Group Theory: Selected Problems 9788173714917, 8173714916

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Group eory Selected Problems

B. Sury

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Universities Press (India) Private Limited Registered Office 3-6-747/1/A & 3-6-754/1, Himayatnagar, Hyderabad 500 029 (A.P.), INDIA e-mail: [email protected] Distributed by Orient Blackswan Private Limited Registered Office 3-6-752 Himayatnagar, Hyderabad 500 029 (A.P.), INDIA e-mail: [email protected] Other Offices Bangalore, Bhopal, Bhubaneshwar, Chennai, Ernakulam, Guwahati, Hyderabad, Jaipur, Kolkata, Lucknow, Mumbai, New Delhi, Noida, Patna © Universities Press (India) Private Limited 2004 First published 2004 eISBN 9978 81 7371 893 9 e-edition:First Published 2013 ePUB Conversion: Techastra Solutions Pvt. Ltd. All rights reserved. No part of this publication may be reproduced, distributed, or transmitted in any form or by any means, including photocopying, recording, or other electronic or mechanical methods,

without the prior written permission of the publisher, except in the case of brief quotations embodied in critical reviews and certain other noncommercial uses permitted by copyright law. For permission requests write to the publisher.

"It is not knowledge, but the act of learning, not possession but the act of getting there, which grants the greatest enjoyment. When I have clarified and exhausted a subject, then I turn away from it, in order to go into darkness again; the never-satisfied man is so strange if he has completed a structure, then it is not in order to dwell in it peacefully, but in order to begin another. I imagine the world conqueror must feel thus, who, aer one kingdom is scarcely conquered, stretches out his arms for others." C.F. Gauss in a letter to Bolyai in 1808

When Horn becomes ⊗ and life seems drab and dreary, join the club of problem-solvers in group theory! My group could be a-belian; what is a mere million? Isomorphic copics could havc hucs of a chamclcon! My matrices – my dear Homs – are oen elementary. In such a case, the group can even have the center {e). Aut I convince any more ? Life's normally N or G!

Contents Preface Acknowledgements Examples and Notations Problems Solutions Bibliography

Preface

ese are problems which the author has callected over two decades. e problems are principally aimed at undergraduate mathematics honours students, but the author expects many of them would be useful even for beginning graduate students. Excepting a few preliminary ones, we have included only problems which are not found in standard texts of Artin, Fraleigh, Herstein, Hungerford and Lang. ese problems are meant only to supplement the existing ones in them and not to substitute for them. e author wishes to acknowledge the enormous pleasure he has derived from two textbooks ‐ one, the wanderful graduate text A course in the eory of Groups by D. J. S. Robinson and the other e eory of Groups by I. D. Macdonald. Only abstract groups are discussed here although many problems on finite groups here can be stated at least for profinite groups. A glaring omission (and perhaps a big disappointment to some readers) is of problems in representations of finite groups. On the one hand, many interesting or useful problems involving representation theory require a knowledge of semisimple rings and modules, and concepts like tensor products etc. On the other hand, the author believes that problems involving representations of finite groups alone require a separate manuscript whose size would go beyand the rest of the topics. As a result, the author has decided, albeit grudgingly, to stick only to some basic problems on representations which involve only linear algebra. Aiso, though we have included a few problems on free groups, we have not really gone into combinatorial group theory. Problems involving wreath products and HNN

extensions, etc. are aiso not discussed. e problems are interspersed with comments putting them in perspective with other problems and with the subject itself. We have adopted a convention whereby we have separated the problems into two categories –S and N– which indicate whether they are standard or not. By standard, we do not mean easy problems but we mean problems which appear in some advanced text as a theorem perhaps. e intention of this problem book is two‐fold; on the one hand, to reinforce the existent knowledge by solving new problems and, on the other hand, to introduce some concepts not usually taught at this level, by means of what we have called standard problems. A final word on acknowledging the source of the problems‐we have mentioned the names of people to whom the result is due. However, these problems have been callected over many years and it has not been always possible to locate a reference where it has appeared either as a theorem or a problem etc. erefore, rather than mention only those references which are known to the author, we adopt the policy of not mentioning a reference but only the name in each case.

Acknowledgements

Over the years, D. S. Nagaraj, Dipendra Prasad, Raja Sridharan, Ravi Rao, Amit Roy and T. N. Venkataramana have been some people with whom I have shared the enthusiasm of encountering these types of problems and solving them. I have aiso enjoyed and benefitted from the works of A. Lubotzky, A. Mann, A. Shalev and L. Pyber for their group‐theoretic essence.

Examples and Notations

Let us start by recalling the principal examples which one encounters and basic results on them; this will also serve the purpose of fixing our notations. /n denotes the group of integers modulo n, under addition mod n. ( /n)× = {a ≤ n: (a,n) = 1} under multiplication mod n. e notations /n and ( /n)* are perhaps less familiar compared to n etc. S1 = {z ∊ C : |z| = 1} under multiplication. All the above are abelian groups. We recall also that a finite abelian group is isomorphic to the direct sum of cyclic p-groups for various primes p dividing its order. Sn, the set of all permutations (i.e., bijections) of n symbols, under the composition of permutations. In fact, for any set X, the set Sym(X) of all bijections on X is a group under composition. e subset SF(X) consisting of all those bijections which move only finitely many elements, is a subgroup. We shall use the convention that the permutation στ is obtained by applying σ first and τ later. GL(n, Q), the set of all n × n rational matrices which have non-zero determinants (the symbol GL stands for 'general linear') is a group under matrix multiplication. Similarly, one can define GL(n, R) and GL(n, C). One has also the 'special linear' groups SL(n, Q), SL(n, R) etc. consisting of the matrices which have determinant 1.

SO(n) ={g ∊ SL(n, R) : ggt = In}, is known as the special orthogonal group of degree n. SU(n) = , is known as the special unitary group of degree n. Sp(n) = is known as the symplectic group of degree n; here 0 and In denote n × n block matrices. ese are nonabelian groups when n ﹥ 1. B(n, Q), the upper triangular rational matrices with nonzero diagonal entries, U(n, Q), the subset consisting of those upper triangular matrices which have all diagonal entries equal to 1 and T(n, Q), the diagonal matrices with all diagonal entries nonzero rational numbers, are groups. Note that T(n, Q), for any n and, U(2, Q) are abelian groups. GL(n, Z) = {g an integral matrix: det(g) = ±1}. GL(n, Z[1/p]) where p is a prime number. Here, Z[1/p]denotes the set of rational numbers whose denominators are divisible only by p and, the above group consists of all matrices whose determinants are ±a power of p. Note that where we have used the notation H ≤ G to denote the fact that H is a subgroup of G. We shall also write G ≥ H at times. Also, H < G for groups would mean that H is a proper subgroup. Let us note that the set of matrices

form a group under matrix multiplication but it is not a subgroup of GL(2, R).

We have where Pσ is the 'permutation matrix' whose rows are the rows of the identity matrix permuted according to σ, i.e., the rows of Pσ are Rσ(1), ... , Rσ(n). Note that trivially while the union of two subgroups may not be a subgroup. Further, if g is an element of a group G, then its centraliser CG,(g) := {x ∊ G : xg = gx} ≤ G. Moreover, for any subset S ⊂ G, the subgroup CG(S) := ∩s∊S CG(s) is the centraliser of S. For S = G, the centraliser CG(G) is usually denoted by Z(G), and is called the center of G. If μ(n) denotes the complex nth roots of unity, we have a chain of subgroups For any subset S ⊂ G, one denotes by < S > the subgroup generated by S. In concrete terms, it consists of all finite products of elements of S and their inverses. For instance, for any group G and any positive integer n, the subset Gn := {gn : g ∊ G} gives a subgroup < Gn >. Note that, in any abelian group G, we have < Gn >= Gn. Evidently, for any G and any n, the subgroup < Gn > is normal in G; we write < Gn > G. Given a subset S ⊂ G, one also defines the normal subgroup generated by S as the smallest normal subgroup < S > N of G which contains S. In concrete terms, it consists of all finite products of conjugates of elements of S and their inverses. One says that a group G is finitely generated if there exists a finite subset S ⊂ G such that G = < S >

For any subset S ⊂ G, we define the normaliser NG(S) = {g ∊ G : gSg–1 = S}. Observe < S > NG(S) ≤ G and NG(S) is the largest subgroup of G containing S in which < S > is normal. For any group G, and x, y, ∊ G, we denote [x,y] = xyx–1y–1. Such elements are called commutators in G. If S is the subset of all commutators of G, one obtains the commutator subgroup of G, denoted by [G,G]. Sometimes, one also denotes [G,G] by D(G) and refers to it as the derived group of G. More generally, for H, K ≤ G, [H,K] denotes the subgroup of G generated by {[h, k] : h ∊ H, k ∊ K}. A group G is said to be solvable – the terminology comes from Galois theory – if the chain G ≥ D1(G) := D(G) ≥ D2(G) := D(D(G)) ≥ …… becomes the trivial group in a finite number of steps. Evidently, any abelian group is solvable. A nonabelian example is the group B(n, Q), the upper triangular rational matrices with nonzero diagonal entries for any n ≥ 2. A complementary concept is that of a perfect group; a group G is said to be perfect if G = [G,G]. A group G is said to be nilpotent if the lower central series G ≥ C1(G) := D(G) ≥ C2(G) ≥ C3(G)… becomes trivial in a finite number of steps where Cn+1(G) = [G, Cn(G)]. It is not only evident that abelian groups are nilpotent but also that nilpotent groups are solvable because it is seen by induction on n that Dn(G) ≤ Cn(G) for any G. e group U(n,Q) is nilpotent but nonabelian when n > 2. In any group G, one defines inductively

[x1,… , xn] := [[x1,… , xn–1], xn]. Fn, the free group of rank n, is defined to be the set of all reduced (finite) words in n symbols x1, … , xn and their 'inverse symbols' denoted by x1–1, … , xn–1. Here, a word is said to be reduced if it does not contain two symbols xi and xi–1 in juxtaposition. e empty word is also counted and is the identity element. e group multiplication is by concatenation of two reduced words and cancelling off all consecutive symbols of the form xixi–1 or xi–1xi. Fn is nonabelian if n ≥ 2. It is evident that any group which can be generated by n elements can be identified upto isomorphism with a quotient group of Fn. A group G is said to be finitely presentable if G ≅ Fn/K for some n where K = < R > N for some finite subset R of Fn. e choice of finite sets X ⊂ G with |X| = n and R ⊂ Fn with Fn / < R > N ≅ G is called a finite presentation of G.

If G = < X|R > and H = < Y|S >, then their free product is defined to be the group < X Y|R ∪ S > and is denoted by G * H. Note that Fn is a free product of Z with itself n times (taking free products is an associative operation). We shall see via some problems that familiar groups like SL(2, Z)/{±I} are free products. For a positive integer n and 1≤ i ≠ j ≤ n, the elementary matrix Ei,j is the n × n matrix whose only nonzero entry 1 is at the (i, j)th place. Let us recall the concept of group actions which makes groups one of the most powerful tools in mathematics as well as in other disciplines. Recall that a group G is said to act on a set S if there is a homomorphism from G to the permutation group Sym(S) of S. e action is said to be faithful if the homomorphism is injective. It is customary to write g.s for the element of S that s ∊ S is sent to by the permutation corresponding to g ∊ G. For s ∊ S, the subset G.s := {g.s : g ∊ G} is called the orbit of s. For any g ∊ G,

the subset Sg := {s ∊ S : g.s = s} is called the set of fixed points under g. For s ∊ S, the subgroup Gs := {g ∊ G : g.s = s} is called the isotropy subgroup at s. Note that for s ∊ S, the map g → g.s is a well-defined bijection from the set G/Gs of le cosets of Gs to the orbit G.s of s. us, orbits under a finite group action have cardinalities which are divisors of the order of the group. In the particular case of a group acting on itself by conjugation, we denote the orbit of an element g (this is the conjugacy class of g) by G(g). One also writes g ~h to mean that g and h are in the same orbit; that is, they are conjugate. One calls an action of G on a set S transitive if each orbit is the whole set S. For instance, Sn acts transitively on {1, 2, … , n}. e group GL(n, R) acts transitively on Rn – (0). A group G is said to act r–transitively on a set S if G acts transitively on the set {(s1, … , sr) ∊ Sτ : gi distinct }. For an element g in a group G, one usually denotes the automorphism x → g–1xg of G as Int (g). Sometimes, one also writes xg for g–1xg. e group Aut (G) of all automorphisms of G is defined by means of the composition in the following order: g(στ) = (g(σ)) τ; this is consistent with our convention of multiplying permutations. In this convention, the set of automorphisms x → Int(x) can be identified with a subgroup of Aut G because g(Int (xy)) = (xy)–1)g(xy) = g (Int(x))(Int(y)). is is in fact, a normal subgroup. It is convenient at times to write xθ instead of x(θ) for an automorphism θ. e last important notion we recall is that of composition series and the Jordan–Holder theorem. Recall that a normal series of a group G is simply a

finite sequence of subgroups Of course, we know that the Gi's need not be normal in the whole of G; they are usually called subnormal subgroups of G. A normal series is said to be a refinement of a normal series if each Gi is some Hj ; it is called a proper refinement if m > n. It is elementary to show that any two normal series have refinements which are isomorphic; that is, refinements in which any successive factor Gi/Gi–1 of one is isomorphic to a successive factor of the other and vice versa. One says that a normal series is a composition series if it has no proper refinements. Note that this is equivalent to the successive factors being simple groups. For instance, any normal series for Z will have a proper refinement because the smallest nontrivial term in such a series is an infinite cyclic group and will have proper nontrivial (normal) subgroups. erefore, Z has no composition series. e theorem of Jordan–Holder asserts that for a finite group, each normal series admits a refinement which is a composition series and that any two composition series are isomorphic. Finally, we recall Zorn's lemma which is equivalent to the axiom of choice. e use of Zorn's lemma becomes unavoidable when we need to prove facts like every vector space has a basis. Let (S, ≤) be any non-empty partially ordered set. is means the three properties : (i) s ≤ s for all s ∊ S; (ii) s ≤ t, t ≤ s ⇒ s = t; and (iii) s ≤ t ≤ u ⇒ s ≤ u.

A chain is a subset T of S in which for any two elements s, t ∊ T, either s ≤ t or t ≤ s. Zorn's lemma asserts that if every chain T in S has an upper bound (that is, an element s0 ∊ S such that t ≤ s0 for all t ∊ T), then S has a maximal element (that is, an element m ∊S which is not ≤ any element of S other than itself). Finally, we recall that a group is said to have exponent d if d is the smallest natural number for which each element g satisfies gd = 1. One last word of convention – a subgroup M of a group G is maximal if M is a proper subgroup which is not contained in any other proper subgroup of G. Also, the identity element is always denoted by 1 unless there is an abelian group written additively when it is denoted by 0 . Also, we write O(g) for the order of an element g of a group G while the order of the group itself is written as |G|. e notation pr||n means that pr is the highest p–power dividing n.

Problems

PROBLEM S-l (i) For any infinite cyclic group G = < g >, the map g 1 is an isomorphism onto Z; for finite cyclic G, g ↦ 1 is an isomorphism onto Z/|G|. Use this to prove that the direct product G × H of nontrivial cyclic groups G, H is cyclic if and only if both G, H are finite and their orders m, n are coprime. (ii) Prove that a finite group is cyclic if and only if it has a unique subgroup of each order dividing the order of the group. Hence deduce that for each natural number n. (iii) Let G be a finite group in which, for every n/|G|, the set {g : gn = e} has at most n elements. en, G is cyclic and the set {g : gn = e} has exactly n elements for each n||G|. In particular, any finite subgroup of K* is cyclic for any field K. ASIDE        Note that for the application in (iii) to field theory, we need only prove the result for abelian G. Later, we wlll discuss this problem again and prove (ii) and (iii) as applications of Sylow's theorems.

PROBLEM S-2 (i)Use the group structure of (Z/p)* to prove Fermat's little theorem asserting ap–1≡1 mod pfor prime p and (a,p)=1. (ii) Prove Euler's congruence asserting aϕ(n)≡1mod n for (a,n)=1.

(iii) Prove the assertion n|ϕ(an – 1) for natural numbers a, n. (iv) Prove Wiison s congruence that for a prime p, one has (p – 1)!≡ –1mod p by looking at the product of all elements in . (v) Let a be a natural number and p be a prime. Suppose that for some n≥1. en, show that ap–1≢1 modulo p2). ASIDE        Later, we shall also discuss problems which give various grouptheoreticol generalisations of Fermat's little theorem, Wilson's theorem etc.

PROBLEM N-3 Let G be any group of order n and let p any prime number. Consider the subset S of G ×. . . × G ( p times) defined by S= {(g1,...,gp): g1g2...gp = e}. (i) Prove that |S|≡ ⌗{g:gp)=e} mod p. (ii) Use (i) to prove Cauchy's theorem which asserts that if G is a finite group and p||G| is a prime, then G has an element of order p. (iii) Use (i) to prove Fermat's little theorem for any prime p. ASIDE    Recall thot a finite-dimensionol division algebra D over a field K is a finite-dimensionol vector space D over K such that D has a multiplication under which the set D* of all non zero elements forms group, the multiplication is also required to satisfy x(y+z)=xy+xz and (x+y)z=xz+yz and, λ(xy)=(λx)y for all λ ∊ K and all x, y, z ∊D. Note that D is a field if and only if D* is an abelian group. Further, D is said to be central over Kif the centre of D* is K*. A familiar example is the space H of Hamilton's quaternions H := {a0 + a1i + a2j + a3k : ai ∊ R} where the multiplication is defined by i2 = j2= k2 = –1, and ij= k= –ji. Note thot H is a 4-dimensional vector space over R and is central over R. A very

similar construction produces many examples of central division algebras over Q.

PROBLEM S-4 (i) Let D be any finite-dimensional central division algebra over a field F of characteristic p > 0. Prove that every finite subgroup of ?D* is again cyclic. (ii) Show that the above Hamilton quaternion algebra H :={a0 + a1i + a2j + a3k : ai ∊ R} is such that H* := H\(0) is a group which contains a finite noncyclic group. (iii) Show that (Z/n)*={a≤n:(a,n)=1} is cyclic if, and only if, n=2,4, pr) or 2pr) for some odd prime p. ASIDE        e proof of (i) does require the application to field theory in Problem S-l (iii). A generator in (iii) here is known in number-theoretlc parlance as a primitive root modulo n. lt is interesting to note thot there is no known algorithm to flnd a primitive root modulo a prime.

PROBLEM N-5 Show that the additive groups Z[1/2] and Z[1/3] are not isomorphic.

PROBLEM N-6 Let p, q be distinct odd primes. Consider the group G = (Z/p)* × (Z/q)* and the subgroup H={(1,1),(–1,–1)}. (i) Prove that

is a set of coset representatives for H in G.

(ii) Using the isomorphism θ : (Z/p)* ×(Z/q)* → (Z/pq)* get a set of coset representatives for θ(H). (iii) Use the two sets to get expressions for the product of elements of G/H and deduce the quadratic reciprocity law

ASIDE    e above observation is due to G. Rousseou.

PROBLEM N-7 Prove that Jordan–Holder theorem implies the fundamental theorem of arithmetic.

PROBLEM N-8 (i) If H≤G has index n in G, then show that H contains a normal subgroup N whose index is a divisor of n!. (ii) Let G be a finite group and H ≤ G be of order n where each prime factor of n is at least [G:H]. en, prove that H is normal. (iii) Let N ⊴ G and suppose x, y∊G such that xy∊N. Prove that xryr∊ N for all integers r.

PROBLEM S-9 Let a finite group G act on a finite set S. Prove:

Conclude that the number of orbits is the 'average number of fixed points' . ASIDE    e above statement is usually known as Burnside's lemma and is effectively used in Polya's theory of enumeration and, in particular, in the combinatorial enumeration of isomers of a chemical compound. In informal parlance, one describes the lemma as saying that when one wants to count cattle, count the number of legs and divide by 4 (!) Actually, there ore mony Burnside lemmata of importance which are of different levels of difficulty and this is the easiest of them all.

PROBLEM N-10 Let G be a finite group and H≤G such that HK is a subgroup of G for all K ≤ G. en, prove that H is a subnormal subgroup of G.

PROBLEM N-11 Let G ≤ Sp act transitively on {1, …, p} where p is a prime. Show that any nontrivial normal subgroup aiso acts transitively on {1, …, p} and contains a p-cycle.

PROBLEM N-12 Let p any prime number of the form 4n+1. Consider the finite set S = {(x,y,z) ∊ N × N × N : x2 + 4yz = p}. Define the Zagier map σ : S → S by mapping (x,y,z) to

Prove that σ is a permutation of order 2 and has a unique fixed point. Hence, conclude that any prime number p of the form 4n+1 is a sum of two squares. (ii) Let p be as before but define S1) = {(x,y,z) ∊ Z × N × N : x2 + 4yz = p} and S2 = {(x,y,z) ∊ S1 : z > x + y}. Consider the Nair maps

and β : S2 → S2;(x,y,z)↦ (–x,y,z) or (x,z,y) according as to whether z > y–x or z < y–x. Show that α is an involution with a unique fixed point and draw the conclusion about p using β. (iii) Let p,S1) and S2) be as above. Consider the subset S3)= {(x,y,z)∊S1): z< x+y}. Prove that if no element of S1) is of the form (x,y,y), then all the elements of S1) can be callected in groups of 4 where exactly two are in S2) and two in S3). Consider the Heathbrown-Nair map

to conclude that |S2| is odd and arrive at a contradiction. ASIDE    Zagier's proof wos published as a 'One sentence proof ' but the other vorlonts due to Heothbrown and Mohon Nolr ore unpublished.

PROBLEM N-13 Let G be a finite simple group and p ||G|. If G has exactly n > 1 p-Sylow subgroups, show that G is isomorphic to a subgroup of An.

PROBLEM N-14 Let p be a prime and let a ≥ b be positive integers. Consider disjoint sets S1, ... , Sa and call their union S. Fix a cyclic permutation of each Si and let G

denote the group of permutations of S generated by these a p-cycles. Considering the action of G on the set T of all pb-element subsets of S, prove that

ASIDE    is ideo of proving the congruence is due to J.H. Smith.

PROBLEM S-15 For any prime p and any natural number n, prove that any p-group P ≤ GL(n, Z/p) is conjugate to a subgroup of the group U=U(n,Z/p) consisting of upper triangular matrices with 1's on the diagonal. Hence, compute the number of p-Sylow subgroups.

PROBLEM N-16 Use the Sylow theorems to solve Problem 1 (ii) and (iii).

PROBLEM S-17 Let G be a finite group and p||G| a prime. (i)Prove that if P is a p-Sylow subgroup of G, then NG(NG(P)) = NG(P). (ii) Let N ⊴ G and let Q ≤ N be a p-Sylow subgroup of N. en, show that G=N NG(Q). ASIDE        e above idea is oen used and is referred to as the Frattini argument.

PROBLEM N-18

Let G be a finite group and pr be a prime power dividing the order of G. en, prove that there exist subgroups of order pr in G and that these are ≡1mod p in number. ASIDE    e above generalisation of Sylow's theorems is due to Frobenius. We have the following even stronger result due to E. Snapper.

PROBLEM N-19 Let G be a finite group and p be a prime such that pn||G|. Suppose H ≤ G has order pm ≤ pn. en, prove that: (i) there exist subgroups of order pn containing H and, (ii) the number of subgroups of order pn which contain H is ≡1 modulo p.

PROBLEM N-20 Show that An is (n – 2)-transitive on {1, 2, ... , n}.

PROBLEM N-21 (i) Prove that any element in An is a commutator xyx–1y–1 in Sn where x is an n-cycle. (ii) In S2n+1, prove that the cycle (1 2 ... 2n+1) is expressible as xyx–1y–1 where x is a n+1-cycle. (iii) In any Sn, show that every element is a product of at the most two cycles. (iv) Let F be a finite field. Prove that Sym(F) is generated by the permutations σ : x → x–1 for x ≠ 0; σ(0) = 0 and τa,b: x → ax+b for a, b∊F. ASIDE   e observation in (iv) was made by L. Carlitz.

PROBLEM N-22

(i) For any n, it is well-known that the permutations σ = (1 2) and τ=(1 2...n) generate the whole of Sn. Prove this. Further, if p is a prime, show that any transposition and any p-cycle generate Sp. (ii) For general n, and for a transposition σ and any n-cycle τ, find a necessary and sufficient candition for Sn to be generated by σ and τ. ASIDE    e proofs of simplicity of the groups An for n ≥ 5 and of PSL (n,F) = SL (n,F)/center for any field F and (n,F) ≠ (2,Z/2), (2,Z/3)are well-known and contained ln all the standard texts. Here are some variants and corallaries.

PROBLEM N-23 Prove that Sn is not isomorphic to a subgroup of An+1 for any n> 1.

PROBLEM N-24 (i) Consider the alternating group A(N) defined as the set of bijections of N which move only finitely many natural numbers and move them as even permutations. Prove that A(N) is simple. (ii) Prove that any finite group is isomorphic to a subgroup of a finite simple group. (iii) Prove that in the infinite group SL (2, F) for any infinite field F, each element is expressible as a finite product of elements of finite order.

PROBLEM N-25 (i) Suppose G is a group and g ∊ G an element conjugate to g-1 such that G = G(g)G(g):= {xy : x,y ∊ G(g)}, where G(g) denotes the conjugacy class of g. en, prove that

G = {xyx–1y–1 : x ∊ G(g),y ∊ G}. (ii) Deduce that the group PSL(2, C)= SL(2, C)/± I is perfect.

PROBLEM N-26 Prove that SU(2) ≅H1), the group of unit quaternions; that is, the quaternions a+bi+cj+dk which satisfy a2+b2+c2+d2=1. Write down explicitly a homomorphism from SU(2) to SO(3) whose kernel is {±I2}.

PROBLEM N-27 For any field K, prove that GL (n,K) is the disjoint union of double cosets BωB over permutation matrices ω. Here B:=B(n,K), the group of invertible upper triangular matrices with entries from K. ASIDE    e above decomposition is known os the Bruhat decomposition. lt has analogues for many matrix groups and proves to be extremely useful.

PROBLEM N-28 (i) Prove that for any odd prime number p, an n×n integral matrix A≠ Id with entries congruent to the corresponding entries of the identity matrix modulo p is of infinite order. In other words, the 'principal congruence subgroup' ker(GL(n,Z) → GL (n,Z/p) is torsion-free for an odd prime p. (ii) Prove (i) using eigenvalues. (iii) For p = 2, show that any matrix A in ker( GL (n,Z) → GL (n,Z/p)) of finite order, has order 1 or 2 and is expressible as M.diag(1,1,…,1,–1,–1,…,–1).M-1 for some M∊ GL (n,Z).

