Theory of Association Schemes (Springer Monographs in Mathematics)

Springer Monographs in Mathematics Paul-Hermann Zieschang Theory of Association Schemes ABC Paul-Hermann Zieschan...

Author: Paul-Hermann Zieschang

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Springer Monographs in Mathematics

Paul-Hermann Zieschang

Theory of Association Schemes

ABC

Paul-Hermann Zieschang Department of Mathematics University of Texas at Brownsville Brownsville, TX 78520 USA E-mail: [email protected]

Library of Congress Control Number: 2005930450 Mathematics Subject Classification (2000): 05E30, 20N99 ISSN 1439-7382 ISBN-10 3-540-26136-2 Springer Berlin Heidelberg New York ISBN-13 978-3-540-26136-0 Springer Berlin Heidelberg New York This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable for prosecution under the German Copyright Law. Springer is a part of Springer Science+Business Media springeronline.com c Springer-Verlag Berlin Heidelberg 2005 Printed in The Netherlands The use of general descriptive names, registered names, trademarks, 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. Typesetting: by the author and TechBooks using a Springer LATEX macro package Cover design: design & production GmbH, Heidelberg Printed on acid-free paper

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543210

Preface

The present text is an introduction to the theory of association schemes. We start with the definition of an association scheme (or a scheme as we shall say briefly), and in order to do so we fix a set and call it X. We write 1X to denote the set of all pairs (x, x) with x ∈ X. For each subset r of the cartesian product X × X, we define r∗ to be the set of all pairs (y, z) with (z, y) ∈ r. For x an element of X and r a subset of X × X, we shall denote by xr the set of all elements y in X with (x, y) ∈ r. Let us fix a partition S of X × X with ∅ ∈ / S and 1X ∈ S, and let us assume that s∗ ∈ S for each element s in S. The set S is called a scheme on X if, for any three elements p, q, and r in S, there exists a cardinal number apqr such that |yp ∩ zq ∗ | = apqr for any two elements y in X and z in yr. The notion of a scheme generalizes naturally the notion of a group, and we shall base all our considerations on this observation. Let us, therefore, briefly look at the relationship between groups and schemes. Let S be a scheme, and let P and Q be nonempty subsets of S. We define P Q to be the set of all elements s in S for which there exist elements p in P and q in Q satisfying 1 ≤ apqs . If P possesses an element p with {p} = P and Q an element q satisfying {q} = Q, we write pq instead of P Q. The set P Q will be called the complex product of P and Q. The associated operation on the set of all nonempty subsets of S will be referred to as the complex multiplication in S. It follows right from the definition of the complex multiplication that, for any two elements p and q of a scheme S, p∗ q is not empty. An element s of S will be called thin if s∗ s contains exactly one element (which then must be the identity element of S). A nonempty subset of S will be called thin if each of its elements is thin. Given a thin scheme S, it is easy to see that the set S γ of all sets {s} with s ∈ S is a group with respect to the restriction of the complex multiplication in S to S γ . Conversely, let G be a group. For each element g in G, we define

VI

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g τ to be the set of all pairs (e, f ) of elements of G satisfying eg = f . Then the set Gτ of all sets g τ with g ∈ G is a thin scheme on G.1 It turns out that, for each thin scheme S, the scheme S γτ is isomorphic to S. Conversely, for each group G, the group Gτ γ is isomorphic to G.2 This establishes a one-to-one correspondence between groups and thin schemes. We call this one-to-one correspondence the group correspondence. The group correspondence allows us to view the class of groups as a distinguished class of schemes, namely as the class of thin schemes. The group correspondence suggests the development of a general structure theory of schemes based on concepts similar to those which have been so successful in group theory. In fact, we shall follow this idea right from the beginning of our text. In particular, we subdivide our theory according to the shape of group theory as it presents itself nowadays. This means that, after three introductory chapters, we focus separately on the decomposition theory of schemes (including their local theory), their representation theory, and their theory of generators. In order to get a first impression about the contents of the twelve individual chapters, we shall now briefly preview each one separately.3 For any three elements p, q, and r of a scheme S, the cardinal number apqr will be called the structure constant of p, q, and r in S. Structure constants, complex products, and the relationship between these two notions form the subject of the first chapter. Most of these results, in particular those of Section 1.4, are fundamental for the further development of the theory. The complex product gives rise to the notion of a so-called closed subset, and it is this concept on which we focus in the second chapter of this monograph. A nonempty subset R of a scheme is called closed if p∗ q ⊆ R for any two elements p and q in R. For each thin scheme S, the group correspondence establishes a one-to-one correspondence between the closed subsets of S and the subgroups of S γ . This means, in particular, that the notion of a closed subset generalizes that of a subgroup. More importantly, closed subsets retain some of the interesting properties of subgroups and transfer them to scheme theory. For instance, the set of all left cosets of a closed subset of a scheme S in S (defined with the help of the complex multiplication in S) is a partition of S.4 Furthermore, 1

2

3

4

The first of these two observations will be proved as Theorem 5.5.1, the second one as Theorem 5.5.2. The first of these two observations will be proved as Theorem 5.5.3, the second one as Theorem 5.5.4. We shall say more about the various sections at the beginning of the individual chapters. This observation enables us to generalize the notion of a ‘coset geometry’ from group theory to scheme theory. In fact, the relationship between ‘buildings’ (in the sense of Jacques Tits) and those schemes to which we shall refer later as

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Richard Dedekind’s ‘modularity law’ holds for closed subsets. The treatment of so-called length functions, which one obtains from generating sets of closed subsets, is another subject which allows one to mimic group theoretic techniques in scheme theory. Finally, the notion of a closed subset leads us to the notion of an involution in scheme theory. A non-identity element of a scheme is called an involution if it is contained in a closed subset of cardinality 2. In the third chapter of this monograph, we look at closed subsets generated by specific subsets. Our investigation leads us to the definitions of commutator subsets and thin residue of a closed subset, as well as to other characteristic closed subsets of schemes. Closed subsets generated by sets of involutions turn out to be an interesting subject, especially if one imposes appropriate extra conditions on the set of the generating involutions. As an example, we introduce constrained sets of involutions; as another example, we look at sets of involutions satisfying the exchange condition. A constrained set of involutions which satisfies the exchange condition will be called a Coxeter set. At the end of the third chapter, we give some basic information about Coxeter sets. A more rigorous treatment of Coxeter sets will be given in the last two chapters of this monograph. The aforementioned exchange condition is defined in such a way that, via the group correspondence, its thin version is equivalent to the well-known group theoretic exchange condition which distinguishes the Coxeter groups within the class of all groups generated by involutions. In addition to the definition of closed subsets and the notion of an involution, the exchange condition is a further example of our general philosophy that, via the group correspondence, each scheme theoretic definition or statement applied to thin schemes has a natural and well-known group theoretic analogue. There is one significant difference between subgroups of groups and closed subsets of schemes. In order for a quotient structure of a group to be a group one has to ‘factor out’ a normal subgroup. In scheme theory (at least as it applies to finite sets) one may factor out any closed subset without leaving the class of schemes. In this manner one obtains a quotient scheme for each closed subset of a scheme of finite valency, irrespective of whether the closed subset is normal or not. As a consequence, there is much more space for inductive reasoning. Quotient schemes are introduced in the fourth chapter of this monograph. Factorization over non-normal closed subsets provides us with a particularly smooth approach to a generalization of Ludwig Sylow’s theorems on finite Coxeter schemes is based on the idea of looking at coset geometries of schemes; cf. [43; Theorem E]. However, in this monograph, we shall ignore this geometric aspect completely.

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groups to scheme theory; cf. Theorem 4.5.3, Theorem 4.5.5, and Theorem 4.5.7. Generalizing Sylow’s theorems on finite groups to scheme theory requires, of course, an appropriate generalization of Sylow subgroups to closed Sylow subsets in scheme theory. It is clear that our definition of closed Sylow subsets is yet another example of our general philosophy that each scheme theoretic definition or result applied to thin schemes has a corresponding group theoretic definition or result. In the fifth chapter, we introduce morphisms and, more specifically, homomorphisms. We establish the natural relationship between quotient schemes and homomorphisms, thereby generalizing Emmy Noether’s group theoretic Homomorphism Theorem and her two Isomorphism Theorems to scheme theory. The Second Isomorphism Theorem enables us to show that, up to isomorphism, schemes of finite valency admit only one so-called composition series; cf. Theorem 5.4.2. This generalizes a well-known theorem on finite groups due to Camille Jordan and Otto H¨ older. Similar to group theory, this theorem gives rise to a scheme theoretic notion of composition factors, as well as emphasizes a notion of simple schemes. We finish the fifth chapter with an investigation of schemes which contain thin composition factors. Morphisms are related to faithful maps, which lead naturally to the notion of a faithfully embedded closed subset. Such subsets provide an appropriate language for an attempt to establish so-called recognition theorems. These theorems deal with the question of which schemes are quotient schemes of thin schemes. We shall come back to recognition theorems and their role in scheme theory later in this preface. In the sixth chapter, we introduce faithful maps and faithfully embedded closed subsets. In particular, we define a closed subset to be schurian if it is faithfully embedded in itself. This chapter is also the place where we prove the first recognition theorems. The seventh chapter is devoted to products. We define direct products of closed subsets of schemes, quasi-direct products of schemes, and, following an approach due to Sejeong Bang, Mitsugu Hirasaka, and Sung-Yell Song, semidirect products of schemes. One of the main results of this chapter is a theorem of Pamela Ferguson and Alexandre Turull on indecomposable schemes. This theorem is similar to the group theoretic theorem of Wolfgang Krull, Erhard Schmidt, and Robert Remak for commutative groups. Semidirect products will play a role in our investigation of spherical Coxeter sets of cardinality 2 (see the last chapter). The eighth chapter of this monograph deals with thin schemes. In this chapter, we collect basic and well-known ring theoretic concepts and facts which are needed for the representation theory of schemes of finite valency. The ninth chapter is an introduction to the basic ideas of representation theory of schemes of finite valency. It is based on the analysis of the previous chapter. Generalizing the notion of a group ring we define scheme rings. We

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give sufficient criteria for a scheme ring to be semisimple (thereby generalizing Heinrich Maschke’s important result on the semisimplicity of the group ring of a finite group). At the end, we apply a few of our representation theoretic results to finite schemes with three elements. In the tenth chapter, we study schemes generated by a set of two involutions. The main result of this chapter is a characterization of schemes which have finite valency and are generated by a Coxeter set of cardinality 2. Referring to results obtained in the previous chapter we obtain, as a consequence, a representation theoretic characterization of such schemes. The flow of the tenth chapter shows in a particularly convincing way how smooth the exchange condition emerges from the general theory of generators in scheme theory. In the final two chapters of this monograph, we focus on constrained sets of involutions which satisfy the exchange condition. In other words, we look at Coxeter sets. We mentioned earlier that the exchange condition generalizes the well-known group theoretic exchange condition which distinguishes the Coxeter groups among the groups generated by involutions. In a strong sense, this justifies our use of the term ‘Coxeter scheme over L’ when referring to any scheme generated by a constrained set L of involutions satisfying the scheme theoretic exchange condition. The eleventh chapter is an outline of a general theory of Coxeter schemes. It turns out that most of the basic observations on Coxeter schemes are natural generalizations of the corresponding ones in the theory of Coxeter groups. In fact, the treatment of so-called parabolic subgroups can be taken over almost word-by-word to parabolic subsets. This close relationship between Coxeter schemes and Coxeter groups has a firm mathematical basis, and it leads us to a general question in scheme theory which we had postponed earlier and would like to briefly address here. The way in which the group correspondence produces thin schemes from groups can be generalized. In fact, each subgroup H of a group G gives rise to a scheme on the set of all left cosets of H in G. For each element g in G, we define g H to be the set of all pairs (eH, f H) where e and f are elements in G satisfying eg = f . It is easy to see (and well known) that the set G//H of all sets g H with g ∈ G is a scheme on the set G/H of all left cosets of H in G. The case {1} = H is one with which we are already familiar from the group correspondence. (In this case, we obtain thin schemes.) The general case leads to quotient schemes of thin schemes in the same way the case {1} = H leads to thin schemes via the group correspondence. It is well known (and easy to see) that schemes are not necessarily isomorphic to quotient schemes of thin schemes. However, it seems that a general scheme theoretic characterization of such quotient schemes is out of reach. It is for this reason that one might instead ask for specific sufficient conditions for a scheme to be isomorphic to a quotient scheme of a thin scheme.

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Recall that in the sixth chapter, we defined a closed subset to be schurian if it is faithfully embedded in itself. From this definition we easily obtain that a scheme is schurian if and only if it is isomorphic to a quotient scheme of a thin scheme. Let us now return, as earlier promised, to a discussion on recognition theorems. Suppose we are given a condition (or a collection of conditions) S under which a scheme S turns out to be schurian. It is then natural to ask for a group theoretic condition G which is satisfied by any two groups H and G such that H is a subgroup of G and G//H satisfies S. If, conversely, G//H satisfies S for any two groups H and G, H a subgroup of G, satisfying G, then we call the theorem which says that S implies G a recognition theorem. As beautifully and completely developed as group theory is nowadays, one of the major goals of scheme theory must be to establish recognition theorems in which large classes of schemes (different from the class of all thin schemes) are recognized as quotients of specific classes of groups. In this way, schurian schemes may well serve as classifying categories for specific classes of groups, so that characterization theorems or classification theorems in group theory would appear as part of a general scheme theory. In the first part of the final chapter, we show that the exchange condition is almost sufficient for a scheme to be schurian. We prove that Coxeter schemes S over a set L of involutions are always schurian if S is finite, if L has at least three elements, and if L does not contain thin elements. More precisely, we show in Theorem 12.3.4 that, for each Coxeter scheme S satisfying the above three conditions, there exists a group G and a so-called Borel subgroup B such that, via the group correspondence, S is isomorphic to G//B. Since we shall conversely show that, for each group G possessing a Borel subgroup B, the quotient scheme G//B is a Coxeter scheme satisfying the above three conditions, Theorem 12.3.4 may be regarded as one of our recognition theorems.5 5

Developing a theory of coset geometries (as mentioned in the previous footnote), one obtains from our recognition theorem of Coxeter schemes Jacques Tits’ main result on buildings of ‘spherical type’. This theorem asserts that each such building is associated with a group if it is ‘thick’ and of ‘rank’ at least 3. In fact, our interest in Coxeter schemes is partially motivated by the fact that such schemes provide an alternate language for Tits’ theory of buildings, a language in which buildings can be treated as algebraic objects simultaneously with other algebraic objects such as groups. The emphasis on Coxeter schemes themselves, rather than on the elements of the underlying set (‘chambers’ in the language of geometers), shows that many of the traditional concepts in the theory of buildings are actually not needed to investigate buildings. In particular, the ‘free monoid’, traditionally associated with each building as a substantial tool in each introduction to buildings (cf., e.g., [38] or [40]), is not part of the theory of Coxeter schemes the way it is developed in the last two chapters of this monograph.

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Theorem 6.4.5 and Theorem 6.5.3 are two further recognition theorems which deal with involutions. Theorem 6.4.5 relates to George Glauberman’s Z ∗ Theorem a specific class of schemes of finite valency generated by elements of valency 2. After having established the recognition theorem for Coxeter schemes, we present a well-known result of Walter Feit and Graham Higman on finite generalized polygons. (We provide a scheme theoretic version of a proof given by Robert Kilmoyer and Louis Solomon.) At the end of the final chapter, we prove the theorems of Stanley Payne and Udo Ott on polarities of finite generalized quadrangles and finite generalized hexagons. They appear here as results on semidirect products. At this point, we conclude our treatment of schemes, although we acknowledge that there are quite a few more aspects of scheme theory which should have their place in a comprehensive textbook but which are not included in our monograph. Among the more theoretical sources of information about schemes (of finite valency), I would like to mention, in particular, the following two research areas. First, there are numerous results on characters of schemes of finite valency which have not found their way into this text, although they provide an impressive insight into the rich and prosperous theory of schemes. In particular, modular representation theory of schemes of finite valency, as it has been developed mainly by Akihide Hanaki, has been excluded completely.6 The other large source of information about schemes (of finite valency) comes from the numerous interesting results on table algebras which have not been considered in this monograph. (Table algebras generalize the notion of schemes of finite valency.) Most of these results are due to Zvi Arad, Harvey Blau, and Mikhail Muzychuk. It is mainly for these two reasons that this monograph should not be considered to be a comprehensive text on scheme theory. It is much more an attempt to present those concepts and results in scheme theory which seem to have the potential to convince the reader that scheme theory is a young theory the florescence of which is still to come. For the remainder of this text, we fix a set and denote it by X. The letter S will always stand for a scheme on X. We shall always write 1 instead of 1X . The condition which gives us the structure constants for a scheme, will be called the regularity condition. Except for a few isolated cases, the letters v, w, x, y, and z are reserved to denote elements in X. The letters p, q, r, s, t, and u will mostly stand for elements of S. (Clearly, the letter p will also often be used to denote prime 6

This excludes one of the most beautiful results on schemes of finite valency, the theorem of Akihide Hanaki and Katsuhiro Uno which says that schemes are commutative if their valency is a prime number; cf. [19; Theorem 3.3].

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numbers.) Involutions of S will be denoted by h, k, or l, a set of involutions mostly by H, K, or L. The letters O, P , Q, and R will usually denote subsets of S, the letters T , U , V , and W closed subsets of S. I would like to close my introductory remarks by thanking two colleagues to whom I owe research stays which allowed me to concentrate on scheme theory over longer periods of times without any academic duties. Ernest Shult gave me the opportunity to stay at Kansas State University, Manhattan (Kansas), during the academic year 1991/92 as a fellow of the Max-Kade-Foundation, New York (U. S. A.). Eiichi Bannai invited me to visit Kyushu University, Fukuoka (Japan), during the academic year 1996/97 as a fellow of the Long-Term Invitation Fellowship Program for Research in Japan rewarded by the Japan Society for the Promotion of Science (JSPS), Tokyo (Japan). Several colleagues have influenced this text directly or indirectly through stimulating correspondence and discussion. I would like to mention Harvey Blau from Northern Illinois University, DeKalb (Illinois), Akihide Hanaki from Shinshu University, Matsumoto (Japan), Mitsugu Hirasaka from Pusan National University, Pusan (Republic of Korea), Mikhail Muzychuk from Netanya Academic College, Netanya (Israel), Sung-Yell Song from Iowa State University, Ames (Iowa), Paul Terwilliger from University of Wisconsin, Madison (Wisconsin), and Andrew Woldar from Villanova University, Villanova (Pennsylvania). It is, in particular, Mitsugu Hirasaka and Mikhail Muzychuk to whom I owe many general insights, as well as a number of small observations of which I was not immediately aware. Waterneverstorf, May 2005

Paul-Hermann Zieschang

Contents

1

Basic Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Structure Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Symmetric Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.3 The Complex Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.4 Complex Products and Valencies . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.5 Complex Products of Subsets of Cardinality 1 . . . . . . . . . . . . . . 14

2

Closed Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Basic Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Dedekind Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Structure Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Maximal Closed Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Normalizer and Strong Normalizer . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Conjugates of Closed Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

17 17 22 23 28 31 35

3

Generating Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Basic Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 The Thin Residue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Elements of Valency 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Closed Subsets Generated by Involutions . . . . . . . . . . . . . . . . . . 3.5 Basic Results on Constrained Sets of Involutions . . . . . . . . . . . . 3.6 Basic Results on Coxeter Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . .

39 40 45 48 51 55 57

4

Quotient Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 General Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Valencies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Hall Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Sylow Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

63 64 69 74 77 78

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Morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Basic Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 The Isomorphism Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Composition Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 The Group Correspondence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Residually Thin Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

83 84 88 90 93 95 98

6

Faithful Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 6.1 Basic Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 6.2 Faithfully Embedded Closed Subsets . . . . . . . . . . . . . . . . . . . . . . 107 6.3 The Schur Group of a Closed Subset . . . . . . . . . . . . . . . . . . . . . . 112 6.4 Elements of Valency 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 6.5 More About Elements of Valency 2 . . . . . . . . . . . . . . . . . . . . . . . . 122 6.6 Constrained Sets of Involutions . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 6.7 Thin Thin Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

7

Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 7.1 Direct Products of Closed Subsets . . . . . . . . . . . . . . . . . . . . . . . . 133 7.2 Quasidirect Products of Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . 138 7.3 Semidirect Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 7.4 A Characterization of Semidirect Products . . . . . . . . . . . . . . . . . 149

8

From Thin Schemes to Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 8.1 Rings and Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 8.2 Integrality in Associative Rings with 1 . . . . . . . . . . . . . . . . . . . . . 161 8.3 Completely Reducibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 8.4 Irreducible Modules over Associative Rings with 1 . . . . . . . . . . 168 8.5 Semisimple Associative Rings with 1 . . . . . . . . . . . . . . . . . . . . . . 172 8.6 Characters of Associative Rings with 1 . . . . . . . . . . . . . . . . . . . . 175 8.7 Roots of Unity in Integral Domains . . . . . . . . . . . . . . . . . . . . . . . 179

9

Scheme Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 9.1 Basic Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 9.2 Algebraically Closed Base Fields . . . . . . . . . . . . . . . . . . . . . . . . . . 192 9.3 Scheme Rings over the Field of Complex Numbers . . . . . . . . . . 195 9.4 Closed Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 9.5 Schemes with at most Five Elements . . . . . . . . . . . . . . . . . . . . . . 204 9.6 Constrained Sets of Involutions . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

10 Dihedral Closed Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 10.1 General Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 10.2 The Spherical Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 10.3 Arithmetic of the Length Function . . . . . . . . . . . . . . . . . . . . . . . . 217 10.4 Two Characteristic Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220

Contents

XV

10.5 The Constrained Spherical Case . . . . . . . . . . . . . . . . . . . . . . . . . . 226 10.6 Dihedral Closed Subsets of Finite Valency . . . . . . . . . . . . . . . . . 228 11 Coxeter Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 11.1 Parabolic Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238 11.2 Direct Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 11.3 Faithful Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 11.4 The Extension Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244 12 Spherical Coxeter Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 12.1 Elements of Maximal Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 12.2 Faithful Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 12.3 The Main Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 12.4 Coxeter Schemes of Finite Valency and Rank 2 . . . . . . . . . . . . . 259 12.5 Valencies and Multiplicities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 12.6 Polarities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281

1 Basic Facts

This introductory chapter deals with the arithmetic of structure constants and its consequences for complex products.

1.1 Structure Constants In this section, we establish a collection of fundamental equations about structure constants. We shall also look at structural consequences derived from these equations. Lemma 1.1.1 For any two elements p and q of S, the following hold. (i) We have a1pq = δpq and ap1q = δpq . (ii) For each element s in S, apqs = aq∗ p∗ s∗ . (iii) For any two elements t and u in S, apqs astu = apsu aqts . s∈S

s∈S

Proof. (i) This follows immediately from the definition of the structure constants of S. (ii) Let y be an element in X, and let z be an element in ys. Then, by definition, |yp ∩ zq ∗ | = apqs and y ∈ zs∗ . From y ∈ zs∗ we obtain |zq ∗ ∩ yp∗∗ | = aq∗ p∗ s∗ . Since p∗∗ = p, the two equations yield apqs = aq∗ p∗ s∗ . (iii) Let y be an element in X, and let z be an element in yu. We count in two different ways the pairs (v, w) in (yp × zt∗ ) such that u ∈ vq. Then the desired equation follows from the definition of the structure constants. Let s be an element in S. We write ns instead of ass∗ 1 and call ns the valency of s. Note that, as s is assumed to be nonempty, 1 ≤ ns .

2

1 Basic Facts

For each nonempty subset R of S, we define nR to be the sum of the cardinal numbers nr with r ∈ R. We call nR the valency of R. Note that |X| = nS . Let P and Q be nonempty subsets of S such that P ⊆ Q. Then nP ≤ nQ . If Q has finite valency, we have nP = nQ if and only if P = Q. Lemma 1.1.2 For each element s in S, the following hold. (i) For each element x in X, |xs| = ns . (ii) We have |s| = nS ns . (iii) If S has finite valency, ns∗ = ns . Proof. (i) Let x be an element in X. Then, as s∗∗ = s, |xs| = ass∗ 1 , so that the claim follows from the definition of ns . (ii) This follows from (i) together with the observation that |X| = nS . (iii) From (ii) we know that |s∗ | = ns∗ nS and that |s| = ns nS . Thus, as |s∗ | = |s|, ns∗ nS = ns nS . Thus, assuming nS to be finite we obtain ns∗ = ns . Lemma 1.1.3 For any two elements p and q of S, the following hold. (i) If 1 ≤ apq∗ 1 , then p = q. (ii) For each element s in S, apsq nq = aqs∗ p np . (iii) We have

apsq = np =

s∈S

asp∗ q .

s∈S

(iv) We have

apqs ns = np nq .

s∈S

Proof. (i) This follows from the fact that, by definition, the intersection of any two different elements of S is empty. (ii) Applying Lemma 1.1.1(iii) to s, q ∗ , and 1 in the role of q, t, and u we obtain apsq nq = np asq∗ p∗ ; cf. (i). From Lemma 1.1.1(ii) we know that asq∗ p∗ = aqs∗ p . Thus, apsq nq = aqs∗ p np . (iii) Let y be an element in X, and let z be an element in yq. Then, as yp is the disjoint union of the sets yp ∩ zs∗ with s ∈ S, apsq = |yp ∩ zs∗ | = |yp| = np ; s∈S

cf. Lemma 1.1.2(i).

s∈S

1.1 Structure Constants

3

The second equation follows from the first one together with Lemma 1.1.1(ii). (iv) From (ii) we obtain

apqs ns =

s∈S

asq∗ p np .

s∈S

Thus, the claim follows from the second equation of (iii). Lemma 1.1.4 For any two elements p and q of S, the following hold. (i) For each element s in S with ns∗ = ns , we have aspq nq = as∗ qp np . (ii) Let t and u be elements in S, and let us assume that nq∗ = nq , that nt∗ = nt , and that nu∗ = nu . Then we have apqs atus ns = ap∗ ts aqu∗ s ns . s∈S

s∈S

(iii) If q has finite valency, aqp∗ q = aqpq . Proof. (i) From Lemma 1.1.3(ii) we know that aspq nq = aqp∗ s ns and that as∗ qp np = apq∗ s∗ ns∗ . From Lemma 1.1.1(ii) we know that aqp∗ s = apq∗ s∗ . Thus, the desired equation follows from our hypothesis that ns∗ = ns . (ii) From nt∗ = nt , (i), and Lemma 1.1.1(ii) we obtain apqs atus ns = apqs at∗ su nu = aq∗ p∗ s∗ as∗ tu∗ nu . s∈S

s∈S

s∈S

From nq∗ = nq and (i) we obtain ap∗ ts aqu∗ s ns = ap∗ ts aq∗ su∗ nu∗ = aq∗ su∗ ap∗ ts nu∗ . s∈S

s∈S

s∈S

Thus, the claim follows from Lemma 1.1.1(iii). (iii) From Lemma 1.1.3(ii) we know that aqp∗ q nq = aqpq nq , so that our claim follows from the hypothesis that q has finite valency. There is a natural way to generalize the notion of a structure constant, which in turn allows generalization of some of our previous results. We shall not make use of such generalizations very often. However, there are a few places where it is convenient to work with these numbers. Let s be an element in S, let n be an integer with 3 ≤ n, and let r1 , . . . , rn be elements in S. We inductively define integers ar1 ...rn s by ar1 ...rn−1 q aqrn s . ar1 ...rn s := q∈S

Our next lemma generalizes Lemma 1.1.1(iii), (ii) and Lemma 1.1.3(iv).

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1 Basic Facts

Lemma 1.1.5 Let n be an integer with 2 ≤ n, and let r1 , . . . , rn be elements in S. (i) Let s be an element in S. Then, if 3 ≤ n, ar1 ...rn s = ar1 qs ar2 ...rn q . q∈S

(ii) For each element s in S, we have ar1 ...rn s = arn∗ ...r1∗ s∗ . (iii) We have

ar1 ...rn s ns = nr1 · · · nrn .

s∈S

Proof. (i) If n = 3, the claim is just a restatement of Lemma 1.1.1(iii) (with r1 , r2 , r3 , and s in the role of p, q, t, and u). Therefore, we assume that 4 ≤ n. Assuming that the claim holds for n − 1, we obtain ar1 ...rn−1 q aqrn s = ( ar1 pq ar2 ...rn−1 p )aqrn s ar1 ...rn s = q∈S

=

ar2 ...rn−1 p

p∈S

=

ar1 qs (

q∈S

q∈S p∈S

ar1 pq aqrn s =

q∈S

ar2 ...rn−1 p

p∈S

ar2 ...rn−1 p aprn q ) =

p∈S

ar1 qs aprn q

q∈S

ar1 qs ar2 ...rn q .

q∈S

(The fourth equation follows from Lemma 1.1.1(iii).) (ii) If n = 2, the claim is just a restatement of Lemma 1.1.1(ii) (with r1 and r2 in the role of p and q). Therefore, we assume that 3 ≤ n. Assuming that the claim holds for n − 1, we obtain ∗ ∗ ...r ∗ s∗ . ar1 ...rn−1 q aqrn s = arn∗ q∗ s∗ arn−1 ar1 ...rn s = ...r1∗ q ∗ = arn 1 q∈S

q∈S

(The second equation follows from Lemma 1.1.1(ii), the last one from (i).) (iii) If n = 2, the claim is just a restatement of Lemma 1.1.3(iv) (with r1 and r2 in the role of p and q). Therefore, we assume that 3 ≤ n. We have ar1 ...rn s ns = ( ar1 ...rn−1 q aqrn s )ns = ar1 ...rn−1 q aqrn s ns s∈S

s∈S q∈S

=

q∈S

ar1 ...rn−1 q (nq nrn ) = (

q∈S

s∈S

ar1 ...rn−1 q nq )nrn = nr1 · · · nrn .

q∈S

(The third equation follows from Lemma 1.1.3(iv).)

1.2 Symmetric Elements

5

Lemma 1.1.6 Assume that S has finite valency. Then we have aq∗ pqs = nS . nq∗ p∈S q∈S

for each element s in S. Proof. For any two elements q and s in S, we have aq∗ pqs aq∗ pr arqs arqs = = ( aq∗ pr ) = arqs = nq∗ . nq∗ nq∗ nq∗ p∈S

p∈S r∈S

r∈S p∈S

r∈S

(The third equation follows from the first equation of Lemma 1.1.3(iii), the last equation follows from the second equation of Lemma 1.1.3(iii).)

1.2 Symmetric Elements An element s in S is called symmetric if s∗ = s. Let us see how information about the structure constants of S relates to this notion of symmetry. Lemma 1.2.1 If S has odd valency, all symmetric elements in S \ {1} have even valency. Proof. Let s be a symmetric element in S \ {1}. Then y ∈ zs for any two elements y and z in X with z ∈ ys. Thus, as 1 = s, |s| is even. On the other hand, we have |s| = ns nS ; cf. Lemma 1.1.2(ii). Thus, as nS is assumed to be odd, ns must be even. For each subset R of S, we define R∗ to be the set of all elements s in S with s∗ ∈ R. Lemma 1.2.2 Assume that each element of S has finite valency and that no element in S \ {1} is symmetric. Then each element in S has odd valency. Proof. By hypothesis, there exists a subset R of S such that {R∗ , R} is a partition of S \ {1}. Thus, for each element s in S, asrs = as1s + 2 asrs = 1 + 2 asrs . ns = r∈S

r∈R

r∈R

(The first equation follows from the first equation of Lemma 1.1.3(iii), the second one from Lemma 1.1.4(iii), and the last one from the second equation of Lemma 1.1.1(i).) The following lemma is a special case of [1; Theorem 2.2(i)].

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1 Basic Facts

Lemma 1.2.3 Assume that S has finite valency, and let k denote the highest 2-power dividing each ns with s ∈ S \ {1}. Assume that 1 = k, and let s be an element in S \ {1} such that ns k −1 is odd. Then s is symmetric. Proof. Let p and q be elements in S \ {1}. Then, by Lemma 1.1.4(ii), app∗ s aqq∗ s ns = ap∗ qs ap∗ qs ns . s∈S

s∈S

Thus, np nq +

app∗ s aqq∗ s ns = δpq n2p∗ +

s∈S\{1}

(ap∗ qs )2 ns .

s∈S\{1}

We are assuming that 1 = k. Thus, np nq k −1 as well as δpq n2p∗ k −1 are even. Thus, modulo 2, app∗ s aqq∗ s ns k −1 ≡ (ap∗ qs )2 ns k −1 . s∈S\{1}

s∈S\{1}

Let us denote by R the set of all elements s in S \ {1} such that ns k −1 is odd. Then, for each element s in S \ R \ {1}, ns k −1 is even. Thus, modulo 2, app∗ r aqq∗ r nr k −1 ≡ app∗ s aqq∗ s ns k −1 ≡ (ap∗ qs )2 ns k −1 r∈R

s∈S\{1}

≡

ap∗ qs ns k

−1

=

s∈S\{1}

ap∗ qs ns k −1 − δpq np∗ k −1 .

s∈S

s∈S\{1}

According to Lemma 1.1.3(iv), the last difference is equal to np∗ nq k −1 − δpq np∗ k −1 . Let us now assume that p ∈ R. From p ∈ R we obtain that np∗ k −1 is odd. From 1 = q we obtain that nq is even. Thus, modulo 2, np∗ nq k −1 − δpq np∗ k −1 = np∗ k −1 (nq − δpq ) ≡ nq − δpq ≡ δpq . Note, finally, that nr∗ k −1 is odd for each element r in R. Thus, modulo 2, app∗ r aqq∗ r ≡ app∗ r aqq∗ r nr k −1 ≡ δpq . r∈R

r∈R

Let us now look at the matrix A := (app∗ q )pq where p and q are elements in R. What we just saw means that A is regular.

1.3 The Complex Product

7

For any two elements p and q in R, we have app∗ q∗ = app∗ q ; cf. Lemma 1.1.1(ii). Moreover, from Lemma 1.1.2(iii), we obtain R∗ = R. Thus, as A is regular, R contains no non-symmetric element. (Otherwise, A would have two equal columns.) A subset R of S is called symmetric if all of its elements are symmetric. Corollary 1.2.4 Assume that S has finite valency. Assume that all elements in S \ {1} have the same valency and that this valency is even. Then S is symmetric. Proof. This is an immediate consequence of Lemma 1.2.3. A closed subset T of S is called commutative if apqr = aqpr for any three elements p, q, and r in T . Lemma 1.2.5 If S is symmetric, S is commutative. Proof. For any three elements p, q, and r in S, we have apqr = aq∗ p∗ r∗ = aqpr ; cf. Lemma 1.1.1(ii).

1.3 The Complex Product Recall that, for any two nonempty subsets P and Q of S, the complex product P Q of P and Q is defined to be the set of all elements s in S such that there exist elements p in P and q in Q satisfying 1 ≤ apqs . Whenever s is an element in S and T a nonempty subset of S, we write sT instead of {s}T and T s instead of T {s}. There are a few basic facts about the complex product. Occasionally, we shall quote these results without further reference. Firstly, for each nonempty subset R of S, we have R1 = R = 1R. Secondly, for any three nonempty subsets P , Q, and R of S with P ⊆ Q, we have RP ⊆ RQ and P R ⊆ QR. From the second observation we obtain R(P ∪ Q) = RP ∪ RQ and (P ∪ Q)R = P R ∪ QR for any three nonempty subsets P , Q, and R of S. A similar observation is that, for any two nonempty subsets P and Q of S, P Q is equal to the union of the sets P q with q ∈ Q as well as to the union of the sets pQ with p ∈ P . From the second observation we also obtain R(P ∩ Q) ⊆ RP ∩ RQ and (P ∩ Q)R ⊆ P R ∩ QR for any three nonempty subsets P , Q, and R of S. However, in general we do not have RP ∩ RQ ⊆ R(P ∩ Q) or P R ∩ QR ⊆ (P ∩ Q)R. In Lemma 2.2.1, we shall give sufficient conditions for these equations to hold.

8

1 Basic Facts

The following lemma is basic for the further development of the theory. Lemma 1.3.1 For any three nonempty subsets P , Q, and R of S, we have (P Q)R = P (QR). Proof. Let us first convince ourselves that (P Q)R ⊆ P (QR). In order to do so, we pick an element s in (P Q)R, and we shall prove that s ∈ P (QR). From s ∈ (P Q)R we obtain elements t in P Q and r in R such that 1 ≤ atrs . Since t ∈ P Q, we find elements p in P and q in Q such that 1 ≤ apqt . It follows that 1 ≤ apqt atrs . Since 1 ≤ apqt atrs , there exists an element u in S such that 1 ≤ apus aqru ; cf. Lemma 1.1.1(iii). From 1 ≤ apus aqru we obtain 1 ≤ apus and 1 ≤ aqru . From 1 ≤ aqru , q ∈ Q, and r ∈ R we obtain u ∈ QR. Thus, as 1 ≤ apus and p ∈ P , s ∈ P (QR). The inclusion P (QR) ⊆ (P Q)R is obtained similarly. Lemma 1.3.2 For any two nonempty subsets P and Q of S, the following hold. (i) We have 1 ∈ P ∗ Q if and only if P ∩ Q is not empty. (ii) If 1 ∈ P Q, 1 ∈ QP . (iii) We have (P Q)∗ = Q∗ P ∗ . Proof. (i) By definition, 1 ∈ P ∗ Q means that there exist elements p in P and q in Q such that 1 ≤ ap∗ q1 . From Lemma 1.1.3(i), we know that 1 ≤ ap∗ q1 if and only if p = q. (ii) Assume that 1 ∈ P Q. Then, by (i), P ∗ ∩ Q is not empty. It follows that Q∗ ∩ P is not empty. Thus, by (i), 1 ∈ QP . (iii) Let s be an element in (P Q)∗ . Then, by definition, s∗ ∈ P Q. Thus, there exist elements p in P and q in Q such that 1 ≤ apqs∗ . Thus, by Lemma 1.1.1(ii), 1 ≤ aq∗ p∗ s . Thus, as q ∗ ∈ Q∗ and p∗ ∈ P ∗ , s ∈ Q∗ P ∗ . Since s has been chosen arbitrarily in (P Q)∗ , we have shown that (P Q)∗ ⊆ Q∗ P ∗ . The inclusion Q∗ P ∗ ⊆ (P Q)∗ is obtained similarly. Lemma 1.3.3 For any three nonempty subsets P , Q, and R of S, we have the following. (i) The set P ∩ QR is empty if and only if Q ∩ P R∗ is empty. (ii) The set P ∩ RQ is empty if and only if Q ∩ R∗ P is empty. (iii) The set P ∗ ∩ RQ is empty if and only if Q∗ ∩ P R is empty. Proof. (i) Assume that P ∩ QR is not empty, and let us pick an element p in P such that p ∈ QR. Since p ∈ QR, there exist elements q in Q and r in R


9

such that 1 ≤ aqrp . Thus, by Lemma 1.1.3(ii), 1 ≤ apr∗ q . Thus, as p ∈ P and r ∈ R, q ∈ P R∗ . It follows that q ∈ Q ∩ P R∗ . (ii) Assume that P ∩ RQ is not empty. Then, by Lemma 1.3.2(iii), P ∗ ∩ Q∗ R∗ is not empty. Thus, by (i), Q∗ ∩ P ∗ R is not empty. Thus, by Lemma 1.3.2(iii), Q ∩ R∗ P is not empty. (iii) By (i), P ∗ ∩ RQ is empty if and only if R ∩ P ∗ Q∗ is empty. By (ii), R ∩ P ∗ Q∗ is empty if and only if Q∗ ∩ P R is empty. We shall often apply Lemma 1.3.3 to three sets each of which contains only one element. In this case (for instance), the first part of the lemma says that, for any three elements p, q, and r in S, p ∈ qr if and only if q ∈ pr∗ .1 Let n be an integer with 3 ≤ n, and let R1 , . . . , Rn be nonempty subsets of S. We inductively define R1 · · · Rn to be the complex product (R1 · · · Rn−1 )Rn . From Lemma 1.3.1 we easily obtain R1 · · · Rn = R1 (R2 · · · Rn ). With the help of Lemma 1.3.2(iii) we similarly obtain (R1 · · · Rn )∗ = Rn∗ · · · R1∗ . Lemma 1.3.4 Let O, P , Q, and R be nonempty subsets of S. Then the set OP ∩ QR is empty if and only if O∗ Q ∩ P R∗ is empty. Proof. From Lemma 1.3.3(i) we know that OP ∩ QR is empty if and only if Q ∩ OP R∗ is empty. However, by Lemma 1.3.3(ii), Q ∩ OP R∗ is empty if and only if P R∗ ∩ O∗ Q is empty. Lemma 1.3.5 Let n be an integer with 2 ≤ n, let s be an element in S, and let R1 , . . . , Rn be nonempty subsets of S. Then the following statements are equivalent. (a) We have s ∈ R1 · · · Rn . (b) There exist elements s0 , . . . , sn in S such that s0 = 1, sn = s, and, for each element i in {1, . . . , n}, si ∈ si−1 Ri . (c) For each element i in {1, . . . , n}, there exists an element ri in Ri such that 1 ≤ ar1 ...rn s . Proof. There is nothing to show if 2 = n. Thus, we assume that 3 ≤ n. (a) ⇒ (b) Let us assume that s ∈ R1 · · · Rn . Then, by definition, there exists an element q in R1 · · · Rn−1 such that s ∈ qRn . From q ∈ R1 · · · Rn−1 we obtain, by induction, elements s0 , . . . , sn−1 in S such that s0 = 1, sn−1 = q, and, for each element i in {1, . . . , n − 1}, si ∈ si−1 Ri . Setting sn := s we obtain from sn−1 = q and s ∈ qRn that sn ∈ sn−1 Rn . (b) ⇒ (c) By induction, there exists, for each element i in {1, . . . , n − 1}, an element ri in Ri such that 1 ≤ ar1 ...rn−1 sn−1 . From s ∈ sn−1 Rn we also obtain 1

Recall that we write pq instead of {p}{q} whenever p and q are elements in S.

10

1 Basic Facts

an element rn in Rn such that 1 ≤ asn−1 rn s . Thus, 1 ≤ ar1 ...rn−1 sn−1 asn−1 rn s ≤ ar1 ...rn s . (c) ⇒ (a) Let us assume that 1 ≤ ar1 ...rn s . Then, by definition, there exists an element q in S such that 1 ≤ ar1 ...rn−1 q and 1 ≤ aqrn s . From 1 ≤ ar1 ...rn−1 q we obtain, by induction, that q ∈ R1 · · · Rn−1 . From 1 ≤ aqrn s and rn ∈ Rn we obtain s ∈ qRn . Thus, as q ∈ R1 · · · Rn−1 , s ∈ R1 · · · Rn . Let s be an element of S, and let R be a nonempty subset of S. Instead of R{s} we write Rs. Similarly, we write sR instead of {s}R. We define Rs := {r ∈ S | sr ⊆ Rs}. Note that sRs ⊆ Rs. The following lemma gives more details about Rs . Lemma 1.3.6 Let s be an element in S, and let R be a nonempty subset of S. Then the following hold. (i) We have Rs ⊆ s∗ Rs. (ii) If 1 ∈ R, 1 ∈ Rs . (iii) Let P and Q be nonempty subsets of S such that P Q ⊆ R. Then we have P s Qs ⊆ Rs . Proof. (i) From sRs ⊆ Rs we obtain Rs ⊆ s∗ sRs ⊆ s∗ Rs. (ii) Let us assume that 1 ∈ R. Then we have s ∈ Rs. Thus, by definition, 1 ∈ Rs . (iii) Let r be an element in P s Qs . Then there exist elements t in P s and u in Qs such that r ∈ tu. Since t ∈ P s , st ⊆ P s. Since u ∈ Qs , su ⊆ Qs. Thus, as r ∈ tu and P Q ⊆ R, sr ⊆ stu ⊆ P su ⊆ P Qs ⊆ Rs, so that sr ∈ Rs. Thus, by definition, r ∈ Rs . Lemma 1.3.7 Let p and q be elements in S, and let R be a nonempty subset of S such that 1 ∈ R and RR ⊆ R. Then, if Rp = Rq, Rp = Rq . Proof. Let s be an element in Rp . Then, by definition, ps ⊆ Rp. Let us assume that Rp = Rq. Then, as we are assuming that 1 ∈ R, we obtain q ∈ Rp, and this implies qs ⊆ Rps.


11

From ps ⊆ Rp and qs ⊆ Rps we obtain qs ⊆ RRp. Thus, as we are assuming that RR ⊆ R, we have that qs ⊆ Rp. Since we are assuming that Rp = Rq, this implies qs ⊆ Rq. Thus, s ∈ Rq . So far, we have seen that Rp ⊆ Rq . The proof for Rq ⊆ Rp is similar. Let us now, at the end of this section, look at the relationship between subsets of X and complex products. In order to do this we fix a nonempty subset of S and call it R. For each element x in X, we define xR to be the union of the sets xr with r ∈ R. Let x be an element in X, and let Q be a subset of S such that Q ∩ R is not empty. Since elements of S are pairwise disjoint, x(Q ∩ R) = xQ ∩ xR. For each nonempty subset Y of X, we define Y R to be the union of the sets yR with y ∈ Y . Let Z be a nonempty subset of X. It is clear that, for each nonempty subset Y of Z, Y R ⊆ ZR. Moreover, we have ZQ ⊆ ZR for each nonempty subset Q of R. Lemma 1.3.8 Let Y be a nonempty subset of X, and let P and Q be nonempty subsets of S. Then (Y P )Q = Y (P Q). Proof. Let x be an element in (Y P )Q. Then, by definition, there exist elements z in Y P and q in Q such that x ∈ zq. Since z ∈ Y P , there exist elements y in Y and p in P such that z ∈ yp. Let us denote by s the uniquely determined element in S which satisfies x ∈ ys. Then, as z ∈ yp ∩ xq ∗ , 1 ≤ apqs . Thus, by definition, s ∈ pq, so that x ∈ ys ⊆ y(pq) ⊆ y(P Q) ⊆ Y (P Q). Since x has been chosen arbitrarily in (Y P )Q, we have shown that (Y P )Q ⊆ Y (P Q). Conversely, let x be an element in Y (P Q). Then, by definition, there exist elements y in Y and s in P Q such that x ∈ ys. Since s ∈ P Q, there exist elements p in P and q in Q such that 1 ≤ apqs . Thus, as x ∈ ys, yp ∩ xq ∗ is not empty. Let z be an element in yp ∩ xq ∗ . Then x ∈ zq ⊆ zQ ⊆ (yp)Q ⊆ (Y P )Q. Since x has been chosen arbitrarily in Y (P Q), we have shown that Y (P Q) ⊆ (Y P )Q. Lemma 1.3.9 Let n be an integer with 2 ≤ n, let s be an element in S, and let R1 , . . . , Rn be nonempty subsets of S. Then the following statements are equivalent.

12

1 Basic Facts

(a) We have s ∈ R1 · · · Rn . (b) Let y be an element in X, and let z be an element in ys. Then there exist elements x0 , . . . , xn in X such that x0 = y, xn = z, and, for each element i in {1, . . . , n}, xi ∈ xi−1 Ri . (c) There exist elements x0 , . . . , xn in X such that xn ∈ x0 s and, for each element i in {1, . . . , n}, xi ∈ xi−1 Ri . Proof. There is nothing to show if 1 = n. Thus, we assume that 2 ≤ n. (a) ⇒ (b) Let us assume that s ∈ R1 · · · Rn . Then, by definition, there exists an element r in R1 · · · Rn−1 such that s ∈ rRn . From z ∈ ys and s ∈ rRn we obtain z ∈ yrRn . Thus, referring to Lemma 1.3.8 we obtain an element w in yr such that z ∈ wRn . From r ∈ R1 · · · Rn−1 and z ∈ wRn we obtain, by induction, elements x0 , . . . , xn−1 in X such that x0 = y, xn−1 = w, and, for each element i in {1, . . . , n − 1}, xi ∈ xi−1 Ri . Setting xn := z we obtain from xn−1 = w and z ∈ wRn that xn ∈ xn−1 Rn . (b) ⇒ (c) This follows from the fact that s is not empty. (c) ⇒ (a) Let us assume that there exist elements x0 , . . . , xn in X such that xn ∈ x0 s and, for each element i in {1, . . . , n}, xi ∈ xi−1 Ri . Let us further denote by r the element in S which satisfies xn−1 ∈ x0 r. From xn ∈ xn−1 Rn and xn−1 ∈ x0 r we obtain xn ∈ x0 rRn . By induction, we have that r ∈ R1 · · · Rn−1 . Thus, we conclude that xn ∈ x0 R1 · · · Rn . Thus, as xn ∈ x0 s, s ∈ R1 · · · Rn .

1.4 Complex Products and Valencies In this section, we establish a few results relating complex products to valencies. The letters P and Q will stand for nonempty subsets of S, each having finite valency. We start with a generalization of Lemma 1.1.3(iv). Lemma 1.4.1 We have ( apqs )ns = nP nQ . s∈P Q p∈P q∈Q

Proof. From Lemma 1.1.3(iv) we know that apqs ns = np nq s∈P Q

1.4 Complex Products and Valencies

13

for any two elements p in P and q in Q. Thus, we have ( apqs ns ) = np nq , p∈P q∈Q s∈P Q

p∈P q∈Q

and that proves the lemma. Lemma 1.4.2 We have nP Q ≤ nP nQ . Proof. For each element s in P Q, we have 1≤ apqs . p∈P q∈Q

Thus, the claim follows from Lemma 1.4.1. Lemma 1.4.3 For each element s in S, we have apqs = nP p∈P q∈Q

if and only if P ∗ s ⊆ Q. Proof. Let s be an element in S. From the first equation of Lemma 1.1.3(iii) we obtain aprs = nP . p∈P r∈S

Thus, the equation in question holds if and only if, for any two elements p in P and r in S \ Q, aprs = 0. For any two elements p in P and r in S, we have aprs = 0 if and only if s∈ / pr. Moreover, by Lemma 1.3.3(ii), s ∈ / pr is equivalent to r ∈ / p∗ s. Thus, the equation in question holds if and only if, for any two elements p in P and r in S \ Q, r ∈ / p∗ s. This means that P ∗ s ⊆ Q. The following two consequences of Lemma 1.4.3 are among the most frequently quoted results in this monograph. Lemma 1.4.4 The following statements hold. (i) We have nQ ≤ nP Q . (ii) We have nQ = nP Q if and only if Q = P ∗ P Q. Proof. From Lemma 1.4.1, together with the first equation of Lemma 1.1.3(iii), we obtain nP nQ = ns ( apqs ) ≤ ns nP = nP Q nP . s∈P Q

p∈P q∈Q

s∈P Q

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1 Basic Facts

However, we are assuming that nP and nQ are finite. Thus, by Lemma 1.4.2, nP Q is finite, so that we have (i). On the other hand, we know from Lemma 1.4.3 that, for each element s in P Q, apqs = nP p∈P q∈Q

if and only if P ∗ s ⊆ Q. Thus, nQ = nP Q if and only if P ∗ P Q ⊆ Q. The following lemma generalizes Lemma 1.4.4. One obtains Lemma 1.4.4 by setting {1} = R in Lemma 1.4.5. Lemma 1.4.5 Let R be a nonempty subset of S, and assume that R has finite valency. Then the following hold. (i) We have nQ ≤ nP QR . (ii) We have nQ = nP QR if and only if Q = P ∗ P QRR∗ . Proof. (i) From Lemma 1.4.4(i) we know that nQ ≤ nP Q and that n(P Q)∗ ≤ nR∗ (P Q)∗ . From Lemma 1.3.2(iii) we know that (P QR)∗ = R∗ (P Q)∗ . Finally, as nP , nQ , and nR are assumed to be finite, we have n(P Q)∗ = nP Q and n(P QR)∗ = nP QR ; cf. Lemma 1.1.2(iii). Thus, nQ ≤ nP QR . (ii) From (i) we know that nQ ≤ nQR ≤ nP QR . Thus, we have nQ = nP QR if and only if nQ = nQR and nQR = nP QR . The first of these two conditions is equivalent to nQ∗ = nR∗ Q∗ ; cf. Lemma 1.3.2(iii). By Lemma 1.4.4(ii), nQ∗ = nR∗ Q∗ is equivalent to Q∗ = RR∗ Q∗ . However, according to Lemma 1.3.2(iii), this is equivalent to Q = QRR∗ . From Lemma 1.4.4(ii) we also know that nQR = nP QR is equivalent to QR = P ∗ P QR. Note finally that the two equations Q = QRR∗ and QR = P ∗ P QR hold if and only if Q = P ∗ P QRR∗ . We shall often apply Lemma 1.4.5 to the case where {1} = P . Corollary 1.4.6 If nQ = nP ∗ Q , then nP P ∗ ≤ nQ . Proof. Let us assume that nQ = nP ∗ Q . Then, by Lemma 1.4.4(ii), Q = P P ∗ Q. In particular, nQ = nP P ∗ Q . On the other hand, by Lemma 1.4.5(i), nP P ∗ ≤ nP P ∗ Q . Thus, nP P ∗ ≤ nQ .

1.5 Complex Products of Subsets of Cardinality 1 Let p and q be elements in S. Recall that we write pq instead of {p}{q}.

1.5 Complex Products of Subsets of Cardinality 1

15

Lemma 1.5.1 An element s in S is thin if and only if 1 = ns . Proof. By definition, an element s of S is thin if and only if {1} = s∗ s. The / s∗ s for each element r in S \ {1}. equation {1} = s∗ s says that r ∈ Let r be an element in S. Then, by Lemma 1.3.3(ii), r ∈ / s∗ s is equivalent to s∈ / sr. Moreover, s ∈ / sr means that asrs = 0. Thus, s is thin if and only if as1s = ns ; cf. the first equation of Lemma 1.1.3(iii). Thus, the claim follows from the second equation of Lemma 1.1.1(i). Lemma 1.5.2 Let p and q be elements in S such that np and nq are finite. Then |p∗ q| is less than or equal to the greatest common divisor of np and nq . Proof. Let us denote by m the greatest common divisor of np and nq . Then there exists an integer n coprime to nq such that np = mn. Let s be an element in S. Then, by Lemma 1.1.3(ii), apsq nq = aqs∗ p np . Thus, as np = mn, n divides apsq nq . Thus, as nq and n are coprime, n divides apsq . Let s be an element in p∗ q. Then, by Lemma 1.3.3(ii), q ∈ ps, and this means that 1 ≤ apsq . Thus, as n divides apsq , n ≤ apsq . Thus, by the first equation of Lemma 1.1.3(iii), |p∗ q|n ≤ apsq ≤ np = mn. s∈p∗ q

This finishes the proof of the lemma. The following lemma is related to Lemma 1.3.3(ii). Lemma 1.5.3 Let p, q and r be elements in S such that {p} = rq. Assume that nq is finite and that np ≤ nq . Then {q} = r∗ p. Proof. We are assuming that {p} = rq. Thus, np = nrq . Thus, assuming that np ≤ nq , we obtain from Lemma 1.4.4(i) that nq = nrq . We are assuming that nq is finite. According to Lemma 1.4.5(i), we also have that nr ≤ nrq = np ≤ nq , so that nr is finite, too. Thus, as nq = nrq , {q} = r∗ rq; cf. Lemma 1.4.4(ii). Thus, our claim follows from {p} = rq. Lemma 1.5.4 Let s be an element of S, and assume that s has finite valency. Assume that there exist elements p and q in S with np ≤ nq and np ≤ ns ≤ aspq . Then ns = ns∗ s . Proof. From the first equation of Lemma 1.1.3(iii) we know that aspq ≤ ns . Thus, as we are assuming that ns ≤ aspq we obtain aspq = ns . Thus, by Lemma 1.4.3, {p} = s∗ q. From p ∈ s∗ q we obtain q ∈ sp; cf. Lemma 1.3.3(ii). From q ∈ sp we obtain nq ≤ ns np ; cf. Lemma 1.1.3(iv). Thus, as np ≤ ns and ns is assumed to be

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finite, nq must be finite. Thus, as we are assuming that np ≤ nq , we obtain from {p} = s∗ q that {q} = sp; cf. Lemma 1.5.3. It follows that {p} = s∗ sp. Thus, by Lemma 1.4.5(i), ns ≤ ns∗ s ≤ ns∗ sp = np , so that our claim follows from our hypothesis that np ≤ ns . Lemma 1.5.5 Let p and q be elements in S with |p∗ q| = 1. Assume that p and q have finite valency and that np∗ = np∗ q and that nq = np∗ q . Then we have pp∗ = qq ∗ , np∗ = npp∗ , and nq = nqq∗ . Proof. We are assuming that np∗ = np∗ q . Thus, by Lemma 1.4.5(ii), {p∗ } = p∗ qq ∗ . It follows that pp∗ = pp∗ qq ∗ . Thus, as qq ∗ ⊆ pp∗ qq ∗ , qq ∗ ⊆ pp∗ . Similarly, one obtains from nq = np∗ q that pp∗ ⊆ qq ∗ , so that pp∗ = qq ∗ . From nq = np∗ q we also obtain npp∗ ≤ nq ; cf. Corollary 1.4.6. On the other hand, we know from Lemma 1.4.5(i) that nq ≤ nqq∗ . Thus, as pp∗ = qq ∗ , nq = nqq∗ . From np∗ = np∗ q = nq , nq = nqq∗ , and pp∗ = qq ∗ one obtains np∗ = npp∗ . Let s be an element in S. According to Lemma 1.5.1, we have {1} = s∗ s if and only if 1 = ns . Let us now see what happens to s∗ s if ns = 2. Lemma 1.5.6 Let s be an element in S with ns = 2 and ns∗ = 2. Then the following hold. (i) There exists a symmetric element r in s∗ s \ {1} such that nr ≤ 2 and {1, r} = s∗ s. (ii) We have ns∗ s = 2 or ns∗ s = 3. Proof. (i) We are assuming that ns = 2. Thus, by Lemma 1.5.1, s is not thin, and that means that 2 ≤ |s∗ s|. On the other hand, as ns = 2, we obtain from Lemma 1.5.2 that |s∗ s| ≤ 2, so that we have |s∗ s| = 2. From Lemma 1.3.2(i) we know that 1 ∈ s∗ s. Thus, as |s∗ s| = 2, there exists an element r in s∗ s \ {1} such that {1, r} = s∗ s. From {1, r} = s∗ s we obtain r∗ = r; cf. Lemma 1.3.2(iii). Applying Lemma 1.1.3(iv) to s∗ and s instead of p and q, we obtain as∗ s1 + as∗ sr nr ≤ ns∗ ns = 4. Thus, as as∗ s1 = ns∗ = 2, nr ≤ 2. (ii) This follows immediately from (i). Let s be an element in S with ns = 2 and ns∗ = 2. Then, by Lemma 1.5.6(ii), ns∗ s = 2 or ns∗ s = 3. Lemma 1.5.4 gives a sufficient condition for ns∗ s = 2.

2 Closed Subsets

As mentioned in the preface of this monograph closed subsets play an important role in scheme theory. Via the group correspondence, closed subsets generalize the notion of a subgroup. They also generalize some of the important properties of subgroups from group theory to scheme theory. The first section of this chapter is a collection of general observations on closed subsets most of which are straightforward generalizations of facts on subgroups. For instance, we show that the set of all ‘double cosets’ of two closed subset of S is a partition of S; cf. Lemma 2.1.3. We also introduce transversals (as they have been introduced in [45]), and, at the end if this section, we show that closed subsets give rise to subschemes. The second section starts with Dedekind’s modularity law for schemes and includes a selection of consequences of this law. The results will be useful in Section 2.4. In the third section of this chapter, we investigate the relationship between closed subsets of S and structure constants of S. In Section 2.4, we apply some of the previously obtained results on closed subsets in order to derive a sufficient condition for a closed subset to be maximal. In Section 2.5, we define the normalizer and the strong normalizer of closed subsets. In the last of the six sections of this chapter, we introduce conjugates of closed subsets. Conjugates are related to normalizers and strong normalizers and will play a role in Section 4.4 when we investigate Sylow subsets.

2.1 Basic Facts Recall that a nonempty subset R of S is called closed if R∗ R ⊆ R. Note that a nonempty subset R of S is closed if and only if p∗ q ⊆ R for any two elements p and q in R.

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There are a few elementary facts about closed subsets which we occasionally shall quote without reference. Let us look at these facts first. Firstly, it is obvious that {1} and S are closed. Secondly, let T be a closed subset of S. Then, by definition, T is not empty. Thus, by Lemma 1.3.2(i), 1 ∈ T ∗ T ⊆ T . It follows that 1 ∈ T . Since 1 ∈ T ,

T ∗ = T ∗ 1 ⊆ T ∗ T ⊆ T.

Thus, T ∗ ⊆ T . Thus, as T ∗∗ = T , T ⊆ T ∗ , whence T ∗ = T . From T ∗ = T and T ∗ T ⊆ T we obtain T T ⊆ T . Note finally that, as each closed subset of S contains 1 as an element, intersections of closed subsets of S are closed. Lemma 2.1.1 Let T and U be closed subsets of S. Then T U is closed if and only if T U = U T . Proof. Let us first assume that T U is closed. Then, we have (T U )∗ = T U . Now recall that, by Lemma 1.3.2(iii), (T U )∗ = U ∗ T ∗ . Thus, T U = U ∗ T ∗ . However, as T and U are assumed to be closed, we have T ∗ = T and U ∗ = U . Thus, T U = U T . Conversely, if T U = U T , (T U )∗ T U = U ∗ T ∗ T U ⊆ U ∗ T U = U ∗ U T ⊆ U T = T U. (The first equation follows from Lemma 1.3.2(iii).) Therefore, T U is closed. Lemma 2.1.2 Let T and U be closed subsets of S. Then we have {1} = T ∩U if and only if, for each element s in T U , there exist uniquely determined elements t in T and u in U such that s ∈ tu. Proof. Let us first assume that {1} = T ∩ U , and let us fix an element s in T U . Then, by definition, there exist elements t in T and u in U such that s ∈ tu. Let us now pick elements t in T and u in U such that s ∈ t u . We have to show that t = t and u = u. From s ∈ tu ∩ t u we obtain that t∗ t ∩ uu∗ is not empty; cf. Lemma 1.3.4. Since T is assumed to be closed, t∗ t ⊆ T . Similarly, as U is assumed to be closed, uu∗ ⊆ U . It follows that t∗ t ∩ uu∗ ⊆ T ∩ U = {1}. Therefore, 1 ∈ t∗ t and 1 ∈ uu∗ . Thus, by Lemma 1.3.2(i), t = t and u = u. Conversely, let s be an element in T ∩ U . Then s ∈ T U and 1s = s = s1. Thus, as 1 ∈ T ∩ U , s = 1. Lemma 2.1.3 For any two closed subsets T and U of S, {T sU | s ∈ S} is a partition of S.

