Hypergroups 3031394887, 9783031394881

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Paul-Hermann Zieschang

Hypergroups

Hypergroups

Paul-Hermann Zieschang

Hypergroups

Paul-Hermann Zieschang School of Mathematical & Statistical Sciences University of Texas Rio Grande Valley Edinburg, TX, USA

ISBN 978-3-031-39489-8 (eBook) ISBN 978-3-031-39488-1 https://doi.org/10.1007/978-3-031-39489-8 © Springer Nature Switzerland AG 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland Paper in this product is recyclable.

Preface

The present text is an introduction to the theory of hypergroups. We start with the definition of a hyperoperation. By a hyperoperation on a set S we mean a map from the cartesian product S × S to the power set of S. The image of a pair (p, q) of elements of a set S under a hyperoperation on S will be called the hyperproduct or simply the product of p and q, and will be denoted by pq. In reference to this terminology and notation, we will often use the term hypermultiplication when we refer to a hyperoperation. Let S be a set endowed with a hypermultiplication, and let P and Q be subsets of S. We write PQ for the union of the products pq with p ∈ P and q ∈ Q, and refer to PQ as the complex product or simply as the product of P and Q. If P contains an element p with P = {p}, we write pQ instead of PQ. Similarly, we write Pq instead of PQ if Q contains an element q with Q = {q}. Now we fix a set S and a hypermultiplication on S. The hypermultiplication on S is called associative if, for any three elements p, q, and r in S, (pq)r = p(qr). An element e of S will be called a neutral element with respect to the hypermultiplication on S if, for each element s in S, se = {s}. A map s 7→ s∗ from S to S will be called an inversion function with respect to the hypermultiplication on S if q ∈ p∗ r and p ∈ rq∗ for any three elements p, q, and r in S satisfying r ∈ pq. A set endowed with an associative hypermultiplication will be called a hypergroup if it contains a neutral element with respect to its hypermultiplication and admits an inversion function with respect to its hypermultiplication.1 In Lemma 1.1.3, we will see that, for each hypergroup H and any two elements a and b of H, the product ab is not empty. In Lemma 1.1.5, we will see that a hypergroup cannot have two distinct neutral elements. This allows us to speak about the neutral element of a hypergroup. After Lemma 1.1.5, but also already in these introductory remarks, the neutral element of a hypergroup will always be denoted by 1. 1Our definition is the same as the one given in [21], [22], [43], [49], [53], [56], and [57], and is derived from Fr´ed´eric Marty’s definition of a hypergroup given in [32], [33], and [34]. V

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In Lemma 1.1.7, we will see that a hypergroup cannot have two distinct inversion functions. This allows us to speak about the inversion function of a hypergroup. After Lemma 1.1.7, but also already in these introductory remarks, the image of an element h of a hypergroup H under the inversion function of H will always be denoted by h∗ . In Lemma 1.1.1, we will see that, for each hypergroup element h, 1 ∈ h∗ h. A hypergroup element h will be called thin if h∗ h = {1}. Subsets of hypergroups will be called thin if all of their elements are thin. In Section 1.5, we will establish a bijective correspondence between groups and thin hypergroups. This correspondence will be referred to as group correspondence and allows us to consider the class of groups as a distinguished class of hypergroups, namely the class of thin hypergroups.2 The fact that the class of groups can be considered as a distinguished class of hypergroups will be the guiding principle of all considerations in this monograph. But it leads us also to a first class of examples besides groups themselves. In fact, for each subgroup H of a group G, the set of all double cosets HgH with g ∈ G forms a hypergroup: for any two elements d and e in G, the product (HdH)(HeH) is the set of all double cosets HgH with g ∈ dHe, the neutral element is the double coset H, and the inversion function maps each double coset HgH to Hg −1 H. (We will look at this class of examples from a more conceptual point of view in Section 3.4.) In Section 1.8, we will see that this class of examples is part of an even larger class of hypergroups, the class of association schemes. More examples arise as the text unfolds. Before we go into a more detailed preview of this monograph, a word of caution would be in order regarding the term hypergroup. In fact, in the literature the term hypergroup appears in many different variations. It seems that the term of a hypergroup first appeared in 1934 in a note by Fr´ed´eric Marty; cf. [32]. Marty must not have been entirely sure about his definition; he varied it in two further articles on hypergroups which he was able to publish before his untimely death in 1940; cf. [33] and [34]. In [32] and [34], he defines a hypergroup to be a set endowed with an associative hypermultiplication (in the above sense) which admits an inversion function (also in the above sense). In [33], he requires associativity and the existence of a neutral element (in the above sense), and he substitutes the existence of an inversion function with a slightly weaker condition. In this connection, he speaks about normal hypergroups or completely regular hypergroups. In the year after the publication of Marty’s third article, the term of a hypergroup appears in the title of a paper by Hubert S. Wall; cf. [52]. Wall’s definition of a hypergroup is similar to Marty’s definitions. He also assigns to any two elements 2To illustrate the extent to which the notion of a hypergroup generalizes the notion of a group, we mention that, up to isomorphism, there are two hypergroups with two elements, ten hypergroups with three elements, 102 hypergroups with four elements, 4412 hypergroups with five elements, and 4886349 hypergroups with six elements. 910337 out of the 4886349 hypergroups with six elements are not commutative (in the sense of Section 1.2). (All these numbers were communicated to me by C. French.)

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of a given set S a subset of S, but this subset comes with multiplicities. Wall refers to Marty’s work, and so do Lois Wilfred Griffiths (a student of Leonard Eugene Dickson) in [23], Howard Herbert Campaigne (a student of Wall) in [8], and Marc Krasner (a student of Jacques Hadamard who met Marty at Hadamard’s Seminar) in [29] in the subsequent years. Follow-up articles published in the 1940s such as [31] by Jean Kuntzmann, [50] by Aleksandr Ivanovich Vikhrov (a student of Aleksandr Gennadievich Kurosh), [30] by Krasner and Kuntzmann, [14] by Robert Croisot, or [47] by Yuzo Utumi are still written in the spirit of Marty’s definition. That was not any longer the case in 1958, when Sigurður Helgason observed (in [26]) that any pair of equivalence classes of unitary irreducible representations of a given compact group gives rise to a measure on the set of all these equivalence classes and subsequently defined a (discrete) hypergroup to be a set S endowed with a map from S × S to the set of all measures on S. Helgason’s definition was broad enough to provide the framework for a theory of topological hypergroups, as developed in the 1970s by Charles F. Dunkl in [18] and [19], by Robert I. Jewett in [28], and by Ren´e Spector in [40], and about which Kenneth A. Ross later (in [38]) rightly says that it has received much attention from harmonic analysts, since its framework is general enough to cover a variety of important examples including double coset spaces, yet concrete enough to develop a reasonable theory. In [37], Ross gives a fairly complete history of this topological approach to hypergroups. Meanwhile, in this context, the monograph [7] by Walter R. Bloom and Herbert Heyer has become a standard reference. Here, hypergroups are considered as vector spaces with bounded Radon measures on a locally compact Hausdorff space satisfying certain conditions aimed at harmonic analysis. The wealth of possibilities offered by the notion of hypergroup in this sense is impressively demonstrated in the anthology [39].3 As full of ideas as the work on the (harmonic) analytic approach to hypergroups was (and still is), the purely algebraic interest in hypergroups was never completely extinguished. It flared up again in 1979 in John R. McMullen’s article [35], where Richard Brauer’s idea of studying character tables of groups from a purely combinatorial point of view is taken up. A clearer return to Marty’s original definition is evident in Stephen D. Comer’s article [10]. Comer speaks about polygroups, but his definition of a polygroup is exactly the definition of a hypergroup which is used in the present monograph. Not surprisingly, Comer already mentions the connection between hypergroups and association schemes, a connection which we will discuss in Section 1.8.4 With the increasing appreciation of the axiomatization of pure mathematics in the seventies of the last century, a more algebraic-combinatorial treatment of Marty’s notion of hypergroups, quite in the spirit of Comer’s work, broke through indepen3On page 97 of this volume Alan L. Schwartz suggests to replace the term hypergroup by the term hypergroup measure algebra, a conceptualization which sums up the nature of the hypergroup in the (harmonic) analytic sense very well. 4Comer speaks about coherent configurations in this context.

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dently of the (harmonic) analytic theory of hypergroups. It is impossible to give a halfway survey of the overwhelming literature in this direction. Representative texts here are probably the monographs [11] by Piergiulio Corsini, [12] by Corsini and Violeta Leoreanu, [51] by Thomas Vougiouklis, [15] by Bijan Davvaz, and [16] by Davvaz and Violeta Leoreanu-Fotea. All five monographs, together with [13], provide a wealth of references and give an impression of the efforts in this direction. The present monograph can certainly be assigned to the algebraic attempts to fathom the concept of the hypergroup. A specific feature of the definition of a hypergroup as it is used in this monograph, however, is the axiomatically assumed existence of a neutral element. (This largely excludes major overlaps with all of the texts mentioned in the previous paragraph.) The reason why we consider hypergroups only with neutral elements lies in the guiding principle of our development already mentioned above, the strict orientation towards the structural theory of groups. Dedekind Modularity, generating sets, conjugation, commutators, centralizers, normalizers, quotients, homomorphisms, composition factors, solvability, involutions, exchange condition, Coxeter groups, buildings, and twin buildings: all these terms point to the guide under which we try to develop the theory. In fact, our interest in hypergroups originally arose from the desire to set the foundations for the algebraic aspect, or, one could also say, the table algebraic aspect (in the sense of [2], [4], or [5]) of scheme theory. From this point of view, the present text can also be seen as a continuation of considerations which have been developed in [55] for association schemes. To explain this in more detail, we now give a brief overview of the content of each chapter of this monograph. The first chapter begins with the uniqueness of the neutral element and the uniqueness of the inversion function in hypergroups. After that, we compile computational rules for products of elements of hypergroups and products of subsets of hypergroups. We also introduce thin hypergroup elements and establish the above-mentioned group correspondence. Three sections on actions of hypergroups are included, where the emphasis is on hypergroups which admit regular actions. Association schemes are introduced as specific regular actions of hypergroups. In the second chapter, we generalize the concept of a subgroup to the theory of hypergroups. We will see that closed subsets (as we will call the generalized subgroups) give rise to cosets, and these cosets have quite a number of properties in common with cosets of subgroups. Closed subsets naturally lead to Dedekind Modularity and to the notion of generating subsets of hypergroups. Similar to group theory, generating sets give rise to length functions, and length functions play as useful a role in the theory of hypergroups as they do in group theory. We then introduce commutators of hypergroup elements, we explain what it means for closed subsets of hypergroups to be conjugate, and we devote a section to the thin radical of a hypergroup which will be our name for the set of all thin elements of a hypergroup. The chapter ends with a section on foldings by which we mean injective maps from closed subsets F of hypergroups H to left cosets of F in H which allow us to distinguish certain cosets of closed subsets from otherwise less controllable cosets. They will play a crucial role in the second half of Chapter 10.

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Chapter 3 goes closely along the lines of the structure theory of groups. We introduce the centralizer, the normalizer, and the strong normalizer of a closed subset of a hypergroup. We continue with quotients which, in contrast to group theory, arise from any closed subset of a hypergroup, not only from normal closed subsets. This provides more flexibility when proofs are conducted by induction. Quotients lead to the notion of a hypergroup homomorphism, and hypergroup homomorphisms entail, as in any algebraic theory, a homomorphism theorem as well as two isomorphism theorems. In Chapter 4, we study subnormal chains. Specific subnormal chains are subnormal series, among them the composition series; cf. Sections 4.2 and 4.3. Composition series give us the oportunity to speak about composition factors of hypergroups in a similar way as one speaks about composition factors in group theory. A particular class of subnormal series, the class of the strong subnormal series, is closely related to the thin residue of a hypergroup, and this closed subset will be introduced in Section 4.4. The thin residue of a hypergroup leads to the notion of a residually thin hypergroup, and residually thin hypergroups are investigated in Section 4.6. In Section 4.7, we restrict ourselves to finite residually thin hypergroups, and a very specific class of finite residually thin hypergroups, the class of solvable hypergroups, is in the focus of Section 4.8. We mentioned earlier that, for each hypergroup element h, 1 ∈ h∗ h. As a consequence, one obtains that h ∈ hh∗ h for each hypergroup element h. A hypergroup element h is called tight if hh∗ h = {h}, and hypergroups will be called tight if all of their elements are tight. Tight hypergroups are in the center of Chapter 5. We will first show that finite tight hypergroups are residually thin; cf. Theorem 5.1.5. (A hypergroup H is called finite if H, as a set, has finite cardinality.) After that, a major part of Chapter 5 is devoted to the construction of an associative ring which we associate to certain finite tight hypergroups and which naturally generalizes the notion of a scheme ring as defined in [55; Section 9.1], and therefore also the notion of a group ring of a finite group. The final two sections of Chapter 5 are devoted to a specific class of tight hypergroups, to hypergroups whose thin residue is contained in their thin radical. In Chapter 6, we introduce involutions of hypergroups. Clearly, involutions of hypergroups generalize group theoretic involutions, and the role of involutions in the theory of hypergroups seem to be as significant as the role of involutions in group theory. In three sections, we collect details about closed subsets generated by an involution. With an eye on group theory, we then look (in Sections 6.5 and 6.6) at hypergroups generated by sets of involutions. In Section 6.7, we introduce two conditions which a set of involutions of a hypergroup may or may not satisfy and which will be of interest in Chapters 8, 9, and 10, dichotomy and the exchange condition. The exchange condition is a genuine generalization of the group theoretic exchange condition which distinguishes the Coxeter groups among the groups generated by involutions. Theorem 6.7.5 clarifies the relationship between these two conditions, it says that the exchange condition implies dichotomy. At the end of the chapter, in

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Section 6.8, we classify all hypergroups in which every element different from the neutral element is an involution. In Chapter 7, we pause for a moment developing the structure theory of hypergroups and turn to the study of hypergroups with a small number of elements. We first describe all hypergroups with one, two, or three elements. It is shown that, with the exception of one isomorphism class, all of these hypergroups arise as quotients of finite thin hypergroups, i.e., quotients of finite groups. After that, we classify all non-symmetric hypergroups with four elements.5 The remaining part of Chapter 7 is then devoted to several classification theorems of hypergroups with six elements. We continue with two chapters on hypergroups generated by distinguished sets of involutions. In Chapter 8, we introduce constrained sets of involutions. These sets will be defined with the help of the length function. In the first two sections we mainly present technical results, but we also show that, for constrained sets of involutions, the converse of Theorem 6.7.5 holds, namely that, for constrained sets of involutions, dichotomy and the exchange conditions are equivalent; cf. Theorem 8.1.5. In Section 8.3, we consider the thin radical of hypergroups generated by a constrained set of involutions, and in the two subsequent sections, we bring constrained sets of involutions together with the two above-mentioned conditions a set of involutions of a hypergroup may satisfy or not, with dichotomy and with the exchange condition. Our study culminates in Andrew Wang’s beautiful Theorem 8.5.5, a theorem which is similar to Theorem 8.1.5 and provides still another necessary and sufficient condition for constrained sets of involutions to be dichotomic. But this time we require the underlying constrained set of involutions not to have thin elements. (In Corollary 6.7.2, we have already proven the necessity of this condition, regardless how the underlying set of involutions looks like.) Together with Theorem 8.1.5, Theorem 8.5.5 characterizes constrained sets of non-thin involutions satisfying the exchange condition without explicitly referring to the length function. In two final sections, constrained sets of involutions are related to foldings. In Chapter 9, we study hypergroups generated by constrained sets of involutions satisfying the exchange condition. These hypergroups will be called Coxeter hypergroups. We will see that Coxeter hypergroups show features which are similar to well-known features of Coxeter groups. In particular, results on finite Coxeter hypergroups are highly reminiscent of results on finite Coxeter groups. Chapter 10 is the place where buildings and twin buildings (in the sense of Jacques Tits) come into the game. We briefly revisit the definitions of buildings and twin buildings and translate a few elementary observations about buildings and twin buildings into the simple language of binary relations. Then we will see that semiregular buildings and regular actions of Coxeter hypergroups are equivalent mathematical objects.6 The precise nature of this equivalence will be described in Theorems 10.2.1, 5Symmetric hypergroups will be defined in Section 1.2. 6The (interlocutory) definition of a semiregular building will be given in Section 10.1. The class of semiregular buildings contains the class of thick buildings.

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10.2.7, 10.3.1, 10.3.2, 10.3.3, and 10.3.4. After that, we will introduce a class of hypergroups (the members of which we will call twin Coxeter hypergroups), and we will see that (modulo a purely building theoretic hypothesis, which we will leave out of this monograph) thick twin buildings and regular actions of twin Coxeter hypergroups over sets of non-thin involutions are equivalent mathematical objects. The precise nature of this equivalence will be described in Theorems 10.6.1, 10.6.7, 10.7.1, 10.7.2, 10.7.3, and 10.7.4. I started working on this monograph during a stay at the Max-Planck-Institut für Mathematik at Bonn (Germany) from January to May 2010; cf. [56]. I gratefully remember the hospitality and pleasant working environment I experienced during those five months in Bonn. Since about the same time, I have been in mathematical exchange with Christopher French from Grinnell College, Grinnell (Iowa). Numerous of his beautiful ideas and results on hypergroups have found their way into the present text, and I would like to thank him here for the years of collaboration. Significant contributions to this monograph also came from Harvey Blau from Northern Illinois University, DeKalb (Illinois). Harvey not only read large parts of the manuscript line by line (thus saving the text from quite a few minor and major errors), with his constant support and stimulating ideas as an expert on table algebras he also has influenced the form and content of several sections (especially of Section 4.7) of this monograph. Merano, September 2022

Paul-Hermann Zieschang

Contents

1

Basic Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Neutral Elements and Inversion Functions . . . . . . . . . . . . . . . . . . . . . 1.2 Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Complex Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Thin Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Groups and Thin Hypergroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Actions of Hypergroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Hypergroups Admitting Regular Actions . . . . . . . . . . . . . . . . . . . . . . 1.8 Association Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 3 6 9 11 13 18 23

2

Closed Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Basic Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Dedekind Modularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Generating Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Commutators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Conjugation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 The Thin Radical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Foldings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

29 30 34 35 39 40 43 45

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Elementary Structure Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Centralizers and Normalizers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Sufficient Conditions for Normality . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Strong Normality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Computations in Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Hypergroup Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 The Isomorphism Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8 Wreath Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9 Hypergroup Isomorphisms and Regular Actions . . . . . . . . . . . . . . . .

49 49 54 57 61 65 68 72 77 86

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Subnormality and Thin Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 4.1 Subnormal Chains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 4.2 Subnormal Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 4.3 Composition Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 4.4 The Thin Residue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 4.5 Thin Residues of Thin Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 4.6 Residually Thin Hypergroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 4.7 Finite Residually Thin Hypergroups . . . . . . . . . . . . . . . . . . . . . . . . . . 110 4.8 Solvable Hypergroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

5

Tight Hypergroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 5.1 Tight Hypergroup Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 5.2 The Set S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 5.3 The Sets a∗ b ∩ Fc and Sa,b (Fc ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 5.4 The Sets b f1 b∗ ∩ Fa and Sb,( f1,..., fn ) (Fa ) . . . . . . . . . . . . . . . . . . . . . . 132 5.5 Structure Constants of Finite Tight Hypergroups . . . . . . . . . . . . . . . 138 5.6 Rings Arising from Finite Tight Hypergroups . . . . . . . . . . . . . . . . . . 142 5.7 Some Arithmetic in Finite Metathin Hypergroups . . . . . . . . . . . . . . . 144 5.8 Finite Metathin Hypergroups and Prime Numbers . . . . . . . . . . . . . . 148

6

Involutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 6.1 Basic Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 6.2 Closed Subsets Generated by an Involution, I . . . . . . . . . . . . . . . . . . 160 6.3 Closed Subsets Generated by an Involution, II . . . . . . . . . . . . . . . . . . 163 6.4 Closed Subsets Generated by an Involution, III . . . . . . . . . . . . . . . . . 165 6.5 Length Functions Defined by Sets of Involutions . . . . . . . . . . . . . . . 170 6.6 Hypergroups Generated by Two Involutions . . . . . . . . . . . . . . . . . . . 175 6.7 Dichotomy and the Exchange Condition . . . . . . . . . . . . . . . . . . . . . . . 179 6.8 Projective Hypergroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182

7

Hypergroups with a Small Number of Elements . . . . . . . . . . . . . . . . . 189 7.1 Hypergroups of Cardinality at Most 3 . . . . . . . . . . . . . . . . . . . . . . . . . 190 7.2 Non-Symmetric Hypergroups of Cardinality 4 . . . . . . . . . . . . . . . . . 197 7.3 Some Hypergroups of Cardinality 6, I . . . . . . . . . . . . . . . . . . . . . . . . 209 7.4 Some Hypergroups of Cardinality 6, II . . . . . . . . . . . . . . . . . . . . . . . . 220 7.5 Non-Normal Closed Subsets Missing Four Elements . . . . . . . . . . . . 234 7.6 A Characterization of H6,1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240

8

Constrained Sets of Involutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 8.1 Basic Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244 8.2 Constrained Sets of Involutions and Cosets . . . . . . . . . . . . . . . . . . . . 248 8.3 Constrained Sets of Involutions and Thin Elements . . . . . . . . . . . . . 250 8.4 Constrained Sets of Involutions and Dichotomy, I . . . . . . . . . . . . . . . 253 8.5 Constrained Sets of Involutions and Dichotomy, II . . . . . . . . . . . . . . 259 8.6 Constrained Sets of Involutions and Foldings, I . . . . . . . . . . . . . . . . . 264 8.7 Constrained Sets of Involutions and Foldings, II . . . . . . . . . . . . . . . . 269

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Coxeter Sets of Involutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 9.1 General Observations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274 9.2 The Sets V1 (U) for Subsets U of Coxeter Sets V . . . . . . . . . . . . . . . . 278 9.3 The Sets V−1 (U) for Subsets U of Coxeter Sets V . . . . . . . . . . . . . . . 285 9.4 Sets of Subsets of Coxeter Sets of Involutions . . . . . . . . . . . . . . . . . . 287 9.5 Spherical Coxeter Sets of Involutions . . . . . . . . . . . . . . . . . . . . . . . . . 290 9.6 Subsets of Spherical Coxeter Sets of Involutions . . . . . . . . . . . . . . . . 295 9.7 Coxeter Sets of Involutions and Foldings . . . . . . . . . . . . . . . . . . . . . . 300 9.8 Coxeter Sets and Their Coxeter Numbers . . . . . . . . . . . . . . . . . . . . . . 304 9.9 Coxeter Sets and Type Preserving Bijections . . . . . . . . . . . . . . . . . . . 310

10 Regular Actions of (Twin) Coxeter Hypergroups . . . . . . . . . . . . . . . . . 319 10.1 Buildings and Binary Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 10.2 Regular Actions of Coxeter Hypergroups . . . . . . . . . . . . . . . . . . . . . . 326 10.3 Coxeter Hypergroups and Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . 336 10.4 Twin Buildings and Binary Relations, I . . . . . . . . . . . . . . . . . . . . . . . 342 10.5 Twin Buildings and Binary Relations, II . . . . . . . . . . . . . . . . . . . . . . . 350 10.6 Regular Actions of Twin Coxeter Hypergroups . . . . . . . . . . . . . . . . . 354 10.7 Twin Coxeter Hypergroups and Twin Buildings . . . . . . . . . . . . . . . . 370 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389

1 Basic Facts

This chapter contains basic results on hypergroups as they were defined at the beginning of the Preface. We will refer to most of these results throughout the text. In Section 1.1, we show that each hypergroup has only one neutral element and only one inversion function. In Section 1.2, we compile results on products of hypergroup elements, and in Section 1.3, we provide computational rules for products of subsets of hypergroups. In Section 1.4, we consider thin hypergroup elements, and in Section 1.5, we establish the group correspondence which was mentioned in the Preface. In Section 1.6, we introduce actions of hypergroups and present a few general results on regular actions, and in Section 1.7, we present conditions a hypergroup needs to satisfy in order to admit a regular action. In Section 1.8, we connect regular actions of hypergroups with association schemes.

1.1 Neutral Elements and Inversion Functions In this section, we show that each hypergroup possesses only one neutral element and only one inversion function. Lemma 1.1.1 Let H be a hypergroup, let e be a neutral element of H, and let h 7→ h∗ be an inversion function of H. Then, for each element h in H, e ∈ h∗ h. Proof. Let h be an element in H. Since e is a neutral element of H, we have h ∈ he. Thus, as h 7→ h∗ is an inversion function of H, e ∈ h∗ h.  Lemma 1.1.2 Let H be a hypergroup, and let h 7→ h∗ be an inversion function of H. Then, for each element h in H, h∗∗ = h. Proof. Let h be an element in H, and let e be a neutral element of H. From Lemma 1.1.1 we know that e ∈ h∗ h. Thus, as h 7→ h∗ is an inversion function of H, we have © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 P. -H. Zieschang, Hypergroups, https://doi.org/10.1007/978-3-031-39489-8_1

1

2

1 Basic Facts

h ∈ h∗∗ e. On the other hand, as e is a neutral element of H, h∗∗ e = {h∗∗ }. It follows that h∗∗ = h.  Occasionally, we will refer to Lemma 1.1.2 without further mention. Lemma 1.1.3 Let H be a hypergroup, and let a and b be elements of H. Then ab is not empty. Proof. Let e be a neutral element of H, and let h 7→ h∗ be an inversion function of H. From Lemma 1.1.1 we know that e ∈ b∗∗ b∗ , from Lemma 1.1.2 that b∗∗ = b. Thus, e ∈ bb∗ . On the other hand, as e is a neutral element of H, a ∈ ae. It follows that a ∈ abb∗ , and that shows that ab is not empty.  Lemma 1.1.4 Let H be a hypergroup, and let e be a neutral element of H. Then, for each element h in H, eh = {h}. Proof. Let h be an element of H. From Lemma 1.1.3 we know that eh is not empty. Let a be an element in eh. We are done when we succeed in showing that a = h. Let h 7→ h∗ be an inversion function of H. Then, since a ∈ eh, e ∈ ah∗ . From this we obtain that h∗ ∈ a∗ e. Since e is a neutral element of H, we also have a∗ e = {a∗ }.  Thus, a∗ = h∗ . It follows that a = a∗∗ = h∗∗ = h; cf. Lemma 1.1.2. Lemma 1.1.5 Each hypergroup possesses exactly one neutral element. Proof. Let H be a hypergroup, and let a and b be neutral elements of H. Since a is a neutral element of H, we have ab = {b}; cf. Lemma 1.1.4. Since b is a neutral element of H, we have ab = {a}. It follows that a = b.  For the remainder of this monograph, the uniquely determined neutral element of a hypergroup will always be denoted by 1. Thus, we have h·1 = {h} = 1· h for each element h of a hypergroup. Lemma 1.1.6 Let H be a hypergroup, let h 7→ h∗ be an inversion function of H, and let a and b be elements in H. Then we have 1 ∈ a∗ b if and only if a = b. Proof. Assume first that 1 ∈ a∗ b. Then, by Lemma 1.1.2, b ∈ a · 1. Thus, since a·1 = {a}, we have a = b. Assume, conversely, that a = b. Then, since a · 1 = {a}, b ∈ a · 1. It follows that  1 ∈ a∗ b.

1.2

Products

3

Lemma 1.1.6 will often be used in conjunction with an unstated application of Lemma 1.1.2. Lemma 1.1.7 Each hypergroup possesses exactly one inversion function. Proof. Let H be a hypergroup, let h 7→ h[ and h 7→ h] be inversion functions of H, and let h be an element in H. We shall see that h[ = h] . Since [ is an inversion function of H, we have 1 ∈ h[ h; cf. Lemma 1.1.1. Thus, as ] is an inversion function of H, we obtain that h[ ∈ 1 · h] . On the other hand, by Lemma 1.1.4, 1· h] = {h] }. It follows that h[ = h] .  For the remainder of this monograph, the uniquely determined inversion function of a hypergroup will always be denoted by ∗ . For each hypergroup element h, the element h∗ will be called the inverse of h. Lemma 1.1.8 In each hypergroup, we have 1∗ = 1. Proof. From Lemma 1.1.1 we know that 1 ∈ 1∗ 1. Thus, as 1∗ 1 = {1∗ }, we obtain that 1∗ = 1. 

1.2 Products In this section, we compile results on products of hypergroup elements. Lemma 1.2.1 Let H be a hypergroup, and let a, b, and c be elements of H. Then the statements c ∈ ab, b ∈ a∗ c, a∗ ∈ bc∗ , c∗ ∈ b∗ a∗ , b∗ ∈ c∗ a, and a ∈ cb∗ are pairwise equivalent. Proof. From c ∈ ab one obtains that b ∈ a∗ c. Similarly, one obtains a∗ ∈ bc∗ from b ∈ a∗ c, c∗ ∈ b∗ a∗ from a∗ ∈ bc∗ , b∗ ∈ c∗ a from c∗ ∈ b∗ a∗ , a ∈ cb∗ from b∗ ∈ c∗ a, and c ∈ ab from a ∈ cb∗ .  Lemma 1.2.2 Let H be a hypergroup, and let a, b, and c be elements of H with ab = {c}. Then b∗ a∗ = {c∗ }. Proof. We are assuming that ab = {c}. This implies that c ∈ ab, so that, by Lemma 1.2.1, c∗ ∈ b∗ a∗ . Let d be an element in b∗ a∗ . Then, by Lemma 1.2.1, d ∗ ∈ ab. Since we are assuming that ab = {c}, this implies that d ∗ = c. It follows that d = c∗ ; cf. Lemma 1.1.2. Since  d has been chosen arbitrarily in b∗ a∗ , this shows that b∗ a∗ ⊆ {c∗ }.

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1 Basic Facts

Let H be a hypergroup, let n be a positive integer, and let h1 , . . ., hn be elements of H. If n = 1, we set h1 · · · hn := {h1 }. If 3 ≤ n, we inductively define h1 · · · hn := (h1 · · · hn−1 )hn . If H contains an element h with h = hi for each element i in {1, . . . , n}, we write h n instead of h1 · · · hn . Lemma 1.2.3 Let H be a hypergroup, let a, b, and c be elements in H, and assume that ab = ac. Then the following hold. (i) We have b ∈ a∗ a if and only if c ∈ a∗ a. (ii) We have a ∈ bb∗ if and only if a ∈ bc∗ . (iii) We have b∗ ∈ a2 if and only if a∗ ∈ ac. Proof. (i) We have b ∈ a∗ a if and only if a ∈ ab. Furthermore, c ∈ a∗ a if and only if a ∈ ac. Since we are assuming that ab = ac, we also have a ∈ ab if and only if a ∈ ac. (ii) We have a ∈ bb∗ if and only if b ∈ ab. Furthermore, a ∈ bc∗ if and only if b ∈ ac. Since we are assuming that ab = ac, we also have b ∈ ab if and only if b ∈ ac. (iii) We are assuming that ab = ac. Thus, we have a∗ ∈ ab if and only if a∗ ∈ ac.  On the other hand, we have a∗ ∈ ab if and only if b∗ ∈ a2 ; cf. Lemma 1.2.1. A hypergroup element h is called symmetric if h∗ = h. Lemma 1.2.4 Let H be a hypergroup, and let a, b, and c be elements of H. Assume that a is symmetric and that c2 = ab. Then the following hold. (i) We have c ∈ c2 if and only if a ∈ bc∗ . (ii) We have a ∈ c2 if and only if b ∈ a2 . Proof. (i) Since we are assuming that c2 = ab, we have c ∈ c2 if and only if c ∈ ab. On the other hand, since a is symmetric, we have c ∈ ab if and only if a ∈ bc∗ ; cf. Lemma 1.2.1. (ii) Since we are assuming that c2 = ab, we have a ∈ c2 if and only if a ∈ ab. Since a is symmetric, we have a ∈ ab if and only if b ∈ a2 .  Lemma 1.2.5 Let H be a hypergroup, and let a, b, c, and d be elements of H. Then the following hold. (i) The set ab ∩ cd is not empty if and only if a∗ c ∩ bd ∗ is not empty.

1.2

Products

5

(ii) The statements a ∈ bcd ∗ , b ∈ adc∗ , c ∈ b∗ ad, and d ∈ a∗ bc are pairwise equivalent. (iii) The set ab ∩ cd is not empty if and only if a ∈ cdb∗ . Proof. (i) Assume first that ab ∩ cd is not empty, and let e be an element in ab ∩ cd. From e ∈ ab we obtain that b ∈ a∗ e. Since e ∈ cd, this implies that b ∈ a∗ cd. Thus, a∗ c contains an element f such that b ∈ f d. From b ∈ f d we obtain f ∈ bd ∗ . It follows that f ∈ a∗ c ∩ bd ∗ , so that a∗ c ∩ bd ∗ is not empty. The converse follows by repeating the above argument with a∗ , c, b, and d ∗ in place of a, b, c, and d. (ii) Assume that a ∈ bcd ∗ . Then bc contains an element e such that a ∈ ed ∗ . From e ∈ bc we obtain that b ∈ ec∗ . From a ∈ ed ∗ we obtain that e ∈ ad. Thus, b ∈ adc∗ . Assume that b ∈ adc∗ . Then ad contains an element e such that b ∈ ec∗ . From b ∈ ec∗ we obtain that c ∈ b∗ e; cf. Lemma 1.2.1. Thus, c ∈ b∗ ad. Assume that c ∈ b∗ ad. Then ad contains an element e such that c ∈ b∗ e. From e ∈ ad we obtain that d ∈ a∗ e. From c ∈ b∗ e we obtain that e ∈ bc. Thus, d ∈ a∗ bc. Assume that d ∈ a∗ bc. Then bc contains an element e such that d ∈ a∗ e. From d ∈ a∗ e we obtain that a ∈ ed ∗ ; cf. Lemma 1.2.1. Thus, a ∈ bcd ∗ . (iii) Assume first that ab ∩ cd is not empty, and let e be an element in ab ∩ cd. From e ∈ ab we obtain that a ∈ eb∗ . Thus, as e ∈ cd, a ∈ cdb∗ . Conversely, assume that a ∈ cdb∗ . Then cd contains an element e such that a ∈ eb∗ .  From a ∈ eb∗ we obtain that e ∈ ab. It follows that e ∈ ab ∩ cd. Recall from the Preface that a hypergroup element h is called tight if hh∗ h = {h}. Lemma 1.2.6 Inverses of tight hypergroup elements are tight. Proof. Let h be a tight hypergroup element, and let a be an element in h∗ h. From a ∈ h∗ h we obtain that a∗ ∈ h∗ h; cf. Lemma 1.2.1. Since h is tight, we also have hh∗ h = {h}. Thus, ha∗ = {h}. Thus, by Lemma 1.2.2, ah∗ = {h∗ }. Since a has been chosen arbitrarily in h∗ h, this shows that h∗ hh∗ = {h∗ }, and that  means that h∗ is tight. Lemma 1.2.7 Let H be a hypergroup, and let a, b, c, d, and e be elements of H with e ∈ ab ∩ cd. Then the following hold. (i) We have ab ∩ cd ⊆ e(e∗ eb∗ b ∩ d ∗ d). (ii) Assume that b, d, and e are tight. Then ab ∩ cd = e(e∗ eb∗ b ∩ d ∗ d). Proof. (i) Let f be an element in ab ∩ cd. We will be done if we succeed in showing that e∗ eb∗ b ∩ d ∗ d contains an element g with f ∈ eg.

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1 Basic Facts

From e ∈ cd we obtain that c ∈ ed ∗ . From f ∈ cd we obtain that c ∈ f d ∗ . It follows that c ∈ ed ∗ ∩ f d ∗ . Thus, we find an element g in e∗ f ∩ d ∗ d; cf. Lemma 1.2.5(i). Since g ∈ e∗ f , f ∈ eg. Thus, as g ∈ d ∗ d, it remains to be shown that g ∈ e∗ eb∗ b. From e ∈ ab we obtain that a ∈ eb∗ . Thus, as f ∈ ab, f ∈ eb∗ b. Since g ∈ e∗ f , we now conclude that g ∈ e∗ eb∗ b. (ii) We are assuming that b, d, and e are tight and that e ∈ ab ∩ cd. Thus, we have e(e∗ eb∗ b ∩ d ∗ d) ⊆ ee∗ eb∗ b ∩ ed ∗ d ⊆ eb∗ b ∩ ed ∗ d ⊆ abb∗ b ∩ cdd ∗ d = ab ∩ cd, so that the desired equation follows from (i).



Let H be a hypergroup. For each element a in H, we define SH (a) := {b ∈ H | ab = {a}}. Note that 1 ∈ SH (h) for each element h in H and SH (1) = {1}. Lemma 1.2.8 Let H be a hypergroup, and let h be an element in H. Then the following hold. (i) We have SH (h) ⊆ h∗ h. (ii) We have SH (h) = h∗ h if and only if h is tight. Proof. (i) Let a be an element in SH (h). Then h ∈ ha. It follows that a ∈ h∗ h. (ii) Assume first that SH (h) = h∗ h. Then ha = {h} for each element a in h∗ h. Thus, hh∗ h ⊆ {h}, so that, by Lemma 1.1.3, hh∗ h = {h} which means that h is tight. Assume, conversely, that h is tight, and let a be an element in h∗ h. Then ha ⊆ hh∗ h = {h}. Thus, a ∈ SH (h). Since a has been chosen arbitrarily in h∗ h, we have shown that h∗ h ⊆ SH (h). Now  we obtain from (i) that SH (h) = h∗ h. A subset of a hypergroup is called symmetric if each of its elements is symmetric. A hypergroup H is called commutative if ab = ba for any two elements a and b in H. Lemma 1.2.9 Symmetric hypergroups are commutative. Proof. Let H be a symmetric hypergroup, let a and b be elements in H, and let c be an element in ab. From c ∈ ab we obtain that c∗ ∈ b∗ a∗ ; cf. Lemma 1.2.1. Thus, as H is symmetric, we have c ∈ ba. This shows that ab ⊆ ba. The reverse containment  follows similarly.

1.3 Complex Products Recall from the Preface that, for any two subsets A and B of a hypergroup, we write AB to denote the union of the products ab with a ∈ A and b ∈ B.

1.3

Complex Products

7

For each subset A of a hypergroup, we set A∗ := {a∗ | a ∈ A}. A subset A of a hypergroup will be called ∗ -invariant if A∗ = A. Clearly, symmetric subsets of hypergroups are ∗ -invariant, but the converse is not true. Lemma 1.3.1 Let H be a hypergroup, and let A and B be subsets of H. Then the following hold. (i) Assume that A ⊆ B. Then A∗ ⊆ B∗ . (ii) We have (AB)∗ = B∗ A∗ . (iii) Assume that AB ⊆ BA. Then B∗ A∗ ⊆ A∗ B∗ . (iv) Assume that A and B are ∗ -invariant and that AB ⊆ BA. Then AB = BA. Proof. (i) Let c be an element in A∗ . Then A contains an element a with c = a∗ . Since a ∈ A, a ∈ B. Thus, a∗ ∈ B∗ . Since c = a∗ , this implies that c ∈ B∗ . (ii) Let c be an element in (AB)∗ . Then AB contains an element d with c = d ∗ . Since d ∈ AB, there exist elements a in A and b in B such that d ∈ ab. Thus, by Lemma 1.2.1, d ∗ ∈ b∗ a∗ . Since a∗ ∈ A∗ and b∗ ∈ B∗ this implies that d ∗ ∈ B∗ A∗ . Since c = d ∗ this implies that c ∈ B∗ A∗ . This shows that (AB)∗ ⊆ B∗ A∗ . Let c be an element in B∗ A∗ . Then there exist elements b in B and a in A such that c ∈ b∗ a∗ . From c ∈ b∗ a∗ we obtain that c∗ ∈ ab; cf. Lemma 1.2.1. Thus, as a ∈ A and b ∈ B, c∗ ∈ AB. It follows that c ∈ (AB)∗ . This shows that B∗ A∗ ⊆ (AB)∗ . (iii) Considering (ii) this follows from (i). (iv) We are assuming that AB ⊆ BA. Thus, by (iii), B∗ A∗ ⊆ A∗ B∗ . Since A and B are assumed to be ∗ -invariant, this implies that BA ⊆ AB.  The first part of the subsequent lemma generalizes Lemma 1.1.2. Lemma 1.3.2 Let A be a subset of a hypergroup. Then the following hold. (i) We have A∗∗ = A. (ii) Assume that A∗ ⊆ A. Then A∗ = A. (iii) The set A∗ A is ∗ -invariant. Proof. (i) This follows from Lemma 1.1.2. (ii) We are assuming that A∗ ⊆ A. Thus, by Lemma 1.3.1(i), A∗∗ ⊆ A∗ . On the other hand, by (i), A∗∗ = A. Thus, A ⊆ A∗ . (iii) From Lemma 1.3.1(ii) we know that (A∗ A)∗ = A∗ A∗∗ , and from (i) we know that A∗∗ = A. Thus, (A∗ A)∗ = A∗ A. 

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1 Basic Facts

Lemma 1.3.3 Let H be a hypergroup, and let A, B, and C be subsets of H. If one of the sets C ∩ AB, B ∩ A∗ C, A∗ ∩ BC ∗ , C ∗ ∩ B∗ A∗ , B∗ ∩ C ∗ A, and A ∩ CB∗ is not empty, so are the other five. Proof. The set C ∩ AB is not empty if and only if there exist elements a in A, b in B, and c in C such that c ∈ ab. Similar observations for the other five sets, together  with Lemma 1.2.1, finish the proof. Lemma 1.3.4 Let H be a hypergroup, and let A, B, and C be subsets of H. Then we have (AB)C = A(BC). Proof. Let f be an element in (AB)C. Then there exist elements d in AB and c in C such that f ∈ dc. Since d ∈ AB, there exist elements a in A and b in B such that d ∈ ab. It follows that f ∈ abc. Thus, bc contains an element e such that f ∈ ae. Since b ∈ B and c ∈ C, bc ⊆ BC. Thus, as e ∈ bc, e ∈ BC. Thus, as a ∈ A and f ∈ ae, f ∈ A(BC). Since f has been chosen arbitrarily among the elements of (AB)C, this shows that  (AB)C ⊆ A(BC). The reverse containment follows similarly. Let H be a hypergroup, let n be a positive integer, and let A1 , . . ., An be subsets of H. If n = 1, we set A1 · · · An := A1 . If 3 ≤ n, we inductively set A1 · · · An := (A1 · · · An−1 )An ; cf. Lemma 1.3.4. If H contains a subset A with A = Ai for each element i in {1, . . . , n}, we write An instead of A1 · · · An . For each subset A of a hypergroup, we set A0 := {1}. The following lemma generalizes Lemma 1.2.5. Lemma 1.3.5 Let H be a hypergroup, and let A, B, C, and D be subsets of H. Then the following hold. (i) The set AB ∩ CD is not empty if and only if A∗ C ∩ BD∗ is not empty. (ii) If one of the sets A ∩ BCD∗ , B ∩ ADC ∗ , C ∩ B∗ AD, and D ∩ A∗ BC is not empty, so are the other three. (iii) The set AB ∩ CD is not empty if and only if A ∩ CDB∗ is not empty. Proof. (i) Assume first that AB ∩ CD is not empty. Then there exist elements a in A, b in B, c in C, and d in D such that ab ∩ cd is not empty. From this we obtain that a∗ c ∩ bd ∗ is not empty; cf. Lemma 1.2.5(i). It follows that A∗ C ∩ BD∗ is not empty. The converse follows by repeating the above argument with A∗ , C, B, and D∗ in place of A, B, C, and D.

1.4

Thin Elements

9

(ii) The set A ∩ BCD∗ is not empty if and only if there exist elements a in A, b in B, c in C, and d in D such that a ∈ bcd ∗ . Similar observations for the other five sets, together with Lemma 1.2.5(ii), finish the proof. (iii) The equivalence is obtained by applying Lemma 1.3.3 to CD, B∗ , and A in place  of A, B, and C. Lemma 1.3.6 Let H be a hypergroup, and let B and C be subsets of H. Assume that H contains a subset A with BA = C A and B∗ A = B∗ . Then the following hold.

∗ -invariant

(i) We have B2 = CB. (ii) The set B ∩ CB∗ is not empty if and only C ∩ CB∗ is not empty. Proof. (i) From B∗ A = B∗ we obtain that A∗ B = B; cf. Lemma 1.3.1(ii). Since A is ∗ -invariant, this implies that AB = B. Thus, as we are assuming that BA = C A, we have B2 = BAB = C AB = CB. (ii) From Lemma 1.3.3 we know that B ∩ CB∗ is empty if and only if C ∩ B2 is empty. Similarly, C ∩ CB∗ is empty if and only if C ∩ CB is empty. Now recall from  (i) that B2 = CB. Thus, B ∩ CB∗ is empty if and only C ∩ CB∗ is empty.

1.4 Thin Elements Recall from the Preface that a hypergroup element h is called thin if h∗ h = {1}. From Lemma 1.1.8 we know that 1∗ = 1. Thus, since 1∗ 1 = {1∗ }, we have 1∗ 1 = {1}, and that means that 1 is thin. Lemma 1.4.1 Thin hypergroup elements are tight. Proof. Assume that h is thin. Then h∗ h = {1}. It follows that hh∗ h = {h}, and that  means that h is tight. Lemma 1.4.2 Let H be a hypergroup, let a and b be elements of H, and assume that ab = {1}. Then a∗ = b and b is thin. Proof. From ab = {1} we obtain that a∗ ab = {a∗ }. Thus, as b ∈ a∗ ab, a∗ = b. From a∗ = b we obtain that a∗∗ = b∗ , and from Lemma 1.1.2 we know that a∗∗ = a.  Thus, a = b∗ . Thus, as ab = {1}, b∗ b = {1}, and that means that b is thin. Lemma 1.4.3 Let H be a hypergroup, and let a and b be elements of H. Then the following hold. (i) If b is thin, |ab| = 1.

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1 Basic Facts

(ii) If a∗ is thin, |ab| = 1. (iii) If a and b both are thin, ab contains a thin element c with ab = {c}. Proof. (i) From Lemma 1.1.3 we know that ab is not empty. Let c be an element in ab. Then a ∈ cb∗ . Thus, if b is thin, ab ⊆ cb∗ b = {c}, and that shows that ab = {c}. (ii) From Lemma 1.1.3 we know that ab is not empty. Let c be an element in ab. Then, by Lemma 1.2.1, c∗ ∈ b∗ a∗ . Now assume that a∗ is thin. Then, by (i), b∗ a∗ = {c∗ }, so that, by Lemma 1.2.2, ab = {c}. (iii) By (i), ab contains an element c such that ab = {c}. Thus, by Lemma 1.2.2, b∗ a∗ = {c∗ }. It follows that c∗ c = b∗ a∗ ab = b∗ b = {1}, which says that c is thin.  Recall from the Preface that a subset of a hypergroup is called thin if each of its elements is thin. Corollary 1.4.4 Let h be a hypergroup element, and assume that h∗ h is thin. Then h is tight. Proof. Let a be an element in h∗ h. Then h ∈ ha. On the other hand, since h∗ h is assumed to be thin, a is thin, whence, by Lemma 1.4.3(i), |ha| = 1. It follows that ha = {h}. Since a has been chosen arbitrarily in h∗ h, this shows that hh∗ h ⊆ {h}. From this  we obtain hh∗ h = {h}, since, by Lemma 1.1.3, hh∗ h is not empty. Lemma 1.4.5 Let H be a hypergroup, and let a and b be elements in H. Then the following hold. (i) Assume that a∗ is thin and that b ∈ a∗ a. Then ab∗ = {a}. (ii) Assume that ab∗ contains a thin element. Then a∗ a ⊆ b∗ b. Proof. (i) Since a∗ is assumed to be thin, |ba∗ | = 1; cf. Lemma 1.4.3(i). Furthermore, from our hypothesis that b ∈ a∗ a we obtain that a∗ ∈ ba∗ . Thus, ba∗ = {a∗ }. It follows that ab∗ = {a}; cf. Lemma 1.2.2. (ii) Let c be a thin element in ab∗ . From c ∈ ab∗ we obtain that a∗ ∈ b∗ c∗ and that a ∈ cb; cf. Lemma 1.2.1. Thus, as c is thin, we conclude that a∗ a ⊆ b∗ c∗ cb = b∗ b, as wanted.  Lemma 1.4.6 Let H be a hypergroup, and let d, e, and f be elements in H. Assume that d is thin and that de ∩ df is not empty. Then e = f . Proof. Since we are assuming that de ∩ df is not empty, we obtain from Lemma 1.2.5(i) that d ∗ d ∩ e f ∗ is not empty. On the other hand, since d is assumed to be  thin, d ∗ d = {1}. Thus, 1 ∈ e f ∗ , so that, by Lemma 1.1.6, e = f .

1.5

Groups and Thin Hypergroups

11

Lemma 1.4.7 Let H be a hypergroup, let h be a thin element of H, and let A be a subset of H. Then the following hold. (i) We have | A| ≤ |hA|. (ii) If h∗ is thin, | A| = |hA|. Proof. (i) Let b and c be elements in A, and assume that hb ∩ hc is not empty. Then, by Lemma 1.4.6, b = c. (ii) Assume that h∗ is thin. Then, for each element a in A, |ha| = 1; cf. Lemma 1.4.3(ii). Thus, for each element a in A, there exists an element ah in ha such that ha = {ah }. From Lemma 1.4.6 we obtain that a 7→ ah is an injective map from A to hA, from  Lemma 1.1.3 that this map is surjective. Thus, we have | A| = |hA|. Lemma 1.4.8 Let H be a hypergroup, let h be a thin element of H, and let A and B be subsets of H. Then h(A ∩ B) = hA ∩ hB. Proof. Since h(A ∩ B) ⊆ hA and h(A ∩ B) ⊆ hB, we have h(A ∩ B) ⊆ hA ∩ hB. To show the reverse containment, we choose an element in hA ∩ hB and denote it by c. Since c ∈ hA, A contains an element a with c ∈ ha. Similarly, as c ∈ hB, B contains an element b with c ∈ hb. It follows that ha ∩ hb is not empty. Thus, by Lemma 1.4.6, a = b. Since a ∈ A and b ∈ B, we now have a ∈ A ∩ B. Thus, as  c ∈ ha, we conclude that c ∈ h(A ∩ B).

1.5 Groups and Thin Hypergroups In the Preface, we mentioned that there is an obvious way to identify groups with thin hypergroups, so that the class of groups can be viewed as a distinguished class of hypergroups. It is the purpose of this section to establish this correspondence between groups and thin hypergroups. Let G be a group. The map which assigns to any two elements d and e in G the set {de} is a hyperoperation on G. We call this hyperoperation the hypermultiplication induced by the group G. Lemma 1.5.1 Let G be a group. Then G is a thin hypergroup with respect to the hypermultiplication induced by the group G and with inversion function g 7→ g −1 . Proof. Let · denote the hypermultiplication induced by the group G. For any three elements d, e, and f in G, we have

12

1 Basic Facts

(d ·e)· f = {de}· f = {de f } = d ·{e f } = d ·(e· f ). Thus, the hypermultiplication induced by the group G is associative. For each element g in G, we have g·1 = {g}. This shows that 1 is a neutral element with respect to the hypermultiplication induced by the group G. For each element g in G, define g ∗ := g −1 . Let d, e, and f be elements in G satisfying f ∈ d · e. From f ∈ d · e we obtain that f ∈ {de}, and then that f = de. It follows that e = d −1 f and d = f e−1 . From e = d −1 f we obtain that e ∈ d ∗ · f , and from d = f e−1 we obtain that d ∈ f ·e∗ . This shows that g 7→ g ∗ is an inversion function with respect to the hypermultiplication induced by the group G. So far, we have seen that G is a hypergroup with respect to the hypermultiplication induced by the group G. For each element g in G, we have g ∗ ·g = {g −1 g} = {1} which shows that G is a thin hypergroup with respect to the hypermultiplication induced by the group G.  The thin hypergroup which one obtains from a group G via Lemma 1.5.1 will be denoted by Gτ . Let H be a thin hypergroup. The map which assigns to any two elements a and b in H the uniquely determined element in ab is an operation on H. We call this operation on H the multiplication induced by the thin hypergroup H. Lemma 1.5.2 Let H be a thin hypergroup. Then H is a group with respect to the multiplication induced by H and with inversion function h 7→ h∗ . Proof. Let · denote the multiplication on H induced by the thin hypergroup H. Let a, b, and c be elements in H. Let d denote the unique element in ab, let e denote the unique element in dc, let d 0 denote the unique element in bc, and let e 0 denote the unique element in ad 0. Then {e} = dc = (ab)c = a(bc) = ad 0 = {e 0 }, so that e = e 0. It follows that (a·b)·c = d ·c = e = e 0 = a·d 0 = a·(b·c). Thus, the multiplication induced by the thin hypergroup H is associative. The definition of the multiplication induced by the thin hypergroup H implies that the neutral element of the thin hypergroup H is a neutral element with respect to the multiplication induced by the thin hypergroup H. Let h be an element in H. Then, since H is thin, h∗ h = {1} = hh∗ . It follows that h∗ · h = 1 = h· h∗ , so that, h∗ is the inverse of h with respect to the multiplication on H induced by the thin hypergroup H. 

1.6

Actions of Hypergroups

13

The group which one obtains from a thin hypergroup H via Lemma 1.5.2 will be denoted by Hγ . Theorem 1.5.3 The following statements hold. (i) For each group G, we have Gτγ = G. (ii) For each thin hypergroup H, we have Hγτ = H. Proof. (i) Let G be a group, let · denote the multiplication on G, and let ·τ denote the hypermultiplication of the hypergroup Gτ , i.e., the hypermultiplication on G induced by the group G. Then, for any two elements d and e in G, d ·τ e = {d ·e}. Let ·τγ denote the multiplication of the group Gτγ , i.e., the multiplication on Gτ induced by the thin hypergroup Gτ . Then, for any two elements d and e in G, d ·τ e = {d ·τγ e}. It follows that, for any two elements d and e in G, d · e = d ·τγ e. As a consequence, we have G = Gτγ . (ii) Let H be a thin hypergroup, let · denote the hypermultiplication on H, and let ·γ denote the multiplication of the group Hγ , i.e., the multiplication on H induced by the thin hypergroup H. Then, for any two elements a and b in H, a · b = {a·γ b}. Let ·γτ denote the hypermultiplication of the hypergroup Hγτ , i.e., the hypermultiplication on Hγ induced by the group Hγ . Then, for any two elements a and b in H, a ·γτ b = {a ·γ b}. It follows that, for any two elements a and b in H, a · b = a ·γτ b. As a consequence,  we have H = Hγτ . The correspondence between groups and thin hypergroups established in Theorem 1.5.3 allows us to identify groups with thin hypergroups. It is called the group correspondence.

1.6 Actions of Hypergroups Let H be a hypergroup, let X be a non-empty set, and let ω be a map from X × H to the power set of X. For any two elements x in X and h in H, we may simply write xh instead of ω(x, h). Let W be a subset of X, and let A be a subset of H. We write W A for the union of the sets wa with w ∈ W and a ∈ A. If W contains an element w with W = {w}, we write w A instead of W A. Similarly, we write W a instead of W A if A contains an element a with A = {a}. We say that ω is a hypergroup action or simply an action of H on (the non-empty set) X or that H acts on X (via ω) if the following conditions hold. O1

For any three elements x in X and a, b in H, we have (xa)b = x(ab).

14

O2

1 Basic Facts

For each element x in X, we have x ·1 = {x}.

If H acts on X, we often write xab instead of (xa)b or x(ab) for elements x in X and a, b in H. Lemma 1.6.1 Let H be a hypergroup, let X be a set, and assume that H acts on X. Let x be an element in X, and let h be an element in H. Then the set xh is not empty. Proof. From Lemma 1.1.1 we know that 1 ∈ h∗∗ h∗ , from Lemma 1.1.2 that h∗∗ = h. Thus, 1 ∈ hh∗ . Now recall from Condition O2 that x ∈ x·1. It follows that x ∈ xhh∗ ,  and that shows that xh is not empty. Let H be a hypergroup, let X be a set, and let ω be an action of H on X. We say that ω is a regular action on X or that H acts regularly on X (via ω) if the following condition holds. O3

For any two elements y and z in X, there exists exactly one element h in H such that z ∈ yh.

(Recall that yh stands for ω(y, h).) Lemma 1.6.2 Let H be a hypergroup, let X be a set, and assume that H acts regularly on X. Let y and z be elements in X, and let h be an element in H with z ∈ yh. Then the following hold. (i) Let A be a subset of H with z ∈ y A. Then h ∈ A. (ii) We have y ∈ zh∗ . Proof. (i) Since z ∈ y A, A contains an element a with z ∈ ya. Thus, as z ∈ yh, Condition O3 forces h = a. Since a ∈ A, this implies that h ∈ A. (ii) By Condition O3, H contains an element b with y ∈ zb. Thus, as z ∈ yh, y ∈ yhb. On the other hand, by Condition O2, y ·1 = {y}. Now, as we have y ∈ y·1 and y ∈ yhb, we obtain from (i) that 1 ∈ hb. It follows that h∗ = b; cf. Lemma 1.1.6. From y ∈ zb and h∗ = b we obtain that y ∈ zh∗ .  Lemma 1.6.3 Let H be a hypergroup, let X be a set, and assume that H acts regularly on X. Let x be an element in X, and let A and B be subsets of H. Then x(A ∩ B) = x A ∩ xB. Proof. Since x(A ∩ B) ⊆ x A and x(A ∩ B) ⊆ xB, we have x(A ∩ B) ⊆ x A ∩ xB. To show the reverse containment, we choose an element in x A ∩ xB and denote it by y. Since y ∈ x A, A contains an element a with y ∈ xa. Similarly, as y ∈ xB, B contains an element b with y ∈ xb. It follows that xa ∩ xb is not empty. Thus, by

1.6

Actions of Hypergroups

15

Condition O3, a = b. Since a ∈ A and b ∈ B, we now have a ∈ A ∩ B. Thus, as  y ∈ xa, we conclude that y ∈ x(A ∩ B). Lemma 1.6.4 Let H be a hypergroup, let X be a set, and assume that H acts regularly on X. Let y and z be elements in X, and let a and b be elements in H with z ∈ yab∗ . Then ya ∩ zb is not empty. Proof. Since z ∈ yab∗ , ya contains an element x with z ∈ xb∗ . From z ∈ xb∗ we obtain that x ∈ zb; cf. Lemma 1.6.2(ii). Now x ∈ ya ∩ zb, so that ya ∩ zb is not empty.  Next we show how to obtain hypergroups from certain partitions of sets. Let X be a set. We define 1X := {(x, x) | x ∈ X }. For each subset s of X × X, we set s∪ := {(y, z) ∈ X × X | (z, y) ∈ s}. For any two subsets q and r of X × X, we define q ◦ r to be the set of all pairs (y, z) in X × X for which there exists an element x in X with (y, x) ∈ q and (x, z) ∈ r.1 Let S be a set of binary relations on a set X. For any two elements q and r in S, we set qr := {s ∈ S | s ⊆ q ◦ r }. The map from S × S to the power set of S which sends each pair (q, r) of elements of S to qr is a hyperoperation on S. We will call this hyperoperation the set theoretic hypermultiplication on S. Lemma 1.6.5 Let X be a non-empty set, and let P be a partition of X × X. Assume that 1X ∈ P and that, for each element p in P, p∪ ∈ P. Assume further that, for any three elements q, r, and s in P, s ⊆ q ◦ r if s ∩ (q ◦ r) is not empty. Then P is a hypergroup with respect to the set theoretic hypermultiplication on P, with neutral element 1X and inversion function p 7→ p∪ . Proof. To verify associativity of the set theoretic hypermultiplication on P we pick elements q, r, and s in P, and we will show that (qr)s = q(r s). Let p be an element in (qr)s. Then qr contains an element t such that p ∈ ts. From p ∈ ts we obtain that p ⊆ t ◦ s, from t ∈ qr we obtain that t ⊆ q ◦ r. It follows that p ⊆ (q ◦ r) ◦ s = q ◦ (r ◦ s). Now P contains an element u such that p ∩ (q ◦ u) and u ∩ (r ◦ s) both are not empty. Since p ∩ (q ◦ u) is not empty, p ⊆ q ◦ u. Similarly, as 1We notice that, for any three subsets p, q, and r of X × X, (p ◦ q) ◦ r = p ◦ (q ◦ r).

16

1 Basic Facts

u ∩ (r ◦ s) is not empty, u ⊆ r ◦ s. From p ⊆ q ◦ u we obtain that p ∈ qu, and from u ⊆ r ◦ s we obtain that u ∈ r s. Now, as p ∈ qu and u ∈ r s, we have p ∈ q(r s). Since p has been chosen arbitrarily in (qr)s, we have shown that (qr)s ⊆ q(r s). The reverse containment follows similarly, so that we have (qr)s = q(r s). For each element p in P, we have p ◦ 1X = p; equivalently, p · 1X = {p}. Since 1X ∈ P, that shows that 1X is a neutral element with respect to the set theoretic hypermultiplication on P. To show that p 7→ p∪ is an inversion function with respect to the set theoretic hypermultiplication on P, we recall that, for each element p in P, p∪ is our notation for the set {(y, z) ∈ X × X | (z, y) ∈ p}. Let q, r, and s be elements in P, and assume that s ∈ qr. Then, by definition, s ⊆ q ◦ r. It follows that none of the two intersections r ∩ (q∪ ◦ s) and q ∩ (s ◦ r ∪ ) is empty. Thus, by hypothesis, r ⊆ q∪ ◦ s and q ⊆ s ◦ r ∪ . It follows that r ∈ q∪ s and q ∈ sr ∪ .  A hypergroup which arises from a partition of X × X for a non-empty set X via Lemma 1.6.5 will be called a set theoretic hypergroup on X. The following lemma shows that each set theoretic hypergroup H on a set X gives rise to a regular action of H on X. Lemma 1.6.6 Let X be a set, and let H be a set theoretic hypergroup on X. For any two elements x in X and h in H, define ω(x, h) := {y ∈ X | (x, y) ∈ h}. Then ω is a regular action of H on X. Proof. To verify Condition O1 we choose an element y in X and elements a and b in H, and we will show that (ya)b = y(ab). To show that (ya)b ⊆ y(ab), we let z be an element in (ya)b. Then ya contains an element x such that z ∈ xb. From x ∈ ya we obtain that (y, x) ∈ a, and from z ∈ xb we obtain that (x, z) ∈ b. It follows that (y, z) ∈ a ◦ b. Since H is assumed to be a set theoretic hypergroup on X, H is a partition of X × X. Thus, H contains an element c with (y, z) ∈ c. From (y, z) ∈ c and (y, z) ∈ a ◦ b we obtain that c ∩ (a ◦ b) is not empty. Since H is a set theoretic hypergroup on X, this forces c ⊆ a ◦ b. It follows that c ∈ ab. On the other hand, as (y, z) ∈ c, we have z ∈ yc. Thus, we conclude that z ∈ y(ab). Since z has been chosen arbitrarily in (ya)b, we have shown that (ya)b ⊆ y(ab). The reverse containment y(ab) ⊆ (ya)b follows similarly. Since H is a set theoretic hypergroup on X, 1X ∈ H. Thus, x ·1X := {y ∈ X | (x, y) ∈ 1X } = {x}, and that shows that Condition O2 holds.

1.6

Actions of Hypergroups

17

Let y and z be elements in X. Since H is a set theoretic hypergroup on X, H is a partition of X × X. Thus, H contains exactly one element h with (y, z) ∈ h. Thus, H contains exactly one element h with z ∈ yh. This shows that Condition O3 holds.  Let X be a set, and let H be a set theoretic hypergroup on X. The regular action of H on X which we found in Lemma 1.6.6 will be called the canonical action of H on X. Recall from the beginning of this section that we often write xh instead of ω(x, h) when ω stands for an action of a hypergroup H on a set X, x for an element in X, and h for an element in H. If H is a set theoretic hypergroup on a set X, this abbreviation will always refer to the canonical action of H on X. In Lemma 1.6.6, we associated to each set theoretic hypergroup H on a set X a regular action of H on X, namely the canonical action of H on X. We will now see that, conversely, each regular action of a hypergroup H on a set X gives rise to a set theoretic hypergroup on X; cf. Lemma 1.6.8. Let H be a hypergroup, let X be a set, and let ω be a regular action of H on X. Then, for any two elements y and z in X, there exists a uniquely determined element h in H such that z ∈ yh; cf. Condition O3. We denote this element by (y, z)σ and call the map which takes each element (y, z) in X × X to (y, z)σ the color defined by ω. From Lemma 1.6.1 one obtains that the color defined by a regular action ω of a hypergroup H on a set X is a surjective map from X × X to H. Lemma 1.6.7 Let H be a hypergroup, let X be a set, let ω be a regular action of H on X, and let σ denote the color defined by ω. Then the following hold. (i) For any three elements y, z in X and h in H, (y, z)σ = h if and only if z ∈ yh. (ii) For any two elements y and z in X, (y, z)σ∗ = (z, y)σ . (iii) For each element x in X, we have (x, x)σ = 1. Proof. (i) This follows right from the definition of σ. (ii) Set h := (y, z)σ . Then, by (i), z ∈ yh. Thus, by Lemma 1.6.2(ii), y ∈ zh∗ . Now, by (i), (z, y)σ = h∗ . Thus, as h = (y, z)σ , we have (y, z)σ∗ = (z, y)σ . (iii) This follows from Condition O2.



Lemma 1.6.8 Let H be a hypergroup, let X be a set, let ω be a regular action of H on X, and let σ denote the color defined by ω. For each element h in H, define hˆ := {(y, z) ∈ X × X | (y, z)σ = h}. Set Hˆ := { hˆ | h ∈ H}. Then Hˆ is a set theoretic hypergroup on X. Proof. From Condition O3 we obtain that Hˆ is a partition of X × X. (Note that, by Lemma 1.6.1, hˆ is not empty for each element h in H.)

18

1 Basic Facts

ˆ From Lemma 1.6.7(i) we obtain that 1ˆ = 1X . Thus, 1X ∈ H. From Lemma 1.6.7(ii) we obtain that, for any three elements y, z in X and h in H, (y, z) ∈ hˆ ∪

⇔

and (y, z) ∈ hˆ∗

⇔

(z, y) ∈ hˆ (y, z)σ = h∗

⇔

⇔

(z, y)σ = h (z, y)σ = h.

Thus, we have hˆ ∪ = hˆ∗ for each element h in H. Since h∗ ∈ H for each element h in H, this implies that hˆ ∪ ∈ Hˆ for each element h in H. For any two elements x in X and h in H, we will now write xh instead of ω(x, h). Then, by Lemma 1.6.7(i), hˆ := {(y, z) ∈ X × X | z ∈ yh} for each element h in H. ˆ is not empty. We have Let a, b, and c be elements in H, and assume that cˆ ∩ (aˆ ◦ b) ˆ to show that cˆ ⊆ aˆ ◦ b. ˆ is not empty, we find elements y and z in X such that Since cˆ ∩ (aˆ ◦ b) ˆ (y, z) ∈ cˆ and (y, z) ∈ aˆ ◦ b. From (y, z) ∈ cˆ we obtain that z ∈ yc. ˆ From From (y, z) ∈ aˆ ◦ bˆ we obtain an element x in X with (y, x) ∈ aˆ and (x, z) ∈ b. ˆ (y, x) ∈ aˆ we obtain that x ∈ ya, from (x, z) ∈ b we obtain that z ∈ xb. From x ∈ ya and z ∈ xb we obtain that z ∈ yab. From z ∈ yc and z ∈ yab we obtain that c ∈ ab; cf. Lemma 1.6.2(i). ˆ In order to show that cˆ ⊆ aˆ ◦ bˆ we now choose elements y and z in X with (y, z) ∈ c. Then z ∈ yc. Thus, as c ∈ ab, z ∈ yab. It follows that ya contains an element x with ˆ z ∈ xb. From x ∈ ya we obtain that (y, x) ∈ a, ˆ from z ∈ xb we obtain that (x, z) ∈ b. ˆ ˆ  From (y, x) ∈ aˆ and (x, z) ∈ b we obtain that (y, z) ∈ aˆ ◦ b. Let H be a hypergroup, let X be a set, and let ω be a regular action of H on X. The set theoretic hypergroup Hˆ on X which we found in Lemma 1.6.8 will be called the ω-shadow of H. We will come back to ω-shadows in Sections 3.9, 10.3, and 10.7.

1.7 Hypergroups Admitting Regular Actions Due to the group correspondence we know that thin hypergroups have exactly one regular action, the right regular action. For non-thin hypergroups this is not necessarily the case. There are non-thin hypergroups that do not admit any regular action,

1.7

Hypergroups Admitting Regular Actions

19

and there are non-thin hypergroups that admit more than one regular action. Hypergroups that do not allow regular actions appear in Theorem 7.2.9(iii). Hypergroups with more than one regular action can be found in the classes of hypergroups listed in Theorem 7.2.9(i). The hypergroups discussed in Sections 10.2, 10.3, 10.6, and 10.7 are further examples of hypergroups with more than one regular action. In this section, we collect a number of conditions a hypergroup needs to satisfy in order to admit a regular action. They are results of a collaboration with C. French. Lemma 1.7.1 Let H be a hypergroup which admits a regular action, and let a and b be elements in H. Assume that b∗ = b, b ∈ a2 , ba ∩ aa∗ = {b}, and a2 ∩ b2 = {a∗ }. Then a = b = 1. Proof. Let X be a set, and let ω be a regular action of H on X. We write xh instead of ω(x, h) when x stands for an element in X and h for an element in H. Let y be an element in X, and let z be an element in yb. Since we are assuming that a∗ ∈ b2 , we also have b∗ ∈ ab; cf. Lemma 1.2.1. Since b∗ = b, this implies that b ∈ ab. Thus, as z ∈ yb, z ∈ yab. It follows that ya contains an element x with z ∈ xb. We are assuming that b ∈ a2 . Thus, a ∈ ba∗ . Since x ∈ ya, this yields x ∈ yba∗ . It follows that yb contains an element v such that x ∈ va∗ . From z ∈ xb and b∗ = b we obtain that x ∈ zb. Thus, our hypothesis that b ∈ a2 yields that x ∈ za2 . Thus, za contains an element w with x ∈ wa. From w ∈ za and z ∈ yb we obtain that w ∈ yba. On the other hand, as w ∈ xa∗ and x ∈ ya, we have w ∈ yaa∗ . It follows that w ∈ yba ∩ yaa∗ , so that, by Lemma 1.6.3, w ∈ y(ba ∩ aa∗ ). Since we are assuming that ba ∩ aa∗ = {b}, this implies that w ∈ yb. From v ∈ xa and x ∈ wa we obtain that v ∈ wa2 . On the other hand, as w ∈ yb and b∗ = b, we have y ∈ wb. Thus, as v ∈ yb, we have v ∈ wb2 . It follows that v ∈ wa2 ∩ wb2 , so that, by Lemma 1.6.3, v ∈ w(a2 ∩ b2 ). Since we are assuming that a2 ∩ b2 = {a∗ }, this implies that v ∈ wa∗ . From v ∈ xa and x ∈ zb we obtain that v ∈ zba. On the other hand, as v ∈ wa∗ and w ∈ za, we have v ∈ zaa∗ . It follows that v ∈ zba ∩ zaa∗ , so that, by Lemma 1.6.3, v ∈ z(ba ∩ aa∗ ). Since we are assuming that ba ∩ aa∗ = {b}, this implies that v ∈ zb. From v ∈ yb and y ∈ zb we obtain that v ∈ zb2 . Thus, as v ∈ zb, b ∈ b2 ; cf. Lemma 1.6.2(i). Now recall that we are assuming that b ∈ a2 and that a2 ∩ b2 = {a∗ }. Thus, b = a∗ . Since b∗ = b, this implies that a = b. From b = a∗ = a, together with ba ∩ aa∗ = {b}, we obtain that b2 = {b}. Since b∗ = b, this implies that b = 1.  Lemma 1.7.2 Let H be a hypergroup which admits a regular action, and let a, b, and c be elements in H. Assume that c∗ = c, a ∈ (a∗ )2 ∩ bc, b∗ a∗ ∩ ca = {b}, ac ∩ a∗ c = {c}, and bc ∩ b∗ c = {c}. Then, c ∈ c2 .

20

1 Basic Facts

Proof. Let X be a set, and let ω be a regular action of H on X. We write xh instead of ω(x, h) when x stands for an element in X and h for an element in H. Let y be an element in X, and let z be an element in ya. Since we are assuming that a ∈ (a∗ )2 , this implies that z ∈ y(a∗ )2 . Thus, ya∗ contains an element x with z ∈ xa∗ . From x ∈ ya∗ we obtain that y ∈ xa, from z ∈ xa∗ we obtain that x ∈ za; cf. Lemma 1.6.2(ii). From y ∈ xa and a ∈ bc we obtain that y ∈ xbc. Thus, xb contains an element v with y ∈ vc. From c ∈ ac and c∗ = c we obtain that a ∈ c2 . Thus, as x ∈ za, we obtain that x ∈ zc2 . Thus, zc contains an element w with x ∈ wc. From v ∈ xb we obtain that x ∈ vb∗ ; cf. Lemma 1.6.2(ii). Thus, as z ∈ xa∗ , z ∈ vb∗ a∗ . On the other hand, as z ∈ ya and y ∈ vc, we have z ∈ vca. It follows that z ∈ vb∗ a∗ ∩ vca, so that, by Lemma 1.6.3, z ∈ v(b∗ a∗ ∩ ca). Since we are assuming that b∗ a∗ ∩ ca = {b}, this implies that z ∈ vb. From w ∈ zc and z ∈ vb we obtain that w ∈ vbc. On the other hand, as x ∈ wc and c∗ = c, we have w ∈ xc; cf. Lemma 1.6.2(ii). Thus, as x ∈ vb∗ , w ∈ vb∗ c. It follows that w ∈ vbc ∩ vb∗ c, so that, by Lemma 1.6.3, w ∈ v(bc ∩ b∗ c). Since we are assuming that bc ∩ b∗ c = {c}, this implies that w ∈ vc. From w ∈ zc and z ∈ ya we obtain that w ∈ yac. On the other hand, as w ∈ xc and x ∈ ya∗ , we have w ∈ ya∗ c. It follows that w ∈ yac ∩ ya∗ c, so that, by Lemma 1.6.3, w ∈ y(ac ∩ a∗ c). Since we are assuming that ac ∩ a∗ c = {c}, this implies that w ∈ yc. From w ∈ yc and y ∈ vc we obtain that w ∈ vc2 . Thus, as w ∈ vc, c ∈ c2 ; cf. Lemma  1.6.2(i). Lemma 1.7.3 Let H be a hypergroup which admits a regular action, and let a and b be elements in H. Assume that ba∗ ∩ a2 = {a} and that (a∗ )2 ∩ a2 = {b}. Then a ∈ b∗ b. Proof. Let X be a set, and let ω be a regular action of H on X. We write xh instead of ω(x, h) when x stands for an element in X and h for an element in H. Let y be an element in X, and let z be an element in yb. Then, as b ∈ (a∗ )2 , z ∈ y(a∗ )2 . Thus, ya∗ contains an element v such that z ∈ va∗ . Similarly, as b ∈ a2 , z ∈ ya2 . Thus, ya contains an element w such that z ∈ wa. From z ∈ wa and a ∈ a2 we obtain that z ∈ wa2 . Thus, wa contains an element x such that z ∈ xa. From z ∈ xa we obtain that x ∈ za∗ ; cf. Lemma 1.6.2(ii). Thus, as z ∈ yb, we obtain that x ∈ yba∗ . On the other hand, as x ∈ wa and w ∈ ya, we have x ∈ ya2 . It follows that x ∈ yba∗ ∩ ya2 , so that, by Lemma 1.6.3, x ∈ y(ba∗ ∩ a2 ). Since we are assuming that ba∗ ∩ a2 = {a}, this implies that x ∈ ya. From x ∈ za∗ and z ∈ va∗ we obtain that x ∈ v(a∗ )2 . On the other hand, as v ∈ ya∗ , we have y ∈ va; cf. Lemma 1.6.2(ii). Thus, as x ∈ ya, x ∈ va2 . It follows that

1.7

Hypergroups Admitting Regular Actions

21

x ∈ v(a∗ )2 ∩ va2 . Thus, by Lemma 1.6.3, x ∈ v((a∗ )2 ∩ a2 ). Since we are assuming that (a∗ )2 ∩ a2 = {b}, this implies that x ∈ vb. From z ∈ wa we obtain that w ∈ za∗ ; cf. Lemma 1.6.2(ii). Thus, as z ∈ va∗ , we have w ∈ v(a∗ )2 . On the other hand, as y ∈ va and w ∈ ya, we have w ∈ va2 . It follows that w ∈ v(a∗ )2 ∩ va2 , so that, by Lemma 1.6.3, w ∈ v((a∗ )2 ∩ a2 ). Since we are assuming that (a∗ )2 ∩ a2 = {b}, this implies that w ∈ vb. From w ∈ vb we obtain that v ∈ wb∗ ; cf. Lemma 1.6.2(ii). Thus, as x ∈ vb, we have x ∈ wb∗ b. Now recall that x ∈ wa. Thus, by Lemma 1.6.2(i), a ∈ b∗ b.  Lemma 1.7.4 Let H be a hypergroup which admits a regular action, and let a, b, c, d, and e be elements in H. Assume that c ∈ a∗ a, a ∈ b2 , b∗ a ∩ bc = {d}, and b∗ b ∩ db∗ = {e}. Then, if c is symmetric, so is e. Proof. Let X be a set, and let ω be a regular action of H on X. We write xh instead of ω(x, h) when x stands for an element in X and h for an element in H. Let y be an element in X, and let z be an element in yc. Then, as c ∈ a∗ a, z ∈ ya∗ a. Thus, ya∗ contains an element x such that z ∈ xa. From x ∈ ya∗ we obtain that y ∈ xa; cf. Lemma 1.6.2(ii). Thus, as a ∈ b2 , y ∈ xb2 , so that xb contains an element v with y ∈ vb. Similarly, xb contains an element w in with z ∈ wb. From v ∈ xb we obtain that x ∈ vb∗ ; cf. Lemma 1.6.2(ii). Thus, as z ∈ xa, we obtain that z ∈ vb∗ a. On the other hand, as z ∈ yc and y ∈ vb, we have z ∈ vbc. It follows that z ∈ vb∗ a ∩ vbc, so that, by Lemma 1.6.3, z ∈ v(b∗ a ∩ bc). Since we are assuming that b∗ a ∩ bc = {d}, this implies that z ∈ vd. From w ∈ xb and x ∈ vb∗ we obtain that w ∈ vb∗ b. On the other hand, since z ∈ wb, we have w ∈ zb∗ ; cf. Lemma 1.6.2(ii). Thus, as z ∈ vd, w ∈ vdb∗ . It follows that w ∈ vb∗ b ∩ vdb∗ , so that, by Lemma 1.6.3, w ∈ v(b∗ b ∩ db∗ ). Since we are assuming that b∗ b ∩ db∗ = {e}, this implies that w ∈ ve. Now assume that c∗ = c. Then we similarly obtain that v ∈ we, so that, by Lemma  1.6.2(ii), e∗ = e. The following lemma will be needed only in the proof of Lemma 1.7.6. Lemma 1.7.5 Let H be a hypergroup, let X be a set, let ω be a regular action of H on X, and let a and b be elements in H. Assume that (a∗ )2 ∩ a2 = {b}, ab ∩ ba∗ = {b}, and b < a∗ a. Let x, y, and z be elements in X with y ∈ ω(x, b), z ∈ ω(y, b), and x ∈ ω(z, b). Then ω(x, a) ∩ ω(y, a∗ ) ∩ ω(z, a) is empty. Proof. Recall that we write xh instead of ω(x, h) whenever x stands for an element in X and h for an element in H.

22

1 Basic Facts

Assume, by way of contradiction, that xa ∩ ya∗ ∩ za is not empty, and let u be an element in xa ∩ ya∗ ∩ za. We are assuming that (a∗ )2 ∩ a2 = {b}. Thus, by Lemma 1.3.1(ii), b∗ = b. From y ∈ xb and b∗ = b we obtain that x ∈ yb; cf. Lemma 1.6.2(ii). Thus, as b ∈ a2 , x ∈ ya2 . Thus, ya contains an element v with x ∈ va. From v ∈ ya we obtain that y ∈ va∗ ; cf. Lemma 1.6.2(ii). Thus, as u ∈ ya∗ , we obtain that u ∈ v(a∗ )2 . On the other hand, as u ∈ xa and x ∈ va, we have u ∈ va2 . It follows that u ∈ v(a∗ )2 ∩ va2 , so that, by Lemma 1.6.3, u ∈ v((a∗ )2 ∩ a2 ). Since we are assuming that (a∗ )2 ∩ a2 = {b}, this implies that u ∈ vb. From x ∈ zb and b∗ = b we obtain that z ∈ xb; cf. Lemma 1.6.2(ii). Thus, as x ∈ va, we obtain that z ∈ vab. On the other hand, as u ∈ za, we have z ∈ ua∗ ; cf. Lemma 1.6.2(ii). Thus, as u ∈ vb, we obtain that z ∈ vba∗ . It follows that z ∈ vab ∩ vba∗ , so that, by Lemma 1.6.3, z ∈ v(ab ∩ ba∗ ). Since we are assuming that ab ∩ ba∗ = {b}, this implies that z ∈ vb. From x ∈ va, z ∈ vb, and v ∈ ya (together with b∗ = b) we obtain that v ∈ zb ∩ ya ∩ xa∗ ; cf. Lemma 1.6.2(ii). Similarly, we find an element w in xb ∩ ya ∩ za∗ . From w ∈ xb and x ∈ va we obtain that w ∈ vab. On the other hand, as w ∈ za∗ and z ∈ vb, we have w ∈ vba∗ . It follows that w ∈ vab ∩ vba∗ , so that, by Lemma 1.6.3, w ∈ v(ab ∩ ba∗ ). Since we are assuming that ab ∩ ba∗ = {b}, this implies that w ∈ vb. From w ∈ ya and y ∈ va∗ we obtain that w ∈ va∗ a. Now, as we have w ∈ vb and w ∈ va∗ a, we conclude from Lemma 1.6.2(i) that  b ∈ a∗ a, contrary to our hypothesis. Lemma 1.7.6 Let H be a hypergroup which admits a regular action, and let a and b be elements in H. Assume that ab = {a∗, b}, (a∗ )2 ∩ a2 = {b}, and b < a∗ a. Then b < b2 . Proof. Let X be a set, and let ω be a regular action of H on X. We write xh instead of ω(x, h) when x stands for an element in X and h for an element in H. We are assuming that (a∗ )2 ∩ a2 = {b}. Thus, by Lemma 1.3.1(ii), b∗ = b. We are assuming that ab = {a∗, b}. Thus, as b∗ = b, ba∗ = {a, b}. On the other hand, since b ∈ a2 and b < a∗ a, we have a∗ , a. Thus, ab ∩ ba∗ = {b}.

1.8

Association Schemes

23

Assume, by way of contradiction, that b ∈ b2 , and choose elements x in X and z in xb. Then there exists an element y in xb with z ∈ yb. Since y ∈ xb and b ∈ a2 , y ∈ xa2 . Thus, xa contains an element v with y ∈ va. From y ∈ va we obtain that v ∈ ya∗ ; cf. Lemma 1.6.2(ii). Thus, as v ∈ xa, we obtain that v < za; cf. Lemma 1.7.5. It follows that z < va∗ ; cf. Lemma 1.6.2(ii). Since z ∈ yb and y ∈ va, we have z ∈ vab. Thus, as ab = {a∗, b} and z < va∗ , z ∈ vb. Since z ∈ xb and b ∈ a2 , we have z ∈ xa2 . Thus, xa contains an element w with z ∈ wa. From z ∈ wa we obtain that w ∈ za∗ ; cf. Lemma 1.6.2(ii). Thus, as w ∈ xa, we obtain that w < ya; cf. Lemma 1.7.5. It follows that y < wa∗ ; cf. Lemma 1.6.2(ii). From z ∈ yb and b∗ = b we obtain that y ∈ zb; cf. Lemma 1.6.2(ii). Thus, as z ∈ wa, y ∈ wab. Thus, as ab = {a∗, b} and y < wa∗ , y ∈ wb. Thus, as b∗ = b, w ∈ yb; cf. Lemma 1.6.2(ii). From w ∈ yb and y ∈ va we obtain that w ∈ vab. On the other hand, as w ∈ za∗ and z ∈ vb, we have w ∈ vba∗ . It follows that w ∈ vab ∩ vba∗ , so that, by Lemma 1.6.3, w ∈ v(ab ∩ ba∗ ). Since we saw that ab ∩ ba∗ = {b}, this implies that w ∈ vb. From w ∈ vb and v ∈ xa we now obtain that w ∈ xab. Thus, as w ∈ xa, Lemma 1.6.2(i) yields a ∈ ab. It follows that b ∈ a∗ a, contrary to our hypothesis. 

1.8 Association Schemes Let X be a set, let H be a set theoretic hypergroup on X, and let ω denote the canonical action of H on X. Recall that we write xh instead of ω(x, h) whenever x stands for an element in X and h for an element in H. The hypergroup H will be called association scheme on X if, for any three elements a, b, and c in H with c ∈ ab, there exists a cardinal number γabc such that, for any two elements y in X and z ∈ yc, |ya ∩ zb∗ | = γabc . If H is an association scheme, the cardinal numbers γabc with {a, b, c} ⊆ H are called the structure constants of H. In this section, we consider association schemes only on finite sets. This implies, of course, that the structure constants of all association schemes in this section are non-negative integers. Our selection of results on association schemes on finite sets presented in this section is quite rudimentary. It is aimed to the proof of Theorem 7.2.9. A systematic approach to association schemes is given in [55]. A useful list of association schemes on sets of small cardinality is given in [25]. We will refer to that list in the proof of Theorem 7.2.9.

24

1 Basic Facts

Lemma 1.8.1 Let H be an association scheme on a finite set, and let a and b be elements in H. Then the following hold. (i) We have γ1ab = δab and γa1b = δab . (ii) For each element c in H, we have γabc = γb∗ a∗ c∗ . (iii) For any two elements c and d in S, we have Õ Õ γabe γecd = γaed γbce . e ∈H

e ∈H

Proof. Let X be a finite set, and assume that H is an association scheme on X. (i) This follows from the fact that H is a partition of X × X. (ii) Let y be an element in X, and let z be an element in yc. Then |ya ∩ zb∗ | = γabc . From z ∈ yc we also obtain that y ∈ zc∗ ; cf. Lemma 1.6.2(ii). Thus, we also have |zb∗ ∩ ya∗∗ | = γb∗ a∗ c∗ . Since a∗∗ = a, the two equations yield γabc = γb∗ a∗ c∗ . (iii) Let y be an element in X, and let z be an element in yd. Counting in two different ways the pairs (v, w) in ya × zc∗ with w ∈ vb one obtains the desired equation from  the definition of γabe , γecd , γaed , and γbce . For each element h of an association scheme, we write νh instead of γhh∗ 1 . We notice that, for each element h of an association scheme, 1 ≤ νh , since h is not empty. Lemma 1.8.2 Let X be a finite set, let H be an association scheme on X, and let h be an element in H. Then the following hold. (i) For each element x in X, we have |xh| = νh . (ii) We have |h| = |X |νh . Proof. (i) Let x be an element in X. Since h∗∗ = h, we have |xh| = γhh∗ 1 . Thus, the claim follows from the definition of νh . (ii) This follows from (i).



Lemma 1.8.3 Let H be an association scheme on a finite set, and let h be an element in H. Then we have νh∗ = νh . Proof. Let X be a finite set, and assume that H is an association scheme on X. Since X is a finite set, h∗ and h are finite sets, too. On the other hand, by Lemma 1.8.2(ii), we have |h∗ | = |X |νh∗ and |h| = |X |νh . Thus, as |h∗ | = |h|, we conclude that νh∗ = νh .



1.8

Association Schemes

25

Lemma 1.8.4 Let H be an association scheme on a finite set, and let a and b be elements in H. Then the following hold. (i) If 1 ≤ γab∗ 1 , then a = b. (ii) For each element c in H, we have γacb νb = γbc∗ a νa . (iii) We have

Õ

γacb = νa =

c ∈H

(iv) We have

Õ

γca∗ b .

c ∈H

Õ

γabc νc = νa νb .

c ∈H

Proof. Let X be a finite set, and assume that H is an association scheme on X. (i) This follows from the fact that H is a partition of X × X. (ii) Applying Lemma 1.8.1(iii) to c, b∗ , and 1 in the role of b, c, and d we obtain that γacb νb = νa γcb∗ a∗ ; cf. (i). From Lemma 1.8.1(ii) we also know that γcb∗ a∗ = γbc∗ a . Thus, γacb νb = γbc∗ a νa . (iii) Let y be an element in X, and let z be an element in yb. The set ya is the union of the sets ya ∩ zc∗ with c ∈ H. Moreover, for any two distinct elements d and e in H, the sets ya ∩ zd ∗ and ya ∩ ze∗ are disjoint. Thus, {ya ∩ zc∗ | c ∈ H} is a partition of ya. It follows that Õ Õ |ya ∩ zc∗ | = |ya| = νa ; γacb = c ∈H

c ∈H

cf. Lemma 1.8.2(i). The second equation follows from the first one together with Lemma 1.8.1(ii). (iv) From (ii) we obtain Õ c ∈H

γabc νc =

Õ c ∈H

γcb∗ a νa = νa

Õ

γcb∗ a .

c ∈H

Thus, the claim follows from the second equation of (iii).



Lemma 1.8.5 Let H be an association scheme on a finite set, and let b be an element in H. Assume that H contains an element a with a∗ a ∩ b2 = {b}. Then b = 1. Proof. Let X be a finite set, and assume that H is an association scheme on X. Let x be an element in X, let z be an element in xa. From b ∈ a∗ a we obtain that a ∈ ab∗ ; cf. Lemma 1.2.1. Thus, as z ∈ xa, z ∈ xab∗ , so that xa contains an element y with z ∈ yb∗ . From z ∈ xa we obtain that x ∈ za∗ ; cf. Lemma 1.6.2(ii). Similarly, as z ∈ yb∗ , y ∈ zb.

26

1 Basic Facts

Let w be an element in xa ∩ yb. From w ∈ xa and x ∈ za∗ we obtain that w ∈ za∗ a. On the other hand, as w ∈ yb and y ∈ zb, we have w ∈ zb2 . It follows that w ∈ za∗ a ∩ zb2 , so that, by Lemma 1.6.3, w ∈ z(a∗ a ∩ b2 ). Since we are assuming that a∗ a ∩ b2 = {b}, this implies that w ∈ zb. Since w has been chosen arbitrarily in xa ∩ yb, we have shown that xa ∩ yb ⊆ xa ∩ zb. On the other hand, as y ∈ xa, |xa ∩ yb| = γab∗ a . Similarly, as z ∈ xa, |xa ∩ zb| = γab∗ a . Thus, as X is a finite set, we conclude that xa ∩ yb = xa ∩ zb. Now recall that y ∈ xa ∩ zb. Thus, y ∈ yb. Since y ∈ y·1, this implies that b = 1.  Lemma 1.8.6 Let H be an association scheme on a finite set, and let b be an element in H. Assume that H contains an element a with a2 = {b} and a∗ a = {1, b}. Then b = 1. Proof. Applying Lemma 1.8.4(iv) to a in place of b, we obtain that Õ γaac νc = νa2 . c ∈H

Since we are assuming that a2 = {b}, we also have Õ γaac νc = γaab νb . c ∈H

Thus,

νa2 = γaab νb .

On the other hand, by Lemma 1.8.4(iii), γaab ≤ νa . Thus, νa ≤ νb . Assume, by way of contradiction, that b , 1. Since we are assuming that a∗ a = {1, b}, we then have Õ νa2 = νa∗ νa = γa∗ ac νc = γa∗ a1 ν1 + γa∗ ab νb = νa∗ + γa∗ ab νb = νa + γa∗ ab νb ; c ∈H

cf. Lemma 1.8.3 and Lemma 1.8.4(iv). Thus, as νa2 = γaab νb , we conclude that (γaab − γa∗ ab )νb = νa . Since νa ≤ νb , this implies that νa = νb . Thus, as γaab νb = νa2 , we obtain that γaab = νa . On the other hand, we are assuming that a2 = {b} and that b , 1. Thus, by Lemma 1.1.1, a∗ , a.

1.8

Association Schemes

27

From γaab = νa and a∗ , a we obtain that νa + γa∗ ab = γaab + γa∗ ab ≤

Õ

γcab = νa∗ = νa ;

c ∈H

cf. Lemma 1.8.4(iii) and Lemma 1.8.3. It follows that γa∗ ab = 0, contrary to our  hypothesis that b ∈ a∗ a. Lemma 1.8.7 Let H be an association scheme on a finite set, and let a and b be elements in H with a∗ , a, b∗ = b, and H = {1, a, a∗, b}. Assume that {a, b} ⊆ ab. Then b ∈ a2 . Proof. Assume, by way of contradiction, that b < a2 . Then γaab = 0 and a < a∗ b. From a < a∗ b we obtain that γa∗ ba = 0. Furthermore, as we are assuming that a∗ , a, we also have γaa1 = 0; cf. Lemma 1.8.4(i). Thus, Õ γaac γcba = γaaa γaba . c ∈H

From b < a2 we also obtain that a∗ < ab∗ ; cf. Lemma 1.2.1. Thus, γab∗ a∗ = 0, and, since b∗ = b, this implies that γaba∗ = 0. From a∗ , a and b∗ = b we obtain that a , b. Thus, γab1 = 0. It follows that Õ γaca γabc = γaaa γaba + γaba γabb . c ∈H

Now recall from Lemma 1.8.1(iii) that Õ Õ γaac γcba = γaca γabc . c ∈H

Thus,

c ∈H

γaaa γaba = γaaa γaba + γaba γabb .

It follows thatγaba γabb = 0, and that contradicts our hypothesis that {a, b} ⊆ ab. 

2 Closed Subsets

A non-empty subset A of a hypergroup is called closed if A∗ A ⊆ A. Notice that a closed subset F of a hypergroup H is a hypergroup with respect to the hyperoperation which one obtains from the hyperoperation on H if one restricts the domain of the hyperoperation on H to F × F and the codomain of this hyperoperation to the power set of F. Each hypergroup is a closed subset of itself. Note also that, for each hypergroup H, the set {1} is a closed subset of H. Hypergroups which have exactly one closed subset are called trivial, those which have have exactly two closed subsets are called primitive. (Thus, a hypergroup H is trivial, if and only if H = {1}, and it is primitive, if and only if H , {1} and H does not have a closed subset different from {1} and H.) Hypergroups of type H3, j (as defined in Section 7.1) are primitive if j ∈ {1, . . . , 6}, they contain a closed subset with two elements if j ∈ {7, . . . , 10}. Hypergroups of type H4,7 , H4,8 , H4,9 , H4,14 , H4,15 , H4,16 , H4,17 , H4,18 , H4,19 , H4,20 , H4,21 , H4,22 , H4,23 , H4,30 , H4,31 , H4,32 , H4,33 , H4,34 , H4,35 , H4,36 , or H4,37 (as defined in Section 7.2) are primitive, while those of type H4,1 , H4,2 , H4,3 , H4,4 , H4,5 , H4,6 , H4,10 , H4,11 , H4,12 , H4,13 , H4,24 , H4,25 , H4,26 , H4,27 , H4,28 , or H4,29 (also defined in Section 7.2) are not. None of the hypergroups of type H6, j with j ∈ {1, . . . , 25} (as defined in Sections 7.3, 7.4, and 7.5) is primitive. They all contain a closed subset of cardinality 2. The closed subsets of a thin hypergroup H correspond, via the group correspondence, to the subgroups of the group which, via the group correspondence, corresponds to H. This shows that the notion of a closed subset is a genuine generalization of the notion of a subgroup. We begin this chapter with basic facts about closed subsets; cf. Section 2.1. Most of them are straightforward generalizations of basic facts about subgroups. The section includes the definitions of a coset and a double coset of a closed subset and the observation that, like in group theory, the set of all double cosets of a closed subset of a hypergroup H is a partition of H. In Section 2.2, we establish © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 P. -H. Zieschang, Hypergroups, https://doi.org/10.1007/978-3-031-39489-8_2

29

30

2 Closed Subsets

Dedekind Modularity for hypergroups, and, in Section 2.3, we define generating subsets of closed subsets. As mentioned already in the Preface, generating sets give rise to length functions, and length functions play as useful a role in the theory of hypergroups as they do in the theory of groups. Generating subsets are related to commutators, and commutators are the subject of Section 2.4. Conjugates of closed subsets will be introduced in Section 2.5, and in Section 2.6, the focus is on the set of all thin elements of a hypergroup. This set is not necessarily closed, but it is closed if it is finite. In a final section, we introduce specific injective maps from closed subsets of hypergroups H to some of their cosets in H. These maps will be called foldings, and their existence allows us to distinguish certain cosets of closed subsets from otherwise less controllable cosets.

2.1 Basic Facts We start with necessary and sufficient conditions for subsets of hypergroups to be closed. Lemma 2.1.1 Let H be a hypergroup, and let A be subset of H. Then the following conditions are equivalent. (a) The set A is a closed subset of H. (b) We have 1 ∈ A, A∗ = A, and AA = A. (c) We have 1 ∈ A, A∗ ⊆ A, and AA ⊆ A. Proof. (a) ⇒ (b) We are assuming that A is a closed subset of H. Thus, by definition, A is not empty. Thus, by Lemma 1.1.1, 1 ∈ A∗ A ⊆ A. From 1 ∈ A we obtain that A∗ = A∗ · 1 ⊆ A∗ A ⊆ A. Thus, by Lemma 1.3.2(ii), A∗ = A. From 1 ∈ A we obtain that A = A · 1 ⊆ AA, and from A∗ = A we obtain that AA = A∗ A ⊆ A. Thus, AA = A. (b) ⇒ (c) This is obvious. (c) ⇒ (a) We are assuming that 1 ∈ A, A∗ ⊆ A, and AA ⊆ A. From 1 ∈ A we obtain  that A is not empty. From A∗ ⊆ A and AA ⊆ A we obtain that A∗ A ⊆ A. Lemma 2.1.2 Let H be a hypergroup, and let A be a non-empty subset of H. Then the following hold. (i) Assume that AA∗ A ⊆ A. Then A∗ A is a closed subset of H. (ii) Let F be a closed subset of H, and assume that AA∗ ⊆ F. Then A∗ F A is a closed subset of H.

2.1

Basic Facts

31

Proof. (i) Since A is not empty, 1 ∈ A∗ A; cf. Lemma 1.1.1. From Lemma 1.3.1(ii) we know that A∗ A is ∗ -invariant, and our hypothesis that AA∗ A ⊆ A implies that A∗ AA∗ A ⊆ A∗ A. Thus, by Lemma 2.1.1, A∗ A is closed. (ii) From AA∗ ⊆ F we obtain that (A∗ F A)∗ A∗ F A = A∗ F AA∗ F A = A∗ F A; cf. Lemma 1.3.1(ii). Thus, A∗ F A is closed.



Recall from the Preface that a hypergroup element h is called tight if hh∗ h = {h}. Lemma 2.1.3 Let H be a hypergroup, and let h be an element of H. Then the following hold. (i) If h is tight, h∗ h is a closed subset of H. (ii) If h∗ h is thin, h∗ h is a closed subset of H. Proof. (i) This follows from Lemma 2.1.2(i). (ii) Assume that h∗ h is thin. Then, by Corollary 1.4.4, h is tight, so that the claim  follows from (i). Throughout this text, we will frequently write Fh instead of h∗ h when h is a tight hypergroup element. Lemma 2.1.4 Let H be a hypergroup, and let F be a non-empty set of closed subsets of H. Then the intersection of the closed subsets which belong to F is a closed subset of H. Proof. Let A denote the intersection of the closed subsets of H which belong to F . From Lemma 2.1.1 we know that 1 ∈ F for each element F in F . Thus, 1 ∈ A, and that shows that A is not empty. For each element F in F , we have A ⊆ F. From this we obtain that, for each element  F in F , A∗ A ⊆ F ∗ F ⊆ F; cf. Lemma 1.3.1(i). It follows that A∗ A ⊆ A. Lemma 2.1.5 Let H be a hypergroup, and let D and E be closed subsets of H. Then the following conditions are equivalent. (a) The set DE is a closed subset of H. (b) The set DE is ∗ -invariant. (c) We have DE = E D. Proof. (a) ⇒ (b) This follows from Lemma 2.1.1. (b) ⇒ (c) Since D is a closed subset of H, D∗ = D; cf. Lemma 2.1.1. Similarly, E ∗ = E. Thus, by Lemma 1.3.1(ii), DE = (DE)∗ = E ∗ D∗ = E D.

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(c) ⇒ (a) Assume that DE = E D. With a reference to Lemma 1.3.1(ii) we then obtain that (DE)∗ DE = E ∗ D∗ DE ⊆ E ∗ DE = E ∗ E D ⊆ E D = DE. Thus, as 1 ∈ DE, DE is a closed subset of H.



Lemma 2.1.6 Let H be a hypergroup, and let D and E be closed subsets of H. Then the following hold. (i) The set {DhE | h ∈ H} is a partition of H. (ii) We have D ∩ E = {1} if and only if, for each element h in DE, there exist exactly one element d in D and exactly one element e in E such that h ∈ de. (iii) Assume that D ∩ E = {1}. Let d be an element in D, and let e be an element in E with |ed| = 1. Then |d ∗ ed ∩ E | ≤ 1. Proof. (i) Since 1 ∈ D and 1 ∈ E, we have h ∈ DhE for each element h in H. Thus, H is equal to the union of the sets DhE with h ∈ H. To show that {DhE | h ∈ H} is a partition of H, we choose elements a and b in H, and we assume that DaE ∩ DbE is not empty. We will see that DaE = DbE. Since DaE ∩ DbE is not empty, we find an element c in DaE ∩ DbE. Since c ∈ DaE, we have DcE ⊆ DaE. From c ∈ DaE we also obtain that a ∈ D∗ cE ∗ ; cf. Lemma 1.3.5(ii). Since D and E both are closed, this implies that DaE ⊆ DcE; cf. Lemma 2.1.1. From DcE ⊆ DaE and DaE ⊆ DcE we obtain that DaE = DcE. Similarly, one shows that DbE = DcE, so that DaE = DbE. (ii) Assume first that D ∩ E = {1}, and let h be an element in DE. Since h ∈ DE, there exist elements d in D and e in E such that h ∈ de. Let b be an element in D, and let c an element in E with h ∈ bc. Then h ∈ de ∩ bc. Thus, by Lemma 1.2.5(i), d ∗ b ∩ ec∗ is not empty. Since D is a closed subset of H, d ∗ b ⊆ D. Similarly, as E is closed, ec∗ ⊆ E. It follows that d ∗ b ∩ ec∗ ⊆ D ∩ E = {1}. Therefore, 1 ∈ d ∗ b and 1 ∈ ec∗ . Thus, by Lemma 1.1.6, b = d and c = e. Assume now that, for each element h in DE, there exist exactly one element d in D and exactly one element e in E such that h ∈ de. Let c be an element in D ∩ E. Then c ∈ DE and c·1 = {c}. Moreover, by Lemma 1.1.4, 1·c = {c}. Thus, as 1 ∈ D ∩ E, c = 1. (iii) Let a and b be elements in d ∗ ed ∩ E. We have to show that a = b. Since we are assuming that |ed| = 1, ed contains an element h with ed = {h}. Then, as a ∈ d ∗ ed, a ∈ d ∗ h. It follows that h ∈ da. Similarly, h ∈ db, so that h ∈ da ∩ db. Since d ∈ D and {a, b} ⊆ E, this implies that a = b; cf. (ii). 

2.1

Basic Facts

33

Let H be a hypergroup, and let F be a closed subset of H. Let h be an element in H. The product hF is called the left coset of F represented by h. We also say that h is a representative of the left coset hF. The product F h is called the right coset of F represented by h, and h is called a representative of the right coset F h. The product F hF is called the double coset of F represented by h. We also say that h is a representative of the double coset F hF. A left coset of F represented by an element of H is said to be a left coset of F in H. Right cosets of F represented by an element in H are called right cosets of F in H, and double cosets of F represented by an element in H are called double cosets of F in H. From Lemma 1.3.1(ii) (together with Lemma 2.1.1) one obtains that the number of left cosets of F in H is equal to the number of right cosets of F in H. For each subset A of H, we define A/F := {aF | a ∈ A}

and

A//F := {FaF | a ∈ A}.

Note that H/F is the set of all left cosets of F in H and H//F is the set of all double cosets of F in H. The cardinal number |H/F | is called the index of F in H. Setting D = {1} and E = F in Lemma 2.1.6(i) one obtains that H/F is a partition of H. Similarly, setting D = F and E = F in Lemma 2.1.6(i) one obtains that H//F is a partition of H. Most of our applications of Lemma 2.1.6(i) refer to these two cases, but we do not explicitly mention this each time. Lemma 2.1.7 Let H be a hypergroup, let F be a closed subset of H, and let a and b be elements in H. Then the following hold. (i) We have aF = bF if and only if a∗ b ∩ F is not empty. (ii) We have Fa = Fb if and only if ab∗ ∩ F is not empty. Proof. (i) Since 1 ∈ F, b ∈ bF. Thus, aF = bF if and only if b ∈ aF; cf. Lemma 2.1.6(i). By Lemma 1.3.3, b ∈ aF if and only if a∗ b ∩ F is not empty. (ii) Since 1 ∈ F, b ∈ Fb. Thus, Fa = Fb if and only if b ∈ Fa; cf. Lemma 2.1.6(i).  By Lemma 1.3.3, b ∈ Fa if and only if ab∗ ∩ F is not empty. Lemma 2.1.8 Let H be a hypergroup, let F be a closed subset of H, and let h be an element in H. Then we have the following. (i) Assume that F h∩ h∗ F is not empty and that hF = {h}. Then h is symmetric. (ii) Assume that h∗ F = {h∗ } and hF = {h}. Then F hF = {h}. Proof. (i) We are assuming that F h ∩ h∗ F is not empty. Thus, as F ∗ = F, we obtain from Lemma 1.3.5(i) that F h∗ ∩ hF is not empty. Since we are assuming that

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hF = {h}, this implies that h ∈ F h∗ . Now, by Lemma 1.3.3, h∗ ∈ hF = {h}, and this shows that h is symmetric. (ii) From h∗ F = {h∗ } we obtain that F h = {h}; cf. Lemma 1.3.1(ii). Thus, as  hF = {h}, we obtain that F hF = {h}.

2.2 Dedekind Modularity The following lemma is a specific case of a general observation due to Richard Dedekind; cf. [17; Satz VIII]. Lemma 2.2.1 Let H be a hypergroup, let F be a closed subset of H, and let A and B be subsets of H. Then we have the following. (i) Assume that A ⊆ F. Then F ∩ AB = A(F ∩ B). (ii) Assume that B ⊆ F. Then F ∩ AB = (F ∩ A)B. Proof. (i) Let f be an element in F ∩ AB. Since f ∈ AB, B contains an element b with f ∈ Ab. Thus, by Lemma 1.3.3, b ∈ A∗ f . Since we are assuming that A ⊆ F and that F is closed, this implies that b ∈ F. Thus, as b ∈ B, we have b ∈ F ∩ B. Since f ∈ Ab, this yields f ∈ A(F ∩ B). Since f has been chosen arbitrarily among the elements in F ∩ AB, we have shown that F ∩ AB ⊆ A(F ∩ B). For the reverse containment we recall that A ⊆ F by hypothesis. Thus, since F is a closed subset of H, we have A(F ∩ B) ⊆ F; cf. Lemma 2.1.1. Since A(F ∩ B) ⊆ AB, we now conclude that A(F ∩ B) ⊆ F ∩ AB. (ii) Setting A = B∗ and B = A∗ in (i) we obtain that F ∩ B∗ A∗ = B∗ (F ∩ A∗ ). Thus, as F ∗ = F, the desired equation follows from Lemma 1.3.1(ii).  Corollary 2.2.2 Let H be a hypergroup, and let F be a closed subset of H. Let A and B be subsets of H, and assume that AF and BF both are closed. Then we have (A ∩ BF)(B ∩ AF) = AF ∩ AB ∩ BF. Proof. Since F is a closed subset of H, we have 1 ∈ F; cf. Lemma 2.1.1. It follows that A ∩ BF ⊆ AF. Thus, as AF is assumed to be closed, we obtain from Lemma 2.2.1(i) that AF ∩ (A ∩ BF)B = (A ∩ BF)(AF ∩ B). Since 1 ∈ F, we also have B ⊆ BF. Thus, as BF is assumed to be closed, Lemma 2.2.1(ii) yields BF ∩ AB = (BF ∩ A)B. The desired equation follows from the above two equations.



2.3

Generating Sets

35

Corollary 2.2.3 Let H be a hypergroup, and let F be a closed subset of H. Let A and B be subsets of H, and assume that B ⊆ F. Then we have the following. (i) If F ⊆ BAB, then B(F ∩ A)B = F. (ii) If A ∪ B = AB ∩ BA, then (F ∩ A) ∪ B = (F ∩ A)B ∩ B(F ∩ A). Proof. (i) Since we are assuming that B ⊆ F, we obtain from Lemma 2.2.1 that B(F ∩ A)B = (F ∩ BA)B = F ∩ BAB. Thus, if F ⊆ BAB, we obtain that B(F ∩ A)B = F, as wanted. (ii) We are assuming that B ⊆ F. Thus, (F ∩ A) ∪ B = F ∩ (A ∪ B). On the other hand, we know from Lemma 2.2.1 that AB ∩ F ∩ BA = (F ∩ A)B ∩ B(F ∩ A). Thus, if A ∪ B = AB ∩ BA, we obtain (F ∩ A) ∪ B = (F ∩ A)B ∩ B(F ∩ A).



Lemma 2.2.4 Let H be a hypergroup, let D and E be closed subsets of H, and let A and B be subsets of H with A ⊆ D and B ⊆ E. Then D ∩ AB ∩ E = (E ∩ A)(B ∩ D). Proof. Applying Lemma 2.2.1(ii) to E in place of F we obtain that E ∩ AB = (E ∩ A)B. Applying Lemma 2.2.1(i) to D and E ∩ A in place of F and A we obtain that D ∩ (E ∩ A)B = (E ∩ A)(D ∩ B). The desired equation follows from the above two equations.



2.3 Generating Sets Let H be a hypergroup, and let A be a subset of H. We define hAi to be the intersection of the closed subsets of H which contain A as a subset. Note that A ⊆ hAi. From Lemma 2.1.4 we also know that hAi is a closed subset of H. If H = hAi, H is said to be generated by A, and A is called a generating set of H. A hypergroup is called finitely generated if it is generated by a finite set. Finitely generated closed subsets of hypergroups will play a role in Section 6.8.

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2 Closed Subsets

Lemma 2.3.1 Let H be a hypergroup, let A and B be subsets of H, and assume that A ⊆ B. Then the following hold. (i) We have hAi ⊆ hBi. (ii) We have A ⊆ hB \ Ai if and only if hB \ Ai = hBi. Proof. (i) By definition, hAi is the intersection of the closed subsets of H which contain A as a subset. Thus, as hBi is a closed subset of H and A ⊆ B ⊆ hBi, we conclude that hAi ⊆ hBi. (ii) Since B \ A ⊆ hB \ Ai, we have A ⊆ hB \ Ai if and only if B ⊆ hB \ Ai. Furthermore, by definition, we have B ⊆ hB \ Ai if and only if hBi ⊆ hB \ Ai. Now recall from (i) that hB \ Ai ⊆ hBi. Thus, we conclude that hBi ⊆ hB \ Ai if and only if hB \ Ai = hBi.  Lemma 2.3.2 Let H be a hypergroup, let A and B be subsets of H, and assume that hAihBi = hBihAi. Then we have hAihBi = hA ∪ Bi. Proof. From Lemma 2.3.1(i) we know that hAi ⊆ hA ∪ Bi and hBi ⊆ hA ∪ Bi. Thus, as hA ∪ Bi is closed, hAihBi ⊆ hA ∪ Bi; cf. Lemma 2.1.1. Now recall that we are assuming that hAihBi = hBihAi. Thus, by Lemma 2.1.5,  hAihBi is closed. Since A ∪ B ⊆ hAihBi, this implies that hA ∪ Bi ⊆ hAihBi. Lemma 2.3.3 Let H be a hypergroup, and let b and c be elements in H. Let A be a ∗ -invariant subset of H, and assume that b ∈ cA. Then h{b} ∪ Ai = h{c} ∪ Ai. Proof. From b ∈ cA we obtain that h{b} ∪ Ai ⊆ h{c} ∪ Ai. Since A is ∗ -invariant, we obtain from b ∈ cA that c ∈ bA; cf. Lemma 1.3.3. From c ∈ bA we obtain, as  before, that h{c} ∪ Ai ⊆ h{b} ∪ Ai. Let H be a hypergroup. Note that, for each subset A of H, hAi = hA∗ ∪ Ai. Thus, it is no loss of information if one restricts the analysis of closed subsets generated by subsets of H to closed subsets generated by ∗ -invariant subsets of H. Lemma 2.3.4 Let H be a hypergroup, and let A be a ∗ -invariant subset of H. Then the following hold. (i) The set hAi is equal to the union of the sets An with n a non-negative integer. (ii) Assume that A is not empty, and let h be an element of H satisfying h∗ Ah ⊆ hAi. Then h∗ hAih ⊆ hAi.

2.3

Generating Sets

37

Proof. (i) Define B to be the union of the sets An with n a non-negative integer. We have to show that hAi = B. Note first that B is not empty, because 1 ∈ A0 ⊆ B. Now let i and j be non-negative integers. Since A is ∗ -invariant, so is Ai ; cf. Lemma 1.3.1(ii). Thus, (Ai )∗ A j = Ai A j = Ai+j ⊆ B. It follows that B is closed. Thus, as A ⊆ B, hAi ⊆ B. Conversely, for each non-negative integer n, we have An ⊆ hAi n ⊆ hAi. Thus, B ⊆ hAi. (ii) From (i) we know that hAi is equal to the union of the sets An with n a nonnegative integer. Thus, it suffices to show that h∗ An h ⊆ hAi for each non-negative integer n. To show this, we proceed by induction. Since A is assumed to be not empty and ∗ -invariant, we have 1 ∈ Ahh∗ A. Thus, as we are assuming that h∗ Ah ⊆ hAi, we conclude that h∗ A0 h = h∗ ·1· h ⊆ h∗ Ahh∗ Ah ⊆ hAi. Now let n be a positive integer, and assume that h∗ An−1 h ⊆ hAi. Then, as 1 ∈ hh∗ and h∗ Ah ⊆ hAi, h∗ An h ⊆ h∗ An−1 hh∗ Ah ⊆ hAi; cf. Lemma 2.1.1.



Let H be a hypergroup, let A be a ∗ -invariant subset of H, and let f be an element in hAi. By Lemma 2.3.4(i), there exists a non-negative integer n with f ∈ An . The smallest non-negative integer n with f ∈ An is called the A-length of f and will be denoted by ` A( f ). The map ` A from hAi to the set of all non-negative integers is called the length function defined by A. Lemma 2.3.5 Let A be a ∗ -invariant subset of a hypergroup, and let f be an element in hAi. Then ` A( f ∗ ) = ` A( f ). Proof. From Lemma 1.3.1(ii) we obtain that l A( f ∗ ) ≤ l A( f ) and l A( f ∗∗ ) ≤ l A( f ∗ ).  From Lemma 1.1.2 we know that f ∗∗ = f . Thus, ` A( f ∗ ) = ` A( f ). Lemma 2.3.6 Let A be a ∗ -invariant subset of a hypergroup, and let c, d, and e be elements in hAi. Assume that e ∈ cd. Then the following hold. (i) We have ` A(e) ≤ ` A(c) + ` A(d). (ii) Assume that ` A(c) ≤ ` A(d). Then ` A(d) − ` A(c) ≤ ` A(e). (iii) Assume that ` A(d) ≤ ` A(c). Then ` A(c) − ` A(d) ≤ ` A(e). Proof. (i) Set i := ` A(c) and j := ` A(d). Then c ∈ Ai and d ∈ A j . Thus, as e ∈ cd, e ∈ Ai A j = Ai+j . It follows that

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2 Closed Subsets

` A(e) ≤ i + j = ` A(c) + ` A(d). (ii) Set i := ` A(c∗ ) and j := ` A(e). Then c∗ ∈ Ai and e ∈ A j . On the other hand, as e ∈ cd, d ∈ c∗ e. Thus, d ∈ Ai A j = Ai+j . It follows that ` A(d) ≤ i + j = ` A(c∗ ) + ` A(e). From Lemma 2.3.5 we also know that ` A(c∗ ) = ` A(c). Thus, ` A(d) − ` A(c) ≤ ` A(e). (iii) The proof is similar to the proof of (ii).



Occasionally, Lemma 2.3.5 and Lemma 2.3.6 will be used implicitly. Lemma 2.3.7 Let A be a ∗ -invariant subset of a hypergroup, and let e be an element in hAi \ {1}. Then the following hold. (i) There exist elements a in A and d in hAi such that e ∈ ad and ` A(e) = ` A(d) + 1. (ii) There exist elements d in hAi and a in A such that e ∈ da and ` A(e) = ` A(d) + 1. Proof. (i) We set n := ` A(e). Then e ∈ An . Since we are assuming that e , 1, 1 ≤ n. From e ∈ An and 1 ≤ n we obtain elements a in A and d in An−1 with e ∈ ad. From d ∈ An−1 we obtain ` A(d) ≤ n−1, so ` A(d)+1 ≤ n. From n = ` A(e) and e ∈ ad we obtain that n ≤ ` A(d) + 1; cf. Lemma 2.3.6(i). It follows that ` A(e) = ` A(d) + 1. (ii) The proof is similar to the proof of (i).



Lemma 2.3.8 Let A be a ∗ -invariant subset of a hypergroup, and let g be an element in hAi. Then the following hold. (i) Let b and f be elements in hAi with g ∈ b f and ` A(g) = ` A(b) + ` A( f ). Let c and e be elements in hAi satisfying f ∈ ce and ` A( f ) = ` A(c) + ` A(e). Then bc contains an element d such that g ∈ de, ` A(d) = ` A(b) + ` A(c), and ` A(g) = ` A(d) + ` A(e). (ii) Let d and e be elements in hAi with g ∈ de and ` A(g) = ` A(d) + ` A(e). Let b and c be elements in hAi satisfying d ∈ bc and ` A(d) = ` A(b) + ` A(c). Then ce contains an element f such that g ∈ b f , ` A( f ) = ` A(c) + ` A(e), and ` A(g) = ` A(b) + ` A( f ). Proof. (i) We are assuming that ` A(g) = ` A(b) + ` A( f ) and ` A( f ) = ` A(c) + ` A(e). Thus, ` A(g) = ` A(b) + ` A(c) + ` A(e). Since g ∈ b f and f ∈ ce, we have g ∈ bce. Thus, bc contains an element d with g ∈ de.

2.4

Commutators

39

Since g ∈ de, ` A(g) ≤ ` A(d) + ` A(e); cf. Lemma 2.3.6(i). Thus, by the above equation, ` A(b) + ` A(c) ≤ ` A(d). On the other hand, since d ∈ bc, we have ` A(d) ≤ ` A(b) + ` A(c); cf. Lemma 2.3.6(i). It follows that ` A(d) = ` A(b) + ` A(c). From the above two equations we obtain that ` A(g) = ` A(d) + ` A(e). (ii) The proof is similar to the proof of (i).



We conclude this section by introducing a notational simplification which will be useful in the remainder of this monograph. If a subset A of a hypergroup contains an element a with A = {a}, we write hai instead of hAi. We will also write hb, ci instead of hAi if b and c are distinct elements of a subset A of a hypergroup with A = {b, c}.

2.4 Commutators Let H be a hypergroup, and let a and b be elements of H. We set [a, b] := a∗ b∗ ab. The subset [a, b] of H is called the commutator of a and b. From Lemma 1.3.1(ii) we know that [a, b]∗ = [b, a] for any two elements a and b in H. For any two closed subsets D and E of a hypergroup H, we define [D, E] to be the closed subset of H generated by the union of the sets [d, e] with d ∈ D and e ∈ E. Lemma 2.4.1 Let H be a hypergroup, and let D and E be closed subsets of H. Then the following hold. (i) We have [D, E] = [E, D]. (ii) Let C be a closed subset of H, and assume that C ⊆ D. Then [C, E] ⊆ [D, E]. Proof. (i) For any two elements d in D and e in E, we have [d, e] = [e, d]∗ ⊆ [E, D]. Thus, [D, E] ⊆ [E, D]. The reverse containment follows similarly. (ii) This follows from Lemma 2.3.1(i).



Lemma 2.4.2 Let H be a hypergroup, and let D and E be closed subsets of H. Then we have E[D, E] = [D, E]E. Proof. Since E and [D, E] are closed subsets of H, E and [D, E] both are ∗ -invariant. Thus, it suffices to show that E[D, E] ⊆ [D, E]E; cf. Lemma 1.3.1(iv).

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2 Closed Subsets

Let d be an element in D, and let e be an element in E. We will first show that E[d, e] ⊆ [D, E]E. Let c be an element in E[d, e]. Then E contains an element b such that c ∈ b[d, e]. It follows that c ∈ bd ∗ e∗ de ⊆ bd ∗ b∗ dd ∗ be∗ de = [b∗, d]d ∗ be∗ de. Thus, be∗ contains an element a such that c ∈ [b∗, d]d ∗ ade. From a ∈ be∗ and {b, e} ⊆ E we obtain that a ∈ E. Thus, c ∈ [b∗, d]d ∗ ade ⊆ [b∗, d]d ∗ ada∗ ae = [d, b∗ ]∗ [d, a∗ ]ae ∈ [D, E]E. Since c has been chosen arbitrarily in E[d, e], this shows that E[d, e] ⊆ [D, E]E. Similarly, one obtains that E[d, e∗ ] ⊆ [D, E]E, so that E[d, e]∗ = Ee∗ d ∗ ed = E d ∗ ed ⊆ E d ∗ ede∗ e = E[d, e∗ ]e ⊆ [D, E]E. Thus, by induction (and with a reference to Lemma 2.3.4(i)), we obtain that E[D, E] ⊆ [D, E]E.  Lemma 2.4.3 Let H be a hypergroup, and let D and E be closed subsets of H. Let A denote the union of the sets E d ∗ E dE with d ∈ D. Then hAi = E[D, E]. Proof. In Lemma 2.4.2, we saw that E[D, E] = [D, E]E. Thus, for any two elements d in D and e in E, E d ∗ e∗ dE = E[d, e]E ⊆ E[D, E]E = E[D, E]. Since A is the union of the sets E d ∗ E dE with d ∈ D, this shows that A ⊆ E[D, E]. Now recall that E[D, E] = [D, E]E also implies that E[D, E] is closed; cf. Lemma 2.1.5. Thus, hAi ⊆ E[D, E]. To show that E[D, E] ⊆ hAi, we note that, for any two elements d in D and e in E, [d, e] = d ∗ e∗ de ⊆ E d ∗ E dE ⊆ A. Thus, by Lemma 2.3.1(i), [D, E] ⊆ hAi. By definition, we also have E ⊆ A ⊆ hAi.  Since hAi is closed, we conclude that E[D, E] ⊆ hAi.

2.5 Conjugation Let H be a hypergroup, and let A be a subset of H. For each element h of H, we define Ah := {b ∈ H | hb ⊆ Ah}. Note that {1}h = SH (h) for each element h of a hypergroup H.

2.5

Conjugation

41

Lemma 2.5.1 Let H be a hypergroup, let A be a subset of H, and let h be an element in H. Then the following hold. (i) If 1 ∈ A, 1 ∈ Ah . (ii) We have Ah ⊆ h∗ Ah. Proof. (i) Assume that 1 ∈ A. Then h·1 = {h} = 1·h ⊆ Ah; cf. Lemma 1.1.4. Thus, by definition, 1 ∈ Ah . (ii) Let b be an element in Ah . Then, hb ⊆ Ah. Thus, b ∈ h∗ hb ⊆ h∗ Ah.



Lemma 2.5.2 Let H be a hypergroup, let A and B be subsets of H, and let h be an element in H. Then the following hold. (i) If A ⊆ B, Ah ⊆ Bh . (ii) We have hA ⊆ Bh if and only if A ⊆ Bh . (iii) We have Ah Bh ⊆ (AB)h . (iv) Let C be a subset of H, and assume that AB ⊆ C. Then Ah Bh ⊆ C h . Proof. (i) Let c be an element in Ah . Then, by definition, hc ⊆ Ah. Now assume that A ⊆ B. Then Ah ⊆ Bh. Thus, hc ⊆ Bh, and that means that c ∈ Bh . (ii) We have hA ⊆ Bh if and only if, for each element a in A, ha ⊆ Bh, and that means that A ⊆ Bh . (iii) Let e be an element in Ah Bh . Then there exist elements c in Ah and d in Bh such that e ∈ cd. Since c ∈ Ah , hc ⊆ Ah. Since d ∈ Bh , hd ⊆ Bh. It follows that he ⊆ hcd ⊆ Ahd ⊆ ABh, so that, by definition, e ∈ (AB)h . (iv) Considering (i) this follows from (iii).



Let H be a hypergroup, and let D and E be closed subsets of H. We say that D is a conjugate of E in H if H contains an element h with Dh = hE. Note that, if D is a conjugate of E in H, E is a conjugate of D in H, since, for each element h in H, Dh = hE implies that E h∗ = h∗ D; cf. Lemma 1.3.1(ii). This allows us to say that D and E are conjugate in H if D is a conjugate of E in H or E is a conjugate of D in H. Lemma 2.5.3 Let H be a hypergroup, let D and E be closed subsets of H, and let h be an element in H. Then D and E are conjugate in H if and only if H contains an element h such ∗ that D ⊆ E h and E ⊆ Dh .