ASIDE        is is originally due to H. Minkowski and now many proofs are known. e proof (ii) using eigenvalues is due to R. Guralnick. is proof for (ii) is due to Y. Kitaoka. For p=2, one has to go modulo 4 to get a torsionfree subgroup of GL (n,Z).

PROBLEM N-29 Let

G=O(3,Q):={g∊GL(3,Q):

ggt)=I}.

Consider

the

sub

group

. Prove that the set of permutation matrices along with H generate only a subgroup of infinite index in G. ASIDE        e above example was observed by J. Delaney. Recall that on abelion group G is said to be divisible if for each g ∊G and each natural number n, there exists h ∊ G satisfying nh=g. Examples are the additive groups of Q, R and the group Q,/Z. Note that the first two groups are uniquely divisible; that is, given g and n, the element h is uniquely determined, however, the last one is not uniquely divisible. Notice that nontrivial divisible groups are infinite.

PROBLEM S-30 (i) Prove that a nontrivial finitely generated abelian group cannot be divisible. (ii) Prove that if S is any set of generators of the additive group of Q, then S contains a proper subset of generators. (iii) Show that Q does not have proper maximal subgroups. ASIDE    Call an abelian group D injective if any homomorphism α : A → D from a subgroup A of an abelian group B extends to a homomorphism from B to D. at this property is equivalent to divisibility is the contention of the next problem which uses Zorn's lemma.

PROBLEM S-31 (i) Prove that an abelian group is divisible if, and only if, it is injective. (ii) If D ≤ G is a divisible subgroup of an abelian group G, show that G=D ⊕ E for some subgroup E of G. (iii) Prove that every abelian group is isomorphic to a subgroup of a divisible group.

PROBLEM N-32 Let G be any divisible group and let S be any proper subset. Prove that the number of different sets {S+g: g∊ G} must be infinite. e above result is due to D. Costa.

PROBLEM N-33 Let θ : G → H be a surjective homomorphism of abelian groups whose elements have order a power of a prime p. Suppose that kerθ is finite. en, for each divisible element h ∊ H, there is a divisible element g∊G such that θ(g)=h. ASIDE    It is clear tht athe multiplicative group R>0 of positive real numbers is isomorphic to the additive group of real numbers by means of the logarithm map. Also, it is clear that for C*, the polar coordinates provide on isomorphism C* ≅ R>0 × S1. ere is general structure theorem for divisible groups from which one can deduce that the product above is isomorphic to S1 itself but there is more elementary proof due to L.R. Duffy which uses Zorn's lemma.

PROBLEM N-34 Use Zorn s lemma to choose appropriate bases for the Q-vector spaces R and R × R and, show that C* ≅ S1.

PROBLEM N-35 Define the integer θ(G) recursively for any finite group G as

Compute θ(G). ASIDE        e above was a problem posed by O. Morphy and solved by E. Andersen.

PROBLEM S-36 Let G be any group. (i) Prove

(ii) For any r≥1, prove

where we have written uv+w to mean uvuw for elements in a group. (iii) Prove the Hall-Witt identity: [a,b,cb][b,c,ac][c,a,ba] = 1. Suppose [G,G] ≤ Z(G). en, prove for any x, y, z ∊ G and integer n that : (iv) (xy)n = xnyn[y,x]n(n–1)/2. (v) Elements of p-power order for any odd prime p, form a subgroup.

PROBLEM N-37

(i) Prove that for any group G, we have [G,G]≤< G2>. Obtain an explicit expression for any commutator as a product of three squares. (ii) If |G/Z(G)|2 < |[G,G]|, prove that there are elements of [G,G] which are not commutators. (iii) Consider the group G of matrices that

where a, b, c ∊ Z/4. Show

is a commutator [X,Y] but that it is not a product of two

squares. ASIDE    e above group in (iii) has order 43 and is the smallest group to give on example as desired. For instance, if one considers the entries of the matrices to be from Z/2 instead of Z/4, it is easily verified that X13 is indeed the square of X12+X23– I. Here Xij denotes the elementary matrix which differs from the identity matrix only at the (i, j)th entry, which is 1. It was proved by Lyndon and Newman that in the free group on two symbols a, b, the commutator [a,b] is not a product of two squares. e example below due to P. J. Cassidy shows that in a general group G, one may require arbitrarily many commutators to express elements of [G,G].

PROBLEM N-38 Consider the group

of matrices where f(x), g(y) are polynomiais in independent variables x, y over rational numbers and h(x,y) ∊ Q[x,y]. Prove that [G,G] ≅ Q[x,y] and that every n≥ 1, there is an element of [G,G] which cannot be expresed as a product of ≤ n commutators.

PROBLEM N-39 Prove that in a finitely generated group G, if [G,G] is finite, then Z(G) has finite index.

PROBLEM N-40 Let θ : G → G be an automorphism of a finite group G such θ(g)=g–1 for more than 3/4ths of the elements of G. Prove that θ(x)=x–1 for all x ∊ G. If θ takes g to g–1 for exactly 3/4ths of the elements of G, prove that G has an abelian subgroup of index 2.

PROBLEM N-41 Suppose G is a nonabelian finite group. en, show that the number of elements (x,y) ∊ G × G such that xy = yx is at the most . ASIDE    We recall that H is said to be characteristic subgroup of a group G if σ(H)=H for any automorphism σ of G. In particular, characteristic subgroups are normal. In any group G, the subgroups [G,G] and for any n are evidently characteristic subgroups.

PROBLEM N-42 (i) Prove that a characteristic subgroup C of a normal subgroup N of a group G is normal in G. (ii) Prove that a subgroup of a characteristic subgroup is a characteristic subgroup. (iii) Give an example of a normal subgroup which is not characteristic.

PROBLEM N-43 Let G be any finite group and let g ∊ G. Write where pi's are distinct primes. en, prove that every element g ∊ G can be uniquely

expressed as a product g = g1 . . .gr, where gi's commute pairwise and for i = 1, . . . , r.

PROBLEM N-44 Suppose M, N are normal subgroups of a finite group G such that G/N ≅ M. If N is simple, prove that G/M ≅ N. ASIDE    e result below is due to .N. Hersteln and R. Teltler.

PROBLEM N-45 Let θ : G → G be a homomorphism of any group into itself. (i) If g ↦ gθ (g) and g ↦ g2θ(g) are also homomorphisms, prove that G must be abelian. (ii) If g ↦ g2θ(g) and g ↦ g4θ(g) are homomorphisms, prove that G must be abelian.

PROBLEM N-46 Let G be any finite group in which lth powers of elements commute among themselves and mth powers of elements commute among themselves. If (l,m)=1, prove that G must be abelian.

PROBLEM N-47 Let G be any group which has no nontrivial abelian normal subgroups. If θ is an automorphism of G with the property that θ(x) commutes with x for all elements x∊ G, prove that θ=Id. ASIDE        e above result is due to T.J. Laffey and generalises a problem posed by I.N. Herstein which asserts the same with the stronger assumption that G is nonabelian, simple.

A well-known elementary exercise (see Herstein's text, for instance) asserts that a finite group is the union of three of its proper subgroups if, and only if it admits quotient isomorphic to Z/2×Z/2. Also, no finite group can be the union of two of its proper subgroups. Note thot a cyclic group has the evident property that it is not the union of its proper subgroups. e fallowing results constitute the appropriate converse to this. ey are due to B.H. Neumann, J. Sonn and Mira Bhargava.

PROBLEM N-48 (i) Show that if G has a quotient isomorphic to a finite, non-cyclic group, then it is the union of finitely many proper subgroups. (ii) If , then show that all the Hi's of infinite index can be dropped without affecting this decomposition. (iii) Deduce that a group G can be written as a finite union of proper subgroups if, and only if G admits a finite non-cyclic group as a quotient. (iv) Show that a group G can be written as a finite union of proper normal subgroups if, and only if G admits of a quotient isomorphic to Z/p × Z/p for some prime p. Recall that a group is said to have a property virtually if there is a subgroup of finite index with that property.

PROBLEM N-49 Let G be a virtually cyclic group. If G is torsion-free that is, has no nontrivial element of finite order, prove that G itself is cyclic. ASIDE        Actually, a much stronger result holds : ony virtually free group which is torsion-free, is free. is is proved by Jeon–Pierre Serre by methods from homological algebra.

ASIDE    Recall that given groups H, N and a homomorphism θ : H → Aut(N), the semidirect product H α N is the group whose underlying set is H × N and the multiplication is defined as (h1, n1)(h2, n2) := (h1h2,θ (h2)–1(n1)(n2). Here, we need to recall our convention of multiplying automorphisms from le to right, that is, g(στ) :=(g(σ))τ.

PROBLEM N-50 (i)Let G be a group, H ≤ G and N ⊴ G such that G= HN and H ⋂ N={1}. Prove that for the action one has G ≅ H α N. Hence, deduce that G ≅ G/N α N in this case. (ii) Let N ⊴ G. Prove that G ≅ G/N α N if, and only if there exists a homomorphism s : G/N → G such that π∘s=Id where π : G → G/N is the natural homomorphism. ASIDE    Recall that a group G is sold to be free abelian of rank n if, and only if it is an abelian group G wlth bosis of n elements g1,…, gn, i.e., G = < {g1,… , gn}> and

implies ai = 0 for all i.

Note that in our additive notation, G = < {g1,… , gn}> means that G={a1g1+ . . . +angn: ai ∊ Z}. e following five problems deal with the structure of finitely generated abelian groups.

PROBLEM S-51 (i) Show that a group G is free abelian of rank n if and only if it is isomorphic to Zn), the set of integral n-tuples under coordinate-wise addition.

(ii) Prove that, if G is free abelian rank n, then the rank n is uniquely determined.

PROBLEM S-52 (i) Prove that if H is a subgroup of a free abelian group G of rank n, then H is free abelian of rank r ≤ n. Further, prove that there are bases {e1, ..., en} of G and {d1e1, ..., drer} of H respectively where di divides di+1 for i 1, the infinite series and the infinite product converge to the same number. ASIDE        e ideaof considering the Dirichlet series to study a number-theoretic sequence ian goes back to Euler, Riemann and Dirichlet, who proved many properties of prime numbers with this tool. For instance, it helps us find bounds for as a function of x when x is large. e above elementary exercise con be used to find an expliclt constant c between 2 and 3 such that tends to zero as x tends to infinity. In other words, on an average, there are c (an explicitly determined number between 2 and 3) abelian groups of a given order on the average.

PROBLEM S-77 (i) Prove that a group G is solvable if, and only if, there exists a series

for some n where Gi are normal subgroups of G and Gi/Gi+1 is abelian. (ii) Show that subgroups and quotient groups of solvable groups are solvable. (iii) If N is a solvable normal subgroup of a group G such that G/N is solvable, then prove that G is solvable as well. ASIDE    One of the most famous, beautiful theorems in finite group theory is the statement that every group of odd order is solvable. e proof by Walter Feit and John ompson occupies a whole volume of the Paciific Journal of Mathematics odd order theorem apart from using many results proved by earlier. A curiosity is that if one can show that there are no primes p≠q for which pq –1 divides qp – 1, a part of the proof can be shortened!

PROBLEM N-78 (i) Let G be a finite group such that all its proper subgroups are abelian. en, prove that G must be solvable. (ii) Let G be a finite group such that all its proper subgroups are nilpotent. en, prove that G must be solvable. (iii) Exhibit a non-solvable finite group all of whose proper subgroups are solvable. ASIDE        Many resu|ts on finite groups are natural consequences of the general representation theory of finite groups. For instance, theorem of Burnside asserts that any group of order ?prqs for primes p, q is solvable. is follows naturally using representation and, although proof is available now which does not need representation theory, it is rather complicated.

PROBLEM N-79

Use the ideas in Problem 73 to prove that in the free product G=Z/2*Z/2, the number of subgroups of index n is either n or n+1 according as whether n is odd or even. ASIDE    One con oiso deol wlth small free products of finite cycllc groups. For instance, one con prove thot for the group PSL(2,Z) (which can be identified wlth Z/2*Z/3), the number of subgroups of index n is odd if and only if n is of the form 2k–3. ASIDE    ere is a very useful notion known as the transfer homomorphism. is is defined as follows. Let H ≤G be subgroup of some finite index n in any group G. If θ : H→A is any homomorphism from H to an abelian group A, one defines the transfer of θ as follows. Write . en, for any g ∈ G, the cosets ore permuted by the right multiplication, that is, Htig=Ht(i)g. Here, i→ (i)g is the permutation defined by g. One defines . Here, we have written the action of θ as a superscript. e main case where this is useful when A=Hab and θ : H→Hab is the natural map. e following few problems give some important results which follow by making use of the transfer when H itself is abelian. Let H ≤ G and θ be as obove. It is easy to check that θ* is a well-defined homomorphism from G which does not depend on the choice of the coset representatives. Further, if the orbits of g ∈ G are , then

is is easy to see using the coset representotlves sigj; i ≤ k,j ≤ li– 1.

PROBLEM N-80 Show that in any group G, if the centre has finite index n, then [G,G] is finite and satisfies ={1}.

ASIDE    e above result is due to I. Schur and hos been generalised by R. Baer and others in a number of ways.

PROBLEM N-81 Let G be any group in which each conjugacy class is finite. Prove that [G,G] is a torsion group. Further, if G is also finitely generated, prove that [G,G] is finite.

PROBLEM N-82 Let G be a finite group and suppose H ≤ G be a p-Sylow subgroup such that H ≤ Z(NG(H)). en, prove that there exists N ⊴ G such that G = NH and N ⋂ H = {1}. ASIDE    is is due to Burnside and is sometimes referred to as the Burnside p-complement theorem, Burnslde p-complement theorem, lt also has generalisations. Two of them are the Schur-Zossenhaus theorems which are dealt with in the following two problems.

PROBLEM N-83 Let G be a finite group with an abelian normal subgroup P. Suppose that the order n of P and its index m are coprime (i) If, for each element x∈G/P, s(x) is any coset representative, show that the elements f(x,y):=s(xy)–1s(x)s(y) ∈ P and satisfy the 'cocycle identity' –1

f(x, yz)f(y, z) = f(xy, z)f(x, y)s(z) . (ii) Let a be an integer such that ma=1mod n. Prove that, for y∈G/P, the element

is well defined and that the set of elements h(x)as(x)–1 forms a subgroup H such that G=PH and P⋂H={1}. (iii) If, in addition, P is contained in Z(G), then prove that G ≌ P × N.

PROBLEM N-84 Let G be a finite group with a normal subgroup N such that N and G/N have coprime orders. Show that there exists a subgroup L of order |G/N| in G. Further, deduce that LN = G and L ⋂ N = {1}.

PROBLEM N-85 Let G be a group such that the set S of all torsion elements of G is a finite set. en, prove that S is a group.

PROBLEM N-86 Let x1,...,xn be torsion elements in a group such that each conjugacy class G(xi),i≤n is finite. Prove that is finite.

PROBLEM N-87 Suppose that G is a nonabelian finite group G in which all subgroups are normal. (i) Prove that if x, y ∈ G do not commute, then H = < x, y > is a finite, nilpotent group with [H, H] ≤ Z(H). (ii) Prove that elements of finite, coprime orders commute. (iii) If x,y ∈ G are such that 0(x)+O(y) is minimal among noncommuting elements, then the group Q = < x, y> is isomorphic to the quaternion group of order 8. (iv) If Q is as in (iii), prove that G = CG(Q)Q is a torsion group and that CG(Q) ≌ E × where E is an elementary abelian 2- group and is an

abelian group in which all elements have odd order. Deduce that G must be a direct product of a quaternion group, an abelian group of odd order and an elementary abelian 2-group.

PROBLEM N-88 Let θ : G → M be a surjective homomorphism of finite groups. Prove that there exists a subgroup H of G such that θ(H)=M and the set of primes dividing |H| is the same as the set of primes dividing |M|.

PROBLEM N-89 (i) If G is a finite group in which there are elements x,y such that xnyn=gn for some g and some n coprime to |G|. Prove that xy and g are in the same coset of [G,G]. (ii) In any group G of odd order, the product of all elements (in any order) lies in [G,G]. (iii) Let G be a finite group whose 2-Sylow subgroups are cyclic. Let j ∈ G be any element of order 2. en, prove that the product of all the elements of G taken in any order is in the set j[G,G] and is nontrivial. Note that the special case G=(Z/p)>* and j = p – 1 gives us Wilson's theorem. ASIDE    e exercise (iii) is due to D.Broline generolislng a problem of S.V. Kanetkar which itself is a generalisation of wilson's theorem. Kanetkar asks for the proof that the product of all the elements of G in any order is nontrivial under the same hypotheses.

PROBLEM S-90 (i) Show that if N ⊴ G and both N, G/N are finitely generated, then G is finitely generated.

(ii) Show that a subgroup of finite index in a finitely generated group is aiso finitely generated. Hence, deduce that any finite group is finitely presentable. (iii) Prove that Sn (for n > 1) has a presentation

PROBLEM S-91 (i) Prove that the matrices

generate a group

isomorphic to F2. Here e is exponential and one may assume that e is not a root of a nonzero integral polynomial. (ii) Prove that Fm ≌ Fn if, and only if, m=n. (iii) Prove that in the free group F(x,y) on two generators x, y, the subgroup H generated by ynxy–n;n∈ Z is not finitely generated. Hence, show that [F2,F2] is not finitely generated. (iv) Prove that Fn/[Fn,Fn] ≌ Zn.

PROBLEM N-92 Let G = F(x, y) × Z. Consider the subgroups H = < (x, 0), (y, 0)> and K = < (x, 0), (y,1) > of G. Prove that H ⋂ K is not finitely generated. ASIDE    Problem 55 shows how the abelianisation of any group with a given presentation may be computed. In fact, oen the only way to know whether a group given by a presentation is infinite or even nontrivial is by showing that the abelianisation itself is infinite, nontrivial, etc.

PROBLEM S-93 (i) Let G = < X | R > be a finite presentation of a group. Show that the abelianisation Gab):= G/[G,G] has the presentation .

(ii) Let G = < X | R > be a finite presentation of a group where |X| > |R|. en, show that Gab (and therefore G itself) is infinite.

PROBLEM N-94 (i) Consider the matrices

in SL(2, Z). Prove

that they generate SL(2, Z). (ii) By considering the images of S and SX in PSL(2, Z), prove that PSL(2, Z) is isomorphic to the free product Z/2*Z/3. (iii) Show that SL(2,Z) =< x, y|x2y–3, x4 >. (iv) Find the abelianisation of SL(2, Z). ASIDE    Recall from problem 54 thot SL(n,Z) is a perfect group when n ≥ 3. is contrasts with the case n=2 above.

PROBLEM S-95 Let K be any field. Consider the subgroups

and

Prove that they are conjugate subgroups in SL(2, K) and generate the whole of SL(2, K).

PROBLEM N-96 Let K be any field with more than 2 elements.

(i) Prove that if K ≠F2, the group SL (2,K[t]) is generated by matrices of the form where f, g ∈ K[t], a∈ K*. (ii) Prove that SL(2, K) and SL(2,K[t]) are perfect groups. ASIDE    Notice that although the generation by 'elementary' matrices fallows for SL(2,K[t]) from the division algorithm for K[t], this group is perfect unlike SL(2,Z) whose commutator subgroup has index 12 as seen in Problem 94. e main reason here is that unlike Z, the polynomial ring contains a field.

PROBLEM N-97 (i) Compute the automorphism group of an elementary abelian p-group of order pr. Deduce that

(ii) For any ∈ > 0;, prove that there exists a prime p and an elementary abelian p-group G such that AutG has a subgroup A of order coprime to |G| and such that |A|>|G|2–∈). ASIDE    e identity in (i) wos observed and generalised by Peter Neumann to show (using other various deep things like the classification of finite simple groups) that the order of the outer outomorphism group of group of order divides the number . e above observation (ii) hos been used by P. Palfy & L.Pyber (again along with the classification of finite simple groups) to prove that for any nontrivial finite group G, and any subgroup A of AutG of order coprime to |G|, one has |A| 1, the cardinality a(C,n) is a multiple of (n|C|,|G|). (ii) If n = n1n2 with n1,n2 > 1 and (n1,n2)=1, then the truth of the assertion for n1 and for n2 implies its truth for n. (iii) If n= pr > 1 for a prime p dividing O(x) with x∈ Z(G), then the assertion holds for x and for this n. (iv) Let Z;={y ∈Z(G):(n,O(y))=1}. en the number a(y,n) of all g satisfying gn = y is the same for each y∈ Z.

(v) e class equation and the previous steps imply Frobenius's theorem.

PROBLEM N-100 For a finite group G and any real number α >0, define

Prove that the cyclic group C of order |G| satisfies aα(C) ≥ aα(G) if α ≤0 and, aα(C) ≤ aα(G) if α ≥ 0.

PROBLEM N-101 Suppose G is a nontrivial, finite group whose only characteristic subgroups are {1} and G itself. Prove that G must be a direct product G1 ×...×Gn of simple groups which are all isomorphic.

PROBLEM N-102 (i) Let G be any finite group and x, y ∈ G such that their conjugacy classes G(x) and G(y) have coprime cardinalities. en, prove : G = CG(x)CG(y), G(xy) = G(x)G(y). (ii) For any (possibly infinite) group G with finite conjugacy classes G(x) and G(y) of coprime cardinalities, prove the same assertion.

PROBLEM N-103 Let G be any group in which each conjugacy class has finitely many elements. Let m(G) be the maximum size. If N ⊴ G, prove that

PROBLEM N-104 (Generalisation of Sylow theorems for solvable groups :) Let G be a solvable group of order mn with (m,n)=1. en, (i) there are subgroups of order m; (ii) any two subgroups of order m are conjugate; (iii) any subgroup of order dividing m is contained in a subgroup of order m.

PROBLEM S-105 Let π : G → F be a surjective homomorphism of a group onto a free group of finite rank. Prove that there exists a subgroup H of G such that G = Hker(π)> and H ⋂ kerπ={1}. ASIDE        In the above problem, the some assertion can be mode without assuming thot π is surjective. To deduce that, one has to use the fact that the image of π, being a subgroup of a free group, is also free. If this image has infinite rank also, the proof is the some as above.

PROBLEM N-106 Prove that a finite group in which all nontrivial maximal subgroups are conjugate, must be a cyclic group of prime-power order.

PROBLEM N-107 Let G be any finite group. Consider the function s2(g) which counts the number of 'square roots' of g in G. en, prove that

where c is the number of conjugacy classes G(x) for which G(x)=G(x–1).

ASIDE        e above assertion has been conjecturally generalised by R. Higgins & D. Ballew. eir conjecture asserts that for any n, k, there is a constant c(n,k) so thot the correspanding sum . e result below is due to R. Freese and generalises a problem posed by M. Marcus.

PROBLEM N-108 For any σ ∈ Sn, denote by c(σ), the number of cycles (including the cycles of order 1); this is just the number of orbits of {1,2... , n} under . Prove the two polynomial identities

Hence, for all N ≥ n conclude that

PROBLEM N-109 Let G be a finite group and α : G → Sn be a homomorphism. For any positive integer a, consider a finite set A with a elements and regard San as the symmetric group on the an elements which are all the n-tuples (a1,..., an) with ai ∈ A. en, consider the homomorphism β : G → San defined by β(g):(a1,...,an) → (aα(g)(1),...,(aα(g)(n)). Given any subgroup H of G, use the fallowing steps to prove that

Here c(K) denotes the number of orbits of K, and the Moebius function μH corresponding to H is defined on subgroups containing H by μH(H)=1 and for any subgroup L ⊃ H. What does the above result give in the special case when G is cyclic of prime power order pr and H is trivial? (i) For H ≤ G, call (a1,... ,an) exactly H-invariant if g(a1,..., an) = (a1,... ,an) for some g ∈ G if and only if g∈ H. Show that a subset S of An) admits a Gbijection with the set G/H of le cosets (that is, there exists a bijection α : S → G/H with α(g.s)=g.α(s) for all g ∈ G) if and only if S = G·(a1,...,an), and (a1,...,an) is exactly H-invariant. (ii) If e(H) denotes the number of exactly H-invariant elements in An, show that the number is a positive integer. (iii) Use these to compute the number of Gem>-orbits. ASIDE    is group-theoretlc generalisation of Fermat's little theorem is due to Strunkov.

PROBLEM N-110 Let G be any group in which one has the cancellation property xn = yn ⇒ x = y for any n ≥ 1 and x, y ∈ G. If Aut (G) is a torsion group, prove that G is abelian. Further, if G is also finitely generated, show that G ≌ Z. ASIDE    e above result is due to Colin D. Fox. e observation below due to M. Auslander can be interpreted in terms of group cohomology.

PROBLEM N-111 Let G be a group and θ ∈ Aut(G) be so that θn= Int (g) for some g ∈ G and some n ≥ 1. If the fixed point set {x ∈ G :x (θ)=x} is contained in the centre Z(G), prove that θn2 = Id. Further, in this case show that θn=Id if and only if g–1g(θ)=x–1 x(θ) for some x ∈ Z(G).

PROBLEM N-112 Let H, K ≤G be subgroups of the same finite index. en, prove that there exist g1,... ,gnsuch that

ASIDE    is wos proved by O.Ore. Even the cose H=K is interesting.

PROBLEM N-113 Use the following steps to prove the fact that a permutation cannot be simultaneously a product of an odd number of transpositions as well as an even number of transpositions. (i) If there exists such a permutation, then the identity has an expression as a product of an odd number of transpositions. (ii) Let k be the least odd number so that Id is a product of k transpositions as in (i). Let Id =(a,b)... be an expression as a product of k transpositions such that a occurs the least number of times among all such expressions of length k. Derive a contradiction by obtaining an expression Id =(a,c)...... as a product of an odd number ≤ k of transpositions where a occurs a fewer number of times.

PROBLEM N-114 (i) Let G be a finite group, let H be a subgroup normalized by an element g ∈ G. Assume that any two distinct elements x, y of H are in distinct le cosets of CG(g). en, prove that every element of H can be expressed as [g,x] where x ∈ H (ii) Let n ≥ 2 and consider any finite field Fq. Let U(n,q) denote the group of all upper triangular matrices with 1's on the diagonal and entries from Fq. Use (i) to prove that any h ∈U(n,q) can be expressed as [g,x] where g is a diagonal matrix with distinct entries from .