2.1 Basic Facts

19

Proof. Let p and q be elements in S such that p ∈ T qU . From p ∈ T qU we obtain T pU ⊆ T qU . (Recall that 1 ∈ T and 1 ∈ U .) Thus, it is enough to show that q ∈ T pU . Since p ∈ T qU , there exists an element s in qU such that p ∈ T s. From s ∈ qU we obtain q ∈ sU ; cf. Lemma 1.3.3(i). From p ∈ T s we obtain s ∈ T p; cf. Lemma 1.3.3(ii). From q ∈ sU and s ∈ T p we obtain q ∈ T pU . For each nonempty subset R of S, we define X/R := {xR | x ∈ X}. For any two nonempty subsets P and Q of S with P ⊆ Q, we define Q/P := {qP | q ∈ Q}. Let x be an element in X, and let R be a nonempty subset of S. Recall that xR is our notation for the union of the sets xr with r ∈ R. Lemma 2.1.4 Let R be a subset of S with 1 ∈ R. Then the following statements are equivalent. (a) The set R is closed. (b) The set X/R is a partition of X. (c) The set S/R is a partition of X. Proof. (a) ⇒ (b) We pick elements y and z in X such that y ∈ zR. We shall be done if we succeed in showing that yR = zR. Since y ∈ zR, yR ⊆ zR. (This follows from Lemma 1.3.8 together with the hypothesis that R is closed.) From y ∈ zR we also obtain z ∈ yR∗ = yR. Thus, we conclude, as before, that zR ⊆ yR. (b) ⇒ (c) We fix elements p and q in S with p ∈ qR. We shall be done if we succeed in showing that pR = qR. From p ∈ qR we obtain an element r in R such that 1 ≤ aqrp . Thus, there exist elements y in X and z in yp such that yq ∩ zr∗ is not empty. Let us pick an element x in yq ∩ zr∗ . Since z ∈ xr, our hypothesis that X/R is a partition yields xR = zR. In order to show that pR ⊆ qR we pick an element s in pR. From s ∈ pR we obtain p ∈ sR∗ ; cf. Lemma 1.3.3(i). Thus, as z ∈ yp, z ∈ ysR∗ . Thus, there exists an element w in ys such that z ∈ wR∗ . From z ∈ wR∗ we obtain w ∈ zR. Thus, as xR = zR, w ∈ xR. Thus, as x ∈ yq, w ∈ yqR. Thus, as w ∈ ys, s ∈ qR. Now, as s has been chosen arbitrarily in pR, we have shown that pR ⊆ qR. The inclusion qR ⊆ pR is obtained similarly.

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(c) ⇒ (a) Let r be an element in R. Then 1 ∈ r∗ r ⊆ r∗ R. However, we are assuming that 1 ∈ R. Thus, as S/R is assumed to be a partition of S, r∗ R = R. Thus, as r has been chosen arbitrarily in R, we have shown that R∗ R ⊆ R. Let T be a closed subset of S. We call the elements in X/T left cosets of T in X. For each closed subset U of S with T ⊆ U , the elements of U/T will be referred to as left cosets of T in U . A subset R of S is called a transversal of T in X if |yR ∩ zT | = 1 for any two elements y and z in X. Let U and V be closed subsets of S, and let us assume that T ⊆ V and U ⊆ V . A subset R of V is called a transversal of T and U in V if |R ∩ T vU | = 1 for each element v in V . Let U be a closed subset of S such that T ⊆ U . A transversal of {1} and T in U is called a left transversal of T in U , a transversal of T and {1} in U is called a right transversal of T in U . Theorem 2.1.5 For each closed subset T of S, we have the following. (i) Let R be a transversal of T in X. Then R is a left transversal of T in S and, for each element r in R, {1} = r∗ r ∩ T . (ii) Let R be a left transversal of T in S. Assume that, for each element r in R, {1} = r∗ r ∩ T . Then, R is a transversal of T in X. Proof. (i) Let R be a transversal of T in X, and let s be an element in S. We shall see that |R ∩ sT | = 1 Let y be an element in X, and let z be an element in ys. Since R is assumed to be a transversal of T in X, yR ∩ zT is not empty. Thus, as z ∈ ys, there exists an element r in R such that s ∈ rT . From s ∈ rT we obtain r ∈ sT ; cf. Lemma 1.3.3(i). Thus, as r ∈ R, 1 ≤ |R ∩ sT |. Let us now show that |R ∩ sT | ≤ 1. In order to do so we fix elements p and q in R ∩ sT . We shall see that p = q. Since p, q ∈ sT , we have s ∈ pT ∩ qT ; cf. Lemma 1.3.3(i). Since s ∈ pT and z ∈ ys, z ∈ ypT . Thus, there exists an element v in yp such that z ∈ vT . Since v ∈ yp and p ∈ R, v ∈ yR. Thus, as z ∈ vT , v ∈ yR ∩ zT . Similarly, as s ∈ qT and z ∈ ys, there exists an element w in yq such that z ∈ wT . Since w ∈ yq and q ∈ R, w ∈ yR. Thus, as z ∈ wT , w ∈ yR ∩ zT . From v ∈ yR ∩ zT and w ∈ yR ∩ zT we obtain v = w. (Recall that R is assumed to be a transversal of T in X.) Thus, as v ∈ yp and w ∈ yq, p = q. So far, we have seen that R is a left transversal of T in S.

2.1 Basic Facts

21

It is clear that 1 ∈ r∗ r ∩ T . In order to show that r∗ r ∩ T ⊆ {1}, we pick an element t in r∗ r ∩ T . We shall see that t = 1. Let y be an element in X, and let z be an element in yt. Then, as t ∈ r∗ r, z ∈ yr∗ r. Thus, there exists an element x in yr∗ such that z ∈ xr. It follows that y ∈ xr ∩ zt∗ ⊆ xr ∩ zT . Thus, as z ∈ xr ∩ zT , y = z. (Recall that R is assumed to be a transversal of T in X.) Thus, as z ∈ yt, we conclude that t = 1. (ii) Let y and z be elements in X. We have to show that |yR ∩ zT | = 1. Let s be the uniquely determined element in S which satisfies z ∈ ys. Since R is assumed to be a left transversal of T in S, there exists an element r in R ∩ sT . From r ∈ sT we obtain s ∈ rT ; cf. Lemma 1.3.3(i). Thus, as z ∈ ys, z ∈ yrT . Thus, there exists an element x in yr with z ∈ xT . It follows that x ∈ yr ∩ zT . In particular, as r ∈ R, 1 ≤ |yR ∩ zT |. We still have to show that |yR ∩ zT | ≤ 1. In order to do this we pick two elements v and w in yR ∩ zT . We shall see that v = w. From v ∈ zT and w ∈ zT we obtain w ∈ vT ; cf. Lemma 2.1.4. Thus, there exists an element t in T such that w ∈ vt. From v ∈ yR we obtain an element q in R with v ∈ yq. From w ∈ vt and v ∈ yq we obtain w ∈ yqt. Thus, there exists an element p in qt such that w ∈ yp. From w ∈ yR and w ∈ yp we obtain p ∈ R. From p ∈ qt and t ∈ T we obtain p ∈ qT . Thus, as |R ∩ qT | ≤ 1 and q ∈ R ∩ qT , q = p. Thus, as p ∈ qt, q ∈ qt. Thus, by Lemma 1.3.3(ii), t ∈ q ∗ q. Thus, as t ∈ T , our hypothesis leads to t = 1. Thus, as w ∈ vt, v = w. Lemma 2.1.6 Let R be a subset of S such that 1 ∈ R and RR ⊆ R. Then, if R has finite valency, R is closed. Proof. We are assuming that RR ⊆ R. Thus, nRR ⊆ nR . On the other hand, R is assumed to have finite valency. Thus, by Lemma 1.4.4(i), nR ⊆ nRR . It follows that nR = nRR . Thus, by Lemma 1.4.4(ii), R = R∗ RR. Thus, as we are assuming that 1 ∈ R, R∗ R ⊆ R. Corollary 2.1.7 Let T be a closed subset of S, and assume that T s has finite valency. Then T s is closed. Proof. From 1 ∈ T we obtain 1 ∈ T s ; cf. Lemma 1.3.6(ii). From T T ⊆ T we obtain T s T s ⊆ T s ; cf. Lemma 1.3.6(iii). Thus, our claim is a consequence of Lemma 2.1.6. We shall now see that closed subsets of S give rise to new schemes, the socalled subschemes. Let Y be a subset of X. For each element s in S, we set

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sY := s ∩ (Y × Y ). For each subset R of S, we define RY to be the set of all sets rY with r ∈ R. The following theorem is obvious. Theorem 2.1.8 Let x be an element in X, and let T be a closed subset of S. Then we have the following. (i) For each element t in T , (txT )∗ = (t∗ )xT . (ii) The set TxT is a scheme on xT . (iii) For any three elements p, q, and r in T , we have apxT qxT rxT = apqr . Let x be an element in X, and let T be a closed subset of S. We call TxT the subscheme of S defined by xT . The first part of the following lemma describes the (obvious) relationship between the complex multiplication in S and the one in TxT . Lemma 2.1.9 Let x be an element in X, let T be a closed subset of S, and let R be a nonempty subset of T . Then we have the following. (i) For each nonempty subset Q of T , QxT RxT = (QR)xT . (ii) The set RxT is closed if and only if R is closed. Proof. (i) Let t be an element in T . We have txT ∈ QxT RxT if and only if there exist elements q in Q and r in R such that 1 ≤ aqxT rxT txT . By Theorem 2.1.8(iii), 1 ≤ aqxT rxT txT is equivalent to 1 ≤ aqrt . Thus, by definition, txT ∈ QxT RxT is equivalent to t ∈ QR. However, we have t ∈ QR if and only if txT ∈ (QR)xT . (ii) From Theorem 2.1.8(i) and (i) we know that (RxT )∗ RxT = (R∗ )xT RxT = (R∗ R)xT . Thus, RxT is closed if and only if (R∗ R)xT ⊆ RxT . However, this is equivalent to R∗ R ⊆ R, and that means that R is closed.

2.2 Dedekind Identities The following lemma is a special case of a general observation made by Richard Dedekind in 1900; cf. [8; Theorem VIII]. Lemma 2.2.1 Let P and Q be nonempty subsets of S, and let T be a closed subset of S. Then we have the following. (i) If P ⊆ T , T ∩ P Q = P (T ∩ Q). (ii) If Q ⊆ T , T ∩ P Q = (T ∩ P )Q.


23

Proof. (i) In order to show that T ∩ P Q ⊆ P (T ∩ Q), we pick an element t in T ∩ P Q. Since t ∈ P Q, there exists an element q in Q such that t ∈ P q. Thus, by Lemma 1.3.3(iii), q ∗ ∈ t∗ P . Thus, as T is assumed to be closed, P ⊆ T yields q ∈ T . Thus, as t ∈ P q, t ∈ P (T ∩ Q). Conversely, as T is assumed to be closed, P ⊆ T yields P (T ∩ Q) ⊆ T . Thus, as P (T ∩ Q) ⊆ P Q, P (T ∩ Q) ⊆ T ∩ P Q. (ii) Applying (i) to Q∗ and P ∗ in the role of P and Q, this follows from Lemma 1.3.2(iii). Corollary 2.2.2 Let P and Q be nonempty subsets of S, and let T be a closed subset of S. Assume that P T and QT are closed. Then we have (P ∩ QT )(Q ∩ P T ) = P Q ∩ QT ∩ P T . Proof. Since T is assumed to be a closed subset of S, we have 1 ∈ T . Thus, P ⊆ P T . Thus, as P T is assumed to be closed, Lemma 2.2.1(i) yields P T ∩ (P ∩ QT )Q = (P ∩ QT )(P T ∩ Q). From 1 ∈ T we also obtain Q ⊆ QT . Thus, as QT is assumed to be closed, Lemma 2.2.1(ii) yields QT ∩ P Q = (QT ∩ P )Q. The desired equation follows easily from the last two equations. Corollary 2.2.3 Let P and Q be nonempty subsets of S, and let T be a closed subset of S such that Q ⊆ T . Then we have the following. (i) If T ⊆ QP Q, Q(P ∩ T )Q = T . (ii) If P ∪ Q = P Q ∩ QP , (P ∩ T ) ∪ Q = (P ∩ T )Q ∩ Q(P ∩ T ). Proof. (i) Since we are assuming that Q ⊆ T , we obtain from Lemma 2.2.1 that Q(P ∩ T )Q = (QP ∩ T )Q = QP Q ∩ T. Thus, if T ⊆ QP Q, Q(P ∩ T )Q = T . (ii) We are assuming that Q ⊆ T . Thus, (P ∩ T ) ∪ Q = (P ∪ Q) ∩ T . On the other hand, we know from Lemma 2.2.1 that P Q ∩ T ∩ QP = (P ∩ T )Q ∩ Q(P ∩ T ). Thus, if P ∪ Q = P Q ∩ QP , (P ∩ T ) ∪ Q = (P ∩ T )Q ∩ Q(P ∩ T ).

2.3 Structure Constants Let s be an element in S, let n be an integer with 2 ≤ n, and let R1 , . . . , Rn be nonempty subsets of S. We define

24

2 Closed Subsets

aR1 ...Rn s :=

...

r1 ∈R1

ar1 ...rn s .

rn ∈Rn

If, for some element i in {1, . . . , n}, the set Ri contains an element ri with {ri } = Ri , we replace the index Ri in aR1 ...Rn s with ri . Lemma 2.3.1 Let p and q be elements in S, let T be a closed subset of S, and assume that T has finite valency. Then the following hold. (i) If q ∈ pT , apT p = apT q . (ii) We have apT T q = apT q nT . Proof. (i) Let x be an element in X, and let z be an element in xq. Then, as q ∈ pT , z ∈ xpT . Thus, there exists an element y in xp such that z ∈ yT . Since z ∈ yT , yT = zT ; cf. Lemma 2.1.4. In particular, xp ∩ yT = xp ∩ zT . Thus, taking into account that {yt∗ | t ∈ T } is a partition of yT and that {zt∗ | t ∈ T } is a partition of zT the equation follows from the fact that y ∈ xp and z ∈ xq. (ii) Referring to Lemma 1.1.1(iii) we obtain apts asuq = apsq atus = apsq atus . apT s asT q = t∈T u∈T

t∈T u∈T

t∈T u∈T

Thus, referring to Lemma 1.1.3(iii) (in order to obtain the third equation) we obtain apT T q = apT s asT q = apsq atus s∈S

=

apsq

t∈S

t∈T

s∈S

nt =

t∈T u∈T

apsq nT = apT q nT .

t∈S

This finishes the proof of (ii). The following lemma generalizes Lemma 2.3.1(i). Lemma 2.3.1(i) is obtained from Lemma 2.3.2 by setting T = {1} and U = T . Lemma 2.3.2 Let p be an element in S, let T and U be closed subsets of S, and let q be an element in T pU . Then, if T and U have finite valency, aT pU p = aT pU q . Proof. Let us set R := T p ∩ qU . Then aT pU q = atpuq = atpr aruq = aT pr arU q . t∈T u∈U

t∈T u∈U r∈R

r∈R

From Lemma 2.3.1(i) together with Lemma 1.1.1(ii) we obtain aT pr = aT pp for each element r in T p. From Lemma 2.3.1(i) we obtain arU q = arU r for each element r in qU . Thus,


aT pU q =

aT pp arU r = aT pp

r∈R

25

arU r .

r∈R

This proves that aT pU q = aT pU q for any two elements q and q in T pU with q ∈ qU . Similarly, one shows that aT pU q = aT pU q for any two elements q and q in T pU such that q ∈ T q. This finishes the proof of the lemma. Lemma 2.3.3 Let s be an element in S, and let T and U be closed subsets of S. Assume that s, T , and U have finite valency. Then we have aT sU s nT sU = nT ns nU . Proof. From Lemma 2.3.2 we obtain aT sU s nT sU = aT sU s nr = aT sU s nr = aT sU r nr . r∈T sU

r∈T sU

r∈T sU

From Lemma 1.1.5(iii) we know that atsur nr = nt ns nu . r∈S

Note also that, for each element r ∈ S \ tsu, atsur = 0; cf. Lemma 1.3.5. Thus, nT ns nU = atsur nr = ( atsur )nr . t∈T u∈U r∈T sU

r∈T sU t∈T u∈U

Thus, the lemma follows from the definition of aT sU r . Lemma 2.3.4 Let s be an element in S, and let T be a closed subset of S. Assume that s and T have finite valency. Then the following hold. (i) The integer nT divides nsT . (ii) We have asT s nsT = ns nT . (iii) The integer (nT )−1 nsT divides ns . Proof. (i) Let y be an element in X. Then we have |ysT | = nsT . On the other hand, we have zT ⊆ ysT and |zT | = nT for each element z in ysT . Thus, the claim follows from Lemma 2.1.4. (ii) This is Lemma 2.3.3 in the case where T = {1} and U = T . (iii) From (ii) we know that asT s nsT = ns nT . Since ns and nT are assumed to be finite, we obtain from Lemma 1.4.2 that nsT is finite. Thus, our claim follows from (i). Lemma 2.3.5 Let p, q, and s be elements of S, and let T be a closed subset of S. Assume that p, q, and T have finite valency. Then we have apT p aqT q auvs = apT qT s . u∈pT v∈qT

26

2 Closed Subsets

Proof. We have aqT q auvs = aqT v auvs = aqT v auvs v∈qT

v∈qT

=

v∈S

aT q∗ v∗ av∗ u∗ s∗ = aT q∗ u∗ s∗ = auqT s .

v∈S

(The first equation follows from Lemma 2.3.1(i), the third and the fifth equation follow from Lemma 1.1.5(ii).) Thus, apT p aqT q auvs = apT p aqT q auvs u∈pT v∈qT

u∈pT

=

u∈pT

v∈qT

apT u auqT s =

apT u auqT s = apT qT s ,

u∈S

and this proves the lemma. For thin schemes, the second part of the following result is associated with the name of Joseph-Louis Lagrange. Lemma 2.3.6 Let T and U be closed subsets of S, and assume that nT and nU are finite. Then the following hold. (i) We have nT nU = nT U nT ∩U . (ii) If T ⊆ U , nT divides nU . (iii) Assume that nS is finite and that (nT )−1 nS and (nU )−1 nS are coprime. Then T U = S. Proof. (i) For any two elements t in T and u in U , we have at1u1 = atu1 = δtu∗ nt ; cf. Lemma 1.1.3(i). Thus, the claim follows from Lemma 2.3.3. (ii) Let R be a left transversal of T in U . Then, for each element r in R, nT divides nrT ; cf. Lemma 2.3.4(i). On the other hand, we know from Lemma 2.1.4 that {rT | r ∈ R} is a partition of U . Thus, nU is the sum of the integers nrT with r ∈ R. Therefore nT divides nU . (iii) From (ii) we know that nT ∩U divides nT . Thus, (nT )−1 nS divides (nT ∩U )−1 nS . Similarly, we obtain that (nU )−1 nS divides (nT ∩U )−1 nS . On the other hand, we are assuming that (nT )−1 nS and (nU )−1 nS are coprime. Thus, (nT )−1 nS (nU )−1 nS ≤ (nT ∩U )−1 nS . From this we obtain nT ∩U nS ≤ nT nU . Thus, by (i), nT ∩U nS ≤ nT ∩U nT U . It follows that nS ≤ nT U , and that means that T U = S.


27

Assume S to have finite valency, and let T be a finite set of closed subsets of S, and let us denote by U the intersection of the elements in T . From Lemma 2.3.6(i) we inductively deduce that (nT )−1 nS . (nU )−1 nS ≤ T ∈T

Let p and q be elements in S. Recall that {p}q stands for the set of all elements s in S which satisfy qs ⊆ pq. In particular, for each element s in S, {1}s is the set of all elements r in S which satisfy {s} = sr. Lemma 2.3.7 Let s be an element in S such that ns∗ and ns are finite. Then we have the following. (i) The set {1}s is closed. (ii) The integer n{1}s divides ns . (iii) Let T be a closed subset of S such that nT is finite and ns and nT are coprime. Then {1} = {1}s ∩ T . Proof. (i) We are assuming that ns∗ and ns are finite. Thus, by Lemma 1.4.2, ns∗ s is finite. From Lemma 1.3.6(i) we know that {1}s ⊆ s∗ s. Thus, {1}s has finite valency. Thus, by Corollary 2.1.7, {1}s is closed. (ii) From (i) we know that {1}s is closed, and from the definition of {1}s we obtain {s} = s{1}s . Thus, by Lemma 2.3.4(i), n{1}s divides ns . (iii) In (i) we saw that {1}s is closed. Thus, by Lemma 2.3.6(ii), n{1}s ∩T divides n{1}s and nT . From (ii) we know that n{1}s divides ns . Thus, as we are assuming ns and nT to be coprime, we conclude that 1 = n{1}s ∩T . It follows that {1} = {1}s ∩ T . Recall that a non-identity element s of S is called an involution if {1, s} is closed. Note that involutions are symmetric. Lemma 2.3.8 For each involution l of S, the following hold. (i) We have alll = nl − 1. (ii) If l is not thin, l ∈ ll. (iii) Let p and q be elements in S such that p = q and {q} = pl. Then aplq = 1. Proof. (i) Since l is assumed to be an involution, we have ll ⊆ {1, l}. Let s be an element in S \{1, l}. Then s ∈ / ll. Thus, by Lemma 1.3.3(i), l ∈ / sl∗ . Thus, by definition, asl∗ l = 0.

28

2 Closed Subsets

Since s has been chosen arbitrarily in S \ {1, l}, we now obtain from the second equation of Lemma 1.1.3(iii) that a1l∗ l + all∗ l = nl . Since l∗ = l, this implies a1ll + alll = nl . But, by the first equation of Lemma 1.1.1(i), a1ll = 1. Therefore, alll = nl − 1. (ii) Since l is assumed to be an involution, we have ll ⊆ {1, l}. Assuming l not to be thin means that {1} = ll. Thus, we must have that l ∈ ll. (iii) Let y be an element in X, let z be an element in yq, and let us assume that 2 ≤ aplq . Then there exist elements v and w in yp ∩ zl such that v = w. From v ∈ yp we obtain y ∈ vp∗ . Thus, as w ∈ yp, w ∈ vp∗ p. From v ∈ zl we obtain z ∈ vl. Thus, as w ∈ zl, w ∈ vll. Thus, as v = w, w ∈ vl. From w ∈ vp∗ p and w ∈ vl we obtain l ∈ p∗ p. Thus, by Lemma 1.3.3(ii), p ∈ pl, contrary to p = q and {q} = pl. Let p be a prime number. An element s in S is called p-valenced if ns is a power of p. A nonempty subset of S is called p-valenced if all of its elements are p-valenced. A nonempty p-valenced subset R of S is called a p-subset of S if nR is a power of p. Lemma 2.3.9 Let p be a prime number, and let T be a closed p-subset of S. Then there exists an element t in T \ {1} such that 1 = nt . Proof. Since T is assumed to be p-valenced, nt is a power of p whenever t is an element in T . By definition, nT is the sum of the integers nt with t ∈ T . Since T is assumed to be a p-subset of S, nT is a power of p. On the other hand, as we are assuming T to be closed, 1 ∈ T . Thus, as n1 = 1, we obtain from the previous paragraph that T \ {t} contains an element t with 1 = nt .

2.4 Maximal Closed Subsets Let T and U be closed subsets of S such that T ⊆ U and T = U . We call T a maximal closed subset of U if T and U are the only closed subsets of U containing T as a subset. In this section, we give a sufficient condition for a closed subset of S to be maximal. Lemma 2.4.1 Let y and z be elements in X, and let T and U be closed subsets of S such that z ∈ yU T U . Then there exists an element x in X such that xT ∩ yU and xT ∩ zU are not empty. Proof. Since z ∈ yU T U , there exists an element x in yU such that z ∈ xT U . From x ∈ yU (and x ∈ xT ) we obtain x ∈ xT ∩ yU .

2.4 Maximal Closed Subsets

29

Since z ∈ xT U , there exists an element w in xT such that z ∈ wU . From z ∈ wU we obtain w ∈ zU ; cf. Lemma 2.1.4. Thus, w ∈ xT ∩ zU . Lemma 2.4.2 Let v, w, y, and z be elements in X. Let T and U be closed subsets of S satisfying T ∪ U = T U ∩ U T . Assume that none of the sets vT ∩ yU , vT ∩ zU , wT ∩ yU , and wT ∩ zU is empty. Then, if yU = zU , vT = wT . Proof. We are assuming that vT ∩ yU is not empty. Thus, there exists an element y in vT ∩yU . Similarly, we find elements v in vT ∩zU , w in wT ∩yU , and z in wT ∩ zU . Since v ∈ vT and y ∈ vT , v ∈ y T ; cf. Lemma 2.1.4. Similarly, as v ∈ zU , z ∈ v U , so that z ∈ y T U . Thus, as z ∈ zU , z ∈ y T U . Similarly, we obtain from w ∈ yU , y ∈ yU , w ∈ wT , and z ∈ wT that z ∈ y U T . From z ∈ y T U and z ∈ y U T we obtain z ∈ y (T U ∩ U T ). Thus, as we are assuming that T ∪ U = T U ∩ U T , we conclude that z ∈ y (T ∪ U ). / y U ; Let us now assume that yU = zU . Then, as z ∈ zU and y ∈ yU , z ∈ cf. Lemma 2.1.4. Thus, as z ∈ y (T ∪ U ), we must have z ∈ y T . Thus, referring to Lemma 2.1.4 once more we obtain from y ∈ vT and z ∈ wT that vT = wT . For the remainder of this section, we assume S to have finite valency. Lemma 2.4.3 Let T , U , and V be closed subsets of S such that V = U T U and T ∪ U = T U ∩ U T . Then ((nU )−1 nV − 1)nT ∩U = ((nT ∩U )−1 nT − 1)nU . Proof. We fix an element x in X, we set τ := (nT ∩U )−1 nT , and we set υ := (nT ∩U )−1 nU . Referring to Lemma 2.1.4 and to Lemma 2.3.6(ii) we find exactly τ elements wU in X/U with ∅ = xT ∩ wU . Similarly, there exist exactly υ elements wT in X/T with ∅ = wT ∩ xU . With the help of Lemma 2.4.1 and Lemma 2.4.2 we count in two different ways the pairs (yT, zU ) satisfying z ∈ xV and yT ∩ xU = ∅ = yT ∩ zU. Then we get (nU )−1 nV − 1 = (τ − 1)υ. From this we obtain the desired equation. Lemma 2.4.4 Let T and U be closed subsets of S such that T = S, U T U = S, and T ∪ U = T U ∩ U T . Then, we have nT ≤ nU .

30

2 Closed Subsets

Proof. There is nothing to show if U = S. Therefore, we assume that U = S. In this case, we obtain elements y and z in X such that yU = zU . Since we are assuming that U T U = S, we obtain from Lemma 2.4.1 an element w in X such that wT ∩ yU = ∅ = wT ∩ zU. Since we are assuming that T = S, there exists an element x in X such that wT ∩ xU is empty; cf. Lemma 2.4.2. We set τ := (nT ∩U )−1 nT and υ := (nT ∩U )−1 nU . Then, there exist τ elements vU in X/U such that ∅ = wT ∩ vU . For each of these elements, there exists an element v T in X/T such that v T ∩ vU = ∅ = v T ∩ xU ; cf. Lemma 2.4.1. Moreover, for any two different elements uU and vU in X/U with wT ∩ uU = ∅ = wT ∩ vU, we have u T = v T ; cf. Lemma 2.4.2. Thus, τ ≤ υ, and from this we conclude that nT ≤ nU . The following theorem is the main result of this section. It is [44; Theorem]. Theorem 2.4.5 Let U be a closed subset of S such that U = S. Assume that there exists a closed subset T of S which satisfies T = S, U T U = S, and T ∪ U = T U ∩ U T . Then U is a maximal closed subset of S. Proof. Let V be a closed subset of S such that U ⊆ V . We shall be done if we succeed in showing that V ∈ {U, S}. We set τ := (nT ∩U )−1 nT and υ := (nT ∩U )−1 nU . Then, by Lemma 2.4.3, nS = nU (1 + (τ − 1)υ). Next, we set δ := (nT ∩U )−1 nT ∩V . Then, by Corollary 2.2.3 and Lemma 2.4.3, nV = nU (1 + (δ − 1)υ). Finally, we set σ := (nV )−1 nS . Then, the two above equations yield 1 + (τ − 1)υ = (1 + (δ − 1)υ)σ. It follows that (τ − 1 − (δ − 1)σ)υ = σ − 1. Let us now first assume that 2 ≤ δ and that 2 ≤ σ. Since 2 ≤ σ, the last equation yields (δ−1)σ ≤ τ −1 and υ ≤ σ−1. Since 2 ≤ δ and (δ−1)σ ≤ τ −1,

2.5 Normalizer and Strong Normalizer

31

σ ≤ τ − 1. Thus, as υ ≤ σ − 1, υ ≤ τ − 1. Thus, nU ≤ nT − 1, contrary to Lemma 2.4.4. This contradiction shows that we have 1 = δ or 1 = σ. Let us now assume that 1 = δ. Then we have nT ∩U = nT ∩V . Since U ⊆ V , this yields T ∩ U = T ∩ V , so that we have T ∩ V ⊆ U . It follows that U = U (T ∩ V )U . On the other hand, as we are assuming that U T U = S, we have that U (T ∩ V )U = V ; cf. Corollary 2.2.3(i). Thus, as U = U (T ∩ V )U , U = V . Note, finally, that 1 = σ is equivalent to V = S.

2.5 Normalizer and Strong Normalizer Let P and Q be subsets of S, with Q nonempty. We set NP (Q) := {p ∈ P | Qp ⊆ pQ}. The set NP (Q) is called the normalizer of Q in P . Let R be a nonempty subset of S. We say that an element s in S normalizes the set R if s ∈ NS (R). Let T and U be closed subsets of S such that T ⊆ U . The closed subset T is said to be a normal closed subset of U if U ⊆ NS (T ). In this case, we also say that T is normal in U . For each nonempty subset R of S, we obviously have 1 ∈ NS (R). Note that we have T ⊆ NS (T ) for each closed subset T of S. However, in general, NS (T ) is not closed. Lemma 2.5.1 For each closed subset T of S, the following hold. (i) Let R be a subset of S, and assume that R ∩ T is not empty. Then NT (R) ⊆ NT (R ∩ T ). (ii) We have T NS (T ) ⊆ NS (T ). Proof. (i) Let t be an element in NT (R). Then (R ∩ T )t ⊆ Rt ∩ T ⊆ tR ∩ T = t(R ∩ T ); cf. Lemma 2.2.1(i). Thus, t ∈ NT (R ∩ T ). (ii) Let s be an element in T NS (T ). Then there exists an element r in NS (T ) such that s ∈ T r. Since r ∈ NS (T ), T r ⊆ rT . Thus, as s ∈ T r, s ∈ rT . Thus, by Lemma 2.1.3, T s = T r and rT = sT . Thus, as T r ⊆ rT , T s ⊆ sT . Thus, by definition, s ∈ NS (T ).

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2 Closed Subsets

Lemma 2.5.2 For each closed subset T of S, we have the following. (i) For each element s in S, s∗ ∈ NS (T ) is equivalent to T ⊆ T s . (ii) Let s be an element in S such that s∗ , s ∈ NS (T ). Then T s = sT . (iii) For each closed subset U of S with T ⊆ NS (U ), T U is closed. (iv) Assume T to be normal in S, and let U be a normal closed subset of S. Then T U is a normal closed subset of S. Proof. (i) Let s be an element in S. We have s∗ ∈ NS (T ) if and only if T s∗ ⊆ s∗ T . According to Lemma 1.3.2(iii), T s∗ ⊆ s∗ T is equivalent to sT ⊆ T s, and this means that T ⊆ T s . (ii) We are assuming that s∗ ∈ NS (T ), and that means that T s∗ ⊆ s∗ T . Thus, by Lemma 1.3.2(iii), sT ⊆ T s. Since we are also assuming that s ∈ NS (T ), we have T s ⊆ sT , too. (iii) Considering Lemma 2.1.1 this is a consequence of (ii). (iv) This follows from (iii). Let P be a subset of S. For each element s in S, we set CP (s) := {p ∈ P | sp = ps}. The set CP (s) is called the centralizer of s in P . For each nonempty subset Q of S, we define CP (Q) to be the intersection of the sets CP (q) with q ∈ Q. The set CP (Q) is called the centralizer of Q in P . Note that we have CP (Q) ⊆ NP (Q) for each nonempty subset Q of S. Lemma 2.5.3 Let T and U be closed subsets of S such that T ⊆ NS (U ), U ⊆ NS (T ), and {1} = T ∩ U . Then T ⊆ CS (U ). Proof. Let t be an element in T , let u an element in U , and let s an element in tu. We shall be done if we succeed in showing that s ∈ ut. From s ∈ tu (together with the hypothesis that U ⊆ NS (T )) we obtain s ∈ uT . Thus, there exists an element p in T such that s ∈ up. From s ∈ up (together with the hypothesis that T ⊆ NS (U )) we obtain s ∈ pU . Thus, there exists an element q in U such that s ∈ pq. Thus, as s ∈ tu, Lemma 2.1.2 yields t = p. Thus, as s ∈ up, s ∈ ut. Lemma 2.5.4 Let p and q be elements of S, let T be a normal closed subset of S, and assume that p, q, and T have finite valency. Then the following hold. (i) We have aT pq = apT q . (ii) For each element s in S, we have apT qs = apqT s .

2.5 Normalizer and Strong Normalizer

33

(iii) For each element s in S, we have apT p aqT q auvs = nT apqw awT w . u∈pT v∈qT

w∈sT

Proof. (i) From Lemma 2.5.2(ii) we know that T s = sT . Thus, our claim follows from Lemma 2.3.3. (ii) From Lemma 1.1.1(iii) we obtain apT r arqs = aprs aT qr apT qs = r∈S

and apqT s =

r∈S

apqr arT s =

r∈S

aprs aqT r .

r∈S

Thus, the claim follows from (i). (iii) From Lemma 2.3.5 we know that apT p aqT q auvs = apT qT s . u∈pT v∈qT

Moreover, from (ii) and Lemma 2.3.1(ii) we easily deduce that apT qT s = apqT T s = nT apqT s . Thus,

apT p aqT q auvs = nT apqT s .

u∈pT v∈qT

On the other hand, we obtain from Lemma 2.3.1(i) that apqw awT w = apqw awT s = apqT s . w∈sT

w∈sT

This finishes the proof of (iii). Let us now introduce a third operator on the set of all nonempty subsets of S. Again, let P and Q be subsets of S, and let us assume Q to be not empty. We set KP (Q) := {p ∈ P | p∗ Qp ⊆ Q}. The set KP (Q) is called the strong normalizer of Q in P . For each nonempty subset R of S, we obviously have 1 ∈ KS (R). Note also that we have T ⊆ KS (T ) for each closed subset T of S. Lemma 2.5.5 For each nonempty subset R of S, we have KS (R) ⊆ NS (R).

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Proof. Let s be an element in KS (R). Then, by definition, s∗ Rs ⊆ R. Thus, as 1 ∈ s∗ s, Rs ⊆ ss∗ Rs ⊆ sR. It follows that Rs ⊆ sR, and that means that s ∈ NS (R). There are cases where strong normalizer and normalizer coincide. Let P be a thin subset of S, and let Q be a nonempty subset of S. Then we obviously have KP (Q) = NP (Q). Lemma 2.5.6 For any two closed subsets T and U of S, the following hold. (i) We have KS (T ) ∩ KS (U ) ⊆ KS (T ∩ U ). (ii) If T ⊆ NS (U ), KS (T ) ∩ NS (U ) ⊆ KS (T U ). (iii) If T ⊆ U , KS (T ) ∩ NS (U ) ⊆ KS (U ). Proof. (i) Let s be an element in KS (T ) ∩ KS (U ). Then s∗ (T ∩ U )s ⊆ s∗ T s ∩ s∗ U s ⊆ T ∩ U. Thus, s ∈ KS (T ∩ U ). (ii) Let s be an element in KS (T ) ∩ NS (U ). Then s∗ T U s ⊆ s∗ T sU ⊆ T U. Thus, s ∈ KS (T U ). (iii) Since U ⊆ NS (U ), this follows from (i). Let T and U be closed subsets of S such that T ⊆ U . The closed subset T is said to be a strongly normal closed subset of U if U ⊆ KS (T ). In this case, we also say that T is strongly normal in U . Lemma 2.5.5 says that, if a closed subset T of S is strongly normal in a closed subset U of S, then T is normal in U . Lemma 2.5.7 Let T and U be closed subsets of S such that T is strongly normal in U . Let V be a closed subset of S such that U ⊆ NS (V ). Then T V is strongly normal in U V . Proof. We are assuming that U ⊆ NS (V ). Thus, by Lemma 2.5.2(iii), U V is closed. Similarly, we obtain from T ⊆ U and U ⊆ NS (V ) that T V is closed. In order to show that T V is strongly normal in U V , we pick an element s in U V . We shall see that s∗ T V s ⊆ T V . Since s ∈ U V , there exist elements u in U and v in V such that s ∈ uv. From u ∈ U and U ⊆ NS (V ) we obtain V u ⊆ uV . From u ∈ U and U ⊆ KS (T ) we obtain u∗ T u ⊆ T . Thus,

2.6 Conjugates of Closed Subsets

35

s∗ T V s ⊆ v ∗ u∗ T V uv ⊆ v ∗ u∗ T uV v ⊆ v ∗ T V v = T V. (Recall that, by Lemma 1.3.2(iii), s∗ ∈ v ∗ u∗ .) Lemma 2.5.8 Let R be a nonempty subset of S, and let T be a closed subset of S. Then the following hold. (i) For any two elements p and q in KT (R), we have pq ⊆ KT (R). (ii) If T has finite valency, KT (R) is closed. Proof. (i) Let p and q be elements in KT (R). Then we have s∗ Rs ⊆ q ∗ p∗ Rpq ⊆ q ∗ Rq ⊆ R for each element s in pq, and this says that s ∈ KT (R). (ii) Considering that 1 ∈ KT (R) this follows from (i) and Lemma 2.1.6. Note that KS ({1}) is the set of the thin elements of S. For each subset R of S, we set Oϑ (R) := {r ∈ R | 1 = nr }, and call it the thin radical of R. Lemma 2.5.9 For each closed subset T of S, the following hold. (i) We have Oϑ (T ) = KT ({1}). (ii) For any two elements p and q in Oϑ (T ), we have pq ⊆ Oϑ (T ). (iii) If T has finite valency, Oϑ (T ) is closed. Proof. (i) Since KT ({1}) is the set of all thin elements of T , our claim is an immediate consequence of Lemma 1.5.1. (ii) From (i) we know that Oϑ (S) = KS ({1}). Thus, our claim is a consequence of Lemma 2.5.8(i). (iii) This follows from (i) together with Lemma 2.5.8(ii).

2.6 Conjugates of Closed Subsets Let P and Q be nonempty subsets of S. We call Q a conjugate of P if there exists an element s in S such that Q = s∗ P s. Lemma 2.6.1 Let s be an element in S, and let T be a closed subset of S such that ss∗ ⊆ T . Then the following hold. (i) We have T s = s∗ T s.

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(ii) The set s∗ T s is closed. (iii) We have KS (T s ) = KS (T )s . Proof. (i) From Lemma 1.3.6(i) we know that T s ⊆ s∗ T s. However, assuming that ss∗ ⊆ T we obtain ss∗ T s ⊆ T s, so that, by definition, s∗ T s ⊆ T s . (ii) Clearly, we have 1 ∈ s∗ T s and (s∗ T s)∗ = s∗ T s. However, assuming that ss∗ ⊆ T we also obtain s∗ T ss∗ T s ⊆ s∗ T s. (iii) According to (i), it is enough to show that KS (s∗ T s) = s∗ KS (T )s. (Recall that T ⊆ KS (T ).) Let q be an element in s∗ KS (T )s. Then there exists an element p in KS (T ) such that q ∈ s∗ ps. Thus, we obtain from ss∗ ⊆ T and p ∈ KS (T ) that q ∗ s∗ T sq ⊆ s∗ p∗ ss∗ T ss∗ ps ⊆ s∗ p∗ T ps ⊆ s∗ T s. Thus, q ∈ KS (s∗ T s). Since q has been chosen arbitrarily in s∗ KS (T )s, we thus have shown that s∗ KS (T )s ⊆ KS (s∗ T s). What we have shown so far also applies to s∗ and s∗ T s instead of s and T . (Note that, by (ii), s∗ T s is closed.) Thus, we have sKS (s∗ T s)s∗ ⊆ KS (ss∗ T ss∗ ) = KS (T ). Thus, KS (s∗ T s) ⊆ s∗ sKS (s∗ T s)s∗ s ⊆ s∗ KS (T )s. Lemma 2.6.2 Let s be an element in S, and let T be a closed subset of S. Assume that s∗ , s, and T have finite valency. Then the following hold. (i) We have ss∗ ⊆ T if and only if nT = ns∗ T s . (ii) Assume that nT = ns∗ T s , and let U be a closed subset of S with T ⊆ U . Then nU = ns∗ U s . Proof. (i) This is an application of Lemma 1.4.5(ii) to {s∗ }, T , and {s} in the role of P , Q, and R. (ii) Assuming nT = ns∗ T s we obtain from (i) that ss∗ ⊆ T . Thus, as we are assuming that T ⊆ U , ss∗ ⊆ U , so that, again by (i), nU = ns∗ U s . Lemma 2.6.3 Let T and V be closed subsets of S such that T ⊆ V . Assume V to have finite valency and set U := KV (T ). Then we have the following. (i) Let p and q be elements in V such that U p = U q. Then pp∗ ⊆ T if and only if qq ∗ ⊆ T . (ii) The number of sets s∗ T s with s ∈ V and ss∗ ⊆ T is equal to the number of sets U s with s ∈ V and ss∗ ⊆ T . Proof. We are assuming that V has finite valency. Thus, by Lemma 2.5.8(ii), U is closed.

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(i) Since we are assuming that U p = U q, there exists an element u in U such that p ∈ uq. Thus, by Lemma 1.3.3(ii), q ∈ u∗ p. Thus, assuming that pp∗ ⊆ T , we obtain qq ∗ ⊆ u∗ pp∗ u ⊆ u∗ T u ⊆ T. Similarly, one obtains pp∗ ⊆ T from qq ∗ ⊆ T . (ii) Let p and q be elements in V , and let us first assume that p∗ T p ⊆ q ∗ T q and that qq ∗ ⊆ T . Then, qp∗ T pq ∗ ⊆ qq ∗ T qq ∗ ⊆ T . It follows that pq ∗ ⊆ U . Thus, by Lemma 1.3.3(i), p ∈ U q. Thus, by Lemma 2.1.4, U p = U q. Let us now assume that U p = U q. Then there exists an element u in U such that p ∈ uq. It follows that p∗ T p ⊆ q ∗ u∗ T uq ⊆ q ∗ T q. Similarly, one shows that q ∗ T q ⊆ p∗ T p, so that p∗ T p = q ∗ T q. Lemma 2.6.4 Let T and U be closed subsets of S such that {1} = T ∩ U . Let t be an element in T , and let u be an element in U such that 1 = |tu|. Then the following hold. (i) The set u∗ tu ∩ T contains at most one element. (ii) If U ⊆ NS (T ), u∗ tu ∩ T contains exactly one element. Proof. (i) We are assuming that tu contains exactly one element. Let us call this element s, and let us pick two elements p and q in u∗ tu ∩ T . From p ∈ u∗ tu and {s} = tu we obtain p ∈ u∗ s. Thus, by Lemma 1.3.3(ii), s ∈ up. Similarly, we obtain s ∈ uq, so that s ∈ up ∩ uq. From s ∈ up ∩ uq we obtain that u∗ u ∩ pq ∗ is not empty; cf. Lemma 1.3.4. Thus, as u∗ u ∩ pq ∗ ⊆ T ∩ U = {1}, {1} = u∗ u ∩ pq ∗ . It follows that 1 ∈ pq ∗ , so that, according to Lemma 1.3.2(i), p = q. (ii) Let us assume that U ⊆ NS (T ). Then tu ⊆ uT . Thus, as tu is assumed to contain only one element, there exists an element r in T such that tu ⊆ ur. Thus, by Lemma 1.3.3(ii), r ∈ u∗ tu. Now the claim follows from (i). Lemma 2.6.5 Let p and q be elements in S such that np∗ and nq are finite and coprime, and let T be a closed subset of S such that q ∈ p∗ T p. Then q ∈ T p. Proof. We are assuming that q ∈ p∗ T p. Thus, by Lemma 1.3.3(ii), p ∈ T pq. However, we are assuming that np∗ and nq are coprime. Thus, by Lemma 1.5.2, |pq| = 1. From p ∈ T pq and |pq| = 1 we obtain pq ⊆ T p; cf. Lemma 1.3.3(ii). Thus, by definition, q ∈ T p .

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Corollary 2.6.6 Let s be an element in S such that ns∗ is finite, and let T and U be closed subsets of S. Assume that, for each element u in s∗ T s ∩ U , nu is finite and ns∗ and nu are coprime. Then T s ∩ U = s∗ T s ∩ U . Proof. We are assuming that, for each element u in s∗ T s ∩ U , ns∗ and nu are coprime. Thus, by Lemma 2.6.5, s∗ T s ∩ U ⊆ T s , so that the desired equation follows from Lemma 1.3.6(i). Let us see what happens if we apply Corollary 2.6.6 to {1} and S in place of T and U . Corollary 2.6.7 Let s be an element in S such that ns∗ is finite. Assume that, for each element r in s∗ s, nr is finite and ns∗ and nr are coprime. Then the following hold. (i) We have {1}s = s∗ s. (ii) The set s∗ s is closed. (iii) We have ns = ns∗ s . Proof. (i) We are assuming that, for each element r in s∗ s, ns∗ and nr are coprime. Thus, by Corollary 2.6.6, {1}s = s∗ s. (ii) From (i) we obtain {s} = ss∗ s. Thus, by Lemma 1.3.2(iii), {s∗ } = s∗ ss∗ . Thus, by Lemma 1.4.5(i), ns ≤ ns∗ . Thus, as ns∗ is assumed to be finite, ns must be finite. Thus, by Lemma 1.4.2, ns∗ s is finite, so that, by (i), {1}s has finite valency. Thus, referring to Corollary 2.1.7 we obtain (ii) from (i). (iii) From (i) we know that {s} = ss∗ s. Thus, by Lemma 1.4.4(ii), ns = ns∗ s . Let s be an element in S. For each subset R of S, we define DR (s) to be the set of all elements r in R such that r∗ r ⊆ s∗ s. Lemma 2.6.8 Assume S to have finite valency. Let s be an element in S such that, for each element r in s∗ s, ns∗ and nr are coprime. Let T be a closed subset of DS (s∗ ). Assume that T ⊆ T s and that, for each element t in T , ns∗ and nt are coprime. Then T ⊆ KS (s∗ s). Proof. Let t be an element in T . We have to show that t ∈ KS (s∗ s). Since t ∈ T , ns∗ and nt are coprime. Thus, by Lemma 1.5.2, |st| = 1. On the other hand, we are assuming that T ⊆ T s . Thus, as t ∈ T , t ∈ T s , and that means that st ⊆ T s. Since |st| = 1 and st ⊆ T s, there exists an element r in T such that st ⊆ rs. From st ⊆ rs we obtain t∗ s∗ st ⊆ s∗ r∗ rs. From r ∈ T and T ⊆ DS (s∗ ) we obtain r ∈ DS (s∗ ). Thus, r∗ r ⊆ ss∗ . Thus, s∗ r∗ rs ⊆ s∗ ss∗ s. But, by Corollary 2.6.7(ii), s∗ s is closed. Thus, s∗ r∗ rs ⊆ s∗ s. Thus, as t∗ s∗ st ⊆ s∗ r∗ rs, t∗ s∗ st ⊆ s∗ s.

3 Generating Subsets

Let R be a subset of S. We define R to be the intersection of all closed subsets of S which contain R. We say that R generates R or that R is generated by R. Recall that the intersection of closed subsets is closed. Therefore R is closed for each subset R of S. Note also that P ⊆ Q for any two subsets P and Q of S with P ⊆ Q. The first section of this chapter is a collection of basic facts about generating sets of closed subsets. We introduce the length function which one obtains from generating subsets, and we establish a connection between generating subsets and commutator subsets. In the second section, we introduce the thin residue of a closed subset. The thin residue is a specific commutator subset. We look at the thin residue of a complex product of two closed subsets, at the complex product of the thin residue and the thin radical, and at multiple thin residues of closed subsets. In the third section of this chapter, we investigate closed subsets of S generated by elements of valency 2. We start with closed subsets of S which are generated by a single symmetric element of valency 2. After that, we shall look at more general cases. Generating sets of elements of S are particularly interesting if they consist of involutions. In fact, a major part of this monograph deals with closed subsets generated by distinguished sets of involutions. Section 3.4 deals with general aspects of closed subsets generated by involutions. In the last two sections of this chapter, we impose specific conditions on sets of involutions. In Section 3.5, we look at constrained sets of involutions, and in the last of the six sections of this chapter, we look at constrained sets of involutions which satisfy the exchange condition. According to what we said in the preface of this monograph, such sets of involutions are called Coxeter sets.

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It might be worth mentioning that, via the group correspondence, most of the results about Coxeter sets which we compile in the last section are natural generalizations of well-known facts on Coxeter groups. Some of the results in the last two sections of this chapter, in particular Theorem 3.6.4 and Theorem 3.6.6, foreshadow the importance of closed subsets generated by involutions in scheme theory.

3.1 Basic Facts Let R be a nonempty subset of S. We set R0 := {1}. For each positive integer n, we inductively define Rn := Rn−1 R. Lemma 3.1.1 For each nonempty subset R of S, we have the following. (i) The set R is the union of the sets (R∗ ∪ R)n with n a non-negative integer. (ii) If R has finite valency, R is the union of the sets Rn with n a nonnegative integer. Proof. (i) We set P := R∗ ∪ R, and we define Q to be the union of all sets P n such that n is a non-negative integer. We have to show that R = Q. From Lemma 1.3.2(iii) we obtain (P n )∗ = (P ∗ )n for each non-negative integer n. Thus, as P ∗ = P , we obtain (P n )∗ = P n for each non-negative integer n. It follows that, for any two non-negative integers l and m, (P l )∗ P m = P l P m = P l+m ⊆ Q. Thus, Q is closed. Thus, as R ⊆ Q, R ⊆ Q. Conversely, for each non-negative integer n, we have P n ⊆ P n ⊆ P = R. Therefore, Q ⊆ R. (ii) Let us denote by Q the union of all sets Rn such that n is a non-negative integer. Then 1 ∈ Q, R ⊆ Q, Q2 ⊆ Q, and, by (i), Q ⊆ R. We are assuming R to have finite valency. Thus, as Q ⊆ R, Q has finite valency. Thus, as 1 ∈ Q and Q2 ⊆ Q, Q is closed; cf. Lemma 2.1.6. Thus, as R ⊆ Q, R ⊆ Q. From Q ⊆ R and R ⊆ Q we obtain R = Q, and that finishes the proof of (ii). Let R be a nonempty subset of S. Let s be an element in R. Then, by Lemma 3.1.1(i), there exists a nonnegative integer n such that s ∈ (R∗ ∪ R)n . In the following, we shall denote by R (s) the smallest non-negative integer n satisfying s ∈ (R∗ ∪ R)n . We call R (s) the length of s with respect to R.

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Since 1 ∈ R0 , R (1) = 0. Note also that we have R (s∗ ) = R (s) for each element s in R. Lemma 3.1.2 Let R be a nonempty subset of S, and let s be an element in

R \ {1}. Then there exist elements q in R and r in R∗ ∪ R such that s ∈ qr and R (s) = R (q) + 1. Proof. We set n := R (s). Then, by definition, s ∈ (R∗ ∪ R)n . On the other hand, we are assuming that 1 = s, so that 1 ≤ n. From s ∈ (R∗ ∪ R)n and 1 ≤ n we obtain elements q in (R∗ ∪ R)n−1 and r in R∗ ∪ R such that s ∈ qr. From q ∈ (R∗ ∪ R)n−1 we obtain R (q) ≤ n − 1. From n = R (s) and s ∈ qr we obtain n ≤ R (q) + 1. Lemma 3.1.3 Let s be an element in S, and let R be a nonempty subset of S with R∗ = R. Then the following hold. (i) Assume that Rs ⊆ s R. Then s ∈ NS ( R). (ii) We have Rs ⊆ Rs . Proof. (i) Let us denote by Q the set of all elements q in R with qs ⊆ s R. By way of contradiction, we assume that Q is not empty. We fix an element q in Q such that R (q) is as small as possible. Since s ∈ s R, 1 = q. Thus, by Lemma 3.1.2, there exist elements p in R and r in R such that q ∈ pr and R (q) = R (p) + 1. Since R (p) = R (q) − 1, p∈ / Q. Thus, as p ∈ R, ps ⊆ s R. Thus, as rs ⊆ s R, qs ⊆ prs ⊆ ps R ⊆ s R, contradiction. (ii) Let us denote by Q the set of all elements q in Rs with sq ⊆ Rs. By way of contradiction, we assume that Q is not empty. We fix an element q in Q such that R (q) is as small as possible. Since s ∈ Rs, 1 = q. Thus, by Lemma 3.1.2, there exist elements p in Rs and r in Rs such that q ∈ pr and R (q) = R (p) + 1. Since R (p) = R (q) − 1, p∈ / Q. Thus, as p ∈ Rs , sp ⊆ Rs. On the other hand, as r ∈ Rs , sr ⊆ Rs. Thus, sq ⊆ spr ⊆ Rsr ⊆ RRs = Rs, contradiction. Lemma 3.1.4 Let s be an element in S, and let R be a nonempty subset of S such that s∗ Rs ⊆ R. Then s ∈ KS ( R).

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Proof. Let us denote by Q the set of all elements q in R with s∗ qs ⊆ R. By way of contradiction, we assume that Q is not empty. We fix an element q in Q such that R (q) is as small as possible. We are assuming that s∗ Rs ⊆ R. Thus, s∗ s ⊆ s∗ Rss∗ R∗ s ⊆ R. This shows that 1 ∈ / Q. In particular, as q ∈ Q, 1 = q. Thus, by Lemma 3.1.2, there exist elements p in R and r in R∗ ∪ R such that q ∈ pr and / Q. Thus, as p ∈ R, R (q) = R (p) + 1. Since R (p) = R (q) − 1, p ∈ s∗ ps ⊆ R. We are assuming that s∗ Rs ⊆ R. Thus, as R is closed, (s∗ Rs)∗ ⊆ R. On the other hand, by Lemma 1.3.2(iii), s∗ R∗ s = (s∗ Rs)∗ . Thus, s∗ R∗ s ⊆ R. It follows that s∗ (R∗ ∪ R)s ⊆ R. Thus, as r ∈ R∗ ∪ R, s∗ qs ⊆ s∗ prs ⊆ s∗ pss∗ rs ⊆ R, contradiction. Lemma 3.1.5 Let R be a subset of S. Then we have the following. (i) The set R∗ ∪ R is thin if and only R is thin. (ii) If R is symmetric, Oϑ (R) ⊆ Oϑ ( R). Proof. (i) Since R∗ ∪ R ⊆ R, R∗ ∪ R is thin if R is thin. Let us assume that R∗ ∪ R is thin and that R is not thin. Assuming R not to be thin we find an element s in R such that s is not thin. Among the non-thin elements of R we fix s in such a way that R (s) is as small as possible. Since s is not thin, 1 = s. Thus, by Lemma 3.1.2, there exist elements q in R and r in R∗ ∪R such that s ∈ qr and R (s) = R (q)+1. Since R (q) = R (s)−1, the minimal choice of s forces q to be thin. Since r ∈ R and R is assumed to be thin, r is thin. Thus, as s ∈ qr, s is thin; cf. Lemma 2.5.9(i), (ii). (ii) We set Q := Oϑ (R). Then Q ⊆ Oϑ ( R). Thus, for each non-negative integer n, Qn ⊆ Oϑ ( R); cf. Lemma 2.5.9(ii). Thus, by Lemma 3.1.1(i), Q ⊆ Oϑ ( R). Theorem 3.1.6 Let p be a prime number, and let R be a subset of S. Assume that nr ≤ p − 1 for each element r in R∗ ∪ R. Then, for each element s in

R, p does not divide ns . Proof. Let us denote by Q the set of all elements s in R such that p divides ns . By way of contradiction, we assume that Q is not empty. We pick an element q in Q such that min R (Q) = R (q).

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Since q ∈ Q, p divides nq . Thus, 1 = q. Thus, by Lemma 3.1.2, there exist elements t in R and r in R∗ ∪ R such that q ∈ tr and R (q) = R (t) + 1. Since r ∈ R∗ ∪ R, nr ≤ p − 1. Moreover, by the second equation of Lemma 1.1.3(iii), aqr∗ t ≤ nr . Thus, aqr∗ t ≤ p−1. Now recall that, by Lemma 1.1.3(ii), atrq nq = aqr∗ t nt . Moreover, as q ∈ tr, 1 ≤ atrq . Thus, as p divides nq and aqr∗ t ≤ p − 1, p must divide nt . / Q. Thus, On the other hand, as R (t) = R (q) − 1 and min R (Q) = R (q), t ∈ as t ∈ R, p does not divide nt . This contradiction finishes the proof. Corollary 3.1.7 Let R be a subset of S, and assume that each element in R∗ ∪ R has valency 2. Then R is 2-valenced. Proof. Let p be an odd prime number, and let s be an element in R. Then, by Theorem 3.1.6, p does not divide ns . Thus, ns is a power of 2. Let T be a closed subset of S such that {1} = T . We define Φ(T ) to be the intersection of all maximal closed subsets of T . Note that Φ(T ) is closed. We call Φ(T ) the Frattini subset of T . Let T and U be closed subsets of S satisfying T U = S and U = S. Then U is called a supplement of T in S. Lemma 3.1.8 Let T be a closed subset of S such that {1} = T . Then the following hold. (i) Let R be a subset of T which satisfies Φ(T ) ∪ R = T . Then, we have

R = T . (ii) The closed subset Φ(T ) does not have a supplement in T . Proof. (i) Suppose, by way of contradiction, that R = T . Then there exists a maximal closed subset U of T such that R ⊆ U . However, by definition of the Frattini subset, we have Φ(T ) ⊆ U . Therefore, we have Φ(T ) ∪ R ⊆ U , contrary to Φ(T ) ∪ R = T . (ii) This follows immediately from (i). For any two elements p and q in S, we define [p, q] := p∗ q ∗ pq and call this set the commutator of p and q. It is obvious that, for any two elements p and q in S, [p, q]∗ = [q, p]. Let P and Q be nonempty subsets of S. We define [P, Q] to be the closed subset of S generated by the union of all sets [p, q] with p ∈ P and q ∈ Q. The set [P, Q] is called the commutator subset of P and Q.

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If there exists an element q in Q with {q} = Q, we write [P, q] instead of [P, Q]. Similarly, if there exists an element p in P with {p} = P , we write [p, Q] instead of [P, Q]. Note that [P, Q] = [Q, P ]. Lemma 3.1.9 Let R be a nonempty subset of S, and let T be a closed subset of S. Then the following hold. (i) We have [R, T ]T = T [R, T ]. (ii) Let Q be the union of the sets T r∗ T rT with r ∈ R. Then Q = [R, T ]T . Proof. (i) Let r be an element in R, and let t be an element in T . We shall show first that [r, t]T ⊆ T [R, T ]. Let s be an element in [r, t]T . Then, as [r, t] = r∗ t∗ rt and t ∈ T , s ∈ r∗ t∗ rT . Thus, there exists an element q in T such that s ∈ r∗ t∗ rq. It follows that s ∈ r∗ t∗ qrr∗ q ∗ rq = r∗ t∗ qr[r, q]. Thus, there exists an element p in t∗ q such that s ∈ r∗ pr[r, q]. From p ∈ t∗ q and t∗ q ⊆ T we obtain p ∈ T . Thus, s ∈ pp∗ r∗ pr[r, q] = p[p, r][r, q] ⊆ T [r, p]∗ [r, q] ⊆ T [R, T ]. Since s has been chosen arbitrarily in [r, t]T , we have shown that, for any two elements r in R and t in T , [r, t]T ⊆ T [R, T ]. From what we have seen so far we also obtain [r, t]∗ T = t∗ r∗ trT = t∗ r∗ trt∗ T ⊆ t∗ [r, t∗ ]T ⊆ t∗ T [R, T ] = T [R, T ] for any two elements r in R and t in T . Thus, by Lemma 3.1.1(i), [R, T ]T ⊆ T [R, T ]. With the help of Lemma 1.3.2(iii) one obtains from [R, T ]T ⊆ T [R, T ] also that T [R, T ] ⊆ [R, T ]T . Thus, [R, T ]T = T [R, T ]. (ii) Let s be an element in Q \ {1}. Then, by Lemma 3.1.1(i), there exists a positive integer n such that s ∈ Qn . Thus, there exist elements r1 , . . . , rn in R such that s ∈ T r1∗ T r1 T · · · T rn∗ T rn T . Thus, there exist elements t1 , . . . , tn in T such that s ∈ T r1∗ t∗1 r1 T · · · T rn∗ t∗n rn T = T [r1 , t1 ]T · · · T [rn , tn ]T ⊆ [R, T ]T ; cf. (i). Since s has been chosen arbitrarily in Q \ {1}, we have shown that

Q ⊆ [R, T ]T . On the other hand, it is clear that, for any two elements r in R and t in T , [r, t] = r∗ t∗ rt ⊆ Q. Thus, as T ⊆ Q, [R, T ]T ⊆ Q. From Q ⊆ [R, T ]T and [R, T ]T ⊆ Q we obtain Q = [R, T ]T .