42

2 Closed Subsets

Proof. By definition, D and E are conjugate in H if H contains an element h with Dh = hE. From Dh = hE we obtain that Dh ⊆ hE and hE ⊆ Dh. From Dh ⊆ hE we obtain that h∗ D ⊆ E h∗ ; cf. Lemma 1.3.1(i), (ii). From Lemma ∗ 2.5.2(ii) we know that h∗ D ⊆ E h∗ if and only if D ⊆ E h . From Lemma 2.5.2(ii) we also know that hE ⊆ Dh if and only if E ⊆ Dh .  Lemma 2.5.4 Let H be a hypergroup, let A be a subset of H, and let h be an element in H. Assume that Ah is ∗ -invariant. Then hAh i ⊆ hAi h . Proof. Let n be a non-negative integer. From Lemma 2.5.2(iii) we obtain that (Ah )n ⊆ (An )h . On the other hand, by Lemma 2.3.4(i), An ⊆ hAi, so that, by Lemma 2.5.2(i), (An )h ⊆ hAi h . Now, as we have (Ah )n ⊆ (An )h and (An )h ⊆ hAi h for each non-negative integer n, we have (Ah )n ⊆ hAi h for each non-negative integer n. This finishes the proof, since, by Lemma 2.3.4(i), hAh i is equal to the union of the sets (Ah )n with n a non-negative integer.  Lemma 2.5.5 Let H be a hypergroup, let F be a closed subset of H, and let h be an element in H. Assume that hh∗ ⊆ F. Then F h = h∗ F h. Proof. From Lemma 2.5.1(ii) we know that F h ⊆ h∗ F h. Thus, we just have to show that h∗ F h ⊆ F h . We are assuming that hh∗ ⊆ F. Thus, by Lemma 2.1.2(ii), h∗ F h is closed. From hh∗ ⊆ F we also obtain that hh∗ F h ⊆ F h. Thus, we may apply Lemma 2.5.2(ii) to  h∗ F h and F in place of A and B. We obtain that h∗ F h ⊆ F h , as wanted. Lemma 2.5.6 Let H be a hypergroup, let F be a closed subset of H, and let a and b be elements in H. Then the following hold. (i) Assume that Fa = Fb. Then F a = F b . (ii) Assume that |ab| = 1 and that b ∈ a∗ Fa. Then b ∈ F a . Proof. (i) Let c be an element in F a . Then ac ⊆ Fa. Since we are assuming that Fa = Fb, we also have b ∈ Fa. Thus, bc ⊆ Fac ⊆ Fa = Fb. It follows that c ∈ F b . Since c has been chosen arbitrarily in F a , we have seen that F a ⊆ F b . The reverse containment is shown similarly. (ii) We are assuming that b ∈ a∗ Fa. Thus, a∗ F contains an element c such that b ∈ ca. From c ∈ a∗ F we obtain that c∗ ∈ Fa; cf. Lemma 1.3.3. From b ∈ ca we obtain that a ∈ c∗ b. Thus, a ∈ Fab. Now recall that we are assuming that |ab| = 1. Thus, as a ∈ Fab, we conclude that  ab ⊆ Fa; cf. Lemma 1.3.3. It follows that b ∈ F a .

2.6

The Thin Radical

43

Lemma 2.5.7 Let H be a hypergroup, let D and E be closed subsets of H, and let h be an element in H. Assume that, for each element e in h∗ Dh ∩ E, |he| = 1. Then Dh ∩ E = h∗ Dh ∩ E. Proof. From Lemma 2.5.1(ii) we know that Dh ⊆ h∗ Dh. Thus, Dh ∩ E ⊆ h∗ Dh ∩ E. To show the reverse containment, we recall that, by hypothesis, |he| = 1 for each  element e in h∗ Dh ∩ E. Thus, by Lemma 2.5.6(ii), h∗ Dh ∩ E ⊆ Dh . Lemma 2.5.8 Let H be a hypergroup, let D and E be closed subsets of H, and let h be an element in H with h∗ h ⊆ E. Then the following hold. (i) If Dh ⊆ hE, Dh ⊆ E. (ii) If Dh = hE, Dh = E. Proof. (i) Let a be an element in Dh . Then ha ⊆ Dh. Now assume that Dh ⊆ hE. Then a ∈ h∗ ha ⊆ h∗ Dh ⊆ h∗ hE. Since we are assuming that h∗ h ⊆ E, this implies that a ∈ E. (ii) From Dh ⊆ hE we obtain that Dh ⊆ E; cf. (i). From hE ⊆ Dh we obtain that E ⊆ Dh ; cf. Lemma 2.5.2(ii). 

2.6 The Thin Radical Let H be a hypergroup. For each subset A of H, we define Oϑ (A) to be the set of all thin elements in A. The set Oϑ (H) will be called the thin radical of H. Lemma 2.6.1 Let H be a hypergroup, and let A be a ∗ -invariant subset of H. Assume that A ⊆ Oϑ (H). Then hAi ⊆ Oϑ (H). Proof. From Lemma 2.3.4(i) we know that hAi is equal to the union of the sets An with n a non-negative integer. Thus, it suffices to show that An ⊆ Oϑ (H) for each non-negative integer n. To show this, we proceed by induction. Note first that A0 ⊆ Oϑ (H), since A0 = {1}. Now assume that An−1 ⊆ Oϑ (H) for some positive integer n, and recall that, by hypothesis, A ⊆ Oϑ (H). Then, by Lemma 1.4.3(iii), An = An−1 A ⊆ Oϑ (H).  Lemma 2.6.2 Let H be a hypergroup, and assume that Oϑ (H) is finite. Then Oϑ (H) is a closed subset of H. Proof. By Lemma 2.6.1, it suffices to show that Oϑ (H) is ∗ -invariant.

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2 Closed Subsets

Let h be an element in Oϑ (H), and assume that Oϑ (H) is finite. Then there exist distinct positive integers i and j such that hi ∩ h j is not empty. Assume, without loss of generality, that i ≤ j − 1. Then, as h is thin, 1 ∈ h j−i ; cf. Lemma 1.3.3. Set n := j − i. Then 1 ≤ n and 1 ∈ h n . Thus, as h is thin, h∗ ∈ h∗ h n = h∗ hh n−1 = h n−1 . Now recall from Lemma 1.4.3(iii) that h n−1 ⊆ Oϑ (H). Thus, h∗ ∈ Oϑ (H).



As a consequence of Lemma 2.6.2 one obtains that the thin radical of a finite hypergroup H is a closed subset of H. It is easy to find an infinite hypergroup the thin radical of which is not closed. Lemma 2.6.3 Let H be a hypergroup, and assume Oϑ (H) to be finite. Let F be a closed subset of H. Then Oϑ (H)F is the disjoint union of |Oϑ (H)/Oϑ (F)| left cosets of F with representatives in Oϑ (H). Proof. From Lemma 2.1.6(i) we obtain that Oϑ (H)F is the union of pairwise disjoint left cosets of F with representatives in Oϑ (H). Let d and e be elements in Oϑ (H). From Lemma 2.1.7(i) we know that dF = eF if and only if d ∗ e ⊆ F. On the other hand, as {d, e} ⊆ Oϑ (H) and Oϑ (H) is assumed to be finite, we have d ∗ e ⊆ Oϑ (H); cf. Lemma 2.6.2. Thus, dF = eF if and only if d ∗ e ⊆ F ∩ Oϑ (H) = Oϑ (F). It follows that F has exactly |Oϑ (H)/Oϑ (F)| left cosets with representatives in Oϑ (H).  The third part of the following lemma is due to Joseph-Louis de Lagrange. Lemma 2.6.4 Let H be a hypergroup, and assume Oϑ (H) to be finite. Let F be a finite closed subset of H. Then the following hold. (i) For each element h in Oϑ (H), we have |F | = |hF |. (ii) We have |Oϑ (H)F | = |Oϑ (H)/Oϑ (F)| · |F |. (iii) Assume that F ⊆ Oϑ (H). Then |Oϑ (H)| = |Oϑ (H)/F | · |F |. Proof. (i) Since Oϑ (H) is assumed to be finite, Oϑ (H) is a closed subset of H; cf. Lemma 2.6.2. Thus, for each thin element h of H, h∗ is thin. Thus, we obtain from Lemma 1.4.7(ii) that |F | = |hF |. (ii) This follows from Lemma 2.6.3 together with (i). (iii) We are assuming that Oϑ (H) is finite. Thus, by Lemma 2.6.2, Oϑ (H) is a closed subset of H. It follows that Oϑ (H)F = Oϑ (H). Thus, the claim follows from (ii).  Lemma 2.6.5 Let H be a hypergroup, and let F be a closed subset of H. Assume that |H \ F | is finite and that Oϑ (H) * F. Then the following hold.

2.7

Foldings

45

(i) The hypergroup H is finite. (ii) For each element h in Oϑ (H), we have |F | = |hF |. Proof. (i) We are assuming that Oϑ (H) * F. Thus, we find an element h in Oϑ (H) with h < F. Since h < F, we obtain from Lemma 2.1.6(i) that hF ⊆ H \ F. Since |H \ F | is assumed to be finite, this implies that |hF | is finite. On the other hand, as h ∈ Oϑ (H), Lemma 1.4.7(i) yields |F | ≤ |hF |. It follows that |F | is finite. Since |F | and |H \ F | both are finite, so is |H|. (ii) In (i), we saw that H is finite. Thus, Oϑ (H) and F both are finite, so that the  claim follows from Lemma 2.6.4(i).

2.7 Foldings Let H be a hypergroup, and let F be a closed subset of H. An injective map ρ from F to H is called folding if, for each element f in F, f ρ f = {1ρ }. Our first result shows that foldings are easy to describe if their domain is thin. Lemma 2.7.1 Let H be a hypergroup, and let F be a thin closed subset of H. A map ρ from F to H is a folding if and only if H contains an element m such that, for each element f in F, m f ∗ = { f ρ }. Proof. Let ρ be a map from F to H, and assume first that ρ is a folding. Then, for each element f in F, 1ρ ∈ f ρ f . It follows that, for each element f in F, f ρ ∈ 1ρ f ∗ . Since F is assumed to be thin, this implies that, for each element f in F, 1ρ f ∗ = { f ρ }; cf. Lemma 1.4.3(i). Setting m := 1ρ this shows that, for each element f in F, m f ∗ = { f ρ }. Conversely, let m be an element in H, and assume that, for each element f in F, m f ∗ = { f ρ }. Then, in particular, m·1∗ = {1ρ }, so that 1ρ = m. To show that ρ is a folding, we now fix an element f in F. Then f ρ ∈ m f ∗ . It follows that m ∈ f ρ f . Since f is thin, this implies that f ρ f = {m}; cf. Lemma 1.4.3(i). Thus, as 1ρ = m, we have f ρ f = {1ρ }.  Lemma 2.7.2 Let H be a hypergroup, let F be a closed subset of H, and let ρ be a folding from F to H. Then F ρ ⊆ 1ρ F. Proof. Let f be an element in F. Then 1ρ ∈ f ρ f . It follows that f ρ ∈ 1ρ f ∗ . Thus, as f ∗ ∈ F, f ρ ∈ 1ρ F. Since f has been chosen arbitrarily among the elements of F, this proves that F ρ ⊆ 1ρ F. 

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2 Closed Subsets

Let H be a hypergroup, let F be a closed subset of H, and let ρ be a folding from F to H. From Lemma 2.1.6(i) one obtains that F = 1ρ F if F ∩ 1ρ F is not empty. Thus, by Lemma 2.7.2, the image of ρ is either contained in its domain or disjoint from its domain. Since ρ is a folding from F to H, ρ is injective. Thus, the map from F to F ρ which takes each element f of F to f ρ is bijective. In the following we write ρ−1 to denote the inverse of this map. Lemma 2.7.3 Let H be a hypergroup, let F be a closed subset of H, and let ρ be a folding from F −1 to H. Assume that F ρ∗ = F ρ . For each element f in F, define f π := f ∗ρ∗ρ . Then π is a permutation of F. Proof. The map which sends each element f in F to f ∗ is a permutation of F. Since ρ is injective, the map from F to F ρ which sends each element f in F to f ρ is bijective. Since we are assuming that F ρ∗ = F ρ , the map which sends each element h of F ρ to h∗ is a permutation of F ρ . The map from F ρ to F which sends each element h in −1  F ρ to hρ is bijective. Let H be a hypergroup, let F be a closed subset of H, and let ρ be a folding from F to H satisfying F ρ∗ = F ρ . The permutation π of F which one obtains from ρ via Lemma 2.7.3 will be called the permutation of F induced by ρ. Permutations induced by foldings will play a role in Sections 3.6, 8.7, 9.5, 9.7, 9.9, 10.6, and 10.7. Lemma 2.7.4 Let H be a hypergroup, let F be a closed subset of H, and let ρ be a folding from F to H. Assume that F ρ∗ = F ρ . Let π denote the permutation of F induced by ρ, and let f be an element in F. Then the following hold. (i) We have f π∗ = f ∗π (ii) We have f π = f

π −1

−1

and f π

−1 ∗

= f ∗π .

if and only if f π∗ = f ∗π .

Proof. (i) By definition, we have f π∗ = f ∗ρ∗ρ −1 −1 obtain that f π ∗ = f ρ∗ρ = f ∗π .

−1 ∗

−1

= f ∗π , and from Lemma 1.1.2 we

(ii) If f π = f π , the second equation of (i) yields f π∗ = f ∗π . Conversely, if −1  f π∗ = f ∗π , the second equation of (i) yields f π = f π∗∗ = f ∗π∗ = f π . −1

Lemma 2.7.5 Let H be a hypergroup, let F be a closed subset of H, and let ρ be a folding from F to H. Assume that 1ρ∗ = 1ρ and that F ρ∗ = F ρ . Let π denote the permutation of F induced by ρ, and let f be an element in F. Then f ·1ρ = 1ρ f π .

2.7

Foldings

47

Proof. By definition, we have f ∗ρ f ∗ = {1ρ } and f ∗ρ∗ f π = {1ρ }. From the first of these two equations (together with 1ρ∗ = 1ρ ) we obtain that f f ∗ρ∗ = {1ρ }; cf. Lemma 1.2.2. Thus, as f ∗ρ∗ f π = {1ρ }, we obtain that f ·1ρ = f f ∗ρ∗ f π = 1ρ f π .  Let H be a hypergroup, let F be a closed subset of H, and let ρ be a folding from F to H. In the previous three lemmas, the set F ρ was assumed to be ∗ -invariant. We continue with a condition under which this hypothesis is satisfied. Lemma 2.7.6 Let H be a hypergroup, let F be a closed subset of H, and let ρ be a folding from F to H. Assume that F , H and that H = F ∪ F ρ . Then the following hold. (i) We have F ρ = 1ρ F = H \ F. (ii) We have F ρ∗ = F ρ . Proof. (i) From Lemma 2.7.2 we know that F ρ ⊆ 1ρ F. From F ρ ⊆ 1ρ F, together with our hypothesis that H = F ∪ F ρ , we obtain that H = F ∪ 1ρ F. Since we are assuming that F , H, this implies that 1ρ F * F. Thus, by Lemma 2.1.6(i), 1ρ F ⊆ H \ F. From H = F ∪ F ρ we obtain that H \ F ⊆ F ρ . (ii) From (i) we know that F ρ = H \ F. Thus, as H \ F is ∗ -invariant, so is F ρ .



Let H be a hypergroup, and let F be a closed subset of H. We say that H is twinned over F if there exists a folding ρ from F to H satisfying 1ρ∗ = 1ρ,

1ρ∗ 1ρ ⊆ F,

and

H = F ∪ Fρ.

Hypergroups which are twinned over some of their closed subsets appear in Sections 3.3, 9.7, 9.9, 10.6, and 10.7. Note that, via the group correspondence, the thin hypergroups H which are twinned over a closed subset of H correspond to groups G which contain subgroups H and K with G = K H, K ∩ H = {1}, and |H| ≤ 2. We conclude this section by showing that, in hypergroups with non-trivial thin radical, foldings never come alone. Lemma 2.7.7 Let H be a hypergroup, let F be a closed subset of H, and let σ be a folding from F to H. Let c be a thin element in F, and assume that c∗ is thin. For each element f in F, define f τ := ( f 0)σ , where f 0 stands for the uniquely determined element in f c. Then τ is a folding from F to H with 1τ = cσ and F τ = F σ . Proof. Note first that τ is injective. In fact, let d and e be elements in F, and assume that d τ = eτ . Let d 0 denote the uniquely determined element in dc, and let e 0 denote the uniquely determined element in ec. Then d τ = (d 0)σ and eτ = (e 0)σ .

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2 Closed Subsets

Since d τ = eτ , we have (d 0)σ = (e 0)σ . Since σ is injective, that implies d 0 = e 0. Thus, as dc = {d 0 } and ec = {e 0 }, we have dc = ec. Since c∗ is assumed to be thin, we obtain that {d} = dcc∗ = ecc∗ = {e}. It follows that d = e. Since 1·c = {c}, we have 1τ = cσ . To show that τ is a folding from F to H, we choose an element f in F and define f 0 to be the uniquely determined element in f c. Then, f τ = ( f 0)σ . Thus, as c∗ is thin and σ a folding from F to H, we obtain that f τ f = ( f 0)σ f = ( f 0)σ f cc∗ = ( f 0)σ f 0 c∗ = 1σ c∗ = cσ cc∗ = 1τ ·1 = {1τ }. It remains to show that F τ = F σ . Let f be an element in F, and define f 0 to be the uniquely determined element in f c. Then, f τ = ( f 0)σ . Since f and c both are in F, so is f 0. Thus, f τ ∈ F σ . Since f has been chosen arbitrarily in F, this shows that F τ ⊆ F σ . Let f 0 be an element in F. Since c∗ is assumed to be thin, f 0 c∗ contains an element f with f 0 c∗ = { f }; cf. Lemma 1.4.3(i). Since c is assumed to be thin, this implies that f c = { f 0 }. Note also that f ∈ F, since { f 0, c} ⊆ F. Thus, f τ = ( f 0)σ . Since f 0  has been chosen arbitrarily in F, this shows that F σ ⊆ F τ . In Lemma 8.6.5(i), we will provide a condition under which the converse of Lemma 2.7.7 holds.

3 Elementary Structure Theory

Centralizers, normalizers, and strong normalizers are basic concepts in the structure theory of hypergroups. Centralizers and normalizers will be introduced and discussed in Section 3.1, strong normalizers in Section 3.3. Sufficient conditions for closed subsets of hypergroups H to be normal in H will be given in Section 3.2. They will be particularly useful in Sections 7.3, 7.4, 7.5, and 7.6 when we investigate hypergroups H which contain a closed subset F satisfying |H \ F | = 4. Each closed subset F of a hypergroup H gives rise to a new hypergroup, the quotient of H over F. Quotients will be defined in Section 3.4, and in Section 3.5, we consider the question to what extent certain concepts and basic properties of hypergroups are inherited by quotients. In Section 3.6, we introduce hypergroup homomorphisms, and in Section 3.7, we prove the Homomorphism Theorem and the two Isomorphism Theorems for hypergroups. These theorems and their proofs are modeled on and generalize the corresponding group theoretic theorems proved by Emmy Noether in 1929; cf. [36; I. §2]. Section 3.7 includes applications of the Isomorphism Theorems some of which will be useful in Chapter 4. In Section 3.8, we introduce the wreath product of hypergroups, and in the final section of this chapter, we study the relationship between hypergroup isomorphisms and regular actions.

3.1 Centralizers and Normalizers Let H be a hypergroup. Let F be a closed subset of H. For each element h in H, we set CF (h) := { f ∈ F | h f = f h} and call this set the centralizer of h in F. Let F be a closed subset of H, and let A be a subset of H. If A is empty, we set CF (A) := F. If A is not empty, we define CF (A) to be the intersection of the sets CF (a) with a ∈ A. The set CF (A) will be called the centralizer of A in F.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 P. -H. Zieschang, Hypergroups, https://doi.org/10.1007/978-3-031-39489-8_3

49

50

3 Elementary Structure Theory

For any two closed subsets D and E of H, we set ND (E) := {d ∈ D | E d ⊆ dE } and call this set the normalizer of E in D. From Lemma 2.1.1 one obtains that, for each closed subset F of H, F ⊆ NH (F). Lemma 3.1.1 Let H be a hypergroup, and let F be a closed subset of H. Then the following hold. (i) We have CH (F) ⊆ NH (F). (ii) We have NH (F) = F NH (F)F. (iii) For each element h in H, we have h∗ ∈ NH (F) if and only if F ⊆ F h . Proof. (i) This follows right from the definition of a centralizer (together with the definition of a normalizer). (ii) Since F is a closed subset of H, 1 ∈ F; cf. Lemma 2.1.1. Thus, NH (F) ⊆ F NH (F)F. To show that F NH (F)F ⊆ NH (F), we choose an element b in F NH (F)F. We will see that b ∈ NH (F). Since b ∈ F NH (F)F, NH (F) contains an element a with b ∈ FaF. Since a ∈ NH (F), we have Fa ⊆ aF. It follows that Fb ⊆ aF. From Fb ⊆ aF, together with 1 ∈ F, we obtain that b ∈ aF. Thus, by Lemma 2.1.6(i), aF = bF. Since Fb ⊆ aF, this implies that Fb ⊆ bF, and that means that b ∈ NH (F). (iii) By definition, we have h∗ ∈ NH (F) if and only if F h∗ ⊆ h∗ F. From Lemma 1.3.1(iii) we know that F h∗ ⊆ h∗ F if and only if hF ⊆ F h. From Lemma 2.5.2(ii) we know that hF ⊆ F h if and only if F ⊆ F h .  Lemma 3.1.2 Let H be a hypergroup, and let D and E be closed subsets of H. Then the following hold. (i) We have D ⊆ CH (E) if and only if E ⊆ CH (D). (ii) We have D ⊆ NH (E) if and only if E ⊆ E d for each element d in D. (iii) Assume that D ⊆ NH (E), E ⊆ NH (D), and D∩E = {1}. Then D ⊆ CH (E). Proof. (i) This follows from the definition of a centralizer. (ii) This follows from Lemma 3.1.1(iii). (iii) Let d be an element in D, let e be an element in E, and let a be an element in ed. Since the hypotheses of the lemma are symmetric in D and E, we shall be done if we succeed in showing that a ∈ de.

3.1

Centralizers and Normalizers

51

We are assuming that D ⊆ NH (E). Thus, as d ∈ D, E d ⊆ dE. Thus, as a ∈ ed, a ∈ dE. Thus, E contains an element c such that a ∈ dc. We are assuming that E ⊆ NH (D). Thus, as c ∈ E, Dc ⊆ cD. Thus, as a ∈ dc, a ∈ cD. Thus, D contains an element b such that a ∈ cb. From a ∈ cb and a ∈ ed we obtain that c = e; cf. Lemma 2.1.6(ii). Thus, as a ∈ dc,  we conclude that a ∈ de. Let H be a hypergroup, and let D and E be closed subsets of H. If D ⊆ CH (E), we say that D centralizes E. If D ⊆ NH (E), we say that D normalizes E. From Lemma 3.1.2(iii) (together with Lemma 3.1.2(i)) we obtain that two closed subsets D and E of a hypergroup centralize each other if they normalize each other and their intersection is {1}. Lemma 3.1.3 Let H be a hypergroup, let F be a closed subset of H, and let h be an element in H. Then the following hold. (i) Assume that h ∈ NH (F). Then h∗ ∈ NH (F) if and only if F h = hF. (ii) Assume that (hF)∗ ∈ H/F. Then h ∈ NH (F). Proof. (i) We are assuming that h ∈ NH (F). Thus, F h ⊆ hF. For the reverse containment we notice that h∗ ∈ NH (F) if and only if F h∗ ⊆ h∗ F and that, by Lemma 1.3.1(iii), F h∗ ⊆ h∗ F if and only if hF ⊆ F h. (ii) We are assuming that (hF)∗ ∈ H/F. Thus, as h∗ ∈ (hF)∗ , we must have h∗ F = (hF)∗ ; cf. Lemma 2.1.6(i). It follows that F h = hF; cf. Lemma 1.3.1(ii).  A closed subset F of a hypergroup H is said to be normal if NH (F) = H. If F is a normal closed subset of a hypergroup H, we say that F is normal in H. Note that, in each hypergroup H, the closed subsets {1} and H are normal in H. A non-trivial hypergroup H is called simple if {1} and H are the only normal closed subsets of H. Thus, a hypergroup H is simple if it has exactly two normal closed subsets, namely {1} and H. Of course, primitive hypergroups are simple, but the converse is not true. Hypergroups of type H3, j (as defined in Section 7.1) are simple if j ∈ {1, . . . , 6}. (In fact, we mentioned already in the introduction to Section 2 that they are primitive.) They are not simple if j ∈ {7, . . . , 10}. Hypergroups of type H4,7 , H4,8 , H4,9 , H4,14 , H4,15 , H4,16 , H4,17 , H4,18 , H4,19 , H4,20 , H4,21 , H4,22 , H4,23 , H4,30 , H4,31 , H4,32 , H4,33 , H4,34 , H4,35 , H4,36 , or H4,37 (as defined in Section 7.2) are simple. (Also here, we mentioned already in the introduction to Section 2 that they all are primitive.) Hypergroups of type H4,1 , H4,2 , H4,3 , H4,4 , H4,5 , H4,6 , H4,10 , H4,11 , H4,12 , H4,13 , H4,24 , H4,25 , H4,26 , H4,27 , H4,28 , or H4,29 (also defined in Section 7.2) are not simple.

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3 Elementary Structure Theory

Hypergroups of type H6, j (as defined in Sections 7.3, 7.4, and 7.5) are simple if j ∈ {6, . . . , 19}, they are not simple if j ∈ {1, . . . , 5}. Simple hypergroups arise naturally in Sections 4.3, 4.7, and 9.4. Lemma 3.1.4 Let H be a hypergroup, let D and E be closed subsets of H, and assume that D ⊆ NH (E). Then the following hold. (i) For each element d in D, we have E d = dE. (ii) We have E D = DE. (iii) The product E D is a closed subset of H. (iv) The closed subset E of H is normal in E D. Proof. (i) This follows from Lemma 3.1.3(i). (ii) This follows from (i). (iii) Considering (ii) this follows from Lemma 2.1.5. (iv) From Lemma 3.1.1(ii) we obtain DE ⊆ E NH (E)E ⊆ NH (E).



Lemma 3.1.5 Let H be a hypergroup, and let D and E be closed subsets of H. Then the following hold. (i) We have ND (E) ⊆ ND (D ∩ E). (ii) Assume that DE is closed. Then NH (D) ∩ NH (E) ⊆ NH (DE). (iii) Assume that D is simple and that D ⊆ NH (E). Then D∩ E = {1} or D ⊆ E. Proof. (i) Let d be an element in ND (E). Then E d ⊆ dE. Thus, by Lemma 2.2.1, (D ∩ E)d ⊆ D ∩ E d ⊆ D ∩ dE = d(D ∩ E), so that, by definition, d ∈ ND (D ∩ E). (ii) For each element h in NH (D) ∩ NH (E), we have DE h ⊆ DhE ⊆ hDE. (iii) We are assuming that D ⊆ NH (E). Thus, by (i), D ⊆ NH (D ∩ E). This means that D ∩ E is normal in D. Since D is assumed to be simple, this implies that D ∩ E = {1} or D ∩ E = D. If D ∩ E = D, D ⊆ E.  The following lemma will be useful in the proofs of Lemma 3.7.4 and Theorem 4.3.2. Lemma 3.1.6 Let H be a hypergroup, and let B, C, D, and E be closed subsets of H with B ⊆ D and C ⊆ E. Assume that B is normal in D and that C is normal in E. Then the following hold. (i) The sets D ∩ BC and D ∩ BE are closed subsets of H.

3.1

Centralizers and Normalizers

53

(ii) We have (D ∩ BC)(D ∩ E) = D ∩ BE. (iii) The closed subset D ∩ C of H is normal in D ∩ E. (iv) The closed subset D ∩ BC of H is normal in D ∩ BE. Proof. (i) Since B ⊆ D, we have D ∩ BC = B(D ∩ C); cf. Lemma 2.2.1(i). On the other hand, as D ∩ C ⊆ NH (B), we obtain from Lemma 3.1.4(iii) that B(D ∩ C) is a closed subset of H. Thus, D ∩ BC is a closed subset of H. From D ∩ E ⊆ NH (B) one similarly obtains that D ∩ BE is a closed subset of H. (ii) In (i), we saw that D ∩ BE is closed. Thus, (D ∩ BC)(D ∩ E) ⊆ D ∩ BE. From Lemma 2.2.1(i) we also know that D ∩ BE = B(D ∩ E). Thus, D ∩ BE ⊆ (D ∩ BC)(D ∩ E). It follows that (D ∩ BC)(D ∩ E) = D ∩ BE. (iii) From Lemma 3.1.5(i) we know that ND (C) ⊆ ND (D ∩ C). Thus, as D ∩ E ⊆ ND (C), we have D ∩ E ⊆ ND (D ∩ C) which shows that D ∩ C is normal in D ∩ E. (iv) Let d be an element in D ∩ BE. We will see that (D ∩ BC)d ⊆ d(D ∩ BC). Since B ⊆ D, we have D∩ BE = B(D∩E); cf. Lemma 2.2.1(i). Thus, as d ∈ D∩ BE, d ∈ B(D ∩ E). It follows that D ∩ E contains an element e with d ∈ Be, and from d ∈ Be we obtain that Bd = Be; cf. Lemma 2.1.6(i). Since D ∩ C ⊆ NH (B), we have B(D ∩ C) = (D ∩ C)B; cf. Lemma 3.1.4(ii). Thus, as Bd = Be, we obtain B(D ∩ C)d = (D ∩ C)Bd = (D ∩ C)Be = B(D ∩ C)e. Since e ∈ D ∩ E, we have e ∈ NH (D ∩ C); cf. (iii). From d ∈ D we obtain that d ∈ NH (B). Thus, as Bd = Be, we obtain B(D ∩ C)e ⊆ Be(D ∩ C) = Bd(D ∩ C) ⊆ dB(D ∩ C). It follows that B(D ∩ C)d ⊆ dB(D ∩ C). On the other hand, as B ⊆ D, we have D ∩ BC = B(D ∩ C); cf. Lemma 2.2.1(i). Thus, we conclude that (D ∩ BC)d ⊆ d(D ∩ BC).  Lemma 3.1.7 Let H be a hypergroup, let A be a ∗ -invariant subset of H, and let h be an element in H. Assume that Ah ⊆ hhAi. Then h ∈ NH (hAi). Proof. From Lemma 2.3.4(i) we know that hAi is equal to the union of the sets An with n a non-negative integer. Thus, it suffices to show that An h ⊆ hhAi for each non-negative integer n. To show this, we proceed by induction. Since A0 = {1}, we have A0 h = {h} ⊆ hhAi.

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Now assume that An−1 h ⊆ hhAi for some positive integer n, and recall that, by hypothesis, Ah ⊆ hhAi. Then An h ⊆ An−1 Ah ⊆ An−1 hhAi ⊆ hhAi, since hAi is a closed subset of H.



Lemma 3.1.8 Let H be a hypergroup, and let D and E be closed subsets of H. Assume that, for each element d in D \ E, dE = {d}. Then D ⊆ NH (E). Proof. Let d be an element in D. If d ∈ E, we have E d = E = dE, and that implies that D ⊆ NH (E). Thus, we assume that d < E. Then, by hypothesis, dE = {d}. Since D and E are closed subsets of H, we obtain from d ∈ D \ E that d ∗ ∈ D \ E. Thus, by hypothesis, d ∗ E = {d ∗ }, so that, by Lemma 1.3.1(ii), E d = {d}. From dE = {d} and E d = {d} we obtain that E d = dE, so d ∈ NH (E).



Let n be an integer with 2 ≤ n, and let F1 , . . . , Fn be pairwise distinct normal closed subsets of H. For each element i in {1, . . . , n}, define Fˆi to be the product of the closed subsets Fj with j ∈ {1, . . . , n} \ {i}. The hypergroup H is called the direct product of the closed subsets F1 , . . . , Fn if, for each element i in {1, . . . , n}, Fi ∩ Fˆi = {1}

and

Fi Fˆi = H.

If H is the direct product of F1 , . . . , Fn , we write H = F1 × · · · × Fn . Lemma 3.1.9 Let H be a hypergroup, let n be an integer with 2 ≤ n, and let F1 , . . . , Fn be pairwise distinct closed subsets of H. Assume that H = F1 × · · · × Fn . Then Fi ⊆ CH (Fj ) for any two distinct elements i and j in {1, . . . , n}. Proof. Let i and j be two distinct elements in {1, . . . , n}. We define Fˆi to be the product of the closed subsets Fk with k ∈ {1, . . . , n} \ {i}. Then Fj ⊆ Fˆi . Thus, as Fi ∩ Fˆi = {1}, we have Fi ∩ Fj = {1}. Now, since Fi and Fj normalize each other, we obtain from that Lemma 3.1.2(iii) that Fi ⊆ CH (Fj ). 

3.2 Sufficient Conditions for Normality In this section, we present sufficient conditions for closed subsets of hypergroups H to be normal in H. Lemma 3.2.1 Let H be a hypergroup, and let F be a closed subset of H. Each of the following conditions implies that F is normal in H.

3.2

Sufficient Conditions for Normality

55

(i) For each element h in H, we have (hF)∗ ∈ H/F. (ii) Each left coset of F in H is ∗ -invariant. (iii) The set H \ F is symmetric. (iv) We have |H/F | ≤ 2. (v) We have |H/F | = |H \ F | + 1. Proof. (i) This follows from Lemma 3.1.3(ii). (ii) This follows from (i). (iii) This follows from (ii). (iv) There is nothing to show if |H/F | = 1. Thus, we assume that |H/F | = 2. In this case, the only left coset of F in H different from F is H \ F. Since F is ∗ -invariant, so is H \ F. Thus, the claim follows from (ii). (v) Assume that |H/F | = |H \ F | + 1. Then hF = {h} for each element h in H \ F.  Thus, the claim follows from (i). Lemma 3.2.2 Let H be a hypergroup, and let F be a closed subset of H. Assume that |H/F | = |H\F | and that H \ F contains exactly two non-symmetric elements. Then F is normal in H. Proof. We are assuming that |H/F | = |H \ F |. Thus, F has exactly one left coset in H which is different from F and has exactly two elements, all other left cosets of F in H which are different from F consist of a single element. Let a and b denote the two elements in the unique left coset of F in H which is different from F and has cardinality 2. Assume first that (aF)∗ , aF. Then a∗ < {a, b} or b∗ < {a, b}. Without loss of generality, we assume that a∗ < {a, b}. Then, since H \ F is assumed to contain exactly two non-symmetric elements, b∗ = b. Thus, as b ∈ aF, b ∈ Fa∗ ∩ aF. Note also that a∗ F = {a∗ }, since a∗ < {a, b}. Thus, applying Lemma 2.1.8(i) to a∗ in place of h we obtain that a is symmetric, contradiction. The contradiction which we obtained shows that (aF)∗ = aF. As a consequence, (aF)∗ ∈ H/F. Now, as any left coset of F in H different from F and aF consists of a single element, we have (hF)∗ ∈ H/F for each element h in H. Thus, by Lemma 3.2.1(i), F is normal in H.  Theorem 3.2.3 Let H be a hypergroup, and let F be a closed subset of H with |H \ F | ≤ 3. Then F is normal in H. Proof. If |H/F | ≤ 2, F is normal in H by Lemma 3.2.1(iv). Thus, we assume that 3 ≤ |H/F |. Then, as |H/F | ≤ |H \ F | + 1, we have |H/F | ∈ {3, 4}. It follows that (|H \ F |, |H/F |) ∈ {(2, 3), (3, 3), (3, 4)}.

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If (|H \ F |, |H/F |) ∈ {(2, 3), (3, 4)}, we are done by Lemma 3.2.1(v). Assume that |H/F | = |H \ F | = 3. If H \ F is symmetric, we are done by Lemma 3.2.1(iii). If H \ F contains a non-symmetric element, H \ F contains exactly two  non-symmetric elements, and we are done by Lemma 3.2.2. Theorem 3.2.4 Let H be a hypergroup, and let F be a non-normal closed subset of H. Assume that |H \ F | = 4. Then |H/F | ∈ {3, 4}. Proof. We are assuming that F is not normal in H. Thus, by Lemma 3.2.1(iv), 3 ≤ |H/F |. We are also assuming that |H \ F | = 4. Thus, |H/F | ≤ 5. It follows that |H/F | ∈ {3, 4, 5}. Assume that |H/F | = 5. Then, we have |H/F | = |H \ F | + 1, since we are assuming that |H \ F | = 4. Thus, by Lemma 3.2.1(v), F is normal in H, contradiction. It follows that |H/F | ∈ {3, 4}, as wanted.  Lemma 3.2.5 Let H be a hypergroup, and let F be a non-normal closed subset of H. Assume that |H \ F | = 4 and that |H/F | = 3. Then H \ F contains pairwise distinct elements a, b, and c such that a∗ = a, b∗ = b, aF = {a, c}, and bF = {b, c∗ }. Proof. Since F is assumed to be not normal in H, H contains an element c with c∗ < cF; cf. Lemma 3.2.1(ii). Assume that cF = {c}. Then |c∗ F | = 3, since |H \ F | = 4 and |H/F | = 3. Thus, c∗ F contains an element d with d ∗ ∈ c∗ F. It follows that d ∈ Fc ∩ c∗ F. Thus, by Lemma 2.1.8(i), c is symmetric, contradiction. This contradiction shows that 2 ≤ |cF |. Similarly, one shows that 2 ≤ |c∗ F |. Thus, as |H \ F | = 4, we obtain that |cF | = 2 = |c∗ F |. Since |cF | = 2, cF \ {c} contains an element a with cF = {a, c}. If a∗ ∈ c∗ F, (cF)∗ = c∗ F. Thus, by Lemma 3.2.1(i), F is normal in H, contradiction. It follows that a∗ ∈ cF, and, since c∗ < cF, a∗ = a. Similarly, one finds an element b in c∗ F such that b∗ = b and c∗ F = {b, c∗ }.  Lemma 3.2.6 Let H be a hypergroup, and let F be a non-normal closed subset of H. Assume that |H \ F | = 4 and that |H/F | = 4. Then H \ F contains two distinct elements a and b such that a∗ F = {a∗ }, aF = {a, b}, and b∗ F = {b∗ }. Proof. We are assuming that F is not normal in H. Thus, by Lemma 3.2.1(iii), H \ F is not symmetric. Furthermore, by Lemma 3.2.2, H \ F does not contain two symmetric and two non-symmetric elements. Thus, none of the elements in H \ F is symmetric.

3.3

Strong Normality

57

From |H \ F | = 4 and |H/F | = 4 we obtain that one of the left cosets of F in H different from F has two elements, each of the remaining two left cosets of F in H different from F consists of a single element. Let a and b denote the two elements in the left coset of F in H different from F which contains two elements. Then, by Lemma 3.2.1(i), a∗ , b. Thus, a∗ F = {a∗ }, aF = {a, b}, and b∗ F = {b∗ }.  Corollary 3.2.7 Let H be a hypergroup with |H| = 6, and let F be a non-normal closed subset of H. Then |F | = 2 and |H/F | ∈ {3, 4}. Proof. From Theorem 3.2.3 we obtain that |F | = 2. From |F | = 2 and |H| = 6 we obtain that |H \ F | = 4. Thus, by Theorem 3.2.4, |H/F | ∈ {3, 4}.  Hypergroups with six elements containing a non-normal closed subset will be considered in more detail in Sections 7.3, 7.4, 7.5 , and 7.6.

3.3 Strong Normality Let H be a hypergroup. For any two closed subsets D and E of H, we set KD (E) := {d ∈ D | d ∗ E d ⊆ E } and call this set the strong normalizer of E in D. Note that KH ({1}) = Oϑ (H). This shows that the strong normalizer generalizes the notion of the thin radical. From Lemma 2.1.1 one obtains that, for each closed subset F of H, F ⊆ KH (F). Lemma 3.3.1 Let H be a hypergroup, and let D and E be closed subsets of H. Then the following hold. (i) We have KD (E) ⊆ ND (E). (ii) We have KD (E)KD (E) ⊆ KD (E). (iii) Let d be an element in ND (E), and assume that d ∗ d ⊆ E. Then d ∈ KD (E). Proof. (i) Let d be an element in KD (E). Then d ∗ E d ⊆ E. It follows that dd ∗ E d ⊆ dE. On the other hand, as 1 ∈ dd ∗ , E d ⊆ dd ∗ E d. Thus, E d ⊆ dE, and that means that d ∈ ND (E). (ii) Let a and b be elements in KD (E). Then, for each element c in ab, c∗ Ec ⊆ b∗ a∗ Eab ⊆ b∗ E b ⊆ E.

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(iii) Since d ∈ ND (E), E d ⊆ dE. It follows that d ∗ E d ⊆ d ∗ dE. Since we are assuming that d ∗ d ⊆ E, this implies that d ∗ E d ⊆ E; cf. Lemma 2.1.1. Thus,  d ∈ KD (E). From Lemma 3.3.1(iii) one obtains that, for each thin hypergroup H and each closed subset F of H, KH (F) = NH (F). The following lemma is similar to Lemma 3.1.4(i). Lemma 3.3.2 Let H be a hypergroup, let D and E be closed subsets of H, and assume that D ⊆ KH (E). Then, for each element d in D, we have d ∗ E d = E. Proof. Let d be an element in D. Then, d ∈ KH (E), so d ∗ E d ⊆ E. Since d ∈ D and D is a closed subset of H, we have d ∗ ∈ D. Thus, as D ⊆ KH (E), d ∗ ∈ KH (E) which means that dE d ∗ ⊆ E. It follows that E ⊆ d ∗ dE d ∗ d ⊆ d ∗ E d.  The first two parts of our next lemma are similar to the corresponding parts of Lemma 3.1.5. Lemma 3.3.3 Let H be a hypergroup, and let D and E be closed subsets of H. Then the following hold. (i) We have KD (E) ⊆ KD (D ∩ E). (ii) Assume that DE is closed. Then NH (D) ∩ KH (E) ⊆ KH (DE). (iii) Assume that E ⊆ D. Then NH (D) ∩ KH (E) ⊆ KH (D). Proof. (i) Let d be an element in KD (E). Then d ∗ E d ⊆ E. Thus, by Lemma 2.2.1, d ∗ (D ∩ E)d ⊆ (D ∩ d ∗ E)d ⊆ D ∩ d ∗ E d ⊆ D ∩ E. This shows that d ∈ KD (D ∩ E). (ii) Since DE is assumed to be closed, DE = E D; cf. Lemma 2.1.5. Thus, for each element h in NH (D) ∩ KH (E), h∗ DE h = h∗ E Dh ⊆ h∗ E hD ⊆ E D = DE. This shows that h ∈ KH (DE). (iii) This follows from (ii).



Let H be a hypergroup, let F be a closed subset of H, and let h be an element in H with hh∗ ⊆ F. From Lemma 2.1.2(ii) we know that h∗ F h is a closed subset of H. Lemma 3.3.4 Let H be a hypergroup, and let F be a closed subset of H. Let h be an element in H, and assume that hh∗ ⊆ F. Then KH (h∗ F h) = h∗ KH (F)h.

3.3

Strong Normality

59

Proof. Let c be an element in KH (h∗ F h). Then hc∗ h∗ F hch∗ ⊆ hh∗ F hh∗ = F, since we are assuming that hh∗ ⊆ F. From this we obtain that hch∗ ⊆ KH (F), so that c ∈ h∗ hch∗ h ⊆ h∗ KH (F)h. Conversely, let c be an element in h∗ KH (F)h. Then KH (F) contains an element b such that c ∈ h∗ bh. Thus, as we are assuming that hh∗ ⊆ F, we conclude that c∗ h∗ F hc ⊆ h∗ b∗ hh∗ F hh∗ bh ⊆ h∗ b∗ Fbh ⊆ h∗ F h. This shows that c ∈ KH (h∗ F h).



Let H be a hypergroup, let h be an element in H, and assume that h∗ h is thin. Then, by Lemma 2.1.3(ii), h∗ h is a closed subset of H. In the following lemma, we look at the strong normalizer of this closed subset. Its second part will be useful in the proof of Lemma 4.6.8(i). Lemma 3.3.5 Let H be a hypergroup, let F be a closed subset of H, and let h be an element in H. Then the following hold. (i) Assume that h is tight, that F ⊆ F h , and that, for each element f in F, |h f | = 1 and f ∗ f ⊆ hh∗ . Then F ⊆ KH (h∗ h). (ii) Assume that F and h∗ h both are thin and that h∗ ∈ NH (F). Then F ⊆ KH (h∗ h). Proof. (i) Let d be an element in F. We have to show that d ∈ KH (h∗ h). Since d ∈ F, |hd| = 1. On the other hand, we are assuming that F ⊆ F h . Thus, as d ∈ F, d ∈ F h , and that means that hd ⊆ F h. Since |hd| = 1 and hd ⊆ F h, there exists an element e in F such that hd ⊆ eh. From e ∈ F we obtain that e∗ e ⊆ hh∗ . Thus, as h is tight, we have d ∗ h∗ hd ⊆ h∗ e∗ eh ⊆ h∗ hh∗ h ⊆ h∗ h, and that means that d ∈ KH (h∗ h). (ii) We are assuming that h∗ h is thin. Thus, by Corollary 1.4.4, h is tight. We are assuming that h∗ ∈ NH (F). Thus, by Lemma 3.1.1(iii), F ⊆ F h . We are assuming that F is thin. Thus, by Lemma 1.4.3(i), |h f | = 1 for each element f in F. Since F is thin, we also have f ∗ f ⊆ hh∗ for each element f in F, so that the claim follows from (i).  Let H be a hypergroup, and let F be a closed subset of H. Recall from Section 3.1 that F is said to be normal if NH (F) = H. The closed subset F is said to be strongly normal if KH (F) = H. If F is a strongly normal closed subset of H, we also say that F is strongly normal in H.

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Hypergroups of type H3,7 or H3,9 (as defined in Section 7.1) contain a strongly normal closed subset, while those of type H3,8 or H3,10 (also defined in Section 7.1) contain a normal closed subset which is not strongly normal. Hypergroups of type H4,1 , H4,2 , H4,3 , H4,4 , H4,5 , H4,6 , H4,10 , H4,11 , H4,12 , H4,13 , H4,24 , H4,25 , H4,26 , or H4,27 (as defined in Section 7.2) all contain a strongly normal closed subset, while those of type H4,28 or H4,29 (also defined in Section 7.2) contain a normal closed subset which is not strongly normal. Hypergroups of type H6,1 , H6,2 , H6,3 , H6,4 , or H6,5 (as defined in Section 7.3) all contain a strongly normal closed subset. Let H be a hypergroup. From Lemma 3.3.1(i) one obtains that closed subsets of H are normal in H if they are strongly normal in H. Lemma 3.3.3(iii) implies that, conversely, normal closed subsets of H are strongly normal in H if they contain a closed subset which is strongly normal in H. Note that {1} is normal in H, but strongly normal in H only if H is thin. Lemma 3.3.6 Let H be a hypergroup, and let F be a non-empty set of strongly normal closed subsets of H. Then the intersection of the closed subsets which belong to F is a strongly normal closed subset of H. Proof. Let E denote the intersection of the closed subsets of H which belong to F . From Lemma 2.1.4 we know that E is closed. Let h be an element in H. Then, for each element F in F , h∗ E h ⊆ h∗ F h ⊆ F. Thus,  h∗ E h ⊆ E, and that means that h ∈ KH (E). Lemma 3.3.7 Let H be a hypergroup, and let C, D, and E be closed subsets of H with D ⊆ E ⊆ NH (C). Assume that D is strongly normal in E. Then CD is strongly normal in CE. Proof. We are assuming that D ⊆ E ⊆ NH (C). Thus, by Lemma 3.1.4(iii), the products CE and CD are closed subsets of H. To show that CD is strongly normal in CE, we choose an element f in CE. We will see that f ∗ CD f ⊆ CD. From f ∈ CE we obtain an element e in E such that f ∈ Ce. From f ∈ Ce we obtain that f ∗ ∈ e∗ C; cf. Lemma 1.3.1(ii). Since D ⊆ NH (C), we have CD = DC; cf. Lemma 3.1.4(ii). Since E ⊆ NH (C), we have Ce∗ = e∗ C; cf. Lemma 3.1.4(i). Thus, as D is assumed to be strongly normal in E, we obtain that f ∗ CD f ⊆ e∗ CDCe = e∗ CDe = Ce∗ De ⊆ CD. Since f has been chosen arbitrarily in CE, we have shown that CD is strongly normal  in CE.

3.4

Quotients

61

Lemma 3.3.8 Let H be a hypergroup, and let F be a closed subset of H. Then the following hold. (i) Let h be an element in H satisfying h∗ h ⊆ F and H = F ∪ hF. Then F is strongly normal in H. (ii) If H is twinned over F, F is strongly normal in H. Proof. (i) From Lemma 3.2.1(iv) we know that F is normal in H. In particular, h ∈ NH (F). Since we are assuming that h∗ h ⊆ F, this implies that h ∈ KH (F); cf. Lemma 3.3.1(iii). From h ∈ KH (F) and F ⊆ KH (F) we obtain that hF ⊆ KH (F); cf. Lemma 3.3.1(ii). Since we are assuming that H = F ∪ hF, this implies that H ⊆ KH (F). Thus, by definition, F is strongly normal in H. (ii) Assume that H is twinned over F. Then, by definition, there exists a folding ρ from F to H satisfying 1ρ∗ 1ρ ⊆ F and H = F ∪ F ρ .1 If F = H, F is obviously strongly normal in H. Therefore, we assume that F , H. From F , H, together with H = F ∪ F ρ , we obtain that F ρ = 1ρ F; cf. Lemma 2.7.6(i). It follows that H = F ∪ 1ρ F, so that the claim follows from (i), 

3.4 Quotients In this section, we will show that each closed subset F of a hypergroup H gives rise to a new hypergroup. This new hypergroup will be called the quotient of H over F. Let H be a hypergroup, let F be a closed subset of H, and let A be a subset of H. Recall that A//F is our notation for the set of all double cosets FaF with a ∈ A. Lemma 3.4.1 Let H be a hypergroup, let F be a closed subset of H, and let A be a subset of H. Then the following hold. (i) We have A//F = F AF//F. (ii) Assume that F ⊆ A and that A2 ⊆ A. Then h ∈ A for each element h in H with F hF ∈ A//F. Proof. (i) Let h be an element in H, and assume first that F hF ∈ A//F. Then A contains an element a such that F hF = FaF. It follows that h ∈ F AF, so F hF ∈ F AF//F. Assume, conversely, that F hF ∈ F AF//F. Then F AF contains an element b such that F hF = FbF. Since b ∈ F AF, A contains an element a such that b ∈ FaF. From b ∈ FaF we obtain that FbF = FaF; cf. Lemma 2.1.6(i). From F hF = FbF and FbF = FaF we obtain that F hF = FaF. Thus, as FaF ∈ A//F, F hF ∈ A//F. 1We also have 1ρ∗ = 1ρ , but we do not need that.

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(ii) Let h be an element in H, and assume that F hF ∈ A//F. Then, A contains an  element a with F hF = FaF. It follows that h ∈ FaF ⊆ F AF ⊆ A. Let H be a hypergroup, and let F be a closed subset of H. We will now define a hyperoperation on the set H//F of all double cosets of F in H and show that, with respect to this hyperoperation, H//F is a hypergroup. Lemma 3.4.2 Let H be a hypergroup, and let F be a closed subset of H. Let a, b, c, and d be elements in H, and assume that FaF = FcF and that FbF = FdF. Then {FeF | e ∈ aFb} = {FeF | e ∈ cFd}. Proof. Let h be an element in H, and assume that F hF ∈ {FeF | e ∈ aFb}. Then aFb contains an element e such that F hF = FeF. It follows that h ∈ FeF ⊆ FaFbF = FcFdF. Thus, cFd contains an element g with h ∈ FgF. From h ∈ FgF we obtain F hF = FgF; cf. Lemma 2.1.6(i). Thus, as g ∈ cFd, F hF ∈ {FeF | e ∈ cFd}. What we have seen so far is that {FeF | e ∈ aFb} ⊆ {FeF | e ∈ cFd}. The reverse containment follows similarly.  Let H be a hypergroup, and let F be a closed subset of H. For each element h of H, we set h F := F hF. From Lemma 3.4.2 we obtain a hyperoperation on the set H//F of all hyperproducts h F with h ∈ H if we define a F bF := {c F | c ∈ aFb} for any two elements a and b in H. We call this hyperoperation on H//F the hypermultiplication on H//F defined by F. We notice that the hypermultiplication on H//F defined by F carries the danger of ambiguity. In fact, the hypermultiplication on H//F defined by F is defined on the same set as the restriction to H//F × H//F of the complex multiplication of H. (By the complex multiplication of H we mean the map which takes each pair (P, Q) of subsets of H to the complex product PQ.) However, while the codomain of the restriction to H//F × H//F of the complex multiplication of H is the power set of H, the codomain of the hypermultiplication on H//F defined by F is the power set of H//F. To say it differently, for any two elements a and b in H, (FaF)(FbF) stands for the union of the double cosets FcF with c ∈ aFb, whereas a F bF is defined to be the set of all products c F with c ∈ aFb, although FaF = a F and FbF = bF . Notice that, by Lemma 1.3.1(ii), (h∗ )F = (h F )∗ for each element h in H.

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63

Theorem 3.4.3 Let H be a hypergroup, and let F be a closed subset of H. Then H//F is a hypergroup with respect to the hypermultiplication on H//F defined by F and with neutral element 1F and inversion function h F 7→ (h∗ )F . Proof. Let a, b, c, and h be elements in H, and assume that h F ∈ (a F bF )c F . We will see that h F ∈ a F (bF c F ). Since h F ∈ (a F bF )c F , H contains an element k with k F ∈ a F bF and h F ∈ k F c F . From k F ∈ a F bF we obtain an element d in aFb with k F = d F . From h F ∈ k F c F we obtain an element e in kFc with h F = e F . It follows that h ∈ FeF ⊆ F kFcF = FdFcF ⊆ FaFbFcF. Thus, bFc contains an element l such that h ∈ FaFlF. Now aFl contains an element g such that h ∈ FgF. It follows that h F = g F ∈ a F l F ⊆ a F (bF c F ). What we have seen so far is that (a F bF )c F ⊆ a F (bF c F ), and, since the reverse containment follows similarly, we have a F (bF c F ) = (a F bF )c F for any three elements a, b, and c in H. This shows that the hypermultiplication on H//F defined by F is associative. For each element h in H, we have h F 1F = {a F | a ∈ hF} = {h F }, and this shows that 1F is a neutral element with respect to the hypermultiplication on H//F defined by F. To show that h F 7→ (h∗ )F is an inversion function with respect to the hypermultiplication on H//F defined by F, we let a, b, and c be elements in H, and we assume that c F ∈ a F bF . Then, by definition, c F = d F for some element d ∈ aFb. From d ∈ aFb we obtain an element f in F with d ∈ a f b. It follows that b ∈ f ∗ a∗ d; cf. Lemma 1.3.3 and Lemma 1.3.1(ii). Thus, a∗ d contains an element e such that b ∈ f ∗ e. From e ∈ a∗ d we obtain that e ∈ a∗ Fd. Thus, e F ∈ (a∗ )F d F = (a∗ )F c F . On the other hand, as b ∈ f ∗ e, b ∈ FeF, whence bF = e F . It follows that bF ∈ (a∗ )F c F . The proof for a F ∈ c F (b∗ )F follows similarly.



The hypergroup which we found in Theorem 3.4.3 will be called the quotient of H over F. A hypergroup is called schurian if it is isomorphic to a quotient of a thin hypergroup. This implies, of course, that thin hypergroups are schurian. The search for sufficient conditions for hypergroups to be schurian is an interesting research topic.

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We will see a number of schurian hypergroups in Section 7.1. Lemma 3.4.4 Let H be a hypergroup, and let F be a closed subset of H. Let h be an element in H, and let A1 , . . ., An be subsets of H. Then we have h F ∈ (A1 //F) · · · (An //F) if and only if h ∈ (F A1 F) · · · (F An F). Proof. If n = 1, the claim follows from Lemma 3.4.1(i). Therefore, we assume that 2 ≤ n. Suppose first that h F ∈ (A1 //F) · · · (An //F). Then there exist elements b and c in H such that bF ∈ (A1 //F) · · · (An−1 //F),

c F ∈ An //F,

and

h F ∈ bF c F .

From h F ∈ bF c F we obtain an element d in bFc such that h F = d F . It follows that h ∈ FdF ⊆ FbFcF. Since bF ∈ (A1 //F) · · · (An−1 //F), induction yields b ∈ (F A1 F) · · · (F An−1 F). It follows that FbF ⊆ (F A1 F) · · · (F An−1 F). Moreover, from c F ∈ An //F we obtain that c ∈ F An F; cf. Lemma 3.4.1(i). Thus, FcF ⊆ F An F. It follows that FbFcF ⊆ (F A1 F) · · · (F An F), and, since we have seen that h ∈ FbFcF, we now conclude that h ∈ (F A1 F) · · · (F An F), as wanted. Suppose, conversely, that h ∈ (F A1 F) · · · (F An F). Then there exist elements b in (F A1 F) · · · (F An−1 F) and c in F An F such that h ∈ bc. From h ∈ bc (together with bc ⊆ bFc) we obtain that h F ∈ bF c F . Since b ∈ (F A1 F) · · · (F An−1 F), induction yields bF ∈ (A1 //F) · · · (An−1 //F). By Lemma 3.4.1(i), c ∈ F An F implies that c F ∈ An //F. It follows that bF c F ⊆ (A1 //F) · · · (An //F). Thus, as h F ∈ bF c F , we have h F ∈ (A1 //F) · · · (An //F), as wanted.



Corollary 3.4.5 Let H be a hypergroup, and let F be a closed subset of H. Let A be a subset of H, and assume that A//F is a closed subset of H//F. Then F AF is a closed subset of H. Proof. Let h be an element in F AF. Then, by Lemma 3.4.4, h F ∈ A//F. Thus, as A//F is assumed to be closed and (h∗ )F is the inverse of h F , (h∗ )F ∈ A//F. Thus, by Lemma 3.4.4, h∗ ∈ F AF. Let c and d be elements in F AF, and let e be an element in cd. Then e ∈ (F AF)(F AF). Thus, by Lemma 3.4.4, e F ∈ (A//F)(A//F). Since A//F is assumed

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65

to be closed, this implies that e F ∈ A//F. Thus, by Lemma 3.4.4, e ∈ F AF. Since e has been chosen arbitrarily in cd, this shows that cd ⊆ F AF.  Theorem 3.4.6 Let H be a hypergroup, and let D be a closed subset of H. Then E 7→ E//D is a bijective map from the set of all closed subsets of H containing D to the set of all closed subsets of H//D. Proof. Let E be a closed subset of H, and assume that D ⊆ E. We first show that E//D is a closed subset of H//D. Since E is closed, we have 1 ∈ E; cf. Lemma 2.1.1. Thus, by definition, 1D ∈ E//D. Since E is closed, we have e∗ ∈ E for each each element e in E; cf. Lemma 2.1.1. Thus, by definition, (e∗ )D ∈ E//D. Let a and b be elements in E, and let c be an element in H with c D ∈ a D bD . From c D ∈ a D bD we obtain that c ∈ DaDbD; cf. Lemma 3.4.4. Thus, as DaDbD ⊆ E, c ∈ E, so that c D ∈ E//D. Since c has been chosen arbitrarily in H with c D ∈ a D bD , this shows that a D bD ⊆ E//D. What we have seen so far is that E//D is a closed subset of H//D; cf. Lemma 2.1.1. Since E has been chosen arbitrarily among the closed subset of H containing D, this shows that E 7→ E//D is a map from the set of all closed subsets of H containing D to the set of all closed subsets of H//D. To show that E 7→ E//D is injective, we let B and C be closed subsets of H with D ⊆ B, D ⊆ C, and B//D = C//D. We will see that B = C. Let b be an element in B. Then bD ∈ C//D. Thus, by Lemma 3.4.1(ii), b ∈ C. Since b has been chosen arbitrarily from B, this shows that B ⊆ C. Similarly, one obtains that C ⊆ B, so that B = C. To show that E 7→ E//D is surjective, we let A be a subset of H, and we assume that A//D is a closed subset of H//D. Then, by Corollary 3.4.5, DAD is a closed subset  of H. On the other hand, by Lemma 3.4.1(i), A//D = DAD//D.

3.5 Computations in Quotients In this section, we will see the ways in which some of the concepts we have developed so far interact with quotients. We will start with conjugation. Lemma 3.5.1 Let H be a hypergroup, and let C, D, and E be closed subsets of H with C ⊆ D ∩ E. Then the following hold. (i) If D and E are conjugate in H, D//C and E//C are conjugate in H//C. (ii) Assume that C is normal in H and that D//C and E//C are conjugate in H//C. Then D and E are conjugate in H.

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Proof. (i) Assume that D and E are conjugate in H. Then, by definition, H contains an element h such that Dh = hE. From Dh = hE and C ⊆ D ∩ E we obtain that (CDC)(ChC) = CDhC = ChEC = (ChC)(CEC). Thus, by Lemma 3.4.4, (D//C)hC = hC (E//C), and that means that D//C and E//C are conjugate in H//C. (ii) Assume that D//C and E//C are conjugate in H//C. Then, by definition, H contains an element h such that (D//C)hC = hC (E//C). Then, by Lemma 3.4.4, (CDC)(ChC) = (ChC)(CEC). On the other hand, since C is assumed to be normal in H and C ⊆ D, we also have (CDC)(ChC) = Dh; cf. Lemma 3.1.4(i). Similarly, as C is normal in H and C ⊆ E, we have (ChC)(CEC) = hE. Thus, Dh = hE, and that means that D and E are  conjugate in H. Now we will see how commutators interact with quotients. Lemma 3.5.2 Let H be a hypergroup, and let F be a closed subset of H. Then we have [H//F, F//F] = F[H, F]//F. Proof. Let A denote the union of the sets F h∗ F hF with h ∈ H. We will see that [H//F, F//F] = hAi//F, so that our claim will follow from Lemma 2.4.3. For each element h in H, we have [h F , 1F ] = (h∗ )F (1∗ )F h F 1F = (h∗ )F h F . Thus, [H//F, F//F] is the closed subset of H//F generated by the union of the sets (h∗ )F h F with h ∈ H. On the other hand, by Lemma 3.4.4, e F ∈ (d ∗ )F d F if and only if e ∈ Fd ∗ FdF for any two elements d and e in H. Thus, the desired equation follows from the definition  of A. Now we will see how tight elements interact with quotients. Lemma 3.5.3 Let H be a hypergroup, let F be a normal closed subset of H, and let h be a tight element in H. Then h F is a tight element of H//F. Proof. Since F is a normal closed subset of H, we have F h = hF; cf. Lemma 3.1.4(i). Thus,

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67

F hF h∗ F hF = F hh∗ hF = F hF, so that, by Lemma 3.4.4, h F (h∗ )F h F = {h F }. This equation says that h F is tight.  The subsequent lemma shows how normalizers and strong normalizers interact with quotients. Lemma 3.5.4 Let H be a hypergroup, let D and E be closed subsets of H, and assume that D ⊆ E. Then we have the following. (i) We have NH (E)//D ⊆ NH//D (E//D). (ii) Assume that D is normal in H. Then NH (E)//D = NH//D (E//D). (iii) We have KH (E)//D = KH//D (E//D). Proof. (i) Let h be an element in NH (E). Then E h ⊆ hE. It follows that E hD ⊆ hE D = hE ⊆ DhE. Thus, by Lemma 3.4.4, (E//D)h D ⊆ h D (E//D), so that h D ∈ NH//D (E//D). (ii) Let h be an element in H, and assume that h D ∈ NH//D (E//D). By (i) we shall be done if we succeed in showing that h D ∈ NH (E)//D. Since h D ∈ NH//D (E//D), we have (E//D)h D ⊆ h D (E//D). Thus, by Lemma 3.4.4, E hD ⊆ DhE. Since we are assuming that D is normal in H, we have Dh = hD; cf. Lemma 3.1.4(i). Thus, as D ⊆ E, E h ⊆ hE, and that means that h ∈ NH (E). It follows that h D ∈ NH (E)//D. (iii) Let h be an element in H. From Lemma 3.4.1(ii) (together with Lemma 3.3.1(ii)) we know that h D ∈ KH (E)//D if and only if h ∈ KH (E). By definition, h ∈ KH (E) means that h∗ E h ⊆ E. From Lemma 3.4.4 we obtain that h∗ E h ⊆ E if and only if  (h∗ )D (E//D)h D ⊆ E//D, and that means that h D ∈ KH//D (E//D). Corollary 3.5.5 Let H be a hypergroup, let D and E be closed subsets of H, and assume that D ⊆ E. Then we have the following. (i) If E is normal in H, E//D is normal in H//D. (ii) Assume that D is normal in H. Then E is normal in H if and only if E//D is normal in H//D. (iii) The closed subset E of H is strongly normal in H if and only if E//D is strongly normal in H//D. Proof. (i) Assume that E is normal in H. Then NH (E) = H. Thus, by Lemma 3.5.4(i), NH//D (E//D) = H//D, and that means that E//D is normal in H//D. (ii) From (i) we know that E//D is normal in H//D if E is normal in H. Thus, we just have to show that E is normal in H if E//D is normal in H//D.

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Assume that E//D is normal in H//D. Then, by definition, NH//D (E//D) = H//D. Since D is assumed to be normal in H, this implies that NH (E)//D = H//D; cf. Lemma 3.5.4(ii). From NH (E)//D = H//D we obtain that NH (E)//D is a closed subset of H//D. Thus, by Corollary 3.4.5, DNH (E)D is a closed subset of H. From NH (E)//D = H//D we also obtain that DNH (E)D//D = H//D; cf. Lemma 3.4.1(i). It follows that DNH (E)D = H; cf. Theorem 3.4.6. Now recall from Lemma 3.1.1(ii) that NH (E) = E NH (E)E. Thus, since DNH (E)D = H and D ⊆ E, NH (E) = H, and that means that E is normal in H. (iii) By definition, E is strongly normal in H if and only if KH (E) = H. From Lemma 3.5.4(iii) (together with Theorem 3.4.6) we obtain that KH (E) = H if and only if KH//D (E//D) = H//D, and KH//D (E//D) = H//D means that E//D is strongly  normal in H//D. The final lemma of this section decribes how thin elements interact with quotients. Lemma 3.5.6 Let H be a hypergroup, and let F be a closed subset of H. Then the following hold. (i) We have KH (F)//F = Oϑ (H//F). (ii) The closed subset F of H is strongly normal in H if and only if H//F is thin. Proof. (i) Applying Lemma 3.5.4(iii) to F in place of D and E we obtain that KH (F)//F = KH//F (F//F) = {h F | (h∗ )F (F//F)h F = F//F}. From F//F = {1F } we obtain further that {h F | (h∗ )F (F//F)h F = F//F} = {h F | (h∗ )F (h F ) = {1F }} = Oϑ (H//F). Thus, KH (F)//F = Oϑ (H//F). (ii) This follows from (i).



Setting F = {1} in Lemma 3.5.6(i) we obtain that KH ({1}) = Oϑ (H). This observation was made already right after the definition of the strong normalizer at the beginning of in Section 3.3.

3.6 Hypergroup Homomorphisms Let H and H 0 be hypergroups, and let φ be a map from H to H 0. For each subset A of H, we set φ(A) := {φ(a) | a ∈ A}. The map φ is called a hypergroup homomorphism or simply a homomorphism if

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Hypergroup Homomorphisms

69

φ(1) = 1 and, for any two elements a and b in H, φ(ab) = φ(a)φ(b). If φ is exponentially written, φ is (of course) a hypergroup homomorphism if 1φ = 1 and, for any two elements a and b in H, (ab)φ = aφ bφ . The following lemma shows that composites of hypergroup homomorphisms are hypergroup homomorphisms. Lemma 3.6.1 Let H, H 0, and H 00 be hypergroups, let χ be a homomorphism from H to H 0, and let ψ be a homomorphism from H 0 to H 00. Then χψ is a homomorphism from H to H 00. Proof. For any two elements a and b in H, we have (ab) χψ = ((ab) χ )ψ = (a χ b χ )ψ = (a χ )ψ (b χ )ψ = a χψ b χψ . This shows that the composite χψ of χ and ψ is a homomorphism from H to H 00.  Lemma 3.6.2 Let H and H 0 be hypergroups, let φ be a homomorphism from H to H 0, and let h be an element in H. Then φ(h∗ ) = φ(h)∗ . Proof. By definition, φ(1) = 1. Furthermore, by Lemma 1.1.1, 1 ∈ h∗ h. Thus, 1 ∈ φ(h∗ h) = φ(h∗ )φ(h). It follows that φ(h∗ ) ∈ 1·φ(h)∗ . On the other hand, by Lemma 1.1.4, 1·φ(h)∗ = {φ(h)∗ }. Thus, φ(h∗ ) = φ(h)∗ .  Let H and H 0 be hypergroups, and let φ be a homomorphism from H to H 0. We define ker(φ) := {h ∈ H | φ(h) = 1} and call this set the kernel of φ. Lemma 3.6.3 Let H and H 0 be hypergroups, and let φ be a homomorphism from H to H 0. Set F := ker(φ), and let a and b be elements in H. Then φ(a) = φ(b) if and only if aF = bF. Proof. From Lemma 3.6.2 we obtain that φ(a)∗ φ(b) = φ(a∗ )φ(b) = φ(a∗ b). Thus, by Lemma 1.1.6, φ(a) = φ(b) if and only if 1 ∈ φ(a∗ b).

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On the other hand, since F = ker(φ), we have 1 ∈ φ(a∗ b) if and only if a∗ b ∩ F is not empty, and, by Lemma 2.1.7(i), a∗ b ∩ F is not empty if and only if aF = bF.  Bijective hypergroup homomorphisms will be called hypergroup isomorphisms or simply isomorphisms. Lemma 3.6.4 Let H and H 0 be hypergroups, and let φ be an isomorphism from H to H 0. Then φ−1 is an isomorphism from H 0 to H. Proof. For any two elements a and b in H, we have φ−1 (φ(a)φ(b)) = φ−1 (φ(ab)) = ab = φ−1 (φ(a))φ−1 (φ(b)). Thus, as φ is surjective, φ−1 is a homomorphism from H 0 to H. Since inverses of bijective maps are bijective, we have shown that φ−1 is an isomorphism from H 0 to H.  Hypergroup isomomorphisms from a hypergroup H to itself are called hypergroup automorphisms of H or simply automorphisms of H. Theorem 3.6.5 The set of all automorphisms of a hypergroup is a group with respect to composition. Proof. Let H be a hypergroup. The identity map 1 H on H is obviously an automorphism of H, and, for each automorphism α of H, we have α·1 H = α = 1 H ·α. Since composites of bijective maps are bijective, we obtain from Lemma 3.6.1 that composites of automorphisms of H are automorphisms of H. From Lemma 3.6.4 we obtain that inverses of automorphisms of H are automorphisms of H.  The group of all automorphisms of a hypergroup H is called the automorphism group of H and is denoted by Aut(H). We shall now see that subgroups of automorphism groups of hypergroups provide a method to obtain new hypergroups from given hypergroups. For this, we let H be a hypergroup, and we let G be a subgroup of Aut(H). For each element h in H, we define hG := {hg | g ∈ G}. This set is called the orbit of h in H under G. Lemma 3.6.6 Let H be a hypergroup, let G be a subgroup of Aut(H), and let a, b, and c be elements in H. Assume that cG ∩ aG bG is not empty. Then cG ⊆ aG bG . Proof. We are assuming that that cG ∩ aG bG is not empty. Without loss of generalization, we may assume that c ∈ aG bG . Thus, G contains elements d and e with

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71

c ∈ a d be . It follows that cg ∈ (a d be )g = (a d )g (be )g = a dg beg ⊆ aG bG for each element g in G. Thus, cG ⊆ aG bG .



Let H be a hypergroup, let G be a subgroup of Aut(H), and set H G := {cG | c ∈ H}. Then

(aG, bG ) 7→ {cG | cG ⊆ aG bG }

is a hyperoperation on H G . We call this hyperoperation the hypermultiplication induced by G. Theorem 3.6.7 Let H be a hypergroup, and let G be a subgroup of Aut(H). Then H G is a hypergroup with respect to the hypermultiplication induced by G and with inversion function hG 7→ (h∗ )G . Proof. For any two elements a and b in H, we define aG ·bG := {cG | cG ⊆ aG bG }. Let a, b, c, and f be elements in H, and assume that f G ∈ (aG ·bG )·cG . We will see that f G ∈ aG ·(bG ·cG ). Since f G ∈ (aG · bG ) · cG , H contains an element d such that d G ∈ aG · bG and f G ∈ d G ·cG . From d G ∈ aG ·bG we obtain that d G ⊆ aG bG , and from f G ∈ d G ·cG we obtain that f G ⊆ d G cG . Thus, f G ⊆ (aG bG )cG . From f G ⊆ (aG bG )cG we obtain that f G ⊆ aG (bG cG ); cf. Lemma 1.3.4. Thus, f ∈ aG (bG cG ). Since f ∈ aG (bG cG ), bG cG contains an element e such that f ∈ aG e. From e ∈ bG cG we obtain that eG ⊆ bG cG ; cf. Lemma 3.6.6. Similarly, f ∈ aG e yields f G ⊆ aG eG . From eG ⊆ bG cG we obtain that eG ∈ bG ·cG , and from f G ⊆ aG eG we obtain that f G ∈ aG ·eG . It follows that f G ∈ aG ·(bG ·cG ). Since f has been chosen arbitrarily in H with f G ∈ (aG · bG )· cG , we have shown that (aG ·bG )·cG ⊆ aG ·(bG ·cG ). The reverse containment follows similarly. For each element h in H, we have hG 1G = hG {1} = hG whence hG ·1G = {hG }. Let a, b, and c be elements in H, and assume that cG ∈ aG ·bG . Then, by definition, cG ⊆ aG bG . It follows that bG ∩ (aG )∗ cG is not empty; cf. Lemma 1.3.3. From Lemma 3.6.2, on the other hand, one obtains that (a∗ )G = (aG )∗ . Thus, bG ∩(a∗ )G cG is not empty, so that, by Lemma 3.6.6, bG ⊆ (a∗ )G cG . This means that bG ∈ (a∗ )G ·cG . Similarly, one sees that aG ∈ cG ·(b∗ )G , so that hG 7→ (h∗ )G is an inversion function  with respect to the hypermultiplication induced by G.

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Lemma 3.6.8 Let H be a hypergroup, let F be a closed subset of H, and let ρ be a folding from F to H. Let H 0 be a hypergroup, let φ be a hypergroup isomorphism from H to H 0, and 0 set F 0 := F φ . For each element f in F, define f φρ := f ρφ . Then the following hold. (i) The map ρ0 is a folding from F 0 to H 0. (ii) Let π denote the permutation of F induced by ρ, and let π 0 denote the permutation of F 0 induced by ρ0. Then φπ 0 = πφ. Proof. (i) Let f be an element in F. Since ρ is a folding from F to H, we have f ρ f = {1ρ }. Thus, f φρ f φ = f ρφ f φ = ( f ρ f )φ = {1ρ } φ = {1ρφ } = {1ρ }. 0

0

Thus, as F 0 = F φ , ρ0 is a folding from F 0 to H 0. (ii) Let f be an element in F. With a reference to Lemma 3.6.2 we obtain f φπ = f φ∗ρ ∗ρ 0

0

0−1

0

= f ∗φρ ∗ρ

0−1

0−1

= f ∗ρφ∗ρ

0−1

= f ∗ρ∗φρ .

Since f π = f ∗ρ∗ρ , we have −1

f πφ = f πφρ ρ

0 0−1

= f πρφρ

0−1

0−1

= f ∗ρ∗φρ .

Thus, f φπ = f πφ . 0



Lemma 3.6.8 will be useful in the proof of Lemma 9.7.4.

3.7 The Isomorphism Theorems For each hypergroup homomorphism φ with domain H, we define im(φ) := φ(H) and call this set the image of φ. Let H and H 0 be hypergroups. From Lemma 3.6.4 we know that the inverse of an isomorphism from H to H 0 is an isomorphism from H 0 to H. The hypergroups H and H 0 are said to be isomorphic if there exists an isomorphism from H to H 0, and one indicates this by writing H  H 0. Theorem 3.7.1 [Homomorphism Theorem] Let H and H 0 be hypergroups, and let φ be a homomorphism from H to H 0. Then the following hold. (i) The set ker(φ) is a normal closed subset of H. (ii) The set im(φ) is a closed subset of H 0.

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73

(iii) We have H//ker(φ)  im(φ). Proof. (i) Since φ(1) = 1, 1 ∈ ker(φ), so ker(φ) is not empty. Let a and b be elements in ker(φ). Then φ(a) = 1 and φ(b) = 1. Thus, by Lemma 3.6.2, φ(a∗ b) = φ(a∗ )φ(b) = φ(a)∗ φ(b) = 1∗ 1 = {1}, and that means that a∗ b ⊆ ker(φ). Since a and b have been chosen arbitrarily in ker(φ), we have shown that ker(φ) is a closed subset of H. To show that ker(φ) is normal in H, we set F := ker(φ), and we choose elements f in F and b in H. We have to show that f b ⊆ bF. Let a be an element in f b. Then, as f ∈ F, φ(a) ∈ φ( f b) = φ( f )φ(b) = 1·φ(b) = {φ(b)}. It follows that φ(a) = φ(b), so that, by Lemma 3.6.3, a ∈ bF. (ii) Since 1 = φ(1) ∈ im(φ), im(φ) is not empty. Let c and d be elements in im(φ). Then H contains elements a and b such that φ(a) = c and φ(b) = d. Thus, referring to Lemma 3.6.2 we obtain that c∗ d = φ(a)∗ φ(b) = φ(a∗ )φ(b) = φ(a∗ b) ⊆ im(φ). Since c and d have been chosen arbitrarily in im(φ), we have shown that im(φ) is a closed subset of H 0. (iii) We set F := ker(φ). From (i) we know that F is normal in H. Thus, by Lemma 3.6.3, φ(a) = φ(b) if and only if a F = bF for any two elements a and b in H. For each element h in H, we define ψ(h F ) := φ(h). Since φ(a) = φ(b) if and only if a F = bF for any two elements a and b in H, ψ is an injective map from H//F to im(φ). The definition of ψ also implies that im(ψ) = im(φ). To show that ψ is a homomorphism, we choose elements a and b in H. Then we have ψ(a F bF ) = ψ({c F | c ∈ aFb}) = {ψ(c F ) | c ∈ aFb} and

{ψ(c F ) | c ∈ aFb} = {φ(c) | c ∈ aFb} = φ(aFb).

Moreover, for each element f in F, φ(a f b) = φ(a)φ( f )φ(b) = φ(a)·1·φ(b) = φ(a)φ(b) = ψ(a F )ψ(bF ). It follows that ψ(a F bF ) = ψ(a F )ψ(bF ).



Theorem 3.7.2 [First Isomorphism Theorem] Let H be a hypergroup, let D and E be closed subsets of H, and assume that D ⊆ E. Then (H//D)//(E//D)  H//E.

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Proof. For each element h in H, we define φ(h E ) := (h D )E//D . Then φ is a surjective map from H//E to (H//D)//(E//D). We will see that φ is a hypergroup isomorphism. From the definition of φ we obtain that φ(1E ) = (1D )E//D . To show that φ is a hypergroup homomorphism, we now let a and b be elements in H. Then φ(a E bE ) = φ({c E | c ∈ aE b}) = {φ(c E ) | c ∈ aE b} = {(c D )E//D | c ∈ aE b}. Since {c D | c ∈ aE b} = {c D | c ∈ DaE bD} for each element c in H, we also have {(c D )E//D | c ∈ aE b} = {(c D )E//D | c ∈ DaE bD}. Thus,

φ(a E bE ) = {(c D )E//D | c ∈ DaE bD}.

On the other hand, we have φ(a E )φ(bE ) = (a D )E//D (bD )E//D = {(c D )E//D | c D ∈ a D (E//D)bD } and, by Lemma 3.4.4, c ∈ DaE bD

⇔

c D ∈ a D (E//D)bD .

Thus, φ(a E bE ) = φ(a E )φ(bE ). To show that φ is injective, we let a and b be elements in H with φ(a E ) = φ(bE ). Then (a D )E//D = (bD )E//D . It follows that (E//D)a D (E//D) = (E//D)bD (E//D), so that, by Lemma 3.4.4, {c D | c ∈ EaE } = {c D | c ∈ E bE }. This implies that EaE = E bE, so that we have a E = bE .



Let H be a hypergroup, let D and E be closed subsets of H, and assume that D ⊆ NH (E). From Lemma 3.1.4(iii) we know that E D is a closed subset of H. In the following theorem, we look at the quotient E D//E. Theorem 3.7.3 [Second Isomorphism Theorem] Let H be a hypergroup, let D and E be closed subsets of H, and assume that D ⊆ NH (E). Then E D//E  D//(E ∩ D). Proof. We claim that

φ : D → E D//E,

d 7→ d E

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75

is a hypergroup homomorphism. From the definition of φ we obtain that φ(1) = 1E . To show that φ is a hypergroup homomorphism, we now let b and c be elements in D. We have to show that φ(bc) = φ(b)φ(c). Note first that φ(bc) = {φ(d) | d ∈ bc} = {d E | d ∈ bc} = bc//E and

φ(b)φ(c) = bE c E = {d E | d ∈ bEc} = bEc//E.

On the other hand, since we are assuming that D ⊆ NH (E), we obtain from Lemma 3.1.4(i) that E bcE = bEc. Thus, by Lemma 3.4.1(i), bc//E = E bcE//E = bEc//E. It follows that φ(bc) = φ(b)φ(c), as wanted. Since D ⊆ NH (E), we have E D = E DE, and from Lemma 3.4.1(i) we know that D//E = E DE//E. Thus, we have E D//E = D//E, and this shows that φ is surjective. Notice finally that ker(φ) = E ∩ D. Thus, we have D//(E ∩ D) = D//ker(φ)  im(φ) = E D//E; 

cf. Theorem 3.7.1(iii). Lemma 3.7.4

Let H be a hypergroup, and let B, C, D, and E be closed subsets of H with B ⊆ D and C ⊆ E. Assume that B is normal in D and that C is normal in E. Then we have (D ∩ BE)//(D ∩ BC)  (D ∩ E)//(D ∩ BC ∩ E). Proof. In Lemma 3.1.6(ii), we saw that (D ∩ BC)(D ∩ E) = D ∩ BE, and from Lemma 3.1.6(iv) we know that D ∩ E ⊆ NH (D ∩ BC). Thus, the claim follows from Theorem 3.7.3 (with D ∩ E in place of D and D ∩ BC in place of E).  The following theorem generalizes a group theoretic result of Hans Zassenhaus; cf. [54; II. §5]. Theorem 3.7.5 Let H be a hypergroup, and let B, C, D, and E be closed subsets of H with B ⊆ D and C ⊆ E. Assume that B is normal in D and that C is normal in E. Then we have (D ∩ BE)//(D ∩ BC)  (E ∩ CD)//(E ∩ CB).

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Proof. In Lemma 3.7.4, we saw that (D ∩ BE)//(D ∩ BC)  (D ∩ E)//(D ∩ BC ∩ E). Applying Lemma 3.7.4 to C, B, E, and D in place of B, C, D, and E we obtain that (E ∩ CD)//(E ∩ CB)  (E ∩ D)//(E ∩ CB ∩ D). From Lemma 2.2.1 (together with the fact that B is normal in D) we obtain that D ∩ BC = B(D ∩ C) = (D ∩ C)B = D ∩ CB. Our claim follows from these three statements.



We conclude this section with two results on finite hypergroups. The first one will be useful in the proof of the second one and in Section 4.7. Lemma 3.7.6 Let H be a finite hypergroup, and let C, D, and E be closed subsets of H with D ⊆ E. Assume that D is strongly normal in E. Then |E//D| = |(E//D)/(D(C ∩ E)//D)| · |(C ∩ E)//(C ∩ D)|. Proof. Since D is assumed to be strongly normal in E, we have C∩E ⊆ NH (D). Thus, by Lemma 3.1.4(iii), D(C ∩ E) is a closed subset of E. It follows that D(C ∩ E)//D is a closed subset of E//D; cf. Theorem 3.4.6. On the other hand, since D is strongly normal in E, Lemma 3.5.6(ii) tells us that E//D is thin. Thus, by Lemma 2.6.4(iii), |E//D| = |(E//D)/(D(C ∩ E)//D)| · |D(C ∩ E)//D|. From C ∩ E ⊆ NH (D) we also obtain that D(C ∩ E)//D  (C ∩ E)//(C ∩ D); cf. Theorem 3.7.3. Thus, |D(C ∩ E)//D| = |(C ∩ E)//(C ∩ D)|, so that the desired equation follows.



Lemma 3.7.7 Let H be a finite hypergroup, and let C, D, and E be closed subsets of H with D ⊆ E. Assume that D is strongly normal in E and that E ⊆ NH (C). Then we have |E//D| = |CE//CD| · |(C ∩ E)//(C ∩ D)|. Proof. Since D ⊆ E ⊆ NH (C), CD is a closed subset of H; cf. Lemma 3.1.4(iii). Since E ⊆ NH (C) ∩ NH (D), E normalizes CD; cf. Lemma 3.1.5(ii). Thus, by Theorem 3.7.3, CE//CD  E//(CD ∩ E).

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From Lemma 3.1.4(ii) we obtain that CD = DC, and from Lemma 2.2.1(i) we know that E ∩ DC = D(E ∩ C). Thus, CD ∩ E = D(C ∩ E). It follows that E//(CD ∩ E) = E//D(C ∩ E). From Theorem 3.7.2 we also know that (E//D)//(D(C ∩ E)//D)  E//D(C ∩ E). Now recall that, by hypothesis, E ⊆ NH (C). Thus, by Lemma 3.1.5(i), E ⊆ NH (C ∩ E). Thus, as E ⊆ NH (D), D(C ∩ E) is normal in E; cf. Lemma 3.1.5(ii). It follows that D(C ∩ E)//D is normal in E//D; cf. Corollary 3.5.5(i). Thus, |(E//D)//(D(C ∩ E)//D)| = |(E//D)/(D(C ∩ E)//D)|. Now we conclude that |CE//CD| = |(E//D)/(D(C ∩ E)//D)|, so that the desired equation follows from Lemma 3.7.6.



3.8 Wreath Products Let H1 and H2 be hypergroups. Assume that H1 ∩ (H2 \ {1}) is empty, and set S := H1 ∪ (H2 \ {1}). For any two elements p and q in S, define pq {q} {p} p ◦ q :=   pq     \ {1}) H ∪ (pq  1        

if if if if if

{p, q} ⊆ H1 p ∈ H1 and q ∈ H2 \ {1} p ∈ H2 \ {1} and q ∈ H1 {p, q} ⊆ H2 \ {1} and p∗ , q {p, q} ⊆ H2 \ {1} and p∗ = q.

Then (p, q) 7→ p ◦ q is a hyperoperation on S. We call this hyperoperation the wreath hypermultiplication on S. Theorem 3.8.1 Let H1 and H2 be hypergroups. Assume that H1 ∩ (H2 \ {1}) is empty, and set S := H1 ∪(H2 \{1}). Then S is a hypergroup with respect to the wreath hypermultiplication. Proof. To show that the wreath hypermultiplication on S is associative, we let p, q, and r be elements in S. If {p, q, r } ⊆ H1 , we have (p ◦ q) ◦ r = pq ◦ r = (pq)r = p(qr) = p ◦ qr = p ◦ (q ◦ r),

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since H1 is assumed to be a hypergroup. If {p, q} ⊆ H1 and r ∈ H2 \ {1}, we have (p ◦ q) ◦ r = pq ◦ r = {r } = p ◦ r = p ◦ {r } = p ◦ (q ◦ r). If {p, r } ⊆ H1 and q ∈ H2 \ {1}, we have (p ◦ q) ◦ r = {q} ◦ r = q ◦ r = {q} = p ◦ q = p ◦ {q} = p ◦ (q ◦ r). The case where {q, r } ⊆ H1 and p ∈ H2 \ {1} can be treated analogously to the case where {p, q} ⊆ H1 and r ∈ H2 \ {1}. Assume that p ∈ H1 and {q, r } ⊆ H2 \ {1}. If q∗ , r, we have q ◦ r = qr. Moreover, by Lemma 1.1.6, qr ⊆ H2 \ {1}. Thus, (p ◦ q) ◦ r = {q} ◦ r = q ◦ r = qr = p ◦ qr = p ◦ (q ◦ r). If q∗ = r, we have q ◦ r = H1 ∪ (qr \ {1}). Since p ∈ H1 and q ∈ H2 we further have p ◦ q = {q}. Thus, (p ◦ q) ◦ r = {q} ◦ r = q ◦ r = H1 ∪ (qr \ {1}). Since p ∈ H1 and qr \ {1} ⊆ H2 \ {1}, we also have H1 ∪ (qr \ {1}) = pH1 ∪ (qr \ {1}) = (p ◦ H1 ) ∪ (p ◦ (qr \ {1})), and with a second reference to q ◦ r = H1 ∪ (qr \ {1}) we obtain that (p ◦ H1 ) ∪ (p ◦ (qr \ {1})) = p ◦ (H1 ∪ (qr \ {1})) = p ◦ (q ◦ r). Thus, we are done also in this case. If q ∈ H1 and {p, r } ⊆ H2 \ {1}, we have (p ◦ q) ◦ r = {p} ◦ r = p ◦ r = p ◦ {r } = p ◦ (q ◦ r). The case where r ∈ H1 and {p, q} ⊆ H2 \ {1} can be treated analogously to the case where p ∈ H1 and {q, r } ⊆ H2 \ {1}. Now assume that {p, q, r } ⊆ H2 \ {1}, and suppose first that p∗ , q,

p∗ < qr,

and

q∗ , r.

From p∗ < qr we obtain that r ∗ < pq; cf. Lemma 1.2.1. Thus, we have p ◦ (q ◦ r) = p ◦ qr = p(qr) = (pq)r = pq ◦ r = (p ◦ q) ◦ r. Next suppose that

3.8

p∗ = q,

p∗ < qr,

and

Wreath Products

79

q∗ , r.

Since p∗ = q, we have p ◦ q = H1 ∪ (pq \ {1}). Thus, (p ◦ q) ◦ r = (H1 ∪ (pq \ {1})) ◦ r = (H1 ◦ r) ∪ ((pq \ {1}) ◦ r). Since r ∈ H2 , we have H1 ◦ r = {r }. Moreover, since p∗ < qr, we have r ∗ < pq; cf. Lemma 1.2.1. Thus, (pq \ {1}) ◦ r = (pq \ {1})r. It follows that (H1 ◦ r) ∪ ((pq \ {1}) ◦ r) = {r } ∪ (pq \ {1})r = {1}r ∪ (pq \ {1})r = (pq)r. Thus, we have

(p ◦ q) ◦ r = (pq)r.

On the other hand, since q∗ , r, we have q ◦ r = qr. Moreover, by Lemma 1.1.6, qr ⊆ H2 \ {1}. Thus, as p∗ < qr, we obtain that p ◦ (q ◦ r) = p ◦ (qr) = p(qr), so that the desired equation holds also in this case. Next suppose that

p∗ , q,

p∗ ∈ qr,

and

q∗ , r.

Since p∗ , q, we have p ◦ q = pq. Furthermore, since p∗ ∈ qr, we obtain from Lemma 1.2.1 that r ∗ ∈ pq. Thus, we have (p ◦ q) ◦ r = pq ◦ r = (r ∗ ◦ r) ∪ ((pq \ {r ∗ }) ◦ r). From r ∗ ∈ pq we also obtain that (r ∗ ◦ r) ∪ ((pq \ {r ∗ }) ◦ r) = H1 ∪ (r ∗ r \ {1}) ∪ ((pq \ {r ∗ }) ◦ r). Since p∗ , q, we have pq ⊆ H2 \ {1}; cf. Lemma 1.1.6. Thus, (pq \ {r ∗ }) ◦ r = (pq \ {r ∗ })r. It follows that H1 ∪ (r ∗ r \ {1}) ∪ ((pq \ {r ∗ }) ◦ r) = H1 ∪ ({r ∗ }r \ {1}) ∪ (pq \ {r ∗ })r. Thus, as

H1 ∪ ({r ∗ }r \ {1}) ∪ (pq \ {r ∗ })r = H1 ∪ ((pq)r \ {1}),

we have The equation

(p ◦ q) ◦ r = H1 ∪ ((pq)r \ {1}). p ◦ (q ◦ r) = H1 ∪ (p(qr) \ {1})

is shown analogously, so that the desired equation holds also in this case. The case where

p∗ , q,

p∗ < qr,

and

q∗ = r

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80

can be treated analogously to the case where p∗ = q, p∗ < qr, and q∗ , r. Next suppose that

p∗ = q,

p∗ ∈ qr,

and

q∗ , r.

Since p∗ = q, we have p ◦ q = H1 ∪ (pq \ {1}). Thus, (p ◦ q) ◦ r = (H1 ∪ (pq \ {1})) ◦ r = (H1 ◦ r) ∪ ((pq \ {1}) ◦ r). From r ∈ H2 \ {1} we obtain that H1 ◦ r = {r }. Thus, (H1 ◦ r) ∪ ((pq \ {1}) ◦ r) = {r } ∪ ((pq \ {1}) ◦ r) From p∗ ∈ qr we further obtain that r ∗ ∈ pq; cf. Lemma 1.2.1. Thus, {r } ∪ ((pq \ {1}) ◦ r) = {r } ∪ (r ∗ ◦ r) ∪ ((pq \ {r ∗, 1}) ◦ r). Since pq \ {r ∗, 1} ⊆ H2 \ {1}, we also have (pq \ {r ∗, 1}) ◦ r = (pq \ {r ∗, 1})r. Thus, {r } ∪ (r ∗ ◦ r) ∪ ((pq \ {r ∗, 1}) ◦ r) = {r } ∪ (r ∗ ◦ r) ∪ (pq \ {r ∗, 1})r. Now, as r ∗ ◦ r = H1 ∪ (r ∗ r \ {1}), we obtain {r } ∪ (r ∗ ◦ r) ∪ (pq \ {r ∗, 1})r = {r } ∪ H1 ∪ (r ∗ r \ {1}) ∪ (pq \ {r ∗, 1})r. Note finally that {r } ∪ H1 ∪ (r ∗ r \ {1}) = H1 ∪ {1}r ∪ ({r ∗ }r \ {1}) and

H1 ∪ {1}r ∪ ({r ∗ }r \ {1}) ∪ (pq \ {r ∗, 1})r = H1 ∪ ((pq)r \ {1}).

Thus,

(p ◦ q) ◦ r = H1 ∪ ((pq)r \ {1}).

We obtain

p ◦ (q ◦ r) = H1 ∪ (p(qr) \ {1})

in exactly the same way as in the case where p∗ , q, p∗ ∈ qr, and q∗ , r. Thus, we have the desired equation also in this case. Next suppose that

p∗ = q,

p∗ < qr,

and

q∗ = r.

Since p∗ = q, we have p ◦ q = H1 ∪ (pq \ {1}). Thus, (p ◦ q) ◦ r = (H1 ∪ (pq \ {1})) ◦ r = (H1 ◦ r) ∪ ((pq \ {1}) ◦ r). From r ∈ H2 \ {1} we obtain that H1 ◦ r = {r }. Thus, (H1 ◦ r) ∪ ((pq \ {1}) ◦ r) = {r } ∪ ((pq \ {1}) ◦ r).

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Since p∗ < qr, we have r ∗ < pq; cf. Lemma 1.2.1. Thus, as r ∈ H2 \ {1}, we conclude that (pq \ {1}) ◦ r = (pq \ {1})r. It follows that {r } ∪ (pq \ {1})r = {1}r ∪ (pq \ {1})r = (pq)r, so that we have

(p ◦ q) ◦ r = (pq)r.

The equation p ◦ (q ◦ r) = p(qr) is shown analogously, so that the desired equation holds also in this case. The case where

p∗ , q,

p∗ ∈ qr,

and

q∗ = r

can be treated analogously to the case where p∗ = q, p∗ ∈ qr, and q∗ , r. Suppose finally that p∗ = q,

p∗ ∈ qr,

and

q∗ = r.

In this case, we obtain (p ◦ q) ◦ r = H1 ∪ ((pq)r \ {1}) in exactly the same way as in the case where p∗ = q, p∗ ∈ qr, and q∗ , r. The equation p ◦ (q ◦ r) = H1 ∪ (p(qr) \ {1}) is shown analogously, so that the desired equation holds also in this case. From the definition of the wreath hypermultiplication we obtain that the neutral element of H1 is a neutral element with respect to the wreath hypermultiplication on S. Thus, we shall be done if we succeed in showing that s 7→ s∗ is an inversion function with respect to the wreath hypermultiplication on S. For this, we let p, q, and r be elements in S, and we assume that r ∈ p ◦ q. We will see that q ∈ p∗ ◦ r Assume first that

and

p ∈ r ◦ q∗ .

{p, q} ⊆ H1 .

Then p ◦ q = pq. Thus, as r ∈ p ◦ q, r ∈ pq. Thus, as pq ⊆ H1 , r ∈ H1 . Now, as H1 is assumed to be a hypergroup, this implies that q ∈ p∗ r and p ∈ rq∗ . Thus, as p∗ ◦ r = p∗ r and r ◦ q∗ = rq∗ , we conclude that q ∈ p∗ ◦ r and p ∈ r ◦ q∗ . Assume now that p ∈ H1

and

q ∈ H2 \ {1}.

Then p ◦ q = {q}. Thus, as r ∈ p ◦ q, we have q = r. From p ∈ H1 and q ∈ H2 \ {1} we obtain that p∗ ◦ q = {q}. It follows that q ∈ p∗ ◦ q. Thus, as q = r, q ∈ p∗ ◦ r.

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From q = r and q ∈ H2 \ {1} we obtain that r ◦ q∗ = H1 ∪ (rq∗ \ {1}). Thus, as p ∈ H1 , we have p ∈ r ◦ q∗ . Next assume that p ∈ H2 \ {1}

and

q ∈ H1 .

Then p ◦ q = {p}. Thus, as r ∈ p ◦ q, we have p = r. From p = r and p ∈ H2 \ {1} we obtain that p∗ ◦ r = H1 ∪ (p∗ r \ {1}). Thus, as q ∈ H1 , we have q ∈ p∗ ◦ r. From q ∈ H1 and r ∈ H2 \ {1} we obtain that r ◦ q∗ = {r }. It follows that r ∈ r ◦ q∗ . Thus, as p = r, p ∈ r ◦ q∗ . Next assume that {p, q} ⊆ H2 \ {1} In this case, we have

p∗

and r ∈ H1 .

= q, since r ∈ p ◦ q.

Since p ∈ H2 \ {1} and r ∈ H1 , we have p∗ ◦ r = {p∗ }. It follows that p∗ ∈ p∗ ◦ r. Thus, as p∗ = q, we have q ∈ p∗ ◦ r. Since q ∈ H2 \ {1} and r ∈ H1 , we have r ◦ q∗ = {q∗ }. It follows that q∗ ∈ r ◦ q∗ . Thus, as p∗ = q, we have p ∈ r ◦ q∗ . Finally, assume that {p, q} ⊆ H2 \ {1}

and r < H1 .

In this case, since r ∈ p ◦ q and p ◦ q ⊆ H1 ∪ (pq \ {1}), we have r ∈ pq. Thus, as H2 is a hypergroup, we have q ∈ p∗ r and p ∈ rq∗ . Suppose that p , r. Then p∗ ◦ r = p∗ r. Thus, as q ∈ p∗ r, q ∈ p∗ ◦ r. Suppose that p = r. Then p∗ ◦ r = H1 ∪ (p∗ r \ {1}). Thus, as q ∈ p∗ r \ {1}, we have q ∈ p∗ ◦ r. Suppose that r , q. Then r ◦ q∗ = rq∗ . Thus, as p ∈ rq∗ , p ∈ r ◦ q∗ . Suppose that r = q. Then r ◦ q∗ = H1 ∪ (rq∗ \ {1}). Thus, as p ∈ rq∗ \ {1}, we have  p ∈ r ◦ q∗ . Let H1 and H2 be hypergroups, and assume that H1 ∩ (H2 \ {1}) is empty. The hypergroup which we found in Theorem 3.8.1 will be denoted by H1 o H2 , and each hypergroup which is isomorphic to H1 o H2 is called a wreath product of H1 and H2 . Lemma 3.8.2 Let H, H1 , and H2 be hypergroups, and assume that H1 ∩ (H2 \ {1}) is empty. Then H is a wreath product of H1 and H2 if and only if H contains a closed subset F such that F  H1 , H//F  H2 , and, for each element h in H \ F, hF = {h}. Proof. Assume first that H is a wreath product of H1 and H2 . Then H is isomorphic to H1 o H2 , and that means that there exists an isomorphism from H to H1 o H2 . We choose one of these isomorphisms and denote it by φ. We define

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Wreath Products

83

F := φ−1 (H1 ). Since c∗ ◦ d = c∗ d for any two elements c and d in H1 , H1 is a closed subset of H1 o H2 . Thus, as φ is an isomorphism from H to H1 o H2 and F = φ−1 (H1 ), F is a closed subset of H and F  H1 . Let h be an element in H \ F. Then φ(h) ∈ (H1 o H2 ) \ H1 = H2 \ {1}. It follows that φ(hF) = φ(h) ◦ φ(F) = φ(h) ◦ H1 = {φ(h)} = φ({h}). Since φ is bijective, this implies that hF = {h}, and it remains to be shown that H//F  H2 . Since we have hF = {h} for each element in H \ F, we have φ(c) = φ(d) for any two elements c and d in H \ F with c F = d F . Set ψ(h F ) := φ(h) for each element h in H \ F and ψ(1F ) := 1. (The 1 on the right hand side stands for the neutral element of H2 .) We claim that ψ is an isomorphism from H//F to H2 . We will first show that ψ is a homomorphism from H//F to H2 . For this, we choose elements c and d in H \ F. We have to show that ψ(c F d F ) = ψ(c F )ψ(d F ). Since ψ(1F ) = 1, we may assume that c , 1 and d , 1. From cF = {c} we obtain that cFd = cd. Thus, c F d F = {e F | e ∈ cFd} = {e F | e ∈ cd}. Assume first that c∗ , d. Then, as c∗ F = {c∗ } and dF = {d}, we have c∗ F , dF. Thus, by Lemma 2.1.7(i), cd ⊆ H \ F, so that ψ(c F d F ) = ψ({e F | e ∈ cd}) = {ψ(e F ) | e ∈ cd} = {φ(e) | e ∈ cd}. Now assume that c∗ = d. Then, by Lemma 1.1.6, 1 ∈ cd. Thus, ψ(c F d F ) = ψ({e F | e ∈ cd}) = {ψ(1F )} ∪ {ψ(e F ) | e ∈ cd \ {1}}. From ψ(1F ) = 1 we also obtain that {ψ(1F )} ∪ {ψ(e F ) | e ∈ cd \ {1}} = {1} ∪ {φ(e) | e ∈ cd \ {1}} = {φ(e) | e ∈ cd}. Thus, as

ψ(c F )ψ(d F ) = φ(c)φ(d) = {φ(e) | e ∈ cd},

we now have ψ(c F d F ) = ψ(c F )ψ(d F ) in both cases. To show that ψ is injective, we let c and d be elements in H, and we assume that ψ(c F ) = ψ(d F ).

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3 Elementary Structure Theory

If {c, d} ∩ F is empty, we have ψ(c F ) = φ(c) and ψ(d F ) = φ(d). Thus, as ψ(c F ) = ψ(d F ), we obtain that φ(c) = φ(d), and, since φ is injective, this implies that c = d. It follows that c F = d F . Assume that c ∈ F and d ∈ H \ F. From c ∈ F we obtain that c F = 1F . Thus, ψ(c F ) = 1. From d ∈ H \ F we obtain that ψ(d F ) = φ(d) ∈ H2 \ {1}. It follows that ψ(c F ) , ψ(d F ), contrary to our hypothesis that ψ(c F ) = ψ(d F ). The case where c ∈ H \ F and d ∈ F can be treated analogously to the case where c ∈ F and d ∈ H \ F. If {c, d} ⊆ F, we have c F = d F . To show that ψ is surjective, we let e be an element in H2 . If e , 1, the surjectivity of φ (together with the definition of F) guarantees an element h in H \ F with φ(h) = e. From h ∈ H \ F we obtain that ψ(h F ) = φ(h), so that we have ψ(h F ) = e. If e = 1, on the other hand, we have ψ(1F ) = e. Assume now, conversely, that H contains a closed subset F such that F  H1,

H//F  H2,

and, for each element h in H \ F, hF = {h}. We have to show that there exists an isomorphism from H to H1 o H2 . Since F  H1 , there exists an isomorphism φ1 from F to H1 . Since H//F  H2 , there exists an isomorphism ψ from H//F to H2 . Since hF = {h} for each element h in H \ F, σ : h 7→ h F is a bijective map from H \ F to (H//F) \ {1F }. We define τ to be the map from (H//F) \ {1F } to H2 \ {1} which coincides on (H//F) \ {1F } with ψ and write φ2 for the composite of σ and τ. Note that we have φ2 (cd) = φ2 (c)φ2 (d) for any two elements c and d in H \ F with cd ⊆ H \ F. We set φ := φ1 ∪ φ2 and claim that φ is an isomorphism from H to H1 o H2 . Let c and d be elements in H. We have to show that φ(cd) = φ(c)φ(d). If {c, d} ⊆ F, we have φ(cd) = φ1 (cd) = φ1 (c)φ1 (d) = φ(c)φ(d), since φ|F = φ1 and φ1 is a homomorphism. Assume that c ∈ F and d ∈ H \ F. From d ∈ H \ F we obtain that d ∗ ∈ H \ F. Thus, d ∗ F = {d ∗ }. It follows that Fd = {d}; cf. Lemma 1.3.1(ii). Thus, as c ∈ F, cd = {d}, so that φ(cd) = φ({d}) = {φ(d)}. On the other hand, since d ∈ H \ F, we obtain that φ(d) ∈ H2 \ {1}. Thus, as φ(c) ∈ H1 , φ(c)φ(d) = {φ(d)}. From φ(cd) = {φ(d)} and φ(c)φ(d) = {φ(d)} we obtain that φ(cd) = φ(c)φ(d), as wanted. The case where c ∈ H \ F and d ∈ F can be treated analogously to the case where c ∈ F and d ∈ H \ F.

3.8

Wreath Products

85

Assume next that {c, d} ⊆ H \ F and c∗ , d. From c∗ , d we obtain that c∗ F , dF, since c∗ F = {c∗ } and dF = {d}. Thus, by Lemma 2.1.7(i), cd ⊆ H \ F. It follows that φ(cd) = φ2 (cd) = φ2 (c)φ2 (d) = φ(c)φ(d), so we are done also in this case. Assume finally that {c, d} ⊆ H \ F and c∗ = d. From c∗ = d we first obtain that F ⊆ d ∗ dF = d ∗ {d} = cd. Thus, φ(cd) = φ(F) ∪ φ(cd \ F) = H1 ∪ {φ(e) | e ∈ cd \ F}. Since φ and φ2 coincide on H \ F, this implies that φ(cd) = H1 ∪ {φ2 (e) | e ∈ cd \ F}. From {c, d} ⊆ H \ F, on the other hand, we obtain that φ(c) = ψ(c F ) and φ(d) = ψ(d F ). Furthermore, since c∗ = d, we have (c∗ )F = d F . Thus, as ψ is a hypergroup homomorphism, we have φ2 (c)∗ = ψ(c F )∗ = ψ(d F ) = φ2 (d); cf. Lemma 3.6.2. Thus, by definition, φ(c)φ(d) = ψ(c F )ψ(d F ) = H1 ∪ ψ(c F d F \ {1F }). Since ψ(c F d F \ {1F }) = {ψ(e F ) | e F ∈ c F d F \ {1F }}, this yields φ(c)φ(d) = H1 ∪ {ψ(e F ) | e F ∈ c F d F \ {1F }}. Now recall that, for each element e in H, e ∈ cd \ F if and only if e F ∈ c F d F \ {1F }. Furthermore, for each element e in H \ F, ψ(e F ) = φ(e). Thus, φ(c)φ(d) = H1 ∪ {φ(e) | e ∈ cd \ F}, so that we have shown φ(cd) = φ(c)φ(d) also in this case.



Let H, H1 , and H2 be hypergroups. Assume that H1 ∩ (H2 \ {1}) is empty and that H is a wreath product of H1 and H2 . Then, by Lemma 3.8.2, H contains a closed subset F such that F  H1 and, for each element h in H \ F, hF = {h}. Note that, by Lemma 3.1.8, this closed subset F of H is normal in H. Corollary 3.8.3 A hypergroup H is a wreath product of two hypergroups if and only if it contains a closed subset F all left cosets of which in H different from F consist of only one element. Proof. This follows from Lemma 3.8.2.



Corollary 3.8.4 A hypergroup H is a wreath product of a hypergroup and a hypergroup with two elements if and only if it contains a closed subset F with |H \ F | = 1.

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3 Elementary Structure Theory

Proof. This follows from Corollary 3.8.3.



Lemma 3.8.5 For any three hypergroups H1 , H2 , and H3 , we have (H1 o H2 ) o H3  H1 o (H2 o H3 ). Proof. Set H := (H1 o H2 ) o H3 . Then, by Lemma 3.8.2, H contains a closed subset E such that E  H1 o H2, H//E  H3, and, for each element h in H \ E, hE = {h}. Since E  H1 o H2 , E contains a closed subset D such that D  H1, E//D  H2, and eD = {e} for each element e in E \ D; cf. Lemma 3.8.2. Since D ⊆ E, we have hD = {h} for each element h in H \ D. Thus, as D  H1 , we shall be done if we succeed in showing that H//D  H2 o H3 ; cf. Lemma 3.8.2. From Theorem 3.7.2 we know that (H//D)//(E//D)  H//E. Thus, as H//E  H3 , (H//D)//(E//D)  H3 . Let h be an element in H \ E, and let e be an element in E. Since h ∈ H \ E, hE = {h}. Thus, hDe ⊆ hE = {h}. It follows that h D e D = {h D }. Since e has been chosen arbitrarily in E, this shows that (h D )E//D = {h D }. Thus, as E//D  H2 and (H//D)//(E//D)  H3 , we obtain from Lemma 3.8.2 that H//D  H2 o H3 . 

3.9 Hypergroup Isomorphisms and Regular Actions The concept of a hypergroup isomorphism is designed to express what it means for two hypergroups to have a similar structure. To express what it means for a regular action ω of a hypergroup H to be similar to a regular action ω 0 of a hypergroup H 0 isomorphic to H we now introduce the concept of an (ω, ω 0)-isomorphism from H to H 0. Let H and H 0 be hypergroups, let X and X 0 be sets, let ω be a regular action of H on X, and let ω 0 be a regular action of H 0 on X 0. A hypergroup isomorphism φ from H to H 0 is called an (ω, ω 0)-isomorphism if there exists a bijective map υ from X to X 0 such that, for any two elements x in X and h in H, ω(x, h)υ = ω 0(x υ, hφ ). The bijective map υ from X to X 0 is called the supporting bijection of φ. The results of this section will play a role in Sections 10.3 and 10.7.

Hypergroup Isomorphisms and Regular Actions

3.9

87

Lemma 3.9.1 Let H and H 0 be hypergroups, let X and X 0 be sets, let ω be a regular action of H on X, and let ω 0 be a regular action of H 0 on X 0. Let φ be a hypergroup isomorphism from H to H 0, and let υ be a bijective map from X to X 0. Then φ is an (ω, ω 0)isomorphism with supporting bijection υ if and only if, for any three elements y and z in X and h in H, the statements z ∈ yh and zυ ∈ y υ hφ are equivalent. Proof. For any three elements y and z in X and h in H, we have zυ ∈ ω(y, h)υ and

z ∈ ω(y, h)

⇔

zυ ∈ ω 0(y υ, hφ )

⇔

⇔

z ∈ yh

zυ ∈ y υ hφ .

Now assume first that φ is an (ω, ω 0)-isomorphism from H to H 0 with supporting bijection υ. Then, for any two elements x in X and h in H, ω(x, h)υ = ω 0(x υ, hφ ). Thus, for any three elements y and z in X and h in H, the statements zυ ∈ ω(y, h)υ and zυ ∈ ω(y, h)υ are equivalent, and, if that is so, the above equivalences show that, for any three elements y and z in X and h in H, the statements z ∈ yh and zυ ∈ y υ hφ are equivalent. 

The converse follows similarly.

Let H be a hypergroup, let X be a set, and let ω be a regular action of H on X. Recall from Section 1.6 that the map which takes each pair (y, z) of elements in X to the unique element h in H satisfying z ∈ yh is called the color defined by ω. Lemma 3.9.2 Let H and H 0 be hypergroups, let X and X 0 be sets, let ω be a regular action of H on X, and let ω 0 be a regular action of H 0 on X 0, let σ denote the color defined by ω, let σ 0 denote the color defined by ω 0, and let φ be an (ω, ω 0)-isomorphism from H to H 0 with supporting bijection υ. Then, for any two elements y and z in X, 0 (y, z)σφ = (y υ, zυ )σ . Proof. Let y and z be elements in X, and set h := (y, z)σ . Then, by Lemma 1.6.7(i), z ∈ yh. Since φ is an implies that

(ω, ω 0)-isomorphism

from H to H 0 with supporting bijection υ, this zυ ∈ y υ hφ ;

cf. Lemma 3.9.1. Referring once more to Lemma 1.6.7(i) we obtain

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3 Elementary Structure Theory

(y υ, zυ )σ = hφ . 0

Thus, as (y, z)σ = h, we have (y, z)σφ = (y υ, zυ )σ . 0



Let σ denote the color defined by ω. For each element h in H, define hˆ := {(y, z) ∈ X × X | (y, z)σ = h}, and set Hˆ := { hˆ | h ∈ H}. In Lemma 1.6.8, we saw that Hˆ is a (set theoretic) hypergroup on X. This hypergroup was called the ω-shadow of H. We will now see that H and its ω-shadow are isomorphic. Theorem 3.9.3 Let H be a hypergroup, let X be a set, let ω be a regular action of H on X, and let σ denote the color defined by ω. For each element h in H, define hˆ := {(y, z) ∈ X × X | (y, z)σ = h}, and set Hˆ := { hˆ | h ∈ H}. Then h 7→ hˆ is a hypergroup ˆ isomorphism from H to H. ˆ We will show that φ is a hypergroup Proof. For each element h in H, define hφ := h. ˆ isomorphism from H to H. Note first that, for any three elements y, z in X and h in H, (y, z) ∈ hφ

⇔

(y, z) ∈ hˆ

⇔

(y, z)σ = h

⇔

z ∈ yh;

cf. Lemma 1.6.7(i). From Condition O2 we obtain that 1φ = 1X , from Condition O3 that φ is a bijective ˆ Thus, we shall be done if we succeed in showing that, for any two map from H to H. elements a and b in H, (ab)φ = aφ bφ . We first show that (ab)φ ⊆ aφ bφ . Let c be an element in ab. We have to show that cφ ∈ aφ bφ . Let y and z be elements in X with (y, z) ∈ cφ . Then z ∈ yc. Since c ∈ ab, this implies that z ∈ yab. Thus, ya contains an element x with z ∈ xb. From x ∈ ya we obtain that (y, x) ∈ aφ , from z ∈ xb we obtain that (x, z) ∈ bφ . Thus, (y, z) ∈ aφ ◦ bφ , and, since y and z have been chosen arbitrarily with (y, z) ∈ cφ , this shows that cφ ⊆ aφ ◦ bφ . Thus, by definition, cφ ∈ aφ bφ . Now we show that aφ bφ ⊆ (ab)φ . Let c be an element in H with cφ ∈ aφ bφ . We are done when we succeed in showing that c ∈ ab. Let y and z be elements in X with (y, z) ∈ cφ . From cφ ∈ aφ bφ we obtain that cφ ⊆ aφ ◦ bφ . Then, as (y, z) ∈ cφ , we obtain that (y, z) ∈ aφ ◦ bφ . Thus, X contains an element x with (y, x) ∈ aφ and (x, z) ∈ bφ . It follows that x ∈ ya and z ∈ xb, so that z ∈ yab.

3.9

Hypergroup Isomorphisms and Regular Actions

89

On the other hand, (y, z) ∈ cφ also implies that z ∈ yc. Thus, we obtain from Lemma 1.6.2(i) that c ∈ ab, as wanted.  Let H be a hypergroup, and let ω be a regular action of H. The hypergroup isomorphism from H to its ω-shadow which we found in Theorem 3.9.3 is called the ω-canonical hypergroup isomorphism. Lemma 3.9.4 Let H be a hypergroup, let X be a set, and let ω be a regular action of H on X. Let Hˆ denote the ω-shadow of H, and let ωˆ denote the canonical action of Hˆ on X. Then ˆ the ω-canonical hypergroup isomorphism from H to Hˆ is an (ω, ω)-isomorphism with supporting bijection 1X . ˆ We will Proof. Let φ denote the ω-canonical hypergroup isomorphism from H to H. see that φ is an (ω, ω)-isomorphism ˆ with supporting bijection 1X . Let h be an element in H, and let σ denote the color defined by ω. Then, as φ denotes ˆ the ω-canonical hypergroup isomorphism from H to H, hφ = {(y, z) ∈ X × X | (y, z)σ = h}. It follows that

(y, z) ∈ hφ

⇔

(y, z)σ = h

for any two elements y and z in X. From Lemma 1.6.7(i) we also know that (x, y)σ = h

⇔

y ∈ xh,

(x, y) ∈ hφ

⇔

y ∈ xhφ

and, by definition, we have

for any two elements y and z in X. Thus, we have y ∈ xh

⇔

y ∈ xhφ

ˆ for any two elements y and z in X. Thus, by Lemma 3.9.1, φ is an (ω, ω)-isomorphism with supporting bijection 1X .  Let H and H 0 be hypergroups, let X and X 0 be sets, let ω be a regular action of H on X, and let ω 0 be a regular action of H 0 on X 0. The hypergroups H and H 0 are called (ω, ω 0)-isomorphic if there exists an (ω, ω 0)-isomorphism from H to H 0. H be a hypergroup, let X be a set, and let ω be a regular action of H on X. Let Hˆ denote the ω-shadow of H, and let ωˆ denote the canonical action of Hˆ on X. Lemma 3.9.4 says that H and Hˆ are (ω, ω)-isomorphic. ˆ

4 Subnormality and Thin Residues

We start this chapter with the definition of subnormal chains. General properties of subnormal chains will be compiled in Section 4.1. In Section 4.2, we restrict our attention to subnormal series. (Subnormal series are subnormal chains which contain {1}.) Maximal subnormal series of hypergroups will be called composition series, and composition series will be in the center of Section 4.3. With the help of the Second Isomorphism Theorem (Theorem 3.7.3) we will see (in Theorem 4.3.2) that any two composition series of a given hypergroup are isomorphic. This leads to the notion of a composition factor of a hypergroup and underscores the importance of simple hypergroups. A particular class of subnormal series is the class of the strong subnormal series. Strong subnormal series are related to the thin residue of a hypergroup, and this important closed subset of a hypergroup will be studied in Sections 4.4 and 4.5. The notion of the thin residue of a hypergroup leads naturally to residually thin hypergroups, and residually thin hypergroups will be discussed in the final three sections of this chapter. General results on residually thin hypergroups will be considered in Section 4.6. In Section 4.7, we study finite residually thin hypergroups. Finite residually thin hypergroups allow structural results that escape the general treatment of hypergroups and are similar to the fundamental theorems [42; Th´eor`emes I, II] of Ludwig Sylow and [24; Theorem] of Philip Hall in finite group theory. A very specific class of finite residually thin hypergroups, the class of solvable hypergroups, will be discussed in a final section of this chapter, in Section 4.8.

4.1 Subnormal Chains Let H be a hypergroup, and let D be a non-empty finite set of closed subsets of H. Assume that, for any two elements A and B in D, A ⊆ B or B ⊆ A. Let C denote the intersection of the elements of D, and define E to be the union of the elements of D. For each element D in D \ {E }, define D D to be the intersection of all elements

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 P. -H. Zieschang, Hypergroups, https://doi.org/10.1007/978-3-031-39489-8_4

91

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4 Subnormality and Thin Residues

of D \ {D} which contain D as a subset. The set D is called a subnormal chain from C to E if, for each element D in D \ {E }, D is normal in D D . Lemma 4.1.1 Let H be a hypergroup, and let B, D, and E be closed subsets of H. Assume that B ⊆ D, and let C be a subnormal chain from B to D. Then the following hold. (i) The set {E ∩ C | C ∈ C} is a subnormal chain from E ∩ B to E ∩ D. (ii) If D ⊆ NH (E), {EC | C ∈ C} is a subnormal chain from E B to E D. Proof. (i) Let C be an element in C \ {D}. Then, by Lemma 3.1.5(i), E ∩ C C ⊆ NE (C) ⊆ NE (E ∩ C). Since E ∩ C ⊆ E ∩ C C , this shows that E ∩ C is normal in E ∩ C C . (ii) Let C be an element in C \ {D}, and assume that D ⊆ NH (E). Then, by Lemma 3.1.5(ii), C C ⊆ NH (E) ∩ NH (C) ⊆ NH (EC). It follows that

EC C ⊆ ECNH (EC) ⊆ NH (EC);

cf. Lemma 3.1.1(ii). Thus, as EC ⊆ EC C , EC is normal in EC C .



A closed subset F of a hypergroup H is said to be a subnormal closed subset of H if there exists a subnormal chain from F to H. If F is a subnormal closed subset of a hypergroup H, we also say that F is subnormal in H. Corollary 4.1.2 Let H be a hypergroup, let D be a subnormal closed subset of H, and let E be a closed subset of H. Then the following hold. (i) The closed subset E ∩ D is subnormal in E. (ii) If D ⊆ E, D is subnormal in E. (iii) If E is subnormal in H, so is E ∩ D. (iv) If D ⊆ NH (E), E D is subnormal in H. Proof. (i) This follows from Lemma 4.1.1(i). (We apply this lemma to D and H in place of B and D.) (ii) This follows from (i). (iii) In (i), we saw that E ∩ D is subnormal in E. Thus, if E is subnormal in H, E ∩ D is subnormal in H. (iv) This follows from Lemma 4.1.1(ii). (We apply this lemma to D and H in place of B and D.) 

4.1

Subnormal Chains

93

Let H be a hypergroup, let D be a subnormal chain of closed subsets of H, and let E denote the union of the elements of D. The subnormal chain D is called strong if, for each element D in D \ {E }, D is strongly normal in D D . The subsequent lemma and its corollary are similar to Lemma 4.1.1 and Corollary 4.1.2. Lemma 4.1.3 Let H be a hypergroup, and let B, D, and E be closed subsets of H. Assume that B ⊆ D, and let C be a strong subnormal chain from B to D. Then the following hold. (i) The set {E ∩ C | C ∈ C} is a strong subnormal chain from E ∩ B to E ∩ D. (ii) If D ⊆ NH (E), {EC | C ∈ C} is a strong subnormal chain from E B to E D. Proof. (i) Let C be an element in C \ {D}. Then, by Lemma 3.3.3(i), E ∩ C C ⊆ KE (C) ⊆ KE (E ∩ C). Since E ∩ C ⊆ E ∩ C C , this shows that E ∩ C is strongly normal in E ∩ C C . (ii) Let C be an element in C \ {D}, and assume that D ⊆ NH (E). Then, by Lemma 3.3.3(ii), C C ⊆ NH (E) ∩ KH (C) ⊆ KH (EC). It follows that EC C ⊆ KH (EC); cf. Lemma 3.3.1(ii). Thus, as EC ⊆ EC C , EC is  strongly normal in EC C . A closed subset F of a hypergroup H is said to be a strongly subnormal closed subset of H if there exists a strong subnormal chain from F to H. If F is a strongly subnormal closed subset of a hypergroup H, we also say that F is strongly subnormal in H. Corollary 4.1.4 Let H be a hypergroup, let D be a strongly subnormal closed subset of H, and let E be a closed subset of H. Then the following hold. (i) The closed subset E ∩ D is strongly subnormal in E. (ii) If D ⊆ E, D is strongly subnormal in E. (iii) If E is strongly subnormal in H, so is E ∩ D. (iv) If D ⊆ NH (E), E D is strongly subnormal in H. Proof. (i) This follows from Lemma 4.1.3(i). (We apply this lemma to D and H in place of B and D.) (ii) This follows from (i). (iii) From (i) we know that E ∩ D is strongly subnormal in E. Thus, if E is strongly subnormal in H, E ∩ D is strongly subnormal in H. (iv) This follows from Lemma 4.1.3(ii). (We apply this lemma to D and H in place  of B and D.)

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4 Subnormality and Thin Residues

We will now see in which way subnormal chains of hypergroups are related to subnormal chains of their quotients. The lemma is reminiscent of the first two parts of Corollary 3.5.5. Lemma 4.1.5 Let H be a hypergroup, let B and C be closed subsets of H, and assume that B ⊆ C. Then the following hold. (i) Let D be a subnormal chain from C to H. Then {D//B | D ∈ D} is a subnormal chain from C//B to H//B. (ii) Assume that B is normal in H, and let D be a set of closed subsets of H. Then D is a subnormal chain from C to H if and only if {D//B | D ∈ D} is a subnormal chain from C//B to H//B. Proof. (i) Assume first that C = H. Then C//B = H//B. Moreover, since D is a subnormal chain from C to H, we obtain that D = {H}. From D = {H} we obtain that {D//B | D ∈ D} = {H//B}. Thus, as C//B = H//B, we conclude that {D//B | D ∈ D} is a subnormal chain from C//B to H//B. Now assume that C , H. Then D , {H}. Thus, as D is a subnormal chain from C to H, D \ {H} contains an element E such that D \ {H} is a subnormal chain from C to E and E is normal in H. Since D \ {H} is a subnormal chain from C to E, we obtain by induction that {D//B | D ∈ D \ {H}} is a subnormal chain from C//B to E//B. Since E is normal in H, Corollary 3.5.5(i) yields that E//B is normal in H//B. It follows that {D//B | D ∈ D} is a subnormal chain from C//B to H//B. (ii) From (i) we know that {D//B | D ∈ D} is a subnormal chain from C//B to H//B if D is a subnormal chain from C to H. Thus, we just have to show that D is a subnormal chain from C to H if {D//B | D ∈ D} is a subnormal chain from C//B to H//B. Assume that {D//B | D ∈ D} a subnormal chain from C//B to H//B. Assume first that C//B = H//B. Then, by Theorem 3.4.6, C = H. Moreover, since {D//B | D ∈ D} is a subnormal chain from C//B to H//B, we obtain that {D//B | D ∈ D} = {H//B}. It follows that D = {H}; cf. Theorem 3.4.6. From C = H and D = {H} we obtain that D is a subnormal chain from C to H. Now assume that C//B , H//B. Then {D//B | D ∈ D} , {H//B}. Thus, as {D//B | D ∈ D} a subnormal chain from C//B to H//B, D contains an element E such that E//B , H//B, {D//B | D ∈ D \ {H}} is a subnormal chain from C//B to E//B, and E//B is normal in H//B. Since E//B , H//B, E , H. Thus, as {D//B | D ∈ D \ {H}} is a subnormal chain from C//B to E//B, we obtain by induction that D \ {H} is a subnormal chain from C to E. Since B is normal in H (by hypothesis) and E//B is normal in H//B, we obtain from Corollary 3.5.5(ii) also that E is normal in H. It follows that D is a  subnormal chain from C to H.

