  ∈ O(V ) invert V1 and centralize V2 ,  = ±. Then γ  induces a graph  We have automorphism of order 2 on K and K.   H := O3 (C (γ  )) ∼ = Ω(V  ) × Ω(V  ) ∼ = A4 × A4 . K

1

2

∼ Aut(K)  Thus, in the notation of [IA , 4.5.1], the image of γ  in Aut(K) = ∼ lies in a coset of Inn(K) labelled ’g’ in that table. Now Out(K) = D8 acts  Denote  [IA , 6.3.1], so γ  induces a reflection on Z(K). faithfully on Z(K)   its eigenspaces on Z(K) by Z and Z . In the same coset is the image of an involution h ∈ O(V ) inducing a reflection on V1 and centralizing V2 . Thus h induces an outer automorphism on Ω(V1 ) and centralizes Ω(V2 ); and its  are again Z and Z  . eigenspaces on Z(K)   or the preimages in K  of Sylow 3-subgroups Either H splits over Z(K)    In the of Ω(V1 ) and Ω(V2 ) have a nontrivial commutator ZH  ≤ Z(K).   latter case, h inverts ZH  , so ZH  = Z or Z ; and without loss we may assume that ZH + = Z. (Note that the image of s interchanges Z and Z  .)  satisfies  + of H + in K Thus the preimage H  + /Z ∼ H = A4 × A4 × Z3 . ∼  As K/Z = 31 U4 (3) by our definition of 31 U4 (3), the lemma is proved.



Lemma 10.10.17. Let K = 3U4 (3) and assume that K has a subgroup H∼ = A4 × A4 . Set E = O2 (H). Then the following conditions hold: (a) NK (E) = EJ × Z(K) with J ∼ = A6 acting irreducibly on E; (b) Let K = K/Z(K). Then CAut(K) (EJ) has odd order. Moreover, F := CAut(K) (E) ∼ = E25 , and there is f ∈ F − E such that CK (f ) ∼ = U4 (2). Proof. We continue with the proof of Lemma 10.10.16. Now E fixes (elementwise) orthogonal frames in V1+ and V2+ , whose union is an orthogonal frame F of V . The stabilizer of F in P O(V ) is then a monomial group

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∼ E25 and Σ = ∼ Σ6 permutes F naturally. The F Σ, where F = CP O(V ) (E) = stabilizer of F in K clearly is NK (E), which must be EJ , where J = [Σ, Σ], because |K|2 = 27 . Since H ∩ Z(K) = 1, J splits over Z(K), which implies (a). Also a reflection f ∈ F satisfies CK (f ) ∼ = Ω5 (3) ∼ = U4 (2). The proof is complete.  Lemma 10.10.18. Let K = 32 U4 (3) and let t ∈ I2 (Aut(K) − Inn(K)) with t inverting Z(K). Then the following conditions hold: (a) E(CK (t)) ∼ = U3 (3) or A6 ; and (b) CK (t) does not contain a copy of Q8 ∗ Q8 . Proof. Since t inverts Z(K), it is enough to show (a) and (b) with K replaced by K := K/Z(K). By Lemma 10.10.13, either t ∈ Inndiag(K) or t is in an Inn(K)-coset labelled g  in [IA , 4.5.1]. That table then proves (a). Suppose that (b) fails. Let L = E(CK (t)). As CK (u) is solvable for all u ∈ I2 (K), CK (L) has odd order, and so Q8 ∗ Q8 embeds in Aut(L). But | Aut(A6 )|2 = 25 and Aut(A6 ) ≥ P GL2 (9) ≥ Z8 , while Q8 ∗ Q8 has exponent 4. Therefore L ∼ = U3 (3), whence t acts on K like a reflection. Therefore CK/Z(K) (t) is the image of CSU4 (3) (t), which induces only innerdiagonal automorphisms on L. So AutK (L) = L and Q8 ∗ Q8 must embed in L. Hence L has Q8 ∗ Q8 Sylow 2-subgroups, whereas it has Z4 Z2 Sylow 2-subgroups, with just two generators, a contradiction.  Lemma 10.10.19. Let F ∗ (X) = K ∼ = 32 U4 (3) and suppose X/Z(K) ∼ = Aut(32 U4 (3)). Let z ∈ I2 (K) and A ∈ E33 (CX (z)). Then the following conditions hold: (a) CCX (z) (A/Z(K))/A ∼ = Z4 ; (b) If t ∈ I2 (CX (z)), t inverts Z(K), |CK ( t, z)|3 = 3 and t normalizes both solvable components of CK (z), then the coset Kt is uniquely determined. All involutions in Kt are K-conjugate and centralize an A6 subgroup of K; and (c) If t ∈ I2 (CX (Z(K))), then either t ∈ K or K t = CX (Z(K)), and in either case, all involutions in Kt are K-conjugate. Proof. Let K = K/Z(K) ∼ = U4 (3). By [IA , 6.3.1], Out(K) ∼ = D8 acts faithfully on the Schur multiplier E32 of K. By [IA , 4.5.1], one of the fourgroups in Out(K) contains cosets of Inn(K) labelled d and g, and in a g-coset  lies an involution u such that O3 (CK (u)) ∼ = P Sp4 (3). The other four-group, likewise, contains cosets of Inn(K) labelled d and g  ; and all the involutions  v in a g  coset are Inndiag(K)-conjugate and satisfy O3 (CK (v)) ∼ = A6 . It is the second four-group that is Out(K), by Lemma 10.10.13. Write A = A1 × A2 where the Ai are long root groups, each in a solvable component of CK (z). Consider a 2-element h ∈ Inndiag(K) centralizing A, or more generally normalizing A1 and A2 . Then h normalizes O3 (NK (A1 ))A2 ∈ Syl3 (K) so

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h lies in a Cartan subgroup H of Inndiag(K). Conversely, we may assume that H normalizes the root groups A1 and A2 . Let V be a natural F9 K-module. We may take V to have basis {vi }4i=1 such that (vi , vj ) = 0 for i ≤ j, except that (v1 , v4 ) = (v2 , v3 ) = 1. Then we may assume that H consists (modulo scalars) of elements h = h(α, β) := diag(α, β, β −3, α−3 ), as α and β range over F× 9 . Here z is represented as h(−1, 1). Now A1 and A2 may be assumed to be represented by upper triangular matrices with nonzero off-diagonal entries in the (2, 3) or (1, 4) place. Thus, [A, h] = 1 if and only if α4 = β 4 = 1. Factoring out scalars, we see that CInndiag(K) ( z, A ) is isomorphic to Z4 and its image in Outdiag(K) has order 2. Since u may  also be taken to centralize P Sp4 (3) and A, it follows that CAut(K) ( z, A ) maps onto the (d, g)-type four-subgroup of Out(K). Therefore any element t ∈ CAut(K) (z) in a coset labelled g  acts nontrivially on AO2 (CK (z))/O2 (CK (z)). In particular, CCX (z) (A/Z(K))/A is the (isomorphic) image of h(i, 1) and hence (a) holds. Moreover, since Ω1 (Outdiag(K)) inverts the Schur multiplier of K, it follows that in Aut(K), one of the cosets of Out(K) labelled g  inverts Z(K) and the other centralizes Z(K). Therefore to complete the proofs of (b) and (c) it suffices to show that in a g  -coset of Inn(K) in Aut(K), all involutions t are K-conjugate. (We know from [IA , 4.5.1] that all such t are Inndiag(K)-conjugate, and that each one centralizes an A6 subgroup of K.) Let t be such an involution. Let R ∈ Syl5 (K). Then NK (R) is a Frobenius group of order 5.4. Hence by a Frattini argument, CAut(K) (R) maps isomorphically onto Out(K). In particular some t ∈ Kt centralizes an involution v mapping onto an involution of Inndiag(K). As elements of order 4 in Outdiag(K) do not even normalize the coset Kt, we get I2 (Kt) = (t )KΩ1 (Inndiag(K)) = (t )vK = (t )K = tK , as desired. The proof is complete.  Lemma 10.10.20. Let V = E34 and X = a1 , a2  ≤ GL(V ) with X ∼ = E32 and CV (X) ∼ = Z3 . Let X = { a ∈ E1 (X) | |[V, a]| = 32 and [V, a, a] = 1}. Then |X| ≤ 2. Proof. We may assume that ai ∈ X, i = 1, 2. Let Vi = CV (ai ) = [V, ai ], i = 1, 2. Since CV (X) ∼ = Z3 , there is a basis {v1 , v2 } of V1 such that [v1 , a2 ] = 0 and [v2 , a2 ] = v1 . Choose v4 ∈ V such that [v4 , a1 ] = v2 and set v3 = [v4 , a2 ]. As commutation with a1 is an a2 -isomorphism from V /V1 to V1 , {vi }4i+1 is a basis of V , and [v3 , a1 ] = v1 . A straighforward matrix calculation shows that a1 a2 and a−1 1 a2 both have a 3 × 3 Jordan block, so  X = { a1  , a2 }, as required.

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10.11. M12 , HJ. Lemma 10.11.1. Let K = M12 and T ∈ Syl2 (K). Let T0 be a nonidentity subgroup of T which is strongly closed in T with respect to K. Then T0 = T . Proof. We use [IA , 5.3b] freely. Suppose that T0 < T . We have T =

I2 (T ) and I2 (K) = z K ∪ tK where z = Z(T ) and t ∈ I2 (T ) − z K . As 1 = T0  T , I2 (T0 ) = z K ∩T . Then |T : T0 | > 2, for otherwise the Thompson transfer lemma yields a contradiction to the simplicity of K. Let U  T with U ∼ = E22 . Then |CT (U )| ≥ 25 > |CK (t)|2 so U # ⊆ z K . Let v ∈ I3 (CK (z)); then as U ≤ O2 (CK (z)) ∼ = 21+4 + and CO2 (CK (z)) (v) = z, v E := U, U  ∼ E and E ≤ T . Notice also that K contains a subgroup 3 = 2 0 ∼ ∼ H = M11 , and H ≥ D = D8 with I2 (D) ⊆ I2 (H) ⊆ z K . Without loss, D ≤ T0 . Thus as |T : T0 | > 2, T0 = ED ∼ = Z2 × D8 . In particular NK (T0 ) normalizes both [T0 , T0 ] ∼ = Z2 and Z(T0 ) ∼ = E22 , so NK (T0 ) is not transitive on Ω1 (Z(T0 ))# = z K ∩ Z(T0 ). But by [IG , 16.9], NK (T0 ) is transitive on  Ω1 (Z(T0 ))# , a contradiction. The lemma follows. Lemma 10.11.2. Let M ∈ C2 be simple of 2-rank 3. Suppose CM (z) ∼ = CM12 (z0 ), where z is a 2-central involution of M and z0 is a 2-central involution of M12 . Assume also that mr (M ) ≤ 3 for all odd primes r. Then M∼ = M12 . Proof. Suppose not. Our hypotheses and Lemma 9.2 imply that M ∼ = 3 L2 (8), U3 (8), or 2B2 (2 2 ). But in all cases, |CM (z)| = |CM12 (z0 )|, a contradiction.  3 ∼ Lemma 10.11.3. Suppose that  K = M12 . Then there is B1 ∈ E∗ (K)  # such that K = CK (b) | b ∈ B1 and K is locally unbalanced with respect to B1 . There also are two conjugacy classes of E32 -subgroups A (interchanged in Aut(K)) such that ΓB,1 (G) is isomorphic to the holomorph of E32 . For each such A, K is locally balanced with respect to A.

∼ Z3 × A4 , and Proof. Refer to [IA , 5.3b]. Let b ∈ I3 (K) with CK (b) = let B1 ∈ Syl3 (CK (b)). Then K is locally unbalanced with respect to B1 , and by [IA , 7.5.5], ΓB1 ,1 (K) = K. On the other hand there are nonconjugate but Aut(K)-conjugate maximal subgroups M1 and M2 isomorphic to the holomorph of E32 . If we put Ai = O3 (Mi ), then Mi = NK (Ai ) and it is clear from the 3-local information in [IA , 5.3b] that Mi = ΓAi ,1 (K). In particular, K is locally balanced with respect to each Ai , by [IA , 7.5.5]. The lemma is proved.  Lemma 10.11.4. Let K = M12 and let P ∈ Syl3 (K) and V ∈ E32 (P ). Then the following hold. (a) If K is locally balanced with respect to V , then there exists an involution t ∈ NK (V ) ∩ NK (P ) inverting Z(P ) and P/V , and centralizing V /Z(P ); and

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(b) If K is not locally balanced with respect to V , then K = ΓV,1 (K). Proof. Let N = NK (P ) = P U with F ∗ (N ) = P and U ∼ = E22 . Therefore U has three orbits on E32 (P ). In the notation of Lemma 10.11.3, these orbits must be represented by A1 , A2 , and B1 . Therefore U normalizes A1 and A2 . Then for i = 1, 2, there exists an involution t ∈ NN (Ai ) centralizing Ai /Z(P ) and inverting A3−i elementwise. This implies (a), and (b) is immediate from Lemma 10.11.3.  Lemma 10.11.5. Suppose that F ∗ (X) ∼ = M12 and A ∈ E32 (X). Then CX (A) = A. Proof. From [IA , 5.3b], |CX (x)|3 ≤ 3 for all 3 -elements x ∈ X of prime order. As Sylow 3-subgroups of X are isomorphic to 31+2 , the result follows.  

Lemma 10.11.6. Let F ∗ (X) = M12 . If c ∈ I3 (X) and O2 (O3 (CX (c))) =

t, u ∼ = A5 , E(CX (t)) u ∼ = Σ5 , and O3 (CX (t)) = = E22 , then E(CX (t)) ∼  ∼

t, t  = E22 or Z2 .  ∼ Lemma 10.11.7. Let K = HJ and let t ∈ I2 (K) with J := E(CK (t)) = A5 . Let x ∈ I5 (J). Then CAut(K) (x) ∼ = CK (x) ∼ = Z5 × A5 . Proof. We use [IA , 5.3g]. We have CK (t) ∼ = J ×E22 . Therefore |CK (x)| is divisible by 4, whence CK (x) ∼ CK (x), then = Z5 × A5 . If CAut(K) (x) ∼ = CAut(K) (x) contains an involution z ∈ Aut(K) − Inn(K). But for such an involution, |CK (z)| is not divisible by 5, contradicting [z, x] = 1. The lemma follows.  Proof. This follows directly from the information in [IA , 5.3b].

Lemma 10.11.8. Let K = HJ or M12 . Then K has no subgroup isomorphic to L2 (24 ) or L± 3 (4). Proof. This follows easily from Lagrange’s theorem and the complete  lists of maximal subgroups of K [IA , 5.3bg]. 10.12. M22 . Lemma 10.12.1. Let K = Aut(M22 ) and let V be a 10-dimensional F2 V module such that dim CV (t) = 4 for t ∈ I3 (K), dim CV (f ) = 2 for f ∈ I5 (K) and, for s ∈ I7 (K), dim CV (s) = 1 and there is a free F2 s-summand on V . Then up to isomorphism, V is determined up to duality. Proof. Note that V is irreducible since 11 divides |K|, and even absolutely irreducible since K contains a Frobenius group of order 9.8. Let ϕ be the Brauer character of V . Elements of K of orders 3, 5, and 11 are rational, so ϕ takes the values 1, 0, and −1 on them. As F2 s-module V is the sum √ of a free module and an irreducible 3-dimensional module, so −1± −7 . There are no elements of composite odd order [IA , 5.3c], ϕ(s) = 2 and since the Brauer character determines the module [V9 , 12.1a], the two choices for ϕ(s) mean that up to duality, V is determined. 

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The next lemma concerns the Golay and Todd modules G10 and T10 for Aut(M22 ). These are defined as follows. Let V24 be the natural 24dimensional F2 M24 -module, with permutation basis  {vi }24 i=1 . Then the Golay code V12 ⊆ V24 is the submodule generated by all O vi , as O ranges over the octads in the Steiner system S(5, 8, 24) preserved by M24 . By abuse of notation we may refer to i∈S vi , where S ⊆ {1, . . . , 24}, just as S; then the vector space addition becomes symmetric difference. We have dim V 12 = 12, 24 and V12 has a 1-dimensional trivial submodule V1 generated by i=1 vi . The elements of V12 − V1 are the octads, their complements, and certain 12-element sets called “dodecads.” Consider the restrictions of all these modules to Aut(M22 ), the fixer of, say, the subset {1, 2}. We use the same notation for the modules over Aut(M22 ). Then V11 = {S ∈ V12 | |S ∩ {1, 2}| is even} is an Aut(M22 )submodule of V11 . We set G10 = V11 /V1 . Of course v1 , v2  is an Aut(M22 )submodule and we set T10 = V24 /(V12 + v1 , v2 ). Notice that if we declare the vi to be an orthonormal basis of V24 , then we have a nondegenerate ⊥ = V , so V /V ∼ ∗ M24 -invariant bilinear form on V24 , and V12 12 24 12 = V12 . Also an element e ∈ I11 (M24 ) has two disjoint 11-cycles, so it acts without fixed points on T10 and G10 , which are therefore irreducible for Aut(M22 ). Therefore T10 ∼ = G∗10 . Lemma 10.12.2. The following conditions hold: (a) T10 and G10 satisfy the conditions of Lemma 10.12.1; and (b) There is a vector t ∈ T10 such that F ∗ (CAut(M22 ) (t)) ∼ = M21 (∼ = L3 (4)). Proof. Let ti be the image of vi in T10 . Then t1 = t2 = 0 and Aut(M22 ) permutes t3 , . . . , t24 naturally. If g ∈ Aut(M22 ) has odd order and has γ cycles on 24 points, including fixed points, then dim CT10 (g) = dim CG10 (g) and so γ = dim CV24 (g) = 2 dim CG10 (g) + 4. Thus dim CT10 (g) = dim CG10 (g) = γ 2 − 2, and (a) is easily checked. Notice that a 7-cycle in the decomposition of any s ∈ I7 (Aut(M22 )) yields a free summand for s on T10 since the minimal nonzero elements of V12 are octads. In (b), the vector t = v3 satisfies the condition. The proof is complete.  10.13. J3 . Lemma 10.13.1. Let S ∈ Syl3 (J3 ) and let M be a maximal subgroup of S. Then M is not abelian. Proof. From [IA , 5.3h], NJ3 (S) acts irreducibly on S/Φ(S) ∼ = E32 . Hence if M existed, there would exist g ∈ NJ3 (S) − NJ3 (M ), and then  |M ∩ M g | = 33 and M ∩ M g ≤ Z(S). But |Z(S)| = 32 , contradiction. Lemma 10.13.2. Let K = 3J3 , K = KZ(K), and Q ∈ Syl3 (K). Then (a) Ω1 (Q) is abelian; (b) Ω1 (Z(Q)) = Z(K); and (c) For every t ∈ I2 (Aut(K)), m3 (CK (t)) = 1.

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Proof. This follows from the information in [IA , 5.3h].

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Lemma 10.13.3. Let K = J3 and let A ∈ E33 (K). Let X = {x ∈ A# | E(CK (x)) ∼ = A6 }. Then K = E(CK (x)) | x ∈ X. Proof. We use [IA , 5.3h] freely. For some P ∈ Syl3 (K), we have A = Ω1 (P ), Z(P ) ∼ = E32 , and X = A − Z(P ). Let L = E(CK (x)) | x ∈ X and N = NK (P ). Then X and L are N -invariant, so it suffices to show that H := LN = K. From [IA , 5.3h] we see that N = NK (Z(P )) = NK (A) ≥ NK (D) for all 1 = D ≤ Z(P ), whence NK (E) ≤ N ≤ H for all E ≤ P with E ∩ Z(P ) = 1. Also by a Frattini argument, for any x ∈ X, NK ( x) ≤ E(CK (x))NK (A) ≤ LN = H. Hence ΓP,1 (K) ≤ H. But by [IA , 7.6.1], K has no strongly 3-embedded subgroup, so H = K. The proof is complete.  10.14. Suz. Lemma 10.14.1. Let K = Suz and P ∈ Syl5 (K). Let X = { x ≤ ∼ K (x)) = A6 } and K0 = E(CK (x)) | x ∈ X. Then K = K0 .

P # | E(C

Proof. From [IA , 5.3o] we see that |X| = 3. Then by L5 -balance, for any x ∈ X, E(CK (x)) has a vertical pumpup J in K0 , i.e., A6 ↑5 J := J/O5 (J). Then J is as in Lemma 13.25. Order considerations rule out all but J ∼ = Suz and A11 . In the latter case |X| = 2, contradiction. Hence ∼ J = K and K = K0 , as required.  Lemma 10.14.2. Let K = Suz and X = Aut(K). Let z ∈ K be 2-central in K. Then for some y ∈ CX (z) − CK (z), CCK (z) (y) ∼ = CAut(HJ) (u), where u is a 2-central involution of HJ. Proof. See [IA , 5.3o]. There is y ∈ I2 (X) − K such that CK (y) ∼ = Aut(HJ). Let z  ∈ CK (y) be 2-central. Since z   = Φ(E) for some E ≤  CK (y) with E ∼ = 21+4 − , z must be 2-central in K, which implies the result.  Lemma 10.14.3. Let K = Suz and X = Aut(K). Let y ∈ I2 (K) be non-2-central, and let P ∈ Syl2 (CX (y)). Let P ≤ T ∈ Syl2 (X). Then [P, P, P, P, P ] = Z(T ) ∩ E(CK (y)) = Z(P ) ∩ E(CK (y)). Proof. We have P = (V × U ) v and T = P w where V ∼ = = E22 ∼ ∼

v, w, V v = D8 , [V, w] = 1, T U ∈ Syl2 (L) where L = E(CK (y)) ∼ = L3 (4), and v and w induce field and graph automorphisms on L, respectively, so that CU (v) ∼ = D8 and CU (w) = Z(U ) and v, w acts freely on U/Z(U ). Hence Z(P ) = (Z(P ) ∩ V ) × (Z(P ) ∩ U ) with both factors of order 2, proving the second claimed equality. Since v, w acts freely on U/Z(U ) and is disjoint from U , [P, P ] = Z(V v) × [U, v, w]. Then [P, P, P ] ≤ U with [P, P, P ]/[U, U ] ∼ = Z2 , so [P, P, P, P, P ] = [Z(U ), v] = Z(P ) ∩ U as desired. 