3.2 The Thin Residue

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3.2 The Thin Residue In this section, S is assumed to have finite valency. The letter T will always stand for a closed subset of S. We define Oϑ (T ) to be the intersection of all strongly normal closed subsets of T and call it the thin residue of T . Note that Oϑ (T ) is closed. Theorem 3.2.1 The following statements hold. (i) The set Oϑ (T ) is strongly normal in T . (ii) We have Oϑ (T ) = [T, 1]. (iii) For each closed subset U of S with T ⊆ U , we have Oϑ (T ) ⊆ Oϑ (U ). Proof. (i) This follows from Lemma 2.5.6(i). (ii) Let us denote by R the union of all sets t∗ t with t ∈ T . We have to show that Oϑ (T ) = R. Let p and q be elements in T . First of all, we shall prove that q ∗ p∗ pq ⊆ R. Let t be an element in pq. Then, by Lemma 1.3.3(i), p ∈ tq ∗ , whence t∗ pq ⊆ t∗ tq ∗ q ⊆ R. Since t has been chosen arbitrarily in pq, this yields q ∗ p∗ pq ⊆

R; cf. Lemma 1.3.2(iii). Since p and q have been chosen arbitrarily in T , we have shown t∗ Rt ⊆ R for each element t in T . Thus, by Lemma 3.1.4, R is strongly normal in T . Thus, by definition, Oϑ (T ) ⊆ R. In order to show that R ⊆ Oϑ (T ) it suffices to show that R ⊆ Oϑ (T ). (This is because Oϑ (T ) is closed.) Let t be an element in T . Then, as 1 ∈ Oϑ (T ), t∗ t ⊆ t∗ Oϑ (T )t. On the other hand, we know from (i) that Oϑ (T ) is strongly normal in T , so that t∗ Oϑ (T )t ⊆ Oϑ (T ). Thus, t∗ t ⊆ Oϑ (T ). Since t has been chosen arbitrarily in T , we have shown that R ⊆ Oϑ (T ). (iii) This is an immediate consequence of (ii). Lemma 3.2.2 Let U be a closed subset of S such that T ⊆ NS (U ). Then we have Oϑ (T )U = Oϑ (T U )U . Proof. From Theorem 3.2.1(iii) we know that Oϑ (T ) ⊆ Oϑ (T U ), and from this we obtain Oϑ (T )U ⊆ Oϑ (T U )U . By Theorem 3.2.1(i), Oϑ (T ) is strongly normal in T . Moreover, we are assuming that T ⊆ NS (U ). Thus, by Lemma 2.5.7, Oϑ (T )U is strongly normal in T U . Thus, the definition of Oϑ (T U ) gives us Oϑ (T U ) ⊆ Oϑ (T )U , and from this we obtain Oϑ (T U )U ⊆ Oϑ (T )U .

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Lemma 3.2.3 Let U be a closed subset of S such that T U is closed and Oϑ (U ) ⊆ T . Then Oϑ (T U ) is the intersection of all strongly normal closed subsets of T which contain Oϑ (U ). Proof. Let us denote by V the set of all strongly normal closed subsets of T which contain Oϑ (U ), and let us define W to be the intersection of all elements of V. We have to show that Oϑ (T U ) = W . Let us first show that Oϑ (T U ) ⊆ W . In order to do so we pick an element s in T U and an element V in V. Since T U is assumed to be closed, T U = U T ; cf. Lemma 2.1.1. Thus, as s ∈ T U , s ∈ U T . Thus, there exist elements t in T and u in U such that s ∈ ut. Since u ∈ U , u∗ u ⊆ Oϑ (U ); cf. Theorem 3.2.1(ii). Since V ∈ V, t∗ V t ⊆ V and Oϑ (U ) ⊆ V . Thus, s∗ s ⊆ t∗ u∗ ut ⊆ t∗ Oϑ (U )t ⊆ V. Now, as s has been chosen arbitrarily in T U , we conclude that Oϑ (T U ) ⊆ V ; cf. Theorem 3.2.1(ii). But also V has been chosen arbitrarily in V. Therefore, we have shown that Oϑ (T U ) ⊆ W . Let us now show that, conversely, W ⊆ Oϑ (T U ). From Oϑ (T U ) ⊆ W and W ⊆ T we obtain Oϑ (T U ) ⊆ T . In particular, Oϑ (T U ) is strongly normal in T . On the other hand, we know from Theorem 3.2.1(iii) that Oϑ (U ) ⊆ Oϑ (T U ). Thus, Oϑ (T U ) ∈ V, so that, by definition, W ⊆ Oϑ (T U ). Corollary 3.2.4 Let U be a thin closed subset of S such that T U is closed. Then we have Oϑ (T ) = Oϑ (T U ). Proof. Since U is assumed to be thin, {1} = Oϑ (U ). Thus, the claim follows from Lemma 3.2.3. Recall that a closed subset U of S is called a supplement of T in S if T U = S and U = S. Referring to this terminology we obtain from Corollary 3.2.4 that, if Oϑ (S) has a supplement in S, Oϑ (S) = S. Lemma 3.2.5 Assume that there exists a prime number p such that T is p-valenced. Then, if {1} = Oϑ (Oϑ (T )), Oϑ (T )Oϑ (T ) = T . Proof. Let t be an element in T . Then, by Theorem 3.2.1(ii), t∗ t ⊆ Oϑ (T ). Thus, by Lemma 1.4.4(ii), nOϑ (T ) = ntOϑ (T ) . We are assuming that T is p-valenced. Thus, assuming that {1} = Oϑ (Oϑ (T )), we obtain that p divides nOϑ (T ) − 1. Thus, as nOϑ (T ) = ntOϑ (T ) , the set Oϑ (T ) ∩ tOϑ (T ) is not empty. Thus, by Lemma 1.3.3(i), t ∈ Oϑ (T )Oϑ (T ). We set (Oϑ )0 (T ) := T . For each positive integer n, we inductively define

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(Oϑ )n (T ) := Oϑ ((Oϑ )n−1 (T )). Note that (Oϑ )n (T ) is closed for each non-negative integer n. Note also that, for each positive integer n, (Oϑ )n (T ) ⊆ (Oϑ )n−1 (T ). Here is a generalization of Theorem 3.2.1(iii). Lemma 3.2.6 Let n be a non-negative integer, and let U be a closed subset of S such that T ⊆ U . Then (Oϑ )n (T ) ⊆ (Oϑ )n (U ). Proof. Our lemma is certainly true if n = 0. By induction, we may assume that (Oϑ )n−1 (T ) ⊆ (Oϑ )n−1 (U ). Then, by Theorem 3.2.1(iii), (Oϑ )n (T ) = Oϑ ((Oϑ )n−1 (T )) ⊆ Oϑ ((Oϑ )n−1 (U )) = (Oϑ )n (U ), and that proves the lemma. The first part of the following lemma generalizes Lemma 3.2.2. Lemma 3.2.7 Let n be a non-negative integer, and let U be a closed subset of S such that T ⊆ NS (U ). Then the following hold. (i) We have (Oϑ )n (T )U = (Oϑ )n (T U )U . (ii) If Oϑ (T U )U = T U , (Oϑ )n (T U )U = T U . Proof. (i) There is nothing to show if n = 0. Therefore, we assume that 1 ≤ n. Assume, by way of contradiction, that there exists a non-negative integer violating the claim. Then 1 ≤ n and (Oϑ )n−1 (T )U = (Oϑ )n−1 (T U )U. Thus, Oϑ ((Oϑ )n−1 (T )U )U = Oϑ ((Oϑ )n−1 (T U )U )U. On the other hand, applying Lemma 3.2.2 to (Oϑ )n−1 (T ) in place of T , we obtain (Oϑ )n (T )U = Oϑ ((Oϑ )n−1 (T ))U = Oϑ ((Oϑ )n−1 (T )U )U. Finally, applying Lemma 3.2.2 to (Oϑ )n−1 (T U ) in place of T , we obtain (Oϑ )n (T U )U = Oϑ ((Oϑ )n−1 (T U ))U = Oϑ ((Oϑ )n−1 (T U )U )U. Thus, we have (Oϑ )n (T )U = (Oϑ )n (T U )U .

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(ii) We are assuming that T ⊆ NS (U ). Thus, we obtain from Lemma 2.5.1(ii) (together with Lemma 2.5.2(iii)) that T U ⊆ NS (U ). In particular, (Oϑ )n−1 (T U ) ⊆ NS (U ). Thus, by Lemma 3.2.2, Oϑ ((Oϑ )n−1 (T U ))U = Oϑ ((Oϑ )n−1 (T U )U )U. On the other hand, we may assume that (Oϑ )n−1 (T U )U = T U and we are assuming that Oϑ (T U )U = T U . Thus, Oϑ ((Oϑ )n−1 (T U )U )U = T U. Thus, as Oϑ ((Oϑ )n−1 (T U )) = (Oϑ )n (T U ), (Oϑ )n (T U )U = T U .

3.3 Elements of Valency 2 In this section, we assume S to have finite valency. We shall first look at closed subsets of S which are generated by a single elements of valency 2. Lemma 3.3.1 Let s be a symmetric element of S, and assume that s has valency 2. Then each element of s has valency at most 2. Proof. Let us write instead of {s} and Q to denote the set of all elements q in s with 3 ≤ nq . By way of contradiction, we assume that Q is not empty. We fix an element q in Q such that min (Q) = (q). Since q ∈ Q, 3 ≤ nq . Thus, 1 = q. Thus, as q ∈ s, there exists an element p in s such that q ∈ ps and (q) = (p) + 1; cf. Lemma 3.1.2. Since (p) = (q) − 1 and min (Q) = (q), p ∈ / Q. Thus, as p ∈ s, np ≤ 2. Now recall that, by Lemma 1.1.3(ii), apsq nq = aqs∗ p np . Moreover, as q ∈ ps, 1 ≤ apsq . Thus, as np ≤ 2 and 3 ≤ nq , 2 ≤ aqs∗ p . We are assuming that ns = 2. Thus, as 3 ≤ nq , q = s. Thus, as q ∈ ps, 1 = p. Thus, as p ∈ s, there exists an element t in s such that p ∈ ts and (p) = (t) + 1; cf. Lemma 3.1.2. Since p ∈ ts and s∗ = s, p ∈ ts∗ . Thus, 1 ≤ ats∗ p . Since (p) = (t) + 1 and (q) = (p) + 1, t = q. Thus, as 2 ≤ aqs∗ p , 3 ≤ ats∗ p + aqs∗ p . On the other hand, by the second equation of Lemma 1.1.3(iii), aqs∗ p + ats∗ p ≤ ns , so that 3 ≤ ns . This contradiction finishes the proof of the lemma. Lemma 3.3.2 Let s be a symmetric element of S, assume that s has valency 2, and let p and q be elements in s with {s} (p) = {s} (q). Then p = q. Proof. Let us write instead of {s} and Q to denote the set of all elements q in s such that there exists an element p in s with (p) = (q) and p = q. By way of contradiction, we assume that Q is not empty, and we fix an element q in Q such that min (Q) = (q).

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Since q ∈ Q, there exists an element p in s such that (p) = (q) and p = q. Clearly, 1 = p. Thus, as p ∈ s, there exists an element t in s such that p ∈ ts and (p) = (t) + 1; cf. Lemma 3.1.2. Similarly, as 1 = q and q ∈ s, there exists an element u in s such that q ∈ us and (q) = (u) + 1. From q ∈ us we obtain u ∈ qs∗ ; cf. Lemma 1.3.3(i). Thus, by definition, 1 ≤ aqs∗ u . From (p) = (t) + 1, (q) = (u) + 1, and (p) = (q) we obtain (t) = (u). Thus, by the (minimal) choice of q, t = u. Thus, as p ∈ ts, p ∈ us. Thus, by Lemma 1.3.3(i), u ∈ ps∗ . It follows that 1 ≤ aps∗ u . Since 1 = u and u ∈ s, there exists an element r in s such that u ∈ rs and (u) = (r) + 1; cf. Lemma 3.1.2. From u ∈ rs and s∗ = s we obtain u ∈ rs∗ . Thus, by definition, 1 ≤ ars∗ u . Note finally that, by the second equation of Lemma 1.1.3(iii), ars∗ u + aps∗ u + aqs∗ u ≤ ns . Thus, as p = q and ns = 2, we must have that r = p or r = q. However, this contradicts (r) ≤ (p) − 1 = (q) − 1. Recall that, for each element r in s, {s} (r∗ ) = {s} (r). Thus, Lemma 3.3.2 says, in particular, that all elements in s are symmetric. Let us now look at the structure of closed subsets of S generated by an arbitrary (not necessarily symmetric) element of valency 2. Lemma 3.3.3 Let s be an element of S, assume that s has valency 2 and that {1} = Oϑ ( s). Then s is symmetric and ns is odd. Proof. From ns = 2 we obtain that s is 2-valenced; cf. Corollary 3.1.7. Thus, assuming that {1} = Oϑ ( s), we obtain that ns is odd. From Lemma 1.5.6(i) we obtain an element r in s∗ s\{1} such that {1, r} = s∗ s. From r ∈ s∗ s we obtain r ⊆ s. Thus, as ns is odd, nr is odd; cf. Lemma 2.3.6(ii). Since s∗ s = {1, r} ⊆ r, nsr = nr ; cf. Lemma 1.4.4(ii). Thus, as nr is odd, nsr is odd. Thus, as s is 2-valenced, we obtain from s r ⊆ s and {1} = Oϑ ( s) that 1 ∈ s r. Thus, by Lemma 2.1.4, s ∈ r. Thus, by Lemma 1.2.3, s is symmetric. Lemma 3.3.4 Let s be an element of S, assume that s has valency 2 and that Oϑ ( s) has odd valency. Then Oϑ ( s)Oϑ ( s) = s. Proof. We are assuming that ns = 2. Thus, by Lemma 1.5.6(i), there exists an element r in s \ {1} such that nr ≤ 2, r∗ = r, and {1, r} = s∗ s. Let v be an element in X, let w be an element in vr. Since w ∈ vr and r ∈ s∗ s, w ∈ vs∗ s. Thus, there exists an element x in vs∗ such that w ∈ xs.

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We are assuming that Oϑ ( s) has odd valency. Thus, as r ∈ s∗ s ⊆ Oϑ ( s),

r has odd valency; cf. Lemma 2.3.6(ii). Thus, as r∗ = r and nr ≤ 2, there exists an element q in r such that nq = 2 and r ∈ q ∗ q. Since w ∈ vr and r ∈ q ∗ q, w ∈ vq ∗ q. Thus, there exists an element y in vq ∗ such that w ∈ yq. Let us denote by t the uniquely determined element in S satisfying y ∈ xt. Then asq∗ t = 2. We wish to show that t ∈ Oϑ ( s). In order to do so we pick an element y in xt. Since asq∗ t = 2, |xs ∩ y q| = 2. On the other hand, xs = {v, w}. Thus, y ∈ vq ∗ ∩ wq ∗ . Thus, as r has odd valency, y = y. Since y has been chosen arbitrarily in xt, xt = {y}. It follows that t ∈ Oϑ (S). Thus, as s ∈ tq and q ∈ r ⊆ Oϑ ( s), s ∈ Oϑ ( s)Oϑ ( s). On the other hand, we know from Theorem 3.2.1(i) and Lemma 2.1.1 that Oϑ ( s)Oϑ ( s) is closed. Thus, Oϑ ( s)Oϑ ( s) = s. Let us now look at the structure of closed subsets of S generated by an arbitrary set of elements of valency 2. Theorem 3.3.5 Let T be a closed subset of S generated by a set of elements of valency 2. Then, we have the following. (i) If Oϑ (T ) has odd valency, Oϑ (T )Oϑ (T ) = T . (ii) If {1} = Oϑ (T ), Oϑ (T ) = T . Proof. (i) We are assuming that T contains a subset R of elements of valency 2 such that R = T . Let r be an element in R. Then, as R ⊆ T , Oϑ ( r) ⊆ Oϑ (T ); cf. Theorem 3.2.1(iii). Thus, as we are assuming Oϑ (T ) to have odd valency, Oϑ ( r) has odd valency; cf. Lemma 2.3.6(ii). Thus, by Lemma 3.3.4, Oϑ ( r)Oϑ ( r) = r. From Oϑ ( r) ⊆ Oϑ (T ) and Oϑ ( r) ⊆ Oϑ (T ) we also obtain Oϑ ( r)Oϑ ( r) ⊆ Oϑ (T )Oϑ (T ). Thus, r ∈ Oϑ (T )Oϑ (T ). Since r has been chosen arbitrarily in R, we have shown that R ⊆ Oϑ (T )Oϑ (T ). Thus, as Oϑ (T )Oϑ (T ) is closed, the desired equation follows from R = T . (ii) We are assuming that T is generated by a set of elements all of which have valency 2. Thus, by Corollary 3.1.7, T is 2-valenced.

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Since T is 2-valenced, our hypothesis that {1} = Oϑ (T ) forces T to have odd valency. Thus, by Lemma 2.3.6(ii), Oϑ (T ) has odd valency, so that our claim follows from (i).

3.4 Closed Subsets Generated by Involutions In this section, the letter L stands for a set of involutions of S. We shall look at L. Instead of L we shall write . Lemma 3.4.1 Let n be a positive integer, and let s0 , . . . , sn be elements in

L such that, for each element i in {1, . . . , n}, si ∈ si−1 L. Assume that (s0 ) + n = (sn ). Then, for each element i in {0, . . . , n}, (s0 ) + i = (si ). Proof. Our first claim is that, for each element i in {0, . . . , n}, si ∈ s0 Li . The claim is obviously true if i = 0. Therefore, we assume that i ∈ {1, . . . , n}. Then, assuming that si−1 ∈ s0 Li−1 we obtain from si ∈ si−1 L that si ∈ s0 Li−1 L = s0 Li . This shows that, for each element i in {0, . . . , n}, (si ) ≤ (s0 ) + i. Let us now prove that, for each element i in {0, . . . , n}, sn ∈ si Ln−i . The claim is obviously true if i = n. Therefore, we assume that i ∈ {0, . . . , n − 1}. Then, assuming that sn ∈ si+1 Ln−(i+1) we obtain from si+1 ∈ si L that sn ∈ si LLn−(i+1) = si Ln−i . This shows that (sn ) ≤ (si ) + n − i for each element i in {0, . . . , n}. Thus, as we are assuming that (s0 ) + n = (sn ), we obtain (s0 ) + i ≤ (si ) for each element i in {0, . . . , n}. From the definition of we obtain (s) ≤ (p) + (q) for any three elements p, q, and s in L with s ∈ pq, . The following lemma focuses on the case where equality holds. Lemma 3.4.2 Let p, q, and r be elements in L satisfying r ∈ pq and (r) = (p) + (q). Let t and u be elements in L satisfying q ∈ tu and (q) = (t) + (u). Then, there exists an element s in pt such that r ∈ su, (s) = (p) + (t), and (r) = (s) + (u). Proof. Since r ∈ pq and q ∈ tu, r ∈ ptu. Thus, there exists an element s in pt such that r ∈ su. Since s ∈ pt, (s) ≤ (p) + (t). Since r ∈ su, (r) ≤ (s) + (u). Thus, as we are assuming that (r) = (p) + (q) and that (q) = (t) + (u), we have

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(r) ≤ (s) + (u) ≤ (p) + (t) + (u) = (r). It follows that (s) = (p) + (t) and (r) = (s) + (u). Let q be an element in L. We define S−1 (q, L) to be the set of all elements r in L such that there exists an element p in L with r ∈ pq and (r) = (p) + (q). By S1 (q, L) we shall denote the set of all elements p in L such that there exists an element r in pq with (r) = (p) + (q). As a consequence of the remark right before Lemma 3.4.2 we obtain that, in both of these definitions, the equation (r) = (p) + (q) can be replaced with the condition that (p) + (q) ≤ (r). Let s be an element in L. For the remainder of this section, we shall write S−1 (s) instead of S−1 (s, L) and S1 (s) instead of S1 (s, L). Note that, for each element s in L \ {1}, S−1 (s) ∩ S1 (s) is empty. Note also that, for each element s in L \ {1}, there exists an element l in L such that s ∈ S−1 (l); cf. Lemma 3.1.2. Note, finally, that, for each element s in L, s ∈ S−1 (s) and 1 ∈ S1 (s). Lemma 3.4.3 For each element s in L, the following hold. (i) We have S−1 (s) ⊆ S1 (s)s. (ii) We have S1 (s) ⊆ S−1 (s)s∗ . Proof. (i) Let q be an element in S−1 (s). Then, by definition, there exists an element p in L such that q ∈ ps and (q) = (p) + (s). Thus, by definition, p ∈ S1 (s). Thus, as q ∈ ps, q ∈ S1 (s)s. (ii) Let p be an element in S1 (s). Then, by definition, there exists an element q in L such that q ∈ ps and (q) = (p) + (s). Thus, by definition, q ∈ S−1 (s). On the other hand, as q ∈ ps, p ∈ qs∗ ; cf. Lemma 1.3.3(i). Thus, p ∈ S−1 (s)s∗ . Lemma 3.4.4 For any two elements p and q in L, we have the following. (i) If p ∈ S1 (q), q ∗ ∈ S1 (p∗ ). (ii) If S−1 (p) ∩ S1 (q) is not empty, p ∈ S1 (q). (iii) If q ∈ S−1 (p), S−1 (q) ⊆ S−1 (p). (iv) If q ∈ S−1 (p), S1 (q ∗ ) ⊆ S1 (p∗ ). Proof. (i) This follows from (p∗ ) = (p) and (q ∗ ) = (q). (ii) Let r be an element in S−1 (p) ∩ S1 (q). Since r ∈ S−1 (p), there exists an element t in L such that r ∈ tp and (r) = (t) + (p). Since r ∈ S1 (q), there exists an element s in L such that s ∈ rq and (s) = (r) + (q). Thus, by Lemma 3.4.2, there exists an element u in pq such that s ∈ tu, (u) = (p) + (q), and (s) = (t) + (u). From u ∈ pq and (u) = (p) + (q) we obtain p ∈ S1 (q).

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(iii) Let us assume that q ∈ S−1 (p), and let us pick an element s in S−1 (q). We shall show that s ∈ S−1 (p). Since s ∈ S−1 (q), there exists an element u in L such that s ∈ uq and (s) = (u) + (q). Since we are assuming that q ∈ S−1 (p), there exists an element t in L such that q ∈ tp and (q) = (t) + (p). Now, by Lemma 3.4.2, there exists an element r in ut such that s ∈ rp, (r) = (u) + (t), and (s) = (r) + (p). From s ∈ rp and (s) = (r) + (p) we obtain s ∈ S−1 (p). (iv) Let us assume that q ∈ S−1 (p), and let us pick an element s in S1 (q ∗ ). We shall show that s ∈ S1 (p∗ ). Since s ∈ S1 (q ∗ ), q ∈ S1 (s∗ ); cf. (i). Thus, as we are assuming that q ∈ S−1 (p), p ∈ S1 (s∗ ); cf. (ii). Thus, by (i), s ∈ S1 (p∗ ). For each subset R of L, we define S−1 (R) to be the intersection of the sets S−1 (r) with r ∈ {1} ∪ R. (Note that S−1 (1) = L.) Recall that, for each element s in L, s ∈ S−1 (s) and 1 ∈ S1 (s). The following lemma deals with the case where s ∈ S−1 (L). Lemma 3.4.5 Let s be an element in S−1 (L). Then the following hold. (i) We have {1} = S1 (s∗ ). (ii) We have {s∗ } = S−1 (s∗ ). Proof. (i) It is clear that 1 ∈ S1 (s∗ ). Let us assume, by way of contradiction, that S1 (s∗ ) contains an element q with 1 = q. From q ∈ S1 (s∗ ) we obtain an element t in qs∗ such that (t) = (q) + (s∗ ). From 1 = q we obtain elements p in L and l in L such that q ∈ pl and (q) = (p) + 1; cf. Lemma 3.1.2. From s ∈ S−1 (L) we obtain an element r in L such that s ∈ rl and (s) = (r) + 1. From t ∈ qs∗ , q ∈ pl, and s ∈ rl we obtain t ∈ pllr ∗ ⊆ pr∗ ∪ plr∗ . In particular, (t) ≤ (p) + 1 + (r∗ ). On the other hand, as (t) = (q) + (s∗ ), (q) = (p) + 1, and (s) = (r) + 1 we conclude that (t) = (p) + 2 + (r), contradiction. (ii) This follows from Lemma 3.4.3(i) together with (i). For each subset R of L, we define S1 (R) to be the intersection of the sets S1 (r) with r ∈ {1} ∪ R. (Note that S1 (1) = L.)

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Lemma 3.4.6 Let p and q be elements in L such that p ∈ S1 (q). Then the following hold. (i) We have S1 (pq) ⊆ S1 (p). (ii) For each element r in S1 (pq), there exists an element s in rp such that s ∈ S1 (q). Proof. We are assuming that p ∈ S1 (q). Thus, by definition, there exists an element u in pq such that (u) = (p) + (q). (i) Let s be an element in S1 (pq). Then, as u ∈ pq, s ∈ S1 (u). Thus, there exists an element r in su such that (r) = (s) + (u). Thus, by Lemma 3.4.2, there exists an element t in sp such that (t) = (s) + (p). It follows that s ∈ S1 (p). (ii) Let r be an element in S1 (pq). Then, as u ∈ pq, r ∈ S1 (u). Thus, there exists an element t in ru such that (t) = (r) + (u). Thus, by Lemma 3.4.2, there exists an element s in rp such that t ∈ sq and (t) = (s) + (q). From t ∈ sq and (t) = (s) + (q) we obtain s ∈ S1 (q). We say that L satisfies the exchange condition if, for any three elements h, k in L and s in S1 (k), h ∈ S1 (s) implies hs ⊆ sk ∪ S1 (k). We shall investigate the exchange condition in Section 3.6 and, in more detail, in the last two chapters of this monograph. For the last two results of this section, we shall now look at a slightly different condition. We assume that, for any three elements h, k in L and s in S1 (k), h ∈ S1 (s) implies hs ⊆ S−1 (k) ∪ S1 (k). Lemma 3.4.7 For each element l in L, S−1 (l) ∪ S1 (l) = L. Proof. Let us assume the claim to be false. Among the elements in L not contained in S−1 (l) ∪ S1 (l) we pick s in such a way that (s) is as small as possible. / S1 (l), 1 = s. Thus, by Lemma 3.1.2, there exist Since 1 ∈ S1 (l) and s ∈ elements k in L and r in L such that s ∈ kr and (s) = 1 + (r). Since (s) = 1 + (r), the (minimal) choice of s yields r ∈ S−1 (l) ∪ S1 (l). / S−1 (l), r ∈ / S−1 (l); Since s ∈ kr and (s) = 1 + (r), s ∈ S−1 (r). Thus, as s ∈ cf. Lemma 3.4.4(iii). Thus, as r ∈ S−1 (l) ∪ S1 (l), r ∈ S1 (l). Note also that, as s ∈ kr and (s) = 1 + (r), k ∈ S1 (r). From k ∈ S1 (r) and r ∈ S1 (l) (together with our hypothesis) we obtain kr ⊆ S−1 (l) ∪ S1 (l). Thus, as s ∈ kr, s ∈ S−1 (l) ∪ S1 (l). This contradiction concludes the proof of our lemma. Lemma 3.4.8 For each subset K of L, S1 (K) K = L.

3.5 Basic Results on Constrained Sets of Involutions

55

Proof. Let us assume that S1 (K) K = L. Among the elements in L not contained in S1 (K) K we pick s such that (s) is as small as possible. / S1 (K). Thus, there exists an element k in K such Since s ∈ / S1 (K) K, s ∈ that s ∈ / S1 (k). Thus, by Lemma 3.4.7, s ∈ S−1 (k). Thus, there exists an element r in L such that s ∈ rk and (s) = (r) + 1. Since (s) = (r) + 1, the (minimal) choice of s yields r ∈ S1 (K) K. Thus, as s ∈ rk and k ∈ K, s ∈ S1 (K) K, contradiction.

3.5 Basic Results on Constrained Sets of Involutions In this section, the letter L stands for a set of involutions of S. For each element s in L, we write S1 (s) instead of S1 (s, L). The set L is called constrained if, for any two elements q in L and p in S1 (q), 1 = |pq|. For the remainder of this section, we assume L to be constrained. Instead of L we shall write . Lemma 3.5.1 For any two elements q and r in L, there exists at most one element p in L such that r ∈ pq and (r) = (p) + (q). Proof. Let us fix an element r in L. We shall denote by Q the set of all elements q in L such that there exist elements p and p in L with r ∈ pq, r ∈ p q, (r) = (p) + (q), (p ) = (p), and p = p. By way of contradiction, we assume that Q is not empty. We pick an element q in Q, and we do this in such a way that (q) is as small as possible. Since 1 ∈ / Q and q ∈ Q, 1 = q. Thus, by Lemma 3.1.2, there exist elements l in L and u in L such that q ∈ lu and (q) = 1 + (u). Thus, as r ∈ pq and (r) = (p) + (q), there exists an element t in pl such that r ∈ tu, (t) = (p) + 1, and (r) = (t) + (u); cf. Lemma 3.4.2. Similarly, we find an element t in p l such that r ∈ t u, (t ) = (p ) + 1, and (r) = (t ) + (u). Since (q) = 1 + (u), the (minimal) choice of q yields u ∈ / Q. Thus, as r ∈ tu, r ∈ t u, (r) = (t) + (u), and (r) = (t ) + (u), t = t. We are assuming that L is constrained. Thus, as t ∈ pl and (t) = (p) + 1, we conclude that {t} = pl. Similarly, {t } = p l. Thus, as t = t, p l = pl. It follows that p ∈ {p, t}. Thus, as t = t and (t ) = (p ) + 1, p = p. This contradiction finishes the proof of the lemma. Lemma 3.5.2 Let p, q, and r be elements in L such that r ∈ pq and (r) = (p) + (q). Then apqr = 1.

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Proof. Let us denote by R the set of the elements r in L such that there exist elements p and q in L with r ∈ pq, (r) = (p) + (q), and 2 ≤ apqr . By way of contradiction, we assume that R is not empty. We pick an element r in R, and we do this in such a way that (r) is as small as possible. Since r ∈ R, there exist elements p and q in L such that r ∈ pq, (r) = (p) + (q), and 1 = apqr . Since r ∈ pq and 1 = apqr , we have 2 ≤ apqr . In particular, 1 = p and 1 = q. Since 1 = q, there exist elements t in L and l in L such that q ∈ tl, and (q) = (t) + 1; cf. Lemma 3.1.2. Since r ∈ pq, (r) = (p) + (q), q ∈ tl, and (q) = (t) + 1, there exists an element u in pt such that r ∈ ul, (u) = (p) + (t), and (r) = (u) + 1; cf. Lemma 3.4.2. We are assuming that L is constrained. Thus, as r ∈ ul and (r) = (u) + 1, {r} = ul. From Lemma 1.1.1(iii) we know that apts aslr = apsr atls . s∈S

s∈S

Since u ∈ pt and (u) = (p) + (t), {u} = pt. Thus, the left hand side of the above equation is equal to aptu aulr . Since q ∈ tl and (q) = (t) + 1, {q} = tl. Thus, the right hand side of the above equation is equal to apqr atlq . The choice of r yields aptu = 1 and atlq = 1. (Recall that 1 = p. Therefore, (q) ≤ (r) − 1.) Thus, aulr = apqr . Thus, as 2 ≤ apqr , 2 ≤ aulr . On the other hand, we have that {r} = ul. Thus, by Lemma 2.3.8(iii), r = u, contrary to (r) = (u) + 1. Let us now look at the thin elements of L. Lemma 3.5.3 Let R be a subset of L such that Oϑ (L) ⊆ R. Assume that R2 ⊆ R and that, for each element r in R, r∗ r ⊆ R. Then R ⊆ L ∩ R. Proof. Let us abbreviate K := L ∩ R. We shall see that R ⊆ K. Suppose, by way of contradiction, that R ⊆ K. Then, R \ K is not empty. We pick an element r in R \ K, and we do this in such a way that (r) is as small as possible. Since r ∈ / K, 1 = r. Thus, by Lemma 3.1.2, there exist elements q in L and l in L such that r ∈ ql and (r) = (q) + 1. Since L is assumed to be constrained, we obtain from r ∈ ql and (r) = (q) + 1 that {r} = ql. Let us first prove that l ∈ R. If 1 = nl , this follows from our hypothesis that Oϑ (L) ⊆ R. If 2 ≤ nl , we obtain from Lemma 2.3.8(ii) that l ∈ ll. Thus,

3.6 Basic Results on Coxeter Sets

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as 1 ∈ q ∗ q and {r} = ql, l ∈ l∗ q ∗ ql = r∗ r. However, we are assuming that r∗ r ⊆ R. Thus, l ∈ R. Since r ∈ ql, q ∈ rl; cf. Lemma 1.3.3(i). Thus, as l ∈ R, q ∈ rR. Thus, as we are assuming that R2 ⊆ R, we obtain q ∈ R. Recall that (r) = (q) + 1. Thus, the (minimal) choice of r yields q ∈ / R\

K. Thus, as q ∈ R, q ∈ K. Thus, as r ∈ ql and l ∈ K, r ∈ K. This contradiction proves the lemma. Corollary 3.5.4 For each closed subset T of L, we have the following. (i) If Oϑ (L) ⊆ T , L ∩ T = T . (ii) If Oϑ (L) is empty, L ∩ T = T . Proof. (i) Considering Lemma 2.5.9(ii) we obtain from Lemma 3.5.3 that T ⊆ L ∩ T . Since T is assumed to be closed we also have that L ∩ T ⊆ T . (ii) This follows from (i). Lemma 3.5.5 The following statements hold. (i) We have Oϑ (L) = Oϑ ( L). (ii) The set L is thin if and only if L is thin. (iii) We have {1} = Oϑ ( L) if and only if Oϑ (L) is empty. Proof. (i) From Lemma 1.5.2 we know that Oϑ ( L)2 ⊆ Oϑ ( L). Moreover, for each element s in Oϑ ( L), we have s∗ s = {1} ⊆ Oϑ ( L); cf. Lemma 1.5.1. Thus, by Lemma 3.5.3, Oϑ ( L) ⊆ Oϑ (L). Conversely, we know from Lemma 3.1.5(ii) that Oϑ (L) ⊆ Oϑ ( L). (ii) If L is thin, so is L. If L is thin, Oϑ (L) = L; cf. Lemma 1.5.1. Thus, by (i), L = Oϑ ( L). From this we obtain that L is thin; cf. Lemma 1.5.1. (iii) This is an immediate consequence of (i). From Lemma 3.5.5(i) we obtain, in particular, that Oϑ ( L) is a closed subset of S.

3.6 Basic Results on Coxeter Sets Throughout this section, the letter L stands for a set of involutions of S. Instead of L we shall write . In accordance with Section 3.4 we shall write, for each element s in L, S1 (s) instead of S1 (s, L). Recall that L is said to satisfy the exchange condition if, for any three elements h, k in L and s in S1 (k), h ∈ S1 (s) implies hs ⊆ sk ∪ S1 (k).

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We call L a Coxeter set if L is constrained and satisfies the exchange condition. We consider Theorem 3.6.4 and Theorem 3.6.6 to be the main results of this section. The equations given in these theorems are crucial for our approach to Coxeter sets. For the remainder of this section, we assume L to be a Coxeter set. Lemma 3.6.1 Let K be a subset of L, and let s be an element in K. Then (s) = K (s). Proof. Assume the claim to be false. Among the elements in K which do not satisfy the equation in question we choose s in such a way that K (s) is as small as possible. Since (s) = K (s), 1 = s. Thus, by Lemma 3.1.2, there exist elements h in K and r in K such that s ∈ hr and K (s) = 1 + K (r). / K. Thus, as s ∈ hr and h ∈ K, 1 = r. Thus, by Since (s) = K (s), s ∈ Lemma 3.1.2, there exist elements p in K and k in K such that r ∈ pk and K (r) = K (p) + 1. Now, by Lemma 3.4.2, there exists an element q in hp such that s ∈ qk, K (q) = 1 + K (p), and K (s) = K (q) + 1. Since K (s) = K (q) + 1, the (minimal) choice of s yields (q) = K (q). Similarly, as K (s) = 1 + K (r) and K (r) = K (p) + 1, (p) = K (p). Thus, as q ∈ hp and K (q) = 1 + K (p), h ∈ S1 (p). Similarly, one obtains p ∈ S1 (k). Thus, as L is assumed to satisfy the exchange condition, we must have hp = pk or hp ⊆ S1 (k). Since s ∈ hpk, the first of these two cases yields s ∈ pkk = {p} ∪ pk, contrary to K (s) = K (p) + 2. Since q ∈ hp, the second case yields q ∈ S1 (k). Thus, as s ∈ qk, (s) = (q) + 1. (Here we use the hypothesis that L is constrained.) Thus, as (q) = K (q) and K (s) = K (q)+1, (s) = K (s). This contradiction finishes the proof of the lemma. Lemma 3.6.2 For each subset K of L, we have the following. (i) The set K is a Coxeter set. (ii) We have K = L ∩ K. Proof. (i) This is an immediate consequence of Lemma 3.6.1. (ii) It is clear that K ⊆ L∩ K. Thus, we just have to show that L∩ K ⊆ K. In order to show that L ∩ K ⊆ K we fix an element l in L ∩ K. Since l ∈ L, (l) = 1. Since l ∈ K, (l) = K (l); cf. Lemma 3.6.1. From (l) = 1 and (l) = K (l) we obtain K (l) = 1, and that means that l ∈ K. We are assuming that L is a Coxeter set. In particular, L is constrained. Under this hypothesis, one can modify the exchange condition, and we shall do this in the following lemma.


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Formally, our modification is a consequence of the exchange condition. However, the exchange condition is obtained from our modification by setting q = 1. Lemma 3.6.3 Let h be an element in L, and let p, t be elements in L such that t ∈ hp and (t) = 1 + (p). Let k be an element in L, and let q, u be elements in L such that u ∈ qk and (u) = (q) + 1. Assume that t ∈ S1 (q) and p ∈ S1 (u). Then we have tq = pu or t ∈ S1 (u). Proof. Assume that t ∈ S1 (q). Then, there exists an element s in tq such that (s) = (t) + (q). Thus, as t ∈ hp and (t) = 1 + (p), there exists an element r in pq such that s ∈ hr, (r) = (p) + (q), and (s) = 1 + (r); cf. Lemma 3.4.2. From r ∈ pq and (r) = (p) + (q) we obtain {r} = pq. From s ∈ hr and (s) = 1 + (r) we obtain h ∈ S1 (r). Similarly, using {r} = pq, we conclude from p ∈ S1 (u) that r ∈ S1 (k). Thus, as L is assumed to satisfy the exchange condition, we may conclude that hr = rk or that hr ⊆ S1 (k). Since {t} = hp, {r} = pq, and {u} = qk, the first case yields tq = pu. Since s ∈ hr, the second case yields s ∈ S1 (k). Thus, by definition, there exists an element s in sk such that (s ) = (s) + 1. Since s ∈ sk and sk ⊆ hrk = hpqk = tu, s ∈ tu. Since (s ) = (s) + 1 and (s) + 1 = 1 + (r) + 1 = 1 + (p) + (q) + 1 = (t) + (u), (s ) = (t) + (u). Thus, t ∈ S1 (u). Recall that, for each element q in L, S−1 (q, L) is our notation for the set of all elements r in L such that there exists an element p in L with r ∈ pq and (r) = (p) + (q). In accordance with Section 3.4 we shall write S−1 (s) instead of S−1 (s, L). Recall that, for each subset R of L, S1 (R) is our notation for the intersection of the sets S1 (r) with r ∈ {1} ∪ R. Theorem 3.6.4 For each subset K of L, we have S1 ( K) = S1 (K). Proof. Let us assume that S1 ( K) = S1 (K). Then, as S1 ( K) ⊆ S1 (K), S1 (K) ⊆ S1 ( K). Among the elements in S1 (K) which are not in S1 ( K) we choose t such that (t) is as small as possible. Since t ∈ / S1 ( K) and 1 ∈ S1 ( K), 1 = t. Thus, by Lemma 3.1.2, there exist elements h in L and p in L such t ∈ hp and (t) = 1 + (p).

60


Since t ∈ hp and (t) = 1 + (p), t ∈ S−1 (p). Thus, as t ∈ S1 (K), p ∈ S1 (K); cf. Lemma 3.4.4(ii). Thus, as (t) = 1 + (p), the (minimal) choice of t yields p ∈ S1 ( K). / S1 (u). Since t ∈ / S1 ( K), there exists an element u in K such that t ∈ Among the elements u in K satisfying t ∈ / S1 (u) we choose u in such a way that (u) is minimal. Since t ∈ / S1 (u), 1 = u. Thus, by Lemma 3.1.2 together with Lemma 3.6.1, there exist elements q in K and k in K such that u ∈ qk and (u) = (q) + 1. Since (u) = (q) + 1 and q ∈ K, the (minimal) choice of u yields t ∈ S1 (q). Since p ∈ S1 ( K) and u ∈ K, p ∈ S1 (u). Thus, by Lemma 3.6.3, tq = pu or t ∈ S1 (u). Thus, by the choice of u, tq = pu. Thus, as q, u ∈ K, t ∈ p K. Thus, there exists an element s in K such that t ∈ ps. Since p ∈ S1 ( K) and s ∈ K, p ∈ S1 (s). Thus, as t ∈ ps, (t) = (p) + (s). (Here we use the hypothesis that L is constrained.) Since (t) = 1 + (p), this means that (s) = 1. Thus, by Lemma 3.6.1, s ∈ K. On the other hand, as / S1 (K), contrary to our t ∈ ps and (t) = (p) + (s), t ∈ S−1 (s). Thus, t ∈ choice of t. Lemma 3.6.5 Let l be an element in L, let p be an element in S1 (l), and let q be the element in pl. Then S−1 (p) ∩ S−1 (l) ⊆ S−1 (q). Proof. Let us define R to be the set of elements in S−1 (p) ∩ S−1 (l) which are not in S−1 (q). By way of contradiction, we assume that R is not empty. We fix an element r in R, and we do this in such a way that (r) is as small as possible. Since r ∈ R, r ∈ S−1 (p). Thus, by definition, there exists an element u in L such that r ∈ up and (r) = (u) + (p). Since r ∈ S−1 (l) and p ∈ S1 (l), r = p. Thus, as r ∈ up, 1 = u. Thus, by Lemma 3.1.2, there exist elements h in L and t in L such that u ∈ ht and (u) = 1 + (t). It follows that there exists an element s in tp such that r ∈ hs, (s) = (t) + (p), and (r) = 1 + (s); cf. Lemma 3.4.2. Since (r) = 1 + (s), (s) = (r) − 1. Thus, the minimal choice of r yields s∈ / R. Moreover, as s ∈ tp and (s) = (t) + (p), s ∈ S−1 (p). / R and s ∈ S−1 (p), s ∈ S−1 (q). On Suppose that s ∈ S−1 (l). Then, as s ∈ the other hand, as r ∈ hs and (r) = 1 + (s), r ∈ S−1 (s). Thus, by Lemma 3.4.4(iii), r ∈ S−1 (q), contrary to r ∈ R. This contradiction yields s ∈ / S−1 (l). Thus, by Lemma 3.4.7, s ∈ S1 (l). Thus, as L is assumed to satisfy the exchange condition, hs ⊆ sl ∪ S1 (l). Thus, as r ∈ hs, r ∈ sl ∪ S1 (l). Thus, as r ∈ S−1 (l), r ∈ sl ⊆ tpl = tq. It follows that, (r) ≤ (t) + (q) ≤ (t) + (p) + 1 = (s) + 1 = (r). This yields (r) = (t) + (q). Thus, as r ∈ tq, r ∈ S−1 (q), contrary to r ∈ R.


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Recall that, for each subset R of L, S−1 (R) is our notation for the intersection of the sets S−1 (r) with r ∈ {1} ∪ R. Theorem 3.6.6 For each subset K of L, we have S−1 ( K) = S−1 (K). Proof. Let us assume that S−1 ( K) = S−1 (K). Then, as S−1 ( K) ⊆ S−1 (K), S−1 (K) ⊆ S−1 ( K). Thus, there exists an element s in S−1 (K) such that s ∈ / S−1 ( K). / S−1 (q). Since s ∈ / S−1 ( K), there exists an element q in K such that s ∈ We pick an element q in K with s ∈ / S−1 (q), and we do this in such a way that K (q) is as small as possible. Since s ∈ / S−1 (q), 1 = q. Thus, Lemma 3.1.2 gives us elements p in K and k in K such that q ∈ pk and K (q) = K (p) + 1. Now, the minimal choice of q yields s ∈ S−1 (p). Thus, as s ∈ S−1 (K) ⊆ S−1 (k), s ∈ S−1 (q); cf. Lemma 3.6.5. This contradiction finishes our proof. Lemma 3.6.7 Let K be a subset of L, let p be an element in K, and let q be an element in S−1 (K). Then the following hold. (i) We have (p) ≤ (q). (ii) If (p) = (q), p = q. Proof. We are assuming that q ∈ S−1 (K). Thus, by Theorem 3.6.6, q ∈ S−1 ( K); cf. Theorem 3.6.6. Thus, as p is assumed to be an element in K, q ∈ S−1 (p). Thus, by definition, there exists an element s in L such that q ∈ sp and (q) = (s) + (p). (i) From (q) = (s) + (p) and 0 ≤ (s) we obtain (p) ≤ (q). (ii) Assume that (q) = (p). Then (s) = 0. Thus, 1 = s. Thus, as q ∈ sp, p = q. The set L is called spherical if S−1 (L) is not empty. The following lemma will be needed in the beginning of Section 12.1. Lemma 3.6.8 The following statements are equivalent. (a) The set L is spherical. (b) The set S−1 (L) has exactly one element. (c) The set of all integers (s) with s ∈ L has a maximal element. Proof. (a) ⇒ (b) Let us assume L to be spherical, and let us pick elements p and q in S−1 (L). We shall see that p = q. From Lemma 3.6.7(i) we obtain (p) = (q). Thus, by Lemma 3.6.7(ii), p = q.

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(a) ⇒ (c) Assume L to be spherical. Then, by definition, S−1 (L) is not empty. Let s be an element in S−1 (L). Then, for each element r in L, (r) ≤ (s); cf. Lemma 3.6.7(i). (c) ⇒ (a) Let us assume that there exists an element s in L such that, for each element r in L, (r) ≤ (s). We shall see that, for each element l in L, s ∈ S−1 (l). Let l be an element in L. Then, by Lemma 3.4.7, s ∈ S−1 (l) ∪ S1 (l). Assume that s ∈ S1 (l). Then, by definition, there exists an element t in sl such that (t) = (s) + 1. From s ∈ L and l ∈ L, we obtain t ∈ L, contrary to the choice of s.

4 Quotient Schemes

A subset R of S is called naturally valenced if each element of R has finite valency. The present chapter starts with the observation that naturally valenced schemes give rise to quotient schemes over finite closed subsets. After the definition of quotient schemes we shall always assume S to be naturally valenced. In the first section, we compute the structure constants of quotient schemes of S in terms of those of S. We relate the complex multiplication in S to the one in quotient schemes of S, and we look at the relationship between subschemes of quotient schemes and quotient schemes of subschemes. In the second section, we shall relate specific closed subsets of S containing a finite closed subset T to the corresponding closed subsets of the quotient scheme of S over T . Among other issues we focus on the relationship between commutators and quotient schemes. This leads naturally to the connection between the thin residue of S and the thin residue of quotient schemes of S. This relationship will be described in Theorem 4.2.8, a result which depends on Lemma 3.2.7. Theorem 4.2.8 turns out to be useful in Section 5.5 where we discuss residually thin schemes. In the third section, we assume S to have finite valency. We investigate the arithmetic between the structure constants of S and the structure constants of the quotient schemes of S. In the last two sections of this chapter, in which S is again assumed to have finite valency, we use the arithmetic of the structure constants of quotient schemes over closed subsets to investigate Hall subsets and Sylow subsets of S. We prove, for instance, that, if there exists a prime number p such that S is p-valenced, S possesses at least one closed subset with valency equal to the highest power of p dividing the valency of S. This is a generalization of the first of the famous theorems of Ludwig Sylow on finite groups. We also include corresponding generalizations of Sylow’s other two theorems on finite groups.

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4.1 Basic Definitions Let T be a closed subset of S. For each element s in S, we define sT := {(yT, zT ) | z ∈ yT sT }. Note that, for each element s in S, sT = {(yT, zT ) | z ∈ ys}; cf. Lemma 2.1.4. Lemma 4.1.1 Let p and q be elements in S, and let T be a closed subset of S. Then the conditions ∅ = pT ∩ q T , T pT = T qT , and pT = q T are pairwise equivalent. Proof. Let us first assume that pT ∩ q T is not empty. Then there exist elements y and z in X such that (yT, zT ) ∈ pT ∩ q T . From (yT, zT ) ∈ pT we obtain z ∈ yT pT . From (yT, zT ) ∈ q T we obtain z ∈ yT qT . Since z ∈ yT pT , there exists an element s in T pT such that z ∈ ys. Thus, as z ∈ yT qT , s ∈ T qT . It follows that s ∈ T pT ∩ T qT , so that, by Lemma 2.1.3, T pT = T qT . Let us now assume that T pT = T qT , and let us pick elements y and z in X. By definition, we have (yT, zT ) ∈ pT if and only if z ∈ yT pT . Similarly, we have (yT, zT ) ∈ q T if and only if z ∈ yT qT . Thus, as we are assuming that T pT = T qT , we have (yT, zT ) ∈ pT if and only if (yT, zT ) ∈ q T . Finally, as pT is not empty, pT = q T implies that pT ∩ q T is not empty. Recall that, for each closed subset T of S, X/T is our notation for the set of all sets xT with x ∈ X. Lemma 4.1.2 For each closed subset T of S, we have the following. (i) We have 1X/T = 1T . (ii) For each element s in S, (sT )∗ = (s∗ )T . Proof. (i) It is obvious that 1X/T ⊆ 1T . Conversely, let y and z be elements in X such that (yT, zT ) ∈ 1T . Then z ∈ yT , so that, by Lemma 2.1.4, yT = zT . This means that (yT, zT ) ∈ 1X/T . (ii) Let y and z be elements in X, and let s be an element in S. We have (yT, zT ) ∈ (sT )∗ if and only if (zT, yT ) ∈ sT , and (zT, yT ) ∈ sT means that y ∈ zT sT . By Lemma 1.3.2(iii), y ∈ zT sT is equivalent to z ∈ yT s∗ T . Finally, z ∈ yT s∗ T means that (yT, zT ) ∈ (s∗ )T . Let T be a closed subset of S. For each nonempty subset R of S, we define R//T := {rT | r ∈ R}.

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For any two nonempty subsets P and Q of S, we write P Q//T instead of (P Q)//T and P ∩ Q//T instead of (P ∩ Q)//T . Moreover, for R a nonempty subset of S and U a closed subset of S, we write R//T ∩U instead of R//(T ∩U ). For the remainder of this section, we shall now assume S to be naturally valenced. Assuming S to be naturally valenced, a closed subset of S is finite if and only if it has finite valency. Theorem 4.1.3 Let T be a finite closed subset of S. Then we have the following. (i) The set S//T is a scheme on X/T . (ii) For any three elements p, q, and r in S, we have auvr . apT qT rT nT = u∈T pT v∈T qT

(iii) For each element s in S, we have nsT nT = nT sT . Proof. (i), (ii) By Lemma 4.1.1, S//T is a partition of X/T × X/T . From Lemma 4.1.2(i) we know that 1X/T ∈ S//T . From Lemma 4.1.2(ii) we know that, for each element s in S, (sT )∗ ∈ S//T . Thus, in order to prove (i), is suffices to verify the regularity condition for S//T . We compute the structure constants explicitly, so that also (ii) will be proved. Let p, q, and r be elements in S, and let y and z be elements in X such that zT ∈ (yT )rT . Lemma 2.1.4 allows us to assume that z ∈ yr. Since we are assuming nT and np to be finite, nT pT is finite; cf. Lemma 2.3.3. Thus, yT pT is finite. Define I := yT pT ∩ zT q ∗ T . Then I is finite. Moreover, as z ∈ yr, auvr . |I| = u∈T pT v∈T qT

On the other hand, with the help of Lemma 4.1.2(ii) we obtain that, for each element x in X, x ∈ I if and only if xT ∈ (yT )pT ∩ (zT )(q T )∗ . Thus, auvr . |(yT )pT ∩ (zT )(q T )∗ |nT = u∈T pT v∈T qT

(iii) This follows from (ii) together with Lemma 1.1.3(i). For each finite closed subset T of S, we call S//T the quotient scheme of S over T . Assume that S is thin. It is easy to see that, in this case, Theorem 4.1.3(i) holds for each closed subset, not only for finite such sets. The proof is almost the same as the one of Theorem 4.1.3(i).

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From Theorem 4.1.3(iii) together with Lemma 2.3.3 we obtain that S//T is naturally valenced for each finite closed subset T of S. The following lemma relates the complex multiplication in S with the complex multiplication in quotient schemes of S over closed subsets of finite valency. Lemma 4.1.4 Let n be a positive integer, let s be an element in S, let R1 , . . . , Rn be nonempty subsets of S, and let T be a finite closed subset of S. Then we have sT ∈ (R1 //T ) · · · (Rn //T ) if and only if s ∈ (T R1 T ) · · · (T Rn T ). Proof. Let us first prove the statement for n = 1. By definition, we have sT ∈ R1 //T if and only if there exists an element r in R1 such that sT = rT . By Lemma 4.1.1, we have sT = rT if and only if T sT = T rT . Thus, we have sT ∈ R1 //T if and only if s ∈ T R1 T . Let us now assume that 2 ≤ n. We shall refer to the case n = 1. Assume first that sT ∈ (R1 //T ) · · · (Rn //T ). Then, by definition, there exist elements p and q in S such that pT ∈ (R1 //T ) · · · (Rn−1 //T ), q T ∈ Rn //T , and 1 ≤ apT qT sT . Since 1 ≤ apT qT sT , there exist elements u in T pT and v in T qT such that 1 ≤ auvs ; cf. Theorem 4.1.3(ii). Thus, by definition, s ∈ (T pT )(T qT ). Since pT ∈ (R1 //T ) · · · (Rn−1 //T ), induction allows us to assume that p ∈ (T R1 T ) · · · (T Rn−1 T ). Moreover, we saw earlier that q T ∈ Rn //T implies q ∈ T Rn T . Therefore, (T pT )(T qT ) ⊆ (T R1 T ) · · · (T Rn T ). Thus, as s ∈ (T pT )(T qT ), s ∈ (T R1 T ) · · · (T Rn T ). Conversely, let us assume that s ∈ (T R1 T ) · · · (T Rn T ). Then there exist elements p in (T R1 T ) · · · (T Rn−1 T ) and q in T Rn T such that s ∈ pq. Since s ∈ pq, 1 ≤ apqs . Thus, by Theorem 4.1.3(ii), 1 ≤ apT qT sT , and that means that sT ∈ pT q T . On the other hand, by induction, we obtain that pT ∈ (R1 //T ) · · · (Rn−1 //T ). Moreover, proof that q ∈ T Rn T is equivalent to q T ∈ (R1 //T ) · · · (Rn //T ). Thus, as sT ∈ pT q T , sT

from p ∈ (T R1 T ) · · · (T Rn−1 T ) we saw in the first part of this Rn //T . Thus, we have pT q T ⊆ ∈ (R1 //T ) · · · (Rn //T ).

Lemma 4.1.5 Let s be an element in S, and let T and U be closed subsets of S such that T ⊆ U . Then sT normalizes U//T if and only if U sT ⊆ T sU . Proof. Let r be an element in S. Then, as we are assuming that T ⊆ U , rT ∈ (U//T )sT if and only if r ∈ U sT ; cf. Lemma 4.1.4. Similarly, rT ∈ sT (U//T ) if and only if r ∈ T sU . Thus, we have (U//T )sT ⊆ sT (U//T ) if and only if U sT ⊆ T sU .

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Lemma 4.1.6 Let P and Q be nonempty subsets of S, and let T be a finite closed subset of S. Then P T Q//T = (P//T )(Q//T ). Proof. Let s be an element in S, and let us first assume that sT ∈ P T Q//T . Then, by definition, there exists an element r in P T Q such that sT = rT . From r ∈ P T Q we obtain r ∈ (T P T )(T QT ). Thus, by Lemma 4.1.4, rT ∈ (P//T )(Q//T ). Thus, as sT = rT , sT ∈ (P//T )(Q//T ). Let us now, conversely, assume that sT ∈ (P//T )(Q//T ). Then, by Lemma 4.1.4, s ∈ T P T QT . Thus, there exists an element r in P T Q such that s ∈ T rT . From r ∈ P T Q we obtain rT ∈ P T Q//T . From s ∈ T rT we obtain sT = rT ; cf. Lemma 4.1.1. Thus, we have that sT ∈ P T Q//T . Lemma 4.1.7 Let R be a nonempty subset of S, and let T be a finite closed subset of S. Then the following hold. (i) Assume that T RT ⊆ R. Then R//T is closed if and only if R is closed. (ii) Assume that R//T is closed, and set U := {s ∈ S | sT ∈ R//T }. Then U is closed, and R//T = U//T . Proof. (i) We are assuming that T RT ⊆ R. Thus, R∗ T R = R∗ R. Thus, by Lemma 4.1.6, R∗ R//T = (R∗ //T )(R//T ). Thus, if R is closed, R//T is closed. Conversely, if R//T is closed, the above equation yields R∗ R//T = R//T. Let s be an element in R∗ R. Then, as R∗ R//T = R//T , sT ∈ R//T . Thus, by Lemma 4.1.4, s ∈ T RT . Thus, as we are assuming that T RT ⊆ R, s ∈ R. Since s has been chosen arbitrarily in R∗ R, we have shown that R∗ R ⊆ R, and that means that R is closed. (ii) Let u be an element in U . Then, by definition, uT ∈ R//T . Thus, as R//T is assumed to be closed, we have that (uT )∗ ∈ R//T . Thus, by Lemma 4.1.2(ii), (u∗ )T ∈ R//T . Thus, by definition, u∗ ∈ U . Let p and q be elements in U , and let s be an element in pq. Since p ∈ U , pT ∈ R//T . Similarly, as q ∈ U , q T ∈ R//T . However, since s ∈ pq, sT ∈ pT q T ; cf. Lemma 4.1.4. Thus, as R//T is assumed to be closed, sT ∈ R//T . Thus, by definition, s ∈ U . That R//T = U//T follows right from the definition of U . The following result describes the relationship between quotient schemes of subschemes of S and subschemes of quotient schemes of S. Note that its first part generalizes Theorem 4.1.3(i).

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Theorem 4.1.8 Let x be an element in X, and let T and U be closed subsets of S such that T ⊆ U and T is finite. Then we have the following. (i) The set UxU //TxU is a scheme on xU/TxU . (ii) The set (U//T )(xT )(U//T ) is a scheme on (xT )(U//T ). (iii) We have xU/TxU = (xT )(U//T ) and UxU //TxU = (U//T )(xT )(U//T ) . Proof. (i) Since U is assumed to be closed, UxU is a scheme on xU ; cf. Theorem 2.1.8(ii). On the other hand, as T is assumed to be closed, TxU is closed; cf. Lemma 2.1.9(ii). Thus, by Theorem 4.1.3(i), UxU //TxU is a scheme on xU/TxU . (Note that, for each element t in U , ntxU is finite, and, as nT is assumed to be finite, nTxU is finite; cf. Theorem 2.1.8(iii).) (ii) According to Theorem 4.1.3(i), S//T is a scheme on X/T . On the other hand, as U is assumed to be closed, U//T is closed; cf. Lemma 4.1.7(i). Thus, the claim follows from Theorem 2.1.8(ii). (iii) Note that xU/TxU = {y(TxU ) | y ∈ xU } = {yT | y ∈ xU }. On the other hand, (xT )(U//T ) is the union of the sets (xT )uT with u ∈ U , and, for each element u in U , (xT )uT = {yT | y ∈ xT uT }. Thus, (xT )(U//T ) = {yT | y ∈ xU }, so that the first equation is proved. In order to establish the second equation, we first mention that, for each element u in U , (uxU )TxU = {(y(TxU ), z(TxU )) ∈ xU/TxU × xU/TxU | z ∈ yTxU uxU TxU } and (uT )(xT )(U//T ) = {(yT, zT ) ∈ (xT )(U//T ) × (xT )(U//T ) | z ∈ yT uT }. Thus, referring to Lemma 2.1.9(i) we obtain from the first equation that (uxU )TxU = (uT )(xT )(U//T ) . Thus, as UxU //TxU = {(uxU )TxU | u ∈ U } and (U//T )(xT )(U//T ) = {(uT )(xT )(U//T ) | u ∈ U }, we conclude that UxU //TxU = (U//T )(xT )(U//T ) . Let x be an element in X, and let T and U be closed subsets of S. Assume that T ⊆ U and that T has finite valency.

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In the following, we shall write (U//T )x to denote the scheme (U//T )(xT )(U//T ) . According to Theorem 4.1.8(iii), (U//T )x then also denotes the scheme UxU //TxU on xU/TxU . If {1} = T , U//{1} is identified with U . Thus, the previous remark implies that we shall write Ux to denote the scheme UxU on xU . If U = S, we shall write U//T instead of (U//T )x .