4.1

Subnormal Chains

95

Lemma 4.1.6 Let H be a hypergroup, let B and C be closed subsets of H, and assume that B ⊆ C. Let D be a set of closed subsets of H. Then D is a strong subnormal chain from C to H if and only if {D//B | D ∈ D} is a strong subnormal chain from C//B to H//B. Proof. Assume first that D is a strong subnormal chain from C to H. Assume first that C = H. Then C//B = H//B. Moreover, since D is a strong subnormal chain from C to H, we obtain that D = {H}. From D = {H} we obtain that {D//B | D ∈ D} = {H//B}. Thus, as C//B = H//B, we conclude that {D//B | D ∈ D} is a strong subnormal chain from C//B to H//B. Now assume that C , H. Then D , {H}. Thus, as D is a strong subnormal chain from C to H, D \ {H} contains an element E such that D \ {H} is a strong subnormal chain from C to E and E is strongly normal in H. Since D \ {H} is a strong subnormal chain from C to E, we obtain by induction that {D//B | D ∈ D \ {H}} is a strong subnormal chain from C//B to E//B. Since E is strongly normal in H, Corollary 3.5.5(iii) yields that E//B is strongly normal in H//B. It follows that {D//B | D ∈ D} is a strong subnormal chain from C//B to H//B. Assume now that, conversely, {D//B | D ∈ D} a strong subnormal chain from C//B to H//B. Assume first that C//B = H//B. Then, by Theorem 3.4.6, C = H. Moreover, since {D//B | D ∈ D} is a strong subnormal chain from C//B to H//B, we obtain that {D//B | D ∈ D} = {H//B}. It follows that D = {H}; cf. Theorem 3.4.6. From C = H and D = {H} we obtain that D is a strong subnormal chain from C to H. Now assume that C//B , H//B. Then {D//B | D ∈ D} , {H//B}. Thus, as {D//B | D ∈ D} a strong subnormal chain from C//B to H//B, D contains an element E such that E//B , H//B, {D//B | D ∈ D \ {H}} is a strong subnormal chain from C//B to E//B, and E//B is strongly normal in H//B. Since E//B , H//B, E , H. Thus, as {D//B | D ∈ D \ {H}} is a strong subnormal chain from C//B to E//B, induction yields that D \ {H} is a strong subnormal chain from C to E. Since E//B is strongly normal in H//B, we obtain from Corollary 3.5.5(ii) also that E is strongly normal in H. It follows that D is a strong subnormal  chain from C to H. The following corollary follows from Lemma 4.1.5 and Lemma 4.1.6 the same way as Corollary 3.5.5 follows from Lemma 3.5.4. Corollary 4.1.7 Let H be a hypergroup, let D and E be closed subsets of H, and assume that D ⊆ E. Then we have the following. (i) If E is subnormal in H, E//D is subnormal in H//D.

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(ii) Assume that D is normal in H. Then E is subnormal in H if and only if E//D is subnormal in H//D. (iii) The closed subset E of H is strongly subnormal in H if and only if E//D is strongly subnormal in H//D. Proof. (i) Assume that E is subnormal in H. Then, by definition, there exists a subnormal chain from E to H. Now, by Lemma 4.1.5(i), there exists a subnormal chain from E//D to H//D, and that means that E//D is subnormal in H//D. (ii) The fact that E is subnormal in H means that there exists a subnormal chain from E to H. By Lemma 4.1.5(ii), this is the case if and only if there exists a subnormal chain from E//D to H//D, and that means that E//D is subnormal in H//D. (iii) The fact that E is strongly subnormal in H means that there exists a strong subnormal chain from E to H. By Lemma 4.1.6, this is the case if and only if there exists a strong subnormal chain from E//D to H//D, and that means that E//D is  strongly subnormal in H//D.

4.2 Subnormal Series Let H be a hypergroup. A subnormal chain from {1} to H is called a subnormal series of H. Each hypergroup H possesses a subnormal series, namely {{1}, H}. Lemma 4.2.1 Let H be a hypergroup, let D be a closed subset of H, and let E be a subnormal series of H. Then the following hold. (i) The set {D ∩ E | E ∈ E} is a subnormal series of D. (ii) If D is normal in H, {DE//D | E ∈ E} is a subnormal series of H//D. Proof. (i) This follows from Lemma 4.1.1(i). (We apply this lemma to {1}, H, D, and E in place of B, D, E, and C.) (ii) Applying Lemma 4.1.1(ii) to {1}, H, D, and E in place of B, D, E, and C we obtain that {DE | E ∈ E} is a subnormal chain from D to H. Now we apply Lemma 4.1.5(i) to D in place of both, B and C, and to {DE | E ∈ E} in place of D. Then we obtain that {DE//D | E ∈ E} is a subnormal series of H//D.  A subnormal series E of a hypergroup H is called strong if, for each element E in E \ {H}, E is strongly normal in E E . Lemma 4.2.2 Let H be a hypergroup, let D be a closed subset of H, and let E be a strong subnormal series of H. Then the following hold. (i) The set {D ∩ E | E ∈ E} is a strong subnormal series of D.

4.3

Composition Series

97

(ii) If D is normal in H, {DE//D | E ∈ E} is a strong subnormal series of H//D. Proof. (i) This follows from Lemma 4.1.3(i). (We apply this lemma to {1}, H, D, and E in place of B, D, E, and C.) (ii) Applying Lemma 4.1.3(ii) to {1}, H, D, and E in place of B, D, E, and C we obtain that {DE | E ∈ E} is a strong subnormal chain from D to H. Then applying Lemma 4.1.6 to D in place of B and C and to {DE | E ∈ E} in place of D, we  obtain that {DE//D | E ∈ E} is a strong subnormal series of H//D.

4.3 Composition Series A maximal subnormal series of a hypergroup H is called a composition series of H. Let H be a hypergroup. If H is trivial, {H} is the only composition series of H. If H is simple, {{1}, H} is the only composition series of H. Hypergroups do not necessarily possess composition series. Lemma 4.3.1 We have the following. (i) Each subnormal series of a finite hypergroup H is contained in a composition series of H. (ii) Each finite hypergroup possesses a composition series. Proof. (i) Let H be a finite hypergroup, and let C be a subnormal series of H. Among the subnormal series of H which contain C, we choose D as large as possible. (This is possible, since H is assumed to be finite.) If D is not a composition series of H, H possesses a subnormal series E which properly contains D. In this case, since C ⊆ D and D ⊆ E, we conclude that C ⊆ E, contrary to the choice of D. (ii) Each hypergroup H possesses a subnormal series, namely {{1}, H}. Thus, the  claim follows from (i). Let H be a hypergroup. Let F be a composition series of H. Recall from Section 4.1 that, for each element F in F \ {H}, F F is our notation for the intersection of all elements of F \ {F} which contain F as a subset. Let D and E be composition series of H. We say that D and E are isomorphic if there exists a bijective map η from D \ {H} to E \ {H} such that, for each element D in D \ {H}, D D //D  Dη E //Dη . The following theorem generalizes and is modeled on a group theoretic theorem which was proved in 1889 by Otto Hölder; cf. [27; §10].

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Theorem 4.3.2 Let H be a hypergroup, and let D and E be composition series of H. Then D and E are isomorphic. Proof. There is nothing to show if H = {1}. Thus, we assume that H , {1}, and we let D be an element in D \ {H}. We define Dµ to be the uniquely determined element E in E \ {H} satisfying D D ⊆ DE E and D D * DE. Since D D ⊆ DDµ E ,

D D = D D ∩ DDµ E .

Thus, by Lemma 3.1.6(iv), D D ∩ DDµ is a normal closed subset of D D . Note also that D ⊆ D D ∩ DDµ . Thus, as D is a composition series of H, we have D = D D ∩ DDµ

or

D D ∩ DDµ = D D .

Since D D * DDµ , D D ∩ DDµ , D D . Thus, D = D D ∩ DDµ . From D D = D D ∩ DDµ E and D = D D ∩ DDµ we obtain that D D //D  (D D ∩ Dµ E )//(D D ∩ DDµ ∩ Dµ E ); cf. Lemma 3.7.4. Now we define a map ν from E \ {H} to D \ {H} analogously to µ. Then Dµ E ⊆ Dµ Dµν D and Dµ E * Dµ Dµν . From the above isomorphism we obtain that Dµ E * DDµ . With a reference to Lemma 1.3.1(ii) this implies that Dµ E * Dµ D. Thus, the definition of ν forces D ⊆ Dµν . On the other hand, as above, the fact that E is a composition series of H yields that Dµ = Dµ E ∩ Dµ Dµν . Therefore, D D ⊆ Dµν would lead to D D ∩ Dµ E ⊆ Dµ , and that contradicts the above isomorphism. Thus, D D * Dµν . From D ⊆ Dµν and D D * Dµν we obtain that Dµν = D. We have shown that, for each element D in D \ {H}, Dµν = D. Similarly, one shows that, for each element E in E \ {H}, E νµ = E. Thus, µ is a bijective map from D \ {H} to E \ {H}, ν is a bijective map from E \ {H} to D \ {H}, and µ and ν are inverses of each other. Now we notice that the above isomorphism also holds for Dµ , E, D, and ν in place of D, D, E, and µ. Thus, we have Dµ E //Dµ  (Dµ E ∩ Dµν D )//(Dµ E ∩ Dµ Dµν ∩ Dµν D ). It follows that D D //D  Dµ E //Dµ .



A hypergroup C is called a composition factor of a hypergroup H if H possesses a composition series F and F \ {H} contains an element F such that C  F F //F.

4.3

Composition Series

99

Assume that a hypergroup H possesses a composition series F . Then |F | − 1 is called the composition length of H. Notice that, by Theorem 4.3.2, the definition of the composition length of H does not depend on the choice of the composition series F. The trivial hypergroup has no composition factor and composition length 0. Lemma 4.3.3 We have the following. (i) Composition factors of hypergroups are simple. (ii) Composition factors of thin hypergroups are thin. Proof. Let H be a hypergroup, and let C be a composition factor of H. Then there exist a composition series F of H and an element F in F \{H} such that C  F F //F. (i) Assume, by way of contradiction, that C is not simple. Then so is F F //F. Thus, by Theorem 3.4.6, F F contains a closed subset E such that F//F ⊆ E//F ⊆ F F //F, F//F , E//F , F F //F, and E//F is normal in F F //F. From F//F , E//F we obtain that F , E; cf. Theorem 3.4.6. Similarly, E//F , F F //F implies that E , F F . Since E//F is normal in F F //F, we obtain from Corollary 3.5.5(ii) that E is normal in F F , contradiction. (ii) Assume that H is thin. Then by Lemma 3.5.6(ii), {1} is strongly normal in H. It follows that {1} is strongly normal in F F . Since F is a composition series of H, F F ⊆ NH (F). Thus, applying Lemma 3.3.7 to F, {1}, and F F in place of C, D, and E we obtain that F is strongly normal in F F . It follows that F F //F is thin; cf. Lemma 3.5.6(ii). Since C  F F //F, this implies that C is thin.  Lemma 4.3.4 Let H be a hypergroup, and let D be a normal closed subset of H. Let E be a composition series of H, and let E be an element in E \ {H}. Then the following hold. (i) We have D ∩ E = D ∩ E E or E E //E  (D ∩ E E )//(D ∩ E). (ii) We have (DE E //D)//(DE//D)  E E //E or DE//D = DE E //D. (iii) We have E E //E  (D ∩ E E )//(D ∩ E) or (DE E //D)//(DE//D)  E E //E. Proof. (i) Since E is normal in E E , D ∩ E E ⊆ NH (E). Thus, by Lemma 3.1.4(iii), E(D ∩ E E ) is a closed subset of H. From E E ⊆ NH (D) we obtain that E E ⊆ NH (D ∩ E E ); cf. Lemma 3.1.5(i). Thus, as E E ⊆ NH (E), E E ⊆ NH (E(D ∩ E E )); cf. Lemma 3.1.5(ii). It follows that E(D ∩ E E ) is normal in E E . Note also that E ⊆ E(D ∩ E E ). Thus, as E is a composition series of H, we have E = E(D ∩ E E ) or

E(D ∩ E E ) = E E .

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In the first case, we obtain that D ∩ E = D ∩ E E . In the second case, we obtain from Theorem 3.7.3 that E E //E  (D ∩ E E )//(D ∩ E). (ii) From Lemma 3.1.5(ii) we know that E E normalizes DE. Thus, by Lemma 3.1.5(i), DE ∩ E E is a normal closed subset of E E . Note also that E ⊆ DE ∩ E E . Thus, as E is a composition series of H, we have E = DE ∩ E E

or

DE ∩ E E = E E .

In the first case, we obtain from Theorem 3.7.3 that DE E //DE  E E //E. On the other hand, by Theorem 3.7.2, (DE E //D)//(DE//D)  DE E //DE. Thus, (DE E //D)//(DE//D)  E E //E. In the second case, we obtain that DE = DE E , so that DE//D = DE E //D. (iii) Assume that E E //E and (D ∩ E E )//(D ∩ E) are not isomorphic. Then, by (i), D ∩ E = D ∩ EE. Assume also that (DE E //D)//(DE//D) and E E //E are not isomorphic. Then, by (ii), DE//D = DE E //D. Now recall from Theorem 3.7.3 that DE//D  E//(D ∩ E) and

DE E //D  E E //(D ∩ E E ).

From this we obtain that E = E E , contradiction.



Lemma 4.3.5 Let H be a hypergroup, let D be a normal closed subset of H, and let E be a composition series of H. Then the following hold. (i) The set {D ∩ E | E ∈ E} is a composition series of D. (ii) The set {DE//D | E ∈ E} is a composition series of H//D. Proof. (i) From Lemma 4.2.1(i) we know that {D ∩ E | E ∈ E} is a subnormal series of D. Let E be an element in E \ {H}. Then, by Lemma 4.3.4(i), D ∩ E = D ∩ EE

or

E E //E  (D ∩ E E )//(D ∩ E).

By Lemma 4.3.3(i), E E //E is simple. Thus, if D ∩ E , D ∩ E E , the above dichotomy forces (D ∩ E E )//(D ∩ E) to be simple. Thus, D ∩ E E does not contain a proper normal closed subset which properly contains D ∩ E; cf. Corollary 3.5.5(i). Since E has been chosen arbitrarily in E\{H}, we have shown that the set {D∩E | E ∈ E} is a composition series of D.

4.4

The Thin Residue

101

(ii) From Lemma 4.2.1(ii) we know that {DE//D | E ∈ E} is a subnormal series of H//D. Let E be an element in E \ {H}. Then, by Lemma 4.3.4(ii), (DE E //D)//(DE//D)  E E //E

or

DE//D = DE E //D.

From Lemma 4.3.3(i) we know that E E //E is simple. Thus, if DE//D , DE E //D, the above dichotomy forces (DE E //D)//(DE//D) to be simple. It follows that DE E //D does not contain a proper normal closed subset which properly contains DE//D; cf. Corollary 3.5.5(i). Now recall that E has been chosen arbitrarily in E \ {H}. Thus, we have shown that  {DE//D | E ∈ E} is a composition series of H//D. Theorem 4.3.6 Let H be a hypergroup, and let D be a normal closed subset of H. Assume that H possesses a composition series. Then the following hold. (i) The closed subset D possesses a composition series, and each composition factor of D is a composition factor of H. (ii) The quotient H//D possesses a composition series, and each composition factor of H//D is a composition factor of H. (iii) Each composition factor of H is a composition factor of D or a composition factor of H//D. Proof. (i) Let E be a composition series of H. Then, by Lemma 4.3.5(i), {D ∩ E | E ∈ E} is a composition series of D. Thus, by Lemma 4.3.4(i), each composition factor of D is a composition factor of H. (ii) Let E be a composition series of H. From Lemma 4.3.5(ii) we know that {DE//D | E ∈ E} is a composition series of H//D. Thus, by Lemma 4.3.4(ii), each composition factor of H//D is a composition factor of H. (iii) Let C be a composition factor of H. Then H possesses a composition series E and E \ {H} contains an element E with C  E E //E. Then, by Lemma 4.3.4(iii), C  (D ∩ E E )//(D ∩ E) or C  (DE E //D)//(DE//D). In the first case, C is a composition factor of D; cf. Lemma 4.3.5(i). In the second  case, C is a composition factor of H//D; cf. Lemma 4.3.5(ii).

4.4 The Thin Residue For each hypergroup H, we define O ϑ (H) to be the intersection of all closed subsets of H which are strongly normal in H. (Note that H contains at least one closed subset which is strongly normal in H, namely H itself.)

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Theorem 4.4.1 Let H be a hypergroup. Then the following statements hold. (i) The set O ϑ (H) is a strongly normal closed subset of H. (ii) The quotient H//O ϑ (H) is thin. (iii) Let F be a closed subset of H, and assume that H//F is thin. Then we have O ϑ (H) ⊆ F. Proof. (i) This follows from Lemma 3.3.6. (ii) From (i) we know that O ϑ (H) is strongly normal in H. Thus, by Lemma 3.5.6(ii), H//O ϑ (H) is thin. (iii) We are assuming that H//F is thin. Thus, by Lemma 3.5.6(ii), F is strongly normal in H. Thus, the definition of O ϑ (H) forces O ϑ (H) ⊆ F.  The strongly normal closed subset O ϑ (H) of a hypergroup H will be called the thin residue of H. Lemma 4.4.2 For each hypergroup H, we have O ϑ (H) = [H, {1}]. Proof. Let H be a hypergroup, let a and b be elements in H, and let c be an element in ab. From c ∈ ab we obtain that a ∈ cb∗ . Thus, c∗ ab ⊆ c∗ cb∗ b = [c, 1][b, 1] ⊆ [H, {1}]. Since c has been chosen arbitrarily among the elements of ab, this yields b∗ a∗ ab ⊆ [H, {1}]. Thus, as b∗ [a, 1]b = b∗ a∗ ab, b∗ [a, 1]b ⊆ [H, {1}]. Since a and b have been chosen arbitrarily among the elements of H, we have shown that [H, {1}] is strongly normal in H; cf. Lemma 2.3.4(ii). Thus, O ϑ (H) ⊆ [H, {1}]. To show the reverse containment, we recall from Theorem 4.4.1(i) that O ϑ (H) is strongly normal in H. Thus, we have [h, 1] = h∗ h ⊆ h∗ O ϑ (H)h ⊆ O ϑ (H) for each element h in H. Thus, as O ϑ (H) is closed, [H, {1}] ⊆ O ϑ (H).



Corollary 4.4.3 Let H be a hypergroup. Then the following statements hold. (i) The hypergroup H is thin if and only if O ϑ (H) = {1}. (ii) For each closed subset F of H, we have O ϑ (F) ⊆ O ϑ (H). Proof. (i) Note that H is thin if and only if [H, {1}] = {1}. Thus, by Lemma 4.4.2, H is thin if and only if O ϑ (H) = {1}.

4.4

The Thin Residue

(ii) Considering Lemma 4.4.2 this is a consequence of Lemma 2.4.1(ii).

103



Lemma 4.4.4 Let H be a hypergroup, and let D and E be closed subsets of H. Assume that E D is a closed subset of H and that O ϑ (E) ⊆ D. Then O ϑ (E D) is equal to the intersection of all strongly normal closed subsets of D which contain O ϑ (E). Proof. Define C to be the set of all strongly normal closed subsets of D which contain O ϑ (E), and let F denote the intersection of all elements of C. We have to show that O ϑ (E D) = F. We first show that O ϑ (E D) ⊆ F. For this, we let b be an element in E D and C an element in C. Since b ∈ E D, there exist elements e in E and d in D such that b ∈ ed. Since e ∈ E, e∗ e ⊆ O ϑ (E); cf. Lemma 4.4.2. From C ∈ C we obtain that d ∗ Cd ⊆ C and that O ϑ (E) ⊆ C. Thus, b∗ b ⊆ d ∗ e∗ ed ⊆ d ∗ O ϑ (E)d ⊆ C. Since b has been chosen arbitrarily among the elements of E D, we have shown that O ϑ (E D) ⊆ C; cf. Lemma 4.4.2. But also C has been chosen arbitrarily among the elements in C. Thus, we have shown that O ϑ (E D) ⊆ F. We now prove that F ⊆ O ϑ (E D). For this, we first note that O ϑ (E D) ⊆ F ⊆ D

and

D ⊆ E D ⊆ KH (O ϑ (E D)).

Thus, O ϑ (E D) is strongly normal in D. On the other hand, by Corollary 4.4.3(ii),  O ϑ (E) ⊆ O ϑ (E D). Thus, O ϑ (E D) ∈ C, so that, by definition, F ⊆ O ϑ (E D). Corollary 4.4.5 Let H be a hypergroup, and let D and E be closed subsets of H. Assume that E D is a closed subset of H and that E is thin. Then we have O ϑ (D) = O ϑ (E D). Proof. Since E is assumed to be thin, O ϑ (E) = {1}; cf. Corollary 4.4.3(i). Thus, the  claim follows from Lemma 4.4.4. Lemma 4.4.6 Let H be a hypergroup, let D and E be closed subsets of H, and assume that D ⊆ NH (E). Then EO ϑ (D) = EO ϑ (E D). Proof. We are assuming that D ⊆ NH (E). Thus, by Lemma 3.1.4(iii), E D is a closed subset of H. Thus, applying Corollary 4.4.3(ii) to E D in place of H we obtain that O ϑ (D) ⊆ O ϑ (E D). It follows that EO ϑ (D) ⊆ EO ϑ (E D). We are assuming that D ⊆ NH (E), and from Theorem 4.4.1(i) we know that O ϑ (D) is strongly normal in D. Thus, by Lemma 3.3.7, EO ϑ (D) is strongly normal in E D. Now the definition of O ϑ (E D) forces O ϑ (E D) ⊆ EO ϑ (D). Thus, as E is closed, we conclude that EO ϑ (E D) ⊆ EO ϑ (D). 

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4 Subnormality and Thin Residues

We continue with three observations about the thin residues of quotients. Lemma 4.4.7 Let H be a hypergroup, and let F be a closed subset of H. Then the following hold. (i) We have O ϑ (H//F) = F[H, F]//F. (ii) We have FO ϑ (H)//F ⊆ O ϑ (H//F). (iii) If F is normal in H, FO ϑ (H)//F = O ϑ (H//F). Proof. (i) From Lemma 4.4.2 we know that O ϑ (H//F) = [H//F, F//F], and in Lemma 3.5.2, we saw that [H//F, F//F] = F[H, F]//F. Thus, we have O ϑ (H//F) = F[H, F]//F. (ii) From Lemma 4.4.2 we know that O ϑ (H) = [H, {1}]. Thus, as [H, {1}] ⊆ [H, F], we obtain that O ϑ (H) ⊆ [H, F]. It follows that FO ϑ (H)//F ⊆ F[H, F]//F, so that the claim follows from (i). (iii) Assume that F is normal in H and recall from Theorem 4.4.1(i) that O ϑ (H) is strongly normal in H. Applying Lemma 3.3.7 to F, O ϑ (H), and H in place of C, D, and E we obtain that FO ϑ (H) is strongly normal in H. Thus, by Corollary 3.5.5(iii), FO ϑ (H)//F is strongly normal in H//F. It follows that O ϑ (H//F) ⊆ FO ϑ (H)//F, so that the desired equation follows from (ii).



Recall from Section 2.1 that we write Fh instead of h∗ h whenever h is a tight hypergroup element. Lemma 4.4.8 Let H be a hypergroup, and let h be a tight element in H. Then the following hold. (i) The set Fh is a closed subset of O ϑ (H). (ii) If h ∈ Fh , h = 1. Proof. (i) We are assuming that h is tight. Thus, by Lemma 2.1.3(i), Fh is a closed subset of H. From Lemma 4.4.2 we know that Fh ⊆ O ϑ (H). (ii) Assume that h ∈ Fh , and recall from (i) that Fh is closed. Then, h∗ ∈ Fh . It follows that hh∗ ⊆ hFh . On the other hand, since h is assumed to be tight, we have hFh = {h}. Thus, we have hh∗ = {h}. Now recall from Lemma 1.1.1 that 1 ∈ hh∗ . Thus, h = 1. 

4.5

Thin Residues of Thin Residues

105

4.5 Thin Residues of Thin Residues Let H be a hypergroup. We set (O ϑ )0 (H) := H. For each positive integer n, we inductively define (O ϑ )n (H) := O ϑ ((O ϑ )n−1 (H)). The first part of the following lemma generalizes Corollary 4.4.3(ii), its second part generalizes Lemma 4.4.7(iii). Lemma 4.5.1 Let H be a hypergroup, let F be a closed subset of H, and let n be a non-negative integer. Then the following hold. (i) We have (O ϑ )n (F) ⊆ (O ϑ )n (H). (ii) If F is normal in H, F(O ϑ )n (H)//F = (O ϑ )n (H//F). Proof. (i) If n = 0, we have (O ϑ )n (F) = F ⊆ H = (O ϑ )n (H), so we are done in this case. If 1 ≤ n and (O ϑ )n−1 (F) ⊆ (O ϑ )n−1 (H), we obtain from Corollary 4.4.3(ii) that (O ϑ )n (F) = O ϑ ((O ϑ )n−1 (F)) ⊆ O ϑ ((O ϑ )n−1 (H)) = (O ϑ )n (H), so that, by induction, we are also done in this case. (ii) If n = 0, we have F(O ϑ )n (H)//F = H//F = (O ϑ )n (H//F), as wanted. Assume that 1 ≤ n and that F is normal in H. Applying Lemma 4.4.6 to (O ϑ )n−1 (H) and F in place of D and E we obtain that FO ϑ ((O ϑ )n−1 (H)) = FO ϑ (F(O ϑ )n−1 (H)). Thus, as O ϑ ((O ϑ )n−1 (H)) = (O ϑ )n (H), F(O ϑ )n (H) = FO ϑ (F(O ϑ )n−1 (H)). It follows that

F(O ϑ )n (H)//F = FO ϑ (F(O ϑ )n−1 (H))//F.

On the other hand, applying Lemma 4.4.7(iii) to F(O ϑ )n−1 (H) in place of H we obtain that FO ϑ (F(O ϑ )n−1 (H))//F = O ϑ (F(O ϑ )n−1 (H)//F). Now assume by induction that F(O ϑ )n−1 (H)//F = (O ϑ )n−1 (H//F). Then

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O ϑ (F(O ϑ )n−1 (H)//F) = O ϑ ((O ϑ )n−1 (H//F)) = (O ϑ )n (H//F). The desired equation now follows from the last three centered equations.



The first part of the following lemma generalizes Lemma 4.4.6. Lemma 4.5.2 Let H be a hypergroup, let D and E be closed subsets of H, and assume that D ⊆ NH (E). Let n be a non-negative integer. Then the following hold. (i) We have E(O ϑ )n (D) = E(O ϑ )n (E D). (ii) If EO ϑ (E D) = E D, E(O ϑ )n (E D) = E D. Proof. (i) If n = 0, we have E(O ϑ )n (D) = E D = E(O ϑ )n (E D), so we are done in this case. Assume that 1 ≤ n. Applying Lemma 4.4.6 to (O ϑ )n−1 (D) in place of D we obtain E(O ϑ )n (D) = EO ϑ ((O ϑ )n−1 (D)) = EO ϑ (E(O ϑ )n−1 (D)). On the other hand, by induction, we may assume that E(O ϑ )n−1 (D) = E(O ϑ )n−1 (E D), so that

EO ϑ (E(O ϑ )n−1 (D)) = EO ϑ (E(O ϑ )n−1 (E D)).

We finally notice that, by Lemma 3.1.4(iv), E D ⊆ NH (E). It follows that (O ϑ )n−1 (E D) ⊆ NH (E). Thus, applying Lemma 4.4.6 to (O ϑ )n−1 (E D) in place of D we obtain E(O ϑ )n (E D) = EO ϑ ((O ϑ )n−1 (E D)) = EO ϑ (E(O ϑ )n−1 (E D)). It follows that E(O ϑ )n (D) = E(O ϑ )n (E D). (ii) There is nothing to show if n = 0. Therefore, we assume that 1 ≤ n. Then, by (i) (together with Lemma 3.1.1(ii)), (O ϑ )n−1 (E D) ⊆ NH (E). Thus, by Lemma 4.4.6, E(O ϑ )n (E D) = EO ϑ ((O ϑ )n−1 (E D)) = EO ϑ (E(O ϑ )n−1 (E D)). Now recall that we are assuming that EO ϑ (E D) = E D. Furthermore, by induction, we may assume that E(O ϑ )n−1 (E D) = E D. Thus, we have EO ϑ (E(O ϑ )n−1 (E D)) = E D. The desired equation now follows from the two centered equations.



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Lemma 4.5.3 Let H be a hypergroup, let D be a closed subset of H, and let E be a strong subnormal chain from D to H. Set n := |E |. Then (O ϑ )n−1 (H) ⊆ D. Proof. If n = 1, we have (O ϑ )n−1 (H) = H = D, and we are done. Assume that 2 ≤ n. Then, by induction, (O ϑ )n−2 (H) ⊆ D E . Thus, by Corollary 4.4.3(ii), (O ϑ )n−1 (H) = O ϑ ((O ϑ )n−2 (H)) ⊆ O ϑ (D E ). Since D is strongly normal in D E , we also have O ϑ (D E ) ⊆ D. Thus, we obtain that (O ϑ )n−1 (H) ⊆ D. 

4.6 Residually Thin Hypergroups A hypergroup H is called residually thin if there exists a non-negative integer n such that (O ϑ )n (H) = {1}. Let H be a residually thin hypergroup. The smallest non-negative integer n satisfying (O ϑ )n (H) = {1} is called the residual depth of H, and will be denoted by dp(H). The trivial hypergroup is the only residually thin hypergroup of residual depth 0. From Corollary 4.4.3(i) we obtain that the residually thin hypergroups of residual depth 1 are exactly the non-trivial thin hypergroups. At the end of this section, we will take a brief look at residually thin hypergroups of residual depth at most 2. We begin this section with a necessary and sufficient condition for a hypergroup to be residually thin. Theorem 4.6.1 A hypergroup is residually thin if and only if it possesses a strong subnormal series. Proof. Let H be a hypergroup, and assume first that H is residually thin. Set n := dp(H). Then, by Theorem 4.4.1(i), {(O ϑ )0 (H), . . . , (O ϑ )n (H)} is a strong subnormal series of H. Assume, conversely, that H possesses a strong subnormal series F , and set n := |F |. Then, by Lemma 4.5.3, (O ϑ )n−1 (H) = {1}.  Residually thin hypergroups do not necessarily possess composition series. Theorem 4.6.2 Let H be a residually thin hypergroup, and let F be a closed subset of H. Then we have the following. (i) The closed subset F of H is residually thin with dp(F) ≤ dp(H). (ii) If F is normal in H, H//F is residually thin with dp(H//F) ≤ dp(H). Proof. (i) This follows from Lemma 4.5.1(i).

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(ii) This follows from Lemma 4.5.1(ii).



Theorem 4.6.3 Let H be a hypergroup, and let D be a closed subset of H. Assume that D and H//D are residually thin. Then H is residually thin with dp(H) ≤ dp(D) + dp(H//D). Proof. Assume first that dp(H//D) = 0. Then D = H. Thus, the assumption that D is residually thin implies that H is residually thin. From D = H we also obtain that dp(D) = dp(H). Thus, as dp(H//D) = 0, we have dp(H) = dp(D) + dp(H//D). Assume now that 1 ≤ dp(H//D). Then, as H//D is assumed to be residually thin, we have O ϑ (H//D) , H//D. By Theorem 3.4.6, H contains closed subsets E such that D ⊆ E and E//D = O ϑ (H//D). Since E//D = O ϑ (H//D), E//D is residually thin with dp(E//D) = dp(H//D) − 1. Thus, by induction, E is residually thin with dp(E) ≤ dp(D) + dp(E//D). By Lemma 4.4.7(ii), DO ϑ (H)//D ⊆ O ϑ (H//D). Thus, as E//D = O ϑ (H//D), we conclude that DO ϑ (H)//D ⊆ E//D. It follows that O ϑ (H) ⊆ E. Define m := dp(E). Then (O ϑ )m (E) = {1}. Thus, as O ϑ (H) ⊆ E, Lemma 4.5.1(i) yields (O ϑ )m+1 (H) = (O ϑ )m (O ϑ (H)) ⊆ (O ϑ )m (E) = {1}. This shows that H is residually thin with dp(H) ≤ m + 1. Thus, since m = dp(E), we  conclude that dp(H) ≤ dp(D) + dp(H//D). Lemma 4.6.4 Let H be a hypergroup, and let D be a closed subset of H. Then the following hold. (i) If H is residually thin and D subnormal in H, H//D is residually thin. (ii) If H//D is residually thin, D is strongly subnormal in H. Proof. (i) There is nothing to show if D = H. Therefore, we assume that D , H. In this case, since D is assumed to be subnormal in H, H contains a normal closed subset E such that E , H and D is subnormal in E. Now assume that H is residually thin. Then, by Theorem 4.6.2(i), E is residually thin. Thus, by induction, E//D is residually thin. Since H is residually thin and E is a normal closed subset of H, we obtain from Theorem 4.6.2(ii) that H//E is residually thin. On the other hand, by Theorem 3.7.2, (H//D)//(E//D)  H//E. Thus, (H//D)//(E//D) is residually thin.

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Now, as both E//D and (H//D)//(E//D) are residually thin, H//D is residually thin; cf. Theorem 4.6.3. (ii) Assume that H//D is residually thin. Then, by Theorem 4.6.1, H//D possesses a strong subnormal series or, what is the same, a strong subnormal chain from D//D to H//D. By definition, that means that D//D is strongly subnormal in H//D. Thus, by Corollary 3.5.5(iii), D is strongly subnormal in H.  Corollary 4.6.5 Let H be a residually thin hypergroup, let D and E be closed subsets of H, and assume that D ⊆ E. Assume that D is subnormal in H and that E//D is subnormal in H//D. Then E is strongly subnormal in H. Proof. Since H is assumed to be residually thin and D to be subnormal in H, we obtain from Lemma 4.6.4(i) that H//D is residually thin. Thus, as E//D is assumed to be subnormal in H//D, E//D is strongly subnormal in H//D; cf. Lemma 4.6.4. Now we obtain from Corollary 3.5.5(iii) that E is strongly subnormal in H.  For the remainder of this section, we restrict our attention to residually thin hypergroups of residual depth at most 2, in other words, we focus on hypergroups H satisfying O ϑ (H) ⊆ Oϑ (H). Hypergroups H satisfying O ϑ (H) ⊆ Oϑ (H) are called metathin. Lemma 4.6.6 The residually thin hypergroups of residual depth at most 2 are exactly the metathin hypergroups. Proof. Let H be a hypergroup, and assume first that H is residually thin and has residual depth at most 2. Then, by definition, (O ϑ )2 (H) = {1}. Now recall that, for each element h in O ϑ (H), h∗ h ⊆ (O ϑ )2 (H); cf. Lemma 4.4.2. Thus, for each element h in O ϑ (H), h∗ h = {1}. This means that O ϑ (H) is thin; equivalently, O ϑ (H) ⊆ Oϑ (H). Assume, conversely, that H is metathin. Then O ϑ (H) ⊆ Oϑ (H), and that means that O ϑ (H) is thin. Thus, applying Corollary 4.4.3(i) to O ϑ (H) in place of H we obtain that (O ϑ )2 (H) = {1}.  Lemma 4.6.7 Metathin hypergroups are tight. Proof. Let H be a metathin hypergroup. Then, O ϑ (H) ⊆ Oϑ (H). It follows that, for each element h in H, h∗ h is thin; cf. Lemma 4.4.2. Thus, for each element h in H, h  is tight; cf. Corollary 1.4.4. This means that H is tight. Recall from Section 2.1 that we write Fh for h∗ h when h is a tight hypergroup element.

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Lemma 4.6.8 Let H be a metathin hypergroup, and let h be an element of H. Then the following hold. (i) The set Fh is a strongly normal closed subset of O ϑ (H). (ii) Let F be a closed subset of O ϑ (H). Then h∗ F h is a closed subset of O ϑ (H), too. Proof. (i) Since H is metathin, H is tight; cf. Lemma 4.6.7. Thus, by Lemma 4.4.8(i), Fh is a closed subset of O ϑ (H). Since H is assumed to be metathin, O ϑ (H) is thin. Thus, as Fh ⊆ O ϑ (H), Fh is thin, too. Since h ∈ H, we have {h∗, h} ⊆ NH (O ϑ (H)); cf. Theorem 4.4.1(i). Thus, applying Lemma 3.3.5(ii) to O ϑ (H) in place of F, we obtain that Fh is strongly normal in O ϑ (H). (ii) Since F is ∗ -invariant, so is h∗ F h; cf. Lemma 1.3.1(ii). From Lemma 4.6.7 we know that H is tight, from (i) we obtain that Fh∗ is normal in O ϑ (H). Thus, (h∗ F h)(h∗ F h) = h∗ FFh∗ F h = h∗ FFh∗ h = h∗ F h. Now, as 1 ∈ h∗ F h, Lemma 2.1.1 yields that h∗ F h is closed. From Theorem 4.4.1(i) we also obtain that h∗ F h ⊆ h∗ O ϑ (H)h ⊆ O ϑ (H).



Lemma 4.6.9 Let H be a metathin hypergroup, let a and b be elements in H, and let d and e be elements in ab. Then Fd = Fe . Proof. From d ∈ ab, we obtain that b ∈ a∗ d. Similarly, since e ∈ ab, b ∈ a∗ e. Thus, b ∈ a∗ d ∩ a∗ e. It follows that aa∗ ∩ de∗ is not empty; cf. Lemma 1.2.5(i). Let c be an element in aa∗ ∩ de∗ . Since c ∈ aa∗ , c ∈ O ϑ (H) ; cf. Lemma 4.4.2. Thus, as O ϑ (H) ⊆ Oϑ (H), c is thin. On the other hand, as c ∈ de∗ , d ∈ ce. Thus, by Lemma 1.2.1, Fd ⊆ e∗ c∗ ce = Fe . The reverse containment Fe ⊆ Fd is shown similarly.



4.7 Finite Residually Thin Hypergroups In this section, we restrict ourselves to finite residually thin hypergroups. We start with two necessary and sufficient conditions for a residually thin hypergroup to be finite. Theorem 4.7.1 Let H be a residually thin hypergroup. Then the following conditions are equivalent. (a) The hypergroup H is finite.

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(b) The hypergroup H possesses a strong composition series, and each composition factor of H is finite. (c) The hypergroup H possesses a strong subnormal series E such that, for each element E in E \ {H}, E E //E is finite. Proof. (a) ⇒ (b) Since H is assumed to be finite, H possesses a composition series; cf. Lemma 4.3.1(ii). Let E be a composition series of H, and let E be an element in E \ {H}. Then E E is a closed subset of H, so that, by Theorem 4.6.2(i), E E is residually thin. Since E E is residually thin and E is normal in E E , E E //E is residually thin; cf. Theorem 4.6.2(ii). On the other hand, by Lemma 4.3.3(i), E E //E is simple. Thus, by Theorem 4.6.1, E E //E is thin. It follows that E is strongly normal in E E ; cf. Lemma 3.5.6(ii). Since E has been chosen arbitrarily in E \ {H}, E is a strong composition series. Since H is assumed to be finite, all composition factors of H are finite. (b) ⇒ (c) Let E be a strong composition series of H, and assume that each composition factor of H is finite. Since E is a strong composition series of H, E is a strong subnormal series of H. From the fact that E is a composition series of H we also obtain that, for each element E in E \ {H}, E E //E is a composition factor of H. Thus, as all composition factors of H are assumed to be finite, E E //E is finite for each element E in E \ {H}. (c) ⇒ (a) There is nothing to show if H is trivial. Therefore, we assume that H is not trivial and possesses a strong subnormal series E such that, for each element E in E \ {H}, E E //E is finite. Let D denote the intersection of all elements in E \ {{1}}. Then E \ {{1}} is a strong subnormal chain from D to H. Thus, by Lemma 4.1.6, {E//D | E ∈ E \ {{1}}} is a strong subnormal series of H//D. For each element E in E \ {{1}, H}, we have (E E //D)//(E//D)  E E //E; cf. Theorem 3.7.2. On the other hand, for each element E in E \ {{1}, H}, E E //E is finite. Thus, for each element E in E \ {{1}, H}, (E E //D)//(E//D) is finite. Now, as {E//D | E ∈ E \ {{1}}} is a strong subnormal series of H//D and, for each element E in E \ {{1}, H}, (E E //D)//(E//D) is finite, induction yields that H//D is finite. From Lemma 2.1.6(i) we know that {DhD | h ∈ H} is a partition of H. Thus, as H//D = {DhD | h ∈ H} and H//D is finite, it suffices to show that, for each element h in H, |DhD| is finite. Let h be an element in H. Since D is thin, we obtain from Lemma 1.4.3(i) that |hD| ≤ |D|. Thus, as |D| is finite, |hD| is finite, too. It follows that, for each element e in D, |ehD| ≤ |hD|; cf. Lemma 1.4.7(i). Thus, we conclude that |DhD| ≤ |D| 2 , so that, since |D| is finite, |DhD| is finite. 

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Let H be a finite residually thin hypergroup, and let F be a strong subnormal series of H. Then, for each element F in F \ {H}, F F //F is finite. If H = {1}, we set n F := 1. If H , {1}, we set Ö |F F //F |. n F := F ∈ F\{H }

We will now see that, for any two strong subnormal series D and E of H, n D = n E . Lemma 4.7.2 Let H be a finite residually thin hypergroup, and let B and C be strong subnormal series of H. Then n B = n C . Proof. We first assume that B ⊆ C. By induction, we may assume that C \ B contains an element C such that B ∪ {C} = C. Then, B \ {H} contains an element B such that B ⊆ C ⊆ B B . Since B is a strong subnormal series of H, B is strongly normal in B B . Thus, by Lemma 3.5.6(ii), B B //B is thin. Thus, as C//B is a closed subset of B B //B, we conclude that |B B //B| = |(B B //B)/(C//B)| · |C//B|; cf. Lemma 2.6.4(iii). Since C is normal in B B we also obtain that C//B is normal in B B //B; cf. Corollary 3.5.5(i). Thus, |(B B //B)//(C//B)| = |(B B //B)/(C//B)|. Now recall from Theorem 3.7.2 that (B B //B)//(C//B)  B B //C. Thus, |(B B //B)//(C//B)| = |B B //C|. It follows that

|B B //B| = |B B //C| · |C//B|,

and this equation implies that n B = n C . Now we drop the extra condition that B ⊆ C. We let D be a composition series which contains B, and we let E be a composition series which contains C; cf. Lemma 4.3.1(i). From Theorem 4.7.1 we know that H possesses a strong composition series. We choose one of the strong composition series of H and denote it by F . By Theorem 4.3.2, D and E both are isomorphic to F . Thus, D and E both are strong. Now the first part of the proof yields n B = n D and n C = n E . Since D and E both are isomorphic to F , D and E are isomophic to each other. Thus, n D = n E . It follows that n B = n C .  Let H be a finite residually thin hypergroup. By Theorem 4.7.1, H possesses a strong subnormal series. We define nH to be the arithmetic mean of the integers n F where F is a strong subnormal series of H. From Lemma 4.7.2 we know that nH = n F for each strong subnormal series F of H.

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Note that each thin hypergroup H satisfies nH = |H|. Thus, the following lemma generalizes part of Lemma 2.6.4(iii). Lemma 4.7.3 Let H be a finite residually thin hypergroup, and let C be a closed subset of H. Then the following hold. (i) The integer nC divides nH . (ii) If C is subnormal in H, nH = nH//C nC . Proof. (i) The claim is obviously true if H = {1}. Therefore, we assume that H , {1}. Since H is a finite residually thin hypergroup different from {1}, H possesses a strong subnormal series E such that, for each element E in E \ {H}, E E //E is finite; cf. Theorem 4.7.1. By definition, we have Ö |E E //E |. nH = E ∈E\{H }

From the fact that E is a strong subnormal series of H, we also obtain that {C∩E | E ∈ E} is a strong subnormal series of C; cf. Lemma 4.2.2(i). Thus, we also have Ö |(C ∩ E E )//(C ∩ E)|. nC = E ∈E\{H }

Now recall from Lemma 3.7.6 that, for each element E in E \{H}, |(C∩E E )//(C∩E)| divides |E E //E |. Thus, by the above two equations, nC divides nH . (ii) There is nothing to show if {1} = C or C = H. Therefore, we assume that {1} , C , H. Since H is a residually thin hypergroup and C a subnormal closed subset of H, C is strongly subnormal in H; cf. Lemma 4.6.4. Thus, we find a strong subnormal chain E from C to H. Thus, by Lemma 4.1.6, {E//C | E ∈ E} is a strong subnormal series of H//C. It follows that Ö nH//C = |(E E //C)//(E//C)|. E ∈E\{H }

On the other hand, by Theorem 3.7.2, (E E //C)//(E//C)  E E //E for each element E in E \ {H}. Thus, for each element E in E \ {H}, |(E E //C)//(E//C)| = |E E //E |. It follows that

nH//C =

Ö E ∈E\{H }

|E E //E |.

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Since E is residually thin, so is C; cf. Theorem 4.6.2(i). Thus, by Theorem 4.6.1, C possesses a strong subnormal series D. Thus, as {1} , C, Ö |D D //D|. nC = D ∈ D\{C }

Note also that D ∪ E is a strong subnormal series of H, so Ö Ö |D D //D| |E E //E |. nH = D ∈D\{C }

E ∈E\{H }

Now the above three equations yield nH = nH//C nC .



Lemma 4.7.4 Let H be a finite residually thin hypergroup, let D and E be closed subsets of H, and assume that D ⊆ NH (E). Then nE nD = nE D nE∩D . Proof. Since D ⊆ NH (E), E D is a closed subset of H; cf. Lemma 3.1.4(iii). Thus, by Theorem 4.6.2(i), E D is residually thin. Now recall from Lemma 3.1.4(iv) that E is normal in E D. Thus, by Theorem 4.6.2(ii), E D//E is residually thin. On the other hand, by Theorem 3.7.3, E D//E  D//(E ∩ D). Thus, D//(E ∩ D) is residually thin and nE D//E = nD//(E∩D) . Since E is normal in E D, we obtain from Lemma 4.7.3(ii) that nE D = nE D//E nE . From Lemma 3.1.5(i) we also obtain that E ∩ D is normal in D, so that, again referring to Lemma 4.7.3(ii), nD = nD//(E∩D) nE∩D . From the latter three equations we now obtain that nE nD = nE nD//(E∩D) nE∩D = nE D//E nE nE∩D = nE D nE∩D, as wanted.



Let π be a set of prime numbers. A positive integer all prime divisors of which belong to π is called a π-number. A positive integer no prime divisor of which belongs to π is called a π 0-number. A closed subset F of a finite residually thin hypergroup H is called a closed π-subset of H if nF is a π-number. Lemma 4.7.5 Let π be a set of prime numbers, and let H be a finite residually thin hypergroup. Assume that H has no non-trivial subnormal closed π-subset. Assume also that, for each element h in H with Fh ⊆ Oϑ (H), Fh is a closed π-subset of H. Then H is thin.

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Proof. Set n := dp(H), and assume first that 2 ≤ n. We set F := (O ϑ )n−2 (H). Then F is residually thin and dp(F) = 2. Thus, by Lemma 4.6.6, F is metathin. Let f be an element in F. Since F is metathin, Ff is a strongly normal closed subset of O ϑ (F); cf. Lemma 4.6.8(i). Thus, as O ϑ (F) is normal in F and F is subnormal in H, Ff is subnormal in H. On the other hand, since Ff ⊆ O ϑ (F) ⊆ Oϑ (F) ⊆ Oϑ (H), one of the hypotheses of our lemma forces Ff to be a closed π-subset of H. Now, as Ff is subnormal in H, the other hypotheses of our lemma forces Ff = {1}, and that means that f is thin. Since f has been chosen arbitrarily from F, we have shown that F is thin. Thus, by Corollary 4.4.3(i), dp(F) ≤ 1, contradiction. Thus, we have n ≤ 1, and that means that H is thin; cf. Corollary 4.4.3(i).



We conclude this section with two results of H. Blau; cf. [6; Theorem 1.2] and [6; Theorem 1.3]. They both are based on the following terminology which was introduced in [6; Definition 1.3]. Let π be a set of prime numbers, and let H be a finite residually thin hypergroup. An element h in H is called π-valenced if, for each subnormal closed π-subset F of H with Fh F ⊆ Oϑ (H//F), Fh F is a closed π-subset of H//F. The hypergroup H is called π-valenced if each element of H is π-valenced. Theorem 4.7.6 Let π be a set of prime numbers, and let H be a finite residually thin hypergroup. Assume that H is π-valenced. Then H contains a strongly normal closed π-subset which contains all subnormal closed π-subsets of H. Proof. Let D be maximal among the subnormal closed π-subsets of H. By hypothesis, H is residually thin. Thus, as D is subnormal in H, Lemma 4.6.4(i) tells us that H//D is residually thin. Let E be a closed subset of H with D ⊆ E, and assume that E//D is a subnormal closed π-subset of H//D. Since D is subnormal in H and E//D is subnormal in H//D, we obtain from Corollary 4.6.5 that E is strongly subnormal in H, and then also subnormal in H. On the other hand, nD as well as nE//D are π-numbers. Thus, by Lemma 4.7.3(ii), nE is a π-number. Thus, E is a closed π-subset of H. Now the choice of D forces D = E, and, since E has been chosen arbitrarily among the closed subsets of H with D ⊆ E and such that E//D is a subnormal closed π-subset of H//D, this shows that H//D has no non-trivial subnormal closed π-subset. Now recall that H is assumed to be π-valenced. This, means that each element of H is π-valenced. It follows that, for each element h in H with Fh D ⊆ Oϑ (H//D), Fh D is a closed π-subset of H//D.

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4 Subnormality and Thin Residues

Considering the last two paragraphs we now obtain from Lemma 4.7.5 that H//D is thin. Thus, by Lemma 3.5.6(ii), D is strongly normal in H, and it remains to show that D contains all subnormal closed π-subsets of H. Let C be a subnormal closed π-subset of H. Then, by Corollary 4.1.2(iv), DC is subnormal in H. On the other hand, by Lemma 4.7.4, nD nC = nDC nD∩C . Thus, as D and C are closed π-subsets of H, DC is a closed π-subset of H, too. Since D ⊆ DC, the choice of D now forces C ⊆ D.  Let π be a set of prime numbers, and let H be a finite residually thin hypergroup. Assume that H is π-valenced. In Theorem 4.7.6, we saw that H contains a strongly normal closed π-subset which contains all subnormal closed π-subsets of H. In the following, this closed subset will be denoted by Oπ (H). 0 A closed π-subset F of H is called a Hall π-subset if nH n−1 F is a π -number.

Theorem 4.7.7 Let π be a set of prime numbers, and let H be a finite residually thin hypergroup. Assume that H is π-valenced. Then we have the following. (i) Each Hall π-subset of H contains Oπ (H). (ii) The map E 7→ E//Oπ (H) is a bijective map from the set of all closed π-subsets of H containing Oπ (H) to the set of all closed π-subsets of H//Oπ (H). (iii) The map E 7→ E//Oπ (H) is a bijective map from the set of all Hall π-subsets of H to the set of all Hall π-subsets of H//Oπ (H). (iv) Each π-subset of H is contained in a Hall π-subset of H if and only if each π-subset of H//Oπ (H) is contained in a Hall π-subset of H//Oπ (H). (v) Any two Hall π-subsets of H are conjugate in H if and only if any two Hall π-subsets of H//Oπ (H) are conjugate in H//Oπ (H). Proof. Set C := Oπ (H). (i) Let E be a Hall π-subset of H. We have to show that C ⊆ E. Since C is strongly normal in H, CE is a closed subset of H; cf. Lemma 3.1.4(iii). Both C and E are closed π-subsets. Thus, by Lemma 4.7.4, CE is a closed π-subset. Since E is a closed subset of CE, we obtain from Lemma 4.7.3(i) that nE divides nCE . Thus, as nCE is a π-number, so is nCE n−1 E . Since CE is a closed subset of H, we obtain from Lemma 4.7.3(i) that nCE divides −1 nH . Thus, as nE divides nCE , nCE n−1 E divides n H nE . Since E is a Hall π-subset of 0 −1 −1 0 H, nH nE is a π -number. Thus, nCE nE is a π -number. 0 −1 Since nCE n−1 E is a π-number and, at the same time, a π -number, nCE nE = 1, and that means that nCE = nE . Thus, as E ⊆ CE, E = CE, whence C ⊆ E.

4.8

Solvable Hypergroups

117

(ii) Let E be a closed subset of H with C ⊆ E, and recall that nC is a π-number. From Lemma 4.7.3(ii) we know that nE = nE//C nC . Thus, E is a closed π-subset of H if and only if E//C is a closed π-subset of H//C. (iii) Let E be a closed subset of H with C ⊆ E. From Lemma 4.7.3(ii) we know that nE = nE//C nC and nH = nH//C nC . Thus, −1 nH n−1 E = n H//C nE//C 0 0 −1 which shows that nH n−1 E is a π -number if and only if n H//C nE//C is a π -number. Now the claim follows from (ii).

(iv) Assume first that each closed π-subset of H is contained in a Hall π-subset of H. We let D be a closed subset of H with C ⊆ D, and we assume that D//C is a closed π-subset of H//C. Since C is strongly normal in H, we obtain from Lemma 4.7.3(ii) that nD = nD//C nC . Since D//C is a closed π-subset of H//C and C is a closed π-subset of H, we obtain from nD = nD//C nC that D is a closed π-subset of H. Thus, by assumption, H contains a Hall π-subset E with D ⊆ E. From (iii) we obtain that E//C is a Hall π-subset of H//C. From D ⊆ E we obtain that D//C ⊆ E//C. Assume now that each closed π-subset of H//C is contained in a Hall π-subset of H//C. We let D be a closed π-subset of H. Since C is strongly normal in H, CD is a closed subset of H; cf. Lemma 3.1.4(iii). It follows that CD//C is a closed subset of H//C; cf. Theorem 3.4.6. From Lemma 4.7.4 we know that nC nD = nCD nC∩D . Thus, as C and D are closed π subsets of H, CD is a closed π subset of H. Thus, by Lemma 4.7.3(ii), CD//C is a closed π subset of H//C. Thus, by assumption, H contains a closed subset E with C ⊆ E such that E//C is a Hall π-subset of H//C with CD//C ⊆ E//C. From (iii) we obtain that E is a Hall π-subset of H. From CD//C ⊆ E//C we obtain that D ⊆ E. (v) Considering (iii) this follows from Lemma 3.5.1.



4.8 Solvable Hypergroups A hypergroup is called solvable if it possesses a composition series and all of its composition factors are thin and have prime order. From Theorem 4.6.1 one obtains that solvable hypergroups are residually thin. Thus, by Theorem 4.7.1, solvable hypergroups are finite. Theorem 4.8.1 Closed subsets of solvable hypergroups are solvable.

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4 Subnormality and Thin Residues

Proof. Let H be a solvable hypergroup, and let D be a closed subset of H. We will see that D is solvable. Let E be a composition series of H. From Lemma 4.2.1(i) we know that {D ∩ E | E ∈ E} is a subnormal series of D. Let E be an element in E \ {H}. Since E is a composition series of H, E E //E is thin and has prime order; cf. Theorem 4.3.2. On the other hand, as E ⊆ E(D ∩ E E ) ⊆ E E , E(D ∩ E E )//E is a closed subset of E E //E; cf. Theorem 3.4.6. Thus, by Lemma 2.6.4(iii), E//E = E(D ∩ E E )//E

or

E(D ∩ E E )//E = E E //E.

In the first case, we obtain that D ∩ E = D ∩ E E . In the second case, we recall from Theorem 3.7.3 that E(D ∩ E E )//E  (D ∩ E E )//(D ∩ E). Thus,

E E //E  (D ∩ E E )//(D ∩ E).

Since E E //E is thin and has prime order, this shows that (D ∩ E E )//(D ∩ E) is thin and has prime order. Since E has been chosen arbitrarily from E \{H}, we have seen that, for each element E of E \ {H}, the quotient (D ∩ E E )//(D ∩ E) is either trivial or thin and of prime order. It follows that {D ∩ E | E ∈ E} is a composition series of D, and then, by Theorem 4.3.2, that all composition factors of D are thin and have prime order. This means that D is solvable.  The following theorem is similar to Theorem 4.6.3. Theorem 4.8.2 Let H be a hypergroup. Assume that H contains a solvable closed subset D such that H//D is solvable. Then H is solvable. Proof. Let D be a solvable closed subset of H, and assume that H//D is solvable. There is nothing to show if D = H. Thus, we assume that D , H. Since D , H, H//D , D//D. Thus, as H//D is solvable, H contains a closed subset E with D ⊆ E such that (H//D)//(E//D) is thin and has prime order; cf. Theorem 3.4.6. Now recall from Theorem 3.7.2 that (H//D)//(E//D)  H//E. Thus, H//E is thin and has prime order. Since H//D is assumed to be solvable, E//D is solvable; cf. Theorem 4.8.1. Thus, as D is a solvable closed subset of E, induction forces E to be solvable. Since E is normal in H, each composition factor of H is a composition factor of E or a composition factor of H//E; cf. Theorem 4.3.6(iii). Since E is solvable, all

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119

composition factors of E are thin and have prime order. Thus, as H//E is thin and has prime order, all composition factors of H are thin and have prime order, and this  means that H is solvable. Statement and proof of the following corollary are similar to statement and proof of Lemma 4.6.4(i). Corollary 4.8.3 Let H be a solvable hypergroup, and let D be a subnormal closed subset of H. Then H//D is solvable. Proof. There is nothing to show if D = H. Therefore, we assume that D , H. In this case, since D is subnormal in H, H contains a normal closed subset E such that E , H and D is subnormal in E. From Theorem 4.8.1 we know that E is solvable. Thus, as D is subnormal in E, induction forces E//D to be solvable. Since H is solvable and E is a normal closed subset of H, H//E is solvable; cf. Theorem 4.3.6(ii). On the other hand, by Theorem 3.7.2, (H//D)//(E//D)  H//E. Thus, (H//D)//(E//D) is solvable. Now, since E//D and (H//D)//(E//D) both are solvable, H//D is solvable; cf. Theorem 4.8.2. 

5 Tight Hypergroups

Recall from the Preface that a hypergroup element h is called tight if hh∗ h = {h}. Recall also that hypergroups are called tight if all of their elements are tight. In this chapter, we study finite tight hypergroups. Key to our analysis is Theorem 5.1.5, which states that finite tight hypergroups are residually thin. This provides us with a useful invariant which can be associated to each finite tight hypergroups, its residual depth, and it is this invariant which determines our study of finite tight hypergroups. Let H be a finite tight hypergroup, and let n denote its residual depth. Our analysis focuses on the set S of all n-tuples ( f1, . . . , fn ) consisting of elements of O ϑ (H) such that, for each element i in {1, . . . , n} with 2 ≤ i, fi ∈ Ffi−1 . Recall from Section 2.1 that Fh stands for the product h∗ h whenever h is a tight hypergroup element. Thus, one obtains from Lemma 4.4.2 that, for each n-tuple ( f1, . . . , fn ) in S and any element i in {1, . . . , n}, fi ∈ (O ϑ )i (H). (We show this in Lemma 5.2.1(i).) In particular fn ∈ (O ϑ )n (H), so that, since n stands for the residual depth of H, fn = 1. The (finite) set S will be studied in Sections 5.2, 5.3, 5.4, 5.5, and 5.6. Under the hypothesis that all closed subsets Fh with h ∈ H are normal in O ϑ (H) it gives rise to an array of non-negative integers which are reminiscent of the structure constants of an association scheme on a finite set; cf. Section 1.8. In Section 5.6 we will see that these structure constants are the structure constants of an associative ring which we associate to each finite tight hypergroup H such that, for every element h in H, Fh is normal in O ϑ (H) and to every commutative ring (with 1). If H is an association scheme, this ring is the scheme ring in the sense of [55; Section 9.1]. If H is thin, this ring coincides with the group ring of the group which, via the group correspondence, corresponds to H.1 While finite tight hypergroups are residually thin, finite residually thin hypergroups are not necessarily tight. (An example is given right after Theorem 5.1.5.) However, in Lemma 4.6.7 (together with Lemma 4.6.6), we saw that residually thin hypergroups 1There is of course no reason to expect that hypergroups in general give rise to associative rings the same way as groups give rise to group rings. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 P. -H. Zieschang, Hypergroups, https://doi.org/10.1007/978-3-031-39489-8_5

121

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5 Tight Hypergroups

are tight if their residual depth is at most 2. We will consider (mostly finite) residually thin hypergroups of residual depth at most 2 in the final two sections of this chapter, in Sections 5.7 and 5.8. Our approach in Sections 5.2, 5.3, 5.4, 5.5, and 5.6 follows [22].

5.1 Tight Hypergroup Elements We begin this section with two results on tight hypergroup elements. After Lemma 5.1.2, we consider tight hypergroups. Recall from Section 2.1 that we write Fh instead of h∗ h whenever h is a tight hypergroup element. Lemma 5.1.1 Let H be a hypergroup, and let a, b, and c be tight elements in H with c ∈ ab. Then b∗ Fa b = Fc Fb . Proof. From c ∈ ab we obtain that b∗ ∈ c∗ a and a ∈ cb∗ ; cf. Lemma 1.2.1. Thus, as a is tight, we have b∗ Fa b ⊆ c∗ aFa b = c∗ ab ⊆ c∗ cb∗ b = Fc Fb . From c ∈ ab we also obtain that c∗ ∈ b∗ a∗ ; cf. Lemma 1.2.1. Thus, as b is tight, we have Fc Fb = c∗ cb∗ b ⊆ c∗ abb∗ b = c∗ ab ⊆ b∗ a∗ ab = b∗ Fa b. Thus, b∗ Fa b = Fc Fb .



Let H be a hypergroup. From Lemma 5.1.1 one obtains that Fd Fb = Fe Fb for any four tight elements a, b, c, and d of H satisfying {d, e} ⊆ ab. Lemma 5.1.2 Let H be a hypergroup, and let a and b be tight elements in H. Assume that ab∗ contains a tight element. Then Fa Fb = Fb Fa . Proof. Let c be a tight element in ab∗ . From c ∈ ab∗ we obtain that a ∈ cb. Thus, applying Lemma 5.1.1 to c and a in place of a and c we obtain that b∗ Fc b = Fa Fb . Since c is tight, Fc is closed; cf. Lemma 2.1.3(i). As a consequence, Fc is ∗ -invariant. It follows that b∗ Fc b is ∗ -invariant; cf. Lemma 1.3.1(ii). Since b∗ Fc b = Fa Fb , this  implies that Fa Fb is ∗ -invariant, so that, by Lemma 2.1.5, Fa Fb = Fb Fa . In the remaining part of this section, we will look at tight hypergroups. Lemma 5.1.3 Let H be a tight hypergroup. Then we have Fa bFa b∗ Fa bFa = Fa bFa for any two elements a and b in H.

5.1

Tight Hypergroup Elements

123

Proof. Let a and b be elements in H. From Lemma 1.1.3 we obtain an element c in ab. Note first that, by Lemma 5.1.1 and Lemma 5.1.2, Fa bFa b∗ Fa bFa = Fa bFa Fc Fb Fa = Fa bFc Fb Fa . From the same references we also obtain that Fa bFc Fb Fa = Fa bb∗ Fa bFa = Fa Fb∗ Fa bFa = Fa Fb∗ bFa . Thus, as Fb∗ b = {b}, we conclude that Fa bFa b∗ Fa bFa = Fa bFa .



Lemma 5.1.4 Let H be a tight hypergroup, and let h be an element of H. Then the following hold. (i) The quotient H//Fh is tight. (ii) We have O ϑ (H)//Fh = O ϑ (H//Fh ). Proof. (i) Let a be an element in H. Then, by Lemma 5.1.3, Fh aFh a∗ Fh aFh = Fh aFh . Thus, by Lemma 3.4.4, a Fh (a Fh )∗ a Fh = a Fh (a∗ )Fh a Fh = {a Fh }, and that says that a Fh is tight. (ii) From Lemma 4.4.8(i) we know that Fh is a closed subset of O ϑ (H). Thus, by Lemma 4.4.7(ii), O ϑ (H)//Fh ⊆ O ϑ (H//Fh ). To show the reverse containment, we let a and b be elements in H with bFh ∈ (a Fh )∗ (a Fh ). Then, by Lemma 3.4.4, b ∈ Fh a∗ Fh aFh . Now choose an element c in ha. Then, by Lemma 5.1.1, a∗ Fh a = Fc Fa . Thus, we obtain that b ∈ Fh Fc Fa Fh , so that, by Lemma 4.4.2, b ∈ O ϑ (H). It follows that bFh ∈ O ϑ (H)//Fh . Since b has been chosen arbitrarily in H with bFh ∈ (a Fh )∗ (a Fh ), we have shown that (a Fh )∗ (a Fh ) ⊆ O ϑ (H)//Fh . Now, since a has been chosen arbitrarily in H, Lemma  4.4.2 yields O ϑ (H//Fh ) ⊆ O ϑ (H)//Fh . The following theorem is [21; Theorem 7.7]. Theorem 5.1.5 Finite tight hypergroups are residually thin. Proof. Let H be a finite tight hypergroup. There is nothing to show if H is thin. Thus, we assume that H is not thin. We choose a non-thin element in H and denote it by h. Since h is not thin, Fh , {1}. From Lemma 5.1.4(i) we know that H//Fh is tight. From Fh , {1} we also obtain that |H//Fh | ≤ |H| − 1. Thus, by induction, H//Fh is residually thin.

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5 Tight Hypergroups

Since h is not thin, h , 1. Thus, by Lemma 4.4.8(ii), h < Fh . Since h ∈ H, this implies that Fh , H. Thus, by Theorem 3.4.6, Fh //Fh , H//Fh . Since H//Fh is residually thin, this implies that O ϑ (H//Fh ) , H//Fh . On the other hand, by Lemma 5.1.4(ii), O ϑ (H)//Fh = O ϑ (H//Fh ). It follows that O ϑ (H)//Fh , H//Fh , whence, by Theorem 3.4.6, O ϑ (H) , H. Now  induction forces O ϑ (H) to be residually thin. Thus, H is residually thin, too. Theorem 5.1.5 implies that finite non-trivial tight hypergroups always have thin elements different from 1. The converse of Theorem 5.1.5 does not hold. Here is an example of a finite residually thin hypergroup which is not tight.

1 a b d c

1 {1} {a} {b} {d} {c}

a {a} {1} {b} {c} {d}

b {b} {b} {1, a} {c, d} {c, d}

c {c} {d} {c, d} {1, b} {a, b}

d {d} {c} {c, d} {a, b} {1, b}

Its residual depth is 3. From Lemma 4.6.6 (together with Lemma 4.6.7) we know that there is no finite residually thin hypergroup of residual depth less than 3 which is not tight. Lemma 5.1.6 Let H be a finite tight hypergroup. Then O ϑ (H) is the product of the closed subsets Fh with h ∈ H. Proof. From Lemma 4.4.2 we know that O ϑ (H) is the smallest closed subset of H containing all closed subset Fh with h ∈ H. From Lemma 5.1.2, together with Lemma 2.1.5, we know that the product of all closed subset Fh with h ∈ H is closed. Thus, O ϑ (H) is the product of all closed subsets Fh with h ∈ H.  Lemma 5.1.7 Let H be a tight hypergroup, let a, b, and c be elements in H, and assume that ca ∩ ab is not empty. Assume further that Fa∗ c ⊆ cFa∗ . Then the following hold. (i) We have ab ⊆ ca. (ii) We have aFb ⊆ Fc a. Proof. (i) We are assuming that ca∩ab is not empty. Thus, by Lemma 1.3.3, b ∈ a∗ ca, and from this we obtain that

5.2

The Set S

125

ab ⊆ Fa∗ ca. Since we are assuming that Fa∗ c ⊆ cFa∗ , we also have Fa∗ ca ⊆ cFa∗ a = ca. As a result, we obtain that ab ⊆ ca. (ii) Since ca ∩ ab is assumed not to be empty, we have b ∈ a∗ ca; cf. Lemma 1.3.3. From this we obtain that Fb ⊆ a∗ c∗ Fa∗ ca. Since we are assuming that Fa∗ c ⊆ cFa∗ , we also have a∗ c∗ Fa∗ ca ⊆ a∗ c∗ cFa∗ a = a∗ Fc a. As a result, we obtain that Fb ⊆ a∗ Fc a. Now recall from Lemma 5.1.2 that Fa∗ Fc = Fc Fa∗ . Thus, we have aFb ⊆ Fa∗ Fc a = Fc Fa∗ a = Fc a.  Lemma 5.1.8 Let H be a tight hypergroup, and let a and b be elements in H. Assume further that a∗ ba contains an element c with Fa c ⊆ cFa . Then, for each element d in a∗ b, a∗ ba = da. Proof. We are assuming that a∗ ba contains an element c with Fa c ⊆ cFa . From c ∈ a∗ ba we obtain that ca∗ ∩ a∗ b is not empty; cf. Lemma 1.2.5(iii). Thus, applying Lemma 5.1.7(i) to a∗ in place of a we obtain that a∗ b ⊆ ca∗ . Let d be an element in a∗ b. Then, as a∗ b ⊆ ca∗ , d ∈ ca∗ . It follows that c ∈ da. Thus, as a∗ b ⊆ ca∗ , a∗ b ⊆ dFa∗ . It follows that a∗ ba ⊆ dFa∗ a = da. The reverse containment follows from d ∈ a∗ b.



5.2 The Set S This section is the beginning of a more thorough study of finite tight hypergroups. We fix a finite tight hypergroup and denote it by H. From Theorem 5.1.5 we know that H is residually thin. We set n := dp(H) and assume that 1 ≤ n. As in Section 5.1, we write Fh instead of h∗ h whenever h stands for an element of H. (This is because H is tight.) For each closed subset F of O ϑ (H), we define S(F) to be the set of all n-tuples ( f1, . . . , fn ) consisting of elements of H satisfying f1 ∈ F and, for each element i in {1, . . . , n} with 2 ≤ i, fi ∈ Ffi−1 . Instead of S(O ϑ (H)) we simply write S.

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5 Tight Hypergroups

Lemma 5.2.1 Let ( f1, . . . , fn ) be an element in S. Then the following hold. (i) For each element i in {1, . . . , n}, we have fi ∈ (O ϑ )i (H). (ii) We have fn = 1. (iii) Assume that 2 ≤ n. Then { f2, . . . , fn } ⊆ Ff1 . Proof. (i) Since ( f1, . . . , fn ) ∈ S, we have f1 ∈ O ϑ (H). Thus, as O ϑ (H) = (O ϑ )1 (H), we have f1 ∈ (O ϑ )1 (H). Let i be an element in {1, . . . , n}, and assume that 2 ≤ i. Suppose, by induction, that fi−1 ∈ (O ϑ )i−1 (H). Then we have Ffi−1 ⊆ O ϑ ((O ϑ )i−1 (H)); cf. Lemma 4.4.2. Thus, as fi ∈ Ffi−1 and O ϑ ((O ϑ )i−1 (H)) = (O ϑ )i (H), we conclude that fi ∈ (O ϑ )i (H). (ii) Since dp(H) = n, we have (O ϑ )n (H) = {1}. Thus, by (i), fn = 1. (iii) Let i be an element in {2, . . . , n}. If i = 2, we obtain fi ∈ Ff1 immediately from the definition of S. Assume that 3 ≤ i and suppose, by induction, that fi−1 ∈ Ff1 . Then Ffi−1 ⊆ Ff1 . Since ( f1, . . . , fn ) ∈ S, we also have fi ∈ Ffi−1 . Thus, fi ∈ Ff1 .  Corollary 5.2.2 Let F be a closed subset of O ϑ (H), and let ( f1, . . . , fn ) be an element in S(F). Then { f1, . . . , fn } ⊆ F. Proof. Since ( f1, . . . , fn ) ∈ S(F), we have f1 ∈ F, and from f1 ∈ F we obtain that Ff1 ⊆ F. Thus, by Lemma 5.2.1(iii), { f2, . . . , fn } ⊆ F.  We set ∂(1) := (1). If 2 ≤ n, we define, for each element ( f1, . . . , fn ) in S, ∂( f1, . . . , fn ) := ( f2, . . . , fn, fn ). We first show that ∂ is a map from S to S. Lemma 5.2.3 For each element ( f1, . . . , fn ) in S, ∂( f1, . . . , fn ) ∈ S. Proof. Let ( f1, . . . , fn ) be an element in S. From Lemma 5.2.1(ii) we know that fn = 1. Set fn+1 := 1. Then fn = fn+1 , so that ∂( f1, . . . , fn ) = ( f2, . . . , fn+1 ). From Corollary 5.2.2 we know that f2 ∈ O ϑ (H). Since ( f1, . . . , fn ) ∈ S, we also have fi ∈ Ffi−1 for each element i in {1, . . . , n} with 3 ≤ i. Note also that fn+1 ∈ Ffn , since fn+1 = 1. Thus, by definition, ( f2, . . . , fn+1 ) ∈ S.  Lemma 5.2.4 Let f be an element in O ϑ (H). Then the restriction of ∂ to the set of all elements ( f1, . . . , fn ) in S with f1 = f is injective and has image S(Ff ). Proof. The statement is obviously true if n = 1. Therefore, we assume that 2 ≤ n.

5.2

The Set S

127

Let ( f1, . . . , fn ) be an element in S with f1 = f . Since ( f1, . . . , fn ) ∈ S, we have ∂( f1, . . . , fn ) ∈ S; cf. Lemma 5.2.3. From ( f1, . . . , fn ) ∈ S we also obtain that f2 ∈ Ff1 . Thus, as f1 = f and ∂( f1, . . . , fn ) = ( f2, . . . , fn, fn ), we conclude that ∂( f1, . . . , fn ) ∈ S(Ff ). Let ( f2, . . . , fn+1 ) be an element in S(Ff ). Then f2 ∈ Ff . Set f1 := f . Then f2 ∈ Ff1 . Since ( f2, . . . , fn+1 ) ∈ S(Ff ), we also have fi ∈ Ffi−1 for each element i in {2, . . . , n} with 3 ≤ i.2 Thus, ( f1, . . . , fn ) ∈ S. Note also that ∂( f1, . . . , fn ) = ( f2, . . . , fn+1 ). So far, we have seen that the restriction of ∂ to the set of all elements ( f1, . . . , fn ) in S with f1 = f has image S(Ff ). To show that the restriction of ∂ to the set of all elements ( f1, . . . , fn ) in S with f1 = f is injective, we choose elements (d1, . . . , dn ) and (e1, . . . , en ) in S, and we assume that d1 = f , e1 = f , and ∂(d1, . . . , dn ) = ∂(e1, . . . , en ). From d1 = f and e1 = f we obtain that d1 = e1 . From ∂(d1, . . . , dn ) = ∂(e1, . . . , en ) we obtain that  (d2, . . . , dn ) = (e2, . . . , en ). Thus, (d1, . . . , dn ) = (e1, . . . , en ). Corollary 5.2.5 For each closed subset F of O ϑ (H), we have Õ |S(F)| = |S(Ff )|. f ∈F

Proof. Let F be a closed subset of O ϑ (H). For each element f in F, we find exactly |S(Ff )| elements ( f1, . . . , fn ) in S(F) with f1 = f ; cf. Lemma 5.2.4. Thus, the equation follows from the fact that, for each element ( f1, . . . , fn ) in S(F), there exists exactly one element f in F with f1 = f .  Let (d1, . . . , dn ) and (e1, . . . , en ) be elements in S, and let b be an element in H. We say that (d1, . . . , dn ) is b-compatible with (e1, . . . , en ) if, for each element i in {1, . . . , n}, the intersection di b ∩ bei is not empty. Lemma 5.2.6 Let (d1, . . . , dn ) and (e1, . . . , en ) be elements in S, and let b be an element in H. Assume that (d1, . . . , dn ) is b-compatible with (e1, . . . , en ). Then ∂(d1, . . . , dn ) is b-compatible with ∂(e1, . . . , en ). Proof. Since (d1, . . . , dn ) is assumed to be b-compatible with (e1, . . . , en ), the intersection di b ∩ bei is not empty for each element i in {2, . . . , n}.3 Thus, as ∂(d1, . . . , dn ) = (d2, . . . , dn, dn ) and ∂(e1, . . . , en ) = (e2, . . . , en, en ), ∂(d1, . . . , dn ) is  b-compatible with ∂(e1, . . . , en ).

2We also have fn+1 ∈ Ffn , but we do not need that. 3We also know that the intersection d1 b ∩ be1 is not empty, but we do not need that.

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5.3 The Sets a∗ b ∩ Fc and Sa,b (Fc ) As in the previous section, the letter H stands for a finite tight hypergroup, the residual depth of H will be denoted by n, and we assume that 1 ≤ n. As in Section 5.1, we write Fh for the product h∗ h whenever h stands for an element of H. Retaining our notation introduced in Section 5.2 we denote, for each closed subset F of O ϑ (H), by S(F) the set of all n-tuples ( f1, . . . , fn ) consisting of elements of H such that f1 ∈ F and, for each element i in {1, . . . , n} with 2 ≤ i, fi ∈ Ffi−1 . As before, we write S instead of S(O ϑ (H)). Let h be an element in H. From Lemma 1.1.3 we know that, for each element a in H, ah is not empty. Thus, there exist maps from H to H which take each element a in H to an element in ah. We choose one of these maps and denote it by ρh . We will write ρh exponentially, so that we have aρh ∈ ah for each element a in H. Given elements h1 , . . ., hm in H, we write ρ(h1,...,hm ) to denote the composite ρh1 · · · ρhm . Lemma 5.3.1 Let a be an element in H, and let ( f1, . . . , fn ) be an element in S. Then, for each element i in {1, . . . , n}, aρ( f1, ..., fi ) ∈ a f1 . Proof. Let i be an element in {1, . . . , n}. If i = 1, we have aρ( f1, ..., fi ) = aρ f1 ∈ a f1 , so we are done in this case. Assume that 2 ≤ i, and suppose, by induction, that aρ( f1, ..., fi−1 ) ∈ a f1 . Since ( f1, . . . , fn ) ∈ S and 2 ≤ i, we also have fi ∈ Ff1 ; cf. Lemma 5.2.1(iii). Thus, we obtain that aρ( f1, ..., fi ) = (aρ( f1, ..., fi−1 ) )ρ fi ∈ aρ( f1, ..., fi−1 ) fi ⊆ a f1 Ff1 = a f1, 

as wanted. (Recall that f1 is tight.) Lemma 5.3.2

Let a, b, and c be elements in H with c ∈ a∗ b, and let ( f1, . . . , fn ) be an element in S(Fb ). Then cρ( f1, ..., fn ) ∈ a∗ b. Proof. By Lemma 5.3.1, cρ( f1, ..., fn ) ∈ c f1 . Since ( f1, . . . , fn ) ∈ S(Fb ), we also have f1 ∈ Fb . Thus, as c ∈ a∗ b, we conclude that cρ( f1, ..., fn ) ∈ a∗ bFb = a∗ b.  Let F be a closed subset of O ϑ (H). For any two elements a and b in H, we define Sa,b (F) to be the set of all elements ( f1, . . . , fn ) in S(F) with aρ( f1, ..., fn ) = b. Lemma 5.3.3 Let a, b, and c be elements in H, and let ( f1, . . . , fn ) be an element in Sa,b (Fc ). Then f1 ∈ a∗ b ∩ Fc .

5.3

The Sets a∗ b ∩ Fc and Sa,b (Fc )

129

Proof. We are assuming that ( f1, . . . , fn ) ∈ Sa,b (Fc ). Thus, aρ( f1, ..., fn ) = b. From Lemma 5.3.1 we also know that aρ( f1, ..., fn ) ∈ a f1 . Thus, b ∈ a f1 . It follows that f1 ∈ a∗ b. Since ( f1, . . . , fn ) ∈ S(Fc ), we also have f1 ∈ Fc .



Lemma 5.3.4 Let a and b be elements in H, and let d and e be elements in Fb . Then Sd,e (Fa ∩ Fb ) = Sd,e (Fa ). Proof. Since Fa ∩ Fb ⊆ Fa , we have Sd,e (Fa ∩ Fb ) ⊆ Sd,e (Fa ). To show the reverse containment, let ( f1, . . . , fn ) be an element in Sd,e (Fa ). From ( f1, . . . , fn ) ∈ Sd,e (Fa ) we obtain that f1 ∈ d ∗ e ∩ Fa ; cf. Lemma 5.3.3. Now recall from Lemma 2.1.3(i) that Fb is closed. Thus, as {d, e} ⊆ Fb , d ∗ e ⊆ Fb . It follows that f1 ∈ Fa ∩ Fb , so that ( f1, . . . , fn ) ∈ S(Fa ∩ Fb ). Since ( f1, . . . , fn ) ∈ Sd,e (Fa ), we also have d ρ( f1, ..., fn ) = e. Thus, we obtain from ( f1, . . . , fn ) ∈ S(Fa ∩ Fb ) that ( f1, . . . , fn ) ∈ Sd,e (Fa ∩ Fb ).  Lemma 5.3.5 Let a, b, and c be elements in H, and let d be an element in a∗ b∩ Fc . Then S(Fb ∩ Fc ) is the disjoint union of the sets Sd,e (Fb ) with e ∈ a∗ b ∩ Fc . Proof. Let ( f1, . . . , fn ) be an element in S(Fb ∩ Fc ). Applying Lemma 5.3.2 to d in place of c we obtain that d ρ( f1, ..., fn ) ∈ a∗ b. Applying the same lemma to c in place of a and b and to d in place of c we obtain that d ρ( f1, ..., fn ) ∈ Fc . Thus, d ρ( f1, ..., fn ) ∈ a∗ b∩Fc , and that shows that ( f1, . . . , fn ) ∈ Sd,e (Fb ∩ Fc ) for some element e in a∗ b ∩ Fc . Notice also that, for each element e in a∗ b ∩ Fc , Sd,e (Fb ∩ Fc ) = Sd,e (Fb ); cf. Lemma 5.3.4.  The following lemma is similar to Lemma 5.2.4. Lemma 5.3.6 Let a, b, and c be elements in H, and let e be an element in a∗ b ∩ Fc . Then the restriction of ∂ to the set of all elements ( f1, . . . , fn ) in Sa,b (Fc ) satisfying f1 = e is injective and has image Sa ρ e ,b (Fe ). Proof. The statement is obviously true if n = 1. Therefore, we assume that 2 ≤ n. Let ( f1, . . . , fn ) be an element in Sa,b (Fc ), and assume that f1 = e. Then aρ( f1, ..., fn ) = b and Set fn+1 := 1. Then and, since a

ρ( f1 , ..., fn )

(aρe )ρ( f2, ..., fn+1 ) = (aρ f1 )ρ( f2, ..., fn+1 ) .

∂( f1, . . . , fn ) = ( f2, . . . , fn+1 ). = b,

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5 Tight Hypergroups

(aρ f1 )ρ( f2, ..., fn+1 ) = aρ( f1, ..., fn+1 ) = (aρ( f1, ..., fn ) )ρ fn+1 = bρ fn+1 ∈ b·1 = {b}. From (aρe )ρ( f2, ..., fn+1 ) = (aρ f1 )ρ( f2, ..., fn+1 ) and (aρ f1 )ρ( f2, ..., fn+1 ) = {b} we obtain that (aρe )ρ( f2, ..., fn+1 ) = {b}. Now recall that f1 = e. Thus, by Lemma 5.2.4, ∂( f1, . . . , fn ) ∈ S(Fe ). Thus, as ∂( f1, . . . , fn ) = ( f2, . . . , fn+1 ), ( f2, . . . , fn+1 ) ∈ S(Fe ). From ( f2, . . . , fn+1 ) ∈ S(Fe ) and (aρe )ρ( f2, ..., fn+1 ) = {b} we now obtain that ( f2, . . . , fn+1 ) ∈ Sa ρ e ,b (Fe ), and, since ∂( f1, . . . , fn ) = ( f2, . . . , fn+1 ), this implies that ∂( f1, . . . , fn ) ∈ Sa ρ e ,b (Fe ). Conversely, let ( f2, . . . , fn+1 ) be an element in Sa ρ e ,b (Fe ), and set f1 := e. Then f1 ∈ Fc , f2 ∈ Ff1 , and, for each element i in {2, . . . , n} with 3 ≤ i, fi ∈ Ffi−1 .4 Thus, ( f1, . . . , fn ) ∈ S(Fc ). From ( f2, . . . , fn+1 ) ∈ Sa ρ e ,b (Fe ) we also obtain that (aρe )ρ( f2, ..., fn+1 ) = b, and in Lemma 5.2.1(ii), we saw that that fn+1 = 1. Thus, (aρe )ρ( f2, ..., fn ) ∈ (aρe )ρ( f2, ..., fn ) ·1 = {(aρe )ρ( f2, ..., fn+1 ) } = {b}. From f1 = e, on the other hand, we obtain that aρ( f1, ..., fn ) = (aρ f1 )ρ( f2, ..., fn ) = (aρe )ρ( f2, ..., fn ) . It follows that aρ( f1, ..., fn ) = b. Thus, as ( f1, . . . , fn ) ∈ S(Fc ), ( f1, . . . , fn ) ∈ Sa,b (Fc ). Note finally that fn = fn+1 , so ∂( f1, . . . , fn ) = ( f2, . . . , fn+1 ). From Lemma 5.2.4 we know that the restriction of ∂ to the set of all elements ( f1, . . . , fn ) in Sa,b (Fc ) with f1 = e is injective.  Lemma 5.3.7 For any three elements a, b, and c in H, we have Õ |Sa ρ e ,b (Fe )|. |Sa,b (Fc )| = e ∈a∗ b∩Fc

4We also have fn+1 ∈ Ffn , but we do not need that.

5.3

The Sets a∗ b ∩ Fc and Sa,b (Fc )

131

Proof. Let a, b, and c be elements in H. In Lemma 5.3.3, we saw that, for each element ( f1, . . . , fn ) in Sa,b (Fc ), f1 ∈ a∗ b ∩ Fc . From Lemma 5.3.6, on the other hand, we obtain that, for each element e in a∗ b ∩ Fc , the set of all elements ( f1, . . . , fn ) in Sa,b (Fc ) with f1 = e has cardinality |Sa ρ e ,b (Fe )|.  Recall that H stands for a finite tight hypergroup. In particular, each closed subset of H is finite and tight. Thus, by Theorem 5.1.5, each closed subset of H is residually thin. This implies that each closed subset of H has a residual depth. Lemma 5.3.8 Let a, b, and c be elements in H, and assume that the intersection a∗ b ∩ Fc is not empty. Then we have |Sa,b (Fc )| = |S(Fb ∩ Fc )|. Proof. We proceed by induction on dp(Fb ) + dp(Fc ) and assume first that dp(Fc ) = 0. In this case, we have Fc = {1}. Since we are assuming that a∗ b ∩ Fc is not empty, this implies that 1 ∈ a∗ b. Thus, by Lemma 1.1.6, a = b. It follows that Sa,b (Fc ) = {(1, . . . , 1)} = S(Fb ∩ Fc ), so that we are done in this case. Now assume that 1 ≤ dp(Fc ). Let e be an element in a∗ b ∩ Fc . From e ∈ a∗ b we obtain that a ∈ be∗ ; cf. Lemma 1.2.1. Thus, aρe ∈ bFe . It follows that (aρe )∗ b ∩ Fe is not empty; cf. Lemma 1.3.3. From e ∈ Fc , on the other hand, we obtain that dp(Fe ) ≤ dp(Fc ) − 1. Thus, by induction |Sa ρ e ,b (Fe )| = |S(Fb ∩ Fe )|. Now we fix an element d in a∗ b ∩ Fc . From d ∈ a∗ b we obtain that a∗ ∈ db∗ . Thus, as e ∈ a∗ b, we conclude that e ∈ dFb . It follows that d ∗ e ∩ Fb is not empty; cf. Lemma 1.3.3. From e ∈ Fc , on the other hand, we obtain that dp(Fe ) ≤ dp(Fc ) − 1. Thus, by induction |Sd,e (Fb )| = |S(Fe ∩ Fb )|. Now recall from Lemma 5.3.7 that |Sa,b (Fc )| =

Õ

|Sa ρ e ,b (Fe )|

e ∈a∗ b∩Fc

and from Lemma 5.3.5 that Õ

|Sd,e (Fb )| = |S(Fb ∩ Fc )|.

e ∈a∗ b∩Fc

Now the desired equation follows from the last four equations.



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5 Tight Hypergroups

5.4 The Sets b f1 b ∗ ∩ Fa and Sb,( f1 ,..., fn ) (Fa ) As in the previous two sections, the letter H stands for a finite tight hypergroup, the residual depth of H will be denoted by n, and we assume that 1 ≤ n. As in Section 5.1, we write Fh for the product h∗ h whenever h stands for an element of H. Retaining our notation introduced in Section 5.2 we denote, for each closed subset F of O ϑ (H), by S(F) the set of all n-tuples ( f1, . . . , fn ) consisting of elements of H such that f1 ∈ F and, for each element i in {1, . . . , n} with 2 ≤ i, fi ∈ Ffi−1 . As before, we write S instead of S(O ϑ (H)). The main goal of this section is Theorem 5.4.8. This theorem lays the foundation for all results in Sections 5.5, 5.6, 5.7, and 5.8. Let (d1, . . . , dn ) and (e1, . . . , en ) be elements in S, and let b be an element in H. Recall from Section 5.2 that (d1, . . . , dn ) is said to be b-compatible with (e1, . . . , en ) if, for each element i in {1, . . . , n}, the intersection di b ∩ bei is not empty. Let a and b be elements in H, and let ( f1, . . . , fn ) be an element in S. We define Sb,( f1,..., fn ) (Fa ) to be the set of all elements in S(Fa ) which are b-compatible with ( f1, . . . , fn ). The first part of the following lemma is similar to Lemma 5.3.3. Lemma 5.4.1 Let a and b be elements in H, and let (e1, . . . , en ) be an element in S. Then the following hold. (i) Let (d1, . . . , dn ) be an element in Sb,(e1,...,en ) (Fa ). Then we have d1 ∈ be1 b∗ ∩ Fa . (ii) Assume that Fa normalizes Fb∗ , and let c be an element in be1 b∗ ∩ Fa . Then there exists an element (d1, . . . , dn ) in Sb,(e1,...,en ) (Fa ) such that d1 = c. Proof. (i) Since (d1, . . . , dn ) ∈ Sb,(e1,...,en ) (Fa ), d1 ∈ Fa and the intersection d1 b∩be1 is not empty. Since d1 b∩be1 is not empty, d1 ∈ be1 b∗ ; cf. Lemma 1.2.5(iii). It follows that d1 ∈ be1 b∗ ∩ Fa . (ii) Set d1 := c. Then, as c ∈ be1 b∗ ∩ Fa , we have d1 ∈ be1 b∗ ∩ Fa . From d1 ∈ be1 b∗ we obtain that d1 b ∩ be1 is not empty; cf. Lemma 1.2.5(iii). Let i be an element in {1, . . . , n}, and assume that 2 ≤ i. Suppose there exist elements d1 , . . ., di−1 in Fa such that d1 = c, dι ∈ Fdι−1 for each element ι in {2, . . . , n} with ι ≤ i − 1, and dι b ∩ beι , ∅ for each element ι in {1, . . . , i − 1}.

5.4

The Sets b f1 b∗ ∩ Fa and Sb,( f1,..., fn ) (Fa )

133

We are assuming that Fa normalizes Fb∗ . Thus, as di−1 ∈ Fa , Fb∗ di−1 ⊆ di−1 Fb∗ . Thus, as di−1 b ∩ bei−1 is not empty, we may apply Lemma 5.1.7(ii) to b, ei−1 , and di−1 in place of a, b, and c. We obtain that bFei−1 ⊆ Fdi−1 b. Since ei ∈ Fei−1 , this implies that bei ⊆ Fdi−1 b. Now Fdi−1 contains an element di such that di b ∩ bei is not empty. From di ∈ Fdi−1 and di−1 ∈ Fa we also obtain that di ∈ Fa . Thus, by induction, we have found elements d1 , . . ., dn in Fa such that d1 = c, di ∈ Fdi−1 for each element i in {2, . . . , n}, and di b ∩ bei , ∅ for each element i in {1, . . . , n}. Since di ∈ Fdi−1 for each element i in {2, . . . , n}, we have (d1, . . . , dn ) ∈ S. Thus, as d1 ∈ Fa , (d1, . . . , dn ) ∈ S(Fa ). Since di b ∩ bei is not empty for each element i in  {1, . . . , n}, this implies that (d1, . . . , dn ) ∈ Sb,(e1,...,en ) (Fa ). The following lemma is similar to Lemma 5.3.6. Lemma 5.4.2 Let a and b be elements in H, let (e1, . . . , en ) be an element in S, and let c be an element in be1 b∗ ∩ Fa . Then the restriction of ∂ to the set of all elements (d1, . . . , dn ) in Sb,(e1,...,en ) (Fa ) with d1 = c is injective and has image Sb,∂(e1,...,en ) (Fc ). Proof. The statement is obviously true if n = 1, Therefore, we assume that 2 ≤ n. Let (d1, . . . , dn ) be an element in Sb,(e1,...,en ) (Fa ), and assume that d1 = c. Then, by Lemma 5.2.4 (together with Lemma 5.2.6), ∂(d1, . . . , dn ) ∈ Sb,∂(e1,...,en ) (Fc ). Conversely, let (d2, . . . , dn+1 ) be an element in Sb,∂(e1,...,en ) (Fc ), and define d1 := c. Then d1 ∈ Fa , d2 ∈ Fd1 , and, for each element i in {2, . . . , n} with 3 ≤ i, di ∈ Fdi−1 .5 Thus, (d1, . . . , dn ) ∈ S(Fa ). From c ∈ be1 b∗ and d1 = c we obtain that d1 ∈ be1 b∗ . Thus, by Lemma 1.2.5(iii), the intersection d1 b ∩ be1 is not empty. From (d2, . . . , dn+1 ) ∈ Sb,∂(e1,...,en ) (Fc ) we also obtain that the intersection di b ∩ bei is not empty for each element i in {2, . . . , n}. Thus, as (d1, . . . , dn ) ∈ S(Fa ), (d1, . . . , dn ) ∈ Sb,(e1,...,en ) (Fa ). Note finally that dn = dn+1 , so 5We also have dn+1 ∈ Fdn , but we do not need that.

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5 Tight Hypergroups

∂(d1, . . . , dn ) = (d2, . . . , dn+1 ). From Lemma 5.2.4 we know that the restriction of ∂ to the set of all elements (d1, . . . , dn ) in Sb,(e1,...,en ) (Fa ) with d1 = c is injective.  The following lemma is similar to Lemma 5.3.7. Lemma 5.4.3 For any three elements a, b in H and ( f1, . . . , fn ) in S, we have Õ |Sb,( f1,..., fn ) (Fa )| = |Sb,∂( f1,..., fn ) (Fc )|. c ∈b f1 b ∗ ∩Fa

Proof. Let a and b be elements in H, and let ( f1, . . . , fn ) be an element in S. In Lemma 5.4.1(i), we saw that, for each element (d1, . . . , dn ) in Sb,( f1,..., fn ) (Fa ), d1 ∈ b f1 b∗ ∩ Fa . From Lemma 5.4.2, on the other hand, we obtain that, for each element c in b f1 b∗ ∩Fa , the set of all elements (d1, . . . , dn ) in Sb,( f1,..., fn ) (Fa ) with d1 = c has cardinality  |Sb,∂( f1,..., fn ) (Fc )|. For the remainder of this section, we consider elements a and b in H such that Fa normalizes Fb∗ . (This condition appeared first in Lemma 5.4.1(ii). Our next lemma is similar to Lemma 5.3.8. Lemma 5.4.4 Let a and b be elements in H, and assume that Fa normalizes Fb∗ . Let ( f1, . . . , fn ) be an element in S, and assume that Sb,( f1,..., fn ) (Fa ) is not empty. Then we have |Sb,( f1,..., fn ) (Fa )| = |S(Fa ∩ Fb∗ )|. Proof. We proceed by induction on dp(Fa ) and assume first that dp(Fa ) = 0. In this case, we have Fa = {1}. Since Sb,( f1,..., fn ) (Fa ) is assumed not to be empty, this implies that Sb,( f1,..., fn ) (Fa ) = {(1, . . . , 1)} = S(Fa ∩ Fb∗ ), so that we are done in this case. Now assume that 1 ≤ dp(Fa ). Since Sb,( f1,..., fn ) (Fa ) is assumed not to be empty, b f1 b∗ ∩ Fa is not empty; cf. Lemma 5.4.1(i). Let c be an element in b f1 b∗ ∩ Fa . From c ∈ Fa we obtain that Fb∗ c ⊆ cFb∗ . (Recall that Fa is assumed to normalize Fb∗ .) Set e := bρ f1 . Then e ∈ b f1 . Thus, as c ∈ b f1 b∗ and Fb∗ c ⊆ cFb∗ , we may apply Lemma 5.1.8 to b∗ , f1 , and e in place of a, b, and d. Then we obtain that b f1 b∗ = eb∗ . Now recall from Lemma 5.4.3 that |Sb,( f1,..., fn ) (Fa )| =

Õ c ∈b f1 b ∗ ∩Fa

|Sb,∂( f1,..., fn ) (Fc )|.

5.4

The Sets b f1 b∗ ∩ Fa and Sb,( f1,..., fn ) (Fa )

135

Now let c be an element in b f1 b∗ ∩ Fa . Then, by Lemma 5.4.1(ii) (together with Lemma 5.4.2), the set Sb,∂( f1,..., fn ) (Fc ) is not empty. From c ∈ Fa we obtain that dp(Fc ) ≤ dp(Fa ) − 1 and that Fc ⊆ Fa . From Fc ⊆ Fa we obtain that Fc normalizes Fb∗ . Thus, by induction, |Sb,∂( f1,..., fn ) (Fc )| = |S(Fc ∩ Fb∗ )|. Now, since b f1 b∗ = eb∗ , |Sb,( f1,..., fn ) (Fa )| =

Õ

|S(Fc ∩ Fb∗ )|.

c ∈eb ∗ ∩Fa

Let c be an element in eb∗ ∩ Fa . From c ∈ eb∗ we obtain that e∗ ∈ b∗ c∗ ; cf. Lemma 1.2.1. It follows that (e∗ )ρc ∈ b∗ Fc . Thus, by Lemma 1.3.3, ((e∗ )ρc )∗ b∗ ∩ Fc is not empty. Applying Lemma 5.3.8 to (e∗ )ρc and b∗ in place of a and b we now obtain that |S(e∗ )ρ c ,b∗ (Fc )| = |S(Fb∗ ∩ Fc )|. Applying Lemma 5.3.7 to e∗ , b∗ , and a in place of a, b, and c we obtain that Õ |Se∗,b∗ (Fa )| = |S(e∗ )ρ c ,b∗ (Fc )|. c ∈eb ∗ ∩Fa

Recall (from further above) that b f1 b∗ ∩ Fa is not empty and that b f1 b∗ = eb∗ . Thus, eb∗ ∩ Fa is not empty. Thus, applying Lemma 5.3.8 to e∗ , b∗ , and a in place of a, b, and c we obtain that |Se∗,b∗ (Fa )| = |S(Fb∗ ∩ Fa )|. Now the desired equation follows from the last four equations.



The following lemma is similar to Lemma 5.3.2. Lemma 5.4.5 Let a and b be elements in H, and assume that Fa normalizes Fb∗ . Let ( f1, . . . , fn ) be an element in S, and assume that Sb,( f1,..., fn ) (Fa ) is not empty. Then, for each element e in ab, eρ( f1, ..., fn ) ∈ ab. Proof. Let (d1, . . . , dn ) be an element in Sb,( f1,..., fn ) (Fa ). Then d1 ∈ Fa , and the intersection d1 b ∩ b f1 is not empty. We are assuming that Fa normalizes Fb∗ . Thus, as d1 ∈ Fa , we have Fb∗ d1 ⊆ d1 Fb∗ . Thus, as d1 b ∩ b f1 is not empty, we may apply Lemma 5.1.7(i) to b, f1 , and d1 in place of a, b, and c. Then we obtain that b f1 ⊆ d1 b, so that, for each element e in ab, eρ( f1, ..., fn ) ∈ e f1 ⊆ ab f1 ⊆ ad1 b ⊆ aFa b = ab; cf. Lemma 5.3.1.



136

5 Tight Hypergroups

Lemma 5.4.6 Let a and b be elements in H, and assume that Fa normalizes Fb∗ . Let ( f1, . . . , fn ) be an element in S, and assume that H contains elements c, d, and e with {c, d} ⊆ ab and ( f1, . . . , fn ) ∈ Sc,d (Fe∗ ). Then Sb,( f1,..., fn ) (Fa ) is not empty. Proof. From ( f1, . . . , fn ) ∈ Sc,d (Fe∗ ) we obtain that f1 ∈ c∗ d; cf. Lemma 5.3.3. Thus, as {c, d} ⊆ ab, f1 ∈ b∗ Fa b. It follows that b f1 b∗ ∩ Fa is not empty; cf. Lemma  1.3.5(ii). Thus, by Lemma 5.4.1(ii), Sb,( f1,..., fn ) (Fa ) is not empty. Lemma 5.4.7 Let a, b, and c be elements in H, and assume that Fa normalizes Fb∗ . Define S ◦ (Fc∗ ) to be the set of all elements ( f1, . . . , fn ) in S(Fc∗ ) such that Sb,( f1,..., fn ) (Fa ) is not empty. Let d be an element in abc, and let g be an element in the intersection ab∩dc∗ . Then S ◦ (Fc∗ ) is the disjoint union of the sets Sg,e (Fc∗ ) with e ∈ ab ∩ dc∗ . Proof. Let ( f1, . . . , fn ) be an element in S ◦ (Fc∗ ). Then Sb,( f1,..., fn ) (Fa ) is not empty. Thus, as g ∈ ab, Lemma 5.4.5 yields g ρ( f1, ..., fn ) ∈ ab. From g ∈ dc∗ and ( f1, . . . , fn ) ∈ S(Fc∗ ) we also obtain that g ρ( f1, ..., fn ) ∈ dc∗ ; cf. Lemma 5.3.2. Thus, we have g ρ( f1, ..., fn ) ∈ ab ∩ dc∗ . In other words, there exists an element e in ab ∩ dc∗ such that ( f1, . . . , fn ) ∈ Sg,e (Fc∗ ). Conversely, let ( f1, . . . , fn ) be an element in Sg,e (Fc∗ ) with e ∈ ab ∩ dc∗ . Then, by  Lemma 5.4.6, Sb,( f1,..., fn ) (Fa ) is not empty, whence ( f1, . . . , fn ) ∈ S ◦ (Fc∗ ). Let a, b, c, and d be elements in H. If d < abc, we set Pabcd := ∅. If d ∈ abc, we define Pabcd to be the set of all pairs ((d1, . . . , dn ), (e1, . . . , en )) in S(Fa ) × S(Fc∗ ) such that (d1, . . . , dn ) is b-compatible with (e1, . . . , en ). Assume that d ∈ abc, let (e1, . . . , en ) be an element in S(Fc∗ ), and let (d1, . . . , dn ) be an element in S(Fa ). Notice that ((d1, . . . , dn ), (e1, . . . , en )) ∈ Pabcd if and only if (d1, . . . , dn ) ∈ Sb,(e1,...,en ) (Fa ). Notice that, by Lemma 1.2.5(i), |Pabcd | = |Pc∗ b∗ a∗ d∗ |. Theorem 5.4.8 Let a, b, and c be elements in H, and assume that Fa normalizes Fb∗ . Then Õ |S(Fa ∩ Fb∗ )| · |S(Fe ∩ Fc∗ )| |Pabcd | = e ∈ab∩dc ∗

for each element d in abc. Proof. Let d be an element in abc. Recall that, for each element ( f1, . . . , fn ) in S(Fc∗ ), Sb,( f1,..., fn ) (Fa ) is the set of all elements (d1, . . . , dn ) in S(Fa ) with ((d1, . . . , dn ), ( f1, . . . , fn )) ∈ Pabcd .

5.4

The Sets b f1 b∗ ∩ Fa and Sb,( f1,..., fn ) (Fa )

137

As in the statement of Lemma 5.4.7, we define S ◦ (Fc∗ ) to be the set of all elements ( f1, . . . , fn ) in S(Fc∗ ) such that Sb,( f1,..., fn ) (Fa ) is not empty. Then S ◦ (Fc∗ ) is the set of all elements ( f1, . . . , fn ) in S(Fc∗ ) such that there exists an element (d1, . . . , dn ) in S(Fa ) with ((d1, . . . , dn ), ( f1, . . . , fn )) ∈ Pabcd . It follows that Pabcd is the disjoint union of the sets {((d1, . . . , dn ), ( f1, . . . , fn )) | (d1, . . . , dn ) ∈ Sb,( f1,..., fn ) (Fa )} with ( f1, . . . , fn ) ∈ S ◦ (Fc∗ ). As a consequence, Õ |Pabcd | = |Sb,( f1,..., fn ) (Fa )|. ( f1,..., fn )∈S ◦ (Fc ∗ )

On the other hand, for each element ( f1, . . . , fn ) in S ◦ (Fc∗ ), |Sb,( f1,..., fn ) (Fa )| = |S(Fa ∩ Fb∗ )|; cf. Lemma 5.4.4. Thus, Õ

|Pabcd | =

|S(Fa ∩ Fb∗ )|.

( f1,..., fn )∈S ◦ (Fc ∗ )

We now fix an element g in ab ∩ dc∗ . Then, by Lemma 5.4.7, S ◦ (Fc∗ ) is the disjoint union of the sets Sg,e (Fc∗ ) with e ∈ ab ∩ dc∗ . It follows that Õ Õ |Pabcd | = |S(Fa ∩ Fb∗ )|. e ∈ab∩dc ∗ ( f1,..., fn )∈Sg, e (Fc ∗ )

It follows that |Pabcd | =

Õ

|Sg,e (Fc∗ )| · |S(Fa ∩ Fb∗ )|.

e ∈ab∩dc ∗

Let e be an element in dc∗ . Since g ∈ dc∗ , we have d ∈ gc. Thus, e ∈ gFc∗ , so that, by Lemma 1.3.3, g ∗ e ∩ Fc∗ is not empty. Thus, by Lemma 5.3.8, |Sg,e (Fc∗ )| = |S(Fe ∩ Fc∗ )|. It follows that |Pabcd | =

Õ

|S(Fa ∩ Fb∗ )| · |S(Fe ∩ Fc∗ )|

e∈ab∩dc ∗

as wanted.



At this point, we mention that Theorem 5.4.8 is the only result from Section 5.3 or Section 5.4 which will be needed later in this monograph, and the only place where we need Theorem 5.4.8 is the proof of Proposition 5.5.6.

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5.5 Structure Constants of Finite Tight Hypergroups As in the previous three sections, the letter H stands for a finite tight hypergroup, the residual depth of H will be denoted by n, and we assume that 1 ≤ n. As in Section 5.1, we write Fh for the product h∗ h whenever h stands for an element of H. Retaining our notation introduced in Section 5.2 we denote, for each closed subset F of O ϑ (H), by S(F) the set of all n-tuples ( f1, . . . , fn ) consisting of elements of H such that f1 ∈ F and, for each element i in {1, . . . , n} with 2 ≤ i, fi ∈ Ffi−1 . For any three elements a, b, and c in H, we set  |S(Fa ∩ Fb∗ )| if c ∈ ab γabc := 0 if c < ab. The elements γabc with a, b, c ∈ H will be called the structure constants of H. Lemma 5.5.1 Let a and b be elements in H. Then the following hold. (i) For any two elements d and e in ab, we have γabd = γabe . (ii) For each element c in H, we have γabc = γb∗ a∗ c∗ . Proof. (i) Let d and e be elements in ab. Then, by definition, γabd = |S(Fa ∩ Fb∗ )| = γabe . (ii) Let c be an element in H, and assume first that c ∈ ab. Then, by Lemma 1.2.1, c∗ ∈ b∗ a∗ . Thus, γabc = |S(Fa ∩ Fb∗ )| = |S(Fb∗ ∩ Fa )| = γb∗ a∗ c∗ . Now assume that c < ab. Then, by Lemma 1.2.1, c∗ < b∗ a∗ . It follows that γabc = 0 = γb ∗ a∗ c ∗ .  For each element h of H, we define νh := γhh∗ 1 and call this positive integer the valency of h. Lemma 5.5.2 For each element h in H, we have νh = |S(Fh )|. Proof. By definition, νh = γhh∗ 1 . On the other hand, by Lemma 1.1.1, 1 ∈ hh∗ . Thus, by definition, γhh∗ 1 = |S(Fh )|. It follows that νh = |S(Fh )|.  The following lemma shows that the structure constants of H which are different from 0 can be expressed in terms of valencies.

5.5

Structure Constants of Finite Tight Hypergroups

139

Lemma 5.5.3 Let a, b, and c be elements in H with c ∈ ab. Then we have Õ γabc = νf . f ∈Fa ∩Fb ∗

Proof. Applying Corollary 5.2.5 to Fa ∩ Fb∗ in place of F we obtain that Õ |S(Fa ∩ Fb∗ )| = |S(Ff )|. f ∈Fa ∩Fb ∗

Let c be an element in ab. Then, by definition, γabc = |S(Fa ∩ Fb∗ )|. On the other hand, for each element f in Fa ∩ Fb∗ , we have ν f = |S(Ff )|; cf. Lemma 5.5.2.  Corollary 5.5.4 For each element h in H, we have νh =

Õ

νf .

f ∈Fh

Proof. The equation is obtained by applying Lemma 5.5.3 to h, h∗ , and 1 in place of a, b, and c.  Lemma 5.5.5 Let a and b be elements in H. Then the following hold. (i) Assume that b ∈ Fa∗ . Then we have νa = γaa∗ b . (ii) Assume that a ∈ Fb . Then we have νa = γba∗ b . Proof. (i) The equation is obtained by applying Lemma 5.5.1(i) to a, a∗ , 1, and b in place of a, b, d, and e. (ii) We are assuming that a ∈ Fb . Thus, by Lemma 1.2.1, b ∈ ba∗ , so that γba∗ b = |S(Fb ∩ Fa )|. From Lemma 2.1.3(i) we know that Fb is a closed subset of H. Thus, as a ∈ Fb , we have Fa ⊆ Fb . From γba∗ b = |S(Fb ∩ Fa )| and Fa ⊆ Fb we obtain that γba∗ b = |S(Fa )|. On the other hand, by Lemma 5.5.2, νa = |S(Fa )|. Thus, νa = γba∗ b .



In the proof of the subsequent proposition, we refer to the sets Pabcd with a, b, c in H and d in abc. These sets were introduced in Section 5.4. Proposition 5.5.6 Let a, b, and c be elements in H. Assume that Fa normalizes Fb∗ and that Fc∗ normalizes Fb . Then, for each element d in abc,

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5 Tight Hypergroups

Õ

Õ

γabe γecd =

γa f d γbc f .

f ∈a∗ d∩bc

e∈ab∩dc ∗

Proof. Let d be an element in abc. Then, by Theorem 5.4.8, Õ |Pabcd | = γabe γecd . e ∈ab∩dc ∗

Since we are assuming that d ∈ abc, we also have d ∗ ∈ c∗ b∗ a∗ ; cf. Lemma 1.3.3 and Lemma 1.2.1. This allows us to apply Theorem 5.4.8 also to c∗ , b∗ , a∗ , and d ∗ in place of a, b, c, and d. Thus, Õ |Pc∗ b∗ a∗ d∗ | = γc ∗ b ∗ f γ f a∗ d ∗ . f ∈c ∗ b ∗ ∩d ∗ a

From Lemma 5.5.1(ii) and Lemma 1.2.1, on the other hand, we have Õ Õ Õ γbc f γa f d . γc ∗ b ∗ f γ f a∗ d ∗ = γbc f ∗ γa f ∗ d = f ∈c ∗ b ∗ ∩d ∗ a

Thus,

f ∈c ∗ b ∗ ∩d ∗ a

|Pc∗ b∗ a∗ d∗ | =

f ∈bc∩a∗ d

Õ

γbc f γa f d .

f ∈bc∩a∗ d

Now notice that, by Lemma 1.2.5(i), |Pabcd | = |Pc∗ b∗ a∗ d∗ |. Thus, we have Õ Õ γa f d γbc f , γabe γecd = e∈ab∩dc ∗

f ∈a∗ d∩bc



as wanted. We now come to the main results of this section. Corollary 5.5.7 Let a and b be elements in H, and assume that Fb∗ normalizes Fa∗ . Then Õ νa = γacb . c ∈a∗ b

Proof. Applying Proposition 5.5.6 to a, a∗ , b, and b in place of a, b, c, and d we obtain that Õ Õ γaa∗ f γ f bb = γacb γa∗ bc . f ∈Fa ∗ ∩Fb ∗

c ∈a∗ b

To express both sides of this equation in a different way, we choose an element d in a∗ b; cf. Lemma 1.1.3. From Lemma 5.5.5(i) and Lemma 5.5.1(ii) we obtain that

5.5

Õ

Structure Constants of Finite Tight Hypergroups

Õ

γaa∗ f γ f bb = νa

γ f bb = νa

f ∈Fa ∗ ∩Fb ∗

f ∈Fa ∗ ∩Fb ∗

Õ

141

γb ∗ f ∗ b ∗ .

f ∈Fa ∗ ∩Fb ∗

From Lemma 5.5.5(ii) and Lemma 5.5.3 we obtain that Õ Õ γb ∗ f ∗ b ∗ = ν f = γa∗ bd . f ∈Fa ∗ ∩Fb ∗

Thus,

f ∈Fa ∗ ∩Fb ∗

Õ

γaa∗ f γ f bb = νa γa∗ bd .

f ∈Fa ∗ ∩Fb ∗

Since d ∈ a∗ b, we obtain from Lemma 5.5.1(i) also that Õ Õ γacb γa∗ bd . γacb γa∗ bc = c ∈a∗ b

It follows that

c ∈a∗ b

νa γa∗ bd =

Note finally that γa∗ bd , 0, since d ∈

Õ

γacb γa∗ bd . c ∈a∗ b a∗ b. Thus,

νa =

Õ

γacb,

c ∈a∗ b



as wanted. Corollary 5.5.8

Let a, b, and c be elements in H with c ∈ ab. Assume that Fa normalizes Fb∗ and that Fc normalizes Fb . Then we have γabc νc = γcb∗ a νa . Proof. Applying Proposition 5.5.6 to c∗ and 1 in place of c and d we obtain that γabc γcc∗ 1 = γaa∗ 1 γbc∗ a∗ . By definition, we have γcc∗ 1 = νc and γaa∗ 1 = νa , and from Lemma 5.5.1(ii) we  know that γbc∗ a∗ = γcb∗ a . Thus, we have γabc νc = νa γcb∗ a . Corollary 5.5.9 Let a and b be elements in H. Assume that Fa normalizes Fb∗ and that, for each element c in ab, Fc normalizes Fb . Then Õ γabc νc . νa νb = c ∈ab

Proof. From Corollary 5.5.7 we know that Õ νb = γbca∗ . c ∈b ∗ a∗

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5 Tight Hypergroups

Multiplying this equation by νa we obtain that Õ Õ Õ Õ γabc∗ νc∗ = γ c ∗ b ∗ a νa = γbca∗ νa = νa νb = γabc νc ; c ∈b ∗ a∗

c ∈b ∗ a∗

c ∈b ∗ a∗

c ∈ab



cf. Lemma 5.5.1(ii), Corollary 5.5.8, and Lemma 1.2.1.

5.6 Rings Arising from Finite Tight Hypergroups Let H be a finite tight hypergroup, let R be a commutative ring (with 1), and let R H denote the free R-module over H. The addition of the module R H will be (temporarily) called the componentwise addition on R H . We will now define a multiplication on R H . For this, we recall from the beginning of Section 5.5 that, for any three elements a, b, and c in H,  |S(Fa ∩ Fb∗ )| if c ∈ ab γabc := 0 if c < ab. For any two subsets {σh | h ∈ H} and {τh | h ∈ H} of R, we define Õ Õ Õ ρh h, σh h, µ( τh h) := h ∈H

where

h ∈H

ρh :=

h ∈H

Õ Õ

σa τb γabh .

a ∈H b ∈H

This definition implies that, for any two elements a and b in H, Õ µ(a, b) = γabc c. c ∈ab

Note that µ is an operation on R H . The operation µ will be called (also only temporarily) the numerical multiplication on R H . Theorem 5.6.1 Let H be a finite tight hypergroup, and assume that, for each element h in H, Fh is normal in O ϑ (H). Let R be a commutative ring (with 1). Then R H is a ring with respect to componentwise addition and numerical multiplication, containing R in its center. Proof. We denote by ∗ the numerical multiplication defined on the free R-module R H over H, and we let a, b, and c be elements in H. Then we have Õ Õ Õ Õ Õ γabe γech h = γabe γech h (a ∗ b) ∗ c = ( γabe e) ∗ c = e∈ab

and

h ∈ec e ∈ab

h ∈H e ∈ab∩hc ∗

5.6

a ∗ (b ∗ c) = a ∗ (

Õ

γbc f f ) =

Rings Arising from Finite Tight Hypergroups

Õ Õ

γbc f γa f h h =

Õ

143

γa f h γbc f h.

h ∈H f ∈a∗ h∩bc

h ∈a f f ∈bc

f ∈bc

Õ

Thus, by Proposition 5.5.6, (a ∗ b) ∗ c = a ∗ (b ∗ c). Since R H is an R-module, this  finishes the proof of the theorem. Let H be a finite tight hypergroup. Assume that, for each element h in H, Fh is normal in O ϑ (H). Let R be a commutative ring (with 1). The ring which we found in Theorem 5.6.1 is called the hypergroup ring of H over R. Assume that H is thin, and let R be a commutative ring (with 1). Then, for each element h of H, Fh = {1}. In particular, for each element h in H, Fh is normal in O ϑ (H). Note also that, for each closed subset F of O ϑ (H), |S(F)| = |F |. Thus, for any three elements a, b, and c of a thin hypergroup H, γabc = 1 if c ∈ ab and γabc = 0 otherwise, so that, in this case, the hypergroup ring of H over R coincides with the group ring over R of the group which, via the group correspondence, corresponds to H. Theorem 5.6.2 Let H be a finite tight hypergroup, and assume that, for each element h in H, Fh is normal in O ϑ (H). Let R be a subfield of the field of complex numbers, and assume that R is invariant under complex conjugation. Then the hypergroup ring of H over R is semisimple. Proof. For each element ρ in R, we define ρ∗ to be the complex conjugate of ρ. For each set {ρh | h ∈ H} of elements in R, we define Õ Õ ( ρ∗h h∗ ρh h)∗ := h ∈H

h ∈H

Let J denote the Jacobson radical of the hypergroup ring of H over R. Then J ∗ = J. Note that the hypergroup ring of H over R is artinian, since it is finitely generated over R. Thus, J is nilpotent, so that there exists a positive integer n with J n = {0}. Among the positive integers n with J n = {0} we choose m as small as possible. Then J m = {0} and J m−1 , {0}. Set I := J m−1 . Then I , {0} and I 2 = {0}. Since I , {0}, we find an element i in I with i , 0. Since i , 0, there exists, for each element h in H an element ρh in R such that {ρh | h ∈ H} , {0} and Õ i= ρh h. h ∈H

Since J ∗ = J, I ∗ = I. Thus, i ∗ ∈ I. Let ρ denote the coefficient of 1 in ii ∗ . Then Õ Õ |S(Fh )| · | ρh | 2 . ρ= γhh∗ 1 | ρh | 2 = h ∈H

Thus, as i , 0, ρ , 0. Thus,

ii ∗

h ∈H

, 0, contrary to ii ∗ ∈ I 2 and I 2 = {0}.



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5 Tight Hypergroups

5.7 Some Arithmetic in Finite Metathin Hypergroups Recall from Section 4.6 that a hypergroup H is called metathin if O ϑ (H) ⊆ Oϑ (H). Metathin hypergroups were in the focus already at the end of Section 4.6, they also played a role in the proof of Lemma 4.7.5. In this section, we restrict our attention to finite metathin hypergroups. Proposition 5.5.6, which played a crucial role already in Section 5.6, is also the key in our study of finite metathin hypergroups. Recall from Section 2.1 that we write Fh instead of h∗ h whenever h is a tight hypergroup element. Theorem 5.7.1 Let H be a finite metathin hypergroup, let a, b, c, and d be elements in H, and assume that d ∈ abc. Then |ab ∩ dc∗ | · |Fa ∩ Fb∗ | · |Fe ∩ Fc∗ | = |a∗ d ∩ bc| · |Fa ∩ Ff ∗ | · |Fb ∩ Fc∗ | for any two elements e in ab ∩ dc∗ and f in a∗ d ∩ bc. Proof. Since H is a finite metathin hypergroup, H is residually thin with dp(H) ≤ 2; cf. Lemma 4.6.6. We set n := dp(H). Then n ≤ 2. There is nothing to show if n = 0. Thus, we assume that n = 1 or n = 2. From Lemma 4.6.7 we also know that H is tight. Retaining our notation introduced in Section 5.2 we denote, for each closed subset F of O ϑ (H), by S(F) the set of all n-tuples ( f1, . . . , fn ) consisting of elements of H such that f1 ∈ F and, for each element i in {1, . . . , n} with 2 ≤ i, fi ∈ Ffi−1 . If n = 1, we have O ϑ (H) = {1}. In this case, we have |S(F)| = 1 and |F | = 1 for each closed subset F of O ϑ (H), so that we have |S(F)| = |F | for each closed subset F of O ϑ (H). If n = 2, we obtain from Lemma 5.2.1(ii) that S(F) = {( f , 1) | f ∈ F} for each closed subset F of O ϑ (H). Thus, also in this case, |S(F)| = |F | for each closed subset F of O ϑ (H). It follows that, for any three elements a, b, and c in H,  |Fa ∩ Fb∗ | if c ∈ ab γabc = if c < ab. 0 On the other hand, since H is metathin, Fh is normal in O ϑ (H) for each element h in H; cf. Lemma 4.6.8(i). Thus, by Proposition 5.5.6, Õ Õ γabe γecd = γa f d γbc f . e∈ab∩dc ∗

f ∈a∗ d∩bc

Now let e and g be elements in ab ∩ dc∗ . From {e, g} ⊆ ab we obtain that γabe = γabg ;

5.7

Some Arithmetic in Finite Metathin Hypergroups

145

cf. Lemma 5.5.1(i). Similarly, since {e, g} ⊆ dc∗ , γdc∗ e = γdc∗ g . Now recall from Corollary 5.5.8 that γecd νd = γdc∗ e νe

and

γgcd νd = γdc∗ g νg .

Furthermore, by Lemma 4.6.9, Fe = Fg , and, by Lemma 5.5.2, νe = |S(Fe )| and νg = |S(Fg )|. Thus, νe = νg , so that γecd = γgcd . From γabe = γabg and γecd = γgcd we obtain that γabe γecd = γabg γgcd, and similarly one shows that γa f d γbc f = γagd γbcg for any two elements f and g in a∗ d ∩ bc. Thus, |ab ∩ dc∗ |γabe γecd = |a∗ d ∩ bc|γa f d γbc f for any two elements e in ab ∩ dc∗ and f in a∗ d ∩ bc.



Corollary 5.7.2 Let H be a finite metathin hypergroup, and let a, b, and c be elements in H with c ∈ ab. Then the following hold. (i) We have |Fa ∩ Fb∗ | · |Fc | = |Fc ∩ Fb | · |Fa |. (ii) We have |Fb | = |ab| · |Fb ∩ Fc |. (iii) We have |Fa | · |Fb | = |ab| · |Fa ∩ Fb∗ | · |Fc |. Proof. (i) Apply Theorem 5.7.1 to c∗ and 1 in place of c and d. (ii) Apply Theorem 5.7.1 to b, b∗ , a∗ , a∗ , 1, and c∗ in place of a, b, c, d, e, and f . (iii) This follows from (i) and (ii).



Theorem 5.7.3 Let H be a finite metathin hypergroup, let F be a closed subset of O ϑ (H), and let h be an element of H. Then the following hold. (i) We have |F | = |hF | · |F ∩ Fh |. (ii) We have |F | · |Fh | = |h∗ F h| · |F ∩ Fh∗ |. Proof. (i) We are assuming that F ⊆ O ϑ (H). Since H is metathin, we also have O ϑ (H) ⊆ Oϑ (H). Thus, F is thin.