ASIDE        e fallowing set of four problems leod to beautiful elementary proof of Frobenius's theorem due to R. Brauer.

PROBLEM N-115 Let N be a finite normal subgroup of a group G. If g ∈G, x (gx)|N| and g|N| are conjugate by an element of N.

∈ N, prove that

PROBLEM N-116 Let H ≤ G be a finite subgroup of a group G. (i) For x, y ∈ G, show that xnH = ynH for all integers n if and only if yxnH = xn+1 H for all integers n. Further, show that

is a subgroup of H normalized by x. (ii) If xn H = yn H for all integers n, show that x|H| and y|H| are conjugate in G.

PROBLEM N-117 Let H ≤G be a finite subgroup of a group G. Call x, y ∈ G equivalent with respect to H if there exists h ∈ H such that x?nH=hynH for all integers n. (i) Show that any y ∈ G which is equivalent to x with respect to H can be uniquely expressed as hxzh–1 where z ∈ Hx, and where h runs through a set of right coset representatives of Hx in H. (ii) Prove that, for any x∈G, there are exactly |H| elements y which are equivalent to x with respect to H. Further, show for such x, y that x|H| and y|H| are conjugate.

PROBLEM N-118

Use the previous problem to prove Frobenius's theorem: In a finite group G, if C is a conjugacy class and n||G|, the number of elements g such that gn ∈ C is a multiple of n. ASIDE    e following results due to I. M. Isaacs and G. R. Robinson give another proof of Frobenius's theorem without recourse to representation theory.

PROBLEM N-119 Let G be a finite group and, for any natural number n, denote by fn(G) the cardinality of the set {x ∈ G : xn = 1}. en, prove: (i) e number of elements of order n is a multiple of ϕ(n). In particular, for a prime p such that pk|||G|, one has fpk (G) ≡ 0 fpr (G) ≡ 0 mod pr for all r < k. More particularly, if fpk(G) ≡ 0 mod pk, then fpr (G) ≡ 0 mod pr for all r < k. (ii) Let pk||n and let Q be a set of representatives of conjugacy classes of k

elements y such that y|G|/p = 1. en, fn(G) =

.

(iii) If pk|||G|, then fp(k ≡ 0 mod pk. (iv) If n/ |G|, then fn (G) ≡ 0 mod n.

PROBLEM N-120 (i) Let G be a finite group which has a normal subgroup N whose order is the smallest prime p dividing O(G). Prove that N ≤ Z(G), the centre of G (ii) Use (i) to prove that a finite group G, with the property that for each normal subgroup N of G, the quotient G/N has normal subgroups of all orders dividing O(G/N), must be nilpotent.

Solutions

SOLUTION 1 (i) If one of G, H (say G is finite and the other (say H is infinite, and if G×H is cyclic, then there is an isomorphism θ : G×H ≅ Z. But then, for two different elements a, b of G, we have |G|θa,0) = θ|G|a,0) = θ0,0) = 0, which means that θa,0) = 0 and, |G|θb,0) = θ|G|b,0) = θ0,0) = 0, which means that θb,0) = 0. us, θ cannot be an isomorphism. If both G, H are infinite, and if G × H is cyclic, generated by some (g,h, consider the isomorphism

So, Image (θ = {(gn,hn : n ∈ Z}. So, if r ≠ s are integers, then clearly we cannot have (gr, hs ∈ Image(θ. Now, if both G = < g > , H = < h > are finite, of orders m, n respectively, G × H has order mn. Consider the isomorphism

So, θ( ) = (gr, hr. en, clearly the LCM [m,n] of m and n is in kerθ. As θ is an isomorphism, this means that in Z/n. us, mn = [m,n]. Hence (m,n = 1 if G × H is to be cyclic. Finally, conversely, if (m,n = 1 and G = < g >, H = < h > are cyclic groups of orders m, n, then the map

is 1–1 since (gr, hr = (1,1) implies that mn|r since (m,n = 1. Being an 1-1 map of finite sets, θ must be onto also. (ii) If G = < g > is of order n, a typical element gr has order as follows.

. e reason is

Now, if d|n, clearly < gn/d > is a subgroup of order d. If two subgroups < gr > and < gs > have the same order, then (n,r = (n,s. Since , there is some integer u such that modulo . Multiplying by (n,r = s (n,s, we get ru ≡ s modulo n. In other words, < g > ≤ < gr >. Since they have the same order, they are equal. Let us prove the converse. Suppose G is a finite group which has a unique subgroup of each order d||G|. Call |G| = n. Now, we claim that G satisfies the hypothesis in (iii) and then we will give a common argument. For each d|n, now . If we show that there are at most ϕ(r) elements of each order r|d, then the le-hand side has cardinality at the most . So, let r|d and suppose x has order r (if there are no such elements, we are through). Now, the cyclic group < x > is the unique subgroup of order r. As before, in this group, there are exactly ϕ(r) elements of order r. If there were any elements y ∉ < x > which have order r, then we would have two distinct subgroups < y > and < x > of order r which

contradicts our hypothesis. us, the only elements of order r are all in < x > and are ϕ(r) in number. is proves our claimed assertion. (iii) Let |G| = n and d|n. Consider Nd = # {g ∈ G : Og = d}. If N(d) ≠ 0, look at some element g with Og = d. As e, g, g2, … gd–1 are distinct and are solutions of xd = e, these are all the solutions of the equation xd = e. As elements of order d in G are among these and are ϕ(d) in number, we have proved that Nd = ϕ(d) if Nd ≠ 0. As every element of G has some order d dividing n, we have . Since , we must have the equality Nd = ϕd for all d|n. In particular, Nn = ϕn ≠ 0.

SOLUTION 2 (i) & (ii) ese are simply a re-statement of Lagrange's theorem for the groups (Z/p* and (Z/n*. (iii) In the group (Z/(an–1))*, the integer a has order n. Once again, Lagrange's theorem implies the divisibility property asserted. (iv) In the product, each element cancels with its inverse except for those elements which are their inverses. e only elements of (Z/p* which are their own inverses are 1 and p–1 because if i2 ≡ 1 mod p, then p|(i–1) or p| (i+1); so the resultant product is p–1. (v) Let θ : (Z/p2)* → (Z/p* be the canonical homomorphism. Considering a as an element of (Z/p2)*, it follows that an ∈ kerθ. Clearly, kerθ has order p and an is a nontrivial element of (Z/p2)* since p2 by hypothesis. erefore, an has order p in (Z/p2)*. is means p divides the order of a in (Z/p2)* and, thus ap–1 ≢ 1 modulo p2.

SOLUTION 3

(i) Evidently, |S| = np–1. For each tuple (g1, … , gp in S, there are exactly p distinct tuples (g2, … , gp, g1) (g3, … , gp, g1, g2) etc. in S unless g1 = g2 = … = gp (here is where we use the fact that p is prime). Note that g1 = g2 = … = gp if and only if = e. us, we have that |S| ≡ # {g : gp = e}mod p. (ii) If p|n, then p divides np–1 = |S| = # {g : gp = e}mod p. In this case (since ep = e, there are at least p – 1 elements of order p in G. is proves Cauchy's theorem. (iii) If n, then one has gp = e for some g if and only if e = g(n,p = g. us, |S| ≡ 1 mod p. is proves Fermat's little theorem.

SOLUTION 4 (i) Let F = Z(D) be the centre of D. Note that F ⊇ Fp, the field of p elements. If G ≤ D* is a finite subgroup, then the set

is a finite-dimensional Fp–vector space. So, it is a finite division algebra, and therefore, by a celebrated theorem of Wedderburn, it must be a field K and G ≤ K*. erefore, by (i), G must be cyclic. (ii) e quaternion group Q = {±1,±i, ±j, ±k} is a non-cyclic subgroup of H*. (iii) If

given by

, then consider the homomorphism

.

Here, of course, we have used to denote elements in different groups. Clearly, θ is 1–1 and, therefore, onto as well as the sets are finite. is is nothing but the usual Chinese remainder theorem.

Now, let us consider the restriction of θ above to the subset (Z/n)* consisting of all with r ≤ n and (r,n = 1. is subset is a subgroup under multiplication modulo n. Its image under the above θ maps into the subset

Note that θ restricted to this subset is actually a group homomorphism under multiplication modulo n. Once again, this is 1–1 as it has trivial kernel because a ≡ 1 mod n if a ≡ 1 mod for all i ≤ r. Hence, being finite a map of finite sets, it is onto as well and thus,

By problem 1 (i), it suffices to check that (Z/pa* for a prime p is cyclic if and only if p is odd or p = 2 and a = 1,2. Of course, (Z/pa * has order pa–1(p–1). Let p be odd first. It suffices to produce elements g and h of the coprime orders pa–1 and p–1 respectively, since gh would then have order pa–1(p–1). By the binomial theorem, if a ≥ 2 and p > 2,

In other words, the element

has order pa–1 if p is odd and a > 1.

We shall now show the existence of an element of order p– 1 in (Z/pa* for p odd and a ≥ 1. First, in the field Z/p of p elements, every nonzero polynomial has at the most its degree number of roots by the remainder theorem. us, for each d, there are at most d elements x of (Z/p* satisfying xd = 1. By what we have proved earlier, the group (Z/p* must be cyclic. If is a generator, then the corresponding element rp–1 in (Z/pa* has order some power pk of p.

erefore, in (Z/pa*, the element rpk has order p – 1. us, the problem is solved when p is odd. For p = 2, a > 2, we see by the binomial theorem that

Hence, if a ≥ 2, the group (Z/2a)* of order 2a–1 is generated by the element – 1 of order 2 and the element 5 of order 2a–2. Now, if –1 were to be a power of 5, the orders would force –1 = 52a–3 in (Z/2a)*. en, we would get –1 ≡ 1 + 2a–1 modulo 2a which is impossible when a > 2. erefore, if a > 2, then (Z/2a)* is isomorphic to the direct product of the group of order 2 and the cyclic group of order 2a–2; so it is not cyclic. Of course, (Z/2)* is trivial and (Z/4)* is the cyclic group of order 2.

SOLUTION 5 Suppose, if possible, there is an isomorphism θ : Z[1/2] → Z[1/3]. If θ1) = with r, s ≥ 0 and a ∈ Z odd, then consider θ1/2r+1) = x say. Now 2r+1x = θ1) = is impossible as a is odd. us, there cannot be any such isomorphism.

SOLUTION 6 (i) Follows because one can get the pairs (i,j for j > (p – i, q – j.

modulo H as (–i, –j =

(ii) Using the isomorphism θ obtained from Chinese reminder theorem, we have the set

of coset representatives in (Z)/pq* for θ(H). Note that for k > , we have used the fact that (k mod p, k mod q = (pq – k mod p, pq – k mod q modulo θ(H). (iii) We obtain the expression for of S. We get

by multiplying all the elements

Now, let us multiply all the elements of T to get

Multiplying the k's in T modulo p, we have the expression

Simplifying this, we get

Similarly, modulo q, the product of elements of T gives

erefore, the product of elements of T maps to the element

For comparison of the two expressions obtained, let us simplify fallows.

as

erefore, we have that

.

is gives us the quadratic reciprocity law for p, q.

SOLUTION 7 We apply the Jordan-Holder theorem to the finite abelian group Z/n. Let {0} = G0 ≤ G1 ≤ … ≤ Gr = Z/n be a composition series for Z/n. en the successive factors Gi/Gi–1 are simple, abelian groups, and therefore, cyclic groups of prime orders. us, the order n of Z/n is the product of the orders of the various Gi/Gi–1. at is, n is a product of prime numbers and one can write where pj's are distinct and there are exactly aj successive factors Gi/Gi–1 of orders pj. Now, if is another factorisation of n into products of primes, then consider the normal series {0} = H0 ≤ H1 ≤ … ≤ Hs = Z/n, where the Hi's are defined as follows. H1, … , Hb1 are the subgroups of Z/n generated, respectively, by

.

Similarly, the next b2 of the Hi's are defined as the subgroups generated, respectively, by . All the Hi's are defined proceeding in this manner. As each Hi/Hi–1 has order p, this is a composition series. By the Jordan-Holder theorem, now comparing the orders of factors, we get uniqueness of prime decomposition up to order.

SOLUTION 8 (i) Let G act on the set X of le cosets of H. is gives θ : G → sym(X ≅ Sn. Take N = kerθ. Note that N ≤ H as N acts trivially on the identity coset and that G/N ≅ θG ≤ Sn. (ii)By (i), ∃N ≤ H ≤ G with N normal in G and [G : N]|p! Since [G : N]||G| as well, [G : N] divides the GCD of |G| and p! which is p since p is the smallest prime dividing |G|. But, H contains N, which implies p|[G : N]. us, [G : N] = p = [G : H]. (iii) Let us prove by induction that for any nonnegative integer n, one has xnyn ∈ N. By hypothesis, we have the assertion for n = 1. Assume that xnyn ∈ N for all 1 ≤ n ≤ r. en, xr+1yr+1 = xxryry = x(xryr)x–1xy ∈ N. us, for all natural numbers n, we have xnyn ∈ N. Now, x–ny–n = (ynxn–1 = (ynxnyny–n–1 ∈ N for all nonnegative integers n.

SOLUTION 9 e set A := {(g,s ∈ G × S : g.s = s}can be written as a union of subsets in two ways – one way is to gather together all g for a particular s and the other is to reverse the roles of g and s. As , we have a coincidence of cardinalities as asserted. erefore,

In the last sum, each orbit is counted its cardinality number of times, so, the right-hand side is the number of orbits.

SOLUTION 10

First, note that HK is a subgroup if and only if HK = KH. Now, if H is not maximal with respect to this property in G, then choose a subgroup H < G1 < G where G1 is maximal with respect to this property. Again, if H is not maximal with respect to this property in G1, then there exists H < G2 < G1 such that G2 is maximal with respect to this property in G1. Proceeding in this manner, since G is finite, we get a chain of subgroups H < Gr … G1 < G, where each group K in the chain has the property that for every subgroup L of the group Q on its immediate right, KL = LK and is maximal with respect to this property. We claim that whenever K ⊴ Q, this would prove that H is subnormal in G. If K ⋪ Q, then there is some conjugate qKq–1 ≠ K. Since KqKq–1) = (qKq–1)K > K and has the asserted property in Q, the maximality of K implies that KqKq–1) = (qKq–1)K = Q. So, if we write q = kʹk with kʹ ∈ qKq–1, we have qKq–1 = kʹk Kk–1(kʹ–1 = kʹKkʹ–1 so that K = (kʹ–1qKq–1kʹ = qkq–1, which is a contradiction.

SOLUTION 11 Write S = {1, … , p}. Suppose N ⊴ G. If s ∈ S and g ∈ G, then Ng.s = (Ng.s = (gN.s = gN.s. us, |N(g.s)| = |N.s|. As G acts transitively, this shows that all N–orbits have the same cardinality, say r. As the union of these orbits is S, r|p. If r = 1, then N fixes each point of S. is means N = {1}. So, if N is nontrivial, r = p; that is, N acts transitively on {1, … ,p}. Moreover, since the orbits under a nontrivial N has cardinality p, it follows that p||N|. But then by Cauchy's theorem, there is an element of order p. An element of order p in Sp has to be a p–cycle.

SOLUTION 12

(i) First, note that in the definition of ∑ one could have taken the < sign wherever ≤ appears; the reason is that x = y – z and x = 2y are impossible to hold in S. Now, it is clear that σ has the unique fixed point (1, 1, n). Now, we note the general fact that for any permutation τ of order 2 on a set, the non-fixed points can be paired off and thus the number of fixed points is of the same parity as |S|. Applying this to σ, we have that |S| must be odd. Turning this around and applying the above observation to the permutation τ : (x, y, z) → (x, z, y), it follows that τ must have an odd number of fixed points. erefore, τ does have at least one fixed point. Any fixed point of τ is a tuple (x, y, y) which means that p = x2 + 4y2. (ii) Evidently, α is an involution with the unique fixed point

.

(iii) Clearly, if no element (x, y, y) exists in S1, then the elements can be grouped in fours as (x, y, z), (x, z, y), (–x, y, z) and (–x, z, y).

SOLUTION 13 Let P1, … , Pn be the various p-Sylow subgroups of G. ey are all conjugates. Since any conjugate of any one of them, say P1, is a p-Sylow subgroup, one can identify the set S := {P1, … , Pn} with the conjugacy class of P1. Now, G acts on S by conjugation; this gives a homomorphism θ from G to Sn = sym(S. As ker θ is a normal subgroup and G is simple, we must

have either ker θ = {1} or kerθ = G. e latter conclusion implies that gP1g–1 = P1 for all g ∈ G; that is, G = NGP1). Hence P1 ⊴ G. Again, by simplicity, P1 = G. But then S = {P1} i.e., n = 1, which is a contradiction of the hypothesis. us, we must have that θ is 1–1. us, θG ≅ G is simple. As An ⊲ Sn, we have that θG ∩ An must be either trivial or the whole of θG. In the former case, θG has order 1 or 2 as θG/(θG ∩ An ≅ Sn/An ≅ Z/2. However, G cannot have order > 2 since it has n > 1 p-Sylow subgroups. us, the latter conclusion holds and θG ≤ An which proves the result.

SOLUTION 14 Now |G| = pa and . Consider any A ∈ T. If A is the union of b of the Si's, then it is evidently fixed by the G-action. On the other hand, if A partially intersects the sets S1, … , Sr say, (that is, if ), then the G-orbit of A has at least pr elements of T. Note that unless A is the union of b of the Si's, it partially intersects at least two of the Si's and then its Gorbit has at least p2 elements in T. erefore, the cardinality of T is modulo p2, the number of A in T which are unions of b of the Si's. In other words, modulo p2.

SOLUTION 15 Let us compute the order of GL (n,Z/p first. e proof of this is quite similar to the proof that a finite field has cardinality, power of a prime. Since this is the group of all invertible linear transformations of the n-dimensional vector space over Z/p, its elements are in bijection with bases of the vector space. is bijection is simply by sending any matrix in this group to the various column vectors. Now, the first column vector can be an arbitrary nonzero vector; there are pn– 1 choices for it. e secand vector can be any vector not in the linear span (that is, not a scalar multiple) of the first one; this gives pn – p choices since there are p scalars. Similarly, by induction, one sees that the order is (pn –1)(pn – p(pn – p2)… (pn – pn–1). e total power of p dividing this number is 1+2+…+(n–1) = nn–1)/2. Now, |U| = pnn–1)/2 which can be seen by induction on n. us, U is a p-Sylow subgroup. erefore, by Sylow's secand theorem, P is conjugate to a subgroup of U. Now, an easy matrix calculation by first principles shows that a matrix A normalises this subgroup if, and only if, it is an upper triangular invertible matrix. Its order is (p–1)npnn–1)/2. Now, the number of p-Sylow

subgroups is, by the secand theorem of Sylow, the index of the normaliser of U. erefore, the number of p-Sylow subgroups is

SOLUTION 16 We are given a finite group G with one of the following two properties: (i) either that for each n||G|, there are at the most n elements g ∈ G satisfying gn = e, (ii) or that for each n||G|, there is a unique subgroup of order n. In particular, in both cases, every subgroup of G is the unique subgroup of that order. erefore, every subgroup is normal. Let P be the p-Sylow subgroup for some prime p dividing |G|. If |P| = pn, then consider the elements of P which are not generators; they have orders pr for some r < n. By the hypothesis, the number of nongenerators of P is ≤ 1+p+p2+ … +pn–1. is is evidently < pn and thus P has a generator. In other words, the Sylow subgroups are cyclic. Let and, let Pi = < gi > be the pi-Sylow subgroup of G. en, Pi, Pj commute elementwise, because if xi ∈ Pi, xj ∈ Pj, then [xi, xj] ∈ Pi ∩ Pj = {e}. erefore, the product g1 … gr must have order |G|.

SOLUTION 17 (i) Let n ∈ NGNGP). en, nPn–1 ≤ nNGPn–1 = NGP. erefore, both P and nPn–1 are p-Sylow subgroups of NGP. Applying the secand Sylow theorem to NGP, there exists an element x ∈ NGP such that nPn–1 = xPx–1. But then n–1x ∈ NGP which gives us that n ∈ NGP. (ii) Let g ∈ G. en, gPg–1 ≤ gNg–1 = N. So, again there are two p-Sylow subgroups P and gPg–1 of the same finite group N. Applying the secand Sylow theorem to N, we have an element n ∈ N such that gPg–l = nPn–1. So, n–1g ∈ NGP which proves that G = NNGP.

SOLUTION 18 Consider the action by le multiplication of G on the set Ω of all subsets of G with pr elements. Let us write |G| = pnd with n = r and p d. e cardinality of Ω is the binomial coefficient . It can be shown by elementary number theory that |Ω| = pn–rd mod pn–r+1 but as we observe below this aiso fallows from some group-theoretic counting. Break up Ω into disjoint orbits, say, , where Si ∈ Ω. We claim that for any S ∈ Ω, G = ∪g∈GgS. To see that this claim holds, take any S ∈ Ω and any s ∈ S. Now, 1 = s–1s∈ s–1S ⊂ G.S. Further, if g ∈ G, then g = g.1 ∈ g.S. us, the claim is true. Hence, in any orbit, the number of subsets is at least pn–rd and divides |G| = pnd. is means that exactly one of the two things happen: either an orbit has exactly pn–rd elements or it has a multiple of pn–r+1 number of elements. Note that |Ω|≢0 mod pn–r+1. Hence, not each orbit can have cardinality a multiple of pn–r+1. is already proves that there are orbits with exactly pn–rd elements. Note that, obviously, the orbit of a set S in Ω has exactly pn–rd elements if, and only if, the corresponding isotropy subgroup has order pr. Suppose the number of such orbits with exactly pn–rd elements is t, then |S| ≡ tpn–rd mod pn–r+1. Although, one can prove by elementary number theory that one could obtain this already from the above sentence as it is valid for any G of order pnd and applying this to the correspanding cyclic group, one gets t = 1. Hence, . us, t ≡ 1 mod p. We note that t is the number of subgroups of order pr be cause an oribit of cardinality pn–rd has exactly one subgroup of order pr in it. is proves the result.

SOLUTION 19 (i) We know that there exists a p-Sylow subgroup P containing H. Let P ≥ C1(P) ≥ C2(P) … ≥ Cr(P) = {1}

be the lower central series for P. But then P = PH ≥ C1(P)H ≥ Cr (P)H = H ≥ {1} is a chain with each term normal in the previous term since CiP/Ci+1(P is contained in the centre of P/Ci+1(P. By Jordan-Holder theorem, one may refine to get a composition series for P. In a composition series with successive factors of orders powers of a prime p, each such factor has to be of order p. us, there are subgroups of P of any order pl for |P| ≥ pl ≥ pm containing H. (ii) Note that if m = 0 (i.e., if H = {1}, the assertion is the above result of Frobenius. Also, if m = n, this is a trivial assertion. Let us first deal separately with the case n = m+1. Now, if G1, … ,Gr are the various subgroups of order pm+1 which contain H, then we know from (i) that r > 0. Now, as the index [Gi: H] = p, which is the smallest (indeed, the only) prime dividing |Gi| it follows that H ⊴ Gi. In other words, Gi ≤ NGH for each i. erefore, Gi/H ; i ≤ r constitute the list of subgroups of NGH/H which have order p. ey are ≡ 1 mod p in number by (i). us, r ≡ 1 mod p and the case n = m+1 is proved for every m. We shall prove the assertion for n = m+l;l ≥1 by induction on l. We have dealt with l = 1 above. Let us take n = m+d+1 where the assertion is assumed to hold for all n ≤ m+d. Let K1, … ,Kr be the list of subgroups of order pm+d which contain H. By the induction hypothesis, we know that r ≡ 1 mod p. If L1,… ,Ls is the list of subgroups of order pm+d+1 which contain H, then we know from (i) that s ≥ 1. We need to show that s ≡ 1 mod p. We form the 'containment' matrix A which has r rows and s columns and the (i,jth entry is 1 or 0 according as to whether Ki≤ Lj or not. Evidently, the sum Ri of the ith row is just the number of Lj's which contain Ki. Similarly, the sum Cj of the jth column gives the number of Ki's which are contained in Lj. Of course, then is the sum of all entries of A. We shall show

that, in fact, each Ri ≡ 1 mod p and that each Cj ≡ 1 mod p. is would prove that r ≡ s mod p and, since r ≡ 1 mod p by the induction hypothesis, we would be through. But, for each i, Ri is just the number of subgroups of order pm+d+1 containing the subgroup Ki of order pm+d. By the first case dealt with separately in the beginning, this number is ≡ 1 mod p. us, each Ri ≡ 1 mod p. Now, look at any Cj. is is the number of Ki's contained in Lj and, therefore, it is aiso the number of subgroups of order pm+d which contain H. is last description shows, by induction hypothesis, that Cj ≡ 1 mod p. us, we have proved that

SOLUTION 20 Let (i1 … in–2) and (j1 … jn–2) be two (n–2)-element subsets of {1,2, … ,n}. Write {1,2,… ,n} \ (i1… in–2) = (a b, {1,2,… ,n} \ (i1… in–2) = (c d. Let σ ∈ Sn map each ir to jr and σa = c, σb = d. Let τ ∈ Sn map each ir to jr and τa = d, τb = c. Clearly, σ–1τ = (c d which means exactly one of σ, τ is an even permutation. at does the job.

SOLUTION 21

(i) We shall show by induction on n that any element σ of An is a product of two n-cycles, say σ = αβ. As n-cycles are all conjugate in Sn, we will have then β = τα–1τ–1 so that σ = [α,τ].

For n = 2, An is trivial and the assertion is clear. Assume n > 2 and suppose that the assertion is known for all m < n. Let σ ∈ An. Write σ1) = t say. Consider τ = (s t 1) with s ≠ t,1, we have that στ fixes 1. Hence, we may view στ as a permutation of Sn–1 = Sym{2,… n}. Since στ is an even permutation, we have by induction hypothesis that στ = σ1τ2 where τ1 τ2 are (n–1)-cycles which fix 1. Now,

where σ1 = τ1(1 t and σ2 = (1 tτ2(1 t(1 s. Evidently, σ1, σ2 are n-cycles. is proves (i). For (ii), merely observe that (a1 … a2n+1) = (a1 … an+1) = (a1 an+2 … a2n+1). Finally (iii) follows from the observation that any element of any Sn can be written as a product of disjoint cycles and these can be re-written as (a11…a1,r1)(a21…a2,r2)…(ak1…ak,,rk = (a11……ak,rk(ak1…a21 a11). (iv) For c ∈ F*, consider θ = τ–c2,0στ1,–cστ1,1/cστ1,–c. Now, θc = 0, τ0) = c and θx = x for all x ≠ 0,c by straightforward calculation. erefore, θ represents the permutation (0 c). Since (a b) = (0 a)(0 b)(0 a) for a, b ∈ F*, all transpositions (x y) with x, y ∈ F are gotten and thus so is sym(F).