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Lemma 10.14.4. Let K = A8 , HS, 2HS, L5 (2), or Sp4 (4). Set H = Aut(K). Let z ∈ I2 (H) with J := E(CH (z)) ∼ = A6 . Let a ∈ I3 (J) with I := E(CH (a)) ∼ = A5 or L3 (2). Then I z ∼ = Aut(I). Proof. Let a ∈ A ∈ Syl3 (J). Then A ∼ = E32 so A ≤ CCH (a) (z). If K/Z(K) ∼ = A5 , and A/ a = HS, then from [IA , 5.3m], z ∈ Inn(K), I ∼ acts faithfully on I. As [z, A] = 1 and CInn(K) (AI) ∼ = Z3 , [z, I] = 1 and so I z ∼ = Sp4 (4), then z induces a nontrivial field automorphism = Σ5 . If K ∼  on K, so it does the same on I = O2 (CH (a)) by [IA , 4.2.3]. So again I z ∼ = L5 (2), I ∼ = L3 (2) is similar, with z inducing graph = Σ5 . The case J ∼ ∼ automorphisms on J and I. If K = A8 , then a is a 3-cycle disjoint from the transposition z, whence the desired result. The lemma is proved.  Lemma 10.14.5. Let K = L± 3 (3) and R ∈ Syl2 (K). Then CAut(K) (R) = Z(R). Proof. In both cases Out(K) is the image of a graph automorphism g of order 2 with CK (g) ∼ = P GL2 (3) [IA , 2.5.12, 4.5.1]. We may assume that g normalizes R and CR (g) ∈ Syl2 (CK (g)), so that CR (g) ∼ = D8 . If the lemma is false, then g induces an inner automorphism on R. But if K = L3 (3) and the lemma is false, then g maps into Inn(R); but R ∼ = SD16 , so CR (CR (g)) = Z(R) and g cannot induce an inner automorphism on R. If K = U3 (3), then K g ∼ = G2 (2), and the Chevalley commutator ∼ formula easily yields Z(R g) = Z2 , whence CAut(K) (R) ≤ R and the lemma follows.  11. Some Other Familiar Groups 11.1. L2 (q), q odd. Lemma 11.1.1. Let K = L2 (q), q odd. Let V ≤ K with V ∼ = E22 , and let R be a 2-subgroup of Aut(K) such that R, V  is a 2-group. (a) Suppose that R ∩ V = 1, and that either [R, V ] = 1 or R ∼ = Z2 m , m > 1. Then any involution of R is a field automorphism on K; (b) If V ≤ R and [R, V ] = 1, then R = V F where F is a group of field automorphisms of K; and Σ∼ (c) Suppose that q = 3n and Σ≤ K with = Σ4 . Then Σ ≤ J ≤ K for some J ∼ = A6 , and J = CJ (z) z ∈ I2 (Σ) . Proof. We have Aut(K) = Inndiag(K)Φ where Φ is a group of field automorphisms of K and Inndiag(K) ∼ = P GL2 (q). All four-subgroups of K are Aut(K)-conjugate and self-centralizing in Inndiag(K) [IA , 4.5.1]. Since CK (Φ) contains L2 (r), where r is the prime of which q is a power, by replacing V by a conjugate, we obtain CAut(K) (V ) = V × Φ. Thus (b) holds. By [IA , 4.9.1], all involutions of V Φ − V are field automorphisms. Hence in proving (a) we may assume that R ∼ = Z2m , m > 1. Expand V, R to S ∈ Syl2 (Aut(K)). Again, by conjugation, we may assume that S is Φinvariant. Set T = S ∩ Inndiag(K). Then T is nonabelian dihedral and has

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a cyclic maximal subgroup T0  S. As T0 ∩ V = 1, we have R ∩ T0 = 1. But S/T0 = T /T0 × (S ∩ Φ)T0 /T0 , so the involution of R lies in (S ∩ Φ)T0 − T0 , and again is a field automorphism by [IA , 4.9.1]. This proves (a). Finally in (c), since |K|2 ≥ 8, n is even. Thus K contains a subgroup J ∼ = Σ4 [IA , 6.5.1], so Σ is determined up to conjugacy = A6 . Also NK (V ) ∼ and we may assume that Σ ≤ J. The final statement follows easily; indeed Σ is maximal in J but CJ (z) ∼  = D8 for all z ∈ I2 (J). 11.2. L2 (27). Lemma 11.2.1. Suppose L2 (27) ∼ = K   X and A ∈ E3k (X), k ≥ 1. Also let P ∈ Syl3 (K). Then the following hold: (a) If AutA (K) ≤ Inn(K), then K = ΓA,k−1 (K); (b) If a ∈ A# and L3 (CK A (a)) = 1, then a centralizes K or does not normalize K; (c) ↑3 (K) = {L2 (39 )}; (d) If A is noncyclic, then unless A normalizes K and faithfully induces inner automorphisms on K, then   A  K = O3 (CK A  (a))L3 (CK A  (a)) a ∈ A# ; in particular, this holds if m3 (A) ≥ 4; (e) Sylow 3-subgroups of Aut(K) are isomorphic to Z3 Z3 ; (f) NX (P ; 3 ) ≤ CX (K); and (g) NK (P ) is a Frobenius group P H, with H ∼ = Z13 permuting transi∼ tively the set of subgroups of P = E33 of order 3. Proof. K is unambiguously in Chev(3), so if K ↑3 L, then L ∈ Chev(3)∪ Spor, by [IA , 7.1.10]. But by inspection of [IA , 5.3], L ∈ Spor. Then (c) follows from [IA , 4.9.1, 4.9.2] and the Borel-Tits theorem. By [IA , 2.5.12], Aut(K) is generated by Inndiag(K) and a field automorphism of order 3. Then (e) and (g) follow from the Chevalley relations, as does the fact that CInndiag(K) (g) = P for all g ∈ I3 (P ). This in turn implies (f). As CK (f ) ∼ = A4 for all field automorphisms f ∈ I3 (Aut(K)), (b) holds. Part (a) is reduced to the case K  X and k = 2 by [V9 , 9.3]. In that case CK (a) ∼ of = A4 for all a ∈ A − K, and so by the classification of subgroups   # K [IA , 6.5.1], K = ΓA,1 (K); indeed K = O3 (CK (a)) | a ∈ A , the latter group being A ∩ K-invariant. In particular (a) holds. Similarly (d) holds if A normalizes K, and then in general by [GL1, I.(22-4)]. This completes the proof.  Lemma 11.2.2. Let L = F ∗ (X) ∼ = L2 (33 ). If A ∈ E32 (X), then either A ≤ L or L = ΓA,1 (L). Proof. This immediate from Lemma 11.2.1a.



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11.3. (S)L3 (q). Lemma 11.3.1. Let K = L3 (q 3 ),  = ±1, q ≡  (mod 3). Let f be a field automorphism of K of order 3. Then CInndiag(K) (f ) ≤ Inn(K). Proof. Let (J, σ) be a σ-setup for SL3 (q 3 ). Then some conjugate of f or f −1 , which we may assume is f is induced by a Steinberg endomorphism τ of J lifted to J/Z(J), where τ 3 = σ. Write Z(J) = z ∼ = Z3 . By the Lang-Steinberg theorem, there is x ∈ J such that xτ = xz. Since z τ = z, xσ = x, so x ∈ CJ (σ) ∼ = SL3 (q 3 ). The image x of x in K is ∼ then not in the image of CJ (τ ) = L3 (q), but it lies in CJ/Z(J) (τ ). Hence CJ/Z(J) (τ ) = CJ (τ ) x /Z(J) ∼ = P GL3 (q) lies in CJ (σ)/Z(J) ∼ = Inn(K), as desired.  Lemma 11.3.2. Let K = L3 (q), q ≡  (mod 3),  = ±1. Then the following hold: (a) Out(K) is 3-closed; and (b) If E ∈ E3 (Aut(K)), then m3 (E ∩ Inn(K)) ≥ m3 (E) − 1. Proof. By [IA , 2.5.12], Outdiag(K) ∼ = Z3 and Out(K)/ Outdiag(K) is abelian with cyclic Sylow 3-subgroups. In particular, (a) holds, and if (b) fails, then E contains e, f  where f is a field automorphism of order 3 and e ∈ I3 (Inndiag(K)) − Inn(K). Hence CInndiag(K) (f ) ≤ Inn(K). This contradicts Lemma 11.3.1 and completes the proof.  9). Then there Lemma 11.3.3. Let K = L3 (q),   = ±1, q ≡  (mod  exists E ∈ E32 (K) such that K = L3 (CK (e)) | e ∈ E # . Proof. Let D be the image in K of a full diagonal subgroup Δ of SL3 (q) (with respect to an orthonormal basis if  = −1). Then O3 (Δ) is homocyclic of exponent 3n , n ≥ 2. Hence E := Ω1 (O3 (D)) ∼ = E32 . Since  r # q ≥ 8, O (CK (e)) = L3 (CK (e)) for all e ∈ E , by [IA , 4.2.2]. Then as E is   diagonal, [IA , 7.3.3] implies that K = ΓrE,1 (K). The lemma follows. Lemma 11.3.4. Let K  X = KA with K ∼ = SL3 (q), q > 2, q ≡  (mod 3),  = ±1, A ∼ = E34 , m3 (X) = 4, m3 (CA (X)) = 2, and no element of A inducing a nontrivial field automorphism on K. Then there is F ≤ A such that the following conditions hold: (a) CA (K) ≤ F ∼ = E33 , I := E(CK (F )) ∼ = SL2 (q), and F ∩ K = Z(K); and (b) For any f ∈ F − CA (K), there is f0 ∈ (A ∩ I) f  − f  such that I0 := E(CK (f0 )) ∼ = I and I, I0  = K. Proof. Expand A ≤ P ∈ Syl3 (X) and set P = P/CP (K). Since no element of A = Ω1 (A) induces a field automorphism on K, the image A of A in Aut(K) lies in Inndiag(K). In particular, m3 (A) ≤ m3 (P ) ≤ 2, by [IA , 4.10.3a], we have m3 (A) = m3 (P ) = m3 (CP (K)) = 2. Hence there is # z ∈ Z(P ) ∩ A . In the proof of Lemma 5.10 it was shown that z is in class

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C1 , and so a preimage z ∈ P ∩K of z is a conjugate of diag(1, ω, ω −1 ), where ω 3 = 1 = ω. Hence z has order 3, and D := CP (z) is a diagonal subgroup (with respect to some frame in the natural K-module V , orthogonal in the unitary case) of index 3 in P . We argue that z ∈ A. As z ∈ A, zb ∈ A for some 1 = b ∈ CP (K). In particular (zb)3 = 1, so as [z, b] = 1, b3 = 1. Then, [CA (K), b] = [CA (K), zb] = 1. As m3 (CA (K)) = m3 (CP (K)) = 2, b ∈ A, so z = (zb)b−1 ∈ A, as asserted. It follows that A ≤ D. But D is abelian of rank 2 so A = Ω1 (D). Then some f ∈Ahas two equal eigenvalues   on V . We let F be the full preimage in A of f . Note that if f ≤ K, then, as shown in the proof of Lemma   5.10, f is the image of an element of K of order 9. Hence as F = f , F ∩ K = 1 and (a) holds. Finally, CP (f ) = D. But I := L3 (CK (f  ))  contains   an element of P of order 3, which must be z. Hence (A ∩ I) f = z, f = Ω1 (D). Thus every element   of z, f − z has two equal eigenvalues on V , and (b) follows as well. The lemma is proved.  Lemma 11.3.5. Let K = L3 (q),  = ±1, q ≡  (mod 3). Suppose that q > 4 and q = 8. Let D ∈ E32 (Inndiag(K)) be the image of a nonabelian # subgroup of GL3 (q). For each  d ∈ D #,let Id = [O3 (CK (d)), Rd ] where Rd ∈ Syl3 (CK (d)). Set K0 = Id | d ∈ D . Then K = K0 . Proof. Using [IA , 4.8.1, 4.8.2, 4.8.4] we find that there are three conjugacy classes of subgroups of H = Inndiag(K) ∼ = P GL3 (q) of order 3. (We regard K ≤ H.) Representatives are

t, x, and z, where z is 3-central with CH (z) an extension of a diagonalizable subgroup Tz ∼ = SL2 (q); and = Zq− × Zq− by Z3 ; t is diagonalizable with E(CH (t)) ∼ x ∈ H − K with CK (x) a Frobenius group with kernel Kx ∼ = Z 1 (q2 +q+1) 3 and complement Hx conjugate to z. As a preimage of t in GL3 (q) is not sheared to Z(SL3 (q)), D contains no conjugate of t. Therefore up to conjugacy, either D = z, z   with z  ∈ Tz and all d ∈ D # conjugate to z, or D = x, z  , with z   a Frobenius complement to Kx and weakly closed in D with respect to H. Set Yd = O3 (CK (d)) = O3 (CH (d)) for every d ∈ D # . Our assumptions on q guarantee that Yd = 1 for all d ∈ D # . Then DYd = d × Fd , where Fd is a Frobenius group with complement of order 3 and kernel Yd . According as d is conjugate to z or x, Yd ∼ = Za × Za or Zb , where a = (q − )3 1 2 and b = 3 (q + q + 1). Assume that K0 < K. Now, K0 is N := NH (D)-invariant and we set K1 = K0 N ∩ K. Then |K1 |3 ≥ 32 and K1 < K. Suppose first that D = z, z   as above. Then Yz preserves a unique unordered frame in the natural GL3 (q)Z-module V , and z  permutes transitively the three lines in this frame, so Yz , Yz   and K1 act irreducibly and primitively on V . Also gcd(q − , q 2 + q + 1) = (q 2 + q + 1)3 = 3 and

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|K1 |3 ≥ 32 , with AutK (D) containing Q8 . By [IA , 6.5.3], these conditions imply that (11A)

η η K1 ≤ K2 < K for some K2 ∼ = L3 (q0 ) or P GL3 (q0 ), where q = q0n , n > 1, and q0 ≡ η (mod 3), η = ±1 = n .

(The case K2 ∼ = M10 would imply Yd = 1 for any d ∈ D # , contradiction.) 3 Since O (N ) ≤ K1 and O3 (N ) permutes E1 (D) transitively, D ≤ K1 and every d ∈ D # is 3-central in K2 . Hence (q − )23 = |Yd | = |CK1 (d)|3 = |CK2 (d)|3 = (q0 − η)23 and thus (q0n − η n )3 = (q0 − η)3 . Hence, (q0n − η n )/(q0 − η) = q0n−1 + ηq0n−1 + q0n−2 + · · · = 3m for some m. As 3 divides q0 − η, it must also divide n. It follows that (q03 − η 3 )3 = (q0 − η)3 . But 3 divides q02 + ηq0 + 1 only to the first power, and so q02 + ηq0 + 1 = 3, whence q0 = 2. This implies that Yd = 1, a contradiction. Therefore D = z, x  with x  conjugate to x. Similar arguments, using [IA , 6.5.3], show that again (11A) holds. Then (q03n −η 3n )/3(q0n −η n ) = η 1 2 1 3 2   3 (q +q +1) = |Yx | divides |L3 (q0 )|3 , hence divides 3 (q0 −η)(q0 −1). Thus 3n 3n n n 3 2 3n (q0 − η ) divides a := (q0 − η )(q0 − η)(q0 − 1). However, q0 − η 3n has a prime divisor not dividing a. (As n > 1, this follows from Zsigmondy’s theorem [IG , 1.1], using the assumption that q0n = 4.) This contradiction completes the proof.  Lemma 11.3.6. Let X = ΓU3 (2) and t ∈ I2 (X − GU3 (2)). Then t is unique up to X-conjugacy, and m3 (CX (t)) = 1. Proof. For some such t, CSU3 (2) (t) ∼ = L2 (2) by [IA , 4.9.2]. As SU3 (2) has Q8 Sylow 2-subgroups, this implies that X has SD16 Sylow 2-subgroups. Hence all involutions in X − GU3 (2) are X-conjugate to t and the result follows.  11.4. L3 (4) and F i22 . Lemma 11.4.1. Let (K, p) = (L3 (4), 3) or (F i22 , 5). Let P ∈ Sylp (K). Then P ∼ = Ep2 and if K is K-proper, then the following hold: (a) AutK (P ) ∼ = Q8 or (Z4 ∗ SL2 (3))#2, according as p = 3 or 5; in particular, NK (P ) contains an involution z inverting P ; (b) If z is as in (a), then K = NK (P ), CK (z) and |CNAut(K) (P ) (z)|p ≤ p; (c) If K   X and A ∈ Epn (X), n ≥ 3, then K ≤ ΓA,n−2 (X); (d) CAut(K) (P ) ∼ = P × Z2 , an abelian group; ∗ (e) If P ∈ Sylp (Aut(K)), then P ∗ ∼ = 31+2 or E52 , respectively; in particular mp (Aut(K))  = 2;  (f) NK (P ; p ) = {1} and NAut(K) (P ; p ) = O2 (CAut(K) (P )) ∼ = Z2 ;

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(g) If u ∈ Ip (Aut(K)) and Op (CK (u)) has a subgroup admitting an automorphism of order p, then p = 3, u is an outer diagonal automorphism, and if u normalizes P , then u [P, u] ∈ S3 (Aut(K))  and O3 (CAut(K) (P u)) = Z(P u) ∼ = Z3 ; 4 (h) Suppose that K ↑p L ∈ Cp ∪ {A7 } ∪ Tp ∪ Tp5 ∪ TGp . Then p = 3 and L ∈ TG3 , but L is not flat; and (i) If B ∈ Ep2 (Aut(K)) and B ≤ Inn(K), then K = ΓB,1 (K). Proof. First suppose that p = 3 and K ∼ = L3 (4). Then (e) holds as ∼ = P GL3 (4). Also K contains U3 (2), and |CK (g)| = 9 for all g ∈ I3 (K), so (a) holds. For the same reason, NK (P ; 3 ) = {1}, proving part of (f). In (b), CK (z) ∈ Syl2 (K), so the assertion follows from Tits’s lemma [IA , 2.6.7]. Part (c) follows (for p = 5 as well) from [V9 , 9.3] and the fact that mp (Aut(K)) = 2. The preimage of P in SL3 (4) is absolutely irreducible, so P = CInndiag(K) (P ). Then (d) follows as CK (f ) ∼ = U3 (2) or L3 (2) for f a graph-field or field automorphism of K, respectively. Also the rest of (f) follows for the same reason. Part (g) holds by the structures  of P ∗ , CAut(K) (P ), and the fact that O3 (Aut(K)) = Inndiag(K) In (h), L ∈ C3 ∪ Alt by [IA , 7.1.10], and obviously L ∈ T34 ∪ T35 , so L ∈ TG3 − Alt. Moreover L ∼ M23 or HS, as can be seen from [IA , 5.3dm]. So if L ∈ TG3 , = then L is a classical group as in Lemma 2.1abc. But then CL (x) has no L3 (4) 3-component, by [IA , 4.8.2, 4.9.1, 4.9.2], contradiction. Finally, [IA , 7.3.4] implies (i). Now suppose that p = 5 and K ∼ = F i22 . Then (a), (d), (e), and (g) follow directly from the information in [IA , 5.3t]. That table also implies that NCK (g) (P ; 5 ) = {1} for all g ∈ P # , which implies (f). Lemma 13.20c implies (h); (i) and the second assertion of (b) are vacuous as | Out(K)| = 2; and (c) has already been noted. So it remains to prove that K = K0 in (b), where K0 = NK (P ), CK (z). Write

t = O2 (CAut(K) (P )) and K1 = CK (t) = ΓP,1 (K) ∼ = Aut(D4 (2)).  O 3 (Aut(K))

We claim that K0 ≤ K1 . Suppose that K0 ≤ K1 . Then |K0 |2 ≤ |K1 |2 = 213 , so z must be of class 2C [IA , 5.3t] and |K0 |2 = 213 . But some, hence any, 2central involution of K1 must be of class 2A, contradiction. (Note that Sylow 2-centers of K1 are of order 2.) Thus, K0 ≤ K1 . Let K01 = K0 ∩ K1 < K0 . Then ΓP,1 (K0 ) ≤ K01 . Then O5 (K0 ) = 1 by (f), so by [IA , 7.6.1], if K0 = K, 1 5 then F ∗ (K0 ) ∼ = L2 (52 ), A10 , 2F4 (2 2 ) , M c, or 2B2 (2 2 ). From the structure 1 of NK (P ) = NK0 (P ) we see that the only possibility is F ∗ (K0 ) ∼ = 2F4 (2 2 ) , 1 1 so K0 embeds in Aut(2F4 (2 2 ) ) ∼ = 2F4 (2 2 ). Thus |CK (z)|2 ≤ |K0 |2 ≤ 212 , a contradiction. The proof is complete.  Lemma 11.4.2. Let K and p be as in Lemma 11.4.1. Suppose H = LX is a group with p-component L and L/Op (L) ∼ = K. Let B ∈ Sylp (L) and assume that X ∈ N∗H (B; 2). Suppose that F ∗ (H) = O2 (H) and set X1 = J(X), X2 = Ω1 (Z(X)), and X3 = Ω1 (Z(J1 (X))), where J1 (X) =

528

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  Em2 (X)−1 (X) . Then two of the three groups NH (Xi ), i = 1, 2, 3, cover L/Op (L). Proof. Let H = H/Op (H). By a Frattini argument we may assume that Op (H) is 2-closed. Also, X ∼ = Z2 with C := CL (X) ∼ = U3 (2) or Aut(D4 (2)), according as p = 3 or 5. Each NH (Xi ) contains C.  HIf p = 3, we can quote [GL1, I.28-8], so assume that p = 5. Let V = X2 ≤ Z(O2 (H)) and suppose first that CH (V ) ≤ O5 (H). If J1 (X) ≤ O2 (H), then NH (Xi ) = H for i = 1 and 3, so assume that there is A ∈ Em2 (X)−1 (X) such that A = 1. Then A = X so |V : CV (X)| ≤ 2|A| = 4. H-conjugates X i of X, 1 ≤ i ≤ 4, such that Y := However, there exist   X 1 , X 2 , X 3 , X 4 contains some y of order 11. (To prove this, observe first that the involution x ∈ X, not being 2-central in H, is H-conjugate to xz for some 2-central z ∈ C; then z, lying in an Ω− 4 (2) subgroup of C, inverts some f ∈ I5 (C), so xz does as well; finally, L contains a Frobenius group of order 11.5, generated by two conjugates of f .) Thus |V : CV (y)| ≤ 28 , which is absurd as 11 does not divide |GL8 (2)|. Suppose then that CH (V ) ≤ O5 (H). As X2 ≤ V , H is covered by NH (X2 ), so we may assume that NH (Xi ) covers H for neither i = 1 nor i = 3. In particular J(X) ≤ O2 (H), and we fix A ∈ E∗ (X) with X = argument (cf. [T2, 5.53]) shows that if we now put S = A. A standard  Em2 (X)−1 and V = Ω1 (Z(CO2 (H) (S))), then V  H, CH (V ) ≤ O5 (H) (as NH (X3 ) does not cover H), and |V : CV (A)| ≤ 2. Again two appropriate H-conjugates X 1 , X 2 of A generate a dihedral group W of order 10, and thus |V : CV ( X1 , X2 )| ≤ 4, a contradiction as 5 does not divide |GL2 (2)|. The proof is complete.  Lemma 11.4.3. Let K, p, and P be as in Lemma 11.4.1. Suppose K ≤ I ∈ C2 , ΓP,1 (I) < I, and |I|p ≤ p4 . Then K = I. Proof. First suppose that I ∈ Chev(2). We apply [IA , 7.3.4]. If p = 5, 1 5 then I ∼ = 2F4 (2 2 ) or 2B2 (2 2 ), so K ≤ I by Lagrange’s theorem, contradiction. Thus p = 3, and a similar argument shows that I/Z(I) must be a classical group. Since U3 (2) ≤ L3 (4) ≤ I and U3 (2) is a Frobenius group of order 9 · 8, any nontrivial representation of I/Z(I) in characteristic 2 has dimension at least 8. From [IA , 7.3.4] we deduce that I = K or I ∼ = Sp8 (2). But the latter is impossible as |I|3 ≤ 34 . So the lemma holds in this case. Suppose that I ∈ C2 ∩ Chev(3) − Chev(2). If p = 3, then as |I|p ≤ p4 , ∼ I = L3 (3), which does not contain L3 (4), contradiction. If p = 5, then K ≤ I by Lagrange’s theorem, contradiction. Now I ∼ = L2 (q), q ∈ F M9, because m2 (I) ≥ m2 (K) > 2. Therefore, by definition of C2 [V11 , 3.1], we may assume that I ∈ Spor. If mp (I) ≤ 2, then equality holds as K ≤ I. Hence [IA , 7.5.5] implies that I ∼ = M c, HS, Ru, or F i22 . In every case = M11 or M12 , with p = 3; or I ∼ except the desired I = K ∼ = F i22 , the inclusion K ≤ I violates Lagrange’s