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5 Tight Hypergroups

Since F is thin, the group which, via the group correspondence, corresponds to F acts on {{h} | h ∈ H} by right-multiplication; cf. Lemma 1.4.3(i). By Lemma 1.2.8(ii) the stabilizer of {h} under the group which, via the group correspondence, corresponds to F is F ∩ Fh . Thus, |F | = |hF | · |F ∩ Fh |. (ii) From Lemma 4.6.8(ii) we know that h∗ F h is a closed subset of O ϑ (H). Thus, by (i), |h∗ F h| = |hh∗ F h| · |h∗ F h ∩ Fh |. From Lemma 4.6.8(i) we know that FFh∗ = Fh∗ F. Thus, hh∗ F h = Fh∗ F h = FFh∗ h = F h. Since |F h| = |h∗ F |, this implies that |hh∗ F h| = |h∗ F |. Note also that Fh ⊆ h∗ F h, whence h∗ F h ∩ Fh = Fh . Thus, we obtain that |h∗ F h| = |h∗ F | · |Fh |. On the other hand, applying (i) to h∗ in place of h we conclude that |F | = |h∗ F | · |F ∩ Fh∗ |. Thus, |F | · |Fh | = |h∗ F | · |F ∩ Fh∗ | · |Fh | = |h∗ F h| · |F ∩ Fh∗ |, 

as wanted. Lemma 5.7.4

Let H be a finite metathin hypergroup, and let h be an element in H. Then we have |Fh∗ | = |Fh |. Proof. From Theorem 4.4.1(i) we know that O ϑ (H) is a strongly normal closed subset of H. Thus, by Lemma 3.3.2, h∗ O ϑ (H)h = O ϑ (H). Applying Theorem 5.7.3(ii) to O ϑ (H) in place of F we now obtain that |O ϑ (H)| · |Fh | = |O ϑ (H)| · |Fh∗ |. Since H is finite, this implies that |Fh∗ | = |Fh |.



Lemma 5.7.5 Let H be a finite metathin hypergroup, and let a, b, and c be elements in H with c ∈ ab. Then the following hold. (i) We have |Fa | · |Fb | = |Fc Fb | · |Fa ∩ Fb∗ |. (ii) We have |Fc Fb | = |ab| · |Fc |. (iii) We have |Fc Fb | · |Fa ∩ Fb∗ | = |Fa Fb | · |Fa ∩ Fb |. Proof. (i) Applying Theorem 5.7.3(ii) to Fa and b in place of F and h we obtain that |Fa | · |Fb | = |b∗ Fa b| · |Fa ∩ Fb∗ |.

5.7

Some Arithmetic in Finite Metathin Hypergroups

147

On the other hand, since c ∈ ab, we have b∗ Fa b = Fc Fb ; cf. Lemma 5.1.1. Thus, we obtain that |Fa | · |Fb | = |Fc Fb | · |Fa ∩ Fb∗ |. (ii) This follows from (i), together with Corollary 5.7.2(iii). (iii) In (i), we saw that |Fa | · |Fb | = |Fc Fb | · |Fa ∩ Fb∗ |, and from Theorem 3.7.3 one obtains that |Fa | · |Fb | = |Fa Fb | · |Fa ∩ Fb |. It follows that |Fc Fb | · |Fa ∩ Fb∗ | = |Fa Fb | · |Fa ∩ Fb |.



Lemma 5.7.6 Let H be a finite metathin hypergroup, and let a, b, and c be elements in H with c ∈ ab. Then we have |Fa Fb∗ | = |Fb Fc | = |Fc∗ Fa∗ |. Proof. From Theorem 3.7.3 one obtains that |Fa | · |Fb∗ | = |Fa Fb∗ | · |Fa ∩ Fb∗ |. On the other hand, by Lemma 5.7.4, |Fb∗ | = |Fb |, so that, by Lemma 5.7.5(i), |Fa | · |Fb∗ | = |Fc Fb | · |Fa ∩ Fb∗ |. Thus, since H is assumed to be finite, we conclude that |Fa Fb∗ | = |Fc Fb |. From Lemma 4.6.8(i) we also obtain that Fc Fb = Fb Fc . Thus, |Fa Fb∗ | = |Fb Fc |. Since c ∈ ab, we also have a∗ ∈ bc∗ ; cf. Lemma 1.2.1. From a∗ ∈ bc∗ we obtain |Fb Fc | = |Fc∗ Fa∗ | in exactly the same way as we obtained |Fa Fb∗ | = |Fb Fc | from c ∈ ab.  Lemma 5.7.7 Let H be a finite metathin hypergroup, and let a, b, and c be elements in H with c ∈ ab. Let k, m, and l be positive integers with m = g.c.d.(|Fa |, |Fb |), |Fa | = km, and |Fb | = ml. Then kl divides |Fc |, and |Fc | divides km2 l. Proof. Since Fa and Fb∗ are thin closed subsets of H, |Fa ∩ Fb∗ | divides |Fa | and |Fb∗ |. From Lemma 5.7.4 we also know that |Fb∗ | = |Fb |. Thus, |Fa ∩ Fb∗ | divides m. On the other hand, by Corollary 5.7.2(i), |Fa ∩ Fb∗ | · |Fc | = |Fc ∩ Fb | · |Fa |. Thus, as |Fa | = km, k divides |Fc |. Similarly, one shows that l divides |Fc |. Now, as k and l both divide |Fc |, we obtain from g.c.d.(k, l) = 1 that kl divides |Fc |. That |Fc | divides km2 l follows from Corollary 5.7.2(i).



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Lemma 5.7.8 Let H be a finite metathin hypergroup, and let a, b, c, d, and e be elements of H with e ∈ ab ∩ dc∗ . Then we have |Fe Fb ∩ Fc∗ | = |ab ∩ dc∗ | · |Fe ∩ Fc∗ |. Proof. Since H is metathin, H is tight; cf. Lemma 4.6.7. Thus, we obtain from Lemma 1.2.7(ii) that ab ∩ dc∗ = e(Fe Fb ∩ Fc∗ ). From Lemma 5.1.2 we know that Fe Fb = Fb Fe . It follows that Fe Fb ∩ Fc∗ is a closed subset of O ϑ (H); cf. Lemma 2.1.5 and Lemma 2.1.4. Now, we obtain from Theorem 5.7.3(i) that |Fe Fb ∩ Fc∗ | = |e(Fe Fb ∩ Fc∗ )| · |Fc∗ ∩ Fe |, 

and we are done. Corollary 5.7.9 Let H be a finite metathin hypergroup, let a, b, and c be elements in H. Then |Fe Fb ∩ Fc∗ | · |Fa ∩ Fb∗ | = |Ff ∗ Fb∗ ∩ Fa | · |Fb ∩ Fc∗ | for any two elements e in ab and f in bc.

Proof. From e ∈ ab we obtain that b ∈ a∗ e, from f ∈ bc that b ∈ f c∗ . It follows that b ∈ a∗ e ∩ f c∗ . Thus, by Lemma 1.2.5(i), a f ∩ ec is not empty. Let d be an element in a f ∩ ec. From d ∈ ec we obtain that e ∈ dc∗ , from d ∈ a f we obtain that f ∈ a∗ d. From d ∈ ec and e ∈ ab we obtain that d ∈ abc. Thus, as e ∈ ab ∩ dc∗ and f ∈ a∗ d ∩ bc, Theorem 5.7.1 yields that |ab ∩ dc∗ | · |Fa ∩ Fb∗ | · |Fe ∩ Fc∗ | = |a∗ d ∩ bc| · |Fa ∩ Ff ∗ | · |Fb ∩ Fc∗ |. Since e ∈ ab ∩ dc∗ , we obtain from Lemma 5.7.8 that |Fe Fb ∩ Fc∗ | = |ab ∩ dc∗ | · |Fe ∩ Fc∗ |. From f ∈ a∗ d ∩ bc we obtain that f ∗ ∈ c∗ b∗ ∩ d ∗ a. Thus, applying Lemma 5.7.8 also to c∗ , b∗ , a∗ , d ∗ , and f ∗ in place of a, b, c, d, and e, we also obtain that |Ff ∗ Fb∗ ∩ Fa | = |c∗ b∗ ∩ d ∗ a| · |Ff ∗ ∩ Fa |. The desired equation is a consequence of the above three equations.



5.8 Finite Metathin Hypergroups and Prime Numbers Let H be a finite metathin hypergroup, and let a, b, and c be elements in H with c ∈ ab. Let k, m, and l be positive integers with m = g.c.d.(|Fa |, |Fb |), |Fa | = km, and |Fb | = ml. From Lemma 5.7.7 we know that kl divides |Fc |, and |Fc | divides km2 l. We will now have a look at the extremal case where |Fc | = kl.

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Lemma 5.8.1 Let H be a finite metathin hypergroup, and let a, b, and c be elements in H with c ∈ ab. Let k, m, and l be positive integers with m = g.c.d.(|Fa |, |Fb |), |Fa | = km, and |Fb | = ml. Assume that |Fc | = kl. Then the following hold. (i) We have |ab| = m and |Fb ∩ Fc | = l. (ii) We have |bc∗ | = l and |Fc∗ ∩ Fa∗ | = k. (iii) We have |c∗ a| = k and |Fa ∩ Fb∗ | = m. Proof. (i) Set j := |Fa ∩ Fb∗ |. Then, as Fa and Fb∗ are thin closed subsets of H, j divides g.c.d.(|Fa |, |Fb |) which means that j divides m. Since c ∈ ab, Corollary 5.7.2(i) yields that |Fa ∩ Fb∗ | · |Fc | = |Fc ∩ Fb | · |Fa |. Since |Fc | = kl and |Fa | = km, this implies that j kl = |Fc ∩ Fb |km. It follows that m divides jl. From c ∈ ab we obtain that c∗ ∈ b∗ a∗ ; cf. Lemma 1.2.1. Thus, a second reference to Corollary 5.7.2(i) yields |Fb∗ ∩ Fa | · |Fc∗ | = |Fc∗ ∩ Fa∗ | · |Fb∗ |. Now recall that, by Lemma 5.7.4, |Fc∗ | = |Fc | and |Fb∗ | = |Fb |. Thus, as |Fc | = kl and |Fb | = ml, we obtain that j kl = |Fc∗ ∩ Fa∗ |ml, so that m divides j k. Now, as m divides j k as well as jl, m divides j. (Recall that g.c.d.(k, l) = 1.) Thus, as j divides m, we conclude that j = m. It follows that |Fb ∩ Fc | = l. From Corollary 5.7.2(ii) we know that |Fb | = |ab| · |Fb ∩ Fc |. Thus, as |Fb | = ml and |Fb ∩ Fc | = l, we conclude that ml = |ab|l, whence |ab| = m. (ii) Since c ∈ ab, we have a∗ ∈ bc∗ ; cf. Lemma 1.2.1. Thus, we may apply (i) to b, c∗ , and a∗ in place of a, b, and c. We obtain that |bc∗ | = l and |Fc∗ ∩ Fa∗ | = k. (iii) Since c ∈ ab, we have b∗ ∈ c∗ a; cf. Lemma 1.2.1. Thus, we may apply (i) to c∗ ,  a, and b∗ in place of a, b, and c. We obtain that |bc∗ | = l and |Fc∗ ∩ Fa∗ | = k. Lemma 5.8.2 Let H be a finite metathin hypergroup, and let a, b, and c be elements in H with c ∈ ab. Assume that g.c.d.(|Fa |, |Fb |) = 1. Then the following hold. (i) We have |Fa | · |Fb | = |Fc |. (ii) We have ab = {c}. (iii) We have Fb ⊆ Fc and Fa∗ ⊆ Fc∗ . Proof. (i) This follows from Lemma 5.7.7. (ii) Considering (i) this follows from Corollary 5.7.2(iii). (iii) Considering (ii) this follows from Corollary 5.7.2(ii).



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For the remainder of this section, we study finite metathin hypergroups H such that, for each element h in H \ O ϑ (H), |Fh | is a prime number. We will first see that, in this case, there exists a prime number p such that, for each element h in H \ O ϑ (H), |Fh | = p. Our goal is then to give an upper bound for |H| in terms of |O ϑ (H)| provided that |O ϑ (H)| , p. Our approach is inspired by C. French’s scheme theoretic investigation [20]. Lemma 5.8.3 Let H be a finite metathin hypergroup, and assume that, for each element h in H \ O ϑ (H), |Fh | is a prime number. Then there exists a prime number p satisfying the following. (i) For each element h in H \ O ϑ (H), |Fh | = p. (ii) We have O ϑ (H) = Oϑ (H). (iii) Via the group correspondence, O ϑ (H) corresponds to an elementary abelian p-group. (iv) For each element h in H \ O ϑ (H), |O ϑ (H)| = |hO ϑ (H)| · p. Proof. (i) This follows from Lemma 5.8.2(i). (ii) Since H is assumed to be metathin, we have O ϑ (H) ⊆ Oϑ (H). Let h be an element in Oϑ (H). Then Fh = {1}. Thus, by hypothesis, h ∈ O ϑ (H). (iii) Considering Lemma 5.1.6 this follows from (i). (iv) Let h be an element in H \ O ϑ (H). Applying Theorem 5.7.3(i) to O ϑ (H) in place of F we obtain that |O ϑ (H)| = |hO ϑ (H)| · |Fh |. From (i) we know that |Fh | = p. Thus, |O ϑ (H)| = |hO ϑ (H)| · p.



Lemma 5.8.4 Let H be a finite metathin hypergroup, and assume that, for each element h in H \ O ϑ (H), |Fh | is a prime number. Let a and b be elements in H \ O ϑ (H), and let c be an element in ab. Then the following hold. (i) If c < O ϑ (H), the equations Fa = Fb∗ , Fb = Fc , and Fc∗ = Fa∗ are pairwise equivalent. (ii) If Fa , Fb∗ , we have c < O ϑ (H) and Fb , Fc . Proof. (i) Assume that c ∈ H \ O ϑ (H). Then, by Lemma 5.8.3(i), |Fa | = |Fb∗ | = |Fb | = |Fc | = |Fc∗ | = |Fa∗ |. Thus, the claim follows from Lemma 5.7.6. (ii) Since {a, b} ⊆ H \O ϑ (H), we have |Fa | = |Fb∗ |; cf. Lemma 5.8.3(i). Now assume that Fa , Fb∗ . Then Fa * Fb∗ . Thus, by Lemma 1.4.5(ii), ab does not contain thin

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elements. Since c ∈ ab, this implies that c is not thin. Thus, H is assumed to be metathin, c < O ϑ (H). From c < O ϑ (H) and Fa , Fb∗ , we obtain that Fb , Fc ; cf. (i) .



Lemma 5.8.5 Let H be a finite metathin hypergroup, and assume that, for each element h in H \ O ϑ (H), |Fh | is a prime number. Let a, b, c, d, and f be elements in H \ O ϑ (H) with f ∈ a∗ d ∩ bc. Assume that Fa = Fb∗ , that Fd = Fc , and that ab ∩ O ϑ (H) is empty. Then Fa = Ff ∗ . Proof. We are assuming that f ∈ a∗ d ∩ bc. Thus, by Lemma 1.2.5(i), we find an element e in ab ∩ dc∗ . Since ab ∩ O ϑ (H) is assumed to be empty, we obtain from e ∈ ab that e < O ϑ (H). We are assuming that Fa = Fb∗ . Thus, as e ∈ ab and e < O ϑ (H), we obtain from Lemma 5.8.4(i) that Fb = Fe . We are also assuming that Fd = Fc . Thus, as e ∈ dc∗ (and e < O ϑ (H)), we obtain from Lemma 5.8.4(i) that Fc∗ = Fe . From Fb = Fe and Fc∗ = Fe we obtain that Fb = Fc∗ . Since f ∈ bc, f < O ϑ (H), and Fb = Fc∗ , we now obtain that Ff ∗ = Fb∗ ; cf. Lemma  5.8.4(i). Since we are assuming that Fa = Fb∗ , this yields that Fa = Ff ∗ . Lemma 5.8.6 Let H be a finite metathin hypergroup, and assume that, for each element h in H \ O ϑ (H), |Fh | is a prime number. Let a, b, c, d, and f be elements of H with f ∈ a∗ d ∩ bc. Assume that {a, b, f } ⊆ H \ O ϑ (H), that Fa = Fb∗ , that Fa , Ff ∗ , and that ab ∩ O ϑ (H) is empty. Then Fd , Fc . Proof. We are assuming that f ∈ bc. Thus, c ∈ b∗ f . Now recall that we are assuming that {b∗ f } ⊆ H \ O ϑ (H), that Fa = Fb∗ , and that Fa , Ff ∗ . From Fa = Fb∗ and Fa , Ff ∗ we obtain that Fb∗ , Ff ∗ . Thus, applying 5.8.4(ii) to b∗ and f in place of a and b, we obtain that c < O ϑ (H). From f ∈ a∗ d we obtain that d ∈ a f . By hypothesis, we also have {a, f } ⊆ H \ O ϑ (H) and Fa , Ff ∗ . Thus, applying 5.8.4(ii) to f and d in place of b and c, we obtain that d < O ϑ (H). So far, we have seen that {c, d} ⊆ H \ O ϑ (H). Thus, as we are assuming that {a, b} ⊆ H \ O ϑ (H), that Fa = Fb∗ , and that Fa , Ff ∗ , we obtain from Lemma 5.8.5  that Fd , Fc . For the remainder of this section, we fix the following notation. Let H be a finite metathin hypergroup, and assume that, for each element h in H \ O ϑ (H), |Fh | is a prime number. We define F := {Fh | h ∈ H \ O ϑ (H)}. For each element F in F , we set

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HF := {h ∈ H | Fh = F}. Note that, for each element F in F , HF is not empty and that H \ O ϑ (H) is the disjoint union of the sets HF with F ∈ F . Lemma 5.8.7 Let H be a finite metathin hypergroup, and assume that, for each element h in H \ O ϑ (H), |Fh | is a prime number. Assume that |O ϑ (H)| is not a prime number. Then, for each element F in F , |HF | ≤ |O ϑ (H)|. Proof. By Lemma 5.8.3(iii), there exists a prime number p such that, via the group correspondence, O ϑ (H) corresponds to an elementary abelian p-group. Let F be an element in F . Then H \ O ϑ (H) contains an element a with Fa = F. Since |O ϑ (H)| is assumed not to be a prime number, Fa , O ϑ (H). Thus, by Lemma 4.4.2, H \ O ϑ (H) contains an element b with Fa , Fb∗ . Let c be an element in ab. Then, since Fa , Fb∗ , we obtain from Lemma 5.8.4(ii) that Fc , {1} and Fb , Fc . Define G := Fc Fb . From Lemma 5.8.3(i) we know that |Fb | = p and |Fc | = p, from Lemma 4.6.8 that Fb and Fc normalize each other. Thus, |G| = p2 . Let G denote the set of the closed subsets of G different from Fb having cardinality p. Then, since |G| = p2 , |G| = p. For each element h in HF , we choose an element h 0 in hb, and we define (h) := Fh0 . (Note that h ∈ HF if and only if Fh = Fa .) From Lemma 4.6.9 we know that (h) does not depend on the choice of h 0 in hb. We will now see that, for each element h in HF , (h) ∈ G. Let h be an element in HF . Then h 0 ∈ hb. Thus, by Lemma 5.1.1, b∗ Fh b = Fh0 Fb . From c ∈ ab we similarly obtain that b∗ Fa b = Fc Fb . Thus, as Fh = Fa and G = Fc Fb , Fh0 ⊆ G. Moreover, since h 0 ∈ hb and Fh , Fb∗ , we obtain from Lemma 5.8.4(ii) that h 0 < O ϑ (H) and Fb , Fh0 . From h 0 < O ϑ (H) we obtain that Fh0 , {1}. It follows that |Fh0 | = p. Thus, as Fh0 ⊆ G and Fb , Fh0 , Fh0 ∈ G. Since (h) = Fh0 , this means that (h) ∈ G. This shows that  is a map from HF to G. Now we fix elements d and e in HF . (Then Fd = F and Fe = F.) Assume first that d ∈ O ϑ (H)e. Then O ϑ (H) contains an element t such that d ∈ te. Since H is assumed to be metathin and t ∈ O ϑ (H), t is thin. From d ∈ te and d 0 ∈ db we obtain that d 0 ∈ teb. Since t is thin, this implies that t ∗ d 0 ⊆ eb. Let k be an element in t ∗ d 0. Then Fk = k ∗ k = (d 0)∗ tt ∗ d 0 = Fd0 . On the other hand, as {k, e 0 } ⊆ eb, Lemma 4.6.9 yields Fk = Fe0 . Thus, (d) = Fd0 = Fe0 = (e).

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Assume now that d < O ϑ (H)e. Then, by Lemma 1.3.3, de∗ ∩ O ϑ (H) is empty. Thus, applying Lemma 5.8.6 to d, e∗ , e 0, and d 0 in place of a, b, c, and d we obtain that (d) = Fd0 , Fe0 = (e). What this shows is that any two elements d and e in HF have the same image under  if and only if O ϑ (H)d = O ϑ (H)e. Now recall from Lemma 5.8.3(iv) that, for each element h in HF , |O ϑ (H)| = |O ϑ (H)h| · p. Thus, each element in the image of  has exactly follows that |HF | ≤ |O ϑ (H)|.

1 ϑ p |O (H)|

preimages under . It 

Theorem 5.8.8 Let H be a finite metathin hypergroup, and assume that, for each element h in H \ O ϑ (H), |Fh | is a prime number. Assume that |O ϑ (H)| is not a prime number. Let p be a prime number, and let n be a positive integer such that |O ϑ (H)| = pn . Then the following hold. (i) We have |H| ≤ pn (pn−1 + pn−2 + · · · + p + 2). (ii) We have |H//O ϑ (H)| ≤ pn + pn−1 + · · · + p + 1. Proof. (i) From Lemma 5.8.3(iii) we know that, via the group correspondence, O ϑ (H) corresponds to an elementary abelian p-group of order pn . Thus, O ϑ (H) has pn − 1 p−1 closed subsets of order p. In particular, |F | ≤

pn − 1 . p−1

From Lemma 5.8.7 we also know that, for each element F in F , |HF | ≤ |O ϑ (H)|. Thus, as |O ϑ (H)| = pn , we have |HF | ≤ pn for each element F in F . Since H \ O ϑ (H) is the disjoint union of the sets HF with F ∈ F , we now conclude that Õ pn − 1 |H| = |O ϑ (H)| + |H \ O ϑ (H)| = pn + |HF | ≤ pn + pn · ; p−1 F ∈F

equivalently |H| ≤ pn (pn−1 + pn−2 + · · · + p + 2). (ii) From Theorem 4.4.1(i) we know that O ϑ (H) is a strongly normal closed subset of H. Thus, we have

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O ϑ (H)hO ϑ (H) = hO ϑ (H) for each element h in H. Now recall from Theorem 5.7.3(i) that, for each element h in H \ O ϑ (H), |O ϑ (H)| = |hO ϑ (H)| · |Fh |. Thus, as |O ϑ (H)| = pn and, for each element h in H \ O ϑ (H), |Fh | = p, we have |hO ϑ (H)| = pn−1 for each element h in H \ O ϑ (H). It follows that |O ϑ (H)hO ϑ (H)| = pn−1 for each element h in H \ O ϑ (H). Thus, the above upper bound of |H| (together with |O ϑ (H)| = pn ) yields |H//O ϑ (H)| ≤ 1 + p·

pn − 1 pn+1 − 1 = ; p−1 p−1

equivalently |H//O ϑ (H)| ≤ pn + pn−1 + · · · + p + 1.



Let H be a finite metathin hypergroup, and let p be a prime number. Assume that |O ϑ (H)| = p2 and that, for each element h in H \O ϑ (H), |Fh | = p. Then, by Theorem 5.8.8, |H| ≤ p3 + 2p2 and |H//O ϑ (H)| ≤ p2 + p + 1. Let k and n be positive integers. We say that an association scheme is of HanakiMiyamoto type H Mk (n) if it is listed as the k th association scheme of order n in the database given in [25]. Both upper bounds are attained (with p = 2) by association schemes of Hanaki-Miyamoto type H M176 (28). In fact, if H is an association scheme of Hanaki-Miyamoto type H M176 (28), we have |H| = 16 and |H//O ϑ (H)| = 7. The second inequality was proven for association schemes by Jung Rae Cho, Mitsugu Hirasaka, and Kijung Kim in [9; Corollary 2.8].

6 Involutions

An element h of a hypergroup H is called involution if h , 1 and {1, h} is a closed subset of H. It follows right from the definition of an involution that, via the group correspondence, the involutions of a thin hypergroup correspond to group theoretic involutions.1 So it should come as no surprise that involutions play a similarly important role in the theory of hypergroups as group theoretic involutions do in group theory. We will see this on several occasions throughout the rest of this monograph. In the first section of this chapter, we compile the most fundamental computational rules about products of involutions. In three subsequent sections, we have a closer look at cosets of closed subsets generated by a single involution of a hypergroup. In Section 6.5, we establish basic facts about the length function defined by a set of involutions. The length function defined by a set T of involutions of a hypergroup is useful in the investigation of the structure of hTi. It gives rise to the three subsets T−1 ( f ), T0 ( f ), and T1 ( f ) of hTi which we define for each element f in hTi and which provide a convenient language to which we will frequently refer in the remainder of this monograph. In Section 6.6, we investigate hypergroups generated by two distinct involutions. Via the group correspondence, these hypergroups correspond to dihedral groups. For any two distinct involutions of a hypergroup, we define the Coxeter number of these two involutions. Coxeter numbers provide information about the structure of closed subsets generated by two distinct involutions. They will be useful in Section 9.8 when we analyze Coxeter sets of involutions. They will also play a role in the proofs of Theorems 10.2.7 and 10.6.7 when we discuss regular actions of Coxeter hypergroups and twin Coxeter hypergroups in connection with buildings and twin buildings. In Section 6.7, we introduce two conditions which a set of involutions of a hypergroup may satisfy or not, dichotomy and the exchange condition. It is not difficult to see that the exchange condition implies dichotomy. We will see this in Theorem 6.7.5. 1This explains of course our terminology. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 P. -H. Zieschang, Hypergroups, https://doi.org/10.1007/978-3-031-39489-8_6

155

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Later, in Theorem 8.1.5, we will prove a partial converse of Theorem 6.7.5. Both, dichotomy and the exchange condition, will be of interest in Chapters 8, 9, and 10. In the final section of this chapter, we consider hypergroups all elements of which different from the neutral element are involutions. These hypergroups will be called projective. Via the group correspondence, thin projective hypergroups correspond to elementary abelian 2-groups. The exposition of this section is based on [57].

6.1 Basic Facts We start this section with four elementary observations about involutions. Lemma 6.1.1 Let H be a hypergroup, and let t be an involution of H. Then the following hold. (i) We have t ∗ = t. (ii) We have hti = {1, t}. (iii) If t is thin, t 2 = {1}. (iv) If t is not thin, t 2 = {1, t}. Proof. (i) Since t is an involution, t , 1 and {1, t} is a closed subset of H. Since {1, t} is closed, t ∗ ∈ {1, t}; cf. Lemma 2.1.1. Since t , 1, t ∗ , 1∗ ; cf. Lemma 1.1.2. Thus, by Lemma 1.1.8, t ∗ , 1. It follows that t ∗ = t. (ii) Since t is an involution of H, {1, t} is a closed subset of H. Thus, by definition, hti ⊆ {1, t}. Conversely, by Lemma 2.1.1, 1 ∈ hti. Furthermore, by definition t ∈ hti. Thus, {1, t} ⊆ hti. (iii) Assume that t is thin. Then, by definition, t ∗ t = {1}. From (i), on the other hand, we know that t ∗ = t. Thus, t 2 = {1}. (iv) From Lemma 1.1.1 we know that 1 ∈ t ∗ t. Assume that t is not thin. Then t ∗ t , {1}. On the other hand, since {1, t} is a closed subset of H, we have t ∗ t ⊆ {1, t}. Thus, t ∗ t = {1, t}, so that, by (i), t 2 = {1, t}.  Occasionally, we shall refer to Lemma 6.1.1 without further mention. Lemma 6.1.2 Let H be a hypergroup, and let t be an involution of H. Let a and b be two distinct elements of H, and assume that ahti = {a, b}. Then the following hold. (i) We have b ∈ at ⊆ {a, b}. (ii) We have at = {b} if and only if t < a∗ a. Proof. (i) From b ∈ ahti and hti = {1, t} we obtain that b ∈ a{1, t} = {a} ∪ at. Since we are assuming that a , b, this implies that b ∈ at. From at ⊆ ahti, together with ahti = {a, b}, we obtain that at ⊆ {a, b}.

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(ii) We are assuming that a , b. Thus, by (i), at = {b} if and only if a < at. Now  the claim follows, since a < at is equivalent to t < a∗ a. Lemma 6.1.3 Let H be a hypergroup, and let t be an involution of H. Let a and b be elements of H, and assume that at = {b}. Then the following hold. (i) If t is thin, bt = {a}. (ii) If t is not thin, bt = {a, b}. Proof. (i) Assume that t is thin. Then, by Lemma 6.1.1(iii), t 2 = {1}. It follows that bt = at 2 = a·1 = {a}, since we are assuming that at = {b}. (ii) Assume that t is not thin. Then, by Lemma 6.1.1(iv), t 2 = {1, t}. It follows that bt = at 2 = a{1, t} = {a} ∪ at = {a, b}, since we are assuming that at = {b}.



Lemma 6.1.4 Let H be a hypergroup, let u and v be involutions of H, and assume that neither u nor v is thin. Let a and b be two distinct elements in H. Assume that H contains an element c with av = {c} = ub. Then a ∈ ua and b ∈ bv. Proof. We are assuming that u is not thin. Thus, by Lemma 6.1.1(iv), u ∈ u2 . Now, as c ∈ ub and av = ub, we conclude that c ∈ uav. Thus, ua contains an element d such that c ∈ dv. It follows that d ∈ cv; cf. Lemma 6.1.1(i). On the other hand, since v is assumed not to be thin, we obtain from av = {c} that cv = {a, c}; cf. Lemma 6.1.3(ii). From d ∈ cv and cv = {a, c} we obtain that d = a or d = c. If d = a, we have a ∈ ua, since d ∈ ua. Assume that d = c. Then c ∈ ua ∩ ub. Thus, by Lemma 1.2.5(i), u∗ u ∩ ab∗ is not empty. Since u∗ u ⊆ {1, u}, this implies that 1 ∈ ab∗ or u ∈ ab∗ . If 1 ∈ ab∗ , we obtain from Lemma 1.1.6 that a = b, contrary to our choice of a and b. Thus, u ∈ ab∗ . It follows that a ∈ ub = {c}, and then a = c. Thus, as d = c and d ∈ ua, we obtain that a ∈ ua. Similarly, one obtains that b ∈ bv.



Lemma 6.1.5 Let H be a hypergroup, and let u and v be involutions of H. Then the following hold. (i) For each element t in uv, ht, vi = hu, vi. (ii) Assume that u , v. Then {u, v} ∩ uv is empty.

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Proof. (i) Recall from Lemma 6.1.1(i) that v ∗ = v. Thus, as t ∈ uv, the desired equation follows from Lemma 2.3.3. (ii) Assume, by way of contradiction, that u ∈ uv. Then v ∈ u2 . On the other hand, by Lemma 6.1.1(iii), (iv), u2 ⊆ {1, u}. Thus, as v , 1, u = v, contradicton. This shows that u < uv. Similarly, one shows that v < uv.



Lemma 6.1.6 Let H be a hypergroup, let u and v be involutions of H, and assume that uv = vu. Then the following hold. (i) We have hu, vi = {1, u, v} ∪ uv. (ii) Assume that u and v are thin and that u , v. Then uv contains a thin involution t such that hu, vi = {1, t, u, v}. (iii) Assume that u is thin and v is not thin. Then uv contains a symmetric element t such that hu, vi = {1, t, u, v}, uv = {t}, t 2 = {1, v}, tu = {v}, and tv = {u, t}. Proof. (i) We are assuming that uv = vu. Thus, huihvi = {1, u, v} ∪ uv = {1, u, v} ∪ vu = hvihui. Now, as huihvi = hvihui, Lemma 2.3.2 yields huihvi = hu, vi. It follows that hu, vi = {1, u, v} ∪ uv. (ii) We are assuming that u and v both are thin. Thus, by Lemma 1.4.3(iii), uv contains a thin element t with uv = {t}. It follows that hu, vi = {1, t, u, v}; cf. (i). Since u , v, t , 1. Furthermore, since uv = {t}, u2 = {1}, and v 2 = {1}, we obtain from Lemma 1.2.1 that t ∗ t ⊆ vu2 v = {1}. Thus, {1, t} is closed, so that t is an involution. (iii) We are assuming that u is thin. Thus, by Lemma 1.4.3(ii), uv contains an element t with uv = {t}. Since we are assuming that uv = vu, we obtain from uv = {t} that t is symmetric. From (i) and uv = {t} we obtain that hu, vi = {1, t, u, v}. We are assuming that v is not thin. Thus, by Lemma 6.1.1(iv), v 2 = {1, v}. Thus, as uv = {t}, uv = vu, and u2 = {1}, we have t 2 = uvuv = vu2 v = {1, v}. From t ∈ uv, t ∗ = t, and v ∗ = v we obtain that v ∈ tu; cf. Lemma 1.2.1. On the other hand, as u is assumed to be thin, |tu| = 1; cf. Lemma 1.4.3(i). Thus, tu = {v}. We are assuming that v is not thin. Thus, as uv = {t}, we obtain from Lemma  6.1.3(ii) that tv = {u, t}. A hypergroup H is called projective if each element of H \ {1} is an involution. Lemma 6.1.7 Let H be a hypergroup, and let u and v be involutions of H. Assume that each element of uv \ {1} is an involution. Then the following hold.

6.1

Basic Facts

159

(i) We have uv = vu. (ii) We have hu, vi = {1, u, v} ∪ uv. (iii) The closed subset hu, vi of H is projective. Proof. (i) We are assuming that each element of uv \ {1} is an involution. In Lemma 6.1.1(i), we saw that involutions are symmetric. Thus, uv is symmetric. In particular, uv is ∗ -invariant, so that, by Lemma 1.3.1(ii), uv = vu. (ii) Considering (i) this follows from Lemma 6.1.6(i). (iii) By hypothesis, each element of uv \ {1} is an involution. Furthermore, in (ii), we saw that hu, vi = {1, u, v} ∪ uv. Thus, hu, vi is projective.  Lemma 6.1.8 Let H be a hypergroup, and let u and v be two distinct involutions of H. Assume that each element of uv \ {1} is an involution. Then the following hold. (i) If one of the involutions u or v is not thin, 5 ≤ |hu, vi|. (ii) If one of the involutions u or v is thin, hu, vi is thin and |hu, vi| = 4. Proof. (i) Assume, without loss of generality, that v is not thin. Then, by Lemma 6.1.1(iv), v ∈ v 2 . Assume, by way of contradiction, that |hu, vi| ≤ 4. Then, by Lemma 6.1.5(ii) (together with Lemma 6.1.7(ii)), |uv | = 1. Thus, H contains an involution t with uv = {t}. Thus, by Lemma 6.1.7(i), uv = vu. It follows that v 2 ⊆ vu2 v = uvuv = t 2 ⊆ {1, t}. Thus, as v ∈ v 2 \ {1}, we conclude that v = t. Since t ∈ uv, this implies that v ∈ uv, contrary to Lemma 6.1.5(ii). This contradiction shows that 2 ≤ |uv |. Thus, by Lemma 6.1.7(ii) (together with Lemma 6.1.5(ii)), 5 ≤ |hu, vi|. (ii) We are assuming that each element of uv \ {1} is an involution. Thus, by Lemma 6.1.7(i), uv = vu. Assume, without loss of generality, that u is thin. Then, by Lemma 6.1.6(ii), (iii), |hu, vi| = 4. Thus, by (i), v is thin. Now we obtain from Lemma 6.1.6(ii) that hu, vi  is thin. Lemma 6.1.9 Let H be a hypergroup, let F be a closed subset of H, and let t be an involution of H. Assume that t ∈ NH (F). Then the following hold. (i) The product F hti is a closed subset of H. (ii) Let h be an element in F hti \ F. Then F hhi = F hti. Proof. (i) We are assuming that t ∈ NH (F). Thus, as 1 ∈ NH (F), hti ⊆ NH (F). Thus, by Lemma 3.1.4(iii), F hti is a closed subset of H.

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(ii) In (i), we saw that F hti is a closed subset of H. Thus, as F ⊆ F hti and h ∈ F hti, F hhi ⊆ F hti. From h ∈ F hti \ F we obtain that h ∈ Ft. Thus, by Lemma 1.3.3, t ∈ F h. Thus, by Lemma 2.1.6(i), Ft = F h, whence F hti = F ∪ Ft ⊆ F hhi. 

6.2 Closed Subsets Generated by an Involution, I In this section, as well as in the following two sections, we study hypergroups containing an involution t and pairwise distinct elements a, b, and c satisfying a∗ = a,

b∗ = b,

ahti = {a, c},

and

bhti = {b, c∗ }.

The present section provides a few general observations about these hypergroups. We will be more specific in the two subsequent sections. The results obtained in this section will be applied only in Sections 6.3, 6.4, 7.3, 7.4, and 7.6. Lemma 6.2.1 Let H be a hypergroup, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H, and assume that a∗ = a, b∗ = b, ahti = {a, c}, and bhti = {b, c∗ }. Then the following hold. (i) We have c∗ , c. (ii) We have t < {a, b, c, c∗ }. (iii) We have t < c2 , t < bc, t ∈ bc∗ , t ∈ ac, t < ac∗ , and t < ab. Proof. (i) Assume that c∗ = c. Then, by Lemma 2.1.6(i), ahti = bhti. It follows that b ∈ {a, c}, contradiction. (ii) Assume, by way of contradiction, that t = a. Then a ∈ ahti∩hti. Thus, by Lemma 2.1.6(i), ahti = hti. It follows that ahti is symmetric, contrary to (i). Similarly, one obtains that t , b. From (i) we also obtain that t < {c, c∗ }. (iii) We are assuming that ahti = {a, c}. Thus, by Lemma 2.1.6(i), chti = {a, c}. From (i) we obtain that c∗ < {a, c}. Thus, c∗ < ct. Since t is an involution of H, we have t ∗ = t; cf. Lemma 6.1.1(i). Thus, by Lemma 1.2.1, t < c2 . From (i) we obtain that c < {b, c∗ }. Thus, as bhti = {b, c∗ }, c < bt. Since b∗ = b, this implies that t < bc. From bhti = {b, c∗ } we obtain that c∗ ∈ bt; cf. Lemma 6.1.2(i). Thus, as b∗ = b, t ∈ bc∗ . From ahti = {a, c} we obtain that c ∈ at; cf. Lemma 6.1.2(i). Thus, as a∗ = a, t ∈ ac. From (i) we obtain that c∗ < {a, c}. Thus, as ahti = {a, c}, c∗ < at. Since a∗ = a, this implies that t < ac∗ . From ahti = {a, c} we obtain that b < at. Thus, as a∗ = a, we have t < ab.



6.2

Closed Subsets Generated by an Involution, I

161

Lemma 6.2.2 Let H be a hypergroup, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H, and assume that a∗ = a, b∗ = b, ahti = {a, c}, and bhti = {b, c∗ }. Then the following conditions are equivalent. (a) We have at = {c}. (b) We have c∗ t = {b}. (c) We have t < a2 . (d) We have t < cc∗ . Proof. (a) ⇒ (b) Assume that at = {c}. Then, by Lemma 1.2.2, t ∗ a∗ = {c∗ }. Thus, as a∗ = a, we have (c∗ t)∗ = t ∗ c = t ∗ at = t ∗ a∗ t = c∗ t which shows that c∗ t is ∗ -invariant. On the other hand, by Lemma 2.1.6(i), c∗ t ⊆ bhti = {b, c∗ }, and from Lemma 6.2.1(i) we know that c∗ , c. Thus, c∗ t = {b}. (b) ⇒ (a) From c ∈ ahti and a∗ = a we obtain that a ∈ htic∗ ; cf. Lemma 1.3.3. From bhti = {b, c∗ }, together with b∗ = b, we obtain that htib = {b, c}; cf. Lemma 1.3.1(ii). Thus, assuming that c∗ t = {b} we obtain that at ⊆ htic∗ t = htib = {b, c}. On the other hand, we have at ⊆ {a, c}. Thus, as a , b, at = {c}. (a) ⇔ (c) By Lemma 6.1.2(ii), we have at = {c} if and only if t < a2 . (b) ⇔ (d) By Lemma 6.1.2(ii), we have c∗ t = {b} if and only if t < cc∗ .



Corollary 6.2.3 Let H be a hypergroup, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H, and assume that a∗ = a, b∗ = b, ahti = {a, c}, and bhti = {b, c∗ }. Then the following conditions are equivalent. (a) We have bt = {c∗ }. (b) We have ct = {a}. (c) We have t < b2 . (d) We have t < c∗ c. Proof. Apply Lemma 6.2.2 to b, a, and c∗ in place of a, b, and c.



Let H, t, a, b, and c be as in the previous three results. From ahti = {a, c} we obtain that either at = {c} or at = {a, c}; cf. Lemma 6.1.2(i). Similarly, bhti = {b, c∗ } implies that either

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bt = {c∗ }

or

bt = {b, c∗ }.

We will now consider the four resulting cases one by one, starting with four preliminary results. The case where at = {c} and bt = {c∗ } will be studied in more detail in the subsequent section. The case where at = {c} and bt = {b, c∗ } is similar to the case where at = {a, c} and bt = {c∗ }, and this case will be in the focus of Section 6.4. The case where at = {a, c} and bt = {b, c∗ } will not be entirely investigated. Lemma 6.2.4 Let H be a hypergroup, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H, and assume that a∗ = a, b∗ = b, at = {c}, and bhti = {b, c∗ }. Then the following hold. (i) We have c2 = ab. (ii) The sets bc and ac∗ both are ∗ -invariant. (iii) The statements c∗ ∈ c2 , a ∈ bc, b ∈ ac∗ are pairwise equivalent. Proof. (i) We are assuming that at = {c}. Thus, by Lemma 6.2.2, c∗ t = {b}. Since b∗ = b, this implies that tc = {b}. It follows that c2 = atc = ab. (ii) We are assuming that at = {c}. Thus, by Lemma 6.2.2, c∗ t = {b}. From c∗ t = {b} we obtain that c∗ tc = bc, and this shows that bc is ∗ -invariant. From at = {c} we obtain that ta∗ = {c∗ }; cf. Lemma 1.2.2. From ta∗ = {c∗ } we obtain that ata∗ = ac∗ , and this shows that ac∗ is ∗ -invariant. (iii) In (i), we saw that c2 = ab. Thus, we have c∗ ∈ c2 if and only if c∗ ∈ ab. From Lemma 1.2.1 we know that c∗ ∈ ab is equivalent to both, a ∈ bc and b ∈ ac∗ .  Corollary 6.2.5 Let H be a hypergroup, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H, and assume that a∗ = a, b∗ = b, ahti = {a, c}, and bt = {c∗ }. Then the following hold. (i) We have c2 = ab. (ii) The sets bc and ac∗ both are ∗ -invariant. (iii) The statements c∗ ∈ c2 , a ∈ bc, b ∈ ac∗ are pairwise equivalent. Proof. Apply Lemma 6.2.4 to b, a, and c∗ in place of a, b, and c.



Lemma 6.2.6 Let H be a hypergroup, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H, and assume that a∗ = a, b∗ = b, ahti = {a, c}, and bhti = {b, c∗ }. Then t is thin if and only if at = {c} and bt = {c∗ }.

6.3

Closed Subsets Generated by an Involution, II

163

Proof. Assume first that t is thin. Then, by Lemma 1.4.3(i), |at| = 1. On the other hand, we know from Lemma 6.1.2(i) that c ∈ at. Thus, at = {c}. Similarly, bt = {c∗ }. Assume, conversely, that at = {c} and bt = {c∗ }. From at = {c} we obtain that c∗ t = {b}; cf. Lemma 6.2.2. From bt = {c∗ } and c∗ t = {b} we obtain that t is thin;  cf. Lemma 6.1.3(ii). Lemma 6.2.7 Let H be a hypergroup, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H, and assume that a∗ = a, b∗ = b, ahti = {a, c}, and bhti = {b, c∗ }. Then the following hold. (i) If c is thin, so are t, a, b, and c∗ . (ii) If c∗ is thin, so are t, a, b, and c. (iii) If a and b are thin, so are t, c, and c∗ . Proof. (i) Assume that c is thin. Then c∗ c = {1}. From Lemma 6.1.2(i) we also know that c∗ ∈ bt. Thus, as b∗ = b, t ∈ bc∗ . It follows that tc ⊆ bc∗ c = {b}, so that, by Lemma 1.1.3, tc = {b}. Now, by Lemma 1.2.2, c∗ t = {b}, so that, by Lemma 6.2.2, at = {c}. From c∗ c = {1} we obtain that t < c∗ c. Thus, as chti = {a, c}, Lemma 6.1.2(ii) yields ct = {a}, so that, by Corollary 6.2.3, bt = {c∗ }. From at = {c} and bt = {c∗ } we now obtain that t is thin; cf. Lemma 6.2.6. From a ∈ ct, together with the fact that c and t both are thin, we obtain that a is thin; cf. Lemma 1.4.3(iii). Similarly, since t and c both are thin, we obtain from b ∈ tc that b is thin. Finally, since b and t both are thin, we obtain from bt = {c∗ } that c∗ is thin. (ii) This follows from (i) by interchanging the roles of a and b and the roles of c and c∗ . (iii) Assume that a is thin. Then a∗ a = {1}. Thus, as ahti = {a, c}, Lemma 6.1.2(ii) yields at = {c}. Similarly, we obtain that bt = {c∗ }. Thus, by Lemma 6.2.6, t is thin. Since at = {c} and a and t both are thin, c is thin; cf. Lemma 1.4.3(iii). Similarly, as b and t both are thin, we obtain from bt = {c∗ } that c∗ is thin. 

6.3 Closed Subsets Generated by an Involution, II We continue our study of hypergroups containing an involution t and pairwise distinct elements a, b, and c satisfying a∗ = a, b∗ = b, ahti = {a, c}, and bhti = {b, c∗ }. We focus on the case where at = {c}

and

bt = {c∗ }.

Our results will be applied only in Section 7.3.

164

6 Involutions

Lemma 6.3.1 Let H be a hypergroup, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H, and assume that a∗ = a, b∗ = b, at = {c}, and bt = {c∗ }. Assume further that both c and c∗ are thin. Then so are t, a, and b. Proof. We are assuming that at = {c} and that bt = {c∗ }. Thus, by Lemma 6.2.6, t is thin. From c ∈ at we obtain that a ∈ ct. Thus, as c and t are thin, so is a; cf. Lemma  1.4.3(iii). Similarly, one obtains from c∗ ∈ bt that b is thin. Lemma 6.3.2 Let H be a hypergroup, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H, and assume that a∗ = a, b∗ = b, at = {c}, and bt = {c∗ }. Then the following hold. (i) We have cc∗ = a2 . (ii) We have c∗ c = b2 . (iii) We have ac = cb. Proof. (i) From at = {c}, together with a∗ = a, we obtain that ta = {c∗ }. From bt = {c∗ } we obtain that ct = {a}; cf. Corollary 6.2.3. Thus, cc∗ = cta = a2 . (ii) This follows by applying (i) to b, a, and c∗ in place of a, b, and c. (iii) From bt = {c∗ }, together with b∗ = b, we obtain that tb = {c}. Thus, as  at = {c}, we have ac = atb = cb. Lemma 6.3.3 Let H be a hypergroup, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H, and assume that a∗ = a, b∗ = b, at = {c}, and bt = {c∗ }. Then the following hold. (i) The statements a ∈ c∗ c, a ∈ c2 , b ∈ c2 , and b ∈ cc∗ are pairwise equivalent. (ii) We have b ∈ c∗ c if and only if a ∈ cc∗ . (iii) The statements c ∈ c∗ c, a ∈ ac∗ , a ∈ bc∗ , and b ∈ bc are pairwise equivalent. Proof. (i) We have a ∈ c∗ c if and only if c∗ ∈ ac∗ . In Lemma 6.2.4(ii), we have seen that ac∗ is ∗ -invariant. Thus, c∗ ∈ ac∗ if and only if c ∈ ac∗ . Note finally that c ∈ ac∗ if and only if a ∈ c2 . By Lemma 1.2.1, a ∈ c2 if and only if c∗ ∈ ac. In Lemma 6.3.2(iii), we have seen that ac = cb. Thus, c∗ ∈ ac if and only if c∗ ∈ cb. Note finally that c∗ ∈ cb if and only if b ∈ c2 ; cf. Lemma 1.2.1.

6.4

Closed Subsets Generated by an Involution, III

165

By Lemma 1.2.1, b ∈ c2 if and only if c∗ ∈ bc. In Lemma 6.2.4(ii), we have seen that bc is ∗ -invariant. Thus, c∗ ∈ bc if and only if c ∈ bc. Note finally that c ∈ bc if and only if b ∈ cc∗ . (ii) We have b ∈ c∗ c if and only if c ∈ cb, and from Lemma 6.3.2(iii) we know that ac = cb. Thus, c ∈ cb if and only if c ∈ ac. Note finally that c ∈ ac is equivalent to a ∈ cc∗ . (iii) We have c ∈ c∗ c if and only if c ∈ cc∗ ; cf. Lemma 1.2.1. In Lemma 6.3.2(i), we have seen that cc∗ = a2 . Thus, c ∈ cc∗ if and only if c ∈ a2 . Finally, as a is symmetric, c ∈ a2 if and only if a ∈ ac∗ ; cf. Lemma 1.2.1. We have a ∈ ac∗ if and only if a ∈ ac. In Lemma 6.3.2(iii), we have seen that ac = cb. Thus, a ∈ ac if and only if a ∈ cb. Finally, as a and b both are symmetric, a ∈ cb if and only if a ∈ bc∗ ; cf. Lemma 1.2.1. We have a ∈ bc∗ if and only if b ∈ ac. In Lemma 6.3.2(iii), we have seen that ac = cb. Thus, we have b ∈ ac if and only if b ∈ cb. Finally, as b is symmetric,  b ∈ cb if and only if b ∈ bc; cf. Lemma 1.2.1.

6.4 Closed Subsets Generated by an Involution, III We continue our study of hypergroups containing an involution t and pairwise distinct elements a, b, and c satisfying a∗ = a, b∗ = b, ahti = {a, c}, and bhti = {b, c∗ }. We now focus on the case where at = {a, c}

and

bt = {c∗ }.

Our results will be applied only in Section 7.4. Lemma 6.4.1 Let H be a hypergroup, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H, and assume that a∗ = a, b∗ = b, at = {a, c}, and bt = {c∗ }. Then the following hold. (i) We have c∗ c = b2 ∪ bc. (ii) We have ac = ab ∪ cb. (iii) We have a2 = ac∗ ∪ cc∗ . Proof. (i) We are assuming that at = {a, c}, and this implies that at , {c}. Thus, by Lemma 6.2.2, c∗ t , {b}. It follows that c∗ t = {b, c∗ }; cf. Lemma 6.1.2(i). Thus, by Lemma 1.3.1(ii), tc = {b, c}. Since we are assuming that bt = {c∗ }, we now conclude that c∗ c = btc = b{b, c} = b2 ∪ bc. (ii) We are assuming that bt = {c∗ }. Thus, by Lemma 1.2.2, tb = {c}. Since we are assuming that at = {a, c}, we now conclude that ac = atb = {a, c}b = ab ∪ cb. (iii) We are assuming that bt = {c∗ }. Thus, by Corollary 6.2.3, ct = {a}. It follows that tc∗ = {a}; cf. Lemma 1.2.2. Since we are assuming that at = {a, c}, we now conclude that a2 = atc∗ = {a, c}c∗ = ac∗ ∪ cc∗ . 

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6 Involutions

Lemma 6.4.2 Let H be a hypergroup, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H, and assume that a∗ = a, b∗ = b, at = {a, c}, and bt = {c∗ }. Then the following hold. (i) The statements a ∈ c∗ c, a ∈ c2 , c∗ ∈ ac, c ∈ ac∗ , c∗ ∈ ac∗ , and b ∈ a2 are pairwise equivalent. (ii) The statements b ∈ c2 , b ∈ cc∗ , c ∈ bc, c∗ ∈ bc, c ∈ bc∗ , and a ∈ b2 are pairwise equivalent. (iii) We have c ∈ c∗ c if and only if b ∈ ac. Proof. (i) We have a ∈ c∗ c if and only if c∗ ∈ ac∗ . By Corollary 6.2.5(ii), ac∗ is ∗ -invariant. Thus, c∗ ∈ ac∗ is equivalent to c ∈ ac∗ . Furthermore, c ∈ ac∗ if and only if a ∈ c2 . Since a∗ = a, a ∈ c2 if and only if c∗ ∈ ac; cf. Lemma 1.2.1. Finally, by Corollary 6.2.5(i), c2 = ab, so that, by Lemma 1.2.4(ii), a ∈ c2 if and only if b ∈ a2 . (ii) From Corollary 6.2.5(ii) we know that bc is ∗ -invariant. Thus, the statements b ∈ cc∗ , c ∈ bc, c∗ ∈ bc, and c ∈ bc∗ are pairwise equivalent. From Lemma 1.2.1 we know that b ∈ c2 if and only if c∗ ∈ bc. From Corollary 6.2.5(i) we know that c2 = ab. It follows that (c∗ )2 = ba; cf. Lemma 1.3.1(ii). Thus, applying Lemma 1.2.4(ii) to b, a and c∗ in place of a, b, and c we obtain that b ∈ (c∗ )2 if and only if a ∈ b2 . Note finally that b ∈ (c∗ )2 if and only if b ∈ c2 . (iii) We have c ∈ c∗ c if and only if c ∈ c2 . From Corollary 6.2.5(i) we know that  c2 = ab. Thus, c ∈ c2 is equivalent to c ∈ ab, and that is equivalent to b ∈ ac. Lemma 6.4.3 Let H be a hypergroup, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H, and assume that a∗ = a, b∗ = b, at = {a, c}, and bt = {c∗ }. Assume further that c < b2 . Then the statements b ∈ c∗ c, b ∈ b2 , and c∗ ∈ bc∗ are pairwise equivalent. Proof. Assume first that b ∈ c∗ c, and recall from Lemma 6.4.1(i) that c∗ c = b2 ∪ bc. Then b ∈ b2 or b ∈ bc. Since we are assuming that c < b2 , we have b < bc. Thus, b ∈ b2 . Assume that b ∈ b2 . Then, as we are assuming that bt = {c∗ }, we obtain that  c∗ ∈ b2 t = bc∗ . Now we are done, since c∗ ∈ bc∗ implies that b ∈ c∗ c. Lemma 6.4.4 Let H be a hypergroup, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H, and assume that a∗ = a, b∗ = b, at = {a, c}, and bt = {c∗ }. Assume further that c∗ c ⊆ {1, a, b}. Then the following hold. (i) We have bc = {a}. (ii) We have b2 = {1, b}.

6.4

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167

(iii) We have c∗ c = {1, a, b}. (iv) We have c2 ∩ {a, b, c, c∗ } = {a, c∗ }. (v) We have cc∗ ∩ {a, b, c, c∗ } = {a}. (vi) We have bc∗ ∩ {a, b, c, c∗ } = {c∗ }. (vii) We have ac ∩ {a, b, c, c∗ } = {a, c, c∗ }. (viii) We have {a, b, c, c∗ } ⊆ ac∗ . (ix) We have {a, b, c, c∗ } ⊆ a2 . Proof. (i) From Lemma 6.4.1(i) we know that c∗ c = b2 ∪ bc. By hypothesis, we have c∗ c ⊆ {1, a, b}. Thus, b2 ⊆ {1, a, b} and bc ⊆ {1, a, b}. From b2 ⊆ {1, a, b} we obtain that c < b2 . Thus, b < bc. Since b , c, we also have 1 < bc; cf. Lemma 1.1.6. Thus, as bc ⊆ {1, a, b}, we have bc = {a}. (ii) From Lemma 6.4.1(i) we know that b2 ⊆ c∗ c. Since we are assuming that c∗ c ⊆ {1, a, b}, this implies that b2 ⊆ {1, a, b}. We first show that a < b2 . Assume, by way of contradiction, that a ∈ b2 . Then b ∈ ab. On the other hand, by Corollary 6.2.5(i), c2 = ab. Thus, b ∈ c2 . It follows that c∗ ∈ bc; cf. Lemma 1.2.1. Now recall from Lemma 6.4.1(i) that bc ⊆ c∗ c. Thus, c∗ ∈ c∗ c, contradicting our hypothesis. So far, we have seen that b2 ⊆ {1, b}. Note that 1 ∈ b2 , since b∗ = b. Thus, we shall be done if we succeed in showing that b ∈ b2 . Assume, by way of contradiction, that b < b2 . Then, as b2 ⊆ {1, b}, b is thin. Thus, by Lemma 1.4.3(i), |ab| = 1. On the other hand, by (i), a ∈ bc, whence c∗ ∈ ab; cf. Lemma 1.2.1. It follows that ab = {c∗ }. Since bc ⊆ c∗ c we obtain from a ∈ bc also that a ∈ c∗ c. Thus, by Lemma 6.4.2(i), a ∈ c2 . Since c2 = ab, this implies that a ∈ ab. From ab = {c∗ } and a ∈ ab we obtain that a = c∗ , contrary to Lemma 6.2.1(i). (iii) Since we are assuming that c∗ c ⊆ {1, a, b}, we just have to show that {1, a, b} ⊆ c∗ c. From Lemma 1.1.1 we know that 1 ∈ c∗ c. Assume, by way of contradiction, that a < c∗ c. Since we are assuming that c∗ c ⊆ {1, a, b}, we then obtain that c∗ c ⊆ {1, b}. From Lemma 6.4.1(i) we know that bc ⊆ c∗ c. Thus, bc ⊆ {1, b}. Since b , c, we have 1 < bc. Thus, by Lemma 1.1.3, b ∈ bc. It follows that c ∈ b2 , contrary to (ii). This contradiction shows that a ∈ c∗ c. From (ii) we know that b ∈ b2 , from Lemma 6.4.1(i) that b2 ⊆ c∗ c. Thus, b ∈ c∗ c. (iv) From (iii) we know that a ∈ c∗ c. Thus, by Lemma 6.4.2(i), a ∈ c2 . From (i) we know that a ∈ bc. Thus, by Lemma 1.2.1, c∗ ∈ ab. Now recall from Corollary 6.2.5(i) that c2 = ab. Thus, c∗ ∈ c2 . From (ii) we know that a < b2 . Thus, by Lemma 6.4.2(ii), b < c2 . From c < c∗ c we obtain that c < c2 .

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(v) In (iii), we saw that b ∈ c∗ c. Thus, c ∈ cb. On the other hand, by Lemma 6.4.1(ii), cb ⊆ ac. Thus, c ∈ ac. It follows that a ∈ cc∗ . In (ii) , we saw that a < b2 . Thus, by Lemma 6.4.2(ii), b < cc∗ . Since c < c∗ c, c < cc∗ ; cf. Lemma 1.2.1. Thus, as cc∗ is ∗ -invariant, c∗ < cc∗ . (vi) From (iii) we know that b ∈ c∗ c. Thus, c∗ ∈ bc∗ . From (iv) we know that c < c2 , from Corollary 6.2.5(i) that c2 = ab. Thus, c < ab. It follows that a < bc∗ ; cf. Lemma 1.2.1. By (ii), c < b2 . Thus, by Lemma 1.2.1, b < bc∗ . From (iv) we know that b < c2 . Thus, c < bc∗ . (vii) From Corollary 6.2.5(i) we know that c2 = ab, from Lemma 6.4.1(ii) that ab ⊆ ac. Thus, c2 ⊆ ac. Thus, by (iv), {a, c∗ } ⊆ ac. From (v) we know that a ∈ cc∗ . Thus, c ∈ ac. From (iv) we know that c < c2 , from Corollary 6.2.5(i) that c2 = ab. Thus, c < ab. It follows that b < ac. (viii) In (iv), we saw that c∗ ∈ c2 , and from Corollary 6.2.5(i) we know that c2 = ab. Thus, c∗ ∈ ab. It follows that b ∈ ac∗ . In (iv), we saw that a ∈ c2 . Thus, c ∈ ac∗ . In (iii), we saw that a ∈ c∗ c. Thus, c∗ ∈ ac∗ . From c ∈ ac∗ we obtain that c ∈ a2 , since we have seen in Lemma 6.4.1(iii) that ac∗ ⊆ a2 . Thus, by Lemma 1.2.1, a ∈ ac∗ . (ix) In (viii), we saw that {a, b, c, c∗ } ⊆ ac∗ , and from Lemma 6.4.1(iii) we know that ac∗ ⊆ a2 . Thus, {a, b, c, c∗ } ⊆ a2 .  Lemma 6.4.5 Let H be a hypergroup, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H, and assume that a∗ = a, b∗ = b, at = {a, c}, and bt = {c∗ }. Assume further that c ∈ c∗ c. Then the following hold. (i) We have {a, c, c∗ } ⊆ cc∗ . (ii) We have {a, b, c} ⊆ ac. (iii) We have {a, c, c∗ } ⊆ a2 . Proof. (i) From c ∈ c∗ c we obtain that c ∈ c2 . From Corollary 6.2.5(i) we know that c2 = ab, and from Lemma 6.4.1(ii) we know that ab ⊆ ac. Thus, c ∈ ac; equivalently, a ∈ cc∗ . From c ∈ c∗ c we also obtain that {c, c∗ } ⊆ cc∗ ; cf. Lemma 1.2.1. (ii) From c ∈ c∗ c we obtain that c ∈ c2 , and from Corollary 6.2.5(i) we know that c2 = ab. Thus, c ∈ ab; equivalently, a ∈ cb. Now recall from Lemma 6.4.1(ii) that cb ⊆ ac. Thus, a ∈ ac. From c ∈ c∗ c we obtain that b ∈ ac; cf. Lemma 6.4.2(iii). From (i) we know that a ∈ cc∗ . Thus, c ∈ ac.

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(iii) From Lemma 6.4.1(iii) we know that cc∗ ⊆ a2 . Thus, the claim follows from  (i). Lemma 6.4.6 Let H be a hypergroup, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H, and assume that a∗ = a, b∗ = b, at = {a, c}, and bt = {c∗ }. Assume further that a < c∗ c. Then the following hold. (i) We have bc ∩ {a, b, c, c∗ } ⊆ {b}. (ii) We have c2 ∩ {a, b, c, c∗ } ⊆ {c}. (iii) We have ac∗ ∩ {a, b, c, c∗ } ⊆ {a}. Proof. (i) By Lemma 6.4.1(i), bc ⊆ c∗ c. Thus, as we are assuming that a < c∗ c, a < bc. Assume, by way of contradiction, that c∗ ∈ bc. Then by Lemma 1.2.1, c∗ ∈ cb. Now recall from Lemma 6.4.1(ii) that cb ⊆ ac. Thus, c∗ ∈ ac. Thus, by Lemma 6.4.2(i), a ∈ c∗ c, contradicting our hypothesis. Thus, c∗ < bc. In Corollary 6.2.5(ii), we saw that bc is ∗ -invariant. Thus, as c∗ < bc, c < bc. (ii) Since a < c∗ c, a < c2 ; cf. Lemma 6.4.2(i). From a < c∗ c we obtain that c∗ < ac; cf. Lemma 6.4.2(i). In Lemma 6.4.1(ii), we saw that cb ⊆ ac. Thus, c∗ < cb, so that, by Lemma 1.2.1, b < c2 . From (i) we know that a < bc. Thus, by Lemma 1.2.1, c∗ < ab. Now recall from Corollary 6.2.5(i) that c2 = ab. Thus, c∗ < c2 . (iii) By (i), a < bc. Thus, b < ac∗ . By (ii), a < c2 . Thus, c < ac∗ . Since a < c∗ c, we  have c∗ < ac∗ . Lemma 6.4.7 Let H be a hypergroup, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H, and assume that a∗ = a, b∗ = b, at = {a, c}, and bt = {c∗ }. Assume further that b < c∗ c. Then the following hold. (i) We have b2 ∩ {a, b, c, c∗ } ⊆ {a}. (ii) We have b < bc. (iii) We have c∗ < bc∗ . (iv) If a < c∗ c, bc ∩ {a, b, c, c∗ } is empty. Proof. (i) Recall from Lemma 6.4.1(i) that c∗ c = b2 ∪ bc. Since we are assuming that b < c∗ c, this implies that b < b2 and b < bc. From b < bc we obtain that c < b2 . Since b2 is ∗ -invariant, we obtain from c < b2 that c∗ < b2 . (ii) Recall from Lemma 6.4.1(i) that bc ⊆ c∗ c. Since we are assuming that b < c∗ c, this implies that b < bc. (iii) From (i) we know that b < b2 and c < b2 . Thus, by Lemma 6.4.3, c∗ < bc∗ .

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(iv) Assume that a < c∗ c. Then, by Lemma 6.4.6(i), bc ∩ {a, b, c, c∗ } ⊆ {b}. In (ii),  we also saw that b < bc. Thus, bc ∩ {a, b, c, c∗ } is empty. Lemma 6.4.8 Let H be a hypergroup, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H, and assume that a∗ = a, b∗ = b, at = {a, c}, and bt = {c∗ }. Assume further that {a, b, c, c∗ } ⊆ c∗ c. Then the following hold. (i) We have {a, c} ⊆ c2 . (ii) We have {a, c∗ } ⊆ bc∗ . (iii) We have {a, b, c, c∗ } ⊆ ac. (iv) We have {a, c, c∗ } ⊆ ac∗ . (v) We have {a, b, c, c∗ } ⊆ a2 . Proof. (i) Since a ∈ c∗ c, a ∈ c2 ; cf. Lemma 6.4.2(i). Since c ∈ c∗ c, c ∈ c2 . (ii) Since c ∈ c∗ c, b ∈ ac; cf. Lemma 6.4.2(iii). From b ∈ ac we obtain that a ∈ bc∗ . Since we are assuming that b ∈ c∗ c, we also have c∗ ∈ bc∗ . (iii) From Lemma 6.4.5(ii) we know that {a, b, c} ⊆ ac. Since we are assuming that a ∈ c∗ c, we also have c∗ ∈ ac; cf. Lemma 6.4.2(i). (iv) From (iii) we know that a ∈ ac. Thus, a ∈ ac∗ . Since we are assuming that a ∈ c∗ c, we also have {c, c∗ } ⊆ ac∗ ; cf. Lemma 6.4.2(i). (v) From Lemma 6.4.5(iii) we know that {a, c, c∗ } ⊆ a2 . Since we are assuming that a ∈ c∗ c, we also have b ∈ a2 ; cf. Lemma 6.4.2(i). 

6.5 Length Functions Defined by Sets of Involutions Let T be a set of involutions of a hypergroup. For each subset A of hTi, we write `T (A) to denote the set of all non-negative integers `T (a) with a ∈ A. Lemma 6.5.1 Let T be a set of involutions of a hypergroup, and let t be an element in T. Let f be an element in hTi, and set n := `T ( f ). Then the set `T ( f t) is equal to one of the sets {n − 1}, {n − 1, n}, {n}, {n, n + 1}, or {n + 1}. Proof. For each element c in f t, we have n − 1 ≤ `T (c) ≤ n + 1; cf. Lemma 2.3.6. Let d and e be elements in f t. Then, by Lemma 2.1.6(i), e ∈ dhti ⊆ {d} ∪ dt, and that implies that `T (e) ≤ `T (d) + 1. Similarly, `T (d) ≤ `T (e) + 1, and the result follows.  Let T be a set of involutions of a hypergroup. In Lemma 2.3.6(i), we saw that, for any three elements c, d, and e in hTi with e ∈ cd,

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`T (e) ≤ `T (c) + `T (d). This observation suggests a focus on elements c, d, and e in hTi satisfying e ∈ cd and `T (e) = `T (c) + `T (d). For each element d in hTi, we define T−1 (d) to be the set of all elements e in hTi such that ed ∗ contains an element c satisfying `T (e) = `T (c) + `T (d). For each element d in hTi, we define T1 (d) to be the set of all elements c in hTi such that cd contains an element e satisfying `T (e) = `T (c) + `T (d). Note that, for each element f in hTi, f ∈ T−1 ( f ) and 1 ∈ T1 ( f ). Lemma 6.5.2 Let T be a set of involutions of a hypergroup, and let d and e be elements in hTi. Then we have the following. (i) If d ∈ T1 (e), e∗ ∈ T1 (d ∗ ). (ii) If T−1 (d) ∩ T1 (e) is not empty, d ∈ T1 (e). (iii) If e ∈ T−1 (d), T−1 (e) ⊆ T−1 (d). (iv) If e ∈ T−1 (d), T1 (e∗ ) ⊆ T1 (d ∗ ). Proof. (i) Assume that d ∈ T1 (e). Then de contains an element f with `T ( f ) = `T (d) + `T (e). From f ∈ de we obtain that f ∗ ∈ e∗ d ∗ ; cf. Lemma 1.2.1. From `T ( f ) = `T (d) + `T (e) we obtain that `T ( f ∗ ) = `T (e∗ ) + `T (d ∗ ); cf. Lemma 2.3.5. Thus, we have e∗ ∈ T1 (d ∗ ). (ii) Let b be an element in T−1 (d) ∩ T1 (e). Since b ∈ T−1 (d), there exists an element a in hTi such that b ∈ ad and `T (b) = `T (a) + `T (d). Since b ∈ T1 (e), there exists an element f in be such that `T ( f ) = `T (b) + `T (e). Thus, by Lemma 2.3.8(ii), de contains an element c such that f ∈ ac, `T (c) = `T (d) + `T (e), and `T ( f ) = `T (a) + `T (c). From c ∈ de and `T (c) = `T (d) + `T (e) we obtain d ∈ T1 (e). (iii) Let f be an element in T−1 (e). Then f e∗ contains an element c such that `T ( f ) = `T (c) + `T (e). Now assume that e ∈ T−1 (d). Then ed ∗ contains an element a such that `T (e) = `T (a) + `T (d). Now, by Lemma 2.3.8(i), ca contains an element b such that f ∈ bd, `T (b) = `T (c) + `T (a), and `T ( f ) = `T (b) + `T (d). From f ∈ bd and `T ( f ) = `T (b) + `T (d) we obtain that f ∈ T−1 (d). (iv) Assume that e ∈ T−1 (d), and let f be an element in T1 (e∗ ). We have to show that f ∈ T1 (d ∗ ). Since f ∈ T1 (e∗ ), e ∈ T1 ( f ∗ ); cf. (i). Thus, as e ∈ T−1 (d), d ∈ T1 ( f ∗ ); cf. (ii). Now,  by (i), f ∈ T1 (d ∗ ). Lemma 6.5.3 Let T be a set of involutions of a hypergroup, and let t be an element in T. Let f be an element in hTi, and set n := `T ( f ). Then the following hold.

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(i) We have f ∈ T−1 (t) if and only if `T ( f t) = {n − 1} or `T ( f t) = {n − 1, n}. (ii) We have f ∈ T1 (t) if and only if `T ( f t) = {n, n + 1} or `T ( f t) = {n + 1}. Proof. (i) Since `T ( f ) = n, we have f ∈ T−1 (t) if and only if f t contains an element e with `T (e) = n − 1. Thus, the claim follows from Lemma 6.5.1. (ii) Since `T ( f ) = n, we have f ∈ T1 (t) if and only if f t contains an element e with `T (e) = n + 1. Thus, the claim follows from Lemma 6.5.1.  Let T be a set of involutions of a hypergroup. Let t be an element in T. Recall from Lemma 1.4.3(i) that, if t is thin, | f t| = 1 for each element f in hTi. Thus, `T ( f t) cannot contain two different elements. Thus, the second and the fourth case in Lemma 6.5.1 do not occur if t is thin. For each element t in T, we define T0 (t) to be the set of all elements d in hTi such that, for each element e in dt, `T (e) = `T (d). We emphasize that T0 (t) is defined only for elements t in T, whereas the sets T−1 ( f ) and T1 ( f ) are defined for any element f in hTi. Lemma 6.5.4 Let T be a set of involutions of a hypergroup, and let u and v be elements in T. Then the statements u ∈ T0 (v), uv ⊆ T, and v ∈ T0 (u) are pairwise equivalent. Proof. Assume first that u ∈ T0 (v). Then, for each element f in uv, `T ( f ) = `T (u). Since u ∈ T, `T (u) = 1. Thus, for each element f in uv, `T ( f ) = 1, and that means that uv ⊆ T. Assume now that uv ⊆ T. Then for each element f in uv, `T ( f ) = 1. Thus, as `T (u) = 1, we have `T ( f ) = `T (u) for each element f in uv, and that means that u ∈ T0 (v). Note also that, by Lemma 6.1.7(i), uv ⊆ T if and only if vu ⊆ T.



Lemma 6.5.5 Let T be a set of involutions of a hypergroup, and let t be an element in T. Then the following hold. (i) The set {T−1 (t), T0 (t), T1 (t)} is a partition of hTi. (ii) For each element f in T0 (t), we have f hti ⊆ T0 (t). (iii) If t is thin, so is T ∩ T0 (t). Proof. (i) Considering Lemma 6.5.1 this follows from Lemma 6.5.3. (ii) Let f be an element in T0 (t), and let e be an element in f hti. We have to show that e ∈ T0 (t). There is nothing to show if e = f . Thus, we assume that e , f . Since e ∈ f hti, this implies that e ∈ f t; cf. Lemma 6.1.1(ii). Thus, as f ∈ T0 (t), `T (e) = `T ( f ). Let d be an element in et. We have to show that `T (d) = `T (e).

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From d ∈ et and e ∈ f hti we obtain that d ∈ f hti. It follows that `T (d) = `T ( f ). Thus, as `T (e) = `T ( f ), `T (d) = `T (e). (iii) Assume that t is thin, and let u be an element in T ∩ T0 (t). Since u ∈ T0 (t), we have ut ⊆ T; cf. Lemma 6.5.4. Since t is thin, this implies that hu, ti is thin; cf. Lemma 6.1.8(ii). As a consequence, u is thin.  The following lemma will play a crucial role in Section 8.4. Lemma 6.5.6 Let T be a set of involutions of a hypergroup, let t be an element in T, and assume that T0 (t) is not empty. Choose an element e in T0 (t) such that `T (e) is as small as possible. Let c and d be elements in hTi such that e ∈ cd and `T (e) = `T (c) + `T (d). Assume that c , 1. Then d ∈ T1 (t). Proof. Assume first that d ∈ T−1 (t). Then, by Lemma 6.5.2(iii), T−1 (d) ⊆ T−1 (t). On the other hand, as e ∈ cd and `T (e) = `T (c) + `T (d), e ∈ T−1 (d). Thus, e ∈ T−1 (t), contrary to the choice of e as an element in T0 (t). This contradiction shows that d < T−1 (t). From `T (e) = `T (c) + `T (d) and c , 1 we obtain that `T (d) ≤ `T (e) − 1. Thus, the minimal choice of e forces d < T0 (t). Now, as d < T−1 (t) and d < T0 (t), we obtain from Lemma 6.5.5(i) that d ∈ T1 (t).  Let T be a set of involutions of a hypergroup, and let A be a subset of hTi. If A is empty, we set T1 (A) := hTi and T−1 (A) := hTi. If A is not empty, we define T1 (A) to be the intersection of the sets T1 (a) with a ∈ A, and T−1 (A) will denote the intersection of the sets T−1 (a) with a ∈ A. Lemma 6.5.7 Let T be a set of involutions of a hypergroup, let e be an element in hTi, and let d be an element in T1 (e). Then T1 (de) ⊆ T1 (d). Proof. Since d ∈ T1 (e), de contains an element f such that `T ( f ) = `T (d) + `T (e). From f ∈ de we obtain that f ∗ ∈ e∗ d ∗ . From `T ( f ) = `T (d) + `T (e) we obtain that `T ( f ∗ ) = `T (e∗ ) + `T (d ∗ ). It follows that f ∗ ∈ T−1 (d ∗ ). Now let c be an element in T1 (de). Then, as f ∈ de, we obtain that c ∈ T1 ( f ). Thus, by Lemma 6.5.2(i), f ∗ ∈ T1 (c∗ ). From f ∗ ∈ T−1 (d ∗ ) ∩ T1 (c∗ ) we obtain that d ∗ ∈ T1 (c∗ ); cf. Lemma 6.5.2(ii). Thus,  by Lemma 6.5.2(i), c ∈ T1 (d). Lemma 6.5.8 Let V be a set of involutions of a hypergroup, and let U be a subset of V. Assume that, for each element u in U, hVi = V−1 (u) ∪ V1 (u). Then the following hold. (i) We have hVi = V1 (U)hUi.

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(ii) For each element e in hVi, V1 (U) ∩ ehUi is not empty. Proof. (i) By definition, V1 (U) ⊆ hVi, and, by Lemma 2.3.1(i), hUi ⊆ hVi. Thus, as hVi is closed, we conclude that V1 (U)hUi ⊆ hVi. To show the reverse containment, we choose an element in hVi and denote it by e. We will see that e ∈ V1 (U)hUi. There is nothing to show if e ∈ V1 (U), so we assume that e < V1 (U). Then U contains an element u with e < V1 (u). Thus, by hypothesis, e ∈ V−1 (u), and that means that eu contains an element d with `V (e) = `V (d) + 1. From d ∈ eu, together with e ∈ hVi and u ∈ U ⊆ V, we obtain that d ∈ hVi. Thus, as `V (e) = `V (d) + 1, induction forces d ∈ V1 (U)hUi. On the other hand, since d ∈ eu, e ∈ du. Thus, as u ∈ U, we conclude that e ∈ V1 (U)hUi. (ii) Let e be an element in hVi. Then, by (i), e ∈ V1 (U)hUi, so that we find an element d in V1 (U) with e ∈ dhUi. From e ∈ dhUi we obtain that d ∈ ehUi; cf. Lemma 2.1.6(i).  Let U and V be sets of involutions of hypergroups. A bijective map φ from hUi to hVi will be called (U, V)-length preserving if, for each element f in hUi, `U ( f ) = `V ( f φ ). (Notice that this implies that U φ = V.) If U = V, we will say U-length preserving instead of (U, V)-length preserving. Lemma 6.5.9 Let U and V be sets of involutions of hypergroups, and let υ be a (U, V)-length preserving bijective map from hUi to hVi. Assume that eυ ∈ cυ d υ for any three elements c, d, and e in hUi with e ∈ cd and `U (e) = `U (c) + `U (d). Then the following hold. (i) If e ∈ U−1 (d), eυ ∈ V−1 (d υ ). (ii) If d ∈ U1 (e), d υ ∈ V1 (eυ ). Proof. (i) Assume that e ∈ U−1 (d). Then ed ∗ contains an element c with `U (e) = `U (c) + `U (d). From c ∈ ed ∗ we obtain that e ∈ cd. Thus, as `U (e) = `U (c) + `U (d), the hypothesis of the lemma forces eυ ∈ cυ d υ . It follows that cυ ∈ eυ (d υ )∗ . On the other hand, since υ is (U, V)-length preserving, we obtain from `U (e) = `U (c) + `U (d) that `V (eυ ) = `V (cυ ) + `V (d υ ). From cυ ∈ eυ (d υ )∗ and `V (eυ ) = `V (cυ ) + `V (d υ ) we obtain that eυ ∈ V−1 (d υ ). (ii) Assume that d ∈ U1 (e). Then de contains an element f with `U ( f ) = `U (d) + `U (e). By hypothesis, we obtain from f ∈ de and `U ( f ) = `U (d) + `U (e) that f υ ∈ d υ eυ . Since υ is (U, V)-length preserving, we obtain from `U ( f ) = `U (d) + `U (e) that `V ( f υ ) = `V (d υ ) + `V (eυ ). From f υ ∈ d υ eυ and `V ( f υ ) = `V (d υ ) + `V (eυ ) we obtain that d υ ∈ V1 (eυ ), as wanted. 

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Let U and V be sets of involutions of hypergroups. Notice that Lemma 6.5.9 can be applied to each (U, V)-length preserving hypergroup isomorphism from hUi to hVi in place of υ.

6.6 Hypergroups Generated by Two Involutions Hypergroups generated by two distinct thin involutions are thin; cf. Lemma 2.6.1. Thus, via the group correspondence, they correspond to groups generated by two distinct involutions, and these groups are exactly the dihedral groups. For the remainder of this section, we fix a hypergroup H and two distinct involutions u and v of H. We set R0 (u, v) := {1}. For each positive integer n, we define Rn (u, v) to be the product t1 · · · tn , where, for each element i in {1, . . . , n}, ti = u if i is odd and ti = v if i is even. (Notice that R1 (u, v) = {u}, R1 (v, u) = {v}, R2 (u, v) = uv, R2 (v, u) = vu, R3 (u, v) = uvu, R3 (v, u) = vuv, etc.) Lemma 6.6.1 Let i and j be positive integers with i + 1 ≤ j, and assume that Ri (u, v) ∩ R j (u, v) is not empty. Then there exists an element n in { j −i, . . . , j +i −1} such that 1 ∈ Rn (u, v) or 1 ∈ Rn (v, u). Proof. From Lemma 6.1.1(iii), (iv) we obtain that 1 ∈ R j (u, v)∗ Ri (u, v) ⊆ R j−1 (v, u)∗ Ri−1 (v, u) ∪ R j−1 (v, u)∗ uRi−1 (v, u). Note that or

R j−1 (v, u)∗ uRi−1 (v, u) = R j+i−1 (u, v) R j−1 (v, u)∗ uRi−1 (v, u) = R j+i−1 (v, u).

Thus, we are done in the case where 1 ∈ R j−1 (v, u)∗ uRi−1 (v, u). Assume that 1 ∈ R j−1 (v, u)∗ Ri−1 (v, u). Then, by Lemma 1.3.3, Ri−1 (v, u) ∩ R j−1 (v, u) is not empty. Suppose first that i = 1. Then Ri−1 (v, u) = {1}. Thus, as Ri−1 (v, u) ∩ R j−1 (v, u) is not empty, 1 ∈ R j−1 (v, u). Since i = 1, this implies that 1 ∈ R j−i (v, u), and we are done. Suppose now that 2 ≤ i. Then i − 1 is a positive integer. Thus, by induction, there exists an element n in { j − i, . . . , j + i − 3} such that 1 ∈ Rn (v, u) or 1 ∈ Rn (u, v).  Lemma 6.6.2 Set T := {u, v}, let f be an element in hTi, and set n := `T ( f ). Then the following hold.

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(i) We have f ∈ Rn (u, v) ∪ Rn (v, u). (ii) If f ∈ T−1 (T), 1 ∈ R2n (u, v). Proof. (i) Assume first that n = 0. Then f = 1 and Rn (u, v) = {1}. Thus, f ∈ Rn (u, v). Assume that n = 1. Then f ∈ {u, v}, Rn (u, v) = {u}, and Rn (v, u) = {v}. It follows that f ∈ Rn (u, v) ∪ Rn (v, u). Assume that 2 ≤ n. From `T ( f ) = n we obtain that f ∈ T n . Thus, T contains elements t1 , . . ., tn such that f ∈ t1 · · · tn . Assume that {1, . . . , n − 1} contains an element i with ti = ti+1 . Then there exist elements d in T i−1 and e in T n−i−1 such that f ∈ dti ti e ⊆ T n−2 ∪ T n−1, contrary to `T ( f ) = n. We have seen that ti , ti+1 for each element i in {1, . . . , n − 1}, and that implies that we have f ∈ Rn (u, v) ∪ Rn (v, u). (ii) Assume that f ∈ T−1 (T). Then, since u ∈ T, f ∈ T−1 (u). Thus, hTi contains an element d with f ∈ du and `T ( f ) = `T (d) + 1. It follows that d ∈ T1 (u). From `T ( f ) = `T (d) + 1 and `T ( f ) = n we obtain that `T (d) = n − 1. Thus, by (i), d ∈ Rn−1 (u, v) or d ∈ Rn−1 (v, u). Since d ∈ T1 (u), we must have d ∈ Rn−1 (u, v) if n is odd and d ∈ Rn−1 (v, u) if n is even. Similarly, we obtain from v ∈ T that hTi contains an element e such that f ∈ ev, e ∈ Rn−1 (v, u) if n is odd, and e ∈ Rn−1 (u, v) if n is even. If n is odd, du ⊆ Rn−1 (u, v)u = Rn (u, v) and ev ⊆ Rn−1 (v, u)v = Rn (v, u). Thus, 1 ∈ f ∗ f ⊆ ud ∗ ev ⊆ Rn (u, v)∗ Rn (v, u) = R2n (u, v). If n is even, du ⊆ Rn−1 (v, u)u = Rn (v, u) and ev ⊆ Rn−1 (u, v)v = Rn (u, v). Thus, 1 ∈ f ∗ f ⊆ ud ∗ ev ⊆ Rn (v, u)∗ Rn (u, v) = R2n (u, v), and we are done in both cases.



Lemma 6.6.3 Assume that there exists a positive integer n satisfying 1 ∈ Rn (u, v). Then the smallest positive integer n satisfying 1 ∈ Rn (u, v) is even. Proof. Let n denote the smallest positive integer satisfying 1 ∈ Rn (u, v). By way of contradiction, we assume that n is odd. Since 1 , u, we have n , 1. Thus, 3 ≤ n. From 1 ∈ Rn (u, v) we obtain that u ∈ Rn−1 (v, u); cf. Lemma 1.3.3. Thus, as n − 1 is even, Rn−2 (v, u) contains an element e with u ∈ eu. It follows that e ∈ u2 . Thus, by Lemma 6.1.1(iii), (iv), e ∈ {1, u}. If e = u, we obtain that 1 ∈ ue ⊆ Rn−1 (u, v), contrary to the minimal choice of n. Thus, e = 1.

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Hypergroups Generated by Two Involutions

177

From e = 1, together with e ∈ Rn−2 (v, u), we obtain that 1 ∈ Rn−2 (v, u). Thus, by Lemma 1.3.3, v ∈ Rn−3 (u, v). Thus, as v , 1, n , 3. Since n is assumed to be odd, this implies that 5 ≤ n. Thus, as n − 3 is even, Rn−4 (u, v) contains an element d with v ∈ dv. It follows that d ∈ v 2 . Thus, by Lemma 6.1.1(iii), (iv), d ∈ {1, v}. If d = v, we obtain that 1 ∈ dv ⊆ Rn−3 (u, v), contrary to the minimal choice of n. Thus, d = 1. From d = 1, together with d ∈ Rn−4 (u, v), we obtain that 1 ∈ Rn−4 (u, v),  contrary to the minimal choice of n. We define C(u, v) to be the set of all positive integers n with 1 ∈ Rn (u, v) and notice that C(u, v) may be empty. In Lemma 6.6.3, we saw that min(C(u, v)) is even if C(u, v) is not empty. We set 1 min(C(u, v)) if C(u, v) , ∅ cu,v := 2 ∞ if C(u, v) = ∅ and call cu,v the Coxeter number of u and v. Since u , v, we have 2 ≤ cu,v ; cf. Lemma 1.1.6. Note also that, by Lemma 1.3.1(ii), Rn (u, v) = Rn (v, u)∗ for each even positive integer n, and that implies that cu,v = cv,u . Lemma 6.6.4 Assume that C(u, v) is not empty, and set T := {u, v}. Let n be an element in {1, . . . , cu,v }, and let f be an element in Rn (u, v). Then `T ( f ) = n. Proof. Set m := `T ( f ). Since f ∈ Rn (u, v), m ≤ n. It remains to show that n ≤ m. From `T ( f ) = m we obtain that f ∈ Rm (u, v) or f ∈ Rm (v, u); cf. Lemma 6.6.2(i). Assume first that f ∈ Rm (u, v). Then f ∈ Rm (u, v) ∩ Rn (u, v). Suppose that m + 1 ≤ n. Since m , 0, we obtain from Lemma 6.6.1 an element k in {n − m, . . . , n + m − 1} such that 1 ∈ Rk (u, v) or 1 ∈ Rk (v, u). From n − m ≤ k ≤ n + m − 1 and m + 1 ≤ n ≤ cu,v we obtain that 1 ≤ k ≤ 2cu,v − 1. From 1 ∈ Rk (u, v) ∪ Rk (v, u) (together with 1 ≤ k), on the other hand, we obtain that 2cu,v ≤ k. This contradiction proves that n ≤ m. Assume now that f ∈ Rm (v, u). Then f ∈ Rm (v, u) ∩ Rn (u, v). From f ∈ Rm (v, u) ∩ Rn (u, v), together with 1 ∈ f ∗ f , we obtain that 1 ∈ Rn+m (u, v) or 1 ∈ Rn+m (v, u). Thus, 2cu,v ≤ n + m. Since n ≤ cu,v , this implies that 2n ≤ n + m. It follows that n ≤ m.  Lemma 6.6.5 Assume that C(u, v) is not empty, and set n := cu,v . Then we have the following. (i) The set Rn (u, v) ∩ Rn (v, u) is not empty. (ii) Set T := {u, v}. Then Rn (u, v) ∩ Rn (v, u) ⊆ T−1 (T). Proof. (i) From n = cu,v we obtain that 1 ∈ R2n (u, v). Thus, by Lemma 1.3.3, Rn (u, v) ∩ Rn (v, u) is not empty.

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(ii) Let f be an element in Rn (u, v) ∩ Rn (v, u). Since f ∈ Rn (u, v), `T ( f ) = n; cf. Lemma 6.6.4. Since f ∈ Rn (u, v), there exists an element e in Rn−1 (u, v) such that f ∈ eu if n is odd and f ∈ ev if n is even. From e ∈ Rn−1 (u, v) we obtain that `T (e) = n − 1; cf. Lemma 6.6.4. It follows that f ∈ T−1 (u) if n is odd and f ∈ T−1 (v) if n is even. From f ∈ Rn (v, u) one obtains similarly that f ∈ T−1 (v) if n is odd and f ∈ T−1 (u) if n is even. Thus, in both cases, f ∈ T−1 (T).  Corollary 6.6.6 Set T := {u, v}. Then C(u, v) is not empty if and only if T−1 (T) is not empty. Proof. From Lemma 6.6.5 we obtain that T−1 (T) is not empty if C(u, v) is not empty. Assume that T−1 (T) is not empty, and let f be an element in T−1 (T). Set n := `T ( f ).  Then, by Lemma 6.6.2(ii), 1 ∈ R2n (u, v). Thus, by definition, 2n ∈ C(u, v). Proposition 6.6.7 Assume that C(u, v) is not empty, and set T := {u, v}. Then we have min({`T ( f ) | f ∈ T−1 (T)}) = cu,v . Proof. Set m := min({`T ( f ) | f ∈ T−1 (T)} and n := cu,v . We shall see that m = n. Since C(u, v) is assumed not to be empty, Rn (u, v) ∩ Rn (v, u) is not empty; cf. Lemma 6.6.5(i). Let f be an element in Rn (u, v)∩ Rn (v, u). Then, by Lemma 6.6.4, `T ( f ) = n, and, by Lemma 6.6.5(ii), f ∈ T−1 (T). It follows that m ≤ n. To show that n ≤ m, we choose an element f in T−1 (T) with `T ( f ) = m. Then 1 ∈ R2m (u, v); cf. Lemma 6.6.2(ii). It follows that 2m ∈ C(u, v), so n ≤ m.  Lemma 6.6.8 Let n be a positive integer with |Rn (u, v)| = 1. Assume that C(u, v) is empty or that C(u, v) is not empty and n ≤ cu,v . Then |Rn−1 (u, v)| = 1. Proof. There is nothing to show if n ∈ {1, 2}. Therefore, we assume that 3 ≤ n. Since |Rn (u, v)| = 1, we find an element f in Rn (u, v) with Rn (u, v) = { f }. Let d and e be elements of Rn−1 (u, v). We will see that d = e. Assume that n is odd. Then du ⊆ Rn (u, v) and eu ⊆ Rn (u, v). Thus, du = { f } = eu. It follows that 1 ∈ f f ∗ = du2 e∗ . Thus, as u2 ⊆ {1, u}, we obtain that 1 ∈ de∗ or 1 ∈ due∗ . Assume that n is even. Then dv ⊆ Rn (u, v) and ev ⊆ Rn (u, v). Thus, dv = { f } = ev. It follows that 1 ∈ f f ∗ = dv 2 e∗ . Thus, as v 2 ⊆ {1, v}, we obtain that 1 ∈ de∗ or 1 ∈ dve∗ . If 1 ∈ de∗ , we obtain from Lemma 1.1.6 that d = e, so we are done in this case. If n is odd and 1 ∈ due∗ , we obtain that 1 ∈ R2n−1 (u, v). (Recall that {d, e} ⊆ Rn−1 (u, v).) Also if n is even and 1 ∈ dve∗ , we obtain 1 ∈ R2n−1 (u, v).

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Dichotomy and the Exchange Condition

179

From 1 ∈ R2n−1 (u, v) we conclude that 2n − 1 ∈ C(u, v). It follows that C(u, v) is not empty and that 2cu,v ≤ 2n − 1. From 2cu,v ≤ 2n − 1 we obtain that cu,v ≤ n − 1,  contradiction.

6.7 Dichotomy and the Exchange Condition A set T of involutions of a hypergroup is called dichotomic if, for each element t in T, hTi = T−1 (t) ∪ T1 (t). Let T be a set of involutions of a hypergroup. Recall from Section 6.5 that, for each element t in T, T0 (t) is our notation for the set of all elements d in hTi such that, for each element e in dt, `T (e) = `T (d). We now define T0 to be the union of the sets T0 (t) with t ∈ T. Recall from Section 6.1 that a hypergroup H is called projective if each element of H \ {1} is an involution. We will now say that a projective hypergroup has positive dimension if it contains more than one involution. Lemma 6.7.1 Let T be a set of involutions of a hypergroup. Then we have the following. (i) The set T is dichotomic if and only if T0 is empty. (ii) The set T does not contain the involutions of a projective closed subset of hTi having positive dimension if and only if T ∩ T0 is empty. Proof. (i) This follows from Lemma 6.5.5(i). (ii) Assume first that the intersection T ∩ T0 is not empty, and let u be an element in T ∩ T0 . Since u ∈ T0 , T contains an element v such that u ∈ T0 (v). From u ∈ T0 (v) we obtain that u , v. From {u, v} ⊆ T and u ∈ T0 (v) we obtain that uv ⊆ T; cf. Lemma 6.5.4. It follows that each element of uv is an involution. Thus, by Lemma 6.1.7(iii), hu, vi is a projective closed subset of hTi. Furthermore, by Lemma 6.1.7(ii), hu, vi = {1, u, v} ∪ uv. This latter equation tells us that hu, vi \ {1} ⊆ T. Assume now that T contains the involutions of a projective closed subset of hTi having positive dimension, and let u and v be two distinct elements of this closed subset. Then uv ⊆ T. Thus, by Lemma 6.5.4, u ∈ T0 (v). It follows that u ∈ T ∩T0 (v) ⊆  T ∩ T0 . Corollary 6.7.2 Let T be a dichotomic set of involutions of a hypergroup. Then T does not contain the involutions of a projective closed subset of hTi having positive dimension. Proof. This follows from Lemma 6.7.1.



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Let T be a set of involutions of a hypergroup. In Corollary 6.7.2, we saw that T does not contain the involutions of a projective closed subset of hTi having positive dimension if T is dichotomic. In Theorem 8.5.5, we will see a condition under which the converse holds. Lemma 6.7.3 Let V be a dichotomic set of involutions of a hypergroup. Then, for each subset U of V, hVi = V1 (U)hUi. Proof. This follows from Lemma 6.5.8(i).



In Corollary 9.2.3, we will prove a result which is similar to Lemma 6.7.3. Let T be a set of involutions of a hypergroup. An element e of hTi is said to have maximal T-length if, for each element d in hTi, `T (d) ≤ `T (e). The following lemma will be useful in Sections 9.3 and 9.6. Lemma 6.7.4 Let T be a dichotomic set of involutions of a hypergroup. Then each element in hTi having maximal T-length belongs to T−1 (T). Proof. Let d be an element in hTi having maximal T-length. Assume, by way of contradiction, that d < T−1 (T). Since d < T−1 (T), T contains an element t with d < T−1 (t). From d < T−1 (t) we obtain that d ∈ T1 (t), because T is assumed to be dichotomic. Now dt contains an element e such that `T (e) = `T (d) + 1. Since e ∈ hTi, this contradicts the choice of d as an element in hTi having maximal T-length.  A set T of involutions of a hypergroup is said to satisfy the exchange condition if, for any three elements u and v in T and f in T1 (v) with u ∈ T1 ( f ), u f ⊆ f v ∪ T1 (v). It is obvious that the empty set satisfies the exchange condition. Note also that each set consisting of a single involution of a hypergroup satisfies the exchange condition. It is easy to see that a set T of thin involutions satisfies the exchange condition if and only if the set of involutions which, via the group correspondence, corresponds to T satisfies the group theoretic Condition (F) as defined in [1; p. 79]. Condition (F) is one of the group theoretic conditions which are equivalent to the group theoretic exchange condition. The full power of the hypergroup theoretic exchange condition comes to light (in Chapters 9 and 10) when it is imposed on constrained sets of involutions of hypergroups. Constrained sets of involutions of hypergroups will be defined in the beginning of Chapter 8. This section will be concluded with two results on the exchange condition without referring to any extra condition.

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Dichotomy and the Exchange Condition

181

Theorem 6.7.5 A set of involutions of a hypergroup is dichotomic if it satisfies the exchange condition. Proof. Let T be a set of involutions of a hypergroup, and assume that T satisfies the exchange condition. Let v be an element in T, and let e be an element in hTi. We have to show that e ∈ T−1 (v) ∪ T1 (v). Since 1 ∈ T1 (v), we assume that e , 1. Then, by Lemma 2.3.7(i), there exist elements u in T and d in hTi such that e ∈ ud and `T (e) = `T (d) + 1. It follows that u ∈ T1 (d) and e ∈ T−1 (d). Assume first that d ∈ T−1 (v). In this case, we obtain from Lemma 6.5.2(iii) that T−1 (d) ⊆ T−1 (v). Thus, as e ∈ T−1 (d), e ∈ T−1 (v), and we are done. Assume now that d < T−1 (v). Then, as `T (e) = `T (d) + 1, induction yields that d ∈ T1 (v). Now recall that u ∈ T1 (d) and that T is assumed to satisfy the exchange condition. Thus, we obtain that ud ⊆ dv ∪ T1 (v). Since e ∈ ud, this implies that e ∈ dv ∪ T1 (v). If e ∈ dv, we obtain from `T (e) = `T (d) + 1 that e ∈ T−1 (v). Thus,  we have e ∈ T−1 (v) ∪ T1 (v). In the subsequent lemma, we look at a specific consequence of the exchange condition. It is an observation which anticipates the essential argument in the proof of Theorem 9.3.1. Lemma 6.7.6 Let T be a set of involutions of a hypergroup, and assume that T satisfies the exchange condition. Let t be an element in T, let d be an element in T1 (t), and let f be an element in T−1 (d) ∩ T−1 (t). Then dt contains an element e with `T (e) = `T (d) + 1 and f ∈ T−1 (e). Proof. Since f ∈ T−1 (d), hTi contains an element b with f ∈ bd and `T ( f ) = `T (b) + `T (d). Since f ∈ T−1 (t) and d ∈ T1 (t), f , d. From f ∈ bd and f , d we obtain b , 1. Thus, there exist elements u in T and a in hTi such that b ∈ ua and `T (b) = `T (a) + 1; cf. Lemma 2.3.7(i). Thus, by Lemma 2.3.8(ii), ad contains an element c such that f ∈ uc, `T (c) = `T (a) + `T (d), and `T ( f ) = `T (c) + 1. Assume first that c ∈ T−1 (t). Since c ∈ ad and `T (c) = `T (a) + `T (d), we also have c ∈ T−1 (d). Thus, as `T ( f ) = `T (c) + 1, induction yields an element e in dt with `T (e) = `T (d) + 1 and c ∈ T−1 (e). From c ∈ T−1 (e) we obtain T−1 (c) ⊆ T−1 (e); cf. Lemma 6.5.2(iii). On the other hand, as f ∈ uc and `T ( f ) = `T (c) + 1, f ∈ T−1 (c). Thus, f ∈ T−1 (e), and we are done in this case. Assume now that c < T−1 (t). Then, by Theorem 6.7.5, c ∈ T1 (t). Thus, as u ∈ T1 (c) and T is assumed to satisfy the exchange condition, uc ⊆ ct ∪ T1 (t). Thus, as f ∈ uc, f ∈ ct ∪ T1 (t). Since f ∈ T−1 (t), this implies that f ∈ ct. Since f ∈ ct, `T ( f ) = `T (c) + 1, c ∈ ad, and `T (c) = `T (a) + `T (d), we obtain an element e in dt with f ∈ ae, `T (e) = `T (d) + 1, and `T ( f ) = `T (a) + `T (e); cf. Lemma 2.3.8(ii).

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From f ∈ ae and `T ( f ) = `T (a) + `T (e) we obtain that f ∈ T−1 (e).



6.8 Projective Hypergroups Recall from Section 6.1 that a hypergroup H is called projective if each element of H \ {1} is an involution. In the introduction to this chapter, we noticed that, via the group correspondence, the thin projective hypergroups correspond to the elementary abelian 2-groups. Thus, the notion of a projective hypergroup generalizes the notion of an elementary abelian 2-group. Via Lemma 6.8.7(iii), we will associate to each projective hypergroup a dimension, in Theorem 6.8.9, we will associate to each projective hypergroup a projective space (projective points, projective lines, and projective planes included). Projective hypergroups will play a role in Section 8.5. Lemma 6.8.1 We have the following. (i) Projective hypergroups are commutative. (ii) Closed subsets of projective hypergroups are projective. (iii) Projective hypergroups which contain a thin involution are thin. Proof. (i) This follows from Lemma 6.1.7(i). (ii) This follows from the definition of projective hypergroups. (iii) This follows from Lemma 6.1.8(ii).



A subset T of the set of the involutions of a projective hypergroup is called independent if, for each element t in T, hT \ {t}i , hTi.2 Lemma 6.8.2 Subsets of independent subsets of projective hypergroups are independent. Proof. Let V be an independent subset of a projective hypergroup, and let U be a subset of V. We will see that U is independent. Since V is independent, hV \ {v}i , hVi for each element v in V. Thus, by Lemma 2.3.1(ii), v < hV \ {v}i for each element v in V. From Lemma 2.3.1(i) we also know that hU \ {u}i ⊆ hV \ {u}i for each element u in U. It follows that u < hU \ {u}i for each element u in U. Thus, by Lemma 2.3.1(ii), hU \ {u}i , hUi for each element u  in U, and that means that U is independent. 2Let T be the set of the involutions of a projective hypergroup. Then, for each element t in T, {t} is independent. Furthermore, for any two distinct elements u and v of T, {u, v} is independent.

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Projective Hypergroups

183

Lemma 6.8.3 Let T be an independent subset of a projective hypergroup, and let v be an element in hTi \ {1}. Then T contains an element u such that h(T \ {u}) ∪ {v}i = hTi. Proof. Since v ∈ hTi\{1}, T contains a non-empty subset U with v ∈ hUi and v < hSi for each proper subset S of U. We choose an element u in U and set S := U \ {u}. Then U = S ∪ {u} and v < hSi. From U = S ∪ {u} we obtain that hUi = hS ∪ {u}i. From Lemma 6.8.1(i) we also know that hTi is commutative, so that, by Lemma 2.3.2, hS ∪ {u}i = hSihui. Thus, we have hUi = hSihui. From v ∈ hUi \ hSi and hUi = hSihui we obtain that v ∈ hSihui \ hSi, so that, by Lemma 6.1.9(ii), hSihvi = hSihui. Thus, as hUi = hSihui, we obtain that hUi = hSihvi. Set R := T \ U. Then, T \ {u} = R ∪ S and T = R ∪ U. Thus, as hUi = hSihvi, h(T \ {u}) ∪ {v}i = hR ∪ S ∪ {v}i = hRihSihvi = hRihUi = hR ∪ Ui = hTi; 

cf. Lemma 2.3.2.

The following theorem is similar to a well-known theorem of Ernst Steinitz; cf. [41; p. 133]. Theorem 6.8.4 Let T and V be independent subsets of a projective hypergroup. Assume that V is finite and that V ⊆ hTi. Then T contains a subset U such that |U| = |V | and h(T \ U) ∪ Vi = hTi. Proof. There is nothing to show if V is empty. Thus, we assume that V is not empty, we choose an element v in V, and we define V 0 := V \ {v}. Since V is independent, so is V 0; cf. Lemma 6.8.2. Thus, by induction, T contains a subset U 0 such that |U 0 | = |V 0 | and h(T \ U 0) ∪ V 0i = hTi. Define R := T \ U 0. Then hR ∪ V 0i = hTi. Thus, by Lemma 2.3.2, hRihV 0i = hTi. (Recall from Lemma 6.8.1(i) that hTi is commutative.) From hRihV 0i = hTi we obtain that hV 0ihRi = hTi. Thus, as v ∈ hTi, hRi contains an element t such that v ∈ hV 0it. On the other hand, V is assumed to be independent, whence hV 0i , hVi. Thus, by Lemma 2.3.1(ii), v < hV 0i. It follows that v ∈ hV 0ihti \ hV 0i. From v ∈ hV 0ihti \ hV 0i we obtain that hV 0ihvi = hV 0ihti; cf. Lemma 6.1.9(ii). Thus, hVi = hV 0 ∪ {v}i = hV 0ihvi = hV 0ihti; cf. Lemma 2.3.2.

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Since v < hV 0i, t , 1. Thus, as R is independent and t ∈ hRi, R contains an element r such that h(R \ {r }) ∪ {t}i = hRi; cf. Lemma 6.8.3. Define R 0 := R \ {r }. Then, hR 0 ∪ {t}i = hRi. Thus, as hVi = hV 0ihti, hR 0ihVi = hR 0ihV 0ihti = hR 0ihtihV 0i = hR 0 ∪ {t}ihV 0i = hRihV 0i; cf. Lemma 2.3.2. Define U := U 0 ∪ {r }. Then T \ U = R 0. Thus, by Lemma 2.3.2, h(T \ U) ∪ Vi = hR 0 ∪ Vi = hR 0ihVi. Now, as hR 0ihVi = hRihV 0i and hRihV 0i = hTi, we have h(T \ U) ∪ Vi = hTi. Since r ∈ R = T \ U 0 and U = U 0 ∪ {r }, we have |U| = |U 0 | + 1. From v ∈ V and V 0 = V \ {v} we also obtain that |V | = |V 0 | + 1. Thus, as |U 0 | = |V 0 |, |U| = |V |.  An independent subset T of a projective hypergroup H is called a basis of H if it satisfies hTi = H. The projective hypergroup {1} has only one basis, namely the empty set. Also projective hypergroups H of cardinality 2 have only one basis, namely H \ {1}. Lemma 6.8.5 Let H be a projective hypergroup. Then the following hold. (i) Each independent subset of H which is maximal among the independent subsets of H is a basis of H. (ii) Each generating subset of H which is minimal among the generating subsets of H is a basis of H. Proof. (i) Let U be an independent subset of H, and assume that U is maximal among the independent subsets of H. We have to show that hUi = H. Assume, by way of contradiction, that hUi , H. Then H contains an element t with t < hUi. Set V := U ∪ {t}. Since t < U, the maximal choice of U implies that V is not independent. Thus, V contains an element v such that hV \ {v}i = hVi. Thus, by Lemma 2.3.1(ii), v ∈ hV \ {v}i. Since t ∈ V, t ∈ hVi. Thus, as t < hUi, hUi , hVi. Since U = V \ {t}, this implies that hV \ {t}i , hVi. Thus, as hV \ {v}i = hVi, v , t. Thus, as v ∈ V = U ∪ {t}, v ∈ U. From v ∈ U and V = U ∪ {t} we obtain that V \ {v} = (U \ {v}) ∪ {t}. Thus, as v ∈ hV \ {v}i, v ∈ h(U \ {v}) ∪ {t}i = hU \ {v}ihti. Since v ∈ U and U is independent, we have hU \ {v}i , hUi. Thus, by Lemma 2.3.1(ii), v < hU \ {v}i.

6.8

Projective Hypergroups

185

From v ∈ hU \ {v}ihti and v < hU \ {v}i we obtain that v ∈ hU \ {v}it. Thus, by Lemma 1.3.3, t ∈ hU \ {v}iv. Since v ∈ U, this implies that t ∈ hUi, contradiction. (ii) Let T be a generating subset of H, and assume that T is minimal among the generating subsets of H. We have to show that T is independent. The choice of T forces hT \ {t}i , H for each element t in T. Since hTi = H, this means that hT \ {t}i , hTi for each element t in T, and that means that T is independent.  Theorem 6.8.6 Each projective hypergroup possesses a basis. Proof. Let H be a projective hypergroup, let S be a set of independent subsets of H, and assume that U ⊆ V or V ⊆ U for any two independent subsets of H belonging to S. We define T to be the union of the independent subsets of H which belong to S and claim that T is independent. Assume that T is not independent. Then T contains an element t with hT \ {t}i = hTi. Thus, by Lemma 2.3.1(ii), t ∈ hT \ {t}i. Thus, T \ {t} contains elements t1 , . . ., tn such that t ∈ t1 · · · tn ; cf. Lemma 2.3.4(i). Since T is the union of the subsets belonging to S, S contains an element S with t ∈ S and {t1, . . . , tn } ⊆ S. Since t ∈ t1 · · · tn and {t1, . . . , tn } ⊆ S \ {t}, t ∈ hS \ {t}i; cf. Lemma 2.3.4(i). Thus, by Lemma 2.3.1(ii), hS \ {t}i = hSi. Since t ∈ S, this implies that S is not independent, contradicting S ∈ S. What we have seen is that each chain of independent subsets of H possesses an upper bound in S. Thus, by Zorn’s Lemma, H contains an independent subset of involutions which is maximal among the independent subsets of involutions of H.  By Lemma 6.8.5(i), this subset is a basis of H. Recall from Section 2.3 that a hypergroup is said to be finitely generated if it possesses a finite generating set. Lemma 6.8.7 Let H be a finitely generated projective hypergroup. Then the following hold. (i) The hypergroup H possesses a finite basis. (ii) Let V be a finite basis of H, and let U be an independent subset of H. Then we have |U| ≤ |V |. (iii) Any two bases of H have the same cardinality. Proof. (i) Since H is finitely generated, H possesses a finite subset V with hVi = H. Among the finite subsets of V which generate H we choose U as small as possible. Then, by Lemma 6.8.5(ii), U is a basis of H.

186

6 Involutions

(ii) Let T be a finite subset of U. Since U is independent, so is T; cf. Lemma 6.8.2. Thus, as T ⊆ hVi, V contains a subset of cardinality |T |; cf. Theorem 6.8.4. It follows that |T | ≤ |V |. What we have seen is that |T | ≤ |V | for each finite subset T of U. Thus, |U| ≤ |V |. (iii) This follows from (i) and (ii).



Let H be a finitely generated projective hypergroup. In Lemma 6.8.7(i), we saw that H possesses a basis with finitely many elements. In Lemma 6.8.7(iii), we saw that any two bases of H have the same number of elements. Thus, there exists a non-negative integer n such that each basis of H has cardinality n. The integer n − 1 is called the dimension of H and will be denoted by dim(H). Lemma 6.8.8 Let H be a finitely generated projective hypergroup, and let F be a closed subset of H. Then the following hold. (i) The closed subset F is finitely generated, and dim(F) ≤ dim(H). (ii) If dim(F) = dim(H), then F = H. Proof. (i) From Lemma 6.8.7(i) we know that H possesses a finite basis. We choose a finite basis of H and denote it by V. From Theorem 6.8.6 we know that F possesses a basis. We choose a basis of F and denote it by U. From Lemma 6.8.7(ii) we obtain that |U| ≤ |V |. Thus, as V is finite, so is U. Since hUi = F, this implies that F is finitely generated. From |U| ≤ |V | we also obtain that dim(F) ≤ dim(H). (ii) Let U be a basis of F, let V be a basis of H, and assume that dim(F) = dim(H).  Then |U| = |V |. Thus, by Theorem 6.8.4, F = hUi = hVi = H. Theorem 6.8.9 Let H be a projective hypergroup. Then the closed subsets of dimension 0 of H together with the closed subsets of dimension 1 of H form a projective space. Proof. From Lemma 6.1.7(ii) (together with Lemma 6.1.5(i)) one obtains that any two distinct closed subsets of dimension 0 of H are contained in exactly one closed subset of dimension 1 of H. Let u, v, w, y, and z be elements in H \ {1} with u , v, y , z, w < {u, v, y, z}, and hwi = hu, vi ∩ hy, zi. We have to find an element x in H \ {1} with hxi = hu, yi ∩ hv, zi. From w ∈ hu, vi ∩ hy, zi and w < {u, v, y, z} we obtain that w ∈ uv ∩ yz; cf. Lemma 6.1.7(ii). Thus, uv ∩ yz is not empty, so that, by Lemma 1.2.5(i), uy ∩ vz is not empty. Let x be an element in uy ∩ vz.

6.8

Projective Hypergroups

187

Assume that x = 1. Then, since x ∈ vz, v = z. In this case, we have v ∈ hu, vi ∩ hy, zi = hwi. It follows that v = w, contradiction. Thus, we have x , 1, and, since we have seen that x ∈ uy ∩ vz, this shows that hxi ⊆ hu, yi ∩ hv, zi. Assume that hu, yi ∩ hv, zi * hxi. Then, since hxi ⊆ hu, yi ∩ hv, zi, Lemma 6.8.8(ii) forces 2 ≤ dim(hu, yi ∩ hv, zi). Thus, as dim(hu, yi) ≤ 2 and dim(hv, zi) ≤ 2, we conclude that hu, yi = hv, zi, again by Lemma 6.8.8(ii). A third application of Lemma 6.8.8(ii) now yields hu, vi = hy, zi. This contradiction shows that hu, yi ∩ hv, zi ⊆ hxi. From hxi ⊆ hu, yi ∩ hv, zi and hu, yi ∩ hv, zi ⊆ hxi we obtain that hxi = hu, yi ∩ hv, zi, as wanted.  Theorem 6.8.9 was already foreshadowed in [48; Theorem] and [5; Theorem 1.1]. It shows that each projective hypergroup gives rise to a projective space. It is easy to associate, conversely, a projective hypergroup to each projective space. To see this, one defines, for any two distinct points q and r of a projective space P, the set qr to be the set of all points of P \ {q, r } which are collinear with q and r. Then (q, r) 7→ qr is a hyperoperation on the set of all points of P, and, with respect to this hyperoperation, the set of all points of P turns out to be a projective hypergroup. The hypermultiplication of the projective hypergroup associated this way to a projective plane of order 3 is described by the following table. 1 a b c d e f g h i j k l m

1 {1} {a} {b} {c} {d} {e} {f } {g} {h} {i} {j} {k } {l} {m}

a {a} {1, a} { f , j} {g, k } {h, l} {i, m} {b, j } {c, k } {d, l} {e, m} {b, f } {c, g} {d, h} {e, i}

b {b} { f , j} {1, b} {d, m} {c, m} {g, l} {a, j } {e, l} {i, k } {h, k } {a, f } {h, i} {e, g} {c, d}

c {c} {g, k } {d, m} {1, c} {b, m} {h, j } {i, l} {a, k } {e, j } { f , l} {e, h} {a, g} { f , i} {b, d}

d {d} {h, l} {c, m} {b, m} {1, d} { f , k} {e, k } {i, j } {a, l} {g, j } {g, i} {e, f } {a, h} {b, c}

e {e} {i, m} {g, l} {h, j } { f , k} {1, e} {d, k } {b, l} {c, j } {a, m} {c, h} {d, f } {b, g} {a, i}

f {f } {b, j } {a, j } {i, l} {e, k } {d, k } {1, f } {h, m} {g, m} {c, l} {a, b} {d, e} {c, i} {g, h}

g {g} {c, k } {e, l} {a, k } {i, j } {b, l} {h, m} {1, g} { f , m} {d, j } {d, i} {a, c} {b, e} { f , h}

h {h} {d, l} {i, k } {e, j } {a, l} {c, j } {g, m} { f , m} {1, h} {b, k } {c, e} {b, i} {a, d} { f , g}

i {i} {e, m} {h, k } { f , l} {g, j } {a, m} {c, l} {d, j } {b, k } {1, i} {d, g} {b, h} {c, f } {a, e}

j {j} {b, f } {a, f } {e, h} {g, i} {c, h} {a, b} {d, i} {c, e} {d, g} {1, j } {l, m} {k, m} {k, l}

k {k } {c, g} {h, i} {a, g} {e, f } {d, f } {d, e} {a, c} {b, i} {b, h} {l, m} {1, k } { j, m} { j, l}

l {l} {d, h} {e, g} { f , i} {a, h} {b, g} {c, i} {b, e} {a, d} {c, f } {k, m} { j, m} {1, l} { j, k }

m {m} {e, i} {c, d} {b, d} {b, c} {a, i} {g, h} { f , h} { f , g} {a, e} {k, l} { j, l} { j, k } {1, m}

Finally, we mention that, if we define, for each projective hypergroup H, by P(H) the projective space as defined in Theorem 6.8.9 and, for each projective space P, by H(P) the projective hypergroup defined as above, then H(P(H)) and H are isomorphic hypergroups and P(H(P)) and P are isomorphic as projective planes.

7 Hypergroups with a Small Number of Elements

In this chapter, we pause developing the structure theory of hypergroups for a moment and turn to the study of hypergroups with a small number of elements. We consider hypergroups of cardinality 1, 2, 3, 4, and 6. Some intermediate results in Section 7.5 will be stated and proven in a more general context. They are about hypergroups H which contain a closed subset F with only a few elements in H \ F. In Section 7.1, we list all hypergroups of cardinality at most 3. Clearly, up to isomorphism, there is only one hypergroup with one element, the trivial hypergroup. There are two isomorphism classes of hypergroups with two elements, and we will see that there exist exactly ten isomorphism classes of hypergroups of cardinality 3. All ten of these classes consist of commutative schurian hypergroups.1 In the Preface, we mentioned that there are 102 different isomorphism classes of hypergroups of cardinality 4. In Section 7.2, we will show that 37 of these classes consist of non-symmetric hypergroups. Among these 37 isomorphism classes we identify five classes consisting of non-commutative hypergroups; cf. Corollary 7.2.8. (Recall from Lemma 1.2.9 that non-commutative hypergroups are not symmetric.) After the classification of the non-symmetric hypergroups of cardinality 4, we turn for three sections to the study of hypergroups of cardinality 6, which contain a non-normal closed subset. From Corollary 3.2.7 we know that a non-normal closed subset of a hypergroup H of cardinality 6 has cardinality 2 and index either 3 or 4 in H. In Sections 7.3 and 7.4, we study hypergroups H of cardinality 6 which contain a non-normal closed subset of index 3 in H. The results of our efforts in these two sections are summarized in Theorems 7.3.8 and 7.4.8. In each of these classification theorems we find one class which consists of Coxeter hypergroups.2 In the former one, it is the class of hypergroups of type H6,1 (as defined in Section 7.3), and in 1Recall from Section 3.4 that a hypergroup is called schurian if it is isomorphic to a quotient of a thin hypergroup. 2We have mentioned Coxeter hypergroups already in the Preface. They will be defined in the introduction to Chapter 9. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 P. -H. Zieschang, Hypergroups, https://doi.org/10.1007/978-3-031-39489-8_7

189

190

7 Hypergroups with a Small Number of Elements

the latter one, it is the class of hypergroups of type H6,11 (as defined in Section 7.4). Hypergroups of type H6,1 are thin and correspond, via the group correspondence, to the symmetric groups of order 6. Hypergroups of type H6,11 correspond, via regular actions, to projective planes; cf. Sections 10.2 and 10.3. In Section 7.5, we study hypergroups H which contain a non-normal closed subset F satisfying |H \ F | = 4 and |H/F | = 4. As a byproduct of our analysis, we obtain a classification of all the hypergroups H of cardinality 6 which contain a non-normal closed subset of index 4 in H; cf. Theorem 7.5.4. We conclude this chapter with a characterization of hypergroups of type H6,1 ; cf. Theorem 7.6.2.

7.1 Hypergroups of Cardinality at Most 3 In this section, we determine all hypergroups containing at most three elements. As mentioned in the introduction to this chapter, the trivial hypergroup is (up to isomorphism) the only hypergroup which consists of a single element. From Lemma 6.1.1(iii) one obtains that any two thin hypergroups with two elements are isomorphic, from Lemma 6.1.1(iv) that any two non-thin hypergroups with two elements are isomorphic. Thin hypergroups with two elements will be called of type H2,1 . Non-thin hypergroups with two elements will be called of type H2,2 . We now classify the hypergroups of cardinality 3. We first present ten isomorphism classes each consisting of hypergroups of cardinality 3. After that, we will see that each hypergroup of cardinality 3 belongs to one of these ten classes. Note first that, via the group correspondence, each cyclic group of order 3 gives rise to a thin hypergroup with three elements. Thin hypergroups with three elements will be called of type H3,1 . Here is a second class of hypergroups with three elements. Let G be the semidirect product of the additive group of the rational numbers and the multiplicative group of the positive rational numbers. Then G is the set of all pairs (a, x) of rational numbers with 0 < x, and, for any four rational numbers b, c, y, and z with 0 < y and 0 < z, we have (b, y)(c, z) = (bz + c, yz). The neutral element of G is (0, 1), the inverse of an element (a, x) in G is (−ax −1, x −1 ). Set g := (1, 1), and define K to be the set of all pairs (a, x) in G with a = 0. Then K is a subgroup of G, and KgK = {(a, x) ∈ G | 0 < a} It follows that

and

Kg −1 K = {(a, x) ∈ G | a < 0}.

G//K = {1K , g K , (g K )∗ }.

7.1

Hypergroups of Cardinality at Most 3

191

Set h := g K and H := G//K. Then the hypermultiplication of the quotient H is described by the following table. H3,2 1 h h∗ 1 {1} h∗ {h∗ } H h {h} {h} H We say that a hypergroup is of type H3,2 if it contains an element h such that its hypermultiplication is described by table H3,2 . A hypergroup element h is called idempotent if h2 = {h} and |hhi| = 3. Note that, for each idempotent hypergroup element h, hhi = {1, h, h∗ }. We now come to a third class of hypergroups with three elements. We let G be a Frobenius group of order 21. Then G contains elements k and l with l 7 = 1, k 3 = 1, and l k = l 2 . Define K := hki. Then {l, l 2, l 4 } ⊆ KlK

and

{l 3, l 5, l 6 } ⊆ Kl 3 K.

Set h := l K and H := G//K. Then h∗ , h and H = {1, h, h∗ }. Since l ∈ (KlK)(KlK), h ∈ h2 ; cf. Lemma 3.4.4. Similarly, as l 3 ∈ (KlK)(KlK), h∗ ∈ h2 . Thus, the hypermultiplication of the quotient H is described by the following table. H3,3 1 h h∗ 1 {1} h∗ {h∗ } H h {h} {h, h∗ } H We say that a hypergroup is of type H3,3 if it contains an element h such that its hypermultiplication is described by table H3,3 . Next let G be a dihedral group of order 10. Then G contains elements k and l with l 5 = 1, k 2 = 1, and l k = l 4 . Define K := hki. Then {l, l 4 } ⊆ KlK

and

{l 2, l 3 } ⊆ Kl 2 K.

Set a := l K , b := (l 2 )K , and H := G//K. Then a , b and H = {1, a, b}. Since l kl = k, (KlK)(KlK) = K ∪ Kl 2 K. Thus, by Lemma 3.4.4, a2 = {1, b}. Similarly, b2 = {1, a}. Thus, the hypermultiplication of the quotient H is described by the following table. H3,4 1 a b

1 a b {1} {a} {1, b} {b} {a, b} {1, a}

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7 Hypergroups with a Small Number of Elements

We say that a hypergroup is of type H3,4 if it contains elements a and b such that its hypermultiplication is described by table H3,4 . Next let G be the group of all permutations of {1, 2, 3, 4, 5}. Let K be the subgroup of G generated by the three transpositions (12), (34), and (45). Then K is a maximal subgroup of G and has order 12. The group K acts on the set C of the ten right cosets of K in G by multiplication from the right hand side. We compute the lengths of the orbits of this action.3 Set d := (13)(24). Then K d = h(12), (34), (25)i. It follows that K d ∩K = h(12), (34)i, so K has an orbit of length 3 in C. Set e := (123). Then K e = h(14), (23), (45)i. It follows that K e ∩ K = h(45)i, so K has an orbit of length 6 in C. Since |C| = 10, this shows that G//K = {K, K dK, KeK }. Set a := d K , b := e K , and H := G//K. Then a , b and H = {1, a, b}. Since d is an involution, a is symmetric. Thus, b is symmetric, too. Since K is a maximal subgroup of G, H is primitive; cf. Theorem 3.4.6. It follows that a and b are not involutions. Thus, as H = {1, a, b}, we conclude that b ∈ a2 and a ∈ b2 . Set k := (12). Then e = keek ∈ KeKeK. Thus, by Lemma 3.4.4, b ∈ b2 . Each element in K dK dK sends 1 to 1 or 2. Thus, d < K dK dK. It follows that a < a2 ; cf. Lemma 3.4.4. Now the hypermultiplication of the quotient H is described by the following table. H3,5 1 a b

1 a b {1} {a} {1, b} {b} {a, b} H

We say that a hypergroup is of type H3,5 if it contains elements a and b such that its hypermultiplication is described by table H3,5 . Next let G be a Frobenius group of order 36. Then G contains elements v, w, and k with vw = wv, v 3 = 1 = w 3 , k 4 = 1, v k = w, and w k = v 2 . Define K := hki. Then {v, w, v 2, w 2 } ⊆ KvK

and

{vw, v 2 w, v 2 w 2, vw 2 } ⊆ KvwK.

Set a := v K , b := (vw)K , and H := G//K. Then a , b and H = {1, a, b}. Since v 3 = 1, v −1 = v 2 . Thus, Kv −1 K = KvK, whence a∗ = a. Similarly, b∗ = b. Notice that KvK = Kv 2 K and v 2 ∈ (KvK)(KvK). Thus, we conclude that KvK ⊆ (KvK)(KvK). Thus, by Lemma 3.4.4, a ∈ a2 . Similarly, b ∈ b2 . Since K is a maximal subgroup of G, H is primitive; cf. Theorem 3.4.6. It follows that a and b are not involutions. Thus, as H = {1, a, b}, we have b ∈ a2 and a ∈ b2 . 3We do this by computing the cardinality of a one-point stabilizer of K. For any two elements k in K and g in G, we have Kgk = Kg if and only if k ∈ K g . Thus, for each element g in G, the stabilizer of Kg in K is K g ∩ K.