4.2 General Facts In this section, S is assumed to be naturally valenced. The letter T stands for a finite closed subset of S. Lemma 4.2.1 Let R be a nonempty subset of S, and let U be a closed subset of S such that T ⊆ U . Then the following hold. (i) We have R ∩ U//T = R//T ∩ U//T . (ii) We have R//T ⊆ U//T if and only if R ⊆ U . Proof. (i) By definition, we have R ∩ U//T ⊆ R//T ∩ U//T . Thus, it suffices to show that R//T ∩ U//T ⊆ R ∩ U//T . In order to prove this, we pick an element s in S such that sT ∈ R//T ∩ U//T . We shall be done if we succeed in showing that sT ∈ R ∩ U//T . Since sT ∈ R//T , there exists an element r in R such that sT = rT . Similarly, as sT ∈ U//T , there exists an element u in U such that sT = uT . It follows that rT = uT , so that, by Lemma 4.1.1, T rT = T uT . We are assuming that T ⊆ U . Thus, we obtain from T rT = T uT that r ∈ U . It follows that r ∈ R ∩ U , so that rT ∈ R ∩ U//T . Thus, as sT = rT , we conclude that sT ∈ R ∩ U//T . (ii) It is clear that, if R ⊆ U , R//T ⊆ U//T . Therefore, we assume, conversely, that R//T ⊆ U//T . In order to show that R ⊆ U , we fix an element r in R. Then, by definition, rT ∈ R//T . Thus, as we are assuming that R//T ⊆ U//T , rT ∈ U//T . Thus, there exists an element u in U such that rT = uT . From rT = uT we obtain T rT = T uT ; cf. Lemma 4.1.1. It follows that r ∈ T uT ⊆ U . Lemma 4.2.2 For each nonempty subset R of S, the following hold. (i) If T ⊆ R, R//T = R//T . (ii) Let U be a closed subset of S such that T ⊆ U and R//T = U//T . Then R ∪ T = U . Proof. (i) Without loss of generalization, we may assume that R∗ = R.

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Let s be an element in S, and let us assume that sT ∈ R//T . Then, by Lemma 3.1.1(i), there exists a non-negative integer n such that sT ∈ (R//T )n . Thus, by Lemma 4.1.4, s ∈ (T RT )n . Now, as we are assuming that T ⊆ R, s ∈ R; cf. Lemma 3.1.1(i). Thus, by definition, sT ∈ R//T . This proves that R//T ⊆ R//T . That R//T ⊆ R//T follows immediately from Lemma 4.1.6 together with Lemma 3.1.1(i). (ii) We are assuming that R//T = U//T . Thus, R ∪ T //T = U//T . Thus, by (i), R ∪ T //T = U//T , so that the claim follows from Lemma 4.2.1(ii). Lemma 4.2.3 For each nonempty subset R of S, we have [R//T, T //T ] = [R, T ]T //T . Proof. Let Q denote the union of the sets T r∗ T rT with r ∈ R. We shall show that [R//T, T //T ] = Q//T , so that our claim will follow from Lemma 4.2.2(i) together with Lemma 3.1.9(ii). Without loss of generalization, we may assume that R∗ = R. Let s be an element in S such that sT ∈ [R//T, T //T ]. Then, by Lemma 3.1.1(i), there exist elements r1 , . . . , rn in R such that sT ∈ (r1T )∗ r1T · · · (rnT )∗ rnT . Thus, for each element i in {1, . . . , n}, there exists an element qi ∈ S such that qiT ∈ (riT )∗ riT and sT ∈ q1T · · · qnT . Let i be an element in {1, . . . , n}. Then, as qiT ∈ (riT )∗ riT , qi ∈ T ri∗ T ri T ⊆ Q; cf. Lemma 4.1.4. Thus, by definition, qiT ∈ Q//T . Thus, as sT ∈ q1T · · · qnT , sT ∈ (Q//T )n . Thus, by Lemma 3.1.1(i), sT ∈ Q//T . Conversely, let s be an element in S such that sT ∈ Q//T . Then there exists a positive integer n such that sT ∈ (Q//T )n ; cf. Lemma 3.1.1(i). Thus, there exist elements q1 , . . . , qn in Q such that sT ∈ q1T · · · qnT . Let i be an element in {1, . . . , n}. Then, as qi ∈ Q, there exists an element ri in R such that qi ∈ T ri∗ T ri T . Thus, by Lemma 4.1.4, qiT ∈ (riT )∗ riT . Thus, as sT ∈ q1T · · · qnT , sT ∈ (r1T )∗ r1T · · · (rnT )∗ rnT ⊆ [R//T, T //T ], and that finishes the proof of the theorem. Recall that KS (T ) is our notation for the set of all elements s in S which satisfy s∗ T s ⊆ T .

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Lemma 4.2.4 Let U and V be closed subsets of S, and let us assume that T ⊆ U . Then the following hold. (i) If T is normal in S, NV //T (U//T ) = NV (U )//T . (ii) We have KV //T (U//T ) = KV (U )//T . Proof. (i) Let s be an element in S. Then, as T is assumed to be normal in S, we have that s∗ , s ∈ NS (T ). Thus, by Lemma 2.5.2(ii), T s = sT . Thus, as T ⊆ U , we have U s = U sT and T sU = sU . From Lemma 4.1.5 we know that sT normalizes U//T if and only if U sT ⊆ T sU . However, s ∈ NS (U ) means that U s ⊆ sU . Thus, as U s = U sT and T sU = sU , we have sT ∈ NS//T (U//T ) if and only if s ∈ NS (U ). So far, we have shown that NS//T (U//T ) = NS (U )//T . Now the claim follows from Lemma 4.2.1(i). (ii) Let s be an element in S. By definition, we have sT ∈ KS//T (U//T ) if and only if (sT )∗ (U//T )sT ⊆ U//T . Since we are assuming that T ⊆ U , we also have that s∗ U s//T = (sT )∗ (U//T )sT ; cf. Lemma 4.1.6. Thus, the latter condition is equivalent to s∗ U s//T ⊆ U//T . According to Lemma 4.2.1(ii), this is equivalent to s∗ U s ⊆ U . However, s∗ U s ⊆ U means that s ∈ KS (U ). Clearly, if s ∈ KS (U ), sT ∈ KS (U )//T . Conversely, if sT ∈ KS (U )//T , there exists an element r in KS (U ) such that sT = rT . From sT = rT we obtain s ∈ T rT ; cf. Lemma 4.1.1. Thus, as T ⊆ U , s ∈ U rU . Thus, as r ∈ KS (U ), s ∈ KS (U ); cf. Lemma 2.5.8(i). So far, we have shown that KS//T (U//T ) = KS (U )//T . Now the claim follows from Lemma 4.2.1(i). Let U be a closed subset of S such that T ⊆ U . Let V be a closed subset of S which normalizes T , and assume that U ⊆ V . Lemma 4.2.4(i) says, in particular, that U//T is normal in V //T if and only if U is normal in V . Recall that T is called strongly normal in U if U ⊆ KS (T ). Let V be closed subsets of S satisfying U ⊆ V . Lemma 4.2.4(ii) says, in particular, that U//T is strongly normal in V //T if and only if U is strongly normal in V . Lemma 4.2.5 Let U be a closed subset of S such that T ⊆ U . Then the following hold. (i) We have Oϑ (U//T ) = KU (T )//T .

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(ii) The subset T is strongly normal in U if and only if U//T is thin. Proof. (i) From Lemma 2.5.9(i) we know that Oϑ (U//T ) = KU//T (T //T ). Thus, the claim follows from Lemma 4.2.4(ii). (ii) By definition, T is strongly normal in U if and only if U ⊆ KU (T ). By (i) (together with Lemma 4.2.1(ii)), the latter condition is equivalent to U//T ⊆ Oϑ (U//T ), and that means that U//T is thin; cf. Lemma 1.5.1. Let us now look at a distinguished series of closed subsets of S. Let U be a closed subset of S such that T ⊆ U . Assume that U has finite valency. We set (KU )0 (T ) := T . For each positive integer n, we inductively define (KU )n (T ) := KU ((KU )n−1 (T )). Since U is assumed to have finite valency, we obtain from Lemma 2.5.8(ii) that (KU )n (T ) is closed for each non-negative integer n. From this we obtain (KU )n−1 (T ) ⊆ (KU )n (T ) for each positive integer n. The following theorem is similar to Lemma 4.2.4(ii). Theorem 4.2.6 Let m be a non-negative integer, let U be a closed subset of S such that T ⊆ U . Assume that U has finite valency, and set V := (KU )m (T ). Then, for each non-negative integer n, (KU//T )n (V //T ) = (KU )m+n (T )//T. Proof. The claim is obvious for n = 0. Therefore, we assume that 1 ≤ n. We set W := (KU )m+n−1 (T ). By induction, we may assume that the claim holds for n − 1. Thus, we have (KU//T )n (V //T ) = KU//T ((KU//T )n−1 (V //T )) = KU//T (W//T ). On the other hand, by Lemma 4.2.4(ii), KU//T (W//T ) = KU (W )//T = (KU )m+n (T )//T, and that finishes the proof. Let us now look how the thin residue works together with quotient schemes. Lemma 4.2.7 Let U be a closed subset of S such that T ⊆ U . Then the following hold. (i) If U//T is thin, Oϑ (U ) ⊆ T .

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(ii) We have Oϑ (U//T ) = [U, T ]T //T . (iii) We have Oϑ (U )T //T ⊆ Oϑ (U//T ). (iv) If T is normal in U , Oϑ (U )T //T = Oϑ (U//T ). Proof. (i) This follows from Lemma 4.2.5(ii) together with the definition of Oϑ (U ). (ii) From Theorem 3.2.1(ii) we know that Oϑ (U//T ) = [U//T, T //T ]. From Lemma 4.2.3 we know that [U//T, T //T ] = [U, T ]T //T . Thus, we have that Oϑ (U//T ) = [U, T ]T //T . (iii) From Theorem 3.2.1(ii) we know that Oϑ (U ) = [U, 1]. Thus, [U, 1] ⊆ [U, T ]. Thus, Oϑ (U )T //T = [U, T ]T //T, so that the claim follows from (ii). (iv) Assume that T is normal in U . Then, by Lemma 2.5.6(ii), Oϑ (U )T is strongly normal in U . Thus, by Lemma 4.2.4(ii), Oϑ (U )T //T is strongly normal in U//T . Thus, by definition, Oϑ (U//T ) ⊆ Oϑ (U )T //T, so that the claim follows from (iii). From Theorem 3.2.1(i) we know that Oϑ (T ) is strongly normal in T . Thus, by Lemma 4.2.5(ii), T //Oϑ (T ) is thin. Thus, we obtain from Lemma 4.2.7(i) that the thin residue Oϑ (T ) of T is the uniquely defined smallest closed subset of S having a thin quotient scheme. From Lemma 4.2.7(ii) we obtain, in particular, that Oϑ (T ) = [T, 1]. Thus, Lemma 4.2.7(ii) is a generalization of Theorem 3.2.1(ii). Theorem 4.2.8 Let n be a non-negative integer, and let U be a closed subset of S such that T ⊆ NS (U ). Then we have (Oϑ )n (T U )U//U = (Oϑ )n (T U//U ). Proof. Let T be a counterexample such that nT is as small as possible. Then 1 ≤ n. We are assuming that T ⊆ NS (U ). Thus, by Lemma 2.5.1(ii), U is normal in T U . Thus, by Lemma 4.2.7(iv), Oϑ (T U )U//U = Oϑ (T U//U ). Let us set V := Oϑ (T U )U. Then U ⊆ V . Thus, by Lemma 2.2.1(ii), (T ∩ V )U = T U ∩ V . Thus, as V ⊆ T U , (T ∩ V )U = V .

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Let us first assume that V = T U . Then T ∩ V = T . Thus, the minimal choice of T yields (Oϑ )n−1 (V )U//U = (Oϑ )n−1 (V //U ). (We write V instead of (T ∩ V )U .) Thus, as V //U = Oϑ (T U//U ), (Oϑ )n−1 (V )U//U = (Oϑ )n−1 (Oϑ (T U//U )) = (Oϑ )n (T U//U ). Applying Lemma 3.2.7(i) to Oϑ (T U ) in place of T , we also obtain (Oϑ )n−1 (Oϑ (T U ))U = (Oϑ )n−1 (Oϑ (T U )U )U. Moreover, the left hand side of this equation is equal to (Oϑ )n (T U )U , and the right hand side is equal to (Oϑ )n−1 (V )U , so that (Oϑ )n (T U )U//U = (Oϑ )n−1 (V )U//U. From this, together with (Oϑ )n−1 (V )U//U = (Oϑ )n (T U//U ), we obtain (Oϑ )n (T U )U//U = (Oϑ )n (T U//U ), contrary to the choice of T . Let us now assume that V = T U . Then, as V = Oϑ (T U )U , Oϑ (T U )U = T U . Thus, by Lemma 3.2.7(ii), (Oϑ )n (T U )U = T U . It follows that (Oϑ )n (T U )U//U = T U//U. However, from V = T U and V //U = Oϑ (T U//U ) we also obtain (Oϑ )n (T U//U ) = T U//U, so that (Oϑ )n (T U )U//U = (Oϑ )n (T U//U ). Again, this contradicts our choice of T . Corollary 4.2.9 Let n be a non-negative integer, and let U be a closed subset of S such that U ⊆ (Oϑ )n (T ). Then we have (Oϑ )n (T )//U = (Oϑ )n (T //U ). Proof. Since we are assuming that U ⊆ (Oϑ )n (T ), U ⊆ T . Thus, T = T U . Thus, (Oϑ )n (T U )U = (Oϑ )n (T ), so that the claim follows from Theorem 4.2.8.

4.3 Valencies In this section, S is assumed to have finite valency. The letter T stands for a closed subset of S. Assuming S to have finite valency, we may speak about the quotient scheme of S over T . We also may apply all results of the previous two sections.

4.3 Valencies

75

Lemma 4.3.1 For each element s in S, we have the following. (i) The integer nsT divides nT ns . (ii) If s∗ , s ∈ NS (T ), nsT divides ns . (iii) If T is thin, ns divides nsT nT . Proof. (i) From Theorem 4.1.3(iii) we know that nsT nT = nT sT , and from Lemma 2.3.3 that nT sT divides nT ns nT . Thus, nsT divides nT ns . (ii) Assume that s∗ , s ∈ NS (T ). Then, by Lemma 2.5.2(ii), T s = sT . From this we obtain T sT = sT . Thus, by Theorem 4.1.3(iii), nsT nT = nsT . It follows that nsT = (nT )−1 nsT , so that the claim follows from Lemma 2.3.4(iii). (iii) Let r be an element in T sT . Then there exist elements p and q in T such that r ∈ psq ∗ . Since r ∈ psq ∗ , nr ≤ npsq∗ . On the other hand, as p and q are thin, {1} = p∗ p and {1} = q ∗ q. Thus, {s} = p∗ psq ∗ q. Thus, by Lemma 1.4.5(ii), ns = npsq∗ . From nr ≤ npsq∗ and ns = npsq∗ we obtain nr ≤ ns . Similarly, we obtain ns ≤ nr , so that we have nr = ns . Thus, as r has been chosen arbitrarily in T sT , we have nT sT = |T sT |ns . In particular, ns divides nT sT , so that the claim follows from Theorem 4.1.3(iii). Assume T to be normal in S, and let s be an element in S. Then, by Lemma 4.3.1(ii), nsT divides ns . In particular, S//T is thin if S is thin. Let π be a set of prime numbers. Recall that a positive integer n is called a π-number if each prime divisor of n is element of π. A nonempty subset R of S is called π-valenced if, for each element r in R, nr is a π-number. Note that, for each prime number p, a {p}-valenced nonempty subset of S is the same as a p-valenced nonempty subset of S. Corollary 4.3.2 Let π be a set of prime numbers, and assume that nT is a π-number. Let U be a closed subset of S such that T ⊆ U . Then we have the following. (i) If U is π-valenced, so is U//T . (ii) Assume T to be thin. Then, if U//T is π-valenced, so is U . Proof. (i) This follows from Lemma 4.3.1(i). (ii) This follows from Lemma 4.3.1(iii). Lemma 4.3.3 Let U be a closed subset S such that T ⊆ U . Then the following hold. (i) We have nU = nU//T nT . (ii) Let V be a closed subset of S with U ⊆ V . Then nV //T = nV //U nU//T .

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Proof. (i) Let R be a transversal of (T, T ) in U . Then, by Lemma 2.1.3, {T rT | r ∈ R} is a partition of S. From this we obtain nU = nT rT r∈R

and, referring to Lemma 4.1.1, that U//T = {rT | r ∈ R}. This last equation implies nU//T = nrT . r∈R

Now the desired equation follows from the fact that, for each element r in R, nrT nT = nT rT ; cf. Theorem 4.1.3(iii). (ii) Applying (i) three times we obtain nV //T nT nU = nV nU = nV //U nU nU//T nT , and that proves (iii). Lemma 4.3.4 Let U be a closed subset of S such that T U is closed. Then nT //T ∩U = nT U//U . Proof. From Lemma 4.3.3(i) we know that nT = nT //T ∩U nT ∩U and that nT U = nT U//U nU . From Lemma 2.3.6(i), we know that nT nU = nT U nT ∩U . Thus, nT U//U = nT //T ∩U . Let π be a set of prime numbers. We shall denote by π the set of all prime numbers not contained in π. Lemma 4.3.5 Let π be a set of prime numbers such that nT is a π-number and nS//T a π -number. Let s be a π-valenced element in S. Then the following hold. (i) We have {1T } = sT (sT )∗ ∩ Oϑ (S//T ). (ii) We have ss∗ ∩ KS (T ) ⊆ T . Proof. (i) From Lemma 2.5.9(iii) we know that Oϑ (S//T ) is closed. Thus, by Corollary 2.6.6, {1T }(s

T ∗

)

∩ Oϑ (S//T ) = sT (sT )∗ ∩ Oϑ (S//T ).

We are assuming that nT and ns are π-numbers. Thus, by Lemma 4.3.1(i), nsT is a π-number, too. From this we obtain that n(sT )∗ is a π-number; cf. Lemma 1.1.2(iii). Now recall that, by Lemma 2.3.7(ii), n{1T }(sT )∗ divides n(sT )∗ . Thus, n{1T }(sT )∗ is a π-number. We are assuming that nS//T is a π -number. Thus, as Oϑ (S//T ) is closed, nOϑ (S//T ) must be a π -number, too; cf. Lemma 2.3.6(ii).

4.4 Hall Subsets

77

Now, as n{1T }(sT )∗ is a π-number and nOϑ (S//T ) is a π -number, {1T } = {1T }(s

T ∗

)

∩ Oϑ (S//T ).

(ii) Let r be an element in ss∗ ∩KS (T ). From r ∈ ss∗ we obtain rT ∈ sT (sT )∗ ; cf. Lemma 4.1.4 (together with Lemma 4.1.2(ii)). From r ∈ KS (T ) we obtain rT ∈ KS (T )//T . Thus, by Lemma 4.2.5(i), rT ∈ Oϑ (S//T ). On the other hand, we know from (i) that {1T } = sT (sT )∗ ∩ Oϑ (S//T ), so that rT = 1T . Thus, by Lemma 4.1.1, r ∈ T .

4.4 Hall Subsets In this section, S is assumed to have finite valency. Recall that, for each set π of prime numbers, π is our notation for the set of all prime numbers not in π. A nonempty π-valenced subset R of S will be called a π-subset of S if nR is a π-number. Let T and U be closed subsets of S such that T ⊆ U . Assume that T is a π-subset of S. We call T a Hall π-subset of U if nU//T is a π -number. Lemma 4.4.1 Let π be a set of prime numbers, and let T , U , and V be closed subsets of S such that T is a Hall π-subset of V , U ⊆ V , and T U = U T . Then T ∩ U is a Hall π-subset of U . Proof. As a subset of T , the set T ∩ U is a closed π-subset of U ; cf. Lemma 2.3.6(ii). Thus, we just have to show that nU//T ∩U is a π -number. Since T U = U T , T U is closed; cf. Lemma 2.1.1. Thus, by Lemma 4.3.3(ii), nT U//T divides nV //T . However, as T is assumed to be a Hall π-subset of V , nV //T is a π -number. Thus, nT U//T is a π -number. Recall, finally, that nU//T ∩U = nT U//T ; cf. Lemma 4.3.4. Thus, nU//T ∩U is a π -number. Lemma 4.4.1 says, in particular, that, whenever a closed subset T of S contains a Hall π-subset of S, then each Hall π-subset of T is a Hall π-subset of S. Lemma 4.4.2 Let π be a set of prime numbers, and let T , U , and V be closed subsets of S. Assume that T is a Hall π-subset of V , that U ⊆ V , and that T ⊆ NS (U ). Then T U//U is a Hall π-subset of V //U . Proof. From Lemma 4.3.3(i) we know that nT = nT //T ∩U nT ∩U . Thus, as nT is assumed to be a π-number, nT //T ∩U is a π-number. Moreover, we are assuming that T ⊆ NS (U ). Thus, by Lemma 2.5.2(iii), T U is a closed subset of S. Thus, by Lemma 4.3.4, nT //T ∩U = nT U//U , so that nT U//U is a π-number.

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4 Quotient Schemes

By Lemma 4.3.3(ii), nV //T U divides nV //T . Moreover, as T is assumed to be a Hall π-subset of V , nV //T is a π -number. Thus, nV //T U is a π -number. It remains to prove that T U//U is π-valenced. In order to do so we pick an element s in T U . We shall see that nsU is a π-number. Since s ∈ T U , there exists an element t in T such that s ∈ tU . From s ∈ tU we obtain U sU = U tU ; cf. Lemma 2.1.3. Thus, by Lemma 4.1.1, sU = tU . Since t is an element in T and T is assumed to be a Hall π-subset of V , nt is a π-number. On the other hand, as t∗ , t ∈ T ⊆ NS (U ), ntU divides nt ; cf. Lemma 4.3.1(ii). Thus, ntU is a π-number. Thus, as sU = tU , nsU is a π-number.

4.5 Sylow Subsets In this section, S is assumed to have finite valency. We fix a prime number and call it p. For each element s in S, we write s instead of {s}. The following lemma on finite groups is commonly referred to Augustin-Louis Cauchy. Lemma 4.5.1 Let T be a thin closed subset of S such that p divides nT . Then T contains a closed subset of valency p. Proof. Assume T to be a minimal counterexample. Then T does not contain closed subsets different from T the valency of which is divisible by p. Let t be an element in T . Then CT (t) is closed. Thus, if CT (t) = T , p does not divide the valency of CT (t). Thus, p divides nT //CT (t) . Let us denote by Z(T ) the intersection of all closed subsets CT (t) with t ∈ T . Then Z(T ) is closed and p divides the valency of Z(T ). Thus, Z(T ) = T . Let U be a closed subsets of T such that {1} = U = T . Then, by Lemma 4.3.3(i), p divides nT //U . Thus, as {1} = U , the minimal choice of T yields a closed subset V of S such that U ⊆ V and nV //U = p. Let t be an element in V \ U . Then t has valency p. Let T be a closed subset of S. Note that a {p}-subset of T is the same as a p-subset of T . We also speak about Sylow p-subsets of T instead of Hall {p}-subsets of T .1 Our notation for the set of all Sylow p-subsets of T will be Sylp (T ). Sylow p-subsets of a closed subset T of S are particularly interesting if T is p-valenced. 1

This is just for historical reasons.

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79

Proposition 4.5.2 Let V be a p-valenced closed subset of S, and let T be closed p-subset of V such that T ∈ / Sylp (V ). Then there exists a closed p-subset U of V such that T ⊆ U ⊆ KV (T ) and pnT = nU . Proof. We are assuming that S is p-valenced and that T is a closed p-subset of V . Thus, by Corollary 4.3.2(i), V //T is p-valenced. We are assuming that T ∈ / Sylp (V ). Thus, as T is a closed p-subset of V , p divides nV //T . Thus, as V //T is p-valenced, p divides nOϑ (V //T ) . Thus, by Lemma 4.5.1, there exists a closed subset U of V such that T ⊆ U , U//T ⊆ Oϑ (V //T ) and nU//T = p. From U//T ⊆ Oϑ (V //T ) we obtain U ⊆ KV (T ); cf. Lemma 4.2.5(i) (together with Lemma 4.2.1(ii)). From nU//T = p and nU = nU//T nT we obtain pnT = nU . The following three theorems were first proved in [24]. A more general approach is given in [4]. The thin case was already proved in 1872 by Ludwig Sylow; cf. [36]. (Note that thin schemes are p-valenced.) Theorem 4.5.3 Each p-valenced closed subset of S possesses at least one Sylow p-subset. Proof. Let T be a p-valenced closed subset of S. Then, Proposition 4.5.2 says that, for each power q of p which divides the valency of T , T possesses a closed p-subset of valency q. In particular, T possesses a Sylow p-subset. Proposition 4.5.4 Let V be a p-valenced closed subset of S, let T be a closed p-subset of V , and let U be a Sylow p-subset of V . Then there exists an element s in V such that s∗ T s ⊆ U . Proof. Let R be a transversal of (T, U ) in V . Then, by Lemma 2.1.3, nV = nT rU . r∈R

Now we are assuming that nT and nU are powers of p. Moreover, for each element r in R, nr is assumed to be a power of p. Thus, for each element r in R, nT rU is a power of p; cf. Lemma 2.3.3. On the other hand, for each element r in R, nU ≤ nT rU ; cf. Lemma 1.4.4(i). Thus, as nU is the highest power of p dividing nV , there exists an element s in R such that nU = nT sU . From nU = nT sU we obtain s∗ T s ⊆ U ; cf. Lemma 1.4.4(ii). Theorem 4.5.5 Let U be a p-valenced closed subset of S, and let T be a Sylow p-subset of U . Then Sylp (U ) = {u∗ T u | u ∈ U, uu∗ ⊆ T }.

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Proof. Let T be a Sylow p-subset of U . Then, by definition, nT = nT . Moreover, by Proposition 4.5.4, there exists an element u in U such that u∗ T u ⊆ T . Thus, nu∗ T u ≤ nT . On the other hand, we know from Lemma 1.4.5(i) that nT ≤ nT u ≤ nu∗ T u . Thus, we have nu∗ T u = nT and nT = nT u . From nu∗ T u = nT and u∗ T u ⊆ T we obtain u∗ T u = T . From nT = nT u we obtain uu∗ ⊆ T ; cf. Lemma 1.4.5(ii). Conversely, let u be an element in U such that uu∗ ⊆ T . Then, by Lemma 2.6.1(ii), u∗ T u is a closed subset of U . Moreover, by Lemma 2.6.2(i), nT = nu∗ T u . Thus, as U is assumed to be p-valenced, u∗ T u ∈ Sylp (U ). Let U be a p-valenced closed subset of S, and let T be a Sylow p-subset of U . From Theorem 4.5.5 together with Lemma 2.6.1(i) we obtain Sylp (U ) = {T s | u ∈ U, uu∗ ⊆ T }. Proposition 4.5.6 Let V be a p-valenced closed subset of S, let T be a Sylow p-subset of V , and set U := KV (T ). Then the following hold. (i) We have |Sylp (V )| = |{U s | s ∈ V, ss∗ ⊆ T }|. (ii) The prime number p divides nV //U − 1. Proof. (i) From Lemma 2.6.3(ii) we know that |{s∗ T s | s ∈ V, ss∗ ⊆ T }| = |{U s | s ∈ V, ss∗ ⊆ T }|, so that our claim is a consequence of Theorem 4.5.5. (ii) We are assuming that V is p-valenced. Thus, as nT is a power of p, V //T is p-valenced; cf. Corollary 4.3.2(i). On the other hand, we know from Lemma 4.2.5(i) that Oϑ (V //T ) = U//T . Thus, p divides nV //T − nU//T . Thus, as nV //T = nV //U nU//T , p divides (nV //U − 1)nU//T . Since we are assuming T to be a Sylow p-subset of V , p does not divide nV //T . Thus, as nU//T divides nV //T , p does not divide nU//T . Thus, as p divides (nV //U − 1)nU//T , p divides nV //U − 1. According to Theorem 4.5.3, each p-valenced closed subset of S possesses at least one Sylow p-subset. Generalizing this theorem we shall now say a little bit more about the number of Sylow p-subsets of such a closed subset of S. Theorem 4.5.7 The number of Sylow p-subsets of a p-valenced closed subset of S is congruent to 1 modulo p. Proof. Let V be a p-valenced closed subset of S, let T be a Sylow p-subset of V , and set U := KV (T ). We fix a right transversal R of U in V . Then nV is the sum of all integers nU r with r ∈ R. Thus, as nV = nV //U nU , nV //U = (nU )−1 nU r . r∈R

4.5 Sylow Subsets

81

Now recall that, for each element r in R, (nU )−1 nr∗ U divides nr ; cf. Lemma 2.3.4(iii). Note also that, for each element r in R, nr∗ U = nU r ; cf. Lemma 1.1.2(iii) and Lemma 1.3.2(iii). Thus, for each element r in R, (nU )−1 nU r divides nr . Thus, as V is assumed to be p-valenced, we conclude that, for each element r in R, (nU )−1 nU r is a power of p. Let us compute the number of elements r in R such that (nU )−1 nU r = 1. From Lemma 1.4.5(ii) we know that, for each element s in V , nU = nU s if and only if ss∗ ⊆ U . Moreover, as V is assumed to be p-valenced, we obtain from Lemma 4.3.5(ii) that, for each element s in V , ss∗ ⊆ U if and only if ss∗ ⊆ T . Thus, by the above equation, p divides nV //U − |{U s | s ∈ V, ss∗ ⊆ T }|. Now our claim follows from Proposition 4.5.6. Let R be a subset of S such that, for each element r in R, nr = 2. Then, by Corollary 3.1.7, the set R is 2-valenced. Thus, the three previous theorems apply to R. For thin schemes, the following lemma is known as a result due to William Burnside. Lemma 4.5.8 Let s be an element in S, let T be a Sylow p-subset of S, and let R be a nonempty subset of T such that ss∗ ⊆ T ⊆ KS (R), and T ⊆ KS (s∗ Rs). Then there exists an element r in KS (T ) such that r∗ Rr ⊆ s∗ Rs. Proof. We are assuming that ss∗ ⊆ T and that T ∈ Sylp (S). Thus, by Theorem 4.5.5, s∗ T s ∈ Sylp (S). Now we have T , s∗ T s ∈ Sylp (KS (s∗ Rs)).2 Thus, by Proposition 4.5.4, there exists an element q in KS (s∗ Rs) such that q ∗ T q ⊆ s∗ T s. From q ∗ T q ⊆ s∗ T s we obtain sq ∗ T qs∗ ⊆ ss∗ T ss∗ ⊆ T. Thus, qs∗ ⊆ KS (T ). Thus, as KS (T ) is closed, sq ∗ ⊆ KS (T ). From q ∈ KS (s∗ Rs) we obtain q ∗ ∈ KS (s∗ Rs). Thus, qs∗ Rsq ∗ ⊆ s∗ Rs. Thus, for each element r in sq ∗ , we have r∗ Rr ⊆ s∗ Rs.

2

Note that, by Lemma 2.5.8(ii), KS (s∗ Rs) is a closed subset of S.

5 Morphisms

˜ be a set, let S˜ be a scheme on X, ˜ let φ be a map from X ∪ S to X ˜ ∪ S, ˜ Let X ˜ ˜ and assume that Xφ ⊆ X and that Sφ ⊆ S. The map φ is called a morphism (or a scheme morphism) if, for any two elements x in X and s in S, (xs)φ ⊆ xφsφ. A morphism φ from X ∪ S will be called a homomorphism (or a scheme homomorphism) if, for any three elements y, z in X and s in S with zφ ∈ yφsφ, there exist elements v in X and w in vs such that vφ = yφ and wφ = zφ. In the first section of this chapter, we compile the most general facts about morphisms and homomorphisms. A bijective homomorphism is called an isomorphism (or a scheme isomorphism). An isomorphism from X ∪ S to X ∪ S is called an automorphism (or a scheme automorphism) of S. The set of all automorphisms of S will be denoted by Aut(S). In the second section, we shall see that the set Aut(S) is a group with respect to composition. We also consider subgroups of Aut(S) and groups which contain Aut(S) as a subgroup. In the third section, we shall prove a Homomorphism Theorem and two Isomorphism Theorems for schemes of finite valency. All three of these results naturally generalize the finite versions of Emmy Noether’s corresponding theorems for groups. In Section 5.4, we define composition series of schemes having finite valency. The main result will be a generalization of a group theoretic theorem due to Camille Jordan and Otto H¨ older. This generalization says that any two composition series of a closed subset of a scheme of finite valency are isomorphic. Composition series lead naturally to the notion of a composition factor, and, according to Lemma 4.2.4(i), composition factors must be simple. Thus, the above theorem on composition series gives reason to consider simple schemes as crucial in scheme theory.

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5 Morphisms

The notion of a homomorphism enables us to establish the group correspondence which has been mentioned in the beginning of the preface. The group correspondence is the content of Section 5.5. In the last of the six sections of this chapter, we look at schemes which have finite valency and only thin composition factors.

5.1 Basic Facts ˜ be a set, let S˜ be a scheme on X, ˜ let φ be a map from X ∪ S to X ˜ ∪ S, ˜ Let X ˜ and that Sφ ⊆ S. ˜ Note that φ is a morphism if and and assume that Xφ ⊆ X only if, for any three elements y in X, s in S, and z in ys, zφ ∈ yφsφ. Lemma 5.1.1 Let φ be a morphism from X ∪ S. Then we have the following. (i) For each element s in S, s∗ φ = (sφ)∗ . (ii) For any two elements p and q in S, (pq)φ ⊆ pφqφ. pq˜)φ−1 . (iii) For any two elements p˜ and q˜ in Sφ, p˜φ−1 q˜φ−1 ⊆ (˜ Proof. (i) Let s be an element in S. In order to show that s∗ φ = (sφ)∗ , it suffices to show that s∗ φ ∩ (sφ)∗ is not empty. Let y be an element in X, and let z be an element in ys. From z ∈ ys we obtain y ∈ zs∗ . Thus, as φ is assumed to be a morphism, yφ ∈ zφs∗ φ. On the other hand, z ∈ ys yields zφ ∈ yφsφ, and that means that yφ ∈ zφ(sφ)∗ . (ii) Let p and q be elements in S, and let s be an element in pq. We have to show that sφ ∈ pφqφ. Let y be an element in X, and let z be an element in ys. Since z ∈ ys and s ∈ pq, z ∈ ypq. Thus, there exists an element x in yp such that z ∈ xq. From x ∈ yp we obtain xφ ∈ yφpφ. From z ∈ xq we obtain zφ ∈ xφqφ. It follows that zφ ∈ yφpφqφ. On the other hand, as z ∈ ys, zφ ∈ yφsφ. Thus, sφ ∈ pφqφ. (iii) Let p˜ and q˜ be elements in Sφ, and let s be an element in p˜φ−1 q˜φ−1 . We have to show that s ∈ (˜ pq˜)φ−1 . Since s ∈ p˜φ−1 q˜φ−1 , there exist elements t in p˜φ−1 and u in q˜φ−1 such that s ∈ tu. Thus, by (ii), sφ ∈ (tu)φ ⊆ tφuφ = p˜q˜, whence s ∈ (˜ pq˜)φ−1 . Let φ be a morphism from X ∪ S, and let R be a subset of S. Considering Lemma 3.1.1(i) we obtain from Lemma 5.1.1(i), (ii) that Rφ ⊆ Rφ. ˜ be a set, let S˜ be a scheme on X, ˜ and let φ be a Lemma 5.1.2 Let X ˜ ∪ S. ˜ Then the following hold. morphism from X ∪ S to X

5.1 Basic Facts

85

(i) We have 1φ = 1X˜ . ˜ Then T˜φ−1 is a closed subset of S. (ii) Let T˜ be a closed subset of S. Proof. (i) Let x be an element in X. Then, as φ is assumed to be a morphism, ˜ xφ ∈ xφ1 ˜ . Thus, 1φ ∩ 1 ˜ is not xφ ∈ xφ1φ. On the other hand, as xφ ∈ X, X X empty, so that 1φ = 1X˜ . (ii) We are assuming that T˜ is closed. Thus, 1X˜ ∈ T˜. Thus, by (i), 1φ ∈ T˜. It follows that 1 ∈ T˜φ−1 . In particular, T˜φ−1 is not empty. Let p and q be elements in T˜φ−1 , and let s be an element in p∗ q. Then, referring to Lemma 5.1.1(ii), (i) we obtain sφ ∈ (p∗ q)φ ⊆ p∗ φqφ = (pφ)∗ qφ ⊆ T˜∗ T˜ ⊆ T˜. It follows that s ∈ T˜φ−1 . ˜ be a set, and let S˜ be a scheme on X. ˜ For each morphism φ from X ∪ S Let X ˜ ∪ S, ˜ we set φX := φ ∩ (X × X). ˜ to X Lemma 5.1.3 Let φ be a morphism from X ∪ S, and assume φX to be surjective. Then the following hold. (i) For any two elements p˜ and q˜ in Sφ, (˜ pq˜)φ−1 = p˜φ−1 q˜φ−1 . (ii) The map φ is surjective. Proof. (i) Let p˜ and q˜ be elements in Sφ, and let s be an element in (˜ pq˜)φ−1 . −1 −1 We shall show that s ∈ p˜φ q˜φ . The claim will then follow from Lemma 5.1.1(iii). Let y be an element in X, and let z be an element in ys. Then zφ ∈ yφsφ ⊆ yφ˜ pq˜. Thus, there exists an element x ˜ in yφ˜ p such that zφ ∈ x ˜q˜. Since φX is assumed to be surjective, there exists an element x in X such that xφ = x ˜. Let p be the element in S which satisfies x ∈ yp, and let q be the element in S with z ∈ xq. From x ∈ yp we obtain xφ ∈ yφpφ, from z ∈ xq we obtain zφ ∈ xφqφ. From xφ = x ˜, xφ ∈ yφpφ, and x ˜ ∈ yφ˜ p we obtain pφ = p˜. Similarly, we obtain from xφ = x ˜, zφ ∈ xφqφ, and zφ ∈ x ˜q˜ that qφ = q˜. Finally, as z ∈ xq and x ∈ yp, z ∈ ypq. Thus, as z ∈ ys, s ∈ pq ⊆ p˜φ−1 q˜φ−1 . ˜ the codomain of φ and by S˜ the codomain of φ. (ii) Let us denote by X ˜ We shall show In order to show that φ is surjective, we fix an element s˜ in S. that there exists an element s in S such that sφ = s˜. ˜ there exist elements y˜ and z˜ in X ˜ such that z˜ ∈ y˜s˜. Since φX is Since s˜ ∈ S, assumed to be surjective, there exist elements y and z in X such that yφ = y˜ and zφ = z˜.

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Let s be the uniquely determined element in S such that z ∈ ys. Then, as φ is assumed to be a morphism, zφ ∈ yφsφ. Thus, as yφ = y˜ and zφ = z˜, z˜ ∈ y˜sφ. Thus, as z˜ ∈ y˜s˜, sφ = s˜. Let φ be a morphism from X ∪ S. We define ker(φ) := {s ∈ S | sφ = 1φ}. The set ker(φ) is called the kernel of φ. From Lemma 5.1.2(ii) one obtains that ker(φ) is closed for each morphism φ from X ∪ S. Lemma 5.1.4 Let φ be a morphism from X ∪ S, and set T := ker(φ). Then we have the following. (i) For any two elements y and z in X, we have yT = zT if and only if yφ = zφ. (ii) Let p and q be elements in S such that pT = q T . Then pφ = qφ. Proof. (i) Let y and z be elements in X, and let us write s to denote the element in S which satisfies z ∈ ys. Since φ is assumed to be a morphism, we obtain from z ∈ ys that zφ ∈ yφsφ. Since z ∈ ys, we have yT = zT if and only if s ∈ T ; cf. Lemma 2.1.4. Since T stands for the kernel of φ, s ∈ T is equivalent to sφ = 1φ. From zφ ∈ yφsφ we obtain that sφ = 1φ if and only if yφ = zφ; cf. Lemma 5.1.2(i). (ii) We are assuming that pT = q T . Thus, by Lemma 4.1.1, T pT = T qT . In particular, q ∈ T pT . Since q is not empty, there exist elements y and z in X such that z ∈ yq. Thus, as q ∈ T pT , z ∈ yT pT . Thus, there exists an element v in yT such that z ∈ wpT . Since z ∈ vpT , there exists an element w in vp such that z ∈ wT . From y ∈ vT we obtain vT = yT ; from z ∈ wT we obtain wT = zT . Since φ is assumed to be a morphism, we obtain from w ∈ vp that wφ ∈ vφpφ. Similarly, we obtain from z ∈ yq that zφ ∈ yφqφ. Since vT = yT , we obtain from (i) that vφ = yφ. Similarly, wT = zT yields wφ = zφ. Thus, pφ = qφ. ˜ be a set, let S˜ be a scheme on X, ˜ let φ be a morphism Lemma 5.1.5 Let X ˜ ˜ from X ∪ S to X ∪ S, and set T := ker(φ). For each element x in X, set (xT )φ¯ := xφ. For each element s in S, set (sT )φ¯ := sφ. ˜ ∪ S. ˜ Then φ¯ is a morphism from X/T ∪ S//T to X Proof. It follows from Lemma 5.1.4 that φ is a (well-defined) map which ˜ and S//T to S. ˜ maps X/T to X

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Let y and z be elements in X, and let s be an element in S such that zT ∈ (yT )(sT ). Then, by definition, z ∈ yT sT . Thus, there exists an element v in yT such that z ∈ vsT . Since z ∈ vsT , there exists an element w in vs such that z ∈ wT . From v ∈ yT we obtain vT = yT , so that vφ = yφ. Similarly, z ∈ wT yields wφ = zφ. Finally, as φ is a morphism, w ∈ vs yields wφ ∈ vφsφ. Thus, we ¯ ¯ T )φ. conclude that zφ ∈ yφsφ, and that means that (zT )φ¯ ∈ (yT )φ(s ˜ be a set, and let S˜ be a scheme on X. ˜ For each morphism φ from X ∪ S Let X ˜ ˜ ˜ to X ∪ S, we set φS := φ ∩ (S × S). Lemma 5.1.6 Let φ be a morphism from X ∪ S. Then we have the following. (i) Assume that φS is injective. Then, for any three elements y, z in X and s in S, zφ ∈ yφsφ implies z ∈ ys. (ii) Assume that, for any three elements y, z in X and s in S, zφ ∈ yφsφ implies z ∈ ys. Then {1} = ker(φ). (iii) If {1} = ker(φ), φX is injective. Proof. (i) Let y and z be elements in X, and let s be an element in S such that zφ ∈ yφsφ. We have to show that z ∈ ys. Let us denote by r the uniquely determined element in S satisfying z ∈ yr. Then zφ ∈ yφrφ. Thus, as zφ ∈ yφsφ, rφ = sφ. Since we are assuming φ to be injective, this yields r = s. Thus, as z ∈ yr, z ∈ ys. (ii) Let s be an element in ker(φ). We have to show that s = 1. Since s ∈ ker(φ), sφ = 1φ. Let y be an element in X, and let z be an element in ys. From z ∈ ys we obtain zφ ∈ yφsφ. Thus, as sφ = 1φ, zφ ∈ yφ1φ. Thus, by hypothesis, y = z. Thus, as z ∈ ys, s = 1. (iii) This is an immediate consequence of Lemma 5.1.4(i). Recall that a morphism φ from X ∪ S is called a homomorphism if, for any three elements y, z in X and s in S with zφ ∈ yφsφ, there exist elements v in X and w in vs such that vφ = yφ and wφ = zφ. Let φ be a morphism from X ∪ S, and assume that φS is injective. Then, by Lemma 5.1.6(i), φ is a homomorphism. Lemma 5.1.7 Let φ be a homomorphism from X ∪ S, and assume that φX is injective. Then φ is injective. Proof. Let p and q be elements in S, and let us assume that pφ = qφ. We have to show that p = q Since p is not empty, there exist elements y and z in X such that z ∈ yp. From z ∈ yp we obtain zφ ∈ yφpφ. Thus, as pφ = qφ, zφ ∈ yφqφ. Thus, as φ

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is assumed to be a homomorphism, there exist elements v in X and w in vq such that vφ = yφ and wφ = zφ. Since we are assuming φX to be injective, we obtain from vφ = yφ that v = y. Similarly, we obtain from wφ = zφ that w = z. From v = y, w = z, and w ∈ vq we obtain z ∈ yq. Thus, as z ∈ yp, p = q. Lemma 5.1.8 A homomorphism φ is injective if and only if {1} = ker(φ). Proof. Considering Lemma 5.1.6 this is a consequence of Lemma 5.1.7.

5.2 Isomorphisms It follows right from the definition of a morphism that the product of two morphisms is a morphism. This is not necessarily the case with homomorphisms. However, we have the following. ˜ be a set, let S˜ be a scheme on X, ˜ let φ be a homoLemma 5.2.1 Let X ˜ ˜ ˜ ˜ ∪ S. ˜ morphism from X ∪ S to X ∪ S, and let φ be a homomorphism from X ˜ Assume φX to be surjective. Then φφ is a homomorphism. Proof. Let y and z be elements in X, and let s be an element in S such that ˜ ∈ y(φφ)s(φ ˜ ˜ z(φφ) φ). We have to show that there exist elements v in X and w in vs such that ˜ = y(φφ) ˜ and w(φφ) ˜ = z(φφ). ˜ v(φφ) ˜ ∈ y(φφ)s(φ ˜ ˜ we obtain From z(φφ) φ) ˜ ˜ (zφ)φ˜ ∈ (yφ)φ(sφ) φ. ˜ Thus, as φ˜ is assumed to be a homomorphism, there exist elements v˜ in X ˜ ˜ ˜ ˜ and w ˜ in v˜(sφ) such that v˜φ = (yφ)φ and w ˜ φ = (zφ)φ. Since φX is assumed to be surjective, there exist elements v in X with v φ = v˜ and w in X with w φ = w. ˜ ˜ and w ˜ ∈ v˜(sφ) we obtain w φ ∈ v φ(sφ). Thus, as φ From v φ = v˜, w φ = w, is assumed to be a homomorphism, there exist elements v in X and w in vs such that vφ = v φ and wφ = w φ. From vφ = v φ, v φ = v˜, and v˜φ˜ = (yφ)φ˜ we obtain ˜ = (vφ)φ˜ = (v φ)φ˜ = v˜φ˜ = (yφ)φ˜ = y(φφ). ˜ v(φφ) ˜ = Similarly, we obtain from wφ = w φ, w φ = w, ˜ and w ˜ φ˜ = (zφ)φ˜ that w(φφ) ˜ z(φφ).

5.2 Isomorphisms

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Recall that an isomorphism from X ∪ S to X ∪ S is called an automorphism of S. Recall also that Aut(S) is our notation of the set of all automorphisms of S. Recall, finally, that, for each element a in Aut(S), aS is our notation for a ∩ (S × S). Let us define Sch(S) to be the set of all elements a in Aut(S) such that aS = {1S }. A bijective map from a set to itself is called a permutation. For each closed subset T of S, we define Stc(T ) to be the set of all permutations σ of T such that, for any three elements p, q, and r in T , apqr = apσqσrσ . Theorem 5.2.2 We have the following. (i) The set Aut(S) is a group with respect to composition. (ii) The set Stc(S) is a group with respect to composition. (iii) For each element a in Aut(S), we define aζ := aS . Then ζ is a group homomorphism from Aut(S) to Stc(S) with kernel Sch(S). Proof. (i) Note that the inverse map of an isomorphism is an isomorphism. Thus, the claim is an immediate consequence of Lemma 5.2.1. (ii) This is an immediate consequence of the definition of Stc(S). (iii) Let a be an automorphism of S. Then aS is a permutation of S. Let p, q, and r be elements in S, and let y and z be elements in X with z ∈ yr. Since a is assumed to be an automorphism, a is a morphism. Thus, as z ∈ yr, za ∈ yara. From z ∈ yr we also obtain |yp∩zq ∗ | = apqr . Thus, as a is assumed to be a bijective map, |yapa ∩ za(qa)∗ | = apqr . It follows that apqr = apaqara . Lemma 5.2.3 For each element σ in Stc(S), the following hold. (i) For any two elements p and q in S, (pq)σ = pσqσ. (ii) We have 1σ = 1. (iii) For each element s in S, s∗ σ = (sσ)∗ . Proof. (i) Let s be an element in (pq)σ. Then there exists an element r in pq such that s = rσ. Since r ∈ pq, 1 ≤ apqr . Thus, as σ ∈ Stc(S), 1 ≤ apσqσrσ . Thus, as s = rσ, s ∈ pσqσ. Conversely, let s be an element in pσqσ. Then 1 ≤ apσqσs . Thus, as σ ∈ Stc(S), 1 ≤ apqsσ−1 . This means that sσ −1 ∈ pq. Thus, by definition, s ∈ (pq)σ. (ii) From (i) we obtain {1σ} = 1σ1σ. Thus, 1σ = 1. (iii) From (ii), Lemma 1.3.2(i), and (i) we obtain 1 = 1σ ∈ (s∗ s)σ = s∗ σsσ. Thus, the claim follows from Lemma 1.3.6(ii).

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5.3 The Isomorphism Theorems In this section, S is assumed to have finite valency. We shall show first that each closed subset of S induces naturally a homomorphism from X ∪ S. Theorem 5.3.1 Let T be a closed subset of S. For each element x in X, we set xφ := xT ; for each element s in S, we set sφ := sT . Then we have the following. (i) The map φ is a surjective homomorphism from X ∪ S to X/T ∪ S//T . (ii) We have T = ker(φ). Proof. (i) Let y be an element in X, let s be an element in S, and let z be an element in ys. From z ∈ ys (and 1 ∈ T ) we obtain z ∈ yT sT . Thus, by definition, zT ∈ (yT )(sT ), and that means that zφ ∈ yφsφ. Since y and z have been chosen arbitrarily in X and s arbitrarily in S, we have shown that φ is a morphism. Let us now show that φ is a homomorphism. In order to do so we fix elements y, z in X and s in S, and we assume that zφ ∈ yφsφ. From zφ ∈ yφsφ we obtain zT ∈ (yT )(sT ), and that means that z ∈ yT sT . Thus, there exists an element v in yT such that z ∈ vsT . Since z ∈ vsT , there exists an element w in vs such that z ∈ wT . By Lemma 2.1.4, v ∈ yT is equivalent to vT = yT , and that means that vφ = yφ. Similarly, we obtain from z ∈ wT that wφ = zφ. Thus, φ is a homomorphism from X ∪ S to X/T ∪ S//T . That φ is surjective follows right from the definition of φ. (ii) Let s be an element in S. By definition, s ∈ ker(φ) if and only if sT = 1T . By Lemma 4.1.1, sT = 1T if and only if s ∈ T . Thus, T = ker(φ). The homomorphism φ in Theorem 5.3.1(i) is called the natural homomorphism from X ∪ S to X/T ∪ S//T . Let φ be a homomorphism from X ∪ S. From Lemma 5.1.2(ii) we know that ker(φ) is closed. Thus, we obtain from Theorem 4.1.3(i) that S//ker(φ) is a scheme on X/ker(φ). The following theorem is called the Homomorphism Theorem for schemes. Theorem 5.3.2 Let φ be a homomorphism from X ∪ S, and set T := ker(φ). For each element x in X, set (xT )φ¯ := xφ. For each element s in S, set (sT )φ¯ := sφ. Then φ¯ is an injective homomorphism from X/T ∪ S//T . Proof. From Lemma 5.1.5 we know that φ¯ is a morphism from X/T ∪ S//T . Thus, according to Lemma 5.1.6(i), it suffices to show that φ¯ is injective.

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That φ¯X is injective follows immediately from Lemma 5.1.4(i). In order to show that φ¯S is injective we fix elements p and q in S, and we ¯ Then, by definition, pφ = qφ. assume that (pT )φ¯ = (pT )φ. Let y and z be elements in X such that zφ ∈ yφpφ. Then, as φ is assumed to be a homomorphism, there exist elements v in X and w in vp such that vφ = yφ and wφ = zφ. Since vφ = yφ, vT = yT . Similarly, as wφ = zφ, wT = zT . From w ∈ vp and vT = yT we obtain w ∈ yT p. Thus, as z ∈ wT , z ∈ yT pT . From zφ ∈ yφpφ and pφ = qφ we also obtain zφ ∈ yφqφ. Thus, replacing p with q in the previous paragraph, we obtain z ∈ yT qT ; too. However, from z ∈ yT pT and z ∈ yT qT we conclude that T pT = T qT ; cf. Lemma 2.1.3. Thus, by Lemma 4.1.1, pT = q T . Recall that a bijective homomorphism is called an isomorphism. ˜ be a set, and let S˜ be a scheme on X. ˜ The schemes S and S˜ are said to Let X ˜ ∪ S. ˜ We shall be isomorphic if there exists an isomorphism from X ∪ S to X write S ∼ = S˜ in order to indicate that S and S˜ are isomorphic. The following theorem is called the First Isomorphism Theorem for schemes. Theorem 5.3.3 Let T and U be closed subsets of S such that T ⊆ U . Then (S//T )//(U//T ) ∼ = S//U . Proof. Let us denote by φ the natural homomorphism from X ∪ S to X/T ∪ S//T and by ψ the natural homomorphism from X/T ∪ S//T to (X/T )/(U//T ) ∪ (S//T )//(U//T ). By Theorem 5.3.1(i) and Lemma 5.2.1, φψ is a surjective homomorphism from X ∪ S to (X/T )/(U//T ) ∪ (S//T )//(U//T ). On the other hand, we know from Theorem 5.3.1(ii) that U//T = ker(ψ). Therefore, we have ker(φψ) = {s ∈ S | sφ ∈ U//T } = {s ∈ S | sT ∈ U//T } = U ; use Lemma 4.2.1(ii) and Lemma 4.1.7(ii) for the last equation. Now the claim follows from Theorem 5.3.2. The following theorem is called the Second Isomorphism Theorem for schemes. Theorem 5.3.4 Let T and U be closed subsets of S, and assume that T ⊆ NS (U ). Then the following hold. (i) The closed set T ∩ U is normal in T . (ii) The closed set U is normal in T U .

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(iii) For each element x in X, we have (T //T ∩ U )x ∼ = (T U//U )x . Proof. (i) This follows from Lemma 2.5.1(i). (ii) This follows from Lemma 2.5.1(ii). (iii) For each element y in xT , we define yφ := yU. For each element t in T , we define (txT )φ := (tU )(xU )(T U//U ) . Then φ is a map from xT ∪ TxT to (xU )(T U//U ) ∪ (T U//U )(xU )(T U//U ) which maps xT to (xU )(T U//U ) and TxT to (T U//U )(xU )(T U//U ) . We claim that φ is a morphism. In order to show this we pick elements y in xT , t in T , and z in yt. From z ∈ yt and 1 ∈ U we obtain z ∈ yU tU . Thus, by definition, zU ∈ (yU )(tU ). From y ∈ xT we similarly obtain yU ∈ (xU )(T U//U ), and from z ∈ xT we obtain zU ∈ (xU )(T U//U ). Thus, zU ∈ (yU )(tU )(xU )(T U//U ) , and that means that zφ = yφ(txT )φ. Let us now show that φ is a homomorphism. In order to do so we pick elements y and z in xT and t in T such that zφ ∈ yφ(txT )φ. From zφ ∈ yφ(txT )φ we obtain zU ∈ (yU )(tU ), whence, by definition, z ∈ yU tU . Since z ∈ yU tU , there exists an element v in yU such that z ∈ vtU . Since z ∈ vtU , there exists an element w in vt such that z ∈ wU . From w ∈ vt and v ∈ yU we obtain w ∈ yU t. However, as we are assuming that T ⊆ NS (U ), we have that U t ⊆ tU . Thus, w ∈ ytU . Thus, there exists an element w in yt such that w ∈ w U . From z ∈ wU and w ∈ w U we obtain w φ = w U = zU = zφ; cf. Lemma 2.1.4. From w ∈ yt we obtain w ∈ ytxT . Thus, φ is a homomorphism. Let us compute ker(φ). For each element t in T , we have (txT )φ = (1xT )φ if and only if tU = 1U . From Lemma 4.1.1 we know that, for each element t in T , tU = 1U is equivalent to t ∈ U . Therefore, ker(φ) = (T ∩ U )xT . For each element z in xT U , there exists an element y in xT such that yU = zU . Therefore, φxT is surjective. Thus, according to Lemma 5.1.3(ii), φ is surjective. Now the claim follows from Theorem 5.3.2.

5.4 Composition Series

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Note that the Homomorphism Theorem and the First Isomorphism Theorem deal with factorizations over arbitrary closed subsets, whereas the Second Isomorphism Theorem deals with factorizations over normal closed subsets. We shall generalize the Second Isomorphism Theorem in the next section. The Homomorphism Theorem and the two Isomorphism Theorems were first proved in [35]. The thin case was already proved in 1929 by Emmy Noether; cf. [32; I. §2]. Let y and z be elements in X, and let T be a closed subset of S. In general, it is not true that Ty ∼ = Tz . However, from Theorem 5.3.4(iii) we immediately obtain the following. Corollary 5.3.5 Let T and U be closed subsets of S such that {1} = T ∩ U and T ⊆ NS (U ). Then, for each element x in X, Tx ∼ = (T U//U )x . Note that, by Lemma 2.5.1(ii), the hypothesis T ⊆ NS (U ) in Corollary 5.3.5 says exactly that U is a normal closed subset of S.

5.4 Composition Series In this short section, S is assumed to have finite valency. We start with a generalization of a theorem of Hans Zassenhaus; cf. [41; II. §5]. Theorem 5.4.1 Let T , T , U , and U be closed subsets of S. Assume that T is normal in T and that U is normal in U . Then the following hold. (i) The set T ∩ T U and T ∩ T U are closed. (ii) The closed set T ∩ T U is normal in T ∩ T U . (iii) For each element x in X, we have (T ∩ U //T ∩ U ∩ T U )x ∼ = (T ∩ T U //T ∩ T U )x . Proof. From Lemma 2.2.1(i) we know that T ∩ T U = T (T ∩ U ) and that

T ∩ T U = T (T ∩ U ).

Since T ∩ U ⊆ NS (T ), T (T ∩ U ) is closed; cf. Lemma 2.5.2(iii). Thus, as T ∩ T U = T (T ∩ U ), T ∩ T U is closed. Similarly, one obtains from T ∩ T U = T (T ∩ U ) that T ∩ T U is closed. Since T ∩ T U is closed, (T ∩ U )(T ∩ T U ) ⊆ T ∩ T U . Conversely, as T ∩ T U = (T ∩ U )T , T ∩ T U ⊆ (T ∩ U )(T ∩ T U ). It follows that

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(T ∩ U )(T ∩ T U ) = T ∩ T U . Note also that, by Lemma 2.5.1(i), T ∩ U ⊆ NT (T U ) ⊆ NS (T ∩ T U ). Thus, the theorem follows from Theorem 5.3.4(i), (iii) (with T ∩ U in place of T and T ∩ T U in place of U ). Let T be a closed subset of S, and let U be a set of closed subsets of T such that {1} ∈ U and T ∈ U. Let us assume that, for any two elements V and W in U, V ⊆ W or W ⊆ V . For each element U in U \ {T }, we define U U to be the intersection of all elements V of U \ {U } which contain U as a subset. (Since S is assumed to have finite valency, we have U U ∈ U.) The set U is called a subnormal series of T if, for each element U in U, U is normal in U U . A maximal subnormal series of T is called a composition series of T . Two composition series U and V of T are called isomorphic, if there exists a bijective map η from U \ {T } to V \ {T } such that, for any two elements x in X and U in U \ {T }, (U U //U )x ∼ = (U ηV //U η )x . The following theorem is [35; Theorem 4.2]. The thin case was already proved in 1889 by Otto H¨ older; cf. [26; §10]. Theorem 5.4.2 Let T be a closed subset of S. Then any two composition series of T are isomorphic. Proof. Let U and V be composition series of T , let us fix an element U in U \{T }, and let us denote by U ν the uniquely determined element V in V \{T } which satisfies U U ⊆ U V V and U U ⊆ U V . Since U U ⊆ U U νV ,

U U = U U ∩ U U νV .

Thus, by Theorem 5.4.1(ii), U U ∩ U U ν is normal in U U . Since U U ⊆ U U ν , U U ∩ U U ν = U U . Moreover, U is normal in U U ∩ U U ν , and U U ∩ U U ν is normal in U U . Thus, as U is assumed to be a composition series of T , we conclude that U = UU ∩ UUν. Using the last two equations we obtain (U U ∩ U νV //U U ∩ U νV ∩ U U ν )x ∼ = (U U //U )x for each element x in X; cf. Theorem 5.4.1(iii).

5.5 The Group Correspondence

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Let us now define a map υ from V \ {T } to U \ {T } analogously to ν. Then U νV ⊆ U ν U νυU and U νV ⊆ U ν U νυ . From the above given isomorphism we obtain U νV ⊆ U U ν . Therefore, U νV ⊆ U ν U . Thus, by definition of υ, U ⊆ U νυ . On the other hand, as above, the fact that V is a composition series of T yields U ν = U νV ∩ U ν U νυ . Therefore, U U ⊆ U νυ would lead to U U ∩ U νV ⊆ U ν , and that contradicts the above given isomorphism. Thus, U U ⊆ U νυ . From U ⊆ U νυ and U U ⊆ U νυ we obtain U = U νυ . Now we have shown that ν and υ are inverse to each other. In particular, ν is a bijective map from U \ {T } to V \ {T }, and υ is a bijective map from V \ {T } to U \ {T }. Thus, the above given isomorphism and its analogue yield that, for each element x in X, (U U //U )x ∼ = (U νV //U ν )x . Let T be a closed subset of S. A scheme S˜ is called a composition factor of T if there exists a composition series U of T , an element U in U \ {T }, and an element x in X such that S˜ ∼ = (U U //U )x . A closed subset T of S is called simple if T has exactly two normal closed subsets, namely {1} and T . It follows immediately from Lemma 4.2.4(i) that composition factors are simple.