SOLUTION 22 (i) Now τ–iστi = (i+1 i+2) ∀ i = 0,1 … ,n–2 by induction on i. But (i i+2) = (i+1 i+2)(i i+1)(i+1 i+2), (i i+3) = (i+1 i+3)(i i+1)(i+1 i+3), etc. us, once again by induction along with the fact that transpositions generate Sn, imply that σ, τ generate Sn. For a prime p, all the p-cycles are powers of each other. us, any p-cycle and any transposition do generate Sp). (ii) We may assume that τ = (1 2 … n and that σ = (i j with i < j. We claim that σ and τ generate Sn if and only if (n, j–i = 1. First, let us assume that (n, j–i = 1. en, σ1 := τj–iστi–j = (j 2j–i. Here, the integer 2j–i is taken modulo n and this obvious convention will be fallowed elsewhere. en, σ2 := σσ1σ1–1 = (i 2j–i. Once again, we consider σ3 := τj–iσ2τi–j which turns out to be (j 3j–2i. In this manner, we define inductively σ2r–1 = τj–iτ2r–2τi–j, σ2r = σσ2r–1σ–1 which turn out inductively to be, respectively, (j (r+1)j–ri and (i (r+1)j–ri).

Since (n, j–i = 1, we may choose r such that rj–i ≡ 1 mod n. is gives, σ2r–1 = (j j+1). We have already seen in (i) that (j j+1) and (1 2 … n generate Sn. Conversely, suppose that (j–i, n = d > 1. We will show that each θ ∈ < σ, τ > satisfies θ (i ≡ θj modulo d. is will show that since d > 1, the transposition (1 2) is not in < σ, τ >. Indeed, it is clear by induction on r that

Hence clearly, θ (i ≡ θj modulo d.

SOLUTION 23 Let, if possible, Sn be identified isomorphically with a subgroup of An+1. Since the index , we must have that n is odd. Further, since A4 is seen to contain no subgroup of order 6 at all, n is also > 3. Now, the action of An+1 on the set of le cosets of Sn gives rise to a homomorphism An+1 → S(n+1)/2. As for any n ≥ 5, it fallows that the above homomorphism has a nontrivial kernel. is contradicts the fact that An is simple for n = 5.

SOLUTION 24 (i) Note that each An ≤ A(N) and, indeed, A(N) = An. Suppose {1} ≠ N ⊴ A(N). en, there exists some σ ≠ 1 in N ∩ An for some n. us, σ ∈ N ∩Ak for every k ≥ n since Ar ≤ Ar+1 etc. erefore, N ∩ Ak is a nontrivial normal subgroup of the simple group Ak for every k ≥ max (n, 5). erefore, Ak ≤ N for every k ≥ max (n,5). Since ∪k≥max(n,5)Ak = ∪r≥1Ar = AN) we have N = AN).

(ii) It suffices to prove this for Sn. But, it is easy to see that Sn ≤ An+2 for any n. (iii) Consider the matrix

. en, clearly M2 = –I and M ≠ I

and M has finite order. Since PSL(2, F := SL(2, F/{±I} is a simple group, any normal subgroup N of SL(2, F maps either to the trivial subgroup or the whole group PSL(2, F. So, any normal subgroup N of SL(2, F is either contained in the centre {±I} or N.{±I} = SL(2, F. Let N be the normal subgroup generated by the matrix M above. Clearly, it is not contained in the centre but contains the centre. us, by the above argument, N = SL(2, F. Note that any element of N is expressible as a finite product of conjugates of M, each of which has finite order.

SOLUTION 25 (i) Write g–1 = aga–1. If x ∈ G is arbitrary, then x ∈ GgGg = GgGg–1). erefore, x = bgb–1cg–1c–1 = b[g,b–1c]b–1 = [bgb–1, cb–1]. (ii) Let t ∈ C be different from 0,1, –1. Consider

.

Since we are dealing with the group PSL(2,C) = SL(2, C)/±I, we shall prove that Gg = Gg–1) and that GgGg ± I = SL(2, C). First, clearly,

us, it suffices to show the secand assertion. Now, any x ∈ SL(2 C) is conjugate to a lower trangular matrix by a matrix y ∈ SL2,C). is can be seen as fOllows. First, it is clear that one can get x to

lower triangular form by some matrix y ∈ SL(2, C). en, one can replace y by with 1/s2 = dety. Evidently SL(2, C) and does the job. Since t ≠ t–1, it follows that the conjugacy class Gg consists of all matrices in SL(2, C) whose eigenvalues are t and t–1. us, it suffices to show that for any lower triangular matrix there is z ∈ Gg such that xz has the eigenvalues t and t–1. If a = 1, then is

implies

. also

that

when

. If a ≠ ±1 and c ≠ 0, then we take

where

With this choice, it is easy to see that xz has trace t+t–1 and determinant 1 which means that the eigenvalues of xz are t and t–1. Finally, if a ≠ ±1 and c = 0, then we choose where

is will complete the proof.

SOLUTION 26 Any element of SU(2) looks like

with a, b,c, d ∈ R such that

a2+b2+c2+d2 = 1. is reminds us of Hamilton's quaternions. erefore, let us define the map

Note that the image is contained in the group H1 of "unit" quaternions; that is, a+ib+cj +dk for which a2+b2+c2+d2 = 1 or, equivalently, (a + ib + cj + dk–1 = a – bi – cj – dk. It is trivial to check that θ is a homomorphism. Also, clearly it is 1–1 as well as onto H1. erefore, SU(2) ≅ H1. Now, as the group of nonzero elements H* acts on the 3-dimensional real vector space V generated by i,j,k by conjugation, we may think of SU(2) as acting on this space. us, we have a homomorphism ρ : SU(2) → GL(V. To explicitly evaluate it, we use the isomorphism θ to view SU(2) as H1. For any , we may compute qiq–1, qjq–1 and qkq–1 where q = θg = a+bi+cj+dk. We arrive at the fallowing matrix with respect to the ordered basis {i,j,k} :

Now, since g–1 = is obtained by changing b, c, d to their negatives, it is clear that ρg–1 = ρgt; that is, ρg∈ O(3,R). Note that if g ∈kerρ, then the correspOnding quaternion q commutes with i,j, k and, hence, with the whole of H. In particular, this gives g ∈ ZSU(2)) = {±I}. Of course, it is clear that –I does indeed belong to kerρ.

e final assertion le is to show that the determinant of ρg is always 1 is proved by direct (but messy) calculation. A better way would be to use a little bit of topology which shows that SU(2) is connected, ρ on SU(2) and the determinant function on GL(3, R) are continuous, that the image of det ∘ρ is a connected subset of R contained in {±1} and containing I.

SOLUTION 27 Write G for GL(n,K. Fix any g ∈ G. For any b ∈ B, consider the matrix bg. Suppose the ith row of bg starts with ai zeroes. Note that for any b, and each i, the numbers ai are between and n–1 since bg is invertible. Let us choose b = b0 so that the sum a1 + … + an for b0g is maximum possible among the matrices bg. We claim that for any such choice of b, the numbers a1, … , an are distinct. Suppose it is not so; let i < j be so that ai = aj. If b1 ∈ B is the matrix which differs from the identity matrix only at the (i,jth place where it has some entry t, then b1b0g has all the ak's(k ≠ i the same as for b0g. Also, the first ai entries of the ith row of b1b0g are also unaffected and, therefore, remain zero. However, the next entry in the ith row is u+tv where u, v are the first nonzero entries of the ith and the jth rows of b0g respectively. Hence, choosing t = – u/v, we have a matrix b1b0 ∈ B for which a1 + … + an increases at least by 1. is is a contradiction of the choice of b0. So, we must have that ai are distinct. Since these are n integers between 0 and n–1, these are the n integers 0 to n–1 in some order. Multiplying by the permutation matrix w corresponding to the permutation that a1, … ,an is of the numbers 0,… , n–1, we get wb0g has ai = i – 1 for all 1 ≤ i ≤ n. In other words, wb0g ∈ B. is proves the decomposition asserted. To show that the double cosets are disjoint, we argue as follows. Suppose w1 = b1w2b2 for some b1, b2 ∈ B and permutation matrices w1, w2. So, b1 = w1b2–1w2–1. Now, for any matrix g ∈ G and any two permutation matrices w, wʹ, it is clear that

where σ,σ are the permutations correspanding to w,wʹ. erefore, in our case

which gives us that

As w1w2–1 is a permutation matrix while b1 is an upper triangular matrix, it follows that w1w2–1 is a diagonal matrix and, therefore, the identity matrix. is proves uniqueness of the decomposition.

SOLUTION 28 (i) For the moment, let p be any prime. Write A = I + pdB where B ∈ Mn, Z) and some entry of B is not a multiple of p. It suffices to show that A does not have prime order because one could apply this to a power of A. Suppose, if possible, A has order a prime q. First, we claim that q = p. For, I = Aq = I + qpdB modulo p2d gives pd divides all entries of B, a contradiction for any d ≥ 1. Moreover, even if q = p is odd, the same equality reads I = Ap = I + pd+1B modulo p2d+1 since pd ≥ 2d+1 and

≡ 0 mod p for 1 ≤ r < p.

us, once again we have a contradiction unless p = 2. (ii) Write A = I + pB, with B ∈ Mn,Z). e characteristic polynomial of A is χAt = det (tI – A = det (t – 1)I – pB. erefore, χAα = 0 if and only if . Now, if A has finite order, its eigenvalues are all roots of unity. Since |α| = 1, we have

us, all the roots of χBt have absolute values < 1. Since χBt is an integral polynomial whose top coefficient is 1, this implies that χBt = tn; that is, all the eigenvalues of B are zero. In other words, all eigenvalues of A are 1. As A has finite order, it must be the identity matrix. (iii) As seen in (ii), A must have order a power of 2 if it has finite order at all. First, suppose A2 = I, A ≠ I. Consider the correspanding homomorphism TA : Zn → Zn; v → vA where v is written as a row. en, A ≡ I mod 2 and A2 = I imply that TAx = x mod 2Zn and TA2 = I. erefore, Zn = {v : TAv = v}⊕{v : TAv = –v} which gives that A is expressible in the form M.diag(1,1,… ,1,–1,–1,…,–1).M–1 for some M ∈ GLn,Z. Finally, let |A| = 2r ≥ 4. We shall prove this is impossible. e identity A2 = I + 2d+1B modulo 22d gives that A2 ≡ I modulo 4. us, if C := A2r–1 = I + 2dB, we have d ≥ 2. en, I = C2 = I + 2d+1B + 22dB2 which implies B ≡ 0 mod 2d–1, a contradiction.

SOLUTION 29

e basic observation here is that for coprime integers a, b, the number a2 + b2 cannot have prime factors of the form 4n+3. us, since matrices in H can be written with a/c and b/c in their lowest terms. Since the determinant is 1, we have a2+b2 = c2 with (a,b) = 1 for, any common prime factor would also have to divide c. erefore, elements of H can have entries with denominators divisible only by 2 and by primes of the form 4n+1. As the permutation matrices are integral, they do not give rise to any additional denominators. Hence, the subgroup generated by H along with the permutation matrices consists of matrices of the form where B is an integral matrix and d is divisible only by 2 and by primes of the form 4n+1. erefore, matrices like are in G and are not in the subgroup generated by H along with the permutation matrices. One can similarly construct examples of elements in G with denominators divisible by arbitrary primes of the form 4n+3. us, the subgroup has infinite index.

SOLUTION 30 (i) If G is an abelian group with a minimal set {a1,…,an} of generators, it follows that G/ < {a2,…,an} is a nontrivial abelian group generated by a single element. Also, as quotients of divisible groups are divisible, it suffices to show that an abelian group generated by a single element is not divisible. is is clear because being infinite, such a group has a generating element a with ra ≠ 0 for any r. But then the element a cannot be divisible by any n ≥ 2. (ii) Suppose S ⊂Q be so that Q = < S >. For any 0 ≠ s ∈ S, we consider the subgroup A = < S \ {s} >. If a ∈ A is any nonzero element, then writing s = u/v and a = b/c, we have bvs = acu ∈ A. Write for some d ∈ A and some integer e. erefore, s = bv(d+es) = bvd + acu ∈ A since d,a ∈ A and b, v, c, u are integers. is means that

Q = A +Zs = A = < S \ {s} >. at is, S \ {s} is aiso a set of generators. (iii) Suppose, if possible, M < Q is a maximal subgroup. Let a/b ∉ M. By maximality,

Also, evidently M ≠ {0} and, let us take some 0 ≠ u/v ∈ M. Consider the rational number integer d, we get

since

writing

for some m ∈ M and

and as b, u, d, a, v are integers.

is contradiction proves that there are no maximal subgroups in Q.

SOLUTION 31 (i) Suppose G is an injective group, g ∈ G and let n be a positive integer. We need to show that g = nh for some h ∈ G. Now, θ : na ↦ ag is a homomorphism from the subgroup nZ of Z into G. By injectivity of G, there is a homomorphism : Z → G extending the above map. erefore, n (1) = (n) = g. Conversely, suppose G is a divisible group. Let A ≤ B be abelian groups and α : A → G be a homomorphism. We need to check that this extends to a homomorphism from B to G. Let us apply Zorn's lemma. Consider pairs (C,αC) where A ≤ C ≤ B and αC extends α. We define a partial order in the set S of such pairs by (C,αC) ≤ (D,αD) if C ⊆ D and αD extends αC. e set S is non-empty, since (A,α) is in S. Also, if (Ci,αCi); i ∈ I is a chain (that is, a totally ordered subset) in S, then take D := ∪ i∈ICi and define αD(a)

= αCi(a) if x ∈ Ci. Clearly, (D, αD)∈ S is an upper bound. us, by Zorn's lemma, there is a maximal element (M,αM) in S. We claim that M = B. If M ≠ B, take any b ∈ B,b ∉ M. If M ∩ Zb = (0), then M + Zb is a subgroup of B properly containing M and the map m + ab → αM(m) is a well-defined homomorphism. is contradicts the maximality of (M, αM). Hence, we must have M ∩ Zb ≠ 0. Let a be the smallest natural number such that ab ∈ M ∩ Zb. erefore, every element of M + Zb has a unique expression as m + db with 0 ≤ d < a. Also, since ab ∈ M, we have an element g = αM (ab). As G is divisible, one can write g = ah for some h ∈ G. Define β : M + Zb → G by m + db → αM(m) + dh ∀ 0 ≤ d < a. en β(m + db) + β(mʹ + dʹb) = αM(m) + dh + αM(mʹ) + dʹh = αM(m + mʹ) + (d + dʹ)h. If d + dʹ = qa + e with 0 ≤ e < a, we have β(m+db+mʹ+dʹb) = β(m+mʹ+qab+eb) = αM(m+mʹ+qab)+eh αM(m+mʹ)+qg+eh since αM(ab) = g be definition. erefore, the last expression on the right side is αM(m + mʹ) + qah + eh = αM(m + mʹ) + (d + dʹ)h. Hence, β is a homomorphism. us, (M + Zb,β) ∈ S and the resultant contradiction proves that M = B. Hence G is injective.

(ii) Let D ≤ G be a divisible subgroup of an abelian group G. We use the fact that D is injective. Look at the identity homomorphism from D to itself. By injectivity of D, this extends to a homomorphism θ from G to D. Note that this means that θ(d) = d for all d ∈ D. Look at E := kerθ. If g ∈ G, then θ(θ(g)) = θ(g) as θ(g) ∈ D. us, θ(θ(g) – g) = 0; that is, θ(g)– g ∈ kerθ = E. Hence, we have G = D + E. If d ∈ D ∩ E, then d = θ(d) = 0; so G = D ⊕ E. (iii) Any abelian group G is a quotient A/B of a free abelian group A. If S is a basis of A, then

and, therefore, D := ⊕s∈SQ is a divisible group containing a group isomorphic to A. As quotients of divisible groups are also clearly divisible, G is isomorphic to a subgroup of the divisible group D/B.

SOLUTION 32 Consider H := {g ∈ : S + g = S}; this is a subgroup which is properly contained in G by the hypothesis. Clearly, the mapping S + g → H + g from the set of all sets of the form S + x to the group G/H, is well-defined and onto. Since G/H is a nontrivial divisible group and, therefore, is infinite. erefore, the set of all sets of the form S+x is infinite as well.

SOLUTION 33 It suffices to check that if h is p-divisible, then there is a p-divisible g mapping to it. Now, write h = pnhn with hn ∈ H. So, if θ(gn) = hn, then θ(pngn) = h. erefore, pg1 – pngn ∈ kerθ for all n. As kerθ is finite, we deduce that infinitely many of the elements pngn give the same element g of G. us, the element g is p-divisible and θ(g) = h.

SOLUTION 34 First of all, the polar coordinates provide an isomorphism C* ≅ R>0 × S1 ≅ R×S1. So, it suffices to show that R×S1 ≅ S1. Start with any Q-vector space basis of R which contains 1. Already, we have used Zorn's lemma here when trying to extend the linearly independent set {1} to a basis. en, the set

is a basis of the Q-vector space R × R. Once again, it is a consequence of Zorn's lemma that and are in bijection. Let θ : → be a bijection which maps (0,1) to 1. en, there is a unique extension of θ to a Q-vector space isomorphism from R × R onto R. Since θ(0,1) = 1 and is a Q-vector space isomorphism, it gives an isomorphism of {0} × Z onto Z. erefore, θ gives (R × R) / ({0}× Z) ≅ R/Z. Evidently, the le hand side above is isomorphic to R/{0} × R/Z. is completes the proof, since t → e2iπt gives an isomorphism from R/Z onto S1.

SOLUTION 35 Computing θ(Z/p) for a prime p, gives us an idea of how to proceed. For this case, it is clearly p–1. Now, for a general n, we get by induction that θ(Z/n) = Φ(n). So, for a general group G, let us define the, a priori, a new function α(G) := Number of generators of G. Evidently, α(G) = 0 if G is not cyclic and α(Z/n) = Φ(n) = θ(Z/n). Now, each element g of G generates a unique cyclic subgroup < g > and the number of elements generating the particular cyclic group is α(< g >), we get

Note that we have used the fact that α(H) = 0 if H is not cyclic. Since α({1}) = 1, and since α is defined by the same recursion as θ, we have θ = α.

SOLUTION 36 Recall the notation first. [x, y] := xyx–1y–1 and xy = yxy–1. Also, x–y = yx–1y–1.

(ii) is is easy to verify by induction on r. Indeed, if [xr, y] = [x, y]xr–1+…+x+1, then

(iii) By putting u = aba–1ca,v = bcb–1ab,w = cac–1bc, we see that the identity reduces to uv–1vw–1wu–1 = 1. (iv) Now, we are in a group with [G,G] ≤ Z(G). So, (ii) gives [x, y]r = [xr, y] for any r; that is, xry = yxr[x, y]r. We now prove the assertion by induction on n.

(1)

using (1) to rewrite ynx in the above expression. Clearly, this is equal to xn+1yn+1[y,x]n(n+1)/2 which proves (iv). (v) Let xp,yp ∈ Gp. We want to show that xpy–p ∈ Gp; that is, that it is pth power. Now, if p is an odd prime, then p(p–1)/2 is a multiple of p and, so

as [G,G] ≤ Z(G).

SOLUTION 37 (i) In G/ < G2 >, every element satisfies x2 = e. But, clearly any group with this property is abelian (for, xy = x2x–1y–1y2 = x–1y–1 = (yx)–1 = (yx)–1(yx)2 = yx). Explicitly, aba–1b–1 = a2(a–1ba)2(a–1b–1)2. (ii) Let . en, the set {[x, y] : x, y ∈ G} of commutators is just the set {[gi,gj]; i,j ≤ n} which has at the most n2 elements. By the hypothesis, n2 < |[G,G]| which means that there are elements in [G,G] which are not commutators. (iii) For i ≠ j, let us write Xij for the elementary matrix which differs from the identity matrix only at the (i,j) th entry, which is 1 here. en, the given matrix is X13 = [X12, X23]. Suppose X13 = X2Y2 where

and

en, comparing entries

2a + 2aʹ = 0 = 2c + 2cʹ, 2b + 2bʹ + ac + aʹcʹ = 1 since 4 = 0 in Z/4. us, aʹ = –a or 2 – a and cʹ = –c or 2 – c, each of which gives the (1, 3)-entry of X2Y2 to be of the form 2d for some d. us, this can never be 1 and the resultant contradiction shows that X13 is not a product of two squares.

SOLUTION 38 Write for simplicity

en, clearly

m(0, yi, 0).

erefore, [G,G] = {m(0,0,h) ∈ Q[x, y]} which shows that [G,G] ≅ Q[x, y]. Now, for any n ≥ 1, let us consider the particular polynomial . We claim that m(0,0,h) is not a product of n commutators. If possible, let this be so. Write

View both sides as polynomiais in x and equate the correspanding coefficients of powers of x. If fjx = , then we have

Consequently, the linearly independent elements 1, y, … y2n+1 of the vector space Q[y] are all in the subspace generated by the 2n elements gj(y), vj(y). is is a manifest contradiction proving thereby that m(0,0,h) is in [G,G] but, for the h(x, y) above, it is not a product of n commutators.

SOLUTION 39 Let us note that the conjugacy class of any element g is finite as it is bijective with the set {xgx–1g–1 : x ∈ G} which is finite by hypothesis. erefore, CG(g) has finite index in G for each element g. Now, let g1,…,gn be generators for G. en, is of finite index.

SOLUTION 40 Let S = {g ∈ G : θ(g) = g–1}. Write |G| = n; we are given that |S| ≥ 3n/4. If x ∈ S, then, xS ∩ S = {xy : θ(y) = y–1,x–1y–1 = (xy)–1} = x(S ∩ CG(x)). erefore, |xS ∩ S| = |S ∩ CG(x)|. Note that for any g ∈ G, g ∉Z(G),|CG(g)| ≤ then

erefore, if x ∈ S, x ∉Z(G),

Hence, if we have strict inequality that |S| > 3n/4, then we would have a contradiction because then |xS\S| + |S| > n. is means that when |S| > 3n/4, each x ∈ S is in Z(G). us, |Z(G)| > 3n/4 which gives |Z(G)| = |G|; that is, G is abelian. If |S| = 3n/4, then, for each x ∈ S \ Z(G), we must have equality everywhere in (A).

In particular, CG(x) is contained in S and has order n/2 for such x. If S ⊂ Z(G), then |Z(G)| ≥ 3n/4 which means that Z(G) = G; that is, G is abelian. en, we would have S to be a group, i.e., S = G, which is a contradiction of our present hypothesis that |S| = 3n/4. us, there does exist an element x ∈ S\Z(G). As observed above, for this x, the centraliser CG(x) is of order n/2 and is contained in S. erefore, θ(y) = y–1 for all y ∈ CG(x). So, for y, z ∈ CG(x), (yz)-1 = θ(yz) = θ(y)θ(z) = y-1z-1. In other words, CG(x) is abelian. is completes the proof.

SOLUTION 41 First, note that [G : Z(G)] ≥ 4 because if G/Z(G) is cyclic, then G must be abelian. Indeed, if gZ(G) generates G/Z(G), then any x ∈ G is of the form gizi for some zi ∈ Z(G). ese clearly commute. us, we have . Now,

We have written G(x) for the conjugacy class of x. As |G(x)| ≥ 2 for each x ∈ Z(G), we get

SOLUTION 42

(i) For any g ∈ G, the inner automorphism Int g is an automorphism of N, as N ⊲ G. As C is characteristic in N, such an automorphism leaves C stable; that is, gCg–1 = C for all g ∈ G. (ii) Let A ⊲ B ⊲ C be such that A is characteristic in B and B is characteristic in C. Any σ ∈ AutC gives an automorphism σB ∈ AutB simply by restriction, as B is characteristic in C. But, then σ(C) = σB(C) = C as C is characteristic in B. (iii) In the abelian group Z × Z, all subgroups are normal but the subgroup Z × {0} is not characteristic. e automorphism which takes any (a, b) to (b, a) does not leave this subgroup stable.

SOLUTION 43 Let

. en, since

Take g2 have orders

we may write

for some integers a, b.

. Of course, g1 g2 = g2g1 = g. We shall check that g1, and n1 respectively.

Now,

. , then we get gbn1d = 1; so O(g)|bn1d. is gives

Moreover, if

however (p1, b) = 1 as seen from . Similarly, .

. us,

so that n1|O(g2). If

which proves that

, then

so that

Hence n1|l as (n1, a) = 1. is implies that O(g2) = n1. Now, we may proceed by induction with g replaced by g2. Ultimately, we will have elements xi, i ≤ r which are powers of g and have orders such that g = x1 … xr.

Finally, we show that such an expression is unique. If g = y1…yr is another expression where yi has order and commute pairwise, then they commute with g. As xi's are powers of g, the xi's and the yj's all commute pairrwise. So,

has order dividing as seen from the le side and at the same time has order dividing as seen from the right side. us, x1 = y1. Proceeding by induction, we get xi = yi for all i.

SOLUTION 44 We shall use the Jordan-Holder theorem. Let {1} = M0 ⊲ M1… ⊲ Mr = M be a composition series for M. Since we are given that M ≅ G/N, we have subgroups G1,…, Gr = G such that Mi ≅ Gi/N under the above isomorphism. So, Gi ⊲ Gi+1 for each i and aiso N ⊲ G1. erefore, {1} = G0 ⊲ N ⊲ G1 ⊲ G2 … ⊲ Gr = G is a normal series for G. We claim that it is a composition series. We need to check that the successive factors are simple. N is given to be simple. Also, G1/N ≅ M1 is simple. For i > 1, is simple. Hence, the above normal series is, indeed, a composition series for G. However, we have a normal series for G, viz. {1} = M0 ⊲ M1 … ⊲ Mr = M ⊲ G. us, by the Jordan-Holder theorem, this admits a refinement which is a composition series. Looking at the length of the composition series of G

written above, it fallows that the last normal series is, indeed, a composition series. But its successive factors are Mi/Mi–1 for 1 ≤ i ≤ r and G/M. Since the earlier composition series has successive factors N, G1/N, Gi/Gi–1 for 1 < i = r, and since Mi/Mi–1 ≅ Gi/Gi–1 for 1 < i ≤ r and M1/M0 = M1 ≅ G1/N, the only possibility le is that G/M ≅ N.