7.1

Hypergroups of Cardinality at Most 3

193

Now the hypermultiplication of the quotient H is described by the following table. H3,6 1 a b

a b 1 {1} {a} H {a} {a, b} H

We say that a hypergroup is of type H3,6 if it contains elements a and b such that its hypermultiplication is described by table H3,6 . Next let G be a dihedral group of order 8. Then G contains elements k and l with l 4 = 1, k 2 = 1, and l k = l 3 . Define K := hki, h := l K , t := (l 2 )K , and H := G//K. Then t , h and H = {1, t, h}. Since l 2 kl 2 = k and l 4 = 1, (Kl 2 K)(Kl 2 K) = K. Thus, by Lemma 3.4.4, t 2 = {1}. Similarly, as l 2 ∈ (KlK)(KlK), t ∈ h2 , and, since l < (KlK)(KlK), h < h2 . Thus, the hypermultiplication of the quotient H is described by the following table. H3,7 1 t h

1 t h {1} {t} {1} {h} {h} {1, t}

We say that a hypergroup is of type H3,7 if it contains elements t and h such that its hypermultiplication is described by table H3,7 . Next let G be a symmetric group of order 24, let K be one of the three cyclic subgroups of order 4 of G, and set H := G//K. The group K acts on the set C of the six right cosets of K in G by multiplication from the right hand side. We compute the lengths of the orbits of this action. Note that NG (K) , G and that, for each element g in G \ NG (K), K g ∩ K = {1}. Thus, K has at least one orbit of length 4 in C. Thus, in its action on C, K has two orbits of length 1 and one orbit of length 4. It follows that |H| = 3. Since K , NG (K) , G, we have {1} , Oϑ (H) , H; cf. Lemma 3.5.6(i). Thus, as |H| = 3, one of the elements of H must be a thin involution, the other one is not thin. Let t denote the thin involution of H, and let h denote the uniquely determined element in H \ {1, t}. Since h , t, 1 < ht; cf. Lemma 1.1.6. Since t is a thin involution, h < t 2 , and that implies that t < ht. Thus, by Lemma 1.1.3, ht = {h}. Since h is symmetric, we have 1 ∈ h2 . Furthermore, as h ∈ ht, we have t ∈ h2 . We will now see that h ∈ h2 . Assume, by way of contradiction, that h < h2 . Then, as {1, t} ⊆ h2 and H = {1, t, h}, we have h2 = {1, t}. It follows that h2 = Oϑ (H). Thus, as {1, t} ⊆ Oϑ (H), Lemma 4.4.2 yields that O ϑ (H) = Oϑ (H).

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7 Hypergroups with a Small Number of Elements

Recall from Theorem 4.4.1(i) that O ϑ (H) is strongly normal in H. Thus, as O ϑ (H) = Oϑ (H), we obtain that Oϑ (H) is strongly normal in H. It follows that Oϑ (G//K) is strongly normal in G//K. On the other hand, by Lemma 3.5.6(i), KG (K)//K = Oϑ (G//K). Thus, KG (K)//K is strongly normal in G//K, so that, by Lemma 3.5.4(iii), the normalizer NG (K) of K in G is normal in G, contradiction. Now the hypermultiplication of the quotient G//K is described by the following table. H3,8 1 t h

1 t h {1} {t} {1} {h} {h} H

We say that a hypergroup is of type H3,8 if it contains elements t and h such that its hypermultiplication is described by table H3,8 . Next let G be the subgroup of the group of all permutations of {1, 2, 3, 4, 5, 6} generated by the elements a, b, c, d, and x, where a := (12),

b := (23),

c := (45),

d := (56),

and

x := (14)(25)(36).

Then |G| = 72. Define L := ha, b, c, di

and

K := ha, b, ci.

Then |L| = 36 and |K | = 12. Furthermore, K d = ha, b, c d i,

K x = ha, c, di,

|K d ∩ K | = 6,

and

|K x ∩ K | = 4.

It follows that |K dK | = 24 and |K xK | = 36. Set t := d K , h := x K , and H := G//K. Then t , h and H = {1, t, h}. Since d ∈ L, t ∈ L//K. Similarly, as x < L, h < L//K. Thus, L//K = {1, t}. It follows that t is an involution, and, since K is not normal in L, t is not thin. Thus, t 2 = {1, t}. From h < t 2 we obtain that t < ht. From t , h we obtain that 1 < ht. Thus, as H = {1, t, h}, we conclude that ht = {h}. From h ∈ ht we obtain that t ∈ h2 . Thus, as h∗ = h, we have {1, t} ⊆ h2 . Assume that h ∈ h2 . From Lemma 4.4.2 we know that O ϑ (H) = [H, {1}]. Thus, h ∈ O ϑ (H). On the other hand, L is normal in G, whence, by Corollary 3.5.5(iii), L//K is strongly normal in G//K. It follows that O ϑ (H) ⊆ L//K. Thus, we have h ∈ L//K, and from this we obtain that x ∈ L. This contradiction shows that h < h2 . Now the hypermultiplication of the quotient H is described by the following table. H3,9 1 t h

1 t h {1} {t} {1, t} {h} {h} {1, t}

Hypergroups of Cardinality at Most 3

7.1

195

We say that a hypergroup is of type H3,9 if it contains elements t and h such that its hypermultiplication is described by table H3,9 . Next let G be the subgroup of the group of all permutations of {1, . . . , 9} generated by the elements a, b, c, d, e, f , y, and z, where a := (12),

b := (23),

d := (56),

c := (45),

e := (78),

f := (89)

and z := (47)(58)(69).

y := (14)(25)(36) and Then |G| = 24 ·34 . Define L := ha, b, c, d, e, f , yi

and

K := ha, b, c, d, e, yi.

Then |L| = 24 ·33 and |K | = 24 ·32 . Furthermore, we have K f = ha, b, c, d, e f , yi, and

|K f ∩ K | = 23 ·32,

K z = ha, b, e, f , c, y z i |K z ∩ K | = 23 ·3.

It follows that |K f K | = 25 ·32 and |K zK | = 25 ·33 . Set t := f K , h := z K , and H := G//K. Then t , h and H = {1, t, h}. Since f ∈ L, t ∈ L//K. Similarly, as z < L, h < L//K. Thus, L//K = {1, t}. It follows that t is an involution, and, since K is not normal in L, t is not thin. Thus, t 2 = {1, t}. From h < t 2 we obtain that t < ht. From t , h we obtain that 1 < ht. Thus, as H = {1, t, h}, we conclude that ht = {h}. From h ∈ ht we obtain that t ∈ h2 . Thus, as h∗ = h, we have {1, t} ⊆ h2 . Assume that h2 , H. Then, as H = {1, t, h} and {1, t} ⊆ h2 , we have h2 = {1, t}. Thus, as t 2 = {1, t}, we conclude that [H, {1}] = {1, t}. Since L//K = {1, t}, this implies that [H, {1}] = L//K. From Lemma 4.4.2, on the other hand, we know that O ϑ (H) = [H, {1}]. Thus, we have O ϑ (H) = L//K, so that, by Theorem 4.4.1(i), L//K is strongly normal in G//K. It follows that L is a normal subgroup of G; cf. Corollary 3.5.5(iii). This contradiction shows that h2 = H. Now the hypermultiplication of the quotient H is described by the following table. H3,10 1 t h

t h 1 {1} {t} {1, t} {h} {h} H

We say that a hypergroup is of type H3,10 if it contains elements t and h such that its hypermultiplication is described by table H3,10 .

196

7 Hypergroups with a Small Number of Elements

This finishes our collection of examples of hypergroups of cardinality 3. We will now see that each hypergroup with three elements is of type H3, j for some element j in {1, . . . , 10}. Theorem 7.1.1 Each non-symmetric hypergroup with three elements is of type H3,1 , H3,2 , or H3,3 . Proof. Let H be a hypergroup with three elements, and assume that H is not symmetric. Then H contains an element h with H = {1, h, h∗ }. If h is thin, so is H; cf. Lemma 2.6.2. In this case, H is of type H3,1 . Assume that h is not thin. Then h∗ h , {1}. Thus, as h∗ h is ∗ -invariant, we obtain that h∗ h = H. Since h is not thin, h∗ is not thin; cf. Lemma 2.6.2. Thus, hh∗ , {1}. Since hh∗ is ∗ -invariant, this implies that hh∗ = H. From h∗ h = H and hh∗ = H we obtain that H is of type H3,2 or of type H3,3 .



Theorem 7.1.2 Each symmetric hypergroup with three elements which does not contain an involution is of type H3,4 , H3,5 , or H3,6 . Proof. Let H be a symmetric hypergroup with three elements, and let a and b denote the two elements in H \ {1}. Assume that neither a nor b is an involution. Since a is not an involution, a2 * {1, a}. Thus, as H = {1, a, b}, we have b ∈ a2 . Since a is symmetric, 1 ∈ a2 . Thus, {1, b} ⊆ a2 . Similarly, {1, a} ⊆ b2 . From b ∈ a2 we obtain that a ∈ ab. Similarly, as a ∈ b2 , b ∈ ab. Thus, as 1 < ab, ab = {a, b}. It follows that H is of type H3,4 , H3,5 , or H3,6 .  Theorem 7.1.3 Each hypergroup with three elements which contains an involution is of type H3,7 , H3,8 , H3,9 , or H3,10 . Proof. Let H be a hypergroup with three elements, let t be an involution of H, and set F = hti. Let h denote the unique element in H \ F. Since t , h, 1 < ht; cf. Lemma 1.1.6. Since h < t 2 , t < ht. Thus, ht = {h}. From h ∈ ht (and h∗ = h) we obtain that t ∈ h2 , so that {1, t} ⊆ h2 . It follows that H is of type H3,7 , H3,8 , H3,9 , or H3,10 .  Recall from Section 3.4 that a hypergroup is called schurian if it is isomorphic to a quotient of a thin hypergroup. Theorem 7.1.4 Each hypergroup with three elements is commutative and schurian.

7.2

Non-Symmetric Hypergroups of Cardinality 4

197



Proof. This follows from Theorems 7.1.1, 7.1.2, and 7.1.3.

Let H be a hypergroup of type H3,7 , H3,8 , H3,9 , or H3,10 . Then H contains an involution t and an element h different from 1 and t such that H = {1, t, h}. Set F := {1, t}. Then we have H \ F = {h} and hF = {h}, so that H is the wreath product of two hypergroups of cardinality 2. We have H3,7 = H2,1 o H2,1 , H3,8 = H2,1 o H2,2 ,

H3,9 = H2,2 o H2,1 , H3,10 = H2,2 o H2,2 .

(By H3,7 = H2,1 o H2,1 we mean that each hypergroup of type H3,7 is a wreath product of a hypergroup of type H2,1 and a hypergroup of type H2,1 . The other three equations are understood correspondingly.)

7.2 Non-Symmetric Hypergroups of Cardinality 4 In this section, we determine all non-symmetric hypergroups with four elements. Let H be a hypergroup with of four elements, and assume that H is not symmetric. Then H contains elements a and b with a∗ , a, b , 1, b∗ = b, and H = {1, a, a∗, b}. We will see that the hypermultiplication of H is described by one of the following 37 tables. H4,1 1 a∗ a b

H4,2 1 a∗ a b

H4,4 1 a∗ a b

1 {1} {a∗ } {a} {b}

a

a∗

1 {1} {a∗ } {a} {b}

a

a∗

{1} {b} {1} {a∗ } {a} {1}

b

{1} {a∗ } {1} {b} {b} {1, a, a∗ }

1 a a∗ b {1} {a∗ } {1, b} {a} {a∗ } {1, b} {b} {a} {a∗ } {1}

b

H4,5 1 a∗ a b

H4,3 1 a∗ a b

1 {1} {a∗ } {a} {b}

a

a∗ b

{1} {a∗ } {1} {b} {b} H

1 a a∗ b {1} {a∗ } {1, b} {a} {a∗ } {1, b} {b} {a} {a∗ } {1, b}

198

7 Hypergroups with a Small Number of Elements

H4,6 1 a∗ a b

b 1 a a∗ {1} {a∗ } {1, b} {a} {b} {1, b} {b} {a, a∗ } {a, a∗ } {1, b}

H4,7 1 a∗ a b

1 a a∗ b {1} {a∗ } {1, b} {a} {b} {1, b} {b} {a, a∗, b} {a, a∗, b} H

H4,8 1 a∗ a b

1 a a∗ b {1} {a∗ } {1, b} {a} {a∗, b} {1, b} {b} {a, a∗, b} {a, a∗, b} {1, a, a∗ }

H4,9 1 a∗ a b

H4,10 1 a∗ a b

1 a a∗ b {1} {a∗ } {1, a, a∗ } {a} {a} {1, a, a∗ } {b} {b} {b} {1, a, a∗ }

H4,11 1 a∗ a b

1 a a∗ b {1} {a∗ } {1, a, a∗ } {a} {a} {1, a, a∗ } {b} {b} {b} H

H4,12 1 a∗ a b

1 a a∗ b {1} {a∗ } {1, a, a∗ } {a} {a, a∗ } {1, a, a∗ } {b} {b} {b} {1, a, a∗ }

H4,13 1 a∗ a b

1 a a∗ b {1} {a∗ } {1, a, a∗ } {a} {a, a∗ } {1, a, a∗ } {b} {b} {b} H

H4,14 1 a∗ a b

H4,15 1 a∗ a b

1 a a∗ b {1} {a∗ } {1, b} {a} {a∗, b} {1, b} {b} {a, a∗, b} {a, a∗, b} H

1 a a∗ b {1} {a∗ } {1, a, a∗ } {a} {a, a∗, b} {1, a, a∗ } {b} {a∗ } {a} {1}

1 a a∗ b {1} {a∗ } {1, a, a∗ } {a} {a, b} {1, a, a∗ } {b} {a∗, b} {a, b} {1, a, a∗ }

H4,16 1 a∗ a b

1 a a∗ b {1} {a∗ } {1, a, a∗ } {a} {a, b} {1, a, a∗ } {b} {a∗, b} {a, b} H

7.2

H4,17 1 a∗ a b

b a∗ a 1 {1} {a∗ } {1, a, a∗ } {a} {a, a∗, b} {1, a, a∗ } {b} {a∗, b} {a, b} {1, a, a∗ }

H4,19 1 a∗ a b

H4,20 1 a∗ a b

Non-Symmetric Hypergroups of Cardinality 4

H4,18 1 a∗ a b

199

1 a a∗ b {1} {a∗ } {1, a, a∗ } {a} {a, a∗, b} {1, a, a∗ } {b} {a∗, b} {a, b} H

1 a a∗ b {1} {a∗ } {1, a, a∗ } {a} {a} H {b} {a, b} {b} H

1 a a∗ b {1} {a∗ } {1, a, a∗ } {a} {a, b} H {b} {a, a∗, b} {a, b} {1, a, a∗ }

H4,21 1 a∗ a b

1 a a∗ b {1} {a∗ } {1, a, a∗ } {a} {a, a∗, b} H {b} {a, a∗, b} {a, b} {1, a, a∗ }

1 a a∗ b {1} {a∗ } {1, a, a∗ } {a} {a, b} H {b} {a, a∗, b} {a, b} H

H4,23 1 a∗ a b

1 a a∗ b {1} {a∗ } {1, a, a∗ } {a} {a, a∗, b} H {b} {a, a∗, b} {a, b} H

H4,22 1 a∗ a b

H4,24 1 a∗ a b

H4,26 1 a∗ a b

1 {1} {1} {a} {b}

1 {1} {a∗ } {a} {b}

a

a∗

b

H {a} H {a} {a∗ } {1}

a

a∗

b

H {a} H {a} {a∗ } {1, b}

H4,25 1 a∗ a b

1 a a∗ b {1} {a∗ } H {a} {a, a∗ } H {b} {a} {a∗ } {1}

H4,27 1 a∗ a b

1 a a∗ b {1} {a∗ } H {a} {a, a∗ } H {b} {a} {a∗ } {1, b}

200

7 Hypergroups with a Small Number of Elements

1 a a∗ b {1} {a∗ } H {a} {a, b} H {b} {a, a∗ } {a, a∗ } {1, b}

H4,29 1 a∗ a b

1 a a∗ b {1} {a∗ } H {a} {a, a∗, b} H {b} {a, a∗ } {a, a∗ } {1, b}

1 a a∗ b {1} {a∗ } H {a} {a} H {b} {a, b} {a∗, b} {1, a, a∗ }

H4,31 1 a∗ a b

1 a a∗ b {1} {a∗ } H {a} {a, a∗ } H {b} {a, b} {a∗, b} {1, a, a∗ }

H4,28 1 a∗ a b

H4,30 1 a∗ a b

1 a a∗ b {1} {a∗ } H {a} {a, b} H {b} {a, a∗, b} {a, a∗, b} {1, a, a∗ }

H4,33 1 a∗ a b

1 a a∗ b {1} {a∗ } H {a} {a, a∗, b} H {b} {a, a∗, b} {a, a∗, b} {1, a, a∗ }

1 a a∗ b {1} {a∗ } H {a} {a} H {b} {a, b} {a∗, b} H

H4,35 1 a∗ a b

1 a a∗ b {1} {a∗ } H {a} {a, a∗ } H {b} {a, b} {a∗, b} H

1 a a∗ b {1} {a∗ } H {a} {a, b} H {b} {a, a∗, b} {a, a∗, b} H

H4,37 1 a∗ a b

1 a a∗ b {1} {a∗ } H {a} {a, a∗, b} H {b} {a, a∗, b} {a, a∗, b} H

H4,34 1 a∗ a b

H4,36 1 a∗ a b

H4,32 1 a∗ a b

7.2

Non-Symmetric Hypergroups of Cardinality 4

201

Let j be an element in {1, . . . , 37}. We say that a hypergroup is of type H4, j if it contains elements a and b such that its hypermultiplication is described by table H4, j . If H is a hypergroup with elements a and b satisfying a∗ , a, b , 1, b∗ = b, and H = {1, a, a∗, b}, then the product a∗ a is equal to one of the sets {1},

{1, b},

{1, a, a∗ },

or

H.

(This is because a∗ a is ∗ -invariant.) We will discuss these four cases one by one and start with the case where a∗ a = {1}. Lemma 7.2.1 Let H be a hypergroup, and let a and b be elements of H satisfying a∗ , a, b , 1, b∗ = b, and H = {1, a, a∗, b}. Assume that a∗ a = {1}. Then H is of type H4,1 , H4,2 , or H4,3 . Proof. We are assuming that a∗ a = {1}, and that means that a is thin. Thus, by Lemma 2.6.2, a∗ is thin, too. It follows that {1, a, a∗ } = Oϑ (H) or Oϑ (H) = H. If Oϑ (H) = H, H is thin. In this case, H is of type H4,1 . Assume that {1, a, a∗ } = Oϑ (H). Since Oϑ (H) is finite, Oϑ (H) is closed. Thus, by Theorem 7.1.1, a2 = {a∗ }. If a ∈ ba, b ∈ aa∗ = {1}, contradiction. If a∗ ∈ ba, b ∈ a2 ; cf. Lemma 1.2.1. Since a2 = {a∗ }, also this is impossible. Note finally that a , b, so 1 < ba; cf. Lemma 1.1.6. Thus, as H = {1, a, a∗, b}, Lemma 1.1.3 forces ba = {b}. From b ∈ ba we obtain that b ∈ ba∗ . On the other hand, as a∗ is thin, |ba∗ | = 1; cf. Lemma 1.4.3(i). Thus, ba∗ = {b}. Since b is symmetric, 1 ∈ bb. Since b ∈ ba, we have a ∈ b2 . Thus, as b2 is ∗ invariant, a∗ ∈ b2 . It follows that b2 = {1, a, a∗ } or b2 = H, and H is of type H4,2 or  of type H4,3 . Next we look at the case where a∗ a = {1, b}. Lemma 7.2.2 Let H be a hypergroup, and let a and b be elements of H satisfying a∗ , a, b , 1, b∗ = b, and H = {1, a, a∗, b}. Assume that a∗ a = {1, b}. Then the following hold. (i) We have aa∗ = {1, b}. (ii) We have a2 ⊆ {a∗, b}. (iii) Assume that b ∈ a2 . Then {a, a∗ } ⊆ ba and {a, a∗ } ⊆ ba∗ . (iv) If a2 = {a∗ }, H is of type H4,4 or H4,5 . (v) If a2 = {b}, H is of type H4,6 or H4,7 . (vi) If a2 = {a∗, b}, H is of type H4,8 or H4,9 .

202

7 Hypergroups with a Small Number of Elements

Proof. (i) Since a < a∗ a, a∗ < aa∗ . Thus, as aa∗ is ∗ -invariant, a < aa∗ . It follows that aa∗ ⊆ {1, b}. Assume, by way of contradiction, that aa∗ = {1}. Then, by Lemma 2.6.2, a∗ a = {1}, contradiction. It follows that a∗ a = {1, b}. (ii) From a < a∗ a we obtain that a < a2 . Thus, as 1 < a2 , a2 ⊆ {a∗, b}. (iii) We are assuming that b ∈ a2 . Thus, by Lemma 1.2.1, a∗ ∈ ba. From (i) we also know that b ∈ aa∗ , and that implies that a ∈ ba. Thus, {a, a∗ } ⊆ ba. From b ∈ a2 we also obtain that a ∈ ba∗ . Now recall that we are assuming that b ∈ a∗ a, whence a∗ ∈ ba∗ . It follows that {a, a∗ } ⊆ ba∗ . (iv) Assume that a2 = {a∗ }. Then, as b ∈ a∗ a, ba ⊆ a∗ a2 = (a∗ )2 = {a}, so ba = {a}. From (i) we know that b ∈ aa∗ . Thus, ba∗ ⊆ a(a∗ )2 = a2 = {a∗ }, and then ba∗ = {a∗ }. From b < ba we obtain that a < b2 , from b < ba∗ that a∗ < b2 . Thus, as H = {1, a, a∗, b}, b2 ⊆ {1, b}. It follows that H is of type H4,4 or H4,5 . (v) Assume that a2 = {b}. Then by (iii), {a, a∗ } ⊆ ba and {a, a∗ } ⊆ ba∗ . Note also that 1 < ba and 1 < ba∗ . Furthermore, we have b ∈ ba if and only if b ∈ ba∗ . From b ∈ a∗ a we obtain that a ∈ ab. Thus, since a2 = {b}, we conclude that b ∈ a2 ⊆ a2 b = b2 . Since {1, b} ⊆ b2 , the statements b ∈ ba and b2 = H are equivalent. It follows that H is of type H4,6 or H4,7 . (vi) Assume that a2 = {a∗, b}. Then by (iii), {a, a∗ } ⊆ ba and {a, a∗ } ⊆ ba∗ . Note also that 1 < ba and 1 < ba∗ . Assume that b < ba. Then ba = {a, a∗ }. Thus, as a∗ ∈ a2 , a ∈ ba∗ ⊆ ba2 = a2 ∪ a∗ a = {1, a∗, b}, contradiction. Thus, b ∈ ba, and then b ∈ ba∗ and {a, a∗ } ⊆ b2 . It follows that H is of type H4,8 or H4,9 .  Now we consider the third of the above four cases, the case where a∗ a = {1, a, a∗ }. Note that, in this case, Lemma 1.2.1 forces {a, a∗ } ⊆ aa∗ . Thus, as 1 ∈ aa∗ , we have {1, a, a∗ } ⊆ aa∗ . Since H = {1, a, a∗, b}, that implies that we have either aa∗ = {1, a, a∗ } or aa∗ = H. We first assume that aa∗ = {1, a, a∗ }. In this case, we will obtain nine different isomorphism classes. After that, we will assume that aa∗ = H in which case we will obtain five different isomorphism classes. The first case will be considered in Lemmas 7.2.3 and 7.2.4, the second case in Lemma 7.2.5.

7.2

Non-Symmetric Hypergroups of Cardinality 4

203

Lemma 7.2.3 Let H be a hypergroup, and let a and b be elements of H satisfying a∗ , a, b , 1, b∗ = b, and H = {1, a, a∗, b}. Assume that a∗ a = {1, a, a∗ } = aa∗ and that b < a2 . Then H is of type H4,10 , H4,11 , H4,12 , or H4,13 . Proof. From b < a2 we obtain that b < (a∗ )2 ; cf. Lemma 1.2.1. Thus, as a∗ a = {1, a, a∗ } = aa∗ , the set {1, a, a∗ } is a closed subset of H. By Theorem 7.1.1, {1, a, a∗ } is a hypergroup of type H3,2 or H3,3 . Since a , b, we have 1 < ba; cf. Lemma 1.1.6. Since b < aa∗ , a < ba, and, since b < a2 , a∗ < ba; cf. Lemma 1.2.1. Thus, as H = {1, a, a∗, b}, ba = {b}. Similarly, ba∗ = {b}. From b ∈ ba we obtain that a ∈ b2 , from b ∈ ba∗ that a∗ ∈ b2 . It follows that {1, a, a∗ } ⊆ b2 . Thus, H is of type H4,10 , H4,11 , H4,12 , or H4,13 .  Lemma 7.2.4 Let H be a hypergroup, and let a and b be elements of H satisfying a∗ , a, b , 1, b∗ = b, and H = {1, a, a∗, b}. Assume that a∗ a = {1, a, a∗ } = aa∗ and that b ∈ a2 . Then the following hold. (i) We have a2 = {a, b} or a2 = {a, a∗, b}. (ii) We have ba = {a∗ } or ba = {a∗, b}. (iii) If ba = {a∗ }, H is of type H4,14 . (iv) If ba = {a∗, b}, H is of type H4,15 , H4,16 , H4,17 , or H4,18 . Proof. (i) By hypothesis, we have a ∈ a∗ a. Thus, a ∈ a2 . We are also assuming that b ∈ a2 . Thus, as 1 < a2 and H = {1, a, a∗, b}, we must have a2 = {a, b} or a2 = {a, a∗, b}. (ii) We are assuming that b < aa∗ Thus, a < ba. Thus, as 1 < ba and H = {1, a, a∗, b}, we have ba ⊆ {a∗, b}. From our hypothesis that b ∈ a2 we also obtain that a∗ ∈ ba; cf. Lemma 1.2.1. Thus we have ba = {a∗ } or ba = {a∗, b}. (iii) Assume that ba = {a∗ }. Then b < ba. Thus, a < b2 . Since b is symmetric, this implies that a∗ < b2 . On the other hand, as we are assuming that b ∈ a2 , we obtain that b2 ⊆ ba2 = a∗ a = {1, a, a∗ }. Thus, b2 = {1}. Assume that a2 = {a, b}. Then, as a∗ a = {1, a, a∗ } and ba = {a∗ }, a ∈ a∗ a = ba2 = ba ∪ b2 = {1, a∗ }. This contradiction shows that a2 = {a, a∗, b}; cf. (i). From ba = {a∗ } and b2 = {1} we obtain that ba∗ = b2 a = {a}, and then that H is of type H4,14 .

204

7 Hypergroups with a Small Number of Elements

(iv) Assume that ba = {a∗, b}. Since b ∈ ba, b ∈ ba∗ . From our hypothesis that b ∈ a2 we obtain that a ∈ ba∗ . Since we are assuming that b < a∗ a, we have a∗ < ba∗ . Thus, as 1 < ba∗ and H = {1, a, a∗, b}, ba∗ = {a, b} From b ∈ ba we also obtain that a ∈ b2 . Since b is symmetric, this implies that a∗ ∈ b2 . Thus, {1, a, a∗ } ⊆ b2 . With a reference to (i) we now obtain that H is of type H4,15 , H4,16 , H4,17 , or H4,18 , as wanted.  Lemma 7.2.5 Let H be a hypergroup, and let a and b be elements of H satisfying a∗ , a, b , 1, b∗ = b, and H = {1, a, a∗, b}. Assume that a∗ a = {1, a, a∗ } and that aa∗ = H. Then the following hold. (i) We have a ∈ a2 . (ii) We have ba = {a, b} or ba = {a, a∗, b}. (iii) We have ba∗ = {b} or ba∗ = {a, b}. (iv) We have {1, a, a∗ } ⊆ b2 . (v) If a2 = {a}, H is of type H4,19 . (vi) If a2 , {a}, H is of type H4,20 , H4,21 , H4,22 , or H4,23 . Proof. (i) Since we are assuming that a ∈ aa∗ , we have a ∈ a2 . (ii) Since we are assuming that b ∈ aa∗ , we have a ∈ ba. Assume, by way of contradiction, that b < ba. Then, by Lemma 1.2.1, b < ab. From b < a∗ a we also obtain that a < ab. Thus, as 1 < ab and H = {1, a, a∗, b}, ab = {a∗ }. Thus, as a ∈ ba, a∗ ∈ bab = ba∗ = {a}; cf. Lemma 1.2.2. This contradiction shows that b ∈ ba. Since a , b, 1 < ba. (iii) From (ii) we know that b ∈ ba. Thus, b ∈ ba∗ . Since b < a∗ a, a∗ < ba∗ , and from a , b we obtain that 1 < ba∗ . (iv) Since b is symmetric, 1 ∈ b2 . From (ii) we know that b ∈ ba. Thus, as b is symmetric, a ∈ b2 . It follows that {1, a, a∗ } ⊆ b2 . (v) Assume that a2 = {a}. Then b < a2 . Thus, by Lemma 1.2.1, a∗ < ba. Thus, by (ii), ba = {a, b}. From b < a2 we also obtain that a < ba∗ . Thus, by (iii), ba∗ = {b}. Since a ∈ ba, a∗ ∈ a∗ b; cf. Lemma 1.2.1. Thus, as ba∗ = {b}, b ∈ ba∗ b = b2 . Thus, by (iv), b2 = H. It follows that H is of type H4,19 . (vi) Assume that a2 , {a}. Then b ∈ a2 or a∗ ∈ a2 , since 1 < a2 . Assume that b < a2 . Then a∗ ∈ a2 . On the other hand, by (ii), a ∈ ba. Then a∗ ∈ ba2 ⊆ ba ∪ ba∗ .

7.2

Non-Symmetric Hypergroups of Cardinality 4

205

However, as b < a2 , a∗ < ba; cf. Lemma 1.2.1. Moreover, as b < a∗ a, a∗ < ba∗ . This contradiction shows that b ∈ a2 . From (i) (together with b ∈ a2 ) we obtain that {a, b} ⊆ a2 . Thus, as 1 < a2 , we have either a2 = {a, b} or a2 = {a, a∗, b}. From b ∈ a2 we also obtain that a∗ ∈ ba; cf. Lemma 1.2.1. On the other hand, by (ii), {a, b} ⊆ ba. Thus, as 1 < ba and H = {1, a, a∗, b}, ba = {a, a∗, b}. From b ∈ a2 we finally obtain that a ∈ ba∗ , and from (iii) we know that b ∈ ba∗ and that a∗ < ba∗ . Thus, as 1 < ba and H = {1, a, a∗, b}, ba∗ = {a, b}. With a reference to (iv) we now obtain that H is of type H4,20 , H4,21 , H4,22 , or H4,23 , as wanted.  We now come to the final case of our analysis. This is the case where a∗ a = H. From a∗ ∈ a∗ a we obtain that a∗ ∈ aa∗ and a ∈ aa∗ ; cf. Lemma 1.2.1. Thus, as 1 ∈ aa∗ , we have {1, a, a∗ } ⊆ aa∗ . If aa∗ = {1, a, a∗ }, we refer to the previous case (interchanging the roles of a and a∗ ). Thus, for the remainder of this section, we assume that aa∗ = H. Lemma 7.2.6 Let H be a hypergroup, and let a and b be elements of H satisfying a∗ , a, b , 1, b∗ = b, and H = {1, a, a∗, b}. Assume that a∗ a = H = aa∗ . Then the following hold. (i) We have a ∈ a2 ⊆ {a, a∗, b}. (ii) We have a ∈ ba ⊆ {a, a∗, b}. (iii) We have a∗ ∈ ba∗ ⊆ {a, a∗, b}. (iv) If b2 = {1}, H is of type H4,24 or H4,25 . (v) If b2 = {1, b}, H is of type H4,26 , H4,27 , H4,28 , or H4,29 . (vi) If b2 = {1, a, a∗ }, H is of type H4,30 , H4,31 , H4,32 , or H4,33 . (vii) If b2 = H, H is of type H4,34 , H4,35 , H4,36 , or H4,37 . Proof. (i) From a ∈ aa∗ we obtain that a ∈ a2 . Since a∗ , a, 1 < a2 . Thus, as H = {1, a, a∗, b}, a2 ⊆ {a, a∗, b}. (ii) From b ∈ aa∗ we obtain that a ∈ ba. From 1 < ba and H = {1, a, a∗, b} we obtain that ba ⊆ {a, a∗, b}. (iii) From b ∈ a∗ a we obtain that a∗ ∈ ba∗ . From 1 < ba∗ and H = {1, a, a∗, b} we obtain that ba∗ ⊆ {a, a∗, b}. (iv) Assume that b2 = {1}. From (ii) we know that a ∈ ba. Thus, ba ⊆ b2 a = {a}, so ba = {a}. Similarly, (iii) yields ba∗ = {a∗ }. From ba∗ = {a∗ } we obtain that b < a2 . Thus, by (i), a2 = {a} or a2 = {a, a∗ }, so that H is of type H4,24 or H4,25 .

206

7 Hypergroups with a Small Number of Elements

(v) Assume that b2 = {1, b}. Then a < b2 , so b < ba and b < ba∗ . From b < ba we obtain that ba = {a} or ba = {a, a∗ }; cf. (ii). From b < ba∗ we obtain that ba∗ = {a∗ } or ba∗ = {a, a∗ }; cf. (iii). Note also that a∗ ∈ ba if and only if a ∈ ba∗ . Moreover, b ∈ a2 if and only if a∗ ∈ ba. Thus, H is of type H4,26 , H4,27 , H4,28 , or H4,29 . (vi), (vii) Assume that a ∈ b2 . Then b ∈ ba and b ∈ ba∗ . From b ∈ ba we obtain that ba = {a, b} or ba = {a, a∗, b}; cf. (ii). From b ∈ ba∗ we obtain that ba∗ = {a∗, b} or ba∗ = {a, a∗, b}; cf. (iii). Note also that a∗ ∈ ba if and only if a ∈ ba∗ . Moreover, b ∈ a2 if and only if a∗ ∈ ba. Thus, if b < b2 , H is of type H4,30 , H4,31 , H4,32 , or H4,33 . Moreover, if b ∈ b2 , H is of type H4,34 , H4,35 , H4,36 , or H4,37 .  Theorem 7.2.7 Each non-symmetric hypergroup with four elements is of type H4, j for some element j in {1, . . . , 37}. Proof. Let H be a non-symmetric hypergroup with four elements. Then H contains elements a and b satisfying a∗ , a, b , 1, b∗ = b, and H = {1, a, a∗, b}. Since a∗ a is ∗ -invariant, a∗ a is equal to one of the sets {1}, {1, b}, {1, a, a∗ }, or H, so that the claim follows from the previous six lemmas.  In a computer search, C. French has shown that, for each element j in {1, . . . , 37}, there exist hypergroups of type H4, j . Corollary 7.2.8 The non-commutative hypergroups with four elements are exactly the hypergroups of type H4,19 , H4,20 , H4,21 , H4,22 , or H4,23 . Proof. Let H be a hypergroup with four elements, and assume that H is not commutative. Since H is not commutative, H is not symmetric; cf. Lemma 1.2.9. Thus, by Theorem 7.2.7, H is of type H4, j for some element j in {1, . . . , 37}. If j < {19, 20, 21, 22, 23}, H is commutative. If j ∈ {19, 20, 21, 22, 23}, a ∈ ba \ ab, so H is not commutative.4  Among the 37 isomorphism classes of non-symmetric hypergroups of cardinality 4 which we found in Theorem 7.2.7, there are twelve isomorphism classes which consist of wreath products of non-trivial hypergroups. In fact, we have H4,2 H4,3 H4,4 H4,5

= = = =

H3,1 o H2,1 , H3,1 o H2,2 , H2,1 o H3,1 , H2,2 o H3,1 ,

H4,10 H4,11 H4,12 H4,13

= = = =

H3,2 o H2,1 , H3,2 o H2,2 , H3,3 o H2,1 , H3,3 o H2,2 ,

4If j ∈ {19, 20, 21, 22, 23}, we also have b ∈ aa∗ \ a∗ a.

H4,24 H4,25 H4,26 H4,27

= = = =

H2,1 o H3,2 , H2,1 o H3,3 , H2,2 o H3,2 , H2,2 o H3,3 .

7.2

Non-Symmetric Hypergroups of Cardinality 4

207

(By H4,2 = H3,1 o H2,1 we mean that each hypergroup of type H4,2 is a wreath product of a hypergroup of type H3,1 and a hypergroup of type H2,1 . The other equations are understood correspondingly.) We conclude this section with a look at regular actions of the non-symmetric hypergroups of cardinality 4 which we found in Theorem 7.2.7. We say that a hypergroup has a realization as an association scheme if it is isomorphic (as a hypergroup) to an association scheme.5 The first part of the following theorem says that among the 37 classes of nonsymmetric hypergroups of cardinality 4 which we found in Theorem 7.2.7, there are 11 classes the members of which have a realization as an association scheme on a finite set. By Lemma 1.6.6, this implies that they admit regular actions. (For most of them we give, in fact, more than one regular action.) In this part, we take advantage of the database [25] of association schemes. The second part of Theorem 7.2.9 says that there are 11 classes of hypergroups among the 37 classes of non-symmetric hypergroups of cardinality 4 which we found in Theorem 7.2.7 the members of which do not have a realization as an association scheme on a finite set. That does not mean that they do not have a realization as an association scheme on an infinite set or that they do not admit a regular action at all. In the proof of this part of the theorem, we refer to results which we obtained in Section 1.8. The third part of Theorem 7.2.9 says that ten out of the 37 classes of non-symmetric hypergroups of cardinality 4 which we found in Theorem 7.2.7 consist of hypergroups which do not admit a regular action. The proof of this part of the theorem relies on the results which we obtained in Section 1.7. Theorem 7.2.9 We have the following. (i) Hypergroups of type H4,1 , H4,2 , H4,3 , H4,4 , H4,5 , H4,12 , H4,13 , H4,14 , H4,25 , H4,27 , and H4,29 have a realization as an association scheme on a finite set. (ii) Hypergroups of type H4,6 , H4,7 , H4,10 , H4,11 , H4,15 , H4,19 , H4,24 , H4,26 , H4,32 , H4,34 , or H4,35 do not have a realization as an association scheme on a finite set. (iii) Hypergroups of type H4,8 , H4,16 , H4,17 , H4,18 , H4,20 , H4,21 , H4,22 , H4,28 , H4,30 , or H4,31 do not admit regular actions. Proof. (i) Hypergroups of type H4,1 are thin, so they all have a realization as a thin association scheme.

5Recall that an association scheme is a set theoretic hypergroup H on a set X such that, for any three elements a, b, and c in H with c ∈ ab, there exists a cardinal number γabc such that, for any two elements y in X and z ∈ yc, |ya ∩ zb∗ | = γabc .

208

7 Hypergroups with a Small Number of Elements

Each hypergroup of type H4,2 has a realization as an association scheme of HanakiMiyamoto type H M4 (6).6 Each hypergroup of type H4,3 has a realization as an association scheme of HanakiMiyamoto type H M4 (9), H M6 (12), H M6 (15), H M6 (18), H M5 (21), H M8 (24), H M379 (27), H M8 (30), and H M4 (33). Each hypergroup of type H4,4 has a realization as an association scheme of HanakiMiyamoto type H M6 (6). Each hypergroup of type H4,5 has a realization as an association scheme of HanakiMiyamoto type H M6 (9), H M13 (12), H M11 (15), H M14 (18), H M10 (21), H M23 (24), H M387 (27), H M31 (30), and H M8 (33). Each hypergroup of type H4,12 has a realization as an association scheme of HanakiMiyamoto type H M6 (14), H M6 (22), H M25 (30), H M26 (30), H M7 (38), H M8 (38), and H M9 (38). Each hypergroup of type H4,13 has a realization as an association scheme of HanakiMiyamoto type H M8 (21), H M14 (28), and H M7 (33). Each hypergroup of type H4,14 has a realization as an association scheme of HanakiMiyamoto type H M6 (8), H M11 (16), H M14 (24), and H M14 (32). Each hypergroup of type H4,25 has a realization as an association scheme of HanakiMiyamoto type H M5 (14), H M5 (22), H M13 (30), H M5 (38), and H M6 (38). Each hypergroup of type H4,27 has a realization as an association scheme of HanakiMiyamoto type H M7 (21), H M72 (28), and H M6 (33). Each hypergroup of type H4,29 has a realization as an association scheme of HanakiMiyamoto type H M18 (16) and H M19 (16). (ii) From Lemma 1.8.6 we obtain that hypergroups of type H4,6 or H4,7 do not have a realization as an association scheme of finite order. Applying Lemma 1.8.5 to a in place of both, a and b we obtain that all hypergroups of type H4,10 , H4,11 , H4,15 , H4,19 , H4,24 , H4,26 , or H4,34 do not have a realization as an association scheme of finite order. To see that hypergroups of type H4,32 do not have a realization as an association scheme of finite order, we apply Lemma 1.8.5 to b and a in place of a and b. By Lemma 1.8.7, hypergroups of type H4,35 do not have a realization as an association scheme of finite order. (iii) From Lemma 1.7.1 we obtain that hypergroups of type H4,8 do not admit a regular action. Applying Lemma 1.7.2 to a∗ in place of b and b in place of c one obtains that no hypergroup of type H4,17 admits a regular action. Applying the same lemma to a in

6Let k and n be positive integers. Recall from Section 5.8 that an association scheme is said to be of Hanaki-Miyamoto type H Mk (n) if it is listed as the k th association scheme of order n in the database given in [25].

Some Hypergroups of Cardinality 6, I

7.3

209

place of b and b in place of c one obtains that no hypergroup of type H4,31 admits a regular action. From Lemma 1.7.3 one obtains that hypergroups of type H4,28 do not admit regular actions. Applying Lemma 1.7.4 to b, a, b, b, and a in place of a, b, c, d, and e one obtains that hypergroups of type H4,16 as well as hypergroups of type H4,18 do not admit a regular action.7 Applying the same lemma to a∗ , b, a, and a∗ in place of a, c, d, and e one obtains that no hypergroup of type H4,20 or of type H4,21 admits a regular action. Applying Lemma 1.7.4 to b, a, and a in place of c, d, and e one obtains that no hypergroup of type H4,30 admits a regular action. From Lemma 1.7.6 one finally obtains that hypergroups of type H4,22 do not admit regular actions. 

7.3 Some Hypergroups of Cardinality 6, I In this section, we start a study of hypergroups of cardinality 6. We limit our investigation to hypergroups of cardinality 6 containing a non-normal closed subset. Let H be a hypergroup with six elements, and assume that H contains a non-normal closed subset. We choose one of the non-normal closed subsets of H and denote it by F. From Corollary 3.2.7 we know that |F | = 2

and

|H/F | ∈ {3, 4}.

In the present section as well as in the subsequent section, we will look at the case where |H/F | = 3. The case where |H/F | = 4 will be considered in Section 7.5. From |H| = 6 and |F | = 2 we obtain that |H \ F | = 4. Thus, assuming now that |H/F | = 3, we find pairwise distinct elements a, b, and c in H \ F with a∗ = a,

b∗ = b,

aF = {a, c}, and bF = {b, c∗ }; cf. Lemma 3.2.5. Since |F | = 2, we further find an element t in F \ {1} with F = {1, t}. Since F is a closed subset of H, this implies that F = hti. Thus, since aF = {a, c} and bF = {b, c∗ }, we have ahti = {a, c}

and

bhti = {b, c∗ }.

From F \ {1} = {t}, together with the fact that F is a closed subset of H, we also obtain that t is an involution. Thus, by Lemma 6.1.2(i), c ∈ at ⊆ {a, c}

and

c∗ ∈ bt ⊆ {b, c∗ }.

7That hypergroups of type H4,16 do not admit regular actions is also obtained from Lemma 1.7.6.

210

7 Hypergroups with a Small Number of Elements

Thus, we distinguish the four cases at at at at

= = = =

{c} {c} {a, c} {a, c}

and and and and

bt bt bt bt

= = = =

{c∗ }, {b, c∗ }, {c∗ }, {b, c∗ }.

In the present section, we completely analyze the first of these four cases, the case where at = {c} and bt = {c∗ }. The second case, the case where at = {c} and bt = {b, c∗ }, follows from the third case by interchanging the roles of a and b and the roles of c and c∗ , and this case will be investigated in Section 7.4. The final case, the case where at = {a, c} and bt = {b, c∗ }, will not be considered in this monograph. Our analysis in this section will show that, if at = {c} and bt = {c∗ }, the hypermultiplication of H is described by one of the following ten tables. H6,1 1 t c∗ c b a

H6,2 1 t c∗ c b a

H6,3 1 t c∗ c b a

1 {1} {t} {c∗ } {c} {b} {a}

1 {1} {t} {c∗ } {c} {b} {a}

1 {1} {t} {c∗ } {c} {b} {a}

t {1} {b} {a} {c∗ } {c}

t {1} {b} {a} {c∗ } {c}

t {1} {b} {a} {c∗ } {c}

c∗

c

c

a

b

{1} {c∗ } {1} {a} {t} {1} {t} {b} {c∗ } {1}

c∗

b

a

{1, a} {a, b} {1, b} {c, c∗ } {t, c} {1, a} {t, c∗ } {c, c∗ } {a, b} {1, b}

c

c∗

b

a

{1, a, b} {a, b} {1, a, b} {c, c∗ } {t, c, c∗ } {1, a, b} {t, c, c∗ } {c, c∗ } {a, b} {1, a, b}

7.3

H6,4 1 t c∗ c b a

1 {1} {t} {c∗ } {c} {b} {a}

H6,5 1 t c∗ c b a

1 {1} {t} {c∗ } {c} {b} {a}

H6,6 1 t c∗ c b a

1 {1} {t} {c∗ } {c} {b} {a}

t

H6,7 1 t c∗ c b a

1 {1} {t} {c∗ } {c} {b} {a}

H6,8 1 t c∗ c b a

1 {1} {t} {c∗ } {c} {b} {a}

{1} {b} {a} {c∗ } {c}

t {1} {b} {a} {c∗ } {c}

t

t

c

Some Hypergroups of Cardinality 6, I

c∗

b

a

{1} {b} {1, c, c∗ } {a} {c} {1, c, c∗ } ∗ {c } {b} {t, a, b} {1, c, c∗ } {c} {t, a, b} {a} {c} {1, c, c∗ }

t {1} {b} {a} {c∗ } {c}

c

c∗

b

a

{1, c, c∗ } {c, c∗ } {1, c, c∗ } {a, b} {t, a, b} {1, c, c∗ } {t, a, b} {a, b} {c, c∗ } {1, c, c∗ }

c

c∗

b

a

{1, a, c, c∗ } {a, b, c} {1, b, c, c∗ } {b, c, c∗ } {t, a, b, c} {1, a, c, c∗ } {t, a, b, c∗ } {a, c, c∗ } {a, b, c} {1, b, c, c∗ }

c

c∗

b

a

{1, a, c, c∗ } {a, b, c, c∗ } {1, b, c, c∗ } {a, b, c, c∗ } {t, a, b, c} {1, a, c, c∗ } {t, a, b, c∗ } {a, b, c, c∗ } {a, b, c, c∗ } {1, b, c, c∗ }

c

c∗

b

a

{1} {b} {1, b, c, c∗ } {a} {c} {1, a, c, c∗ } ∗ {c } {b} {t, a, b, c∗ } {1, b, c, c∗ } {c} {t, a, b, c} {a} {c} {1, a, c, c∗ }

211

212

7 Hypergroups with a Small Number of Elements

H6,9 1 t c∗ c b a

H6,10 1 t c∗ c b a

1 {1} {t} {c∗ } {c} {b} {a}

1 {1} {t} {c∗ } {c} {b} {a}

t

c

{1} {b} {a} {c∗ } {c}

c∗

b

a

H \ {t} {a, b, c} H \ {t} {b, c, c∗ } H \ {1} H \ {t} H \ {1} {a, c, c∗ } {a, b, c} H \ {t}

t

c

c∗

b

a

{1} {b} H \ {t} {a} {a, b, c, c∗ } H \ {t} {c∗ } {a, b, c, c∗ } H \ {1} H \ {t} {c} H \ {1} {a, b, c, c∗ } {a, b, c, c∗ } H \ {t}

Since c∗ c is ∗ -invariant and H = {1, t, a, b, c, c∗ }, c∗ c is equal to one of the sets {1}, {1, c, c∗ },

{1, a}, {1, a, c, c∗ },

{1, b},

{1, a, b},

{1, b, c, c∗ },

H \ {t}.

We will disregard the first case for the moment and go through the remaining seven cases one by one. Lemma 7.3.1 Let H be a hypergroup with six elements, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H \ hti, and assume that a∗ = a, b∗ = b, at = {c}, and bt = {c∗ }. Assume further that c∗ c = {1, a}. Then the following hold. (i) We have c2 = {a, b}. (ii) We have cc∗ = {1, b}. (iii) We have bc = {c, c∗ }. (iv) We have bc∗ = {t, c}. (v) We have b2 = {1, a}. (vi) We have ac = {t, c∗ }. (vii) We have ac∗ = {c, c∗ }. (viii) We have ab = {a, b}. (ix) We have a2 = {1, b}. Proof. (i) From Lemma 6.2.1(i) we know that c∗ , c. Thus, by Lemma 1.1.6, 1 < c2 . Furthermore, by Lemma 6.2.1(iii), t < c2 .

7.3

Some Hypergroups of Cardinality 6, I

213

Our assumption that a ∈ c∗ c implies that {a, b} ⊆ c2 ; cf. Lemma 6.3.3(i). Since we are assuming that c < c∗ c, we also have c < c2 . Now assume, by way of contradiction, that c∗ ∈ c2 . Then c2 = {a, b, c∗ }. It follows that (c2 )c = {a, b, c∗ }c = ac ∪ bc ∪ c∗ c. Since c < c∗ c, a < bc∗ and b < bc; cf. Lemma 6.3.3(iii). From a < bc∗ we obtain that b < ac. Thus, as b < c∗ c, b < (c2 )c. On the other hand, assuming c∗ ∈ c2 we also obtain that cc∗ ⊆ c(c2 ), and, since a ∈ c∗ c, we obtain from Lemma 6.3.3(i) that b ∈ cc∗ . Thus, b ∈ c(c2 ), a contradiction which shows that c∗ < c2 . (ii) By Lemma 1.1.1, 1 ∈ cc∗ . Furthermore, by Lemma 6.2.2, t < cc∗ . Since b < c∗ c, a < cc∗ ; cf. Lemma 6.3.3(ii). Since a ∈ c∗ c, b ∈ cc∗ ; cf. Lemma 6.3.3(i). From c < c∗ c we obtain that c < cc∗ and that c∗ < cc∗ ; cf. Lemma 1.2.1. (iii) Since b , c, 1 < bc. Furthermore, by Lemma 6.2.1(iii), t < bc. In (i), we saw that c∗ < c2 . Thus, by Lemma 6.2.4(iii), a < bc. Since c < c∗ c, b < bc; cf. Lemma 6.3.3(iii). From (ii) we know that b ∈ cc∗ . Thus, c ∈ bc. Now recall from Lemma 6.2.4(ii) that bc is ∗ -invariant. Thus, as c ∈ bc, c∗ ∈ bc. (iv) Since b , c, 1 < bc∗ . Furthermore, by Lemma 6.2.1(iii), t ∈ bc∗ . Since we are assuming that c < c∗ c, we have a < bc∗ and b < bc; cf. Lemma 6.3.3(iii). From b < bc we obtain that b < bc∗ . In (i), we saw that b ∈ c2 . Thus, c ∈ bc∗ . Finally, since we are assuming that b < c∗ c, we have c∗ < bc∗ . (v) From Lemma 6.3.2(ii) we know that c∗ c = b2 . Thus, the claim follows from our hypothesis that c∗ c = {1, a}. (vi) From Lemma 6.3.2(iii) we know that ac = cb. Thus, the claim follows from (iv). (vii) Since a , c, 1 < ac∗ . Furthermore, by Lemma 6.2.1(iii), t < ac∗ . Since c < c∗ c, a < ac∗ ; cf. Lemma 6.3.3(iii). In (i), we saw that c∗ < c2 . Thus, by Lemma 6.2.4(iii), b < ac∗ . From a ∈ c∗ c we obtain that c∗ ∈ ac∗ . Recall from Lemma 6.2.4(ii) that ac∗ is ∗ -invariant. Thus, as c∗ ∈ ac∗ , c ∈ ac∗ . (viii) From Lemma 6.2.4(i) we know that c2 = ab, so the claim follows from (i). (ix) From Lemma 6.3.2(i) we know that cc∗ = a2 , so the claim follows from (ii).  Lemma 7.3.2 Let H be a hypergroup with six elements, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H \ hti, and assume that a∗ = a, b∗ = b, at = {c}, and bt = {c∗ }. Then c∗ c , {1, b}. Proof. From Lemma 6.2.1(i) we know that c∗ , c. Thus, by Lemma 1.1.6, 1 < c2 . Furthermore, by Lemma 6.2.1(iii), t < c2 . Now assume, by way of contradiction, that c∗ c = {1, b}. Then, a < c∗ c. Thus, by Lemma 6.3.3(i), a < c2 and that b < c2 . From c < c∗ c, we also obtain that c < c2 . Thus, as H = {1, t, a, b, c, c∗ }, we have c2 = {c∗ }, and, since b ∈ c∗ c, this yields b ∈ (c2 )c.

214

7 Hypergroups with a Small Number of Elements

On the other hand, since a < c∗ c, we obtain from Lemma 6.3.3(i) that b < cc∗ . Thus, as c2 = {c∗ }, b < c(c2 ). This contradiction shows that c∗ c , {1, b}.  Lemma 7.3.3 Let H be a hypergroup with six elements, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H \ hti, and assume that a∗ = a, b∗ = b, at = {c}, and bt = {c∗ }. Assume further that c∗ c = {1, a, b}. Then the following hold. (i) We have c2 = {a, b}. (ii) We have cc∗ = {1, a, b}. (iii) We have bc = {c, c∗ }. (iv) We have bc∗ = {t, c, c∗ }. (v) We have b2 = {1, a, b}. (vi) We have ac = {t, c, c∗ }. (vii) We have ac∗ = {c, c∗ }. (viii) We have ab = {a, b}. (ix) We have a2 = {1, a, b}. Proof. (i) From Lemma 6.2.1(i) we know that c∗ , c. Thus, by Lemma 1.1.6, 1 < c2 . Furthermore, by Lemma 6.2.1(iii), t < c2 . Our assumption that a ∈ c∗ c implies that {a, b} ⊆ c2 ; cf. Lemma 6.3.3(i). Since we are assuming that c < c∗ c, we also have c < c2 . Now assume, by way of contradiction, that c∗ ∈ c2 . Then, since a ∈ c2 , a ∈ (c∗ )2 , so a ∈ (c2 )c∗ . On the other hand, we have c(cc∗ ) = c{1, a, b} = {c} ∪ ca ∪ cb. Since c < c∗ c, we have a < ac∗ and a < bc∗ . Thus, as a is symmetric, a < ca and a < cb. It follows that a < c(cc∗ ). This contradiction shows that c∗ < c2 . (ii) By Lemma 1.1.1, 1 ∈ cc∗ . Furthermore, by Lemma 6.2.2, t < cc∗ . Since b ∈ c∗ c, a ∈ cc∗ ; cf. Lemma 6.3.3(ii). Since a ∈ c∗ c, b ∈ cc∗ ; cf. Lemma 6.3.3(i). From c < c∗ c we obtain that c < cc∗ and that c∗ < cc∗ ; cf. Lemma 1.2.1. (iii) Since b , c, 1 < bc. Furthermore, by Lemma 6.2.1(iii), t < bc. Since c∗ < c2 , we have a < bc; cf. Lemma 6.2.4(iii). Since c < c∗ c, b < bc; cf. Lemma 6.3.3(iii). From (ii) we know that b ∈ cc∗ . Thus, c ∈ bc. Now recall from Lemma 6.2.4(ii) that bc is ∗ -invariant. Thus, as c ∈ bc, c∗ ∈ bc. (iv) Since b , c, 1 < bc∗ . Furthermore, by Lemma 6.2.1(iii), t ∈ bc∗ . Since we are assuming that c < c∗ c, we have a < bc∗ and b < bc; cf. Lemma 6.3.3(iii). From b < bc we obtain that b < bc∗ . In (i), we saw that b ∈ c2 . Thus, c ∈ bc∗ . Finally, since we are assuming that b ∈ c∗ c, we have c∗ ∈ bc∗ . (v) From Lemma 6.3.2(ii) we know that c∗ c = b2 . Thus, the claim follows from our hypothesis that c∗ c = {1, a, b}.

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(vi) From Lemma 6.3.2(iii) we know that ac = cb. Thus, the claim follows from (iv). (vii) Since a , c, 1 < ac∗ . Furthermore, by Lemma 6.2.1(iii), t < ac∗ . Since c < c∗ c, a < ac∗ ; cf. Lemma 6.3.3(iii). Since c∗ < c2 , we have b < ac∗ ; cf. Lemma 6.2.4(iii). From a ∈ c∗ c we obtain that c∗ ∈ ac∗ . Recall from Lemma 6.2.4(ii) that ac∗ is ∗ -invariant. Thus, as c∗ ∈ ac∗ , c ∈ ac∗ . (viii) From Lemma 6.2.4(i) we know that c2 = ab, so the claim follows from (i). (ix) From Lemma 6.3.2(i) we know that cc∗ = a2 , so the claim follows from (ii).  Lemma 7.3.4 Let H be a hypergroup with six elements, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H \ hti, and assume that a∗ = a, b∗ = b, at = {c}, and bt = {c∗ }. Assume further that c∗ c = {1, c, c∗ }. Then the following hold. (i) We have {c} ⊆ c2 ⊆ {c, c∗ }. (ii) We have cc∗ = {1, c, c∗ }. (iii) We have {b} ⊆ bc ⊆ {a, b}. (iv) We have bc∗ = {t, a, b}. (v) We have b2 = {1, c, c∗ }. (vi) We have ac = {t, a, b}. (vii) We have {a} ⊆ ac∗ ⊆ {a, b}. (viii) We have {c} ⊆ ab ⊆ {c, c∗ }. (ix) We have a2 = {1, c, c∗ }. Proof. (i) From Lemma 6.2.1(i) we know that c∗ , c. Thus, by Lemma 1.1.6, 1 < c2 . Furthermore, by Lemma 6.2.1(iii), t < c2 . Our assumption that a < c∗ c implies that a < c2 and that b < c2 ; cf. Lemma 6.3.3(i). Since we are assuming that c ∈ c∗ c, we also have c ∈ c2 . (ii) By Lemma 1.1.1, 1 ∈ cc∗ . Furthermore, by Lemma 6.2.2, t < cc∗ . Since b < c∗ c, a < cc∗ ; cf. Lemma 6.3.3(ii). Since a < c∗ c, b < cc∗ ; cf. Lemma 6.3.3(i). From c ∈ c∗ c we obtain that c ∈ cc∗ and that c∗ ∈ cc∗ ; cf. Lemma 1.2.1. (iii) Since b , c, 1 < bc. Furthermore, by Lemma 6.2.1(iii), t < bc. Since c ∈ c∗ c, b ∈ bc; cf. Lemma 6.3.3(iii). From (ii) we know that b < cc∗ . Thus, c < bc. Now recall from Lemma 6.2.4(ii) that bc is ∗ -invariant. Thus, as c < bc, c∗ < bc. (iv) Since b , c, 1 < bc∗ . Furthermore, by Lemma 6.2.1(iii), t ∈ bc∗ . Since we are assuming that c ∈ c∗ c, we have a ∈ bc∗ and b ∈ bc; cf. Lemma 6.3.3(iii). From b ∈ bc we obtain that b ∈ bc∗ . In (i), we saw that b < c2 . Thus, c < bc∗ . Since we are assuming that b < c∗ c, we have c∗ < bc∗ .

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(v) From Lemma 6.3.2(ii) we know that c∗ c = b2 . Thus, the claim follows from our hypothesis that c∗ c = {1, c, c∗ }. (vi) From Lemma 6.3.2(iii) we know that ac = cb. Thus, the claim follows from (iv). (vii) Since a , c, 1 < ac∗ . Furthermore, by Lemma 6.2.1(iii), t < ac∗ . Since c ∈ c∗ c, a ∈ ac∗ ; cf. Lemma 6.3.3(iii). From a < c∗ c we obtain that c∗ < ac∗ . Recall from Lemma 6.2.4(ii) that ac∗ is ∗ -invariant. Thus, as c∗ < ac∗ , c < ac∗ . (viii) From Lemma 6.2.4(i) we know that c2 = ab, so the claim follows from (i). (ix) From Lemma 6.3.2(i) we know that cc∗ = a2 , so the claim follows from (ii).  Lemma 7.3.5 Let H be a hypergroup with six elements, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H \ hti, and assume that a∗ = a, b∗ = b, at = {c}, and bt = {c∗ }. Assume further that c∗ c = {1, a, c, c∗ }. Then the following hold. (i) We have {a, b, c} ⊆ c2 ⊆ {a, b, c, c∗ }. (ii) We have cc∗ = {1, b, c, c∗ }. (iii) We have {b, c, c∗ } ⊆ bc ⊆ {a, b, c, c∗ }. (iv) We have bc∗ = {t, a, b, c}. (v) We have b2 = {1, a, c, c∗ }. (vi) We have ac = {t, a, b, c∗ }. (vii) We have {a, c, c∗ } ⊆ ac∗ ⊆ {a, b, c, c∗ }. (viii) We have {a, b, c} ⊆ ab ⊆ {a, b, c, c∗ }. (ix) We have a2 = {1, b, c, c∗ }. Proof. (i) From Lemma 6.2.1(i) we know that c∗ , c. Thus, by Lemma 1.1.6, 1 < c2 . Furthermore, by Lemma 6.2.1(iii), t < c2 . Our assumption that a ∈ c∗ c implies that {a, b} ⊆ c2 ; cf. Lemma 6.3.3(i). Since we are assuming that c ∈ c∗ c, we also have c ∈ c2 . (ii) By Lemma 1.1.1, 1 ∈ cc∗ . Furthermore, by Lemma 6.2.2, t < cc∗ . Since b < c∗ c, a < cc∗ ; cf. Lemma 6.3.3(ii). Since a ∈ c∗ c, b ∈ cc∗ ; cf. Lemma 6.3.3(i). From c ∈ c∗ c we obtain that c ∈ cc∗ and that c∗ ∈ cc∗ ; cf. Lemma 1.2.1. (iii) Since b , c, 1 < bc. Furthermore, by Lemma 6.2.1(iii), t < bc. Since c ∈ c∗ c, b ∈ bc; cf. Lemma 6.3.3(iii). From (ii) we know that b ∈ cc∗ . Thus, c ∈ bc. Now recall from Lemma 6.2.4(ii) that bc is ∗ -invariant. Thus, as c ∈ bc, c∗ ∈ bc. (iv) Since b , c, 1 < bc∗ . Furthermore, by Lemma 6.2.1(iii), t ∈ bc∗ . Since we are assuming that c ∈ c∗ c, we have a ∈ bc∗ and b ∈ bc; cf. Lemma 6.3.3(iii). From b ∈ bc we obtain that b ∈ bc∗ . In (i), we saw that b ∈ c2 . Thus, c ∈ bc∗ . Since we are assuming that b < c∗ c, we have c∗ < bc∗ .

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(v) From Lemma 6.3.2(ii) we know that c∗ c = b2 . Thus, the claim follows from our hypothesis that c∗ c = {1, a, c, c∗ }. (vi) From Lemma 6.3.2(iii) we know that ac = cb. Thus, the claim follows from (iv). (vii) Since a , c, 1 < ac∗ . Furthermore, by Lemma 6.2.1(iii), t < ac∗ . Since c ∈ c∗ c, a ∈ ac∗ ; cf. Lemma 6.3.3(iii). From a ∈ c∗ c we obtain that c∗ ∈ ac∗ . Recall from Lemma 6.2.4(ii) that ac∗ is ∗ -invariant. Thus, as c∗ ∈ ac∗ , c ∈ ac∗ . (viii) From Lemma 6.2.4(i) we know that c2 = ab, so the claim follows from (i). (ix) From Lemma 6.3.2(i) we know that cc∗ = a2 , so the claim follows from (ii).  Lemma 7.3.6 Let H be a hypergroup with six elements, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H \ hti, and assume that a∗ = a, b∗ = b, at = {c}, and bt = {c∗ }. Assume further that c∗ c = {1, b, c, c∗ }. Then the following hold. (i) We have c2 = {c}. (ii) We have cc∗ = {1, a, c, c∗ }. (iii) We have {b} ⊆ bc ⊆ {a, b}. (iv) We have bc∗ = {t, a, b, c∗ }. (v) We have b2 = {1, b, c, c∗ }. (vi) We have ac = {t, a, b, c}. (vii) We have {a} ⊆ ac∗ ⊆ {a, b}. (viii) We have {c} ⊆ ab ⊆ {c, c∗ }. (ix) We have a2 = {1, a, c, c∗ }. Proof. (i) From Lemma 6.2.1(i) we know that c∗ , c. Thus, by Lemma 1.1.6, 1 < c2 . Furthermore, by Lemma 6.2.1(iii), t < c2 . Our assumption that a < c∗ c implies that a < c2 and that b < c2 ; cf. Lemma 6.3.3(i). Since we are assuming that c ∈ c∗ c, we also have c ∈ c2 . Now assume, by way of contradiction, that c∗ ∈ c2 . Then c2 = {c, c∗ }. It follows that (c2 )c = {c, c∗ }c = c2 ∪ c∗ c = {c, c∗ } ∪ {1, b, c, c∗ }. In particular, a < (c2 )c. On the other hand, since b ∈ c∗ c, a ∈ cc∗ ; cf. Lemma 6.3.3(ii). Thus, as we are assuming that c∗ ∈ c2 , we obtain a ∈ c(c2 ). This contradiction shows that c∗ < c2 . (ii) By Lemma 1.1.1, 1 ∈ cc∗ . Furthermore, by Lemma 6.2.2, t < cc∗ . Since b ∈ c∗ c, a ∈ cc∗ ; cf. Lemma 6.3.3(ii). Since a < c∗ c, b < cc∗ ; cf. Lemma 6.3.3(i). From c ∈ c∗ c we obtain that c ∈ cc∗ and that c∗ ∈ cc∗ ; cf. Lemma 1.2.1. (iii) Since b , c, 1 < bc. Furthermore, by Lemma 6.2.1(iii), t < bc.

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7 Hypergroups with a Small Number of Elements

Since c ∈ c∗ c, b ∈ bc; cf. Lemma 6.3.3(iii). From (ii) we know that b < cc∗ . Thus, c < bc. Now recall from Lemma 6.2.4(ii) that bc is ∗ -invariant. Thus, as c < bc, c∗ < bc. (iv) Since b , c, 1 < bc∗ . Furthermore, by Lemma 6.2.1(iii), t ∈ bc∗ . Since we are assuming that c ∈ c∗ c, we have a ∈ bc∗ and b ∈ bc; cf. Lemma 6.3.3(iii). From b ∈ bc we obtain that b ∈ bc∗ . In (i), we saw that b < c2 . Thus, c < bc∗ . Since we are assuming that b ∈ c∗ c, we have c∗ ∈ bc∗ . (v) From Lemma 6.3.2(ii) we know that c∗ c = b2 . Thus, the claim follows from our hypothesis that c∗ c = {1, b, c, c∗ }. (vi) From Lemma 6.3.2(iii) we know that ac = cb. Thus, the claim follows from (iv). (vii) Since a , c, 1 < ac∗ . Furthermore, by Lemma 6.2.1(iii), t < ac∗ . Since c ∈ c∗ c, a ∈ ac∗ ; cf. Lemma 6.3.3(iii). From a < c∗ c we obtain that c∗ < ac∗ . Recall from Lemma 6.2.4(ii) that ac∗ is ∗ -invariant. Thus, as c∗ < ac∗ , c < ac∗ . (viii) From Lemma 6.2.4(i) we know that c2 = ab, so the claim follows from (i). (ix) From Lemma 6.3.2(i) we know that cc∗ = a2 , so the claim follows from (ii).  Lemma 7.3.7 Let H be a hypergroup with six elements, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H \ hti, and assume that a∗ = a, b∗ = b, at = {c}, and bt = {c∗ }. Assume further that c∗ c = H \ {t}. Then the following hold. (i) We have {a, b, c} ⊆ c2 ⊆ {a, b, c, c∗ }. (ii) We have cc∗ = H \ {t}. (iii) We have {b, c, c∗ } ⊆ bc ⊆ {a, b, c, c∗ }. (iv) We have bc∗ = H \ {1}. (v) We have b2 = H \ {t}. (vi) We have ac = H \ {1}. (vii) We have {a, c, c∗ } ⊆ ac∗ ⊆ {a, b, c, c∗ }. (viii) We have {a, b, c} ⊆ ab ⊆ {a, b, c, c∗ }. (ix) We have a2 = H \ {t}. Proof. (i) From Lemma 6.2.1(i) we know that c∗ , c. Thus, by Lemma 1.1.6, 1 < c2 . Furthermore, by Lemma 6.2.1(iii), t < c2 . Our assumption that a ∈ c∗ c implies that {a, b} ⊆ c2 ; cf. Lemma 6.3.3(i). Since we are assuming that c ∈ c∗ c, we also have c ∈ c2 . (ii) By Lemma 1.1.1, 1 ∈ cc∗ . Furthermore, by Lemma 6.2.2, t < cc∗ . Since b ∈ c∗ c, a ∈ cc∗ ; cf. Lemma 6.3.3(ii). Since a ∈ c∗ c, b ∈ cc∗ ; cf. Lemma 6.3.3(i). From c ∈ c∗ c we obtain that c ∈ cc∗ and that c∗ ∈ cc∗ ; cf. Lemma 1.2.1. (iii) Since b , c, 1 < bc. Furthermore, by Lemma 6.2.1(iii), t < bc.

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Since c ∈ c∗ c, b ∈ bc; cf. Lemma 6.3.3(iii). From (ii) we know that b ∈ cc∗ . Thus, c ∈ bc. Now recall from Lemma 6.2.4(ii) that bc is ∗ -invariant. Thus, as c ∈ bc, c∗ ∈ bc. (iv) Since b , c, 1 < bc∗ . Furthermore, by Lemma 6.2.1(iii), t ∈ bc∗ . Since we are assuming that c ∈ c∗ c, we have a ∈ bc∗ and b ∈ bc; cf. Lemma 6.3.3(iii). From b ∈ bc we obtain that b ∈ bc∗ . In (i), we saw that b ∈ c2 . Thus, c ∈ bc∗ . Since we are assuming that b ∈ c∗ c, we have c∗ ∈ bc∗ . (v) From Lemma 6.3.2(ii) we know that c∗ c = b2 . Thus, the claim follows from our hypothesis that c∗ c = H \ {t}. (vi) From Lemma 6.3.2(iii) we know that ac = cb. Thus, the claim follows from (iv). (vii) Since a , c, 1 < ac∗ . Furthermore, by Lemma 6.2.1(iii), t < ac∗ . Since c ∈ c∗ c, a ∈ ac∗ ; cf. Lemma 6.3.3(iii). From a ∈ c∗ c we obtain that c∗ ∈ ac∗ . Recall from Lemma 6.2.4(ii) that ac∗ is ∗ -invariant. Thus, as c∗ ∈ ac∗ , c ∈ ac∗ . (viii) From Lemma 6.2.4(i) we know that c2 = ab, so the claim follows from (i). (ix) From Lemma 6.3.2(i) we know that cc∗ = a2 , so the claim follows from (ii).  Let j be an element in {1, . . . , 10}. We say that a hypergroup is of type H6, j if it contains elements t, a, b, and c such that its hypermultiplication is described by table H6, j . Theorem 7.3.8 Let H be a hypergroup with six elements. Assume that H contains a thin non-normal closed subset which has index 3 in H. Then {1, . . . , 10} contains an element j such that H is of type H6, j . Proof. Let F be a thin non-normal closed subset of H, and assume that |H/F | = 3. Since |H| = 6 and F is not normal in H, we have |F | = 2; cf. Corollary 3.2.7. It follows that |H \ F | = 4. Thus, as |H/F | = 3, H \ F contains pairwise distinct elements a, b, and c such that a∗ = a,

b∗ = b,

aF = {a, c}, and bF = {b, c∗ }; cf. Lemma 3.2.5. Since |F | = 2, F \ {1} contains an element t with F = {1, t}. Since F is a closed subset of H, this implies that F = hti. Thus, as aF = {a, c} and bF = {b, c∗ }, we obtain that ahti = {a, c} and bhti = {b, c∗ }. From F \ {1} = {t}, together with the fact that F is a closed subset of H, we also obtain that t is an involution. Moreover, since F is assumed to be thin, so is t. Thus, by Lemma 6.2.6, at = {c} and bt = {c∗ }. It follows that

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7 Hypergroups with a Small Number of Elements

c∗ t = {b}

and

ct = {a};

cf. Lemma 6.2.2 and Corollary 6.2.3. Since c∗ c is ∗ -invariant and H = {1, t, a, b, c, c∗ }, c∗ c is equal to one of the sets {1}, {1, c, c∗ },

{1, a}, {1, a, c, c∗ },

{1, b},

{1, a, b},

{1, b, c, c∗ },

H \ {t}.

Assume that c∗ c = {1}. Then c is thin, whence, by Lemma 2.6.2, c∗ is thin. Thus, as H = {1, t, a, b, c, c∗ }, we obtain from Lemma 6.3.1 that H is thin. Now, as |H| = 6, H corresponds, via the group correspondence, to a group of order 6. Since H has a non-normal closed subset, this implies that H is of type H6,1 . Assume that c∗ c = {1, a}. Then, by Lemma 7.3.1, H is of type H6,2 . From Lemma 7.3.2 we know that c∗ c , {1, b}. Assume that c∗ c = {1, a, b}. Then, by Lemma 7.3.3, H is of type H6,3 . Assume that c∗ c = {1, c, c∗ }. If c∗ < c2 , we obtain from Lemma 6.2.4(iii) that a < bc and b < ac∗ . Thus, by Lemma 7.3.4, H is of type H6,4 . If c∗ ∈ c2 , we obtain from Lemma 6.2.4(iii) that a ∈ bc and b ∈ ac∗ . Thus, by Lemma 7.3.4, H is of type H6,5 . Assume that c∗ c = {1, a, c, c∗ }. If c∗ < c2 , we obtain from Lemma 6.2.4(iii) that a < bc and b < ac∗ . Thus, by Lemma 7.3.5, H is of type H6,6 . If c∗ ∈ c2 , we obtain from Lemma 6.2.4(iii) that a ∈ bc and b ∈ ac∗ . Thus, by Lemma 7.3.5, H is of type H6,7 . Assume that c∗ c = {1, b, c, c∗ }. Then, by Lemma 7.3.6, H is of type H6,8 . Assume that c∗ c = H \ {t}. If c∗ < c2 , we obtain from Lemma 6.2.4(iii) that a < bc and b < ac∗ . Thus, by Lemma 7.3.7, H is of type H6,9 . If c∗ ∈ c2 , Lemma 6.2.4(iii)  yields that a ∈ bc and b ∈ ac∗ . Thus, by Lemma 7.3.7, H is of type H6,10 . In a computer search, C. French has shown that, for each element j in {1, . . . , 10}, there exist hypergroups of type H6, j .

7.4 Some Hypergroups of Cardinality 6, II We continue our study of hypergroups of cardinality 6 containing a non-normal closed subset which we started in Section 7.3. More precisely, we will study hypergroups H with six elements containing an involution t and pairwise distinct elements a, b, and c in H \ hti satisfying a∗ = a,

b∗ = b,

at = {a, c},

and

bt = {c∗ }.

(The only difference from the conditions in Section 7.3 is that we assume at = {a, c} instead of at = {c}.) The main result of this section is Theorem 7.4.8. This theorem says that the hypermultiplication of hypergroups H containing an involution t and pairwise distinct

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221

elements a, b, and c satisfying the above four equations is described by one of the following nine tables. Similarly, to Theorem 7.3.8, Theorem 7.4.8 will be stated without an explicit reference to the above four equations. H6,11 1 t c∗ c b a

H6,12 1 t c∗ c b a

H6,13 1 t c∗ c b a

1 {1} {t} {c∗ } {c} {b} {a}

1 {1} {t} {c∗ } {c} {b} {a}

1 {1} {t} {c∗ } {c} {b} {a}

H6,14 1 t c∗ c b a

t

c∗

c

b

a

{1, t} {b, c∗ } {1, a, b} {a} {a, c∗ } {1, t, a} ∗ {c } {a} {t, c∗ } {1, b} ∗ {a, c} {t, a, c, c } {a, b, c, c∗ } {a, c∗ } H

t

c∗

c

{1, t} {b, c∗ } {a} {c∗ } {a, c}

b

a

{1, a, c, c∗ } {a, b, c, c∗ } H {a, c, c∗ } {t, a, c} {1, a} H \ {1} {a, b, c, c∗ } {a, b, c, c∗ } H

t

c∗

c

b

a

{1, t} {b, c∗ } {1, b, c, c∗ } {a} {c} H \ {b} {c∗ } {b} {t, a, b, c∗ } {1, b, c, c∗ } {a, c} {t, a, b, c} {a} {c} H \ {b}

1 {1} {t} {c∗ } {c} {b} {a}

t {1, t} {b, c∗ } {a} {c∗ } {a, c}

c

c∗

b

a

H \ {t} {a, c, c∗ } H \ {b} {a, b} {t, a, b, c∗ } {1, b, c, c∗ } H \ {1} {a, b, c, c∗ } {a, c, c∗ } H

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7 Hypergroups with a Small Number of Elements

H6,15 1 t c∗ c b a

1 {1} {t} {c∗ } {c} {b} {a}

H6,16 1 t c∗ c b a

{1, t} {b, c∗ } {a} {c∗ } {a, c}

1 {1} {t} {c∗ } {c} {b} {a}

H6,17 1 t c∗ c b a

1 {1} {t} {c∗ } {c} {b} {a}

H6,18 1 t c∗ c b a

1 {1} {t} {c∗ } {c} {b} {a}

H6,19 1 t c∗ c b a

1 {1} {t} {c∗ } {c} {b} {a}

t

c∗

c

t

b

a

H \ {t} {a, b, c} H {b, c, c∗ } H \ {1} {1, a, c, c∗ } H \ {1} {a, c, c∗ } {a, b, c} H

t {1, t} {b, c∗ } {a} {c∗ } {a, c}

c

c∗

b

a

H \ {t} {a, b, c} H {b, c, c∗ } H \ {1} H \ {t} H \ {1} {a, c, c∗ } {a, b, c} H

c

c∗

b

a

{1, t} {b, c∗ } H \ {t} {a} {a, b, c, c∗ } H {c∗ } {a, c, c∗ } {t, a, c, c∗ } {1, a, b} {a, c} H \ {1} {a, b, c, c∗ } {a, b, c, c∗ } H

t

c

c∗

b

a

{1, t} {b, c∗ } H \ {t} {a} {a, b, c, c∗ } H {c∗ } {a, b, c, c∗ } H \ {1} {1, a, c, c∗ } {a, c} H \ {1} {a, b, c, c∗ } {a, b, c, c∗ } H

t

c

c∗

b

a

{1, t} {b, c∗ } H \ {t} {a} {a, b, c, c∗ } H {c∗ } {a, b, c, c∗ } H \ {1} H \ {t} {a, c} H \ {1} {a, b, c, c∗ } {a, b, c, c∗ } H

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The following seven lemmas provide the necessary information for the proof of Theorem 7.4.8. Lemma 7.4.1 Let H be a hypergroup with six elements, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H \ hti, and assume that a∗ = a, b∗ = b, at = {a, c}, and bt = {c∗ }. Assume further that c < c∗ c. Then the following hold. (i) We have c∗ c = {1, a, b}. (ii) We have c2 = {a, c∗ }. (iii) We have cc∗ = {1, t, a}. (iv) We have bc = {a}. (v) We have bc∗ = {t, c∗ }. (vi) We have b2 = {1, b}. (vii) We have ac = {t, a, c, c∗ }. (viii) We have ac∗ = {a, b, c, c∗ }. (ix) We have ab = {a, c∗ }. (x) We have a2 = H. Proof. (i) By Lemma 1.1.1, 1 ∈ c∗ c. Since we are assuming that bt = {c∗ }, we obtain from Corollary 6.2.3 that t < c∗ c. Since c∗ c is ∗ -invariant, we obtain from c < c∗ c that c∗ < c∗ c. Thus, as H = {1, t, a, b, c, c∗ }, we have c∗ c ⊆ {1, a, b}, so that the desired equation follows from Lemma 6.4.4(iii). (ii) Since c∗ , c, 1 < c2 . Furthermore, by Lemma 6.2.1(iii), t < c2 . From (i) we know that c∗ c = {1, a, b}. Thus, we may apply Lemma 6.4.4(iv), and we obtain that c2 ∩ {a, b, c, c∗ } = {a, c∗ }. (iii) From Lemma 1.1.1 we know that 1 ∈ cc∗ . Furthermore, since we are assuming that at = {a, c}, we have t ∈ cc∗ ; cf. Lemma 6.2.2. From (i) we know that c∗ c = {1, a, b}. Thus, by Lemma 6.4.4(v), cc∗ ∩ {a, b, c, c∗ } = {a}. (iv) From (i) we know that c∗ c = {1, a, b}. Thus, the desired equation follows from Lemma 6.4.4(i). (v) Since b , c, we have 1 < bc∗ , and from Lemma 6.2.1(iii) we know that t ∈ bc∗ . From (i) we know that c∗ c = {1, a, b}. Thus, by Lemma 6.4.4(vi), bc∗ ∩ {a, b, c, c∗ } = {c∗ }. (vi) From (i) we know that c∗ c = {1, a, b}. Thus, the desired equation follows from Lemma 6.4.4(ii). (vii) Since a , c, 1 < ac. Furthermore, by Lemma 6.2.1(iii), t ∈ ac.

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From (i) we know that c∗ c = {1, a, b}. Thus, we may apply Lemma 6.4.4(vii), and we obtain that ac ∩ {a, b, c, c∗ } = {a, c, c∗ }. (viii) Since a , c, 1 < ac∗ . Furthermore, by Lemma 6.2.1(iii), t < ac∗ . From (i) we know that c∗ c = {1, a, b}. Thus, we obtain from Lemma 6.4.4(viii) that {a, b, c, c∗ } ⊆ ac∗ . (ix) Since we are assuming that bt = {c∗ }, we have c2 = ab; cf. Corollary 6.2.5(i). Thus, the claim follows from (ii). (x) From Lemma 6.4.1(iii) we know that a2 = ac∗ ∪ cc∗ . Thus, the desired equation  follows from (viii) and (iii). Lemma 7.4.2 Let H be a hypergroup with six elements, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H \ hti, and assume that a∗ = a, b∗ = b, at = {a, c}, and bt = {c∗ }. Assume further that c ∈ c∗ c and b < c∗ c. Then the following hold. (i) We have c∗ c = {1, a, c, c∗ }. (ii) We have c2 = {a, b, c, c∗ }. (iii) We have cc∗ = H. (iv) We have bc = {a, c, c∗ }. (v) We have bc∗ = {t, a, c}. (vi) We have b2 = {1, a}. (vii) We have ac = H \ {1}. (viii) We have ac∗ = {a, b, c, c∗ }. (ix) We have ab = {a, b, c, c∗ }. (x) We have a2 = H. Proof. (i) By Lemma 1.1.1, 1 ∈ c∗ c. Since we are assuming that bt = {c∗ }, we obtain from Corollary 6.2.3 that t < c∗ c. Since b , c, 1 < bc. From Lemma 6.2.1(iii) we also know that t < bc. Thus, as H = {1, t, a, b, c, c∗ }, we obtain that bc ∩ {a, b, c, c∗ } is not empty; cf. Lemma 1.1.3. Since we are assuming that b < c∗ c, this implies that a ∈ c∗ c, cf. Lemma 6.4.7(iv). Since we are assuming that c ∈ c∗ c, we obtain that c∗ ∈ c∗ c; cf. Lemma 1.2.1. (ii) Since c∗ , c, 1 < c2 . Furthermore, by Lemma 6.2.1(iii), t < c2 . From (i) we know that a ∈ c∗ c. Thus, by Lemma 6.4.2(i), a ∈ c2 . Recall from Lemma 6.4.1(i) that c∗ c = b2 ∪ bc. Thus, as b < c∗ c, b < bc. It follows that c∗ < b2 ; cf. Lemma 1.2.1. Since c∗ ∈ c∗ c and c∗ c = b2 ∪ bc, this implies that c∗ ∈ bc. Thus, by Lemma 1.2.1, b ∈ c2 . From our hypothesis that c ∈ c∗ c we obtain that c ∈ c2 .

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From Corollary 6.2.5(i) we know that c2 = ab. Thus, as {b, c} ⊆ c2 , {b, c} ⊆ ab. From c ∈ ab we obtain that a ∈ cb. Thus, as b ∈ ab, we have b ∈ cb2 . From Lemma 6.4.7(i), on the other hand, we know that b2 ∩ {a, b, c, c∗ } ⊆ {a}. Thus, as chti = {a, c}, we have b ∈ {a, c} ∪ ca. It follows that b ∈ ca. Thus, by Lemma 1.2.1, c∗ ∈ ab, and, as c2 = ab, this implies that c∗ ∈ c2 . (iii) From Lemma 1.1.1 we know that 1 ∈ cc∗ . Furthermore, since we are assuming that at = {a, c}, we have t ∈ cc∗ ; cf. Lemma 6.2.2. From c ∈ c∗ c we obtain that c ∈ c2 , from Corollary 6.2.5(i) we know that c2 = ab, and, in Lemma 6.4.1(ii), we saw that ab ⊆ ac. Thus, we have c ∈ ac; equivalently, a ∈ cc∗ . From (ii) we know that b ∈ c2 . Thus, by Lemma 6.4.2(ii), b ∈ cc∗ . We are assuming that c ∈ c∗ c. Thus, by Lemma 1.2.1, {c, c∗ } ⊆ cc∗ . (iv) Since b , c, 1 < bc. Furthermore, by Lemma 6.2.1(iii), t < bc. From (ii) we know that c∗ ∈ c2 , from Corollary 6.2.5(i) that c2 = ab. Thus, c∗ ∈ ab. It follows that a ∈ bc; cf. Lemma 1.2.1. We are assuming that b < c∗ c. Thus, by Lemma 6.4.7(ii), b < bc. From (ii) we know that b ∈ c2 . Thus, by Lemma 6.4.2(ii), {c, c∗ } ⊆ bc. (v) Since b , c, we have 1 < bc∗ , and from Lemma 6.2.1(iii) we know that t ∈ bc∗ . By hypothesis, c ∈ c∗ c. Thus, by Lemma 6.4.2(iii), b ∈ ac; equivalently, a ∈ bc∗ . From (iv) we know that b < bc. Thus, b < bc∗ . From (ii) we know that b ∈ c2 . Thus, c ∈ bc∗ . From Lemma 6.4.7(iii) we know that c∗ < bc∗ . (vi) Since b∗ = b, 1 ∈ b2 . Furthermore, since we are assuming that bt = {c∗ }, we have t < b2 ; cf. Corollary 6.2.3. From Lemma 6.4.7(i) we know that that b2 ∩ {a, b, c, c∗ } ⊆ {a}. It remains to be shown that a ∈ b2 . From (ii) we know that b ∈ c2 , from Corollary 6.2.5(i) that c2 = ab. Thus, b ∈ ab; equivalently, a ∈ b2 . (vii) Since a , c, 1 < ac. Note also that, by Lemma 6.2.1(iii), t ∈ ac. From Lemma 6.4.5(ii) we know that {a, b, c} ⊆ ac. Furthermore, since we have seen in (i) that a ∈ c∗ c, we obtain from Lemma 6.4.2(i) that c∗ ∈ ac. (viii) Since a , c, 1 < ac∗ . Furthermore, by Lemma 6.2.1(iii), t < ac∗ . From (vii) we know that a ∈ ac; equivalently, a ∈ ac∗ . From (iv) we know that a ∈ bc; equivalently, b ∈ ac∗ . From (i) we know that a ∈ c∗ c. Thus, by Lemma 6.4.2(i), {c, c∗ } ⊆ ac∗ . (ix) From Corollary 6.2.5(i) we know that c2 = ab. Thus, the claim follows from (ii). (x) From Lemma 6.4.1(iii) we know that cc∗ ⊆ a2 . Thus, the desired equation follows from (iii).  Lemma 7.4.3 Let H be a hypergroup with six elements, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H \ hti, and assume that a∗ = a, b∗ = b,

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at = {a, c}, and bt = {c∗ }. Assume further that c ∈ c∗ c and a < c∗ c. Then the following hold. (i) We have c∗ c = {1, b, c, c∗ }. (ii) We have c2 = {c}. (iii) We have cc∗ = H \ {b}. (iv) We have bc = {b}. (v) We have bc∗ = {t, a, b, c∗ }. (vi) We have b2 = {1, b, c, c∗ }. (vii) We have ac = {t, a, b, c}. (viii) We have ac∗ = {a}. (ix) We have ab = {c}. (x) We have a2 = H \ {b}. Proof. (i) By Lemma 1.1.1, 1 ∈ c∗ c. Since we are assuming that bt = {c∗ }, we obtain from Corollary 6.2.3 that t < c∗ c. Since b , c, 1 < bc. From Lemma 6.2.1(iii) we also know that t < bc. Thus, as H = {1, t, a, b, c, c∗ }, we obtain that bc ∩ {a, b, c, c∗ } is not empty; cf. Lemma 1.1.3. Since we are assuming that a < c∗ c, this implies that b ∈ c∗ c, cf. Lemma 6.4.7(iv). Since we are assuming that c ∈ c∗ c, we obtain that c∗ ∈ c∗ c; cf. Lemma 1.2.1. (ii) Since c∗ , c, 1 < c2 . Furthermore, by Lemma 6.2.1(iii), t < c2 . From Lemma 6.4.6(ii) we know that c2 ∩ {a, b, c, c∗ } ⊆ {c}. Thus, since H = {1, t, a, b, c, c∗ }, c2 = {c}; cf. Lemma 1.1.3. (iii) From Lemma 1.1.1 we know that 1 ∈ cc∗ . Furthermore, since we are assuming that at = {a, c}, we have t ∈ cc∗ ; cf. Lemma 6.2.2. We are assuming that c ∈ c∗ c. Thus, by Lemma 6.4.5(i), {a, c, c∗ } ⊆ cc∗ . From Lemma 6.4.6(i) we know that c < bc. Thus, b < cc∗ . (iv) Since b , c, 1 < bc. Furthermore, by Lemma 6.2.1(iii), t < bc. From Lemma 6.4.6(i) we know that bc ∩ {a, b, c, c∗ } ⊆ {b}. Thus, as H = {1, t, a, b, c, c∗ }, bc = {b}; cf. Lemma 1.1.3. (v) Since b , c, we have 1 < bc∗ , and from Lemma 6.2.1(iii) we know that t ∈ bc∗ . From (ii) we know that c ∈ c2 , from Corollary 6.2.5(i) that c2 = ab. Thus, c ∈ ab. It follows that a ∈ bc∗ ; cf. Lemma 1.2.1. Furthermore, in (iv) we saw that b ∈ bc, so b ∈ bc∗ . From (ii) we know that b < c2 , which implies that c < bc∗ . From (i) we know that b ∈ c∗ c; equivalently, c∗ ∈ bc∗ . (vi) Since b∗ = b, 1 ∈ b2 . Furthermore, since it is assumed that bt = {c∗ }, we have t < b2 ; cf. Corollary 6.2.3. From (ii) we know that b < c2 , from Corollary 6.2.5(i) that c2 = ab. Thus, b < ab, whence a < b2 .

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Now assume, by way of contradiction, that b < b2 . Then c(b2 ) = c{1, c, c∗ } = {c} ∪ c2 ∪ cc∗ = H \ {b}. On the other hand, by (v), {b, c} ⊆ cb, whence b ∈ (cb)b. This contradiction shows that b ∈ b2 . From (iv) we know that b ∈ bc. Thus, c ∈ b2 . Since b2 is ∗ -invariant, this implies that c∗ ∈ b2 , too. (vii) Since a , c, 1 < ac. From Lemma 6.4.1(ii) we know that cb ⊆ ac. Thus, by (v) {t, a, b, c} ⊆ ac. From (ii) we know that a < c2 . Thus, by Lemma 1.2.1, c∗ < ac. (viii) Since a , c∗ , 1 < ac∗ . Furthermore, from Lemma 6.2.1(iii) we know that t < ac∗ . In Lemma 6.4.6(iii), we saw that ac∗ ∩ {a, b, c, c∗ } ⊆ {a}. Thus, as H = {1, t, a, b, c, c∗ }, ac∗ = {a}; cf. Lemma 1.1.3. (ix) From Corollary 6.2.5(i) we know that c2 = ab. Thus, the claim follows from (ii). (x) From Lemma 6.4.1(iii) we know that a2 = ac∗ ∪ cc∗ . Thus, the desired equation follows from (viii) and (iii).  Lemma 7.4.4 Let H be a hypergroup with six elements, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H \ hti, and assume that a∗ = a, b∗ = b, at = {a, c}, and bt = {c∗ }. Assume further that {a, b, c} ⊆ c∗ c and that b < cc∗ . Then the following hold. (i) We have c∗ c = H \ {t}. (ii) We have c2 = {a, c, c∗ }. (iii) We have cc∗ = H \ {b}. (iv) We have bc = {a, b}. (v) We have bc∗ = {t, a, b, c∗ }. (vi) We have b2 = {1, b, c, c∗ }. (vii) We have ac = H \ {1}. (viii) We have ac∗ = {a, b, c, c∗ }. (ix) We have ab = {a, c, c∗ }. (x) We have a2 = H. Proof. (i) By Lemma 1.1.1, 1 ∈ c∗ c. Since we are assuming that bt = {c∗ }, we obtain from Corollary 6.2.3 that t < c∗ c. We are assuming that {a, b, c} ⊆ c∗ c, and from c ∈ c∗ c we obtain that c∗ ∈ c∗ c, since c∗ c is ∗ -invariant. (ii) Since c∗ , c, 1 < c2 . Furthermore, by Lemma 6.2.1(iii), t < c2 .

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In (i), we saw that {a, b, c, c∗ } ⊆ c∗ c. Thus, by Lemma 6.4.8(i), {a, c} ⊆ c2 . Moreover, as we are assuming that b < cc∗ , we obtain from Lemma 6.4.2(ii) that b < c2 . From a ∈ c2 we obtain that c∗ ∈ ac; cf. Lemma 1.2.1. Since are assuming that b < cc∗ , we have c < bc∗ ; cf. Lemma 6.4.2(ii). From c < bc∗ we obtain that c∗ < cb; cf. Lemma 1.2.1. Now recall that, by Lemma 6.4.1(ii), ac = ab ∪ cb. Thus, c∗ ∈ ab. It follows that c∗ ∈ c2 ; cf. Corollary 6.2.5(i). (iii) From Lemma 1.1.1 we know that 1 ∈ cc∗ . Furthermore, since we are assuming that at = {a, c}, we have t ∈ cc∗ ; cf. Lemma 6.2.2. We are assuming that c ∈ c∗ c. Thus, by Lemma 6.4.5(i), {a, c, c∗ } ⊆ cc∗ . (iv) Since b , c, 1 < bc. Furthermore, by Lemma 6.2.1(iii), t < bc. From (ii) we know that c∗ ∈ c2 , from Corollary 6.2.5(i) we know that c2 = ab. Thus, c∗ ∈ ab, so that, by Lemma 1.2.1, a ∈ bc. We are assuming that c ∈ c∗ c. Thus, by Lemma 6.4.2(iii), b ∈ ac. On the other hand, by (ii), b < c2 , and, by Corollary 6.2.5(i), c2 = ab. Thus, b < ab. Now recall that, by Lemma 6.4.1(ii), ac = ab ∪ cb. Thus, b ∈ cb, so that, by Lemma 1.2.1, b ∈ bc. By hypothesis, b < cc∗ . Thus, by Lemma 6.4.2(ii), c < bc and c∗ < bc. (v) Since b , c, we have 1 < bc∗ , and from Lemma 6.2.1(iii) we know that t ∈ bc∗ . From Lemma 6.4.8(ii) we know that {a, c∗ } ⊆ bc∗ . By (i), c∗ ∈ c∗ c, and, by Lemma 6.4.1(i), c∗ c = b2 ∪ bc. By (ii), b < c2 , whence c∗ < bc. Thus, c∗ ∈ b2 ; equivalently, b ∈ bc∗ . Note finally that, since b < c2 , c < bc∗ . (vi) Since b∗ = b, 1 ∈ b2 . Furthermore, since it is assumed that bt = {c∗ }, we have t < b2 ; cf. Corollary 6.2.3. From (ii) we know that b < c2 . Thus, by Lemma 6.4.2(ii), a < b2 . Now assume, by way of contradiction, that b < b2 . Then c(b2 ) = c{1, c, c∗ } = {c} ∪ c2 ∪ cc∗ = H \ {b}. On the other hand, by (v), {b, c} ⊆ cb, whence b ∈ (cb)b. This contradiction shows that b ∈ b2 . From (iv) we know that b ∈ bc. Thus, c ∈ b2 , and, since b2 is ∗ -invariant, c ∈ b2 implies that c∗ ∈ b2 . (vii) Since a , c, 1 < ac, and, by Lemma 6.2.1(iii), t ∈ ac. Now recall from (i) that {a, b, c, c∗ } ⊆ c∗ c. Thus, by Lemma 6.4.8(iii), {a, b, c, c∗ } ⊆ ac. (viii) Since a , c∗ , 1 < ac∗ . Furthermore, from Lemma 6.2.1(iii) we know that t < ac∗ . In (i), we have seen that {a, b, c, c∗ } ⊆ c∗ c. Thus, by Lemma 6.4.8(iv), {a, c, c∗ } ⊆ ac∗ . From (iv) we know that a ∈ bc. Thus, b ∈ ac∗ . (ix) From Corollary 6.2.5(i) we know that c2 = ab. Thus, the claim follows from (ii). (x) From Lemma 6.4.1(iii) we know that a2 = ac∗ ∪ cc∗ . Thus, the desired equation follows from (viii) and (iii). 

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Lemma 7.4.5 Let H be a hypergroup with six elements, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H \ hti, and assume that a∗ = a, b∗ = b, at = {a, c}, and bt = {c∗ }. Assume further that {a, b, c} ⊆ c∗ c, that b ∈ cc∗ , and that c∗ < ab. Then the following hold. (i) We have c∗ c = H \ {t}. (ii) We have c2 = {a, b, c}. (iii) We have cc∗ = H. (iv) We have bc = {b, c, c∗ }. (v) We have bc∗ = H \ {1}. (vi) We have {1, a, c, c∗ } ⊆ b2 ⊆ H \ {t}. (vii) We have ac = H \ {1}. (viii) We have ac∗ = {a, c, c∗ }. (ix) We have ab = {a, b, c}. (x) We have a2 = H. Proof. (i) By Lemma 1.1.1, 1 ∈ c∗ c. Since we are assuming that bt = {c∗ }, we obtain from Corollary 6.2.3 that t < c∗ c. We are assuming that {a, b, c} ⊆ c∗ c, and from c ∈ c∗ c we obtain that c∗ ∈ c∗ c, since c∗ c is ∗ -invariant. (ii) Since c∗ , c, 1 < c2 . Furthermore, by Lemma 6.2.1(iii), t < c2 . In (i), we saw that {a, b, c, c∗ } ⊆ c∗ c. Thus, by Lemma 6.4.8(i), {a, c} ⊆ c2 . Since we are assuming that b ∈ cc∗ , we obtain from Lemma 6.4.2(ii) that b ∈ c2 . Since we are assuming that c∗ < ab, we obtain from Corollary 6.2.5(i) that c∗ < c2 . (iii) From Lemma 1.1.1 we know that 1 ∈ cc∗ . Furthermore, since we are assuming that at = {a, c}, we have t ∈ cc∗ ; cf. Lemma 6.2.2. Since b ∈ c∗ c, c ∈ cb. On the other hand, by Lemma 6.4.1(ii), cb ⊆ ac, so that c ∈ ac; equivalently, a ∈ cc∗ . From (ii) we know that c ∈ c2 . Thus, c ∈ cc∗ , and, since cc∗ is ∗ -invariant, c ∈ cc∗ implies that c∗ ∈ cc∗ . (iv) Since b , c, 1 < bc. Furthermore, by Lemma 6.2.1(iii), t < bc. By hypothesis, c∗ < ab. Thus, by Lemma 1.2.1, a < bc. By hypothesis, b ∈ cc∗ . Thus, by Lemma 6.4.2(ii), {c, c∗ } ⊆ bc. Assume, by way of contradiction, that b < bc. Then c < b2 , and, since b2 is ∗ invariant, c < b2 implies that c∗ < b2 . Thus, as H = {1, t, a, b, c, c∗ }, we have b2 ⊆ {1, a, b}. (Notice that t < b2 , cf. Corollary 6.2.3.) It follows that b3 ⊆ b2 ·b ⊆ {1, a, b}b = {b} ∪ ab ∪ b2 ⊆ {1, a, b, c}.

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(Recall from Corollary 6.2.5(i) that c2 = ab, so that, by (ii), ab = {a, b, c}.) Since b3 is ∗ -invariant, this implies that b3 ⊆ {1, a, b}. On the other hand, by (ii), c ∈ c2 = ab, and, by Lemma 6.4.2(ii), a ∈ b2 . Thus, c ∈ b3 , contradiction. (v) Since b , c, we have 1 < bc∗ , and from Lemma 6.2.1(iii) we know that t ∈ bc∗ . From Lemma 6.4.8(ii) we know that {a, c∗ } ⊆ bc∗ . From (iv) we know that b ∈ bc, whence b ∈ bc∗ . From (ii) we know that b ∈ c2 , whence c ∈ bc∗ . (vi) Since b∗ = b, 1 ∈ b2 . Furthermore, since it is assumed that bt = {c∗ }, we have t < b2 ; cf. Corollary 6.2.3. From (ii) we know that b ∈ c2 . Thus, by Lemma 6.4.2(ii), a ∈ b2 . From (iv) we know that b ∈ bc. Thus, c ∈ b2 . Since b2 is ∗ -invariant, we obtain from c ∈ b2 that c∗ ∈ b2 . (vii) Since a , c, 1 < ac, and, by Lemma 6.2.1(iii), t ∈ ac. Now recall from (i) that {a, b, c, c∗ } ⊆ c∗ c. Thus, by Lemma 6.4.8(iii), {a, b, c, c∗ } ⊆ ac. (viii) Since a , c∗ , 1 < ac∗ . Furthermore, from Lemma 6.2.1(iii) we know that t < ac∗ . In (i), we have seen that {a, b, c, c∗ } ⊆ c∗ c. Thus, by Lemma 6.4.8(iv), {a, c, c∗ } ⊆ ac∗ . From (iv) we know that a < bc. Thus, b < ac∗ . (ix) From Corollary 6.2.5(i) we know that c2 = ab. Thus, the claim follows from (ii). (x) From Lemma 6.4.1(iii) we know that cc∗ ⊆ a2 . Thus, the desired equation follows  from (iii). Lemma 7.4.6 Let H be a hypergroup with six elements, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H \ hti, and assume that a∗ = a, b∗ = b, at = {a, c}, and bt = {c∗ }. Assume further that {a, b, c} ⊆ c∗ c, b ∈ cc∗ , c∗ ∈ ab, and b < bc. Then the following hold. (i) We have c∗ c = H \ {t}. (ii) We have c2 = {a, b, c, c∗ }. (iii) We have cc∗ = H. (iv) We have bc = {a, c, c∗ }. (v) We have bc∗ = {t, a, c, c∗ }. (vi) We have b2 = {1, a, b}. (vii) We have ac = H \ {1}. (viii) We have ac∗ = {a, b, c, c∗ }. (ix) We have ab = {a, b, c, c∗ }. (x) We have a2 = H.

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Proof. (i) By Lemma 1.1.1, 1 ∈ c∗ c. Since we are assuming that bt = {c∗ }, we obtain from Corollary 6.2.3 that t < c∗ c. We are assuming that {a, b, c} ⊆ c∗ c, and from c ∈ c∗ c we obtain that c∗ ∈ c∗ c, since c∗ c is ∗ -invariant. (ii) Since c∗ , c, 1 < c2 . Furthermore, by Lemma 6.2.1(iii), t < c2 . In (i), we saw that {a, b, c, c∗ } ⊆ c∗ c. Thus, by Lemma 6.4.8(i), {a, c} ⊆ c2 . Since we are assuming that b ∈ cc∗ , we obtain from Lemma 6.4.2(ii) that b ∈ c2 . Since we are assuming that c∗ ∈ ab, we obtain from Corollary 6.2.5(i) that c∗ ∈ c2 . (iii) From Lemma 1.1.1 we know that 1 ∈ cc∗ . Furthermore, since we are assuming that at = {a, c}, we have t ∈ cc∗ ; cf. Lemma 6.2.2. Since b ∈ c∗ c, c ∈ cb. On the other hand, by Lemma 6.4.1(ii), cb ⊆ ac, so that c ∈ ac; equivalently, a ∈ cc∗ . From (ii) we know that c ∈ c2 . Thus, c ∈ cc∗ , and, since cc∗ is ∗ -invariant, c ∈ cc∗ implies that c∗ ∈ cc∗ . (iv) Since b , c, 1 < bc. Furthermore, by Lemma 6.2.1(iii), t < bc. By hypothesis, c∗ ∈ ab. Thus, by Lemma 1.2.1, a ∈ bc. By hypothesis, b ∈ cc∗ . Thus, by Lemma 6.4.2(ii), {c, c∗ } ⊆ bc. (v) Since b , c, we have 1 < bc∗ , and from Lemma 6.2.1(iii) we know that t ∈ bc∗ . From Lemma 6.4.8(ii) we know that {a, c∗ } ⊆ bc∗ . Since we are assuming that b < bc, we have b < bc∗ . In (ii) we saw that b ∈ c2 . Thus, c ∈ bc∗ . (vi) Since b∗ = b, 1 ∈ b2 . Furthermore, since it is assumed that bt = {c∗ }, we have t < b2 ; cf. Corollary 6.2.3. From (ii) we know that b ∈ c2 . Thus, by Lemma 6.4.2(ii), a ∈ b2 . We are assuming that b ∈ c∗ c. In Lemma 6.4.1(i), we saw that c∗ c = b2 ∪ bc, and, in (iv), we saw that b < bc. Thus, we have b ∈ b2 . We are assuming that that b < bc. Thus, c < b2 , and, since b2 is ∗ -invariant, this implies that c∗ < b2 . (vii) Since a , c, 1 < ac, and, by Lemma 6.2.1(iii), t ∈ ac. Now recall from (i) that {a, b, c, c∗ } ⊆ c∗ c. Thus, by Lemma 6.4.8(iii), {a, b, c, c∗ } ⊆ ac. (viii) Since a , c∗ , 1 < ac∗ . Furthermore, from Lemma 6.2.1(iii) we know that t < ac∗ . In (i), we have seen that {a, b, c, c∗ } ⊆ c∗ c. Thus, by Lemma 6.4.8(iv), {a, c, c∗ } ⊆ ac∗ . From (iv) we know that a ∈ bc. Thus, b ∈ ac∗ . (ix) From Corollary 6.2.5(i) we know that c2 = ab. Thus, the claim follows from (ii). (x) From Lemma 6.4.1(iii) we know that cc∗ ⊆ a2 . Thus, the desired equation follows  from (iii). Lemma 7.4.7 Let H be a hypergroup with six elements, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H \ hti, and assume that a∗ = a, b∗ = b,

232

7 Hypergroups with a Small Number of Elements

at = {a, c}, and bt = {c∗ }. Assume further that {a, b, c} ⊆ c∗ c, b ∈ cc∗ , c∗ ∈ ab, and b ∈ bc. Then the following hold. (i) We have c∗ c = H \ {t}. (ii) We have c2 = {a, b, c, c∗ }. (iii) We have cc∗ = H. (iv) We have bc = {a, b, c, c∗ }. (v) We have bc∗ = H \ {1}. (vi) We have {1, a, c, c∗ } ⊆ b2 ⊆ H \ {t}. (vii) We have ac = H \ {1}. (viii) We have ac∗ = {a, b, c, c∗ }. (ix) We have ab = {a, b, c, c∗ }. (x) We have a2 = H. Proof. (i) By Lemma 1.1.1, 1 ∈ c∗ c. Since we are assuming that bt = {c∗ }, we obtain from Corollary 6.2.3 that t < c∗ c. We are assuming that {a, b, c} ⊆ c∗ c, and from c ∈ c∗ c we obtain that c∗ ∈ c∗ c, since c∗ c is ∗ -invariant. (ii) Since c∗ , c, 1 < c2 . Furthermore, by Lemma 6.2.1(iii), t < c2 . In (i), we saw that {a, b, c, c∗ } ⊆ c∗ c. Thus, by Lemma 6.4.8(i), {a, c} ⊆ c2 . Since we are assuming that b ∈ cc∗ , we obtain from Lemma 6.4.2(ii) that b ∈ c2 . Since we are assuming that c∗ ∈ ab, we obtain from Corollary 6.2.5(i) that c∗ ∈ c2 . (iii) From Lemma 1.1.1 we know that 1 ∈ cc∗ . Furthermore, since we are assuming that at = {a, c}, we have t ∈ cc∗ ; cf. Lemma 6.2.2. Since b ∈ c∗ c, c ∈ cb. On the other hand, by Lemma 6.4.1(ii), cb ⊆ ac, so that c ∈ ac; equivalently, a ∈ cc∗ . From (ii) we know that c ∈ c2 . Thus, c ∈ cc∗ , and, since cc∗ is ∗ -invariant, c ∈ cc∗ implies that c∗ ∈ cc∗ . (iv) Since b , c, 1 < bc. Furthermore, by Lemma 6.2.1(iii), t < bc. By hypothesis, c∗ ∈ ab. Thus, by Lemma 1.2.1, a ∈ bc. By hypothesis, b ∈ cc∗ . Thus, by Lemma 6.4.2(ii), {c, c∗ } ⊆ bc. (v) Since b , c, we have 1 < bc∗ , and from Lemma 6.2.1(iii) we know that t ∈ bc∗ . From Lemma 6.4.8(ii) we know that {a, c∗ } ⊆ bc∗ . Since we are assuming that b ∈ bc, we have b ∈ bc∗ . In (ii) we saw that b ∈ c2 . Thus, c ∈ bc∗ . (vi) Since b∗ = b, 1 ∈ b2 . Furthermore, since it is assumed that bt = {c∗ }, we have t < b2 ; cf. Corollary 6.2.3. From (ii) we know that b ∈ c2 . Thus, by Lemma 6.4.2(ii), a ∈ b2 . From (iv) we know that b ∈ bc. Thus, c ∈ b2 . Since b2 is ∗ -invariant, we obtain from c ∈ b2 that c∗ ∈ b2 .

Some Hypergroups of Cardinality 6, II

7.4

233

(vii) Since a , c, 1 < ac, and, by Lemma 6.2.1(iii), t ∈ ac. Thus, the claim holds, since we have seen in Lemma 6.4.8(iii) that {a, b, c, c∗ } ⊆ ac. (viii) Since a , c∗ , 1 < ac∗ . Furthermore, from Lemma 6.2.1(iii) we know that t < ac∗ . In (i), we have seen that {a, b, c, c∗ } ⊆ c∗ c. Thus, by Lemma 6.4.8(iv), {a, c, c∗ } ⊆ ac∗ . From (iv) we know that a ∈ bc. Thus, b ∈ ac∗ . (ix) From Corollary 6.2.5(i) we know that c2 = ab. Thus, the claim follows from (ii). (x) From Lemma 6.4.1(iii) we know that cc∗ ⊆ a2 . Thus, the desired equation follows from (iii).  Let j be an element in {11, . . . , 19}. We say that a hypergroup is of type H6, j if it contains elements t, a, b, and c such that its hypermultiplication is described by table H6, j . Theorem 7.4.8 Let H be a hypergroup with six elements, and let F be a non-thin non-normal closed subset of H which has index 3 in H. Assume that H \ F contains an element h such that, for each element f in F, |h f | = 1. Then {11, . . . , 19} contains an element j such that H is of type H6, j . Proof. Since |H| = 6 and F is not normal in H, we have |F | = 2; cf. Corollary 3.2.7. It follows that |H \ F | = 4. Thus, as |H/F | = 3, H \ F contains pairwise distinct elements a, b, and c such that a∗ = a,

b∗ = b,

aF = {a, c}, and bF = {b, c∗ }; cf. Lemma 3.2.5. Since |F | = 2, F \ {1} contains an element t with F = {1, t}. Since F is a closed subset of H, this implies that F = hti. Thus, as aF = {a, c} and bF = {b, c∗ }, we obtain that ahti = {a, c} and bhti = {b, c∗ }. From F \ {1} = {t}, together with the fact that F is a closed subset of H, we also obtain that t is an involution. Thus, by Lemma 6.1.2(i), c ∈ at ⊆ {a, c}

and

c∗ ∈ bt ⊆ {b, c∗ }.

⇔

c∗ t = {b, c∗ },

In Lemma 6.2.2 we also saw that at = {a, c}

and, in Corollary 6.2.3, we, similarly, saw that bt = {c∗ }

⇔

ct = {a}.

234

7 Hypergroups with a Small Number of Elements

Now recall that we are assuming that H \ hti contains an element h with |ht| = 1. Thus, as H \ hti = {a, b, c, c∗ }, we cannot have at = {a, c} and bt = {b, c∗ } at the same time. On the other hand, since t is not thin, we have at = {a, c}

or

bt = {b, c∗ };

cf. Lemma 6.2.6. Thus, we must have either at = {c} and bt = {b, c∗ } or at = {a, c} and bt = {c∗ }. Without loss of generalization, we assume that at = {a, c} Then

c∗ t = {b, c∗ }

and

and

bt = {c∗ }. ct = {a}.

To finish the proof we now apply the preceding seven lemmas. Assume first that c < c∗ c. Then, by Lemma 7.4.1, H is of type H6,11 . If c ∈ c∗ c and b < c∗ c, we obtain from Lemma 7.4.2 that H is of type H6,12 . If c ∈ c∗ c and a < c∗ c, we obtain from Lemma 7.4.3 that H is of type H6,13 . If {a, b, c} ⊆ c∗ c and b < cc∗ , we obtain from Lemma 7.4.4 that H is of type H6,14 . If {a, b, c} ⊆ c∗ c, b ∈ cc∗ , and c∗ < ab, we obtain from Lemma 7.4.5 that H is of type H6,15 or of type H6,16 . If {a, b, c} ⊆ c∗ c, b ∈ cc∗ , c∗ ∈ ab, and b < bc, we obtain from Lemma 7.4.6 that H is of type H6,17 . If {a, b, c} ⊆ c∗ c, b ∈ cc∗ , c∗ ∈ ab, and b ∈ bc, we obtain from Lemma 7.4.7 that H  is of type H6,18 or of type H6,19 . In a computer search, C. French has shown that, for each element j in {11, . . . , 19}, there exist hypergroups of type H6, j .

7.5 Non-Normal Closed Subsets Missing Four Elements In Theorem 3.2.4, we saw that a non-normal closed subset F of a hypergroup H with |H \ F | = 4 has index either 3 or 4 in H. In this section, we consider the second of these two cases. We will apply our results to hypergroups H of cardinality 6 having a non-normal closed subset of index 4 in H. Besides Theorems 7.3.8 and 7.4.8, this will be the third of our classification theorems of hypergroups with six elements having a non-normal closed subset; cf. Theorem 7.5.4. Let H be a hypergroup, let F be a closed subset of H, and assume that F is not normal in H. Assume that |H \ F | = 4 and that |H/F | = 4. Then, by Lemma 3.2.6, H \ F contains two distinct elements a and b such that a∗ F = {a∗ },

aF = {a, b},

and

b∗ F = {b∗ }.

7.5

Non-Normal Closed Subsets Missing Four Elements

235

We first compute the products of the elements a, a∗ , b, and b∗ . Lemma 7.5.1 Let H be a hypergroup, let F be a closed subset of H, and assume that |H \ F | = 4. Let a and b be two distinct elements of H, and assume that a∗ F = {a∗ }, aF = {a, b}, and b∗ F = {b∗ }. Then the following hold. (i) We have a2 = ba and b2 = ab. (ii) We have aa∗ = H and bb∗ = H. Proof. (i) Applying Lemma 1.3.6(i) to F, {a}, and {b} in place of A, B, and C we obtain that a2 = ba. Applying the same lemma to F, {b}, and {a} in place of A, B, and C we obtain that b2 = ab. (ii) We are assuming that a∗ F = {a∗ }, and from this we obtain that F ⊆ aa∗ . Thus, in order to prove that aa∗ = H, it suffices to show that H \ F ⊆ aa∗ . Recall from (i) that a2 = ba. Thus, a∗ b∗ = (a∗ )2 . Thus, applying Lemma 1.2.3(i) to a∗ , b∗ , and a∗ in place of a, b, and c we obtain that b∗ ∈ aa∗ if and only if a∗ ∈ aa∗ . Since aa∗ is ∗ -invariant, we also have a ∈ aa∗ if and only if a∗ ∈ aa∗ and, similarly, b ∈ aa∗ if and only if b∗ ∈ aa∗ . Thus, as H \ F = {a, a∗, b, b∗ }, we have aa∗ ⊆ F or H \ F ⊆ aa∗ . Thus, it suffices to show that aa∗ * F. Assume, by way of contradiction, that aa∗ ⊆ F. Then b∗ aa∗ ⊆ b∗ F = {b∗ }, so that b∗ aa∗ = {b∗ }. Assume first that a ∈ ba∗ . Then a∗ ∈ b∗ a. Thus, as b∗ aa∗ = {b∗ }, (a∗ )2 = {b∗ }. It follows that a2 = {b}, and then that ab = a3 = ba. On the other hand, by (i), a2 = ba. Thus, ab = {b}. Now, as aF = {a, b}, we obtain that a ∈ bF = a2 F = a2 ∪ ab = {b}. This contradiction shows that a < ba∗ . Applying Lemma 1.3.6(ii) to F, {a}, and {b} in place of A, B, and C we obtain from a < ba∗ that b < ba∗ . Assume next that b∗ ∈ ba∗ . Then b ∈ b∗ a. Thus, as b∗ aa∗ = {b∗ }, ba∗ = {b∗ }. It follows that ab∗ = {b}. Now, as aF = {a, b}, we obtain that {b} = ab∗ = ab∗ F = bF = {a, b}. This contradiction shows that b∗ < ba∗ . From b∗ < ba∗ we obtain that b < ab∗ . Thus, applying Lemma 1.3.6(ii) to F, {b}, and {a} in place of A, B, and C we obtain that a < ab∗ ; equivalently, a∗ < ba∗ . We have seen that none of the elements a, a∗ , b, or b∗ belongs to ba∗ . Since H \ F = {a, a∗, b, b∗ }, this implies that ba∗ ⊆ F. It follows that a∗ ∈ b∗ ba∗ F ⊆ b∗ F = {b∗ },

236

7 Hypergroups with a Small Number of Elements

contradiction. The equation bb∗ = H follows similarly.



Lemma 7.5.2 Let H be a hypergroup, let F be a closed subset of H, and assume that |H \ F | = 4. Let a and b be two distinct elements of H, and assume that a∗ F = {a∗ }, aF = {a, b}, and b∗ F = {b∗ }. Then the following hold. (i) We have ba∗ = H \ F. (ii) We have a2 = {a, b} or a2 = H \ F. (iii) We have ab = ba. Proof. (i) We are assuming that a∗ F = {a∗ } and that b∗ F = {b∗ }. Thus, as a , b, a∗ F , b∗ F. It follows that ba∗ ∩ F is empty; cf. Lemma 2.1.7(i). Thus, ba∗ ⊆ H \ F, and it remains to show that H \ F ⊆ ba∗ . Assume, by way of contradiction, that a < ba∗ . Then, applying Lemma 1.3.6(ii) to F, {a}, and {b} in place of A, B, and C we obtain that b < ba∗ . Thus, as ba∗ ⊆ H \ F and H \ F = {a, a∗, b, b∗ }, ba∗ ⊆ {a∗, b∗ }. On the other hand, applying Lemma 1.3.6(ii) to F, {b}, and {a} in place of A, B, and C we have b ∈ ab∗ if and only if a ∈ ab∗ . Thus, b∗ ∈ ba∗ if and only if a∗ ∈ ba∗ . It follows that ba∗ = {a∗, b∗ }, and then ab∗ = {a, b}. In Lemma 7.5.1(ii), we saw that bb∗ = H. Thus, a ∈ bb∗ . It follows that b∗ ∈ b∗ a. From Lemma 7.5.1(i) we also know that a2 = ba. Thus, as ab∗ = {a, b}, b ∈ ab∗ ⊆ ab∗ a = a2 ∪ ba = a2 . It follows that a ∈ ba∗ , and this contradiction shows that a ∈ ba∗ . Applying Lemma 1.3.6(ii) to F, {a}, and {b} in place of A, B, and C we obtain from a ∈ ba∗ that b ∈ ba∗ . In the same way as we have shown that {a, b} ⊆ ba∗ one proves that {a, b} ⊆ ab∗ , and from this latter containment one obtains that {a∗, b∗ } ⊆ ba∗ . Now, as H \ F = {a, a∗, b, b∗ }, we have H \ F ⊆ ba∗ . (ii) By hypothesis, a < a∗ F. Thus, by Lemma 1.3.3, F∩a2 is empty. Thus, a2 ⊆ H\F. In Lemma 7.5.1(ii), we saw that aa∗ = H. Thus, a ∈ aa∗ ; equivalently, a ∈ a2 . From (i) we know that a ∈ ba∗ . Thus, b ∈ a2 . It follows that {a, b} ⊆ a2 . From Lemma 7.5.1(i) we know that a2 = ba. Thus, a∗ ∈ a2 if and only if a∗ ∈ ba. Furthermore, by Lemma 1.2.1, we have a∗ ∈ ba if and only if b∗ ∈ a2 , so that we have a∗ ∈ a2 if and only if b∗ ∈ a2 . Now the claim follows from H \ F = {a, a∗, b, b∗ }. (iii) Considering Lemma 7.5.1(i) we obtain from (ii) that ba = {a, b} or ba = H \ F. Similarly, ab = {a, b} or ab = H \ F. On the other hand, by Lemma 1.2.1, a∗ ∈ ab if and only if a∗ ∈ ba. Thus, ab = ba. 

7.5

Non-Normal Closed Subsets Missing Four Elements

237

Lemma 7.5.3 Let H be a hypergroup, let F be a closed subset of H, and assume that |H \ F | = 4. Let a and b be two distinct elements of H, and assume that a∗ F = {a∗ }, aF = {a, b}, and b∗ F = {b∗ }. Then the following hold. (i) Set A := { f ∈ F | a f = {b}}. Then a∗ a = H \ A. (ii) Set B := { f ∈ F | b f = {b}}. Then b∗ a = H \ B. Proof. (i) From Lemma 7.5.2(ii) we know that a ∈ a2 . Thus, a ∈ a∗ a. Since a∗ a is this implies that {a, a∗ } ⊆ a∗ a.

∗ -invariant,

From Lemma 7.5.1(i) we know b2 = ab. Thus, (b∗ )2 = b∗ a∗ . From Lemma 7.5.2(i) we know that b∗ ∈ ba∗ . Thus, by Lemma 1.2.1, b∗ ∈ a∗ b. Now applying Lemma 1.2.3(ii) to b∗ , a∗ , and b∗ in place of a, b, and c we obtain that b∗ ∈ a∗ a. Since a∗ a is ∗ -invariant, this implies that {b, b∗ } ⊆ a∗ a. Since H \ F = {a, a∗, b, b∗ }, we have shown that H \ F ⊆ a∗ a. Let f be an element in F. We have f ∈ a∗ a if and only if a ∈ a f . Since aF = {a, b}, we have a ∈ a f if and only if f ∈ F \ A. Thus, F ∩ a∗ a = F \ A. From H \ F ⊆ a∗ a and F ∩ a∗ a = F \ A we obtain that a∗ a = H \ A. (ii) In Lemma 7.5.1(ii), we saw that aa∗ = H. Thus, a ∈ aa∗ ; equivalently, a ∈ a2 . From Lemma 7.5.1(i) we know that a2 = ba. Thus, a ∈ ba. It follows that a ∈ b∗ a. From Lemma 7.5.2(i) we also know that a ∈ ba∗ , so a∗ ∈ b∗ a. Thus, we have {a, a∗ } ⊆ b∗ a. Similarly, one obtains {b, b∗ } ⊆ a∗ b which is equivalent to {b, b∗ } ⊆ b∗ a. Now recall that H \ F = {a, a∗, b, b∗ }. Thus, H \ F ⊆ b∗ a. Let f be an element in F. We have f ∈ b∗ a if and only if a ∈ b f . Since aF = {a, b}, we have a ∈ b f if and only if f ∈ F \ B. Thus, F ∩ b∗ a = F \ B. From H \ F ⊆ b∗ a and F ∩ b∗ a = F \ B we obtain that b∗ a = H \ B.



We now apply our results to hypergroups H of cardinality 6 containing a non-normal closed subset of index 4 in H. We will see that the hypermultiplication of these hypergroups is described by one of the following six tables.