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theorem. Thus, we may assume that mp (I) ≥ 3, whence mp (I) = 3 or 4. If p = 5, the only possibility for I is then Co1 , using [IA , 5.6.1] and the fact that |I|p ≤ p4 . But F i22 does not embed in Co1 (they have Sylow 3-subgroups of the same order, but of different ranks). Thus, p = 3. As m3 (I) = 3 or 4, and I ∈ C2 with |I|3 ≤ 34 , we find from [IA , 5.6.1] that I cannot lie in Spor. The proof is complete.  Lemma 11.4.4. Suppose I ↑p J ∈ Kp , with (I, p) = (L3 (4), 3) or (F i22 , 5). Then J ∈ Cp ∪ Tp4 ∪ Tp5 , and if p = 3, then J is not a flat TG3 -group or isomorphic to A7 . Proof. We have Tp4 = {I}. By Lemma 7.13, J ∈ Cp . If J ∈ Tp4 ∪ Tp5 , then by Lemma 8.3, J ∼ = I, a contradiction. As |A7 | < |I|, we may assume that J is a flat TG3 -group and p = 3. From [IA , 5.3d, 5.3m, 5.2.9], J is not as in Lemma 2.1d. If J is as in Lemma 2.1c, then from [IA , 4.8.2] we see that q(I) = 2, which is not the case. Hence J ∼ = Sp4 (q) or L− n (q), n = 4 or 5, q ≡  (mod 3),  = ±1. Therefore a Sylow 3-subgroup of J is abelian of rank 2, and with [IA , 4.8.1], there are two J-conjugacy classes of subgroups of J of order 3. But if J ≥ I ∼ = L3 (4), then J, like I, would have just one such conjugacy class, contradiction. The proof is complete.  11.5. A9 . Lemma 11.5.1. Let K = A9 , t1 ∈ I3 (K), and ti ∈ I2 (K), 2 ≤ i ≤ 7. Suppose that t1 t2 ∈ I3 (K), t1 ti ∈ I2 (K) (3 ≤ i ≤ 7), ti ti+1 ∈ I3 (K) (3 ≤ i < 7), ti tj ∈ I2 (K) (2 ≤ i < j − 1 < 7). Then (t2 t3 )3 = 1. Proof. We have Σ := t1 , t2  ∼ = A4 , H := t4 , t5 , t6 , t7  ∼ = Σ5 , and ∼ J := t3 , H = Σ6 . Moreover, ti inverts t1 for all 3 ≤ i ≤ 7, so [[J, J], t1 ] = 1 and ti | i = 1, 3, 4, 5, 6, 7 = NK ( t1 ) ∼ = Σ3 Y J. Hence without loss we may assume that t1 = (123) and ti = ui (i + 1, i + 2) for 3 ≤ i ≤ 7, where ui ∈ Σ{1,2,3} is a transposition. As [ti , tj ] = 1 for all 3 ≤ i < j − 1 < 7, u3 = u4 = u5 = u6 = u7 and we may assume that u3 = (12). Now [t2 , tj ] = 1 for all 4 ≤ j ≤ 7, so t2 centralizes [H, H] = A{5,6,7,8,9} and hence lies in CΣ (t4 ). Thus, t2 = (12)(34). As t3 = (12)(45), the lemma holds.  Lemma 11.5.2. Let X = Σ9 and X0 = O2 (X). Let z be a 2-central involution of X0 , S = O2 (CX (z)), and S0 = S ∩ X0 . Let O be the set of all transpositions in S. Then S0 is the unique subgroup of S isomorphic to Q8 ∗ Q8 , S0 acts transitively by conjugation on each of O and Oz, and I2 (S − S0 ) = O ∪ Oz. Proof. We have z = z1 z2 z3 z4 , where the zi are disjoint transpositions, ∼ ∼ and CX (z)  = AH = Z2 Σ4 with H = Σ4 permuting the four zi ’s and H A = z1 . Thus, S = AV , where V is the Klein four-subgroup of Σ.

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Moreover, S0 = (A ∩ S0 )V = Ω1 (S0 ) is easily checked to be isomorphic to Q8 ∗ Q8 . Let t, u ∈ I2 (S) with t ∈ A and u ∈ S0 . Then [t, zi ] = 1 for each i, so t ∈ z X ⊆ S0 . Consequently u ∈ A,so u = zi or zi z for some i, and the last two assertions follow. Also uS = A, which is not embeddable in Q8 ∗ Q8 , so u lies in no subgroup R such that Q8 ∗ Q8 ∼ = R ≤ S (note that |S : R| = 2). So S0 = Ω1 (S0 ) is unique, as claimed. The proof is complete.  11.6. A10 . Lemma 11.6.1. Let K = A10 and P ∈ Syl3 (K). Then P ∼ = Z3 Z3 and (a) If t ∈ P and K is not locally balanced with respect to t, then t ∈ J(P ) and t has four fixed points in the natural representation of K on 10 letters; and (b) Let E ≤ P with E ∼ = E32 . If E contains no 3-cycle, then it contains a product of three disjoint 3-cycles. Proof. Let ti ∈ I3 (K) be the product of i disjoint 3-cycles, i = 1, 2, 3. Every element of I3 (K) is conjugate to some ti . We have P ∼ = Z3 Z3 with every element of P − J(P ) conjugate to t3 . Also CAut(K) (t1 ) ∼ = Z3 × Σ7 and CAut(K) (t3 ) ∼ = Z3 Σ3 with respect to the natural action of Σ3 on three letters. Hence O3 (CK (ti )) = O3 (CAut(K) (ti )) = 1 for i = 1 and 3, which implies (a). If (b) is false, then every element of E # is conjugate to t2 , and by Burnside’s lemma, the number of orbits of E on 10 letters is 19 (10 + 8 · 4), which is not an integer, an absurdity.  12. Sylow Subgroups Lemma 12.1. Let K ∈ T3 with Z(K) = 1. Let P ∈ Syl3 (K). Then CAut(K) (P ) is a 3 -group. Proof. We have K ∼ = 3A6 or SL3 (q), q ≡  (mod 3),  = ±1. In the first case P ∼ = = 31+2 and P/Z(K) is self-centralizing in K/Z(K) ∼  3 O (Aut(K)), whence the desired conclusion. In the second case P is absolutely irreducible on the natural module, which implies that its centralizer in Inndiag(K) ∼ = P GL3 (q) is trivial. Hence if the result failed then P would be centralized by a field automorphism of order 3. But this is impossible as 1  |SL3 (q)|3 > |SL3 (q 3 )|. The lemma follows. Lemma 12.2. Let K ∈ Cp , p ≤ 5, and Q ∈ Ep2 (Aut(K)) ∪ Sylp (K). Then the following hold: 1 (a) If K ∼ = A6 , U3 (3), L2 (8), or M11 , with p = 3, or 2F4 (2 2 ) or 5 2B (2 2 ), with p = 5, then C 2 Aut(K) (Q) is a p-group; and ∼ (b) If K = L3 (4) with p = 3, or F i22 with p = 5, then |CAut(K) (Q)| = 2p2 .

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∼ A6 , then Aut(K) = Aut(A1 (9)), then CAut(K) (Q) = Q Proof. If K = by [IA , 2.6.5e]. The same reference will handle K ∼ = U3 (3) = 2 A2 (3) and 1 Aut(K) ∼ = 2 G2 (3 2 ) once we remark that a Sylow 3-subgroup = Aut(L2 (8)) ∼ P of Aut(K) containing Q is extraspecial of width 1, so O3 (CAut(K) (Q)) = O 3 (CAut(K) (P )) = 1. Next, the assertions about M11 and F i22 follow from [IA , 5.3a, 5.3t (see Note 3)]. If K ∼ = L3 (4), then again for Q ≤ P ∈ Syl3 (Aut(K)), P ∼ = 31+2 , so O3 (CAut(K) (Q)) = O3 (CAut(K) (P )). By [IA , 2.5.12], Out(K) ∼ = Z2 × Σ3 , and by Lemma 12.1, CK (P ∩ K) = P ∩ K. 3 Hence |O (CK (P ))| ≤ 2, and equality holds as some graph-field involutory automorphism centralizes U3 (2) and hence P ∩ K. 1  If K ∼ = 2F4 (2 2 ) , then thanks to [IA , 4.8.7, 7.6.2e], O2 (CAut(K) (Q) = 1, 5 so CAut(K) (Q) = CK (Q) = Q. Finally, if K ∼ = 2B2 (2 2 ), then P is extraspecial of order 53 . With [IA , 7.7.11d], CAut(K) (Z(P )) = P , which implies the desired conclusion. The proof is complete.  Lemma 12.3. Let K ∈ Chev(p) have twisted Lie rank 1 and p-rank at least 3, p an odd prime. Let P ∈ Sylp (K). Then NK (P )/P acts irreducibly on P/Φ(P ). Proof. Here NG (P ) is a Borel subgroup of K. This lemma is a routine, if messy, consequence of the Chevalley relations [IA , Sec. 2.4], in particular the structure of P [IA , Table 2.4]; the definition of elements of the Cartan subgroup [IA , Table 2.4.7]; and the action of the Cartan subgroup on P [IA , 1.12.1g]. Note that in every case, P/Φ(P ) is parametrized by t ∈ Fα , and in the case of the Ree groups, the assumption mp (P ) ≥ 3 is needed to guarantee this parametrization.  Lemma 12.4. A Sylow 3-subgroup of K = Aut(3D4 (2)) has exponent 9. Proof. By [IA , 2.5.12], for a suitable graph automorphism f ∈ I3 (K), CK (f ) ∼ = Z3 × P GU3 (2) [IA , 4.9.2], and if S ∈ Syl3 (CK (f )), S ∼ = Z3 × 31+2 has exponent 3. Let SK := S ∩ [K, K]. Then CSK f  (f ) is normal in SK f  of index 3. So SK f  has exponent at most 9. As K contains L2 (8), there  are elements of order 9 in SK , and the equality holds. 13. Pumpups and Subcomponents Lemma 13.1. Let X be a K-group with O3 (X) = 1, let x ∈ I3 (X), and let L be a quasi-3-component of CX (x), so that L/O3 (L) ∼ = SU3 (2). Set  L0 = O3 (L) and let K be the subnormal closure of L0 in X. Then one of the following holds: (a) (b) (c) (d)

K K K K

≤ O3 (X); ∈ Spor − TG3 ; ∈ Chev(2), and either K has level q(K) = 2 or K ∼ = SU3 (8); or ∈ Chev(3), but K ∈ Alt ∪ Chev(r) for any r = 3.

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Proof. Assume that (a) fails. By solvable L3 -balance [IG , 13.8], L0 lies in a component of X, which must be K. Then as L0  L, L normalizes K. We may pass to KL x /O3 (KL x) and assume that X = KL x. There exists a 2-subgroup Q ≤ L such that L = O3 (L)L0 Q, and we may pass to KL0 Q, x and assume that X = K Q, x, with Q/Q ∩ O3 (CX (x)) ∼ = Q8 . Since L   CX (x), Q centralizes every Q-invariant abelian 3-subgroup of CX (x), and consequently [Q, Z(K)] = 1 by the A × B-lemma. Thus [O 3 (X), Z(K)] = 1. Now Z(K) is a 3-group, and so in X := X/Z(K) we have O 3 (CX (x)) = O3 (CX (x)). As L   O3 (CX (x)) it follows that L   CX (x). Hence we may assume that either Z(K) ∩ L = 1 or Z(K) = 1.  Furthermore, as O3 (L) = L0 ≤ K, KL/K is a 3 -group. But also KL  KL x = X, so O3 (KL) = 1 and hence F ∗ (KL) = K. Suppose that K ∈ Alt. If Z(K) = 1 then K ∼ = 3A6 or 3A7 , so L 1+2 ∼ contains R ∈ Syl3 (K) with R = 3 . But then CAut(K) (R) = 1 and so x centralizes K, a contradiction. Thus Z(K) = 1 and K ∼ = A3p+q , where x acts as the product of p 3-cycles, and with q fixed points. As L has nonabelian Sylow 3-subgroups, O3 (CΣ3p+q (x)) is nonabelian. But CΣ3p+q (x) is the direct product of Σq with an extension of E3p by Σp . In particular, O3,2 (CK (x)) has a normal self-centralizing E3p -subgroup. This is inconsistent with the structure of L, however. Thus, K ∈ Alt. Suppose next that K ∈ Chev(r) − Alt for some prime r. Suppose that r = 3. If x induces a field or graph-field automorphism on K with C := CK (x) nonsolvable, then by [IA , 4.9.1,7.1.4c], CInn(K) ([C, C]) = 1, a contradiction since [C, C, L0 ] = 1. Thus if x induces a field or graphfield automorphism on K, C is solvable, forcing K ∼ = SU3 (8), so (c) holds. Therefore we may assume in this case that x induces an inner-diagonal or graph automorphism on K. Inspection of [IA , 4.7.3A] shows that it must  be inner-diagonal. If r = 2, then as O2 (CK (x)) is the central product of Lie components with level at least as high as that of K, by [IA , 4.2.2],  (c) must hold. If r > 3, then again Kx := Or (CK (x)) is the commuting product of quasisimple groups in Chev(r), and CK (x)/Kx has an abelian normal r-complement. But then [L, L] is a subnormal r -subgroup of Kx , so [L, L] ≤ Or (CK (x)) ≤ Z(CK (x)). This is a contradiction as [L, L] is not abelian. Therefore r = 3 and (d) holds. It remains to rule out the possibility K ∈ Spor ∩ TG3 = {M23 , M24 , HS, J4 , He, Ru, 3ON }. Suppose that this occurs and let P ∈ Syl3 (K). As | Out(K)| ≤ 2, x induces an inner automorphism on K and so we may assume that x O3 (L) ≤ P .  If x ∈ O3 (L), the only possibility is K ∼ = 3ON , but then O3 (CK (x)) ∼ = 3A6 × Z3 , so O3 (CK (x)) is abelian, a contradiction. Hence x ∈ O3 (L). In particular |K|3 ≥ 3 and in case of equality x is 3-central in K. This forces K∼ = J4 , He, or Ru, but in every case O3 (CK (x)) = x, a final contradiction. The lemma is proved. 

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Lemma 13.2. Suppose that K ↑2 L ∈ C2 via x ∈ I2 (Aut(L)). Suppose also that mr (L) ≤ 3 for all odd primes r. Then the following hold. (a) If K ∼ = A5 , then L/Z(L) ∼ = L2 (24 ), L± 3 (4), M12 , or HJ, with Z(L) elementary abelian; (b) If K ∼ = L3 (4), or HJ; = L2 (7), then L/Z(L) ∼ (c) If K ∼ = J3 ; = L2 (17), then L ∼ (d) If K ∼ = L4 (2), L5 (2), Sp4 (4), HS, 2HS, or U4 (2); = A6 , then L ∼ ∼ (e) If K = Sp4 (8), then L ∼ = Sp4 (26 ) or L± 4 (8); (f) If K ∼ = 3D4 (22 ); and = 3D 4 (2), then L ∼ 5 (g) K ∼ 2F 4 (2 2 ). = Proof. Part (a) follows from [III11 , 1.6] and the definition of C2 [V11 , 3.1]. Part (d) similarly follows from [V8 , 3.10n], with a few groups ruled out that violate the condition m3 (L) ≤ 3 [IA , 3.3.3, 4.10.3]. Given the remaining possibilities for K, we have L ∈ Chev(2) ∪ Spor, by [IA , 7.1.10], [IA , 2.2.10], and the definition of C2 . If L ∈ Spor, we survey [IA , 5.3] to verify that (b) or (c) holds with L ∼ = HJ or J3 . Now assume that L ∈ Chev(2). If x is a field or graph-field automor5 phism, then L is as in (b), (e), or (f), or K ∼ = F4 (25 ). The = 2F4 (2 2 ) and L ∼ last of these satisfies m3 (L) > 3, however, contrary to assumption. So, by [IA , 4.9.1, 4.9.2] and the Borel-Tits theorem, we may assume that x is a ± graph automorphism, whence K ∼ = L± = Sp4 (8) and L ∼ 4 (8) or L5 (8). In the last case, m3 (L) > 3 or m7 (L) > 3, contrary to assumption. The proof is complete.  2 Lemma 13.3. Let K = L± 4 (3) and t ∈ U ∈ E2 (Aut(K)) with E(CK (t)) = ∼ E(CK (U )) = A6 . Then the following conditions hold: (a) t ∈ Inndiag(K); and (b) E(CK (u)) ∼ = U4 (2) for all u ∈ U − t.

Proof. We write K = L4 (3) and use [IA , 4.5.1]. Note that Out(K) is dihedral of order 2(3 − ) and Outdiag(K) is cyclic of order 3 − . Since [E(CK (t)), U ] = 1, CInndiag(K) (E(CK (t))) = 1. Hence t is in the class t2 of [IA , 4.5.1]. In particular, (a) holds. Moreover, for any u ∈ U − t,  u induces a graph automorphism on K. Then E(CK (u)) = O3 (CK (u)) contains E(CK (t)) ∼ = A6 , so u is of class γ1 or γ2 in [IA , 4.5.1]. But if u  were of class γ2 , then E(CK (u)) = E(CK (t)) so [E(CK (u)), t] = 1, contrary to the table, which would specify CInndiag(K) (E(CK (u))) = 1. Therefore u  is of class γ1 and so (b) holds. The proof is complete. Lemma 13.4. Suppose that K ∈ K, O2 (K) = 1, U4 (2) ↑3 K/O3 (Z(K)) or L± 4 (3) ↑3 K/O3 (Z(K)), and Sp6 (2) ↑2 K/O2 (Z(K)). Suppose also that m2,3 (K) ≤ 3. Then K ∼ = U6 (2), SU6 (2), or Ω+ 8 (2). Proof. Using [IA , 5.3] we see that the condition Sp6 (2) ↑2 K/O2 (Z(K)) forces K ∈ Spor. Likewise K ∈ Chev(r) ∪ Alt for any odd r, by [IA , 7.1.10]. So K ∈ Chev(2). By the Borel-Tits theorem, CK (g) ∼ = Sp6 (2) for some

534

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involutory field, graph, or graph-field automorphism g of K. By [IA , 4.9.1, 4.9.2], either K is of one of the claimed isomorphism types, or K ∼ = Sp6 (4), (2). In the U (2) case, m (K) ≥ m (U U7 (2), L6 (2), L7 (2), or Ω− 7 2,3 3 5 (2)) = 4, 8 contradiction. In the other cases m3 (Aut(K)) = 3 [IA , 4.10.3, 2.5.12]. The condition K/O3 (Z(K)) ↓3 U4 (2) or L± 4 (3), however, forces m3 (Aut(K)) ≥ 4, ruling these cases out. The lemma follows.  5 Lemma 13.5. Suppose that 2B2 (2 2 ) ↑2 K ∈ C2 . Then K ∼ = Sp4 (25 ).