5.5 The Group Correspondence Let G be a set. A map from G × G to G is called an operation on G. Let us fix an operation on G and call it µ. For any two elements e and f in G, we write ef instead of (e, f )µ. The operation µ is called associative if, for any three elements d, e, and f in G, d(ef ) = (de)f . An element n of G is called a neutral element of G if, for each element g in G, gn = g = ng. Assume that n and n are neutral elements of G. Then, by definition, n = n n = n. Thus, G possesses at most one neutral element. Let us assume that G possesses a neutral element n, and let e and f be elements of G. The element f is called an inverse of e if ef = n = f e. The set G is called a group (with respect to µ) if µ is associative, if G possesses a neutral element and, for each of its elements, an inverse. Let G be a group. Then, by definition, there exists an operation µ on G. If, for any two elements e and f in G, we write ef instead of (e, f )µ (as we did so far), then we say that G is written multiplicatively. In this case, the

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neutral element of G is denoted by 1, and it is also called the identity element of G. Moreover, for each element g in G, the inverse of g is denoted by g −1 . Usually, we write groups multiplicatively. However, sometimes we write, for any two elements e and f in G, e + f instead of (e, f )µ. In this case, we say that G is written additively, we denote the neutral element of G by 0, and, for each element g in G, the inverse of g will be denoted by −g. Assume that an element d in G has inverse elements e and f . Then, by definition, e = e1 = e(df ) = (ed)f = 1f = f. Thus, d possesses at most one inverse element. Recall that, if S is thin, S γ is our notation for the set of all sets {s} with s ∈ S. Theorem 5.5.1 Assume S to be thin. Then S γ is a group with respect to the restriction of the complex multiplication in S to S γ and with {1} as identity element. Proof. From Lemma 2.5.9(i), (ii) together with Lemma 1.5.2 we obtain that the restriction of the complex multiplication in S to S γ is an operation on S γ . From Lemma 1.3.1 we know that this operation is associative. It follows right from the definition of the complex multiplication in S that {1} is an identity element of S γ . Finally, as S is assumed to be thin, we deduce from Lemma 1.3.2(i) that, for each element s in S, s∗ s = {1} = ss∗ , so that {s}−1 = {s∗ }. This finishes the proof of the theorem. Let G be a group. Recall that, for each element g in G, g τ is our notation for the set of all pairs (e, f ) of elements of G satisfying eg = f . Recall also that Gτ is our notation for the set of all sets g τ with g ∈ G. Theorem 5.5.2 Let G be a group. Then Gτ is a thin scheme on G. Proof. It is clear that 1τ = 1G and that, for each element g in G, (g τ )∗ = (g −1 )τ . In order to verify the regularity condition we pick elements b, c, d, e, and f in G. We assume that (b, c) ∈ f τ . Thus, by definition, we have that bf = c. It follows that f = b−1 c. Let us now assume that bdτ ∩ c(eτ )∗ is not empty. Then, as (eτ )∗ = (e−1 )τ , bdτ ∩ c(e−1 )τ is not empty. Thus, as {bd} = bdτ and {ce−1 } = c(e−1 )τ , bd = ce−1 . From bd = ce−1 we obtain de = b−1 c.

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From f = b−1 c and de = b−1 c we obtain de = f . Note, finally, that, for each element g in G, {1τ } = (g τ )∗ g τ . Thus, Gτ is thin. Theorem 5.5.3 If S is thin, S γτ ∼ = S. Proof. We set G := S γ , and we fix an element v in X. For each element x in X, we define xφ to be the uniquely determined element g in G which satisfies x ∈ vg. For each element s in S, we set sφ := {s}τ . Since S is assumed to be thin, φ is a bijective map from X ∪ S to G ∪ Gτ . Let y and z be elements in X, and let s be the uniquely defined element in S satisfying z ∈ ys. By Lemma 5.1.6(i), we shall be done if we succeed in showing that zφ ∈ yφsφ. Let p be the uniquely defined element in S such that y ∈ vp, and let q be the uniquely defined element in S satisfying z ∈ vq. From z ∈ ys and y ∈ vp we obtain z ∈ vps. Thus, as z ∈ vq, q ∈ ps. Thus, by Lemma 1.5.2, {q} = {p}{s}. Thus, by definition, {q} ∈ {p}{s}τ . On the other hand, as y ∈ vp, {p} = yφ. Similarly, as z ∈ vq, {q} = zφ. Thus, as sφ = {s}τ , zφ ∈ yφsφ. ˜ be groups. A map from G to G ˜ is called a group homomorphism or Let G and G simply a homomorphism if, for any two elements e and f in G, (ef )φ = eφf φ. ˜ are called isomorphic if there exists a bijective group Two groups G and G ˜ If two groups G and G ˜ are isomorphic, one homomorphism from G to G. ∼ ˜ writes G = G. Theorem 5.5.4 For each group G, Gτ γ ∼ = G. Proof. For each element g in G, we define gφ := {g τ }. Then φ is a bijective map from G to Gτ γ . Let e and f be elements in G, and set g := ef . We shall be done if we succeed in showing that gφ = eφf φ. First of all, since g = ef , g ∈ ef τ . Moreover, we also have e ∈ 1eτ and g ∈ 1g τ . From g ∈ ef τ and e ∈ 1eτ we obtain g ∈ 1eτ f τ . Thus, as g ∈ 1g τ , g τ ∈ eτ f τ . But, by Theorem 5.5.2, 1 = neτ and 1 = nf τ . Thus, by Lemma 1.5.2, {g τ } = eτ f τ , and that means that {g τ } = {eτ }{f τ }. It follows that gφ = eφf φ. The four theorems which have been proved in this section show that we may view thin schemes as groups. In particular, the class of all groups can be viewed as a subclass of the class of all schemes.

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5.6 Residually Thin Schemes Throughout this section, S is assumed to have finite valency. The letter T will stand for a closed subset of S. We define OΘ (T ) to be the intersection of the sets (Oϑ )n (T ) with n a nonnegative integer. Since S is assumed to have finite valency, there exists a non-negative integer m such that, for each integer n with m ≤ n, OΘ (T ) = (Oϑ )n (T ). We call T residually thin if {1} = OΘ (T ). Theorem 5.6.1 The set T is residually thin if and only if each composition factor of T is thin. Proof. Define T to be the set of all sets (Oϑ )n (T ) such that n is a nonnegative integer. Then, by Theorem 3.2.1(i) and Lemma 2.5.5, {{1}} ∪ T is a subnormal series of T . Let U be a composition series of T which contains {{1}} ∪ T . Let us first assume T to be residually thin, and let us pick an element U in U \ {T }. Since T is assumed to be residually thin, {1} = OΘ (T ). Thus, as U ∈ U, there exists a non-negative integer n such that (Oϑ )n+1 (T ) ⊆ U and U U ⊆ (Oϑ )n (T ).1 By Lemma 2.5.6(iii), this implies that U is strongly normal in U U . Thus, by Lemma 4.2.5(ii), U U //U is thin. Thus, for each element x in X, (U U //U )x is thin; cf. Theorem 2.1.8(iii). Thus, as U has been chosen arbitrarily in U \ {T }, we have proved that each composition factor of T is thin; cf. Theorem 5.4.2. Let us now assume that T is not residually thin. We shall show that T has a composition factor which is not thin. Since T is assumed not to be residually thin, {1} = OΘ (T ). Thus, as OΘ (T ) ∈ U, there exists an element U in U such that U U = OΘ (T ). Since U U = OΘ (T ), Oϑ (U U ) = U U . Therefore, by Lemma 4.2.5(ii), U U //U is not thin. Thus, for each element x in X, (U U //U )x is not thin; cf. Theorem 2.1.8(iii). Lemma 5.6.2 Let U be a closed subsets of S such that T ⊆ U . Then, if U is residually thin, so is T . Proof. This is an immediate consequence of Lemma 3.2.6. Lemma 5.6.3 For each closed subset U of S, the following hold. 1

Recall that U U stands for the intersection of all elements V of U \ {U } which contain U .

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(i) Assume that T ⊆ NS (U ) and that OΘ (T U ) ⊆ U . Then T U//U is residually thin. (ii) Assume that T is normal in U . Then, if U is residually thin, so is U//T . Proof. (i) This is an immediate consequence of Theorem 4.2.8. (ii) This is a particular case of (i). Lemma 5.6.2 and Lemma 5.6.3(ii) say that the property of being residually thin is inherited by closed subsets and by quotient schemes over normal closed subsets. Here is a partial converse. Theorem 5.6.4 Let U be a closed subset of S such that T ⊆ NS (U ). Then, if T and U are residually thin, so is T U . Proof. We are assuming that T ⊆ NS (U ). Thus, by Theorem 5.3.4(i), T ∩ U is normal in T . Let us assume that T is residually thin. Then, as T ∩ U is normal in T , we obtain from Lemma 5.6.3(ii) that T //T ∩ U is residually thin. On the other hand, from our hypothesis that T ⊆ NS (U ) we also obtain (T //T ∩ U )x ∼ = (T U//U )x for each element x in X; cf. Theorem 5.3.4(iii). Thus, as T //T ∩U is residually thin, T U//U is residually thin. Since T U//U is residually thin, OΘ (T U ) ⊆ U ; cf. Theorem 4.2.8. Thus, as U is assumed to be residually thin, Lemma 5.6.2 yields {1} = OΘ (T U ), and that means that T U is residually thin. From Theorem 5.6.4 we obtain that S possesses a uniquely determined biggest residually thin normal closed subset. Let us now introduce a second operator on the set of all closed subsets of S. This operator is dual to OΘ . Let T be a closed subset of S. We shall write OΘ (T ) to denote the union of the sets (KT )n ({1}) such that n is a non-negative integer. Note that, by Lemma 2.5.8(ii), OΘ (T ) is a closed subset of T . Note also that there exists a non-negative integer m such that, for each integer n with m ≤ n, (KT )n ({1}) = OΘ (T ). (Recall that S is assumed to have finite valency.) Lemma 5.6.5 The following statements hold. (i) We have Oϑ (T ) ⊆ OΘ (T ). (ii) We have OΘ (T //Oϑ (T )) = OΘ (T )//Oϑ (T ).

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Proof. (i) From Lemma 2.5.9(i) we know that Oϑ (S) = KS ({1}). Thus, the claim is an immediate consequence of the definition of OΘ (T ). (ii) We fix a non-negative integer and call it n. Applying Theorem 4.2.6 to Oϑ (T ), T , and 1 instead of T , U , and m, we obtain (KT //Oϑ (T ) )n (KT (Oϑ (T ))//Oϑ (T )) = (KT )n (Oϑ (T ))//Oϑ (T ). On the other hand, by Lemma 4.2.4(ii), KT //Oϑ (T ) (Oϑ (T )//Oϑ (T )) = KT (Oϑ (T ))//Oϑ (T ), so that (KT //Oϑ (T ) )n (KT //Oϑ (T ) (Oϑ (T )//Oϑ (T ))) = (KT )n (Oϑ (T ))//Oϑ (T ). Now recall that, by Lemma 2.5.9(i), KT ({1}) = Oϑ (T ). Thus, (KT //Oϑ (T ) )n+1 (Oϑ (T )//Oϑ (T )) = (KT )n+1 ({1})//Oϑ (T ). Now the claim follows from the fact that n has been chosen arbitrarily. Theorem 5.6.6 The set OΘ (T ) is residually thin. Proof. We set U := OΘ (T ). Then, as S is assumed to have finite valency, there exists a positive integer n such that (KT )n ({1}) = U . It follows that (Oϑ )0 (U ) = (KT )n ({1}). Let us now prove that, for each element i in {1, . . . , n}, (Oϑ )i (U ) ⊆ (KT )n−i ({1}). By induction, we may (and shall) assume that (Oϑ )i−1 (U ) ⊆ (KT )n−i+1 ({1}). By definition, (KT )n−i ({1}) is strongly normal in (KT )n−i+1 ({1}). Thus, referring to Theorem 3.2.1(iii) we obtain (Oϑ )i (U ) = Oϑ ((Oϑ )i−1 (U )) ⊆ Oϑ ((KT )n−i+1 ({1})) ⊆ (KT )n−i ({1}). It follows that {1} = (Oϑ )n (U ), and that means that U is residually thin. Theorem 5.6.7 Let p be a prime number, and let us assume that nT is a power of p. Then the following conditions are equivalent. (a) We have OΘ (T ) = T . (b) The set T is residually thin.

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(c) The set T is a closed p-subset of S. Proof. (a) ⇒ (b) This follows from Theorem 5.6.6. (b) ⇒ (c) Let T be a counterexample of minimal valency. Then T is residually thin. Thus, there exists a positive integer n such that {1} = (Oϑ )n+1 (T ) and {1} = (Oϑ )n (T ). We set U := (Oϑ )n (T ). Then {1} = U ⊆ Oϑ (T ) and, by Corollary 4.2.9, U//U = (Oϑ )n (T //U ). The latter equation implies that T //U is residually thin. On the other hand, we are assuming that nT is a power of p. Thus, by Lemma 4.3.3(i), nT //U is a power of p. Thus, the (minimal) choice of T forces T //U to be p-valenced. Thus, as U is thin, T is p-valenced; cf. Corollary 4.3.2(ii). (c) ⇒ (a) Let T be a counterexample of minimal valency. Then T is a closed p-subset of S. Thus, by Lemma 2.3.9, {1} = Oϑ (T ). On the other hand, the hypothesis that T is p-valenced implies also that T //Oϑ (T ) is p-valenced; cf. Corollary 4.3.2(i). Moreover, as nS is assumed to be a power of p, nT //Oϑ (T ) is a power of p; cf. Lemma 4.3.3(i). Thus, the (minimal) choice of T leads to OΘ (T //Oϑ (T )) = T //Oϑ (T ). From Lemma 5.6.5(ii) we also know that OΘ (T //Oϑ (T )) = OΘ (T )//Oϑ (T ). Thus, by Lemma 4.2.1(ii), OΘ (T ) = T , contrary to the choice of T . Corollary 5.6.8 Let p be a prime number. Then T is a closed p-subset of S if and only if each composition factor of T is thin and has valency p. Proof. Assume first that T is a closed p-subset of S. Then, T is p-valenced and nT is a power of p. Thus, by Theorem 5.6.7, T is residually thin. Thus, by Theorem 5.6.1, each composition factor of S is thin. Conversely, let us assume that each composition factor of T is thin and has valency p. We may assume that {1} = T . Then T possesses a thin closed subset U of valency p. Thus, by induction, T //U is a closed p-subset of S//U . Thus, by Corollary 4.3.2(i), T is a closed p-subset of S. Corollary 5.6.9 Let p be a prime number, assume that T is a closed p-subset of S, and let U be a closed p-subset of S such that T ⊆ NS (U ). Then T U is a closed p-subset of S. Proof. By hypothesis, nT and nU are powers of p. Thus, by Lemma 2.3.6(i), nT U is a power of p.

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We assume T to be a closed p-subset of S. Thus, by Theorem 5.6.7, T is residually thin. Similarly, as U is assumed to be a closed p-subset of S, U is residually thin. Thus, by Theorem 5.6.4, T U is residually thin. Thus, as nT U is a power of p, T U is a closed p-subset of S; cf. Theorem 5.6.7.

6 Faithful Maps

A map χ from a subset Y of X to X is called faithful if, for any three elements v, w in Y and s in S, w ∈ vs implies wχ ∈ vχs. The first section of this chapter is a collection of basic facts about faithful maps. Let Y and Z be subsets of X such that Y ⊆ Z, let χ be a faithful map from Y to X, and let χ ¯ be a faithful map from Z to X. We say that χ extends faithfully to χ ¯ if, for each element y in Y , yχ = y χ. ¯ Let T and U be closed subsets of S, and assume that T ⊆ U . We say that T is faithfully embedded in U if, for any two elements y in X and z in yT , each faithful map χ from {y, z} to yU extends faithfully to a bijective (faithful) map from yT to yχT . The second section of this chapter deals with faithfully embedded closed subsets of S. We mainly discuss the question to which extent the property of being faithfully embedded is inherited from given quotient schemes of closed subsets of S to other quotient schemes of closed subsets of S. A closed subset of S will be called schurian if it is faithfully embedded in itself. Let x be an element in X, and let T be a closed subset of S. It is easy to see that the set of all bijective faithful maps from xT to xT is a group with respect to composition. We call this group the Schur group of T with respect to x. If T = S, we just speak about the Schur group of S. Recall that Sch(S) is defined to be the set of all elements a in Aut(S) such that aS = {1S }. Referring to this notation it is clear that the Schur group of S is just the set of all elements aS with a ∈ Sch(S). The Schur group is the subject of the third section of this chapter. We shall see, in particular, that S is schurian if and only if S is isomorphic to a quotient scheme of a thin scheme, and this is the case if and only if, modulo the group

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correspondence, S is isomorphic to the quotient scheme of its own Schur group over one of the ‘one-point stabilizers’; cf. Theorem 6.3.1. In Section 6.4, we use previously obtained results about Schur groups in order to establish a recognition theorem for certain schemes of finite valency all elements of which have valency at most 2. The theorem is related to one of the most significant results in finite group theory, to George Glauberman’s Z ∗ -Theorem. In Section 6.5, we assume S to have finite valency. We shall prove that closed subsets of S which are generated by a single symmetric element of valency 2 are faithfully embedded in S. We also establish the corresponding recognition theorem. After that we shall look at closed subsets of S in which each nonidentity element has valency 2. Faithful maps are particularly interesting in connection with constrained sets. Section 6.6 provides a few results about this relationship. They will turn out to be useful in Chapter 11. In the last of the seven sections of this chapter, we investigate closed subsets T of S which have finite valency and satisfy Oϑ (T ) ⊆ Oϑ (T ). We shall give a sufficient criterion for T to be faithfully embedded in S.

6.1 Basic Facts Recall that a map χ from a subset Y of X to X is called faithful if, for any three elements v, w in Y and s in S, w ∈ vs implies wχ ∈ vχs. Note that faithful maps are necessarily injective. Lemma 6.1.1 Let x be an element in X, let T be a closed subset of S, and let χ be a map from xT to X. Then the following hold. (i) The map χ is faithful if and only if, for any two elements w in xT and t in T , wtχ ⊆ wχt. (ii) Assume T to have finite valency. Then χ is faithful if and only if, for any two elements w in xT and t in T , wtχ = wχt. (iii) Assume T to have finite valency and χ to be faithful. Then χ is bijective map from xT to xχT , and its inverse is faithful. Proof. (i) This is just the definition of a faithful map rewritten under the hypothesis that the domain of χ contains wt. (ii) Assume χ to be faithful. Then χ is injective. Thus, as wt and wχt have the same (finite) number of elements, χ|wt is a bijective map from wt to wχt. (iii) This follows from (ii).

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Lemma 6.1.2 Let x be an element in X, let T and U be closed subsets of S such that T ⊆ U , and let χ be a faithful map from xU to X. For each element w ∈ xU , we define (wT )χT := (wχ)T. Then χT is a faithful map from (xT )(U//T ) to X/T . Proof. Let v and w be elements in xU such that vT = wT . Since vT = wT , there exists an element t in T such that w ∈ vt. Thus, as χ is assumed to be faithful, wχ ∈ (vχ)t. In particular, (vχ)T = (wχ)T . Thus, by definition, (vT )χT = (wT )χT . Thus, as v and w have been chosen arbitrarily in xU such that vT = wT , we have shown that χT is a map. In order to show that χT is faithful, we pick two elements y and z in xU , and we denote by u the uniquely determined element in S which satisfies z ∈ yu. Since z ∈ yu, zT ∈ (yT )uT . On the other hand, as χ is assumed to be faithful, z ∈ yu yields zχ ∈ (yχ)u. From zχ ∈ (yχ)u we obtain (zχ)T ∈ ((yχ)T )uT . Thus, by definition, (zT )χT ∈ (yT )χT uT . Thus, as y and z have been chosen arbitrarily in xU , we have shown that χT is faithful. Lemma 6.1.3 Let y and z be elements in X, and let T and U be closed subsets of S. Assume that T is strongly normal in U . Let χ be a faithful map from {y, z} to X. Then, if z ∈ yU and yχ ∈ yT , zχ ∈ zT . Proof. Let us assume that z ∈ yU . Then there exists an element u in U such that z ∈ yu. From u ∈ U and U ⊆ KS (T ) we obtain u∗ T u ⊆ T . Since z ∈ yu, y ∈ zu∗ and, as χ is assumed to be faithful, zχ ∈ yχu. Thus, assuming that yχ ∈ yT , we obtain zχ ∈ yT u ⊆ zu∗ T u ⊆ zT, and that finishes the proof of the lemma. Lemma 6.1.4 Let T , U , and V be closed subsets of S such that T ⊆ U ⊆ V . Assume that T is normal in V and that, for each element s in V \U , {s} = sT . Let x be an element in X, and let χ be a map from xV to X such that, for each element w in xV , χ|wU is faithful and wχ ∈ wT . Then χ is faithful. Proof. Let y and z be elements in xV . Then, by Lemma 2.1.4, z ∈ yV . Thus, there exists an element s in V such that z ∈ ys. We have to show that zχ ∈ yχs. Since χ|yU is assumed to be faithful, there is nothing to show if z ∈ yU . Thus, we may assume that z ∈ / yU . From z ∈ / yU and z ∈ ys we obtain s ∈ V \ U . Thus, by hypothesis, {s} = sT . On the other hand, we are assuming that T

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is normal in V . Thus, by Lemma 2.5.2(ii), T s = sT . Thus, we obtain from {s} = sT that {s} = T sT . We are assuming that yχ ∈ yT . Thus, by Lemma 2.1.4, y ∈ yχT . Thus, as zχ ∈ zT and z ∈ ys, zχ ∈ ysT ⊆ yχT sT. Thus, as {s} = T sT , zχ ∈ yχs. The following lemma shows in which way generating subsets of S and faithful maps work together. Lemma 6.1.5 Let R be a subset of S, and assume that R has finite valency. Assume that p = q for any two elements p and q in R with R (p) = R (q). Then the following hold. (i) There exists a symmetric element r in R such that {r} = R. (ii) Let x be an element in X, let r be the element in R, and let χ be an injective map from x R to X such that, for each element y in x R, yrχ ⊆ yχr. Then χ is faithful. Proof. (i) This follows immediately from the hypothesis that R contains only one element of length 1 (with respect to R). (ii) Let us denote by Q the set of all elements s in R for which there exists an element y in x R with ysχ ⊆ yχs. By way of contradiction, we assume that Q is not empty. Let us write instead of R , and let us fix an element q in Q with min (Q) = (q). Since 1 ∈ / Q and q ∈ Q, 1 = q. Thus, referring to (i) and Lemma 3.1.2 we find an element p in R such that q ∈ pr and (q) = (p) + 1. Since q ∈ Q, there exists an element y in x R with yqχ ⊆ yχq. Thus, there exists an element z in yq with zχ ∈ / yχq. Since z ∈ yq and q ∈ pr, z ∈ ypr. Thus, there exists an element w in yp such that z ∈ wr. Since (p) = (q) − 1, the (minimal) choice of q forces p ∈ / Q. Thus, as p ∈ R and y ∈ x R, ypχ ⊆ yχp. Thus, as w ∈ yp, wχ ∈ yχp. On the other hand, we obtain from z ∈ wr and wrχ ⊆ wχr that zχ ∈ wχr. Thus, zχ ∈ yχpr, so that there exists an element t in pr with zχ ∈ yχt. From zχ ∈ yχt and zχ ∈ / yχq we obtain t = q. Since t ∈ pr, (t) ≤ (p) + 1. Thus, as (q) = (p) + 1, (t) ≤ (q).

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From t = q and our hypothesis we obtain (t) = (q). Thus, as (t) ≤ (q), we must have that (t) ≤ (q) − 1. Thus, the (minimal) choice of q forces ytχ ⊆ yχt. However, we are assuming χ to be injective. Thus, as |yt| = |yχt|, ytχ = yχt. From zχ ∈ yχt and ytχ = yχt we obtain zχ ∈ ytχ. Thus, there exists an element z in yt such that zχ = z χ. Thus, as χ is assumed to be injective, we obtain z = z . Thus, as z ∈ yt, z ∈ yt. Thus, as z ∈ yq, q = t, contradiction.

6.2 Faithfully Embedded Closed Subsets Let Y and Z be subsets of X such that Y ⊆ Z, let χ be a faithful map from Y to X, and let χ ¯ be a faithful map from Z to X. Recall that we say that χ extends faithfully to χ ¯ if, for each element y in Y , yχ = y χ. ¯ Let T and U be closed subsets of S such that T ⊆ U . Recall that T is said to be faithfully embedded in U if, for any two elements y in X and z in yT , each faithful map χ from {y, z} to yU extends faithfully to a bijective map from yT to yχT . Lemma 6.2.1 Thin closed subsets of S are faithfully embedded in S. Proof. Let us fix a thin closed subset of S and call it T . Let y be an element in X, let z be an element in yT , and let χ be a faithful map from {y, z} to X. We have to show that χ extends faithfully to a bijective map from yT to yχT . For each element x in yT , we define xχ ¯ to be the uniquely defined element in yχt, where t stands for the uniquely defined element in T with x ∈ yt. Then χ ¯ is a bijective map from yT to yχT satisfying yχ = y χ ¯ and zχ = z χ. ¯ We claim that χ ¯ is faithful. In order to show this we pick two elements v and w in yT , and we denote by t the uniquely determined element in T satisfying w ∈ vt. We have to show that wχ ¯ ∈ v χt. ¯ Let us denote by p the element in T which satisfies v ∈ yp and by q the element in T which satisfies w ∈ yq. From v ∈ yp we obtain y ∈ vp∗ . Thus, as w ∈ yq, w ∈ vp∗ q. Thus, as w ∈ vt, t ∈ p∗ q. However, we know from Lemma 1.5.2 that |p∗ q| = 1. Thus, {t} = p∗ q. From v ∈ yp we obtain v χ ¯ ∈ y χp, ¯ and from w ∈ yq we obtain wχ ¯ ∈ y χq. ¯ ¯ ∈ v χt. ¯ Thus, wχ ¯ ∈ v χp ¯ ∗ q. Thus, as {t} = p∗ q, wχ The following two propositions show that the property of being faithfully embedded is inherited to closed subsets and to quotient schemes.

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Proposition 6.2.2 Let T , U , V , and W be closed subsets of S such that T ⊆ V , U ⊆ W , T ⊆ U , and V ⊆ W . Then, if U is faithfully embedded in W , T is faithfully embedded in V . Proof. Let y be an element in X, let z be an element in yT , and let χ be a faithful map from {y, z} to yV . We have to show that there exists a bijective faithful map χ ¯ from yT to yχT such that yχ = y χ ¯ and zχ = z χ. ¯ Since z ∈ yT and T ⊆ U , z ∈ yU . Since V ⊆ W and χ is a faithful map from {y, z} to yV , χ is a faithful map from {y, z} to yW . Thus, as U is assumed to be faithfully embedded in W , there exists a bijective faithful map χ ¯ from yU to yχU such that yχ = y χ ¯ and zχ = z χ. ¯ Since χ ¯ is faithful, we have (yT )χ ¯ ⊆ y χT ¯ . Thus, as yχ = y χ, ¯ (yT )χ ¯ ⊆ yχT . ¯ is a bijective Thus, as χ| ¯ yT is a faithful map from yT to yχT . Finally, as χ from yU to yχU , χ| ¯ yT is a bijective from yT to yχT . Proposition 6.2.3 Let T , U , and V be closed subsets of S such that T ⊆ U ⊆ V . If U is faithfully embedded in V , U//T is faithfully embedded in V //T . Proof. Let y and z be elements in X such that z ∈ yU , and let φ be a faithful map from {yT, zT } to (yT )(V //T ). We have to show that there exists a bijective faithful map φ¯ from (yT )(U//T ) to (yT )φ(U//T ) such that (yT )φ = ¯ (yT )φ¯ and (zT )φ = (zT )φ. Since z ∈ yU , there exists an element u in U such that z ∈ yu. In particular, zT ∈ (yT )uT . Thus, as φ is assumed to be faithful, (zT )φ ∈ (yT )φuT . Thus, by definition, (zT )φ ⊆ (yT )φT uT . Thus, there exist elements v in (yT )φ and w in (zT )φ such that w ∈ vu. Thus, as U is assumed to be faithfully embedded in V , we obtain from v ∈ yV a bijective faithful map χ from yU to vU such that yχ = v and zχ = w. For each element x ∈ yU , we define (xT )φ¯ := (xχ)T. By Lemma 6.1.2, φ¯ is a faithful map from (yT )(U//T ) to (vT )(U//T ). Moreover, we have that (yT )φ = vT = (yχ)T = (yT )φ¯ ¯ and, similarly, (zT )φ = (zT )φ. Since χ is a bijective map from yU to vU , we obtain from (yT )φ = vT that φ¯ is a bijective map from (yT )(U//T ) to (yT )φ(U//T ). For the remainder of this section, we shall assume S to have finite valency. Proposition 6.2.2 says that the property of being faithfully embedded is inherited to closed subsets. Given closed subsets T and U in S with T ⊆ U , we shall now ask ourselves under which hypotheses the property of being faithfully embedded in S can be lifted from T to U .


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Lemma 6.2.4 Let T , U , and V be closed subsets of S such that T ⊆ U ⊆ V ⊆ KS (T ) and, for each element s in V \ U , {s} = sT . Assume that U is faithfully embedded in S. Let y, z be elements in X such that z ∈ yV , and let χ be a faithful map from {y, z} to X such that yχ ∈ yT and zχ ∈ zT . Then χ extends faithfully to a bijective map from yV to yχV . Proof. Since U is assumed to be faithfully embedded in S, there exists a bijective faithful map χ ¯ from yU ∪ zU to yχU ∪ zχU such that χ|yU and χ|zU are faithful. (It might be that yU = zU , it also might be that yU = zU .) Thus, for each element x in yU ∪ zU , xχ ∈ xT ; cf. Lemma 6.1.3. For each element x in yV \ (yU ∪ zU ) we define xχ := x. Then, by Lemma 6.1.4, χ must be faithful. Lemma 6.2.5 Let T , U , and V be closed subsets of S such that T ⊆ U ⊆ V ⊆ KS (T ) and, for each element s in V \ U , {s} = sT . Assume that U is faithfully embedded in S. Let x be an element in X. Then, for each faithful map ψ from (xT )(V //T ) to X/T , there exists a faithful map φ from xV to X such that, for each element w in xV , wφ ∈ (wT )ψ. Proof. By Lemma 2.1.4, we find a subset Z of xV such that, for each element w in xV , |Z ∩ wU | = 1. Let ψ be a faithful map from (xT )(V //T ) to X/T . For each element u in Z, we pick an element in (uT )ψ and call it uφ. Since U is assumed to be faithfully embedded in S, there exists, for each element u in Z, a bijective faithful map φu from uU to uφU such that uφu = uφ. We extend φ from Z to xV by setting wφ := wφu for any two elements u in Z and w in uU . Let us first prove that, for each element w in xV , wφ ∈ (wT )ψ. In order to do so we pick an element w in xV , and we denote by u the uniquely determined element in Z ∩wU . Since u ∈ wU , w ∈ uU ; cf. Lemma 2.1.3. Thus, there exists an element p in U such that w ∈ up. Thus, as φ is faithful on uU , wφ ∈ uφp. On the other hand, uφ is defined to be an element of (uT )ψ. Thus, wφ ∈ (uT )ψp. By Lemma 4.2.5(ii), V //T is thin. Thus, there exists an element v in X such that {vT } = (uT )ψpT . Thus, by Lemma 4.1.4, vT = (uT )ψT pT . Thus, as wφ ∈ (uT )ψp, wφ ∈ vT. From w ∈ up we also obtain wT ∈ (uT )pT . Thus, as ψ is assumed to be faithful, we must have

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(wT )ψ ∈ (uT )ψpT . Thus, as {vT } = (uT )ψpT , (wT )ψ = vT . Thus, as wφ ∈ vT , wφ ∈ (wT )ψ. Let us now show that φ is faithful. In order to do so we pick two elements y and z in xV . From y, z ∈ xV we obtain z ∈ yV ; cf. Lemma 2.1.4. Thus, there exists an element q in V such that z ∈ yq. We have to show that zφ ∈ yφq. If q ∈ U , we are done by the definition of φ. Thus, we may assume that q ∈ / U. Since z ∈ yq, zT ∈ (yT )q T . Thus, as ψ is assumed to be faithful, we must have (zT )ψ ∈ (yT )ψq T . From yφ ∈ (yT )ψ we obtain yφT = (yT )ψ. Similarly, as zφ ∈ (zT )ψ, zφT = (zT )ψ. Thus, as (zT )ψ ∈ (yT )ψq T , zφT ∈ (yφT )q T . By Lemma 4.1.4, this means that zφ ∈ yφT qT . On the other hand, we are assuming that q ∈ / U . Thus, by hypothesis, {q} = qT . Thus, as we are assuming that V ⊆ KS (T ), we obtain from Lemma 2.5.5 and Lemma 2.5.2(ii) that T q = qT . Thus, as we are assuming that {q} = qT , {q} = T qT . Thus, as zφ ∈ yφT qT , zφ ∈ yφq. The following theorem is a partial inverse of Proposition 6.2.2. It is similar to [25; Theorem A]. Theorem 6.2.6 Let T , U , and V be closed subsets of S such that T ⊆ U ⊆ V , V ⊆ KS (T ), and, for each element s in V \ U , {s} = sT . Then, if U is faithfully embedded in S, so is V . Proof. Let y, z be elements in X such that z ∈ yV , and let χ be a faithful map from {y, z} to X. We have to show that χ extends faithfully to a bijective map from yV to yχV . Since z ∈ yV , there exists an element r in V such that z ∈ yr. Since χ is assumed to be faithful, we have zχ ∈ yχr. Since z ∈ yr, zT ∈ (yT )(rT ). Similarly, as zχ ∈ yχr, zχT ∈ (yχT )(rT ). Thus, there exists a faithful map ψ from (yT )(V //T ) to X/T with (yT )ψ = yχT and (zT )ψ = zχT ; cf. Lemma 6.2.1 together with Lemma 4.2.5(ii). Thus, by Lemma 6.2.5, there exists a faithful map φ from yV to X such that, for each element x in yV , xφ ∈ (xT )ψ. From yφ ∈ (yT )ψ and (yT )ψ = yχT we obtain yφ ∈ yχT . Thus, by Lemma 2.1.4, yχ ∈ yφT . Similarly, we obtain from zφ ∈ (zT )ψ and (zT )ψ = zχT that zχ ∈ zφT . Finally, as yχ ∈ yφT and zχ ∈ zφT , there exists a faithful map φ from yχV to X such that yφφ = yχ and zφφ = zχ; cf. Lemma 6.2.4. Let T and U be closed subsets of S such that T ⊆ U .


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Proposition 6.2.3 says that the property of being faithfully embedded is inherited from U to U//T . We shall now ask ourselves under which hypotheses the property of being faithfully embedded in S can be transferred from U//T to U . Let x be an element in X, and let us denote by F (respectively F T ) the set of all faithful maps from xU (respectively (xT )(U//T )) to X (respectively X/T ). For each element χ in F, we define λx (χ) := χT . Then by Lemma 6.1.2, λx is a map from F to F T . We shall say that U covers T if, for each element x in X, the map λx is surjective. Lemma 6.2.7 Let T and U be closed subsets of S such that T ⊆ U . Assume that U contains a closed subset V of S which covers T . Assume further that {u ∈ U | 1 = |T uT |} ⊆ V . Then U covers T . Proof. Let x be an element in X, let φ be a faithful map from (xT )(U//T ) to X/T . We have to find a faithful map χ from xU to X such that χT = φ. Let Y be a subset of xU such that, for each element y in xU , |Y ∩yV | = 1. Let w be an element in Y . Then φ|(wT )(V //T ) is faithful. Thus, as V is assumed to cover T , there exists a faithful map χw from wV to X such that (χw )T = φ|(wT )(V //T ) . For each element y in xU , we define yχ := yχw , where w denotes the uniquely determined element w in Y which satisfies y ∈ wV . Then χ is a map from xU to X satisfying χT = φ. We claim that χ is faithful. In order to prove that χ is faithful, we pick two elements y and z in xU , and we denote by u the uniquely determined element in U which satisfies z ∈ yu. We have to show that zχ ∈ yχu. Let w denote the uniquely determined element in Y ∩ yV . Then, if u ∈ V , z ∈ yu ⊆ yV = wV . Thus, in this case, zχ = zχw ∈ yχw u = yχu. Thus, we may (and we shall) assume that u ∈ / V. Since w ∈ Y ∩ yV , (yT )φ = (yT )φ|(wT )(V //T ) = (yT )(χw )T = (yχw )T = (yχ)T. Similarly, (zT )φ = (zχ)T . From z ∈ yu we obtain zT ∈ (yT )uT . Thus, as φ is assumed to be faithful, (zT )φ ∈ (yT )φuT . Thus, the above two equations yield (zχ)T ∈ ((yχ)T )uT . Thus, by definition, zχ ∈ yχT uT . On the other hand, we are assuming that u ∈ / V . Thus, by hypothesis, {u} = T uT . Thus, zχ ∈ yχu. In our last result, we do not need the full strength of the notion of a faithfully embedded closed subset. In order to weaken this concept we fix closed subsets

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T and U such that T ⊆ U . We say that U is faithfully T -embedded in U if, for any two elements y in X and z in yU , each faithful map χ from {y, z} to X with yχ ∈ yT and zχ ∈ zT extends faithfully to a bijective (faithful) map from yU to yχU . Here is a partial inverse of Proposition 6.2.3. Theorem 6.2.8 Let T and U be closed subsets of S such that T ⊆ U . Assume that U contains a closed subset V of S which covers T . Assume further that {h ∈ U | 1 = |T hT |} ⊆ V and that V is faithfully T -embedded in S. Then, if U//T is faithfully embedded in S//T , U is faithfully embedded in S. Proof. Let y and z be elements in X such that z ∈ yU , and let χ be a faithful map from {y, z} to X. We have to show that χ extends faithfully to a bijective map from yU to (yχ)U . Since z ∈ yU , there exists an element u in U such that z ∈ yu. Thus, as χ is assumed to be faithful, zχ ∈ yχu. From z ∈ yu, we obtain zT ∈ (yT )uT . Similarly, as zχ ∈ yχu, (zχ)T ∈ (yχ)T uT . Thus, as U//T is assumed to be faithfully embedded in S//T , there exists a faithful map φ from (yT )(U//T ) to X/T such that (yχ)T = (yT )φ and (zχ)T = (zT )φ. We are assuming that V covers T and that {u ∈ U | 1 = |T uT |} ⊆ V . Thus, by Lemma 6.2.7, U covers T . Thus, by definition, there exists a faithful map ζ from yU to X such that ζ T = φ. From ζ T = φ we obtain (yζ)T = (yT )ζ T = (yT )φ = (yχ)T. In particular, yζ ∈ yχT . Similarly, zζ ∈ zχT . Set yζη := yχ and zζη := zχ. Then yζη ∈ yζT and zζη ∈ zζT . Moreover, as zχ ∈ yχu and zζ ∈ yζu, η is faithful. Thus, as we are assuming that U is faithfully T -embedded in S, η extends faithfully to a faithful map χ¯ from (yζ)U to (yζη)U . Thus, as ζ is a faithful map from yU to (yζ)U , χ ¯ is faithful. This finishes the proof, because we have χ| ¯ {y,z} = χ.

6.3 The Schur Group of a Closed Subset Let x be an element in X, and let T be a closed subset of S. Recall that the group of all bijective faithful maps from xT to xT is called the Schur group of T with respect to x. The identity on xT is the identity element of the Schur group of T with respect to x. Let G be a subgroup of the Schur group of T with respect to x. Let Y be a subset of xT . Assume that, for any two elements y in Y and g in G, yg ∈ Y . One says that the group G acts transitively on Y if, for any two elements v and w in Y , there exists an element g in G such that vg = w.

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We set Gx := {g ∈ G | xg = x}. The set Gx is called the stabilizer of x in G. Recall that a closed subset of S is called schurian if it is faithfully embedded in itself. Theorem 6.3.1 Let x be an element in X, and let T be a closed subset of S. Then the following statements are equivalent. (a) The Schur group G of T with respect to x acts transitively on xT and, for each element t in T , Gx acts transitively on xt. (b) Let G be the Schur group of T with respect to x. Then Tx ∼ = Gτ //(Gx )τ . (c) The scheme Tx is a quotient scheme of a thin scheme. (d) The scheme Tx is schurian. Proof. (a) ⇒ (b) We set H := Gx . We pick an element w in xT . For any two elements e and f in G with w = xe−1 and w = xf −1 , we have xe−1 f = x. That means that e−1 f ∈ H. Thus, by Lemma 1.3.3(ii) and Lemma 2.1.4, eH = f H, and this implies eH τ = f H τ . For any two elements w in xT and g in G with w = xg −1 , we define wφ := gH τ . Let e and f be elements in G such that xe−1 ∈ xt and xf −1 ∈ xt. Then, by (a), there exists an element h in H such that xe−1 h = xf −1 . It follows that e−1 hf ∈ H, so that e ∈ Hf H. Thus, HeH = Hf H, and this yields τ τ (eτ )H = (f τ )H . For any two elements t in T and g in G with xg −1 ∈ xt, we define τ

(txT )φ := (g τ )H . Note that φ is a map from xT ∪ TxT to G/H τ ∪ Gτ //H τ which maps xT to G/H τ and TxT to Gτ //H τ . Let us now show that φ is an isomorphism from xT ∪ TxT to G/H τ ∪ Gτ //H τ . Let w be an element in xT , and let t be an element in T such that w ∈ xt. Since G is assumed to act transitively on xT , we shall be done if we succeed in showing that wφ ∈ xφtxT φ. Since G acts transitively on xT , there exists an element g in G such that w = xg −1 . Thus, by definition, wφ = gH τ . From w = xg −1 and w ∈ xt we obtain xg −1 ∈ xt. Thus, by definition, τ (txT )φ = (g τ )H . τ

Since g ∈ 1H τ g τ H τ , gH τ ∈ (1H τ )(g τ )H . Thus, as xφ = 1H τ , wφ = gH τ τ and (txT )φ = (g τ )H , wφ ∈ xφtxT φ.

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(b) ⇒ (c) This is obvious. (c) ⇒ (d) Considering Lemma 6.2.1 this is an immediate consequence of Proposition 6.2.3. (d) ⇒ (a) Let y and z be elements in xT . Since Tx is assumed to be schurian, TxT is faithfully embedded in TxT . Thus, there exists a bijective faithful map g from xT to xT such that yg = z. Let t be an element in T , and let y and z be elements in xt. Then, as TxT is assumed to be faithfully embedded in TxT , there exists a bijective faithful map from xT to xT such that yg = z. From this we obtain (a). Corollary 6.3.2 The scheme S is schurian if and only if S is a quotient scheme of a thin scheme. Proof. This follows immediately from Theorem 6.3.1. Theorem 6.3.3 A closed subset T of S is schurian if and only if, for each element x in X, the Schur group G of T with respect to x acts transitively on xT and, for each element t in T , Gx acts transitively on xt. Proof. Let T be a closed subset of S, and let us first assume T to be schurian. We fix an element x in X. In order to show that the Schur group G of T with respect to x acts transitively on xT , we fix elements y and z in xT . The map χ from {y} to {z} with yχ := z is faithful. Thus, by definition, there exists an element g in G such that yg = z. This shows that G acts transitively on xT . Let t be an element in T , and let y and y be elements in xt. Define xχ := x and yχ := y . Then χ is a faithful map from {x, y} to xT . Thus, as T is assumed to be schurian, there exists an element g in Gx such that yg = y . Let us now assume that, for each element x in X, the Schur group G of T with respect to x acts transitively on xT and that, for each element t in T , Gx acts transitively on xt. We have to show that T is faithfully embedded in T. Let y be an element in X, let z be an element in yT , and let χ be a faithful map from {y, z} to yT . Since z ∈ yT , there exists an element t in T such that z ∈ yt. Thus, as χ is assumed to be faithful, zχ ∈ yχt. We are assuming that G acts transitively on yT . Thus, there exists an element e in G such that ye = yχ. We are assuming that Gye acts transitively on yet. Thus, there exists an element f in Gye such that zef = zχ. Theorem 6.3.1, together with Proposition 6.2.2, also shows that, if S is schurian, each closed subset of S is faithfully embedded in S. Lemma 6.2.1 says that thin closed subsets are faithfully embedded in S. The set of all closed

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subsets of S which are faithfully embedded in S can be viewed as a measure how close S is to be schurian. The following lemma will be needed in Section 6.5 and in Section 12.3. Lemma 6.3.4 Let x be an element in X, let R be a subset of S, and let G be a subgroup of the Schur group of R with respect to x. Assume that, for any two elements y and z in x R with z ∈ y(R∗ ∪ R), there exists an element g in G with yg = z. Then G acts transitively on x R. Proof. This follows from Lemma 3.1.1(i) by induction. We shall now look at closed subsets T of S in which each element has valency at most 2. The following lemma is [23; Lemma 3.5]. Note that we do not assume S to have finite valency. Lemma 6.3.5 Let x be an element in X, let T be a closed subset of S, and let G denote the Schur group of T with respect to x. Assume that T is not thin and that, for each element t in T , nt ≤ 2. Then 2 ≤ |Gx |. Proof. Let x be an element in X, and let w be an element in xT . Since w ∈ xT , there exists an element t in T such that w ∈ xt. If nt = 1, we set wh := w. If nt = 2, we define wh to be the uniquely determined element in xt \ {w}. Note that h ∈ Gx . Note also that h is a bijective map from xT to xT . Since T is assumed to be not thin, 1 = h. Thus, we just have to show that h is faithful. In order to show that h is faithful we pick elements y and z in xT . From y ∈ xT and z ∈ xT we obtain z ∈ yT ; cf. Lemma 2.1.4. Thus, there exists an element t in T such that z ∈ yt. We now have to show that zh ∈ yht. Let us denote by p the uniquely determined element in T such that y ∈ xp. Then, as z ∈ yt, z ∈ xpt. Thus, there exists an element q in pt such that z ∈ xq. From z ∈ xq we obtain zh ∈ xq. Let us first assume that np = 1. In this case, we have {y} = xp and yh = y. Thus, as zh ∈ xq and q ∈ pt, zh ∈ xpt = yt = yht, so that we are done in this case. The case where nq = 1 is treated similarly. Let us, therefore, assume that np = 2 and that nq = 2. Since y ∈ xp and z ∈ xq ∩ yt, 1 ≤ aqt∗ p . Thus, as yh ∈ xp, there exists an element w in xq ∩ yht. Since w ∈ xq = {z, zh} we now must have w = z or w = zh.

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If w = zh, w ∈ yht leads to zh ∈ yht, and we are done. If w = z, {y, yh} ⊆ xp ∩ zt∗ . Thus, as z ∈ xq, 2 ≤ aptq . Thus, as zh ∈ xq, 2 ≤ |xp∩zht∗ |. Thus, as xp = {y, yh}, {y, yh} ⊆ zht∗ . It follows that zh ∈ yht; again, we are done. The following lemma is [23; Theorem 3.7(iv)]. Note that, again, we do not require S to have finite valency. Lemma 6.3.6 Let x be an element in X, let T be a closed subset of S, and let G denote the Schur group of T with respect to x. Assume that Oϑ (T ) does not have valency 2 and that, for each element t in T , nt ≤ 2. Then |Gx | ≤ 2. Proof. Assume, by way of contradiction, that 3 ≤ |Gx |. Then there exists an element h in Gx \ {1} which fixes an element in x(T \ Oϑ (T )). We define P to be the set of all elements t in T \ Oϑ (T ) such that h fixes none of the two elements in xt, and we set Q := T \ P \ Oϑ (T ). Since 1 = h, P is not empty. Recall also that h fixes at least one element in x(T \ Oϑ (T )). Thus, Q is not empty. Thus, by Lemma 1.5.5, we shall be done if we succeed in showing that, for any two elements p in P and q in Q, |p∗ q| = 1. (Recall that we are assuming that Oϑ (T ) does not have valency 2.) Let p be an element in P , let q be an element in Q, let v be an element in xp, and let w be an element in xq. Then vh = v and wh = w. Since v ∈ xp, x ∈ vp∗ . Thus, as w ∈ xq, w ∈ vp∗ q. From w ∈ vp∗ q we obtain an element r in p∗ q such that w ∈ vr. It follows that v ∈ xp ∩ wr∗ . Since v ∈ xp and xh = x, vh ∈ xph ⊆ xhp = xp. Similarly, as wh = w and v ∈ wr∗ , vh ∈ wr∗ . Thus, vh ∈ xp ∩ wr∗ . From {vh, v} ⊆ xp ∩ wr∗ , w ∈ xq, and vh = v, we conclude that 2 ≤ aprq . Thus, as np = 2, aprq = np . Thus, by Lemma 1.4.3, {r} = p∗ q.

6.4 Elements of Valency 2 In Lemma 6.3.5 and Lemma 6.3.6 we investigated the Schur group of closed subsets of S in which each element has valency at most 2. From Lemma 1.5.6(ii) we know that, for each element s in S with ns = 2, ns∗ s ∈ {2, 3}. In this section, we focus on closed subsets T of S in which each element t satisfies nt ≤ 2 and nt∗ t = 2. We shall first prove that a closed subset T of S is schurian if each element t of T satisfies nt ≤ 2 and nt∗ t = 2. After that, we shall see which finite groups are Schur groups of these closed subsets. This way we shall establish a recognition theorem for these schemes.


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The following proposition is [31; (4.1)]. Note that we do not require the closed subset T of S to have finite valency. Proposition 6.4.1 Let T be a closed subset of S, and assume that, for each element t in T , nt ≤ 2 and nt∗ t = 2. Then T is faithfully embedded in S. Proof. Let y and y be elements in X, let s be an element in T , let z be an element in ys, and let z be an element in y s. We have to show that there exists a bijective faithful map χ from yT to y T such that yχ = y and zχ = z . Let us first assume that ns = 2. In this case, we define χ in the following way. Let x be an element in yT . Then there exist elements p and q in T such that x ∈ yp ∩ zq ∗ . Thus, as z ∈ ys, 1 ≤ apqs . On the other hand, we are assuming that np ≤ 2, nq ≤ 2, ns = 2, and np∗ p = 2. Thus, by Lemma 1.5.4, apqs ≤ 1, so that apqs = 1. Thus, as z ∈ y s, y p ∩ z q ∗ contains exactly one element. We define xχ to be the uniquely determined element in y p ∩ z q ∗ . It follows from the definition of χ that χ is a surjective map from yT to y T , that yχ = y , and that zχ = z . In order to show that χ is faithful we fix elements v in yT and r in T . We have to show that vrχ ⊆ vχr; cf. Lemma 6.1.1(i). Let us assume, by way of contradiction, that vrχ ⊆ vχr. Then there exists an element w in vr such that wχ ∈ / vχr. Since wχ ∈ / vχr, there exists an element r in T such that r = r and wχ ∈ vχr . Let p and q be elements in S such that v ∈ yp ∩ zq ∗ , and let t and u be elements in S such that w ∈ yt ∩ zu∗ . From the definition of χ we obtain vχ ∈ yχp ∩ zχq ∗ and wχ ∈ yχt ∩ zχu∗ . Since wχ ∈ vχr and vχ ∈ yχp, wχ ∈ yχpr . Thus, as wχ ∈ yχt, t ∈ pr . Thus, as w ∈ yt, w ∈ ypr . Thus, there exists an element v in yp such that w ∈ v r . From w ∈ v r , w ∈ vr, and r = r we obtain v = v. Similarly, as zχ ∈ wχu and wχ ∈ vχr , zχ ∈ vχr u. Thus, as zχ ∈ vχq, q ∈ r u. Thus, as z ∈ vq, z ∈ vr u. Thus, there exists an element w in vr such that z ∈ w u. From w ∈ vr , w ∈ vr, and r = r we obtain w = w. Let us denote by t the uniquely determined element in T satisfying w ∈ yt . If t = t, {w, w } ⊆ yt ∩ zu∗ . Thus, by Lemma 1.5.4, w = w, contradiction. Thus, t = t. From w ∈ yt we obtain w χ ∈ yχt . Thus, as wχ ∈ yχt and t = t, w χ = wχ. From w ∈ zu∗ we obtain w χ ∈ zχu∗ . Thus, as wχ ∈ zχu∗ , w χ = wχ and nu∗ ≤ 2, zχu∗ = {wχ, w χ}. Since z ∈ vq and w ∈ vr ∩ zu∗ , 1 ≤ aruq . Thus, as zχ ∈ vχq, there exists an element x in vχr ∩ zχu∗ . From x ∈ vχr, wχ ∈ vχr , and r = r we obtain x = wχ. Thus, as x ∈ zχu∗ = {wχ, w χ}, x = w χ. Thus, as x ∈ vχr, w χ ∈ vχr.

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Since w χ ∈ yχt and vχ ∈ yχp ∩ w χr∗ , 1 ≤ aprt . Thus, as w ∈ yt , there exists an element x in yp ∩ w r∗ . From x ∈ w r∗ , v ∈ w r∗ , and r = r we obtain x = v. Thus, as x ∈ yp = {v, v }, x = v . Thus, as w ∈ xr, w ∈ v r. Since v ∈ yp and w ∈ yt ∩ vr, 1 ≤ atr∗ p . Thus, as v ∈ yp, there exists an element x in yt ∩ v r. Similarly, as v ∈ yp and w ∈ yt ∩ v r , 1 ≤ atr∗ p . Thus, as v ∈ yp, there exists an element x in yt ∩ vr . Since x ∈ v r, w ∈ v r , and r = r, x = w. Since x ∈ vr , w ∈ vr, and r = r, x = w. Thus, as {w, x, x } ⊆ yt, x = x. Thus, as x ∈ vr , x ∈ vr . It follows that {x, w } ⊆ vr ∩ v r. Since v ∈ yp and y ∈ vp∗ , v ∈ vp∗ p. Thus, as v = v, there exists an element m in p∗ p \ {1} such that v ∈ vm. Since m ∈ p∗ p \ {1}, {1, m} = p∗ p; cf. Lemma 1.5.6(i). Thus, as we are assuming that np∗ p = 2, nm = 2. Thus, by Lemma 1.5.4, ar r∗ m = 1. From ar r∗ m = 1, v ∈ vm, and {w , x} ⊆ vr ∩ v r we obtain w = x. Thus, as w ∈ yt and x ∈ yt, t = t, contradiction. This contradiction proves that χ is faithful. Let us now assume that ns = 1. By Lemma 6.2.1, we may assume that T is not thin. Thus, there exists an element r in T such that nr = 2. Let x be an element in yr∗ . Then, by the previous case (applied to x and y instead of y and z), we obtain a bijective faithful map χ from yT to y T such that yχ = y . Since s is assumed to be thin, zχ = z . Let p and q be elements in S. If S is thin, one usually writes q p instead of {q}p . Recall that, for each element s in S, we write s instead of {s}. The following lemma shows in which way thin schemes provide the situation required in Proposition 6.4.1. Lemma 6.4.2 Assume S to be thin, let s be an element in S, and let l be an involution of S. Then the following hold. (i) We have nsl ≤ 2. (ii) Assume that nsl = 2. Then n(sl )∗ sl = 2 if and only if lls = ls l. Proof. (i) Since l is assumed to be an involution, nlsl ≤ 4. On the other hand, we know from Theorem 4.1.3(iii) that nsl nl = nlsl . Thus, as nl = 2, nsl ≤ 2. (ii) Let us assume that nsl = 2. Then, by Lemma 1.5.6(i), there exists an element r in S such that 1l = rl and {1l , rl } = (sl )∗ sl . / l; cf. Lemma 4.1.1. From 1l = rl we obtain r ∈


119

From rl ∈ (sl )∗ sl we obtain r ∈ ls∗ ls l; cf. Lemma 4.1.4. Thus, as r ∈ S \ l, r ∈ lls l. Thus, lr = rl if and only if lls = ls l. However, we have lr = rl if and only if nrl = 1; cf. Lemma 4.2.5(i). Moreover, as {1l , rl } = (sl )∗ sl , this is the case if and only if n(sl )∗ sl = 2. Assume S to be thin, and let l be an involution of S. If S has finite valency, the local condition that, for each element s in S \ CS (l), lls = ls l can be expressed globally.1 This fact, namely that condition (c) of the following theorem follows from condition (a) or condition (b), is due to George Glauberman; cf. [15; Theorem 1]. The proof of Glauberman’s theorem requires Richard Brauer’s Second Main Theorem on modular representations of finite groups; cf. [5]. A proof of Glauberman’s theorem is beyond the scope of this monograph. From Lemma 2.5.2(iv) and Lemma 2.3.6(i) we obtain that a thin scheme possesses a uniquely determined maximal normal closed subset of odd valency. We shall denote this closed subset by O(S). Theorem 6.4.3 Assume S to be thin and to have finite valency. Then, for each involution l of S, the following conditions are equivalent. (a) For each element s in S \ CS (l), ls l has odd order. (b) For each element s in S \ CS (l), lls = ls l. (c) We have [O(S), l]CS (l) = S. Proof. (a) ⇒ (b) Let us assume, by way of contradiction, that S \ CS (l) contains an element s with lls = ls l. Then (ls l)2 = ls lls l = ls ls ll = 1. Thus, / CS (l), ls l has order 2. ls l has order at most 2. Thus, as s ∈ (b) ⇒ (a) Let us assume, by way of contradiction, that S \ CS (l) contains an element s such that ls l has even order. Then there exists an integer n such that (ls l)n is an involution. Thus, (lls )n = (ls l)n . Set k := (ls l)n−1 ls . Then s s m lk = kl. Moreover, if n is even and 2m + 2 = n, k = ll (ll ) . If n is odd and s m 2m + 1 = n, then k = (ls )(ll ) . (a) ⇒ (c) This is the content of [15; Theorem 1]. (c) ⇒ (a) Assume that [O(S), l]CS (l) = S, and let s be an element in S \CS (l). Then there exist elements p in [O(S), l] and q in CS (l) such that qp = s. Then lp = ls . Thus, ls l = lp l = p−1 pl ∈ O(S), so that ls l has odd order. Assume that S is simple and satisfies condition (c) in Theorem 6.4.3. Then

l = S or [O(S), l] = S. In the second case, O(S) = S, and that means that S has odd valency. However, l has even valency, so that, according to Lemma 2.3.6(ii), S cannot have odd valency. This contradiction shows that S cannot 1

Recall that CS (l) is an abbreviation for CS ({l}).

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be simple if S satisfies one of the three conditions (a), (b), or (c) of Theorem 6.4.3. The following theorem is the main result of this section. It was first proved in [31; (5.1)]. Theorem 6.4.4 Assume that S is not thin and that S has finite valency. Assume that, for each element s in S, ns ≤ 2 and ns∗ s = 2. Then there exists ¯ ¯l]CS¯ (¯l) = S¯ a finite thin scheme S¯ and an involution ¯l in S¯ such that [O(S), ∼ ¯ ¯ and S = S// l. Proof. Let us denote by G the Schur group of S, and let us fix an element x in X. Then, by Lemma 6.3.5 and Lemma 6.3.6, there exists an element h in G \ {1} such that {1, h} = Gx . Let g be an element in G \ CG (h). We shall prove that hhg = hg h. Let us denote by s the uniquely determined element in S satisfying xg ∈ xs. Then xgh ∈ xsh ⊆ xhs = xs, so that {xg, xgh} ⊆ xs. Assume that xgh = xg. Then xghg −1 = x. Thus, ghg −1 ∈ Gx . Thus, as {1, h} = Gx , ghg −1 = h. It follows that g ∈ CG (h), contrary to the choice of g. This contradiction shows that xgh = xg. From {xg, xgh} ⊆ xs and xgh = xg we obtain {xg, xgh} = xs. In particular, xgs∗ ∪ xghs∗ = xss∗ . Since xg ∈ xs, x ∈ xsg −1 ⊆ xg −1 s. Thus, xg −1 ∈ xs∗ , so that xhg = xg −1 hg ∈ xs∗ hg ⊆ xhgs∗ = xgs∗ . On the other hand, we obtain from xg ∈ xs also that x ∈ xgs∗ . Thus, as x = xhg , {x, xhg } = xgs∗ . From this we obtain

{x, xhg h} = xghs∗ .