H6,20 1 t a∗ b∗ a b

1 {1} {t} {a∗ } {b∗ } {a} {b}

t {1} {a∗ } {b∗ } {b} {a}

a

b

a∗

b∗

H \ {t} H \ {1} H \ {t} {a, b} {a, b} H {a, b} {a, b} H \ hti H

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7 Hypergroups with a Small Number of Elements

H6,21 1 t a∗ b∗ a b

1 {1} {t} {a∗ } {b∗ } {a} {b}

H6,22 1 t a∗ b∗ a b

1 {1} {t} {a∗ } {b∗ } {a} {b}

H6,23 1 t a∗ b∗ a b

1 {1} {t} {a∗ } {b∗ } {a} {b}

H6,24 1 t a∗ b∗ a b

1 {1} {t} {a∗ } {b∗ } {a} {b}

H6,25 1 t a∗ b∗ a b

1 {1} {t} {a∗ } {b∗ } {a} {b}

t

a

{1} {a∗ } {b∗ } {b} {a}

{1, t} {a∗ } {b∗ } {b} {a, b}

{1, t} {a∗ } {b∗ } {b} {a, b} t

a∗

b∗

H \ {t} H \ {1} H \ {t} H \ hti H \ hti H H \ hti H \ hti H \ hti H

t

t

b

a

b

a∗

b∗

H \ {t} H \ {1} H {a, b} {a, b} H {a, b} {a, b} H \ hti H a

b

a∗

b∗

H \ {t} H \ {1} H H \ hti H \ hti H H \ hti H \ hti H \ hti H a

b

a∗

b∗

{1, t} {a∗ } H {b∗ } H \ {1} H {a, b} {a, b} {a, b} H {a, b} {a, b} {a, b} H \ hti H t

a

b

a∗

b∗

{1, t} {a∗ } H {b∗ } H \ {1} H {a, b} H \ hti H \ hti H {a, b} H \ hti H \ hti H \ hti H

7.5

Non-Normal Closed Subsets Missing Four Elements

239

Let j be an element in {20, . . . , 25}. We say that a hypergroup is of type H6, j if it contains elements t, a, and b such that its hypermultiplication is described by table H6, j . Theorem 7.5.4 Let H be a hypergroup with six elements. Assume that H contains a non-normal closed subset of index 4 in H. Then {20, . . . , 25} contains an element j such that H is of type H6, j . Proof. We are assuming that H contains a non-normal closed subset of index 4 in H. We choose one of these closed subsets of H and denote it by F. Since |H| = 6 and F is not normal in H, we have |F | = 2; cf. Corollary 3.2.7. It follows that |H \ F | = 4. Since |H/F | = 4, this allows us to apply Lemma 3.2.6, so that we find two distinct elements a and b in H \ F satisfying a∗ F = {a∗ },

aF = {a, b},

and

b∗ F = {b∗ }.

We are now in the position to take advantage of Lemmas 7.5.1, 7.5.2, and 7.5.3. From Lemma 7.5.1(i) (together with Lemma 7.5.2(iii)) we obtain that a2 = ba = ab = b2, and in Lemma 7.5.2(ii), we saw that or

a2 = H \ F.

and

bb∗ = H,

a2 = {a, b} In Lemma 7.5.1(ii), we saw that aa∗ = H

and from Lemma 7.5.2(i), we know that ba∗ = H \ F. To determine b∗ a we recall that |F | = 2. Thus, F \ {1} contains an element t with F = {1, t}. Since F is a closed subset of H, this implies that F = hti. Thus, as aF = {a, b}, we have ahti = {a, b}. From F \ {1} = {t}, together with the fact that F is a closed subset of H, we obtain that t is an involution. Thus, as ahti = {a, b}, we obtain from Lemma 6.1.2(i) that bt , {b}. Thus, by Lemma 7.5.3(ii), b∗ a = H \ {1}. Next we determine a∗ a. From Lemma 6.1.2(ii) we know that at = {b} if and only if t < a∗ a. Thus, by Lemma 7.5.3(i), we have

240

7 Hypergroups with a Small Number of Elements

a∗ a = H \ {t} and

Similarly, we have and

a∗ a = H

if

b∗ b = H \ {t} b∗ b = H

if

if

at = {b}

at = {a, b}.

if

bt = {a}

bt = {a, b}.

If t is thin, Lemma 1.4.3(i) (together with Lemma 6.1.2(i)) forces at = {b} and bt = {a}. Thus, in this case, H is of type H6,20 or H6,21 . If t is not thin, Lemma 6.1.3(ii) leaves us with three cases. Either we have at = {b} and bt = {a, b}, or we have at = {a, b} and bt = {a}, or we have at = {a, b} and bt = {a, b}. In the first case, H is of type H6,22 or H6,23 . In the second case, we obtain the same isomorphism classes as in the first case, and in the last case, H is of type H6,24 or of type H6,25 .  As before, in a computer search, C. French has shown that, for each element j in {20, . . . , 25}, there exist hypergroups of type H6, j .

7.6 A Characterization of H6,1 In this section, we look, for a last time, at hypergroups H having a non-normal closed subset F with |H \ F | = 4. We will see that the structure of H is severely restricted if we assume that Oϑ (H) * F. Lemma 7.6.1 Let H be a hypergroup with six elements, and let t be an involution of H. Let a, b, and c be pairwise distinct elements of H \ hti, and assume that a∗ = a, b∗ = b, ahti = {a, c}, and bhti = {b, c∗ }. Assume that one of the elements a and b is thin. Then H is thin. Proof. Without loss of generalization, we assume that a is thin. Since c ∈ ahti and a , c, c ∈ at. Since a is assumed to be thin, a∗ is thin, too; cf. Lemma 2.6.2. Thus, by Lemma 1.4.3(ii), |at| = 1. It follows that at = {c}. From this we obtain that

c∗ t = {b};

cf. Lemma 6.2.2. From at = {c} we also obtain that ac∗ is ∗ -invariant; cf. Lemma 6.2.4(ii). On the other hand, as a∗ is thin, |ac∗ | = 1; cf. Lemma 1.4.3(ii). Thus, we have ac∗ = {t}, ac∗ = {a}, or ac∗ = {b}. (Recall from Lemma 6.2.1(i) that c∗ , c.)

7.6

A Characterization of H6,1

241

Since c∗ t = {b} and a , b, a < c∗ t. Thus, by Lemma 1.2.1, t < ac∗ . From Lemma 6.2.1(i) we also obtain that c∗ , 1. Thus, as a is thin, ac∗ , {a}. Now we are left with the third case which means that ac∗ = {b}. Since a is thin, this implies that ab = {c∗ }. From ac∗ = {b}, c∗ t = {b}, and ab = {c∗ } we obtain that bt = {c∗ }. From at = {c} and bt = {c∗ } we obtain that t is thin; cf. Lemma 6.2.6. Since a and t both are thin, we obtain from at = {c} that c is thin; cf. Lemma 1.4.3(iii). Now, by Lemma 2.6.2, c∗ is thin. Thus, as c∗ t = {b}, b is thin; cf. Lemma  1.4.3(iii). Theorem 7.6.2 Let H be a hypergroup, and let F be a non-normal closed subset of H. Assume that |H \ F | = 4 and that Oϑ (H) * F. Then H is of type H6,1 . Proof. We are assuming that F is a non-normal closed subset of H and that |H\F | = 4. Thus, by Theorem 3.2.4, |H/F | ∈ {3, 4}. Assume first that |H/F | = 4. Then, by Lemma 3.2.6, H \ F contains two distinct elements a and b such that a∗ F = {a∗ },

aF = {a, b},

and

b∗ F = {b∗ }.

In particular, |h∗ F | , |hF | for each element h in H \ F, so that, by Lemma 2.6.5(ii), Oϑ (H) ⊆ F, contradiction. Assume now that |H/F | = 3. Then, by Lemma 3.2.5, H \ F contains pairwise distinct elements a, b, and c such that a∗ = a, b∗ = b, aF = {a, c}, and bF = {b, c∗ }. Since Oϑ (H) * F, one of the elements a, b, c, or c∗ must be thin. If a is thin, |F | ≤ |aF |; cf. Lemma 1.4.7(i). Thus, as |aF | = 2, |F | = 2. (Recall that F is not normal in H.) Similarly, we obtain that |F | = 2 if one of the elements b, c, or c∗ is thin. From |F | = 2 and |H \ F | = 4 we obtain that |H| = 6. Thus, if c is thin, H is thin by Lemma 6.2.7(i), and if c∗ is thin, H is thin by Lemma 6.2.7(ii). If one of the elements a and b is thin, H is thin by Lemma 7.6.1. Now, as H is a thin hypergroup with six elements, H corresponds, via the group correspondence, to a group of order 6. Thus, as H has a non-normal closed subset, H is of type H6,1 . 

8 Constrained Sets of Involutions

Let T be a set of involutions of a hypergroup. Recall from Section 6.5 that, for each element d in hTi, T1 (d) stands for the set of all elements c in hTi such that cd contains an element e satisfying `T (e) = `T (c) + `T (d). A set T of involutions of a hypergroup is said to be constrained if, for any two elements t in T and f in T1 (t), | f t| = 1. Notice that sets consisting of a single involution of a hypergroup are constrained. From Lemma 1.4.3(i) one obtains further that each thin set of involutions of a hypergroup is constrained. In this chapter, we outline an abstract approach to constrained sets of involutions of hypergroups. More concrete aspects of constrained sets of involutions of hypergroups will be presented in the final two chapters of this monograph when we consider constrained sets of involutions of hypergroups satisfying the exchange condition. In Section 8.1, we start with basic facts about constrained sets of involutions of hypergroups. We first show that |de| = 1 whenever T stands for a constrained set of involutions of a hypergroup, e for an element in hTi and d for an element in T1 (e). This gives more flexibility in the investigation of constrained sets of involutions. A major result of this first section is Theorem 8.1.5, where we show that constrained sets of non-thin involutions of hypergroups are dichotomic if and only if they satisfy the exchange condition. (In Theorem 6.7.5, we saw already that the exchange condition implies dichotomy, regardless how the underlying set of involutions looks like.) In Section 8.2, we study (left) cosets of closed subsets generated by a single involution which belongs to a constrained set of involutions of a hypergroup, and in Section 8.3, we collect information about the thin radical of hypergroups generated by a constrained set of involutions. In Section 8.4, we investigate constrained sets of involutions which are not dichotomic, and in Section 8.5, we study, more specifically, constrained sets of involutions which are not dichotomic and which do not contain thin elements. Our study culminates in Theorem 8.5.5, a theorem which is similar to Theorem 8.1.5

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 P. -H. Zieschang, Hypergroups, https://doi.org/10.1007/978-3-031-39489-8_8

243

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8 Constrained Sets of Involutions

and provides a necessary and sufficient condition for constrained sets of non-thin involutions to be dichotomic. In the final two sections of this chapter, we consider foldings the domains of which are generated by a constrained set of involutions. The results show that the definition of a folding works well with the definition of a constrained set of involutions of a hypergroup. Much of the content of Sections 8.1, 8.4, and 8.5 was proved by Wang in his doctoral dissertation [53]. In particular, Theorem 8.1.5 is a consequence of [53; Theorem 2.3.1], and the content of Theorem 8.5.5 is developed implicitly in [53; Section 2.3].

8.1 Basic Results We begin with two necessary and sufficient conditions for a set of involutions of a hypergroup to be constrained. Lemma 8.1.1 Let T be a set of involutions of a hypergroup. Then the following conditions are equivalent. (a) The set T is constrained. (b) For any two elements e in hTi and d in T1 (e), we have |de| = 1. (c) For any three elements c, d, and e in hTi with e ∈ cd and `T (e) = `T (c) + `T (d), we have cd = {e}. Proof. (a) ⇒ (b) Let e be an element in hTi, and let d be an element in T1 (e). We will see that |de| = 1. If e = 1, we have de = {d}, so we are done in this case. Therefore, we assume that e , 1. In this case, we find elements b in hTi and t in T with e ∈ bt and `T (e) = `T (b) + 1; cf. Lemma 2.3.7(ii). Since d ∈ T1 (e), de contains an element f with `T ( f ) = `T (d) + `T (e). Thus, as e ∈ bt and `T (e) = `T (b) + 1, db contains an element c such that f ∈ ct, `T (c) = `T (d) + `T (b), and `T ( f ) = `T (c) + 1; cf. Lemma 2.3.8(i). From f ∈ ct and `T ( f ) = `T (c) + 1 we obtain that c ∈ T1 (t). Thus, as T is assumed to be constrained, |ct| = 1. From c ∈ db and `T (c) = `T (d) + `T (b) we obtain that d ∈ T1 (b). Thus, as `T (e) = `T (b) + 1, induction yields |db| = 1. Since c ∈ db, this implies that db = {c}. From e ∈ bt and `T (e) = `T (b) + 1 we obtain that b ∈ T1 (t). Thus, as T is assumed to be constrained, |bt| = 1. Since e ∈ bt, this implies that bt = {e}. From db = {c} and bt = {e} we obtain that de = ct. Thus, as |ct| = 1, |de| = 1. (b) ⇒ (c) Let c, d, and e be elements in hTi, and assume that e ∈ cd and that `T (e) = `T (c) + `T (d). Then, by definition, c ∈ T1 (d). Thus, by (b), |cd| = 1. Thus, as e ∈ cd, cd = {e}.

8.1

Basic Results

245

(c) ⇒ (a) Let t be an element in T, and let d be an element in T1 (t). Then, by definition, dt contains an element e such that `T (e) = `T (d) + 1. Now (c) yields  dt = {e}. It follows that |dt| = 1. Lemma 8.1.2 Let T be a constrained set of involutions of a hypergroup, and let e be an element in hTi. Then the following hold. (i) For each element d in hTi, there exists at most one element c in hTi with e ∈ cd and `T (e) = `T (c) + `T (d). (ii) For each element c in hTi, there exists at most one element d in hTi with e ∈ cd and `T (e) = `T (c) + `T (d). Proof. (i) Let d be an element in hTi, let c and c 0 be elements in hTi, and assume that e ∈ cd,

e ∈ c 0 d,

`T (e) = `T (c) + `T (d),

and `T (e) = `T (c 0) + `T (d).

We have to show that c = c 0. If d = 1, we have c = e = c 0, so we are done in this case. Therefore, we assume that d , 1. In this case, we find elements t in T and b in hTi with d ∈ tb and `T (d) = `T (b) + 1; cf. Lemma 2.3.7(i). Thus, as e ∈ cd and `T (e) = `T (c) + `T (d), ct contains an element a such that e ∈ ab, `T (a) = `T (c) + 1, and `T (e) = `T (a) + `T (b); cf. Lemma 2.3.8(i). Similarly, we find an element a 0 in c 0t such that e ∈ a 0 b, `T (a 0) = `T (c 0) + 1, and `T (e) = `T (a 0) + `T (b). Since e ∈ ab,

e ∈ a 0 b,

`T (e) = `T (a) + `T (b),

`T (e) = `T (a 0) + `T (b),

and `T (d) = `T (b) + 1, induction yields a 0 = a. Now recall that T is constrained. Thus, as a ∈ ct and `T (a) = `T (c) + 1, we obtain from Lemma 8.1.1 that ct = {a}. Similarly, we obtain that c 0t = {a 0 }. Thus, as a = a 0, ct = c 0t. It follows that c ∈ {c 0, a 0 }; cf. Lemma 6.1.1(ii). Since a = a 0 and `T (a) = `T (c) + 1, we cannot have c = a 0. Thus, c = c 0. (ii) The proof is similar to the proof of (i).



Lemma 8.1.3 Let T be a constrained set of involutions of a hypergroup. Let a, b, and c be elements in hTi with c ∈ ab and `T (c) = `T (a) + `T (b). Let d be an element in hTi, and assume that c ∈ T1 (d). Then bd contains an element e such that bd = {e}, ae = cd, and a ∈ T1 (e). Proof. Fom c ∈ T1 (d) we obtain an element f in cd such that `T ( f ) = `T (c) + `T (d). Thus, as c ∈ ab and `T (c) = `T (a) + `T (b), bd contains an element e such that f ∈ ae, `T (e) = `T (b) + `T (d), and `T ( f ) = `T (a) + `T (e); cf. Lemma 2.3.8(ii).

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Now recall that T is constrained. Thus, as e ∈ bd and `T (e) = `T (b) + `T (d), we obtain from Lemma 8.1.1 that bd = {e}. Similarly, we conclude that ae = { f } and cd = { f }, so that ae = cd. From f ∈ ae and `T ( f ) = `T (a) + `T (e) we also obtain that a ∈ T1 (e).



Lemma 8.1.4 Let T be a constrained set of involutions of a hypergroup. Let u and v be elements in T, and let c be an element in T1 (v). Assume that u ∈ T1 (c), that uc ⊆ T−1 (v), and that v is not thin. Then uc = cv. Proof. Since u ∈ T1 (c), uc contains an element d such that `T (d) = `T (c) + 1. Since T is constrained, this implies that uc = {d}; cf. Lemma 8.1.1. Since c ∈ T1 (v), cv contains an element e with `T (e) = `T (c) + 1. Since T is constrained, this implies that cv = {e}; cf. Lemma 8.1.1. From uc = {d} and cv = {e} we obtain that dv = ue. We are assuming that uc ⊆ T−1 (v). Thus, as d ∈ uc, d ∈ T−1 (v). This means that dv contains an element b with `T (d) = `T (b) + 1. Since T is constrained, this implies that bv = {d}; cf. Lemma 8.1.1. Thus, as v is assumed to be not thin, we obtain from Lemma 6.1.3(ii) that dv = {b, d}. From b ∈ dv and dv = ue we obtain that b ∈ ue. It follows that e ∈ ub. Note also that `T (e) = `T (c) + 1 = `T (d) = `T (b) + 1. Thus, u ∈ T1 (b), and, since T is constrained, this implies that |ub| = 1; cf. Lemma 8.1.1. Since e ∈ ub, this implies that ub = {e}. Thus, by Lemma 6.1.1(iii), (iv), ue = u2 b ⊆ {b} ∪ ub = {b, e}. Now, as dv = ue and dv = {b, d}, we have {b, d} ⊆ {b, e}. It follows that d = b or d = e. Since `T (d) = `T (b) + 1, we cannot have d = b. Thus, d = e. Since uc = {d} and cv = {e}, this implies that uc = cv.  Recall from Section 6.7 that a set T of involutions of a hypergroup is called dichotomic if, for each element t in T, hTi = T−1 (t) ∪ T1 (t). In the same section we said that a set T of involutions of a hypergroup satisfies the exchange condition if, for any three elements u and v in T and f in T1 (v) with u ∈ T1 ( f ), u f ⊆ f v ∪ T1 (v).

8.1

Basic Results

247

In Theorem 6.7.5, we proved that sets of involutions of hypergroups are dichotomic if they satisfy the exchange condition. The following theorem shows that, for constrained sets of non-thin involutions of hypergroups, the converse of that theorem holds. Theorem 8.1.5 Let T be a constrained set of involutions of a hypergroup, and assume that none of the elements in T is thin. Then T is dichotomic if and only if T satisfies the exchange condition. Proof. Regarding Theorem 6.7.5 we just need to show that T satisfies the exchange condition if T is dichotomic. Let u and v be elements in T, and let d be an element in T1 (v) with u ∈ T1 (d). We have to show that ud ⊆ dv ∪ T1 (v). Since u ∈ T1 (d), ud contains an element e with `T (e) = `T (d) + 1. Thus, as T is assumed to be constrained, we obtain from Lemma 8.1.1 that ud = {e}. Now assume that T is dichotomic. Then we have e ∈ T−1 (v) or e ∈ T1 (v). If e ∈ T−1 (v), we obtain from ud = {e} that ud ⊆ T−1 (v). Thus, by Lemma 8.1.4,  ud = dv. If e ∈ T1 (v), we obtain from ud = {e} that ud ⊆ T1 (v). Lemma 8.1.6 Let T be a constrained set of involutions of a hypergroup, let f be an element in hTi, and set n := `T ( f ). Let t1 , . . ., tn be elements in T such that f ∈ t1 · · · tn . Then t1 · · · tn = { f }. Proof. Assume first that n = 1. Then, by definition, t1 · · · tn = {t1 }. Thus, as f ∈ t1 · · · tn , f ∈ {t1 }; equivalently, f = t1 . It follows that t1 · · · tn = { f }. Assume now that 2 ≤ n. Then, as f ∈ t1 · · · tn , the product t1 · · · tn−1 contains an element e with f ∈ etn . From f ∈ etn we obtain that `T ( f ) ≤ `T (e) + 1; cf. Lemma 2.3.6(i). Thus, as `T ( f ) = n, we have n ≤ `T (e) + 1; equivalently, n − 1 ≤ `T (e). From e ∈ t1 · · · tn−1 , on the other hand, we obtain that `T (e) ≤ n − 1; cf. Lemma 2.3.6(i). Thus, we have `T (e) = n − 1. This equation has two consequences. First, since e ∈ t1 · · · tn−1 , it allows us to conclude (by induction) that t1 · · · tn−1 = {e}. Second, it yields `T ( f ) = `T (e) + 1 which, together with f ∈ etn , allows us to conclude that etn = { f }; cf. Lemma 8.1.1. From t1 · · · tn−1 = {e} and etn = { f } we obtain that t1 · · · tn = { f }.  Lemma 8.1.7 Let u and v be two distinct involutions of a hypergroup. Assume that C(u, v) is not empty and that {u, v} is constrained. Then, for each element n in {1, . . . , cu,v }, |Rn (u, v)| = 1.

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8 Constrained Sets of Involutions

Proof. Let n be an element in {1, . . . , cu,v }, let f be an element in Rn (u, v), and set T := {u, v}. Then, by Lemma 6.6.4, `T ( f ) = n. It follows that Rn (u, v) = { f }; cf.  Lemma 8.1.6. We conclude this section by showing that subsets of constrained sets of involutions of hypergroups are not necessarily constrained. Let H be a projective hypergroup of dimension 1 and of cardinality 5. Then the hypermultiplication of H is given by the following table. 1 v w x y

1 {1} {v} {w} {x} {y}

v {v} {1, v} {x, y} {w, y} {w, x}

w {w} {x, y} {1, w} {v, y} {v, x}

x {x} {w, y} {v, y} {1, x} {v, w}

y {y} {w, x} {v, x} {v, w} {1, y}

Set U := {v, w, x, y}. Then `U (u) = 1 for each element u in U. Thus, U1 (u) = {1} for each element u in U. It follows that U is constrained. Set T := {v, w}. Since x ∈ vw and x < T, `T (x) = 2. It follows that v ∈ T1 (w). Thus, as |vw| = 2, {v, w} is not constrained.

8.2 Constrained Sets of Involutions and Cosets In this section, we look at cosets f hti, where t is an element of a constrained set T of involutions of a hypergroup and f ∈ hTi. Lemma 8.2.1 Let T be a constrained set of involutions of a hypergroup, let t be an element in T, and let f be an element in hTi. Assume that hTi = T−1 (t) ∪ T1 (t). Then f hti contains elements d and e with f hti = {d, e} and `T (e) = `T (d) + 1. Proof. From Lemma 6.5.8(ii) we know that T1 (t) ∩ f hti is not empty. Let d be one of the elements in T1 (t) ∩ f hti. Since d ∈ T1 (t), dt contains an element e with `T (e) = `T (d) + 1. Since T is constrained, this implies that dt = {e}. From Lemma 6.1.1(ii) we obtain that dhti = {d} ∪ dt. Thus, as dt = {e}, we have dhti = {d, e}. On the other hand, as d ∈ f hti, Lemma 2.1.6(i) yields dhti = f hti. Thus, f hti = {d, e}, as wanted.



8.2

Constrained Sets of Involutions and Cosets

249

Let T be a set of involutions of a hypergroup. Recall from Section 6.5 that, for each element t in T, T0 (t) is our notation for the set of all elements d in hTi such that, for each element e in dt, `T (e) = `T (d). Lemma 8.2.2 Let T be a constrained set of involutions of a hypergroup, let t be an element in T, and let f be an element in T0 (t). Then the following hold. (i) We have f < f t. (ii) We have | f t| = 1 if and only if t is thin. Proof. (i) Since 1 ∈ T1 (t) and f ∈ T0 (t), f , 1. Thus, as f ∈ hTi, there exist elements u in T and e in hTi such that f ∈ ue and `T ( f ) = `T (e) + 1; cf. Lemma 2.3.7(i). From f ∈ ue and `T ( f ) = `T (e) + 1 we obtain that f ∈ T−1 (e). Thus, as f < T−1 (t), T−1 (e) * T−1 (t), so that, by Lemma 6.5.2(iii), e < T−1 (t). From e < T−1 (t) we obtain that e ∈ T0 (t) or e ∈ T1 (t); cf. Lemma 6.5.5(i). If e ∈ T0 (t), induction yields that e < et. If e ∈ T1 (t), we also have e < et; this time, since T is assumed to be constrained. Thus, we have e < et From f ∈ ue and `T ( f ) = `T (e)+1 (together with the hypothesis that T is constrained) we obtain that ue = { f }; cf. Lemma 8.1.1. It follows that u f = u2 e ⊆ {e} ∪ ue = {e, f }, and then that u f t ⊆ et ∪ f t. Now recall that e < et. Since f ∈ T0 (t) and `T ( f ) = `T (e) + 1, we also have e < f t. Thus, as u f t ⊆ et ∪ f t, e < u f t. Since e ∈ u f , this implies that f < f t. (ii) Assume first that | f t| = 1, and let e denote the uniquely determined element in f t. From f ∈ T0 (t) and e ∈ f t we obtain that e ∈ T0 (t); cf. Lemma 6.5.5(ii). Thus, by (i), e < et. It follows that t is thin; cf. Lemma 6.1.3(ii). The converse, that | f t| = 1 if t is thin, is known from Lemma 1.4.3(i).



Corollary 8.2.3 Let T be a constrained set of involutions of a hypergroup, let t be an element in T, let d be an element in T0 (t), and let e be an element in dt. Then the following hold. (i) We have dhti \ {d, e} = dt \ {e}. (ii) We have dt \ {e} = et \ {d}. Proof. (i) Since d ∈ T0 (t), d < dt; cf. Lemma 8.2.2(i). Thus, as dhti = {d} ∪ dt, we obtain that dhti \ {d} = dt. It follows that dhti \ {d, e} = dt \ {e}. (ii) In (i) we saw that

dhti \ {d, e} = dt \ {e}.

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8 Constrained Sets of Involutions

From d ∈ T0 (t) and e ∈ dt we obtain that e ∈ T0 (t); cf. Lemma 6.5.5(ii). Since e ∈ dt, we also have d ∈ et. Thus, by (i), ehti \ {d, e} = et \ {d}. From e ∈ dt we also obtain that dhti = ehti; cf. Lemma 2.1.6(i). Thus, dhti \ {d, e} = ehti \ {d, e}, so that the desired equation follows.



Corollary 8.2.4 Let T be a constrained set of involutions of a hypergroup, let t be an element in T, let d be an element in T0 (t), and let e be an element in dt. Then dhti = {d, e} if and only if t is thin. Proof. By Corollary 8.2.3(i), dhti = {d, e} if and only if dt = {e}. Since e ∈ dt, we have dt = {e} if and only if |dt| = 1, and in Lemma 8.2.2(ii), we saw that |dt| = 1 if and only if t is thin. 

8.3 Constrained Sets of Involutions and Thin Elements In this section, we look at the thin radical of hypergroups generated by a constrained set of involutions. Lemma 8.3.1 Let T be a constrained set of involutions of a hypergroup, and let A be a subset of hTi. Assume that Oϑ (T) ⊆ A, that A2 ⊆ A, and that, for each element a in A, a∗ a ⊆ A. Then A ⊆ hT ∩ Ai. Proof. Let e be an element in A. We will see that e ∈ hT ∩ Ai. If `T (e) = 0, e = 1, so we have e ∈ hT ∩ Ai by Lemma 2.1.1. Assume that e , 1. Then, by Lemma 2.3.7(ii), there exist elements d in hTi and t in T such that e ∈ dt and `T (e) = `T (d) + 1. Since T is constrained, we obtain that dt = {e}; cf. Lemma 8.1.1. Note that t ∈ A. In fact, if t is thin, this follows from our hypothesis that Oϑ (T) ⊆ A, and if t is not thin, we obtain from Lemma 6.1.1(iv) (together with dt = {e}) that t ∈ t 2 ⊆ td ∗ dt = e∗ e ⊆ A. Since e ∈ dt, d ∈ et. Thus, as e ∈ A and t ∈ A, we have d ∈ A2 . Since we are assuming that A2 ⊆ A, this implies that d ∈ A. Thus, as `T (e) = `T (d) + 1, induction yields that d ∈ hT ∩ Ai. Now, since e ∈ dt and t ∈ T ∩ A, we obtain that e ∈ hT ∩ Ai, as wanted. 

8.3

Constrained Sets of Involutions and Thin Elements

251

Corollary 8.3.2 Let T be a constrained set of involutions of a hypergroup, let F be a closed subset of hTi, and assume that Oϑ (T) ⊆ F. Then hT ∩ Fi = F. Proof. Since F is a closed subset of hTi, we have hT ∩ Fi ⊆ F. From Lemma 8.3.1 we obtain that F ⊆ hT ∩ Fi. Thus, hT ∩ Fi = F.  Corollary 8.3.3 Let T be a constrained set of involutions of a hypergroup. Then the following hold. (i) We have hOϑ (T)i = Oϑ (hTi). (ii) The set Oϑ (T) is empty if and only if Oϑ (hTi) = {1}. Proof. (i) Recall from Lemma 6.1.1(i) that T is symmetric. Thus, Oϑ (T) is a thin subset of hTi. It follows that hOϑ (T)i ⊆ Oϑ (hTi); cf. Lemma 2.6.1.

∗ -invariant

To show the reverse containment, we first notice that, by Lemma 1.4.3(iii), Oϑ (hTi)2 ⊆ Oϑ (hTi). Furthermore, for each element f in Oϑ (hTi), f ∗ f = {1} ⊆ Oϑ (hTi). Thus, by Lemma 8.3.1, Oϑ (hTi) ⊆ hOϑ (T)i. (ii) This is a consequence of (i).



Let T be a constrained set of involutions of a hypergroup. As a consequence of Corollary 8.3.3(i) one obtains that Oϑ (hTi) is a closed subset of hTi. (Recall from Section 2.6 that, in general, the thin radical of a hypergroup is not closed.) Corollary 8.3.3(ii) says that Oϑ (hTi) = {1} for each constrained set T of non-thin involutions. Hypergroups generated by constrained sets of non-thin involutions will be in the focus of Sections 8.5, 8.6, 9.1, 9.2, 9.4, 9.7, 9.9, 10.6, and 10.7. Recall from Section 6.7 that a set T of involutions of a hypergroup is called dichotomic if, for each element t in T, hTi = T−1 (t) ∪ T1 (t). Let U and V be sets of involutions of hypergroups. Recall from Section 6.5 that a bijective map φ from hUi to hVi is called (U, V)-length preserving if, for each element f in hUi, `U ( f ) = `V ( f φ ). Theorem 8.3.4 Let U and V be constrained sets of involutions of hypergroups. Assume that U is dichotomic. Let υ be a (U, V)-length preserving bijective map from hUi to hVi satisfying Oϑ (U)υ = Oϑ (V). Assume that eυ ∈ cυ d υ for any three elements c, d, and e in hUi with e ∈ cd and `U (e) = `U (c) + `U (d). Then υ is a hypergroup isomorphism from hUi to hVi. Proof. Since υ is (U, V)-length preserving, we have `U (1) = `V (1υ ). Thus, as `U (1) = 0, we have `V (1υ ) = 0. It follows that 1υ = 1. Let d and e be elements in hUi. We have to show that (de)υ = d υ eυ . If e = 1, eυ = 1υ . Thus, as 1υ = 1,

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(de)υ = (d ·1)υ = {d}υ = {d υ } = d υ 1 = d υ 1υ = d υ eυ, and we are done in this case. Therefore, we assume that e , 1. From e , 1 we obtain elements u in U and b in hUi with e ∈ ub and `U (e) = `U (b)+1; cf. Lemma 2.3.7(i). Since U is constrained, Lemma 8.1.1 yields ub = {e}. From e ∈ ub and `U (e) = `U (b) + 1 we also obtain that eυ ∈ uυ bυ , and, since υ is assumed to be (U, V)-length preserving, `U (e) = `U (b)+1 yields `V (eυ ) = `V (bυ )+1. Thus, as V is constrained, Lemma 8.1.1 yields uυ bυ = {eυ }. Assume first that d ∈ U1 (u). Then du contains an element a such that `U (a) = `U (d) + 1. As before, Lemma 8.1.1 yields du = {a}

and

d υ uυ = {aυ }.

It follows that (de)υ = (dub)υ = (ab)υ = aυ bυ = d υ uυ bυ = d υ eυ . (The third equality holds by induction, since `U (e) = `U (b) + 1.) Thus, we are done also in this case. Assume now that d < U1 (u). Then, as U is assumed to be dichotomic, d ∈ U−1 (u). Thus, du contains an element a with `U (d) = `U (a) + 1. As before, Lemma 8.1.1 yields au = {d} and aυ uυ = {d υ }. Suppose first that u is thin. Then, as a ∈ du, Lemma 1.4.3(i) yields du = {a}. Now recall that we are assuming that Oϑ (U)υ = Oϑ (V). Thus, as u is thin, so is uυ . Since u ∈ U and `U (u) = `V (uυ ), we also have uυ ∈ V. In particular, uυ is an involution. Thus, as aυ uυ = {d υ }, d υ uυ = {aυ }. It follows that (de)υ = (dub)υ = (ab)υ = aυ bυ = d υ uυ bυ = d υ eυ . (The third equality holds by induction, since `U (e) = `U (b) + 1.) Thus, we are done also in this case. Suppose now that u is not thin. Then, as au = {d}, Lemma 6.1.3(ii) yields

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du = {a, d}. It follows that (de)υ = (dub)υ = ({a, d}b)υ = (ab ∪ db)υ = (ab)υ ∪ (db)υ . We are assuming that Oϑ (U)υ = Oϑ (V). Thus, as u is not thin, uυ is not thin. Since u ∈ U and `U (u) = `V (uυ ), we also have uυ ∈ V. In particular, uυ is an involution. Thus, as aυ uυ = {d υ }, Lemma 6.1.3(ii) yields d υ uυ = {aυ, d υ }. It follows that d υ eυ = d υ uυ bυ = {aυ, d υ }bυ = aυ bυ ∪ d υ bυ . Now recall that `U (e) = `U (b) + 1. Thus, by induction, (ab)υ = aυ bυ and (db)υ =  d υ bυ . It follows that (de)υ = d υ eυ also in this case. Theorem 8.3.4 will be applied in the proofs of Lemma 9.5.9(ii), Lemma 9.7.2, and Theorem 9.9.2.

8.4 Constrained Sets of Involutions and Dichotomy, I Let T be a set of involutions of a hypergroup. Recall from Section 6.5 that, for each element t in T, T0 (t) is our notation for the set of all elements d in hTi such that, for each element e in dt, `T (e) = `T (d). In Section 6.7, we defined T0 to be the union of the sets T0 (t) with t ∈ T. In Lemma 6.7.1(i), we saw that a set T of involutions of a hypergroup is dichotomic if and only if T0 is empty. In this section, we begin a study of constrained sets of involutions of hypergroups which are not dichotomic. In other words, we investigate constrained sets T of involutions of hypergroups and assume T0 not to be empty. Clearly, the focus will be on the elements in T0 which have smallest possible Tlength. Lemma 8.4.1 Let T be a constrained set of involutions of a hypergroup, let v be an element in T, and assume that T0 (v) is not empty. Choose an element c in T0 (v) such that `T (c) is as small as possible. Let u be an element in T, and let b be an element in hTi such that c ∈ ub and `T (c) = `T (b) + 1. Let d be an element in bv. Then the following hold. (i) We have bv = {d}. (ii) We have cv = ud. (iii) We have `T (c) = `T (d).

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(iv) We have d ∗ ∈ T0 (u). Proof. (i) From Lemma 6.5.6 we know that b ∈ T1 (v). Thus, as T is assumed to be constrained, |bv | = 1. Since d ∈ bv, this implies that bv = {d}. (ii) From c ∈ ub and `T (c) = `T (b) + 1 we obtain that ub = {c}; cf. Lemma 8.1.1. (Recall that T is assumed to be constrained.) In (i), we saw that bv = {d}. Thus, cv = ud. (iii) From Lemma 6.5.6 we know that b ∈ T1 (v), from (i) that bv = {d}. Thus, `T (d) = `T (b) + 1. Since we are assuming that `T (c) = `T (b) + 1, this implies that `T (c) = `T (d). (iv) Since c ∈ T0 (v), each element in cv has T-length `T (c). Thus, by (ii) and (iii), each element in ud has T-length `T (d). It follows that each element in d ∗ u has T-length `T (d ∗ ). (We refer to Lemmas 1.3.1(ii) and 2.3.5.) Thus, d ∗ ∈ T0 (u).  Lemma 8.4.2 Let T be a constrained set of involutions of a hypergroup, let v be an element in T, and assume that T0 (v) is not empty. Choose an element c in T0 (v) such that `T (c) is as small as possible. Let u be an element in T, and let b be an element in hTi such that c ∈ ub and `T (c) = `T (b) + 1. Let d be an element in bv, and let e be an element in cv. Then (chvi \ {c, e})∗ = d ∗ hui \ {d ∗, e∗ }. Proof. Since c ∈ T0 (v) and e ∈ cv, we have chvi \ {c, e} = cv \ {e}; cf. Corollary 8.2.3(i). From Lemma 8.4.1(ii) we know that cv = ud. Thus, as e ∈ cv, e ∈ ud. It follows that e∗ ∈ d ∗ u; cf. Lemma 1.2.1. In Lemma 8.4.1(iv), we also saw that d ∗ ∈ T0 (u). Thus, by Corollary 8.2.3(i), d ∗ hui \ {d ∗, e∗ } = d ∗ u \ {e∗ }. Now we have (chvi \ {c, e})∗ = (cv \ {e})∗ = (ud \ {e})∗ = d ∗ hui \ {d ∗, e∗ }. (Once more, we refer to the equation cv = ud.)



We notice that, in Lemma 8.4.2, the set chvi \ {c, e} may be empty. In fact, according to Corollary 8.2.4, chvi \ {c, e} is empty if and only if v is thin. Lemma 8.4.3 Let T be a constrained set of involutions of a hypergroup, let v be an element in T, and assume that T0 (v) is not empty. Choose an element c in T0 (v) such that `T (c) is as small as possible. Let u be an element in T, and let b be an element in hTi such that c ∈ ub and `T (c) = `T (b) + 1. Then vc∗ ⊆ T0 (u).

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Proof. Let d be an element in bv. Then, by Lemma 8.4.1(iv), d ∗ ∈ T0 (u). It follows that d ∗ u ⊆ T0 (u); cf. Lemma 6.5.5(ii). Now recall from Lemma 8.4.1(ii) that cv = ud.  Thus, by Lemma 1.3.1(ii), vc∗ = d ∗ u. It follows that vc∗ ⊆ T0 (u). Lemma 8.4.4 Let T be a constrained set of involutions of a hypergroup, let t be an element in T, and assume that T0 (t) is not empty. Choose an element d in T0 (t) such that `T (d) is as small as possible. Let e be an element in dt. Let u and v be elements in T, and let b and c be elements in hTi such that d ∈ ub, e ∈ vc, `T (d) = `T (b) + 1, and `T (e) = `T (c) + 1. Then u , v. Proof. Applying Lemma 8.4.3 to d and t in place of c and v we obtain that td ∗ ⊆ T0 (u). On the other hand, as e ∈ dt, e∗ ∈ td ∗ ; cf. Lemma 1.2.1. Thus, e∗ ∈ T0 (u). Since e ∈ vc and `T (e) = `T (c) + 1, we have e∗ ∈ c∗ v and `T (e∗ ) = `T (c∗ ) + 1. Thus, e∗ ∈ T−1 (v). From e∗ ∈ T0 (u) and e∗ ∈ T−1 (v) we obtain that u , v.



Let T be a constrained set of involutions of a hypergroup, and let t be an element in T. So far, we were assuming that T0 (t) is not empty, and our focus was on elements in T0 (t) which have smallest possible T-length among the elements in T0 (t). Now we will assume that T0 is not empty, and we will look at elements in T0 which have smallest possible T-length among the elements in T0 . (Recall from Section 6.7 that T0 stands for the union of the sets T0 (t) with t ∈ T.) Of course, all results which we have obtained so far in this section apply to these elements in T0 . Lemma 8.4.5 Let T be a constrained set of involutions of a hypergroup, and assume that T0 is not empty. Choose an element d in T0 such that `T (d) is as small as possible. Let v be an element in T with d ∈ T0 (v), and let e be an element in dv. Let u and v 0 be elements in T, and let b and c be elements in hTi such that d ∈ ub, e ∈ cv 0, `T (d) = `T (b) + 1, and `T (e) = `T (c) + 1. Then the following hold. (i) We have b ∈ T1 (v) and u ∈ T1 (c). (ii) Let d 0 denote the element in bv, and let e 0 denote the element in uc. Then we have d 0 ∈ T0 (v 0), e 0 ∈ d 0 v 0, and dhvi \ {d, e} = d 0 hv 0i \ {d 0, e 0 }. Proof. (i) Since d has the smallest possible T-length among the elements in T0 and d in T0 (v), d has the smallest possible T-length among the elements in T0 (v). Thus, applying Lemma 6.5.6 to v, d, u, and b in place of t, e, c, and d we obtain that b ∈ T1 (v). Similarly, applying Lemma 8.4.3 to d in place of c we obtain that vd ∗ ⊆ T0 (u). From e ∈ dv we also obtain that e∗ ∈ vd ∗ . Thus, e∗ ∈ T0 (u). Since d ∈ T0 (v) and e ∈ dv, `T (e) = `T (d). Thus, by Lemma 2.3.5, `T (e∗ ) = `T (d). Thus, as d has the smallest possible T-length among the elements in T0 , e∗ has

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the smallest possible T-length among the elements in T0 (u). Thus, applying Lemma 6.5.6 to u, e∗ , v 0, and c∗ in place of t, e, c, and d we obtain that c∗ ∈ T1 (u). Now Lemma 6.5.2(i) yields u ∈ T1 (c). (ii) Applying Lemma 8.4.1(ii) to d and d 0 in place of c and d we obtain that dv = ud 0 . Thus, as e ∈ dv, e ∈ ud 0. It follows that e∗ ∈ (d 0)∗ u; cf. Lemma 1.2.1. On the other hand, by Lemma 8.4.1(iv), (d 0)∗ ∈ T0 (u). Thus, Corollary 8.2.3(ii) yields that (d 0)∗ u \ {e∗ } = e∗ u \ {(d 0)∗ }, so that, by Lemma 1.3.1(ii), ud 0 \ {e} = ue \ {d 0 }. From (d 0)∗ ∈ T0 (u) and e∗ ∈ (d 0)∗ u we obtain that e∗ ∈ T0 (u); cf. Lemma 6.5.5(ii). Thus, we may apply Lemma 8.4.1(ii) also to u, e∗ , v 0, c∗ , and (e 0)∗ in place of v, c, u, b, and d. Then we obtain that e∗ u = v 0(e 0)∗ , so that, by Lemma 1.3.1(ii), ue = e 0 v 0 . Since e ∈ ud 0, we have d 0 ∈ ue, and from this, together with ue = e 0 v 0, we obtain that d 0 ∈ e 0 v 0. It follows that e0 ∈ d 0v 0. Since e∗ ∈ T0 (u), we may apply also Lemma 8.4.1(iv) to u, e∗ , v 0, c∗ , and (e 0)∗ in place of v, c, u, b, and d. Then we obtain that e 0 ∈ T0 (v 0). Thus, as d 0 ∈ e 0 v 0, Lemma 6.5.5(ii) yields that d 0 ∈ T0 (v 0), and Corollary 8.2.3(i) yields that e 0 hv 0i \ {e 0, d 0 } = e 0 v 0 \ {d 0 }. Similarly, as d ∈ T0 (v) and e ∈ dv, Corollary 8.2.3(i) yields that dhvi \ {d, e} = dv \ {e}. Thus, as dv = ud 0, dhvi \ {d, e} = ud 0 \ {e} = ue \ {d 0 } = e 0 v 0 \ {d 0 } = e 0 hv 0i \ {e 0, d 0 }. Now we are done, since d 0 hv 0i = e 0 hv 0i.



Lemma 8.4.6 Let T be a constrained set of involutions of a hypergroup, and assume that T0 is not empty. Choose an element d in T0 such that `T (d) is as small as possible. Let t be an element in T with d ∈ T0 (t), and let e be an element in dt. Set k := `T (d), let t1 , . . ., tk be elements in T such that d ∈ t1 · · · tk , let tk+2 , . . ., t2k+1 be elements in T such that e∗ ∈ tk+2 · · · t2k+1 , and set tk+1 := t. For each element i in {1, . . . , 2k + 1}

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and each positive integer n, set tn(2k+1)+i := ti . Let n be a positive integer. Then there exist elements dn in T0 (tn+k ) and en in dn tn+k such that dn ∈ tn · · · tn+k−1 , e∗n ∈ tn+k+1 · · · tn+2k , and dhti \ {d, e} = dn htn+k i \ {dn, en }. Proof. By hypothesis, we have d ∈ T0 (t). From d ∈ T0 (t) and e ∈ dt we obtain that e ∈ T0 (t); cf. Lemma 6.5.5(ii). By hypothesis, we also have d ∈ t1 · · · t k

and

e∗ ∈ tk+2 · · · t2k+1 .

Thus, as t = tk+1 , we are done if n = 1. Let n be an integer with 2 ≤ n, and assume that there exist elements dn−1 in T0 (tn+k−1 ) and en−1 in dn−1 tn+k−1 such that dn−1 ∈ tn−1 · · · tn+k−2, and

e∗n−1 ∈ tn+k · · · tn+2k−1,

dhti \ {d, e} = dn−1 htn+k−1 i \ {dn−1, en−1 }.

From dn−1 ∈ tn−1 · · · tn+k−2 we obtain that `T (dn−1 ) ≤ k. On the other hand, since `T (d) = k and dn−1 ∈ T0 (tn+k−1 ), the minimal choice of d forces k ≤ `T (dn−1 ). Thus, we have `T (dn−1 ) = k. Since dn−1 ∈ tn−1 · · · tn+k−2 , we find an element a in tn · · · tn+k−2 such that dn−1 ∈ tn−1 a. From dn−1 ∈ tn−1 a we obtain that `T (dn−1 ) ≤ `T (a) + 1. On the other hand, since a ∈ tn · · · tn+k−2 , we have `T (a) ≤ k − 1 = `T (dn−1 ) − 1, so that `T (a) + 1 ≤ `T (dn−1 ). It follows that `T (dn−1 ) = `T (a) + 1. From e∗n−1 ∈ tn+k · · · tn+2k−1 we also find an element b in tn+k+1 · · · tn+2k−1 such that en−1 ∈ b∗ tn+k

and `T (en−1 ) = `T (b) + 1.

Apply Lemma 8.4.5(i) to dn−1 , tn+k−1 , en−1 , tn−1 , tn+k , a, and b∗ in place of d, v, e, u, v 0, b, and c. Then a ∈ T1 (tn+k−1 ) and tn−1 ∈ T1 (b∗ ). Let dn be an element in atn+k−1 , and let en be an element in tn−1 b∗ . We apply Lemma 8.4.5(ii) to dn−1 , tn+k−1 , en−1 , tn−1 , tn+k , a, b∗ , dn , and en in place of d, v, e, u, v 0, b, c, d 0, and e 0. Then dn ∈ T0 (tn+k ), and

en ∈ dn tn+k ,

dn−1 htn+k−1 i \ {dn−1, en−1 } = dn htn+k i \ {dn, en }.

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From dn ∈ atn+k−1 and a ∈ tn · · · tn+k−2 we obtain that dn ∈ tn · · · tn+k−1, and from e∗n ∈ btn−1 and b ∈ tn+k+1 · · · tn+2k−1 we obtain that e∗n ∈ tn+k+1 · · · tn+2k . From dhti \ {d, e} = dn−1 htn+k−1 i \ {dn−1, en−1 } and dn−1 htn+k−1 i \ {dn−1, en−1 } = dn htn+k i \ {dn, en } we obtain that dhti \ {d, e} = dn htn+k i \ {dn, en }.



Lemma 8.4.7 Let T be a constrained set of involutions of a hypergroup, and assume that T0 is not empty. Choose an element d in T0 such that `T (d) is as small as possible. Let t be an element in T with d ∈ T0 (t), and let e be an element in dt. Then dhti \ {d, e} is ∗ -invariant. Proof. Set k := `T (d), let t1 , . . ., tk be elements in T such that d ∈ t1 · · · tk , let tk+2 , . . ., t2k+1 be elements in T such that e∗ ∈ tk+2 · · · t2k+1 , and set tk+1 := t. For each element i in {1, . . . , 2k + 1} and each positive integer n, set tn(2k+1)+i := ti . From Lemma 8.4.6 we obtain, for each positive integer n, elements dn in T0 (tn+k ) and en in dn tn+k such that dn ∈ tn · · · tn+k−1, and

e∗n ∈ tn+k+1 · · · tn+2k ,

dhti \ {d, e} = dn htn+k i \ {dn, en }.

In particular, we have d2 ∈ t2 · · · tk+1

and

dk+2 ∈ tk+2 · · · t2k+1,

and, since t2k+1+i = ti for each element i in {1, . . . , 2k + 1}, we also obtain that ek+2 ∈ dk+2 t1, and

∗ ek+2 ∈ t2 · · · tk+1,

dhti \ {d, e} = dk+2 ht1 i \ {dk+2, ek+2 }.

Since d ∈ t1 · · · tk , there exists an element b in t2 · · · tk such that d ∈ t1 b. From b ∈ t2 · · · tk we obtain that `T (b) ≤ k − 1, from d ∈ t1 b we obtain that `T (d) ≤ `T (b) + 1. Thus, as `T (d) = k, we have `T (d) = `T (b) + 1. Thus, applying Lemma 8.4.2 to t, d, t1 , and d2 in place of v, c, u, and d we obtain that (dhti \ {d, e})∗ = d2∗ ht1 i \ {d2∗, e∗ }.

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(Recall that tk+1 = t.) Now we shall be done if we succeed in showing that d2∗ ht1 i \ {d2∗, e∗ } = dk+2 ht1 i \ {dk+2, ek+2 }. We first show that d2∗ = ek+2 and e∗ = dk+2 .

Since d2 ∈ T0 (tk+2 ) ⊆ T0 , the minimal choice of d forces k ≤ `T (d2 ). On the other hand, since d2 ∈ t2 · · · tk+1 , we have `T (d2 ) ≤ k. Thus, `T (d2 ) = k, so that, by Lemma ∗ ∗ . It 8.1.6, t2 · · · tk+1 = {d2 }. Thus, as ek+2 ∈ t2 · · · tk+1 , we conclude that d2 = ek+2 ∗ follows that d2 = ek+2 .

From d in T0 (t) and e ∈ dt we obtain that `T (e) = `T (d). Thus, as `T (d) = k, we have `T (e) = k. Now, by Lemma 2.3.5, `T (e∗ ) = k. Thus, as e∗ ∈ tk+2 · · · t2k+1 , Lemma 8.1.6 yields that tk+2 · · · t2k+1 = {e∗ }. Since dk+2 ∈ tk+2 · · · t2k+1 , this implies that e∗ = dk+2 . Recall finally that ek+2 ∈ dk+2 t1 . Thus, ek+2 ht1 i = dk+2 ht1 i. Since d2∗ = ek+2 , this  implies that d2∗ ht1 i = dk+2 ht1 i.

8.5 Constrained Sets of Involutions and Dichotomy, II In the previous section, we considered constrained sets of involutions of hypergroups which are not dichotomic. In other words, we investigated constrained sets T of involutions of hypergroups where T0 is not empty; cf. Lemma 6.7.1(i). The focus was on those elements in T0 which have the smallest possible T-length. In Lemma 8.4.7, we proved that, for each such element d in T0 and for any two elements t in T with d ∈ T0 (t) and e in dt, the difference set dhti \ {d, e} is ∗ -invariant. A little earlier, in Corollary 8.2.4, we saw that these difference sets are not empty if and only if t is not thin. It is the combination of these two results which suggests a closer look at constrained sets of involutions which do not contain thin elements, and these constrained sets are in the focus of the present section. Lemma 8.5.1 Let T be a constrained set of involutions of a hypergroup, and assume that none of the elements in T is thin. Assume that T0 is not empty, and choose an element f in T0 such that `T ( f ) is as small as possible. Then f is symmetric. Proof. Since f ∈ T0 , T contains an element t with f ∈ T0 (t). Since T is assumed not to contain thin elements and t ∈ T, t is not thin. Thus, by Lemma 8.2.2(ii), 2 ≤ | f t|. Let d and e be two distinct elements in f t. Since f ∈ T0 (t) and d ∈ f t, d ∈ T0 (t); cf. Lemma 6.5.5(ii). Since {d, e} ⊆ f t and d , e, e ∈ dt. On the other hand, as f ∈ T0 (t) and d ∈ f hti, `T (d) = `T ( f ). Thus, the minimal choice of f forces dhti \ {d, e} to be ∗ -invariant; cf. Lemma 8.4.7. Since d ∈ f t, we also have dhti = f hti. Thus, f hti \ {d, e} is ∗ -invariant.

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By Lemma 8.2.2(i), f < {d, e} for any two distinct elements d and e in f t. Thus, { f } is the intersection of the sets f hti \ {d, e} with {d, e} ⊆ f t and d , e. Since each of these sets is ∗ -invariant, so is { f }, and that implies that f is symmetric.  Corollary 8.5.2 Let T be a constrained set of involutions of a hypergroup, and assume that none of the elements in T is thin. Assume that T0 is not empty, and choose an element f in T0 such that `T ( f ) is as small as possible. Then the following hold. (i) Each element in T0 which has the same T-length as f is symmetric. (ii) Let t be an element in T with f ∈ T0 (t). Then f t is symmetric. Proof. (i) We are assuming that f has the smallest possible T-length among the elements in T0 . Thus, each element in T0 which has the same T-length as f has the smallest possible T-length among the elements in T0 . Thus, by Lemma 8.5.1, each element in T0 which has the same T-length as f is symmetric. (ii) Let e be an element in f t. Then, as f ∈ T0 (t), we obtain that e ∈ T0 (t); cf. Lemma 6.5.5(ii). In particular, e ∈ T0 . From f ∈ T0 (t) and e ∈ f t we also obtain that `T (e) = `T ( f ). Thus, by (i), e is symmetric.  Lemma 8.5.3 Let T be a constrained set of involutions of a hypergroup, and assume that none of the elements in T is thin. Assume that T0 is not empty, and choose an element d in T0 such that `T (d) is as small as possible. Let t be an element in T with d ∈ T0 (t). Let u be an element in T, and let b be an element in hTi such that d ∈ ub and `T (d) = `T (b) + 1. Then the following hold. (i) The set dt is symmetric. (ii) The set bt is symmetric. (iii) For each element e in dt, we have bt ⊆ eu. (iv) For each element e in dt, we have dhti \ {d, e} ⊆ eu. (v) We have dt ⊆ T0 (u). Proof. (i) Among the elements in T0 we have chosen d in such a way that `T (d) is as small as possible. Thus, as d ∈ T0 (t), we obtain from Corollary 8.5.2(ii) that dt is symmetric. (ii) Let a be an element in bt. Applying Lemma 8.4.1(iv) to t, d, and a in place of v, c, and d we obtain that a∗ ∈ T0 (u) ⊆ T0 . Applying Lemma 8.4.1(iii) to t, d, and a in place of v, c, and d we obtain that `T (d) = `T (a), whence, by Lemma 2.3.5, `T (a∗ ) = `T (d). From a∗ ∈ T0 and `T (a∗ ) = `T (d) we obtain that a∗ is symmetric; cf. Corollary 8.5.2(i). Now, as a∗ is symmetric, so is a. (iii) Let a be an element in bt. Applying Lemma 8.4.1(ii) to t, d, and a in place of v, c, and d we obtain that dt = ua. Now let e be an element in dt. Then e ∈ ua.

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261

Since e ∈ dt, e is symmetric; cf. (i). Since a ∈ bt, a is symmetric; cf. (ii). Thus, as e ∈ ua, we obtain from Lemma 1.2.1 that a ∈ eu. (iv) Let e be an element in dt, and let f be an element in dhti \ {d, e}. We will see that f ∈ eu. From d ∈ T0 (t), e ∈ dt, and f ∈ dhti \ {d, e} we obtain that f ∈ dt \ {e}; cf. Corollary 8.2.3(i). From f ∈ dt we now obtain that bt ⊆ f u; cf. (iii). Let a be an element in bt. Then a ∈ f u. Thus, f ∈ au. From a ∈ bt we also obtain that a ∈ eu; cf. (iii). Thus, f ∈ eu2 ⊆ ehui, and, since e , f , this implies that f ∈ eu. (v) Let e be an element in dt, and let a be an element in bt. Then, by (iii), a ∈ eu. It follows that e ∈ au. On the other hand, by Lemma 8.4.1(iv), a ∈ T0 (u). Thus, by  Lemma 6.5.5(ii), e ∈ T0 (u). Lemma 8.5.4 Let T be a constrained set of involutions of a hypergroup, and assume that none of the elements in T is thin. Assume that T0 is not empty, and choose an element d in T0 such that `T (d) is as small as possible. Let t be an element in T with d ∈ T0 (t), and let e be an element in dt. Let u and v be elements in T, and let b and c be elements in hTi such that d ∈ ub, e ∈ vc, `T (d) = `T (b) + 1, and `T (e) = `T (c) + 1. Then the following hold. (i) For each element g in cu, we have eu = vg. (ii) For each element f in dhti \ {d, e}, we have vb ⊆ f t. (iii) We have e ∈ vb. Proof. (i) Let g be an element in cu, and recall from Lemma 8.5.3(v) that e ∈ T0 (u). Thus, we may apply Lemma 8.4.1(ii) to u, e, v, c, and g in place of v, c, u, b, and d. We obtain that eu = vg. (ii) Let f be an element in dhti \ {d, e}. Then, by Lemma 8.5.3(iv), f ∈ eu. Let g be an element in cu. Then, by (i), eu = vg. Thus, as f ∈ eu, f ∈ vg. Let a be an element in bt. Then, by Lemma 8.5.3(iii), a ∈ eu. On the other hand, by (i), eu = vg. Thus, a ∈ vg. It follows that g ∈ va. Thus, as f ∈ vg, f ∈ v 2 a, so f ∈ hvia. Since f has been chosen from dhti \ {d, e}, we have f ∈ dt; cf. Corollary 8.2.3(i). Thus, applying Lemma 8.4.1(ii) to t, d, and a in place of v, c, and d we obtain that f ∈ ua. Since f ∈ dt, f is symmetric; cf. Lemma 8.5.3(i). Since a ∈ bt, a is symmetric; cf. Lemma 8.5.3(ii). Thus, as f ∈ ua, f ∈ au.

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Applying Lemma 8.4.1(iv) to t, d, and a in place of v, c, and d we obtain that a∗ ∈ T0 (u). Since a is symmetric, this implies that a ∈ T0 (u). Thus, as f ∈ au, a , f ; cf. Lemma 8.2.2(i). Thus, as f ∈ hvia, we have f ∈ va. From f ∈ va and a ∈ bt we obtain that f ∈ vbt. Thus, vb contains an element h such that f ∈ ht. It follows that h ∈ f t. From h ∈ f t and f ∈ dt we obtain that h ∈ dt 2 ⊆ dhti. Thus, as d ∈ T0 (t), `T (h) = `T (d). Since we are assuming that `T (d) = `T (b) + 1, this implies that `T (h) = `T (b) + 1. Thus, as h ∈ vb, vb = {h}; cf. Lemma 8.1.1. Since h ∈ f t this implies that vb ⊆ f t. (iii) We are assuming that none of the elements in T is thin. Since t ∈ T, this implies that t is not thin. Thus, by Lemma 8.2.2(ii), dhti \ {d, e} is not empty. We fix an element f0 in dhti \ {d, e}. Then, by (ii), vb ⊆ f0 t. Since f0 ∈ dhti, we also have f0 t ⊆ dhti. Thus, vb ⊆ dhti. Let f be an element in dhti \ {d, e}. Then, by (ii), vb ⊆ f t. Since f ∈ T0 (t), we have f < f t. Thus, f < vb. Thus, as vb ⊆ dhti, vb ⊆ dhti \ { f }. Since f has been chosen arbitrarily in dhti \ {d, e}, this shows that vb ⊆ {d, e}. Assume that d ∈ vb. Then, as d ∈ ub and `T (d) = `T (b) + 1, Lemma 8.1.2(i) yields u = v, contrary to Lemma 8.4.4. It follows that vb ⊆ {e}, and, since vb is not empty,  this implies that e ∈ vb. Theorem 8.5.5 Let T be a constrained set of involutions of a hypergroup, and assume that none of the elements in T is thin. Then T is dichotomic if and only if T does not contain the involutions of a projective closed subset of hTi having positive dimension. Proof. Regarding Corollary 6.7.2 we just need to show that T is dichotomic if T does not contain the involutions of a projective closed subset of hTi having positive dimension. Assume, by way of contradiction, that T is not dichotomic. Then, by Lemma 6.7.1(i), T0 is not empty. We choose an element d in T0 , and we do it in such a way that `T (d) is as small as possible. Since d ∈ T0 , there exists an element t in T with d ∈ T0 (t). Let e be an element in dt. Then, as d ∈ T0 (t), `T (d) = `T (e). Let u and v be elements in T, and let b and c be elements in hTi such that d ∈ ub,

e ∈ vc,

`T (d) = `T (b) + 1,

Then, by Lemma 8.5.4(iii), e ∈ vb.

and `T (e) = `T (c) + 1.

8.5

Constrained Sets of Involutions and Dichotomy, II

263

From `T (d) = `T (e) and `T (d) = `T (b) + 1 we obtain that `T (e) = `T (b) + 1. Thus, as e ∈ vb, e ∈ vc, and `T (e) = `T (c) + 1, Lemma 8.1.2(ii) yields b = c. We are assuming that T does not contain the involutions of a projective closed subset of hTi having positive dimension. Thus, by Lemma 6.7.1(ii), T ∩ T0 is empty. Since d ∈ T0 , this implies that d < T. It follows that 2 ≤ `T (d). Thus, as `T (d) = `T (b) + 1, 1 ≤ `T (b); equivalently, b , 1. Since b , 1, we obtain from Lemma 2.3.7(ii) elements a in hTi and w in T with b = aw

and `T (b) = `T (a) + 1.

Thus, as d ∈ ub and `T (d) = `T (b) + 1, there exists an element y in ua such that d ∈ yw

and `T (d) = `T (y) + 1;

cf. Lemma 2.3.8(i). Since b = c, we obtain from b = aw and `T (b) = `T (a) + 1 also that c = aw

and `T (c) = `T (a) + 1.

Thus, as e ∈ vc and `T (e) = `T (c) + 1, there exists an element z in va such that e ∈ zw

and `T (e) = `T (z) + 1;

cf. Lemma 2.3.8(i). From d ∈ yw and `T (d) = `T (y) + 1 we obtain that d ∈ w y∗

and `T (d) = `T (y ∗ ) + 1.

(By Lemma 8.5.1, d ∗ = d.) Similarly, as e ∈ zw and `T (e) = `T (z) + 1, we have e ∈ wz∗

and `T (e) = `T (z ∗ ) + 1.

Thus, by Lemma 8.4.4, e < dt, contradiction.



Three remarks about Theorem 8.5.5. First, the hypothesis that none of the elements of T is thin can be expressed by the equation Oϑ (hTi) = {1}; cf. Corollary 8.3.3(i). Second, by Lemma 6.7.1(i), T is dichotomic if and only if T0 is empty. Finally, by Lemma 6.7.1(ii), T does not contain the involutions of a projective closed subset of hTi having positive dimension if and only if T ∩ T0 is empty. Referring to these three observations we obtain a shorter but more technical way to state Theorem 8.5.5. The theorem says that, for each constrained set T of involutions of a hypergroup with Oϑ (hTi) = {1}, T0 is empty if T ∩ T0 is empty.

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8 Constrained Sets of Involutions

Theorem 8.5.5 is interesting in connection with Theorem 8.1.5, since the combination of these two theorems implies that constrained sets of non-thin involutions of a hypergroup satisfy the exchange condition if and only if they do not contain the involutions of a projective closed subset of hTi having positive dimension. We refer to this observation in the following corollary. Corollary 8.5.6 Any constrained set of involutions of a hypergroup consisting of one, two, or three non-thin involutions satisfies the exchange condition. Proof. Let T be a constrained set of involutions of a hypergroup. Assume that none of the elements in T is thin and that T does not satisfy the exchange condition. Then, by Theorem 8.1.5, T is not dichotomic. Thus, by Theorem 8.5.5, T contains the involutions of a projective closed subset F of hTi having positive dimension. Since T is assumed not to contain thin elements, this implies that 5 ≤ |F |; cf. Lemma  6.1.8(i). It follows that 4 ≤ |T |. The constrained set of four non-thin involutions of a hypergroup given at the end of Section 8.1 shows that constrained sets of non-thin involutions of hypergroups do not necessarily satisfy the exchange condition any longer if they contain more than three elements.

8.6 Constrained Sets of Involutions and Foldings, I Recall from Section 2.7 that an injective map ρ from a closed subset F of a hypergroup H to H is called a folding if, for each element f in F, f ρ f = {1ρ }. In this section, we consider foldings the domains of which are generated by constrained sets of involutions. Our results will be useful in Sections 8.7, 9.7, and 10.6. Lemma 8.6.1 Let H be a hypergroup, let T be a constrained set of involutions of H, and let ρ be a folding from hTi to H. Let f be an element in hTi, let h be an element in H, and assume that h f = {1ρ }. Then h = f ρ . Proof. The statement is obviously true if f = 1. Therefore, we assume that f , 1. From f , 1 we obtain elements t in T and e in hTi such that f ∈ te and `T ( f ) = `T (e) + 1; cf. Lemma 2.3.7(i). From f ∈ te and `T ( f ) = `T (e) + 1 (together with the hypothesis that T is constrained) we obtain that te = { f }; cf. Lemma 8.1.1. Since ρ is a folding, f ρ f = {1ρ }. Thus, as te = { f }, f ρ te = {1ρ }. It follows that, for each element d in f ρ t, de = {1ρ }. Thus, by induction, d = eρ for each element d in f ρ t. It follows that f ρ t = {eρ }.

8.6

Constrained Sets of Involutions and Foldings, I

265

Now recall that we are assuming that h f = {1ρ }. Thus, as te = { f }, hte = {1ρ }. It follows that, for each element d in ht, de = {1ρ }. Thus, by induction, d = eρ for each element d in ht. It follows that ht = {eρ }. From f ρ t = {eρ } and ht = {eρ } we obtain that f ρ t = ht. Thus, f ρ ∈ hhti = {h} ∪ ht = {h, eρ }. If f ρ = eρ , the injectivity of ρ forces f = e, and that contradicts `T ( f ) = 1 + `T (e). Thus, we have f ρ = h.  Lemma 8.6.2 Let H be a hypergroup, let T be a constrained set of involutions of H, and let ρ be a folding from hTi to H. Let c, d, and e be elements in hTi, and assume that e ∈ cd and that `T (e) = `T (c) + `T (d). Then e∗ρ d ∗ = {c∗ρ }. Proof. We are assuming that e ∈ cd and `T (e) = `T (c) + `T (d). Thus, as T is constrained, cd = {e}; cf. Lemma 8.1.1. It follows that d ∗ c∗ = {e∗ }; cf. Lemma 1.2.2. Thus, for each element f in e∗ρ d ∗ , f c∗ ⊆ e∗ρ d ∗ c∗ = e∗ρ e∗ = {1ρ }, whence, by Lemma 8.6.1, f = c∗ρ .



Corollary 8.6.3 Let H be a hypergroup, let T be a constrained set of involutions of H, and let ρ be a folding from hTi to H. Let d and e be elements in hTi, and let t be an element in T. Assume that e ∈ dt and `T (e) = `T (d) + 1. Then the following hold. (i) If t is thin, d ∗ρ t = {e∗ρ }. (ii) If t is not thin, d ∗ρ t = {e∗ρ, d ∗ρ }. Proof. From e ∈ dt and `T (e) = `T (d) + 1 we obtain that e∗ρ t = {d ∗ρ }; cf. Lemma 8.6.2. Thus, the corollary follows from Lemma 6.1.3.  The following proposition refers to Lemma 2.7.2. Proposition 8.6.4 Let H be a hypergroup, let T be a constrained set of involutions of H, and let ρ be a folding from hTi to H. Then hTi ρ = 1ρ hTi. Proof. From Lemma 2.7.2 we know that hTi ρ ⊆ 1ρ hTi. Assume, by way of contradiction, that 1ρ hTi * hTi ρ . Then hTi contains an element f such that 1ρ f * hTi ρ . Among the elements f in hTi with 1ρ f * hTi ρ we choose f such that `T ( f ) is as small as possible.

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8 Constrained Sets of Involutions

Since 1ρ f * hTi ρ , 1ρ f contains an element e with e < hTi ρ . Define A to be the set of all elements a in hTi such that there exists an element d in hTi with f ∈ ad, `T ( f ) = `T (a) + `T (d), and e ∈ a∗ρ d. Since e ∈ 1ρ f , 1 ∈ A. Thus, A is not empty. From e < hTi ρ we obtain that f < A. Among the elements in A we choose a such that `T (a) is as large as possible. Since a ∈ A, hTi contains an element d with f ∈ ad,

`T ( f ) = `T (a) + `T (d),

and

e ∈ a∗ρ d.

From a ∈ A and f < A we obtain that a , f . Thus, d , 1. Thus, by Lemma 2.3.7(i), there exist elements t in T and c in hTi such that d ∈ tc and `T (d) = `T (c) + 1. Thus, as f ∈ ad and `T ( f ) = `T (a) + `T (d), at contains an element b such that f ∈ bc, `T (b) = `T (a) + 1, and `T ( f ) = `T (b) + `T (c); cf. Lemma 2.3.8(i). From `T (b) = `T (a) + 1, together with the maximal choice of a, we obtain that b < A. Thus, as f ∈ bc and `T ( f ) = `T (b) + `T (c), we conclude that e < b∗ρ c; equivalently, b∗ρ < ec∗ . From e ∈ a∗ρ d we obtain that d ∗ ∈ e∗ a∗ρ ; cf. Lemma 1.2.1. Similarly, as d ∈ tc, d ∗ ∈ c∗ t. Thus, d ∗ ∈ e∗ a∗ρ ∩ c∗ t, so that, by Lemma 1.2.5(i), ec∗ ∩ a∗ρ t , ∅. From b ∈ at and `T (b) = `T (a) + 1 we obtain that a∗ρ t ⊆ {b∗ρ, a∗ρ }; cf. Corollary 8.6.3. Thus, as b∗ρ < ec∗ and ec∗ ∩ a∗ρ t , ∅, a∗ρ ∈ ec∗ ; equivalently, e ∈ a∗ρ c. Since 1ρ ∈ a∗ρ a∗ , we have a∗ρ ∈ 1ρ a. Thus, as e ∈ a∗ρ c, e ∈ 1ρ ac. Thus, ac contains an element g such that e ∈ 1ρ g. From g ∈ ac, `T (d) = `T (c) + 1, and `T ( f ) = `T (a) + `T (d) we obtain `T (g) ≤ `T (a) + `T (c) = `T (a) + `T (d) − 1 = `T ( f ) − 1. Thus, the minimal choice of f forces 1ρ g ⊆ hTi ρ . Thus, as e ∈ 1ρ g, e ∈ hTi ρ . This  contradiction finishes the proof. Let H be a hypergroup, and let F be a closed subset of H. In Lemma 2.7.7, we saw how to obtain foldings from F to H from given foldings from F to H and certain thin elements in F. The following lemma shows that, if F is generated by a constrained set of involutions, then all foldings from F to H having the same image arise from a given folding from F to H the way described in Lemma 2.7.7. Lemma 8.6.5 Let H be a hypergroup, and let T be a constrained set of involutions of H. Let σ and τ be foldings from hTi to H, and assume that hTi σ = hTiτ . Then the following hold. (i) The set hTi contains a thin element c such that, for each element f in hTi, { f τ } = ( f c)σ . (In particular, 1τ = cσ .)

8.6

Constrained Sets of Involutions and Foldings, I

267

(ii) Assume that none of the elements in T is thin. Then σ = τ. Proof. (i) We are assuming that hTi σ = hTiτ . Thus, 1σ ∈ hTiτ , so that we find an element b in hTi with 1σ = bτ . It follows that 1σ b = {1τ }. Similarly, we obtain an element c in hTi such that 1τ c = {1σ }. From 1σ b = {1τ } and 1τ c = {1σ } we obtain that 1σ bc = {1σ }. Let d be an element in bc. Then 1σ d = {1σ }. Thus, by Lemma 8.6.1, 1σ = d σ . On the other hand, as σ is a folding, σ is injective. Thus, 1 = d, and, since d has been chosen arbitrarily in bc, this shows that bc = {1}. It follows that c is thin; cf. Lemma 1.4.2. Let f be an element in hTi. Since c is thin, f c contains an element e such that f c = {e}; cf. Lemma 1.4.3(i). It follows that f τ e = f τ f c = 1τ c = {1σ }. Thus, by Lemma 8.6.1, f τ = eσ . Thus, as f c = {e}, { f τ } = ( f c)σ . (ii) We are assuming that none of the elements in T is thin. Thus, by Corollary 8.3.3(ii), 1 is the only thin element of hTi, so that the desired equation follows from  (i). Lemma 8.6.6 Let H be a hypergroup, let T be a constrained set of involutions of H, and let ρ be a folding from hTi to H. Assume that H = hTi ∪ hTi ρ . Then the following hold. (i) We have hTi ρ = hTi or hTi ρ = H \ hTi. (ii) We have hTi1ρ hTi = hTi ρ . (iii) The set hTi ρ is is ∗ -invariant. Proof. (i) Recall from Proposition 8.6.4 that hTi ρ = 1ρ hTi. Since we are assuming that H = hTi ∪ hTi ρ , this implies that H = hTi ∪ 1ρ hTi. Thus, by Lemma 2.1.6(i), 1ρ hTi = hTi

or

1ρ hTi = H \ hTi.

Since hTi ρ = 1ρ hTi, this implies that hTi ρ = hTi or hTi ρ = H \ hTi. (ii) From Lemma 2.1.6(i) we know that hTi1ρ hTi = hTi

or

hTi1ρ hTi ⊆ H \ hTi.

Assume that 1ρ ∈ hTi. Then hTi1ρ hTi = hTi and, by (i), hTi ρ = hTi. Thus, we have hTi1ρ hTi = hTi ρ . Assume that 1ρ ∈ H \ hTi. Then hTi1ρ hTi ⊆ H \ hTi and, by (i), hTi ρ = H \ hTi. Thus, hTi1ρ hTi ⊆ hTi ρ . On the other hand, by Proposition 8.6.4, hTi ρ = 1ρ hTi.

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8 Constrained Sets of Involutions

Thus, as 1ρ hTi ⊆ hTi1ρ hTi, we have hTi ρ ⊆ hTi1ρ hTi. Again we obtain that hTi1ρ hTi = hTi ρ . (iii) Since hTi is ∗ -invariant, this follows from (i).



Lemma 8.6.7 Let H be a hypergroup, let T be a constrained set of involutions of H, and assume that none of the elements in T is thin. Then there exists at most one folding ρ from hTi to H satisfying H = hTi ∪ hTi ρ . Proof. Let σ and τ be foldings from hTi to H, and assume that H = hTi ∪ hTi σ

and

H = hTi ∪ hTiτ .

Then, by Lemma 8.6.6(i), hTi σ = hTi or hTi σ = H \ hTi and

hTiτ = hTi or hTiτ = H \ hTi;

cf. Lemma 2.1.6(i). Assume that hTi σ = hTi. Then H = hTi which implies that hTiτ , H \ hTi. Thus, we have hTiτ = hTi. Similarly, hTiτ = hTi implies that hTi σ = hTi, and this shows that hTi σ = hTiτ . Now recall that T is assumed not to contain thin elements. Thus, by Lemma 8.6.5(ii), σ = τ.  Lemma 8.6.8 Let H be a hypergroup, let T be a constrained set of involutions of H, and let ρ be a folding from hTi to H. Assume that H = hTi ∪ hTi ρ , that 1ρ∗ 1ρ ⊆ hTi, and that H acts regularly on a set X. Then X = yhTi ∪ zhTi for any two elements y in X and z in y1ρ . Proof. Let y be an element in X, let z be an element in y1ρ , and let x be an element in X \ yhTi. We shall be done if we succeed in showing that x ∈ zhTi. We are assuming that H = hTi ∪ hTi ρ , and from Proposition 8.6.4 we know that hTi ρ = 1ρ hTi. Thus, we have H = hTi ∪ 1ρ hTi. On the other hand, since H is assumed to act regularly on X, we have x ∈ yH. It follows that x ∈ yhTi

or

x ∈ y1ρ hTi.

Since x has been chosen from X \ yhTi, this implies that x ∈ y1ρ hTi. Now recall that z ∈ y1ρ . Thus, by Lemma 1.6.2(ii), y ∈ z1ρ∗ . Thus, as x ∈ y1ρ hTi, we have x ∈ z1ρ∗ 1ρ hTi. Since we are assuming that 1ρ∗ 1ρ ⊆ hTi, this implies that x ∈ zhTi.



8.7

Constrained Sets of Involutions and Foldings, II

269

8.7 Constrained Sets of Involutions and Foldings, II In this section, we continue studying foldings the domains of which are generated by a constrained set of involutions. Additionally, we assume the constrained sets of involutions to be dichotomic. Lemma 8.7.1 Let H be a hypergroup, let T be a dichotomic constrained set of involutions of H, and let ρ be a folding from hTi to H. Let t be an element in T, and let d and e be elements in hTi. Then we have dhti = ehti if and only if d ∗ρ hti = e∗ρ hti. Proof. There is nothing to show if d = e. Thus, we assume that d , e. Assume first that dhti = ehti. Since T is dichotomic, we obtain from Lemma 8.2.1 that dhti = {d, e}, and, without loss of generalization, `T (e) = `T (d) + 1. Thus, as e ∈ dt, we obtain from Lemma 8.6.2 that e∗ρ t = {d ∗ρ }, so that, by Lemma 2.1.6(i), d ∗ρ hti = e∗ρ hti. Assume, conversely, that d ∗ρ hti = e∗ρ hti. Since T is dichotomic, we have d ∈ T−1 (t) or d ∈ T1 (t). Thus, dt contains an element c such that `T (d) = `T (c) + 1

or `T (c) = `T (d) + 1.

From c ∈ dt we also obtain that d ∈ ct. Thus, in both cases, d ∗ρ hti = {c∗ρ, d ∗ρ }; cf. Corollary 8.6.3. Since we are assuming that d ∗ρ hti = e∗ρ hti, this implies that e∗ρ ∈ {c∗ρ, d ∗ρ }. From e∗ρ ∈ {c∗ρ, d ∗ρ } we obtain that e ∈ {c, d}, since ρ is injective. Thus, since d , e, e = c. Since c ∈ dt, this implies that e ∈ dt, so that, by Lemma 2.1.6(i),  dhti = ehti. Lemma 8.7.2 Let H be a hypergroup, let T be a dichotomic constrained set of involutions of H, and let ρ be a folding from hTi to H. Let d and e be elements in hTi, and assume that d ∗ρ ∈ 1ρ e. Then `T (d) ≤ `T (e) − 1 or d = e. Proof. If e = 1, our hypothesis that d ∗ρ ∈ 1ρ e implies that d ∗ρ = 1ρ . Thus, as ρ is injective, d ∗ = 1. It follows that d = 1, and we are done. Assume that e , 1. Then, by Lemma 2.3.7(ii), there exist elements c in hTi and t in T such that e ∈ ct and `T (e) = `T (c) + 1. We are assuming that d ∗ρ ∈ 1ρ e. Thus, 1ρ ∈ d ∗ρ e∗ . From e ∈ ct we also obtain that e∗ ∈ tc∗ ; cf. Lemma 1.2.1. Thus, 1ρ ∈ d ∗ρ tc∗ . Now d ∗ρ t contains an element b with 1ρ ∈ bc∗ . From 1ρ ∈ bc∗ we obtain that b ∈ 1ρ c. Thus, as c ∈ hTi, b ∈ 1ρ hTi. It follows that b ∈ hTi ρ ; cf. Proposition 8.6.4. Thus, hTi contains an element a with b = a∗ρ .

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8 Constrained Sets of Involutions

From b ∈ d ∗ρ t and b = a∗ρ we obtain that a∗ρ ∈ d ∗ρ t, so that, by Lemma 2.1.6(i), a∗ρ hti = d ∗ρ hti. It follows that ahti = dhti; cf. Lemma 8.7.1. Now `T (d) ≤ `T (a) + 1. From 1ρ ∈ bc∗ and b = a∗ρ we obtain that 1ρ ∈ a∗ρ c∗ . Thus, a∗ρ ∈ 1ρ c, so that, by induction, `T (a) ≤ `T (c) − 1 or a = c. Assume that `T (e) ≤ `T (d). Then, since `T (e) = `T (c) + 1 and `T (d) ≤ `T (a) + 1, we conclude that `T (c) ≤ `T (a). Thus, the above dichotomy forces a = c, and, since ahti = dhti, this implies that d ∈ chti. It follows that d ∈ {c} ∪ ct. Since `T (e) = `T (c) + 1 and `T (e) ≤ `T (d), we also have d , c. Thus, d ∈ ct. Now recall that T is constrained. Thus, as e ∈ ct and `T (e) = `T (c) + 1, we conclude that ct = {e}. Thus, as d ∈ ct, we obtain that d = e, as wanted.  Let H be a hypergroup, let F be a closed subset of H, and let ρ be a folding from F to H. Assume that F ρ∗ = F ρ . In Lemma 2.7.3, we saw that, in this case, the map from F to F which sends each −1 element f of F to f ∗ρ∗ρ is a permutation of F. In Section 2.7, we called this map the permutation of F induced by ρ. Lemma 8.7.3 Let H be a hypergroup, let T be a dichotomic constrained set of involutions of H, and let ρ be a folding from hTi to H. Assume that 1ρ∗ = 1ρ and that hTi ρ∗ = hTi ρ . Let π denote the permutation of hTi induced by ρ. Then the following hold. (i) We have π 2 = 1 hT i . (ii) For each element f in hTi, we have f π∗ = f ∗π . Proof. (i) Let f be an element in hTi. We shall see that f π = f . 2

From our assumption that 1ρ∗ = 1ρ we obtain that 1π = 1, and then that 1π = 1. 2

Since f ∈ hTi, f π ∈ hTi. Thus, 1ρ ∈ f π ρ f π . It follows that f π ρ∗ ∈ f π ·1ρ ; cf. −1 −1 Lemma 1.2.1. Now recall from Lemma 2.7.5 that f π ·1ρ = 1ρ f . Thus, f π ρ∗ ∈ 1ρ f . −1

−1

−1

−1

−1

From the definition of π one obtains that f π ρ∗ = f π ∗ρ . Thus, as f π ρ∗ ∈ 1ρ f , −2 −2 we have f π ∗ρ ∈ 1ρ f , and we may apply Lemma 8.7.2 to f π and f in place of d and e. Then −2 −2 `T ( f π ) ≤ `T ( f ) − 1 or f π = f . −1

−1

−2

If `T ( f π ) ≤ `T ( f ) − 1, induction yields ( f π )π = f π , contrary to `T ( f π ) ≤ −2 2 `T ( f ) − 1. Thus, f π = f , and that implies that f π = f . −2

−2

2

(ii) Considering (i) this follows from Lemma 2.7.4(ii).

−2

−2



8.7

Constrained Sets of Involutions and Foldings, II

271

The results of this section will be applied in the proofs of Lemma 9.7.1, Lemma 9.7.2, Lemma 9.9.8, and Theorem 10.6.1.

9 Coxeter Sets of Involutions

Recall from the introduction to Chapter 8 that a set T of involutions of a hypergroup is called constrained if, for any two elements t in T and f in T1 (t), | f t| = 1. In Section 6.7, we said that a set T of involutions of a hypergroup satisfies the exchange condition if, for any three elements u and v in T and f in T1 (v) with u ∈ T1 ( f ), u f ⊆ f v ∪ T1 (v). In this chapter, we investigate constrained sets of involutions of hypergroups which satisfy the exchange condition. These sets will be called Coxeter sets, and a hypergroup H which is generated by a Coxeter set T of involutions of H will be called a Coxeter hypergroup over T. If a hypergroup is generated by a Coxeter set of involutions and we do not specify this Coxeter set, we simply speak about a Coxeter hypergroup. In the introduction to Chapter 8, we mentioned that thin sets of involutions of hypergroups are constrained, and in Section 6.7 (right after the definition of the exchange condition), we noticed that a set T of thin involutions satisfies the exchange condition if and only if the set of involutions which, via the group correspondence, corresponds to T satisfies the group theoretic exchange condition. Thus, the thin Coxeter hypergroups are exactly those hypergroups which, via the group correspondence, correspond to Coxeter groups. Hypergroups of type H6,11 (as defined in Section 7.4) give examples of non-thin Coxeter hypergroups. In Sections 10.2 and 10.3, we will see that Coxeter hypergroups (thin or not thin) are closely related to buildings in the sense of [44] or [45]. The present chapter is an introduction to Coxeter hypergroups. In Section 9.1, we discuss basic properties of Coxeter sets of involutions of hypergroups. We present some useful observations about length functions defined by Coxeter sets and study closed subsets of Coxeter hypergroups. In Section 9.2, we look at the sets V1 (U) where V is a Coxeter set of involutions of a hypergroup and U a subset of V, and in Section 9.3, we look at the sets V−1 (U) where, again, V is a Coxeter set of involutions of a hypergroup and U a subset of V. (The sets V1 (U) and V−1 (U) were introduced in Section 6.5.)

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 P. -H. Zieschang, Hypergroups, https://doi.org/10.1007/978-3-031-39489-8_9

273

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9 Coxeter Sets of Involutions

In Section 9.4, we study intersections and direct products of closed subsets generated by Coxeter sets of involutions of a hypergroup, and in Sections 9.5 and 9.6, we restict our considerations to finite Coxeter hypergroups. In Section 9.7, we study foldings the domains of which are generated by Coxeter sets of involutions, in other words, we consider foldings the domains of which are Coxeter hypergroups. In Section 9.8, the focus is on the Coxeter numbers of involutions (as defined in Section 6.6) belonging to a Coxeter set. Bijective maps between Coxeter sets of involutions which respect the Coxeter numbers will be called type preserving, and these maps are the subject of Section 9.9. They will be related to hypergroup isomorphisms between Coxeter hypergroups.

9.1 General Observations Our first lemma is similar to Lemma 8.1.1. It gives two necessary and sufficient conditions for constrained sets of involutions of a hypergroup to satisfy the exchange condition. Lemma 9.1.1 Let T be a constrained set of involutions of a hypergroup. Then the following conditions are equivalent. (a) The set T satisfies the exchange condition. (b) For any three elements u and v in T and f in T1 (v) with u ∈ T1 ( f ), we have u f = f v or u f ⊆ T1 (v). (c) For any six elements u and v in T, a and b in hTi, d ∈ ua, and e ∈ bv with `T (d) = `T (a) + 1, `T (e) = `T (b) + 1, d ∈ T1 (b) and a ∈ T1 (e), we have db = ae or d ∈ T1 (e). Proof. (a) ⇒ (b) Let u and v be elements in T, and let f be an element in T1 (v) with u ∈ T1 ( f ). Since T is assumed to satisfy the exchange condition, we have u f ⊆ f u ∪ T1 (v). Since u ∈ T1 ( f ) and T is constrained, we have |u f | = 1; cf. Lemma 8.1.1. Similarly, as f ∈ T1 (v), | f v | = 1. Thus, as u f ⊆ f u ∪ T1 (v), we have u f = f v or u f ⊆ T1 (v). (b) ⇒ (c) Let u be an element in T, let a be an element in hTi, and let d be an element in ua with `T (d) = `T (a) + 1. Let b be an element in hTi, let v be an element in T, and let e be an element in bv with `T (e) = `T (b) + 1. Assume that d ∈ T1 (b) and a ∈ T1 (e). From d ∈ ua and `T (d) = `T (a) + 1 we obtain that ua = {d}; cf. Lemma 8.1.1. (Recall that T is assumed to be constrained.) Similarly, bv = {e}. Since d ∈ T1 (b), db contains an element f with `T ( f ) = `T (d) + `T (b). Thus, as d ∈ ua and `T (d) = `T (a) + 1, ab contains an element c such that f ∈ uc, `T (c) = `T (a) + `T (b), and `T ( f ) = `T (c) + 1; cf. Lemma 2.3.8(ii).

9.1

General Observations

275

From c ∈ ab and `T (c) = `T (a) + `T (b) (together with the hypothesis that T is constrained) we obtain that ab = {c}; cf. Lemma 8.1.1. From f ∈ uc and `T ( f ) = `T (c) + 1 we obtain that u ∈ T1 (c). Similarly, using ab = {c}, we conclude from a ∈ T1 (e) that c ∈ T1 (v). Now recall that T is assumed to satisfy the exchange condition. Thus, as u ∈ T1 (c) and c ∈ T1 (v), we have uc = cv or uc ⊆ T1 (v). If uc = cv, we obtain from ua = {d}, ab = {c}, and bv = {e} that db = ae. If uc ⊆ T1 (v), we obtain from f ∈ uc that f ∈ T1 (v). Thus, by definition, f v contains an element g such that `T (g) = `T ( f ) + 1. From g ∈ f v and f v ⊆ ucv = uabv = de we obtain that g ∈ de. From `T (g) = `T ( f ) + 1 and `T ( f ) + 1 = `T (c) + 2 = `T (a) + `T (b) + 2 = `T (d) + `T (e) we obtain that `T (g) = `T (d) + `T (e). It follows that d ∈ T1 (e). (c) ⇒ (a) The exchange condition is the case b = 1.



Lemma 9.1.2 Let V be a Coxeter set of involutions of a hypergroup, and let U be a subset of V. Then, for each element f in hUi, `U ( f ) = `V ( f ). Proof. Assume the assertion to be false. Among the elements f in hUi with `U ( f ) , `V ( f ) we choose an element e with smallest possible U-length. From `U (e) , `V (e) we obtain that e , 1. Thus, by Lemma 2.3.7(i), there exist elements x in U and d in hUi such that e ∈ xd and `U (e) = `U (d) + 1. From `U (e) , `V (e), together with U ⊆ V, we obtain that e < U. Thus, as e ∈ xd and x ∈ U, d , 1. Thus, by Lemma 2.3.7(ii), there exist elements b in hUi and y in U such that d ∈ by and `U (d) = `U (b) + 1. Now, by Lemma 2.3.8(i), xb contains an element c such that e ∈ cy, `U (c) = `U (b) + 1, and `U (e) = `U (c) + 1. Since `U (e) = `U (c) + 1, the minimal choice of e forces `U (c) = `V (c). From `U (e) = `U (d) + 1 and `U (d) = `U (b) + 1 we obtain `U (e) = `U (b) + 2. Thus, the minimal choice of e forces `U (b) = `V (b). From `U (c) = `U (b) + 1, together with `U (c) = `V (c) and `U (b) = `V (b), we now obtain that `V (c) = `V (b) + 1. Thus, as c ∈ xb, x ∈ V1 (b). Similarly, one obtains that b ∈ V1 (y). Thus, as V is assumed to satisfy the exchange condition, we have xb ⊆ by ∪ V1 (y). Since c ∈ xb, this implies that c ∈ by

or

c ∈ V1 (y).

Assume that c ∈ by. Then, as e ∈ cy, we obtain that e ∈ by 2 ⊆ {b} ∪ by, contrary to `U (e) = `U (b) + 2.

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Assume that c ∈ V1 (y). Then, as e ∈ cy, we have `V (e) = `V (c) + 1. (Recall that V is assumed to be constrained.) Thus, as `U (e) = `U (c) + 1 and `U (c) = `V (c), we have  `U (e) = `V (e), contrary to the choice of e. Corollary 9.1.3 Let V be a Coxeter set of involutions of a hypergroup, let U be a subset of V, and let d be an element in hUi. Then the following hold. (i) We have U1 (d) = hUi ∩ V1 (d). (ii) We have U−1 (d) = hUi ∩ V−1 (d). Proof. (i) Let c be an element in U1 (d). Then c ∈ hUi and cd contains an element e such that `U (e) = `U (c) + `U (d). Thus, by Lemma 9.1.2, `V (e) = `V (c) + `V (d). (Note that {c, d, e} ⊆ hUi.) Since c ∈ hVi, this implies that c ∈ V1 (d). Conversely, let c be an element in hUi ∩ V1 (d). Since c ∈ V1 (d), the product cd contains an element e such that `V (e) = `V (c) + `V (d). Thus, by Lemma 9.1.2, `U (e) = `U (c) + `U (d). (Note that {c, d, e} ⊆ hUi.) It follows that c ∈ U1 (d). (ii) Let e be an element in U−1 (d). Then e ∈ hUi and ed ∗ contains an element c such that `U (e) = `U (c) + `U (d). Thus, by Lemma 9.1.2, `V (e) = `V (c) + `V (d). (Note that {c, d, e} ⊆ hUi.) Since e ∈ hVi, this implies that e ∈ V−1 (d). Conversely, let e be an element in hUi ∩ V−1 (d). Since e ∈ V−1 (d), the product ed ∗ contains an element c such that `V (e) = `V (c) + `V (d). Thus, by Lemma 9.1.2, `U (e) = `U (c) + `U (d). (Note that {c, d, e} ⊆ hUi.) It follows that e ∈ U−1 (d).  Corollary 9.1.4 Let V be a Coxeter set of involutions of a hypergroup, and let U be a subset of V. Then U = V ∩ hUi. Proof. Since U ⊆ V ∩ hUi, it suffices to show that V ∩ hUi ⊆ U. Let v be an element in V ∩ hUi. Since v ∈ V, we have `V (v) = 1, and from v ∈ hUi we obtain that `U (v) = `V (v); cf. Lemma 9.1.2. It follows that `U (v) = 1, and that means that v ∈ U.  Recall from Section 8.1 that subsets of constrained sets of involutions of a hypergroup are not necessarily constrained. The following theorem shows that the situation is different for Coxeter sets of involutions of hypergroups. Theorem 9.1.5 Let V be a Coxeter set of involutions of a hypergroup, and let U be a subset of V. Then U is a Coxeter set. Proof. We first show that U is constrained. For this, we fix elements u in U and f in U1 (u). We have to show that | f u| = 1.

9.1

General Observations

277

Since u ∈ hUi, U1 (u) ⊆ V1 (u); cf. Corollary 9.1.3(i). Thus, as f ∈ U1 (u), f ∈ V1 (u). Now recall that V is a Coxeter set. In particular, V is constrained. Thus, as u ∈ V, | f u| = 1. To show that U satisfies the exchange condition, we let x and y be elements in U, and we let f be an element in U1 (y) with x ∈ U1 ( f ). We have to show that x f ⊆ f y ∪ U1 (y). Since y ∈ hUi, U1 (y) ⊆ V1 (y); cf. Corollary 9.1.3(i). Thus, as f ∈ U1 (y), f ∈ V1 (y). Similarly, as f ∈ hUi, U1 ( f ) ⊆ V1 ( f ). Thus, as x ∈ U1 ( f ), x ∈ V1 ( f ). Now recall that V is a Coxeter set. In particular, V satisfies the exchange condition. Thus, we have x f ⊆ f y ∪ V1 (y). On the other hand, as x ∈ U and f ∈ hUi, x f ⊆ hUi. Similarly, f y ⊆ hUi. Thus, x f ⊆ hUi ∩ ( f y ∪ V1 (y)) = (hUi ∩ f y) ∪ (hUi ∩ V1 (y)) = f y ∩ U1 (y); cf. Corollary 9.1.3(i).



Lemma 9.1.6 Let T be a Coxeter set of involutions of a hypergroup, let U and V be subsets of T, and assume that U ⊆ C hT i (V). Then the following hold. (i) We have hUi ⊆ C hT i (hVi). (ii) We have hUihVi = hU ∪ Vi. (iii) The closed subset hUi is normal in hU ∪ Vi. Proof. (i) Let d be an element in hUi, and let e be an element in hVi. We have to show that ed = de. There is nothing to show if d = 1 or e = 1. Thus, we assume that d , 1 and e , 1. From e ∈ hVi \ {1} we obtain elements v in V and c in hVi such that e ∈ vc and `V (e) = `V (c) + 1; cf. Lemma 2.3.7(i). It follows that v ∈ V1 (c). Thus, as c ∈ hVi, v ∈ T1 (c); cf. Corollary 9.1.3(i). Since T is a Coxeter set, this implies that vc = {e}. From d ∈ hUi \ {1} we obtain elements b in hUi and u in U such that d ∈ bu and `U (d) = `U (b) + 1; cf. Lemma 2.3.7(ii). It follows that b ∈ U1 (u). Thus, as u ∈ hUi, b ∈ T1 (u); cf. Corollary 9.1.3(i). Since T is a Coxeter set, this implies that bu = {d}. Since `U (d) = `U (b) + 1, induction yields cb = bc and vb = bv. Since `V (e) = `V (c) + 1, induction yields cu = uc. Since we are assuming that U ⊆ C hT i (V), we have vu = uv. Thus, as vc = {e} and bu = {d}, we have ed = vcbu = vbcu = bvuc = buvc = de. (ii) From (i), we know that hUi ⊆ C hT i (hVi). Thus, hVi ⊆ C hT i (hUi). It follows that hVi ⊆ N hT i (hUi); cf. Lemma 3.1.1(i). Now Lemma 3.1.4(ii) yields hUihVi = hVihUi. Thus, by Lemma 2.3.2, hUihVi = hU ∪ Vi. (iii) From (i), we know that hUi ⊆ C hT i (hVi). Thus, hVi ⊆ C hT i (hUi). It follows that hVi ⊆ N hT i (hUi); cf. Lemma 3.1.1(i). Thus, by Lemma 3.1.1(ii),

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hUihVi ⊆ hUiN hT i (hUi) = N hT i (hUi). Now recall from (ii) that hUihVi = hU ∪ Vi. Thus, hU ∪ Vi ⊆ N hT i (hUi). Since  hUi ⊆ hU ∪ Vi, this says that hUi is normal in hU ∪ Vi. In the remaining two results of this section, we look at Coxeter sets consisting of non-thin involutions. Theorem 9.1.7 Let V be a Coxeter set of involutions of a hypergroup, and assume that none of the elements in V is thin. Then U 7→ hUi is a bijective map from the power set of V to the set of all closed subsets of hVi. Proof. We are assuming that Oϑ (V) is empty. Thus, for each closed subset F of hVi, hV ∩ Fi = F; cf. Corollary 8.3.2. This shows that U 7→ hUi is a surjective map from the power set of V to the set of all closed subsets of hVi. From Corollary 9.1.4 we obtain that X = V ∩ hXi = V ∩ hY i = Y for any two subsets X and Y of V with hXi = hY i. Thus, U 7→ hUi is an injective map from the power  set of V to the set of all closed subsets of hVi. Corollary 9.1.8 Let T be a Coxeter set of involutions of a hypergroup, and assume that none of the elements in T is thin. Then, for any two subsets U and V of T, we have hU ∩ Vi = hUi ∩ hVi. Proof. Let U and V be subsets of T. Then, by Lemma 2.1.4, hUi ∩ hVi is a closed subset of hTi. Thus, by Theorem 9.1.7, T contains a subset W with hWi = hUi ∩ hVi. From Corollary 9.1.4 we know that U = T ∩ hUi. Thus, as W ⊆ T ∩ hUi, we have W ⊆ U. Similarly, W ⊆ V, so W ⊆ U ∩ V. From Corollary 9.1.4 we know that W = T ∩ hWi. Thus, as U ∩V ⊆ T ∩ hUi ∩ hVi = T ∩ hWi, we have U ∩ V ⊆ W. From W ⊆ U ∩ V and U ∩ V ⊆ W we obtain that W = U ∩ V. It follows that hWi = hU ∩ Vi. Thus, as hWi = hUi ∩ hVi, we have hU ∩ Vi = hUi ∩ hVi. 

9.2 The Sets V1 (U) for Subsets U of Coxeter Sets V In this section, we have a look at the sets V1 (U) where V is a Coxeter set of involutions of a hypergroup and U a subset of V. Theorem 9.2.1 Let V be a Coxeter set of involutions of a hypergroup, and let U be a subset of V. Then V1 (U) = V1 (hUi).

9.2

The Sets V1 (U) for Subsets U of Coxeter Sets V

279

Proof. Assume, by way of contradiction, that V1 (U) , V1 (hUi). Then, since V1 (hUi) ⊆ V1 (U), we have V1 (U) * V1 (hUi). Among the elements in V1 (U) which are not in V1 (hUi) we choose c such that `V (c) is as small as possible. Since c < V1 (hUi), hUi contains an element d such that c < V1 (d). Among the elements d in hUi satisfying c < V1 (d) we choose d such that `V (d) is as small as possible. Since c < V1 (hUi) and 1 ∈ V1 (hUi), c , 1. Thus, by Lemma 2.3.7(i), there exist elements v in V and a in hVi such that c ∈ va

and `V (c) = `V (a) + 1.

Since c ∈ va and `V (c) = `V (a) + 1, we have c ∈ V−1 (a). Thus, as c ∈ V1 (U), a ∈ V1 (U); cf. Lemma 6.5.2(ii). Thus, as `V (c) = `V (a) + 1, the minimal choice of c forces a ∈ V1 (hUi). Thus, as d ∈ hUi, we have a ∈ V1 (d). Since c < V1 (d) and c ∈ V1 (1), d , 1. Thus, by Lemma 2.3.7(ii), there exist elements b in hUi and u in U such that d ∈ bu

and `U (d) = `U (b) + 1.

From `U (d) = `U (b) + 1 and {b, d} ⊆ hUi we obtain that `V (d) = `V (b) + 1; cf. Lemma 9.1.2. Thus, as b ∈ hUi, the minimal choice of d forces c ∈ V1 (b). Now we apply Lemma 9.1.1 to V, v, c, d, and u in place of T, u, d, e, and v. Then cb = ad

or

c ∈ V1 (d).

Since c < V1 (d), this implies that cb = ad. Now recall that {b, d} ⊆ hUi. Thus, c ∈ ahUi; cf. Lemma 2.1.6(i). It follows that hUi contains an element e with c ∈ ae. Since a ∈ V1 (hUi) and e ∈ hUi, a ∈ V1 (e). Thus, as c ∈ ae, `V (c) = `V (a) + `V (e). (Here we use the hypothesis that V is constrained, and we apply Lemma 8.1.1.) Since `V (c) = `V (a) + 1, this implies that `V (e) = 1. As a consequence, e ∈ V. From e ∈ V ∩ hUi we now obtain that e ∈ U; cf. Corollary 9.1.4. On the other hand, as c ∈ ae and `V (c) = `V (a) + `V (e), we have c ∈ V−1 (e). Thus, c < V1 (U), contrary  to our choice of c. Corollary 9.2.2 Let V be a Coxeter set of involutions of a hypergroup, and let U be a subset of V. Then hUi ⊆ V1 (hV \ Ui). Proof. Let v be an element in V \ U. Since V is a Coxeter set of involutions of a hypergroup, V satisfies the exchange condition. Thus, v ∈ V1 (U). It follows that v ∈ V1 (hUi); cf. Theorem 9.2.1. Thus, by Lemma 6.5.2(i), hUi ⊆ V1 (v). Since v has been chosen arbitrarily among the elements of V \ U, we have shown that hUi ⊆ V1 (V \ U). Thus, by Theorem 9.2.1, hUi ⊆ V1 (hV \ Ui).  The following corollary is similar to Lemma 6.7.3.

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Corollary 9.2.3 Let T be a Coxeter set of involutions of a hypergroup, and let U and V be subsets of T. Then hUi = (hUi ∩ T1 (V))hU ∩ Vi. Proof. Since T is a Coxeter set, T satisfies the exchange condition. Thus, T is dichotomic; cf. Theorem 6.7.5. Thus, by Lemma 6.7.3, hTi = T1 (U ∩ V)hU ∩ Vi. It follows that hUi = hUi ∩ T1 (U ∩ V)hU ∩ Vi = (hUi ∩ T1 (U ∩ V))hU ∩ Vi; cf. Lemma 2.2.1(ii). On the other hand, by Corollary 9.2.2, hUi ⊆ T1 (V \ U). Thus, hUi ∩ T1 (U ∩ V) = hUi ∩ T1 (V). It follows that hUi = (hUi ∩ T1 (V))hU ∩ Vi.



We will now see that, given a Coxeter set V of involutions of a hypergroup and a subset U of V, each left coset of hUi in hVi contains exactly one element of V1 (U). Corollary 9.2.4 Let V be a Coxeter set of involutions of a hypergroup, and let U be a subset of V. Then, for each element f in hVi, |V1 (U) ∩ f hUi| = 1. Proof. Let f be an element in hVi. By Corollary 9.2.3, V1 (U) contains an element c such that f ∈ chUi. It follows that c ∈ f hUi; cf. Lemma 1.3.3. Thus, we have c ∈ V1 (U) ∩ f hUi. Let b be an element in V1 (U) ∩ f hUi. We will see that b = c. From {b, c} ⊆ f hUi we obtain that c ∈ bhUi; cf. Lemma 2.1.6(i). Thus, hUi contains an element d such that c ∈ bd. From b ∈ V1 (U) we obtain that b ∈ V1 (hUi); cf. Theorem 9.2.1. Thus, as d ∈ hUi, b ∈ V1 (d). Thus, bd contains an element e with `V (e) = `V (b) + `V (d). Since V is assumed to be constrained, we obtain from b ∈ V1 (d) also that |bd| = 1; cf. Lemma 8.1.1. Thus, as {c, e} ⊆ bd, c = e. Thus, as `V (e) = `V (b) + `V (d), `V (c) = `V (b) + `V (d). Similarly, one obtains that `V (b) = `V (c) + `V (d ∗ ). It follows that `V (d) = 0, so d = 1. Thus, as c ∈ bd, b = c.  The following theorem will be useful in the proof of Theorem 9.4.4. Theorem 9.2.5 Let V be a Coxeter set of involutions of a hypergroup, and let U be a subset of V. Then the following conditions are equivalent.

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The Sets V1 (U) for Subsets U of Coxeter Sets V

281

(a) We have U ⊆ C hV i (V \ U). (b) We have hUi ⊆ C hV i (hV \ Ui). (c) The closed subset hUi is normal in hVi. Proof. (a) ⇔ (b) This follows from Lemma 9.1.6(i). (a) ⇒ (c) This follows from Lemma 9.1.6(iii). (c) ⇒ (a) Let u be an element in U and let v be an element in V \ U. We have to show that uv = vu. Since u ∈ U and v ∈ V \ U, u ∈ V1 (v); cf. Corollary 9.2.2. Thus, by definition, uv contains an element e with `V (e) = `V (u) + `V (v) = 2. On the other hand, as V is a Coxeter set of involutions of a hypergroup, V is constrained. Thus, we obtain from u ∈ V1 (v) that |uv | = 1; cf. Lemma 8.1.1. Since e ∈ uv, this implies that uv = {e}. We are assuming that hUi is normal in hVi. Thus, we have uv ⊆ v hUi. Since e ∈ uv, this implies that e ∈ v hUi. Thus, hUi contains an element d with e ∈ vd. Since v ∈ V \ U and d ∈ hUi we obtain from Corollary 9.2.2 that v ∈ V1 (d). Since V is constrained, we obtain from v ∈ V1 (d) that |vd| = 1; cf. Lemma 8.1.1. Since e ∈ vd, this implies that vd = {e}. From v ∈ V1 (d) and vd = {e} we obtain that `V (e) = `V (d) + 1. Thus, as `V (e) = 2, we obtain that `V (d) = 1; equivalently, d ∈ V. From uv = {e} and vd = {e} we obtain that uv = vd. It follows that d ∈ hu, vi. From d ∈ V ∩ hu, vi we obtain that d ∈ {u, v}; cf. Corollary 9.1.4. Since v ∈ V1 (d),  d , v. Thus, d = u. Now, as uv = vd, we have uv = vu. Lemma 9.2.6 Let T be a Coxeter set of involutions of a hypergroup, and assume that none of the elements in T is thin. Then each subnormal closed subset of hTi is normal in hTi. Proof. Let D be a subnormal closed subset of hTi. By induction, we may assume that hTi contains a normal closed subset E such that D is normal in E. Since none of the elements in T is thin, T contains a subset V such that hVi = E; cg. Theorem 9.1.7. Since E is normal in hTi, this means that hVi is normal in hTi. As a consequence, hT \ Vi normalizes hVi. Thus, by Lemma 3.1.4(ii), hVihT \ Vi = hT \ VihVi. Now Lemma 2.3.2 yields that hVihT \ Vi = hTi. Let f be an element in hTi. We have to show that D f ⊆ f D. Since f ∈ hTi and hVihT \ Vi = hTi, we have f ∈ hVihT \ Vi. Thus, there exist elements b in hVi and c in hT \ Vi such that f ∈ bc. From b ∈ hVi and c ∈ hT \ Vi

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we obtain that b ∈ T1 (c); cf. Corollary 9.2.2. Thus, as f ∈ bc and T is constrained, we obtain from Lemma 8.1.1 that bc = { f }. Since D is normal in E and b ∈ E, we have Db ⊆ bD. Since hVi is normal in hTi, we have hVi ⊆ C hT i (hT \ Vi); cf. Theorem 9.2.5. It follows that hT \ Vi ⊆ C hT i (hVi). Thus, as hVi = E, hT \ Vi ⊆ C hT i (E). Thus, as c ∈ hT \ Vi and D ⊆ E, we have c ∈ C hT i (D). As a consequence, Dc = cD. From bc = { f }, Db ⊆ bD, and Dc ⊆ cD we obtain that D f = Dbc ⊆ bDc = bcD = f D, 

as wanted. Lemma 9.2.7

Let V be a Coxeter set of involutions of a hypergroup, and let U be a subset of V. Let d be an element in V1 (U), and let e be an element in dhUi. Then the following hold. (i) We have e∗ ∈ V−1 (d ∗ ). (ii) Let v be an element in V ∩ V1 (e). Then vehUi = ehUi or vd ⊆ V1 (U). Proof. (i) We are assuming that e ∈ dhUi. Thus, hUi contains an element c with e ∈ dc. We are assuming that d ∈ V1 (U). Thus, by Theorem 9.2.1, d ∈ V1 (hUi). Since c ∈ hUi, this implies that d ∈ V1 (c). Now, as V is constrained, we have |dc| = 1. Since e ∈ dc, that implies that dc = {e} and `V (e) = `V (d) + `V (c). From e ∈ dc, we obtain that e∗ ∈ c∗ d ∗ ; cf. Lemma 1.2.1. From `V (e) = `V (d)+`V (c) we obtain that `V (e∗ ) = `V (c∗ ) + `V (d ∗ ); cf. Lemma 2.3.5. Thus, e∗ ∈ V−1 (d ∗ ). (ii) From (i) we know that e∗ ∈ V−1 (d ∗ ). On the other hand, we are assuming that v ∈ V1 (e), and that implies e∗ ∈ V1 (v); cf. Lemma 6.5.2(i). Thus, by Lemma 6.5.2(ii), d ∗ ∈ V1 (v), so that, again by Lemma 6.5.2(i), v ∈ V1 (d). Now recall that V is assumed to satisfy the exchange condition. Thus, we obtain from v ∈ V1 (d) and d ∈ V1 (U) that vd ⊆ dU

or

vd ⊆ V1 (U).

In the first case, we obtain that ve ⊆ vdhUi ⊆ dhUi, and then, by Lemma 2.1.6(i),  that vehUi = ehUi. (Notice that, by Lemma 8.1.1, |ve| = 1.) The following lemma generalizes Lemma 8.1.2(ii) under the hypothesis that the underlying set of involutions is not just constrained but even a Coxeter set. Lemma 9.2.8 Let V be a Coxeter set of involutions of a hypergroup, and let U be a subset of V. Let d and e be elements in hVi, and let f be an element in V1 (d) ∩ V1 (e). Then we have f dhUi = f ehUi if and only if dhUi = ehUi.

9.2

The Sets V1 (U) for Subsets U of Coxeter Sets V

283

Proof. It is obvious that f dhUi = f ehUi follows from dhUi = ehUi. To show the converse, we assume that f dhUi = f ehUi. There is nothing to show if f = 1. Thus, we assume that f , 1. In this case, we find elements v in V and c in hVi such that f ∈ vc and `V ( f ) = `V (c) + 1; cf. Lemma 2.3.7(i). Applying Lemma 8.1.3 to V, v, c, and f in place of T, a, b, and c we obtain an element a in cd such that cd = {a},

va = f d,

and

v ∈ V1 (a).

Applying the same lemma to V, v, c, f , and e in place of T, a, b, c, and d we obtain an element b in ce such that ce = {b},

vb = f e,

and

v ∈ V1 (b).

Now recall that we are assuming that f dhUi = f ehUi. Thus, as va = f d and vb = f e, we have vahUi = vbhUi. It follows that a ∈ bhUi

or

a ∈ vbhUi;

cf. Lemma 6.1.1(ii). Assume first that a ∈ bhUi. Then, ahUi = bhUi. Since cd = {a} and ce = {b}, this implies that cdhUi = cehUi. From f ∈ vc and `V ( f ) = `V (c) + 1 we obtain that f ∈ V−1 (c). Thus, as f ∈ V1 (d), we obtain from Lemma 6.5.2(ii) that c ∈ V1 (d). Similarly, as f ∈ V1 (e), we obtain that c ∈ V1 (e). Thus, as `V ( f ) = `V (c) + 1, induction yields dhUi = ehUi, so that we are done in this case. Assume now that a < bhUi. Then a ∈ vbhUi. From Lemma 6.7.3 (together with Theorem 6.7.5) we know that hVi = V1 (U)hUi. Thus, since b ∈ hVi, there exists an element l in V1 (U) such that b ∈ l hUi. Now recall that v ∈ V1 (b). Thus, applying Lemma 9.2.7(ii) to l and b in place of d and e we obtain that vbhUi = bhUi or vl ⊆ V1 (U). In the first case, we obtain that a ∈ bhUi, contradiction. Thus, vl ⊆ V1 (U). Since b ∈ l hUi, hUi contains an element g such that b ∈ lg. From l ∈ V1 (U) and g ∈ hUi, we obtain that l ∈ V1 (g). Thus, as V is constrained, |lg| = 1; cf. Lemma 8.1.1. Since b ∈ lg, we now have lg = {b}. Thus, as l ∈ V1 (g), we obtain from Lemma 6.5.7 that V1 (b) ⊆ V1 (l). Now, as v ∈ V1 (b), we have v ∈ V1 (l). Similarly, we find an element k in V1 (U) such that

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a ∈ k hUi,

v k ⊆ V1 (U),

and

v ∈ V1 (k).

From a ∈ k hUi and b ∈ l hUi we obtain that v k hUi = vahUi = f dhUi = f ehUi = vbhUi = vl hUi. Since v ∈ V1 (k) and v ∈ V1 (l), we have that |v k | = 1 and |vl | = 1. Thus, as v k ⊆ V1 (U) and vl ⊆ V1 (U) (and v k hUi = vl hUi), we conclude that v k = vl; cf. Corollary 9.2.4. From v ∈ V1 (k), v ∈ V1 (l), and v k = vl we obtain that k = l; cf. Lemma 8.1.2(ii). It follows that a ∈ k hUi = l hUi = bhUi, 

contradiction. The following lemma will be useful in the proof of Lemma 9.6.3. Lemma 9.2.9

Let V be a Coxeter set of involutions of a hypergroup, and let U be a subset of V. Let d be an element in V1 (U). Let x and y be elements in U, and let e be an element in hUi such that e∗ ∈ U−1 (x) ∩ U−1 (y). Assume that dxe = dye. Then x = y. Proof. We are assuming that e∗ ∈ U−1 (x). Thus, xe contains an element a such that `U (e) = `U (a) + 1. Now recall that V is assumed to be a Coxeter set of involutions of a hypergroup. Thus, U is a Coxeter set, too; cf. Theorem 9.1.5. In particular, U is constrained. Thus, as e ∈ xa and `U (e) = `U (a) + 1, we have xa = {e}; cf. Lemma 8.1.1. It follows that hxia = {a, e}, and then, by Lemma 2.1.6(i), hxie = {a, e}. Similarly, we find an element b in ye with `U (e) = `U (b) + 1 and hyie = {b, e}. From dxe = dye we obtain that dhxie = dhyie, and then da ∪ de = d{a, e} = d{b, e} = db ∪ de. Now recall that d ∈ V1 (U). Thus, by Theorem 9.2.1, d ∈ V1 (hUi). Since e ∈ hUi, this implies that d ∈ V1 (e). Thus, as V is constrained, de consists of a single element, and that element has V-length `V (d) + `V (e). Thus, as `U (e) = `U (a) + 1, da ∩ de is empty. (Here we refer to Lemma 9.1.2.) Thus, as da ∪ de = db ∪ de, we obtain that da ⊆ db. The reverse containment follows similarly, so that we have da = db. From d ∈ V1 (hUi) and a ∈ hUi we obtain that d ∈ V1 (a). Thus, da contains an element f such that `V ( f ) = `V (d) + `V (a). Since d ∈ V1 (hUi) and b ∈ hUi, we also have d ∈ V1 (b). Thus, as da = db, we have `V ( f ) = `V (d) + `V (b) and f ∈ db. Now, by Lemma 8.1.2(ii), a = b. It follows that e ∈ xa ∩ ya. Now Lemma 8.1.2(i) yields x = y. 

9.3

The Sets V−1 (U) for Subsets U of Coxeter Sets V

285

9.3 The Sets V−1 (U) for Subsets U of Coxeter Sets V In this section, we have a look at the sets V−1 (U) where V is a Coxeter set of involutions of a hypergroup and U a subset of V. Our first result is similar to Theorem 9.2.1. Theorem 9.3.1 Let V be a Coxeter set of involutions of a hypergroup, and let U be a subset of V. Then V−1 (U) = V−1 (hUi). Proof. Assume, by way of contradiction, that V−1 (U) , V−1 (hUi). Then, since V−1 (hUi) ⊆ V−1 (U), we have V−1 (U) * V−1 (hUi). Thus, V−1 (U) contains an element e such that e < V−1 (hUi). Since e < V−1 (hUi), hUi contains an element d such that e < V−1 (d). Among the elements d in hUi with e < V−1 (d) we choose d such that `V (d) is as small as possible. Since e < V−1 (d) and e ∈ V−1 (1), d , 1. Thus, as d ∈ hUi, there exist elements c in hUi and u in U such that d ∈ cu and `U (d) = `U (c) + 1; cf. Lemma 2.3.7(ii). From `U (d) = `U (c) + 1 we obtain that `V (d) = `V (c) + 1; cf. Lemma 9.1.2. Since `V (d) = `V (c) + 1 and c ∈ hUi, the minimal choice of d yields e ∈ V−1 (c). From e ∈ V−1 (U) and u ∈ U we also obtain that e ∈ V−1 (u). From d ∈ cu and `V (d) = `V (c) + 1 (together with the fact that V is constrained) we obtain that  cu = {d}. Thus, by Lemma 6.7.6, e ∈ V−1 (d), contradiction. Lemma 9.3.2 Let V be a Coxeter set of involutions of a hypergroup, and let U be a subset of V. Let d be an element in hUi, and let e be an element in V−1 (U). Then the following hold. (i) We have `V (d) ≤ `V (e). (ii) If `V (d) = `V (e), then d = e. Proof. From e ∈ V−1 (U) we obtain that e ∈ V−1 (hUi); cf. Theorem 9.3.1. Thus, as d ∈ hUi, e ∈ V−1 (d). Thus, by definition, hVi contains an element c such that e ∈ cd and `V (e) = `V (c) + `V (d). (i) From `V (e) = `V (c) + `V (d) we obtain that `V (d) ≤ `V (e). (ii) Assume that `V (e) = `V (d). Then, since `V (e) = `V (c) + `V (d), we conclude that  `V (c) = 0. It follows that c = 1. Thus, as e ∈ cd, d = e. Let T be a Coxeter set of involutions of a hypergroup. Recall from Section 6.7 that an element e of hTi is said to have maximal T-length if, for each element d in hTi, `T (d) ≤ `T (e). Corollary 9.3.3 Let T be a Coxeter set of involutions of a hypergroup. Then we have the following. (i) The set T−1 (T) contains at most one element.

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9 Coxeter Sets of Involutions

(ii) The set hTi contains at most one element having maximal T-length. Proof. (i) Let d and e be elements in T−1 (T). Applying Lemma 9.3.2(i) to T in place of U and V we obtain that `T (d) = `T (e). Thus, by Lemma 9.3.2(ii), d = e. (ii) From Theorem 6.7.5 we know that Coxeter sets are dichotomic. Thus, the claim follows from (i), together with Lemma 6.7.4.  Lemma 9.3.4 Let V be a Coxeter set of involutions of a hypergroup, let f be an element in hVi, and set U := {v ∈ V | f ∈ V−1 (v)}. Then hUi is a finite set. Proof. For each element u in U, we have f ∈ V−1 (u). Thus, f ∈ V−1 (U). It follows that f ∈ V−1 (hUi); cf. Theorem 9.3.1. Thus, for each element e in hUi, f ∈ V−1 (e). As a consequence, we have `V (e) ≤ `V ( f ) for each element e in hUi. Since `V (e) ≤ `V ( f ) for each element e in hUi, |U| ≤ `V ( f ); cf. Corollary 9.2.2. We shall now use induction to show that, for each positive integer n with n ≤ `V ( f ), hUi contains only finitely many elements of V-length n. Let n be a positive integer with n ≤ `V ( f ), and let d be an element in hUi with `V (d) = n. Then, by Lemma 2.3.7(ii), there exist elements c in hUi and u in U such that d ∈ cu and `V (d) = `V (c) + 1. Thus, as V is assumed to be constrained, cu = {d}; cf. Lemma 8.1.1. By induction, we have only finitely many choices for c.  Thus, as U is finite, hUi contains only finitely many elements of V-length n. Theorem 9.3.5 Let T be a Coxeter set of involutions of a hypergroup. Then the following statements are equivalent. (a) The set hTi is finite. (b) The set hTi contains at least one element having maximal T-length. (c) The set hTi contains exactly one element having maximal T-length. (d) The set T−1 (T) is not empty. (e) The set T−1 (T) contains exactly one element. Proof. (a) ⇒ (b) This follows from the definition of elements in hTi having maximal T-length. (b) ⇒ (d) This follows from Lemma 6.7.4, since, by Theorem 6.7.5, Coxeter sets are dichotomic. (d) ⇒ (a) Assume that T−1 (T) is not empty, and let f be an element in T−1 (T). Then f ∈ T−1 (t) for each element t in T. Thus, by Lemma 9.3.4, hTi is finite. (b) ⇔ (c) This follows from Corollary 9.3.3(ii). (d) ⇔ (e) This follows from Corollary 9.3.3(i).



9.4

Sets of Subsets of Coxeter Sets of Involutions

287

Let T be a Coxeter set of involutions of a hypergroup, and assume that hTi is finite. Then, by Lemma 6.7.4, the uniquely determined element of hTi having maximal T-length which we found in Theorem 9.3.5 is the same as the uniquely determined element in T−1 (T) which we found in Theorem 9.3.5.

9.4 Sets of Subsets of Coxeter Sets of Involutions In this section, we consider sets of subsets of Coxeter sets of involutions of hypergroups. We have results on intersections and products. Lemma 9.4.1 Let V be a Coxeter set of involutions of a hypergroup, and let U be a non-empty set of subsets of V. Then we have Ù Ù h hUi. Ui = U ∈U

U ∈U

Proof. Define T to be the intersection of the subsets belonging to U, and let F denote the intersection of the sets hUi with U ∈ U. We will show that hTi = F. From Lemma 2.3.1(i) we know that, for each element U in U, hTi ⊆ hUi. Thus, hTi ⊆ F, and it remains to be shown that F ⊆ hTi. Let f be an element in F. We will see that f ∈ hTi. If f = 1, we obtain f ∈ hTi from Lemma 2.1.1. Thus, we assume that f , 1. In this case, we find elements e in hVi and v in V such that f ∈ ev and `V ( f ) = `V (e) + 1; cf. Lemma 2.3.7(ii). It follows that f ∈ V−1 (v). Thus, as f ∈ F, v ∈ T; cf. Corollary 9.2.2. Since f ∈ ev and f ∈ F, this implies that e ∈ F. Now, as `V ( f ) = `V (e) + 1,  induction yields e ∈ hTi. Thus, as f ∈ ev and v ∈ T, f ∈ hTi. The following lemma generalizes Lemma 9.4.1. Lemma 9.4.2 Let V be a Coxeter set of involutions of a hypergroup, and let U be a non-empty set of subsets of V. Then Ù Ù fh Ui = f hUi U ∈U

U ∈U

for each element f in hVi. Proof. Let f be an element in hVi, define T to be the intersection of the subsets belonging to U, and let A denote the intersection of the cosets f hUi with U ∈ U. We will see that f hTi = A. From Lemma 2.3.1(i) we obtain that, for each element U in U, f hTi ⊆ f hUi. Thus, f hTi ⊆ A, and it remains to be shown that A ⊆ f hTi.

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9 Coxeter Sets of Involutions

Let a be an element in A. We have to show that a ∈ f hTi. If f = 1, we are done by Lemma 9.4.1. Thus, we assume that f , 1. In this case, we find elements v in V and e in hVi with f ∈ ve and `V ( f ) = `V (e) + 1; cf. Lemma 2.3.7(i). Since V is constrained, we obtain from f ∈ ve and `V ( f ) = `V (e) + 1 that ve = { f }; cf. Lemma 8.1.1. Assume first that v ∈ V1 (a). Then, as V is constrained, we obtain from Lemma 8.1.1 that |va| = 1. Thus va contains an element b such that va = {b}. Let U be an element in U. Then, as a ∈ A, we have a ∈ f hUi. Thus, by Lemma 2.1.6(i), f ∈ ahUi. Since f ∈ ve, we also have e ∈ v f . Thus, as va = {b}, we obtain that e ∈ bhUi. It follows that b ∈ ehUi; cf. Lemma 2.1.6(i). Thus, as U has been chosen arbitrarily in U and `V ( f ) = `V (e) + 1, induction yields b ∈ ehTi. From b ∈ va we obtain that a ∈ vb. Thus, as b ∈ ehTi and ve = { f }, we obtain that a ∈ f hTi, so we are done also in this case. Assume now that v < V1 (a). Then, by Lemma 6.5.2(i), a∗ < V1 (v). Thus, by Theorem 6.7.5, a∗ ∈ V−1 (v), and that means that hVi contains an element b such that a∗ = b∗ v and `V (a∗ ) = `V (b∗ ) + 1. It follows that a ∈ vb and `V (a) = `V (b) + 1; cf. Lemma 1.2.1 and Lemma 2.3.5. Now we have v ∈ V1 (b). Since f ∈ ve and `V ( f ) = `V (e) + 1, we also have v ∈ V1 (e). Since V is constrained, we obtain from v ∈ V1 (b) that |vb| = 1; cf. Lemma 8.1.1. Since a ∈ vb, that implies that vb = {a}. Thus, as ve = { f }, we have, for each element U in U, vbhUi = ahUi = f hUi = vehUi. Now we are in the position to apply Lemma 9.2.8 to v and b in place of f and d. We obtain that, for each element U in U, bhUi = ehUi. It follows that, for each element U in U, b ∈ ehUi. Thus, as `V ( f ) = `V (e) + 1, induction yields b ∈ ehTi. From a ∈ vb, b ∈ ehTi and ve = { f }, we obtain that a ∈ f hTi, so that we are done also in this case.  Corollary 9.4.3 Let V be a Coxeter set of involutions of a hypergroup, and let U be a non-empty set of subsets of V. Then Ù Ù hUi f h Ui f = U ∈U

for each element f in hVi.