Proof. As in the proof of Lemma 13.4, we reduce to the case K ∈ 5 Chev(2), with 2B2 (2 2 ) a component of CK (g) for some field, graph-field, or graph automorphism g of K This time [IA , 4.9.1, 4.9.2] yield only K ∼ =  Sp4 (25 ), as claimed. 1

Lemma 13.6. Let K = 2F4 (2 2 ) , M11 , M12 , 2M12 , HJ, 2HJ, [2 × 2]L3 (4), L3 (3), HS, 2HS, Ru, or 2Ru. Then the following hold: (a) There is no J ∈ Co2 such that K ↑2 J and mr (J) ≤ 3 for all odd primes r; and (b) In the cases K = 2M12 , 2HJ, and 2Ru, there is no J ∈ Co2 such that K ↑2 J and m2,r (J) ≤ 3 for all odd primes r. Proof. Suppose J exists satisfying the condition in (a) or (b). By [IA , 7.1.10], J ∈ Chev(2) ∪ Spor or K = L3 (3). Suppose J ∈ Spor. If K = 2F (2 21 ) , then by the Borel-Tits theorem and [I , 4.9.1, 4.9.2], J ∼ F (2), = 4 4 A so m3 (J) > 3 [IA , 4.10.3]. If K = [2 × 2]L3 (4), then for the same reasons, Z(J) must be noncyclic and J/Z(J) ∼ = L3 (42 ), contradicting [IA , 6.1.4]. If K = L3 (3), then for the same reasons J ∈ Chev(3) ∩ Co2 , so J/Z(J) ∼ = L± 4 (3) and m3 (J) = 4. Therefore J ∈ Spor, and we use [IA , 5.3] freely, as well as the rank 2Ru, and J ∼ of J from [IA , 5.6.1]. In (b), K ∼ = = 2Suz, whence m2,3 (J) = m3 (J) = 5. In (a), the only other cases that arise are M12 ↑2 Suz, HJ ↑2 Suz, [2×2]L3 (4) ↑2 He, and 2HS ↑2 F5 . But m3 (Suz) = 5, and Co2 contains neither He nor F5 . The lemma follows.  Lemma 13.7. Let p = 5, 7, or 17, and K = L2 (p). Let L be as in Lemma 13.2abc, respectively. Suppose that K = E(CL (t)) for some t ∈ I2 (Aut(L)). Assume L/Z(L) ∼ M12 or HJ. Then p does not divide |CAut(L) (K t)|. = Proof. If L ∈ Spor, this holds by inspection of [IA , 5.3]. If L ∼ = L2 (42 ) 2 or L3 (2 ), this holds by [IA , 7.1.4c]. Otherwise t is a graph automorphism of K ∼ = L3 (4) or U3 (4), and CInn(K) (K t) = 1 by [IA , 4.9.2]. Moreover, t  inverts O 2 (Out(K)), and the lemma follows. Lemma 13.8. Suppose that K ∈ C2 , L ∈ K2 , and K ↑2 L via x ∈ I2 (Aut(L)). If (a), (b), or (c) holds, then K is semirigid in L, and CAut(L) (K x) = x : 5 (a) K ∼ = L2 (8) or 2B2 (2 2 );

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∼ HJ or M12 or Ru; or (b) K = ∼ (c) K = A5 , and L/Z(L) ∼ = L2 (24 ) or L± 3 (4). Proof. By [IA , 7.1.10], L ∈ Alt. Suppose next that L ∈ Spor. Then surveying [IA , 5.3] we see that L/Z(L) ∼ = Suz and K ∼ = HJ or M12 ; moreover, CAut(L) ( x K) = x, as required. Finally, assume that L ∈ Chev(r), so that K ∈ Chev(r) by [IA , 7.1.10]. 1 If r = 2, then K ∼ = L2 (8) = 2 G2 (3 2 ) , L ∼ = G2 (3), and K ∗ := CL (x) ∼ = 1 2 G (3 2 ) ∼ Aut(K). Therefore C ∗ (K) = 1, so C (K

x) embeds = 2 Aut(K ) Aut(L) in CAut(L) (K ∗ ), which equals x by [IA , 7.1.4c], as x is a graph-field automorphism in this case. Therefore we may assume that r = 2, whence by the Borel-Tits theorem and [IA , 4.9.1, 4.9.2], x induces a field, graph-field, or  graph automorphism on L. Moreover, K = O2 (CL (x)) in every case (note that L ∼ = L2 (82 ) or Sp4 (25 ) in (a)). Again [IA , 7.1.4c] implies the desired conclusion unless possibly x is a graph automorphism. This only occurs in (c) with L/Z(L) ∼ = L± 3 (4). By [IA , 4.9.2], CInn(L) ( x K) ≤ Z(K) = 1 when we regard K as a subgroup of Inn(L). Moreover, by [IA , 4.9.2c], COutdiag(L) (x) ∼ = CZ (x) = 1 where Z = Z(Lu ), Lu being the universal version of L. Using [IA , 2.5.12], we see that Out(L)/ Outdiag(L) ∼ = E22 or ∼ Z2 according as L/Z(L) = L3 (4) or U3 (4). Hence if the lemma fails, then L/Z(L) ∼ = L3 (4) and CAut(L) (K) contains a field automorphism f . However, CL/Z(L) (f ) ∼ = L3 (2), so [K, f ] = 1, a contradiction. The proof is complete.  Lemma 13.9. Let K = He and p = 5 or 7. Then there is no L ∈ Kp such that K ↑p L and L/Z(L) ∈ A. Proof. Clearly, L ∈ Spor if L exists, and the result follows by inspection of [IA , 5.3].  Lemma 13.10. ↑2 (J4 ) = ∅. Proof. If J4 ↑2 K, then K ∈ Spor and the result follows by inspection  of the tables [IA , 5.3]. n

Lemma 13.11. ↑2 (2F4 (2 2 ) ) = {F4 (2n )}, for n = 2k + 1 ≥ 1. n

Proof. Suppose 2F4 (2 2 ) ↑2 K. By inspection of [IA , 5.3], K ∈ Spor, n and then as 2F4 (2 2 ) is unambiguously in Chev(2), we have K ∈ Chev(2). By the Borel-Tits theorem, and then by [IA , 4.9.1, 4.9.2], K ∼ = F4 (2n ) (and 2F (2 n  4 2 ) = E(CK (φ)) for a graph-field automorphism φ of K of order 2.)  Lemma 13.12. Let p = 7 or 11, and let L ∈ Kp with Ap ↑p L. Let I be a covering group of L. Then m2 (L) ≥ 5 and e(I) > 3. Proof. By [IA , 7.1.10], L ∈ Alt∪Spor. If L ∈ Alt, then L contains A14 , and I contains A14 or 2A14 . As A14 contains A4 × A4 × A4 and E33 × A4 , the lemma holds in this case. So suppose that L ∈ Spor. From [IA , 5.3], p = 7 and L ∼ = Co1 or F i24 . The information in [IA , 5.6.1, 5.6.2] on ranks and 2-local ranks finishes the proof. 

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Lemma 13.13. Let K ∈ Co2 with A5 ↑2 K and e(G) ≤ 3. Let x ∈ Ip (Aut(K)), p ∈ {7, 11}. Then CK (x) has no composition factor isomorphic to A7 . Proof. Let L ∼ = A7 be a composition factor of CK (x). Since p > 5, ∩ Chev(p) = ∅. If K ∈ Chev(r), then r = p, and so by [IA , 4.2.2, 4.9.1], L ∈ Chev(r), contradiction. If K ∈ Alt, then K/Z(K) ∼ = An for some n ≡ 7 (mod p), so K ∈ Co2 . Hence, K ∈ Spor. As A5 ↑2 K and K ∈ Co2 , K/Z(K) ∼ = M12 or HJ, so 7p does not divide |K|, a final contradiction. 

Co2

Lemma 13.14. Let p be an odd prime. a+1 {L2 (pp )}.

a

For a ≥ 1, ↑p (L2 (pp )) =

a

Proof. Let L2 (pp ) ↑p K via x ∈ Ip (Aut(K)). With [IA , 2.2.10, 4.9.6, 5.2.9, 5.3], K ∈ Chev(p). By the Borel-Tits theorem, x ∈ Inn(K). Then by  [IA , 4.9.1, 4.9.2], x is a field automorphism, completing the proof. Lemma 13.15. Let K = L3 (q), q ≡  (mod 3),  = ±1, q > 2. Let x ∈ I3 (K) be 3-central in K. Then CAut(K) (x) has no SU3 (2)-subgroup. Proof. Suppose false, so that there is a F = P Q ≤ Aut(K) such that P ∼ = Q8 , and F/Z(P ) is a Frobenius group. In partic= 31+2 , Q ∼ ular Q is irreducible on P/Z(P ). As a result, since m3 Outdiag(K)) = m3 (Out(K)/ Outdiag(K)) = 1, P ≤ Inn(K). Therefore |K|3 > 9, so q ≡  (mod 9). Let C1 and C2 be the K-conjugacy classes of subgroups of K of order 3 (see the proof of Lemma 5.9). By that lemma, if U ≤ P with U ∼ = E32 , all subgroups of U of order 3 are in class C1 . Indeed, if u ∈ U − Z(P ) and

u ∈ C1 , then U = C1 ∩ P   F , which is not the case. And if Z(P ) ∈ C2 , then CK (Z(P )) has abelian Sylow 3-subgroups, whereas P is nonabelian. Hence, every subgroup of P of order 3 lies in C1 . On the other hand, a Sylow 3-subgroup S of K containing P has a maximal diagonalizable subgroup D of index 3, and Ω1 (D) ∼ = E32 contains only one subgroup of class C1 . But 2  |D ∩ P | ≥ 3 , a contradiction. Lemma 13.16. Suppose p is an odd prime, K ≤ L ∈ Kp , and A ∈ If K is a component of CK (a) for all a ∈ A# , then K ∈ Tp .

Ep2 (Aut(L)).

Proof. This follows directly from [III11 , 1.16] and the fact that if p = 3, L∼ = A6 , then = 3Suz, K = E(CL (a)) for some a ∈ A# , and KZ(L)/Z(L) ∼ ∼  K = A6 as well [IA , 5.3o]. Lemma 13.17. Let X be a group with O3 (X) = 1. Let K ∈ CTG3 be a component of X such that one of the following holds: (a) For some J0 ∈ CTG3 such that J0 /Z(J0 ) is not locally balanced with respect to p = 3, J0 2. If K is a C3 -group, T3 -group, or flat TG3 -group such that L ↑3 K, then K ∼ = L3 (q 3 ) ∈ T32 . Proof. Write q = ra where r is prime. Then r = 3. Also, L ∈ Alt ∪ Chev(s) for any prime s = r, so K ∈ Chev(r) ∪ Spor. Also, our assumptions imply that m3 (Aut(K)) ≥ 3. If K is a flat TG3 -group, we conclude that K ∼ = Sp8 (2), D4− (2), or D5 (2); but then by [IA , 4.8.2], L has level q = 2, contrary to assumption. If K ∈ T3 , the only possibility besides the desired K ∼ = G2 (8); but then = L3 (q 3 ) is K ∼  ∼ L = L2 (8), SU3 (8), or G2 (2) by [IA , 4.7.3A, 4.9.1], contradiction. Suppose finally that K ∈ C3 . If K ∈ Spor, then L3 (q) ∈ ↓3 (K) by examination of [IA , 5.3], so K is one of the eight groups in C3 lying in Chev(2) − Chev(3) [V11 , 3.1]. As q > 2, we see from [IA , 4.8.2, 4.7.3A] that none of these eight  lies in ↑3 (L3 (q)). The proof is complete.

538

17. PROPERTIES OF K-GROUPS

Lemma 13.19. Let p be an odd prime and suppose that K ∈ Tpi for some i, x ∈ Ip (Aut(K)), and I is a component of CK (x) that is not locally 2balanced for p. Assume that if i = 1, then K ∼ = HJ. Then i = 2 and x is a field automorphism. Proof. Refer to [IA , 7.7.7c] to verify local 2-balance. If K ∼ = HJ, then by [IA , 5.3g], I ∼ = 3A6 , which is locally 2-balanced for p = 3. Therefore i > 1. If K ∼ = L3 (q), q ≡  (mod 3),  = ±1, q > 4, then by [IA , 4.8.2, 4.8.4, 4.9.1], either I ∼ = L2 (q), which is locally 2-balanced, or x is a field automorphism, in which case q ≡  (mod 9) so i = 2. Finally, if K ∼ = A7 , M12 , or M22 , L3 (4) (with p = 3) or F i22 (with p = 5), or if mp (K) = 1, then mp (I) = 1, if indeed I exists at all [IA , 5.2.9, 5.3bct, 4.8.2, 4.8.4] and again I is locally 2-balanced for p, a final contradiction.  Lemma 13.20. The following conditions hold: (a) ↑p (F1 ) = ∅ for all p dividing |F1 |; (b) ↑3 (HJ) = ∅; (c) ↑5 (F i22 ) = ∅; (d) ↑5 (HS) = ∅; (e) ↑5 (Ru) = ∅; and (f) ↑2 (L3 (3)) ∩ Co2 = {L4 (3)}. Proof. Except for the last assertion, by [IA , 4.9.6, 5.2.9], each set consists of sporadic groups. The result follows by inspection of [IA , 5.3]. In the case of the last assertion, one must also inspect [IA , 4.5.1] for pumpups in  Chev(3) of L3 (3) lying in Co2 . Lemma 13.21. Suppose that K ∼ = D4− (2) or D5 (2), and b ∈ I3 (Aut(K)). Then no two components of E(CK (b)) are isomorphic. Proof. This follows from the description of CK (b) given in [IA , 4.8.2].  Lemma 13.22. Let K = P Sp4 (q), q ≡ 0 (mod 3), or L− n (q), n ∈ {4, 5}, q ≡  (mod 3),  = ±1. Let u ∈ I3 (K). Then O3 (CK (u)) is cyclic. Proof. If K = P Sp4 (2) or L4 (2), then O3 (CK (u)) = 1, so we may exclude these cases from our argument. Let V be the natural module for J := Sp4 (q) or SL− n (q) and t ∈ J be a preimage of u ∈ K. Let L = E(CJ (t)). First suppose dim[V, t] = 2. Then L ∼ = SL2 (q) or SL− n−2 (q), respectively, and CJ (t) respects the decomposition V = [V, t] ⊕ [V, L] (orthogonal sum if  = 1), with [V, L] = CV (t) a natural L-module. As 3 divides |L|, O3 (CJ (t)) induces scalars on [V, L]. Also t is orthogonally indecomposable on [V, t], so O3 (CJ (t)) induces a cyclic group Z on [V, t]. The kernel W of the restriction mapping ρ[V,t] : g → g|[V,t] on O3 (CJ (t)) consists of scalars on [V, L] of determinant 1. Hence W = 1 if n = 5 and W ∼ = Z2 otherwise. Thus ρ[V,t] induces an isomorphism O3 (CK (t)) ∼ = Z, as desired.

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Likewise if dim[V, t] = 4, the only other possibility, we have L ∼ = SL2 (q), 2 2 ∼ SL2 (q ), or SL2 (q ), respectively. Suppose that K = P Sp4 (q). If q is even, then application of a graph-field automorphism reduces us to the previous case, so we may assume that q is odd. Then using the isomorphism K ∼ = Ω5 (q), we see easily that O3 (CK (u)) embeds in CΩ2 (q) (u ) ∼ = Zq− for some u ∈ I3 (Ω2 (q)), so O3 (CK (u)) is cyclic, as desired. Finally suppose that J ∼ = SL− n (q), so that V = [V, t] ⊕ CV (t) with dim CV (t) ≤ 1, [V, L] = [V, t], and CV (L) = CV (t). Again [O3 (CL (t)), L] = 1. Since all elements of J have determinant 1, the restriction map on O3 (CL (t)) into the isometry group of [V, L] is injective, and so we are re− duced to the case n = 4. Interpreting K = L− 4 (q) as P Ω6 (q), we again see that O3 (CK (t)) embeds in the centralizer of an element of order 3 in O2 (q), and hence is cyclic. The proof is complete.  Lemma 13.23. Let K = L3 (q), q ≡  (mod 3),  = ±1, and suppose that 1 f is a field automorphism of K of order 3. Then CK (f ) ∼ = P GL3 (q 3 ) and CInndiag(K) (f ) ≤ Inn(K). 1  Proof. Write q0 = q 3 . Let K0 = Or (CK (f )) ∼ = L3 (q0 ), where q is a power of the prime r. By [IA , 4.9.1b], CInndiag(K) (f ) ∼ = P GL3 (q0 ), so it is enough to show that CK (f ) > K0 . Let (K, σq ) be a σ-setup for K,   and identify K and K0 with Or (CK (σq )) and Or (CK (σq0 )), respectively. Let L be the universal version of K, φ : L → K a covering, and write

z = Z(L) ∼ = Z3 . Then by the Lang-Steinberg theorem, z = g −1 σq0 (g) for some g ∈ L. Then φ(g) ∈ CK (σq0 ). However, the full preimage of K0 in L is CL (σq0 ), which does not contain g. So φ(g) ∈ CK (f ) − K0 , and the lemma follows. 

Lemma 13.24. Suppose that K ↑p L ∈ Cp , where (K, p) = (L2 (8), 3), 5 (L2 (4), 5), (2B2 (2 2 ), 5), or (L3 (2), 7). Also assume that L ∈ A. Then L ∼ = 3 2 p ∼ Co3 or G2 (3 2 ), both with p = 3 and m2,3 (L) = 3; or L = L2 (p ), p = 5 or 7; or L ∈ Chev(2); or mp (Aut(L)) = 2. Proof. We may assume that L ∈ Chev(2). If L ∈ Chev(r) for some r, then K ∈ Chev(r) and so r = p [IA , 2.2.10]. Then by the Borel-Tits 3 theorem and [IA , 4.9.1, 4.9.2], L ∼ = 2 G2 (3 2 ) or L2 (pp ), as asserted. Note that 3 3 m2,3 (2 G2 (3 2 )) = 3 since any involution centralizer in 2 G2 (3 2 ) is isomorphic to Z2 × L2 (33 ) [IA , 4.5.1]. So we may assume that L ∈ Chev. If L ∈ Alt, then K ∈ Alt, whence K ∼ = An with = A5 and p = 5. As L ∈ A, L ∼ n ≤ 12 so m5 (Aut(L)) = m5 (Σn ) = 2. Finally suppose that L ∈ Spor. Then mp (Aut(L)) ≥ 2 as K ↑p L, and as | Out(L)| ≤ 2, we may assume that mp (L) > 2. Using [IA , 5.6.1, 5.6.2] and the tables [IA , 5.3], we find that  L∼ = Co3 and m2,3 (L) = 3. The lemma is proved. Lemma 13.25. Let A6 ↑5 K ∈ K5 and suppose that K ∈ A. Then K ∼ = A11 , L2 (310 ), U3 (9), Suz, Sp4 (25 ), Sp8 (2), or F4 (2).

540

17. PROPERTIES OF K-GROUPS

Proof. Note that | Aut(K)|5 ≥ 52 . From [IA , 5.3] we see that if K ∈ Spor, then K ∼ = Suz. If K ∼ = An , then by [IA , 5.2.6] and as K ∈ A, n = 11. If J ∈ Chev(2) (possible as A6 ∼ = Sp4 (2) ), then the Dynkin diagram of J must contain a double bond, so J ∼ = Sp4 (25 ), Sp2n (2), n ≥ 4, or F4 (2). Again as K ∈ A, the desired conclusion holds. Finally if K ∈ Chev(3), then by order considerations K is as in Def. 1.1ab, and the result follows easily. This completes the proof.  Lemma 13.26. Let I ↑5 K ∈ K5 with I ∈ S := {A7 , A8 , A9 , A12 }, and suppose that K ∈ A. Then K ∈ S, or K ∈ Chev(2) is a pumpup of A8 . Proof. If K ∈ Chev(r), then I ∈ Chev(r) by [IA , 4.9.6], so the last possibility occurs. If K ∈ Spor ∪ Alt − S, then by [IA , 5.3, 5.2.6], K ∈ A, contradiction, proving the lemma.  Lemma 13.27. ↑3 (3A6 ) ∩ Chev = ∅. Proof. Suppose that K ∈ Chev(r), x ∈ I3 (Aut(K)), and L   CK (x) with L ∼ = 3A6 . If r = 3, then Z(K) ≤ Z(L) by the Borel-Tits theorem, so K/Z(K) ∼ = U4 (3), G2 (3), or Ω7 (3) by [IA , 6.1.4]. In these cases Out(K) is a 3 -group, so the Borel-Tits theorem is contradicted. Thus, r = 3. We may pass to K/O3 (K), whence K = [I, I] for some I ∈ Lie(r) (usually K = I), and x extends to an automorphism of I of order 3, by [IA , 2.5.15]. Then L is a component of CI (x) and hence by [IA , 4.2.2], L = [J, J] for some J ∈ Lie(r). But L ∈ Lie(r), so L < J. Hence J/Z(J) ∼ = Sp4 (2). But then ∼ L = A6 , a contradiction. The proof is complete.  ∗ Lemma 13.28. Suppose that K = F (X) ∼ = HS or 2HS, u ∈ I2 (X), and ∼ m3 (CX (u)) = 2. Then E(CX (u)) = A6 or A8 , and CX (E(CX (u)))/Z(X) is elementary abelian. Proof. See [IA , 5.3m]. Note that if E(CX (u)) ∼ = A6 , then there is an  involution v ∈ CAut(K) (E(CX (u))) such that E(CX (v)) ∼ = A8 . Lemma 13.29. Suppose that X is a K-group, m3 (E(X)) = 2, A ≤ X with A ∼ = E32 , and every component of E(X) lies in C2 . Suppose that for each a ∈ A# , A ∈ Syl3 (CX (a)) and E(CE(X) (a)) ∼ = A5 . Then E(X) ∼ = Sp4 (4), HS, 2HS, or M23 . Proof. Since A ∈ Syl3 (CX (a)) for all a ∈ A# , A ∈ Syl3 (X). Let K be a component of X of order divisible by 3. Then A ∩ K = 1. If there is a second component L in E(X), then L ≤ CE(X) (A ∩ K) so L ∼ = A5 and # A ∩ L = 1. But then for any a ∈ A − K − L, E(CE(X) (a)) is a 3 -group, contrary to assumption. Therefore E(X) = K. b Likewise the assumption E(CE(X) (a)) ∼ = A5 forces A ≤ K and O3 (K) = 1. Thanks to our knowledge of Schur multipliers [IA , 6.1.4], we may pass modulo O3 (X) and assume that K is simple. If K ∈ Spor, the conditions E(CK (a)) ∼ = A5 and A ∈ Syl3 (K) yield K ∼ = HS or M23 [IA , 5.3]. If K ∼ = An , then 6 ≤ n ≤ 8 and the hypothesis fails for

13. PUMPUPS AND SUBCOMPONENTS

541

a being the product of two disjoint 3-cycles. So assume K ∈ Chev(r) − Alt for some r. By [IA , 4.9.6], r ∈ {2, 5}, as A5 ∈ Chev(r) for r = 2 or 5 only. Since K ∈ C2 , we may take r = 2. n # is 3-central and E(C (a)) = 1, If K ∼ = L± K 3 (2 ), then some a ∈ A n ), so K is 3(2 contradiction. As A ∈ Syl3 (K), K does not involve L± 6 saturated. It follows from Steinberg’s connectedness theorem that for a σ-setup (K, σ) of K, A lies in a maximal torus of K. Hence by [IA , 7.4.1a],  A acts 2-exactly on K [IA , 7.2.5]. Note that for any a ∈ A# , O2 (CK (a)) has no solvable Lie components; as A ∈ Syl3 (K), such Lie components would 1 be impossible whether they had order divisible by 3 or K ∼ = 2F4 (2 2 ) . Thus |CK (a)|2 = 22 , and 2-exactness implies that |K|2 = 28 . The only possibility  is K ∼ = Sp4 (4) (L2 (28 ) has 3-rank 1). The proof is complete. Lemma 13.30. Let I ∈ C2 , and let u ∈ Aut(I) with u3 = 1. Let I1 be a component of CI (u) such that I1 /O3 (I1 ) ∼ = A8 . If m3 (I) = 2, and L3 (CI (t)/O3 (CI (t))) ∼ = A5 for every 3-element t ∈ I1 mapping on a 3-cycle in I1 /O3 (I1 ), then I1 = I. Proof. With [IA , 4.9.6, 5.2.6, 5.3] we have I/Z(I) ∼ = A8+3k , k > 0, or I ∈ Chev(2). But A8+3k ∈ C2 only for k = 0, so I ∈ Chev(2). As 2A8 ∈ C2 , we may assume that u = 1 and it suffices to argue to a contradiction. Since m3 (I) = 2 = m3 (A8 ), u induces a non-inner automorphism on I. Because of the hypothesis on t, q(I) = 2 or 4, by [IA , 4.2.2e], and I ∼ 3D4 (q) for any q, by [IA , 4.7.3A]. Therefore I has no field, graph, = or graph-field automorphism of order 3, so I1 u embeds in Inndiag(I). Hence, m3 (I) = 2 < m3 (Inndiag(I)). Using [IA , 2.5.12] we conclude that I ∼ = L3 (4); but = L3n (4), U3n (2), E6 (4), or 2 E6 (2). As m3 (I) = 2, I ∼ then m3 (Inndiag(I)) = m3 (P GL3 (4)) = 2, a contradiction. The lemma follows.  Lemma 13.31. Suppose that L ↑3 J ∈ K3 with L ∈ Chev(r), r = 3, and 3A6 and J ∈ Chev(3) ∪ Spor ∪ m3 (L) − m3 (Z(L)) = 1. Assume that L ∼ = Alt − ∪s =3 Chev(s). Assume also that J ∈ A and m3 (Aut(J)) > 2. Then 3 (L, J) ∼ = (A5 , A11 ), (L2 (8), Co3 ), or (L2 (8), 2 G2 (3 2 )). Proof. If J ∈ Chev(3), then L ∈ Chev(3) by [IA , 4.9.6]. Hence by 3 [IA , 2.2.10], and as m3 (L) − m3 (Z(L)) = 1, L ∼ = L2 (8). Then J ∼ = 2 G2 (3 2 ), as desired, by the Borel-Tits theorem and [IA , 4.9.1, 4.9.2]. If J ∈ Spor, then the information in [IA , 5.3] shows that the only possibility is (L, J) ∼ = (L2 (8), Co3 ). Finally if J/Z(J) ∼ = An , then L ∈ Alt so L ∼ = A5 . As J ∈ A, n ≤ 12, so since m3 (Aut(J)) > 2 and n ≡ 5 (mod 3), n = 11. The proof is complete.  Lemma 13.32. Let K ∈ K be simple and let b ∈ I3 (Aut(K)) and L  CK (b) with L ∼ = A6 . ∼ (a) If K = J3 , then L ≤ L∗ ≤ CK (b) with L∗ ∼ = Σ6 . (b) If K ∼ = Σ6 . = Sp6 (2) or Sp4 (8), then L ≤ L∗ ≤ CK (b) with L∗ ∼

17. PROPERTIES OF K-GROUPS

542

Proof. For J3 , m3 (CAut(K) (t)) = 1 for any t ∈ I2 (Aut(K)) (Lemma 10.13.2c). For K = Sp6 (2), CK (b) ∼ = Z3 × Sp4 (2) [IA , 4.8.2], and for K =  Sp4 (8), CK (b) ∼ = Sp4 (2) [IA , 4.9.1]. Lemma 13.33. Let A ≤ K where K ∼ = E32 . Then for = Sp4 (8) and A ∼ # ∼ any x ∈ A , E(CK (x)) = L2 (8). Proof. See [IA , 4.8.1, 4.8.2, 4.8.4].