Thus, as xgs∗ ∪ xghs∗ = xss∗ , {x, xhg , xhg h} = xss∗ . Thus, as we are assuming that nss∗ = 2, we conclude that xhg = xhg h. In particular, we have hhg = hg h. We set S¯ := Gτ and ¯l := hτ . For each element g in G \ CG (h), we have hhg = hg h. Thus, for each element ¯ ¯ ¯l]CS¯ (¯l) = S. s¯ in S¯ \ CS¯ (¯l), ¯l¯ls¯ = ¯ls¯¯l. Thus, by Theorem 6.4.3, [O(S), ¯ ¯l. Note finally that, by Theorem 6.3.1 and Proposition 6.4.1, S ∼ = S//


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Note that each element s in S satisfies nss∗ = 2 if Oϑ (Oϑ (S)) has odd valency. Thus, the conclusion of Theorem 6.4.4 remains valid if we assume that all elements of S have valency at most 2 and Oϑ (S) has odd valency. Here is the converse of Theorem 6.4.4. It makes Theorem 6.4.4 to be one of our recognition theorems. Theorem 6.4.5 Assume S to be thin and to have finite valency. Let l be an involution of S such that [O(S), l]CS (l) = S. Then S// l has finite valency and, for each element s in S, nsl ≤ 2 and n(sl )∗ sl = 2. Proof. Considering Theorem 6.4.3 we obtain the claim from Lemma 6.4.2. Note that Theorem 6.4.4, in the form as we stated it, requires the abovementioned theorem of George Glauberman. In order to avoid the use of Glauberman’s theorem one could, of course, replace the condition that ¯ ¯l]CS¯ (¯l) = S¯ [O(S), with one of the equivalent conditions (a) or (b) in Lemma 6.4.3. Contrary to this, the following corollary of Theorem 6.4.4 relies essentially on Glauberman’s theorem. Corollary 6.4.6 Assume that S has finite valency and that, for each element s in S, ns ≤ 2 and nss∗ = 2. Then Oϑ (S)Oϑ (S) = S, and Oϑ (S) has odd valency. Proof. The statement is obvious if S is thin. If S is not thin, we obtain from Theorem 6.4.4 a finite thin scheme S¯ and an involution ¯l in S¯ such that ¯ ¯l]CS¯ (¯l) = S¯ and S ∼ ¯ ¯l. [O(S), = S//

¯ ¯l]CS¯ (¯l) = S¯ we obtain From [O(S), ¯ ¯l; ¯ ¯l]// ¯l)(CS¯ (¯l)// ¯l) = S//

([O(S), cf. Lemma 4.1.6. ¯ ¯l). On the other hand, we ¯ ¯l we obtain Oϑ (S) ∼ From S ∼ = Oϑ (S//

= S//

ϑ ¯ ¯ ¯ ¯l] ¯l// ¯l. Thus, as [S, ¯ ¯l] = know from Lemma 4.2.3 that O (S// l) = [S, ¯ ¯ [O(S), l], ¯ ¯l] ¯l// ¯l. Oϑ (S) ∼ = [O(S), ¯ ¯l). Moreover, by Lemma ¯ ¯l we also obtain Oϑ (S) ∼ From S ∼ = Oϑ (S//

= S//

¯ ¯ ¯ ¯ 4.2.5(i), Oϑ (S// l) = CS¯ (l)// l. Thus, Oϑ (S) ∼ = CS¯ (¯l)// ¯l. ¯ ¯l. Now Oϑ (S)Oϑ (S) = S follows from S ∼ = S//

ϑ ¯ ¯l] ¯l// ¯l. That O (S) has odd valency follows from Oϑ (S) ∼ = [O(S),

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6.5 More About Elements of Valency 2 In this section, S is assumed to have finite valency. We shall first prove that a closed subset of S which is generated by a single symmetric element of valency 2 is faithfully embedded in S. After that we shall look at closed subsets in which each non-identity element has valency 2. Lemma 6.5.1 Let s be a symmetric element of S, assume that s has valency 2, and let y and z be elements in X. Then there exists a bijective faithful map χ from y s to z s such that yχ = z. Proof. By Lemma 3.3.2 and Lemma 6.1.5(ii), we shall be done if we succeed in showing that there exists a map χ from y s to X such that yχ = z and ysχ ⊆ yχs. (As for the bijectivity we refer to Lemma 6.1.1(iii).) Let us write instead of {s} . For each non-negative integer n, we define Rn to be the set of all elements r in s such that (r) ≤ n. Let us assume, by way of contradiction, that there does not exist a map χ from y s to X with yχ = z and ysχ ⊆ yχs. Then, by Lemma 3.1.1(i), there exists a non-negative integer n such that there does not exist a map χ from yRn to X such that yχ = z and, for any two elements u and v in yRn , v ∈ us implies vχ ∈ uχs. We pick n as small as possible. Clearly, 1 ≤ n. Thus, the minimal choice of n gives us a map χ from yRn−1 to X such that yχ = z and, for any two elements u and v in yRn−1 , v ∈ us implies vχ ∈ uχs. From Lemma 3.3.2 we know that |Rn \ Rn−1 | = 1. Let us denote by q the uniquely determined element in Rn \ Rn−1 . From q ∈ Rn we obtain q ∈ s. Thus, by Lemma 3.3.1, nq ≤ 2. Thus, there exist elements v and w in yq such that {v, w} = yq. Let us first assume that v = w. Since q ∈ Rn \ Rn−1 , (q) = n. Thus, there exists an element p in s such that q ∈ ps and (q) = (p) + 1; cf. Lemma 3.1.2. Since v ∈ yq and q ∈ ps, v ∈ yps. Thus, there exists an element v in yp such that v ∈ v s. Similarly, we find an element w in yp such that w ∈ w s. We define vχ to be the unique element in v χs ∩ yχRn . Similarly, we define wχ to be the unique element in w χs ∩ yχRn . Note that v is the only element in vs ∩ yRn−1 . Otherwise we would have 2 ≤ apsq . Then, as apsq nq = aqs∗ p np , nq = 1, and then v = w. Also, if w ∈ vs, wχ ∈ vχs. Thus, we are done in this case. Let us now assume that v = w. In this case, we have 1 = |yRn \ yRn−1 |, so that nq = 1. Thus, |yχRn \ yχRn−1 | = 1. We denote by vχ the uniquely determined element in yχRn \ yχRn−1 .

6.5 More About Elements of Valency 2

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Since nq = 1, there exist elements t and u in vs ∩ yRn−1 such that t = u. Thus, apsq = 2. Thus, vχ ∈ tχs and vχ ∈ uχs. Again, we are done. Proposition 6.5.2 Let s be a symmetric element of S, and assume that s has valency 2. Then s is faithfully embedded in S. Proof. Let y be an element in X, let z be an element in y s, and let χ be a faithful map from {y, z} to X. We have to show that χ extends faithfully to a bijective map from y s to yχ s. From Lemma 6.5.1 we obtain a bijective faithful map φ from y s to X such that yφ = yχ. Since z ∈ y s, there exists an element q in s such that z ∈ yq. Since χ is faithful, we obtain from z ∈ yq that zχ ∈ yχq. Since φ is faithful, we obtain from z ∈ yq that zφ ∈ yφq. Thus, as yφ = yχ, zφ ∈ yχq. By Lemma 3.3.1, each element in s has valency at most 2. Thus, by Lemma 6.3.5, there exists a (bijective) faithful map ψ from yχ s to yχ s such that (yχψ = yχ and) zφψ = zχ. Now φψ is a faithful map from y s to yχ s which coincides with χ on {y, z}. For any two elements p and q in S, we write p, q instead of {p, q}. Theorem 6.5.3 Let s be a symmetric element of S such that ns = 2 and ¯ and

s = S. Then there exists a finite thin scheme S¯ containing involutions h ∼ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ k with hk = k h, h, k = S, and S = S// k. Proof. Let us denote by G the Schur group of S, and let us fix an element x in X. It is easy to see that our claim holds if nss∗ = 2. Let us, therefore, assume that nss∗ = 2. Then Oϑ (S) does not have valency 2. Thus, referring to Lemma 3.3.1 we obtain from Lemma 6.3.5 and Lemma 6.3.6 that G possesses an involution k such that k = Gx . We set S¯ := Gτ and k¯ := k τ . Then S¯ is a finite thin scheme, and, by Theorem ¯ ¯ k. 6.3.1 and Proposition 6.5.2, S ∼ = S//

¯ ¯ and s = S we obtain an element h ¯ in S¯ such that h ¯ k ¯ k has From S ∼ = S//

¯ k ¯ ¯ ¯ valency 2 and h = S// k. ¯ ¯ ¯ k¯ = k¯h; ¯ cf. Lemma 4.2.5(i). Since h ¯ k ¯ ¯ k ¯ k, has valency 2, h = S//

Since h ¯ ¯ ¯

h, k = S; cf. Lemma 4.2.2(i). Here is the converse of Theorem 6.5.3. It makes Theorem 6.5.3 to be one of our recognition theorems.

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Theorem 6.5.4 Assume S to be thin. Let h and k be involutions of S such that hk = kh and h, k = S. Then hk is symmetric, hk has valency 2, and we have hk = S// k. Proof. Applying Lemma 4.2.2(i) to {h, k} instead of R, we obtain hk = S// k. From Lemma 6.4.2(i) we know that nhk ≤ 2. Since hk = kh, 2 ≤ nhk ; cf. Lemma 4.2.5(i). Thus, nhk = 2. From Lemma 4.1.2(ii) we obtain that hk is symmetric. Note the similarity between Theorem 6.4.4 and Theorem 6.5.3. The first theorem says that groups with a ‘Glauberman involution’ provide the only examples for non-thin schemes of finite valency in which each element s satisfies ns ≤ 2 and nss∗ = 2. The second theorem says that finite dihedral groups provide the only examples for schemes of finite valency which are generated by a single symmetric element. We now list a few consequences of Theorem 6.5.3. Corollary 6.5.5 Let s be a symmetric element of S, and assume that s has valency 2. Set n := ns . Then the following hold. (i) If n is odd, {1} = Oϑ ( s) and Oϑ ( s) = s. (ii) Assume that n is even. Then Oϑ ( s) = s and there exists an element r in s such that {s} (r) = max {s} ( s) and {1, r} = Oϑ ( s). (iii) Assume that n is even and that 4 does not divide n. Then we have {1} = Oϑ ( s) ∩ Oϑ ( s) and Oϑ ( s)Oϑ ( s) = s. (iv) If 4 divides n, Oϑ ( s) ⊆ Oϑ ( s). Proof. All these statements follow from Theorem 6.5.3. The second part of the following theorem is due to Mitsugu Hirasaka. It is [22; Theorem 4.9]. Theorem 6.5.6 Let T be a closed subset of S in which each element has valency at most 2. Then we have the following. (i) The set Oϑ (T ) is schurian. (ii) If each element in T \ {1} has valency 2, T is schurian. Proof. (i) Let x be an element in X, set U := Oϑ (T ), and let us write G to denote the Schur group of U with respect to x.

6.6 Constrained Sets of Involutions

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Each element in T is assumed to have valency at most 2. Thus, as U ⊆ T , each element in U has valency at most 2. Thus, we obtain from Lemma 6.3.5 that, for any two elements w in xU and u in U , Gw acts transitively on (the two elements of) wu. Thus, we just have to show that G acts transitively on xU ; cf. Theorem 6.3.1 and Theorem 6.3.3. In order to show this latter condition we fix elements y and z in xU . According to Lemma 6.3.4, we may assume that there exists an element t in T such that z ∈ yt∗ t. Since z ∈ yt∗ t, there exists an element w in yt∗ such that z ∈ wt. From y ∈ wt and z ∈ wt we now obtain an element g in the Schur group of T with respect to x which (fixes w) and maps y to z. From yg = z and z ∈ yU we obtain that the restriction of g to xU is in G. (ii) Referring to Theorem 3.3.5 this follows immediately from (i).

6.6 Constrained Sets of Involutions Let L be a set of involutions, and let us write instead of L . Recall that, for each element q in L, S1 (q, L) is our notation for the set of all elements p in L such that pq contains an element r with (r) = (p) + (q). In accordance with Section 3.4 we shall write, for each element s in L, S1 (s) instead of S1 (s, L). Recall that L is called constrained if, for any two elements q in L and p in S1 (q), 1 = |pq|. Throughout this section, the letter L stands for a constrained set of involutions. Lemma 6.6.1 Let q be an element in L, and let p be an element in S1 (q). Let x be an element in X, let y be an element in xp, and let z be an element in yq. Let χ be a map from {x, y, z} to X. Then, if χ|{x,y} and χ|{y,z} are faithful, so is χ. Proof. Let us denote by s the uniquely determined element in S which satisfies z ∈ xs. We have to prove that zχ ∈ xχs. Since z ∈ yq and y ∈ xp, z ∈ xpq. Thus, as z ∈ xs, s ∈ pq. Thus, as we are assuming that p ∈ S1 (q) and that L is constrained, {s} = pq. Assume that χ|{x,y} is faithful. Then, as y ∈ xp, yχ ∈ xχp. Assume that χ|{y,z} is faithful. Then, as z ∈ yq, zχ ∈ yχq. From zχ ∈ yχq and yχ ∈ xχp we obtain zχ ∈ xχpq. Thus, as {s} = pq, zχ ∈ xχs. Let y and z be elements in X, and let n be the smallest non-negative integer n with z ∈ yLn . We write D(y, z) to denote the union of the sets yLi ∩ zLj which satisfy i + j = n.

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Lemma 6.6.2 Let y be an element in X, let z be an element in y L, let v be an element in D(y, z), and let w be an element in D(v, z). Let χ be a map from {v, w, y, z} to X. Then, if χ|{y,v,z} and χ|{y,w,z} are faithful, so is χ. Proof. Let us denote by p the uniquely determined element in S which satisfies v ∈ yp, by t the one which satisfies w ∈ vt. Then, we have w ∈ ypt. Thus, there exists an element s in pt such that w ∈ ys. Since w ∈ ys, wχ ∈ yχs. (We are assuming that χ|{y,w,z} is faithful.) Thus, as s ∈ pt, wχ ∈ yχpt. Thus, there exists an element x in yχp such that wχ ∈ xt. Let us denote by u the uniquely determined element in S which satisfies z ∈ wu. Then, as w ∈ vt, z ∈ vtu. Thus, there exists an element q in tu such that z ∈ vq. From q ∈ tu and w ∈ D(v, z) we obtain (q) = (t) + (u). Thus, t ∈ S1 (u). Thus, as L is assumed to be constrained, we obtain from q ∈ tu that {q} = tu. Since z ∈ wu, zχ ∈ wχu. (Again, we use the hypothesis that χ|{y,w,z} is faithful.) Thus, as wχ ∈ xt, zχ ∈ xtu. Thus, as {q} = tu, zχ ∈ xq. From z ∈ vq and v ∈ yp we obtain z ∈ ypq. Thus, there exists an element r in pq such that z ∈ yr. From r ∈ pq and v ∈ D(y, z) we obtain (r) = (p) + (q). Thus, by Lemma 3.5.2, apqr = 1. Since z ∈ yr, zχ ∈ yχr. Since v ∈ yp ∩ zq ∗ , vχ ∈ yχp ∩ zχq ∗ . (This time, we use that χ|{y,v,z} is assumed to be faithful.) On the other hand, we also have x ∈ yχp ∩ zχq ∗ . Thus, as apqr = 1, vχ = x. Thus, as wχ ∈ xt, wχ ∈ vχt. Lemma 6.6.3 Let x be an element in X, and let χ be a map from x L to X. Assume that, for any two elements w in x L and l in L, wlχ ⊆ wχl. Then χ is faithful. Proof. Let us assume that χ is not faithful, and let us define R to be the set of all elements s in L such that xsχ ⊆ xχs. Then R is not empty. We fix an element r in R such that (r) is as small as possible. Among the elements in R we pick r as small as possible. Then 1 = r. Thus, by Lemma 3.1.2, there exist elements q in L and l in L such that r ∈ ql and (r) = (q) + 1. From q ∈ L and l ∈ L we obtain q ∈ L. Thus, as (q) = (r) − 1, the minimal choice of r yields q ∈ / R. That means that xqχ ⊆ xχq. By hypothesis, we have that, for each element w in x L, wlχ ⊆ wχl. Since L is assumed to be constrained, we also have {r} = ql. Thus, xrχ = x(ql)χ = (xq)lχ ⊆ (xqχ)l ⊆ (xχq)l = xχ(ql) = xχr, contradiction.


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Let x be an element in X, and let G be a subgroup of the Schur group of L with respect to x. For any two elements y and z in X, we define Gyz to be the intersection of Gy and Gz . Lemma 6.6.4 Let x be an element in X, and let G be a subgroup of the Schur group of L with respect to x. Let q be an element in L, let p be an element in S1 (q). Then we have the following. (i) If Gx acts transitively on xpq, Gx acts transitively on xp. (ii) Assume that Gx acts transitively on xp and that, for each element w in xp, Gxw acts transitively on wq. Then Gx acts transitively on xpq. Proof. (i) Let y and z be elements in xp. We shall prove that there exists an element g in Gx such that yg = z. We are assuming that p ∈ S1 (q). Thus, as L is assumed to be constrained, there exists an element r in pq such that {r} = pq. Let v be an element in yq, and let w be an element in zq. From v ∈ yq and y ∈ xp we obtain v ∈ xpq. Thus, as {r} = pq, v ∈ xr. Similarly, we obtain from w ∈ zq and z ∈ xp that w ∈ xr. Thus, as we are assuming that Gx acts transitively on xpq, there exists an element g in Gx such that vg = w. Since y ∈ xp ∩ vq ∗ , yg ∈ xp ∩ vgq ∗ = xp ∩ wq ∗ . Thus, as {r} = pq, yg = z, cf. Lemma 3.5.2. (ii) Let y and z be elements in xpq. We shall prove that there exists an element g in Gx such that yg = z. Since y ∈ xpq, there exists an element v in xp such that y ∈ vq. Since z ∈ xpq, there exists an element w in xp such that z ∈ wq. We are assuming that Gx acts transitively on xp. Thus, as v, w ∈ xp, there exists an element e in Gx such that ve = w. We are also assuming that Gxw acts transitively on wq. Thus, there exists an element f in Gxw such that yef = z. Finally, as e ∈ Gx and f ∈ Gxw , ef ∈ Gx . Lemma 6.6.5 Let x be an element in X, and let G be the Schur group of

L with respect to x. Assume that, for any four elements l in L, s in S1 (l), y in x L, and z in ys, Gyz acts transitively on zl. Then, Lx is schurian. Proof. That G acts transitively on x L follows from the hypothesis with s = 1. From Lemma 6.6.4(ii) we obtain by induction that, for each element s in L, Gx acts transitively on xs. Thus, the claim follows from Theorem 6.3.1.

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6.7 Thin Thin Residues In this section, the letter T stands for a closed subset of S which has finite valency and satisfies Oϑ (T ) ⊆ Oϑ (T ). We shall prove that, under a certain additional condition, T is faithfully embedded in S. Recall from Section 2.6 that, for each element s in S, DT (s) is our notation for the set of all elements t in T satisfying t∗ t ⊆ s∗ s. Lemma 6.7.1 For each element t in T , the following hold. (i) We have t∗ t ⊆ Oϑ (T ). (ii) We have Oϑ (T ) ⊆ DT (t). (iii) We have {t} = tt∗ t. (iv) The set t∗ t is closed. (v) We have Oϑ (T ) ⊆ KS (t∗ t). (vi) We have nt = nt∗ t . Proof. (i) From Theorem 3.2.1(ii) we know that, for each element t in T , t∗ t ⊆ Oϑ (T ). Thus, as we are assuming that Oϑ (T ) ⊆ Oϑ (T ), we must have t∗ t ⊆ Oϑ (T ). (ii) It follows right from the definition of DT (t) that thin elements of T belong to DT (t). Thus, by Lemma 1.5.1, Oϑ (T ) ⊆ DT (t). Thus, the claim follows from our hypothesis that Oϑ (T ) ⊆ Oϑ (T ). (iii) From (i) we know that t∗ t ⊆ Oϑ (T ). Thus, by Corollary 2.6.6 (applied to t, {1} and T in the roles of s, T and U ), 1t = t∗ t. Thus, by definition, tt∗ t ⊆ {t}. It follows that {t} = tt∗ t. (iv) From (iii) we obtain t∗ tt∗ t ⊆ t∗ t, and this implies that t∗ t is closed. (v) From (i) we know that t∗ t ⊆ Oϑ (T ), from (ii) we know that Oϑ (T ) ⊆ DT (t∗ ). Clearly, we also have that tOϑ (T ) ⊆ tOϑ (T )t∗ t ⊆ Oϑ (T )t. Thus, Oϑ (T ) ⊆ Oϑ (T )t . Thus, by Lemma 2.6.8, Oϑ (T ) ⊆ KS (t∗ t). (vi) Considering Lemma 1.4.4(ii) this follows from (iii). Corollary 6.7.2 Let π be a set of primes such that, for each element t in T , nt is a π-number. Then the valency of Oϑ (T ) is a π-number. Proof. From Theorem 3.2.1(ii) we know that Oϑ (T ) is generated by the subsets t∗ t with t ∈ T . From Lemma 6.7.1(iv), (v) we know that, for each

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element t in T , t∗ t is a normal closed subset of T , and Lemma 6.7.1(vi) says that, for each element t in T , nt = nt∗ t . Thus, as we are assuming that, for each element t in T , nt is a π-number, Oϑ (T ) is generated by normal closed subsets of valency a π-number. Thus, considering Lemma 2.1.1 the claim follows from Lemma 2.3.6(i). Assume that there exists a prime p such that nt = p for each element t in T . Then, via the group correspondence, Oϑ (T ) is an elementary abelian p-group; cf. Corollary 6.7.2. We shall now focus on a condition which turns out to be sufficient for T to be faithfully embedded in S. We shall assume that, for any two elements p and q in T , p∗ p ⊆ q ∗ q or q ∗ q ⊆ p∗ p. Lemma 6.7.3 Assume that {t∗ t | t ∈ T } is linearly ordered with respect to set theoretic inclusion. Then, for each element t in T , tt∗ = t∗ t. Proof. We are assuming that T has finite valency. Thus, by Lemma 1.1.2(iii), nt∗ = nt . Thus, by Lemma 6.7.1(vi), ntt∗ = nt∗ t . Thus, as we are assuming that t∗ t ⊆ tt∗ or tt∗ ⊆ t∗ t, tt∗ = t∗ t. Lemma 6.7.4 Assume that {t∗ t | t ∈ T } is linearly ordered with respect to set theoretic inclusion. Then, for each element t in T , DT (t) is closed. Proof. Let p and q be elements in DT (t), and let s be an element in p∗ q. We have to show that s ∈ DT (t). Thus, we shall be done if we succeed in showing that s∗ s ⊆ t∗ t. Since {t∗ t | t ∈ T } is assumed to be linearly ordered with respect to set theoretic inclusion, we have pp∗ ⊆ qq ∗ or qq ∗ ⊆ pp∗ . Let us first assume that pp∗ ⊆ qq ∗ . Then, as s ∈ p∗ q, s∗ s ⊆ q ∗ pp∗ q ⊆ q ∗ qq ∗ q. However, according to Lemma 6.7.1(iv), q ∗ q is closed, so that q ∗ qq ∗ q ⊆ q ∗ q. Thus, s∗ s ⊆ q ∗ q. On the other hand, we have picked q in DT (t), and, therefore, we have q ∗ q ⊆ t∗ t. It follows that s∗ s ⊆ t∗ t. Let us now assume that qq ∗ ⊆ pp∗ . Then, as s ∈ p∗ q, ss∗ ⊆ p∗ qq ∗ p ⊆ p∗ pp∗ p. However, we know from Lemma 6.7.1(iv) that p∗ p is closed. Thus, ss∗ ⊆ p∗ p. Thus, as p has been chosen from DT (t), p∗ p ⊆ t∗ t. It follows that ss∗ ⊆ t∗ t. Thus, by Lemma 6.7.3, s∗ s ⊆ t∗ t. The following theorem is [25; Theorem B]. Theorem 6.7.5 Assume that {t∗ t | t ∈ T } is linearly ordered with respect to set theoretic inclusion. Then, T is faithfully embedded in S.

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Proof. Let us abbreviate V := Oϑ (T ). By Lemma 6.2.1, we may assume that T is not thin. Thus, by Lemma 4.2.5(ii), {1} = V . In particular, there exists an element t in T such that t∗ t = V . Let us denote by U the set of all elements t in T which satisfy t∗ t = V . Since we are assuming {t∗ t | t ∈ T } to be linearly ordered with respect to set theoretic inclusion, there exists an element s in T such that U = DT (s). Thus, by Lemma 6.7.4, U is closed. Note also that U = T . We are assuming that Oϑ (T ) ⊆ Oϑ (T ). Thus, by Theorem 3.2.1(iii), we also have Oϑ (U ) ⊆ Oϑ (U ). Thus, by induction, U is faithfully embedded in S. Recall that V ⊆ Oϑ (T ) and U = DT (s). Thus, as Oϑ (T ) ⊆ DT (s), V ⊆ U . According to the definition of U , we have, for each element t in T \U , t∗ t = V . Thus, for each element t in T \ U , we have {t} = tV ; cf. Lemma 6.7.1(iii). Thus, by Theorem 6.2.6, T is faithfully embedded in S. Corollary 6.7.6 If Oϑ (T ) is simple, T is faithfully embedded in S. Proof. From Lemma 6.7.1(iv), (v) we know that, for each element t in T , t∗ t is a normal closed subset of Oϑ (T ). Thus, as we are assuming Oϑ (T ) to be simple, we have that, for each element t in T , {1} = t∗ t or t∗ t = Oϑ (T ). Thus, the claim follows from Theorem 6.7.5. Lemma 6.7.7 Assume that {t∗ t | t ∈ T } is linearly ordered with respect to set theoretic inclusion. Then the following hold. (i) For any two elements p and q in T , we have p∗ p ⊆ q ∗ q if and only if DT (p) ⊆ DT (q). (ii) The set {DT (t) | t ∈ T } is linearly ordered with respect to set theoretic inclusion. (iii) Let p and q be elements in T such that q ∈ / DT (p). Then |p∗ q| = 1. Proof. (i) Let p and q be elements in T , and let us first assume that p∗ p ⊆ q ∗ q. Let s be an element in DT (p). Then, by definition, s∗ s ⊆ p∗ p. Thus, as we are assuming that p∗ p ⊆ q ∗ q, we conclude that s∗ s ⊆ q ∗ q. Thus, by definition, s ∈ DT (q). Since s has been chosen arbitrarily in DT (p), we have shown that DT (p) ⊆ DT (q). Let us now, conversely assume that DT (p) ⊆ DT (q). Then p ∈ DT (q). Thus, by definition, p∗ p ⊆ q ∗ q. (ii) This is an immediate consequence of (i). (iii) Let t and u be elements in p∗ q. We shall see that t = u. Assume that t ∈ DT (p). From Lemma 6.7.4 we know that DT (p) is closed. Thus, as p ∈ DT (p), pt ⊆ DT (p).

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On the other hand, we obtain from t ∈ p∗ q that q ∈ pt; cf. Lemma 1.3.3(ii). Thus, q ∈ DT (p), contradiction. Thus, t ∈ / DT (p), so that, according to (i), p ∈ DT (t). Thus, by definition, p∗ p ⊆ t∗ t. Thus, by Lemma 6.7.3, p∗ p ⊆ tt∗ . From u ∈ p∗ q and q ∈ pt we obtain u ∈ p∗ pt. From u ∈ p∗ pt and p∗ p ⊆ tt∗ we obtain u ∈ tt∗ t. Now recall that, by Lemma 6.7.1(iii), {t} = tt∗ t. Thus, t = u.

7 Products

In this chapter we investigate various types of products arising naturally in scheme theory. We define direct products of closed subsets of S, direct products of schemes, quasi-direct products of schemes, and semidirect products of schemes.

7.1 Direct Products of Closed Subsets Let n be a positive integer, let T1 , . . . , Tn be closed subsets of S, and let us write T to denote the closed subset generated by the union of the sets Ti with i ∈ {1, . . . , n}. For each element i in {1, . . . , n}, we define Tî to be the closed subset generated by the union of the sets Tj with j ∈ {1, . . . , n} \ {i}. Assume that, for each element i in {1, . . . , n}, {1} = Ti ∩ Tî . We call T the direct product of the closed subsets T1 , . . . , Tn if, for any two elements i and j in {1, . . . , n}, Ti ⊆ NS (Tj ). We shall write T1 × . . . × Tn = T in order to indicate that the closed subset T is the direct product of the closed subsets T1 , . . . , Tn . Assume that T1 × . . . × Tn = T . Then, by Lemma 2.1.1, T1 · · · Tn = T . Moreover, for any two elements i and j in {1, . . . , n} with i = j, Ti ⊆ CS (Tj ); cf. Lemma 2.5.3. Theorem 7.1.1 Let n be a positive integer, and let T1 , . . . , Tn be closed subsets of S such that, for any two elements i and j in {1, . . . , n}, Ti ⊆ NS (Tj ). Then, for each closed subset T of S, the following conditions are equivalent. (a) We have T1 × . . . × Tn = T .

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(b) Let t be an element in T . Then there exists, for each element i in {1, . . . , n}, a uniquely determined element ti in Ti such that t ∈ t1 · · · tn . Proof. The claim is obvious for n = 1. Therefore, we assume that 2 ≤ n. (a) ⇒ (b) Since t ∈ Tˆn Tn , there exist uniquely determined elements s in Tˆn and tn in Tn such that t ∈ stn ; cf. Lemma 2.1.2. By induction, there exists, for each element i in {1, . . . , n − 1}, a uniquely determined element ti in Ti such that s ∈ t1 · · · tn−1 . From t ∈ stn and s ∈ t1 · · · tn−1 we obtain t ∈ t1 · · · tn . (b) ⇒ (a) We have to show that, for each element i in {1, . . . , n}, {1} = Ti ∩ Tî . We are assuming that, for any two elements i and j in {1, . . . , n}, Ti ⊆ NS (Tj ). Thus, for any two elements i and j in {1, . . . , n}, Ti Tj = Tj Ti . Therefore it suffices to prove that {1} = Tn ∩ Tˆn . Let s be an element in Tn ∩ Tˆn . Since Tˆn = T1 · · · Tn−1 , we obtain from s ∈ Tˆn that s ∈ T1 · · · Tn−1 . Thus, there exists, for each element i in {1, . . . , n − 1}, an element ti in Ti such that s ∈ t1 · · · tn−1 . On the other hand, we have s ∈ Tn . Thus, the uniqueness assumed in (b) yields s = 1. Lemma 7.1.2 Let n be a positive integer, let T1 , . . . , Tn be closed subsets of S, and let T be a closed subset of S such that T1 × . . . × Tn = T . Then we have the following. (i) For each element t in T , t ∈ (T1 ∩ tTˆ1 ) · · · (Tn ∩ tTˆn ). (ii) Let U be a closed subset of S such that U ⊆ T . Assume that (T1 ∩ U Tˆ1 ) · · · (Tn ∩ U Tˆn ) = T. Then, for each element i in {1, . . . , n}, U Tî = T . Proof. (i) Let t be an element in T . Since we are assuming that T1 · · · Tn = T , there exists, for each element i in {1, . . . , n}, an element ti in Ti such that t ∈ t1 · · · tn . Let i be an element in {1, . . . , n}. Then, for each element j in {1, . . . , n} \ {i}, {1} = Ti ∩ Tj , Ti ⊆ NS (Tj ), and Tj ⊆ NS (Ti ). Thus, for each element j in {1, . . . , n} \ {i}, ti tj = tj ti ; cf. Lemma 2.5.3. Therefore, t ∈ ti Tî . Thus, by Lemma 2.1.4, ti ∈ tTî . It follows that ti ∈ Ti ∩ tTî . Since i ∈ {1, . . . , n} has been chosen arbitrarily, we now obtain from t ∈ t1 · · · tn that t ∈ (T1 ∩ tTˆ1 ) · · · (Tn ∩ tTˆn ). (ii) Let i be an element in {1, . . . , n}, and let t be an element in Ti . Then t ∈ T , so that, by hypothesis, t ∈ (T1 ∩ U Tˆ1 ) · · · (Tn ∩ U Tˆn ).

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Since t ∈ (T1 ∩ U Tˆ1 ) · · · (Tn ∩ U Tˆn ), there exists, for each element j in {1, . . . , n}, an element tj in Tj ∩ U Tˆj such that t ∈ t1 · · · tn . Thus, as t ∈ Ti , t = ti ; cf. Theorem 7.1.1. It follows that t ∈ U Tî . Since t ∈ Ti has been chosen arbitrarily, we have proved that Ti ⊆ U Tî . It follows that T = Ti Tî ⊆ U Tî . A closed subset T of S is called decomposable if there exist closed subsets U and V of S such that {1} = U , {1} = V , and U × V = T . A closed subset of S different from {1} is called indecomposable if it is not decomposable. Theorem 7.1.3 Let T be a closed subset of S such that {1} = T . Then we have the following. (i) Let U and U be closed subsets of S such that U × U = T . Let V and W be closed subsets of S such that V × W = U . Then U × V × W = T . (ii) If S has finite valency, T is the direct product of indecomposable closed subsets of S. Proof. (i) Let s be an element in V ∩ U W . Since s ∈ U W , there exist elements u in U and w in W such that s ∈ uw. Thus, by Lemma 1.3.3(i), u ∈ sw∗ ⊆ V W . It follows that u ∈ U ∩ U = {1}. Thus, as s ∈ uw, s = w. Thus, we obtain from s ∈ V , w ∈ W , and V ∩ W = {1} that s = 1. Since s has been chosen arbitrarily in V ∩ U W , we have shown that {1} = V ∩ U W . Similarly, one obtains {1} = W ∩ U V . From V ×W = U we obtain V ⊆ NS (W ) and W ⊆ NS (V ). Since U ×U = T , V W = U ⊆ NS (U ). Referring to Lemma 2.5.3 we also obtain U ⊆ CS (U ) ⊆ CS (V ) ⊆ NS (V ). Similarly, we obtain U ⊆ NS (W ). (ii) This follows from (i) by induction. Proposition 7.1.4 Let T be a closed subset of S. Assume that T has finite valency and that, for any two elements p and q in T , qp = pq. Let m and n be positive integers, and let U1 , . . . , Um and V1 , . . . , Vn be indecomposable closed subsets of S such that U1 × . . . × Um = T = V1 × . . . × Vn . Then, for each element i in {1, . . . , m}, there exists an element j in {1, . . . , n} î . such that Ui × Vˆj = T = Vj × U Proof. Let i be an element in {1, . . . , m}, and set ˜i := (V1 ∩ Ui Vˆ1 ) · · · (Vn ∩ Ui Vˆn ). U

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For each element j in {1, . . . , n}, we define ˆ1 ) · · · (Um ∩ Vj U ˆm ). V˜j := (U1 ∩ Vj U ˜i = T . Let us first assume that U ˜i . Thus, by Lemma 2.2.1(i), Ui (U ˜i ∩ U î ) = U ˜i . On By Lemma 7.1.2(i), Ui ⊆ U ˆ ˜ ˆ the other hand, as Ui ∩ Ui = {1}, Ui ∩ (Ui ∩ Ui ) = {1}. Thus, ˜i ∩ U î ) = U ˜i . Ui × (U (Recall that we are assuming that, for any two elements p and q in T , qp = pq.) By definition, we also have ˜i . (V1 ∩ Ui Vˆ1 ) × . . . × (Vn ∩ Ui Vˆn ) = U With the help of Theorem 7.1.3(i) we may refine both of the above represen˜i to direct products of indecomposable closed subsets of S. Thus, tations of U ˜i = T , we obtain, by induction, an element j in as we are assuming that U {1, . . . , n} and an indecomposable closed subset W of Vj ∩ Ui Vˆj such that î ) = U ˜i . ˜i ∩ U W × (U ˜i and W ∩ (U ˜i ∩ U î ) = {1} we conclude that W ∩ U î = {1}. On From W ⊆ U the other hand, we have that î = Ui (U ˜i ∩ U î )U î = U ˜i U ˜i ∩ U î )U î = T. î = W (U WU Therefore, î = T. W ×U î = {1} we conclude that W ∩ (Vj ∩ U î ) = {1}. From W ∩ U ˆ i ) = Vj ∩ W U î = Vj ; cf. î = T and W ⊆ Vj we obtain W (Vj ∩ U From W U Lemma 2.2.1(i). Thus, ˆ i ) = Vj . W × (Vj ∩ U However, by hypothesis, Vj is indecomposable. Therefore, Vj = W . Thus, as î = T , W ×U î = T. Vj × U From Vj = W and W ⊆ Ui Vˆj we obtain Vj ⊆ Ui Vˆj . Thus, T = Vj Vˆj ⊆ Ui Vˆj . It follows that Ui Vˆj = T . ˆ i = T = Vj × U î , Lemma 2.3.6(i) yields nU = On the other hand, as Ui × U i nVj . (Recall that T is assumed to have finite valency.) Therefore, referring to Lemma 2.3.6(i) once again, we obtain from Ui Vˆj = T that

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Ui × Vˆj = T. ˜i = T . This proves the proposition in the case where U ˜i = T . Then, for each element j in {1, . . . , n}, Let us now assume that U ˆ Ui Vj = T ; cf. Lemma 7.1.2(ii). We shall now prove that there exists an element î = T . j in {1, . . . , n} such that Vj U î = Assume, by way of contradiction, that, for each element j in {1, . . . , n}, Vj U ˜ T . Then, for each element j in {1, . . . , n}, Vj = T ; cf. Lemma 7.1.2(ii). Thus, similar to the first case, we find, for each element l in {1, . . . , n}, an element k in {1, . . . , m} such that ˆk . Uk × Vˆl = T = Vl × U Thus, if k = i, we are done. Let us, therefore, assume that k = i. From Uk × Vˆl = T we obtain V1 Uk //Uk × . . . × Vl−1 Uk //Uk × Vl+1 Uk //Uk × . . . × Vn Uk //Uk = T //Uk . Thus, as we are assuming that k = i, we find, by induction, an element j in {1, . . . , n} such that Vj Uk //Uk × Ui Uk //Uk = T //Uk . From U1 //Uk × . . . × Uk−1 //Uk × Uk+1 //Uk × . . . × Um //Uk = T //Uk we obtain

î ∩ U ˆk )Uk //Uk . Uk //Uk = (U Ui

From this we obtain î ∩ U ˆk ) = T. Vj Uk (U î . Thus, by Lemma 2.2.1(i), On the other hand, as k = i, Uk ⊆ U î = Uk (U î ∩ U ˆk ). U î = T , contrary to our hypothesis. It follows that Vj U Thus, we have shown that there exists an element j in {1, . . . , n} such that î . Ui Vˆj = T = Vj U Similar to the reasoning in the first case we obtain from this that î ; Ui × Vˆj = T = Vj × U cf. Lemma 2.3.6(i).

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The following theorem is due to Pamela Ferguson and Alexandre Turull; cf. [11; Theorem 3.11]. Theorem 7.1.5 Assume that S has finite valency and that, for any two elements p and q in S, qp = pq. Let m and n be positive integers, and let T1 , . . . , Tm and U1 , . . . , Un be indecomposable closed subsets of S satisfying T1 × . . . × Tm = S = U1 × . . . × Un . Then the following hold. (i) We have m = n. (ii) There exists a permutation π of the set {1, . . . , n} such that, for any two elements x in X and i in {1, . . . , n}, (Ti )x ∼ = (Uiπ )x . Proof. Let i be a positive integer such that i ≤ m. Then, by Proposition 7.1.4, there exists an element j in {1, . . . , n} such that ˆj = S = Uj × Tî . Ti × U Thus, as Ti × Tî = S,

ˆj = Ti × Tî Ti × U

and Ti × Tî = Uj × Tî . From the second of these equations we obtain (Ti )x ∼ = S//Tî ∼ = (Uj )x for each element x in X; cf. Corollary 5.3.5. Thus, setting iπ := j we obtain (Ti )x ∼ = (Uiπ )x . From the first of the above two equations we obtain ˆj )x (Tî )x ∼ = S//Ti ∼ = (U for each element x in X; cf. Corollary 5.3.5. Thus, we are done by induction.

7.2 Quasidirect Products of Schemes Let n be a positive integer. For each an element i in {1, . . . , n}, we fix a nonempty set Xi and a scheme Si on Xi . For each element (s1 , . . . , sn ) in S1 × . . . × Sn , we define (s1 , . . . , sn )ς to be the set of all pairs ((y1 , . . . , yn ), (z1 , . . . , zn )) which satisfy, for each element i in {1, . . . , n}, zi ∈ yi si .

7.2 Quasidirect Products of Schemes

139

Lemma 7.2.1 Let n be a positive integer. For each element i in {1, . . . , n}, we fix a nonempty set Xi and a scheme Si on Xi . Then the following hold. (i) The set (S1 × . . . × Sn )ς is a partition of X1 × . . . × Xn . (ii) We have (1X1 , . . . , 1Xn )ς = 1X1 ×...×Xn . Proof. For each element i in {1, . . . , n}, we fix elements yi and zi in Xi . (i) For each element i in {1, . . . , n}, there exists a uniquely determined element si in Si such that zi ∈ yi si . Thus, (s1 , . . . , sn )ς is the uniquely determined element in (S1 × . . . × Sn )ς which satisfies ((y1 , . . . , yn ), (z1 , . . . , zn )) ∈ (s1 , . . . , sn )ς. (ii) We have ((y1 , . . . , yn ), (z1 , . . . , zn )) ∈ (1X1 , . . . , 1Xn )ς if and only if, for each element i in {1, . . . , n}, yi = zi . The latter condition is equivalent to (y1 , . . . , yn ) = (z1 , . . . , zn ). Thus, (1X1 , . . . , 1Xn )ς = 1X1 ×...×Xn . Lemma 7.2.2 Let n be a positive integer. For each element i in {1, . . . , n}, we fix a nonempty set Xi , a scheme Si on Xi , and an element si in Si . Then the following hold. (i) We have ((s1 , . . . , sn )ς)∗ = (s∗1 , . . . , s∗n )ς. (ii) Let us fix, for each element i in {1, . . . , n}, elements yi and zi in Xi and elements pi and qi in Si . Then, if ((y1 , . . . , yn ), (z1 , . . . , zn )) ∈ (s1 , . . . , sn )ς, |(y1 , . . . , yn )(p1 , . . . , pn )ς ∩ (z1 , . . . , zn )((q1 , . . . , qn )ς)∗ | = ap1 q1 s1 · · · apn qn sn . Proof. (i) We have ((y1 , . . . , yn ), (z1 , . . . , zn )) ∈ (s1 , . . . , sn )ς if and only if, for each element i in {1, . . . , n}, zi ∈ yi si . The latter condition is equivalent to the fact that, for each element i in {1, . . . , n}, yi ∈ zi s∗i . This means that ((z1 , . . . , zn ), (y1 , . . . , yn )) ∈ (s∗1 , . . . , s∗n )ς. (ii) For each element i in {1, . . . , n}, we fix an element xi in Xi . Then, by definition, (x1 , . . . , xn ) ∈ (y1 , . . . , yn )(p1 , . . . , pn )ς ∩ (z1 , . . . , zn )((q1 , . . . , qn )ς)∗ if and only if, for each element i in {1, . . . , n}, xi ∈ yi pi ∩ zi qi∗ . Therefore, the sets (y1 , . . . , yn )(p1 , . . . , pn )ς ∩ (z1 , . . . , zn )((q1 , . . . , qn )ς)∗ and

y1 p1 ∩ z1 q1∗ × . . . × yn pn ∩ zn qn∗

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are equal. Thus, the claim follows from the fact that, for each element i in {1, . . . , n}, |yi pi ∩ zi qi∗ | = api qi si . Theorem 7.2.3 Let n be a positive integer. For each element i in {1, . . . , n}, we fix a nonempty set Xi and a scheme Si on Xi . Then the following hold. (i) The set (S1 × . . . × Sn )ς is a scheme on X1 × . . . × Xn . (ii) Let us fix, for each element i in {1, . . . , n}, elements pi , qi , and ri in Si . Then a(p1 ,...,pn )ς(q1 ,...,qn )ς(r1 ,...,rn )ς = ap1 q1 r1 · · · apn qn rn . Proof. This follows from Lemma 7.2.1 and Lemma 7.2.2. Let n be a positive integer. For each element i in {1, . . . , n}, we fix a nonempty set Xi and a scheme Si on Xi . The scheme (S1 × . . . × Sn )ς will be called the direct product of the schemes S1 , . . . , Sn . ˜ := X1 × . . . × Xn and S˜ := (S1 × . . . × Sn )ς. We set X For each element i in {1, . . . , n}, we define S˜i to be the set of all elements (s1 , . . . , sn )ς in S˜ such that, for each element j in {1, . . . , n} \ {i}, sj = 1Xj . We call S a quasi-direct product of the schemes S1 , . . . , Sn if there exists a ˜ ∪ S˜ such that φX is bijective and, for any two morphism φ from X ∪ S to X sφ−1 | = 1. elements i in {1, . . . , n} and s˜ in S˜i , |˜ Note that the morphism φ in the definition of quasi-direct products is necessarily surjective; cf. Lemma 5.1.3(ii). Note also that, if the morphism φ in the definition of quasi-direct products is injective, the second condition of the definition, namely that, for any two sφ−1 | = 1, is necessarily satisfied. In elements i in {1, . . . , n} and s˜ in S˜i , |˜ particular, direct products are quasi-direct products. The relationship between quasi-direct products of schemes and direct products of closed subsets of S (as defined in the previous section) is described in the following theorem which generalizes [11; Proposition 3.13]. Theorem 7.2.4 Let n be a positive integer. For each element i in {1, . . . , n}, we fix a nonempty set Xi and a scheme Si on Xi . Then the following conditions are equivalent. (a) The scheme S is a quasi-direct product of the schemes S1 , . . . , Sn . (b) The scheme S possesses closed subsets T1 , . . . , Tn with T1 ×. . .×Tn = S such that, for any two elements i in {1, . . . , n} and x in X, Si ∼ = (Ti )x . ˜ := X1 × . . . × Xn and S˜ := (S1 × . . . × Sn )ς. Proof. (a) ⇒ (b) We set X


141

For each element i in {1, . . . , n}, we define S˜i to be the set of all elements (s1 , . . . , sn )ς in S˜ such that, for each element j in {1, . . . , n} \ {i}, sj = 1Xj . Since S is assumed to be a quasi-direct product of the schemes S1 , . . . , Sn , ˜ ∪ S˜ such that φX is bijective and, there exists a morphism φ from X ∪ S to X sφ−1 | = 1. for any two elements i in {1, . . . , n} and s˜ in S˜i , |˜ For each element i in {1, . . . , n}, we define Ti := S˜i φ−1 . Since S˜i is closed, Ti is closed; cf. Lemma 5.1.2(ii). Since φX is bijective, we obtain from Lemma 5.1.3(i) that T1 · · · Tn = S. Recall that, for each element i in {1, . . . , n}, Tî is our notation for the closed subset generated by the union of the sets Tj with j ∈ {1, . . . , n} \ {i}. Let i be an element in {1, . . . , n}. Then ˆ˜ = {1φ}. (Ti ∩ Tî )φ ⊆ Ti φ ∩ Tî φ ⊆ S˜i ∩ S i Thus, as {1} = (1φ)φ−1 , Ti ∩ Tî = {1}. Let i and j be elements in {1, . . . , n}. We shall show that Ti ⊆ NS (Tj ). Let s be an element in Ti , and set s˜ := sφ. Then s˜ ∈ S˜i . From Lemma 7.2.2(ii) we obtain S˜j s˜ = s˜S˜j . On the other hand, we have {s} = s˜φ−1 . Thus, by Lemma 5.1.3(i), Tj s = S˜j φ−1 s˜φ−1 = (S˜j s˜)φ−1 = (˜ sS˜j )φ−1 = s˜φ−1 S˜j φ−1 = sTj . Thus, as s has been chosen arbitrarily in Ti , we have shown that Ti ⊆ NS (Tj ). So far, we have proved that T1 × . . . × Tn = S. Let us now fix an element i in {1, . . . , n} and an element x in X. We set ˜ i := xφS˜i . We have to show that Si ∼ X = (Ti )x . For each element w in xTi , we set wψ := wφ. For each element s in Ti , we set (sxTi )ψ := sφ. ˜ i ∪ S˜i . Since φ is a morphism, ψ is a morphism from xTi ∪ (Ti )xT to X i

Let p and q be elements in Ti with (pxTi )ψ = (qxTi )ψ. Then pφ = qφ. But, as qφ ∈ S˜i , |(qφ)φ−1 | = 1. Therefore, p = q. This proves that ψ(Ti )xTi injective. Thus, by Lemma 5.1.6(i), ψ is an injective homomorphism. Let us now show that ψ is surjective. According to Lemma 5.1.3(ii), it suffices to show that ψxTi is surjective. ˜ i . Then, ˜ in X In order to show that ψxTi is surjective we fix an element w by definition, w ˜ ∈ xφS˜i . Therefore, there exists an element s˜ in S˜i such that w ˜ ∈ xφ˜ s.

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Since φ is surjective, there exists an element w in X such that wφ = w. ˜ Let us write s to denote the element in S which satisfies w ∈ xs. Since φ is a morphism, we obtain from w ∈ xs that wφ ∈ xφsφ. Thus, as wφ = w, ˜ w ˜ ∈ xφsφ. Thus, as w ˜ ∈ xφ˜ s, sφ = s˜ ∈ S˜i . Thus, by definition, s ∈ Ti . Thus, as w ∈ xs, w ∈ xTi . From w ∈ xTi we obtain wψ = wφ. ˜ i , we have shown that ψxT is surjecSince w ˜ has been chosen arbitrarily in X i tive. ˜ i ∪ S˜i . It follows that ψ is an isomorphism from xTi ∪ (Ti )xT to X i

On the other hand, it is clear that S˜i ∼ = Si . (b) ⇒ (a) Let x be an element in X. We shall be done if we succeed in showing that S is a quasi-direct product of the schemes (T1 )x , . . . , (Tn )x . We define ˜ := xT1 × . . . × xTn . X and S˜ := ((T1 )xT1 × . . . × (Tn )xTn )ς .

Let y be an element in X, let i be an element in {1, . . . , n}. From Ti × Tî = S we obtain y ∈ xTi Tî and Ti ∩ Tî = {1}. Thus, |xTi ∩ y Tî | = 1. Let yi be the element in xTi ∩ y Tî . We set yφ := (y1 , . . . , yn ). Let s be an element in S. Then we find, for each element i in {1, . . . , n}, a uniquely determined element ti in Ti such that s ∈ t1 · · · tn ; cf. Theorem 7.1.1. We set sφ := ((t1 )xT1 , . . . , (tn )xTn )ς. ˜ ∪ S. ˜ Then φ is an injective map from X ∪ S to X For each element i in {1, . . . , n}, we define Ui to be the set of all elements ((s1 )xT1 , . . . , (sn )xTn )ς such that, for each element j in {1, . . . , n} \ {i}, (sj )xTj = 1xTj . It follows from the definition of φ that, for any two elements i in {1, . . . , n} and f in Ui , |f φ−1 | = 1. Therefore, we shall be done if we succeed in showing that φ is a morphism. To this end, let y be an element in X, let s be an element in S, and let z be an element in ys. ˜ be such that Let (y1 , . . . , yn ) ∈ X yφ = (y1 , . . . , yn ), ˜ be such that and let (z1 , . . . , zn ) ∈ X zφ = (z1 , . . . , zn ).


143

Then, for each element i in {1, . . . , n}, yi ∈ xTi and zi ∈ xTi . Thus, for each element i in {1, . . . , n}, there exists an element ti in Ti such that zi ∈ yi ti ; cf. Lemma 2.1.4. It follows that (yφ, zφ) = ((y1 , . . . , yn ), (z1 , . . . , zn )) ∈ ((t1 )xT1 , . . . , (tn )xTn )ς. Now we shall be done if we succeed in showing that sφ = ((t1 )xT1 , . . . , (tn )xTn )ς. Let i be an element in {1, . . . , n}. From the definition of yi we obtain y ∈ yi Tî ⊆ zi t∗i Tî = zi Tî t∗i . Therefore, zi Tî ∩ yti is not empty. Let w be an element in zi Tî ∩ yti . From the definition of zi we obtain z ∈ zi Tî . Thus, as w ∈ zi Tî , Lemma 2.1.4 yields z ∈ wTî ⊆ yti Tî . It follows that s ∈ ti Tî . Since i has been chosen arbitrarily in {1, . . . , n}, we obtain s ∈ t1 · · · tn . Thus, by definition, sφ = ((t1 )xT1 , . . . , (tn )xTn )ς. The first part of the following theorem generalizes the first part of [11; Theorem 3.17]. Its second part is due to Pamela Ferguson and Alexandre Turull; it is the second part of [11; Theorem 3.17]. Theorem 7.2.5 Assume S to have finite valency. Then the following hold. (i) If {1} = S, S is a quasi-direct product of indecomposable schemes. (ii) Let m and n be positive integers, and let T1 , . . . , Tm and U1 , . . . , Un be indecomposable schemes. Assume that S is a quasi-direct product of T1 , . . . , Tm as well as a quasi-direct product of U1 , . . . , Un . Assume that, for any two elements p and q in S, qp = pq. Then m = n and there exists a permutation π of the set {1, . . . , n} such that, for each element i in {1, . . . , n}, Ti ∼ = Uiπ . Proof. (i) By Theorem 7.1.3(ii), we find a positive integer n and closed subsets T1 , . . . , Tn of S such that T1 × . . . × Tn = S and, for each element i in {1, . . . , n}, Ti is indecomposable. Let x be an element in X. Then, by Theorem 7.2.4, S is a quasi-direct product of the schemes (T1 )x , . . . , (Tn )x . For each element i in {1, . . . , n}, Ti is indecomposable. Thus, for each such i, the scheme (Ti )x is indecomposable. (ii) The scheme S is assumed to be a quasi-direct product of the schemes T1 , such that . . . , Tm . Thus, S possesses closed subsets T1 , . . . , Tm

144

7 Products T1 × . . . × Tm =S

and, for any two elements i in {1, . . . , m} and x in X, Ti ∼ = (Ti )x ; see Theorem 7.2.4. Similarly, S possesses closed subsets U1 , . . . , Un such that U1 × . . . × Un = S and, for any two elements j in {1, . . . , n} and x in X, Uj ∼ = (Uj )x . For each element i in {1, . . . , m}, Ti is assumed to be indecomposable. Thus, for each such i, the scheme (Ti )x is indecomposable. Similarly, we see that, for each element j in {1, . . . , n}, (Uj )x is indecomposable. Now the claim follows from Theorem 7.1.5.

7.3 Semidirect Products In this section, we define the semidirect product of two schemes which has been mentioned in the preface of these monograph. We follow an approach which, in this generality, was given first by Sejeong Bang, Mitsugu Hirasaka, and Sung-Yell Song in [3]. Let W be a finite set, and let A be a scheme on W . Since W is assumed to be a finite set, A has finite valency. Thus, Oϑ (A) is defined and, according to Theorem 4.1.3(i), A//Oϑ (A) is a scheme on W/Oϑ (A). Let us denote by π the natural homomorphism from W ∪ A to W/Oϑ (A) ∪ A//Oϑ (A). According to Theorem 3.2.1(i), Oϑ (A) is strongly normal in A. Thus, by Lemma 4.2.5(ii), A//Oϑ (A) is thin. Thus, (A//Oϑ (A))γ is a group. From Theorem 5.2.2(ii) we also know that Stc(S) is a group with respect to composition. Let ζ be a group homomorphism from (A//Oϑ (A))γ to Stc(S), and let w be an element in W . For any two elements s in S and a in A, we define sζ,w a to be the set of all pairs ((y, u), (z, v)) with y, z ∈ X and u, v ∈ W satisfying v ∈ ua and z ∈ ys(dπζ), where d denotes the uniquely determined element in A which satisfies

7.3 Semidirect Products

145

w ∈ ud. For any two subsets R of S and B of A, we define Rζ,w B to be the set of all elements rζ,w b where r is an element in R and b an element in B. Lemma 7.3.1 Let W be a set, let A be a scheme on W , let ζ be a group homomorphism from (A//Oϑ (A))γ to Stc(S), and let w be an element in W . Then we have the following. (i) The set Sζ,w A is a partition of X × W . (ii) We have 1X×W = 1ζ,w 1. Proof. Let y, z be elements in X, and let u, v be elements in W . Let d be the uniquely determined element in A with w ∈ ud, let π be the natural homomorphism from W ∪ A to W/Oϑ (A) ∪ A//Oϑ (A), and set σ := dπζ. Then σ ∈ Stc(S). (i) Let s be the uniquely determined element in S with z ∈ ys, and let a be the uniquely determined element in A with v ∈ ua. Since σ ∈ Stc(S), σ is bijective. Thus, by definition, we must have that ((y, u), (z, v)) ∈ (sσ −1 )ζ,w a. Now let p, q be elements in S, and let b, c be elements in A such that ((y, u), (z, v)) ∈ pζ,w b ∩ qζ,w c. Then, by definition, v ∈ ub ∩ uc and z ∈ ypσ ∩ yqσ. From v ∈ ub ∩ uc we obtain b = c. From z ∈ ypσ ∩ yqσ we obtain pσ = qσ. Thus, as σ is bijective, p = q. It follows that pζ,w b = qζ,w c. (ii) Since σ ∈ Stc(S), 1σ = 1; cf. Lemma 5.2.3(ii). Thus, ((y, u), (z, v)) ∈ 1ζ,w 1 if and only if (u, v) ∈ 1 and (y, z) ∈ 1. This is the case if and only if (y, u) = (z, v). Lemma 7.3.2 Let W be a set, let A be a scheme on W , let π denote the natural homomorphism from W ∪ A to W/Oϑ (A) ∪ A//Oϑ (A), let ζ be a group homomorphism from (A//Oϑ (A))γ to Stc(S), let w be an element in W , let s be an element in S, and let a be an element in A. (i) We have (sζ,w a)∗ = s∗ (aπζ)ζ,w a∗ . (ii) Let y, z be elements in X, and let u, v be elements in W such that (z, v) ∈ (y, u)sζ,w a. Then, for any four elements p, q in S and b, c in A, we have |(y, u)pζ,w b ∩ (z, v)(qζ,w c)∗ | = apq(b∗ πζ)s abca . Proof. (i) Let y, z be elements in X, and let u, v be elements in W such that ((y, u), (z, v)) ∈ sζ,w a. Then, by definition, v ∈ ua and z ∈ ys(dπζ), where d denotes the uniquely determined element in A which satisfies w ∈ ud.

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From z ∈ ys(dπζ) we obtain y ∈ z(s(dπζ))∗ . On the other hand, as dπζ ∈ Stc(S), s∗ (dπζ) = (s(dπζ))∗ ; cf. Lemma 5.2.3(iii). Thus, y ∈ zs∗ (dπζ). Let us denote by d the uniquely determined element in A which satisfies w ∈ vd . From w ∈ vd and v ∈ ua we obtain w ∈ uad . Thus, as w ∈ ud, d ∈ ad . Thus, by Lemma 5.1.1(ii), dπ ∈ aπd π, and then dπζ = (aπζ)(d πζ). Thus, as y ∈ zs∗ (dπζ), y ∈ zs∗ (aπζ)(d πζ). Thus, as u ∈ va∗ and w ∈ vd , ((z, v), (y, u)) ∈ s∗ (aπζ)ζ,w a∗ . (ii) Let x be an element in X, and let t be an element in W such that (x, t) ∈ (y, u)pζ,w b ∩ (z, v)(qζ,w c)∗ . From (x, t) ∈ (y, u)pζ,w b we obtain t ∈ ub and x ∈ yp(dπζ), where d denotes the uniquely determined element in A which satisfies w ∈ ud. From (z, v) ∈ (x, t)qζ,w c we obtain v ∈ tc and z ∈ xq(d πζ), where d denotes the uniquely determined element in A which satisfies w ∈ td . Since w ∈ ud and u ∈ tb∗ , w ∈ tb∗ d. Thus, as w ∈ td , d ∈ b∗ d. Thus, by Lemma 5.1.1(ii), d π ∈ b∗ πdπ, and then d πζ = (b∗ πζ)(dπζ). Thus, as z ∈ xq(d πζ), z ∈ xq(b∗ πζ)(dπζ). Thus, as x ∈ yp(dπζ), x ∈ yp(dπζ) ∩ z(q(b∗ πζ)(dπζ))∗ . Thus, as t ∈ ub ∩ vc∗ , (x, t) ∈ (yp(dπζ) ∩ z(q(b∗ πζ)(dπζ))∗ ) × (ub ∩ vc∗ ). Thus, as (x, t) has been chosen arbitrarily in (y, u)pζ,w b ∩ (z, v)(qζ,w c)∗ , (y, u)pζ,w b ∩ (z, v)(qζ,w c)∗ = (yp(dπζ) ∩ z(q(b∗ πζ)(dπζ))∗ ) × (ub ∩ vc∗ ). On the other hand, we are assuming that (z, v) ∈ (y, u)sζ,w a. Thus, by definition, z ∈ ys(dπζ) and v ∈ ua. Thus, |ye(dπζ) ∩ z(q(b∗ πζ)(dπζ))∗ | = ap(dπζ)q(b∗ πζ)(dπζ)s(dπζ) and

|ub ∩ vc∗ | = abca .

Finally recall that, as dπζ ∈ Stc(S), ap(dπζ)q(b∗ πζ)(dπζ)s(dπζ) = apq(b∗ πζ)s . Thus,

|(y, u)pζ,w b ∩ (z, v)(qζ,w c)∗ | = apq(b∗ πζ)g abca .

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147

This finishes the proof of Lemma 7.3.2. Theorem 7.3.3 Let W be a set, let A be a scheme on W , and let ζ be a group homomorphism from (A//Oϑ (A))γ to Stc(S). (i) For each element w in W , Sζ,w A is a scheme on X × W . (ii) For any two elements u and v in W , Sζ,u A = Sζ,v A. Proof. (i) This follows from Lemma 7.3.1 together with Lemma 7.3.2. (ii) Let p and q be elements in S, let v and w be elements in W , and let b and c be elements in A such that pζ,v b ∩ qζ,w c is not empty. We shall show that pζ,v b = qζ,w c. Assuming that pζ,v b ∩ qζ,w c is not empty we find elements y and z in X and t and u in W such that ((y, t), (z, u)) ∈ pζ,v b ∩ qζ,w c. Let d be the uniquely defined element in A satisfying v ∈ td, and let e be the uniquely defined element in A satisfying w ∈ te. Let us denote by π the natural homomorphism from W ∪ A to W/Oϑ (A) ∪ A//Oϑ (A). From ((y, t), (z, u)) ∈ pζ,v b and v ∈ td we obtain z ∈ yp(dπζ). Similarly, as ((y, t), (z, u)) ∈ qζ,w b and w ∈ te, z ∈ yq(eπζ). Thus, as S is a scheme, p(dπζ) = q(eπζ). Note also that u ∈ tb ∩ tc. Thus, as A is assumed to be a scheme, b = c. Let y and z be elements in X, and let t and u be elements in W such that ((y , t ), (z , u )) ∈ pζ,v b. We shall be done if we succeed in showing that ((y , t ), (z , u )) ∈ qζ,w c. From ((y , t ), (z , u )) ∈ pζ,v b we obtain u ∈ t b and z ∈ y p(dπζ), where v ∈ t d. From z ∈ y p(dπζ) and p(dπζ) = q(eπζ) we obtain z ∈ y q(eπζ). Let e be the uniquely defined element in A satisfying w ∈ t e , and let r be the element in A satisfying t ∈ tr. From w ∈ t e and t ∈ tr we obtain w ∈ tre. Thus, as w ∈ te, e ∈ re . From t ∈ vd∗ and v ∈ td we obtain t ∈ tdd∗ . Thus, as t ∈ tr, r ∈ dd∗ . Now recall that, by Theorem 3.2.1(ii), dd∗ ⊆ Oϑ (A), so that r ∈ Oϑ (A). From e ∈ re and r ∈ Oϑ (A) we obtain e ∈ Oϑ (A)e . It follows that eπ = e π. Thus, as z ∈ y q(eπζ), z ∈ y q(e πζ).