U ∈U

9.4

Sets of Subsets of Coxeter Sets of Involutions

289

Proof. Let f be an element in hVi, define T to be the intersection of the subsets belonging to U, and let B denote the intersection of the sets hUi f with U ∈ U. We will see that hTi f = B. From Lemma 2.5.2(i) we know that, for each element U in U, hTi f ⊆ hUi f . Thus, hTi f ⊆ B, and it remains to be shown that B ⊆ hTi f . Let b be an element in B. Then, for each element U in U, b ∈ hUi f . Thus, by definition, f b ⊆ hUi f for each element U in U. It follows that, for each element U in U, b∗ f ∗ ⊆ f ∗ hUi; cf. Lemma 1.3.1(ii). Thus, applying Lemma 9.4.2 to f ∗ in place of f we obtain that b∗ f ∗ ⊆ f ∗ hTi. By Lemma 1.3.1(ii), this implies that f b ⊆ hTi f , so that, again by definition, b ∈ hTi f .  Recall from Corollary 8.3.3(ii) that a Coxeter set T of involutions of a hypergroup does not contain thin elements if and only if Oϑ (hTi) = {1}. We now shall see that, if a non-empty Coxeter set T of involutions of a hypergroup is finite and does not contain thin elements, then hTi is the direct product of simple closed subsets each of which is generated by the elements of T which it contains. In other words, we will see that each non-trivial hypergroup generated by a finite Coxeter set of non-thin involutions is the direct product of simple Coxeter hypergroups. Theorem 9.4.4 Let V be a finite and non-empty Coxeter set of involutions of a hypergroup, and assume that none of the elements in V is thin. Then V possesses a partition {U1, . . . , Un } such that hVi = hU1 i × . . . × hUn i and, for each element i in {1, . . . , n}, hUi i is a simple Coxeter hypergroup. Proof. Let F1 , . . ., Fn denote the (pairwise distinct) normal closed subsets of hVi which are minimal (with respect to set theoretic inclusion) among the non-trivial normal closed subsets of hTi. From Theorem 9.1.7 we know that V contains nonempty subsets U1 , . . ., Un such that, for each element i in {1, . . . , n}, hUi i = Fi . We first show that {U1, . . . Un } is a partition of V. From Lemma 9.2.6 we obtain that, for each element i in {1, . . . , n}, Fi is simple. Thus, for any two distinct elements i and j in {1, . . . , n}, we have Fi ∩ Fj = {1}; cf. Lemma 3.1.5(iii). Equivalently, we have hUi i ∩ hU j i = {1} for any two distinct elements i and j in {1, . . . , n}. Now we obtain from Corollary 9.1.8 that, for any two distinct elements i and j in {1, . . . , n}, Ui ∩ U j = ∅. To show that V = U1 ∪ . . . ∪ Un , we define T := V \ (U1 ∪ . . . ∪ Un ) and assume, by way of contradiction, that T is not empty. From Lemma 3.1.5(ii) (together with Lemma 2.3.2) we obtain that U1 ∪ . . . ∪ Un is normal in H. Thus, by Theorem 9.2.5, hTi is normal in hVi. Since T is assumed to be not empty, we also have hTi , {1}. Thus, {1, . . . , n} contains an element i with

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9 Coxeter Sets of Involutions

hUi i ⊆ hTi. Now Corollary 9.1.4 forces Ui ⊆ T, and this contradiction shows that T is empty. It follows that V = U1 ∪ . . . ∪ Un, and we have shown that {U1, . . . , Un } is a partition of V. To show that hVi = hU1 i × . . . × hUn i, we fix an element i in {1, . . . , n}. Define Fˆi to be the product of the closed subsets hU j i with j ∈ {1, . . . , n} \ {i}, and set Uˆ i := V \ Ui . Then, by Lemma 2.3.2, hUˆ i i = Fˆi . Since Ui ∩ Uˆ i is empty, we obtain from Corollary 9.1.8 that hUi i ∩ hUˆ i i = {1}. It follows that Fi ∩ Fˆi = {1}. Since Fi is normal in hVi, so is hUi i. Thus, by Theorem 9.2.5, Ui ⊆ C hV i (Uˆ i ). It follows that hUi ihUˆ i i = hVi; cf. Lemma 9.1.6(ii). Thus, Fi Fˆi = hVi. This shows that hVi = F1 × . . . × Fn ; equivalently, hVi = hU1 i × . . . × hUn i. Let i be an element in {1, . . . , n}. Then, Fi is simple; equivalently, hUi i is simple. From Theorem 9.1.5 we also know that Ui is a Coxeter set, whence hUi i is a Coxeter  hypergroup. The hypothesis in Theorem 9.4.4 that V does not contain thin elements is essential. In fact, hypergroups of type H6,1 (as defined in Section 7.3) are not direct products of simple hypergroups, although they are Coxeter hypergroups. Hypergroups of type H6,11 (as defined in Section 7.4) are examples of simple Coxeter hypergroups. Theorem 9.4.4 tells us that, in order to investigate Coxeter sets without thin elements, it is enough to look at simple Coxeter hypergroups.

9.5 Spherical Coxeter Sets of Involutions A Coxeter set T of involutions of a hypergroup is called spherical if the set hTi is finite. In Theorem 9.3.5, various necessary and sufficient conditions were given for Coxeter sets of involutions of hypergroups to be spherical. In this section, we have a closer look at the structure of hypergroups generated by a spherical Coxeter set of involutions. Clearly, the empty set is spherical. Also, each set consisting of a single involution of a hypergroup is a spherical Coxeter set. More generally, for each Coxeter set T of involutions of a hypergroup and any element f in hTi, the set {t ∈ T | f ∈ T−1 (t)} is a spherical Coxeter set. (This is an immediate consequence of Theorem 9.1.5 and Lemma 9.3.4).

9.5

Spherical Coxeter Sets of Involutions

291

Recall from Theorem 9.3.5 that, for each spherical Coxeter set T of involutions of a hypergroup, hTi contains exactly one element having maximal T-length. In the course of this section, this element will be denoted by mT , so that we have T−1 (T) = {mT } for each spherical Coxeter set T of involutions of a hypergroup. Lemma 9.5.1 Let T be a spherical Coxeter set of involutions of a hypergroup. Then mT is symmetric. Proof. Since mT is the only element in hTi having maximal T-length in hTi, this  follows from (mT )∗ ∈ hTi together with `T ((mT )∗ ) = `T (mT ). Let T be a spherical Coxeter set of involutions of a hypergroup. Then mT ∈ T−1 (T). Thus, by Theorem 9.3.1, mT ∈ T−1 ( f ) for each element f in hTi. Let e be an element in hTi. From mT ∈ T−1 (e) we obtain an element d in hTi such that mT ∈ de and `T (mT ) = `T (d) + `T (e). Since T is constrained, we obtain from Lemma 8.1.2(i) that hTi contains at most one such element. Thus, hTi contains exactly one element d with mT ∈ de and `T (mT ) = `T (d) + `T (e). In the following, this element will be denoted by e(T ) . Thus, we have mT ∈ f (T ) f

and `T (mT ) = `T ( f (T ) ) + `T ( f )

for each element f in hTi. Lemma 9.5.2 Let T be a spherical Coxeter set of involutions of a hypergroup. Then the following hold. (i) We have 1(T ) = mT . (ii) The map hTi → hTi, f 7→ f (T ) is bijective. Proof. (i) By definition, we have mT ∈ 1(T ) ·1 = {1(T ) }. (ii) Let d and e be elements in hTi, and assume that d (T ) = e(T ) . Then mT ∈ e(T ) d

and `T (mT ) = `T (e(T ) ) + `T (d).

On the other hand, we have mT ∈ e(T ) e

and `T (mT ) = `T (e(T ) ) + `T (e).

Thus, by Lemma 8.1.2(ii), d = e. Since d and e have been chosen arbitrarily in hTi, this shows that f 7→ f (T ) is an injective map from hTi to hTi. The surjectivity follows from the injectivity, together with the fact that hTi is assumed to be finite. 

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9 Coxeter Sets of Involutions

Lemma 9.5.3 Let T be a spherical Coxeter set of involutions of a hypergroup, and let f be an element in hTi. Then the following hold. (i) We have f (T ) f = {mT }. (ii) We have f f ∗(T )∗ = {mT }. (iii) We have f ∗(T )∗(T ) = f . (iv) We have f (T )∗(T )∗ = f . Proof. (i) By definition, mT ∈ f (T ) f and `T (mT ) = `T ( f (T ) ) + `T ( f ). Thus, as T is constrained, Lemma 8.1.1 yields f (T ) f = {mT }. (ii) From (i) we know that f ∗(T ) f ∗ = {mT }. Thus, by Lemma 1.3.1(ii), f f ∗(T )∗ = {(mT )∗ }. Now recall from Lemma 9.5.1 that mT is symmetric. Thus, f f ∗(T )∗ = {mT }. (iii) By definition, we have mT ∈ f ∗(T )∗(T ) f ∗(T )∗ . From (ii) we know that mT ∈ f f ∗(T )∗ . From Lemma 2.3.5 we also obtain that `T ( f ) = `T ( f ∗(T )∗(T ) ) and `T (mT ) = `T ( f ) + `T ( f ∗(T )∗ ). Thus, by Lemma 8.1.2(i), f ∗(T )∗(T ) = f . (iv) Applying (iii) to f ∗ in place of f we obtain that f (T )∗(T ) = f ∗ . Thus, by Lemma  1.1.2, f (T )∗(T )∗ = f . Lemma 9.5.4 Let T be a spherical Coxeter set of involutions of a hypergroup, and let c, d, and e be elements in hTi with e ∈ cd and `T (e) = `T (c) + `T (d). Then we have d (T ) ∈ e(T ) c and `T (d (T ) ) = `T (e(T ) ) + `T (c). Proof. By definition, we have mT ∈ e(T ) e and `T (mT ) = `T (e(T ) ) + `T (e). By hypothesis, we have e ∈ cd and `T (e) = `T (c) + `T (d). Thus, by Lemma 2.3.8(i), e(T ) c contains an element f such that mT ∈ f d, `T ( f ) = `T (e(T ) ) + `T (c), and `T (mT ) = `T ( f ) + `T (d). From mT ∈ f d and `T (mT ) = `T ( f ) + `T (d) we obtain f = d (T ) . Thus, the claim follows from f ∈ e(T ) c and `T ( f ) = `T (e(T ) ) + `T (c).  Let T be a spherical Coxeter set of involutions of a hypergroup. For each element f in hTi, we write f [T ] instead of f (T )(T ) . Thus, we have mT ∈ f [T ] f (T )

and `T (mT ) = `T ( f [T ] ) + `T ( f (T ) )

for each element f in hTi. Lemma 9.5.5 Let T be a spherical Coxeter set of involutions of a hypergroup, and let c, d, and e be elements in hTi with e ∈ cd and `T (e) = `T (c) + `T (d). Then we have e[T ] ∈ c[T ] d [T ] and `T (e[T ] ) = `T (c[T ] ) + `T (d [T ] ).

9.5

Spherical Coxeter Sets of Involutions

Proof. This follows from Lemma 9.5.4. (Apply this lemma three times.)

293



Lemma 9.5.6 Let T be a spherical Coxeter set of involutions of a hypergroup, and let f be an element in hTi. Then the following hold. (i) We have `T ( f ) = `T ( f [T ] ). (ii) We have f = f [T ][T ] . (iii) We have mT f = f [T ] mT . (iv) The element f is thin if and only if f [T ] is thin. Proof. (i) We have `T (mT ) = `T ( f (T ) ) + `T ( f ) and `T (mT ) = `T ( f [T ] ) + `T ( f (T ) ). Thus, `T ( f ) = `T ( f [T ] ). (ii) By definition, mT ∈ f [T ](T ) f [T ] , and, by Lemma 9.5.1, (mT )∗ = mT . Thus, mT ∈ f [T ]∗ f [T ](T )∗ ; cf. Lemma 1.2.1. Since mT ∈ f (T ) f , this implies that f (T ) f ∩ f [T ]∗ f [T ](T )∗ is not empty. Thus, by Lemma 1.2.5(i), f (T )∗ f [T ]∗ ∩ f f [T ](T ) is not empty. On the other hand, by Lemma 9.5.3(i), we have f [T ] f (T ) = {mT }. It follows that f (T )∗ f [T ]∗ = {mT }; cf. Lemma 1.2.2. Thus, mT ∈ f f [T ](T ) . Now recall from (i) that `T ( f [T ][T ] ) = `T ( f ). Thus, by definition, f = f [T ][T ] . (iii) Referring two times to Lemma 9.5.3(i) we obtain that mT f = f [T ] f (T ) f = f [T ] mT . (iv) By (ii), it suffices to show that f [T ] is thin if f is thin. Thus, we assume that f is thin, and we will show that f [T ] is thin. Since f is thin and f ∈ hTi, f ∈ Oϑ (hTi). Thus, by Corollary 8.3.3(i), f ∈ hOϑ (T)i. If f = 1, we obtain from (i) that f [T ] = 1. In particular, f [T ] is thin, and we are done in this case. Thus, we assume that f , 1. In this case, Lemma 2.3.7(i) provides elements t in Oϑ (T) and e in hOϑ (T)i such that f ∈ te and `T ( f ) = `T (e) + 1. From f ∈ te and `T ( f ) = `T (e) + 1 we obtain that f [T ] ∈ t [T ] e[T ] ; cf. Lemma 9.5.5. Since `T ( f ) = `T (e) + 1, induction yields that e[T ] is thin. Thus, by Lemma 1.4.3(iii), it suffices to show that t [T ] is thin. Since t ∈ Oϑ (T), Lemma 1.4.3(i) yields that |mT t| = 1. Thus, by (iii), |t [T ] mT | = 1. Since t [T ] and mT both are symmetric, this implies that |mT t [T ] | = 1. Assume, by way of contradiction, that t [T ] is not thin, and recall from Lemma 9.5.3(i) that t [T ](T ) t [T ] = {mT }. Then, by Lemma 6.1.3(ii), mT t [T ] = {t [T ](T ), mT }.

294

9 Coxeter Sets of Involutions

Since t [T ](T ) , mT , this implies that |mT t [T ] | = 2, and this contradiction shows that t [T ] is thin.  Lemma 9.5.7 Let T be a spherical Coxeter set of involutions of a hypergroup, and let f be an element in hTi. Then the following hold. (i) We have f [T ]∗(T ) = f (T )∗ . (ii) We have f [T ](T )∗ = f ∗(T ) . (iii) We have f [T ]∗ = f ∗[T ] . Proof. (i) Applying Lemma 9.5.3(iv) to f (T ) in place of f , we obtain that f [T ]∗(T )∗ = f (T ) . Thus, by Lemma 1.1.2, f [T ]∗(T ) = f (T )∗ . (ii) From Lemma 9.5.6(ii) we know that f = f [T ][T ] . Thus, applying (i) to f [T ] in place of f , we obtain that f [T ](T )∗ = f ∗(T ) . (iii) From (i) we obtain that f [T ]∗[T ] = f (T )∗(T ) , and from Lemma 9.5.3(iv) we obtain that f (T )∗(T ) = f ∗ . Thus, f [T ]∗[T ] = f ∗ , so that the desired equation follows from  Lemma 9.5.6(ii). Lemma 9.5.8 Let T be a spherical Coxeter set of involutions of a hypergroup. Then the map ρ : hTi → hTi, f 7→ f (T ) is a folding. Proof. In Lemma 9.5.2(ii) we saw that f 7→ f ρ is a bijective map from hTi to hTi. Let f be an element in hTi. Then, by Lemma 9.5.3(i), f (T ) f = {mT }. From Lemma 9.5.2(i) we also know that 1(T ) = mT . Thus, we have f (T ) f = {1(T ) }, and, since f (T ) = f ρ and 1(T ) = 1ρ , we conclude that f ρ f = {1ρ }.  In Lemma 9.5.8, we saw that a Coxeter hypergroup H admits a folding from H to H if it is finite. In Theorem 9.7.5, we will see that the converse is true for Coxeter hypergroups H with Oϑ (H) = {1}. Let T be a spherical Coxeter set of involutions of a hypergroup. The bijective folding from hTi to hTi which we found in Lemma 9.5.8 will be called the spherical folding on hTi. Referring to this terminology we obtain from Lemma 9.5.8 that spherical foldings are bijective. As a consequence, one obtains that the image of a spherical folding is ∗ -invariant, so that each spherical folding induces a permutation on hTi. (Permutations induced by foldings were introduced in Section 2.7.) We shall now look at permutations induced by spherical foldings. Lemma 9.5.9 Let T be a spherical Coxeter set of involutions of a hypergroup, let ρ denote the spherical folding on hTi, and let π denote the permutation induced by ρ. Then the following hold.

9.6

Subsets of Spherical Coxeter Sets of Involutions

295

(i) For each element f in hTi, we have f π = f [T ] . (ii) The map π is a T-length preserving hypergroup automorphism of hTi. (iii) We have π 2 = 1 hT i . Proof. (i) Let f be an element in hTi. Then, by Lemma 9.5.7(ii), f πρ = f ∗ρ∗ = f ∗(T )∗ = f [T ](T ) = f [T ]ρ, so that the desired equation follows from the injectivity of ρ. (ii) By definition, π is a bijective map from hTi to hTi, and from Lemma 9.5.6(i) (together with (i)) we obtain that π is T-length preserving. From Lemma 9.5.6(iv) we know that an element t in T is thin if and only if t π is thin, and from Lemma 9.5.5 (together with (i)) we obtain that e π ∈ c π d π for any three elements c, d, and e in hTi with e ∈ cd and `T (e) = `T (c) + `T (d). Now we obtain from Theorem 8.3.4 that π is a hypergroup automorphism of hTi. (iii) Considering (i) this follows from Lemma 9.5.6(ii).



9.6 Subsets of Spherical Coxeter Sets of Involutions In this section, we look at subsets of spherical Coxeter sets of involutions of hypergroups. From Theorem 9.1.5 we know that each subset of a spherical Coxeter set of involutions of a hypergroup is a Coxeter set, from Lemma 2.3.1(i) we obtain that each such subset is spherical. Following our notation introduced in the previous section, we denote, for each spherical Coxeter set T of involutions of a hypergroup, by mT the uniquely determined element in hTi which has maximal T-length. Lemma 9.6.1 Let V be a spherical Coxeter set of involutions of a hypergroup, and let U be a subset of V. Then the following hold. (i) We have V1 (mU ) = V1 (hUi). (ii) We have (mU )(V ) ∈ V1 (hUi). (iii) We have (mU )(V ) ∈ mV hUi. Proof. (i) By Lemma 6.7.4, mU ∈ U−1 (U). Thus, by Corollary 9.1.3(ii), mU ∈ V−1 (U). On the other hand, by Lemma 9.5.1, (mU )∗ = mU . Thus, as U is symmetric, we obtain from Lemma 6.5.2(iv) that V1 (mU ) ⊆ V1 (U). It follows that V1 (mU ) ⊆ V1 (hUi); cf. Theorem 9.2.1. The reverse containment follows from mU ∈ hUi. (ii) This follows from (i), since (mU )(V ) ∈ V1 (mU ).

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9 Coxeter Sets of Involutions

(iii) By definition, mV ∈ (mU )(V ) mU . Thus, as mU ∈ hUi, mV ∈ (mU )(V ) hUi, so that the claim follows from Lemma 2.1.6(i).  Let T be a spherical Coxeter set of involutions of a hypergroup. For each subset A of hTi, we define A[T ] to be the set of all elements a[T ] with a ∈ A. Lemma 9.6.2 Let V be a spherical Coxeter set of involutions of a hypergroup, and let U be a subset of V. Then the following hold. (i) We have hUi [V ] = hU [V ] i. (ii) We have (mU )[V ] = mU [V ] . (iii) We have (mU )(V )∗ = (mU [V ] )(V ) . Proof. (i) Let e be an element in hUi. We first show that e[V ] ∈ hU [V ] i. If e = 1, e[V ] = 1; cf. Lemma 9.5.6(i). Thus, in this case, e[V ] ∈ hU [V ] i. Assume that e , 1. Then, by Lemma 2.3.7(ii), there exist elements d in hUi and u in U such that e ∈ du and `U (e) = `U (d) + 1. From `U (e) = `U (d) + 1 we obtain that `V (e) = `V (d) + 1; cf. Lemma 9.1.2. Thus, by Lemma 9.5.5, e[V ] ∈ d [V ] u[V ] . Since d ∈ hUi and `V (e) = `V (d) + 1, induction yields d [V ] ∈ hU [V ] i. By definition, we also have u[V ] ∈ U [V ] . Thus, as e[V ] ∈ d [V ] u[V ] , we conclude that e[V ] ∈ hU [V ] i. Since e has been chosen arbitrarily in hUi, we have seen that hUi [V ] ⊆ hU [V ] i. Similarly, one shows that hU [V ] i [V ] ⊆ hU [V ][V ] i, and from this one obtain that hU [V ] i [V ][V ] ⊆ hU [V ][V ] i [V ] . Now Lemma 9.5.6(ii) yields hU [V ] i ⊆ hUi [V ] . (ii) Let e be an element in hU [V ] i. We will see that `U [V ] (e) ≤ `U [V ] ((mU )[V ] ). Since e ∈ hU [V ] i, e ∈ hUi [V ] ; cf. (i). Thus, hUi contains an element d such that e = d [V ] . From d ∈ hUi we obtain that `U (d) ≤ `U (mU ), and then, by Lemma 9.1.2, that `V (d) ≤ `V (mU ). Thus, by Lemma 9.5.6(i), `V (d [V ] ) ≤ `V ((mU )[V ] ). Since e = d [V ] , this yields `V (e) ≤ `V ((mU )[V ] ). Thus, by Lemma 9.1.2, `U [V ] (e) ≤ `U [V ] ((mU )[V ] ). Since e has been chosen arbitrily from hU [V ] i, the definition of mU [V ] now forces (mU )[V ] = mU [V ] . (iii) From (ii) we obtain that (mU )[V ] (mU )(V ) = mU [V ] (mU )(V ) . Thus, as mV ∈ (mU )[V ] (mU )(V ) , we have mV ∈ mU [V ] (mU )(V ) . Now recall from Lemma 9.5.1 that mV and mU [V ] are symmetric. Thus, by Lemma 1.2.1, mV ∈ (mU )(V )∗ mU [V ] . On the other hand, with a reference to Lemma 9.5.6(i) we obtain from (ii) that `V (mU ) = `V ((mU )[V ] ) = `V (mU [V ] ),

9.6

and that implies that

Subsets of Spherical Coxeter Sets of Involutions

297

`V ((mU )(V ) ) = `V ((mU [V ] )(V ) ).

Now recall from Lemma 2.3.5 that `V ((mU )(V )∗ ) = `V ((mU )(V ) ), so that `V ((mU )(V )∗ ) = `V ((mU [V ] )(V ) ). Thus, as mV ∈ (mU )(V )∗ mU [V ] , we obtain that (mU )(V )∗ = (mU [V ] )(V ) , as wanted.  Lemma 9.6.3 Let V be a spherical Coxeter set of involutions of a hypergroup, and let U be a subset of V. Let f be an element in hUi. Then the following hold. (i) We have (mU )(V ) f [U] = f [V ] (mU )(V ) . (ii) We have f [U][V ] = f [V ][U

[V ] ]

.

Proof. (i) Assume first that f = 1. Then, by Lemma 9.5.6(i), f [U] = 1 and f [V ] = 1. Thus, (mU )(V ) f [U] = {(mU )(V ) } = f [V ] (mU )(V ), so we are done in this case. Assume now that f , 1. In this case, we obtain from Lemma 2.3.7(ii) elements e in hUi and u in U such that f ∈ eu

and `U ( f ) = `U (e) + 1.

Our first goal is to show that (mU )(V ) u[U] = u[V ] (mU )(V ) . From Lemma 9.6.1(ii) we know that (mU )(V ) ∈ V1 (hUi). Thus, as u[U] ∈ hUi, (mU )(V ) ∈ V1 (u[U] ). Since V is constrained, this implies that |(mU )(V ) u[U] | = 1; cf. Lemma 8.1.1. Thus, (mU )(V ) u[U] contains an element c with (mU )(V ) u[U] = {c}. From Lemma 9.6.1(ii) (together with Lemma 9.6.2(i)) we obtain that (mU [V ] )(V ) ∈ V1 (hUi [V ] ). Thus, as u[V ] ∈ hUi [V ] , (mU [V ] )(V ) ∈ V1 (u[V ] ). It follows that u[V ] ∈ V1 ((mU )(V ) ); cf. Lemma 6.5.2(i) and Lemma 9.6.2(iii). Since V is constrained, this implies that |u[V ] (mU )(V ) | = 1; cf. Lemma 8.1.1. Thus, u[V ] (mU )(V ) contains an element d with u[V ] (mU )(V ) = {d}. From Lemma 9.5.6(iii) we know that mU u = u[U] mU and that mV u = u[V ] mV . Thus,

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9 Coxeter Sets of Involutions

(mU )(V ) u[U] mU = (mU )(V ) mU u = mV u = u[V ] mV = u[V ] (mU )(V ) mU . It follows that cmU = dmU , so d ∈ chUi. Since c ∈ (mU )(V ) hUi, this implies that d ∈ (mU )(V ) hUi. Thus, hUi contains an element b such that d ∈ (mU )(V ) b. From Lemma 9.6.1(ii) we know that (mU )(V ) ∈ V1 (hUi). Thus, as b ∈ hUi, (mU )(V ) ∈ V1 (b). Since V is constrained, this implies that |(mU )(V ) b| = 1; cf. Lemma 8.1.1. Thus, as d ∈ (mU )(V ) b, we have (mU )(V ) b = {d} From u[V ] ∈ V1 ((mU )(V ) ) together with u[V ] (mU )(V ) = {d} we obtain that `V (d) = `V ((mU )(V ) ) + 1. Similarly, since (mU )(V ) ∈ V1 (b) and (mU )(V ) b = {d}, `V (d) = `V ((mU )(V ) ) + `V (b). Thus, `V (b) = 1. It follows that b ∈ V. Thus, as b ∈ hUi, Corollary 9.1.4 forces b ∈ U. On the other hand, since (mU )(V ) u[U] = {c} and (mU )(V ) b = {d}, we have (mU )(V ) u[U] mU = cmU = dmU = (mU )(V ) bmU . Thus, applying Lemma 9.2.9 to (mU )(V ) , u[U] , b, and mU in place of d, x, y, and e we conclude that u[U] = b. It follows that (mU )(V ) u[U] = (mU )(V ) b = {d} = u[V ] (mU )(V ), and we have achieved our first goal. Now recall that f ∈ eu and `U ( f ) = `U (e) + 1. Thus, by Lemma 9.5.5, f [U] ∈ e[U] u[U] Thus, by Lemma 8.1.1, Similarly,

and `U ( f [U] ) = `U (e[U] ) + 1.

e[U] u[U] = { f [U] }. e[V ] u[V ] = { f [V ] }.

Since `U ( f ) = `U (e) + 1, induction yields that (mU )(V ) e[U] = e[V ] (mU )(V ) . Thus, as (mU )(V ) u[U] = u[V ] (mU )(V ) , we obtain that

9.6

Subsets of Spherical Coxeter Sets of Involutions

299

(mU )(V ) e[U] u[U] = e[V ] (mU )(V ) u[U] = e[V ] u[V ] (mU )(V ) . Thus, as e[U] u[U] = { f [U] } and e[V ] u[V ] = { f [V ] }, we obtain that (mU )(V ) f [U] = f [V ] (mU )(V ), as wanted. (ii) Assume first that f = 1. Then, by Lemma 9.5.6(i), f [U][V ] = 1 and f [V ][U so we are done in this case.

[V ] ]

= 1,

Assume now that f , 1. In this case, we obtain from Lemma 2.3.7(ii) elements e in hUi and u in U such that and `U ( f ) = `U (e) + 1.

f ∈ eu

Our first goal is to show that u[U][V ] = u[V ][U

[V ] ]

.

From Lemma 9.6.2(ii) (together with Lemma 9.5.9(ii)) we obtain that mU [V ] = (mU )[V ] ∈ (u[U] u(U) )[V ] = u[U][V ] u(U)[V ] . Thus, as `U [V ] (u[U][V ] ) = 1 = `U [V ] (u(U)[V ](U

[V ] )

), u[U][V ] = u(U)[V ](U

[V ] )

. Similarly,

mU [V ] = (mU )[V ] ∈ (u(U) u)[V ] = u(U)[V ] u[V ] . Thus, as `U [V ] (u(U)[V ] ) = `U [V ] (mU [V ] ) − 1 = `U [V ] (u[V ](U From

u[U][V ]

=

[V ] u(U)[V ](U )

u(U)[V ]

and

u[U][V ] = u[V ](U

=

[V ] u[V ](U )

[V ] )(U [V ] )

[V ] )

), u(U)[V ] = u[V ](U

we obtain that

= u[V ][U

[V ] ]

,

and we have achieved our first goal. Now recall that f ∈ eu and `U ( f ) = `U (e) + 1. Thus, by Lemma 9.5.5, f [U] ∈ e[U] u[U] Thus, by Lemma 8.1.1,

and `U ( f [U] ) = `U (e[U] ) + 1.

e[U] u[U] = { f [U] }.

Similarly,

e[V ] u[V ] = { f [V ] }.

Since `U ( f ) = `U (e) + 1, induction yields that e[U][V ] = e[V ][U Thus, as u[U][V ] = u[V ][U

[V ] ]

[V ] ]

.

, we obtain that

(e[U] u[U ])[V ] = e[U][V ] u[U][V ] = e[V ][U

[V ] ]

u[V ][U

[V ] ]

= (e[V ] u[V ] )[U

[V ] ]

.

[V ] )

.

300

9 Coxeter Sets of Involutions

(The first and the last equality follow from Lemma 9.5.9(i), (ii).) Thus, since e[U] u[U] = { f [U] } and e[V ] u[V ] = { f [V ] }, we obtain that f [U][V ] = f [V ][U

[V ] ]

, 

as wanted. Lemma 9.6.4

Let V be a spherical Coxeter set of involutions of a hypergroup, and let U be a subset of V. Then the following hold. (i) We have (mU )(V ) hUi = hU [V ] i(mU )(V ) . (V )

(ii) We have hUi ⊆ hU [V ] i (mU ) . Proof. (i) From Lemma 9.6.2(i) we know that hUi [V ] = hU [V ] i. Thus, the desired equation follows from Lemma 9.6.3(i). (ii) Considering (i) this follows from Lemma 2.5.2(ii).



9.7 Coxeter Sets of Involutions and Foldings In this section, we consider foldings the domains of which are generated by Coxeter sets of involutions. We assume that none of the elements of these Coxeter sets are thin. In other words, we consider foldings the domains of which are Coxeter hypergroups with a thin radical equal to {1}. The first two results of this section will play a crucial role in the proof of Theorem 10.6.1. The proofs of the results of this section depend on results that we have obtained in Sections 8.6 and 8.7. Lemma 9.7.1 Let H be a hypergroup, let T be a Coxeter set of involutions of H, and assume that none of the elements in T is thin. Let ρ be a folding from hTi to H satisfying 1ρ∗ = 1ρ and hTi ρ∗ = hTi ρ . Then the permutation of hTi induced by ρ is T-length preserving. Proof. Let π denote the permutation of hTi induced by ρ. From Lemma 8.7.3(i) we know that π 2 = 1 hT i . Thus, it suffices to show that, for each element f in hTi, `T ( f π ) ≤ `T ( f ). Let f be an element in hTi. If f = 1, we obtain from 1ρ∗ = 1ρ that f π = 1. Thus, `T ( f π ) = `T ( f ), and we are done in this case. Assume that f , 1, and set

e := f π .

Since π is injective and 1π = 1, we have e , 1. Thus, we find elements c in hTi and v in T such that e ∈ cv and `T (e) = `T (c) + 1;

9.7

Coxeter Sets of Involutions and Foldings

301

cf. Lemma 2.3.7(ii). By hypothesis, none of the elements in T is thin. Thus, as v ∈ T, v is not thin, so that, by Corollary 8.6.3(ii), c∗ρ ∈ c∗ρ v. Since 1ρ ∈ c∗ρ c∗ , we also have c∗ρ ∈ 1ρ c. Thus, c∗ρ ∈ 1ρ cv. Now recall that T is assumed to be a Coxeter set. In particular, T is constrained. Thus, as e ∈ cv and `T (e) = `T (c) + 1, we have cv = {e}; cf. Lemma 8.1.1. Since c∗ρ ∈ 1ρ cv, this implies that c∗ρ ∈ 1ρ e. From Lemma 2.7.5 we also know that f ·1ρ = 1ρ e. Thus, c∗ρ ∈ f ·1ρ . Set

d := c π .

Then d ρ∗ = c∗ρ . Thus, as c∗ρ ∈ f ·1ρ , we conclude that d ρ∗ ∈ f ·1ρ . It follows that d ρ ∈ 1ρ f ∗ ; cf. Lemma 1.2.1. Applying Lemma 8.7.2 to d ∗ and f ∗ in place of d and e we now obtain that `T (d ∗ ) ≤ `T ( f ∗ ) − 1

or

d∗ = f ∗.

Assume that d ∗ = f ∗ . Then, as π 2 = 1 hT i , c π = d = f = e π . Since π is injective, this implies that c = e, contrary to `T (e) = `T (c) + 1. Thus, the above dichotomy forces `T (d ∗ ) ≤ `T ( f ∗ ) − 1. It follows that `T (d) ≤ `T ( f ) − 1, so that, by induction, `T (d π ) ≤ `T (d). Thus, as d π = c, `T (c) ≤ `T (d). Now recall that `T (e) = `T (c) + 1 and `T (d) ≤ `T ( f ) − 1. Thus, we conclude that `T (e) ≤ `T ( f ). Since e = f π , this implies that `T ( f π ) ≤ `T ( f ), as wanted.  Lemma 9.7.2 Let H be a hypergroup, let T be a Coxeter set of involutions of H, and assume that none of the elements in T is thin. Let ρ be a folding from hTi to H satisfying 1ρ∗ = 1ρ and hTi ρ∗ = hTi ρ . Then the permutation of hTi induced by ρ is a hypergroup automorphism of hTi. Proof. Let π denote the permutation of hTi induced by ρ. The claim is that π is a hypergroup automorphism of hTi. In Lemma 9.7.1 we saw that π is T-length preserving. Furthermore, since none of the elements in T is thin, we have Oϑ (T)π = Oϑ (T). Let c, d, and e be elements in hTi with e ∈ cd and `T (e) = `T (c) + `T (d). We will see that e π ∈ c π d π . From e ∈ cd and `T (e) = `T (c) + `T (d) (together with the fact that T is constrained) we obtain that cd = {e}; cf. Lemma 8.1.1. Thus, by Lemma 2.7.5, 1ρ e π = e·1ρ = cd ·1ρ = c·1ρ d π = 1ρ c π d π . On the other hand, since 1ρ ∈ e π∗ρ e π∗ , we also have e π∗ρ ∈ 1ρ e π . Thus, e π∗ρ ∈ 1ρ c π d π . Now c π d π contains an element f such that e π∗ρ ∈ 1ρ f . Thus, applying Lemma 8.7.2 to e π and f in place of d and e, we obtain that

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9 Coxeter Sets of Involutions

`T (e π ) ≤ `T ( f ) − 1 or

eπ = f .

Now recall that `T (e) = `T (c) + `T (d) and that π is T-length preserving. Thus, we have `T (e π ) = `T (c π ) + `T (d π ). Since f ∈ c π d π , on the other hand, we also have `T ( f ) ≤ `T (c π ) + `T (d π ). Thus, `T ( f ) ≤ `T (e π ), so that, by the above dichotomy, e π = f . From e π = f and f ∈ c π d π we obtain that e π ∈ c π d π . So far we have seen that π satisfies all conditions required from υ in Theorem 8.3.4. It follows that π is a hypergroup automorphism of hTi.  Recall from Section 2.7 that a hypergroup H is said to be twinned over a closed subset F of H if there exists a folding ρ from F to H satisfying 1ρ∗ = 1ρ,

1ρ∗ 1ρ ⊆ F,

and

H = F ∪ Fρ.

Let H be a hypergroup, and let T be a set of involutions of H. We say that H is a twin Coxeter hypergroup over T if T is a Coxeter set and H is twinned over hTi. By a twin Coxeter hypergroup we mean a hypergroup H which is a twin Coxeter hypergroup over a set of involutions of H. Lemma 9.7.3 Let H be a hypergroup, let T be a set of involutions of H, and assume that H is a twin Coxeter hypergroup over T. Assume that none of the elements in T is thin. Then there exists exactly one folding ρ from hTi to H satisfying H = hTi ∪ hTi ρ . Proof. We are assuming that H is a twin Coxeter hypergroup over T. This means that T is a Coxeter set of involutions of H and H is twinned over hTi. Since H is twinned over hTi, there exists a folding ρ from hTi to H satisfying H = hTi ∪ hTi ρ . Since T is a Coxeter set of involutions of H and none of the elements in T is thin, there is at most one folding ρ from hTi to H satisfying  H = hTi ∪ hTi ρ ; cf. Lemma 8.6.7. Let H be a hypergroup, let T be a set of involutions of H, and assume that H is a twin Coxeter hypergroup over T. Assume that none of the elements in T is thin. The folding which we found in Lemma 9.7.3 will be called the canonical folding defined by H and T. Canonical foldings defined by twin Coxeter hypergroups H and sets of non-thin involutions of H will play a role in Sections 10.6 and 10.7. Lemma 9.7.4 Let H and H 0 be hypergroups, let T be a set of involutions of H, let T 0 be a set of involutions of H 0, assume that H is a twin Coxeter hypergroup over T and that H 0 is a twin Coxeter hypergroup over T 0. Assume that none of the elements in T or T 0 is thin. Let ρ denote the canonical folding defined by H and T, and let ρ0 denote the canonical folding defined by H 0 and T 0. Let φ be a hypergroup isomorphism from H to H 0 with T φ = T 0. Then the following hold.

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Coxeter Sets of Involutions and Foldings

303

(i) For each element f in hTi, we have f φρ = f ρφ . 0

(ii) Let π denote the permutation of hTi induced by ρ, and let π 0 denote the permutation of hT 0i induced by ρ0. Then φπ 0 = πφ. Proof. (i) For each element f in hTi, define f φρ := f ρφ . Applying Lemma 3.6.8(i) to hTi in place of F we obtain that ρ00 is a folding from hT 0i to H 0. Note also that 00 H 0 = hT 0i ∪ hT 0i ρ . Thus, we have ρ0 = ρ00, since ρ0 stands for the canonical folding 0 defined by H 0 and T 0. It follows that f φρ = f ρφ for each element f in hTi. 00

(ii) Considering (i) this follows from Lemma 3.6.8(ii).



Theorem 9.7.5 Let H be a Coxeter hypergroup, and assume that Oϑ (H) = {1}. Then the following statements are equivalent. (a) There exists exactly one folding from H to H. (b) There exists a folding from H to H. (c) The set H is finite. Proof. (a) ⇒ (b) Here is nothing to show. (b) ⇒ (c) Since H is a Coxeter hypergroup, H is generated by a Coxeter set T. Since we are assuming that Oϑ (H) = {1}, T does not contain thin elements. Let ρ be a folding from H to H. Then ρ is a folding from hTi to hTi. Let t be an element in T with 1ρ ∈ T1 (t). Since t ∈ T and T does not contain thin elements, t is not thin. Thus, by Lemma 6.1.1(iv), t 2 = {1, t}. Thus, as t ρ t = {1ρ }, 1ρ t = t ρ t 2 = {t ρ } ∪ t ρ t = {t ρ, 1ρ }. On the other hand, as T is assumed to be a Coxeter set, T is constrained. Thus, as 1ρ ∈ T1 (t), Lemma 8.1.1 yields |1ρ t| = 1. Since 1ρ t = {t ρ, 1ρ }, this implies that t ρ = 1ρ . Thus, as ρ is injective, t = 1, contrary to t ∈ T. This contradiction shows that, for each element t in T, 1ρ < T1 (t). Now recall that T is a Coxeter set. Thus, T satisfies the exchange condition, so that, by Theorem 6.7.5, T is dichotomic. Thus, we have 1ρ ∈ T−1 (t) for each element t in T, and that means that that 1ρ ∈ T−1 (T). Now, by Theorem 9.3.5, hTi is a finite set. Thus, as H = hTi, H is a finite set. (c) ⇒ (a) From Lemma 9.5.8 we know that there exists a folding from hTi to hTi, the spherical folding. From Proposition 8.6.4 one obtains that each folding from hTi to hTi is surjective. Thus, considering the fact that T does not contain thin elements, we obtain from Lemma 8.6.5(ii) that there do not exist two distinct foldings from hTi to hTi.  As a consequence of Theorem 9.7.5 one obtains that a Coxeter hypergroup which is twinned over itself is finite if its thin radical is trivial.

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9 Coxeter Sets of Involutions

9.8 Coxeter Sets and Their Coxeter Numbers In this section, we work with the free monoid over sets of involutions of hypergroups. In the first two results of this section, the sets of involutions under consideration are not subject to any restrictions. In the remaining four results, they need to be Coxeter sets. All results obtained in this section will be used in Section 9.9, but not later. Let T be a set of involutions of a hypergroup. By F(T) we denote the free monoid over T. The operation of F(T) will be denoted by ∗, the neutral element by 1. Let t1 , . . ., tn be elements in T. If f denotes the product t1 ∗ · · · ∗ tn , we write f ∗ for the product tn ∗ · · · ∗ t1 . Since F(T) is a free monoid, there exists exactly one monoid homomorphism from F(T) to the additive monoid of the non-negative integers which sends each element of T to 1. This monoid homomorphism will be denoted by λT . (Note that λT (1) = 0.) Since F(T) is a free monoid, there exists exactly one monoid homomorphism from F(T) to the monoid of all non-empty subsets of hTi which sends each element t of T to {t}. This monoid homomorphism will be denoted by ρT . (Note that ρT (1) = {1}.) The first result in this section establishes a relationship between the monoid homomorphisms λT and ρT . Lemma 9.8.1 Let T be a set of involutions of a hypergroup, and let f be an element in hTi. Then the following hold. (i) Let f be an element in F(T) with f ∈ ρT (f). Then `T ( f ) ≤ λT (f). (ii) There exists an element f in F(T) such that f ∈ ρT (f) and `T ( f ) = λT (f). Proof. (i) Assume first that f = 1. Then λT (f) = 0 and ρT (f) = {1}. From f ∈ ρT (f) and ρT (f) = {1} we obtain that f = 1, and from f = 1 we obtain that `T ( f ) = 0. Thus, as λT (f) = 0, we have `T ( f ) = λT (f). Assume now that f , 1. Then there exist elements e in F(T) and t in T with f = e ∗ t. Thus, as f ∈ ρT (f), f ∈ ρT (e ∗ t) = ρT (e)t. Now ρT (e) contains an element e such that f ∈ et. Since e ∈ ρT (e), induction yields `T (e) ≤ λT (e). From f ∈ et we also obtain that `T ( f ) ≤ `T (e) + 1; cf. Lemma 2.3.6(i). Thus, `T ( f ) ≤ λT (e) + 1 = λT (e ∗ t) = λT (f). (ii) Assume first that f = 1. Then f ∈ ρT (1) and `T ( f ) = λT (1). Assume now that f , 1. Then, by Lemma 2.3.7(ii), there exist elements e in hTi and t in T such that f ∈ et and `T ( f ) = `T (e) + 1. By induction, F(T) contains an element e such that e ∈ ρT (e) and `T (e) = λT (e). Set f := e ∗ t. Then, as f ∈ et and e ∈ ρT (e), f ∈ ρT (e)t = ρT (e ∗ t) = ρT (f).

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Coxeter Sets and Their Coxeter Numbers

305

From `T ( f ) = `T (e) + 1 and `T (e) = λT (e) we also obtain that `T ( f ) = λT (e) + 1 = λT (e ∗ t) = λT (f), 

as wanted. Lemma 9.8.2

Let T be a set of involutions of a hypergroup, let f be an element in hTi, and let f be an element in F(T) with f ∈ ρT (f) and `T ( f ) = λT (f). Let e be an element in F(T), and let t be an element in T such that f = e ∗ t. Then ρT (e) contains an element e with f ∈ et and `T ( f ) = `T (e) + 1. Proof. From f = e ∗ t and f ∈ ρT (f) we obtain that f ∈ ρT (e ∗ t) = ρT (e)t. Thus, ρT (e) contains an element e such that f ∈ et. Since f ∈ et, we have `T ( f ) ≤ `T (e) + 1; cf. Lemma 2.3.6(ii). From e ∈ ρT (e), on the other hand, we obtain that `T (e) ≤ λT (e); cf. Lemma 9.8.1(i). Moreover, since f = e ∗ t, we have λT (f) = λT (e) + 1. Thus, as `T ( f ) = λT (f), we  conclude that `T (e) + 1 ≤ `T ( f ). Let T be a set of involutions of a hypergroup, and let u and v be two distinct elements of T. For each positive integer n, we define fn (u, v) := t1 ∗ · · · ∗ tn, where, for each element i in {1, . . . , n}, ti = u if i is odd and ti = v if i is even. Notice that, for each positive integer n, fn (u, v) ∈ F(T) and recall from Section 6.6 that, for each positive integer n, Rn (u, v) is our notation for the product t1 · · · tn , where, for each element i in {1, . . . , n}, ti = u if i is odd and ti = v if i is even. Thus, we have ρT (fn (u, v)) = Rn (u, v) for each positive integer n. Recall from Section 6.6 that C(u, v) stands for the set of all positive integers n with 1 ∈ Rn (u, v) and cu,v for the Coxeter number of u and v. Let d and e be elements in F(T). The elements d and e are called elementary T-homotopic if they are equal or if there exist elements a, b in F(T) and u, v in T such that u , v, C(u, v) is not empty, d = a ∗ fn (u, v) ∗ b,

and

e = a ∗ fn (v, u) ∗ b,

where n := cu,v . The elements d and e are called T-homotopic if F(T) contains elements f0 , . . ., fn such that f0 = d, fn = e, and, for each element i in {1, . . . , n}, the elements fi−1 and fi are elementary T-homotopic.

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9 Coxeter Sets of Involutions

Note that being T-homotopic is an equivalence relation on F(T). The equivalence classes of this equivalence relation will be called T-homotopy classes of F(T). Let f be an element of F(T). We notice that d ∗ f and e ∗ f are T-homotopic if d and e are T-homotopic. Similarly, f ∗ d and f ∗ e are T-homotopic if d and e are T-homotopic. For the remainder of this section, the sets of involutions which we consider will always be Coxeter sets. Lemma 9.8.3 Let T be a Coxeter set of involutions of a hypergroup, and let d and e be T-homotopic elements of F(T). Then we have ρT (d) = ρT (e). Proof. There is nothing to show if d = e. Thus, we assume that d , e. By induction, we may assume that d and e are elementary T-homotopic. Then there exist elements a, b in F(T) and u, v in T such that u , v, C(u, v) is not empty, d = a ∗ fn (u, v) ∗ b,

and

e = a ∗ fn (v, u) ∗ b,

where n := cu,v . Since C(u, v) is not empty, Rn (u, v) ∩ Rn (v, u) is not empty; cf. Lemma 6.6.5(i). On the other hand, since T is a Coxeter set, we obtain from Lemma 8.1.7 (together with Theorem 9.1.5) that |Rn (u, v)| = 1 and |Rn (v, u)| = 1. Thus, we have ρT (fn (u, v)) = Rn (u, v) = Rn (v, u) = ρT (fn (v, u)), and from this we obtain that ρT (d) = ρT (e).



Lemma 9.8.4 Let T be a Coxeter set of involutions of a hypergroup, and let d and e be elements in F(T). Assume that ρT (d) ∩ ρT (e) contains an element f with λT (d) = `T ( f ) = λT (e). Then d and e are T-homotopic. Proof. Assume first that d = 1. Then λT (d) = 0. Thus, as we are assuming that λT (d) = λT (e), λT (e) = 0. It follows that e = 1, so we are done in this case. Assume now that d , 1. Then there exist elements d0 in F(T) and u in T such that d = d0 ∗ u. Thus, as f ∈ ρT (d) and `T ( f ) = λT (d), ρT (d0) contains an element d with f ∈ du and `T ( f ) = `T (d) + 1; cf. Lemma 9.8.2. It follows that `T (d) = λT (d0) and

f ∈ T−1 (u).

Similarly, there exist elements e0 in F(T), v in T, and e in ρT (e0) such that e = e0 ∗ v, f ∈ ev,

`T ( f ) = `T (e) + 1,

`T (e) = λT (e0),

and

f ∈ T−1 (v).

9.8

Coxeter Sets and Their Coxeter Numbers

307

Assume that u = v. Then f ∈ eu. Thus, as f ∈ du, `T ( f ) = `T (d) + 1, and `T ( f ) = `T (e) + 1, we obtain from Lemma 8.1.2(i) that d = e. It follows that d ∈ ρT (e0) and `T (d) = λT (e0). Thus, as d ∈ ρT (d0), λT (d0) = `T (d), and λT (d) = λT (d0) + 1, induction yields that d0 and e0 are T-homotopic. Thus, as u = v, d and e are Thomotopic, too, and we are done also in this case. Assume now that u , v, and set W := {u, v}. Since T is a Coxeter set and W ⊆ T, W is a Coxeter set; cf. Theorem 9.1.5. On the other hand, as f ∈ T−1 (u) and f ∈ T−1 (v), hWi is finite; cf. Lemma 9.3.4. Thus, by Theorem 9.3.5, W−1 (W) contains an element m with W−1 (W) = {m}.1 Since W−1 (W) is not empty, C(u, v) is not empty; cf. Corollary 6.6.6. Thus, as W−1 (W) = {m}, we obtain from Proposition 6.6.7 that `W (m) = cu,v . Now recall from Lemma 9.1.2 that `W (m) = `T (m). Thus, `T (m) = cu,v . Since f ∈ T−1 (u) and f ∈ T−1 (v), f ∈ T−1 (W). Thus, by Theorem 9.3.1, f ∈ T−1 (hWi). Thus, as m ∈ hWi, f ∈ T−1 (m), and that means that hTi contains an element a such that f ∈ am and `T ( f ) = `T (a) + `T (m). Since m ∈ W−1 (u), hWi contains an element b with m ∈ bu and `T (m) = `T (b) + 1. From f ∈ am, `T ( f ) = `T (a) + `T (m), m ∈ bu, and `T (m) = `T (b) + 1 we obtain an element c in ab such that f ∈ cu, `T (c) = `T (a) + `T (b), and `T ( f ) = `T (c) + 1; cf. Lemma 2.3.8(i). Now recall that f ∈ du and `T ( f ) = `T (d)+1. Thus, as f ∈ cu and `T ( f ) = `T (c)+1, we conclude that c = d; cf. Lemma 8.1.2(i). It follows that d ∈ ab and `T (d) = `T (a) + `T (b). Set

n := `T (m).

Then `T (b) = n − 1. Thus, by Lemma 9.1.2, `W (b) = n − 1. Thus, by Lemma 6.6.2(i), b ∈ Rn−1 (u, v) or b ∈ Rn−1 (v, u). It follows that b ∈ ρT (fn−1 (u, v)) or b ∈ ρT (fn−1 (v, u)). Set  f (u, v) if n is odd b := n−1 fn−1 (v, u) if n is even. Then, as m ∈ bu and n = `T (b) + 1, we have b ∈ ρT (b) and `T (b) = λT (b). Now recall that a ∈ hTi. Thus, F(T) contains an element a such that a ∈ ρT (a) and `T (a) = λT (a); 1In Section 9.5 and Section 9.6, we would have written mW instead of m.

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cf. Lemma 9.8.1(ii). From d ∈ ab, a ∈ ρT (a), and b ∈ ρT (b) we obtain that d ∈ ρT (a)ρT (b) = ρT (a ∗ b). From `T (d) = `T (a) + `T (b), `T (a) = λT (a), and `T (b) = λT (b) we obtain that `T (d) = λT (a) + λT (b) = λT (a ∗ b). Thus, as d ∈ ρT (d0), λT (d0) = `T (d), and λT (d) = λT (d0) + 1, induction yields that d0 and a ∗ b are T-homotopic. It follows that d and a ∗ b ∗ u are T-homotopic, too. Similarly, e and a ∗ c ∗ v are T-homotopic, where  f (v, u) if n is odd c := n−1 fn−1 (u, v) if n is even. Now observe that

if n is odd and

fn (v, u) = c ∗ v

fn (u, v) = b ∗ u

and

fn (u, v) = c ∗ v

and fn (v, u) = b ∗ u

if n is even. On the other hand, as n = cu,v , fn (u, v) and fn (v, u) are elementary T-homotopic. Thus, in both cases, b ∗ u and c ∗ v are elementary T-homotopic. It follows that a ∗ b ∗ u and a ∗ c ∗ v are elementary T-homotopic. Now recall that d and a∗b∗u are T-homotopic and that e and a∗c∗v are T-homotopic.  Thus, d and e are T-homotopic, as wanted. Lemma 9.8.4 will be needed only in the proof of Lemma 9.8.5 and in the proof of Proposition 9.8.6(iii). Let T be a Coxeter set of involutions of a hypergroup. An element of F(T) is called T-reduced if it is not T-homotopic to an element a ∗ t ∗ t ∗ b with a, b ∈ F(T) and t ∈ T. Note that T-homotopy classes of F(T) which contain a T-reduced element of F(T) cannot contain elements of F(T) that are not T-reduced. Lemma 9.8.5 Let T be a Coxeter set of involutions of a hypergroup, and let f be an element in F(T). Then f is T-reduced if and only if ρT (f) contains an element f with `T ( f ) = λT (f). Proof. Assume first that f is T-reduced. If f = 1, λT (f) = 0 and ρT (f) = {1}. Thus, setting f := 1, we obtain that f ∈ ρT (f) and `T ( f ) = λT (f).

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309

Assume now that f , 1. Then we find a T-reduced element e in F(T) and an element t in T such that f = e ∗ t. Since f = e ∗ t, we have λT (f) = λT (e) + 1. Thus, by induction, ρT (e) contains an element e such that `T (e) = λT (e). Suppose that e ∈ T−1 (t). Then hTi contains an element d such that e ∈ dt and `T (e) = `T (d) + 1. Since d ∈ hTi, F(T) contains an element d such that d ∈ ρT (d) and `T (d) = λT (d); cf. Lemma 9.8.1(ii). From e ∈ dt and d ∈ ρT (d) we obtain that e ∈ ρT (d)t = ρT (d ∗ t). Thus, since e ∈ ρT (e), e ∈ ρT (d ∗ t) ∩ ρT (e). From `T (d) = λT (d) and `T (e) = `T (d) + 1 we obtain that λT (d ∗ t) = `T (e). Thus, as `T (e) = λT (e), we obtain from Lemma 9.8.4 that the elements d ∗ t and e are T-homotopic. Since f = e ∗ t, this implies that f and d ∗ t ∗ t are T-homotopic, contrary to the hypothesis that f is T-reduced. This contradiction forces e < T−1 (t). Thus, by Theorem 6.7.5, e ∈ T1 (t). This means that et contains an element f with `T ( f ) = `T (e) + 1. From f ∈ et and e ∈ ρT (e) we obtain that f ∈ ρT (e)t = ρT (e ∗ t) = ρT (f). From `T ( f ) = `T (e) + 1 and `T (e) = λT (e) we obtain that `T ( f ) = λT (e) + 1 = λT (e ∗ t) = λT (f). Assume now that f is not T-reduced. Then f is T-homotopic to an element a ∗ t ∗ t ∗ b with a, b ∈ F(T) and t ∈ T. It follows that ρT (f) = ρT (a ∗ t ∗ t ∗ b) = ρT (a)tt ρT (b); cf. Lemma 9.8.3. Thus, for each element f in ρT (f), f ∈ ρT (a)tt ρT (b) ⊆ ρT (a)ρT (b) ∪ ρT (a)t ρT (b) = ρT (a ∗ b) ∪ ρT (a ∗ t ∗ b). From this we obtain that, for each element f in ρT (f), `T ( f ) ≤ λT (f) − 1; cf. Lemma 9.8.1(i).  The third part of the following proposition is a partial converse of Lemma 9.8.3. Proposition 9.8.6 Let T be a Coxeter set of involutions of a hypergroup. Then the following hold. (i) For each T-reduced element f of F(T), we have | ρT (f)| = 1.

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(ii) For each element f in hTi, there exists a T-reduced element f in F(T) such that ρT (f) = { f }. (iii) Any two T-reduced elements d and e of F(T) satisfying ρT (d) = ρT (e) are T-homotopic. Proof. (i) Let f be a T-reduced element in F(T). If f = 1, ρT (f) = {1}, so that we are done in this case. Assume that f , 1. Then we find a T-reduced element e in F(T) and an element t in T such that f = e ∗ t. Since f = e ∗ t, we have λT (f) = λT (e) + 1. Thus, by induction, ρT (e) contains an element e with ρT (e) = {e}. From e ∈ ρT (e) we obtain that `T (e) ≤ λT (e); cf. Lemma 9.8.1(i). Since f is T-reduced, ρT (f) contains an element f with `T ( f ) = λT (f); cf. Lemma 9.8.5. From f ∈ ρT (f), f = e ∗ t, and ρT (e) = {e} we obtain that f ∈ ρT (e ∗ t) = ρT (e)t = et. From `T ( f ) = λT (f), f = e ∗ t, and `T (e) ≤ λT (e) we obtain that `T (e) + 1 ≤ λT (e) + 1 = λT (e ∗ t) = `T ( f ). From f ∈ et and `T (e) + 1 ≤ `T ( f ) we now obtain that e ∈ T1 (t). Since T is constrained, this implies that et = { f }. Thus, as f = e ∗ t and ρT (e) = {e}, we have ρT (f) = { f }. (ii) Let f be an element in hTi. Then, by Lemma 9.8.1(ii), F(T) contains an element f such that f ∈ ρT (f) and `T ( f ) = λT (f). Since f ∈ ρT (f) and `T ( f ) = λT (f), f is T-reduced; cf. Lemma 9.8.5. Thus, by (i), | ρT (f)| = 1. Since f ∈ ρT (f), this implies that ρT (f) = { f }. (iii) Let d and e be T-reduced elements in F(T), and assume that ρT (d) = ρT (e). Since d and e are T-reduced, hTi contains elements d and e with ρT (d) = {d} and ρT (e) = {e}; cf. (i). Thus, by Lemma 9.8.5, `T (d) = λT (d) and `T (e) = λT (e). From ρT (d) = ρT (e), ρT (d) = {d}, and ρT (e) = {e} we obtain that d = e. Set f := d. Then f ∈ ρT (d) ∩ ρT (e) and λT (d) = `T ( f ) = λT (e). Thus, by Lemma 9.8.4,  d and e are T-homotopic. Let T be a Coxeter set of involutions of a hypergroup. Proposition 9.8.6 (together with Lemma 9.8.3) shows that ρT induces a bijective map from hTi to the set of all T-homotopy classes of F(T) consisting of T-reduced elements.

9.9 Coxeter Sets and Type Preserving Bijections Let U and V be Coxeter sets of involutions of hypergroups. A bijective map ι from U to V is called a type preserving bijection if, for any two distinct elements y and z in U, cy,z = cy ι,z ι . (We follow our notation introduced in Section 6.6 and denote by

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cy,z the Coxeter number of y and z. Similarly, cy ι,z ι stands for the Coxeter number of the elements y ι and z ι in V.) Let ι be a type preserving bijection from U to V. Since ι is a map from U to V, there exists exactly one monoid homomorphism from F(U) to F(V) which sends each element u of U to uι . We denote this monoid homomorphism by κ. Since ι is bijective, so is κ. Since κ is a monoid homomorphism we have, for any two distinct elements y and z in U and each positive integer n, fn (y, z)κ = fn (y ι, z ι ). (Remember that fn (y, z) stands for the product t1 ∗ · · · ∗ tn , where, for each element i in {1, . . . , n}, ti = y if i is odd and ti = z if i is even.) This observation has two consequences. First, any two elements d and e of F(U) are U-homotopic if and only if dκ and eκ are V-homotopic. This follows from the fact that ι is assumed to be type preserving. Second, an element f in F(U) is U-reduced if and only if f κ is V-reduced. This follows from the first observation. We will now establish a map from hUi to hVi. For this, we choose an element in hUi and denote it by f . From Proposition 9.8.6(ii) we obtain a U-reduced element f in F(U) with ρU (f) = { f }. Since f is U-reduced, f κ is V-reduced. It follows that | ρV (f κ )| = 1; cf. Proposition 9.8.6(i). We define f τ to be the uniquely determined element in ρV (f κ ). Note that f τ ∈ hVi. Assume that F(U) contains U-reduced elements d and e with ρU (d) = { f } and ρU (e) = { f }. Then, by Proposition 9.8.6(iii), d and e are T-homotopic. Thus, dκ and eκ are V-homotopic, so that, by Lemma 9.8.3, ρV (dκ ) = ρV (eκ ). This shows that f τ does not depend on the choice of the U-reduced element f in F(U) satisfying ρU (f) = { f }. What we have seen so far is that τ is a map from hUi to hVi. In the following, the map τ will be called the signature induced by ι. Theorem 9.9.1 Let U and V be Coxeter sets of involutions of hypergroups, let ι be a type preserving bijection from U to V, and let τ denote the signature induced by ι. Then the following hold. (i) We have τ|U = ι. (ii) The map τ is a bijective map from hUi to hVi. (iii) The map τ is (U, V)-length preserving. (iv) For any three elements c, d, and e in hUi with e ∈ cd and `U (e) = `U (c) + `U (d), we have eτ ∈ cτ d τ .

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(v) For each element f in hUi, we have f τ∗ = f ∗τ . (vi) For any two elements d and e in hUi with e ∈ U−1 (d), we have eτ ∈ V−1 (d τ ). (vii) For any two elements c and d in hUi with c ∈ U1 (d), we have cτ ∈ V1 (d τ ). Proof. Let κ denote the uniquely determined monoid isomorphism from F(U) to F(V) such that, for each element u of U, uκ = uι . Since τ stands for the signature induced by ι, we have ρV (f κ ) = { f τ } for each element f in hUi and each reduced element f in F(U) with ρU (f) = { f }. (i) Let u be an element in U. Then uι ∈ V, whence ρV (uι ) = {uι }. Thus, as ρV (uκ ) = {uτ } and uκ = uι , we have uτ = uι . (ii) To show that τ is injective, we choose elements d and e in hUi and assume that d τ = eτ . Since d ∈ hUi, F(U) contains a U-reduced element d with ρU (d) = {d}; cf. Proposition 9.8.6(ii). Similarly, since e ∈ hUi, F(U) contains a U-reduced element e with ρU (e) = {e}. It follows that

ρV (dκ ) = {d τ }

and

ρV (eκ ) = {eτ }.

Since we are assuming that d τ = eτ , we obtain that ρV (dκ ) = ρV (eκ ). Note also that dκ and eκ are V-reduced, because d and e are U-reduced. Thus, by Proposition 9.8.6(iii), dκ and eκ are V-homotopic. It follows that d and e are U-homotopic, so that, by Lemma 9.8.3, ρU (d) = ρU (e). Since ρU (d) = {d} and ρU (e) = {e}, this implies that d = e. To show that τ is surjective, we choose an element e in hVi. By Proposition 9.8.6(ii), F(V) contains a V-reduced element e such that ρV (e) = {e}. On the other hand, since κ is a surjective map from F(U) to F(V), F(U) contains an element d with dκ = e. Thus, ρV (dκ ) = {e}. Since e is V-reduced and dκ = e, d is U-reduced. Thus, by Proposition 9.8.6(i), hUi contains an element d with ρU (d) = {d}. It follows that ρV (dκ ) = {d τ }. Thus, as ρV (dκ ) = {e}, we obtain that d τ = e. (iii) Let f be an element in hUi. Then F(U) contains a U-reduced element f with ρU (f) = { f }; cf. Proposition 9.8.6(ii). Since f is U-reduced and ρU (f) = { f }, we have `U ( f ) = λU (f); cf. Lemma 9.8.5. From the fact that f is U-reduced we also obtain that f κ is V-reduced. Also, since f is U-reduced and ρU (f) = { f }, we have ρV (f κ ) = { f τ }. Thus, again by Lemma 9.8.5, `V ( f τ ) = λV (f κ ).

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From `U ( f ) = λU (f), λU (f) = λV (f κ ), and `V ( f τ ) = λV (f κ ) we obtain that `U ( f ) = `V ( f τ ). (iv) Let c, d, and e be elements in hUi with e ∈ cd and `U (e) = `U (c) + `U (d). We will see that eτ ∈ cτ d τ . From e ∈ cd and `U (e) = `U (c) + `U (d) (together with the fact that U is constrained) we obtain that cd = {e}; cf. Lemma 8.1.1. From Proposition 9.8.6(ii) we obtain U-reduced elements c, d, and e such that ρU (c) = {c},

ρU (d) = {d},

and

ρU (e) = {e}.

These three equations imply that ρV (cκ ) = {cτ },

ρV (dκ ) = {d τ },

and

ρV (eκ ) = {eτ }.

Since c and d are U-reduced, we obtain from ρU (c) = {c} and ρU (d) = {d} that `U (c) = λU (c) and `U (d) = λU (d); cf. Lemma 9.8.5. From e ∈ cd (together with ρU (c) = {c} and ρU (d) = {d}) we obtain that e ∈ ρU (c)ρU (d) = ρU (c ∗ d). From `U (e) = `U (c) + `U (d) (together with `U (c) = λU (c) and `U (d) = λU (d)) we obtain that `U (e) = λU (c) + λU (d) = λU (c ∗ d). Thus, by Lemma 9.8.5, c ∗ d is U-reduced. On the other hand, also e is U-reduced, and, since ρU (c) = {c}, ρU (d) = {d}, cd = {e}, and ρU (e) = {e}, we have ρU (c ∗ d) = ρU (c)ρU (d) = cd = ρU (e). Thus, by Proposition 9.8.6(iii), c ∗ d and e are U-homotopic. It follows that (c ∗ d)κ and eκ are V-homotopic. Thus, by Lemma 9.8.3, ρV ((c ∗ d)κ ) = ρV (eκ ). Thus, as ρV (eκ ) = {eτ }, ρV ((c ∗ d)κ ) = {eτ }. From ρV (cκ ) = {cτ } and ρV (dκ ) = {d τ } we also obtain that cτ d τ = ρV (cκ )ρV (dκ ) = ρV (cκ ∗ dκ ) = ρV ((c ∗ d)κ ). Thus, we have eτ ∈ cτ d τ .

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(v) Let f be an element in hUi. Then F(U) contains a U-reduced element f with ρU (f) = { f }; cf. Proposition 9.8.6(ii). From the definition of τ, together with ρU (f) = { f }, we obtain that ρV (f κ ) = { f τ }. It follows that ρV (f κ∗ ) = { f τ∗ }. On the other hand, since ρU (f) = { f }, we have ρU (f ∗ ) = { f ∗ }, whence ρV (f ∗κ ) = { f ∗τ }. Thus, as f κ∗ = f ∗κ , we conclude that f τ∗ = f ∗τ . (vi) From (iii) we know that τ is (U, V)-length preserving, from (ii) that τ is bijective. In (iv), we saw that eτ ∈ cτ d τ for any three elements c, d, and e in hUi with e ∈ cd and `U (e) = `U (c) + `U (d). Thus, the claim follows from Lemma 6.5.9(i). (vii) This follows from Lemma 6.5.9(ii) in exactly the same way as (vi) follows from Lemma 6.5.9(i).  The following theorem provides a sufficient condition for the signature induced by a type preserving bijection to be a hypergroup isomorphism. Theorem 9.9.2 Let U and V be Coxeter sets of involutions of hypergroups, let ι be a type preserving bijection from U to V, and let τ denote the signature induced by ι. Assume that Oϑ (U)τ = Oϑ (V). Then τ is a hypergroup isomorphism from hUi to hVi. Proof. We are assuming that U and V are Coxeter sets of involutions of hypergroups. Thus, by definition, both U and V are constrained. Since U is a Coxeter set, U also satisfies the exchange condition, whence, by Theorem 6.7.5, U is also dichotomic. In Theorem 9.9.1(iii) we saw that τ is (U, V)-length preserving, and from Theorem 9.9.1(ii) we know that τ is a bijective map from hUi to hVi. Recall also that we are assuming that Oϑ (U)τ = Oϑ (V). Finally, in Theorem 9.9.1(iv), we saw that eτ ∈ cτ d τ for any three elements c, d, and e in hUi with e ∈ cd and `U (e) = `U (c) + `U (d). Now, by Theorem 8.3.4, τ is a hypergroup isomorphism from hUi to hVi.



Let U and V be Coxeter sets of involutions of hypergroups, let ι be a type preserving bijection from U to V, and let τ denote the signature induced by ι. As a consequence of Theorem 9.9.2 one obtains that τ is a hypergroup isomorphism from hUi to hVi if both U and V are thin or if both U and V do not contain thin elements. The following lemma shows that the signature induced by a type preserving bijection is characterized by the first four conditions of Theorem 9.9.1. Lemma 9.9.3 Let U and V be Coxeter sets of involutions of hypergroups, and let ι be a type preserving bijection from U to V. Let υ be a (U, V)-length preserving bijective map from hUi to hVi satisfying υ|U = ι. Assume that eυ ∈ d υ uυ for any three elements d

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and e in hUi and u in U with e ∈ du and `U (e) = `U (d) + 1. Then υ is the signature induced by ι. Proof. Let τ denote the signature induced by ι, and let f be an element in hUi. We have to show that f υ = f τ and proceed by induction on `U ( f ). Assume that `U ( f ) = 0. Then, as υ is (U, V)-length preserving, `V ( f υ ) = 0. It follows that f υ = 1. Also τ is (U, V)-length preserving; cf. Theorem 9.9.1(iii). Thus, similarly, f τ = 1. From f υ = 1 and f τ = 1 we obtain that f υ = f τ , and we are done in this case. Assume that f , 1. Then, by Lemma 2.3.7(ii), there exist elements e in hUi and u in U with f ∈ eu and `U ( f ) = `U (e) + 1. From f ∈ eu and `U ( f ) = `U (e) + 1 we also obtain that f υ ∈ eυ uυ . (This is by hypothesis.) Since υ is assumed to be (U, V)-length preserving, we have `V ( f υ ) = `V (eυ ) + 1. Thus, as V is a assumed to be a Coxeter set of involutions, Lemma 8.1.1 yields eυ uυ = { f υ }. From f ∈ eu and `U ( f ) = `U (e) + 1 we obtain that f τ ∈ eτ uτ ; cf. Theorem 9.9.1(iv). From `U ( f ) = `U (e) + 1 we also obtain that `V ( f τ ) = `V (eτ ) + 1; cf. Theorem 9.9.1(iii). Thus, as V is a assumed to be a Coxeter set of involutions, Lemma 8.1.1 yields eτ uτ = { f τ }. By induction, eυ = eτ , and, by hypothesis, uυ = uτ . Thus, as eυ uυ = { f υ } and  eτ uτ = { f τ }, we obtain that f υ = f τ . Let U and V be Coxeter sets of involutions of hypergroups, and let ι be a type preserving bijection from U to V. Notice that Lemma 9.9.3 can be applied to each (U, V)-length preserving hypergroup isomorphism φ from hUi to hVi satisfying φ|U = ι. Lemma 9.9.4 Let U and V be Coxeter sets of involutions of hypergroups, and let φ be a (U, V)length preserving hypergroup isomorphism from hUi to hVi. Then the following hold. (i) The map φ|U is a type preserving bijection from U to V. (ii) The signature induced by φ|U is φ. Proof. (i) Since φ be assumed to be a (U, V)-length preserving hypergroup isomorphism from hUi to hVi, φ|U is a bijective map from U to V. To show that φ|U is type preserving, we let y and z be two distinct elements of U.

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Assume first that cy,z is finite. Then C(y, z) is not empty. Set k := cy,z . Then 2k = min(C(y, z)). Thus, 1 ∈ R2k (y, z). It follows that 1 ∈ R2k (y, z)φ = R2k (y φ, z φ ), and then that cy φ ,z φ ≤ cy,z . Set l := cy φ ,z φ . Then 2l = min(C(y φ, z φ )). Thus, 1φ = 1 ∈ R2l (y φ, z φ ) = R2l (y, z)φ . Since φ is injective, this implies that 1 ∈ R2l (y, z). It follows that cy,z ≤ cy φ ,z φ . This shows that cy,z = cy φ ,z φ if cy,z is assumed to be finite. Similarly, this equation is obtained under the assumption that cy φ ,z φ is finite. (ii) For each element u in U, define uι := uφ . Then, by (i), ι is a type preserving bijection from U to V and φ|U = ι. Thus, by Lemma 9.9.3, φ is the signature induced by φ|U .  Lemma 9.9.5 Let U, V, and W be Coxeter sets of involutions of hypergroups, let η be a type preserving bijection from U to V, and let θ be a type preserving bijection from V to W. Then the following hold. (i) The composite ηθ is a type preserving bijection from U to W. (ii) Let χ denote the signature induced by η, and let ψ denote the signature induced by θ. Then χψ is the signature induced by ηθ. Proof. (i) Since η is a type preserving bijection from U to V, we have cy,z = cy η ,z η for any two distinct elements y and z in U. Since θ is a type preserving bijection from V to W, we have cy η ,z η = cy η θ ,z η θ for any two distinct elements y and z in U. It follows that cy,z = cy η θ ,z η θ for any two distinct elements y and z, and that shows that ηθ is a type preserving bijection from U to W. (ii) Let λ denote the uniquely determined monoid homomorphism from F(U) to F(V) which sends each element u of U to uη , and let µ denote the uniquely determined monoid homomorphism from F(V) to F(W) which sends each element v of V to v θ . Then λµ is the uniquely determined monoid homomorphism from F(U) to F(W) which sends each element u of U to uηθ . Let f be an element in hUi, and let ξ denote the signature induced by ηθ. We are done when we succeed in showing that f χψ = f ξ . By Proposition 9.8.6(ii), F(U) contains a U-reduced element f with ρU (f) = { f }. Then, by definition, ρV (f λ ) = { f χ }, and this implies that ρW (f λµ ) = { f χψ }. Since λµ is the uniquely determined monoid homomorphism from F(U) to F(W) which sends each element u of U to uηθ and ξ is the signature induced by ηθ, ρW (f λµ ) = { f ξ }. From ρW (f λµ ) = { f χψ } and ρW (f λµ ) = { f ξ } we obtain that f χψ = f ξ .



Lemma 9.9.6 Let U and V be Coxeter sets of involutions of hypergroups, and let ι be a type preserving bijection from U to V. Then the following hold. (i) The map ι−1 is a type preserving bijection from V to U.

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(ii) Let τ denote the signature induced by ι. Then τ −1 is the signature induced by ι−1 . Proof. (i) This follows immediately from the definition of a type preserving bijection. (ii) Let φ denote the signature of ι−1 . Then, by Lemma 9.9.5(ii), τφ is the signature induced by the composite ιι−1 . On the other hand, since ιι−1 = 1U , the signature induced by the composite ιι−1 is 1 hU i ; cf. Lemma 9.9.4. Thus, τφ = 1 hU i , and that means that φ = τ −1 .  Lemma 9.9.7 Let U and V be Coxeter sets of involutions of hypergroups, let ι be a type preserving bijection from U to V, and let τ denote the signature induced by ι. Let α be a U-length preserving hypergroup automorphism of hUi, and assume that V is thin. Then the composite τ −1 ατ is a V-length preserving automorphism of hVi. Proof. Since α is a U-length preserving hypergroup automorphism of hUi, α|U is a type preserving permutation of U; cf. Lemma 9.9.4(i). Furthermore, by Lemma 9.9.4(ii), α is the signature induced by α|U . Since ι is a type preserving bijection from U to V, ι−1 is a type preserving bijection from V to U; cf. Lemma 9.9.6(i). Thus, as α|U is a type preserving permutation of U and ι is a type preserving bijection from U to V, the composite ι−1 α|U ι is a type preserving permutation of V; cf. Lemma 9.9.5(i). Since τ −1 is the signature induced by ι−1 (by Lemma 9.9.6(ii)), α is the signature induced by α|U , and τ is the signature induced by ι, we obtain from Lemma 9.9.5(ii) that τ −1 ατ is the signature induced by ι−1 α|U ι. Since τ −1 ατ is the signature induced by ι−1 α|U ι, we now obtain from Theorem 9.9.1(iii) that τ −1 ατ is V-length preserving. Since τ −1 ατ is the signature induced by ι−1 α|U ι and V is thin, we obtain from Theorem 9.9.2 that τ −1 ατ is an automorphism  of hVi. Let H be a hypergroup, let F be a closed subset of H, and let ρ be a folding from F to H satisfying F ρ∗ = F ρ . Recall from Lemma 2.7.3 that the map from F to F −1 which takes each element f in F to f ∗ρ∗ρ is bijective. As before, we will refer to this permutation of F as the permutation of F induced by the folding ρ. Lemma 9.9.8 Let H be a hypergroup, let U be a Coxeter set of involutions of H, and assume that none of the elements in U is thin. Let ρ be a folding from hUi to H satisfying 1ρ∗ = 1ρ and hUi ρ∗ = hUi ρ , and let π denote the permutation of hUi induced by ρ. Let V be a Coxeter set of involutions of a thin hypergroup, let ι be a type preserving bijection from U to V, and let τ denote the signature induced by ι. Then τ −1 πτ is a V-length preserving automorphism of hVi having order 1 or 2.

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Proof. From Lemma 9.7.1 we know that π is U-length preserving, from Lemma 9.7.2 that π is a hypergroup automorphism of hUi. Thus, applying Lemma 9.9.7 to π in place of α we obtain that τ −1 πτ is a V-length preserving automorphism of hVi. From Lemma 8.7.3(i) we know that π has order 1 or 2. Thus, the composite τ −1 πτ  has order 1 or 2, too. Corollary 9.9.9 Let H be a hypergroup, let U be a set of involutions of H, and assume that H is a twin Coxeter hypergroup over U. Assume that none of the elements in U is thin. Let ρ denote the canonical folding defined by H and U, and let π denote the permutation of hUi induced by ρ. Let V be a Coxeter set of involutions of a thin hypergroup, let ι be a type preserving bijection from U to V, and let τ denote the signature induced by ι. Then τ −1 πτ is a V-length preserving automorphism of hVi having order 1 or 2. Proof. This follows from Lemma 9.9.8.



The V-length preserving automorphism of hVi having order 1 or 2 which one obtains via Corollary 9.9.9 from a twin Coxeter hypergroup H over a set U of non-thin involutions of H together with a type preserving bijection ι from U to V will be called the automorphism of hVi induced by ι.2

2Note that, although the automorphism of hVi induced by ι is defined with the help of τ and π, it really depends only on τ, since the permutation π of hUi is induced by the canonical folding defined by H and U, and the canonical folding defined by H and U is, in fact, the only folding ρ from hUi to H satisfying H = hTi ∪ hTi ρ .

10 Regular Actions of (Twin) Coxeter Hypergroups

In this chapter we will show that both, the notion of a building in the sense of [44] or [45] and (modulo a certain building theoretic hypothesis) the notion of a twin building in the sense of [46], can be considered in a natural way as components of a theory of hypergroups. In Sections 10.2 and 10.3, we will see that semiregular buildings (as they will be defined in Section 10.1) and regular actions of Coxeter hypergroups are equivalent mathematical objects. The precise nature of this equivalence will be described in Theorems 10.2.1, 10.2.7, 10.3.1, 10.3.2, 10.3.3, and 10.3.4. In Sections 10.6 and 10.7, we will see that (modulo the above-mentioned hypothesis) thick twin buildings and regular actions of twin Coxeter hypergroups over sets of nonthin involutions are equivalent mathematical objects. The precise nature of this equivalence will be described in Theorems 10.6.1, 10.6.7, 10.7.1, 10.7.2, 10.7.3, and 10.7.4. The first of the above series of main results will be prepared in Section 10.1, the second one in Sections 10.4 and 10.5.

10.1 Buildings and Binary Relations Let W be a group, and let I be a set of involutions of W. The pair (W, I) is called a Coxeter system if W is generated by I and I satisfies the (group theoretic) exchange condition.1 Note that, according to the terminology introduced in the introduction to Chapter 9, (W, I) is a Coxeter system if and only if, via the group correspondence, W corresponds to a thin Coxeter hypergroup over I. Let (W, I) be a Coxeter system. For each element w in W, the I-length of w will be denoted by `I (w). For each element u in W, we write I1 (u) to denote the set of all elements t in W satisfying `I (tu) = `I (t) + `I (u). (Via the group correspondence, this notation is compatible with the quite similar notation which we introduced in Section 6.5. This allows us to apply all results on the sets T1 (d) which we have established 1Here, we consider Condition (F) as defined in [1; p. 79] as the group theoretic exchange condition. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 P. -H. Zieschang, Hypergroups, https://doi.org/10.1007/978-3-031-39489-8_10

319

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for sets T of involutions of hypergroups and elements d in hTi to sets I1 (w) with w an element in W.) Now we are ready to give the definition of a building. Our definition is taken from [1; p. 218]. Let (W, I) be a Coxeter system, let X be a non-empty set, and let δ be a map from X × X to W. The pair (X, δ) is called a building of type (W, I) if, for any three elements y, z in X and w in W with δ(y, z) = w, the following conditions hold. B1

We have w = 1 if and only if y = z.

B2

For any two elements i in I and x in X with δ(z, x) = i, we have δ(y, x) = wi or δ(y, x) = w. If, in addition, w ∈ I1 (i), δ(y, x) = wi.

B3

For each element i in I, there exists an element x in X with δ(y, x) = wi and δ(z, x) = i.

Let (W, I) be a Coxeter system, and let (X, δ) be a building of type (W, I). For each element w in W, we define hw := {(y, z) ∈ X × X | δ(y, z) = w}. The purpose of this section is to compile several results on the sets hw with w ∈ W. Lemma 10.1.1 Let (W, I) be a Coxeter system, and let (X, δ) be a building of type (W, I). Then h1 = 1 X . Proof. Considering the definition of h1 this follows from Condition B1.



Let (W, I) be a Coxeter system, and let (X, δ) be a building of type (W, I). Since the sets hw with w ∈ W are subsets of X × X, we may adopt the following notation from Section 1.6. For each element w in W, we write hw∪ to denote the set of all pairs (y, z) in X × X with (z, y) ∈ hw . For any two elements u and v in W, we write hu ◦ hv to denote the set of all pairs (y, z) in X × X for which we can find an element x in X with (y, x) ∈ hu and (x, z) ∈ hv . Lemma 10.1.2 Let (W, I) be a Coxeter system, let (X, δ) be a building of type (W, I), and let i be an element in I. Then the following statements hold. (i) We have hi∪ = hi . (ii) We have h1 ⊆ hi ◦ hi ⊆ h1 ∪ hi . Proof. (i) Let y and z be elements in X, and assume that (y, z) ∈ hi∪ . Then (z, y) ∈ hi , and that means that δ(z, y) = i. Thus, by Condition B3, there exists an element x in X with δ(z, x) = 1 and δ(y, x) = i. From δ(z, x) = 1 we obtain that z = x; cf. Condition B1. Thus, as δ(y, x) = i, δ(y, z) = i. It follows that (y, z) ∈ hi .

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321

The reverse containment follows similarly. (ii) Let y and z be elements in X, and assume that (y, z) ∈ h1 . Then, by Condition B3, there exists an element x in X with δ(y, x) = i and δ(z, x) = i. From δ(y, x) = i we obtain that (y, x) ∈ hi , and from δ(z, x) = i we obtain that (z, x) ∈ hi . From (z, x) ∈ hi we obtain that (x, z) ∈ hi ; cf. (i). From (y, x) ∈ hi and (x, z) ∈ hi we obtain that (y, z) ∈ hi ◦ hi . Let y and z be elements in X, and assume that (y, z) ∈ hi ◦ hi . Then we find an element x in X such that (y, x) ∈ hi and (x, z) ∈ hi . From (y, x) ∈ hi we obtain that δ(y, x) = i, and from (x, z) ∈ hi we obtain that δ(x, z) = i. Thus, by Condition B2, δ(y, z) = 1 or δ(y, z) = i. In the first case, we have (y, z) ∈ h1 , in the second case, we have (y, z) ∈ hi . Thus, (y, z) ∈ h1 ∪ hi .  Let (W, I) be a Coxeter system. Let (X, δ) be a building of type (W, I), and let i be an element in I. In Lemma 10.1.2(ii), we saw that h1 ⊆ hi ◦ hi ⊆ h1 ∪ hi . In the following, we will say that hi is of first type if h1 = hi ◦ hi , and we will say that hi is of second type if hi ◦ hi = h1 ∪ hi . Of course, hi may be neither of first nor of second type. Recall that, for each element w in W, `I (w) is our notation for the I-length of w. For each element u in W, we will write I−1 (u) to denote the set of all elements v in W such that `I (v) = `I (vu−1 ) + `I (u). (Via the group correspondence, this notation is compatible with the quite similar notation which we introduced in Section 6.5. This allows us to apply all results on the sets T−1 (d) which we have established for sets T of involutions of a hypergroup and elements d in hTi to sets I−1 (w) with w an element in W.) Lemma 10.1.3 Let (W, I) be a Coxeter system, let (X, δ) be a building of type (W, I), let w be an element in W, and let i be an element in I. Then the following hold. (i) If w ∈ I1 (i), hw ◦ hi = hwi . (ii) Assume that w ∈ I−1 (i) and that hi is of first type. Then hw ◦ hi = hwi . (iii) Assume that w ∈ I−1 (i) and that hi is of second type. Then hw ◦hi = hwi ∪hw . Proof. (i) Let y and z be elements in X, and assume first that (y, z) ∈ hw ◦ hi . Then X contains an element x such that (y, x) ∈ hw and (x, z) ∈ hi . It follows that δ(y, x) = w and δ(x, z) = i. Now assume that w ∈ I1 (i). Then, by Condition B2, δ(y, z) = wi, and that means that (y, z) ∈ hwi . Assume, conversely, that (y, z) ∈ hwi . Then δ(y, z) = wi. Now, by Condition B3, X contains an element x such that δ(y, x) = w and δ(z, x) = i. From δ(y, x) = w we obtain that (y, x) ∈ hw . From δ(z, x) = i we obtain that (z, x) ∈ hi , and then, by Lemma 10.1.2(i), that (x, z) ∈ hi . From (y, x) ∈ hw and (x, z) ∈ hi we obtain that (y, z) ∈ hw ◦ hi . (ii), (iii) We are assuming that w ∈ I−1 (i), and that is equivalent to wi ∈ I1 (i). Thus, by (i), hwi ◦ hi = hw . It follows that

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hw ◦ hi = hwi ◦ hi ◦ hi . If hi is of first type, we have hi ◦ hi = h1 , so that, by the above equation, hw ◦ hi = hwi . If hi is of second type, we have hi ◦ hi = h1 ∪ hi , so that, by the above equation, hw ◦ hi = hwi ◦ (h1 ∪ hi ) = (hwi ◦ h1 ) ∪ (hwi ◦ hi ) = hwi ∪ hw, 

as wanted. Corollary 10.1.4

Let (W, I) be a Coxeter system, and let (X, δ) be a building of type (W, I). Let u and v be elements in W with u ∈ I1 (v). Then hu ◦ hv = huv . Proof. We proceed by induction on `I (v). If `I (v) = 0, v = 1. In this case, we have hu ◦ hv = hu ◦ h1 = hu = huv , so that we are done in this case. Assume that 1 ≤ `I (v). Then v , 1, so that we find elements t in W and i in I with v = ti

and `I (v) = `I (t) + 1.

Since we are assuming that u ∈ I1 (v), we also have `I (uv) = `I (u) + `I (v). Thus, by Lemma 2.3.8(i), `I (u) + `I (t) = `I (ut) and `I (uv) = `I (ut) + 1. From the first of these two equations we obtain that u ∈ I1 (t). Thus, since `I (v) = `I (t) + 1, induction yields hu ◦ ht = hut . From the second equation (together with v = ti) we obtain that ut ∈ I1 (i). Thus, as v = ti, Lemma 10.1.3(i) yields hut ◦ hi = huv . From v = ti and `I (v) = `I (t) + 1 we obtain that t ∈ I1 (i). Thus, as v = ti, Lemma 10.1.3(i) also yields ht ◦ hi = hv . From ht ◦ hi = hv , hu ◦ ht = hut , and hut ◦ hi = huv we now obtain hu ◦ hv = huv , as wanted.  Lemma 10.1.5 Let (W, I) be a Coxeter system, and let (X, δ) be a building of type (W, I). Then, for each element w in W, hw∪ = hw −1 . Proof. We proceed by induction on `I (w).

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323

If `I (w) = 0, w = 1. In this case, we have hw∪ = h1∪ = h1 = h1−1 = hw −1 , so that we are done in this case. Assume that 1 ≤ `I (w). Then w , 1, so that we find elements v in W and i in I with w = vi

and `I (w) = `I (v) + 1.

From these two equations we obtain that v ∈ I1 (i). Thus, as w = vi, Lemma 10.1.3(i) yields hv ◦ hi = hw . It follows that hw∪ = hi∪ ◦ hv∪ . Now recall from Lemma 10.1.2(i) that hi∪ = hi . Furthermore, since `I (w) = `I (v)+1, induction yields hv∪ = hv−1 . Thus, we have hi∪ ◦ hv∪ = hi ◦ hv−1 . From v ∈ I1 (i) we obtain that i ∈ I1 (v −1 ); cf. Lemma 6.5.2(i). Thus, as iv −1 = w −1 , Corollary 10.1.4 yields hi ◦ hv−1 = hw −1 . From hw∪ = hi∪ ◦ hv∪ , hi∪ ◦ hv∪ = hi ◦ hv−1 , and hi ◦ hv−1 = hw −1 we obtain that  hw∪ = hw −1 , as wanted. We now generalize Lemma 10.1.3. In fact, Lemma 10.1.3 is the case (u, v) = (w, i) of the subsequent lemma. Lemma 10.1.6 Let (W, I) be a Coxeter system, let (X, δ) be a building of type (W, I), let u and v be elements in W, and let i be an element in I. Assume that v −1 ∈ I−1 (i). Then the following hold. (i) If u ∈ I1 (i), hu ◦ hv = hui ◦ hiv . (ii) Assume that u ∈ I−1 (i) and that hi is of first type. Then hu ◦ hv = hui ◦ hiv . (iii) Assume that u ∈ I−1 (i) and that hi is of second type. Then hu ◦ hv = (hui ◦ hiv ) ∪ (hu ◦ hiv ). Proof. We are assuming that v −1 ∈ I−1 (i), and that is equivalent to v −1 i ∈ I1 (i). Thus, by Lemma 6.5.2(i), i ∈ I1 (iv). It follows that hi ◦ hiv = hv ; cf. Corollary 10.1.4. (i) Assume that u ∈ I1 (i). Then, by Lemma 10.1.3(i), hu ◦hi = hui . Since hi ◦hiv = hv , this implies that hu ◦ hv = hui ◦ hiv .

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10 Regular Actions of (Twin) Coxeter Hypergroups

(ii) We are assuming that u ∈ I−1 (i) and that hi is of first type. Thus, by Lemma 10.1.3(ii), hu ◦ hi = hui . Since hi ◦ hiv = hv , this implies that hu ◦ hv = hui ◦ hiv . (iii) We are assuming that u ∈ I−1 (i) and that hi is of second type. Thus, by Lemma 10.1.3(iii), hu ◦ hi = hui ∪ hu . Since hi ◦ hiv = hv , this implies that hu ◦ hv = hu ◦ hi ◦ hiv = (hui ∪ hu ) ◦ hiv = (hui ◦ hiv ) ∪ (hu ◦ hiv ), 

as wanted.

Let (W, I) be a Coxeter system, and let (X, δ) be a building of type (W, I). The building (X, δ) will be called semiregular if, for each element i in I, hi is either of first type or of second type. The building (X, δ) is called thick if, for each element i in I, hi is of second type.2 Notice that the class of semiregular buildings contains the class of thick buildings. Lemma 10.1.7 Let (W, I) be a Coxeter system, and let (X, δ) be a semiregular building of type (W, I). Let t, u, and v be elements in W, and assume that hv ∩ (ht ◦ hu ) is not empty. Then hv ⊆ ht ◦ hu . Proof. We proceed by induction on `I (u). Assume first that `I (u) = 0. Then u = 1, so that hu = h1 . It follows that hv ∩ (ht ◦ hu ) = hv ∩ (ht ◦ h1 ) = hv ∩ ht . Since hv ∩ (ht ◦ hu ) is assumed not to be empty, this implies that hv ∩ ht is not empty; equivalently, hv = ht . It follows that hv = ht ◦ h1 = ht ◦ hu , so we are done in this case. Now assume that 1 ≤ `I (u). Then u = 1, so that we find an element i in I with `I (u) = `I (iu) + 1. It follows that

u−1 ∈ I−1 (i).

Assume first that t ∈ I−1 (i), and recall that (X, δ) is assumed to be semiregular. Then, since u−1 ∈ I−1 (i), we may apply Lemma 10.1.6(ii), (iii) to t and u in place of u and v. We obtain that ht ◦ hu = hti ◦ hiu

or

ht ◦ hu = (hti ◦ hiu ) ∪ (ht ◦ hiu ).

2The definition of a thick building is common in the literature, the one of a semiregular building is not.

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325

Suppose that ht ◦ hu = hti ◦ hiu . Then, as hv ∩ (ht ◦ hu ) is assumed not to be empty, hv ∩ (hti ◦ hiu ) is not empty. Thus, as `I (u) = `I (iu) + 1, induction yields hv ⊆ hti ◦ hiu . Since we are assuming that ht ◦ hu = hti ◦ hiu , this implies that hv ⊆ ht ◦ hu , so that we are done also in this case. Suppose that ht ◦ hu = (hti ◦ hiu ) ∪ (ht ◦ hiu ). Since hv ∩ (ht ◦ hu ) is assumed not to be empty, this implies that one of the two intersections hv ∩ (hti ◦ hiu ) and

hv ∩ (ht ◦ hiu )

is not empty. Thus, as `I (u) = `I (iu) + 1, induction yields hv ⊆ hti ◦ hiu

or

hv ⊆ ht ◦ hiu .

Since we are assuming that ht ◦ hu = (hti ◦ hiu ) ∪ (ht ◦ hiu ), this implies that hv ⊆ ht ◦ hu , so that we are done also in this case. Assume now that t ∈ I1 (i). Then, since u−1 ∈ I−1 (i), we may apply Lemma 10.1.6(i) to t and u in place of u and v. We obtain that ht ◦ hu = hti ◦ hiu . Then, as hv ∩ (ht ◦ hu ) is assumed not to be empty, hv ∩ (hti ◦ hiu ) is not empty. Thus, as `I (u) = `I (iu) + 1, induction yields hv ⊆ hti ◦ hiu . Since ht ◦ hu = hti ◦ hiu ,  this implies that hv ⊆ ht ◦ hu . The final lemma of this section is independent of all previous results in this section and will be needed only in Lemma 10.4.2, in Lemma 10.4.3, and in Theorem 10.6.1. Lemma 10.1.8 Let (W, I) be a Coxeter system, let (X, δ) be a building of type (W, I), and let α be an I-length preserving automorphism of W. For any two elements y and z in X, define (y, z) := δ(y, z)α . Then (X, ) is a building of type (W, I). Proof. Let y and z be elements in X, and let w be an element in W with (y, z) = w. From (y, z) = w we obtain that δ(y, z) = w α . −1

Since α is an automorphism of W, we have w = 1 if and only if w α = 1. Since −1 −1 (X, δ) is a building of type (W, I) and δ(y, z) = w α , we have w α = 1 if and only if y = z. This shows that w = 1 if and only if y = z, so that (X, ) satisfies Condition B1. −1

To show that (X, ) satisfies Condition B2, we choose elements i in I and x in X with (z, x) = i. We have to show that (y, x) = wi or (y, x) = w. Moreover, we have to show that (y, x) = wi if w ∈ I1 (i). From (z, x) = i we obtain that

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10 Regular Actions of (Twin) Coxeter Hypergroups

δ(z, x) = i α . −1

Since α is assumed to be I-length preserving, we have `I (i) = `I (i α ). Thus, as i ∈ I, −1 i α ∈ I. −1

Now recall that (X, δ) is assumed to be a building of type (W, I). Thus, since −1 −1 −1 δ(y, z) = w α , i α ∈ I, and δ(z, x) = i α , we have δ(y, x) = w α i α −1

If, in addition, w α

−1

−1

δ(y, x) = w α . −1

or

∈ I1 (i α ), δ(y, x) = w α i α . −1

−1

−1

If δ(y, x) = w α i α , we obtain that δ(y, x) = (wi)α , so that (y, x) = wi. If −1 δ(y, x) = w α , we obtain that (y, x) = w. −1

−1

−1

If, in addition, w ∈ I1 (i), we obtain from Theorem 9.9.1(vii) (together with Lemma −1 −1 −1 −1 −1 9.9.4) that w α ∈ I1 (i α ). Thus, δ(y, x) = w α i α . It follows that δ(y, x) = (wi)α , and then (y, x) = wi. To show that (X, ) satisfies Condition B3, we choose an element i in I. We have to find an element x in X with (y, x) = wi and (z, x) = i. Since α is assumed to be I-length preserving, we have `I (i) = `I (i α ). Thus, as i ∈ I, −1 i α ∈ I. −1

Now recall that (X, δ) is assumed to be a building of type (W, I). Thus, as δ(y, z) = −1 −1 w α and i α ∈ I, X contains an element x with δ(y, x) = w α i α −1

δ(z, x) = i α . −1

−1

and

From δ(y, x) = w α i α we obtain that δ(y, x) = (wi)α , and then (y, x) = wi. −1  From δ(z, x) = i α we obtain that (z, x) = i. −1

−1

−1

10.2 Regular Actions of Coxeter Hypergroups Let (W, I) be a Coxeter system. In this section, we first show that each regular action of a Coxeter hypergroup H over a set T of involutions of H together with a type preserving bijection from T to I gives rise to a semiregular building of type (W, I).3 After that, we will see that, conversely, each semiregular building of type (W, I) gives rise to a regular action of a Coxeter hypergroup H over a set T of involutions of H and a type preserving bijection from T to I. 3The assumption of the existence of a type preserving bijection from T to I is not at all a restriction of the class of all Coxeter hypergroups. In [3; Theorem 1.1.2], it was shown that (up to isomorphism) there is a bijective correspondence between Coxeter matrices (as defined in [3]) and Coxeter systems. Thus, each Coxeter set T of involutions of a hypergroup comes with a Coxeter system (W, I) and a type preserving bijection from T to I.

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327

To prove the first statement we introduce the following notation. Let H be a hypergroup, let T be a set of involutions of H, and assume that H is a Coxeter hypergroup over T. Let X be a set, and let ω be a regular action of H on X. The color defined by ω will be denoted by σ. Let (W, I) be a Coxeter system, and let ι be a type preserving bijection from T to I. The signature induced by ι will be denoted by τ. From Lemma 1.6.1 we obtain that σ is a surjective map from X × X to H, and in Theorem 9.9.1(ii), we saw that τ is a bijective map from H to W. Thus, the composite στ of σ and τ is a surjective map from X × X to W. This surjective map will be denoted by δ, so that, for any two elements y and z in X, δ(y, z) = (y, z)στ . We call δ the W-distance function induced by ω and ι. Theorem 10.2.1 Let H be a hypergroup, let T be a set of involutions of H, and assume that H is a Coxeter hypergroup over T. Let X be a set, and let ω be a regular action of H on X. Let (W, I) be a Coxeter system, let ι be a type preserving bijection from T to I, and let δ denote the W-distance function induced by ω and ι. Then (X, δ) is a semiregular building of type (W, I). Proof. Let σ denote the color defined by ω, and let τ denote the signature induced by ι. Since δ is the W-distance function induced by ω and ι, δ is the composite of σ and τ. To show that (X, δ) is a building of type (W, I), we choose elements y and z in X and w in W, and we assume that δ(y, z) = w. From Theorem 9.9.1(ii) we know that τ is a bijective map from H to W. Thus, as w ∈ W, H contains an element d with d τ = w. Since δ(y, z) = w, this implies that δ(y, z) = d τ . Thus, as δ(y, z) = (y, z)στ , we obtain that (y, z)στ = d τ , and, since τ is injective, this implies that (y, z)σ = d. Thus, by Lemma 1.6.7(i), z ∈ yd. To show that (X, δ) satisfies Condition B1, we recall that, by Theorem 9.9.1(iii), τ is (T, I)-length preserving. Thus, as d τ = w, we have d = 1 if and only w = 1. Since z ∈ yd, this implies that w = 1 if and only if y = z. To show that (X, δ) satisfies Condition B2, we choose elements i in I and x in X, and we assume that δ(z, x) = i. We have to show that δ(y, x) = wi or δ(y, x) = w. Moreover, we have to show that δ(y, x) = wi if w ∈ I1 (i). From Theorem 9.9.1(i) we know that τ|T = ι. Thus, as ι is a bijection from T to I and i ∈ I, T contains an element t with

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t τ = i. Since δ(z, x) = i, this implies that δ(z, x) = t τ . Thus, as δ(z, x) = (z, x)στ , we obtain that (z, x)στ = t τ , and, since τ is injective, this implies that (z, x)σ = t. Thus, by Lemma 1.6.7(i), x ∈ zt. From x ∈ zt and z ∈ yd, we obtain that x ∈ ydt. Set

c := (y, x)σ

and

v := cτ .

Then, since δ(y, x) = (y, x)στ , δ(y, x) = v. Thus, we have to show that v = wi or v = w. Moreover, we have to show that v = wi if w ∈ I1 (i). From (y, x)σ = c we obtain that x ∈ yc; cf. Lemma 1.6.7(i). Thus, as x ∈ ydt, c ∈ dt; cf. Lemma 1.6.2(i). It follows that `T (c) ∈ {`T (d) − 1, `T (d), `T (d) + 1}. cf. Lemma 2.3.6(i), (iii). Suppose that `T (c) = `T (d)−1. Then `T (d) = `T (c)+1. On the other hand, as c ∈ dt, d ∈ ct; cf. Lemma 6.1.1(i). Thus, by Theorem 9.9.1(iv), d τ = cτ t τ . Since d τ = w, cτ = v, and t τ = i, this implies that w = vi; equivalently, v = wi. Suppose next that `T (c) = `T (d). Then, as chti = dhti, we obtain from Lemma 8.2.1 that c = d. Since cτ = v and d τ = w, this implies that v = w. Suppose now that `T (c) = `T (d) + 1. Then, as c ∈ dt, we obtain from Theorem 9.9.1(iv) that cτ = d τ t τ . Since cτ = v, d τ = w, and t τ = i, this implies that v = wi. Assume that w ∈ I1 (i). Then, as d τ = w and t τ = i, d τ ∈ I1 (t τ ). Thus, by Theorem 9.9.1(vi), d ∈ T1 (t). Since c ∈ dt, this implies that `T (c) = `T (d) + 1. From c ∈ dt and `T (c) = `T (d) + 1 we obtain that cτ = d τ t τ ; cf. Theorem 9.9.1(iv). Thus, as cτ = v, d τ = w, and t τ = i, we obtain v = wi, as wanted. To show that (X, δ) satisfies Condition B3, we choose an element i in I. We have to show that X contains an element x with δ(y, x) = wi and δ(z, x) = i. Recall that τ is a bijective map from H to W. Thus, as wi ∈ W, H contains an element c with cτ = wi. Recall also that τ|T = ι. Thus, as ι is a bijection from T to I and i ∈ I, T contains an element t with t τ = i.

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329

We will now see that d ∈ ct. Assume first that d ∈ T−1 (t). Then, by definition, dt contains an element b with `T (d) = `(b) + 1. From b ∈ dt we obtain that d ∈ bt; cf. Lemma 6.1.1(i). Thus, as `T (d) = `T (b) + 1, we obtain from Theorem 9.9.1(iv) that d τ = bτ t τ . Since d τ = w and t τ = i, that means that w = bτ i. It follows that bτ = wi. Thus, as cτ = wi, bτ = cτ . It follows that b = c, since τ is injective. From b = c and d ∈ bt we obtain that d ∈ ct. Assume now that d < T−1 (t). Then, by Theorem 6.7.5, d ∈ T1 (t). Thus, dt contains an element e such that `T (e) = `T (d) + 1. From e ∈ dt and `T (e) = `T (d) + 1 we obtain that eτ = d τ t τ ; cf. Theorem 9.9.1(iv). Since d τ = w and t τ = i, this means that eτ = wi. Thus, as cτ = wi, cτ = eτ . It follows that c = e, since τ is injective. From c = e and e ∈ dt we obtain that c ∈ dt, so that we have d ∈ ct also in this case. From z ∈ yd and d ∈ ct we now obtain that z ∈ yct. Thus, by Lemma 1.6.4, yc ∩ zt is not empty. We choose an element in yc ∩ zt and denote it by x. From x ∈ yc we obtain that (y, x)σ = c; cf. Lemma 1.6.7(i). Thus, δ(y, x) = (y, x)στ = cτ = wi. From x ∈ zt we obtain that (z, x)σ = t, whence δ(z, x) = (z, x)στ = t τ = i. So far, we have seen that (X, δ) is a building of type (W, I). We still have to show that (X, δ) is semiregular. To show this, we recall from Section 10.1 that, for each element w in W, hw is our notation for the set of all pairs (y, z) of elements in X with δ(y, z) = w. We will see that, for each element i in I, hi is either of first type or of second type. Let i be an element in I. As before, we find an element t in T with t τ = i. It follows that, for any two elements y and z in X, (y, z) ∈ hi

⇔

δ(y, z) = t τ

⇔

(y, z)σ = t

⇔

z ∈ yt;

cf. Lemma 1.6.7(i). Assume first that t is thin, and let y and z be elements in X satisfying (y, z) ∈ hi ◦ hi . Then X contains an element x such that (y, x) ∈ hi and (x, z) ∈ hi . It follows that x ∈ yt and z ∈ xt. Thus, we have z ∈ yt 2 . Since t is assumed to be thin, t 2 = {1}. Thus, z ∈ y ·1, so that (y, z) ∈ h1 . Since y and z have been chosen arbitrarily in X with (y, z) ∈ hi ◦ hi , we have seen that hi ◦ hi ⊆ h1 , and that means that hi is of first type. Now assume that t is not thin, and let y and z be elements in X with (y, z) ∈ hi . From (y, z) ∈ hi we obtain that z ∈ yt. On the other hand, as t is not thin, t ∈ t 2 ; cf. Lemma 6.1.1(iv). Thus, z ∈ yt 2 , so that yt contains an element x with z ∈ xt. From

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10 Regular Actions of (Twin) Coxeter Hypergroups

x ∈ yt we obtain that (y, x) ∈ hi , from z ∈ xt we obtain that (x, z) ∈ hi . It follows that (y, z) ∈ hi ◦ hi . Since y and z have been chosen arbitrarily in X with (y, z) ∈ hi , we have seen that  hi ⊆ hi ◦ hi , and that says that hi is of second type. Let (W, I) be a Coxeter system. Theorem 10.2.1 shows that each regular action of a Coxeter hypergroup H over a set T of involutions of H together with a type preserving bijection from T to I gives rise to a semiregular building of type (W, I). The semiregular building of type (W, I) which one obtains via Theorem 10.2.1 from a regular action ω of a Coxeter hypergroup H over a set T of involutions of H together with a type preserving bijection ι from T to I will be called the building induced by ω and ι. We will now see that, conversely, each semiregular building of type (W, I) gives rise to a regular action of a Coxeter hypergroup H over a set T of involutions of H and a type preserving bijection from T to I. Lemma 10.2.2 Let (W, I) be a Coxeter system, and let (X, δ) be a semiregular building of type (W, I). Define hw := {(y, z) ∈ X × X | δ(y, z) = w} for each element w in W, and set H := {hw | w ∈ W }. Then H is a set theoretic hypergroup on X. Proof. The definition of H implies that H is a partition of X × X. From Lemma 10.1.1 we know that h1 = 1X . Thus, as h1 ∈ H, 1X ∈ H. From Lemma 10.1.5 we know that, for each element w in W, hw∪ = hw −1 . Thus, we have hw∪ ∈ H for each element w in W. From Lemma 10.1.7 we know that, for any three elements t, u, and v in W, hv ⊆ ht ◦hu if hv ∩ (ht ◦ hu ) is not empty. Now, by Lemma 1.6.5, H is a set theoretic hypergroup on X.



The hypergroup which one obtains from a semiregular building (X, δ) via Lemma 10.2.2 will be called the hypergroup induced by (X, δ). Lemma 10.2.3 Let (W, I) be a Coxeter system, let (X, δ) be a semiregular building of type (W, I), and let H denote the hypergroup induced by (X, δ). Define T := {hi | i ∈ I}. Then the following hold. (i) Each element in T is an involution of H. (ii) We have H = hTi. Proof. (i) Let t be an element in T. We have to show that t is an involution. For this, it is enough to show that t ∗ = t and that t 2 ⊆ {1, t}. Since t ∈ T, I contains an element i such that t = hi .

10.2

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331

Since H is a set theoretic hypergroup on X, hi∗ = hi∪ . On the other hand, by Lemma 10.1.5, hi∪ = hi . Thus, hi∗ = hi . Since t = hi , this means that t ∗ = t. Since (X, δ) is assumed to be semiregular, hi is either of first or of second type. Thus, as i ∈ I−1 (i), we obtain from Lemma 10.1.3(ii), (iii) that hi ◦ hi ⊆ h1 ∪ hi . Thus, the definition of the set theoretic hypermultiplication implies that hi2 ⊆ {h1, hi }. Since t = hi , this implies that t 2 ⊆ {1, t}. (ii) Let h be an element in H. We have to show that h ∈ hTi. Since h ∈ H, W contains an element w such that h = hw . Thus, we have to show that hw ∈ hTi. We proceed by induction on `I (w). If `I (w) = 0, w = 1. Then hw = h1 . Thus, as h1 ∈ hTi, hw ∈ hTi, as wanted. Assume that 1 ≤ `I (w). Then w , 1, so that we find elements v in W and i in I with w = vi

and `I (w) = `I (v) + 1.

From w = vi and `I (w) = `I (v) + 1 we obtain that v ∈ I1 (i). Thus, by Lemma 10.1.3(i), hv ◦ hi = hw . It follows that hw ∈ hv hi . On the other hand, since `I (w) = `I (v) + 1, induction yields that hv ∈ hTi. Thus, as hi ∈ T, hv hi ⊆ hTi. From hw ∈ hv hi and hv hi ⊆ hTi we obtain that hw ∈ hTi, as wanted.



Let (W, I) be a Coxeter system, let (X, δ) be a semiregular building of type (W, I), and let H denote the hypergroup induced by (X, δ). The set T of involutions which we found in Lemma 10.2.3(i) will be called the set of involutions induced by (X, δ). Lemma 10.2.4 Let (W, I) be a Coxeter system, let (X, δ) be a semiregular building of type (W, I), and let T denote the set of involutions induced by (X, δ). Then hw 7→ w is a (T, I)-length preserving bijective map from hTi to W. Proof. By Lemma 10.2.3(ii), hw 7→ w is a bijective map from hTi to W. To show that hw 7→ w is (T, I)-length preserving, we fix an element in W and denote it by w. We have to show that `T (hw ) = `I (w) and proceed by induction on `I (w). If `I (w) = 0, w = 1. Then hw = h1 , so that `T (hw ) = 0. It follows that `T (hw ) = `I (w), so that we are done in this case. Assume that 1 ≤ `I (w). Then w , 1, so that we find elements v in W and i in I with w = vi

and `I (w) = `I (v) + 1.

From w = vi and `I (w) = `I (v) + 1 we obtain that v ∈ I1 (i). Thus, by Lemma 10.1.3(i), hv ◦ hi = hw ; equivalently,

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10 Regular Actions of (Twin) Coxeter Hypergroups

hv hi = {hw }. From w , 1 we also obtain that hw , h1 . Thus, by Lemma 2.3.7(ii) (together with Lemma 10.2.3(ii)), there exist elements u in W and i in I such that hw ∈ hu hi

and `T (hw ) = `T (hu ) + 1.

From hv hi = {hw } and hw ∈ hu hi we obtain that hv hi ⊆ hu hi . Now recall from Lemma 10.2.3(i) that hi is an involution of hTi. Thus, by Lemma 6.1.1(iii), (iv), hi hi ⊆ {h1, hi }. It follows that hv ∈ {hu } ∪ hu hi . Assume that hv ∈ hu hi . Then hu ∈ hv hi = {hw }. It follows that hu = hw , contrary to `T (hw ) = `T (hu ) + 1. Thus, as hv ∈ {hu } ∪ hu hi , we have hv = hu . From hv = hu , together with `T (hw ) = `T (hu ) + 1, we obtain that `T (hw ) = `T (hv ) + 1. On the other hand, since `I (w) = `I (v) + 1, induction yields `T (hv ) = `I (v). From `T (hw ) = `T (hv ) + 1, `T (hv ) = `I (v), and `I (w) = `I (v) + 1 we obtain that  `T (hw ) = `I (w), as wanted. Corollary 10.2.5 Let (W, I) be a Coxeter system, let (X, δ) be a semiregular building of type (W, I), and let T denote the set of involutions induced by (X, δ). Let w be an element in W, and let i be an element in I. Then the following hold. (i) We have w ∈ I−1 (i) if and only if hw ∈ T−1 (hi ). (ii) We have w ∈ I1 (i) if and only if hw ∈ T1 (hi ). Proof. Since W = I−1 (i) ∪ I1 (i) and T−1 (hi ) ∩ T1 (hi ) is empty, it suffices to show that hw ∈ T−1 (hi ) follows from w ∈ I−1 (i) and that hw ∈ T1 (hi ) follows from w ∈ I1 (i). (i) Assume that w ∈ I−1 (i). Then wi ∈ I1 (i). Thus, by Lemma 10.1.3(i), hwi ◦hi = hw . It follows that hw ∈ hwi hi . On the other hand, as wi ∈ I1 (i), `I (w) = `I (wi) + 1. Thus, by Lemma 10.2.4, `T (hw ) = `T (hwi ) + 1. From hw ∈ hwi hi and `T (hw ) = `T (hwi ) + 1 we obtain that hw ∈ T−1 (hi ). (ii) Assume that w ∈ I1 (i). Then, by Lemma 10.1.3(i), hw ◦ hi = hwi . It follows that hwi ∈ hw hi . On the other hand, as w ∈ I1 (i), `I (wi) = `I (w) + 1. Thus, by Lemma 10.2.4,

Regular Actions of Coxeter Hypergroups

10.2

333

`T (hwi ) = `T (hw ) + 1. From hwi ∈ hw hi and `T (hwi ) = `T (hw ) + 1 we obtain that hw ∈ T1 (hi ).



Lemma 10.2.6 Let (W, I) be a Coxeter system, let (X, δ) be a semiregular building of type (W, I), and let T denote the set of involutions induced by (X, δ). Then T is a Coxeter set. Proof. We first show that T is constrained. For this, we choose an element t in T and an element h in T1 (t). We have to show that |ht| = 1. Since h ∈ hTi, W contains an element w with h = hw . Similarly, since t ∈ T, t = hi for some element i in I. Thus, as h ∈ T1 (t), hw ∈ T1 (hi ). Now Corollary 10.2.5(ii) yields w ∈ I1 (i). It follows that hw ◦ hi = hwi ; cf. Lemma 10.1.3(i). From hw ◦ hi = hwi we obtain that hw hi = {hwi }. Thus, as h = hw and t = hi , |ht| = 1, as wanted. Now we show that T satisfies the exchange condition. For this, we fix elements p and q in T and an element c in T1 (q) with p ∈ T1 (c). Since we have seen already that T is constrained, it suffices to show that pc = cq

or

pc ⊆ T1 (q);

cf. Lemma 9.1.1. Since p ∈ T1 (c), |pc| = 1; cf. Lemma 8.1.1. Thus, pc contains an element d with pc = {d}. Similarly, cq contains an element e with cq = {e}. Since {c, d, e} ⊆ hTi and {p, q} ⊆ T, there exist elements t, u, v in W and j, k in I such that c = h t , d = hu , e = h v , p = h j , q = h k . From p ∈ T1 (c), p = h j , and c = ht , we obtain that h j ∈ T1 (ht ). Thus, by Corollary 10.2.5(ii), j ∈ I1 (t). It follows that h j ◦ ht = h jt ; cf. Corollary 10.1.4. Thus, by definition, h j ht = {h jt }. Since p = h j and c = ht , this implies that pc = {h jt }. Now recall that pc = {d} and that d = hu . Thus, h jt = hu ; equivalently, jt = u. Similarly, one obtains from c ∈ T1 (q), c = ht , q = hk , and cq = {e} that t ∈ I1 (k)

and tk = v.

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10 Regular Actions of (Twin) Coxeter Hypergroups

Now recall that (W, I) is a Coxeter system. Thus, as j ∈ I1 (t) and t ∈ I1 (k), we have jt = t k

or

jt ∈ I1 (k).

Assume first that jt = tk. Then, u = v, because we saw earlier that jt = u and tk = v. Thus, pc = {d} = {hu } = {hv } = {e} = cq, so that we are done in this case. Now assume that jt ∈ I1 (k). Then, as jt = u, u ∈ I1 (k). Then, by Corollary 10.2.5(ii), hu ∈ T1 (hk ). Since d = hu and q = hk this implies that d ∈ T1 (q). Thus, as pc = {d},  we have that pc ⊆ T1 (q), so that we are done also in this case. We are now ready to state and prove the second main result of this section. Theorem 10.2.7 Let (W, I) be a Coxeter system, let (X, δ) be a semiregular building of type (W, I), let H denote the hypergroup induced by (X, δ), and let T denote the set of involutions induced by (X, δ). Then the following hold. (i) The set H is a Coxeter hypergroup over T. (ii) The map (x, hw ) 7→ {y ∈ X | (x, y) ∈ hw } is a regular action of H on X. (iii) The map hi 7→ i is a type preserving bijection from T to I. Proof. (i) In Lemma 10.2.6, we saw that T is a Coxeter set of involutions of H, and from Lemma 10.2.3(ii) we know that H = hTi. Thus, H is a Coxeter hypergroup over T. (ii) The map (x, hw ) 7→ {y ∈ X | (x, y) ∈ hw } is the canonical action of H on X. Thus, by Lemma 1.6.6, it is a regular action of H on X. (iii) It follows from the definition of T that hi 7→ i is a bijective map from T to I. Thus, it remains to be shown that, for any two elements k and l in I, chk ,hl = ck,l . Let k and l be elements in I, and set J := {k, l} and U := {hk , hl }. Recall from the definition of Coxeter numbers that chk ,hl is finite if and only if C(hk , hl ) is not empty. From Corollary 6.6.6 we also know that C(hk , hl ) is not empty if and only if U−1 (U) is not empty. Similarly, ck,l is finite if and only if J−1 (J) is not empty. Now recall that, by Corollary 10.2.5(i), U−1 (U) is not empty if and only if J−1 (J) is not empty. Thus, chk ,hl is finite if and only if ck,l is finite, and we may (and will) assume that chk ,hl and ck,l both are finite. Since chk ,hl is assumed to be finite, C(hk , hl ) is not empty. Thus, by Proposition 6.6.7, min({`U (hw ) | hw ∈ U−1 (U)}) = chk ,hl . Similarly,

10.2

Regular Actions of Coxeter Hypergroups

335

min({`J (w) | w ∈ J−1 (J)}) = ck,l . Let w be an element in J−1 (J) with `J (w) = ck,l . Since w ∈ J−1 (J), hw ∈ U−1 (U); cf. Corollary 10.2.5(i). Thus, chk ,hl ≤ `U (hw ). Now recall from Lemma 10.2.4 that `T (hw ) = `I (w), and, since T and I both are Coxeter sets, that implies that `U (hw ) = `J (w); cf. Lemma 9.1.2. Thus, chk ,hl ≤ ck,l . Let w be an element in W with hw ∈ U−1 (U) and `U (hw ) = chk ,hl . Since hw ∈ U−1 (U), w ∈ J−1 (J); cf. Corollary 10.2.5(i). Thus, ck,l ≤ `J (w). Now recall from Lemma 10.2.4 that `T (hw ) = `I (w), and, since T and I both are Coxeter sets, that implies that `U (hw ) = `J (w); cf. Lemma 9.1.2. Thus, ck,l ≤ chk ,hl . From chk ,hl ≤ ck,l and ck,l ≤ chk ,hl we obtain that chk ,hl = ck,l .



Let (W, I) be a Coxeter system, and let (X, δ) be a semiregular building of type (W, I). With an eye on Theorem 10.2.7(i), we will now call the hypergroup which we obtained from (X, δ) via Lemma 10.2.2 the Coxeter hypergroup induced by (X, δ). The regular action which we found in Theorem 10.2.7(ii) will be called the regular action induced by (X, δ). The type preserving bijection from T to I which we found in Theorem 10.2.7(iii) will be called the type preserving bijection induced by (X, δ). We emphasize that the Coxeter hypergroup induced by a semiregular building is always a set theoretic hypergroup. Lemma 10.2.8 Let (W, I) be a Coxeter system, let (X, δ) be a semiregular building of type (W, I), υ := w for each let ι denote the type preserving bijection induced by (X, δ), and set hw element w in W. Then υ is the signature induced by ι. Proof. Let T denote the set of involutions induced by (X, δ). Then, by Lemma 10.2.4, υ is a (T, I)-length preserving bijective map from hTi to W. Since ι stands for the type preserving bijection induced by (X, δ), we have hiι = i for each element i in I. Thus, υ|T = ι. Let u and v be elements in W, and let i be an element in I with h v ∈ hu hi

and `T (hv ) = `T (hu ) + 1.

From hv ∈ hu hi we obtain that hv ⊆ hu ◦ hi . On the other hand, by Lemma 10.1.3, hu ◦ hi ⊆ hui ∪ hu . Thus, hv ⊆ hui ∪ hu . It follows that hv = hui or hv = hu , so that we have v = ui or v = u.

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10 Regular Actions of (Twin) Coxeter Hypergroups

From `T (hv ) = `T (hu ) + 1 we obtain that `I (v) = `I (u) + 1, since υ is (T, I)-length preserving. Thus, v , u. It follows that v = ui; equivalently, hvυ = huυ hiυ . We have seen that υ is a (T, I)-length preserving bijective map from hTi to W with υ|T = ι satisfying hvυ = huυ hiυ for any three elements u and v in W and i in I with hv ∈ hu hi and `T (hv ) = `T (hu )+1. Thus, by Lemma 9.9.3, υ is the signature induced  by ι.

10.3 Coxeter Hypergroups and Buildings In this section, we will see that regular actions of Coxeter hypergroups and semiregular buildings are equivalent mathematical objects. For this, we need to define what it means for two regular actions of Coxeter hypergroups to be isomorphic and what it means for two buildings to be isomorphic. We first define what it means for two regular actions of Coxeter hypergroups to be isomorphic. Let H and H 0 be hypergroups, let T be a set of involutions of H, and let T 0 be a set of involutions of H 0. Assume that H is a Coxeter hypergroup over T and that H 0 is a Coxeter hypergroup over T 0. Recall from Section 6.5 that a bijective map φ from H to H 0 is called (T, T 0)-length preserving if, for each element h in H, `T (h) = `T 0 (hφ ). Let X and X 0 be sets, let ω be a regular action of H on X, and let ω 0 be a regular action of H 0 on X 0. Recall from Section 3.9 that a hypergroup isomorphism φ from H to H 0 is called an (ω, ω 0)-isomorphism if there exists a bijective map υ from X to X 0 such that, for any two elements x in X and h in H, ω(x, h)υ = ω 0(x υ, hφ ). In Section 3.9, we called the bijective map υ from X to X 0 the supporting bijection of φ. Notice that the inverse of a (T, T 0)-length preserving (ω, ω 0)-isomorphism from H to H 0 is a (T 0, T)-length preserving (ω 0, ω)-isomorphism from H 0 to H. The regular actions ω and ω 0 will be called isomorphic if there exists a (T, T 0)-length preserving (ω, ω 0)-isomorphism from H to H 0. Next we define what it means for two buildings to be isomorphic. Let (W, I) and (W 0, I 0) be Coxeter systems, let (X, δ) be a building of type (W, I), and let (X 0, δ 0) be a building of type (W 0, I 0). A bijective map υ from X to X 0 is called a building isomorphism from (X, δ) to (X 0, δ 0) if there exists an (I, I 0)-length preserving group isomorphism τ from W to W 0 such that, for any two elements y and z in X, δ(y, z)τ = δ 0(y υ, zυ ).