SOLUTION 45 (i) For any x, y ∈ G, we have xyθ(x)θ(y) = xθ(x)yθ(y) so that yθ(x) = θ(x)y which means that θ(x) is in Z(G) for any x. Using the fact that g ↦ gβ(g) is a homomorphism where β(g) = gθ(g), we get that β(x) = xθ(x) is in Z(G). Hence x ∈ Z(G) for any x. (ii) Put α(g) = g2θ(g); then for any x, y, xyxyθ(x)θ(y) = α(xy) = θ(x)θ(y) = xxθ(x)yyθ(y). at is, yxyθ(x) = xθ(x)y2∀ xy.

(1)

Notice that, in particular, for x = y, we get x2θ(x) = θ(x)x2 ∀x. Now g ↦ g4θ(g) = g2α(g) is a homomorphism. is gives us yxyα(x) = xθ(x)y2 ∀ x, y. at is, yxyx2θ(x) = x3θ(x)y2.

(2)

Cancelling off yxy from the le-hand sides of (1) and (2), we get θ(x)-1 x2θ(x) = y-2θ(x)-1x2θ(x)y2. since we observed that g2 and θ(g) commute. us, the above equality reduces to x2 = y–2x2y2. at is, squares of elements commute. is gives immediately from (1) that xy = yx for all x, y.

SOLUTION 46 bm

Write al + = 1 for some integers a, b. As any g ∈ G is expressible as g = (ga)l (gb)m, it suffices to prove that xl ym = ym xl for all x, y ∈ G. Now,

Similarly, bm

(xlym)

= (ymxl)bm.

Multiplying the two equalities, we get the asserted equality.

SOLUTION 47 For x, y ∈ G, we have xyθ(x)θ(y) = θ(x)θ(y)xy which gives

yθ(x)θ(y)y–1 = x–1θ(x)θ(y)x. us, yθ(x)y–1 = θ(x)x–1θ(y)xθ(y)–1 = θ(x)[x–1, θ(y)].

(1)

Applying (1) to x–1, we have yθ(x)–1y–1 = θ(x)–1 [x,θ(y)]. Taking inverses, we have yθ(x)y–1 = [θ(y),x]θ(x).

(2)

From the right-hand sides of (1) and (2), we have that x–1θ(x) commutes with θ(y)xθ(y)–1. Hence x–1θ(x) commutes with yθ(x)y–1 also. As this holds for all y ∈ G, it fallows that x–1θ(x) commutes elementwise with the normal subgroup N(x) := < x, θ(x) >N by x and θ(x). erefore, x–1θ(x) ∈ Z(N(x)) which is an abelian normal subgroup of G. As Z(N(x)) is trivial by hypothesis, it fallows that θ = Id.

SOLUTION 48 (i) If G/N is a finite, non-cyclic group, then it is the union of its proper cyclic subgroups. e inverse images of these in G are proper subgroups and cover G. (ii) Step I: Whenever a group the Hi's is of finite index.

, then we claim that at least one of

We prove this by induction on the number r of different Hi's. If r = 1, then H1 = … = Hn so that which gives [G : H1] = ≤ n.

Suppose now that r >1 and that the assertion of step I holds when the number of different Hi's is less than r. Renaming, we may assume that H1 = H2 = … = Hk and that this is different from each of Hk+1,…, Hn. e number of different subgroups among Hk+1,…, Hn is r–1. Now,

If words,

, then [G : H1] = ≤ k. erefore, we suppose there is some . us, evidently H1g does not intersect . In other

Hence, for each j ≤ k, we have

erefore,

As the number of different subgroups occurring is r–1, we have [G : Hi] < ∞ for some i between k+1 and n. erefore, by induction on r, we have step 1. Now, we proceed with the proof. Without loss of generality, we may assume that with some Hi's having infinite index; otherwise, we are through. Rename so that H1,…, Hr have infinite index and the others — Hr+1,…, Hn have finite index.

Note that if , then [G : H0] < ∞. Also, each right coset of any of Hr+1,…, Hn is a finite union of right cosets of H0. So, we may write

for some elements xj of G. If

, then we may restore the right cosets of the subgroups Hr+1,…, Hn and the assertion is proved. Suppose not. en, once again, we argue as in step 1; namely, choose some x ∈ G such that . en, . In other words,

Hence,

for all j ≤ s. us,

is is a contradiction of step I since each Hi for i ≤ r has infinite index in G. is completes the proof. (iii) By (i), we need only check that if G is a union of finitely many proper subgroups, it has a finite, non-cyclic group as a quotient. Let where Hi are proper subgroups. By (ii), we may assume that each Hi has finite index in G. But then is of finite index and, therefore, contains a normal subgroup N of finite index in G. As , and Hi/N are proper subgroups of the finite group G/N, G/N cannot be cyclic. (iv) By (ii), if for proper normal subgroups Hi, then we may assume that each Hi is of finite index. Further, we may assume that the decomposition is irredundant, i.e., we cannot drop any Hi. Consider G/H1. en, the images of Hi, i ≥ 2 are proper normal subgroups whose union is

the whole of G/H1. So, if we prove the assertion for finite groups, it would fallow that G/H1 admits of a quotient isomorphic to Z/p×Z/p for some prime p. us, this would be true for G itself. Hence, we have reduced the proof to finite groups G. Now, we may assume that Hi are maximal normal subgroups. Again, since the quotient group G/(∩iHi) admits a covering by proper normal subgroups (the images of Hi's), it suffices to prove that G/(∩iHi) admits of a quotient isomorphic to Z/p×Z/p for some prime p. us, we may assume that ∩iHi is trivial. Let I ⊂ {1,2,…,n} be a maximal subset such that ∩i∈IHi ≠ {1}. Renaming, we may assume that I = {1,2,…,k} for some k < n. We first observe that k ≠ n – 1. Otherwise, we may choose some x ∈ Hn which is not in ∪i 1 and that the theorem holds for m < n. Correspanding to any basis of G, there is a positive integer with the property that it is the smallest positive integer that occurs as a coefficient in the expression of elements of H in terms of this basis. is positive integer can depend on the basis and let l1 be the smallest such with respect to all

bases of G. Let v1,…, vn be a correspanding basis for G such that . Dividing all the ai by l1, we have ai = qil1 + ri with 0 ≤ ri < l1. Evidently, and is another basis of G. By the minimality of l1, we must have ri = 0 for all i ≥ 2. us, writing w1 for . Look at the subset H0 of H which have coefficients of w1 to be zero in terms of the basis w1, v2,…, vn of G. Clearly, H0 is a subgroup of H such that H0 ∩ Zv = {0}. Also, if h ∈ H, write . Once again, dividing the bi's by l1, say, bi = mil1 + si with 0 ≤ si < l1, we have . us, by the minimality of l1 we get s1 = 0 i.e., h – m1v ∈ H0. us, H = H0 ⊕ Zv. Now, H0 is contained in the subgroup . By induction hypothesis, G0 has a basis w2,…, wn and there exists r ≤ n such that H0 has a basis of the form d2w2,…, drwr with d2|d3|…|dn. Clearly, therefore, H itself has rank r ≤ n and l1w1, d2w2,…, dnwn is a basis for H. We have only to show that l1|d2. Once again, writing d2 = cl1 + d with 0 ≤ d < l1, we notice l1w1 + d2w2 = l1(w1 + cw2) + dw2sub> ∈ H where w1 + cw2, w2,…, wn is a basis of G. us, minimality of l1 forces d = 0 i.e., l1|d2. e proof is complete.

SOLUTION 53 Suppose the matrix statement holds. Let H be a subgroup of a free abelian group G of rank n. en, H is aiso free abelian of rank m ≤ n (this we are assuming known through other arguments). Let α : Zm → H and β : G → Zn be isomorphisms. If i : H ≤ G denotes the inclusion map, we have that the composite map β ∘ i ∘ α correspands to a matrix A ∈ Mn,m (Z) with respect to the canonical ordered bases of Zm and Zn. e matrix statement gives us P ∈ GL (n, Z) and Q ∈ rmGL (m, Z) such that

where di|di+1. Hence, the bases {v1,…,vn} = {Pe1,…,Pensub>} of Zn and {w1,…,wm} = {Qe1,…,Qem} of Zm are so that {β–1 (v1),…,β–1 (vn)} is a basis for G and {α(w1),…,α(wm)} is a basis for H. Now, note that the matrix identity above implies that AQ (ei) = P (diei) where ei on the le side are in Zm and those on the right side are in Zn. at is, βα(Qei) = diP(ei). So, we have βα(wi) = divi, which means that the bases {β–1 (v1),…,β– (vn)} of G and {α(w1),…,α(wm)} of H are as asserted in the invariant factor theorem. Conversely, let us assume that the invariant factor theorem holds. Consider any A ∈ Mn,m (Z) of rank max(m,n). Without loss of generality, we shall take m ≤ n for, otherwise, we could take the transpose. Now, A defines a homomorphism

TA : Zm → Zn ; v ↦ Av. Now the image of TA is a free abelian group generated by the n vectors Ae1, …, Aem. Since the matrix A has rank m, the vectors Ae1,…, Aem are linearly independent vectors over Q. erefore, they are linearly independent over Z aiso. In other words, Image TA is free abelian subgroup of Zn of rank m. By the invariant factor theorem, let us choose bases {v1,…, vn} of Zn and {d1v1,…,dmvm} of Image TA such that di|d+1. Call Awi = divi for all i ≤ m. Let P ∈ GL (n, Z) denote the matrix effecting the change of basis from the canonical basis to the vi's. Similarly, let Q ∈ GL (m, Z) be the matrix effecting the change of basis from the canonical basis to the wi's. en, P–1 AQ(ei) = divi for all i ≤ m. In other words, P–1 AQ has the form asserted.

SOLUTION 54 e proof of the invariant factor theorem clearly shows the generation of GL (n, Z) by the above matrices. Also, as the matrices diag (±1,…,±1) normalise the groups < I + Eij > for each i ≠ j, we can write each element of GL (n, Z) as a product of matrices of the form Xi,j, i ≠ j fallowed by a single matrix of the form diag (±1,…,±1). Now, if g ∈ SL(n, Z), then the matrix diag (±1,…,±1) occurring at the end in the above expression has an even number of –1's. So, it suffices to check that the above diagonal matrices with two –1's are in < Xijsub>; i ≠ j >. is can be seen from the fallowing identities in the correspanding 2 × 2 matrices:

Finally, if n ≥ 3, then Xij = [Xik, Xkj] for i, j, k distinct. is proves that SL(n, Z) is perfect for n ≥ 3.

SOLUTION 55 (i) We know that GL(n, Z) is generated by the matrices of the form Xij = I + Eij; i ≠ j and the matrices diag (±1,…,±1). eisewhere. We shall check for each r that hr(AXij) = hr(XijA) for all i ≠ j ≤ n. By the previous problem, we need to consider only A of the 'diagonal' form with nonzero entries d1,…, dm with di|di+1. erefore, it is clear that hr(AD) = hr(DA) for D = diag(±1,…, ±1). Now, for such A, we have, if i > m that AXij = A and, if i ≤ m, AXij = A + A ´ where A´ is a matrix whose only nonzero entry is di at the (i,j)th place. Clearly, hr(AXij) = hr(A). Similarly, we see aiso that hr(XijA) = hi(A). erefore, we have (i). For (ii), we merely note that for 'diagonal' matrices A as above, with di|di+1, the numbers hi(A) = d1…di. us, the invariant factors are successive quotients of the hi's.

SOLUTION 56 (i) If G is nilpotent, then by our definition, the lower central series

G ≥ C1 (G) ≥ C2(G) ≥ …Cn (G) = {1} for some n. Evidently, Ci+1(G) ⊲ Ci(G) and Ci(G)/Ci+1(G) is contained in the centre of G/Ci(G) by the very definition. Conversely, let G = G0 ⊇ G1 ⊇ … ⊇ Gn = {1} for some n where each Gi is normal in G and Gi–1/Gi is contained in the centre of G/Gi. Note that C0 (G) = G = G0. We prove by induction that for any r ≤ n, Cr (G) ≤ Gr. If we assume Cm(G) ≤ Gm for m < n, then Cm+1(G) = [G,Cm(G)] ≤ [G,Gm] ≤ Gm+1 as Gm/Gm+1 is contained in the centre of G/Gm+1. erefore, the inductive proof fallows, and shows that Cn(G) = {1}. (ii) e penultimate term Cn–1 (G) in the lower central series G ≥ C1(G) ≥ C2(G) ≥ …Cn(G) = {1} is contained in the centre. (iii) Note first that any p-group G has a nontrivial centre Z (this fallows by using the conjugation action of the group on itself). Applying induction, we may assume that = G/Z is nilpotent. Evidently, Ci(G) ≤ i for all i where Ci (G/Z) = i/Z. If Cn( i) = 1, then we have Cn(G) ≤ Z; so Cn+1(G) = 1. us, G is nilpotent. (iv) It is clear that for any H ≤ G, we have Cr (H) ≤ Cr(G) for every r, by induction. us, subgroups of nilpotent groups are aiso nilpotent. Let N ⊲ G and let Cn(G) = {1}. Consider the series G/N ≥ C1(G)N/N ≥ …Cn(G)N/N = {1}.

Clearly, each Ci(G)N/N ⊲ G/N and let us check that any successive factor is contained in the centre of . Now, [G/N, Ci(G)N/N] = [G,Ci(G)]N/N ≤ Ci+1(G)N/N. us, the above series shows that G/N is nilpotent by (i). (v) Call B = B(n, Q) and U = U(n, Q) for simplicity. An easy matrix computation shows that [U, U] ≤ U and that [U,U] consists of matrices which have the entries (1, 2), (2, 3),…, (n … 1, n) to be zero. By induction, one can easily show that Cr(U) consists of matrices whose (i,j) th entries for i < j ≤ i + r – 1 are all zero. Hence, Cn(U) = {1}. Clearly, B = TU with T ∩ U = {1} where T ≤ B is the subgroup of diagonal matrices. Also, U = ker(det : B → Q*) is a normal subgroup. Since T is abelian, it is nilpotent. However, B is not nilpotent since [B,B] = U and [B,U] = U by the same computation as above. (vi) We already know that quotient groups of nilpotent groups are nilpotent. So, we assume that G/Z(G) is nilpotent and show that G must be nilpotent. Using (i) for G/Z(G), we have normal subgroups Gi of G containing Z(G) such that G/Z(G) = G0/Z(G) ⊇ G1/Z(G)…⊇Gn/Z(G) = {1} where Gi–1/Gi is contained in the centre of G/Gi. Consider the series G = G0 ⊇ G1 … ⊇ Gn = Z(G) ⊇ Gn+1 = {1}. Evidently, Gn/Gn+1 is contained in (in fact, equal to) the centre of G/Gn+1. us, G is nilpotent.

SOLUTION 57

Let G be nilpotent and H ≤ G. Suppose G ≥ C1 ≥ C2 ≥ … Cn = {1} be its lower central series. Since, we have that Ci/Ci+1 is contained in the centre of G/Ci+1, it fallows that G = GH ≥ C1H ≥ …CnH = H is a chain of subgroups where each term is normal in the previous one. us, H is subnormal in G; that is, (i) implies (ii). Now, suppose (ii) holds. Let H < G. We have a series H = H0 ⊲ H1 ⊲ … Hr = G. If i is the smallest number for which Hi > H, it fallows that Hi ≤ NG(H) and Hi H. is proves (ii) implies (iii). For a maximal subgroup, M ≠ NG(M) shows that NG(M) = G i.e., M is normal. us (iii) implies (iv). Assume (iv). If P is a p-Sylow subgroup which is not normal, then NG(P) ≤ M for some maximal subgroup M. Since M is normal, we get for any g ∈ G, gPg–1 is again a p-Sylow subgroup of gMg–1 = M. us, by Sylow's secand theorem applied to M, there is some m ∈ M with gPg–1 = mPm–1, i.e., m–1g ∈ NG(P) ≤ M. Hence g ∈ M, a contradiction, since a maximal subgroup, by definition, is a proper subgroup. Hence (iv) implies (v). Assume (v) holds; that is, the Sylow subgroups are normal. We shall show first that elements of orders powers of different primes commute. Let O(x) = pr and O(y) = qs where p ≠ q are primes. If P, Q are the unique p-Sylow subgroup and the unique q-Sylow subgroup, then x ∈ P and y ∈ Q. As xyx–1 y–1 ∈ P ∩ Q = {1}, we have xy = yx.

Now, we deal with the general case. We saw in Problem 43 that in any finite group, every element can be written as commuting elements of prime power orders dividing the order of the element. Let g, h ∈ G be elements in our group which have coprime orders m, n. en, g = x1…xr, h = y1…, ys where xi's commute among themselves, yj's commute among themselves and each of them has prime power order. Also O(xi)|O(g) = m and O(yj)|O(h) = n which are coprime. us, each xi and each yj commute. Hence gh = hg. us, (v) implies (vi). Suppose now that elements of coprime orders commute. We write

Let Pi be any pi-Sylow subgroup of G ; 1 ≤ i ≤ r. en, since [Pi, Pj] = {1} for i ≠ j, the product P1 … Pr is a group. As the orders of Pi's are pairwise coprime to each other, |P1…Pr| = |P1|…|Pr| = |G|. Hence G = P1…Pr and

, which means that G ≅ P1 × … × Pr.

So, we have proved that (vi) implies (vii). Finally, since p-groups are nilpotent, (vii) clearly implies that P1 × … Pr is nilpotent; that is, (i) fallows.

SOLUTION 58 (i) Let G = < S > and s ∈ S ∩ Φ(G). It suffices to show that s ∈< S\{s} >. Now, if H = < S\{s} > < G, then there is some maximal subgroup M of G containing H. But then S\{s} is contained in M as well as s ∈ M. is means that G = < S > ≤ M, which is a contradiction. us, we have shown that elements of Φ(G) can be dropped from any generating set for G.

Conversely, suppose g ∈ G be such that whenever < S ∪ {g} > = G, we have < S > = G. Let, if possible, M be a maximal subgroup not containing g. en, < M ∪ {g} > = G. By the hypothesis, this gives < M > = M = G, a contradiction. Hence all maximal subgroups contain g and (i) is proved. (ii) Since Φ(G) is the intersection of all maximal subgroups of G, and since any automorphism permutes maximal subgroups of G, it fallows that Φ(G) is a characteristic subgroup of G. (iii) By the previous problem, it suffices to prove that the p-Sylow subgroups of Φ(G), for any p, are normal in it. By the Frattini argument (problem 17 (ii)), we have G = Φ(G)NG(P). In particular, NG(P) ∪ Φ(G) generate G which implies that NG(P) = < NG(P) > = G; that is, P ⊲ G. In particular, P ⊴ Φ(G). (iv) Let M1…, Mr be the (proper) maximal subgroups of G. Consider their images in the elementary abelian group G/[G,G] < Gp > . ey are maximal subgroups and every maximal subgroup of G/[G,G]< Gp > is the image of one of these. Of course, two different Mi's may have the same image. Note that the intersection of all maximal subgroups in an elementary abelian pgroup is trivial. is is so because an elementary abelian p-group is isomorphic to Z/p×…Z/p, and in this group, the subproducts with one factor being the trivial groups are maximal subgroups and intersect in the identity. erefore,

Conversely, note that in p-groups, all maximal subgroups are of index p. is means that in our p-group G, < Gp > ≤ Φ(G). Also, since maximal subgroups M in G are necessarily normal, G/M is an abelian group (of order p) and so, [G,G] ≤ M. us, [G,G] ≤ Φ(G) as well. is proves that

Φ(G) = [G,G]< Gp > . (v) By (iv), it fallows that the Frattini subgroup of an abelian p-group A is simply Ap. erefore, clearly A/Φ(A) = A/Ap is an elementary abelian pgroup. If A/Ap has order pr, then it can be generated by r elements x1Ap,…, xrAp say. erefore, A = < Ap, x1,…, xr > . However, Ap is the set of nongenerators and, therefore, A = < x1,…,xr > which implies that r ≥ d(A). However, it is obvious that for any abelian pgroup A, the elementary abelian p-group A/Ap is a direct sum of at the most d(A) copies of the group of order p; so |A/Ap| = pr ≤ pd(A). erefore, we have d(A) = r.

SOLUTION 59 (i) Let S be a finite set of generators of a nilpotent group G. Consider the sets S0 = S, Si+1 = {aba–1b–1 : a ∈ S, b ∈ Si} for all i ≥ 0. Of course, each Si is a finite set. Since G is nilpotent, there is some n so that Sr = 1 for all r ≥ n. For each i ≥ 0, consider the subgroup Gi of G generated by the set ∪r≥iSr . As Sr = 1 for r ≥ n, each Gi is finitely generated. It is clear that [G,Gi] = Gi+1 since S, Si, Si+1 generate G, Gi and Gi+1 and Si+1 = {aba–1b–1 : a ∈ S, b ∈ Si}. In particular, each Gi is normal in G and the subgroup Gi/Gi+1 is central in G/Gi+1. Now, if H is any subgroup of G, then (H ∩ Gi)/(H ∩ Gi+1), being a subgroup of the finitely generated abelian group Gi/Gi+1, is finitely generated. Since H ∩ Gn = 1, this gives inductively that each H ∩ Gi is finitely generated. In particular, H = H ∩ G0 = H ∩ G is finitely generated. (ii) Look at the lower central series of G. As Cn(G)/Cn+1(G) is a subgroup of the finitely generated, nilpotent group G/Cn+1(G), this subgroup is finitely generated as well. But, this is an abelian, torsion group and is, hence, finite. us, G itself is finite.

SOLUTION 60 We have already seen that the Frattini subgroup Φ(P) = [P,P] < Pp > for any p-group P. By our hypothesis, Φ(G) = < Gp > . We have aiso seen earlier that for any group, a subset is a generating set if and only if it generates the group modulo its Frattini subgroup. erefore, since we are interested in the question of whether a certain subset of < Gp > generates it, we may quotient out by the Frattini subgroup, and assume that Φ(< Gp >) = {1}. In particular, [Gp, Gp] = {1} = . Hence [G,G]p ≤ . us, generate < Gp > if a1, …,an generate G.

SOLUTION 61 We show that any element of K having prime order must have order p. is would show by Cauchy's theorem that p is the only prime dividing the order of K. Let θ ∈ K have some prime order q. Note that θ ∈ K means by definition that θ(x)Φ(G) = xΦ(G) ∀xΦ(G) ∈ G/Φ(G). us, for each x ∈ G, we have θ(x) ∈ xΦ(G). Fix generators g1,…, gn of G. Consider the set Ω:= {(g1f1,…gnfn) : fi ∈ Φ(G)}. en (θ(g1)θ(f1),…,θ(gn)θ(fn)) ∈ Φ for any (g1 f1,…gnfn) ∈ Ω. Note that this gives an action of the group < θ > on Ω. Moreover, if (g1f1,…gnfn) ∈ Ω is fixed by θ, then we have that θ(gi) = gi for all i since a subset X of G generates

it if, and only if, its image modulo Φ(G) generates G/Φ(G). us, < θ > is a group of order q acting without fixed points; hence each orbit has order q. In other words, |Ω| is a multiple of q. us q divides |Φ(G)|n; that is, q = p.

SOLUTION 62 (i) Let θ ∈ Aut (G) be a nontrivial element; suppose θ(g) ≠ g for some g ∈ G. en, the element x := θ(g)g–1 ≠ 1. Since G is residually finite, there exists H ≤ G of finite index such that x ∉ H. As G is finitely generated, it has only finitely many sub-groups of any given finite index. In particular, the set of sub-groups {σ(H) : σ ∈ Aut (G)} is a finite set since each of them has index [G : H]. Letting

it follows that C is a characteristic subgroup of finite index in G such that x ∉ C. Consider the homomorphism Aut (G) → Aut (G/C. We claim that the finite quotient group Aut (G/C) of Aut (G), is such that θ maps to a nontrivial element of it. is would prove that Aut (G) is residually finite. Now, if θ were to map to the identity, this precisely means that θ(h)h–1 ∈ C for every h ∈ G. As this does not hold for h = x, we have the assertion. (ii) As G is virtually residually-p, there is a subgroup G' of finite index in G which is residually-p. By taking the intersection of all the finitely many subgroups σ(G') as σ runs through Aut G, we get a characteristic subgroup G0 of finite index in G which is residually-p. If Aut (G0) is virtually residually-p, so is Aut (G) as seen by pulling back via the restriction homomorphism from Aut (G) to Aut (G0). Without loss of generality, we,

therefore, assume that G itself is residually-p. Let H be any characteristic subgroup of p-power index in G. Consider the p-group P = G/H. Now, where Φ(P) denotes the Frattini subgroup of P. Moreover, by problem 58, the number r of copies of Z/p is bounded independently of H; indeed, r ≤ n, where n is the minimal number of generators of G. Now, by the previous problem, ker Aut(P) → Aut(P/Φ(P))) is a p-group. Note that Aut(P/Φ(P)) ≅ GL(r, Z/p, and that there are only finitely many homomorphisms from Aut(G) to GL(n,Z/p), since there are only finitely many subgroups of index bounded by the order of GL(n,Z/p) in the finitely generated group Aut G. Call to be the intersection of the kernels of all the homomorphisms from Aut(G) to GL(n, Z/p. We claim that is residually-p. Let σ ∈ , σ ≠ id. So, there is g such that σ(g)g–1 ≠ id. Let H be a characteristic subgroup of p-power index in G such that σ(g)g–1 ∉ H; then σ ∉ ker(Aut(G) → Aut (G/H). Call P = G/H. Consider, now, the composite

By the choice of , the image goes into N := ker(Aut(P) → Aut (P/Φ(P))), a p-group. Since the image of σ is nontrivial in Aut(P), ker( → N) is normal, of p-power index in , and does not contain σ. is completes the proof.

SOLUTION 63 (i) Let p be any group. Let g ∈ F be a nontrivial element and suppose where xi are basis elements (not necessarily distinct), each ai ≠ 0 and xi ≠ xi+1. Let d be an integer bigger than the power of p dividing the product a1… an. We consider the (finite) p-group P consisting of all (n + 1) × (n + 1) upper triangular matrices with entries from Z/pd and all diagonal entries 1.