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Thus, as w ∈ t e , u ∈ t b, and b = c, ((y , t ), (z , u )) ∈ qζ,w c. Let W , A, and ζ be as before, and let w be an element in W . Theorem 7.3.3(ii) allows us to write Sζ A instead of Sζ,w A. Similarly, for any two elements s in S and a in A, we may write sζ a instead of sζ,w a. We call Sζ A the semidirect product of S and A with respect to ζ. The scheme S is called the kernel of the semidirect product Sζ A, A its complement. Occasionally, we do not need to specify the group homomorphism ζ which defines the semidirect product Sζ A. In this case, we shall just write SA instead of Sζ A. Corollary 7.3.4 Let A be a scheme, let π denote the natural homomorphism from W ∪ A to W/Oϑ (A) ∪ A//Oϑ (A), and let ζ be a group homomorphism from (A//Oϑ (A))γ to Stc(S). Then, for any six elements p, q, r in S and a, b, c in A, the following hold. (i) We have a(pζ b)(qζ c)(rζ a) = apq(b∗ πζ)r abca . (ii) We have rζ a ∈ (pζ b)(qζ c) if and only if r ∈ pq(b∗ πζ) and a ∈ bc. Proof. (i) This follows from Theorem 7.3.3(i) together with Lemma 7.3.2(ii). (ii) This follows from (i). Theorem 7.3.5 Let A be a scheme, and let ζ be a group homomorphism from (A//Oϑ (A))γ to Stc(S). (i) The sets Sζ {1} and {1}ζ A are closed subsets of Sζ A. (ii) For any two elements s in S and a in A, {sζ a} = (sζ 1)(1ζ a). (iii) We have {1ζ 1} = Sζ {1} ∩ {1}ζ A and (Sζ {1})({1}ζ A) = Sζ A. (iv) The closed subset Sζ {1} is normal in Sζ A. Proof. (i) From Lemma 7.3.2(i) we obtain (sζ 1)∗ = (s∗ )ζ 1 ∈ Sζ 1 for each element s in S. From Corollary 7.3.4(ii) we obtain (pζ 1)(qζ 1) = {sζ 1 | s ∈ pq} for any two elements p and q in S. Thus, Sζ {1} is a closed subset of Sζ A. From Lemma 5.2.3(ii) we obtain 1(aπζ) = 1 for each element a in A. Thus, for each element a in A, (1ζ a)∗ = 1ζ a∗ ∈ 1ζ A; cf. Lemma 7.3.2(i). For any two elements b and c in A, (1ζ b)(1ζ c) = {1ζ a | a ∈ bc}; cf. Corollary 7.3.4(ii). Thus, we have shown that {1}ζ A is a closed subset of Sζ A. (ii) This follows immediately from Corollary 7.3.4(ii). (iii) The first claim is obvious, the second claim follows from (i). (iv) According to Lemma 2.5.1(ii), we just have to show that {1}ζ A normalizes Sζ {1}. In order to do so we fix elements s in S and a in A. We have to show that (sζ 1)(1ζ a) ⊆ (1ζ a)(Sζ {1}).

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From (ii) we know that {sζ a} = (sζ 1)(1ζ a). From Corollary 7.3.4(ii) we also obtain sζ a ∈ (1ζ a)(s(aπζ)ζ 1) ⊆ (1ζ a)(Sζ 1). (Note that a∗ πζ = (aπζ)−1 .) Thus, (sζ 1)(1ζ a) ⊆ (1ζ a)(Sζ {1}). Note that, according to Lemma 2.5.1(ii), the last condition in Theorem 7.3.5 is equivalent to {1}ζ A ⊆ NSζ A (Sζ {1}).

7.4 A Characterization of Semidirect Products It is the purpose of this (short) section to show that the four conditions given in Theorem 7.3.5 are sufficient to identify a scheme as a semidirect product. Throughout this section, the letters T and U will stand for closed subsets of S satisfying {1} = T ∩ U and U ⊆ NS (T ). We shall always assume that, for any two elements t in T and u in U , 1 = |tu|. The latter hypothesis implies that, for any two elements t in T and u in U , u∗ tu contains exactly one element which is in T ; cf. Lemma 2.6.4(ii). In the following, we shall denote this element by tσu . Lemma 7.4.1 We have the following. (i) For any two elements t in T and u in U , tu = utσu . (ii) For any three elements p, q, and r in T with r ∈ pq, we have σp σq = σr . (iii) For any two elements t in T and u in Oϑ (U ), tσu = t. (iv) For each element u in U , σu ∈ Stc(T ). Proof. (i) Let t be an element in T , and let u be an element in U . Then, by definition, tσu ∈ u∗ tu. Thus, there exists an element s in tu such that tσu = u∗ s. From s ∈ tu and 1 = |tu| we obtain {s} = tu. From tσu = u∗ s we obtain s ∈ utσu ; cf. Lemma 1.3.3(ii). Thus, as 1 = |utσu |, {s} = utσu . It follows that tu = utσu . (ii) Let t be an element in T , and let p, q be elements in U . Then, by (i), tp = ptσp and tσp q = qtσp σq . Thus, tpq = ptσp q = pqtσp σq . Thus, for each element r in pq, tσr ∈ r∗ tr ∩ T ⊆ q ∗ p∗ tpq ∩ T = q ∗ p∗ pqtσp σq ∩ T. Thus, there exists an element u in q ∗ p∗ pq such that tσr ∈ utσp σq ∩ T . From tσr ∈ utσp σq we obtain u ∈ T , from u ∈ q ∗ p∗ pq we obtain u ∈ U . Thus, u ∈ T ∩ U . However, we are assuming that {1} = T ∩ U . Thus, u = 1. Thus, as tσr ∈ utσp σq , tσr = tσp σq . Thus, as t has been chosen arbitrarily in T , we have shown that σp σq = σr .

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(iii) Let u be an element in Oϑ (U ). Then, by Theorem 3.2.1(ii), there exist elements u1 , . . . , un in U such that u ∈ u∗1 u1 · · · u∗n un . Thus, the claim follows from (ii). (iv) Let u be an element in U . Then, by (ii), σu∗ σu = σ1 . Thus, σu is bijective. In order to show that σu ∈ Stc(T ), we now fix three elements p, q, and r in T . We shall see that apσu qσu rσu = apqr . Let y and z be elements in X such that z ∈ yr. Since rσu ∈ u∗ ru, r ∈ urσu u∗ . Thus, as z ∈ yr, z ∈ yurσu u∗ . Thus, there exists an element v in yu such that z ∈ vrσu u∗ . Since z ∈ vrσu u∗ , there exists an element w in vrσu such that z ∈ wu∗ . Let x be an element in yp ∩ zq ∗ . Since x ∈ yp, y ∈ xp∗ . Thus, as v ∈ yu, v ∈ xp∗ u. On the other hand, we know from (i) that p∗ u = up∗ σu . Thus, v ∈ xup∗ σu . Thus, there exists an element x in xu such that v ∈ x p∗ σu . From p∗ σu = (pσu )∗ and v ∈ x p∗ σu we obtain x ∈ vpσu . Let x be an element in vpσu ∩ xu. Then x ∈ x (u∗ u ∩ T ) = {x }. Thus, we have a bijective map from yp ∩ zq ∗ to vqσu ∩ w(qσu )∗ . It follows that apσu qσu rσu = apqr . Thus, as p, q, and r have been chosen arbitrarily in T , we have shown that σu ∈ Stc(T ). Theorem 7.4.2 For any two elements y in X and z in yT U , Ty ∼ = Tz . Proof. We are assuming that z ∈ yT U . Thus, there exist elements t in T and u in U such that z ∈ ytu. Since z ∈ ytu, there exists an element s in ut such that z ∈ ys. Since s ∈ tu and 1 = |tu|, {s} = tu. Thus, as u∗ tu ∩ T is assumed to be not empty, u∗ s ∩ T is not empty. Thus, there exists an element t in T such that t ∈ u∗ s. Since t ∈ u∗ s, s ∈ ut ; cf. Lemma 1.3.3(ii). Thus, as z ∈ ys, z ∈ yuT . Thus, there exists an element z in yu such that z ∈ z T . Thus, we may assume that z = z. Thus, z ∈ yu. For each element x in yT , we define xφ to be the uniquely determined element in xu ∩ zT . For each element t in T , we define tφ to be the uniquely defined element in u∗ tu ∩ T . Then φ is an isomorphism from yT ∪ TyT to zt ∪ TzT . The following theorem is the converse of Theorem 7.3.5. Theorem 7.4.3 For each element x in X, there exists a group homomorphism ζ from (UxU )//Oϑ (UxU ) to Stc(TxT ) such that (T U )x ∼ = (Tx )ζ (Ux ). Proof. Let us write π to denote the natural homomorphism from xU ∪ UxU to xU/Oϑ (UxU )xU ∪ UxU //Oϑ (UxU ). Then, for any two elements t in T and u in U ,

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txT (uxU πζ) = (tσu )xT ; cf. Lemma 7.4.1(ii), (iii). Let w be an element in xT U . Since we are assuming that {1} = T ∩ U , there exists a uniquely determined element wT in xT such that w ∈ wT U . From w ∈ xT U we also obtain w ∈ xU T . Thus, there exists a uniquely determined element wU in xU such that w ∈ wU T . We set wχ := (wT , wU ). Let s be an element in T U . Then there exist uniquely determined elements t in T and u in U such that {s} = tu; cf. Lemma 2.1.2. We define sφ := (txT )ζ,x (uxU ) and claim that φ is an isomorphism with respect to χ. Let us first show that χ is a bijective map from X to xT × xU . Let u and v be elements in X such that uχ = vχ. Then, by definition, uT = vT and uU = vU . Since u ∈ uT U and v ∈ vT U , uT = vT yields v ∈ uU . Similarly uU = vU yields v ∈ uT . Thus, v ∈ u(T ∩ U ). Thus, as we are assuming that {1} = T ∩ U , u = v. This shows that χ is injective. Let y be an element in xT , and let z be an element in xU . From y ∈ xT we obtain x ∈ yT . Thus, as z ∈ xU , z ∈ yT U . Thus, as T U ⊆ U T , z ∈ yU T . Thus, there exists an element w in yU such that z ∈ wT . Since y ∈ xT and w ∈ yU , y = wT . Similarly, as z ∈ xU and w ∈ zT , z = wU . Thus, wχ = (y, z). Thus, we have shown that χ is a bijective map from X to xT × xU . That φ is a bijective map from T UxT U to (TxT )ζ (UxU ) follows immediately from the definition of φ. Let us now show that φ is a homomorphism with respect to χ. In order to do this, we fix elements y and z in X, and we call s the uniquely determined element in S which satisfies z ∈ ys. We have to show that zχ ∈ yχsφ. Let t and u be the uniquely determined elements in T and U such that {s} = tu. We have to show that (zT , zU ) ∈ (yT , yU )(txT )ζ,x (uxU ). Thus, we have to show that zU ∈ yU u and that zT ∈ yT txT (dxU πζ), where d denotes the uniquely determined element in U which satisfies x ∈ yU d. Since z ∈ ys ⊆ ytu, there exists an element w in yt such that z ∈ wu. From w ∈ yt and y ∈ yU T we obtain w ∈ yU T . Thus, as z ∈ wu, z ∈ yU T u. Thus, z ∈ yU uT . Thus, there exists an element z in yU u such that z ∈ z T . From z ∈ yU u and yU ∈ xU we obtain z ∈ xU . Thus, as z ∈ zT , z = zU . It follows that zU ∈ yU u.

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By definition, x ∈ yU d and yU ∈ yT . Thus, x ∈ yT d. But as d ∈ U , T d ⊆ dT . Thus, x ∈ ydT . Thus, there exists an element y in yd such that x ∈ y T . From yT ∈ yU ∩ xT we, therefore, obtain y = yT . Thus, as y ∈ yd, yT ∈ yd. Similarly, we obtain zT ∈ wd. From y ∈ yT d∗ , w ∈ yt, and zT ∈ wd we obtain zT ∈ yT (d∗ td∩T ). Thus, as tσd is our notation for the only element in d∗ td ∩ T , zT ∈ yT tσd . Thus, as zT ∈ xT and tσd ∈ T , zT ∈ yT (tσd )xT . Thus, by the above, zT ∈ yT txT (dxU πζ).

8 From Thin Schemes to Modules

In this chapter, we shall develop some of the fundamental aspects of the representation theoretic part of scheme theory. Representations of (finite) schemes reflect the arithmetic structure of schemes. They are useful in cases where the structure constants underly extreme constraints. The central notion in representation theory of finite schemes is the one of an associative ring. Rings give rise to modules. The concept of a ring goes back to Richard Dedekind; cf. [7]. The first abstract definition of a ring was given by Adolf Fraenkel; cf. [12]. The use of modules in the theory of finite groups is due to Emmy Noether [32; III]. The idea to generalize the relationship between modules and finite groups to a fruitful relationship between modules and schemes of finite valency had been recognized first by Donald Higman; cf. [20]. Since rings as well as modules are built from commutative groups and groups are identified with thin schemes (via the group correspondence) one may view the notion of a ring and the one of a module as part of ‘thin scheme theory’. The first section of this chapter provides general observations on modules over associative rings with 1. The collection includes the Homomorphism Theorem and the Isomorphism Theorem for modules over associative rings with 1. In the second section, we shall look at commutative associative rings with 1. We prove that the set of all elements of a commutative associative ring D with 1 which are integral over a unitary subring of D forms a ring. In Section 8.3, we focus on completely reducible modules over associative rings with 1. Section 8.4 deals with irreducible modules over these rings. In Section 8.5, we combine the results obtained in these two sections to obtain the famous (and complete) description of semisimple associative rings with 1 which was first given by Joseph Wedderburn and Emil Artin. The interest in semisimple rings is based on Theorem 9.1.5(ii), a result in which we shall see that each scheme of finite valency gives rise to semisimple rings, the so-called scheme rings.

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In Section 8.6, we shall provide a few basic facts on characters of semisimple rings. In the last section of this chapter, in Section 8.7, we present identities about roots of unity in integral domains. The results will be useful in Section 12.4 where we shall investigate Coxeter sets of cardinality 2.

8.1 Rings and Modules Let M be an additively written group (with neutral element 0). For each element m in M with m + m = m, one has m = m + 0 = m + (m + (−m)) = (m + m) + (−m) = m + (−m) = 0. The group M is called commutative if, for any two elements k and l in M , k + l = l + k. Let us now assume M to be commutative, and let D be a further additively written commutative group. A map from M × D to M is called an operation of D on M . Let us fix an operation of D on M . For any two elements m in M and d in D, we shall write md to denote the image of (m, d) under this operation. The (commutative) group M is called a group over D or a D-group if, for any three elements k, l in M and d in D, (k + l)d = kd + ld and, for any three elements m in M and b, c in D, m(b + c) = mb + mc. Let us assume M to be a D-group. For each element m in M , we have m0 + m0 = m(0 + 0) = m0. Thus, the above observation tells us that, for each element m in M , m0 = 0. Similarly, one obtains 0d = 0 for each element d in D. Let m be an element in M and d an element in D. Then we have md + (−m)d = (m + (−m))d = 0d = 0, so that (−m)d = −(md). Similarly, one obtains m(−d) = −(md). The D-group M is called unital if there exists an element d in D \ {0} such that, for each element m in M , md = m.

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An (additively written) group is called a ring if it is a group over itself. Let us assume D to be a ring. Assuming D to be a ring we have an operation of D on D or, what is the same, an operation on D in the sense of Section 5.5. This operation on D is called the multiplication of D, and one stays with the convention that bc denotes the image of (b, c) under the multiplication on D whenever b and c are elements in D. A subgroup C of the additive group of D is called a subring if, with respect to the restriction of the multiplication on D to C × C, C is a C-group. We define Z(D) to be the set of all elements z in D such that, for each element d in D, dz = zd. Note that 0 ∈ Z(D). Moreover, for any two elements d in D and z in Z(D), we have d(−z) = −(dz) = −(zd) = (−z)d. Finally, for any three elements d in D and x, y in Z(D), one has d(x + y) = dx + dy = xd + yd = (x + y)d. Thus, Z(D) is a subgroup of the additive group D. It is called the center of D. The ring D is called commutative if Z(D) = D. Since Z(D) is a subgroup of the additive group D, the additive group D is a Z(D)-group with respect to the restriction of the multiplication of D to D × Z(D). Let x and y be elements in Z(D)\{0}, and let us assume that, for each element d in D, dx = d and dy = d. Then x = xy = yx = y. This shows that, if the Z(D)-group D is unital, then there exists at most one element z in Z(D) \ {0} such that, for each element d in D, dz = d. In this case, we shall denote this unique element by 1. (Thus, by definition, 0 = 1.) The ring D is called a ring with 1 if the Z(D)-group D is unital. The D-group M is called a module over D or a D-module if, for any three elements m in M and b, c in D, m(bc) = (mb)c. Assume M to be a D-module. Then, for each subring C of D, M is a Cmodule. A ring is called associative if it is a module over itself. If D is an associative ring with 1 and M a D-module, we always assume M to be unital. For the remainder of this section, we shall assume D to be an associative ring with 1.

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Assuming D to be associative we obtain d(xy) = (dx)y = (xd)y = x(dy) = x(yd) = (xy)d. for any three elements d in D and x, y in Z(D). Thus, Z(D) is a subring of the ring D. Let d be an element in D. An element c in D is called a multiplicative inverse of d if dc = 1 = cd. Let d be an element in D, and let b and c be multiplicative inverses of d. Then b = b1 = b(dc) = (bd)c = 1c = c. Thus, each element in D has at most one multiplicative inverse. An element d in D is called a unit if it has at least one multiplicative inverse. The multiplicative inverse of a unit d of D is denoted by d−1 . The set of all units of D is denoted by U (D). Recall that, for each element d in D, 0d = 0 = d0. Thus, as 0 = 1, U (D) ⊆ D \ {0}. The ring D is called a division ring if U (D) = D \ {0}. A commutative division ring is called a field. A commutative subring C of D is called a subfield if, for each element c in C \ {0}, c ∈ U (D) and c−1 ∈ C. A (unital) module over a field C is called a vector space over C. Let us now assume M to be a (unital) D-module. Since M is assumed to be unital, D possesses an element e such that, for each element m in M , me = m. It follows that, for each element m in M , m1 = (me)1 = m(e1) = me = m. Let L be a subset of M , and let C be a subset of D. If one of the two sets C or L is empty, we set LC := {0}. If both C and L are not empty, we define LC to be the set of all finite sums of products lc with l ∈ L and c ∈ C. If there exists an element l in L with {l} = L, we write lC instead of LC. Similarly, if there exists an element c in C with {c} = C, we write Lc instead of LC. For any two nonempty subsets C of D and L of M , the set LC is called the complex product of L and C. We also speak of the complex multiplication between M and D. A subgroup L of the additive group M is called a submodule of M if LD ⊆ L. Note that a subgroup L of M is a submodule of M if, for any two elements l in L and d in D, ld ∈ L. Note also that {0} and M are submodules of M . Moreover, for any two submodules K and L of M , K ∩ L is a submodule of M . A submodule L of the D-module D is called an ideal of D if DL ⊆ L.


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It is obvious that {0} and D are ideals of D. If, conversely, these two ideals are the only ideals of D and {0} = D, then the ring D is called simple. Let K and L be nonempty subsets of M . We define K + L to be the set of all sums k + l with k ∈ K and l ∈ L. If there exists an element k in K with {k} = K, we write k + L instead of K + L. Similarly, if there exists an element l in L with {l} = L, we write K + l instead of K + L. For any two nonempty subsets K and L of M , the set K + L is called the complex sum of K and L. The associated operation on the set of all nonempty subsets of M will be referred to as the complex addition in M . It is easy to see that the complex sum of two submodules of M is a submodule of M . Let us now fix a submodule L of M . We define M/L to be the set of all sets L + m with m ∈ M . It is easy to see that the restriction of the complex addition to M/L is an operation on M/L. Moreover, one verifies easily that, with respect to this addition, M/L is a commutative group. For any two elements m in M and d in D, we define (L + m)d := L + md.1 It is easy to see that, for any three elements h, k in M and d in D, ((L + h) + (L + k))d = (L + h)d + (L + k)d. Moreover, for any three elements m in M and b, c in D, one has (L + m)(b + c) = (L + m)b + (L + m)c. Thus, M/L is a D-group. Note also that, for any three elements m in M and b, c in D, (L + m)(bc) = ((L + m)b)c. Thus, M/L is a D-module. Finally, for each element m in M , we have (L + m)1 = L + m, so that the D-module M/L is unital. In fact, all four of these equations follow from the corresponding equation for the D-group M . The unital D-module M/L is called the factor module of M over L. 1

This is not the complex multiplication between M and D.

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˜ be a D-module, and let φ be a map from M to M ˜. Let M The map φ is called a homomorphism if, for any two elements k and l in M , (k + l)φ = kφ + lφ and, for any two elements m in M and d in D, (md)φ = (mφ)d. Sometimes, in order to emphasize the underlying ring, one says D-module homomorphism or D-homomorphism instead of homomorphism. We define ker(φ) to be the set of all elements m in M satisfying mφ = 0. It is easy to see that ker(φ) is a submodule of M and that M φ is a submodule ˜ . (By M φ we mean, as is common, the set of all elements mφ with of M m ∈ M .) ˜ are called isomorphic, if there exists a bijective The D-modules M and M ˜ . Sometimes, we shall write M ∼ ˜ in order to homomorphism from M to M =M ˜ indicate that M and M are isomorphic. The following theorem is called the Homomorphism Theorem for modules over associative rings with 1. Theorem 8.1.1 Let φ be a homomorphism from M . Then ker(φ) is a submodule of M , and M φ is an D-module with M/ker(φ) ∼ = M φ. Proof. It is obvious that ker(φ) is a submodule of M . For any two elements k and l in M , we have kφ + lφ = (k + l)φ ∈ M φ. Also, for any two elements m in M and d in D, we have (mφ)d = (md)φ ∈ M φ. Therefore, M φ is an D-module. Let us abbreviate L := ker(φ). For any two elements h and k in M , we have hφ = kφ if and only if (h − k)φ = 0, and this is equivalent to h − k ∈ L. Note also that h − k ∈ L is equivalent to L + h = L + k. This gives us a bijective map ι from M/L to M φ which, for each element m in M , maps L + m to mφ. For any two elements h and k in M , we have ((L + h) + (L + k))ι = (L + (h + k))ι = (h + k)φ = hφ + kφ = (L + h)ι + (L + k)ι. For any two elements m in M and d in D, we have ((L + m)d)ι = (L + md)ι = (md)φ = (mφ)d = ((L + m)ι)d. Thus, ι is a D-homomorphism.


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The following theorem is the Isomorphism Theorem for modules over associative rings with 1. Theorem 8.1.2 For any two submodules K and L of M , we have K/K ∩L ∼ = (K + L)/L. Proof. For each element k in K, we define kφ := L+k. Then φ is a surjective map from K to (K + L)/L. For any two elements h and k in M , we have (h + k)φ = L + (h + k) = (L + h) + (L + k) = hφ + kφ. For any two elements k in K and d in D, we have (kd)φ = L + kd = (L + k)d = (kφ)d. Therefore, φ is a homomorphism. Note that ker(φ) = K ∩ L. Thus, the claim follows from Theorem 8.1.1. A group homomorphism from the additive group M to itself is called a (group) endomorphism of M . Let us write End(M ) to denote the set of all endomorphisms of the additive group M . Let c and d be elements in End(M ). For each element m in M , we define m(c + d) := mc + md. It is straightforward to see that, as c, d ∈ End(M ), c + d ∈ End(M ). This shows that componentwise addition is an operation on End(M ). Note also that the composition of elements in End(M ) is an operation on End(M ). It is easy to see that, with respect to componentwise addition and composition (as multiplication), End(M ) is a ring. Note also that M is a module over End(M ). A D-homomorphism from M to M is called a D-endomorphism of M . We define EndD (M ) to be the set of all D-endomorphisms of M . Note that EndD (M ) is a subring of End(M ). In particular, M is a module over EndD (M ). For each element d in D, we define dM : M → M, m → md.

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Note that, for each element d in D, dM ∈ End(M ). We set DM := {dM | d ∈ D}. Lemma 8.1.3 Set E := EndD (M ). Then DM ⊆ EndE (M ). Proof. For any three elements m in M , d in D, and e in E, we have (mdM )e = (md)e = (me)d = (me)dM . Thus, dM ∈ EndE (M ). For each element d in D, we define BM (d) to be the kernel of dM . For each subset C of D, we define BM (C) to be the intersection of the sets BM (c) with c ∈ C. Lemma 8.1.4 Set E := EndD (M ). Then, for each subset C of D, BM (C) is a submodule of the E-module M . Proof. Considering that intersections of submodules are submodules, this follows from Theorem 8.1.1 together with Lemma 8.1.3. Let m be an element in M . We define mD : D → M, d → md. Then mD is a D-homomorphism from the D-module D to M . For each element m in M , we define AD (m) to be the kernel of mD . For each subset L of M , we define AD (L) to be the intersection of the sets AD (l) with l ∈ L. Lemma 8.1.5 For each subset L of M , AD (L) is a submodule of the Dmodule D. Proof. Considering that intersections of submodules are submodules, this follows from Theorem 8.1.1. Lemma 8.1.6 We have the following. (i) For each subset C of D, CD ⊆ AD (BM (C)). (ii) Set E := EndD (M ). Then, for each subset L of M , LE ⊆ BM (AD (L)). Proof. (i) Let c be an element in C. Then, for each element m in BM (C), mc = 0. Thus, by definition, c ∈ AD (BM (C)). Since c has been chosen arbitrarily in C, we have shown that C ⊆ AD (BM (C)). Thus, the claim follows from Lemma 8.1.5.

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(ii) Let l be an element in L. Then, for each element d in AD (L), ld = 0. Thus, by definition, l ∈ BM (AD (L)). Since l has been chosen arbitrarily in L, we have shown that L ⊆ BM (AD (L)). Thus, the claim follows from Lemma 8.1.4. A subring C of D is called unitary if 1 ∈ C. Note that the intersection of any two unitary subrings of D is a unitary subring of D. In particular, D possesses a uniquely determined smallest unitary ˘ subring. Let us (for the moment) denote this unitary subring by D. ˘ is finite, we call |D| ˘ the characteristic of D. Otherwise, we say that D If |D| has characteristic 0.

8.2 Integrality in Associative Rings with 1 Throughout this section, the letter D stands for an associative ring with 1. Let C be a unitary subring of D. Let A be a subset of D such that, for any two elements c in C and a in A, ca = ac. Then we shall denote by C[A] the smallest subring of D containing C and A as subsets. If there exists an element a in A with {a} = A, we write C[a] instead of C[A]. Let us now assume D to be commutative. An element d in D is called integral over C if there exist elements c0 , . . . , cn−1 in C such that dn + cn−1 dn−1 + . . . + c1 d + c0 = 0. We define ID (C) to be the set of all elements in D which are integral over C. The ring D is called finite over C if there exists a finite subset A of D such that C[A] = D. A D-module M is called finitely generated if it possesses a finite subset L with LD = M . Occasionally, we shall refer to the elements in L as generators of M. Lemma 8.2.1 Let C be a unitary subring of D such that D is finite over C and integral over C. Then the C-module D is finitely generated. Proof. Since D is assumed to be finite over C, D possesses a finite subset A such that C[A] = D. Since D is assumed to be integral over C, each element of A is integral over C. Thus, for each element a in A, there exists a positive integer na together with elements ca,0 , . . . , ca,na −1 in C such that ana + ca,na −1 ana −1 + . . . + ca,1 a + ca,0 = 0.

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Let a be an element in A. We set Ba := {1, a, . . . , ana −1 }. Then ana ∈ Ba C. Thus, for each element d in Ba C, ad ∈ Ba C, and that means that aBa C ⊆ Ba C. We define B to be the set of all products ba a∈A

such that, for each element a in A, ba ∈ Ba . Then, as A is finite, so is B. Recall that, for each element a in A, aBa C ⊆ Ba C. Thus, for each element a in A, aBC ⊆ BC. Thus, as C[A] = D, DBC ⊆ BC. Thus, as 1 ∈ BC, BC = D. Lemma 8.2.2 Let B be a finite subset of D. For any two elements a and b in B, let dab be an element in D such that, for each element b in B, adab = 0. a∈B

Then, for each element b in B, bdet(dab ) = 0. Proof. Set n := |B|. Then there exist elements b1 , . . . , bn in B such that {b1 , . . . , bn } = B. Set A := (b1 , . . . , bn ). For any two elements i and j in {1, . . . , n}, we set dij := dbi bj . We set M := (dij ). Then AM = 0. For any two elements i and j in {1, . . . , n}, we set nij := (−1)i+j det(Mkl )k=i,l=j . We set N := (nij ). Then M N = det(M )I. From AM = 0 and M N = det(M )I we obtain Adet(M )I = 0. Thus, for each element b in B, bdet(M ) = 0. Lemma 8.2.3 Let B be a finite subset of D, and let C be a unitary subring of D such that BC is a subring of D. Then BC ⊆ ID (C). Proof. Let d be an element in BC. We have to show that d ∈ ID (C). Let b be an element in B. Then, as d ∈ BC and BC is assumed to be a subring of D, bd ∈ BC. Thus, there exists, for each element a in B, an element cab in C such that acab . bd = a∈B

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For any two elements a and b in B such that a = b, we set dab := −cab . For each element b in B, we set dbb := d − cbb . Then, for each element b in B, adab = 0. a∈B

Thus, for each element b in B, bdet(dab ) = 0; cf. Lemma 8.2.2. Recall that D is assumed to be a ring with 1, and BD is assumed to be a unitary subring of D. Thus, 1 ∈ BC. Thus, as, for each element b in B, bdet(dab ) = 0, 1det(dab ) = 0. It follows that det(dab ) = 0. However, det(dab ) is a monic polynomial in d with coefficients in C. In particular, d is integral over C. The following result is due to Richard Dedekind; cf. [7; §160]. Theorem 8.2.4 For each unitary subring C of D, ID (C) is a subring of D. Proof. We set I := ID (C) and pick an element d in I 2 . We have to show that d ∈ I. Since d ∈ I 2 , there exists a finite subset A of I such that d ∈ A2 . Since A is a finite subset of I, C[A] possesses a finite subset B such that C[A] = BC; cf. Lemma 8.2.1. Since A ⊆ BC and BC is a subring of D, d ∈ A2 ⊆ (BC)2 ⊆ BC. On the other hand, as B is finite and BC a subring of D, Lemma 8.2.3 yields that BC ⊆ I. Thus, d ∈ I. Lemma 8.2.5 Assume D to be a field of characteristic 0. Let us denote by Q the smallest subfield of D and by Z the smallest unitary subring of D. Then Z = IQ (Z). Proof. Let q be an element in IQ (Z). We have to show that q ∈ Z. Since q ∈ Q, there exist elements d in Z and c in Z \ {0} such that qc = d. Since D is assumed to have characteristic 0, we may assume that d and c are coprime in Z. We are assuming that q ∈ IQ (Z). Thus, there exist elements z0 , . . . , zn−1 in Z such that q n + zn−1 q n−1 + . . . + z1 q + z0 = 0. Multiplying this equation by cn we obtain dn = −(zn−1 dn−1 + zn−2 dn−2 c + . . . + z1 dcn−2 + r0 cn−1 )c. Thus, c divides d. Thus, as that d and c are assumed to be coprime in Z, c−1 ∈ Z. It follows that q = dc−1 ∈ Z.

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Assume D to be a field, let C be a subfield of D, and let d be an element in D. Instead of saying that d is integral over C one also calls d algebraic over C. If C = ID (C), one says that C is algebraically closed in D. One says that D is algebraically closed if, for each field E which contains D as a subfield, D is algebraically closed in E. Lemma 8.2.6 Assume D to be a division ring, and let C be a subfield of D with C ⊆ Z(D). Assume C to be algebraically closed and D to be a finitely generated C-module. Then C = D. Proof. Let d be an element in D. Since we are assuming that C ⊆ Z(D), C[d] is a field. We are also assuming that D is finitely generated as a C-module, so that C[d] is finitely generated as a C-module. Thus, as we are assuming C to be algebraically closed, we must have C = C[d], and that means that d ∈ C. Since d has been chosen arbitrarily in D, we have shown that C = D.

8.3 Completely Reducibility Throughout this section, the letter D stands for an associative ring with 1, the letter M for a D-module. We shall focus on the structure of M . A submodule L of the D-module M different from M is called maximal if L and M are the only submodules of M containing L (as a subset). The D-module M is called irreducible if {0} is a maximal submodule of M . Note that {0} is not an irreducible D-module. Let K and L be submodules of M . If {0} = K ∩ L, we write K ⊕ L instead of K + L. Lemma 8.3.1 Let K be a submodule of M , and let L be an irreducible submodule of M such that L ⊆ K. Then K is a maximal submodule of M if and only if K ⊕ L = M . Proof. Since L is assumed to be irreducible and L ⊆ K, {0} = K ∩ L. Since L ⊆ K, K = K + L. Thus, assuming that K is maximal we obtain K + L = M. Let us now assume that K + L = M , and let H be a submodule of M such that K ⊆ H. Then, referring to the group correspondence we obtain from Lemma 2.2.1 that H = H ∩ (K + L) = K + (H ∩ L). However, L is assumed to be irreducible. Thus, {0} = H ∩ L or H ∩ L = L. In the first case, we obtain K = H, in the second one, we obtain H = M .

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The module M is called completely reducible if, for each submodule L of M , there exists a submodule K of M such that K ⊕ L = M . Note that {0} is completely reducible. For the remainder of this section, we shall look at the case where M is completely reducible. Lemma 8.3.2 Assume that M is completely reducible. Then so is each submodule of M . Proof. Let L be a submodule of M , and let K be a submodule of L. Since M is assumed to be completely reducible, there exists a submodule H of M such that H ⊕ K = M . From H + K = M and K ⊆ L we obtain (L ∩ H) + K = L; cf. Lemma 2.2.1(ii) (together with the group correspondence). From {0} = H ∩ K, we obtain {0} = (L ∩ H) ∩ K. Thus, (L ∩ H) ⊕ K = L. Lemma 8.3.3 Assume that M is completely reducible and that {0} = M . Then M contains an irreducible submodule. Proof. Let m be an element in M \ {0}. Then, by Zorn’s Lemma, there exists a maximal element L in the set of all submodules of M not containing m. Since M is assumed to be completely reducible, M possesses a submodule K such that K ⊕ L = M . Since m ∈ / L, L = M . Thus, as K + L = M , {0} = K. We claim that K is irreducible, and in order to prove this we assume, by way of contradiction, that K is not irreducible. Assuming K to be not irreducible we obtain from {0} = K a submodule J of K with {0} = J = K. Since M is assumed to be completely reducible and K is a submodule of M , K must be completely reducible; cf. Lemma 8.3.2. Thus, as J is a submodule of K, K possesses a submodule I such that I ⊕ J = K The maximal choice of L yields m ∈ I ⊕ L and m ∈ J ⊕ L. However, as {0} = K ∩ L and {0} = I ∩ J, L = (I ⊕ L) ∩ (J ⊕ L). Thus, m ∈ L, contrary to the choice of L. This contradiction shows that K is irreducible. Let H be a set of submodules of the D-module M . We say that M is the sum of the elements in H if each element of M is sum of finitely many elements in the union of the elements in H. (The sum of zero elements in M is 0.)

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Assume that M is equal to the sum of the elements in H. We say that M is ˆ the direct sum of the elements in H if, for each element H in H, {0} = H ∩ H, ˆ denotes the sum of the submodules in H different from H. where H Proposition 8.3.4 The following statements are equivalent. (a) The module M is sum of irreducible submodules of M . (b) The module M is completely reducible. (c) The module M is a direct sum of irreducible submodules of M . Proof. (a) ⇒ (b) In order to do this we assume that there exists a set H of irreducible submodules of M such that M is the sum of the elements in H. Let L be a submodule of M , and let us define K to be the set of all submodules K of M with {0} = K ∩ L. By Zorn’s Lemma, K has a maximal element. We pick such a maximal element and call it K. Assume that K + L = M . Then there exists an element H in H such that H ⊆ K + L. As H is irreducible, H ⊆ K + L yields {0} = H ∩ (K + L). From {0} = H ∩ (K + L) we obtain (H + K) ∩ L ⊆ K. However, we have {0} = K ∩ L. Thus, {0} = (H + K) ∩ L, contrary to the (maximal) choice of K. (b) ⇒ (c) We define L to be a maximal direct sum of irreducible submodules of M . According to Zorn’s Lemma, we find such a submodule. Then, by (b), there exists a submodule K of M such that K ⊕ L = M . Let us assume that {0} = K. Since M is assumed to be completely reducible and K is a submodule of M , K is completely reducible; cf. Lemma 8.3.2. Thus, by Lemma 8.3.3, K possesses an irreducible submodule H. It follows that H ⊕ L is a direct sum of irreducible submodules and L = H ⊕ L, contrary to the choice of L. Thus, we have shown that {0} = K, so that L = M . (c) ⇒ (a) This is obvious. Lemma 8.3.5 Let H be a set of irreducible submodules of M such that M is the sum of the elements in H, and let L be an irreducible submodule of M . Then there exists an element H in H such that H ∼ = L. Proof. Since L is assumed to be a submodule of M , M possesses a submodule K such that K ⊕ L = M ; cf. Proposition 8.3.4. Since L is assumed to be irreducible, K is a maximal submodule of M ; cf. Lemma 8.3.1. In particular, K = M . Thus, there exists an element H in H such that H ⊆ K. Since K is a maximal submodule of M and H an irreducible submodule of M with H ⊆ K, H ⊕ K = M ; cf. Lemma 8.3.1.

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From H ⊕ K = M and K ⊕ L = M we obtain H ∼ = M/K ∼ = L; cf. Theorem 8.1.2. We call M homogeneous if M is a sum of irreducible submodules of M which are isomorphic in pairs. Note that, as a consequence of Lemma 8.3.3 and Lemma 8.3.5, each completely reducible D-module has maximal homogeneous submodules. Theorem 8.3.6 Assume that M is completely reducible, and let H denote the set of all maximal homogeneous submodules of M . Then the following hold. (i) The module M is the direct sum of the elements in H. (ii) For each irreducible submodule L of M , there exists exactly one element H in H such that L ⊆ H. (iii) Let K and L be irreducible submodules of M . Then K ∼ = L if and only if there exists an element H in H such that K ⊆ H and L ⊆ H. Proof. (i) We are assuming M to be completely reducible. Thus, by Proposition 8.3.4, M is sum of irreducible submodules of M . In particular, M is sum of the elements in H. ˆ to be the sum of Let us fix an element H in H, and define (as we did earlier) H ˆ the submodules in H different from H. Then, by Lemma 8.3.5, {0} = H ∩ H. Thus, as H has been chosen arbitrarily in H, M is direct sum of the elements in H. (ii) Let L be an irreducible submodule of M , and let H be the sum of all submodules of M isomorphic to L. Then L ⊆ H ∈ H. That L is not subset of two different elements of H follows from Lemma 8.3.5. (iii) This follows from (ii). Assume that M is completely reducible and that {0} = M . From Lemma 8.3.3 we know that M contains an irreducible submodule L. Since M is assumed to be completely reducible, there exists a submodule K of M such that K ⊕ L = M . According to Lemma 8.3.1, K is a maximal submodule of M , so that M has at least one maximal submodule. We define J(M ) to be the intersection of all maximal submodules of M and call this submodule the Jacobson radical of M . The module M is called artinian if each set of submodules of M has minimal elements. Theorem 8.3.7 Assume that {0} = M . Then we have the following. (i) If M is completely reducible, {0} = J(M ). (ii) If M is artinian and {0} = J(M ), M is completely reducible.

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Proof. (i) We set J := J(M ). Then J is a submodule of M . Thus, as M is assumed to be completely reducible, there exists a submodule L of M such that L ⊕ J = M . Let us assume, by way of contradiction, that {0} = J. Then L = M , so that M possesses a maximal submodule K with L ⊆ K. Since K is a maximal submodule of M , J ⊆ K. Thus, M = L ⊕ J ⊆ K, contradiction. (ii) Let K be minimal among the submodules of M with completely reducible factor module. We shall be done if we succeed in showing that K is contained in each maximal submodule of M . Let us assume, by way of contradiction, that M possesses a maximal submodule L with K ⊆ L. Then K + L = M . From this we conclude that (K/K ∩ L) ⊕ (L/K ∩ L) = M/K ∩ L. With the help of Theorem 8.1.2 we obtain from K + L = M also that K/K ∩ L ∼ = M/L and that

L/K ∩ L ∼ = M/K.

Now the choice of K yields K ∩ L = K. This means that K ⊆ L, contrary to the choice of L. Theorem 8.3.6 tells us that the structure of a completely reducible D-module depends on the structure of its irreducible submodules. This is the reason why we shall now have a closer look at irreducible D-modules.

8.4 Irreducible Modules over Associative Rings with 1 Throughout this section, the letter D stands for an associative ring with 1, the letter M for an irreducible D-module. Recall that, for each element m in M , AD (m) is our notation for the set of all elements d in D such that md = 0. Recall that, for each element m in M , AD (m) is a submodule of the D-module D; cf. Lemma 8.1.5. Lemma 8.4.1 For each element m in M \ {0}, D/AD (m) ∼ = M. Proof. Let m be an element in M \ {0}. Since 1mD = m and 0 = m, {0} = DmD . Thus, as DmD is a submodule of M and M is assumed to be irreducible, we must have that DmD = M . Thus, looking at the definition of AD (m) the claim follows from Theorem 8.1.1.

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Lemma 8.4.2 We have {0} = M J(D). Proof. We are assuming that M is irreducible. Thus, for each element m in M \ {0}, AD (m) is a maximal submodule of the D-module D. Thus, by definition, J(D) ⊆ AD (M ), and that means that {0} = M J(D). Recall that EndD (M ) is our notation for the ring of all D-endomorphisms of M . (We saw that EndD (M ) is a ring with respect to componentwise addition and composition as multiplication.) The following lemma is often referred to as Schur’s Lemma. Lemma 8.4.3 The ring EndD (M ) is a division ring. Proof. Let e be an element in EndD (M ) such that 0 = e. We have to show that e possesses a multiplicative inverse in EndD (M ). From 0 = e we obtain ker(e) = M . Thus, as ker(e) is a submodule of M and M is assumed to be irreducible, we conclude that {0} = ker(e). It follows that e is injective. From 0 = e we also obtain {0} = M e. Thus, as M e is a submodule of M and M is assumed to be irreducible, we conclude that M e = M , and that means that e is surjective. Let us now see that e−1 ∈ EndD (M ). In order to see this, we fix elements m in M and d in D. Then ((md)e−1 )e = (md)(e−1 e) = (m(e−1 e))d = ((me−1 )e)d = ((me−1 )d)e. Thus, as e is injective, (md)e−1 = (me−1 )d. Thus, as m has been chosen arbitrarily in M and d arbitrarily in D, we have shown that e−1 ∈ EndD (M ). For the remainder of this section, we set E := EndD (M ). According to Lemma 8.4.3, the ring E is a division ring. Lemma 8.4.4 For each finite subset L of M , we have LE = BM (AD (L)). Proof.2 From Lemma 8.1.6(ii) we know that LE ⊆ BM (AD (L)). Thus, we just have to show that BM (AD (L)) ⊆ LE. We set A := AD (L). If L is empty, A = D. From A = D (together with {0} = M and 1 ∈ D) we obtain {0} = BM (A). It follows that BM (A) ⊆ LE. Let us now assume that L is not empty. In this case, we fix an element l in L, and we set L := L \ {l} and A := AD (L ). Then, by induction, 2

Our proof follows the one given for [28; Theorem 13.15].

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BM (A ) ⊆ L E If A = A , BM (A) = BM (A ). Thus, as L E ⊆ LE, BM (A) ⊆ LE. Let us now assume that A = A . Then A ⊆ AD ({l}), so that {0} = lA . Thus, as lA is a submodule of M and M is assumed to be irreducible, lA = M . Let m be an element in BM (A). We shall be done if we succeed in showing that m ∈ LE. Let b and c be elements in A such that lb = lc. Then l(b − c) = 0. Thus, b − c ∈ A ∩ AD (l) = A. Thus, as we are assuming that {0} = mA, we conclude that m(b − c) = 0. It follows that mb = mc. For each element a in A , we define (la)e := ma. Since lA = M , we just saw that e is a map from M to M . It is obvious that e is a group endomorphism of M . However, for any two elements a in A and d in D, we have ((la)d)e = (l(ad))e = m(ad) = (ma)d = ((la)e)d. Thus, e ∈ E. Note, finally, that, for each element a in A , (m − le)a = ma − (le)a = ma − (la)e = 0. This means that m − le ∈ BM (A ). Thus, as BM (A ) ⊆ L E, m − le ∈ L E. Thus, as e ∈ E, m ∈ L E + lE = LE. This proves the lemma. The following theorem, due to Nathan Jacobson, is often referred to as Jacobson Density Theorem. Theorem 8.4.5 Let L be a finite subset of M linearly independent with respect to E. Then, for each element e in EndE (M ), there exists an element d in D such that, for each element l in L, ld = le. Proof. We may assume that L is not empty. We fix an element l in L, and we set L := L \ {l} and A := AD (L ). By induction, there exists an element d in D such that, for each element m in L , md = me. / L E. Thus, by Lemma 8.4.4, l ∈ / BM (A ). Thus, as M is Since l ∈ / L , l ∈ assumed to be irreducible, lA = M . Thus, there exists an element a in A such that la = le − ld .

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Set d := a + d . Then ld = le and, for each element m in L , md = m(a + d ) = ma + md = me. This proves the theorem. The ring D is called artinian if the D-module D is artinian. Lemma 8.4.6 If D is artinian, M is a finitely generated E-module. Proof. Define L to be the set of all finite subsets of M , and define A to be the set of all sets AD (L) with L ∈ L. Since D is assumed to be artinian, L possesses an element L such that AD (L) is minimal in A. We set A := AD (L). Let m be an element in M . If {0} = mA, A ⊆ AD (L ∪ {m}). Thus, as AD (L ∪ {m}) ⊆ A, this contradicts the (minimal) choice of A. Thus, we have that {0} = mA, and that means that m ∈ BM (A). Thus, by Lemma 8.4.4, m ∈ LE. Since m has been chosen arbitrarily in M , we have shown that M = LE. Recall that DM is our notation for the set of all elements dM with d ∈ D. In Lemma 8.1.3 we saw that DM = EndE (M ). The following lemma is often referred to as Double Centralizer. Lemma 8.4.7 Assume D to be artinian. Then DM = EndE (M ). Proof. According to Lemma 8.1.3, it is enough to show that EndE (M ) ⊆ DM . In order to show this, we pick an element in EndE (M ) and call it e. We have to show that e ∈ DM . According to Lemma 8.4.6, there exists a finite set L such that M = LE. In particular, we may assume L to be linearly independent with respect to E. Thus, according to Theorem 8.4.5, there exists an element d in D such that, for each element l in L, ld = le. For each element m in M , we define mdM = md. Then dM ∈ EndE (M ), and dM coincides with e on L. It follows that e = dM . In particular, e ∈ DM . ˜ be a ring, and let φ be a map from D to D. ˜ Let D The map φ is called a ring homomorphism or simply a homomorphism if, for any two elements b and c in D, (b + c)φ = bφ + cφ and

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(bc)φ = bφcφ. ˜ is called isomorphic to D if there exists a bijective ring homomorA ring D ˜ We shall write D ∼ ˜ in order to indicate that D and phism from D to D. =D ˜ are isomorphic rings. D Theorem 8.4.8 Assume D to be artinian and simple. Then, D ∼ = EndE (M ). Proof. The set AD (M ) is an ideal of D. Moreover, 1 ∈ / AD (M ). Thus, as we are assuming D to be simple, {0} = AD (M ). It follows that D ∼ = DM . Thus, our claim follows from Lemma 8.4.7.

8.5 Semisimple Associative Rings with 1 An associative ring D with 1 is called semisimple if the D-module D is completely reducible. In this section, we shall combine our results about completely reducible modules obtained in Section 8.3 with those about irreducible modules (and artinian simple rings) obtained in Section 8.4 in order to give a complete description of semisimple rings. For the remainder of this section, we shall now fix an associative ring with 1 and call it D. Lemma 8.5.1 Let L be a set of submodules of the D-module D such that D is the sum of the elements in L. Then there exists a finite subset K of L such that D is the sum of the elements in K. Proof. We are assuming that D is a ring with 1. Since D is the sum of the elements in L, there exists a finite subset K of L and, for each element K in K, an element dK in K such that dK . 1= K∈K

Thus, for each element d in D, d = 1d = (

K∈K

dK )d =

dK d.

K∈K

For each element K in K, we have dK d ∈ K. Thus, as d has been chosen arbitrarily, we have shown that D is the sum of the elements in K. Recall that a subgroup L of a D-module M is called a submodule of M if LD ⊆ L. Recall that a submodule L of the D-module D is called an ideal of D if DL ⊆ L. It is obvious that {0} and D are ideals of D.

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Let C be an ideal of D such that {0} = C. We call C a minimal ideal of D if {0} and C are the only ideals of D contained (as subsets) in C. Proposition 8.5.2 Assume D to be semisimple. Then each maximal homogeneous submodule of the D-module D is a minimal ideal of D. Proof. Let H be a maximal homogeneous submodule of the D-module D, let d be an element in D, and let K be an irreducible submodule of H. Since K is a submodule of the D-module D, so is dK. Let K be an element in K. Then dK = KdD is a submodule of D. Therefore, as K is irreducible, we have either dK ∼ = K or {0} = dK; cf. Theorem 8.1.1. Thus, dK ⊆ H. Since d has been chosen arbitrarily in D, we conclude that DK ⊆ H. Since K has been chosen arbitrarily among the irreducible submodules of H, we conclude that DH ⊆ H. (Recall that H is homogeneous. In particular, H is the sum of its irreducible submodules.) This means that H is an ideal of D. Since H is an ideal of D different from {0}, H contains a minimal ideal C of D. We shall prove that C = H, and in order to see that we assume, by way of contradiction, that C = H. From C = H we obtain an irreducible submodule K of H such that {0} = K ∩ C. It follows that {0} = KC. Let L be an irreducible submodule of C. Then, as H is homogeneous, there exists a bijective D-homomorphism φ from K to L. Thus, LC = (Kφ)C = (KC)φ. Thus, as {0} = KC, {0} = LC. Since L has been chosen arbitrarily, we have shown that {0} = C 2 . Since the D-module D is assumed to be semisimple, D possesses a submodule K such that K ⊕ C = D. Then {0} = KC. Thus, as D is assumed to be a ring with 1 and {0} = C 2 , C = DC = (K + C)C = {0}, contrary to the choice of C as minimal ideal of D. This contradiction shows that C = H, so that H is a minimal ideal of D. The following two theorems are the main theorems about semisimple associative rings with 1. They are due to Emil Artin; cf. [2]. Less general versions have been given earlier by Joseph Wedderburn; cf. [39; Theorem 10] and [39; Theorem 17]. Theorem 8.5.3 Assume D to be semisimple, and let H denote the set of all maximal homogeneous submodules of the D-module D. Then we have the following. (i) The set H is finite. (ii) The D-module D is the direct sum of the elements in H.

174


(iii) For any two elements K and L in H, we have {0} = KL. (iv) Each element in H is a direct sum of finitely many of its (pairwise isomorphic) irreducible submodules. Proof. (i) The D-module D is sum of the elements in H. Thus, by Lemma 8.5.1, H is a finite set. (ii) This is an application of Theorem 8.3.6(i) to the D-module D. (iii) Let K and L be elements in H. According to Proposition 8.5.2, K is an ideal of D. Thus, KL ⊆ K. Similarly, as L is an ideal of D, KL ⊆ L. However, we know from (ii) that {0} = K ∩ L. (iv) Let H be an element in H. Then H is a submodule of the D-module D. Since D is assumed to be semisimple, the D-module D is completely reducible. Thus, by Lemma 8.3.2, H is completely reducible. Thus, by Proposition 8.3.4, H is a direct sum of irreducible submodules. Now, considering that H is a ring with 1, the claim follows from Lemma 8.5.1. Theorem 8.5.4 Assume D to be semisimple, let H be a maximal homogeneous submodule of the D-module D, let M be an irreducible submodule of the D-module D contained in H, and set E := EndD (M ). Then we have the following. (i) We have H ∼ = EndE (M ). (ii) Let C be an algebraically closed subfield of Z(E) such that E is a finitely generated vector space over C. Then C = E and H ∼ = EndC (M ). Proof. (i) Let C be an ideal of H. For each element K in H \ {H}, we have {0} = HK. Thus, C is an ideal of D. Thus, by Proposition 8.5.2, {0} = C or C = H. This proves that H is a simple ring. That H is artinian follows from Theorem 8.5.3(iv). Thus, the claim follows from Theorem 8.4.8. (ii) Applying Lemma 8.2.6 to E in place of D, we obtain C = E. Thus, the claim follows from (i). Corollary 8.5.5 Assume D to be semisimple, and let C be an algebraically closed subfield of Z(D). The ring D is commutative if and only if, for each irreducible D-module M of D, dimC (M ) = 1. Proof. This follows from Theorem 8.5.3(iii) together with Theorem 8.5.4(ii). Theorem 8.5.3 together with Theorem 8.5.4 gives a complete picture about semisimple rings. The following theorem gives a complete picture about the collection of all modules over semisimple rings.

8.6 Characters of Associative Rings with 1

175

Theorem 8.5.6 If D is semisimple, we have the following. (i) Each D-module is completely reducible. (ii) Each irreducible D-module is isomorphic to a submodule of the D-module D. Proof. (i) Let M be a D-module, and let m be an element in M . Since D is assumed to be a ring with 1, there exists a finite set K of irreducible submodules of the D-module D such that 1 is element of the sum C of the elements of K. It follows that m = m1 ∈ mC. Note that mC is the sum of the submodules mK with K ∈ K. We shall now see that, for each element K in K, mK is contained in an irreducible submodule of M . Let K be an element in K. Then mK = KmD is a submodule of M . Thus, as K is irreducible, we must have mK ∼ = K or mK = {0}; cf. Theorem 8.1.1. We have seen that m is element of a sum of irreducible submodules of M . Thus, as m has been chosen arbitrarily, we have shown that M is sum of irreducible submodules of M . Thus, the claim follows from Proposition 8.3.4. (ii) Let M be an irreducible D-module, and let m be an element in M \ {0}. Then, by Lemma 8.4.1, D/AD (m) ∼ = M. Since D is assumed to be semisimple, there exists a submodule K of the D-module D such that K ⊕ AD (m) = D. Thus, by Theorem 8.1.2, D/AD (m) ∼ = K. It follows that M ∼ = K.

8.6 Characters of Associative Rings with 1 Characters of associative rings D with 1 arise from D-modules M when D contains a subfield C in its center such that M is a finitely generated vector space over C. In this section, we shall look at this situation. Throughout this section, the letter D stands for an associative ring with 1. The letter C stands for a subfield of D with C ⊆ Z(D). Let M be a D-module. Then, as C ⊆ D, M is a vector space over C. We assume that M is a finitely generated vector space over C. Recall that, for each element d in D, dM is our notation for the map from M to M which maps each element m in M to md. We saw earlier that, for each element d in D, dM ∈ End(M ). Since C ⊆ Z(D), we now also have that, for any three elements m in M , c in C, and d in D,

176


(mc)dM = (mc)d = m(cd) = m(dc) = (md)c = (mdM )c. Thus, for each element d in D, dM ∈ EndC (M ). For each element d in D, we shall denote by tr(dM ) the trace of the Cendomorphism dM , and we set χM (d) := tr(dM ). Then χM is a homomorphism from D to C. This map is called the C-character of D afforded by M . If there is no danger of ambiguity, we shall simply speak about characters instead of C-characters. The integer χM (σ1 ) (= dimC (M )) is called the degree of χM . Character of degree 1 are called linear. Note that the sum of two characters is a character. ˜ be D-modules finitely generated as vector Theorem 8.6.1 Let M and M ∼ ˜ spaces over C. Then, if M = M , χM = χM˜ . Proof. Let d be an element in D, and let B be a basis of M . Then, for any two elements k and l in B, there exists an element ckl in C such that, for each element k in B, kd = lckl . l∈H

In particular, χM (d) =

cbb .

b∈B

˜ are assumed to be isomorphic, there exists On the other hand, since M and M ˜ . In particular, Bφ is a basis of a bijective D-homomorphism φ from M to M ˜ M , and, for each element k in B, (kφ)d = (kd)φ = ( lckl )φ = (lckl )φ = (lφ)ckl . l∈H

It follows that χM˜ (d) =

l∈H

l∈H

cbb .

b∈B

Thus, as d has been chosen arbitrarily in D, we have shown that χM = χM˜ . A character of D afforded by an irreducible D-module is called irreducible. Let us denote by Irr(D) the set of all irreducible characters of D. Looking at Theorem 8.5.3(i), (iv) and Theorem 8.5.6(i) one has a complete picture about the set of all irreducible modules over a semisimple ring. Thus, if all these irreducible modules are finitely generated vector spaces over C,

8.6 Characters of Associative Rings with 1

177

one can expect a clear picture about the set of all irreducible characters of a semisimple ring, too. For the remainder of this section, we shall now assume D to be semisimple and to be finitely generated as a vector space over C. Assuming D to be semisimple we obtain from Theorem 8.5.6(i) and Proposition 8.3.4 that each character of D is the sum of irreducible characters of D. Let us, therefore, look at the set of all irreducible characters of D. As earlier (in Section 8.3 and Section 8.5), we shall denote by H the set of the maximal homogeneous submodules of the D-module D. For each element H in H, D possesses an irreducible character ψH such that, for each irreducible submodule K of H, χK = ψH ; cf. Lemma 8.3.5 together with Theorem 8.6.1. Theorem 8.6.2 The following hold. (i) We have {ψH | H ∈ H} = Irr(D). (ii) For any two different elements K and L in H, we have ψK (L) = {0}. Proof. (i) Let χ be an element in Irr(D). Then, by definition, there exists an irreducible D-module M such that χ = χM . By Theorem 8.5.6(ii), there exists an irreducible submodule K of the Dmodule D such that M ∼ = K. From M ∼ = K we obtain χM = χK ; cf. Theorem 8.6.1. By Theorem 8.3.6(ii) there exists exactly one element H in H such that K ⊆ H. Since K is a submodule of H, χK = ψH . It follows that χ = ψH . (ii) Let K and L be elements in H. Then, by Theorem 8.5.3(iii), {0} = KL. Let l be an element in L, and let J be an irreducible submodule of K. Since J ⊆ K, χJ = ψK . From J ⊆ K we also obtain JlJ = Jl = 0. Thus, tr(lJ ) = 0, so that, by definition, χJ (l) = 0. Thus, as χJ = ψK , ψK (l) = 0. For each element H in H, there exists a uniquely determined element 1H in H such that 1H ; 1= H∈H

see Theorem 8.5.3(ii). Note that 1H is the multiplicative neutral element of the ring H. Lemma 8.6.3 For any two elements H in H and d in D, we have ψH (d) = ψH (1H d). Proof. Using Theorem 8.6.2(ii) for the last equation we obtain

178


ψH (d) = ψH (1d) = ψH (

1K d) =

K∈H

ψH (1K d) = ψH (1H d);

K∈H

note that, for each element K in K, 1K d ∈ K. Theorem 8.6.2(i) says that the map Ψ : H → Irr(D), H → ψH is surjective. In the first part of the following theorem, we shall see that Ψ is bijective if the characteristic of C does not divide any of the degrees of the irreducible characters of D. Theorem 8.6.4 Assume that, for each irreducible character χ of D, the characteristic of C does not divide χ(σ1 ). Then we have the following. (i) The set Irr(D) is a linearly independent subset of HomC (D, C). (ii) We have |H| = |Irr(D)|. (iii) If C is algebraically closed, χD =

χ(σ1 )χ.

χ∈Irr(D)

Proof. (i) From Lemma 8.6.3 we know that, for each element H in H, ψH (1) = ψH (1H ). From Theorem 8.6.2(ii) we know that, for any two different elements K and L in H, ψK (1L ) = 0. Thus, the claims are consequences of Theorem 8.6.2(i). (ii) This follows from (i) together with Theorem 8.6.2(i). (iii) Assume C to be algebraically closed, and let H be an element in H. Since 1 ∈ C, 1H C is a subfield of H, so that we may apply Theorem 8.5.4(ii). ∼ Let M be an irreducible H-module. Then, by Theorem 8.5.4(ii), H = EndC (M ). Thus, H is the direct sum of χ(σ1 ) copies of M , so that the desired equation follows from (i). Corollary 8.6.5 Assume that, for each irreducible character χ of D, the characteristic of C does not divide χ(σ1 ). Then we have χ(σ1 )2 . dimC (D) = χ∈Irr(D)

Proof. This follows immediately from Theorem 8.6.4(iii).