10.3

Coxeter Hypergroups and Buildings

337

The (I, I 0)-length preserving group isomorphism τ from W to W 0 is called the supporting group isomorphism of υ. It follows right from the definition of a building isomorphism that the inverse of a building isomorphism from (X, δ) to (X 0, δ 0) is a building isomorphism from (X 0, δ 0) to (X, δ). The buildings (X, δ) and (X 0, δ 0) are called isomorphic if there exists a building isomorphism from (X, δ) to (X 0, δ 0). Now we are in the position to explain what we mean by saying that regular actions of Coxeter hypergroups and semiregular buildings are equivalent mathematical objects. For this we let (W, I) be a Coxeter system. In Theorem 10.2.1, we have seen that each regular action of a Coxeter hypergroup H over a set T of involutions of H together with a type preserving bijection from T to I gives rise to a semiregular building of type (W, I). In Theorem 10.3.1, we will see that isomorphic regular actions of Coxeter hypergroups give rise to isomorphic semiregular buildings. In Theorem 10.2.7, we have seen that each semiregular building of type (W, I) gives rise to a regular action of a Coxeter hypergroup H over a set T of involutions of H and a type preserving bijection from T to I. In Theorem 10.3.2, we will see that isomorphic semiregular buildings give rise to isomorphic regular actions of Coxeter hypergroups. Theorem 10.2.1 (together with Theorem 10.3.1) assigns to each isomorphism class of regular actions of Coxeter hypergroups an isomorphism class of semiregular buildings. Theorem 10.2.7 (together with Theorem 10.3.2) assigns to each isomorphism class of semiregular buildings an isomorphism class of regular actions of Coxeter hypergroups. In Theorems 10.3.3 and 10.3.4, we will see that these two assignments are inverses of each other. Theorem 10.3.1 Let H and H 0 be hypergroups, let T be a set of involutions of H, and let T 0 be a set of involutions of H 0. Assume that H is a Coxeter hypergroup over T and that H 0 is a Coxeter hypergroup over T 0. Let X and X 0 be sets, let ω be a regular action of H on X, and let ω 0 be a regular action of H 0 on X 0. Let (W, I) and (W 0, I 0) be Coxeter systems, let ι be a type preserving bijection from T to I, and let ι0 be a type preserving bijection from T 0 to I 0. Assume that ω and ω 0 are isomorphic. Then the building induced by ω and ι is isomorphic to the building induced by ω 0 and ι0. Proof. Let δ denote the W-distance function induced by ω and ι, and let δ 0 denote the W 0-distance function induced by ω 0 and ι0. We are assuming that ω and ω 0 are isomorphic, and we have to find a building isomorphism from (X, δ) to (X 0, δ 0). Since ω and ω 0 are assumed to be isomorphic, there exists a (T, T 0)-length preserving (ω, ω 0)-isomorphism from H to H 0. We fix one of these (T, T 0)-length preserving (ω, ω 0)-isomorphisms from H to H 0 and denote it by φ. The supporting bijection of φ will be denoted by υ.

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10 Regular Actions of (Twin) Coxeter Hypergroups

Since φ is a (T, T 0)-length preserving hypergroup isomorphism from H to H 0, φ|T is a type preserving bijection from T to T 0; cf. Lemma 9.9.4(i). Now recall that ι is a type preserving bijection from T to I and that ι0 is a type preserving bijection from T 0 to I 0. Thus, we obtain from Lemma 9.9.5(i) (together with Lemma 9.9.6(i)) that the composite ι−1 φ|T ι0 is a type preserving bijection from I to I 0. From Lemma 9.9.4(ii) we know that φ is the signature induced by φ|T . Let τ denote the signature induced by ι and τ 0 the signature induced by ι0. Then, by Lemma 9.9.5(ii) (together with Lemma 9.9.6(ii)) the composite τ −1 φτ 0 is the signature induced by ι−1 φ|T ι0. Thus, by Theorem 9.9.1(iii), τ −1 φτ 0 is (I, I 0)-length preserving. Moreover, by Theorem 9.9.2, τ −1 φτ 0 is a group isomorphism from W to W 0.4 We claim that υ is a building isomorphism from (X, δ) to (X 0, δ 0) with supporting group isomorphism τ −1 φτ 0. In other words, we claim that υ is a bijective map from X to X 0 such that, for any two elements y and z in X, δ(y, z)τ

−1 φτ 0

= δ 0(y υ, zυ ).

To prove this claim we let y and z be elements in X, and we let σ denote the color defined by ω. Then δ(y, z) = (y, z)στ . Similarly, we let σ 0 denote the color defined by ω 0, so that δ 0(y υ, zυ ) = (y υ, zυ )σ τ . 0 0

Since φ is an (ω, ω 0)-isomorphism from H to H 0 with supporting bijection υ, we obtain from Lemma 3.9.2 that (y, z)σφ = (y υ, zυ )σ . 0

We obtain that δ(y, z)τ

−1 φτ 0

= (y, z)σφτ = (y υ, zυ )σ τ = δ 0(y υ, zυ ), 0

0 0



as wanted. Theorem 10.3.2

Let (W, I) and (W 0, I 0) be Coxeter systems, let (X, δ) be a semiregular building of type (W, I), and let (X 0, δ 0) be a semiregular building of type (W 0, I 0). Let ω denote the regular action induced by (X, δ), and let ω 0 denote the regular action induced by (X 0, δ 0). Assume that (X, δ) and (X 0, δ 0) are isomorphic. Then ω and ω 0 are isomorphic. Proof. For each element w in W, define hw := {(y, z) ∈ X × X | δ(y, z) = w}. 4We say group isomorphism here, because W and W 0 are groups.

10.3

Set

H := {hw | w ∈ W }

Coxeter Hypergroups and Buildings

339

and T := {hi | i ∈ I}.

Then H is the Coxeter hypergroup induced by (X, δ), and T is the set of involutions induced by (X, δ). Similarly, define

hw0 0 := {(y, z) ∈ X 0 × X 0 | δ 0(y, z) = w 0 }

for each element w 0 in W 0, and set H 0 := {hw0 0 | w 0 ∈ W 0 }

and T 0 := {hi00 | i 0 ∈ I 0 }.

Then H 0 is the Coxeter hypergroup induced by (X 0, δ 0), and T 0 is the set of involutions induced by (X 0, δ 0). We are assuming that (X, δ) and (X 0, δ 0) are isomorphic, and we have to find a (T, T 0)-length preserving (ω, ω 0)-isomorphism from H to H 0. Since (X, δ) and (X 0, δ 0) are assumed to be isomorphic, there exists a building isomorphism from (X, δ) to (X 0, δ 0). We fix one of these building isomorphisms and denote it by υ. Its supporting group isomorphism will be denoted by τ. Then we have δ(y, z)τ = δ 0(y υ, zυ ) for any two elements y and z in X. For each element w in W, set

φ

hw := hw0 τ .

We claim that φ is a (T, T 0)-length preserving (ω, ω 0)-isomorphism from H to H 0. First we show that φ is a hypergroup isomorphism from H to H 0. The bijectivity of φ follows from the bijectivity of τ. Now let t, u, and v be elements in W with φ φ φ hv ∈ ht hu . We have to show that hv ∈ ht hu . From hv ∈ ht hu we obtain that hv ⊆ ht ◦ hu . Thus, we find elements x, y, and z in X with (y, x) ∈ ht ,

(x, z) ∈ hu,

and

(y, z) ∈ hv .

From (y, x) ∈ ht we obtain that δ(y, x) = t. It follows that δ 0(y υ, x υ ) = t τ , so that (y υ, x υ ) ∈ ht0τ . Similarly,

(x υ, zυ ) ∈ hu0 τ

and

(y υ, zυ ) ∈ hv0 τ .

We obtain that hv0 τ ∩ (ht0τ ◦ hu0 τ ) is not empty. Thus, we have hv0 τ ⊆ ht0τ ◦ hu0 τ , and φ φ φ then hv0 τ ∈ ht0τ hu0 τ ; equivalently, hv ∈ ht hu , as wanted.

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10 Regular Actions of (Twin) Coxeter Hypergroups

Next we show that φ is (T, T 0)-length preserving. For this, we let η denote the type preserving bijection induced by (X, δ), and we let χ denote the signature of η. Then, χ by Lemma 10.2.8, hw = w for each element w in W. Moreover, by Theorem 9.9.1(ii), χ is a bijective map from H to W, and from Theorem 9.9.1(iii) we know that χ is (T, I)-length preserving. Similarly, we define η 0 to be the type preserving bijection induced by (X 0, δ 0) and χ 0 0 to be the signature of η 0. Then (hw0 0 ) χ = w 0 for each element w 0 in W 0, and χ 0 is a (T 0, I 0)-length preserving bijective map from H 0 to W 0. Since τ is an (I, I 0)-length preserving group isomorphism from W to W 0, we now obtain that the composite χτ( χ 0)−1 is (T, T 0)-length preserving. Note also that φ = χτ( χ 0)−1 . Thus, we have shown that φ is (T, T 0)-length preserving. We still have to show that φ is an (ω, ω 0)-isomorphism from H to H 0. For this, we choose elements y and z in X and w in W with (y, z) ∈ hw . We have z ∈ yhw and

zυ ∈ y υ hw0 τ

⇔ ⇔

(y, z) ∈ hw (y υ, zυ ) ∈ hw0 τ

⇔ ⇔

δ(y, z) = w δ 0(y υ, zυ ) = w τ .

φ

Thus, as hw = hw0 τ and υ is a building isomorphism from (X, δ) to (X 0, δ 0) with φ supporting group isomorphism τ, we have z ∈ yhw if and only if zυ ∈ y υ hw , so that, by Lemma 3.9.1, φ is an (ω, ω 0)-isomorphism from H to H 0.  Theorem 10.3.3 Let H be a hypergroup, let T be a set of involutions of H, and assume that H is a Coxeter hypergroup over T. Let X be a set, and let ω be a regular action of H on X. Let (W, I) be a Coxeter system, and let ι be a type preserving bijection from T to I. Let (X, δ) denote the building induced by ω and ι, and let ω 0 denote the regular action induced by (X, δ). Then ω and ω 0 are isomorphic. Proof. Let σ denote the color defined by ω, and let τ denote the signature induced by ι. Since (X, δ) stands for the building induced by ω and ι, we have δ(y, z) = (y, z)στ for any two elements y and z in X. For each element w in W, we define hw := {(y, z) ∈ X × X | δ(y, z) = w}. By H 0 we denote the Coxeter hypergroup induced by (X, δ). Then H 0 = {hw | w ∈ W }. For each element h in H, we define

Coxeter Hypergroups and Buildings

10.3

341

hˆ := {(y, z) ∈ X × X | (y, z)σ = h}. By Hˆ we denote the ω-shadow of H. Then Hˆ = { hˆ | h ∈ H}. ˆ Our first claim is that H 0 = H. Let h 0 be an element in H 0. Then W contains an element w such that h 0 = hw . It follows that h 0 = {(y, z) ∈ X × X | δ(y, z) = w} = {(y, z) ∈ X × X | (y, z)στ = w}. Now recall from Theorem 9.9.1(ii) that τ is a bijective map from H to W. Thus, as w ∈ W, H contains an element h such that hτ = w. It follows that ˆ h 0 = {(y, z) ∈ X × X | (y, z)στ = hτ } = {(y, z) ∈ X × X | (y, z)σ = h} = h. ˆ It follows that h 0 ∈ H. ˆ Now notice that Since h 0 has been chosen arbitrarily in H 0, this shows that H 0 ⊆ H. ˆ as claimed. H 0 as well as Hˆ are partitions of X × X. Thus, H 0 = H, Let T 0 denote the set of involutions induced by (X, δ), and let ψ denote the ωcanonical hypergroup isomorphism from H to Hˆ (defined in Section 3.9). We claim that ψ is a (T, T 0)-length preserving (ω, ω 0)-isomorphism from H to H 0. We first show that ψ is (T, T 0)-length preserving. For this, we let ι0 denote the type preserving bijection induced by (X, δ), and we let τ 0 denote the signature induced by ι0. We will see that ψ = τ(τ 0)−1 . Let h be an element in H, and set w := hτ . Then, for any two elements y and z in X, (y, z)σ = h

⇔

δ(y, z) = hτ

⇔

δ(y, z) = w.

It follows that hψ = {(y, z) ∈ X × X | (y, z)σ = h} = {(y, z) ∈ X × X | δ(y, z) = w} = hw . Now recall from Lemma 10.2.8 that (hw )τ = w. Thus, as hψ = hw and w = hτ , we 0 have hψτ = hτ , and, since h has been chosen arbitrarily from H, this shows that ψτ 0 = τ. It follows that ψ = τ(τ 0)−1 . 0

From Theorem 9.9.1(iii), (ii) we know that τ is a (T, I)-length preserving bijective map from H to W and τ 0 is a (T 0, I)-length preserving bijective map from H 0 to W. Thus, τ(τ 0)−1 is a (T, T 0)-length preserving bijective map from H to H 0. Thus, as ψ = τ(τ 0)−1 , ψ is a (T, T 0)-length preserving bijective map from H to H 0. Since ω 0 stands for the regular action induced by (X, δ), ω 0 is the canonical action of ˆ this means that ω 0 is the canonical action of Hˆ on X. Thus, H 0 on X. Since H 0 = H, ˆ Thus, as H 0 = H, ˆ ψ is by Lemma 3.9.4, ψ is an (ω, ω 0)-isomorphism from H to H. 0 0 an (ω, ω )-isomorphism from H to H .

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We have seen that ψ is a (T, T 0)-length preserving (ω, ω 0)-isomorphism from H to  H 0. Thus, ω and ω 0 are isomorphic. Theorem 10.3.4 Let (W, I) be a Coxeter system, and let (X, δ) be a semiregular building of type (W, I). Let ω denote the regular action induced by (X, δ), and let ι denote the type preserving bijection induced by (X, δ). Then the building induced by ω and ι is equal to (X, δ). Proof. Let H denote the Coxeter hypergroup induced by (X, δ). Then H = {hw | w ∈ W }, where, for each element w in W, hw := {(y, z) ∈ X × X | δ(y, z) = w}. Let σ denote the color defined by ω, let τ denote the signature induced by ι, and let  denote the composite of σ and τ. We shall be done when we succeed in showing that, for any two elements y and z in X, (y, z) = δ(y, z). Let y and z be elements in X, and set w := δ(y, z). From δ(y, z) = w we obtain that (y, z) ∈ hw . Since ω stands for the regular action induced by (X, δ), ω is the canonical action of H on X. Thus, we have z ∈ ω(y, hw ). Since σ is the color defined by ω, this implies that (y, z)σ = hw . It follows that (y, z) = (y, z)στ = hτw . On the other hand, as ι is the type preserving bijection induced by (X, δ) and τ the signature induced by ι, hτw = w; cf. Lemma 10.2.8. From (y, z) = hτw , hτw = w, and w = δ(y, z) we now obtain that (y, z) = δ(y, z).  Theorems 10.3.3 and 10.3.4 (together with Theorems 10.2.1, 10.2.7, 10.3.1, and 10.3.2) show that (the isomorphism classes of) semiregular buildings can be identified with (the isomorphism classes of) regular actions of Coxeter hypergroups. The purpose of the remaining sections of this chapter is to establish (modulo Hypothesis 10.5.6) a similar theorem for thick twin buildings.

10.4 Twin Buildings and Binary Relations, I We begin with the definition of a twin building. Our definition is taken from [1; p. 266].

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Let (W, I) be a Coxeter system, let (X−, δ− ) and (X+, δ+ ) be buildings of type (W, I), and let δ◦ be a map from (X− × X+ ) ∪ (X+ × X− ) to W. We assume that X− ∩ X+ is empty. The triple ((X−, δ− ), (X+, δ+ ), δ◦ ) is called a twin building of type (W, I) if, for any four elements  in {−, +}, y in X− , z in X , and w in W with δ◦ (y, z) = w, the following conditions hold. T1

We have δ◦ (z, y) = w −1 .

T2

For any two elements i in I with w ∈ I−1 (i) and x in X with δ (z, x) = i, we have δ◦ (y, x) = wi.

T3

For each element i in I, there exists an element x in X with δ◦ (y, x) = wi and δ (z, x) = i.

Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. For each element w in W, we define fw− := {(y, z) ∈ X− × X− | δ− (y, z) = w π¯ } and

fw+ := {(y, z) ∈ X+ × X+ | δ+ (y, z) = w},

and we set fw := fw− ∪ fw+ . The purpose of this section is to compile several results on the sets fw with w ∈ W. We begin with a lemma which is similar to Lemma 10.1.1. Lemma 10.4.1 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a twin building of type (W, I), let π¯ be an I-length preserving automorphism of W having order 1 or 2, and set X := X− ∪ X+ . Then f1 = 1X . Proof. Considering the definition of f1 this follows from Condition B1.



Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a twin building of type (W, I), let π¯ be an I-length preserving automorphism of W having order 1 or 2, and set X := X− ∪ X+ . Since the sets fw with w ∈ W are subsets of X × X (that are contained in (X− × X− ) ∪ (X+ × X+ )), we may adopt the following notation from Section 1.6. For each element w in W, we write fw∪ to denote the set of all pairs (y, z) in X × X with (z, y) ∈ fw . For any two elements u and v in W, we write fu ◦ fv to denote the set of all pairs (y, z) in X × X for which we can find an element x in X with (y, x) ∈ fu and (x, z) ∈ fv . The following lemma is similar to Lemma 10.1.2. Lemma 10.4.2 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Let i be an element in I. Then the following statements hold.

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(i) We have fi∪ = fi . (ii) We have f1 ⊆ fi ◦ fi ⊆ f1 ∪ fi . Proof. Recall that (X−, δ− ) is a building of type (W, I). For any two elements y and z in X− , define − (y, z) := δ− (y, z)π¯ . Since π¯ is an I-length preserving automorphism of W, (X−, − ) is a building of type (W, I); cf. Lemma 10.1.8. Note also that fw− = {(y, z) ∈ X− × X− | − (y, z) = w} for each element w in W. (i) Since (X−, − ) is a building of type (W, I), ( fi− )∪ = fi− ; cf. Lemma 10.1.2(i). Similarly, as (X+, δ+ ) is a building of type (W, I), ( fi+ )∪ = fi+ . Thus, we have fi∪ = ( fi− )∪ ∪ ( fi+ )∪ = fi− ∪ fi+ = fi . (ii) Since (X−, − ) is a building of type (W, I), we have f1− ⊆ fi− ◦ fi− ⊆ f1− ∪ fi− ; cf. Lemma 10.1.2(ii). Similarly, as (X+, δ+ ) is a building of type (W, I), f1+ ⊆ fi+ ◦ fi+ ⊆ f1+ ∪ fi+ . Thus, we have f1 = f1− ∪ f1+ ⊆ ( fi− ◦ fi− ) ∪ ( fi+ ◦ fi+ ) = ( fi− ∪ fi+ ) ◦ ( fi− ∪ fi+ ) = fi ◦ fi and fi ◦ fi = ( fi− ◦ fi− ) ∪ ( fi+ ◦ fi+ ) ⊆ ( f1− ∪ fi− ) ∪ ( f1+ ∪ fi+ ) = f1 ∪ fi, as wanted.



Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. In the following, we will say that fi is of first type if f1 = fi ◦ fi and that fi is of of second type if fi ◦ fi = f1 ∪ fi . Of course, fi may be neither of first nor of second type. The following lemma is similar to Lemma 10.1.3. Also the proofs of its last two parts are similar to the proofs of the last two parts of Lemma 10.1.3. Lemma 10.4.3 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Let w be an element in W, and let i be an element in I. Then the following hold.

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(i) If w ∈ I1 (i), fw ◦ fi = fwi . (ii) Assume that w ∈ I−1 (i) and that fi is of first type. Then fw ◦ fi = fwi . (iii) Assume that w ∈ I−1 (i) and that fi is of second type. Then fw ◦ fi = fwi ∪ fw . Proof. (i) Recall that (X−, δ− ) is a building of type (W, I). For any two elements y and z in X− , define − (y, z) := δ− (y, z)π¯ . Since π¯ is an I-length preserving automorphism of W, (X−, − ) is a building of type (W, I); cf. Lemma 10.1.8. Note also that fw− := {(y, z) ∈ X− × X− | − (y, z) = w} for each element w in W. Thus, by Lemma 10.1.3(i), − fw− ◦ fi− = fwi .

Similarly, as (X+, δ+ ) is a building of type (W, I), + . fw+ ◦ fi+ = fwi

It follows that + − ∪ fwi fw ◦ fi = ( fw− ∪ fw+ ) ◦ ( fi− ∪ fi+ ) = ( fw− ◦ fi− ) ∪ ( fw+ ◦ fi+ ) = fwi = fwi,

as wanted. (ii), (iii) We are assuming that w ∈ I−1 (i), and that is equivalent to wi ∈ I1 (i). Thus, by (i), fwi ◦ fi = fw . It follows that fw ◦ fi = fwi ◦ fi ◦ fi . If fi is of first type, we have fi ◦ fi = f1 , so that, by the above equation, fw ◦ fi = fwi . If fi is of second type, we have fi ◦ fi = f1 ∪ fi , so that, by the above equation, fw ◦ fi = fwi ◦ ( f1 ∪ fi ) = ( fwi ◦ f1 ) ∪ ( fwi ◦ fi ) = fwi ∪ fw, as wanted.



The following result is similar to Corollary 10.1.4. Also its proof is similar to the proof of Corollary 10.1.4. Corollary 10.4.4 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Let u and v be elements in W with u ∈ I1 (v). Then fu ◦ fv = fuv . Proof. We proceed by induction on `I (v).

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10 Regular Actions of (Twin) Coxeter Hypergroups

If `I (v) = 0, v = 1. In this case, we have fu ◦ fv = fu ◦ f1 = fu = fuv , as wanted. Assume that 1 ≤ `I (v). Then v , 1, so that we find elements t in W and i in I with v = ti

and `I (v) = `I (t) + 1.

Since we are assuming that u ∈ I1 (v), we also have `I (uv) = `I (u) + `I (v). Thus, by Lemma 2.3.8(i), `I (u) + `I (t) = `I (ut) and `I (uv) = `I (ut) + 1. From the first of these two equations we obtain that u ∈ I1 (t). Thus, as `I (v) = `I (t) + 1, induction yields fu ◦ ft = fut . From the second equation (together with v = ti) we obtain that ut ∈ I1 (i). Thus, as v = ti, Lemma 10.4.3(i) yields fut ◦ fi = fuv . From v = ti and `I (v) = `I (t) + 1 we obtain that t ∈ I1 (i). Thus, as v = ti, Lemma 10.4.3(i) also yields ft ◦ fi = fv . From ft ◦ fi = fv , fu ◦ ft = fut , and fut ◦ fi = fuv we now obtain fu ◦ fv = fuv , as  wanted. The following lemma is similar to Lemma 10.1.5. Also its proof is similar to the proof of Lemma 10.1.5. Lemma 10.4.5 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Then, for each element w in W, fw∪ = fw −1 . Proof. We proceed by induction on `I (w). If `I (w) = 0, w = 1. In this case, we have fw∪ = f1∪ = f1 = f1−1 = fw −1 , as wanted.

Assume that 1 ≤ `I (w). Then w , 1, so that we find elements v in W and i in I with w = vi

and `I (w) = `I (v) + 1.

From these two equations we obtain that v ∈ I1 (i). Thus, as w = vi, Lemma 10.4.3(i) yields fv ◦ fi = fw . It follows that fw∪ = fi∪ ◦ fv∪ . Now recall from Lemma 10.4.2(i) that fi∪ = fi . Furthermore, since `I (w) = `I (v)+1, induction yields fv∪ = fv−1 . Thus, we have

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fi∪ ◦ fv∪ = fi ◦ fv−1 . From v ∈ I1 (i) we obtain that i ∈ I1 (v −1 ); cf. Lemma 6.5.2(i). Thus, as iv −1 = w −1 , Corollary 10.4.4 yields fi ◦ fv−1 = fw −1 . From fw∪ = fi∪ ◦ fv∪ , fi∪ ◦ fv∪ = fi ◦ fv−1 , and fi ◦ fv−1 = fw −1 we obtain that fw∪ = fw −1 ,  as wanted. Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. The twin building ((X−, δ− ), (X+, δ+ ), δ◦ ) will be called semiregular if, for each element i in I, fi is either of first type or of second type. It is called thick if, for each element i in I, fi is of second type.5 Note that the twin building ((X−, δ− ), (X+, δ+ ), δ◦ ) is semiregular if and only if the buildings (X−, δ− ) and (X+, δ+ ) both are semiregular. Similarly, the twin building ((X−, δ− ), (X+, δ+ ), δ◦ ) is thick if and only if the buildings (X−, δ− ) and (X+, δ+ ) both are thick. Lemma 10.4.6 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a semiregular twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Then fv− ∩ ( ft− ◦ fu− ) is not empty if and only if fv+ ∩ ( ft+ ◦ fu+ ) is not empty. Proof. (i) We proceed by induction on `I (v). Assume first that `I (v) = 0. Then v = 1, so that fv− = f1− and fv+ = f1+ .

Suppose that fv− ∩ ( ft− ◦ fu− ) is not empty. Then, as fv− = f1− , f1− ∩ ( ft− ◦ fu− ) is not empty. It follows that ( ft− )∪ = fu− . Now recall from Lemma 10.1.5 that ( ft− )∪ = ft−−1 . Thus, we obtain that ft−−1 = fu− ; equivalently, t −1 = u. From t −1 = u we obtain that ft+−1 = fu+ . It follows that f1+ ∩ ( ft+ ◦ fu+ ) is not empty. Thus, as fv+ = f1+ , we obtain that fv+ ∩ ( ft+ ◦ fu+ ) is not empty.

Similarly, one obtains that fv− ∩ ( ft− ◦ fu− ) is not empty if fv+ ∩ ( ft+ ◦ fu+ ) is not empty, so we are done in the case where `I (v) = 0. Now assume that 1 ≤ `I (v). Then v , 1, so that we find elements v 0 in W and i in I with v = v 0i and `I (v) = `I (v 0) + 1. From v = v 0i and `I (v) = `I (v 0) + 1 we obtain that v 0 ∈ I1 (i). Thus, as v = v 0i, Lemma 10.1.3(i) yields

5The definition of a thick twin building is common in the literature, the one of a semiregular twin building is not.

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10 Regular Actions of (Twin) Coxeter Hypergroups

fv−0 ◦ fi− = fv−

and

fv+0 ◦ fi+ = fv+ .

Suppose that fv− ∩ ( ft− ◦ fu− ) is not empty. Then, as fv−0 ◦ fi− = fv− , ( fv−0 ◦ fi− ) ∩ ( ft− ◦ fu− ) is not empty. It follows that (( ft− )∪ ◦ fv−0 ) ∩ ( fu− ◦ fi− ). is not empty. (Recall from Lemma 10.1.2(i) that ( fi− )∪ = ( fi− ).) Let u 0 be an element in W such that fu−0 ∩ (( ft− )∪ ◦ fv−0 ) and fu−0 ∩ ( fu− ◦ fi− ) both are not empty. Since fu−0 ∩ (( ft− )∪ ◦ fv−0 ) is not empty, fv−0 ∩ ( ft− ◦ fu−0 ) is not empty. Thus, since `I (v) = `I (v 0) + 1, induction yields that fv+0 ∩ ( ft+ ◦ fu+0 ) is not empty. It follows that fu+0 ∩ (( ft+ )∪ ◦ fv+0 ) is not empty. Suppose that u ∈ I1 (i). Then, by Lemma 10.1.3(i), fu− ◦ fi− = fui−

and

fu+ ◦ fi+ = fui+ .

Since fu−0 ∩ ( fu− ◦ fi− ) is not empty, we obtain from fu− ◦ fi− = fui− that fu−0 ∩ fui− is not empty. In this case, we have fu−0 = fui− , and we obtain that u 0 = ui. Now fu+0 = fui+ , and, since fu+ ◦ fi+ = fui+ , this yields that fu+ ◦ fi+ = fu+0 . Thus, as fu+0 ∩ (( ft+ )∪ ◦ fv+0 ) is not empty, (( ft+ )∪ ◦ fv+0 ) ∩ ( fu+ ◦ fi+ ) is not empty in this case. Suppose that u ∈ I−1 (i) and that fi is of first type. Then, by Lemma 10.1.3(ii), fu− ◦ fi− = fui−

and

fu+ ◦ fi+ = fui+ .

Since fu−0 ∩ ( fu− ◦ fi− ) is not empty, we obtain from fu− ◦ fi− = fui− that fu−0 ∩ fui− is not empty. In this case, we have fu−0 = fui− , and we obtain that u 0 = ui. Now fu+0 = fui+ , and, since fu+ ◦ fi+ = fui+ , this yields that fu+ ◦ fi+ = fu+0 . Thus, as fu+0 ∩ (( ft+ )∪ ◦ fv+0 ) is not empty, (( ft+ )∪ ◦ fv+0 ) ∩ ( fu+ ◦ fi+ ) is not empty also in this case. Suppose that u ∈ I−1 (i) and that fi is not of first type. Then, as ((X−, δ− ), (X+, δ+ ), δ◦ ) is assumed to be semiregular, fi is of second type. Thus, by Lemma 10.1.3(iii), fu− ◦ fi− = fui− ∪ fu−

and

fu+ ◦ fi+ = fui+ ∪ fu+ .

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Since fu−0 ∩ ( fu− ◦ fi− ) is not empty and fu− ◦ fi− = fui− ∪ fu− , fu−0 ∩ ( fui− ∪ fu− ) is not empty. In this case, we have fu−0 = fui− or fu−0 = fu− , and we obtain that u 0 = ui or u 0 = u. It follows that fu+0 = fui+ or fu+0 = fu+ , and, since fu+ ◦ fi+ = fui+ ∪ fu+ , this yields that fu+0 ⊆ fu+ ◦ fi+ . Thus, as fu+0 ∩ (( ft+ )∪ ◦ fv+0 ) is not empty, (( ft+ )∪ ◦ fv+0 ) ∩ ( fu+ ◦ fi+ ) is not empty also in this case. Now, as we have seen that, in all cases, ( ft+ )∪ ◦ fv+0 ) ∩ ( fu+ ◦ fi+ ) is not empty, we obtain that ( ft+ ◦ fu+ ) ∩ ( fv+0 ◦ fi+ ) is not empty. Thus, as fv+0 ◦ fi+ = fv+ (by Lemma 10.1.3(i)), we obtain that fv+ ∩ ( ft+ ◦ fu+ ) is not empty, as wanted. Similarly, we obtain that fv− ∩( ft− ◦ fu− ) is not empty if the intersection fv+ ∩( ft+ ◦ fu+ ) is not empty.  Lemma 10.4.7 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a semiregular twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Let t, u, and v be elements in W, and assume that fv ∩ ( ft ◦ fu ) is not empty. Then fv ⊆ ft ◦ fu . Proof. We are assuming that fv ∩ ( ft ◦ fu ) is not empty. Since fw := fw− ∪ fw+ for each element w in W, this means that ( fv− ∪ fv+ ) ∩ (( ft− ∪ ft+ ) ◦ ( fu− ∪ fu+ )) is not empty. It follows that ( fv− ∪ fv+ ) ∩ (( ft− ◦ fu− ) ∪ ( ft+ ◦ fu+ )) is not empty, and that means that one of the two intersections fv− ∩ ( ft− ◦ fu− ) and

fv+ ∩ ( ft+ ◦ fu+ )

is not empty. Now, by Lemma 10.4.6, none of the two intersections is empty. Now, by Lemma 10.1.7, fv− ⊆ ft− ◦ fu−

and

fv+ ⊆ ft+ ◦ fu+ .

It follows that fv = fv− ∪ fv+ ⊆ ( ft− ◦ fu− ) ∪ ( ft+ ◦ fu+ ) = ( ft− ∪ ft+ ) ◦ ( fu− ∪ fu+ ) = ft ◦ fu, as wanted.



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10.5 Twin Buildings and Binary Relations, II Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. For each element w in W, define rw := {(y, z) ∈ X− × X+ | δ◦ (y, z) = w −1 } ∪ {(y, z) ∈ X+ × X− | δ◦ (y, z) = (w −1 )π¯ }. The purpose of this section is to compile several results on the sets rw with w ∈ W. We begin with a lemma which is similar to Lemma 10.4.5. Lemma 10.5.1 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Then, for each element w in W, rw∪ = r(w −1 ) π¯ . Proof. Let w be an element in W, and let y and z be elements in X. We will see that (y, z) ∈ rw∪ if and only if (y, z) ∈ r(w −1 ) π¯ . Assume first that y ∈ X− and z ∈ X+ . Then, as π¯ is an automorphism of W, we have (y, z) ∈ rw∪ ⇔ (z, y) ∈ rw ⇔ δ◦ (z, y) = (w −1 )π¯ ⇔ δ◦ (z, y) = (w π¯ )−1 . From Condition T1, on the other hand, we obtain that (y, z) ∈ r(w −1 ) π¯ ⇔ (y, z) ∈ r(w π¯ )−1 ⇔ δ◦ (y, z) = w π¯ ⇔ δ◦ (z, y) = (w π¯ )−1 . Thus, we are done in this case. Assume now that y ∈ X+ and z ∈ X− . Then, by Condition T1, (y, z) ∈ rw∪ ⇔ (z, y) ∈ rw ⇔ δ◦ (z, y) = w −1 ⇔ δ◦ (y, z) = w. Since π¯ is an automorphism of W having order 1 or 2, we also have (y, z) ∈ r(w −1 ) π¯ ⇔ (y, z) ∈ r(w π¯ )−1 ⇔ δ◦ (y, z) = w π¯

2

⇔ δ◦ (y, z) = w.

This proves the desired equation also in this case.



The following lemma is similar to Lemma 10.1.3 and Lemma 10.4.3. Lemma 10.5.2 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Let w be an element in W, and let i be an element in I. Then the following hold. (i) If w −1 ∈ I−1 (i), rw ◦ fi = riw . (ii) Assume that w −1 ∈ I1 (i) and that fi is of first type. Then rw ◦ fi = riw .

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351

(iii) Assume that w −1 ∈ I1 (i) and that fi is of second type. Then rw ◦ fi = riw ∪rw . Proof. (i) Set X := X− ∪ X+ , and let y and z be elements in X. Assume first that (y, z) ∈ rw ◦ fi . We will see that (y, z) ∈ riw . Suppose first that y ∈ X− and z ∈ X+ . In this case, X+ contains an element x such that (y, x) ∈ rw and (x, z) ∈ fi . From (y, x) ∈ rw we obtain that δ◦ (y, x) = w −1 . From (x, z) ∈ fi we obtain that

δ+ (x, z) = i.

Now assume that ∈ I−1 (i). Then, by Condition T2, δ◦ (y, z) = w −1i = (iw)−1 . Thus, by definition, (y, z) ∈ riw , as wanted. w −1

Suppose now that y ∈ X+ and z ∈ X− . In this case, X− contains an element x such that (y, x) ∈ rw and (x, z) ∈ fi . From (y, x) ∈ rw we obtain that δ◦ (y, x) = (w −1 )π¯ . From (x, z) ∈ fi we obtain that δ− (x, z) = i π¯ . Now assume that w −1 ∈ I−1 (i). Then, as π¯ is an I-length preserving automorphism of W, we obtain from Lemma 6.5.9(i) that (w −1 )π¯ ∈ I−1 (i π¯ ). Thus, by Condition T2, δ◦ (y, z) = (w −1 )π¯ i π¯ = (w −1i)π¯ = ((iw)−1 )π¯ , so that, by definition, (y, z) ∈ riw , as wanted. Assume now, conversely, that (y, z) ∈ riw . We will see that (y, z) ∈ rw ◦ fi . Suppose first that y ∈ X− and z ∈ X+ . In this case, δ◦ (y, z) = (iw)−1 = w −1i. Thus, by Condition T3, X+ contains an element x such that δ◦ (y, x) = w −1 and δ+ (z, x) = i. From δ◦ (y, x) = w −1 we obtain that (y, x) ∈ rw . From δ+ (z, x) = i we obtain that (z, x) ∈ fi , and then, by Lemma 10.4.2(i), that (x, z) ∈ fi . It follows that (y, z) ∈ rw ◦ fi , as wanted. Suppose now that y ∈ X+ and z ∈ X− . In this case, δ◦ (y, z) = ((iw)−1 )π¯ = (w −1i)π¯ = (w −1 )π¯ i π¯ . Thus, by Condition T3, X− contains an element x such that δ◦ (y, x) = (w −1 )π¯ and δ− (z, x) = i π¯ . From δ◦ (y, x) = (w −1 )π¯ we obtain that (y, x) ∈ rw . From δ− (z, x) = i π¯ we obtain that (z, x) ∈ fi , and then, by Lemma 10.4.2(i), that (x, z) ∈ fi . It follows that (y, z) ∈ rw ◦ fi , as wanted.

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10 Regular Actions of (Twin) Coxeter Hypergroups

(ii), (iii) We are assuming that w −1 ∈ I1 (i), and that is equivalent to w −1 i ∈ I−1 (i) or, what is the same, (iw)−1 ∈ I−1 (i). Thus, by (i), riw ◦ fi = rw . It follows that rw ◦ fi = riw ◦ fi ◦ fi . If fi is of first type, we have fi ◦ fi = f1 , so that, by the above equation, rw ◦ fi = fiw . If fi is of second type, we have fi ◦ fi = f1 ∪ fi , so that, by the above equation, rw ◦ fi = riw ◦ ( f1 ∪ fi ) = (riw ◦ f1 ) ∪ (riw ◦ fi ) = riw ∪ rw, 

as wanted.

We now generalize Lemma 10.5.2. In fact, Lemma 10.5.2 is the case (u, v) = (w, i) of the subsequent lemma. The lemma is similar to Lemma 10.1.6. Lemma 10.5.3 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Let u and v be elements in W, and let i be an element in I. Assume that v −1 ∈ I−1 (i). Then the following hold. (i) If u−1 ∈ I−1 (i), ru ◦ fv = riu ◦ fiv . (ii) Assume that u−1 ∈ I1 (i) and that fi is of first type. Then ru ◦ fv = riu ◦ fiv . (iii) Assume that u−1 ∈ I1 (i) and that fi is of second type. Then ru ◦ fv = (riu ◦ fiv ) ∪ (ru ◦ fiv ). Proof. We are assuming that v −1 ∈ I−1 (i), and that is equivalent to v −1i ∈ I1 (i). Thus, by Lemma 6.5.2(i), i ∈ I1 (iv). It follows that fi ◦ fiv = fv ; cf. Corollary 10.4.4. (i) Assume that u−1 ∈ I−1 (i). Then, by Lemma 10.5.2(i), ru ◦ fi = riu . Thus, since fi ◦ fiv = fv , we obtain that ru ◦ fv = riu ◦ fiv . (ii) We are assuming that u−1 ∈ I1 (i) and that fi is of first type. Thus, by Lemma 10.5.2(ii), ru ◦ fi = riu . Since fi ◦ fiv = fv , this implies that ru ◦ fv = riu ◦ fiv . (iii) We are assuming that u−1 ∈ I1 (i) and that fi is of second type. Thus, by Lemma 10.5.2(iii), ru ◦ fi = riu ∪ ru . Since fi ◦ fiv = fv , this implies that ru ◦ fv = ru ◦ fi ◦ fiv = (riu ∪ ru ) ◦ fiv = (riu ◦ fiv ) ∪ (ru ◦ fiv ), as wanted.



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353

Corollary 10.5.4 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Then, for each element w in W, rw ◦ fw = r1 . Proof. We proceed by induction on `I (w). If `I (w) = 0, w = 1. In this case, we have rw ◦ fw = r1 ◦ f1 = r1 , as wanted. Assume that 1 ≤ `I (w). Then we find an element i in I with `I (w) = `I (iw) + 1. It follows that w −1 ∈ I−1 (i), so that we may apply Lemma 10.5.3(i) to w in place of both, u and v. We obtain that rw ◦ fw = riw ◦ fiw . On the other hand, since `I (w) = `I (iw) + 1, induction yields riw ◦ fiw = r1 . Thus, we have rw ◦ fw = r1 .  The subsequent lemma is similar to Lemmas 10.1.7 and 10.4.7. Lemma 10.5.5 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a semiregular twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Let t, u, and v be elements in W. Then the following statements hold. (i) Assume that rv ∩ (rt ◦ fu ) is not empty. Then rv ⊆ rt ◦ fu . (ii) Assume that rv ∩ ( fu ◦ rt ) is not empty. Then rv ⊆ fu ◦ rt . Proof. (i) We proceed by induction on `I (u). Assume first that `I (u) = 0. Then u = 1, so that fu = f1 . It follows that rv ∩ (rt ◦ fu ) = rv ∩ (rt ◦ f1 ) = rv ∩ rt . Since rv ∩ (rt ◦ fu ) is assumed not to be empty, this implies that rv ∩ rt is not empty; equivalently, rv = rt . It follows that rv = rt ◦ f1 = rt ◦ fu , so we are done in this case. Now assume that 1 ≤ `I (u). In this case, we find an element i in I with `I (u) = `I (iu) + 1. It follows that

u−1 ∈ I−1 (i).

Assume first that t −1 ∈ I−1 (i). Then, since u−1 ∈ I−1 (i), we may apply Lemma 10.5.3(i) to t and u in place of u and v. We obtain that rt ◦ fu = rit ◦ fiu . Since rv ∩ (rt ◦ fu ) is assumed not to be empty, this implies that rv ∩ (rit ◦ fiu ) is not empty. Thus, as `I (u) = `I (iu) + 1, induction yields rv ⊆ rit ◦ fiu . Since rt ◦ fu = rit ◦ fiu , this implies that rv ⊆ rt ◦ fu , so that we are done also in this case.

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10 Regular Actions of (Twin) Coxeter Hypergroups

Assume now that t −1 ∈ I1 (i), and recall that we have u−1 ∈ I−1 (i). Recall also that ((X−, δ− ), (X+, δ+ ), δ◦ ) is semiregular, by hypothesis. Thus, applying Lemma 10.5.3(ii), (iii) to t and u in place of u and v, we obtain that rt ◦ fu = rit ◦ fiu

or rt ◦ fu = (rit ◦ fiu ) ∪ (rt ◦ fiu ).

Suppose first that rt ◦ fu = rit ◦ fiu . Then, as rv ∩ (rt ◦ fu ) is assumed not to be empty, rv ∩ (rit ◦ fiu ) is not empty. Thus, as `I (u) = `I (iu) + 1, induction yields rv ⊆ rit ◦ fiu . Since we are assuming that rt ◦ fu = rit ◦ fiu , this implies that rv ⊆ rt ◦ fu , so that we are done also in this case. Suppose now that rt ◦ fu = (rit ◦ fiu ) ∪ (rt ◦ fiu ). Then, as rv ∩ (rt ◦ fu ) is assumed not to be empty, one of the two intersections rv ∩ (rit ◦ fiu )

and rv ∩ (rt ◦ fiu )

is not empty. Thus, as `I (u) = `I (iu) + 1, induction yields rv ⊆ rit ◦ fiu

or rv ⊆ rt ◦ fiu .

Since we are assuming that rt ◦ fu = (rit ◦ fiu )∪(rt ◦ fiu ), this implies that rv ⊆ rt ◦ fu . (ii) The proof is similar to the proof of (i).



Hypothesis 10.5.6 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a thick twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Let t, u, and v be elements in W. Assume that fv ∩ (rt ◦ ru ) is not empty. Then fv ⊆ rt ◦ ru . There is some evidence that Hypothesis 10.5.6 can be proven. Of course, if `I (v) = 1, the hypothesis follows right from the definition of a twin building. C. French proved the validity of Hypothesis 10.5.6 also in the case where `I (v) ≤ 5, and in a completely different approach, he verified Hypothesis 10.5.6 under the assumption that |I | = 2. Note also that the hypothesis is valid if ((X−, δ− ), (X+, δ+ ), δ◦ ) admits a strongly transitive group of automorphisms in the sense of [1; Definition 6.1]. From [1; Theorem 8.27] (in combination with [1; Proposition 8.19]) one obtains a selection of conditions which guarantee that a given building admits a strongly transitive group of automorphisms.

10.6 Regular Actions of Twin Coxeter Hypergroups Let (W, I) be a Coxeter system. In this section, we first show that each regular action of a twin Coxeter hypergroup H over a set T of non-thin involutions of H together with a type preserving bijection from T to I gives rise to a thick twin building of type (W, I). In Corollary 9.9.9, we saw already that each type preserving bijection from

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355

T to I gives rise to an I-length preserving automorphism of W having order 1 or 2. After that, we will see that, conversely (and modulo Hypothesis 10.5.6), each thick twin building of type (W, I) together with an I-length preserving automorphism of W having order 1 or 2 gives rise to a regular action of a twin Coxeter hypergroup H over a set T of non-thin involutions of H and a type preserving bijection from T to I. To prove the first statement we introduce the following notation. Let H be a hypergroup, let T be a set of involutions of H, and assume that H is a twin Coxeter hypergroup over T. Assume that none of the elements in T is thin. The letter ρ will stand for the canonical folding defined by H and T (which was introduced in Section 9.7). Thus, we have 1ρ∗ = 1ρ,

1ρ∗ 1ρ ⊆ hTi,

and

H = hTi ∪ hTi ρ .

Let X be a set, and let ω be a regular action of H on X. Following our notation introduced at the beginning of Section 1.6 we write xh instead of ω(x, h) whenever x stands for an element in X and h for an element in H. We fix an element x− in X and an element x+ in x− 1ρ , and we set X− := x− hTi

and

X+ := x+ hTi.

From 1ρ∗ 1ρ ⊆ hTi and H = hTi ∪ hTi ρ we obtain that X = X− ∪ X+ ; cf. Lemma 8.6.8. The pair (X−, X+ ) will be called a pair of opposite sets of ω.6 Now we write σ for the color defined by ω and π for the permutation of hTi induced by ρ. Let (W, I) be a Coxeter system, and let ι be a type preserving bijection from T to I. The signature induced by ι will be denoted by τ. From Lemma 1.6.1 we obtain that the restriction of σ to X− × X− is a surjective map from X− × X− to hTi. From Theorem 9.9.1(ii) we also know that τ is a bijective map from hTi to W. Thus, as π is a permutation of hTi, the composite of the restriction of σ to X− × X− , π, and τ is a surjective map from X− × X− to W. This surjective map will be denoted by δ− , so that we have δ− (y, z) = (y, z)σπτ for any two elements y and z in X− . For any two elements y and z in different opposite sets of X, we define ( −1 (y, z)σρ ∗τ if y ∈ X− and z ∈ X+ δ◦ (y, z) := −1 (y, z)σ∗ρ τ if y ∈ X+ and z ∈ X−, 6Any different choice of elements x−0 in X and x+0 in x−0 1ρ yields a pair (X−0 , X+0 ) of opposite sets for which we then either have X−0 = X− and X+0 = X+ , or X−0 = X+ and X+0 = X− .

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10 Regular Actions of (Twin) Coxeter Hypergroups

and we notice that δ◦ is a surjective map from (X− × X+ ) ∪ (X+ × X− ) to W. In fact, choose elements y in X− and z in X+ . Since x+ ∈ x− 1ρ , z ∈ x+ hTi ⊆ x− 1ρ hTi ⊆ yhTi1ρ hTi. On the other hand, by Lemma 8.6.6(ii), we have hTi1ρ hTi = hTi ρ . Thus, z ∈ yhTi ρ ; equivalently, (y, z)σ ∈ hTi ρ . This shows that the restriction of σ to X− × X+ is a map from X− × X+ to hTi ρ . From Lemma 1.6.1 we obtain that this restriction is surjective. Now observe that ρ is a bijective map from hTi to hTi ρ . Recall further from Theorem 9.9.1(ii) that τ is a bijective map from hTi to W. Thus, the restriction of δ◦ to X− × X+ is a surjective map from X− × X+ to W. Similarly, the restriction of δ◦ to X+ × X− is a surjective map from X+ × X− to W. From Lemma 1.6.1 we obtain that the restriction of σ to X+ × X+ is a surjective map from X+ × X+ to hTi. Thus, as τ is a bijective map from hTi to W, the composite of the restriction of σ to X+ × X+ and τ is a surjective map from X+ × X+ to W. This surjective map will be denoted by δ+ , so that we have δ+ (y, z) = (y, z)στ for any two elements y and z in X+ . We call δ− and δ+ the W-distance functions induced by ω and ι. The map δ◦ will be called the W-codistance function induced by ω and ι. The letters which we have introduced so far are H, T, ρ, π,

X, X−, X+, ω, σ,

W, I, ι, τ,

and

δ−, δ◦, δ+ .

We are now ready to state and prove the first main result of this section. Theorem 10.6.1 Let H be a hypergroup, let T be a set of involutions of H, and assume that H is a twin Coxeter hypergroup over T. Assume that none of the elements in T is thin. Let X be a set, and let ω be a regular action of H on X. Let (W, I) be a Coxeter system, and let ι be a type preserving bijection from T to I. Let (X−, X+ ) denote a pair of opposite sets of ω, let δ− and δ+ denote the W-distance functions induced by ω and ι, and let δ◦ denote the W-codistance function induced by ω and ι. Then ((X−, δ− ), (X+, δ+ ), δ◦ ) is a thick twin building of type (W, I). Proof. Let σ denote the color defined by ω, let ρ denote the canonical folding defined by H and T, let π denote the permutation of hTi induced by ρ, and let τ denote the signature induced by ι. Since δ− stands for the W-distance function on X− × X− induced by ω and ι, we have δ− (y, z) = (y, z)σπτ for any two elements y and z in X− .

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357

Since δ◦ stands for the W-codistance function induced by ω and ι, we have ( −1 (y, z)σρ ∗τ if y ∈ X− and z ∈ X+ ◦ δ (y, z) = −1 (y, z)σ∗ρ τ if y ∈ X+ and z ∈ X− for any two elements y and z in X. Since δ+ stands for the W-distance function on X+ × X+ induced by ω and ι, we have δ+ (y, z) = (y, z)στ for any two elements y and z in X+ . For any two elements y and z in X− , define γ− (y, z) := (y, z)στ . Then, by Theorem 10.2.1, (X−, γ− ) is a building of type (W, I). Now recall that none of the elements in T is thin. Thus, by Lemma 9.9.8, the composite τ −1 πτ is an I-length preserving automorphism of W. Thus, since δ− (y, z) = (y, z)σπτ = (y, z)σττ

−1 πτ

= γ− (y, z)τ

−1 πτ

,

(X−, δ− ) is a building of type (W, I); cf. Lemma 10.1.8. That (X+, δ+ ) is a building of type (W, I) follows immediately from Theorem 10.2.1. We are now ready to show that the triple ((X−, δ− ), (X+, δ+ ), δ◦ ) satisfies Conditions T1, T2, and T3. For this, we let  be an element in {−, +}, y an element in X− , z an element in X , w an element in W, and we assume that δ◦ (y, z) = w. To show that ((X−, δ− ), (X+, δ+ ), δ◦ ) satisfies Condition T1, we first assume that  = +. Then we have δ◦ (z, y) = (z, y)σ∗ρ

−1 τ

= (y, z)σρ

−1 τ

= ((y, z)σρ

−1 ∗τ

)−1 = δ◦ (y, z)−1 ;

cf. Lemma 1.6.7(ii) and Theorem 9.9.1(v). If  = −, we similarly obtain that δ◦ (z, y) = (z, y)σρ

−1 ∗τ

= ((z, y)σρ

−1 τ

)−1 = ((y, z)σ∗ρ

−1 τ

)−1 = δ◦ (y, z)−1 .

Thus, as δ◦ (y, z) = w, we have δ◦ (z, y) = w −1 in both cases, and Condition T1 holds. To show that ((X−, δ− ), (X+, δ+ ), δ◦ ) satisfies Condition T2, we choose elements i in I with w ∈ I−1 (i) and x in X with δ (z, x) = i. We have to show that δ◦ (y, x) = wi. From Theorem 9.9.1(ii) we know that τ is a bijective map from hTi to W. Thus, as w ∈ W, hTi contains an element d such that d τ = w. Since δ◦ (y, z) = w, this implies that δ◦ (y, z) = d τ . Similarly, since ι is a bijective map from T to I and i ∈ I, T contains an element t such that t ι = i. By Theorem 9.9.1(i), on the other hand, we have τ|T = ι. Thus,

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10 Regular Actions of (Twin) Coxeter Hypergroups

t τ = i, and, since δ (z, x) = i, that implies that δ (z, x) = t τ . From w ∈ I−1 (i) (together with d τ = w and t τ = i) we obtain that d τ ∈ I−1 (t τ ). Thus, by Theorem 9.9.1(vii), d ∈ T−1 (t), and that means that hTi contains an element c with d ∈ ct and `T (d) = `T (c) + 1. From d ∈ ct and `T (d) = `T (c) + 1 we obtain that d τ = cτ t τ ; cf. Theorem 9.9.1(iv). Thus, as d τ = w and t τ = i, w = cτ i; equivalently, cτ = wi. Assume first that  = −. In this case, we have δ◦ (y, z) = (y, z)σ∗ρ τ . Thus, since −1 δ◦ (y, z) = d τ , we have (y, z)σ∗ρ τ = d τ , and, since τ is injective, this implies that −1 (y, z)σ∗ρ = d. It follows that (y, z)σ = d ρ∗ , so that, by Lemma 1.6.7(i), z ∈ yd ρ∗ . On the other hand, from the definition of π we obtain that d ρ∗ = d ∗πρ , and from Lemma 8.7.3(ii) we know that d π∗ = d ∗π . Thus, −1

z ∈ yd π∗ρ . Since  = −, we also have δ (z, x) = (z, x)σπτ . Thus, as δ (z, x) = t τ , (z, x)σπτ = t τ , and, since τ is injective, this implies that (z, x)σπ = t. It follows that (z, x)σ = t π ; cf. Lemma 8.7.3(i). Thus, by Lemma 1.6.7(i), x ∈ zt π . From x ∈ zt π and z ∈ yd π∗ρ we obtain that x ∈ yd π∗ρ t π . Recall that none of the elements in T is thin. Thus, as d ∈ ct, we obtain from Lemma 9.7.2 that d π ∈ c π t π . Furthermore, as `T (d) = `T (c) + 1, we obtain from Lemma 9.7.1 that `T (d π ) = `T (c π ) + 1. Now, by Lemma 8.6.2, d π∗ρ t π = {c π∗ρ }. Thus, as x ∈ yd π∗ρ t π , x ∈ yc π∗ρ . It follows that (y, x)σ = c π∗ρ = c∗πρ = cρ∗ ; cf. Lemma 8.7.3(ii). Thus, −1 (y, x)σ∗ρ = c. Thus, as δ◦ (y, x) = (y, x)σ∗ρ

−1 τ

and cτ = wi, we obtain that δ◦ (y, x) = wi,

as wanted.

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Regular Actions of Twin Coxeter Hypergroups

359

Now assume that  = +. In this case, we have δ◦ (y, z) = (y, z)σρ ∗τ . Thus, since −1 δ◦ (y, z) = d τ , we have (y, z)σρ ∗τ = d τ , and, since τ is injective, this implies that −1 (y, z)σρ ∗ = d. It follows that (y, z)σ = d ∗ρ , so that, by Lemma 1.6.7(i), −1

z ∈ yd ∗ρ . Since  = +, we also have δ (z, x) = (z, x)στ . Thus, as δ (z, x) = t τ , (z, x)στ = t τ , and, since τ is injective, this implies that (z, x)σ = t. Thus, by Lemma 1.6.7(i), x ∈ zt. From x ∈ zt and z ∈ yd ∗ρ we obtain that x ∈ yd ∗ρ t. From d ∈ ct and `T (d) = `T (c) + 1 we obtain that d ∗ρ t = {c∗ρ }; cf. Lemma 8.6.2. Thus, as x ∈ yd ∗ρ t, x ∈ yc∗ρ . It follows that (y, x)σ = c∗ρ ; equivalently, (y, x)σρ Thus, as δ◦ (y, x) = (y, x)σρ

−1 ∗τ

−1 ∗

= c.

and cτ = wi, we obtain that δ◦ (y, x) = wi

also in this case, as wanted. To show that ((X−, δ− ), (X+, δ+ ), δ◦ ) satisfies Condition T3, we choose an element i in I. We have to show that X contains an element x with δ◦ (y, x) = wi and δ (z, x) = i. From Theorem 9.9.1(ii) we know that τ is a bijective map from hTi to W. Thus, as w ∈ W, hTi contains an element d such that d τ = w. Since δ◦ (y, z) = w, this implies that δ◦ (y, z) = d τ . Similarly, since ι is a bijective map from T to I and i ∈ I, T contains an element t such that t ι = i. By Theorem 9.9.1(i), on the other hand, we have τ|T = ι. Thus, t τ = i. Assume first that d ∈ T−1 (t). Then, by Theorem 9.9.1(vi), d τ ∈ I−1 (t τ ); equivalently, w ∈ I−1 (i). Suppose first that  = −. By Lemma 1.6.1, zt π is not empty. Let x be an element in zt π . Then, by Lemma 1.6.7(i), (z, x)σ = t π . It follows that (z, x)σπ = t; cf. Lemma

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10 Regular Actions of (Twin) Coxeter Hypergroups

8.7.3(i). Since  = −, we also have δ (z, x) = (z, x)σπτ . Thus, as t τ = i, we obtain that δ (z, x) = i, as wanted. Suppose now that  = +. By Lemma 1.6.1, zt is not empty. Let x be an element in zt. Then, by Lemma 1.6.7(i), (z, x)σ = t. Since  = +, we also have δ (z, x) = (z, x)στ . Thus, as t τ = i, we obtain also in this case δ (z, x) = i, as wanted. Now, as we have w ∈ I−1 (i), δ (z, x) = i, and δ◦ (y, z) = w, we obtain from Condition T2 that δ◦ (y, x) = wi, as wanted. Assume now that d ∈ T1 (t). Then hTi contains an element e such that e ∈ dt

and `T (e) = `T (d) + 1.

From e ∈ dt and `T (e) = `T (d) + 1 we obtain that eτ = d τ t τ ; cf. Theorem 9.9.1(iv). Thus, as d τ = w and t τ = i, eτ = wi. Suppose first that  = −. In this case, we have δ◦ (y, z) = (y, z)σ∗ρ τ . Thus, −1 as δ◦ (y, z) = d τ , (y, z)σ∗ρ τ = d τ , and, since τ is injective, this implies that −1 (y, z)σ∗ρ = d. It follows that (y, z)σ = d ρ∗ , so that, by Lemma 1.6.7(i), z ∈ yd ρ∗ . On the other hand, from the definition of π we obtain that d ρ∗ = d ∗πρ , and from Lemma 8.7.3(ii) we know that d π∗ = d ∗π . Thus, −1

z ∈ yd π∗ρ . Recall that none of the elements in T is thin. Thus, as e ∈ dt, we obtain from Lemma 9.7.2 that e π ∈ d π t π . Furthermore, as `T (e) = `T (d) + 1, we obtain from Lemma 9.7.1 that `T (e π ) = `T (d π ) + 1. Now, by Lemma 8.6.2, e π∗ρ t π = {d π∗ρ }. Thus, as z ∈ yd π∗ρ , z ∈ ye π∗ρ t π , so that ye π∗ρ contains an element x with z ∈ xt π . From x ∈ ye π∗ρ we obtain that (y, x)σ = e π∗ρ = e∗πρ = eρ∗ ; cf. Lemma 8.7.3(ii). It −1 −1 follows that (y, x)σ∗ρ = e. Thus, as δ◦ (y, x) = (y, x)σ∗ρ τ , we have δ◦ (y, x) = eτ . From this, together with eτ = wi, we obtain that δ◦ (y, x) = wi, as wanted. From z ∈ xt π we obtain that x ∈ zt π ; cf. Lemma 1.6.2(ii). It follows that (z, x)σ = t π ; cf. Lemma 1.6.7(i). Thus, as t τ = i, (z, x)σπτ = i. On the other hand, as  = −, δ (z, x) = (z, x)σπτ . Thus,

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361

δ (z, x) = i, as wanted. Suppose now that  = +. In this case, we have δ◦ (y, z) = (y, z)σρ ∗τ . Thus, −1 as δ◦ (y, z) = d τ , (y, z)σρ ∗τ = d τ , and, since τ is injective, this implies that −1 (y, z)σρ ∗ = d. It follows that (y, z)σ = d ∗ρ , so that, by Lemma 1.6.7(i), −1

z ∈ yd ∗ρ . From e ∈ dt and `T (e) = `T (d) + 1 we also obtain that e∗ρ t = {d ∗ρ }; cf. Lemma 8.6.2. Thus, as z ∈ yd ∗ρ , z ∈ ye∗ρ t, so that ye∗ρ contains an element x with z ∈ xt. From x ∈ ye∗ρ we obtain that (y, x)σ = e∗ρ ; cf. Lemma 1.6.7(i). It follows that −1 −1 (y, x)σρ ∗ = e. Thus, as δ◦ (y, x) = (y, x)σρ ∗τ , we have δ◦ (y, x) = eτ . From this, τ together with e = wi, we obtain that δ◦ (y, x) = wi, as wanted. From z ∈ xt we obtain that x ∈ zt; cf. Lemma 1.6.2(ii). It follows that (z, x)σ = t; cf. Lemma 1.6.7(i). Thus, as t τ = i, (z, x)στ = i. On the other hand, as  = +, δ (z, x) = (z, x)στ . Thus, we have δ (z, x) = i, as wanted. So far, we have seen that ((X−, δ− ), (X+, δ+ ), δ◦ ) is a twin building of type (W, I). We still have to show that ((X−, δ− ), (X+, δ+ ), δ◦ ) is thick. For each element w in W, we define fw to be the set of all pairs (y, z) of elements of X satisfying either {y, z} ⊆ X−

and

or {y, z} ⊆ X+

and

δ− (y, z) = w τ

−1 πτ

δ+ (y, z) = w.

We have to show that, for each element i in I, fi is of second type which means that fi ◦ fi = f1 ∪ fi . Let i be an element in I. From Lemma 10.4.3(ii), (iii) we know that f1 ⊆ fi ◦ fi , from Lemma 10.4.3 that fi ◦ fi ⊆ f1 ∪ fi . Thus, it remains to show that fi ⊆ fi ◦ fi . To show that fi ⊆ fi ◦ fi , we fix elements y and z in X with (y, z) ∈ fi . We will see that (y, z) ∈ fi ◦ fi . Since i ∈ I and ι is a bijective map from T to I, T contains an element t such that t ι = i. From Theorem 9.9.1(i), on the other hand, we know that τ|T = ι. Thus, t τ = i.

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10 Regular Actions of (Twin) Coxeter Hypergroups

If {y, z} ⊆ X− , we have δ− (y, z) = (y, z)σπτ . Now recall that π is bijective and from Theorem 9.9.1(ii) that τ is bijective. Thus, we have (y, z) ∈ fi

⇔

δ− (y, z) = t πτ

⇔

(y, z)σ = t

⇔

z ∈ yt.

If {y, z} ⊆ X+ , we have δ+ (y, z) = (y, z)στ , and from this we obtain similarly that (y, z) ∈ fi

⇔

δ+ (y, z) = t τ

⇔

(y, z)σ = t

⇔

z ∈ yt.

This shows that we have z ∈ yt in both cases. Now recall that, by hypothesis, none of the elements in T is thin. Thus, as t ∈ T, t is not thin, so that, by Lemma 6.1.1(iv), t ∈ t 2 . Since z ∈ yt, this implies that z ∈ yt 2 , so that yt contains an element x with z ∈ xt. From x ∈ yt we obtain that (y, x) ∈ fi . From z ∈ xt we obtain that (x, z) ∈ fi . It  follows that (y, z) ∈ fi ◦ fi . Let (W, I) be a Coxeter system. Theorem 10.6.1 shows that each regular action of a twin Coxeter hypergroup H over a set T of non-thin involutions of H together with a type preserving bijection from T to I gives rise to a thick twin building of type (W, I). The thick twin building of type (W, I) which one obtains via Theorem 10.6.1 from a regular action ω of a twin Coxeter hypergroup H over a set T of non-thin involutions of H together with a type preserving bijection ι from T to I will be called the twin building induced by ω and ι. We will now see that, conversely, (modulo Hypothesis 10.5.6) each thick twin building of type (W, I) together with an I-length preserving automorphism of W having order 1 or 2 gives rise to a regular action of a twin Coxeter hypergroup H over a set T of non-thin involutions of H and a type preserving bijection from T to I. The statements and proofs of the remaining results of this sections are similar to the statements and proofs of the corresponding results of Section 10.2. Lemma 10.6.2 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a thick twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Assume that Hypothesis 10.5.6 holds. For each element w in W, define fw := {(y, z) ∈ X− × X− | δ− (y, z) = w π¯ } ∪ {(y, z) ∈ X+ × X+ | δ+ (y, z) = w}, and rw := {(y, z) ∈ X− × X+ | δ◦ (y, z) = w −1 } ∪ {(y, z) ∈ X+ × X− | δ◦ (y, z) = (w −1 )π¯ }. Define further F := { fw | w ∈ W },

R := {rw | w ∈ W },

and

H := F ∪ R.

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363

Then H is a set theoretic hypergroup on X− ∪ X+ . Proof. Set X := X− ∪ X+ . The definition of H implies that H is a partition of X × X. From Lemma 10.4.1 we know that f1 = 1X , and that implies that 1X ∈ H. From Lemma 10.4.5 we know that, for each element w in W, fw∪ = fw −1 . Thus we have fw∪ ∈ H for each element w in W. From Lemma 10.5.1 we know that, for each element w in W, rw∪ = r(w −1 ) π¯ . Thus, we also have rw∪ ∈ H for each element w in W. In Lemma 10.4.7, we saw that, for any three elements t, u, and v in W, fv ⊆ ft ◦ fu if fv ∩ ( ft ◦ fu ) is not empty. From Lemma 10.5.5 we know that, for any three elements t, u, and v in W, rv ⊆ rt ◦ fu if rv ∩ (rt ◦ fu ) is not empty and rv ⊆ fu ◦ rt if rv ∩ ( fu ◦ rt ) is not empty.7 Assuming Hypothesis 10.5.6 we obtain that, for any three elements t, u, and v in W, fv ⊆ rt ◦ ru if fv ∩ (rt ◦ ru ) is not empty. Now, by Lemma 1.6.5, H is a set theoretic hypergroup on X.



Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a thick twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Assume that Hypothesis 10.5.6 holds. The hypergroup which one obtains from ((X−, δ− ), (X+, δ+ ), δ◦ ) and π¯ via Lemma 10.6.2 will be called the hypergroup induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π. ¯ Lemma 10.6.3 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a thick twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Assume that Hypothesis 10.5.6 holds. Let H denote the hypergroup induced by ¯ and set T := { fi | i ∈ I} and F := { fw | w ∈ W }. Then ((X−, δ− ), (X+, δ+ ), δ◦ ) and π, the following hold. (i) Each element in T is a non-thin involution of H. (ii) We have F = hTi. (iii) The map ρ : F → H, fw 7→ rw is a folding from F to H and satisfies ρ∗ ρ ρ ρ∗ f1 = f1 , f1 f1 ⊆ F, and H = F ∪ F ρ . Proof. (i) Let t be an element in T. We have to show that t is a non-thin involution. For this, it is enough to show that t ∗ = t and that t 2 = {1, t}. Since t ∈ T, I contains an element i such that t = fi . Since H is a set theoretic hypergroup on X, fi∗ = fi∪ . On the other hand, by Lemma 10.4.5, fi∪ = fi . Thus, fi∗ = fi . Since t = fi , this means that t ∗ = t. Since ((X−, δ− ), (X+, δ+ ), δ◦ ) is assumed to be thick, fi is of second type. Thus, as i ∈ I−1 (i), we obtain from Lemma 10.4.3(iii) that fi ◦ fi = f1 ∪ fi . Thus, the definition 7In Lemma 10.4.7 and Lemma 10.5.5, we needed only semiregularity, not thickness of the underlying twin building.

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10 Regular Actions of (Twin) Coxeter Hypergroups

of the set theoretic hypermultiplication implies that fi2 = { f1, fi }. Since t = fi , this implies that t 2 = {1, t}. (ii) It follows from the definition of F that F is a closed subset of H. Thus, it suffices to show that F ⊆ hTi. Let f be an element in F. We have to show that f ∈ hTi. Since f ∈ F, W contains an element w such that f = fw . Thus, we have to show that fw ∈ hTi. We proceed by induction on `I (w). If `I (w) = 0, w = 1. Then fw = f1 . Thus, as f1 ∈ hTi, fw ∈ hTi. as wanted. Assume that 1 ≤ `I (w). Then w , 1, so that we find elements v in W and i in I with w = vi

and `I (w) = `I (v) + 1.

From w = vi and `I (w) = `I (v) + 1 we obtain that v ∈ I1 (i). Thus, by Lemma 10.4.3(i), fv ◦ fi = fw . It follows that fw ∈ fv fi . On the other hand, since `I (w) = `I (v) + 1, induction yields that fv ∈ hTi. Thus, as fi ∈ T, fv fi ⊆ hTi. From fw ∈ fv fi and fv fi ⊆ hTi we obtain that fw ∈ hTi, as wanted. (iii) Let w be an element of W. Then, by Corollary 10.5.4, rw ◦ fw = r1 . By definition, ρ ρ fw = rw and f1 = r1 . Thus, ρ ρ fw ◦ fw = f1 . Now the definition of the set theoretic hypermultiplication implies that ρ

ρ

fw fw = { f1 }. Since w has been chosen arbitrarily from W, this shows that ρ is a folding from F to H. From Lemma 10.5.1 we know that r1∪ = r1 , and, since H is a set theoretic hypergroup ρ ρ∗ ρ on X, we have r1∗ = r1∪ . Thus, r1∗ = r1 . Thus, as f1 = r1 , we have f1 = f1 . ρ

ρ∗ ρ

That r1∗ r1 ⊆ F follows from the definition of r1 . Thus, as f1 = r1 , f1 f1 ⊆ F.

From the definition of ρ we obtain that F ρ = R, and, by definition, we have H = F∪R.  Thus, H = F ∪ F ρ . Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a thick twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Assume that Hypothesis 10.5.6 holds. The set of non-thin involutions which we found in Lemma 10.6.3(i) will be called the set of involutions induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π. ¯

The folding from F to H which we found in Lemma 10.6.3(iii) will be called the ¯ folding induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π.

Regular Actions of Twin Coxeter Hypergroups

10.6

365

Lemma 10.6.4 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a thick twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Assume that Hypothesis 10.5.6 holds. Let T denote the set of involutions ¯ Then fw 7→ w is a (T, I)-length preserving induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π. bijective map from hTi to W. Proof. By Lemma 10.6.3(ii), fw 7→ w is a bijective map from hTi to W. To show that fw 7→ w is (T, I)-length preserving, we fix an element in W and denote it by w. We have to show that `T ( fw ) = `I (w) and proceed by induction on `I (w). If `I (w) = 0, w = 1. Then fw = f1 , so that `T ( fw ) = 0. It follows that `T ( fw ) = `I (w), so that we are done in this case. Assume that 1 ≤ `I (w). Then w , 1, so that we find elements v in W and i in I with w = vi

and `I (w) = `I (v) + 1.

From w = vi and `I (w) = `I (v) + 1 we obtain that v ∈ I1 (i). Thus, by Lemma 10.4.3(i), fv ◦ fi = fw ; equivalently, fv fi = { fw }. From w , 1 we also obtain that fw , f1 . Thus, by Lemma 2.3.7(ii) (together with Lemma 10.6.3(ii)), there exist elements u in W and i in I such that fw ∈ fu fi

and `T ( fw ) = `T ( fu ) + 1.

From fv fi = { fw } and fw ∈ fu fi we obtain that fv fi ⊆ fu fi . Now recall from Lemma 10.6.3(i) that fi is an involution of hTi. Thus, by Lemma 6.1.1(iii), (iv), fi fi ⊆ { f1, fi }. It follows that fv ∈ { fu } ∪ fu fi . Assume that fv ∈ fu fi . Then fu ∈ fv fi = { fw }. It follows that fu = fw , contrary to `T ( fw ) = `T ( fu ) + 1. Thus, as fv ∈ { fu } ∪ fu fi , we have fv = fu . From fv = fu together with `T ( fw ) = `T ( fu ) + 1, we obtain that `T ( fw ) = `T ( fv ) + 1. On the other hand, since `I (w) = `I (v) + 1, induction yields `T ( fv ) = `I (v). From `T ( fw ) = `T ( fv ) + 1, `T ( fv ) = `I (v), and `I (w) = `I (v) + 1 we obtain that `T ( fw ) = `I (w), as wanted.  Corollary 10.6.5 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a thick twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order

366

10 Regular Actions of (Twin) Coxeter Hypergroups

1 or 2. Assume that Hypothesis 10.5.6 holds. Let T denote the set of involutions induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π. ¯ Let w be an element in W, and let i be an element in I. Then the following hold. (i) We have w ∈ I−1 (i) if and only if fw ∈ T−1 ( fi ). (ii) We have w ∈ I1 (i) if and only if fw ∈ T1 ( fi ). Proof. Since W = I−1 (i) ∪ I1 (i) and T−1 ( fi ) ∩ T1 ( fi ) is empty, it suffices to show that fw ∈ T−1 ( fi ) follows from w ∈ I−1 (i) and that fw ∈ T1 ( fi ) follows from w ∈ I1 (i). (i) Assume that w ∈ I−1 (i). Then wi ∈ I1 (i). Thus, by Lemma 10.4.3(i), fwi ◦ fi = fw . It follows that fw ∈ fwi fi . On the other hand, as wi ∈ I1 (i), `I (w) = `I (wi) + 1. Thus, by Lemma 10.6.4, `T ( fw ) = `T ( fwi ) + 1. From fw ∈ fwi fi and `T ( fw ) = `T ( fwi ) + 1 we obtain that fw ∈ T−1 ( fi ). (ii) Assume that w ∈ I1 (i). Then, by Lemma 10.4.3(i), fw ◦ fi = fwi . It follows that fwi ∈ fw fi . On the other hand, as w ∈ I1 (i), `I (wi) = `I (w) + 1. Thus, by Lemma 10.6.4, `T ( fwi ) = `T ( fw ) + 1. From fwi ∈ fw fi and `T ( fwi ) = `T ( fw ) + 1 we obtain that fw ∈ T1 ( fi ).



Lemma 10.6.6 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a thick twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Assume that Hypothesis 10.5.6 holds. Let T denote the set of involutions ¯ Then T is a Coxeter set. induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π. Proof. We first show that T is constrained. For this, we choose an element t in T and an element f in T1 (t). We have to show that | f t| = 1. Since f ∈ hTi, W contains an element w with f = fw ; cf. Lemma 10.6.3(ii). Since t ∈ T, t = fi for some element i in I. Thus, as f ∈ T1 (t), fw ∈ T1 ( fi ). Now Corollary 10.6.5(ii) yields w ∈ I1 (i). It follows that fw ◦ fi = fwi ; cf. Lemma 10.4.3(i). From this we obtain that fw fi = { fwi }. Since f = fw and t = fi , this implies that | f t| = 1, as wanted. Now we show that T satisfies the exchange condition. For this, we fix elements p and q in T and an element c in T1 (q) with p ∈ T1 (c). Since we have seen already that T is constrained, it suffices to show that pc = cq

or

pc ⊆ T1 (q);

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Regular Actions of Twin Coxeter Hypergroups

367

cf. Lemma 9.1.1. Since p ∈ T1 (c), we have |pc| = 1; cf. Lemma 8.1.1. Thus, pc contains an element d with pc = {d}. Similarly, cq contains an element e with cq = {e}. Since {c, d, e} ⊆ hTi and {p, q} ⊆ T, there exist elements t, u, v in W and j, k in I such that c = ft , d = fu, e = fv, p = f j , q = fk . From p ∈ T1 (c), p = f j , and c = ft , we obtain that f j ∈ T1 ( ft ). Thus, by Corollary 10.6.5(ii), j ∈ I1 (t). It follows that f j ◦ ft = f jt ; cf. Corollary 10.4.4. Thus, by definition, f j ft = { f jt }. Since p = f j and c = ft , this implies that pc = { f jt }. Now recall that pc = {d} and that d = fu . Thus, f jt = fu ; equivalently, jt = u. Similarly, one obtains from c ∈ T1 (q), c = ft , q = fk , and cq = {e} that t ∈ I1 (k)

and t k = v.

Now recall that (W, I) is a Coxeter system. Thus, as j ∈ I1 (t) and t ∈ I1 (k), we have jt = tk

or

jt ∈ I1 (k).

Assume first that jt = t k. Then, u = v, because we saw that jt = u and tk = v. Thus, pc = {d} = { fu } = { fv } = {e} = cq, and we are done. Now assume that jt ∈ I1 (k). Then, as jt = u, u ∈ I1 (k). Thus, by Corollary 10.6.5(ii), fu ∈ T1 ( fk ). Since d = fu and q = fk this implies that d ∈ T1 (q). Thus, as pc = {d},  we have that pc ⊆ T1 (q), so that we are done also in this case. We are now ready to state and prove the second main result of this section. Theorem 10.6.7 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a thick twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Assume that Hypothesis 10.5.6 holds. Let H denote the hypergroup induced

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10 Regular Actions of (Twin) Coxeter Hypergroups

by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π, ¯ and let T denote the set of involutions induced by ¯ Then the following hold. ((X−, δ− ), (X+, δ+ ), δ◦ ) and π. (i) The set H is a twin Coxeter hypergroup over T. (ii) Set X := X− ∪ X+ . Then (x, fw ) 7→ {y ∈ X | (x, y) ∈ fw }, (x, rw ) 7→ {y ∈ X | (x, y) ∈ rw } is a regular action of H on X. (iii) The map fi 7→ i is a type preserving bijection from T to I. (iv) The folding induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π¯ is the canonical folding defined by H and T. (v) Let ρ denote the folding induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π, ¯ and let π denote the permutation induced by ρ. Then, for each element w in W, fw π¯ = fwπ . Proof. (i) In Lemma 10.6.6, we saw that T is a Coxeter set of involutions of H, and from Lemma 10.6.3(ii), (iii) we know that H is twinned over hTi. Thus, H is a twin Coxeter hypergroup over T. (ii) The map (x, fw ) 7→ {y ∈ X | (x, y) ∈ fw },

(x, rw ) 7→ {y ∈ X | (x, y) ∈ rw }

is the canonical action of H on X. Thus, by Lemma 1.6.6, it is a regular action of H on X. (iii) It follows from the definition of T that fi 7→ i is a bijective map from T to I. Thus, it remains to be shown that, for any two elements k and l in I, c fk , fl = ck,l . Let k and l be elements in I, and set J := {k, l} and U := { fk , fl }. Recall from the definition of Coxeter numbers that c fk , fl is finite if and only if C( fk , fl ) is not empty. From Corollary 6.6.6 we also know that C( fk , fl ) is not empty if and only if U−1 (U) is not empty. Similarly, ck,l is finite if and only if J−1 (J) is not empty. Now recall that, by Corollary 10.6.5(i), U−1 (U) is not empty if and only if J−1 (J) is not empty. Thus, c fk , fl is finite if and only if ck,l is finite, and we may (and will) assume that c fk , fl and ck,l both are finite. Since c fk , fl is assumed to be finite, C( fk , fl ) is not empty. Thus, by Proposition 6.6.7, min({`U ( fw ) | fw ∈ U−1 (U)}) = c fk , fl . Similarly,

min({`J (w) | w ∈ J−1 (J)}) = ck,l .

Let w be an element in J−1 (J) with `J (w) = ck,l . Since w ∈ J−1 (J), fw ∈ U−1 (U); cf. Corollary 10.6.5(i). Thus, c fk , fl ≤ `U ( fw ). Now recall from Lemma 10.6.4 that

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369

`T ( fw ) = `I (w), and, since T and I both are Coxeter sets, that implies that `U ( fw ) = `J (w); cf. Lemma 9.1.2. Thus, c fk , fl ≤ ck,l . Let w be an element in W with fw ∈ U−1 (U) and `U ( fw ) = c fk , fl . Since fw ∈ U−1 (U), w ∈ J−1 (J); cf. Corollary 10.6.5(i). Thus, ck,l ≤ `J (w). Now recall from Lemma 10.6.4 that `T ( fw ) = `I (w), and, since T and I both are Coxeter sets, that implies that `U ( fw ) = `J (w); cf. Lemma 9.1.2. Thus, ck,l ≤ c fk , fl . From c fk , fl ≤ ck,l and ck,l ≤ c fk , fl we obtain that c fk , fl = ck,l , as wanted. (iv) This follows from the definition of the canonical folding defined by H and T. (v) From Lemma 10.6.2 we know that H is a set theoretic hypergroup. Thus, fw∪ = fw∗

and rw∪−1 = rw∗ −1 .

From Lemma 10.4.5 we know that fw∪ = fw −1 . Thus, as fw∪ = fw∗ , we conclude that fw −1 = fw∗ . From Lemma 10.5.1 we know that rw∪−1 = rw π¯ . Thus, as rw∪−1 = rw∗ −1 , we conclude that rw π¯ = rw∗ −1 . ρ−1

∗ρ−1

ρ∗ρ−1

It follows that fw π¯ = rw π¯ = rw −1 = fw −1

∗ρ∗ρ−1

= fw

= fwπ .



Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a thick twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Assume that Hypothesis 10.5.6 holds. With an eye on Theorem 10.6.7(i), we will now call the set which we obtained from ((X−, δ− ), (X+, δ+ ), δ◦ ) and π¯ via Lemma 10.6.2 the twin Coxeter hypergroup induced ¯ by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π. The regular action which we found in Theorem 10.6.7(ii) will be called the regular ¯ action induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π. The bijection from T to I which we found in Theorem 10.6.7(iii) will be called the type preserving bijection induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π. ¯ We emphasize that the twin Coxeter hypergroup induced by a thick twin building of type (W, I) and an I-length preserving automorphism π¯ of W having order 1 or 2 is always a set theoretic hypergroup. Lemma 10.6.8 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a thick twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1

370

10 Regular Actions of (Twin) Coxeter Hypergroups

or 2. Assume that Hypothesis 10.5.6 holds. Let ι denote the type preserving bijection induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π, ¯ and set fwυ := w for each element w in W. Then υ is the signature induced by ι. Proof. Let T denote the set of involutions induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π. ¯ Then, by Lemma 10.6.4, υ is a (T, I)-length preserving bijective map from hTi to W. ¯ we Since ι is the type preserving bijection induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π, have fiι = i for each element i in I. Thus, υ|T = ι. Let u and v be elements in W, and let i be an element in I with fv ∈ fu fi

and `T ( fv ) = `T ( fu ) + 1.

From fv ∈ fu fi we obtain that fv ⊆ fu ◦ fi . On the other hand, by Lemma 10.4.3, fu ◦ fi ⊆ fui ∪ fu . Thus, fv ⊆ fui ∪ fu . It follows that fv = fui or fv = fu , so that we have v = ui or v = u. From `T ( fv ) = `T ( fu ) + 1 we obtain that `I (v) = `I (u) + 1, since υ is (T, I)-length preserving. Thus, v , u. It follows that v = ui; equivalently, fvυ = fuυ fiυ . We have seen that υ is a (T, I)-length preserving bijective map from hTi to W with υ|T = ι satisfying fvυ = fuυ fiυ for any three elements u and v in W and i in I with fv ∈ fu fi and `T ( fv ) = `T ( fu ) + 1. Thus, by Lemma 9.9.3, υ is the signature induced  by ι.

10.7 Twin Coxeter Hypergroups and Twin Buildings In this section, we will see that (modulo Hypothesis 10.5.6) regular actions of twin Coxeter hypergroups over sets of non-thin involutions and thick twin buildings (endowed with a length preserving automorphism having order 1 or 2) are equivalent mathematical objects. For this, we need to define what it means for two regular actions of twin Coxeter hypergroups to be isomorphic and what it means for two thick twin buildings (each of them endowed with a length preserving automorphism having order 1 or 2) to be isomorphic. We first define what it means for two regular actions of twin Coxeter hypergroups to be isomorphic. Let H and H 0 be hypergroups, let T be a set of involutions of H, and let T 0 be a set of involutions of H 0. Assume that H is a twin Coxeter hypergroup over T and that H 0 is a twin Coxeter hypergroup over T 0.8 Recall from Section 6.5 that a bijective map φ from hTi to hT 0i is called (T, T 0)-length preserving if, for each element h in hTi, `T (h) = `T 0 (hφ ). Let X and X 0 be sets, let ω be a regular action of H on X, and let ω 0 be a regular action of H 0 on X 0. 8At this point, we do not assume that the elements of T or T 0 are not thin.

10.7

Twin Coxeter Hypergroups and Twin Buildings

371

Recall from Section 3.9 that a hypergroup isomorphism φ from H to H 0 is called an (ω, ω 0)-isomorphism if there exists a bijective map υ from X to X 0 such that, for any two elements x in X and h in H, ω(x, h)υ = ω 0(x υ, hφ ). In Section 3.9, we called the bijective map υ from X to X 0 the supporting bijection of φ. Notice that the inverse of an (ω, ω 0)-isomorphism from H to H 0 the restriction of which to hTi is (T, T 0)-length preserving is an (ω 0, ω)-isomorphism from H 0 to H the restriction of which to hT 0i is (T 0, T)-length preserving. The regular actions ω and ω 0 will be called isomorphic if there exists an (ω, ω 0)isomorphism from H to H 0 the restriction of which to hTi is (T, T 0)-length preserving. Next we define what it means for two thick twin buildings (each of them endowed with a length preserving automorphism having order 1 or 2) to be isomorphic. Let (W, I) and (W 0, I 0) be Coxeter systems, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a thick twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Similarly, let ((X−0 , δ−0 ), (X+0 , δ+0 ), δ 0◦ ) be a thick twin building of type (W 0, I 0), and let π¯ 0 be an I 0-length preserving automorphism of W 0 having order 1 or 2. A bijective map υ from X− ∪ X+ to X−0 ∪ X+0 is called a twin building isomorphism from ((X−, δ− ), (X+, δ+ ), δ◦ ) to ((X−0 , δ−0 ), (X+0 , δ+0 ), δ 0◦ ) if X−υ = X−0

and

X+υ = X+0 ,

and there exists an (I, I 0)-length preserving group isomorphism τ from W to W 0 with πτ ¯ = τ π¯ 0 such that

δ− (y, z)τ = δ−0 (y υ, zυ )

for any two elements y and z in X− , δ◦ (y, z)τ = δ 0◦ (y υ, zυ ) for any two elements y in X− and z in X+ or y in X+ and z in X− , and δ+ (y, z)τ = δ+0 (y υ, zυ ) for any two elements y and z in X+ . The (I, I 0)-length preserving group isomorphism τ from W to W 0 is called the supporting group isomorphism of υ. It follows right from the definition of a twin building isomorphism that the inverse of a twin building isomorphism from ((X−, δ− ), (X+, δ+ ), δ◦ ) to ((X−0 , δ−0 ), (X+0 , δ+0 ), δ 0◦ )

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10 Regular Actions of (Twin) Coxeter Hypergroups

is a twin building isomorphism from ((X−0 , δ−0 ), (X+0 , δ+0 ), δ 0◦ ) to

((X−, δ− ), (X+, δ+ ), δ◦ ).

The twin buildings ((X−, δ− ), (X+, δ+ ), δ◦ ) and ((X−0 , δ−0 ), (X+0 , δ+0 ), δ 0◦ ) are called isomorphic if there exists a twin building isomorphism from ((X−, δ− ), (X+, δ+ ), δ◦ ) to

((X−0 , δ−0 ), (X+0 , δ+0 ), δ 0◦ ).

Now we are in the position to explain what we mean by saying that regular actions of twin Coxeter hypergroups over sets of non-thin involutions and thick buildings are equivalent mathematical objects. For this we let (W, I) be a Coxeter system. In Theorem 10.6.1, we have seen that each regular action of a twin Coxeter hypergroup H over a set T of non-thin involutions of H together with a type preserving bijection from T to I gives rise to a thick twin building of type (W, I). In Corollary 9.9.9, we saw already that each type preserving bijection from T to I gives rise to an I-length preserving automorphism of W having order 1 or 2. In Theorem 10.7.1, we will see that isomorphic regular actions of twin Coxeter hypergroups over sets of non-thin involutions give rise to isomorphic thick twin buildings. In Theorem 10.6.7, we have seen that (modulo Hypothesis 10.5.6) each thick twin building of type (W, I) together with an I-length preserving automorphism of W having order 1 or 2 gives rise to a regular action of a twin Coxeter hypergroup H over a set T of non-thin involutions of H and a type preserving bijection from T to I. In Theorem 10.7.2, we will see that (modulo Hypothesis 10.5.6) isomorphic thick twin buildings give rise to isomorphic regular actions of twin Coxeter hypergroups. Theorem 10.6.1 (together with Theorem 10.7.1) assigns to each isomorphism class of regular actions of twin Coxeter hypergroups over sets of non-thin involutions together with a type preserving bijection from T to I an isomorphism class of thick buildings. Theorem 10.6.7 (together with Theorem 10.7.2) assigns (modulo Hypothesis 10.5.6) to each isomorphism class of thick buildings an isomorphism class of regular actions of twin Coxeter hypergroups over sets of non-thin involutions. In Theorems 10.7.3 and 10.7.4, we will see that (modulo Hypothesis 10.5.6) these two assignments are inverses of each other. Theorem 10.7.1 Let H and H 0 be hypergroups, let T be a set of involutions of H, and let T 0 be a set of involutions of H 0. Assume that H is a twin Coxeter hypergroup over T and that H 0 is a twin Coxeter hypergroup over T 0. Assume further that none of the elements in T or T 0 is thin. Let X and X 0 be sets, let ω be a regular action of H on X, and let ω 0 be a regular action of H 0 on X 0. Let (W, I) and (W 0, I 0) be Coxeter systems, let ι be a type preserving bijection from T to I, and let ι0 be a type preserving bijection from T 0 to I 0. Assume that ω and ω 0 are isomorphic. Then the twin building induced by ω and ι is isomorphic to the twin building induced by ω 0 and ι0. Proof. Let δ− and δ+ denote the W-distance functions induced by ω and ι, and let δ◦ denote the W-codistance function induced by ω and ι. Let δ−0 and δ+0 denote the

10.7

Twin Coxeter Hypergroups and Twin Buildings

373

W 0-distance functions induced by ω 0 and ι0, and let δ 0◦ denote the W 0-codistance function induced by ω 0 and ι0. We are assuming that ω and ω 0 are isomorphic, and we have to find a twin building isomorphism from ((X−, δ− ), (X+, δ+ ), δ◦ ) to

((X−0 , δ− ), (X+0 , δ+ ), δ 0◦ ).

Since ω and ω 0 are assumed to be isomorphic, there exists an (ω, ω 0)-isomorphism from H to H 0 the restriction of which to hTi is (T, T 0)-length preserving. We fix one of these (ω, ω 0)-isomorphisms from H to H 0 (with a (T, T 0)-length preserving restriction to hTi) and denote it by φ. The supporting bijection of φ will be denoted by υ. Since φ| hT i is a (T, T 0)-length preserving hypergroup isomorphism from hTi to hT 0i, φ|T is a type preserving bijection from T to T 0; cf. Lemma 9.9.4(i). Now recall that ι is a type preserving bijection from T to I and ι0 is a type preserving bijection from T 0 to I 0. Thus, we obtain from Lemma 9.9.5(i) (together with Lemma 9.9.6(i)) that the composite ι−1 φ|T ι0 is a type preserving bijection from I to I 0. From Lemma 9.9.4(ii) we know that φ| hT i is the signature induced by φ|T . Let τ denote the signature induced by ι and τ 0 the signature induced by ι0. Then, by Lemma 9.9.5(ii) (together with Lemma 9.9.6(ii)) the composite τ −1 φ| hT i τ 0 is the signature induced by ι−1 φ|T ι0. Thus, by Theorem 9.9.1(iii), τ −1 φ| hT i τ 0 is (I, I 0)length preserving, and, by Theorem 9.9.2, τ −1 φ| hT i τ 0 is a group isomorphism from W to W 0.9 We will now show that υ is a twin building isomorphism from ((X−, δ− ), (X+, δ+ ), δ◦ ) to

((X−0 , δ− ), (X+0 , δ+ ), δ 0◦ )

with supporting group isomorphism τ −1 φ|T τ 0. Let ρ denote the canonical folding defined by H and T, let π denote the permutation of hTi induced by ρ, let ρ0 denote the canonical folding defined by H 0 and T 0, and let π 0 denote the permutation of hT 0i induced by ρ0. Then τ −1 πτ is the automorphism of W induced by ι and τ 0−1 π 0 τ 0 is the automorphism of W 0 induced by ι0, and, since φπ 0 = πφ (by Lemma 9.7.4(ii)), we have (τ −1 πτ)(τ −1 φ|T τ 0) = τ −1 πφ|T τ 0 = τ −1 φ|T π 0 τ 0 = (τ −1 φ|T τ 0)(τ 0−1 π 0 τ 0). It remains to be shown that X−υ = X−0

and

X+υ = X+0

and (since τ −1 φ| hT i τ 0 = τ −1 φτ 0) that δ− (y, z)τ

−1 φτ 0

= δ−0 (y υ, zυ )

9We say group isomorphism here, because W and W 0 are groups.

374

10 Regular Actions of (Twin) Coxeter Hypergroups

for any two elements y and z in X− with δ− (y, z) ∈ I, δ◦ (y, z)τ

−1 φτ 0

= δ 0◦ (y υ, zυ )

for any two elements y in X− and z in X+ or y in X+ and z in X− with δ◦ (y, z) ∈ I, and 0 −1 δ+ (y, z)τ φτ = δ+0 (y υ, zυ ) for any two elements y and z in X+ with δ− (y, z) ∈ I. Let σ denote the color defined by ω, and let σ 0 denote the color defined by ω 0. Since φ is an (ω, ω 0)-isomorphism from H to H 0 with supporting bijection υ, we obtain from Lemma 3.9.2 that 0 (y, z)σφ = (y υ, zυ )σ . Let y and z be elements in X− . Then δ− (y, z) = (y, z)σπτ

δ−0 (y υ, zυ ) = (y υ, zυ )σ π τ . 0 0 0

and

Since φ is an isomorphism from H to H 0 with T φ = T 0, we obtain from Lemma 9.7.4(ii) that 0 (y, z)σπφ = (y, z)σφπ . Thus, we obtain that δ− (y, z)τ

−1 φτ 0

= (y, z)σπφτ = (y, z)σφπ τ = (y υ, zυ )σ π τ = δ−0 (y υ, zυ ), 0

0 0

0 0 0

as wanted. Now let y be an element in X− , and let z be an element in X+ . Then, since δ◦ stands for the W-codistance function induced by ω and ι, δ◦ (y, z) = (y, z)σρ

−1 ∗τ

.

It follows that δ◦ (y, z)τ

−1 φτ 0

= (y, z)σρ

−1 ∗φτ 0

= (y, z)σρ

−1 φ∗τ 0

= (y, z)σρ

−1 φρ0 ρ0−1 ∗τ 0

.

Since δ 0◦ stands for the W 0-codistance function induced by ω 0 and ι0, we also have δ 0◦ (y υ, zυ ) = (y υ, zυ )σ ρ

0 0−1 ∗τ 0

= (y, z)σφρ

0−1 ∗τ 0

= (y, z)σρ

−1 ρφρ0−1 ∗τ 0

.

Since φ is an isomorphism from H to H 0 with T φ = T 0, we obtain from Lemma 9.7.4(i) that −1 −1 0 (y, z)σρ ρφ = (y, z)σρ φρ . Together we obtain that δ◦ (y, z)τ as wanted.

−1 φτ 0

= δ 0◦ (y υ, zυ ),

10.7

Twin Coxeter Hypergroups and Twin Buildings

A similar argument shows that δ◦ (y, z)τ in X+ and z in X− .

−1 φτ 0

375

= δ 0◦ (y υ, zυ ) holds also for elements y

Now let y and z be elements in X+ . Then δ+ (y, z) = (y, z)στ

δ+0 (y υ, zυ ) = (y υ, zυ )σ τ . 0 0

and

Thus, as (y, z)σφ = (y υ, zυ )σ , we obtain that 0

δ+ (y, z)τ

−1 φτ 0

= (y, z)σφτ = (y υ, zυ )σ τ = δ+0 (y υ, zυ ), 0

0 0



as wanted. Theorem 10.7.2

Let (W, I) and (W 0, I 0) be Coxeter systems, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a thick twin building of type (W, I), let π¯ be an I-length preserving automorphism of W having order 1 or 2, let ((X−0 , δ−0 ), (X+0 , δ+0 ), δ 0◦ ) be a thick twin building of type (W 0, I 0), and let π¯ 0 be an I 0-length preserving automorphism of W 0 having order 1 or 2. Assume that Hypothesis 10.5.6 holds. Let ω denote the regular ac¯ and let ω 0 denote the regular action tion induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π, 0 0 0 0 0◦ 0 induced by ((X−, δ− ), (X+, δ+ ), δ ) and π¯ . Assume that ((X−, δ− ), (X+, δ+ ), δ◦ ) and ((X−0 , δ−0 ), (X+0 , δ+0 ), δ 0◦ ) are isomorphic. Then ω and ω 0 are isomorphic. Proof. For each element w in W, define fw := {(y, z) ∈ X− × X− | δ− (y, z) = w π¯ } ∪ {(y, z) ∈ X+ × X+ | δ+ (y, z) = w} and rw := {(y, z) ∈ X− × X+ | δ◦ (y, z) = w −1 } ∪ {(y, z) ∈ X+ × X− | δ◦ (y, z) = (w −1 )π¯ }. Define further F := { fw | w ∈ W },

R := {rw | w ∈ W },

H := F ∪ R,

and T := { fi | i ∈ I}. Then H is the twin Coxeter hypergroup induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π, ¯ and T is the set of involutions induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π. ¯ For each element w 0 in W 0, write fw0 0 for the union {(y, z) ∈ X−0 × X−0 | δ−0 (y, z) = w 0π¯ } ∪ {(y, z) ∈ X+0 × X+0 | δ+0 (y, z) = w 0 } 0

and rw0 0 for the union {(y, z) ∈ X−0 × X+0 | δ 0◦ (y, z) = w 0−1 } ∪ {(y, z) ∈ X+0 × X−0 | δ 0◦ (y, z) = (w 0−1 )π¯ }. 0

Define further

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10 Regular Actions of (Twin) Coxeter Hypergroups

F 0 := { fw0 0 | w 0 ∈ W 0 }, and

R 0 := {rw0 0 | w 0 ∈ W 0 },

H 0 := F 0 ∪ R 0,

T 0 := { fi0 | i ∈ I}.

Then H 0 is the twin Coxeter hypergroup induced by ((X−0 , δ−0 ), (X+0 , δ+0 ), δ 0◦ ) and π¯ 0, and T 0 is the set of involutions induced by ((X−0 , δ−0 ), (X+0 , δ+0 ), δ 0◦ ) and π¯ 0. We are assuming that ((X−, δ− ), (X+, δ+ ), δ◦ ) and ((X−0 , δ−0 ), (X+0 , δ+0 ), δ 0◦ ) are isomorphic, and we have to find an (ω, ω 0)-isomorphism from H to H 0 the restriction of which to hTi is (T, T 0)-length preserving. Since ((X−, δ− ), (X+, δ+ ), δ◦ ) and ((X−0 , δ−0 ), (X+0 , δ+0 ), δ 0◦ ) are isomorphic, there exists a twin building isomorphism from ((X−, δ− ), (X+, δ+ ), δ◦ ) to ((X−0 , δ−0 ), (X+0 , δ+0 ), δ 0◦ ). We fix one of these twin building isomorphisms and denote it by υ. Its supporting group isomorphism will be denoted by τ. Then we have ¯ = τ π¯ 0 . πτ Furthermore, we have X−υ = X−0 , X+υ = X+0 , δ− (y, z)τ = δ−0 (y υ, zυ ) for any two elements y and z in X− , δ◦ (y, z)τ = δ 0◦ (y υ, zυ ) for any two elements y in X− and z in X+ or y in X+ and z in X− , and δ+ (y, z)τ = δ+0 (y υ, zυ ) for any two elements y and z in X+ . For each element w in W, set φ

fw := fw0 τ

φ

and rw := rw0 τ .

We will now show that φ is an (ω, ω 0)-isomorphism from H to H 0 and that φ| hT i is (T, T 0)-length preserving. We first show that φ is a hypergroup isomorphism from H to H 0. The bijectivity of φ follows from the bijectivity of τ. Now let y and z be elements in X, and let w be an element in W. Assume first that (y, z) ∈ fw . Then {y, z} ⊆ X− or {y, z} ⊆ X+ . ¯ . Suppose that {y, z} ⊆ X− . Then δ− (y, z) = w π¯ . It follows that δ−0 (y υ, zυ ) = w πτ 0 0 ¯ υ 0 υ τ π υ υ Thus, as πτ ¯ = τ π¯ , we obtain that δ− (y , z ) = w . It follows that (y , z ) ∈ fw0 τ . φ Since fw = fw0 τ , this implies that φ

(y υ, zυ ) ∈ fw .

10.7

Twin Coxeter Hypergroups and Twin Buildings

377

Suppose that {y, z} ⊆ X+ . Then δ+ (y, z) = w. It follows that δ+0 (y υ, zυ ) = w τ , and φ then that (y υ, zυ ) ∈ fw0 τ . Since fw = fw0 τ , this implies that φ

(y υ, zυ ) ∈ fw . Assume now that (y, x) ∈ rw . Then (y, z) ∈ X− × X+ or (y, z) ∈ X+ × X− . Suppose that (y, z) ∈ X− × X+ . Then δ◦ (y, z) = w −1 . It follows that δ 0◦ (y υ, zυ ) = φ (w −1 )τ = (w τ )−1 , and then that (y υ, zυ ) ∈ rw0 τ . Since rw = rw0 τ , this implies that φ

(y υ, zυ ) ∈ rw . Suppose that (y, z) ∈ X+ × X− . Then δ◦ (y, z) = (w −1 )π¯ . It follows that δ 0◦ (y υ, zυ ) = 0 ¯ . Thus, as πτ (w −1 )πτ ¯ = τ π¯ 0, we obtain that δ 0◦ (y υ, zυ ) = (w −1 )τ π¯ . It follows that φ υ 0 0 υ 0 (y , z ) ∈ r((w −1 )τ )−1 = rw τ . Since rw = rw τ , this implies that φ

(y υ, zυ ) ∈ rw . What we have seen is that φ

(y υ, zυ ) ∈ fw and that

φ

(y υ, zυ ) ∈ rw

if

(y, x) ∈ fw

if

(y, x) ∈ rw .

if

fv ∈ ft fu,

if

fv ∈ rt ru,

if

rv ∈ ft ru,

if

rv ∈ rt fu .

From this we obtain that φ

φ φ

fv ∈ ft fu that

φ

φ φ

φ

φ φ

φ

φ φ

fv ∈ rt ru that

rv ∈ ft ru and that

rv ∈ rt fu

This shows that φ is a hypergroup isomorphism from H to H 0. Next we show that φ| hT i is (T, T 0)-length preserving. For this, we let η denote the ¯ and we let χ type preserving bijection induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π, χ denote the signature of η. Then, by Lemma 10.6.8, fw = w for each element w in W. Moreover, by Theorem 9.9.1(ii), χ is a bijective map from hTi to W, and from Theorem 9.9.1(iii) we know that χ is (T, I)-length preserving. Let η 0 be the type preserving bijection induced by ((X−0 , δ−0 ), (X+0 , δ+0 ), δ 0◦ ) and π¯ 0, and 0 let χ 0 be the signature of η 0. Then ( fw0 0 ) χ = w 0 for each element w 0 in W 0, and χ 0 is a (T 0, I 0)-length preserving bijective map from hT 0i to W 0.

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10 Regular Actions of (Twin) Coxeter Hypergroups

Since τ is an (I, I 0)-length preserving group isomorphism from W to W 0, we now obtain that the composite χτ( χ 0)−1 is (T, T 0)-length preserving. Note also that φ| hT i = χτ( χ 0)−1 . Thus, we have shown that φ| hT i is (T, T 0)-length preserving. We still have to show that φ is an (ω, ω 0)-isomorphism from H to H 0. For this, we choose elements y and z in X and w in W. Assume first that (y, z) ∈ fw . Then {y, z} ⊆ X− or {y, z} ⊆ X+ . Suppose that {y, z} ⊆ X− . Then z ∈ y fw

(y, z) ∈ fw

⇔

δ− (y, z) = w π¯

⇔

and zυ ∈ y υ fw0 τ

(y υ, zυ ) ∈ fw0 τ

⇔

¯ δ−0 (y υ, zυ ) = w τ π¯ = w πτ . 0

⇔

φ

φ

Thus, as fw = fw0 τ , we have z ∈ y fw if and only if zυ ∈ y υ fw . Suppose that {y, z} ⊆ X+ . Then ⇔

z ∈ y fw and

zυ ∈ y υ fw0 τ

(y, z) ∈ fw (y υ, zυ ) ∈ fw0 τ

⇔

δ+ (y, z) = w

⇔

δ+0 (y υ, zυ ) = w τ .

⇔

φ

φ

Thus, as fw = fw0 τ , we have z ∈ y fw if and only if zυ ∈ y υ fw also in this case. Assume now that (y, x) ∈ rw . Then (y, z) ∈ X− × X+ or (y, z) ∈ X+ × X− . Suppose that (y, z) ∈ X− × X+ . Then

and

zυ ∈ y υ rw0 τ

(y, z) ∈ rw

⇔

δ◦ (y, z) = w −1

(y υ, zυ ) ∈ rw0 τ

⇔

δ 0◦ (y υ, zυ ) = (w τ )−1 .

⇔

z ∈ yrw ⇔

φ

Thus, as (w τ )−1 = (w −1 )τ and rw = rw0 τ , we obtain that z ∈ yrw

⇔

φ

zυ ∈ y υ rw .

Suppose that (y, z) ∈ X+ × X− . Then z ∈ yrw

(y, z) ∈ rw

⇔

δ◦ (y, z) = (w −1 )π¯

(y υ, zυ ) ∈ rw0 τ

⇔

δ 0◦ (y υ, zυ ) = ((w τ )−1 )π¯ .

⇔

and zυ ∈ y υ rw0 τ

⇔

0

φ

¯ and r = r 0 , we obtain Thus, as ((w τ )−1 )π¯ = (w −1 )τ π¯ = (w −1 )πτ w wτ 0

0

z ∈ yrw

⇔

φ

zυ ∈ y υ rw

Twin Coxeter Hypergroups and Twin Buildings

10.7

379

also in this case, and Lemma 3.9.1 now yields that φ is an (ω, ω 0)-isomorphism from H to H 0.  Theorem 10.7.3 Let H be a hypergroup, let T be a set of involutions of H, and assume that H is a twin Coxeter hypergroup over T. Assume that none of the elements in T is thin. Let X be a set, and let ω be a regular action of H on X. Let (W, I) be a Coxeter system, and let ι be a type preserving bijection from T to I. Let ((X−, δ− ), (X+, δ+ ), δ◦ ) denote the twin building induced by ω and ι, and let π¯ denote the automorphism of W induced by ι. Assume that Hypothesis 10.5.6 holds, and let ω 0 denote the regular action induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π. ¯ Then ω and ω 0 are isomorphic. Proof. Let σ denote the color defined by ω, let ρ denote the canonical folding defined by H and T, let π denote the permutation of hTi induced by ρ, and let τ denote the signature induced by ι. Since ((X−, δ− ), (X+, δ+ ), δ◦ ) stands for the twin building induced by ω and ι, we have δ− (y, z) = (y, z)σπτ for any two elements y and z in X− , ( −1 (y, z)σρ ∗τ if y ∈ X− and z ∈ X+ ◦ δ (y, z) = −1 (y, z)σ∗ρ τ if y ∈ X+ and z ∈ X− for any two elements y and z in X, and δ+ (y, z) = (y, z)στ for any two elements y and z in X+ . Since π¯ stands for the automorphism of W induced by ι, we have π¯ = τ −1 πτ. For each element w in W, define fw := {(y, z) ∈ X− × X− | δ− (y, z) = w π¯ } ∪ {(y, z) ∈ X+ × X+ | δ+ (y, z) = w} and rw := {(y, z) ∈ X− × X+ | δ◦ (y, z) = w −1 } ∪ {(y, z) ∈ X+ × X− | δ◦ (y, z) = (w −1 )π¯ }. ¯ Let H 0 be the twin Coxeter hypergroup induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π. Then H 0 = { fw | w ∈ W } ∪ {rw | w ∈ W }. For each element h in H, we define hˆ := {(y, z) ∈ X × X | (y, z)σ = h}. By Hˆ we denote the ω-shadow of H. Then

380

10 Regular Actions of (Twin) Coxeter Hypergroups

Hˆ = { hˆ | h ∈ H}. ˆ Our first claim is that H 0 = H. ˆ For this, we let h 0 be an element in H 0. Then W contains We first show that H 0 ⊆ H. an element w such that h 0 = fw or h 0 = rw . Suppose first that h 0 = fw . Then h 0 = {(y, z) ∈ X− × X− | δ− (y, z) = w π¯ } ∪ {(y, z) ∈ X+ × X+ | δ+ (y, z) = w}, and, since π¯ = τ −1 πτ, this implies that h 0 = {(y, z) ∈ X− × X− | δ− (y, z) = w τ

−1 πτ

} ∪ {(y, z) ∈ X+ × X+ | δ+ (y, z) = w}.

Thus, as δ− (y, z) = (y, z)σπτ and δ+ (y, z) = (y, z)στ , we obtain that h 0 = {(y, z) ∈ X− × X− | (y, z)σπτ = w τ

−1 πτ

} ∪ {(y, z) ∈ X+ × X+ | (y, z)στ = w}.

Now recall from Theorem 9.9.1(ii) that τ is a surjective map from hTi to W. Thus, as w ∈ W, hTi contains an element h such that hτ = w. Then, since π and τ both are injective, we obtain that ˆ h 0 = {(y, z) ∈ X− × X− | (y, z)σ = h} ∪ {(y, z) ∈ X+ × X+ | (y, z)σ = h} = h. ˆ It follows that h 0 ∈ H. Suppose now that h 0 = rw . Then h 0 = {(y, z) ∈ X− × X+ | δ◦ (y, z) = w −1 } ∪ {(y, z) ∈ X+ × X− | δ◦ (y, z) = (w −1 )τ

−1 πτ

}.

(Recall that π¯ = τ −1 πτ.) Now recall from Theorem 9.9.1(ii) that τ is a bijective map from hTi to W. Thus, δ◦ (y, z) = w −1

⇔

(y, z)σρ

−1 ∗τ

= w −1

⇔

(y, z)σ = (w −1 )τ

−1 ∗ρ

for any two elements y in X− and z in X+ and δ◦ (y, z) = (w −1 )τ

−1 πτ

⇔ (y, z)σ∗ρ

−1 τ

= (w −1 )τ

−1 πτ

⇔ (y, z)σ = (w −1 )τ

−1 ∗ρ

for any two elements y in X+ and z in X− . Now recall from Theorem 9.9.1(ii) that τ is a surjective map from hTi to W. Thus, as w ∈ W, hTi contains an element h such that hτ = w. From this we obtain that −1 (w −1 )τ ∗ρ = hρ . Thus, h 0 = {(y, z) ∈ X− × X+ | (y, z)σ = hρ } ∪ {(y, z) ∈ X+ × X− | (y, z)σ = hρ } = hˆρ . ˆ It follows that h 0 ∈ H.

10.7

Twin Coxeter Hypergroups and Twin Buildings

381

ˆ Now notice that Since h 0 has been chosen arbitrarily in H 0, this shows that H 0 ⊆ H. 0 0 ˆ ˆ H as well as H are partitions of X × X. Thus, H = H, as claimed. Let T 0 denote the set of involutions induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π, ¯ and let ψ denote the ω-canonical hypergroup isomorphism from H to Hˆ (defined in Section 3.9). We claim that ψ is a (ω, ω 0)-isomorphism from H to H 0 the restriction of which to hTi is (T, T 0)-length preserving. We first show that ψ| hT i is (T, T 0)-length preserving. For this, we denote by ι0 the ¯ and we denote by type preserving bijection induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π, τ 0 the signature induced by ι0. We will see that ψ| hT i = τ(τ 0)−1 . Let f be an element in hTi, and set w := f τ . Then we have (y, z)σ = f

⇔

δ− (y, z) = f πτ = f τ π¯

⇔

δ− (y, z) = w π¯

for any two elements y and z in X− and (y, z)σ = f

⇔

δ+ (y, z) = f τ

⇔

δ+ (y, z) = w

for any two elements y and z in X+ . Thus, as f ψ = {(y, z) ∈ X × X | (y, z)σ = f } and fw = {(y, z) ∈ X− × X− | δ− (y, z) = w π¯ } ∪ {(y, z) ∈ X+ × X+ | δ+ (y, z) = w}, we have f ψ = fw . Now recall from Lemma 10.6.8 that ( fw )τ = w. Thus, as f ψ = fw and w = f τ , we 0 have f ψτ = f τ , and, since f has been chosen arbitrarily from hTi, this shows that ψ| hT i τ 0 = τ. It follows that ψ| hT i = τ(τ 0)−1 . 0

From Theorem 9.9.1(iii), (ii) we know that τ is a (T, I)-length preserving bijective map from hTi to W and τ 0 is a (T 0, I)-length preserving bijective map from hT 0i to W. Thus, τ(τ 0)−1 is a (T, T 0)-length preserving bijective map from hTi to hT 0i. Thus, as ψ| hT i = τ(τ 0)−1 , ψ| hT i is a (T, T 0)-length preserving bijective map from hTi to hT 0i. Since ω 0 stands for the regular action induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π, ¯ ω 0 is 0 0 0 ˆ this means that ω is the canonical the canonical action of H on X. Since H = H, action of Hˆ on X. Thus, by Lemma 3.9.4, ψ is an (ω, ω 0)-isomorphism from H to ˆ Thus, as H 0 = H, ˆ ψ| hT i is an (ω, ω 0)-isomorphism from H to H 0. H. We have seen that ψ is an (ω, ω 0)-isomorphism from H to H 0 the restriction of which to hTi is (T, T 0)-length preserving. By definition, that means that ω and ω 0  are isomorphic. Statement and proof of our final result are based on an idea suggested by C. French.

382

10 Regular Actions of (Twin) Coxeter Hypergroups

Theorem 10.7.4 Let (W, I) be a Coxeter system, let ((X−, δ− ), (X+, δ+ ), δ◦ ) be a thick twin building of type (W, I), and let π¯ be an I-length preserving automorphism of W having order 1 or 2. Assume that Hypothesis 10.5.6 holds. Let ω denote the regular action induced by ¯ and let ι denote the type preserving bijection induced ((X−, δ− ), (X+, δ+ ), δ◦ ) and π, by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π. ¯ Then the twin building induced by ω and ι is equal (up to a choice of opposite sets) to ((X−, δ− ), (X+, δ+ ), δ◦ ), and the automorphism of W induced by ι is equal to π. ¯ Proof. For each element w in W, define fw := {(y, z) ∈ X− × X− | δ− (y, z) = w π¯ } ∪ {(y, z) ∈ X+ × X+ | δ+ (y, z) = w} and rw := {(y, z) ∈ X− × X+ | δ◦ (y, z) = w −1 } ∪ {(y, z) ∈ X+ × X− | δ◦ (y, z) = (w −1 )π¯ }. Set

F := { fw | w ∈ W },

R := {rw | w ∈ W },

H := F ∪ R,

and T := { fi | i ∈ I}. ¯ and Then H is the twin Coxeter hypergroup induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π, T is the set of involutions induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π. ¯ Let ρ denote the canonical folding defined by H and T. Then, by Lemma 10.6.7(iv), ρ fw = rw for each element w in W. ρ

From f1 = r1 we obtain that (X−, X+ ) is a pair of opposite sets of ω.

Let ((X−, − ), (X+, + ),  ◦ ) denote the twin building induced by ω and ι. Our claim is that ((X−, δ− ), (X+, δ+ ), δ◦ ) = ((X−, − ), (X+, + ),  ◦ ) To establish this equation it suffices to show that − = δ−,

 ◦ = δ◦,

and

+ = δ+ .

Let σ denote the color defined by ω, let π denote the permutation (of hTi) induced by ρ, and let τ denote the signature induced by ι. Since ((X−, − ), (X+, + ),  ◦ ) stands for the twin building induced by ω and ι, we have − (y, z) = (y, z)σπτ for any two elements y and z in X− , ( −1 (y, z)σρ ∗τ if y ∈ X− and z ∈ X+ ◦  (y, z) = −1 (y, z)σ∗ρ τ if y ∈ X+ and z ∈ X−

10.7

Twin Coxeter Hypergroups and Twin Buildings

383

for any two elements y and z in X, and + (y, z) = (y, z)στ for any two elements y and z in X+ . Let y and z be elements in X− , and set w := δ− (y, z)π¯ . Then δ− (y, z) = w π¯ , so that we have (y, z) ∈ fw . Now recall that ω stands for the regular action induced ¯ and σ for the color defined by ω. Thus, we obtain by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π, that (y, z)σ = fw . Since π stands for the permutation induced by ρ, Theorem 10.6.7(v) yields that fw π¯ = fwπ . Since τ denotes the signature induced by ι, we obtain from Lemma 10.6.8 that fwτπ¯ = w π¯ . Thus, the definition of − forces − (y, z) = (y, z)σπτ = fwπτ = fwτπ¯ = w π¯ = δ− (y, z). Let y be an element in X− , let z be an element in X+ , and set w := δ◦ (y, z)−1 . Then δ◦ (y, z) = w −1, so that we have (y, z) ∈ rw . Now recall that ω stands for the regular action induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π, ¯ and σ for the color defined by ω. Thus, we obtain that (y, z)σ = rw . Since ρ stands for the canonical folding defined by H and T, we have ρ−1

fw = rw . From Lemma 10.4.5 we know that fw∪ = fw −1 . Since H is a set theoretic hypergroup, we have fw∪ = fw∗ . Thus, fw −1 = fw∗ . Since τ denotes the signature induced by ι, we obtain from Lemma 10.6.8 that fwτ−1 = w −1 . Thus, the definition of  ◦ forces  ◦ (y, z) = (y, z)σρ

−1 ∗τ

ρ−1 ∗τ

= rw

= fw∗τ = fwτ−1 = w −1 = δ◦ (y, z).

384

10 Regular Actions of (Twin) Coxeter Hypergroups

Let y be an element in X+ , and let z be an element in X− . Then, with a reference to the previous case and to Condition T1, we obtain that δ◦ (y, z) = δ◦ (z, y)−1 =  ◦ (z, y)−1 =  ◦ (y, z). Let y and z be elements in X+ , and set w := δ+ (y, z). Then (y, z) ∈ fw . Now recall that ω stands for the regular action induced by ((X−, δ− ), (X+, δ+ ), δ◦ ) and π, ¯ and σ for the color defined by ω. Thus, we obtain that (y, z)σ = fw . Since τ denotes the signature induced by ι, we obtain from Lemma 10.6.8 that fwτ = w. Thus, the definition of + forces + (y, z) = (y, z)στ = fwτ = w = δ+ (y, z). So far, we have seen that ((X−, δ− ), (X+, δ+ ), δ◦ ) = ((X−, − ), (X+, + ),  ◦ ). We still have to show that π¯ is the automorphism of W induced by ι. Since τ stands for the signature induced by ι and π for the permutation induced by ρ, τ −1 πτ is the automorphism of W induced by ι. Thus, we have to show that π¯ = τ −1 πτ. Let w be an element in W. Then, by Theorem 10.6.7(v), fw π¯ = fwπ . From Lemma 10.6.8 we also know that fwτπ¯ = w π¯ and fwτ = w. Thus, w π¯ = fwτπ¯ = fwπτ = w τ

−1 πτ

.

Since w has been chosen arbitrarily from W, we have shown that π¯ = τ −1 πτ.



Under the Hypothesis 10.5.6, Theorems 10.7.3 and 10.7.4 (together with Theorems 10.6.1, 10.6.7, 10.7.1, and 10.7.2) show that (the isomorphism classes of) thick twin buildings can be identified with (the isomorphism classes of) regular actions of twin Coxeter hypergroups over sets of non-thin involutions.

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Index

acting regularly via (an action), 14 acting via (an action), 13 action (hypergroup action) of a hypergroup, 13 association scheme, 23 associative hypermultiplication, V automorphism (hypergroup automorphism), 70 automorphism group of a hypergroup, 70 automorphism induced by a type preserving bijection, 318 basis of a projective hypergroup, 184 building induced by a regular action of a Coxeter hypergroup and a type preserving bijection, 330 building of type (W, I), 320 canonical (ω-canonical) hypergroup isomorphism, 89 canonical action of a set theoretic hypergroup, 17 canonical folding defined by a twin Coxeter hypergroup and a set of non-thin involutions, 302 centralize, 51 centralizer of a hypergroup element, 49 centralizer of a subset of a hypergroup, 49 closed π-subset, 114 closed subset, 29 codistance (W-codistance) function induced by a regular action of a twin Coxeter hypergroup and a type preserving bijection, 356

color defined by a regular action, 17 commutative hypergroup, 6 commutator, 39 compatible (b-compatible) n-tuple, 127 complex multiplication, 62 complex product, V composition factor, 98 composition length, 99 composition series, 97 conjugate closed subsets, 41 conjugate of a closed subset, 41 constrained set of involutions, 243 Coxeter hypergroup induced by a semiregular building, 335 Coxeter hypergroup over a set of involutions, 273 Coxeter number, 177 Coxeter set of involutions, 273 Coxeter system, 319 dichotomic set of involutions, 179 dimension of a finitely generated projective hypergroup, 186 direct product of closed subsets, 54 distance (W-distance) function induced by a regular action of a Coxeter hypergroup and a type preserving bijection, 327 distance (W-distance) function induced by a regular action of a twin Coxeter hypergroup and a type preserving bijection, 356 double coset of a closed subset, 33 elementary T-homotopic, 305

© Springer Nature Switzerland AG 2023 P. -H. Zieschang, Hypergroups, https://doi.org/10.1007/978-3-031-39489-8

389

390

Index

exchange condition, 180 finite hypergroup, IX finitely generated hypergroup, 35 first type, 321 folding, 45 folding induced by a thick twin building and an I-length preserving automorphism, 364 generating set of a hypergroup, 35 group correspondence, 13 Hall π-subset, 116 Hanaki-Miyamoto type, 154 homomorphism (hypergroup homomorphism), 68 homotopic (T-homotopic), 305 homotopy (T-homotopy) class, 306 hypergroup, V hypergroup induced by a semiregular building, 330 hypergroup induced by a thick twin building and an I-length preserving automorphism, 363 hypergroup of type H2,1 , 190 hypergroup of type H3,1 , 190 hypergroup of type H4, j , 201 hypergroup of type H6, j , 219 hypergroup ring of a finite tight hypergroup, 143 hypermultiplication, V hypermultiplication defined by a closed subset, 62 hypermultiplication induced by a group of automorphisms, 71 hyperoperation, V hyperoperation induced by a group, 11 hyperproduct, V idempotent hypergroup element, 191 image of a hypergroup homomorphism, 72 independent set of involutions of a projective hypergroup, 182 index of a closed subset in a hypergroup, 33 invariant (∗ -invariant) subset of a hypergroup, 7 inverse of a hypergroup element, 3 inversion function, V involution, 155

isomorphic ((ω, ω 0 )-isomorphic), 89 isomorphic buildings, 337 isomorphic composition series, 97 isomorphic hypergroups, 72 isomorphic regular actions of Coxeter hypergroups, 336 isomorphic regular actions of twin Coxeter hypergroups over sets of non-thin involutions, 371 isomorphic twin buildings, 372 isomorphism ((ω, ω 0 )-isomorphism), 86 isomorphism (building isomorphism), 336 isomorphism (hypergroup isomorphism), 70 isomorphism (twin building isomorphism), 371 kernel of a hypergroup homomorphism, 69 left coset of a closed subset, 33 length, 37 length ((U, V)-length, U-length) preserving, 174 length function, 37 maximal T-length, 180 metathin hypergroup, 109 multiplication induced by a thin hypergroup, 12 neutral element, V normal closed subset, 51 normal in, 51 normalize, 51 normalizer of a closed subset, 50 number (π-number, π 0 -number), 114 orbit of a hypergroup element, 70 pair of opposite sets, 355 permutation induced by a folding, 46 positive dimension, 179 primitive hypergroup, 29 product, V projective hypergroup, 158 quotient of a hypergroup, 63 realization as an association scheme, 207 reduced (T-reduced), 308 regular action, 14 regular action induced by a semiregular building, 335

Index regular action induced by a thick twin building and an I-length preserving automorphism, 369 representative of a double coset, 33 representative of a left coset, 33 representative of a right coset, 33 residual depth, 107 residually thin hypergroup, 107 right coset of a closed subset, 33 schurian hypergroup, 63 second type, 321 semiregular building, 324 semiregular twin building, 347 set of involutions induced by a semiregular building, 331 set of involutions induced by a thick twin building and an I-length preserving automorphism, 364 set theoretic hypergroup, 16 set theoretic hypermultiplication, 15 shadow (ω-shadow) of a hypergroup, 18 signature induced by a type preserving bijection, 311 simple hypergroup, 51 solvable hypergroup, 117 spherical Coxeter set of involutions, 290 spherical folding, 294 strong normalizer of a closed subset, 57 strong subnormal chain, 93 strong subnormal series, 96 strongly normal closed subset, 59 strongly normal in, 59 strongly subnormal closed subset, 93 strongly subnormal in, 93 structure constant of a finite tight hypergroup, 138 structure constants of an association scheme, 23 subnormal chain, 92 subnormal closed subset, 92 subnormal in, 92 subnormal series, 96

391

supporting bijection of a hypergroup isomorphism, 86 supporting group isomorphism of a building isomorphism, 337 supporting group isomorphism of a twin building isomorphism, 371 symmetric hypergroup element, 4 symmetric subset of a hypergroup, 6 thick building, 324 thick twin building, 347 thin hypergroup element, VI thin radical, 43 thin residue, 102 thin subset of a hypergroup, VI tight hypergroup, IX tight hypergroup element, IX trivial hypergroup, 29 twin building induced by a regular action of a twin Coxeter hypergroup and a type preserving bijection, 362 twin building of type (W, I), 343 twin Coxeter hypergroup induced by a thick twin building and an I-length preserving automorphism, 369 twin Coxeter hypergroup over a set of involutions, 302 twinned hypergroup, 47 type preserving bijection, 310 type preserving bijection induced by a semiregular building, 335 type preserving bijection induced by a thick twin building and an I-length preserving automorphism, 369 valenced (π-valenced) hypergroup element, 115 valenced (π-valenced) hypergroup), 115 valency of an element of a finite tight hypergroup, 138 wreath hypermultiplication, 77 wreath product of hypergroups, 82