Lemma 13.34. There is no K ∈ C3 ∩ ↑3 (L2 (8)) such that Z(K) = 1, K/Z(K) ∈ A, and e(K) ≤ 3. Proof. Suppose for a contradiction that K satisfies the stated con1 ditions. If K ∈ Chev(3), then since L2 (8) ∼ = 2 G2 (3 2 ) ↑3 K, we have K ∼ = 2 G (3 32 ); but then Z(K) = 1 by [I , 6.1.4], contradiction. If K ∈ Alt∪Spor− 2 A Chev(3), then since L2 (8) ↑3 K, and as L2 (8) ∈ Alt, K ∼ = Co3 [IA , 5.3], so again Z(K) = 1, contradiction. Therefore K ∈ C3 ∩Chev(2)−Chev(3)−Alt. As Z(K) = 1, the only possibility is K ∼ = SU6 (2); but then e(K) ≥  m2,3 (K) = 4, a final contradiction. Lemma 13.35. Let K ∈ K be of restricted even type, with e(K) = 3 and ∩ Gp = ∅ for some odd prime p. Then K ∼ = U5 (2n ) for some n ≥ 3, n and p divides 2 + 1. Lop (K)

Proof. Since Lop (K) = ∅, mp (K) ≥ 4, and K ∈ Chev(p) by the BorelTits theorem. By Theorem 1.3, K ∈ A, as in Definition 1.1. Using [IA , 4.10.3] we easily check that if K ∈ Chev, then K ∼ = U5 (2n ), n ≥ 3, as desired, or K ∼ = Sp8 (2), D4 (2), or F4 (2), all with p = 3. In the latter case, with [IA , 4.8.2, 4.7.3A], Lo3 (K) ⊆ {Sp6 (2), U4 (2), A6 } ⊆ C3 , contradicting Lo3 (K) ∩ G3 = ∅. Hence, we may assume that K ∈ Alt ∪ Spor. Again as K ∈ A and mp (K) ≥ 4, and using [IA , 5.6.1], we have p = 3 and K ∼ = A12 ,  o Co2 , M c, Suz, Ly, O N , F5 , or F3 . Using [IA , 5.3] we see that L3 (K) ⊆ C3 , again contrary to hypothesis. This completes the proof.  14. Cross-Characteristic “Pumpups” Lemma 14.1. Let X ≥ J y where J  X, y ∈ I2 (X), [J, y] = J ∈ Cp for some odd prime p, and Y := E(CJ (y)) = 1. Suppose further that e(X) ≤ 3, p divides |CX (J)|, and some component of Y is a Co2 -group. Then one of the following holds: (a) p = 3 and Y and J/Z(J) are as follows: ± (1) Y ∼ = L2 (34 ), U4 (2), L± = A6 and J/Z(J) ∼ 3 (9), or L4 (3);   (2) Y ∼ = L3 (9) or L4 (3); = L3 (3),  = ±1, and J/Z(J) ∼ ± ∼ ∼ (3) Y = U4 (2) and J/Z(J) = L4 (3); (4) Y ∼ = G2 (3); = L2 (8) and J/Z(J) ∼ 3 2 ∼ (5) Y = B2 (2 2 ) and J/Z(J) ∼ = P Sp4 (8); (6) Y ∼ = A9 ; = A5 and J/Z(J) ∼ (7) Y ∼ = J3 ; or = L2 (17) and J/Z(J) ∼

14. CROSS-CHARACTERISTIC “PUMPUPS”

(b)

(c)

(d) (e)

543

∼ M11 and J/Z(J) ∼ (8) Y = = M c; p = 5 and Y and J/Z(J) are as follows: (1) Y ∼ = L2 (52 ), HJ, L± = A5 and J/Z(J) ∼ 3 (5), or P Sp4 (5); ∼ (5); (2) Y = A5 × A5 and J/Z(J) ∼ = L± 4 (3) Y ∼ = L2 (7) and J/Z(J) ∼ = HJ; (4) Y ∼ = A10 or HS; = A6 or A8 and J/Z(J) ∼ 2B (2 32 ) and J/Z(J) ∼ Ru; (5) Y ∼ = = 2 (6) Y ∼ = He; = [2 × 2]L3 (4) and J/Z(J) ∼ (7) Y ∼ = 2Sp6 (2) or M12 and J ∼ = Co3 ; (8) Y ∼ = M c; or = M11 and J ∼ ∼ (9) Y ∼ 2HS and J = = F5 ; p = 7 and Y and J/Z(J) are as follows: (1) Y ∼ = L2 (72 ), L± = L2 (7) and J/Z(J) ∼ 3 (7), or P Sp4 (7); (2) Y ∼ = L± = L2 (7) × L2 (7) and J/Z(J) ∼ 4 (7); ∼ ∼ (3) Y = A5 and J/Z(J) = A7 ; or (4) Y ∼ = [Z2 × Z2 ]L3 (4) and J/Z(J) ∼ = He; p = 11 and Y and J/Z(J) are as follows: (1) Y ∼ = A11 ; or = A5 and J/Z(J) ∼ p = 17 and Y and J/Z(J) are as follows: (1) Y ∼ = L2 (172 ), L± = L2 (17) and J/Z(J) ∼ 3 (17), or P Sp4 (17); or ∼ (2) Y = L2 (17) × L2 (17) and J/Z(J) ∼ = L± 4 (17).

Proof. Since m2,p (X) ≤ e(X) = 3 and p divides |CX (J)|, m2,p (J) − mp (Z(J)) ≥ 2. Also, clearly mp (CJ (y)) ≤ 3. Suppose first that J ∈ Spor ∪ Alt. If J ∈ Spor and p = 3, then as e(X) ≤ 3 (see Theorem 1.3), m2,p (J) − mp (Z(J)) ≤ 2 (see [IA , 5.6.2]), and J ∈ C3 (see [V11 , 3.1]), we have J ∼ = M11 , J3 , M c, or O N . Then from [IA , 5.3], since Y ↑2 J/O2 (J) and Y ∈ Co2 , we get conclusions (a7) and (a8) only. If p = 5, similar reasoning shows that J ∼ = HJ, Co2 , Co3 , HS, M c, Ly, Ru, or He, and from [IA , 5.3] we get conclusions (b1), (b3), (b7), (b4), (b8), (b5), or (b9) only. Likewise if p ≥ 7, J ∼ = He, O  N , or J4 , and we get conclusion (c4) only. On the other hand, if J ∈ Alt, then as e(X) ≤ 3 and X ∈ Cp , J ∼ = An , where n = kp ≤ 12, k ≤ 3. Then Y ∼ = Am , m < n, m ≡ n (mod 2), m = 5, 6, or 8. This yields conclusions (a6), (b4), (c3), and (d1) only. Next, if J ∈ Chev(p), then Y has a component Y0 ∈ Chev(p) ∩ Co2 . If p > 3, we must have Y0 ∼ = L2 (p), p = 5, 7, or 17. In that case either J∼ = L2 (p2 ), giving conclusions (b1), (c1), and (e1), or y maps into Aut0 (J). ± In the latter case by [IA , 4.5.1], J/Z(J) ∼ = L± 3 (p) or P Ωn (p), n = 5 or 6, the upper bound on n because m2,p (Ω7 (p)) ≥ 4 (from an Ω6 (p) subgroup for appropriate  = ±1). This yields conclusions (b1,2), (c1,2), and (e1,2) only. Thus we may assume that p = 3, whence from (14A) and [IA , 2.2.10] and

(14A)

544

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1 2 ∼ 2  [V11 , 3.1], Y0 ∼ = L2 (32 ), L± 3 (3), P Sp4 (3), G2 (3 ) , or G2 (3). (Y0 /Z(Y0 ) = ± L4 (3) is not possible as m3 (Y0 ) ≤ e(X) ≤ 3.) We again use [IA , 4.5.1, 4.9.1] to determine the possible J’s. If y does not map into Aut0 (J), then J ∼ = 4 L2 (3 ), L3 (9), P Sp4 (9), G2 (3), or G2 (9), respectively. As m2,3 (P Sp4 (9)) ≥ 4 ≤ m2,3 (G2 (9)), the only possibilities are in conclusions (a1), (a2), and  (a4). If y maps into Aut0 (J), then the structure of O3 (CJ (y)) is given in ± a+1 [IA , 4.5.1]. We see that either J contains one of L ∼ ), = SL± 4 (3), L4 (3 3D (3a ), (P )Sp (3), (P )Sp (3a+1 ) (a > 0) or a covering group of Ω (3) (all 4 6 4 7 of which imply the contradiction m2,3 (X) ≥ 4 > e(G)), or the existence of Y0 forces J/Z(J) to be L± 4 (3) or P Sp4 (3), whence case (a1), (a2), or (a3) occurs. Thus we may assume that

J ∈ Chev − Chev(p) − Alt. As J ∈ Cp , it follows from the definition of Cp [V11 , 3.1] that J ∈ Chev(2). By the Borel-Tits theorem, y induces an outer automorphism on J, given by [IA , 4.9.1, 4.9.2]. The existence of such an automorphism forces p = 3 and J/Z(J) ∼ = U5 (2), U6 (2), D4 (2), F4 (2), or Sp4 (8). All but Sp4 (8) satisfy m2,3 (J) − m3 (Z(J)) ≥ 3, contradiction, and case (a5) results. The proof is complete.  Lemma 14.2. Let p be an odd prime and let Y ∈ C2 ∩ Cp . Then Y is simple, and if e(Y ) ≤ 3 and m2,p (Y ) ≤ 2, one of the following holds: 1 3 2 2  (a) p = 3 and Y ∼ = L2 (8), A6 , L± 3 (3), M11 , U4 (2), D4 (2), F4 (2 ) , P Sp4 (8), Sp6 (2), or J3 ; 5 1 5 (b) p = 5 and Y ∼ = A5 , 2B2 (2 2 ), 2F4 (2 2 ) , 2F4 (2 2 ), HJ, HS, or Ru; (c) p ≥ 7 and Y ∼ = L2 (p); or (d) p = 11 and Y ∼ = J4 . Proof. We refer to the definitions of C2 and Cp [V11 , 3.1]. As Y ∈ C2 ∩ Cp , Y is quasisimple and Z(Y ) = O2 (Y ) = Op (Y ) = 1, so Y is simple. If Y ∼ = An , then n ∈ {5, 6, 8} ∩ {p, 2p, 3p} by the above definitions and [IA , 2.2.10], and the result holds in this case. If Y ∈ Spor, the above definitions are routine to check, with the ranks given in [IA , 5.6.1, 5.6.2]. If Y ∈ Chev(2) ∩ Chev(p) − Alt, see [IA , 2.2.10] and note P Sp4 (3) ∼ = U4 (2). If Y ∈ Chev(p) − Chev(2) or Y ∈ Chev(2) − Chev(p), the above definitions, together with the rank results [IA , 3.3.3, 4.10.3] suffice to complete the proof.  Lemma 14.3. Let Y and p be as in Lemma 14.2. Suppose that mp (Y ) + m2,p (Y ) ≤ 2. 5 Then (p, Y ) ∼ = L2 (p), p ∈ F M, p ≥ 5. = (3, L2 (8)) or (5, 2B2 (2 2 )), or Y ∼

Proof. If Y ∈ Spor, this is easily checked from [IA , 5.6.1, 5.6.2]. If ∼ A6 , then m3 (Y ) = 2 and m2,3 (Y ) = 1. So if Y ∼ Y = = An , then n = p = 5

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∼ L2 (p), as desired. If Y ∈ Chev(p) − Alt but Y ∼ and Y = L2 (p), then = ± ∼ ∼ p = 3 and Y = L2 (8), as desired, since Y = L3 (3) and U4 (2) contain A4 and satisfy m3 (Y ) > 1. Finally, suppose that Y ∈ Chev(2) − Chev(p) − Alt. 5 Then unless p = 5 and Y ∼ = 2B2 (2 2 ), some parabolic subgroup of Y has order divisible by p mand mp (Y ) > 1 by [IA , 4.10.3], contradicting the assumed inequality. The lemma follows.  Lemma 14.4. Let L ∈ C3 , L  Y , O3 (Y ) = 1, and e(Y ) ≤ 3. Let y be an involutory automorphism of L with y 2 = 1, and K a component of CL (y). Suppose further that K ∈ C2 , m3 (K) = 2 and Aut(K) contains a subgroup 1 3 2 2  isomorphic to P GU3 (2). Then K ∼ = L± 3 (3), D4 (2), or F4 (2 ) . Proof. Refer to [V11 , 3.1] for the definition of the set C3 , which consists of most groups in Chev(3), A9 , several sporadic groups and covering groups of them, and a handful of groups in Chev(2). If K ∈ Chev(3)−{L± 3 (3)}, then ± ∼ as K ∈ C2 (see also [V11 , 3.1]), K = L4 (3), G2 (3), P Sp4 (3), or L2 (8), none of which satisfies m3 (K) = 2 [IA , 3.3.3], or A6 , for which | Aut(K)|3 = 32 < |P GU3 (2)|3 . So we may assume that K ∈ Chev(3), whence L ∈ Chev(3). If L ∈ Spor, the condition e(Y ) ≤ 3 rules out L/Z(L) ∼ = Suz, and there are no further examples. If L ∼ = A6 , which has been ruled out = A9 , then again K ∼ already. Finally, suppose L ∈ Chev(2) but K ∼ A6 . If y centralizes L, then = 1 3 2 ∼ K = L and as m3 (K) = 2, K = D4 (2) or F4 (2 2 ) , as desired. Otherwise, by the Borel-Tits theorem, y is a graph, field, or graph-field automorphism of L. The possibilities, from [IA , 4.9.1, 4.9.2], are (L, K) = ((S)U6 (2) or D4 (2), Sp6 (2)), (F4 (2), 2F4 (2 2 ) ), and (Sp4 (8), 2B2 (2 2 )). 1

3

1

Only K = 2F4 (2 2 ) satisfies the condition m3 (K) = 2 [IA , 4.10.3]. The lemma is proved.  Lemma 14.5. Suppose p is an odd prime. Let L ∈ Chev(2) have twisted Lie rank 1 or 2. Suppose L  X, Op (X) = 1, CX (L) = 1, and for some u ∈  Ip (X), u induces a nontrivial field automorphism on L with Op (CL (u)) ∈ Cp . If mp (X) = 3, then (L, p) is one of the following pairs: (L2 (83 ), 3), 52

(L2 (45 ), 5), (2B2 (2 2 ), 5), or (L3 (27 ), 7).  Proof. Let Lu = Op (CL (u)). If Z(L) = 1, then L ∼ = SL3 (q) with ∼ p = 3 dividing q − , or L = SU5 (q) with p = 5 dividing q + 1. In neither case is Lu ∈ Cp , contrary to assumption. So Z(L) = 1, whence mp (CL (u)) = mp (Lu ) = 1. Then Lu ∈ Cp ∩ Lie(2). With the definition of Cp [V11 , 3.1], 5 this implies that (Lu , p) ∼ = (2B2 (2 2 ), 5) or Lu ∈ Chev(p) ∩ Lie(2). In the 2 5 former case L ∼ = 2B2 (2 2 ), as desired, so assume the latter case. Then as mp (Lu ) = 1, we see from [IA , 2.2.10] that (L0 , p) = (L2 (8), 3), (L2 (4), 5), or  (L3 (2), 7), whence the conclusion of the lemma. Lemma 14.6. Suppose L ∼ = A8 , A10 , Sp4 (4), HS, 2HS, L5 (2), or U4 (2). Then Aut(L) does not contain Z5 × A6 .

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Proof. In all cases Out(L) is a 5 -group, so we may assume that there is x ∈ I5 (L) such that CL (x) contains a copy of A6 . In particular, m5 (L) ≥ 2, which rules out L = A8 , L5 (2) and U4 (2). For the other groups, CL (x)(∞) ∼ =  1orA5 , by [IA , 5.2.6, 4.8.2, 5.3m, 5.3m], respectively. 15. Connecting p-Groups Lemma 15.1. Let X be a K-group and p an odd prime with mp (X) ≥ 3. Assume that F ∗ (X) is a direct product of simple C2 -groups of orders divisible by p. Let Y0 = {z ∈ I2 (X) | mp (CX (z)) > 0}, and for each i = 0, 1, 2, . . . , let Yi+1 = Yi ∪ {z ∈ I2 (X) A# ⊆ Yi for some A ∈ E22 (CX (z))}; Then Yn = I2 (X) for some n. Proof. Let the components of K := F ∗ (X) be K1 , K2 , . . . , Km , with mp (K1 ) ≥ mp (K2 ) ≥ · · · ≥ mp (Km ). Note that if z ∈ I2 (X) and Kiz = Ki for some i, then z ∈ Y0 . It therefore suffices to show that any involution z ∈ X0 := ∩m i=1 NX (Ki ) lies in some Yj . Obviously m2 (CK (z)) ≥ m. i Set K = CK (Ki ) and X i = CX (Ki ), i = 1, . . . , m. Then if m ≥ 2, we have I2 (X i ) ⊆ Y0 for each i = 1, . . . , m. Suppose that m ≥ 2. Since I2 (K1 ) ⊆ Y0 , the decomposition K = K1 × K 1 then shows that I2 (K) ⊆ Y1 . Moreover, for any z ∈ I2 (X0 ), m2 (CK (z)) ≥ m ≥ 2, so z ∈ Y2 . Hence the lemma holds in this case, and we can assume m = 1, whence K is simple and X0 = X. Next, suppose that some f ∈ Ip (X) induces a nontrivial field automorphism on K. As p is odd and K ∈ C2 , K ∈ Chev(2). There is then an f -invariant long root subgroup U such that |U | ≥ 23 , CU (f ) = 1, and U # is completely fused in K. Thus, U # ⊆ Y0 . Moreover, U ≤ Z(P ) for some P ∈ Syl2 (K), so as |U | > 2, I2 (K) ⊆ Y1 . Now let z ∈ I2 (X) be arbitrary. We may assume that z induces a graph, field, or graph-field automorphism on K. In the graph case, [z, U ] = 1 by [IA , 4.9.2d], so z ∈ Y1 . In the field and graph-field cases, we can take z to centralize f in its action on K, which forces |CU (z)| ≥ 23 as f has odd order. Thus z ∈ Y2 , as required. So we may assume that no such element f exists. We likewise can rule out the possibility that Ki ∼ = 3D4 (q) for some q = 2a , and some f ∈ I3 (X) induces a graph automorphism on some Ki . Namely, replacing f by a suitable element of Kf , we may assume that CK (f ) ∼ = G2 (q). Let V ∼ = Eq4 be the subgroup of an f -invariant Sylow 2-subgroup T of X0 generated by the highest long and highest short root subgroups of K. Then V  T . Moreover, under conjugation by suitable elements of K (in particular, using the Cartan subgroup), each element of V can be brought into CK (f ) and hence lies in Y0 . As m2 (V ) > 2, m2 (CV (z)) ≥ 2 for all z ∈ I2 (T ), so I2 (T ) ⊆ Y1 , as asserted. Hence, we may assume that no such f ’s exist, and it follows that mp (K) ≥ 3. ± n n (Note that for p = 3, m3 (P GL± 3 (2 )) ≤ 2 and m3 (E6 (2 )) ≥ 3, while for n  n 2 ≡  (mod p) and np ≥ 5, mp (Lnp (2 )) ≥ 3.)