Of course, the matrices I + Eij for i < j generate P where Eij has only the one nonzero entry 1 at the (i, j)th place. Of course, to define a homomorphism from F to P, we need to specify only the images of the basis elements but we also need for our purpose to ensure that the element g maps to a nontrivial matrix. erefore, it is natural to gather together, for each basis element x of F, all the i's such that xi = x and define

In the last expression, the factors on the right side commute since consecutive xi and xi+1 are unequal and Ei,i+1Ej,j+1 = 0 unless i + 1 = j. erefore, for a basis element x occurring in g,

Let us check whether θ(g) = I is possible. Note that

We see that E1,2, E2,3,…, En,n+1 occur in that order and their coefficients are precisely the integers a1,…an. erefore, since Ei,jEj,k = Ej,k = Ei,k, the coefficient of E1,n+1 is the product a1…an which is not zero in Z/pd by the choice of d. Hence θ(g) ≠ I and so, F is residually-p. (ii) Let G be any finitely generated, residually finite group. Suppose θ : G → G be a surjective homomorphism which has a nontrivial element g in the kernel.

Let N ⊲ G be of finite index such that g ∉ N. Since G is finitely generated, there are only finitely many different homomorphisms θ1,…,θn from G to G/N. Note that the composite map α = β ∘ π : G → G, where π : G → G/Ker(θ) is the natural map and β : G/Ker(θ) → G is induced by θ, is such that α maps g maps to the identity. Also, since θi ∘ α : G → G/N;i ≤ n are distinct, these are just the θi in some order. is means that every homomorphism from G to G/N maps g to the identity. is contradicts the fact that the natural homomorphism maps g to a nontrivial element. erefore, θ must be an automorphism.

SOLUTION 64 Let a, b, c, d ∈ Z such that ad – bc ≡ 1 mod n. Write ad – bc = 1 + qn. Note that (c,d,n) = 1. We would like to change a, b, c, d mod n so that ad – bc becomes 1. First, let us suppose that (c, d) = 1. Consider aʹ = a + un and bʹ = b + vn where u, v are to be chosen such that aʹd – bʹc = 1. Now aʹd – bʹc = ad – bc + (ud – vc)n = 1 + (q + ud – vc)n. Since (c,d) = 1, we may choose u, v with q – vc – ud; this gives aʹd – bʹc = 1 and we are done. So, we only have to prove that the above supposition holds. Let p1,…,pr be the set of all primes which divide d. If each pi|n, then clearly, none of the pi's divide c since (c, d, n) = 1. In such a case, evidently, (c, d) = 1. So, let us suppose that some of the pi's do not divide n; let p1,…,pk be the subset of these primes. We note that (n, p1,…pk = 1. By the Chinese remainder theorem, choose an integer x ≡ c cmod n and x ≡ 1 mod p1…pk. en, clearly p1,…pk x.

Also, writing x = c + ln, we have that the other pi's which divide n cannot divide c + ln as (c, d, n) = 1. Hence (c + ln, d) = 1 and we are done.

SOLUTION 65 (i) Let K be any field. If A, B ∈ SL2(K) have the same trace ≠ ±2, then they have the same order. is follows from the fact that A and B have the same characteristic polynomial and, hence, the same eigenvalues λ, λ–1. Since these eigenvalues are necessarily distinct, A and B are conjugate in GL2 over an algebraic closure of K. Let us start with l, m, n ≥ 2. Let K = Fp with p a prime ≡ 1 modulo 2lmn. Consider G = PSL2 (K). Let λ ∈ K* be a primitive 2lth root of unity. Note that since 2l ≥ 4, one has λ ≠ λ–1 so that the matrix ∈ SL2(K) has order 2l, for any arbitrary α ∈ K. In particular, has order 2l in SL2(K). Similarly, there exists μ ∈ K* of order has order 2m for any t in K. Now, . Look at its trace λμ + λ–1 μ–1 + t. One can obtain any given element θ of K as trace(AB) by solving for t ∈ K. Choosing θ to be the trace of an element of order 2n ≥ 4 in SL2(K), we have got hold of A and B in SL2(K) such that AB has order 2n in SL2(K). Evidently, the images of A, B and AB in G = PSL2(K) have orders l, m, n respectively. (ii) Note that W can be taken to be of the form Ar1Bs1…ArkBsk with 0 < ri < l and 0 < si < m. We proceed as before by starting with a prime p ≡ 1 mod 2lmn. Once again, we choose of order 2l in SL2(K) and which is of order 2m ∈ SL2(K) for any t in K. Let us write and that

for any d > 0. en, an easy calculation shows

where we have denoted by O(td) a polynomial of degree at most d over Fp. us trace W(A, B) is a nonconstant polynomial in t over Fp since M(si) ≠ 0

≠ L(rj) as μ2si ≠ 1 λ2rj. We choose any root of this polynomial in . en, λ, μ, t lie in a finite field Fp and A, B, W(A, B) have orders l, m, n respectively in PSL2(Fp).

SOLUTION 66 We note first that if we write A = ⊕pAp and B = ⊕pBp where Ap and Bp are p-subgroups for various primes p, then Hom(A,B) ≅ ⊕pHom(Ap,Bp). Also End (A) ≅ ⊕p End(Ap). erefore, it suffices to prove the assertion for p-groups A, B for any prime p. Moreover, note that if an abelian p-group A is the sum of r cyclic pgroups, then A/pA is a vector space over Z/p of dimension r. If B is the sum of s cyclic groups, then

To prove our assertion for p-groups, we shall argue by induction on |A|. If either A or B is trivial, the assertion is obvious. So, we assume that neither is trivial. Look at the subgroups pA and pB and observe that we have homomorphisms

Here α is defined by sending any homomorphism θ : A/pA → B to the homomorphism θ ∘ π where π : A → A/pA is the natural homomorphism.

Also, β is defined by sending any homomorphism θ : A → B to the homomorphism which maps any pa to pθ(a). Clearly, the composite map β ∘ α is the zero map. In other words, Image α ≤ kerβ. We claim that they are actually equal. Indeed, let θ ∈ kerβ. en, pθ(a) = 0 in B for all a ∈ A. consider the homomorphism ɣ : A/pA → B defined as a + pA ↦ θ(a). By the above property of θ, this is a well-defined homomorphism and clearly, α(ɣ) = θ. erefore, we have proved that Image α = kerβ. Hence, the isomorphism Hom(A,B)/ker(β) ≅ Image(β) implies that

and, therefore, divides

As the order of a homomorphic image of a group divides the order of the group, the above number divides

If a, b denote, respectively, the number of cyclic summands of A and B, then

and

SOLUTION 67

(i) e identity is easily verified and the conclusion about groups of exponent 3 is obvious from it. (ii) We first note that from the statement that groups of exponent 3 are nilpotent, it follows from Problem 59 that if the group is also finitely generated, it must be finite. erefore, we proceed to prove the first assertion. For convenience, for any x ∈ G, let us write Cx for the set consisting of all products of conjugates of x and x-1. Note from (i) that any two elements of Cx commute. Moreover, C x is stable under conjugation by any element. Finally, observe that for any x1,…,xn ∈ G, we have [x1,…,xn] ∈ C(xi) for all i ≤ n. In particular, for any g ∈ C(xi) for some i, we have

We shall now consider the elements of the form [x, y, z]; these evidently generate C3(G) in any group. e idea is to prove that these are contained in the centre; that is, we shall show that [x,y,z,w] = 1 ∀ x, y, z, w. First, look at any [x, y, z]; we know that [x,yz] = [x,y][x,z]y (in any group) by problem 36. Similarly, [xy,z] = [y,z]x[x,z].

erefore

Since [x,z] (and therefore [x, z]y) are in C(z), the above expression becomes [x,yz,z] = [x,y,z]

(1)

Similarly,

Since [x,y] ∈ C(y), as we observed in the beginning of the proof,

that is, [x,yz,y] = [x,z,y]

(2)

Now, for each x, y, z,

by (1) and (2). But, notice that [x,y,z] ∈ C(y). Hence 1 = [x,z,y] [x,y,z] But,

(3)

by the first observation.

From this and (3), we have [x,y,z] = [z,x,y]. Changing the roles of x, y, z, we have [x,y,z] = [z,x,y] = [y,z,x]

(4)

Let us use this repeatedly to look at [x,y,z,w].

Let us note the final identity [x,y,z,w] = [x,y,w,z]-1

(5)

erefore, using (5) and repeatedly using (4) as well as the basic identity [u,v] = [v,u]-1, we have

erefore, C3(G) ≤ Z(G) and G is nilpotent. Note also that [[x,y],[z,w]] = 1 which means that [G, G] is abelian. is proves the assertion.

SOLUTION 68 (ii) We first deduce this from (i). We proceed by induction on the minimal number n of generators needed. Trivially, if n = 1, then |G|≤4. Suppose n > 1

generators g1,…,gn generate a group G of exponent 4 and that the (n - 1)generated group H = < g1,…,gn-1 > is finite. Consider

.

From (i), we can conclude that K is finite and so G is finite. (i) As x2 ∈ L, any element of M can be written as

with li ∈ L nontrivial for 1 < i < n. We shall show that if n is minimal for such an expression, then n ≤ |H| + 2. e idea is to get many expressions of length n for g and deduce that if |L| is small compared to n, two such expressions coincide and give rise to a cancellation within the expression and that this would yield for g an expression of smaller length. For any l ∈ L, we have 1 = (xl)4 = xlxlxlxl which implies

where

.

us, we see that in the expression

we may replace any

which leads to the expression

In other words, li-1 has got replaced by could repeat this as follows.

but the length remains n. We

Change l2 to .

, then l3 to

so that l2 has now got changed to

Following this with the change l4 to We can change until ln-1 to changed to any one of

changes l2 to

.

and so, inductively we see that l2 can be

or

according as to whether n is odd or even and 2d or 2d + 1 can increase until n. Note that each expression has n elements from L and that as varies from d = 2 onwards, these are n - 2 elements of L in number. If n - 2 >|L|, two of these expressions are equal. If

with d' > d, then we may cancel off common terms from both sides and conclude that a nontrivial expression of the form

to be the identity element. But, since it is possible to get an expression for g of the same length n where l2d+2 is changed to the above expression representing the trivial element, it means that this part can be cancelled off and we can get an expression of smaller length. is contradiction shows that n - 2 ≤ |L|. erefore, |M| ≤ |L||L|+2.

SOLUTION 69 (i) Now, x–1T(x) = y–1T(y) for some x, y if and only if T fixes yx–1; that is, when x = y. us, the map x ↦ x–1T(x) is 1 - 1. Being a map of finite sets, this is onto as well. (ii) From (i), an arbitrary element of g can be written as g = x–1T(x) so that we get gT(g)…T(g)p-1 = 1. (vii) Taking p = 2, this gives T(g) = g–1 for all g. As T is an automorphism, G is abelian. (iii) Consider the subgroup Z of Aut (G) generated by T. is is a cyclic group of order p acting on G. Each orbit has cardinality dividing p and, as T fixes only the identity element, each orbit other than that of 1 has p elements. So, |G| has 1 mod p number of elements. (iv) Look at the action of Z on the set S of all q-Sylow subgroups of G. Again, the orbits which are not fixed points under Z, have order p. So, if there is no fixed point in S, then the number of elements in S would be a multiple of p. But, we know that the number of elements in S (i.e. the number of qSylow subgroups) is a divisor of |G|. us, p would divide |G|, a contradiction of (iii). (v) Note that if T(Q) = Q, then T(N) = N where N is the normaliser of Q in G. Now, if T fixes another qSylow subgroup gQg–1, then rewriting T(gQg–1 = gQg–1, we have g–1T(g) belongs to N. But, applying (i) to N itself, any element of N is of the form x–1T(x) for some x in N. So, g–1T(g) = x–1T(x), which gives g = x belongs to N. So, gQg–1 = Q. (vi) Let R be any q-group in G which is fixed by T. Let S be a q-group which contains R and is maximal with respect to this property (i.e., S is fixed by T, contains R and is not contained in a strictly larger qgroup which is T-fixed).

We claim that R = Q. For this, first look at the normaliser N(S) of S in G. Since (S) = S, we have also T(N(S)) = N(S), clearly by (v) applied to the subgroup M of N(S) which is Tfixed. Now, S is a qsubgroup of N(S), say, yMy–1 where y is in N(S). But then S = y–1Sy is contained in M. By maximality property of S, we get S = M i.e., S is a q-Sylow subgroup of N(S). But, any Sylow subgroup of a subgroup in any group is the intersection of the subgroup with a Sylow subgroup of the big group. So, we have a qSylow subgroup L of G such that S = N(S) intersection L. In other words, = NL(S), the normaliser of S in L. But, in a qgroup (indeed, in any nilpotent group), the normaliser of a proper subgroup contains the subgroup properly. So, we have S = L. In other words, S is a q-Sylow subgroup of G itself. By uniqueness of the Tfixed qSylow subgroup of G itself. us, we have shown that R is contained in = P. (viii) Now, we prove the result for p = 3. Now x T(x)T2(x) = e for any x in G. So, T(x–1 x–1 = (x T(x))-1 = T2 (x) from the above. Putting y = x–1, we get T(y)y = T2(y–1) = (T2 (y))-1. us, both yT(y) and T(y)y are equal to (T2(y))-1. In other words, every element x in G commutes with the element Tx. Similarly, x commutes with T2(x also. Now, to prove the result, let us consider any prime q and the unique qSylow subgroup Q of G which is Tfixed. Let g be any element of G which has qpower order. Consider the subgroup Q' is generated by the three elements g, T(g) and T2 (g). Clearly, Q' is Tfixed. Also, since g, T(g), T2(g) commute among themselves, Q' is a q-group. erefore, by (vi), Q' is contained in Q. us, Q contains all elements of G of q-power order. Hence, Q is the unique q-Sylow subgroup of G. So, it is normal. is proves the result for p = 3.

SOLUTION 70 (i) We note first that σ : (a,b) ↦ (a2,b4)) is an automorphism of A. e reason is that both a ↦ a2 and b ↦ b4 are isomorphisms on a group of order

7 as 7 is coprime to 2 as well as to 4. en, B ↦ Aut(A);xi ↦ σi is a 1-1 homomorphism as (a,b) ↦ (a2,b4) is of order 3. (ii) is clear as B normalises A. (iii) We see that

since b7 = 1. For any two elements g = xiajbk and h = xuavbw of G, let us simplify gh = xiajbkxuavbw for u = 0,1,2 and write it in the form in (ii).

by using (1). us, once again θ(gh) = θ(g)θ(h). Finally, let u = 2 and consider θ(gh).

and

erefore, θ is a homomorphism which is clearly 1-1 since θ(g) = 1 if and only if i = j = k = 0. Hence, θ is an automorphism because it is a 1-1 map on a finite set to itself. Also, θ2 (xiaj bk) = θ(x-ibja-k) = xi b-ka-j = xia-jb-k for any xi ajbk ∈ G. So, clearly, θ4 is the identity map. erefore, θ has order dividing 4. Since θ2 (a)

= a–1 ≠ a, we have that θ2 cannot be the identity map. erefore, θ has order 4. (iv) We shall show that G has trivial centre. If za = az gives .

is in the centre, then

erefore, x-iaxi = a which forces xi = 1 (we know that xax–1 = a2 ≠ a and x2ax-2 = a4 ≠ a). So, z = ajbk. Now, xz = zx gives

Hence a3j = 1 = b15k; that is, aj = 1 = bk as 3, 15 are coprime to their order 7. Hence z = 1. In other words, Z(G) = {1}, which shows that G is not nilpotent.

SOLUTION 71 Clearly, the first assertion implies the second since it shows that any unimodular vector can be transformed to (1,0,…,0) by an element of SL (n, Z). Let us prove the first assertion. Start with any unimodular vector a := (a1,…,an) Consider the subgroup H

:= < a >≤ Zn, the group of row vectors. Let {v1,…,vn} be a basis of Zn and {d1v1} be a basis of H as given by the invariant factor theorem. As < d1v1 > = H = < a >, we have d1v1 = ±a. is means that d1 divides each ai which implies by the unimodularity of a that d1 = ±1. Changing the sign of v1 if necessary, we may suppose that d1 = 1. If g ∈ GL(n,Z) corresponds to the change of ordered bases from the standard basis {e1,…,en} to {v1,…,vn}, then ge1 = v1 = a; that is, the first column of g is a. If det g = -1, we just change the signs of entries in the second column of g to get a matrix g1 in SL(n,Z) whose first column is a.

SOLUTION 72 If (a, b, c ) = 1, let P ∈ SL(3, Z) be a matrix whose first row is

SOLUTION 73 (i) If ρ : G → Sn is any transitive representation, then the subset H := {g → G :

チ (g)(1) = 1} is a subgroup of G, of index n. Conversely, if H ≤ G is a subgroup of index n in G, the set G/H of le cosets has n elements and G acts transitively on it. ere are (n-1)! ways to identify G/H with the set 1, 2, , n} where the coset H is identified with 1. us, an = tn/(n-1)!. (ii) For each 1 ≤ k ≤ n, there are

ways to choose the orbit of 1, tk

transitive actions on it, and hn-k actions on its complement. is proves the relation

Rewriting the relation in terms of the an, one has

(iii) is immediate from (ii). (iv) If G is d-generated, then

. Hence

.

(v) Now, any subgroup H of Z2 is of the form gZ2 for some g ∈ M(2,Z). It is of finite index if g ∈ GL(2, Q). Note that H = gZ2 = gxZ2 for any x ∈ GL(2, Z). We claim that x can be chosen so that gx is of the form where a > 0,0 ≤ c < d. Now, if

such that au + bv = (a,b). en, and gx is of the form

matrices like

. By multiplying by

, we may assume that gx is of the form

with

a1, d1 > 0. Now, multiplying on the right by a matrix of the form get

, we

where we may assume that 0 ≤ c1 + d1l < d1. erefore, we

have shown that any subgroup H of finite index is of the form gZ2 where , with a, d > 0 and 0 ≤ c < d. Clearly, the index of H is ad. We claim finally that gZ2 = hZ2 for another matrix

,

with a1, d1 > 0 and 0 ≤ c1 < d1 if and only if g = h. To see this, suppose g–1 h ∈ GL(2, Z). is gives a1 = a, d1 = d and d|(c1 - c). As 0 ≤ c1, c < d, we get c1 = c as well. erefore, the number of subgroups of index n is the number of matrices of the form , with d|n and 0 ≤ c < d. us, for each divisor d|n, there are exactly d subgroups of index. erefore,

.

(vi) We claim that hn(Z2) = n!p(n). e reason is as follows. x can be arbitrarily chosen in Sn, and y chosen in its centraliser CSn)(x), so that . is yields us the

identity

SOLUTION 74 If H ≤ Zr, [Zr : H] < ∞, then we have seen earlier (in Problem 53) that H = gZr, for some g ∈ Mr(Z) ∩ GLr(Q). Moreover, . erefore, detg |-s where C is the set . One can take for C the set of lower triangular matrices where the entries aij are nonnegative integers satisfying aij j, and with aii ≥ 1. A simple counting gives the expression in the problem.

SOLUTION 75 Note that

.

If A is a finite abelian p-group and d(G) (the minimal number of generators needed to generate A, then we have already seen (in Problem 58) that A/Ap is an elementary abelian group of order pd(A). If d(A) = r, then one can write A = G/H where G ≅ Zr. But, clearly . erefore, A ≤ Gp. us, finite abelian groups A with d(A) = r correspond to subgroups H of G with G/H a p-group and H ≤ Gp. Hence, the number of these H ≤ Gp which have index pr+k in G is the number of subgroups of index pk in Gp. Since Gp ≅ G, this last number is just apk as in the problem above. Let us write . Decomposing finite abelian groups into prime-power order groups, it follows that has an Euler product expansion and fp(s) is its ppart, i.e., . Using the computation of the zeta function in the theorem,

Now, one can count this in another way. We claim that given an abelian pgroup A with d(A) = r, the number of subgroups H ≤ G with G/H ≅ A is given by the number . Here Epi (G,A) denotes the set of

epimorphisms and thus, Epi (G,A)| is the number of ordered rtuples which generate A. To verify the above assertion, look at the mapping which associates to each epimorphism α : G → A, its kernel H. Suppose α, β have the same kernel H. en, evidently, they define isomorphisms from G/H onto A. us, one has an automorphism θ : A → A where . Note that β = θ ∘ α. Conversely, for any automorphism θ ∈ Aut(A), and any epimorphism α : G → A, the epimorphism β = θ ∘ α has the same kernel. is proves the claim made above. As mentioned above, an rtuple generates A if, and only if, it generates it modulo Ap. As A/Ap is elementary abelian of order pr, the number of rtuples generating it is (pr - 1)(pr - p)…(pr - pr-1). If |A| = pr+k, then |Ap| = pk and so, the number of r tuples in A generating it is (pk)r(pr - 1)(pr - p)…(pr pr-1). erefore, we have

where the sum is over all abelian A of order pr+k with d(A) = r. Sum over all k, to get

where the sum on the right is over all abelian p-groups A with d(A) = r. erefore, we have

Now, Euler's identity above shows that the right side is just completes the solution.

SOLUTION 76

. is

If G is an abelian group of order mn where (m,n) = 1, then consider the various Sylow subgroups of G. If Pi (with i ≤ r are Sylow subgroups corresponding to the primes dividing m and Qj (with j ≤ s are Sylow subgroups corresponding to the primes dividing n, let us consider H = P1… Pr and K = Q1…Qs. Clearly, H, K ≤ G have orders m, n respectively. Since H ∩ K = {1} and HK = G, we have G ≅ H × K. Conversely, if A, B are abelian groups of coprime orders m, n, then G := A × B is an abelian group of order mn in which the corresponding H and K can be identified with the subgroups A × {1} and {1}× B respectively. erefore, a(mn) = a(m)a(n) for (m,n) = 1. Now,

e product above is over all primes. is is further equal to

where pk(n) denotes the number of partitions of n in which each part is at most k. us, the limit is

Also,

where p(r) denotes the number of partitions of r and we take p(0) = 1 for convenience.

is proves the result.

SOLUTION 77 (i) Evidently, the derived series is a series as asserted and, so any solvable group has such a series. Conversely, suppose G admits a finite series as above. We prove by induction that Dr(G) ≤ Gr for all r. at would prove Dn(G) = {1}. e assertion holds for r = 0. Supposing Dr(G) ≤ Gr, we have

as Gr/Gr+1 is abelian. is proves (i). (ii) & (iii) Suppose G is solvable. Start with a series for G as in (i). e subgroups Hr = GrN/N form a series for G/N as in (i), and since Hn = {1}, it follows that G/N is solvable. Of course, any subgroup K of G must be solvable as the terms of its derived series are contained in the corresponding term of the derived series for G. Conversely, suppose N ⊴ G and G/N are solvable. Let

and

be series for N and G/N as in (i). But then

is a series for G as in (i). us, G is solvable.

SOLUTION 78 (iii) A5 is not solvable but all its proper subgroups are.

(i) Let, if possible, G be a minimal counterexample. In other words, G is not solvable while all its proper subgroups are abelian and there are no groups of smaller order with this property. en |G| > 1 obviously. If M ≠ N are maximal proper subgroups, then the subgroup < M, N > = G. So, if M ∩ N ≠ {1}, then any common element commutes with both M and N, i.e., is contained in Z(G). But, then G/(M ∩ N) would be a smaller counterexample. erefore, we suppose that M ∩ N = e for all maximal, proper subgroups M ≠ N. We claim that there is a maximal subgroup M which is normal in G. Suppose NG(M) ≠ G, i.e., NG(M) = M for all maximal proper subgroups M. Writing |G| = l and |M| = m, we have that the number of conjugate subgroups gMg–1 is [G : NG(M)] = l/m. As any two of them intersect trivially, we get that the union of the conjugate subgroups xMx–1 has cardinality (m 1)l/m + 1 < l. us, there are elements of G not in any xMx–1 and there is a maximal proper subgroup of G different from any xMx–1. If N is any such maximal subgroup (of order n say), then the various conjugate subgroups gNg–1 being maximal subgroups, intersect the conjugates of M trivially. In other words, the number of elements in the union of all the conjugate subgroups of M and of N are (m - 1)l/m + (n - 1)l/n + 1 = 2l - l/m - l/n + 1 > l as m, n ≥ 2. is contradiction proves that there is indeed some maximal proper subgroup M which is normal. But then G/M is a smaller counterexample contradicting the minimality of G. erefore, there are no counterexamples and the proof is complete. (ii) Again, let G be a minimal counter-example. Evidently |G| 1. If G has a nontrivial proper nontrivial normal subgroup N, then G/N and N would both be solvable by minimality. erefore, G would be solvable, a contradiction of the assumption. So, we suppose that G is simple. G is nonabelian by the assumption. By the above counting argument, any two distinct maximal subgroups must intersect. If M, N are two different maximal subgroups, then they are nilpotent and, so is M ∩ N ≠ {1}.

erefore, M ∩ N has a nontrivial centre. However, G = < M, N > , which implies that Z(M ∩ N) is a nontrivial normal (even central) subgroup of G. is is a contradiction. Hence there are no counter-examples.

SOLUTION 79 Recall from problem 73 that if hn denotes the number of homomorphisms from a finitely generated group G into Sn and if an is the number of subgroups of index n in G, then an + b1an-1 + … + bn-1a1 = nbn, where br := hr/r!. It is also convenient to take b0 := 1 and rewrite this in terms of the generating functions as :

(is formulation is known as Dey's formula). Now, hn(Z/2*Z/2) = hn(Z/2)2 since hn (H * K) = hn(H)hn(K) for any H, K. Now, it is easy to see by induction on n that

for any prime p. So, for p = 2, we get

which can be rewritten in terms of ncn = cn-1 + cn-2

as (2)

Now,

e recursion (2) for the cn's gives

On the other hand, cn + cn-1 = (n+1)cn+1 implies, by squaring, an expression for cncn-1. Eliminating cncn-1 from the two equations (and recalling that , we get

is yields for the generating function

that

e above generating functions in (1) show clearly that an = n or n + 1 according as n is odd or even.

SOLUTION 80 Let H = Z(G) and consider the transfer of the identity map on H. Since

and since each term is in H, we have

As

, we get

us, η : G → Z(G);g ↦ gn is a homomorphism. As Z(G) is abelian, [G,G] ≤ ker(η). Now, if , then [G,G] is generated by the elements [ti,tj]. Since [G,G]/([G,G]∩Z(G)) ≤ G/Z(G), [G,G]∩Z(G) is of finite index. erefore, it is an abelian group which is finitely generated. As this is contained in the kernel of η, we have that it is finite and satisfies ( [G,G]∩Z(G))n = {1}. Hence [G,G] is also finite and satisfies [G,G]n = {1}.