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Now, if ± K ∈ C2 − Chev(2) ⊆ (Chev(2) ∩ Spor) ∪ {L± 3 (3), L4 (3), G2 (3)},

then it is easily verified, using [IA , 4.5.1, 4.9.1, 5.6.1, 5.3], that every element of I2 (Aut(K)) centralizes an element of Ip (K), so that I2 (X) ⊆ Y0 . Thus, we may assume that K ∈ Chev(2). Let T ∈ Syl2 (X) and P = T ∩ K ∈ Syl2 (K), and let V be the high root subgroup of P . Set U = V if |V | > 2, in which case by Lemma 17.5, p divides |CK (U )|, so U # ⊆ Y0 . Hence if |V | > 2, then I2 (P ) ⊆ Y1 . Moreover, graph automophisms centralize V , and m2 (CK (z)) ≥ 2 if z is a field or graph-field automorphism. (Otherwise, K ∼ = L2 (4) or L3 (4), and mp (K) < 3, contradiction.) Hence we are done if |V | > 2. If |V | = 2, on the other hand, then it suffices to find U  P with U ∼ = E22 such that U # ⊆ Y0 . Indeed then I2 (P ) ⊆ Y1 by the Thompson transfer lemma, and it will follow as in the case |V | > 2 that I2 (T ) ⊆ Y2 . By Lemma 17.3, either I2 (P ) ⊆ Y0 or such a subgroup U exists, so the proof is complete.  Lemma 15.2. Let p be an odd prime, X a group, Op (X) = 1, K  X with K ∈ Tp , mp (K) > 1, and Z(K) = 1. Let Q ∈ Sylp (CX (K)), P ∈ Sylp (K), and S ∈ Sylp (X) with P × Q ≤ S. Let A ∈ Ep∗ (S), B = Ep∗ (P Q) and B ∈ B. Set r = min(mp (B) − 1, 4). Suppose V ≤ P with Ep2 ∼ = V  S and U ≤ Q with Ep2 ∼ = U  S, so that r = 3 or 4. Then there exists a chain A = A0 , A1 , . . . , An−1 , An = B of elementary abelian p-groups such that for all 1 ≤ i ≤ n, (a) mp (Ai−1 ∩ Ai ) ≥ r; (b) Ai ∈ Ep∗ (AP Q) ∪ B; and (c) If B ∩ P = V , then Ai ∩ P = V . Proof. Without loss S = AP Q. Our hypotheses on K imply that mp (P ) = 2 and Z(K) = 1. The first steps in the chain are to A1 = V CA (V ) ∈ Ep∗ (S) and A2 = U CA1 (U ) ∈ Ep∗ (S). Since mp (A) ≥ r + 1, (a) and (b) hold for i = 1, 2, as does Ai ∩ P = V . Let E = B ∩ Q ∈ Ep∗ (Q) and E ∗ = U CE (U ) ∈ Ep∗ (Q). We continue the chain thus: A2 , V E ∗ , V E, the two new terms being in B. If B∩P = V , the chain is complete. If B∩P = V , we extend it by adding B = (B ∩ P )E and note that since mp (P ) = 2, B ∩ P ∩ V = 1. The desired conditions are then easily checked.  16. Small Representations Lemma 16.1. Let X/O2 (X) ∼ = U4 (2) and O2 (X) ∼ = E28 with O2 (X) a natural 4-dimensional module for X/O2 (X) over F4 . Let x ∈ I3 (X). Then |CX (x)|2 ≤ 25 .

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∼ SU4 (2). Then CO (X) (x) has even F2 Proof. Note that X/O2 (X) = 2 dimension not equal to 6. According as CO2 (X) (x) has F2 -dimension 0, 2,  or 4, O2 (CX/O2 (X) (xO2 (X))) ∼ = SL2 (2) × SL2 (2), SU3 (2), or SL2 (2), so 0+2 2+3 4+1  |CX (x)|2 = 2 , 2 , or 2 , as required. Lemma 16.2. Let K = U4 (2) and let Vnat be the natural 4-dimensional F4 K-module, considered as F2 K-module. Let R = 21+8 and suppose that K acts faithfully on R. Let A ∈ E33 (K) and assume that CR (A) = Z(R). Let x1 ∈ I3 (K) and suppose that Vnat = [Vnat , x1 ]. Then |CR/Z(R) (x1 )| = 4 2 . Proof. Without loss, x1 ∈ A. Set R = R/Z(R) and N = NK (A). Then N acts irreducibly on A, and since Vnat = [Vnat , x1 ], | x1 N | = 3. Write x1 N = { xi  | 1 ≤ i ≤ 3}, a set permuted doubly transitively by N . Let Ri = CR (xi ), 1 ≤ i ≤ 3, and assume by way of contradiction that |Ri | = 24 for each i. If Ri ∩ Rj = Z(R) for some i = j, then it holds for all i = j by double transitivity, but then R3 = [R, x1 ] = R2 , a contradiction. Likewise if Ri = Rj for some i = j, then R1 = R2 = R3 ≤ CR (A) = Z(R), contradiction. Therefore Ri ∩ Rj ∼ = Q8 for each i = j. Again as CR (A) = 1, Ri = Ri ∩ Rj , Ri ∩ Rk  whenever {i, j, k} = {1, 2, 3}. Thus R123 := R1 , R2 , R3  ∼ = Q8 ∗ Q8 ∗ Q8 is N -invariant, as are Ro = CR (R123 ) o o and A = CA (R ). By irreducibility of N on A, Ao = A and Ro ≤ CR (A) = Z(R), a final contradiction.  Lemma 16.3. Let V be a 4-dimensional vector space over Fp , p ∈ {3, 5}. Let X be a K-subgroup of GL(V ) containing a subgroup D such that V+ := CV (D) has dimension 2, and D has a normal Q8 -subgroup Q, whose center

z is the unique minimal normal subgroup of D. Assume that D  CX (v) for all 0 = v ∈ V+ , but D  X. Then some 2-element of X interchanges V+ and V− := [V, z]. Proof. Obviously z is weakly closed in D with respect to X. Suppose that z has a conjugate z  = z g = z such that [z, z  ] = 1. Let V+ = CV (z  ) and V− = [V, z  ]. As [z, z  ] = 1, V+ = (V+ ∩ V+ ) ⊕ (V+ ∩ V− ). If the first summand V0 is nonzero then by assumption D  CX (V0 ), so D g normalizes D. Then D g normalizes V− . Hence by irreducibility of D g on V− , V− = V− or V+ and accordingly z = z  or −z  . Since z = z  , V± = V∓ and g interchanges these two subspaces, as desired. So we may assume that no such conjugate z  of z exists, whence z ∈ ∗ Z (X) by the Z ∗ -theorem [IG , 15.3]. If z ∈ Z(X), then D  X, contrary to hypothesis, so [z, O2 (X)] = 1. Choose an odd prime r such that [z, Or (X)] = 1 and consider P D ≤ X, where P = Or (X). If r = p, then CV (P ) = CV+ (P ) ⊕ CV− (P ), a nonzero D-invariant subspace. In this case if CV+ (P ) = 0, then by assumption D  DP , whence [P, z] = 1, contradiction. So CV (P ) ≤ C− and as CV (P ) is D-invariant, CV (P ) = V− . But because of the absolute irreducibility, every element of an irreducible D-submodule of

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P is a transvection, therefore normalizes D and then centralizes D, contradiction. So r = p. But as p ≤ 5 and r ∈ {2, p}, all r-subgroups of GL4 (p) are abelian of rank at most 2, and indeed cyclic unless p = 5 and r = 3. As D is nonabelian, and not embeddable in GL2 (3) if p = 5, we have a contradiction, and the proof is complete.  Lemma 16.4. Let N ≤ GL3 (3) have Sylow 3-subgroup x, where x is a  transvection. Then either x  N or O3 (N ) ∼ = SL2 (3). Proof. Since x is a transvection, it cannot act nontrivially on an E22 or Z13 -subgroup of N . In particular N has a normal 13-complement by Burnside’s theorem, so N is solvable and then a {2, 3}-group. As m2 (SL3 (3)) = 2,   O 3 (N ) = [O2 (N ), x] x ∼ = SL2 (3). Lemma 16.5. Suppose that H < SL3 (3), H is irreducible on the natural SL3 (3)-module, and H is maximal with this property. Then H ∼ = F13.3 or Σ4 . 

Proof. This follows easily from [IA , 6.5.3]. 17. Elements of Order 2p

5 Lemma 17.1. Let K ∼ = 2F4 (2 2 ) with = 3D4 (2) or Sp4 (8) with p = 3, or K ∼ p = 5. Then there is a 2-central involution t of K such that mp (CK (t)) > 0.

Proof. In the first two cases, the normalizer of a high long root subgroup is a parabolic subgroup with Levi factor L2 (8). In the third case, one can check from the commutator formula [IA , 2.4.5] that yi (in the notation 1  of that result) is 2-central and NK ( y5 ) contains x1 , x9  ∼ = 2B2 (2 2 ). Lemma 17.2. Let K ∈ C2 . Let p be an odd prime such that mp (K) > 1 but K has no element of order 2p. Then the following conditions hold. (a) If p = 3, then K ∼ = A6 or L3 (2n ),  = +1 with n ≡ ±2 (mod 6) or  = −1 with n ≡ ±1 (mod 6); and (b) p = 5. Proof. Obviously Z(K) has odd order, so K is simple. If K ∼ = L2 (q), q ∈ F M9, then since mp (K) > 1, p = 3 and K ∼ = A6 . If K ∼ = L± 3 (3), ± P Sp4 (3), L4 (3), or G2 (3), then m5 (K) < 2 and K contains SL2 (3). If K ∈ Spor, the conclusions hold by inspection of [IA , 5.3, 5.6.1]. Finally, we may assume K ∈ Chev(2) − {A6 , U4 (2), G2 (2) }. Let E ≤ K with E ∼ = Ep2 .  2 By assumption ΓE,1 (K) = 1, so K is given by [IA , 7.3.3]. As mp (K) > 1, ∼ L (q), q = 2n ≡  (mod p),  = ±1, or 2F4 (2 21 ) with p = 5. In the K = p

latter group, the centralizer of a 2-central involution contains O2 (2B2 (2 2 )) ∼ = Z5 , so assume K ∼ = Lp (q). If p = 5, K contains Z5 × SL3 (q), so p = 3. Finally, if the asserted congruences on n fail, then q ≡  (mod 9), so K  contains Z3 × SL2 (q). This completes the proof. 1

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Lemma 17.3. Let K ∈ Chev(2) be simple with level q(K) = 2 and assume for some odd prime p that mp (K) ≥ 3. Let P ∈ Syl2 (K) and let X = {z ∈ I2 (K) p divides |CK (z)|}. Then there exists U  P such that U ∼ = E22 and U # ⊆ X. Proof. Since mp (K) ≥ 3, K has untwisted rank at least 3, and K ∼ = 3D (q) or 2F (2 n  ∼ ∼ 2 ) . If K = U4 (2) = P Sp4 (3), then p = 3 and this holds 4 4 by the fact that P is not of maximal class, [IG , 10.11], and the fact [IA , 4.5.1] that X = I2 (P ). If K ∼ = Un (2), n ≥ 5, then CK (Z(P )) = QL ∗(n−2) ∼ ∼ and L = GUn−2 (2) acts naturally on V := Q/Z(P ). where Q = Q8 With this identification, involutions of Q map to isotropic vectors in V , whose centralizers in L contain GUn−4 (2). As mp (K) ≥ 3 and p is odd, p divides |GUn−4 (2)|. But Q contains a normal four-subgroup of P by [IG , 10.11], so the lemma holds for the unitary groups. If K ∼ = Sp2n (2), n ≥ 3 (resp. F4 (2)), then we take U = Z(P ) ∼ = E22 , whose centralizer contains Sp2n−4 (2) (resp. Sp4 (2)) and so contains elements of order p. In all other cases, P contains a long root subgroup V0 of height one less than that of the central root subgroup Z(P ). We take U = Z(P )V0  P and observe that since Z(P ) and V0 are conjugate and V0 ≤ Z(P ), every element of U # is conjugate into Z(P ) and so is centralized by an element of order p. This completes the proof.  Lemma 17.4. Let K ∈ Tp with mp (K) > 1. If K/Z(K) has no element of order 2p, then p = 3 and K/Z(K) ∼ = L3 (2n ), n ≡ ±2 (mod 6), or U3 (2n ), n ≡ ±1 (mod 6), or A6 . Proof. Since K ∈ Tp and mp (K) > 1, p ∈ {3, 5}. By Lemma 17.2, we F i22 , so p = 3, and K ∼ may assume that K ∈ C2 . Hence K ∼ = = A7 , which  contains Z3 × A4 , contradiction. The lemma is proved. Lemma 17.5. Let K ∈ Chev(2) be simple and let p be an odd prime such that mp (K) ≥ 3. With standard notation [IA , 2.10], let X be the root group of K for the highest root. Then p divides |CK (X)|. Proof. We assume the discussion of normalizers of long root groups in [IA , Example 3.2.6]. Note also that the assumption mp (K) ≥ 3 implies that  K ∈ Lie(2) and K is not a Suzuki or Ree group. Let N = O2 (NK (X)). Then N ≤ CK (X) and N/O2 (N ) is the commuting product of groups in Lie(2), and it suffices to show that p divides |N/O2 (N )|. The untwisted Dynkin diagram ΔN of N/O2 (N ) is the subdiagram of the extended untwisted Dynkin diagram of K consisting of all roots orthogonal to the high root. Since mp (K) ≥ 3, the diagram Δ of K has at least 3 nodes [IA , 4.10.3], and of these, at most 2 are not orthogonal to the high root. Hence ΔN is not empty, and in particular we are done if p divides q 2 − 1, where q = q(K) is the level of K. So assume that the order m0 of q modulo p is at least 3. If K has exceptional type, then (Δ, ΔN , m0 ) = (E6 , A5 , 3), (E7 , D6 , 3), (E8 , E7 , 3), or (E8 , E7 , 4) (see [III11 , Table 13.1]). As |N/O2 (N )| is divisible

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by q 6 − 1 in the first three cases and q 8 − 1 in the fourth, p divides |N |, as desired. Finally, assume that K is a classical group. Then (K, N/O2 (N )) = (Ln (q), SLn−2 (q)), (Sp2n (q), Sp2n−2 (q)), or (Ωn (q), Ωn−4 (q)). Since m0 > 2, the dimension of an orthogonally indecomposable representation of Zp of the appropriate type (linear, unitary, etc.) over Fq (or Fq2 in the unitary case) must be at least 2, so mp (N ) ≥ mp (K) − 42 > 0. See [IA , 4.8.1, 4.8.2]. The proof is complete.  Lemma 17.6. Let K = U5 (2), U6 (2), Sp6 (2), Sp8 (2), D4 (2), or F4 (2). Let t ∈ I2 (Aut(K)). Then t centralizes an element of I3 (K). Proof. If t ∈ Inn(K) and K = D4 (2), then by [IA , 4.9.1, 4.9.2], K = 1 U5 (2), U6 (2), or F4 (2) and CK (t) involves Σ6 , Σ6 , or 2F4 (2 2 ), so the result holds in these cases. So we may assume that t ∈ Inn(K) or t ∈ O8+ (2) − Ω+ 8 (2). We claim first that (17A)

t normalizes a subgroup y of K of order 3.

Suppose false and let K be a minimal counterexample. By the BaerSuzuki theorem, t inverts some x ∈ Ip (K) for some odd prime p > 3. If K = F4 (2) and p = 13, then it is easy to check that x lies in no proper parabolic subgroup of K, so CK (x) has odd order by the Borel-Tits theorem. Also from the order formula, x ∈ Syl13 (K). But NK ( x)−CK (x) contains 1 an element u of order 3 since L3 (3) embeds in 2F4 (2 2 ) , which embeds in K [IA , 7.3.3]. Hence t centralizes some element of I3 (NG ( x)), contrary to assumption. Likewise if K = F4 (2) and p = 17, then again by the Borel-Tits theorem, CK (x) has odd order, so K has a unique conjugacy class of D2.17 subgroups. But K has a subsystem subgroup Sp8 (2) ∼ = L, and by Burnside’s transfer theorem, L contains a copy of D2.17 . Hence the minimality of K is contradicted. Suppose that p = 11 and K = U6 (2). Let VU6 be the natural SU6 (2)module. Then for a preimage ξ ∈ SU (VU6 ) of x, a preimage of t normalizes [VU6 , ξ], so we are reduced to the case K = U5 (2), contradicting minimality again. Similarly if K = U5 (2), let VU5 be the natural K-module. Then since dim(CVU5 (t)) ≥ 3, t centralizes a nonisotropic 1-space W . Therefore, t acts on CK (W ) ∼ = U4 (2) ∼ = P Sp4 (3). But then by [IA , 4.5.1], t centralizes an element of I3 (CK (W )). Hence we need no longer consider K = U5 (2). Next suppose that p = 7. If K = F4 (2), then by [IA , 4.8.6c], t acts on E(CK (x)) ∼ = L3 (2) and hence normalizes a subgroup of E(CK (x)) of order 3, contradiction. If K = Sp2n (2), then on the natural module VSp2n , dim[VSp2n , x] = 6 so by minimality n = 3. Then x ∈ Syl7 (K), CK (x) has odd order by examination of the parabolic subgroups of K, and 3 divides | AutK ( x)| since K contains L3 (2). So t normalizes a subgroup of NK ( x) of order 3, contradiction. Similar arguments show that K = D4 (2) and K = U6 (2) are impossible. So p = 7.

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Suppose then that p = 5. If K = F4 (2), then by [IA , 4.8.6b], t acts on  L = O 2 (CK (x)) ∼ = B2 (2) ∼ = Σ6 , inducing an inner-diagonal automorphism by [IA , 4.2.2]. So t normalizes a subgroup of L of order 3, contradiction.  If K = U6 (2), Sp6 (2), Sp8 (2), or D4 (2), then by [IA , 4.8.2], O2 (CK (x)) ∼ = L2 (2), Sp4 (2), or A5 (not respectively), with t inducing an inner-diagonal automorphism, and we reach the usual contradiction. The only remaining case is K = Sp8 (2) and p = 17. Let V be the natural F2 K-module. Then CV (x) = 0 so t is free on V . If (v, v t ) = 1 for some v ∈ V , then t normalizes the nondegenerate subspace W = F2 v + F2 v t . In that case t normalizes CK (W ⊥ ) ∼ = Σ3 and hence normalizes a subgroup of order 3. So we may assume that (v, v t ) = 0 for all v ∈ V . Thus (v, [v, t]) = 0. But

v + X, w + X := (v, [w, t]) defines a nondegenerate form on V /X, which therefore is alternating. Let {v + X, w + X} be a hyperbolic pair with respect to , . Then the subspace Y of V generated by v, [v, t], w, [w, t] is t-invariant, 4-dimensional, and nondegenerate. Thus t ∈ NK (Y ) ∼ = Sp(Y ) × Sp(Y ⊥ ) ∼ = Σ6 × Σ6 , and again t normalizes a subgroup of order 3. Thus (17A) is established. Now suppose that K is a minimal counterexample to the lemma, and y is as in (17A). Then t inverts y. Also, CO3 (CK (y)) (t) = 1 so O3 (CK (y)) is abelian. In particular, if K = F4 (2), then from [IA , 4.7.3A], E(CK (y)) ∼ = Sp6 (2) and we are done by minimality. In the remaining cases let V be the natural K-module. Assume for the moment that K = U6 (2). If [V, y] = V ,  then O 2 (CK (y)) ∼ = SU3 (2) or U4 (2) ∼ = P Sp4 (3). In the first case O3 (CK (y)) is nonabelian, and in the second, as noted above, t centralizes an element of I3 (CK (y)). So [V, y] < V . Now t acts on C := CK ([V, y]); if it normalizes a subgroup y   of C of order 3, then it inverts y  , so it inverts yy  . But then as [V, yy  ] = V we reach a contradiction again. So t normalizes no subgroup of C of order 3. But C ∼ = B2 (2) with t inducing an inner-diagonal = Σ3 , Σ6 ∼ + ∼ automorphism, Sp6 (2), Ω4 (2), or Ω− 6 (2) = U4 (2), and in every case by minimality and inspection, t does centralize a subgroup of C of order 3. We are finally reduced to K = U6 (2). Let y be a preimage of y in SU6 (2) = SU (V ). Since O3 (CK (y)) is abelian, y has no eigenspace on V of dimension 3. Hence dim CV ( y ) = 2 or 4. In any case t normalizes a U4 (2)-subgroup of K, and so centralizes an element of order 3 by [IA , 4.5.1]. The proof is complete.  18. p-Ranks Lemma 18.1. Let K = L2 (q), q ∈ F M9, or L3 (4). Let X = Aut(K), and let z ∈ I2 (X) − Inn(K) be a diagonal or graph-field automorphism, respectively. Then r2 (CX (z)) ≤ 3, where r2 denotes sectional 2-rank. Proof. Let S ∈ Syl2 (CK (z)) and S z ≤ T ∈ Syl2 (CX (z)). Then |T : S z | ≤ 2 in both cases. In the first case, |S| = 2 so |T | ≤ 8. In the second case S ∼ = Q8 , and for a graph automorphism g ∈ T , |CS (g)| ≤

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|CK ( g, z)|2 = 2, so S g ∼ = SD16 . Hence T has a normal cyclic subgroup of index 4, so the result follows.  Lemma 18.2. Let K = M12 or HJ. Then r2 (Aut(K)) < 6. Proof. As | Out(K)| = 2 it suffices to show that r2 (K) < 5. Suppose for a contradiction that T  S ≤ K with S/T ∼ = E25 and |T | = 2a . By [IA , 5.6.1], a > 0. If a = 1 then |CG (T )|2 ≥ |S| = 26 so by [IA , 5.3bg], T = Z(P ) for some S ≤ P ∈ Syl2 (K). But from the structure of CG (T ) given in [IA , 5.3], m2 (P/T ) = 4, a contradiction. Therefore a > 1, |K|2 ≥ 27 , K∼ = HJ, and S ∈ Syl2 (K). By [IA , 9.16], |CS (z)| ≥ 25 for every z ∈ I2 (S), contradicting [IA , 5.3g]. The proof is complete.  Lemma 18.3. mI3 (U6 (2)) ≥ 3. Proof. U6 (2) has a parabolic subgroup M with O2 (M ) ∼ = 21+8 and ∼  M/O2 (M ) = U4 (2). Lemma 18.4. Let K = 3U4 (3) or Z3 × U4 (3). Let R ≤ K with R ∼ = E32 . # Then either m3 (CK (x)) = 5 for some x ∈ R , or m3 (CK (R)) = 4. Proof. Let R ≤ P ∈ Syl3 (K). Then with [IA , 6.4.4], A := J(P ) ∼ = E35 . We may assume that R ∩ A = 1, and show that m3 (CK (R)) = 4. Choose x ∈ R# . Then x has the same action on A as an element x of a Levi factor L of the parabolic subgroup NK (A) = AL, L ∼ = A6 . In particular x ∈ H ∼ = A4  for some H ≤ L, so x and x each have a free submodule on A/Z(K). Hence x has a free submodule A0 on A with A0 ∩ Z(K) = 1. Clearly [A0 , x, x] = [A, x, x] as |A| = 35 . Let A1 = [A, x, x]. Then B := RA1 Z(K)  is elementary abelian and since R ∩ A = 1, m3 (B) = 4, as required. Lemma 18.5. Suppose K ∈ K3 is quasisimple with Z(K) = 1. If K/Z(K) ∼ = G2 (3), U4 (3), Ω7 (3), J3 , M c, or Suz, then m3 (K) ≥ 4. Proof. For 3G2 (3), see [V8 , 1.5]. For 3U4 (3) and 3Ω7 (3), see [IA ,  6.4.4a]. And for the sporadic groups, see [IA , 5.6.1]. Lemma 18.6. Let K ∈ K3 with K/Z(K) ∼ = U4 (2), L± 4 (3), or G2 (3), but ∼ K = 3G2 (3). Then m2,3 (K) ≥ 2 + m3 (Z(K)). Proof. Let z ∈ I2 (K) be 2-central. Then C := CK/Z(K) (z) contains SL2 (3)∗SL2 (3), and by Lemma 10.10.9, a Sylow 3-subgroup of C splits over Z(K) (the only case with Z(K) = 1 is possibly K/Z(K) ∼ = U4 (3)). This implies the lemma.  19. p-Components Lemma 19.1. Let p be an odd prime and X a K-group with Op (X) = 1. Let K be a component of X. Then the following conditions hold: (a) mp (NX (K)) = mp (X); and (b) If K is simple, then every element of Ep∗ (X) normalizes K.