SOLUTION 81 We first assume that G is finitely generated. Let x1,…,xn be generators. en, as the conjugacy classes G(xi) are finite, the centralisers CG(xi) are of finite index in G for i ≤ n. Clearly, then is of finite index in G. We saw above that [G,G] is finite. Now, let G be a general group with the property that all conjugacy classes are finite. Let x ∈ [G,G]. en, evidently since only finitely many elements of G are involved in the expression for x, we have a finitely generated subgroup F of G such that x ∈ [F,F]. As F also must have the property that all conjugacy classes are finite, the subgroup [F, F] is finite as proved above. erefore, the order of x is finite.

SOLUTION 82 As in the previous problem, we consider the transfer corresponding to the identity map on H. As before, we have

and each term is in H. Now, we claim that si and xli commute. Now, and, so commutes with H. Hence, gli commutes with . is means that both H and are p-Sylow subgroups of CG(gli). ey are, therefore, conjugate in CG(gli); say . is gives . As H ≤ Z(NG(H)), we have that commutes with H. Hence csi and, hence si itself, commutes with gli ∈ H. is proves the claim above and shows that the transfer is simply the homomorphism g ↦ gn where [G : H] = n. If m is chosen to satisfy mn ≡ 1 mod |H|, it follows that the homomorphism fixes H. en N := kerτ satisfies the assertion.

SOLUTION 83 (i) For x, y ∈ G/P, the elements s(x)s(y) and s(xy) represent the same coset; therefore, s(xy)-1s(x)s(y) = f(x, y) ∈ P. Also, associativity implies

and

e right-hand sides give us the cocycle identity. (ii) Note that h(x) is well-defined as P is abelian and the orders of the terms does not matter. In the cocycle identity, if we vary x ∈ G/P and multiply all the m expressions, we get

Raising this to the ath power and observing that all terms commute, we get (since |P| = n

at is,

is can be rewritten as

is proves that the set N of all the elements h(x)as(x)-1 forms a group H. Moreover, the map x ↦ h(x)as(x)-1 from G/P to H is an isomorphism because if x is in the kernel, then s(x) = h(x)a ∈ P which means that the coset x corresponding to s(x) is the identity coset. us, H ≅ G/P. Also H ∩ P is trivial as P, H have coprime orders. us, PH is a group with P ∩ H = {1} and PH has mn = |G| elements thereby proving that G = PH. (iii) is immediate from (ii) since elements of P and N commute.

SOLUTION 84 Note first that the last statement follows from the first since the subgroup L ∩ N must be trivial (as its order divides two coprime numbers) and |L||N| = |G|. us, let us prove the first statement. Write n = |N| and m = |G/N|. We shall apply induction on |G|. At some point, we shall reduce the assertion to the previous result where the normal subgroup is abelian. e assertion is evident if n = 1. Suppose n > 1 and that the assertion holds for all groups G' of order < |G| with a normal subgroup N' such that |G'/N'| is coprime to |N'|. Let p be any prime divisor of n. Now, any p-Sylow subgroup P of N is a p-Sylow subgroup of G as |N| is coprime to

|G/N|. In fact, we claim that the p-Sylow subgroups of G are already contained in N. To see this, look at NG (P). As N ⊲ G, all the G-conjugates of P are contained in N. erefore, the numbers of p-Sylow subgroups in G and in N are equal. In other words, [G : NG(P)] = [N : N ∩ NG (P)]. Writing the le-hand side as [G : N][N : N ∩ NG (P)] = [G : NG (P)][NG (P): N ∩ NG (P)] we obtain [G : N] = [NG (P) : N ∩ NG (P)] = m. As , we have that NG (P))/P has order less than |G| and applying the induction hypothesis to it for the normal subgroup N ∩ NG (P))/P, we conclude that NG (P) has a subgroup H containing P such that |H/P| is of order m = [NG ( P): N ∩ NG (P)]. Look at the centre Z(P) of P. As it is a characteristic subgroup of P ⊲ NG (P), it follows that Z(P)⊲ NG (P). Note that Z(P) ⊂ H; so Z(P) ⊲ H. As , one can apply the induction hypothesis to the group H/Z(P) and obtain in it a subgroup K/Z(P) of order m. Finally, if we consider the group K and its abelian normal subgroup Z(P), the previous result applies to yield a subgroup L ≤ K of order m. us, the proof is complete.

SOLUTION 85 Consider H = < S >, the subgroup of G generated by S. is is finitely generated. Note that S = S–1. Also, if h = s1. . .sn ∈ H, then the centraliser CG (h) contains the subgroup CG (s1)∩. . .∩CG(sn) which is of finite index in G as each CG(si) is (the last statement is because each conjugacy class G(si) – being a subset of S – is finite). Now, the centre Z(H) of H is the intersection of all CH(s), s ∈ S. So, Z(H) has finite index in H and so, [H,H] is finite (indeed, it has been shown in an earlier problem that if Z(H) has index n,

then < [H,H ]n > is trivial). Further, the group H/[H,H] is a finitely generated abelian group, generated by torsion elements. Hence it is finite. us, H itself is finite. So, if a, b ∈ S, the element ab ∈ H has finite order and so, ab ∈ S. Hence S = H.

SOLUTION 86 Let G = < x1,. . ., xn >. Suppose d is the LCM of the orders of x1,. . ., xn. en, xd = 1 for all conjugates of all the xi's. Each element of G is a word in the xi's. e trick here is to also include the finitely many conjugates of the xi's in G and regard this appended set y1, ... ,ym as a set of generators for G. e idea is that the minimal length of any element of G in terms of the yi's will be bounded in terms of m and d. We claim that any element has an expression in the yi's of length ≤ m(d – 1). Start with any x = g1 . . . gr where each gi is some yj. Now, if each yi occurs less than d times, then r < m(d – 1). Hence, we suppose r > m(d – 1) and conclude that some yi occurs at least d times in the expression above. If gj is the first term in the expression which equals yi, we consider hk = yi1 g y for all k < j. en, we have k i x = yi h1 h2. . .hj–1 gj+1. . .gr. Using this repeatedly, we can bring d of the yi's one by one to the le. Each time, the expression has the same length r as before. erefore, at the dth step, since yid=1, we have an expression for x which has length r-1. erefore, if we had an expression of minimal length in the yi's to start with, it must be that all the yi's occur less than d times. Hence, we have proved the claim. is shows, in fact, that |G| ≤ mm(d–1).

SOLUTION 87 Before starting, let us recall the basic notation that in any group [x, y] denotes xyx–1 y–1 and xy denotes yxy–1. Also, recall from Problem 36 that in a group A with [A,A] ≤ Z(A), we have, for every n, (xy)n = xn yn [y,x] n(n–1)/2. Let us start the proof now. (i) Since X =< x >, Y =< y > are normal subgroups, the commutator c = [x,y] ∈ X ∩ Y. So, c = xa = yb for some a, b. Hence c commutes with both x and y. Note also that c ≠ 1 implies a, b ≠ 0, ±1. Hence c ∈ Z(H) where H =< x, y >. Moreover, xyx–1 = cy ⇒ xlyx–l = cly, xlym x–l = clmym. erefore, [H,H] =< c > ≤ Z(H). is means that H is a nilpotent group. Moreover, since ca = [x,y]a = [xa,y] = [yb,y] = 1, we have that c is of finite order. Also, then x, y have finite order as c = xa = yb. By Problem 59 (ii), H must be finite as well. (ii) If x, y do not commute, then as above c = xa = yb. But, if x, y have coprime orders, the orders of xa and yb are coprime as well. is means that c = 1, which is a contradiction. (iii) Let x, y be noncommuting elements for which O(x)+O(y) is minimal. Let p be any prime dividing O(x). en, O(xp) < O(x) which means that xp and y must commute by the minimal choice of x, y. But, if c = [x,y], we know that c commutes with x, y and so cp = [x, y]p = [xp,y] = 1. is means that O (c) = p. Analogously, for an arbitrary prime q dividing O(y), we have O(c) = q. us, it follows that O(x), O(y) are powers of a prime p and O(c) = p.

Now, c = xupr = yvps for some r, s > 1 and u, v coprime to p. Writing x1 = xu and y1 = yv, we have O(x1) = O(x) and O(y1) = O(y) since these are powers of p. Note [x1,y1] = [x, y]uv = cuv has order p. Also, O (x) = O(x1) = pr+1 and O(y) = O(y1) = ps+1 from the fact that O(c) = p. To fix notation, suppose r ≥ s without loss of generality. Consider recalled above,

. en, by the commutator identity in Problem 36

e exponent of c above is a multiple of p if p is odd. If p = 2, the exponent is again a multiple of 2 unless r = s = 1. Since O(c) = p, we have that y2ps = 1 unless p = 2 and r = s = 1. But, [x,y2] = [x,y1]–1 = ([x, y]–1 )v = (c–1)v ≠ 1. erefore, x, y2 do not commute and we must have O(y2) ≥ O(y) by the choice of x, y. Hence O(y2) ≥ O(y) = ps+1 which gives y2ps ≠ 1. erefore, we must have p = 2 and r = s = 1. at is, O(x) = O(x1) = pr+1 = 4 = O(y1) = O(y). Also, c = x12 = y12.

Note that [x1,y1] ≠ 1 means that x1 y1 x1–1, which must be in the group < y1 >of order 4. So, x1y1x1–1, must be y1–1 as it has the same order as y1. In other words, < x1, y1 > is isomorphic to the quaternion group of order 8. Finally, < x, y > is a homomorphic image of < x1, y1 > since x1 = xu and y1 = yv with u, v odd. As < x, y > is nonabelian, it must be isomorphic to the quaternion group of order 8.

(iv) Write C = CG (Q) and suppose that g ∈ G\CQ. en g does not commute with at least one of x and y. Let us say yg ≠ y. Since |y| = 4, we must have yg = y–1; therefore gx commutes with y. us gx cannot commute with x; for otherwise gx ∈ C. e same argument shows that gxy commutes with x: but clearly gxy also commutes with y; so gxy ∈ C and g ∈ CQ. It follows that G = CQ. If g ∈ C, then [x,gy] = [x,y] ≤ 1 and by (i), gy has finite order. erefore, g has finite order and G is a torsion group. Now, we show that the elements of C of order a power of 2 form an elementary abelian 2-group. Suppose that g ∈ C has order 4. en [x,gy] ≠ 1 and (gy)4 = 1, which implies that (gy)x = (gy)–1. us [gy,x] = (gy)–2 = g–2 y–2. But [gy,x] = [y,x] = y–2; so g2 = 1, which is a contradiction. We have shown that C has no elements of order 4; that is, the order of any element of 2-power order, is 2. Hence, the elements of C of 2-power order form an elementary abelian 2group E. Also, as we have already shown, the elements of odd order commute among themselves and elements of coprime orders commute. us, C = E × where is the set of elements of odd order in C. Hence G = CQ = (QE) × O. Since E is elementary abelian, we can write E = (Q ∩ E) × E1 for some subgroup E1. us G = (QE)× = Q × E1 × .

SOLUTION 88 Suppose H ≤ G is minimal with respect to the property that θ(H) = M. We first observe that H ∩ kerθ ≤ Φ(H). Suppose this is not so, and let M0 be a maximal subgroup of H not containing H ∩kerθ. en, H =M0(H ∩ kerθ) by the maximality of M0 in H. But then θ(M0) = θ(H) = M, which contradicts the choice of H as a minimal subgroup of G with this property. erefore, indeed H ∩ kerθ ≤ Φ(H). Now, suppose that p is a prime dividing |H| but not dividing O(M), if possible. If P is any p-Sylow subgroup of H, then P ≤ H ∩ kerθ which means that P ≤ Φ(H), which is a nilpotent group. erefore, P is the unique p-

Sylow subgroup of H; thus, it is normal in H. By the Schur–Zassenhaus theorem, there is a subgroup K of H such that KP = H and K ∩ P = {1}. But then θ(K) = M, which is a contradiction of the minimal choice of H.

SOLUTION 89 (i) We need to show that the images of xy and g in the abelian group G/[G,G] are equal. Write an+b|G| = 1. en, in the abelian group G/[G,G], we have xy = (xy)an+b|G| = (xy)an = ((xy)n)a = (xn yn )a = (gn )a = gan = gan+|G| = g. (ii) Let z be the product of all elements of G in some order. As before, let us compute in the abelian group G/[G,G] and keep in mind that the order in which elements are multiplied is irrelevant. Each element cancels with its inverse and there are no nontrivial elements cancelling off since G and, a fortiori, G/[G,G] has odd order. us, z is trivial in G/[G,G]. (iii) Let S be a 2-Sylow subgroup containing j. Note that j is the only element of order 2 (also called an involution) in S as S is cyclic. Similarly, each 2Sylow subgroup (being cyclic) contains a unique involution and the conjugacy of 2-Sylow subgroups gives us the conjugacy of all involutions of G. Hence, the set I of all involutions of G has cardinality [G : CG (j)]. As S ≤ CG (j), the index n: = [G : CG (j)] is a divisor of [G : S] and, is odd. Now, if g is the product of all elements of G in some order, then g[G,G] is an element of the abelian group Gab = G/[G,G], which does not depend on the order of the factors defining g. Now,

where J : = {x ∈ G : O(x) ≠ 2}. Since x ∈ J ⇒ x–1 ≠ x ∈ J, it follows that Further, yjy–1 [G,G] = j[G,G]. Hence,

since j has n conjugates and

as n is odd. Finally, we show that j ∉ [G,G]; this will imply that 1 ∉ j[G,G] and that [G,G] ≠ j[G,G]. We know from an earlier exercise that there is a normal subgroup N of G such that G = NS and N ∩ S = {1}. As G/N ≅ S is abelian, [G,G] ≤ N. is shows that j ∉ [G,G] since j ∈ S and N ∩ S = {1}.

SOLUTION 90 (i) If x1,. . ., xr ∈ N generate N and y1N,. . . , ysN generate G/N, then clearly x1,. . . , xr, y1,. . ., ys generate G. Indeed, if g ∈ G, then which means that for some words w, w0. us, g is a word in the xi's and the yj's and their inverses. (ii) Let H ≤ G be of finite index and write G = ⊔ni = 1 H gi where g1 = 1. en, for each g ∈ G, there is a corresponding permutation i ↦ (i)g of {1, 2, , n} such that Hgig = Hg(i)g. Here, we have denoted by (i)g the action of g on i; this is convenient because we have adopted the convention of applying g first in a product gg1. Now, we have gig = h(i,g)g(i)g for some h(i,g) ∈ H. Let X be a finite set of generators for G. We claim that the elements h(i,x), x ∈ X ∪ X–1 generate H. Let h ∈ H. Write h = x1. . .xr with xi ∈ X ∪ X–1. Now, h = g1h = g1 x1. . .xr = h(1,x1)g (1)x1 x2 . . .xr = h(1,x1)h((1)x1,x2)g (1)x1x2 x3. . .,xr

= h(1,x1)h((1)x1,x2). . .h((1)x1. . .xr–1,xr)g (1)x1. . .xr = h(1,x1)h((1)x1,x2). . .h((1)x1. . .xr–1,xr)g (1)h. Note that h = g1h ∈ Hg(1)h implies that Hg (1)h = H; that is, g (1)h = g1 = 1. is proves the assertion that H is finitely generated. To deduce the assertion that any finite group G is finitely presentable, we write a surjective homomorphism θ : F → G from a free group F of finite rank. Since F is finitely generated and ker θ is of finite index |G| in F, what we proved shows that ker θ is finitely generated as well. us, if R is a finite set of generators for ker θ and X a finite basis of F, then G = < X|R >. (iii) Let G be a group defined by the presentation stated. If πi = (i i+1) ∈ Sn, then clearly the πi's generate Sn and satisfy the relations stated. erefore, there is a surjective homomorphism from G onto Sn. So, |G| ≥ n!. So, it suffices to show that |G|≤ n! for this would show that the above homomorphism has to be an isomorphism. We shall show that |G| ≤ n! by induction on n > 1. is is clear for n = 2. Assume n > 2 and that the subgroup H of G generated by x1,. . ., xn-2 has order ≤ (n-1)!. We shall show that [G:H] ≤ n. Consider the right cosets H, Hxn-1, Hxn-1 xn-2,. . ., Hxn-1 . . . x1. We claim that right multiplication by any xl permutes these right cosets. is would imply that for any word w in the xi's is in one of these cosets and since any element of G is a word in the xi's, we would have [G:H] ≤ n. Let us consider (H xn–1 xn–2. . .xk)xl for different k, l. If k ≥ l +2, then xk xl = (xk xl )-1 = xl xk and, we know that all the x's in the bracket commute with xl. erefore, if ≥ l +2, we get (Hxn-1 xn-2. . .xk)xl = Hxl xn-l xn-2. . .xk = Hxn-l xn-2 xk since xl ∈ H for all l ≤ n-2.

If k < l, then since xl commutes with all the xm for |m-l|>1, we can move xl to the le and get (Hxn-1 xn-2 ...xk)xl = Hxn-1 xn-2. . .xl xl-1 xl xl-2 . . .xk. Now, we rewrite the relation (xl xl-1 )3 = 1 to get xl xl-1 xl = xl-1 xl xl-1 since all xi's have order 2. Using this, the above equality reduces to (Hxn-1 xn-2 . . .xk)xl = Hxn-1 xn-2. . .xl+1 xl xl-1 xl xl-2 . . .xk = Hxn-l xn-2. . .xl+1xl-1 xl xl-1 xl-2. . .xk = Hxl-lxn-l xn-2. . .xl+1 xl. . .xk = Hxn-1 xn-2. . . . . .xk as xl-1 ∈ H for l ≤ n-1. Now, we are le with the cases k = l and = l +1. In the first case, (Hxn-1xn-2 . . .xk) xk = Hxn-l xn-2 xk+1. In the second case, (Hxn-1 xn-2 . . .xk)xk-1 = Hxn-l xn-2 xk xk-1. erefore, the proof that [G:H] ≤ n is complete.

SOLUTION 91 (i) We need to show that any nontrivial reduced word in X, Y does not give the identity matrix. We shall show by induction that any matrix of the form Xa1 Yb1 . . .Xar Ybr

with ai, bi nonzero integers, has (1, 1)th entry and (2, 1)th entry q(e), where p(e), q(e) are (possibly constant) polynomials in e with integer coefficients and degrees strictly less than r. As e does not satisfy any nonzero integer polynomial, the above inductive assertion can easily be seen to prove that nontrivial reduced words in X and Y do not give the identity matrix. Indeed, if a word of the form YbO Xa1 Yb1 . . .Xar Ybr with r > 0 , is I, then Xa1 ) Yb1 . . .Xar Ybr = Y-bO which is impossible since the le side does not have 1 at the (1, 1)th place. Similarly, the other words are also dealt with. Now, let us prove the inductive assertion. Clearly,

which shows the assertion holds for r = 1.

Suppose n > 1 and that Xa1 Yb1 . . .Xan Ybn has (1, 1)th entry and (2, 1)th entry q(e) where p(e), q(e) are (possibly constant) polynomials in e with integer coefficients and degrees strictly less than n. en,

where p1, q1 are integral polynomials in e with degrees less than n+1. is proves the inductive assertion.

(ii) e proof is the same as for free abelian groups in Problem 51; here, we look at the set S(m) := Hom(Fm, /2) of all homomorphisms from Fm to /2. Once again, it has cardinality 2m since each basis element can be sent to either element of /2 and define different homomorphisms. Similarly, S(n) has cardinality 2n. Fixing an isomorphism θ : Fm → Fn, it is clear that the map αnm :S(n) → S(m); ϕ ↦ ϕ 0 θ is a bijection. us 2m = 2n and hence m = n. (iii) Clearly, each element of the subgroup H is uniquely expressible as a reduced word in the elements ynxy-n;n ∈ . us, H is not finitely generated. (iv) Write Fn = < x1,. . ., xn| θ > , and notice that [Fn,Fn] = < [xi,xj ]; i ≠ j > N. us,

SOLUTION 92 Now, one can write the presentation < x, y, z|[x,z], [y,z] > for G. One can identify H with the subgroup of G generated by x and y and K with the subgroup of G generated by x and yz. As z commutes with x, y in G, a word w in x, y, z is in the intersection of the subgroups H and K if and only if the sum of the exponents of z in w is zero. However, note that for any word in K, the sum of the exponents of y and of z are equal. Hence H ∩ K is the subgroup of G generated by all ynxy–n;n ∈ Z. is was seen in the last problem to be infinitely generated.

SOLUTION 93 (i) Let θ : F(X) → G with kerθ = < R >N. Consider α = π ∘ θ : F(X) → Gab where π : G → Gab is the natural map. Note that R ∪ {[x,y];x,y ∈ X} ⊂ kerα.

If f ∈ kerα, then θ(f) ∈ kerπ = [G,G] = < ([x,y]);x, y ∈ X > . erefore, f–1w ∈ kerθ where w ∈< [x,y];x, y ∈ X >. Hence, f–1w is a product of conjugates in F(X) of elements of R. erefore, f is a product of conjugates of elements of R and elements of the form [x,y] as x, y ∈ X. is proves that kerα = < R ∪ {[x,y];x,y ∈ X} >. (ii) Let X = {x1,…,xr}, and R = {w1,…,ws}. If the sum of the exponents of xj in the word wi is the integer aij, then a presentation of Gab is

In other words

where A = (aij) is the As A is a rectangular matrix, this is infinite. In fact, if d1,…,ds are the invariant factors of A, then is infinite since r > s.

SOLUTION 94 (i) We shall prove, equivalently, that S and A := S–1 X–1 generate SL(2, Z). Note that . Now, start with any

. We shall show that le and right

multiplications by powers of X and Y lead to ±I by the usual Euclidean division algorithm. For any integer l, we have

. is shows us that one can

divide a by c and replace a by its residue mod c.

Similarly, one can see that by le multiplication by some Yl, one can reduce c mod a. Repeating these finitely many times, the division algorithm implies that one of a and c becomes zero; the other has to be ±1 as the determinant is always 1.

and g2 = Xb or –X–b Since –I = S2, the assertion follows. (ii) Since S2 = –I also represents the identity element in PSL(2, Z), the image s of S has order 2 in PSL(2, Z). Also, the image b of

in PSL(2, Z) has order 3 as (SX)3 = –

I. We know that the elements s, b generate the whole group; so, we need only show that no matrix SBa1 SBa2…SBar ) with each ai either 1 or 2, can be the matrices I, –I. Since SB = –X and SB2 = Y, it follows that any word in the positive powers of SB and SB2 is a matrix in which a, b, c, d are of the same sign. erefore, if b ≠ 0 , then the corresponding entry –b –d of SBg and b of SB2g are non-zero as well. Similarly, if c ≠ 0, the corresponding entries of SBg and SB2g are nonzero. Since SB and SB2 have the property that either the (1,2)th entry or the 2, 1)th entry is nonzero, any word g in their positive powers has this property; hence g can not be the identity matrix. (iii) follows from (ii) by sending x to S and y to SX.

(iv) Now,

e invariant factors of the subgroup < 2e – 3f, 4e > above are the invariant factors of the matrix . e latter is computed by computing h1(A) = 1 = d1 and h2(A) = 12 = d1d2. Clearly, d1 = 1 and d2 = 12. erefore, the abelianisation of SL(2, Z) is the cyclic group of order 12.

SOLUTION 95 Let

If a = 0, then we see that

Since

it follows that g ∈ < U+, U– > in this case. On the other hand, if c = 0, then we see that

Since

it follows in this case also that g ∈ < U+, U– >. Finally, if ac ≠ 0, then

Since this is of the forms considered earlier, < U+, U– > = SL(2,K). Finally, to show that U+, U– are conjugate, write, for any t ∈ K,

Note that Y(–t) = wX(t)w–1 where

SOLUTION 96 (i) e fact that the above matrices generate SL(2,K[t]) easily follows by using the division algorithm for K[t]. Let us carry it out.

erefore, when either a(t) = 0 or c(t) = 0, we are done.

Suppose a(t)c(t) ≠ 0. Let us write a(t) = q(t)c(t) + r(t) with either r(t) the zero polynomial or degr < deg c. en,

If r(t) is not the zero polynomial, one can write c(t) = q1(t)r(t) + r1(t) with either r1(t) zero or of degree less than deg r(t). But then

In this manner, we could keep dividing the (1, 1)th entry by the (2,1)th entry and vice versa, accomplished by le multiplication by a matrix of one of the above forms. Aer finitely many steps, we arrive at a matrix

where either a1(t) = 0 or c1(t) = 0. We have dealt with both these cases above. us, we have proved the assertion on matrices of certain special forms generating SL(2,K[t]). (ii) By the previous problem, SL(2, K) is generated by U+ and its conjugate subgroup U–. Hence, if we prove that U+ ≤ [SL(2,K),SL(2,K)], it will prove that SL(2, K) is perfect. As |K| > 2, there exists a ∈ K* with a2 ≠ 1. en,

proving that SL(2, K) is perfect. Finally, we prove that SL(2,K[t]) is perfect. For a ∈ K* with a2 ≠ 1, note that

Since

the matrices

are commutators

too.

SOLUTION 97 (i) If G = ⊕Z/p is an elementary abelian p-group, then every group automorphism is also Z/plinear. So, if |G| = pn, then Aut G ≅ GL(n,Z/p). To show the identity above, let us note that (pa – 1)(pa – p)…(pa – pa–1) is just the order of the automorphism group of the elementary abelian pgroup of order pa. Let us denote this by θ(a). Now, consider GL(a+b,Z/p). e subset of all those matrices whose top right a × b blocks consist of zeroes, is a subgroup H. Its order is clearly pab times the product of the orders of GL(a,Z/p) and GL (b,Z/p). us, the fact that this order divides the order of GL(a+b,Z/p) gives us the identity.

(ii) Let p be any prime to start with. Let G be an elementary abelian p-group of order pp–1. en, Aut G ≅ GL(p – 1,Z/p). If A ≤ AutG is the subgroup of monomial matrices (that is, matrices with each row and each column having exactly one non-zero entry), then |A| = (p – 1)p–1(p – 1)! which is coprime to |G|. Now, if p is large enough, then

erefore, since log |G| = (p – 1) log p, we get

for large enough p.