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Proof. Let A ∈ Ep∗ (K). IfA ≤ NX (K), then (a) holds. If not, then  A by [IG , 8.7(iii)], p = 3 and K = K1 K2 K3 where the Ki are components permuted transitively by some a ∈ A# , and K = K1 ∼ = 3A6 , 3A7 , 3M22 , or SL3 (q),  = ±1, q ≡ 4 or 7 (mod 9). In every case m3 (Aut(K)) = 2 (see [IA , 5.2.1, 5.3c, 2.5.12]; the congruences on q prevent  the existence of field automorphisms of order 3). Hence m3 (AutA ( K A )) ≤ 1 + m3 (Aut(K)) = 3. On the other hand, there exist elements  xi ∈ Ki − Z(Ki ) of order 3, i = 1, 2,3, by [IG , 16.11]. SetB = CA ( K A ) x1, x2 , x3 . Then m3 (B) = m3 (CA ( K A )) + 3 ≥ m3 (CA ( K A )) + m3 (AutA ( K A )) ≥ m3 (A), so B ∈  E3∗ (X). As B normalizes K, (a) follows. For (b), see [IG , 8.8]. 20. Miscellaneous Lemma 20.1. Let R = (Rd × RK ) c be a p-group with Rd abelian of p-rank ≤ 2 and c ∈ Ip (R). Suppose that p is odd, and either (a) RK is homocyclic abelian of p-rank 2 and [RK , c] = 1 (RK ); or (b) p = 3, RK = AK w with AK ∼ = Z3 × Z9 and w ∈ I3 (CRK (c)) and [RK , c] = 1 (RK ). Then R has no quotient isomorphic to Zp Zp . Proof. Suppose on the contrary that S  R with R := R/S ∼ = Zp Zp . If RK is abelian, let RK = AK . In either case, let A= Rd × AK and   note that A  R. Suppose that h ∈ A of order p2 . Then h = CR (h) = A  R, which is not the case. Hence A has exponent p. In particular, [RK , c] = 1 and R is the product of the subgroups X := Rd c and Y := RK , which commute elementwise. Passing to R/Z(R) ∼ = p1+2 , we see that either X or Y must cover R/Z(R). Hence either RK or Rd c covers R, since Z(R) ≤ Φ(R). In all cases, it follows that R has a maximal subgroup which is abelian of rank  at most 2, which is impossible as R ∼ = Zp Zp . The proof is complete. Lemma 20.2. Let R = Q1 ∗ Q2 ∗ D with Q1 ∼ = D8 , = Q2 ∼ = Q8 and D ∼ so that R is extraspecial. Suppose that H is a K-group with R = F ∗ (H). Suppose that t ∈ D is an involution with RA/R = F ∗ (CH/R (tZ(R))), where A ≤ H, A ∼ = E32 , [R, A] = Q1 Q2 , and |CH/R (tZ(R)) : RA/R| = 2 or 4. Suppose also that there is w ∈ H interchanging Q1 and Q2 . Then t ∈ [H, H]. Proof. Let H = H/R. As R = F ∗ (H), O2 (H) = 1. If O3 (H) = 1, then since w ∈ H interchanges Q1 and Q2 , we must have [R, O2,3 (H)] = Q1 Q2 , so D  H (as D/Z(R) = CR/Z(R) (O2,3 (H))). Let Z be the unique cyclic subgroup of order 4 in D. Then Z  H. As t ∈ CH (Z), t ∈ [H, H], as claimed. Hence, we may assume that O3 (H) = 1. It follows easily that F (H) = 1 and then J := F ∗ (H) is simple. If J is a {2, 3, p}-group for some prime p, then by [III17 , 9.9], J ∼ = A6 . As A6 has no subgroup of index 8, A8 has a unique class of subgroups isomorphic to A6 , and so H is contained in the stabilizer of a nonsingular line of V = R/Z(R). But then we again have Z  H and t ∈ [H, H], as claimed.

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It follows that |J| is divisible by 35. Let V = R/Z(R) and v = tZ(R) ∈ V . Then v is an isotropic vector in V and so |v H | ≤ 35. As |CH (t)| = 2a · 32 with a ≤ 2, we must have |v H | = 35 and then J = H with |H| = 22 · 32 · 5 · 7, but this violates Sylow’s theorem for p = 7, a final contradiction.  Lemma 20.3. Let J ∈ Co2 , and suppose that u ∈ I2 (Aut(J)) and K is a component of CJ (u). Assume that m3 (J) ≤ 2, B ≤ CAut(J) (u) with B ∼ = E32 , and the image of KB in Aut(J) is isomorphic to A6 , M11 or Aut(L2 (8)) ∼ = P ΓL2 (8). Then for some b ∈ B # , CJ (b) contains A5 and in particular is not solvable. Proof. First suppose that J ∈ Chev(2). By [IA , 4.9.6], K ∈ Chev(2), so K ∼ M11 . Since K ≤ E(CJ (u)), u is a field, graph-field or graph auto= morphism of J by [IA , 3.1.4]. Indeed by [IA , 4.9.1, 4.9.2], either J ∼ = Lm (2), m = 4 or 5,  = ±1, with K ∼ = A6 and u a graph automorphism, or J ∼ = Sp4 (4) or J ∼ = L2 (82 ), with u a field automorphism. In the last case, since KB ∼ = P ΓL2 (8), we may take as b some element of B # which induces a field automorphism on J. If J ∼ = Sp4 (4), then any b ∈ B # satisfies the desired property by [IA , 4.8.2]. If J ∼ = Lm (2), then  = +1 by the hypothesis m3 (J) ≤ 2. Then any b ∈ B with a four-dimensional commutator space on the natural J-module has the property that we want. We may therefore assume that J ∈ Co2 − Chev(2), and we have K ↑2 J. If J ∈ Chev(3) − Chev(2), or J ∼ = L2 (q), q ∈ F M9, then as m3 (J) ≤ 2, J ∼ = L3 (3) [IA , 2.2.10, 3.3.3]. But then K ∈ ↓2 (J) = ∅ [IA , 4.5.1], a contradiction. By definition of Co2 [V11 , 3.1], therefore, J ∈ Spor ∩ C2 . As m3 (J) ≤ 2, J/Z(J) is a Mathieu group, HJ, J4 , HS, or Ru [IA , 5.6.1]. By inspection of [IA , 5.3], the only possibility is HS; but then CJ (b) contains  A5 for every b ∈ I3 (Aut(J)), as claimed. The proof is complete. Lemma 20.4. Let K ∈ Chev(2) with m3 (K) = 2, and let B ≤ Inndiag(K) with B ∼ = E32 . Suppose that CK (b) has cyclic Sylow 2-subgroups for each 1 # b ∈ B . Then K ∼ = A6 , U3 (3), 2F4 (2 2 ) , or L3 (2n ),  = ±1, 2n ≡  (mod 3). 1

Proof. First of all, notice that A6 , U3 (3), 2F4 (2 2 ) and L3 (2n ),  = ±1, n 2 ≡  (mod 3) all satisfy the hypotheses of the lemma. Now, take K to be a group satisfying the hypotheses of this lemma, but 1 not isomorphic to A6 ∼ = G2 (2), or 2F4 (2 2 ) . We may suppose = B2 (2), U3 (3) ∼  that K = d L(2m ). By [IA , 4.2.2, 4.7.3A, 4.9.1], for each b ∈ B # , O2 (CK (b)) is the central product of groups of the form db Lb (2mb ), with m dividing mb . Since Sylow 2-subgroups of CK (b) are cyclic, there is at most one factor in 1 1  this central product, and O2 (CK (b)) ∼ = A1 (2), 2B2 (2 2 ) or 1. In the 2B2 (2 2 ) case, we would have m = 12 , so this case cannot occur by [IA , 4.7.3A]. Hence m = 1 or CK (b) has odd order. Let K be the algebraic group overlying K; 3 then b acts on K as conjugation by some b ∈ K with b ∈ Z(K), and the connected component CK (b)o = T L where T is a maximal torus and L is either trivial or isomorphic to A1 . By [IA , 4.7.1, 4.8.2], we have L = A1 ,

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A2 , or B2 . If m = 1, then the only simple choice for K is A2 (2), which is impossible as m3 (K) = 2. Thus m > 1, and L = A1 or A2 . Since m3 (K) = 2, K ∼  = L3 (2n ), 2n ≡  (mod 3). The proof is complete. Lemma 20.5. Let J ∼ = L2 (5m ) for some m ≥ 2. Let u ∈ I2 (Aut(J)) inducing an inner or diagonal automorphism of J. Then O2 (CJ (u)) = 1. Proof. CJ (u) contains a normal cyclic subgroup of order 5 2−1 or 5 2+1 . As 5m − 1 is divisible by 4, we see that 5m + 1 = 2km with km odd and m km > 1. Hence the claim holds if the subgroup has order 5 2+1 . So assume m the subgroup has order 5 2−1 . If m is even, then 5m − 1 is divisible by 52 − 1 = 24, and again the lemma holds. Finally, if m is odd and m ≥ 3, then 5m − 1 = (5 − 1)(5m−1 + · · · + 5 + 1) = 4sm with sm odd and sm > 1. Hence the lemma holds in all cases.  m

m

Lemma 20.6. Let H be a K-group with O2 (H) = 1, m2 (H) ≤ 3, and m5 (H) ≥ 2. Let Q ∈ Syl2 (H) contain an involution u such that CH (u) is a {2, 3}-group. Assume that one of the following holds: (a) Q is dihedral or semidihedral; (b) Q is abelian; or (c) u is the unique involution in [Q, Q] ∩ Z(Q). Then F ∗ (H) ∼ = L2 (5m ) for some m ≥ 2 and u induces an inner or diagonal automorphism on F ∗ (H). Moreover, O2 (CH (u)) = 1. Proof. Let B < H with B ∼ = E52 . If Q is dihedral or semidihedral, then as [B, O2 (H)] = 1, B acts faithfully on E(H). Then, by [IA , 4.10.5], we see that either F ∗ (H) ∼ = L2 (5m ), m ≥ 2, and every involution acts as an  inner or diagonal automorphism on F ∗ (H), as claimed, or O2 (H) ∼ = L3 (q) ∼ or U3 (q). In the latter case, u ∈ E(H) and E(CH (u)) = SL2 (q), whence q = 3, since CH (u) is a {2, 3, }-group. But this contradicts m5 (H) = 2. If F ∗ (H) := J ∼ = L2 (5m ) with m ≥ 2, then O2 (CH (u)) = 1 by Lemma 20.5, as claimed. Hence we may assume that either Q is abelian or u is the unique involution in [Q, Q] ∩ Z(Q). Also, we may assume that J := E(H) ∼ L2 (q), = L3 (q), or U3 (q) for any odd q. Suppose first that F ∗ (H) = O2 (H) := R. Then as u ∈ Z(Q), u ∈ V := Ω1 (Z(O2 (H))) and as CB (u) = 1, B acts faithfully on V , whence m2 (V ) > 3 = m2 (H), a contradiction. Hence E(H) = 1. Let J be a component of H. If J ∈ Chev(q), q odd, then using [IA , 4.5.1, 4.9.1], we reduce to the case J ∼ = G2 (3). If J ∈ Alt − Chev(q), then J ∼ = A7 . If ∼ J ∈ Spor, then J = M11 , M12 , J1 , 2HJ or O N . As CJ (u) is a {2, 3}group and B acts faithfully on J, these cases are all ruled out. Finally, if J ∈ Chev(2), then using [IA , 3.3.3], we reduce to the cases J ∼ = L2 (8), U3 (4), or Sz(8), which are likewise ruled out by the faithful action of B. Then Lemma 20.5 completes the proof.  Lemma 20.7. Suppose p is an odd prime, and X ≤ K = GL2p (p). Suppose that b ∈ Ip (X) and CX (b) has a subgroup H of index at most 2

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such that H = ( b × Y ) a HZ , where HZ is a group of scalar mappings, Y ∼ = Zm0 , m0 = (pp − 1)/(p − 1), a ∈ Ip (X), and Y a is a Frobenius group with Frobenius kernel Y . Suppose also that X is involved in a group G with e(G) ≤ 3. Then [E(X), Y ] = 1. Proof. Suppose false and let J be a component of E(X) not centralized by Y . For any y0 ∈ Y # , CK (y0 ) = CK (Y ) (cf. [V9 , (13A)]). Hence CY (J) = 1. Since CX (b) is solvable, b acts nontrivially on every component of E(X). If J is a p -group, then mp (Aut(J)) ≤ 1 by [IA , 7.1.2], so a cannot normalize J. But then b induces a field automorphism on each component of J a , so CJ a (b) has sectional 2-rank at least p. However, HZ is cyclic, so CX (b) has sectional 2-rank at most 2, contradiction. Thus p divides |J|. As b = Op (CX (b)), Z(J) is a p -group. Then since |CX (b)|p = p2 and |Y | = m0 is odd, Y normalizes J. We conclude that

b × Y a acts faithfully on J. Now a, b ∈ Sylp (CX (b)). Suppose that a, b induces inner automorphisms on J, or even inner-diagonal automorphisms should J ∈ Chev. Then the image of Y in Aut(J) is of odd p order ≥ 13 and is invariant under CInn(J) (b). As Y = [a, Y ], J is not weakly locally balanced with respect to a, b, and Y induces inner or inner-diagonal automorphisms on J. By  where J is the [IA , 7.7.6], J ∈ Chev(r) for some r = p, and p divides |Z(J)|, universal version of J. As J/Z(J) ∈ A, ∼ U5 (2n ). ∼ L (rn ), or p = 5 with J/Z(J) = p = 3 with J/Z(J) = 3

Here  = ±1, ≡  (mod 3), and n ≡ 2 (mod 4). Since p does not divide |Y |, Y maps into CInn(J) (b). Hence for every prime divisor s of |Y |, Os (CJ (b)) has s-rank exactly 1. But from [IA , 4.8.2, 4.8.4] we see that this implies that CJ/Z(J) (b) either has a component or is a Frobenius group with complement of order p and kernel cyclic of order r2n + rn + 1 or (25n + 1)/(2n + 1), according as p = 3 or 5. As CX (b) is solvable, the Frobenius group alternative holds. Thus, p = 3 and r2n + rn + 1 = (33 − 1)/3 − 1 = 13, or p = 5 and (25n + 1)/(2n + 1) = (55 − 1)/(5 − 1) = 781, which equations are impossible (recall that rn ≡  (mod 3)). In particular, J ∈ Chev. If either x = a or x = b induces a graph or graph-field automorphism on J, then p = 3 and as J ∈ A, the only possibilities are J ∼ = D4 (2) or 3D4 (2n ). As a, b ∈ Syl3 (CX (b)), a, b ∩ J = 1. But m3 (CJ (x)) > 1 [IA , 4.7.3A], so m3 (CX (b)) > 2, contradiction. Therefore x = a or x = b induces a field automorphism on J. If b ∈ J, then b ∈ Sylp (CJ (b)) so b ∈ Sylp (J). But J admits a field automorphism of order p, so from the order formula for J, |J|p > p, contradiction. Hence b induces a field automorphism on J. But this is clearly impossible given the  structure of CX (b). This contradiction completes the proof. rn

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Lemma 20.8. Let K ∈ Chev(r), r = 3, and let x ∈ I3 (Aut(K)) − Inndiag(K). Then CK (x) does not contain a Sylow 3-subgroup of K, unless n K∼ = 2B2 (2 2 ) for some n. Proof. See [IA , 4.9.1, 4.9.2]. If x is a graph automorphism, then K ∼ = D4 (q) or 3D4 (q), and in either case CK (x) ∼ = G2 (q) or P GL± 3 (q). Comparison of the order formulae shows that 3 divides |K : CK (x)|. A similar comparison handles the graph-field automorphism, since 3 divides |D4 (q 3 )|/|3D4 (q)|. So we may assume that x is a field automorphism. We also may assume that K is the adjoint version. Suppose first that K is not of Suzuki or Ree type. Say K = d L(q 3 ). Let L = d L(q), the adjoint version. Then L ≤ CK (x) ≤ Inndiag(L). Write the order formula for |K| as  1 (q 3 )N Φi (q 3 )ni , |K| = | Outdiag(K)| i

where all ni ≥ 0, almost all ni = 0, and the Φ are the cyclotomic polynomials (cf. [IA , (4.10.1)]). Similarly,  Φi (q)ni , | Inndiag(L)| = q N i

Let i0 = 1 or 2 according as q ≡ 1 or −1 (mod 3). Then ni0 > 0, and the cyclotomic values Φi (q 3 ) and Φi (q) divisible by 3 are those for which i/i0 ∈ 3Z. Indeed if |i/i0 | = 3ai bi with (bi , 3) = 1 and ai > 0, then the 3-shares of Φi (q 3 ) and Φi (q) are both 3ai . Moreover, Φi0 (q 3 ) = 3Φi0 (q). Thus |K|3 |K|3 3ni0 is an integer multiple of = . |CK (x)|3 | Inndiag(L)|3 | Outdiag(K)|3 Unless L = A3n−1 or E6 , | Outdiag(K)|3 = 1. Then since ni0 > 0, |K|3 > |L|3 , as desired. If L = E6 , then ni0 ≥ 4 and | Outdiag(K)|3 ≤ 3, so again |K|3 > |L|3 . If L = Am−1 with m = 3a b, (3, b) = 1, a ≥ 1, then a | Outdiag(K)|3 = (3a , Φi0 (q 3 )), while ni0 = m − 1. But 3m−1 ≥ 33 −1 > 3a , so again we are done. A similar calculation in the Ree group cases finishes the proof.  Lemma 20.9. Let K ∈ K. Unless K ∼ = D4 (q), q odd, Out(K) is 2nilpotent. Proof. See [III11 , 2.1].



Lemma 20.10. Suppose K is a component of X, and K ∼ = 3A6 , 3A7 ,  3M22 , or SL3 (q), q ≡  (mod 3),  = ±1. Let P ∈ Syl3 (X). Then [K, Ω1 (Z(P ))] = 1. 3O  N ,

Proof. See [V8 , 9.5].



20. MISCELLANEOUS

559

Lemma 20.11. Let X be a K-group with O3 (X) = 1 = O3 (X). Let K  X with K ∈ C3 . Assume the following: (a) m3 (X) ≥ 5; (b) m2 (K) ≥ 4 and X has an A4 ×A4 subgroup A such that CA (K) = 1; (c) m2,3 (X) ≤ 3; and (d) For any D ∈ E33 (X) such that CK (D) has even order, L3 (CK (D0 )) = 1 for every D0 ∈ E32 (D). Then K ∼ = M c, 3M c, L± 4 (3), or 3U4 (3). Proof. Suppose first that m3 (CX (K)) > 1, and let D0 ∈ E32 (CX (K)). Then (d) implies that there is no x ∈ I3 (K) such that CK (x) has even order. By Lemma 7.17, K ∼ = L2 (8) or L2 (3n ) for some n ≥ 2. But then m2 (K) < 4, contradicting (b). We conclude that m3 (CX (K)) = 1, since O3 (X) = 1 by assumption. Thus by (a), (20A)

m3 (Aut(K)) ≥ 4.

∼ An , then n ≥ 12 by (20A), so K ∈ C3 [V11 , 3.1], contraIf K/Z(K) = diction. If K ∈ Spor, then using [IA , 5.6.1, 5.6.2], (20A), and (c), we find that K ∼ = O N , then L3 (CK (D)) = 1 for = M c, 3M c, or O  N . But if K ∼ 3 some D ∈ E3 (X), contradicting (d). Thus the lemma holds in these cases. If K ∈ Chev(2) − Chev(3), then as K ∈ C3 and m2,3 (KO3 (X)) ≤ 3 with 1 O3 (X) = 1, we have K ∼ = Sp6 (2), 3D4 (2), 2F4 (2 2 ) , or Sp4 (8). But then m3 (Aut(K)) ≤ 3, contradicting (20A). Hence we may assume that K ∈ Chev(3). First suppose that K is of n exceptional Lie type. Since m2 (G2 (3n )) = 2 and m2 (2 G2 (3 2 ) ) = 3, and 3D (3n ) contains SL (3n ) ∗ SL (33n ), K is not one of these three groups. 4 2 2 + n n n n n The inclusions Spin± 4 (3 ) ∗ Spin4 (3 ) ≤ Spin9 (3 ) ≤ F4 (3 ) ≤ Em (3 ) n show that the other exceptional types violate (c), as does Dm (3 ) for m ≥ n 4. Likewise for m ≥ 3, Bm (3n ) contains Ω± 6 (3 ), at least one of which n violates (c). Bm (3 ), m ≥ 2, violates (c) via SL2 (3n ) ∗ Bm−1 (3n ) unless m = 2 and n = 1, in which case m3 (Aut(P Sp4 (3))) = 3, violating (20A). n Finally, consider K = A± m (3 ). If m ≤ 2, then m2 (K) = 2, violating (b). n If m ≥ 4, then K contains SL± 4 (3 ), violating (c). If m = 3 and n > 1, n n then K contains SL2 (3 ) ∗ SL2 (3 ), violating (c). Thus we are left with  K/Z(K) ∼ = L± 4 (3), and as m3 (CX (K)) = 1, the lemma follows. Lemma 20.12. Let F ∗ (X) = K ∈ K be simple with |X|2 = |K|2 = 26 or 29 , and with R ∈ Syl3 (X) satisfying R ∼ = E32 . Suppose that X has a 2-local subgroup M of odd index such that M ≤ K, F ∗ (M ) = O2 (M ), M/O2 (M ) ∼ = L3 (2), and every chief factor of M within O2 (M ) has order 3 2 . Assume also that X acts faithfully on a 2-group S, and |SX|2 = 217 . Then X ∼ = L4 (2). Proof. We have |S| = 28 or 211 . If K ∈ Chev(2), then M is a parabolic subgroup of K, and the restriction on |K|2 yields only X ∼ = L4 (2) or X ∼ = 2 Sp6 (2). Only the former is possible as |X|3 = 3 .