SOLUTION 98 (i) Now, Int G is a normal subgroup of Aut G and is isomorphic to G since Z(G) = {1}. We will show that any element θ ∈ AutG commuting with every element of Int G must be trivial. Recall that x(Int(g)) = g–1xg for all x, g. If θ = Int(g)θInt(g)–1, then for any x ∈ G, we have x(θ) = gθ(g–1xg)g–1. So, y = x(θ) satisfies y(Int(g)) = y(Int(θ(g)g–1)). As y = x(θ) can be any element of G, we get that θ(g)g–1 ∈ Z(G) = {1}. Since this holds for all g ∈ G, θ is the identity automorphism. (ii) Consider the chain

A0 := G ⊴ A1 := Aut(G) ⊴ A2 = Aut(A1) ⊴…… Now, by (i), we have that CAi+1(Ai) = {1} for i ≥ 0. We show that CAi(A0) =

{1} for all i. In fact, we show that CAi(Aj) = {1} for all i > j by induction on i – j. By hypothesis, this assertion holds for i – j = 1.

Suppose i – j > 1 and consider CAi(Aj). By induction hypothesis, CAi–1(Aj) = CAi(Aj+1) = {1}. Now, as Aj+1 normalises Aj, it normalises also CAi(Aj). at is, [Aj+1,CAi(Aj) ≤ CAi(Aj). Also, as Aj+1 ≤ Ai–1, we have

Hence,

In other words,

Hence, we have shown by induction that

In particular, CAi(A0) = {1}. By Wielandt's theorem, since G = A0 is a

subnormal subgroup of each Ai, the orders of all the Ai's are bounded by |A0| = |G|. Hence, the chain of automorphism groups stabilises.

SOLUTION 99

Before proceeding, note that induction is trivially started with the trivial group. (i) Suppose C = C(x) has cardinality > 1; then x ∉ Z(G). Also, then CG(x) ≠ G. We note that for any y = hxh–1 ∈ C, the number a(y,n) = a(x,n) because gn = x ⇔ (hgh–1)n = y. erefore, a(C,n) = |C|a(x,n). Now, let us apply the induction hypothesis to CG(x) and its corresponding conjugacy class of x,

which is simply {x}. Note that if g ∈ G with gn = x, then g ∈ CG (x) as g commutes with gn. erefore, the induction hypothesis gives that a(x,n) is a multiple of (n,|CG(x)|) = (n,|G|/|C|). us, we have that a(C,n) is a multiple of (n|C|,|G|). (ii) If D := {g ∈ G : gn1 ∈ C}, then {g ∈ G : gn ∈ C} = {g ∈ G : gn2 ∈ D}. As n1, n2 > 1, we get by induction hypotheses that a(D, n2) ≡ 0 mod (n2|D|,|G|) and |D| = a(C,n1) ≡ 0 mod (n1|C|,|G|). is clearly implies that a(C,n) ≡ 0 mod (n|C|,|G|). (iii) With n = pr > 1 and x ∈ Z(G) with p|O(x), we claim that if gn = x, then O(g) = prO(x) and that |{gi : gin = x}| = n. e first of these assertions would imply that prǁG| and, so (n,|G|) = n; the second one implies that n divides a(x,n). So, let us prove the claimed assertions. Now, the second one follows from the first because gin = x ⇔ xi–1 = 1; that is, i ≡ 1 mod O(x). Now, if gi = gj for some i = 1+rO(x), j = 1+sO(x), then gi–j = 1 so that i–j is a multiple of O(g) = nO(x). is means that r–s is a multiple of n. Hence, the set of powers gi with gin = x consists of all gi with 0 ≤ i < n and its cardinality is, therefore, n. So, it suffices to prove the first assertion that O(g) = nO(x). Now, O(g)|nO(x) since gnO(x) = xO(x) = 1. Write O(g) = psd where p d. Now,

As p|O(x), we must have s > r and O(x) = ps-rd. erefore, O(g) = prO(x) which proves the assertion. (iv) Let y ∈ Z; hence (n,O(y)) = 1. erefore, the map g ↦ gn from < y > to itself is a homomorphism without kernel. Hence it is onto also. One can write y = ymn for some m < O(y). Now, {g : gn = y} = {g : gn = ymn} = {g : (y–mg)n = 1} = ym{t : t ∈ G, tn = 1}. erefore, a(y,n) = |{t : t ∈ G, tn = 1}|. In other words, this number is independent of y ∈ Z. (v) Finally, for each n, we write the group into a union of sets of elements g for which gn belongs to its various conjugacy classes. Let, as before, Z = {y ∈ Z(G) : (O(y),n) = 1}. erefore,

Using induction on |G|+n and the steps (i) to (iv), we have that (n|Ci|,|G|)|a(Ci,n) for each Ci ⊈ Z. Hence, from the above class equation, we must have (n,|G|)|a(Z,n) as (n,|G|) divides the le hand side as well as each term in the sum. But, we observed in (iv) that a(Z,n) = |Z|a(y,n) for each y ∈ Z. Hence, (n,|G|)||Z|a(y,n) for each y ∈ Z. Now, since each element of Z has order coprime to n, Z forms a group because, if y, z ∈ Z, then yz ∈ Z(G) and O(yz)|O(y)O(z). Moreover, Z has order coprime to n for, if not, there is a prime p|n and an element y ∈ Z of order p by Cauchy's theorem. Hence, we get (n,|G|)|a(y,n) for each y ∈ Z also. e proof is complete.

SOLUTION 100 We know from Frobenius's result above that in any finite group G, the number of elements g satisfying gr = 1 is a multiple of (r,|G|). Let us use this.

In any finite group G, we shall produce for each rǁG|, a set Sr of ϕ(r) elements g satisfying gr = 1 and such that the various Sr are disjoint. If we do this, then we can map the elements of Sr bijectively onto the set of elements of order r in the cyclic group of order |G|. Since O(g) ≤ r for each g ∈ Sr, we will get that aα(G) ≤ aα(C) if α ≤ 0 and aα(G) ≥ aα(C) if α ≥ 0 where C is the cyclic group of order |G|. To produce Sr, we work by induction on r. Firstly, choose S1 = {1}. Suppose the sets Sd for all divisors d < r of |G| have been chosen. Now, the above fact implies that the set {x ∈ G : xr = 1} has at least r elements. e sets Sd;d|r, d < r have been chosen from this set and contribute elements. us, there are ϕ(r) elements from the set {x ∈ G : xr = 1} which are still le over and Sr can be chosen. is completes the proof.

SOLUTION 101 If G is simple, we are through; so assume that there is a nontrivial normal subgroup N of minimal possible order. Let us consider all possible subgroups of G which are isomorphic to a group of the form N1 × … × Nr where Ni ⊲ G and Ni ≅ N for some r ≥ 1. Let M be such a subgroup where r is maximum possible. Now, since Ni are normal and their images generate M, we have M ⊴ G. We claim that M is actually a characteristic subgroup of G. To see this, start with any σ ∈ AutG. Evidently, σ(Ni) ⊲ G and σ(Ni) ≅ N. erefore, if some σ(Ni) ⊈ M, then σ(Ni)∩M is a normal subgroup of G with order less than that of σ(Ni) ≅ N. is forces σ(Ni)∩M = {1}. But then < M, σ(Ni) > ≅ M × σ(Ni) is a group of the same form as M but with r + 1 factors. is is a contradiction of the choice of M. us, we must have that σ(Ni) ≤ M for each i. erefore, σ(M) ≤ M and, so M is a characteristic subgroup of G. As M is nontrivial, this gives by the hypothesis that M = G. Finally, N

must be simple because any normal subgroup of N is also normal in N1 × … × Nr = G.

SOLUTION 102 (i) Let p be any prime dividing |G| and let pn be the highest power of p dividing it. If p /| |G(x)|, then pnǁCG(x)|. If p| |G(x)|, then p |G(y)| and, therefore, pn||CG(y)|. Hence G = CG(x)CG(y). Now, clearly G(xy) ⊆ G(x)G(y). Conversely, let axa–1byb–1 ∈ G(x)G(y). en,

axa–1byb–1 = acxc–1 a–1 bdyd–1 b–1 for any c ∈ CG(x), d ∈ CG(y). Since G = CG(x)CG(y), we may write a–1b = cd– 1 for some c ∈ C (x), d ∈ C (y). But then c–1a–1bd = 1 and thus, axa–1byb–1 G G = ac(xy)c–1a–1 ∈ G(xy).

(ii) If G is infinite and G(x), G(y) are finite, we have that CG(x, y) := CG(x) ∩ CG(y) is of finite index in G. So, we can choose a normal subgroup N of finite index in G which is contained in CG(x, y). Working in the finite group G/N, we have the same assertions.

SOLUTION 103 Let n ∈ N have m(N) conjugates and let gN ∈ G/N have m(G/N) conjugates. en,

Also, if we denote G/N by

us,

and gN by , we have

Taking x = g or gn, we get

SOLUTION 104 We use induction on |G|. Let P be a minimal normal subgroup. en, it is a p-group for some prime p. Write |P| = q say. First assume that p|m. en, evidently q|m. By the induction hypothesis, G/P has a subgroup H/P of order m/q. en, of course, H has order m. Let K be a subgroup of order d dividing m. en, KP is a subgroup of G of order dividing qd. In particular, (|KP|,n) = 1. So, |KP|/P divides m/q. By the induction hypothesis, there exists x ∈ G such that KP/P ≤ xP(H/P)(xP)–1 = xHx–1/P. us, KP ≤ xHx–1 and both (ii) and (iii) follow. Now, we suppose that p m. en p|n. By the induction hypothesis, G/P contains a subgroup L/P of order m. en, of course, |L| = mq. Look at P as a normal subgroup of L. Its order is coprime to its index m in L. us, by the Schur-Zassenhaus theorem, L contains a subgroup M of order m. erefore (i) is proved. Now, if Q is a subgroup of G of order r dividing m, then QP/P ≅ Q is a subgroup of order r in G/P. By induction hypothesis applied to G/P, there is y ∈ G so that yP(QP/P) (yP)–1 ≤ L/P. us, yQy–1 ≤ L and has order r. But yQy–1 is conjugate to a subgroup of a complement Mʹ of P in L. In other words, Q is conjugate to a subgroup of Mʹ (which has order m. erefore, (ii) and (iii) are also proved.

SOLUTION 105 Let {x1,…,xn} be a basis of F. Choose arbitrary 'lis' gi in G of the xi's; that is, let π(gi) = xi. en, the mapping xi ↦ gi extends to a unique homomorphism s : F ↦ G as F is free on the xi's. Consider the subgroup s(F) of G. Note that π ∘ s is the identity homomorphism from F to itself as it is identity on the xi's.

erefore, for g ∈ G, the elements g and s ∘ π(g) have the same images under π. So, g = hn for some n ∈ kerπ and h ∈ s(F). erefore, G = s(F)kerπ. If g = s(f) ∈ s(F) ∩ kerπ, then 1 = π(g) = π(s(f)) = f which gives g = s(f) = 1.

SOLUTION 106 If G is a nontrivial group which has no nontrivial maximal subgroups, it is a cyclic group of prime order. So, let us assume that G does have nontrivial maximal subgroups. By hypothesis, if H is one such, then all nontrivial maximal subgroups are of the form gHg–1. Trivially, each element of G which is not a generator of G generates a proper subgroup and, must lie in some maximal subgroup. Hence, if G is not cyclic, we would have G = ∪g∈GgHg–1. is is easily seen to be impossible. Indeed, if we write , then for each g ∈ G, there is some i for which g ∈ giH; so, , which means . In other words, if , then . However,

as the conjugates always have the common element identity. erefore, G must indeed be cyclic. If p ≠ q are two distinct primes dividing |G|, then the subgroups of orders |G|/p and |G|/q are both maximal subgroups which are not conjugate. Hence |G| is the power of a single prime.

SOLUTION 107 Note that s2(g) ≠ 0 if, and only if, g = x2. erefore, we need to compute .

If x1,…,xn ∈ G are all the different 'square roots' of x2, then . In other words, the sum we want to compute is simply the number of pairs (not ordered) y, z in G such that y2 = z2. Now, if y2 = z2, then y2z–1 = z; that is, y(yz–1)y–1 = zy–1. us, the element yz–1 is conjugate to its inverse. Calling w = yz–1, the various elements t for which twt–1 = w–1 are the elements of the coset xCG(w). erefore,

Here, we have written w ~ w–1 to denote that w is conjugate to w–1. Now, for a conjugacy class G(w) with w ~ w–1, if G(w) = {w1,…,wn}, then |CG(wi)| = |CG(w)| for all i. Hence,

where c is the number of conjugacy classes G(w) in G for which w ~ w–1. is proves the assertion.

SOLUTION 108 Note that the binomial identity follows simply by putting x = N. So, let us prove the polynomial identities. We shall apply induction on n. ey are evident for n = 1. Assume that they are true for n ≥ 1 and let us prove it for n + 1. Look at the first identity for n. We have

en,

Now,

Where

Note that each τ ∈ In+1 restricts to a unique σ ∈ Sn and c(σ) = c(τ) – 1. Moreover, for each r < n+1, as τ runs over Ir, the elements σ := τ(n+1 r) run over Sn. Note that c(σ) = c(τ). erefore,

us, we have proved the first identity. e second is exactly similar.

SOLUTION 109 (i) Our first aim is to count the number of subsets S of An which admit a Gbijection with the set G/H of le cosets of H. In the language of representation theory of finite groups, this amounts to counting the multiplicity of the representation induced on G by the trivial representation of H in the representation An of G. Let S ⊂ An be such a subset and let π : G/H → S be a G-bijection. As G acts transitively on G/H (i.e., any two le cosets are permuted by some element of G, this must hold for the subset S also. is means that S is simply a G-orbit in An, say, S = G · (a1,…,an). We may assume that π(H) = (a1,…,an) where H is the identity coset. Now, π(gH) = g(a1,…,an) for all g ∈ G. Since π is well-defined, it follows that (a1,…,an) is

an H-invariant element. Furthermore, if g(a1,…,an) = (a1,…,an) for some g ∈ G, then gH = H i.e., g ∈ H. In other words, (a1,…,an) is exactly Hinvariant. Conversely, suppose that (a1,…,an) is exactly an H-invariant. en, if we define

then θ is well-defined as (a1,…,an) is H-invariant and it is 1–1 since (a1, …,an) is exactly H-invariant. Evidently, θ is onto. is proves (i). (ii) We wish to count the number of these G-orbits. us, we have to scrutinise the exactly H-invariant elements and decide when two such give the same G-orbit. Suppose, (a1,–,an) and (b1,…,bn) = x(a1,…,an) are both exactly H-invariant for some x ∈ G. Since gH ↦ g(b1,…,bn) = gx(a1,…,an) is well-defined, we must have ghx(a1,…,an) = gx(a1,…,an) for all h ∈ H. In other words, x–1hx fixes (a1,…,an) for any h ∈ H. As (a1,…,an) is exactly Hinvariant, one obtains x–1hx ∈ H ∀ h ∈ H, i.e., x–1Hx = x, i.e., x ∈ NG(H), the normaliser of H in G. Of course, for x, y ∈ NG(H), we have x(a1,…,an) =

y(a1,…,an) if and only if y ∈ xH. Hence, the number of subsets of An which admit a G-bijection with G/H is where e(H) denotes the number of exactly H-invariant elements in An. In particular, the number positive integer.

is a

is could give us useful information if we can compute e(H) explicitly. First, note that since each H-invariant element is exactly K-invariant for a unique subgroup K containing H, one has , the total number of H-invariant elements. Also i(G) = e(G). Now, it is easy to compute i(K) for any subgroup K of G as follows. If the orbits of K in An correspond to the sets of subscripts I1, I2,…,Ic(K), then and c(K) is the number of K-orbits. Clearly, then (a1,…,an) ∈ An is K-invariant if and only if for each j, all the ai's for i ∈ Ij are equal. us, i(K) = |A|c(K) = ac(K).

Now, the expressions for any H, yield, by the inclusionexclusion principle that e(H) = , where the Moebius function μH is defined on subgroups containing H by μH(H) = 1 and for any subgroup L ⊃ H. erefore, we have obtained that

Combining this with the observation above that the number positive integer, we obtain:

is a

e particular case when G ≅ Z/>pr and H = {1} gives us the identity

SOLUTION 110 Note first that if xyn = ynx for some n ≥ 1, then yn = xynx–1 = (xyx–1)n; so y = xyx–1. Now, let x, y ∈ G. Since Aut(G) is a torsion group, the element Int(y) has finite order, say n ≥ 1. en, yn ∈ Z(G) and, therefore, xyn = ynx. By the observation above, xy = yx. us, G is abelian. If G is also finitely generated, then it must be free abelian since there are no nontrivial torsion elements. Writing G ≅ Zr, we have Aut(G) ≅ GL(r,Z). Clearly, this is a torsion group if, and only if, r = 1.

SOLUTION 111 Now, for any x ∈ G, we have x(θn) = x(Int(g)) = g–1xg. So, x(θn+1) = (g–1xg) (θ) = g(θ)–1x(θ)g(θ). In other words, for y = x(θ), we have

y(Int(g)) = y(θn) = y(Int(g(θ))). As x runs through G, so does y and, we get g–1g(θ) ∈ Z(G). Call this element z. So, we have g(θ) = gz. e subgroup H generated by Z(G) and g is abelian and le stable by θ as Z(G) is a characteristic subgroup of G. As θn is inner conjugation by g, we have g(θn) = g. at is, the element gg(θ)g(θ2)…g(θn–1) is fixed by θ. By hypothesis, this is an element of Z(G). On the other hand, gg(θ)g(θ2)…g(θn–1) = gnw 2

for some w ∈ Z(G). Hence, gn ∈ Z(G) as well. is just means that θn ) = Id. Further, suppose θn itself is the identity. en, g ∈ Z(G). Conversely, if g–1g(θ) = x–1x(θ) for some x ∈ Z(G), then xg–1 is fixed by θ. is gives by hypothesis that xg–1 ∈ Z(G). erefore, Id = Int(x) = Int(g) = θn.

SOLUTION 112 Writing G as a union of the double cosets KgH, it is necessary and sufficient to write each double coset in the form asserted since each double coset KgH is a union of right cosets of K in G and also a union of le cosets of H in G. We note the two bijections:

Here, the notation is explained as follows. If B is a subgroup of a group A and if S is a subset of A which is a union of le cosets of B in A, then one has written S/B to denote the set of le cosets of B contained in S. Similarly, if T is a subset of A which is a union of right cosets of B, then one has written B\T to denote the set of right cosets of B contained in T. It is easy to verify that the above are bijections. Note that the right sides of these bijections count cosets of subgroups in groups (and not merely sets). If there is a bijection,

where i = 1,…,r, then we see that

erefore, we only have to establish a bijection between K/(K∩ gHg–1) and (H ∩ g–1Kg)\H. Now, we note that K/(K ∩ gHg–1) is in bijection with g–1Kg/(g–1Kg ∩ H). Now, since H, K have the same finite index in G, so do the subgroups H and g–1Kg of G. us, it suffices to show that the subgroup g–1Kg∩H has finite index. But, this is true since both H and g–1Kg do.

SOLUTION 113 (i) Let σ be such a permutation. en, clearly σ–1 is another such. So, if we take an expression of σ as a product of an even number of transpositions and

an expression of σ–1 as a product of an odd number of transpositions, we have an expression of the identity permutation as a product of an odd number of transpositions. (ii) Look at the second occurrence of a. Since (c,d)(a,e) = (a,e)(c,d) for disjoint transpositions and since (c,d)(a,c) = (a,d)(c,d), one can bring a next to (a,b) without changing either k or the number of times a occurs. Now, we have Id = (a,b)(a,d)……for some d. If d = b, we have an expression of Id as an odd number k – 2 < k of transpositions which is a contradiction of the choice of k. So b ≠ d and we get Id = (a,b)(a,d)… = (a,d)(b,d)… which has length k and a occurs a fewer number of times. us, we have a contradiction and the result follows.

SOLUTION 114 (i) Consider the mapping θ : H → H given by h ↦ [g,h]. is is 1–1 because ghg–1h–1 = gkg–1k–1 implies that h–1k ∈ CG(g) which, in turn, implies that k = h by the hypothesis. Since H is finite, this is onto as well. (ii) Take G = GL(n, Fq) and g = diag(t1,…,tn), where ti ∈ are distinct. en, it is seen that CG(g) = {diag(s1,…,sn) : si ∈ }. If x, y ∈ U(n,q) lie in the same le coset of CG(g), then x–1y would be a diagonal matrix in U(n,q) which implies that x = y. By (i), we get the result.

SOLUTION 115 Put |N| = n and fix g ∈ G, x ∈ N. We need h ∈ N so that h–1(gx)nh = gn i.e., so that h–1 (gx)nhg–n = 1.

e first easy observation we shall make is that there is at least some natural number r for which (gx)r and gr are conjugate under an element of N. In fact, it will be easily seen that every element h of N conjugates some power (gx)r to gr.

To see this, simply consider all the elements of the form h–1(gx)ihg–i for i ≥ 0. As (gx)ig–i ∈ N for all i, we have h–1(gx)ihg–i = h–1(gx)ig–i(gihg–i) ∈ N. As N is finite, there exist i > j ≥ 0 so that h–1(gx)ihg–i = h–1(gx)jhg–j; that is, h–1(gx)i–jh = gi–j. Let r = r(h) be the smallest natural number for which h–1(gx)rh = gr. erefore, (gx)rhg–r = h. We see that

e le-hand side has cardinality mr(h1) while the right-hand side is precisely the set of elements h1z where z ∈ CN(gr(h1)). As CN(gr(h1)) is a subgroup of N, it has for cardinality a divisor of n. us, r(h1)|n. As , we also have . e proof is complete.

SOLUTION 116 (i) Suppose xn+1H = yxnH for all n ∈ Z. en, x–n+1yxn ∈ H for all n ∈ Z. Now,

is means that xnH = ynH for all n. Conversely, suppose xnH = ynH for all n ∈ Z. en,

which means that xn+1H = yxnH for all n. Now, it is evident that Hx is a subgroup of H normalized by x. (ii) Now, we assume that x, y are related as above and show that their |H|th powers are conjugate. As Hx is a subgroup of H which is normalized by x, < Hx, x> >⊴ G. Now, as

y = xx–1y and x–1y ∈ Hx, we conclude by the previous problem that x|H| and y|H| are conjugate.

SOLUTION 117 (i) Let x ∈ G. We shall show that any y ∈ G which is equivalent to x with respect to H can be uniquely expressed as hxzh–1 where z ∈ Hx and where h runs through a set of right coset representatives of Hx in H.

Let us show that any y equivalent to x has an expression as asserted above. Now, there is h ∈ H such that xnH = hynH = (hyh–1)nH for all n. By the previous problem, we have xn+1H = (hyh–1)xnH for all integers n. So, (hyh–1 )–1x ∈ Hx; that is, hyh–1 ∈ xHx. is gives an expression as asserted. For uniqueness, suppose that hxz1h–1 = xz2. en, x–1hx = z2hz1 ∈ HxhHx. As x normalises Hx, we get x–nhxn ∈ H for all n. In other words, hxnH = xnH for all n; that is, h ∈ Hx. us, the uniqueness is also proved. (ii) By (i), there are clearly |H| elements which are equivalent to x with respect to H. If y is equivalent to x, then there exists h ∈ H with xnH = hynH = (hyh–1)nH for all n ∈ Z. By the previous problem, x|H| and (hyh–1)|H| are conjugate, which means that x|H| and y|H| are conjugate. is completes the proof.

SOLUTION 118 Write that the

and let Pi be a pi-Sylow subgroup. If x ∈ G, and i ≤ r, note elements y which are equivalent to x with respect to Pi are such

that and are conjugate. erefore, xn and yn are also conjugate. In other words, for each i ≤ r, the set {x : xn ∈ C} is a union of equivalence classes with respect to Pi and, therefore, has cardinality a multiple of . us, this cardinality is a multiple of n.

SOLUTION 119 (i) e relation x ~ y ⇔ 〈x〉 = 〈y〉 is an equivalence relation. e equivalence class of x looks like {xi/(i, O(x)) = 1}; it has ϕ(O(x)) elements. Writing the set of elements of order n as a union of equivalence classes (of elements of order

n), the assertion follows. Further, if pk/|G| and r < k, then fk(G) – fpr(G) is the

number of elements of orders pr+1,…,pk in G. As ϕ(pr+1) = pr(p – 1), ϕ(pr+2) = pr+1 (p – 1), etc., we have that the number of elements of orders pr+1,…,pk is certainly a multiple of pr. e last assertion is obvious. (ii) By Problem 43, we have each g = xy uniquely where xy = yx and O(x) is a power of p and (O(y),p) = 1. us, g ↦ (x, y) gives a bijection between G and ordered pairs (x, y) for which (O(y),p) = 1 and x ∈ CG(y) has p-power order. So, . Hence

is constant where y varies through its conjugacy k

class, and as this conjugacy class has [G : CG(y)] elements, yn/p = 1 we have CG(y)]fpk (CG(y)). (iii) e proof is by induction on |G|. It is vacuously true for G = {1}. Assume for all proper subgroups H that fpr(H) ≡ 0 mod pr where prǁ|H|. en, by (i), fpl(H) ≡ 0 mod pl if pl/|H|. Now, apply (ii) with n = |G|. We get

Look at any of the proper subgroups CG(y) for y ∈ Q\Z(G). If p ℓ ǁ|CG(y)|, with ℓ ≤ k, then [G : CG (y)] ≡ 0 mod pk–ℓ and, p ℓ |fp ℓ (CG (y)) by induction hypothesis. Note fp ℓ (CG(y)) = fpk (CG(y)). erefore each term in the sum above is a

multiple of pk. As pkǁG| we have pkǁQ∩Z(G)|fpk (G). But p | ∩ Z(G)| for, if it did then it would have an element y of order p which would also satisfy . is is a contradiction, as . erefore pk|fpk)(G).

(iv) Let nǁG|. We write for distinct primes pi. en, it suffices to show that . Now, by (ii), fn(G) is expressible as a sum of terms of the form for subgroups H. By (iii), each such H (including G itself) satisfies where . erefore divides . us divides the sum which is fn(G).

SOLUTION N-120 (i) e inner conjugation action of G on N gives a homomorphism θ from G to Aut N. Since N has order p, Aut N ≅ (Z/p/Z)* which has order p – l. erefore, since Imθ has order dividing O(G) as well as dividing p – l, it must be trivial by the choice of p. us, N ≤ Z(G). (ii) e hypothesis gives that G itself has normal subgroups of all orders dividing O(G). If p is the smallest prime divisor of O(G), this means there is a normal subgroup N of order p. By (i), this is central. erefore, Z(G) ≠ {1}. As G/Z(G) satisfies the same hypothesis that G does, it follows by induction on the order that G/Z(G) is nilpotent. By Problem 56 (vi), G itself must be nilpotent.

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