560

17. PROPERTIES OF K-GROUPS

Note that all r-subgroups of X are abelian for primes r > 3 by the faithful action on S (and for r = 3 by assumption). Hence if K ∈ Chev(r), then K ∼ = L2 (rn ) for some n. As K has odd index in X, X has dihedral Sylow 2-subgroups, hence sectional 2-rank 2, which is impossible by the assumed chief factor structure. If K ∈ Alt − Chev, then K ∼ = A7 by the 3-structure, yielding the same contradiction. Finally if K ∈ Spor, then by the 3-structure, K ∼ = M11 or HS. But HS has a nonabelian 5-subgroup, and M11 has sectional 2-rank 2, so they are both impossible. The proof is complete.  Lemma 20.13. Suppose X is a K-group, O2 (X) = 1, z ∈ I2 (X), and CX (z) = F ∗ (CX (z))Az , with F ∗ (CX (z)) = O2 (CX (z)) ∼ and Az ∼ = 21+4 = A5 − ∼ or SL2 (5). Then either z ∈ Z(X) or X = HJ or J3 . Proof. Let C = CX (z), Q = O2 (C), and T ∈ Syl2 (C). By assumption, C/Q ∼ = A5 . Clearly Z(T ) = z so T ∈ Syl2 (X). It is also clear that C acts irreducibly on Q/ z. Hence O2 (X) = 1, z, or Q, and in the last two cases z ∈ Z(X). So assume that O2 (X) = 1, whence F ∗ (X) = E(X). As |T | = 27 , E(X) has at most three components, so since C = [C, C], C normalizes them all, and then C ≤ E(X). By a Frattini argument X = E(X)C = E(X). As Z(T ) is cyclic and O2 (X) = 1, it follows that X is simple. If X ∼ = An then n = 10 or 11 and from the structure of C, given by [IA , 5.2.8a1], O2 (C) ∼ = 21+4 − , contradiction. If X ∈ Chev(r) − Chev(2), then  ∼ obviously X = L2 (q) for any q, and as Or (C) has no normal subgroup in Chev(r), [IA , 4.2.2] is contradicted. If X ∈ Chev(2) then |X|2 = 27 leads to a contradiction. (If X ∼ = d L(2a ), then a = 1 as Z(T ) = z, but no L has exactly 7 positive roots.) Therefore X ∈ Spor, and the result follows by  inspection of [IA , 5.3]. Lemma 20.14. Let X be a K-group with one conjugacy class of involutions. Let z ∈ I2 (X) and suppose that CX (z) ∼ = D12 . Suppose also that |X|3 = 3. Then X ∼ = L2 (11) or L2 (13). Proof. Let T ∈ Syl (X). Then T ∼ = E22 and 2

T ≤ [NG (T ), NG (T )]  by Burnside’s theorem and the assumed fusion. Next, O2 (X/O2 (X)) ∼ = L2 (q) for some q ≡ ±3 (mod 8), by [IA , 5.6.4]. If Y := O2 (CX (z)) ≤ O2 (X), then since T ≤ [NG (T ), NG (T )] and by a Frattini argument, [T, Y ] = 1, a contradiction. Thus CO2 (X) (z) = 1 and the same holds for all involutions of T , so O2 (X) = 1. Then q±1 = 12, so q is prime and as T ∈ Syl2 (X), the result follows.  ∼ Lemma 20.15. Suppose that X is a K-group, T ∈ Syl2 (X), T = D8 , P ∈ Syl3 (X), P = CX (P ) ∼ = Z3 , and a central involution of T inverts P . ∼ Then X = L3 (2).

Proof. Since Z(T ) = [T, T ], P  X. First assume that O3 (X) = 1. Then F ∗ (X) = E(X) and since P = CX (P ), E(X) is simple and contains

20. MISCELLANEOUS

561

P . Then NX (P ) > CX (P ) so NG (P ) ≤ E(X) and thus X = E(X). By [IA , 5.6.3], X ∼ = L2 (q), q odd, or A7 . As P = CX (P ) and T is not abelian, q = 7, as desired. Finally, if O3 (X) = 1, set X = X/O3 (X). The hypotheses go over to X, so X ∼ = L3 (2), which contains Frobenius groups of order 7.3 and 22 .3. Hence CO3 (X) (P ) = 1 by [IG , 9.12], contradiction. The proof is complete.  Lemma 20.16. Either let K = L4 (2) and let V be a natural F2 K-module, or let K = Sp6 (2) and let V be the spin F2 K-module. Then H 1 (K, V ) = 0. Proof. (Alperin-Gorenstein) In either case let x ∈ I3 (K) with [V, x] = V . Thus E(CK (x)) ∼ = A5 or Sp4 (2) in the two cases, and we can choose x ∈ xK ∩ E(CK (x)), so that [V, x ] = V and CK (x), CK (x ) = K by [IA , 7.5.1, 7.3.2] (and the fact that L4 (2) ∼ = A8 ). Now if W is an F2 Kmodule containing V and such that dim(W/V ) = 1, CW (x) complements V and is invariant under CK (x), x , and then CK (x ) and K. This proves the lemma.  Lemma 20.17. Let P ∈ Syl3 (3D4 (2)). Then P has a unique normal subgroup of order 9, and it is elementary abelian. Proof. Refer to [IA , 4.7.3A]. Let C = CG (t1 ). We may assume t1 ∈ Z(P ). Then P/ t1  ∼ = P ∩ E(C) ∼ = 31+2 , which imply that P is of maximal class and both assertions follow.  Lemma 20.18. There is no L ∈ C2 such that (a) |Z(L)| is even; (b) 27 ≤ |L|2 ≤ 29 ; and (c) For some z ∈ I2 (Aut(L)) and z0 ∈ I2 (J3 ), O2 (CL (z)) ∼ = CJ3 (z0 ). Proof. Suppose such an L exists. Conditions (a) and (b) rule out L ∈ Alt (note that 2An ∈ C2 for any n ≥ 5). Since Z(L) is a 2-group, O 2 (CL (z)) maps onto C := O2 (CL/Z(L) (z)), which by (c) is therefore a split or E24 by A5 , with O2 (C)/Φ(O2 (C)) the core of extension of O2 (C) = 21+4 − a natural permutation module for A5 . If L ∈ Spor, this forces L ∼ = 2HJ or 2HS, and so by (b), L ∼ = 2HJ, whence z ∈ Inn(L) is 2-central. But then by (c), C must split over Z(C). However, involutions in C − O2 (C) do not so split, by [IA , 5.3g], contradiction. Therefore L ∈ Chev(r) for some prime r. Suppose that r > 2. As L ∈ C2 ,  L/Z(L) ∼ L2 (ra ) for any a, so r = 3 and by [IA , 4.2.2], O3 (CL/Z(L) (z)) is = the central product of Lie components of characteristic 3, contradiction. Hence, K ∈ Chev(2). By [IA , 6.1.4] and the definition of C2 [V11 , 3.1], and 8 taking (b) into account, L/Z(L) ∼ = L3 (4) or 2B2 (2 2 ). But then C is simple or solvable, a contradiction. The proof is complete.  Lemma 20.19. Let K = G2 (4) and f ∈ I2 (Aut(K)). If m3 (CK (f )) > 1, then CK (f ) ∼ = G2 (2).

562

17. PROPERTIES OF K-GROUPS

Proof. All maximal parabolic subgroups of K have 3-rank 1. Hence by the Borel-Tits theorem, f ∈ Inn(K). Then by [IA , 2.5.12], f is a field automorphism. The lemma follows.  Lemma 20.20. Let K ∈ S = {J3 , U4 (2), Sp6 (2), 2Sp6 (2)}. Then the following conditions hold: (a) If L ∈ C3 and K ↑3 L, then L ∈ S; and (b) If K = F ∗ (X) and T ∈ Syl2 (X), then m2 (Z(T )) ≤ 2. Proof. As U4 (2) ∼ = P Sp4 (3) ∈ Chev(3) and | Out(U4 (2))|3 = 1, ↓ 3 (U4 (2)) = ∅. Using [IA , 5.3h] and [IA , 4.8.2] for L = J3 and L/Z(L) = Sp6 (2), respectively, we have ↓3 (L) ⊆ {A6 }, and (a) follows. Part (b) holds for K = Sp6 (2) by [IA , 3.3.1c], and for K = 2Sp6 (2) by Lemma 10.2.4ac. In the other two cases, | Out(K)| = 2 by [IA , 2.5.12, 5.3h], so it is enough to show that Z(T ∩ K) is cyclic. But this holds because for some z ∈ I2 (K), F ∗ (CK (z)) ∼ = 21+4 [IA , 4.5.1, 5.3h]. The lemma is proved.  Lemma 20.21. Let K ∈ Kp , p odd. Assume that K is a component of a group X in which a Sylow p-subgroup P satisfies Ω1 (P ) = Ω1 (Z(P )) ∼ = Ep3 . Then Z(K) = 1. Proof. If Z(K) = 1, then the possibilities for K are given in [IA , 6.1.4]. If K/Z(K) ∼ = An , then n = 6 or 7 and p = 3, so P = Ω1 (P ), which is nonabelian by [IG , 15.12]. If K ∈ Spor, then from mp (K) ≤ 3 we get K ∼ = 3M22 or 3O N , and again P = Ω1 (P ) [IA , 5.6.1, 5.3cs], giving the same contradiction. If K ∈ Chev(p) − Alt, then Lemma 18.5 gives the contradiction mp (K) > 3. Finally suppose that K ∈ Chev(r) − Chev(p), r = p. As p − 1 ≤ mp (K) ≤ 3 (see [IA , 4.10.3]) and Z(K) = 1, the only possibility is p = 3 and K ∼ = SL3 (q), q ≡ 1 (mod 3),  = ±1. But then a monomial subgroup of K contains 31+2 so Ω1 (P ) is not abelian, a final contradiction.  Lemma 20.22. Let K ∈ K be simple, let p be an odd prime, and let P ∈ Sylp (K) with mp (K) = 2. If Ω1 (P ) is abelian, then NK (P ) acts irreducibly on Ω1 (P ). Proof. By [GL1, I-12.1], the result holds if P is abelian. So assume that P is not abelian. Note that as p is odd, L = Lp (q) and L = SLp (q), q ≡  (mod p),  = ±1, have Ω1 (Q) nonabelian for Q ∈ Sylp (L). This is visible in a monomial subgroup of L, and shows that K has no subgroup L of either of these two types. It follows easily, using [IA , 2.5.12], that if K ∈ Chev(r), r = p, then p does not divide | Outdiag(K)|. In that case, using [IA , 4.10.2ac, 4.10.3a], we see that P has an complemented abelian normal subgroup of rank mp (P ). As Ω1 (P ) is abelian, this implies that P is abelian, contradiction.

20. MISCELLANEOUS

563

∼ L± (p) If K ∈ Chev(p), then as mp (K) = 2 and P is not abelian, K = 3 [IA , 3.3.3], but then Ω1 (P ) = P is nonabelian. If K ∼ = An , then P is elementary abelian or P contains Zp Zp , impossible as Ω1 (P ) is abelian. And if K ∈ Spor, either P is abelian or Ω1 (P ) is nonabelian (except for K = J3 and p = 3, for which mp (K) > 2), from the information in [IA , 5.3, 5.6.1]. 

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565

566

[GLS1]

[I2 ] [I1 ] [IG ]

[IA ]

[GLS4]

[II1 ] [II3 ] [II2 ]

[GLS5]

[III2 ] [IIIK ] [III1 ] [III3 ] [GLS6]

[IVK ] [IV9 ] [GLS7]

BIBLIOGRAPHY

D. Gorenstein, R. Lyons, and R. Solomon, The classification of the finite simple groups, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, Providence, RI, 1994, DOI 10.1090/surv/040.1. MR1303592 D. Gorenstein, R. Lyons, and R. Solomon, Outline of Proof, ch. 2 in [GLS1], American Mathematical Society, 1994. D. Gorenstein, R. Lyons, and R. Solomon, Overview, ch. 1 in [GLS1], American Mathematical Society, 1994. D. Gorenstein, R. Lyons, and R. Solomon, The Classification of the Finite Simple Groups. Number 2. Part I, Chapter G, General group theory, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, Providence, RI, 1996. MR1358135 (96h:20032) D. Gorenstein, R. Lyons, and R. Solomon, The Classification of the Finite Simple Groups. Number 3. Part I, Chapter A: Almost simple K-groups, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, Providence, RI, 1998. MR1490581 (98j:20011) D. Gorenstein, R. Lyons, and R. Solomon, The Classification of the Finite Simple Groups. Number 4. Part II, Chapters 1–4: Uniqueness theorems, with errata: The classification of the finite simple groups. Number 3. Part I. Chapter A [Amer. Math. Soc., Providence, RI, 1998; MR1490581 (98j:20011)], Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, Providence, RI, 1999, MR1675976 (2000c:20028) D. Gorenstein, R. Lyons, and R. Solomon, General Lemmas, ch. 1 in [GLS4], American Mathematical Society, 1999. D. Gorenstein, R. Lyons, and R. Solomon, p-Component Uniqueness Theorems, ch. 3 in [GLS4], American Mathematical Society, 1999. D. Gorenstein, R. Lyons, and R. Solomon, Strongly Embedded Subgroups and Related Conditions on Involutions, ch. 2 in [GLS4], American Mathematical Society, 1999. D. Gorenstein, R. Lyons, and R. Solomon, The Classification of the Finite Simple Groups. Number 5. Part III, Chapters 1–6: The generic case, stages 1– 3a, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, Providence, RI, 2002. MR1923000 (2003h:20028) D. Gorenstein, R. Lyons, and R. Solomon, General Group-Theoretic Lemmas, ch. 2 in [GLS5], American Mathematical Society, 2002. D. Gorenstein, R. Lyons, and R. Solomon, Properties of K-Groups, ch. 6 in [GLS5], American Mathematical Society, 2002. D. Gorenstein, R. Lyons, and R. Solomon, Theorem C7 : General Introduction, ch. 1 in [GLS5], American Mathematical Society, 2002. D. Gorenstein, R. Lyons, and R. Solomon, Theorem C∗7 : Stage 1, ch. 3 in [GLS5], American Mathematical Society, 2002. D. Gorenstein, R. Lyons, and R. Solomon, The Classification of the Finite Simple Groups. Number 6. Part IV: The special odd case, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, Providence, RI, 2005. MR2104668 (2005m:20039) D. Gorenstein, R. Lyons, and R. Solomon, Preliminary Properties of KGroups, ch. 10 in [GLS6], American Mathematical Society, 2005. D. Gorenstein, R. Lyons, and R. Solomon, Theorem C3 : Stages 2 and 3, ch. 9 in [GLS6], American Mathematical Society, 2005. D. Gorenstein, R. Lyons, and R. Solomon, The Classification of the Finite Simple Groups, Number 7. Part III, Chapters 7–11: The generic case, stages 3b and 4a, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, Providence, RI, 2017.

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[III8 ] [III11 ] [III7 ] [GLS8]

[III17 ] [III13 ] [G1] [Hu1]

[Is2]

[KMa1]

[McL2] [N1]

[SmF] [T2]

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Index

flat TG3 -group, 37

2-obstruction, 9 31 U4 (3), 513 32 U4 (3), 513

G, 1 γ ∗ , 358 ΓA,k (Y ), 20 ΓA,k (Y ), 20 Gilman-maximal, 15 Glauberman, G., xiii Golay module, 520 Gomi, K., 273 Gr¨ un, O., 1

A, 152, 409 A9 –Sp6 (2) setup, 311 A9 -Sp6 (2) setup, 147 AB , 1 Aschbacher, M., xiii, 144, 152 B0 (T ), 45, 55 B3 -group, 31 B13 , 32 Background Results, 334, 364 k + 12 -balance, 8 balance, k-balance, weak balance, 8 balance, strong local, 188 balancing ordered pair, 168 BT , 181 B(T ), 45, 55 BT∗ , 187

H, 146, 179 H∗ (G), 297 Hall, Philip, 272 Hcons , 146 Hinv , 146 Ipo -terminal, 38 J (A), 81 κc (G), κu (G), 145 κ(G), 145 K(G), 204 KM -singularity, 147 k-obstruction, 9 k + 12 -obstruction, 9 Kp , 3

C2 , C2 -group, 148 Co2 , Co2 -group, 149 C(A, K), 199 centralizer of involution pattern, 364 connected, 39 connected, r-connected, 8 Cp , Cp -group, 148 CTGp , 37, 473 CTGp , 37

LA , 291 Lemmas [V17 , 14.1] and [V17 , 14.2], application of, 4 λ(G), λu (G), 144 local k-balance, weak local k-balance, local k + 12 -balance, 8 L∗p -balance, 14 L∗p (X), 14 LTp -type, 29

d0 , 33 d0 = 1a, 1g, 1j, 35 ΔG (D), ΔD , 6 dT,p , 31 e(X), 3 Epn (X), 2 exceptional situation, 33 extraspecial setup, 147, 311

m2,p (X), 3 m  2,p (J), 3 569

570

mB , 55 mp (X), 3 mIp (X), 179 m  p (J), 3, 32 non-isolated ordered pair, 168 normal situation, 33 N (y, J), 39 obstruction, k-obstruction, k + 12 -obstruction, 9 Op (X), Op (X), Oπ (X), Oπ (X),  Op (X), Op (X), 1 outer well-generated, 17

INDEX

TJp (G), 34 TJp (G), 34 TJop (G), 34 Todd module, 520 Tp , 30 Tp (G), 37, 473 Tp -group, 30 Tpi , 1 ≤ i ≤ 5, 30 Tpi (G), 37, 473 U6 (2)-type, 351 U(G; M ; p), 25

P , 44

W (D4 )∗ , 22 weak balance, weak k-balance, 8 weak local balance, 8

Q, 44

(x, K), 43

R, 44 r, 66 restricted even type, 149, 239 rp (X), 11 Signalizer Functor theorem, consequence of, 7 Sp , 2, 152 standard component, of centralizer of involution, 17 standard plan, 173 T , 44 T1 , 44 T33+ , T33− , 68 T(B), 181 terminal component, of centralizer of involution, 17 TG3 , 37 TGp , TGp -group, 32 TGp (G), 37, 473 Θ2 , Θ3 , Θ3/2 , Θ5/2 , 6 Θ3 , Θ3/2 , Θ5/2 , 67 ΘB , ΘB (G; B), 67 Theorem C∗4 , 144 Theorem C∗4 , Case A, 145 Theorem C∗4 , Case B, 146 Theorem C∗4 : Stage A1, 146 Theorem C∗4 : Stage A2, 147 Theorem C∗4 : Stage A3, 147 Theorem C∗4 : Stage A4, 148 Theorem C∗4 : Stage A5, 148 Theorem C∗∗ 6 , 29 Thompson dihedral lemma, 3 Thompson transfer lemma, 1 Tiep, xiii

Selected Published Titles in This Series 40.10 Inna Capdeboscq, Daniel Gorenstein, Richard Lyons, and Ronald Solomon, The Classification of the Finite Simple Groups, Number 10, Part V, Chapters 9-17: Theorem C6 and Theorem C4∗ , Case A, 2023 40.9 Inna Capdeboscq, Daniel Gorenstein, Richard Lyons, and Ronald Solomon, The Classification of the Finite Simple Groups, Number 9, Part V, Chapters 1-8: Theorem C5 and Theorem C6 , Stage 1, 2021 273 Gennadiy Feldman, Characterization of Probability Distributions on Locally Compact Abelian Groups, 2023 272 Tadashi Ochiai, Iwasawa Theory and Its Perspective, Volume 1, 2023 271 Hideto Asashiba, Categories and Representation Theory, 2022 270 Leslie Hogben, Jephian C.-H. Lin, and Bryan L. Shader, Inverse Problems and Zero Forcing for Graphs, 2022 269 Ralph S. Freese, Ralph N. McKenzie, George F. McNulty, and Walter F. Taylor, Algebras, Lattices, Varieties, 2022 268 Ralph S. Freese, Ralph N. McKenzie, George F. McNulty, and Walter F. Taylor, Algebras, Lattices, Varieties, 2022 267 Dragana S. Cvetkovi´ c Ili´ c, Completion Problems on Operator Matrices, 2022 266 Kate Juschenko, Amenability of Discrete Groups by Examples, 2022 265 264 263 262

Nabil H. Mustafa, Sampling in Combinatorial and Geometric Set Systems, 2022 J. Scott Carter and Seiichi Kamada, Diagrammatic Algebra, 2021 Vugar E. Ismailov, Ridge Functions and Applications in Neural Networks, 2021 Ragnar-Olaf Buchweitz, Maximal Cohen–Macaulay Modules and Tate Cohomology, 2021

261 Shiri Artstein-Avidan, Apostolos Giannopoulos, and Vitali D. Milman, Asymptotic Geometric Analysis, Part II, 2021 260 Lindsay N. Childs, Cornelius Greither, Kevin P. Keating, Alan Koch, Timothy Kohl, Paul J. Truman, and Robert G. Underwood, Hopf Algebras and Galois Module Theory, 2021 259 William Heinzer, Christel Rotthaus, and Sylvia Wiegand, Integral Domains Inside Noetherian Power Series Rings, 2021 258 Pramod N. Achar, Perverse Sheaves and Applications to Representation Theory, 2021 257 Juha Kinnunen, Juha Lehrb¨ ack, and Antti V¨ ah¨ akangas, Maximal Function Methods for Sobolev Spaces, 2021 256 Michio Jimbo, Tetsuji Miwa, and Fedor Smirnov, Local Operators in Integrable Models I, 2021 255 Alexandre Boritchev and Sergei Kuksin, One-Dimensional Turbulence and the Stochastic Burgers Equation, 2021 254 Karim Belabas and Henri Cohen, Numerical Algorithms for Number Theory, 2021 253 Robert R. Bruner and John Rognes, The Adams Spectral Sequence for Topological Modular Forms, 2021 252 Julie D´ eserti, The Cremona Group and Its Subgroups, 2021 251 David Hoff, Linear and Quasilinear Parabolic Systems, 2020 250 Bachir Bekka and Pierre de la Harpe, Unitary Representations of Groups, Duals, and Characters, 2020 249 Nikolai M. Adrianov, Fedor Pakovich, and Alexander K. Zvonkin, Davenport–Zannier Polynomials and Dessins d’Enfants, 2020 248 Paul B. Larson and Jindrich Zapletal, Geometric Set Theory, 2020

For a complete list of titles in this series, visit the AMS Bookstore at www.ams.org/bookstore/survseries/.

This book is the tenth in a series of volumes whose aim is to provide a complete proof of the classification theorem for the finite simple groups based on a fairly short and clearly enumerated set of background results. Specifically, this book completes our identification of the simple groups of bicharacteristic type begun in the ninth volume of the series (see SURV/40.9). This is a fascinating set of simple groups which have properties in common with matrix groups (or, more generally, groups of Lie type) defined both over fields of characteristic 2 and over fields of characteristic 3. This set includes 11 of the celebrated 26 sporadic simple groups along with several of their large simple subgroups. Together with SURV/40.9, this volume provides the first unified treatment of this class of simple groups.

For additional information and updates on this book, visit www.ams.org/bookpages/surv-40

SURV/40.10