8.7 Roots of Unity in Integral Domains

179

8.7 Roots of Unity in Integral Domains A commutative associative ring D with 1 is called an integral domain if the product of any two elements in D \ {0} is in D \ {0}. Throughout this section, the letter D stands for an integral domain. Lemma 8.7.1 Let d be an element in D \ {1}, and let n be a positive integer such that dn = 1. Then we have n−1

di = 0.

i=0

Proof. We have

n−1

di (d − 1) = dn − 1.

i=0

Thus, our claim follows from the hypotheses that d − 1 = 0 and dn − 1 = 0. Lemma 8.7.2 Let d be an element in D \ {1}, and let n be a positive integer such that d2n+1 = 1. Then we have n

(di + d−i )2 = 2n + 3

i=0

and

n

(di+1 − d−(i+1) + di − d−i )2 = −(2n + 1)(d + d−1 + 2).

i=0

Proof. We are assuming that d2n+1 = 1. Thus, by Lemma 8.7.1, 2n

di = 0.

i=0

Since we are assuming that d2n+1 = 1, this implies n

(d2i + d−2i ) =

i=0

n n (d2i + d2(n−i)+1 ) = (d2i + d2i+1 ) = 1. i=0

i=0

It follows that n

(di + d−i )2 =

i=0

and that is our first claim.

n i=0

(d2i + 2 + d−2i ) = 2(n + 1) + 1,

180


From

n

(d2i + d−2i ) = 1

i=0

we also obtain

n

(d2i+1 + d−(2i+1) ) = 1

i=0

and

n

(d2(i+1) + d−2(i+1) ) = d + d−1 − 1.

i=0

Thus, as (d

i+1

−d

−(i+1)

+ di − d−i )2 is equal to

(d2(i+1) + d−2(i+1) ) + (d2i + d−2i ) + 2(d2i+1 + d−(2i+1) ) − 2(d + d−1 + 2), we conclude that n

(di+1 − d−(i+1) + di − d−i )2 = −(2n + 1)(d + d−1 + 2),

i=0

and that is our second claim. Lemma 8.7.3 Let d be an element in D \ {−1, 1}, and let n be a positive integer such that d2n = 1. Then we have n−1

(di+1 − d−(i+1) )2 + (di − d−i )2 = −4n

i=0

and

n−1

(di+1 − d−(i+1) )(di − d−i ) = −n(d + d−1 ).

i=0

Proof. We are assuming that d2n = 1. Thus, we obtain from Lemma 8.7.1 that n−1 2n−1 2i d (d + 1) = di = 0. i=0

i=0

Thus, as d + 1 = 0 and D is assumed to be an integral domain, n−1

d2i = 0.

i=0

From this we obtain n−1 i=0

(di − d−i )2 =

n−1 i=0

d2i − 2 + d−2i = −2n.

8.7 Roots of Unity in Integral Domains

181

Multiplying n−1

d2i = 0

i=0 2

by d we obtain n−1

d2(i+1) = 0.

i=0

Thus, n−1

(di+1 − d−(i+1) )2 =

i=0

so that

n−1

d2(i+1) − 2 + d−2(i+1) = −2n,

i=0 n−1

(di+1 − d−(i+1) )2 + (di − d−i )2 = −4n,

i=0

and that is our first claim. Multiplying n−1

d2i = 0

i=0

by d we obtain n−1

d2i+1 = 0.

i=0

Therefore, we also obtain n−1

(di+1 −d−(i+1) )(di −d−i ) =

i=0

n−1

(d2i+1 −d−d−1 +d−(2i+1) ) = −n(d+d−1 ),

i=0

and that is our second claim. Lemma 8.7.4 Assume that D has characteristic 0, and let us denote by Z the smallest unitary subring of D. Let d be an element in D for which there exists a positive integer n with dn = 1. Then, if d+d−1 ∈ Z, d4 = 1 or d6 = 1. Proof. For each integer z, we have (dz + d−z )(d + d−1 ) = (dz+1 + d−(z+1) ) + (dz−1 + d−(z−1) ). Thus, as we are assuming that d + d−1 ∈ Z, induction yields that, for each integer z, dz + d−z ∈ Z. (Note that d0 + d−0 ∈ Z.) We are assuming that there exists a positive integer n with dn = 1. Thus, the set of all sums dz + d−z with integral exponent z is finite. Let m be an integer such that dm + d−m is maximal (with respect to the natural ordering of Z).

182


Since dn = 1, d−n = 1. Thus, dn + d−n = 2. Thus, 2 ≤ dm + d−m . From (d + d−1 )2 (dm + d−m ) = (dm+2 + d−(m+2) ) + (dm−2 + d−(m−2) ) + 2(dm + d−m ), we also obtain

(d + d−1 )2 (dm + d−m ) ≤ 4(dm + d−m ).

Thus, as dm + d−m is positive, we obtain (d + d−1 )2 ≤ 4. Assume that (d + d−1 )2 = 4. Then d2 − 2 + d−2 = 0. It follows that (d2 − 1)2 = d4 − 2d2 + 1 = 0. Thus, as D is assumed to be an integral domain, d2 − 1 = 0, and that means that d2 = 1. Assume that (d + d−1 )2 = 1. Then d2 + 1 + d−2 = 0. It follows that d6 − 1 = (d4 + d2 + 1)(d2 − 1) = 0, whence d6 = 1. Assume that d+d−1 = 0. Then d2 +1 = 0. Thus, d4 −1 = (d2 +1)(d2 −1) = 0, whence d4 = 1.

9 Scheme Rings

In this chapter, S is assumed to have finite valency. With the help of S we shall define, for each field C, an associative ring with 1. We shall denote this ring by CS and call it the scheme ring of S over C. In the first section of this chapter, we shall prove that CS is semisimple if the characteristic of the field C does not divide any of the integers |s∗ | with s ∈ S. This enables us to refer to some of the results about semisimple rings which we obtained in Section 8.5. We shall also see that C ⊆ Z(CS), so that we may speak about characters of CS and refer to results about characters which we obtained in Section 8.6. The results of the first section include the orthogonality relations for fields the characteristic of which does not divide any of the integers |s∗ | with s ∈ S. In the second section, we derive the Schur relations in the case where the characteristic of the underlying base field does not divide any of the integers |s∗ | with s ∈ S and is algebraically closed. In Section 9.3, we assume the underlying base field C to be the field of the complex numbers. We establish a relationship between closed subsets of S and characters of CS. In Section 9.4, we obtain a few general results about closed subsets. In Section 9.5, we shall apply the orthogonality relations in order to investigate the case where |S| ≤ 5. In the last of the six sections of this chapter, we shall briefly look at scheme rings of S under the assumption that S is generated by a constrained set of involutions.

9.1 Basic Facts In this section, the letter C stands for a field. Let us fix, for each element x in X, an element cx in C. We write

184

9 Scheme Rings

cx x

x∈X

to denote the map from X to C which maps each element x in X to cx . The set CX of all maps from X to C is a vector space over C with respect to componentwise addition and componentwise multiplication with elements of C. Each element x in X can be identified with the map from X to C which maps x to 1 and each element different from x to 0. Thus, X can be viewed as a subset of CX. Recall that S is assumed to have finite valency. Thus, X is finite, so that X is a basis of the vector space CX. Let s be an element in S. Then, as X is a basis of CX, there exists a uniquely defined element σs in EndC (CX) such that, for each element x in X, xσs = y. y∈xs

For each nonempty subset R of S, we define CR to be the set of all finite sums of products cσr with c ∈ C and r ∈ R. Note that, for each nonempty subset R of S, CR is a vector space over C with respect to componentwise addition and componentwise multiplication with elements of C. The set {σr | r ∈ R} is a basis of CR. Lemma 9.1.1 The following statements hold. (i) For any two elements p and q in S, we have σp σq = apqs σs . s∈S

(ii) Let n be an integer with 2 ≤ n, and let r1 , . . . , rn be elements in S. Then σ r1 · · · σ rn = ar1 ...rn s σs . s∈S

Proof. (i) For each element x in X, we have x(σp σq ) = (xσp )σq = ( y)σq = yσq = z y∈xp

y∈xp

y∈xp z∈yq

and x(

s∈S

apqs σs ) =

s∈S

apqs xσs =

s∈S

apqs

z∈xs

z=

s∈S z∈xs

apqs z.

9.1 Basic Facts

185

This proves the desired equation. (ii) Assume that 3 ≤ n. Referring to (i) induction yields σ r1 · · · σ rn = ( ar1 ...rn−1 q σq )σrn = ar1 ...rn−1 q (σq σrn ) q∈S

=

ar1 ...rn−1 q

q∈S

=

q∈S

aqrn s σs =

s∈S

( ar1 ...rn−1 q aqrn s )σs s∈S q∈S

ar1 ...rn s σs ,

s∈S

and that proves the claim. From Lemma 9.1.1(i) we obtain that CS is a subring of EndC (CX). The associative ring CS is a ring with 1. We call CS the scheme ring of S over C. The field C is called the base field of CS. Since CS is a subring of EndC (CX), CX is a CS-module. We call CX the standard module of CS. Since 1 ∈ S, σ1 ∈ CS. Thus, the elements of C can be identified with the multiples of σ1 . Thus, C can be viewed as a subfield of Z(CS), the center of CS. The fact that C ⊆ Z(CS) enables us to define characters for each CS-module which is finitely generated over C. Recall that the standard module of CS is finitely generated over C. The character of CS afforded by the standard module is called the standard character of CS. We shall denote the standard character of CS by χCX . In the following lemma, we collect information about the standard character. Lemma 9.1.2 The following statements hold. (i) We have χCX (σ1 ) = nS . (ii) For each element s in S \ {1}, we have χCX (σs ) = 0. (iii) For any two elements p and q in S, χCX (σp∗ σq ) = δpq |p∗ |. (iv) For each element s in S, let cs be an element in C. Set cs σs . σ := s∈S

Then, for each element s in S, χCX (σs∗ σ) = cs |s∗ |. Proof. (i) This equation follows immediately from the definition of χCX . (ii) These equations follow immediately from the definition of χCX . (iii) Let p and q be elements in S. Referring to Lemma 9.1.1(i) we obtain from (i) and (ii) that

186

9 Scheme Rings

χCX (σp∗ σq ) = χCX (

ap∗ qs σs ) =

s∈S

ap∗ qs χCX (σs ) = ap∗ q1 nS .

s∈S

Thus, the claim follows from the definition of ns∗ together with Lemma 1.1.2(ii) and Lemma 1.1.3(i). (iv) This follows from (iii) because CS is distributive. The standard module possesses an irreducible submodule which induces a character of CS all values of which can be computed explicitly. In order to introduce this module we (temporarily) set x. j := x∈X

Note that, for each element s in S, jσs = y = ns∗ x = ns∗ j. x∈X y∈xs

x∈X

Thus, Cj is a submodule of the CS-module CX.1 We call Cj the principal module of CS. The character afforded by the principal module is called the principal character of CS. Instead of χCj we shall write 1CS . The above equation tells us that, for each element s in S, 1CS (σs ) = ns∗ . Having introduced the standard module and the principal module of CS we shall now introduce a third module of CS, the CS-module CS. This module is defined by the multiplication of CS. The CS-module CS is usually called the regular module of CS. Let M be a CS-module. For each element φ in EndC (M ), we define φζM :=

1 σs∗ φσs . ns

s∈S

It is obvious that ζM is a C-module endomorphism of EndC (M ) to EndC (M ). The following lemma says that, in fact, the image of ζM is a subset of EndCS (M ). Lemma 9.1.3 Let M be a CS-module, and let φ be an element in EndC (M ). Then φζM ∈ EndCS (M ). Proof. For any two elements s in S and φ in EndC (M ), we have 1

Recall that Cj denotes the set of all elements cj with c ∈ C.

9.1 Basic Facts

σs (φζM ) = σs

187

1 1 σp∗ φσp = σs σp∗ φσp np np

p∈S

p∈S

1 1 = asp∗ q σq φσp = aps∗ q∗ σq φσp np np p∈S

q∈S

q∈S p∈S

1 1 = aq∗ sp σq φσp = σq φ aq∗ sp σp nq∗ nq∗ q∈S p∈S

q∈S

1 = σq φσq∗ σs = (φζM )σs . nq∗

p∈S

q∈S

(The third and the seventh equations follow from Lemma 9.1.1(i), the fourth from Lemma 1.1.1(ii), and the fifth from Lemma 1.1.3(ii).) The second part of the following theorem gives the transformation matrix of ζCS with respect to the basis {σs | s ∈ S} of CS. In Lemma 1.1.6, this matrix has been proven to be ‘stochastic’ in the sense that each of its column sums is equal to nS . Theorem 9.1.4 Writing ζ instead of ζCS we have the following. (i) We have (CS)ζ ⊆ Z(CS). (ii) For each element p in S, we have as∗ psq ( )σq . σp ζ = ns q∈S s∈S

(iii) We have σ1 ζ ∈ COϑ (S). (iv) There exists an element γ in EndC (CS) such that γζ = σ1 . Proof. (i) This is an application of Lemma 9.1.3 to the CS-module CS. (ii) Let p be an element in S. Then, by Lemma 9.1.1(ii), 1 1 as∗ psq σ s∗ σ p σ s = as∗ psq σq = ( )σq , σp ζ = ns ns ns s∈S

s∈S

q∈S

q∈S s∈S

and that proves the claim. (iii) From the second equation of Lemma 1.1.1(i) we obtain as∗ 1sq = as∗ sq for any two elements q and s in S. Thus, by (ii), as∗ sq σ1 ζ = ( )σq . ns q∈S s∈S

For each element q in S, we define cq :=

as∗ sq s∈S

ns

.

188

9 Scheme Rings

Then σ1 ζ =

cq σq .

q∈S

Let q be an element in S such that cq = 0. Then there exists an element s in S such that 1 ≤ as∗ sq . This means that q ∈ s∗ s. Thus, by Theorem 3.2.1(ii), q ∈ Oϑ (S). Since q ∈ S has been chosen arbitrarily with 0 = cq , we conclude that 1ζ ∈ COϑ (S). (iv) Let us denote by γ the uniquely determined element of EndC (CS) which, for each element s in S, satisfies σs γ = δ1s σs . Then, for each element q in S, we have 1 1 σs∗ γσs = σq σs∗ γσs ns ns s∈S s∈S 1 aqs∗ 1 = aqs∗ p σp γσs = σs = σq . ns ns

σq (γζ) = σq

s∈S

p∈S

s∈S

(The last equation follows from Lemma 1.1.3(i).) Let us now look at the Jacobson radical J(CS) of the scheme ring CS. The third part of the following theorem was first proved by Heinrich Maschke; cf. [30]. Theorem 9.1.5 We have the following. (i) Let us denote by R the set of all elements s in S such that the characteristic of C divides |s∗ |. Then, J(CS) ⊆ CR. (ii) Assume that, for each element s in S, the characteristic of C does not divide |s∗ |. Then CS is semisimple. (iii) Assume S to be thin. If the characteristic of C does not divide the valency of S, CS is semisimple. Proof. (i) Let σ be an element in J(CS). Then σ ∈ CS. Thus, there exists, for each element s in S, an element cs in C such that cs σs . σ= s∈S

Let s be an element in S such that cs = 0. We shall be done if we succeed in showing that the characteristic of C divides |s∗ |. Let

9.1 Basic Facts

189

{0} = V0 ⊆ V1 ⊆ . . . ⊆ Vn = CX be a series of submodules of CX such that, for each element i in {1, . . . , n}, Vi /Vi−1 is irreducible. Then, for each element i in {1, . . . , n}, Vi σs∗ σ ⊆ Vi−1 ; cf. Lemma 8.4.2. Therefore, χCX (σs∗ σ) = 0. On the other hand, by Lemma 9.1.2(iv), χCX (σs∗ σ) = cs |s∗ |. From χCX (σs∗ σ) = 0 and χCX (σs∗ σ) = cs |s∗ | we obtain cs |s∗ | = 0. Thus, as cs = 0, |s∗ | = 0. This means that the characteristic of C divides |s∗ |. (ii) This follows from (i) and Theorem 8.3.7(ii). (iii) Referring to Lemma 1.5.1 this follows from (ii). For the remainder of this section, we assume that, for each element s in S, the characteristic of C does not divide |s∗ |. According to Theorem 9.1.5(ii), this implies that CS is semisimple so that we may refer to results obtained in Section 8.5 and Section 8.6. Let χ be an irreducible character of CS. Since CS is semisimple, we obtain from Theorem 8.6.2(i) together with Theorem 8.6.4(ii) that there exists exactly one maximal homogeneous submodule Hχ of the CS-module CS such that χ = ψHχ . Set χ := 1Hχ . Then 1=

χ .

χ∈Irr(CS)

From Theorem 8.5.6(i) we know that there exists, for each irreducible character χ of CS, a non-negative integer mχ such that mχ χ. χCX = χ∈Irr(CS)

For each irreducible character χ of CS, the element mχ is called the multiplicity of χ. Lemma 9.1.6 Let χ be an irreducible character of CS. Then the following hold. (i) For each element σ in Hχ , we have σ=

mχ χ(σs∗ σ) σs . nS ns∗ s∈S

(ii) We have χ =

mχ χ(σs∗ ) σs . nS ns∗ s∈S

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9 Scheme Rings

(iii) The characteristic of C does not divide mχ . Proof. (i) Let σ be an element in Hχ . Then σ ∈ CS. Thus, there exists, for each element s in S, an element cs in C such that σ= cs σs . s∈S

Let s be an element in S. Then, as σ ∈ Hχ , σs∗ σ ∈ Hχ ; cf. Proposition 8.5.2. Therefore, by Theorem 8.6.2(ii), χCX (σs∗ σ) = mχ χ(σs∗ σ). On the other hand, by Lemma 9.1.2(iv), χCX (σs∗ σ) = cs |s∗ |. Therefore, cs =

mχ χ(σs∗ σ) . |s∗ |

Thus, the claim follows from Lemma 1.1.2(ii). (ii) This follows from (i) and Lemma 8.6.3. (iii) This follows from (ii), because χ = 0. The equations in the following theorem are called the orthogonality relations for schemes with finite valency. Theorem 9.1.7 For any two irreducible characters φ and ψ of CS, the following hold. (i) For each element s in S, we have 1 as∗ pq φ(σs∗ ) φ(σp∗ )ψ(σq ) = δφψ . nS np∗ mφ p∈S q∈S

(ii) We have 1 1 φ(σ1 ) φ(σs∗ )ψ(σs ) = δφψ . nS ns∗ mφ s∈S

Proof. (i) From Lemma 9.1.6(ii) and Lemma 9.1.1(i) we obtain φ ψ =

mφ φ(σp∗ ) mψ ψ(σq∗ ) ( σp ) ( σq ) nS np∗ nS nq∗ p∈S

q∈S

mφ mψ φ(σp∗ )ψ(σq∗ ) σp σq = n2S np∗ nq∗ p∈S q∈S

mφ mψ φ(σp∗ )ψ(σq∗ ) = ( apqs )σs . n2S np∗ nq∗ s∈S p∈S q∈S

9.1 Basic Facts

191

On the other hand, as φ ∈ Hφ and ψ ∈ Hψ , φ ψ = δφψ φ . Referring once again to the equation mφ φ(σs∗ ) φ = σs nS ns∗ s∈S

we, therefore, obtain mφ mψ φ(σp∗ )ψ(σq∗ )apqs mφ φ(σs∗ ) = δφψ nS np∗ nq∗ ns∗ p∈S q∈S

for each element s in S. Recall that, by Lemma 9.1.6(iii), 1 ≤ mφ and 1 ≤ mψ . Therefore we conclude that, for each element s in S, 1 apq∗ s ns∗ φ(σs∗ ) φ(σp∗ )ψ(σq ) = δφψ . nS np∗ nq mφ p∈S q∈S

Finally, recall that, by Lemma 1.1.1(ii) and Lemma 1.1.3(ii), apq∗ s ns∗ = aqp∗ s∗ ns∗ = as∗ pq nq . (ii) This follows from (i) by setting s∗ = 1; use the first equation of Lemma 1.1.1(i). Lemma 9.1.8 The following statements hold. (i) For each non-principal irreducible character χ of CS, we have

χ(σs ) = 0.

s∈S

(ii) We have m1CS = 1. Proof. (i) Set ψ = 1CS in Theorem 9.1.7(ii). (ii) Set φ = 1CS = ψ in Theorem 9.1.7(ii). Let M be a CS-module. Recall from Lemma 9.1.3 that, for each element φ in EndC (M ), φζM ∈ EndCS (M ). Here is an analog of Theorem 9.1.4(iv). Theorem 9.1.9 Let M be a CS-module. Then there exists an element φ in EndC (M ) such that φζM is the identity on M . Proof. We are assuming that, for each element s in S, the characteristic of C does not divide |s∗ |. Thus, by Theorem 9.1.5(ii), CS is semisimple. Let M be an irreducible CS-module. Then, by Theorem 8.5.6(i), we may assume that M is a submodule of the CS-module CS. Since CS is semisimple, CS possesses a submodule L such that L ⊕ M = CS; cf. Proposition 8.3.4.

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9 Scheme Rings

For any two elements l in L and m in M , we define (l + m)π := m. Then, as M is a submodule of CS, we obtain πσs = σs π for each element s in S. From Theorem 9.1.4(iv) we know that there exists an element γ in EndC (CS) such that γζCS = σ1 . Set φ := πγπ. Then φ ∈ EndC (M ) and 1 1 σs∗ φσs = σs∗ πγπσs ns ns s∈S s∈S 1 = π( σs∗ γσs )π = πγζCS π = πσ1 π = πσ1 . ns

φζM =

s∈S

It follows that, for each element m in M , mφζM = m.

9.2 Algebraically Closed Base Fields In this section, the letter C stands for a field. We assume that, for each element s in S, the characteristic of C does not divide |s∗ |, so that, according to Theorem 9.1.5(ii), CS is semisimple. In addition, we assume C to be algebraically closed. Recall that, for each element φ in EndC (CS), φζCS :=

1 σs∗ φσs . ns

s∈S

The following result is due to Donald Higman; cf. [20; (7.1)]. Theorem 9.2.1 Writing ζ instead of ζCS , we have (CS)ζ = Z(CS). Proof. From Theorem 9.1.4(i) we know that (CS)ζ ⊆ Z(CS). Thus, we just have to prove that Z(CS) ⊆ (CS)ζ. Referring to Theorem 8.5.3(ii) and Theorem 8.5.4(ii) we shall be done if we succeed in showing that χ ∈ (CS)ζ. Let K be an irreducible submodule of the CS-module CS such that K ⊆ Hχ . Then, by Theorem 8.5.4(ii), Hχ ∼ = EndC (K). Thus, by Theorem 9.1.9, there exists an element σ in Hχ such that, for each element k in K, k(σζ) = k. Since Hχ is an ideal of CS, σζ ∈ Hχ . Thus, σζ = χ . Corollary 9.2.2 The matrix (

as∗ psq

s∈S

has rank |Irr(CS)|.

ns

)pq

9.2 Algebraically Closed Base Fields

193

Proof. From Theorem 8.6.4(ii) (together with Theorem 8.5.3(ii) and Theorem 8.5.4(ii)) we conclude that dimC (Z(CS)) = |Irr(CS)|. Thus, the claim follows from Theorem 9.1.4(ii) together with Theorem 9.2.1. We shall now see that the assumption that C is algebraically closed gives rise to a generalization of the orthogonality relations. Let χ be an irreducible character of CS, and let M be an irreducible CSmodule such that χM = χ. Fix a basis Bχ of M . Then, for any three elements t, u in Bχ and σ in CS, there exists an element cχtu (σ) in C such that, for each element t in Bχ , χ tσ = ctu (σ)u. u∈Bχ

Let v and w be elements in Bχ . Then, by Theorem 8.5.4(ii), there exists χvw ∈ Hχ such that, for any two elements t and u in Bχ , cχtu (χvw ) = δtw δuv . In other words, we have wχvw = v and, for each element s in Bχ \ {w}, we have sχvw = 0. Lemma 9.2.3 Let χ be an irreducible character of CS, and let t and u be elements in Bχ . Then the following hold. (i) For each element σ in CS, we have χ(σχtu ) = cχtu (σ). (ii) We have χtu =

mχ cχtu (σs∗ ) σs . nS ns∗ s∈S

Proof. (i) For each element v in Bχ , vσχtu = cχvw (σ)wχtu = cχvw (σ)δwu t = cχvu (σ)t. w∈Bχ

w∈Bχ

Therefore, χ(σχtu ) = cχtu (σ). (ii) By definition, χtu ∈ Hχ . Thus, the desired equality follows from (i) together with Lemma 9.1.6(i).

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9 Scheme Rings

In the case where S is thin, the equations of the following theorem are known as Schur relations. Theorem 9.2.4 Let φ and ψ be irreducible characters of CS, let t and u be elements in Bφ , and let v and w be elements in Bψ . Then the following hold. (i) For each element s in S, we have cφvu (σs∗ ) 1 as∗ pq φ ctu (σp∗ )cψ . vw (σq ) = δφψ δtw nS np∗ mφ p∈S q∈S

(ii) We have 1 1 φ 1 ctu (σs∗ )cψ . vw (σs ) = δφψ δtw δuv nS ns∗ mφ s∈S

Proof. (i) From Lemma 9.2.3(ii) and Lemma 9.1.1(i) we obtain φtu ψ vw =

mφ cφtu (σp∗ ) mψ cψ vw (σq ∗ ) ( σp ) ( σq ) nS np∗ nS nq∗ p∈S

q∈S

mφ mψ cφtu (σp∗ )cψ vw (σq ∗ ) σp σq = n2S np∗ nq∗ p∈S q∈S

=

mφ mψ cφtu (σp∗ )cψ vw (σq ∗ ) ( apqs )σs . n2S np∗ nq∗ s∈S p∈S q∈S

φ ψ φ But, by definition, φtu ∈ Hφ and ψ vw ∈ Hψ . Thus, tu vw = δφψ δtw vu . Referring once again to the equation

φvu =

mφ cφvu (σs∗ ) σs nS ns∗ s∈S

we, therefore, obtain mφ cφvu (σs∗ ) mφ mψ cφtu (σp∗ )cψ vw (σq ∗ )apqs = δφψ δtw nS np∗ nq∗ ns∗ p∈S q∈S

for each element s in S. Recall that, by Lemma 9.1.6(iii), mφ = 0 = mψ . Therefore, for each element s in S, we must have 1 apq∗ s ns∗ φ cφvu (σs∗ ) ctu (σp∗ )cψ . vw (σq ) = δφψ δtw nS np∗ nq mφ p∈S q∈S

Finally, recall that, by Lemma 1.1.1(ii) and Lemma 1.1.3(ii),

9.3 Scheme Rings over the Field of Complex Numbers

195

apq∗ s ns∗ = aqp∗ s∗ ns∗ = as∗ pq nq . (ii) This equation follows from (i) by setting s∗ = 1; use the first equation of Lemma 1.1.1(i). Lemma 9.2.5 Let χ be an irreducible character of CS, and let s be an element in S. Then χ(σs ) is integral over the smallest unitary subring of C. Proof. Let us denote by Z the smallest unitary subring of C. The C-module CS is finitely generated. Thus, σs is zero of a monic polynomial over Z. Thus, each characteristic root of σs is integral over Z. Thus, as χ(σs ) is a sum of characteristic roots of σs , the claim follows from Theorem 8.2.4.

9.3 Scheme Rings over the Field of Complex Numbers In this section, the letter C stands for the field of complex numbers. The scheme ring over the field of complex numbers provides us with an additional tool to work with, the norm function of C. Lemma 9.3.1 Let s be an element in S, and let c be a characteristic root of σs . Then |c| ≤ ns . Proof. Since c is a characteristic root of σs , there exists an element m in CX such that m = 0 and mσs = cm. Since m ∈ CX, there exists, for each element x in X, an element cx in C such that cx x. m= x∈X

Thus, as (

x∈X

cx x)σs =

cx xσs =

x∈X

c(

cx x) =

x∈X

y=

y∈xs

x∈X

and

cx

( cy )x x∈X y∈xs∗

ccx x,

x∈X

we obtain from mσs = cm that ccx =

cy

y∈xs∗

for each element x in X. Let us now pick an element z in X such that, for each of the finitely many elements x in X, |cx | ≤ |cz |. Then, by Lemma 1.1.2(i),

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9 Scheme Rings

|c||cz | = |ccz | = |

cy | ≤

y∈zs∗

|cy | ≤ ns∗ |cz |.

y∈zs∗

But cz = 0. Therefore, |c| ≤ ns∗ . Recall, finally, that we are assuming S to have finite valency. Thus, ns∗ = ns . Thus, as |c| ≤ ns∗ , |c| ≤ ns . Corollary 9.3.2 Let s be an element in S, and let χ be a character of CS. Then the following hold. (i) We have |χ(σs )| ≤ ns χ(σ1 ). (ii) Let M be an irreducible CS-module such that χM = χ. Then χ(σs ) = ns χ(σ1 ) if and only if, for each element m in M , mσs = ns m. Proof. (i) Since C is assumed to be the field of complex numbers, C is algebraically closed. Thus, χ(σs ) is the sum of χ(σ1 ) characteristic roots of σs . Thus, the claim follows from Lemma 9.3.1. (ii) Assume that, for each element m in M , mσs = ns m. Then we obviously have that χ(σs ) = ns χ(σ1 ). Let us now assume that χ(σs ) = ns χ(σ1 ). We set e := χ(σ1 ), and we define c1 , . . . , ce to be the characteristic roots of σs . Then, referring to Lemma 9.3.1 we obtain |χ(σs )| = |c1 + . . . + ce | ≤ |c1 | + . . . + |ce | ≤ ns e = ns χ(σ1 ) = χ(σs ) = |χ(σs )|. This proves that, for each element i in {1, . . . , e}, ci = ns . The following two results are due to Akihide Hanaki; cf. [17; Theorem 3.4] and [17; Theorem 3.5]. Lemma 9.3.3 Let T be a normal closed subset of S. Then there exists a ring homomorphism φ from CS to C(S//T ) such that, for each element s in S, φ(σs ) = asT s σsT . Proof. Let p and q be elements in S. Referring to Lemma 9.1.1(i) and to Theorem 4.1.3(ii) we obtain φ(σp )φ(σq ) = apT p σpT aqT q σqT = apT p aqT q apT qT sT σsT sT ∈S//T

= apT p aqT q =

sT ∈S//T

auvs (nT )−1 σsT

sT ∈S//T u∈pT v∈qT

(

u∈pT v∈qT

apT p aqT q auvs (nT )−1 )σsT

9.3 Scheme Rings over the Field of Complex Numbers

197

and φ(σp σq ) = φ(

apqs σs ) =

s∈S

=

s∈S

apqs asT s σsT =

apqs φ(σs )

(

apqw awT w )σsT .

sT ∈S//T w∈sT

s∈S

Thus, the claim follows from Lemma 2.5.4(iii). Lemma 9.3.4 Let T be a normal closed subset of S, and let M be a C(S//T )module. Then M is a CS-module such that, for any two elements m in M and s in S, mσs = asT s mσsT . Proof. Let m be an element in M , and let p and q be elements in S. We have to show that m(σp σq ) = (mσp )σq . Referring to Lemma 9.1.1(i) we obtain apqs σs ) = apqs mσs m(σp σq ) = m( s∈S

=

s∈S

apqs asT s mσsT = m(

s∈S

apqs asT s σsT ).

s∈S

Thus, as (mσp )σq = (apT p mσpT )σq = aqT q apT p mσpT σqT = m(aqT q apT p σpT σqT ), the desired equation follows from Lemma 9.3.3. Let T be a normal closed subset of S. The following corollary shows how to obtain characters of CS from characters of C(S//T ). Corollary 9.3.5 Let T be a normal closed subset of S, let χ be a character of C(S//T ). For each element s in S, we define χ(σ ¯ s ) := asT s χ(σsT ). Then χ ¯ is a character of CS. Proof. This follows immediately from Lemma 9.3.4. For each character χ of CS, we define T (χ) := {s ∈ S | χ(σs ) = ns χ(σ1 )}. The following two results are due to Akihide Hanaki; cf. [18; Theorem 3.2] and [18; Theorem 3.5].

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9 Scheme Rings

Lemma 9.3.6 For each character χ of CS, T (χ) is a closed subset of S. Proof. Let p and q be elements in T (χ). Then, by definition, χ(σp ) = np χ(σ1 ) and χ(σq ) = nq χ(σ1 ). According to Lemma 2.1.6, we shall be done if we succeed in showing that pq ⊆ T (χ). From Lemma 9.1.1(i), Corollary 9.3.2(ii), and Lemma 1.1.3(iv) we obtain apqs χ(σs ) = χ( apqs σs ) = χ(σp σq ) = np nq χ(σ1 ) = apqs ns χ(σ1 ). s∈S

s∈S

s∈S

Now recall from Corollary 9.3.2(i) that, for each element s in S, |χ(σs )| ≤ ns χ(σ1 ). Moreover, for each element s in S, 0 ≤ apqs . Thus, for each element s in S with 1 ≤ apqs , we must have χ(σs ) = ns χ(σ1 ). This means that, for each element s in pq, χ(σs ) = ns χ(σ1 ). Thus, for each element s in pq, s ∈ T (χ). Lemma 9.3.7 For each normal closed subset T of S, there exists a character χ of S such that T = T (χ). Proof. Let T be a normal closed subset of S. For any two elements s in S and χ in Irr(C(S//T )), we define χ(σ ¯ s ) := asT s χ(σsT ). Then, by Corollary 9.3.5, χ ¯ is a character of CS. We define ζ :=

mχ χ, ¯

χ∈Irr(C(S//T ))

and we claim that T = T (ζ). For each element t in T , we have atT t = nt . Thus, for each element t in T , ζ(σt ) = mχ χ(σ ¯ t) = mχ atT t χ(σtT ) χ∈Irr(C(S//T ))

= nt

χ∈Irr(C(S//T ))

mχ χ(σ1T ) = nt χC(X/T ) (σ1T ) = nt nX/T .

χ∈Irr(C(S//T ))

(The last equation follows from Lemma 9.1.2(i).) In particular, ζ(σ1 ) = nX/T . Thus, for each element t in T , ζ(σt ) = nt ζ(σ1 ), and that means that T ⊆ T (ζ). In order to show that T (ζ) ⊆ T we fix an element s in T (ζ). We shall be done if we succeed in showing that s ∈ T .

9.4 Closed Subsets

199

Since s ∈ T (ζ),

ζ(σs ) = ns ζ(σ1 ) = ns

mχ χ(σ ¯ 1 ) = 0.

χ∈Irr(C(S//T ))

On the other hand,

ζ(σs ) =

mχ χ(σ ¯ s)

χ∈Irr(C(S//T ))

= asT s

mχ χ(σsT ) = asT s χC(X/T ) (σsT ).

χ∈Irr(C(S//T ))

Thus, χC(X/T ) (σsT ) = 0. From χC(X/T ) (σsT ) = 0 we obtain 1T = sT ; cf. Lemma 9.1.2(ii). Thus, by Lemma 4.1.1, s ∈ T . Since s in T (ζ) has been chosen arbitrarily, we have shown that T (ζ) ⊆ T .

9.4 Closed Subsets In this section, the letter C stands for a field. We shall look at the relationship between representations of CS and closed subsets of S. For each subset R of S, we denote by C[R] the smallest unitary subring of CS which contains C and {σr | r ∈ R}. Theorem 9.4.1 Let T be a closed subset of S, and let R be a subset of T such that C[R] = CT . Then R = T . Proof. Let t be an element in T . Then σt ∈ CT . Thus, as we are assuming that C[R] = CT , there exist elements r1 , . . . , rm in R and positive integers e1 , . . . , em such that m ∈ / C(T \ {t}). σre11 · · · σrem Thus, setting n :=

m

ei

i=1

we obtain from Lemma 9.1.1(ii) that t ∈ Rn . Since t has been chosen arbitrarily in T , the claim follows from Lemma 2.3.2. Let R be a nonempty subset of S. We define σr . σR := r∈R

(Recall that nR is our notation for the sum of the integers nr with r ∈ R.)

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9 Scheme Rings

Lemma 9.4.2 A nonempty subset R of S is closed if and only if (σR )2 = nR σR . Proof. Let R be a nonempty subset of S. Then, by Lemma 9.1.1(i), (σR )2 = ( σp )( σq ) = apqs σs = ( apqs )σs . p∈R

q∈R

p∈R q∈R s∈R

s∈R p∈R q∈R

From Lemma 1.4.3 we know that, for each element s in S, apqs = nR p∈R q∈R

if and only if R∗ s ⊆ R. Thus, (σR )2 = nR σR if and only if R is closed. Lemma 9.4.3 For each closed subset T of S, the following hold. (i) The closed subset T is normal in S if and only if σT ∈ Z(CS). (ii) Let M be a CS-module, and assume that the characteristic of C does not divide nT . Then M = ker(σT ) ⊕ {m ∈ M | mσT = nT m}. Proof. (i) Let us assume that T is normal in S. Then, for each element s in S, T s = sT ; cf. Lemma 2.5.2(ii). On the other hand, we have σt σs = atsr σr = atsr σr = aT sr σr σT σs = t∈T

t∈T r∈T s

and, similarly, σs σT =

r∈T s t∈T

r∈T s

asT r σr .

r∈sT

Thus, our claim follows from Lemma 2.5.4(i). (ii) Let m be an element in M . Set t := nT m − mσT and u := mσT . Then nT m = t + u. On the other hand, with the help of Lemma 9.4.2, one obtains tσT = (nT m − mσT )σT = nT mσT − mσT2 = m(nT σT − σT2 ) = 0 and

uσT = mσT2 = nT mσT = nT u.

Now the claim follows from the hypothesis that the characteristic of C does not divide nT . Corollary 9.4.4 For each involution l of S, the following hold. (i) We have σl2 = nl σ1 + (nl − 1)σl . (ii) For each CS-module M , we have

9.4 Closed Subsets

201

M = {m ∈ M | mσl = −m} ⊕ {m ∈ M | mσl = nl m}. Proof. Since l is assumed to be an involution of S, {1, l} is closed. Thus, (i) follows from Lemma 9.4.2, and (ii) follows from Lemma 9.4.3(ii). Lemma 9.4.5 Let L be a set of two different involutions. Assume that C is algebraically closed and that C[L] = CS. Then, for each irreducible CSmodule M , dimC (M ) ≤ 2. Proof. Let M be an irreducible CS-module, and let us denote by h and k the elements in L. Since C is assumed to be algebraically closed, there exists an element m in M such that m = 0 and Cmσh σk = Cm. Let us define U := Cm + Cmσh . Then {0} = U . From Corollary 9.4.4(i) we obtain (Cmσh )σh = Cmσh2 ⊆ Cm + Cmσh = U, whence U σh ⊆ U . Similarly, Corollary 9.4.4(i) yields U σk ⊆ U . Recall that we are assuming that C[L] = CS. Therefore, U is a submodule of M . But {0} = U and M is assumed to be irreducible. Therefore, U = M . It follows that dimC (M ) ≤ 2. Proposition 9.4.6 Assume that C has characteristic 0. Let s be an element in S, and let T be a closed subset of S. (i) The vector space C(sT ) is a CT -module. (ii) For each element t in T , we have χC(sT ) (σt ) =

artr .

r∈sT

(iii) The character 1CT has multiplicity 1 in χC(sT ) . Proof. Let p be an element in sT , and let t be an element in T . Then, by Lemma 9.1.1(i), aptq σq . σp σt = q∈S

(i) Let q be an element in S such that 1 ≤ aptq . Then, by definition, q ∈ pt ⊆ pT . But, as p ∈ sT , pT = sT ; cf. Lemma 2.1.4. Therefore, q ∈ sT . Since q has been chosen arbitrarily in S satisfying 1 ≤ aptq , the above equality yields that σp σt ∈ C(sT ). (ii) This claim follows from the above equality together with the fact that {σq | q ∈ sT } is a basis of C(sT ).

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9 Scheme Rings

(iii) From Lemma 2.1.1 we know that, for each element r in sT , arT r = arT s . Thus, artr = artr = arts = arts . t∈T r∈sT

r∈sT t∈T

r∈sT t∈T

t∈T r∈sT

Thus, referring to (ii) and to the second equation of Lemma 1.1.3(iii) we obtain χC(sT ) (σt ) = artr = arts = arts = nt∗ = nT . t∈T

t∈T r∈sT

t∈T r∈sT

t∈T r∈S

t∈T

Now recall that, for each element χ in Irr(CT ) \ {1CT }, χ(σt ) = 0; t∈T

cf. Lemma 9.1.8(i). Thus, the claim follows from 1CT (σt ) = nT t∈T

together with the hypothesis that C has characteristic 0. For each element s in S, we define Cs := {σ ∈ CS | σσs = ns σ}. For each nonempty subset R of S, we define CR to be the intersection of the sets Cr such that r ∈ R. Recall that, for each closed subset T of S, S/T is our notation for the set of all left cosets of T in S. Theorem 9.4.7 Assume that C has characteristic 0. Then, for each closed subset T of S, we have |S/T | = dimC (CT ). Proof. Let R be a left transversal of S. Then, by Lemma 2.1.4, {rT | r ∈ R} is a partition of S. In particular, CS is a direct sum of the submodules C(rT ) with r ∈ R. Let σ be an element in CT . Since σ ∈ CS, there exists, for each element r in R, an element σ (r) in C(rT ) such that σ= σ (r) . r∈R

Let t be an element in T . Then, as σ ∈ CT , σ (r) σt = σσt = nt σ = nt σ (r) = nt σ (r) . r∈R

r∈R

r∈R

9.4 Closed Subsets

203

But, for each element r in R, nt σ (r) ∈ C(rT ) and, by Proposition 9.4.6(i), σ (r) σt ∈ C(rT ). Thus, as CS is a direct sum of the submodules C(rT ) with r ∈ R, we conclude that, for each element r in R, σ (r) σt = nt σ (r) . Thus, for each element r in R, σ (r) ∈ Ct . Since t ∈ T has been chosen arbitrarily, we obtain σ (r) ∈ CT for each element r in R. It follows that (C(rT ) ∩ CT ). CT = r∈R

Thus, the claim follows from Proposition 9.4.6(iii). Let us now see what we can say about characters of semidirect products of S and specific thin schemes. Let A be a subgroup of Aut(S). For each element a in A, we define aζ := aS . (Recall that, for each element a in Aut(S), aS is our notation for a ∩ (S × S).) From Lemma 5.2.2(iii) we know that ζ is a group homomorphism from A to Stc(S). Thus, we may consider the semidirect product Sζ A of S and A with respect to ζ; cf. Section 7.3. In the following theorem, we shall see that the standard module CX of CS is a module also for the scheme ring C(Sζ A) of Sζ A over C. We shall refer to this fact in the proof of Proposition 12.6.3. Theorem 9.4.8 Let A be a subgroup of Aut(S). For each element a in A, we define aζ := aS . For any three elements x in X, s in S, and a in A, we set y. xσsζ a := y∈xasa

Then we have the following. (i) For any two elements x in X and s in S, xσsζ 1 = xσs . (ii) For any two elements x in X and a in A, xσ1ζ a = xa. (iii) For any two subsets {cx | x ∈ X} and {c(s,a) | (s, a) ∈ S × A} of C, we define cx x)( c(s,a) σsζ a ) := cx c(s,a) xσsζ a . ( x∈X

(s,a)∈S×A

x∈X (s,a)∈S×A

Then CX is an C(Sζ A)-module. (iv) Let φ denote the character of C(Sζ A) afforded by CX. Let s be an element in S, and let a be an element in A. Then φ(σsζ a ) = |{x ∈ X | x ∈ xas}|. Proof. (i) This follows immediately from the definition of xσsζ 1 .

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9 Scheme Rings

(ii) This follows immediately from the definition of xσ1ζ a . (iii) Let p and q be elements in S, and let b and c be elements in A. We shall prove first that, for each element x in X, x(σpζ b σqζ c ) = (xσpζ b )σqζ c . First of all, with the help of Lemma 9.1.1(i) and Corollary 7.3.4(i), we obtain σpζ b σqζ c = a(pζ b)(qζ c)(sζ a) σsζ a = apqb∗ s σsζ bc . (s,a)∈S×A

s∈S

Therefore, for each element x in X, x(σpζ b σqζ c ) = x apqb∗ s σsζ bc = apqb∗ s xσsζ bc s∈S

=

s∈S

=

apqb∗ s

s∈S

y=

a(pbc)(qc)(sbc) y =

s∈S y∈(xbc)(sbc)

=

y

y∈xbcsbc

a(pbc)(qc)s y.

s∈S y∈(xbc)s

On the other hand, for each element x in X, (xσpζ b )σqζ c = yσqζ c = y∈(xb)(pb)

a(pbc)(qc)(sbc)

s∈S

y∈(xbc)(sbc)

z

y∈(xb)(pb) z∈(yc)(qc)

z=

yc∈(xbc)(pbc) z∈(yc)(qc)

a(pbc)(qc)s z.

s∈S z∈(xbc)s

Thus, we have proved the desired equation. Now looking at the definition of an C(Sζ A)-module, we see that the above equation suffices in order to show that CX is an C(Sζ A)-module. (iv) This follows immediately from the definition of xσsζ a , where x ∈ X, s ∈ S, and a ∈ A. Let a be an element in A, and set FixX (a) := {x ∈ X | xa = x}. Then, by Theorem 9.4.8, φ(σ1ζ a ) = |FixX (a)|.

9.5 Schemes with at most Five Elements In this section, we shall see how some of the representation theoretic results of the previous sections can be applied successfully if |S| is small.

9.5 Schemes with at most Five Elements

205

We start with two results due to Donald Higman; cf. [20; (4.2)]. Lemma 9.5.1 Let C be an algebraically closed field of characteristic 0. Then, if |Irr(CS)| = 2, |S| = 2. Proof. We are assuming that |Irr(CS)| = 2. Thus, there exists an irreducible character χ of CS such that {1CS , χ} = Irr(CS). Let s be an element in S \ {1}. Then, as 1CS (σs ) = ns∗ , we obtain from Lemma 9.1.8(ii) that ns∗ + mχ χ(σs ) = m1CS 1CS (σs ) + mχ χ(σs ) = χCX (σs ) = 0. It follows that χ(σs ) =

−ns∗ . mχ

Let us write Q to denote the smallest subfield of C, and let Z denote the smallest unitary subring of C. Then χ(σs ) ∈ Q. On the other hand, by Lemma 9.2.5, χ(σs ) is integral over Z. Thus, as C is assumed to have characteristic 0, χ(σs ) ∈ Z. It follows that χ(σs ) ≤ −1. Now recall that, by Corollary 8.6.5, |S| = 1 + χ(σ1 )2 . Thus, χ(σs ) = χ(σ1 ) + χ(σs ) ≤ χ(σ1 ) − (|S| − 1) = χ(σ1 ) − χ(σ1 )2 . s∈S

s∈S\{1}

Thus, by Lemma 9.1.8(i), χ(σ1 ) = χ(σ1 )2 . It follows that χ(σ1 ) = 1. Thus, as |S| = 1 + χ(σ1 )2 , |S| = 2. Theorem 9.5.2 If |S| ≤ 5, S is commutative. Proof. This is an immediate consequence of Lemma 9.5.1 and Corollary 8.6.5. The second part of the following theorem is due to Paul Erd¨ os, Alfred Rényi, and Vera S´ os; cf. [9; Theorem 6]. Its third part is a result of Alan Hoffman and Robert Singleton; cf. [27]. Theorem 9.5.3 Assume that S \ {1} contains symmetric elements p and q such that p = q and {1, p, q} = S. (i) The rational number (a − appq )(nS − 1) + 2np ppp (appp − appq )2 + 4(np − appq ) is an integer.

206

9 Scheme Rings

(ii) If appp = 1, then appq = 1. (iii) If appp = 0 and appq = 1, (np , nS ) ∈ {(2, 5), (3, 10), (7, 50), (57, 3250)}. Proof. (i) Let C be an algebraically closed field of characteristic 0, and let χ be a non-principal irreducible character of CS. Then, as |S| = 3, χ is linear; cf. Corollary 8.6.5. Thus, by Lemma 9.1.1(i), χ(σp )2 = χ(σp2 ) = χ(np σ1 + appp σp + appq σq ) = np + appp χ(σp ) + appq χ(σq ). But, by Lemma 9.1.8(i), χ(σq ) = −χ(σp ) − 1. (Recall that, by hypothesis, {1, p, q} = S.) Therefore, χ(σp )2 = np + appp χ(σp ) + appq (−χ(σp ) − 1), so that

χ(σp )2 − (appp − appq )χ(σp ) − (np − appq ) = 0.

From Corollary 8.6.5 we know that there exist elements φ and ψ in Irr(CS) \ {1CS } such that φ = ψ and {1CS , φ, ψ} = Irr(CS). Without loss of generality, we may assume that 1 φ(σp ) = (appp − appq + (appp − appq )2 + 4(np − appq )) 2 and 1 ψ(σp ) = (appp − appq − (appp − appq )2 + 4(np − appq )). 2 From Lemma 9.1.8(ii) we know that m1CS = 1. Therefore, 1 + m φ + mψ = mχ χ(σ1 ) = χCX (σ1 ) = nS . χ∈Irr(CS)

From 1CS (σp ) = np∗ and p∗ = p we obtain np + mφ φ(σp ) + mψ ψ(σp ) = mχ χ(σp ) = χCX (σp ) = 0. χ∈Irr(CS)

Now we have φ(σp )(nS − 1) + np φ(σp ) − ψ(σp ) 1 (appp − appq )2 + 4(np − appq ))(nS − 1) + np 2 (appp − appq + = (appp − appq )2 + 4(np − appq ) 1 (appp − appq )(nS − 1) + 2np = (nS − 1 + ). 2 (appp − appq )2 + 4(np − appq )

mψ =


207

Thus, the claim follows from the fact that mψ is an integer. (ii) Assume, by way of contradiction, that appp = 1 = appq . Then, by (i), np − 1 divides np . Therefore, np = 2. On the other hand, as we are assuming p to be symmetric, 1 ≤ appq yields 1 ≤ aqpp ; cf. Lemma 1.1.3(ii). Moreover, by the first equation of Lemma 1.1.1(i), we also have that 1 ≤ a1pp . Thus, by the second equation of Lemma 1.1.3(iii), 3 ≤ np∗ = np , contradiction. (iii) Let us assume that appp = 0 and that appq = 1. From the first equation of Lemma 1.1.1(i) we know that a1pp = 1. Thus, as appp = 0, the second equation of Lemma 1.1.3(iii) yields aqpp = np∗ − 1. Since p∗ = p, this means that aqp∗ p = np − 1. Now, as appq = 1, Lemma 1.1.3(ii) yields nq = (np − 1)np . Thus, as {1, p, q} = S, nS = n2p + 1. On the other hand, we know from (i) that 4np − 3 divides nS − 1 − 2np . Therefore, 4np − 3 divides np (np − 2). Assume that np = 2, and set m := Then np =

4np − 3.

1 2 (m + 3) 4

and m divides

1 (m2 + 3)(m2 − 5). 16 In particular, m divides 15, whence m ∈ {1, 3, 5, 15}. Assume that 1 = np . Then, by Lemma 1.5.1, s is thin. However, as we are assuming that appq = 1, we have q ∈ pp, contradiction. Thus, m ∈ {3, 5, 15}. Now the claim follows from 4np = m2 + 3 and nS = n2p + 1. It seems to be a still open problem whether nS = 3250 can actually occur in Theorem 9.5.3(iii).

9.6 Constrained Sets of Involutions Let L be a set of involutions, and let us write instead of L . Recall that, for each element q in L, S1 (q, L) is our notation for the set of all elements p in L such that pq contains an element r with (r) = (p) + (q). In accordance with Section 3.4 we shall write, for each element s in L, S1 (s) instead of S1 (s, L).

208

9 Scheme Rings

Recall that L is called constrained if, for any two elements q in L and p in S1 (q), 1 = |pq|. In this short section, the letter L stands for a constrained set of involutions. Lemma 9.6.1 Assume that L = S and that S has finite valency. Let p, q, and r be elements in S such that r ∈ pq and L (r) = L (p) + L (q). Then, we have the following. (i) For each element s in S, apqs = δsr . (ii) We have σp σq = σr . Proof. (i) From r ∈ pq and (r) = (p) + (q) we obtain apqr = 1; cf. Lemma 3.5.2. On the other hand, as L is assumed to be constrained, r ∈ pq yields {r} = pq. Thus, for each element s in S \ {r}, apqs = 0. (ii) From (i) and Lemma 9.1.1(ii) we obtain apqs σs = δsr σs = σr . σp σq = s∈S

s∈S

This proves (ii). From Theorem 9.4.1 we know that, for each subset R of S with C[R] = CS,

R = S. The following corollary says that the converse of this theorem holds if R is constrained. Corollary 9.6.2 The set Oϑ (L) is empty if and only if {1} = Oϑ ( L). Proof. Let us assume, by way of contradiction, that {1} = Oϑ ( L), let us write instead of L , and let us fix an element s in Oϑ ( L) \ {1} such that (s) is as small as possible. Since 1 = s, there exist elements r in L and l in L such that s ∈ rl and (s) = (r) + 1; cf. Lemma 3.1.2. From (s) = (r) + 1 and the minimal choice of s we obtain 1 = r or r ∈ / Oϑ ( L). If 1 = r, s ∈ L, contradiction. Thus, r ∈ / Oϑ ( L). From s ∈ rl and (s) = (r) + 1 we obtain arlq = δqs for each element q in

L; cf. Lemma 9.6.1(i). Thus, by Lemma 1.1.3(iv), arlq nq = nr nl . ns = q∈S

/ Oϑ ( L). contrary to s ∈ Oϑ ( L) and r ∈ Corollary 9.6.3 Assume that L = S and that S has finite valency. Then, for each field C we have C[L] = CS. Proof. This is an immediate consequence of Lemma 9.6.1(ii).

10 Dihedral Closed Subsets

A closed subset of S is called dihedral if it is generated by a set of two involutions of S. In this chapter, we investigate dihedral closed subsets of S. The letter L stands for a set of two (different) involutions of S. We shall look at L. As earlier, we shall write instead of L . Recall that, for each element q in L, S−1 (q, L) is our notation for the set of all elements r in L such that there exists an element p in L with r ∈ pq and (r) = (p) + (q). Keeping the notation introduced in Section 3.4 and used in Section 3.5 and Section 3.6 we shall write, for each element s in L, S−1 (s) instead of S−1 (s, L). In accordance with Section 3.5 we shall denote by S−1 (L) the intersection of the two sets S−1 (l) with l ∈ L. It may happen that S−1 (L) is empty. In this case, the structure of L is easy to describe. More challenging is it to look at the case where S−1 (L) is not empty, and it is this case on which we shall focus in the present chapter. In fact, we are mainly interested in the case where S−1 (L) contains exactly one element. Later in this chapter, in Section 10.6, we shall assume additionally that L has finite valency. It is one of the main goals of this chapter to show that (in this case) L is a Coxeter set (defined in Section 3.6) or a Moore set. (The definition of a Moore set will be given in Section 10.6.) This goal will be achieved in Theorem 10.6.6. A general theory of Coxeter sets will be developed in the last two chapters of this monograph. A more detailed investigation of the first case of Theorem 10.6.6 will be part of this more general approach. If L = S and S has finite valency, the condition that S−1 (L) has exactly one element is related to a natural condition about the scheme ring of S over an algebraically closed field of characteristic 0. In fact, we shall see that S−1 (L)

210


possesses exactly one element if there exists an algebraically closed field C of characteristic 0 such that C[L] = CS. The results in this section show that the notion of a Coxeter set emerges naturally from the theory of dihedral closed subsets. On this occasion, it might be worthwhile to recall that the theory of Coxeter sets is equivalent to the theory of buildings in the sense of Jacques Tits. In this chapter, we deal only with closed subsets generated by a set of two involutions. However, our approach indicates that the concepts and techniques developed in this chapter generalize meaningfully to closed subsets generated by an arbitrary finite set of involutions.

10.1 General Remarks Let l be an element in L. We set R0 (l) := {1}. Let j be a positive integer, let l1 , . . . , lj be elements in L, and let us assume that, for each integer i with 1 ≤ i ≤ j, li = l if and only if i is odd. In this case, we set Rj (l) := l1 · · · lj . The following lemma is crucial for the remainder of this section. Lemma 10.1.1 Let i and j be positive integers such that i ≤ j, let h and k be elements in L such that h = k, and let s be an element in Rj (h)∗ Ri (h). Then the following hold. (i) If j is odd, there exists an integer n such that j − i ≤ n ≤ j + i − 1 and s ∈ Rn (h). (ii) If j is even, there exists an integer n such that j − i ≤ n ≤ j + i − 1 and s ∈ Rn (k). Proof. Let us first look at the case where i = 1. In this case, our assumption says that s ∈ Rj (h)∗ h. Thus, if j is odd, s ∈ Rj (h)h ⊆ Rj−1 (h) ∪ Rj (h). Similarly, if j is even, s ∈ Rj (k)h ⊆ Rj−1 (k) ∪ Rj (k), so that we are done in the case where i = 1. Let us now assume that 2 ≤ i. Then Rj−1 (k)∗ hhRi−1 (k) ⊆ Rj−1 (k)∗ Ri−1 (k) ∪ Rj−1 (k)∗ hRi−1 (k).

10.1 General Remarks

211

Thus, as s ∈ Rj (h)∗ Ri (h) and Rj (h)∗ Ri (h) = Rj−1 (k)∗ hhRi−1 (k), s ∈ Rj−1 (k)∗ Ri−1 (k) ∪ Rj−1 (k)∗ hRi−1 (k). If j is odd, If j is even,

Rj−1 (k)∗ hRi−1 (k) = Rj+i−1 (h). Rj−1 (k)∗ hRi−1 (k) = Rj+i−1 (k).

Thus, we may assume that s ∈ Rj−1 (k)∗ Ri−1 (k). If j is odd, j − 1 is even. Thus, by induction, there exists an integer n such that j − i ≤ n ≤ j + i − 3 and s ∈ Rn (h). If j is even, j − 1 is odd. Thus, by induction, there exists an integer n such that j − i ≤ n ≤ j + i − 3 and s ∈ Rn (k). Corollary 10.1.2 Let i and j be positive integers, let h and k be elements in L such that h = k. Then we have the following. (i) If Ri (h) ∩ Rj (k) is not empty, 1 ∈ Ri+j (h) ∪ Ri+j (k). (ii) Assume that Ri (h) ∩ Rj (h) is not empty and that i ≤ j. Then there exists an integer n with j − i ≤ n ≤ j + i − 1 and 1 ∈ Rn (h) ∪ Rn (k). Proof. (i) Let us assume that Ri (h) ∩ Rj (k) is not empty, and let s be an element in Ri (h) ∩ Rj (k). Referring to Lemma 1.3.2(iii) we then obtain 1 ∈ s∗ s ⊆ Ri (h)∗ Rj (k) ⊆ Ri+j (h) ∪ Ri+j (k). (ii) We are assuming that Ri (h) ∩ Rj (h) is not empty. Let s be an element in Ri (h) ∩ Rj (h). Then 1 ∈ s∗ s ⊆ Rj (h)∗ Ri (h), so that the claim follows from Lemma 10.1.1. Lemma 10.1.3 Let n be a non-negative integer, and let s be an element in Ln . Then there exist elements i in {0, . . . , n} and l in L such that s ∈ Ri (l). Proof. The claim holds obviously if n = 0. Let us, therefore, assume that 1 ≤ n. Since s ∈ Ln , there exists an element r in Ln−1 such that s ∈ rL. From r ∈ Ln−1 we obtain, by induction, an integer i with 1 ≤ i ≤ n and an element l in L such that r ∈ Ri−1 (l). If 1 = i, we obtain from r ∈ Ri−1 (l) that r ∈ R0 (l). It follows that 1 = r. Thus, as s ∈ rL, s ∈ L, and we are done. If 2 ≤ i, we obtain from s ∈ rL and r ∈ Ri−1 (l) that s ∈ Ri−1 (l)L ⊆ Ri−2 (l) ∪ Ri−1 (l) ∪ Ri (l), and that finishes our proof.

212


Corollary 10.1.4 Let s be an element in L, and let l be an element in L. Then s ∈ S−1 (l) if and only if s∗ ∈ R(s) (l) \ {1}. Proof. We set j := (s), and we assume first that s ∈ S−1 (l). Then 1 = s and, by definition, there exists an element r in L such that s ∈ rl and j = (r) + 1. From j = (r) + 1 we obtain (r) = j − 1. Thus, r ∈ Lj−1 . It follows that r∗ ∈ Lj−1 . Thus, there exist elements i in {1, . . . , j − 1} and k in L such that r∗ ∈ Ri (k); cf. Lemma 10.1.3. From r∗ ∈ Ri (k) we obtain r ∈ Ri (k)∗ . Thus, as s ∈ rl, s ∈ Ri (k)∗ l. If i = 0, we obtain from s ∈ Ri (k)∗ l that s = l, and we are done. Thus, we may assume that 1 ≤ i. If k = l, we obtain from s ∈ Ri (k)∗ l that s ∈ Ri (k)∗ k. Thus, there exists a non-negative integer n such that n ≤ i and s ∈ Ln ; cf. Lemma 10.1.1. It follows that (s) ≤ n ≤ i ≤ j − 1, contradiction. If k = l, we obtain from s ∈ Ri (k)∗ l that s ∈ Ri+1 (l)∗ . Thus, (s) ≤ i + 1 ≤ j, so that (s) = i+1. It follows that s ∈ R(s) (l)∗ , and this implies s∗ ∈ R(s) (l). Let us now, conversely, assume that s∗ ∈ R(s) (l) \ {1}. Then, as j = (s), s∗ ∈ Rj (l) \ {1}. Thus, there exists an element r in Lj−1 such that s ∈ rl. From r in Lj−1 we obtain (r) ≤ j − 1. Thus, (s) = (r) + 1. Thus, as s ∈ rl, s ∈ S−1 (l). Lemma 10.1.5 Assume that there exists a positive integer j and an element l in L such that 1 ∈ Rj (l). Let j be the smallest positive integer for which there exists an element l in L with 1 ∈ Rj (l). Then there exists an element s in S−1 (L) such that j = 2(s). Proof. From 1 ∈ Rj (l) we obtain 4 ≤ j and 1 ∈ R2 (l)Rj−2 (l). By Lemma 1.3.2(ii), the latter condition implies 1 ∈ Rj−2 (l)R2 (l). Let us assume, by way of contradiction, that j is odd. Then, j − 2 is odd, so that Rj−2 (l)∗ = Rj−2 (l). Thus, as 1 ∈ Rj−2 (l)R2 (l) and 2 ≤ j − 2, there exists an integer n with j − 4 ≤ n ≤ j − 1 and 1 ∈ Rn (l); cf. Lemma 10.1.1(i). This contradicts our (minimal) choice of j. Thus, j is even, and that means that there exists an integer d such that j = 2d. It follows that 1 ∈ R2d (h). Thus, there exists an element s in Rd (h) such that 1 ∈ sRd (k)∗ . It follows that s ∈ Rd (h) ∩ Rd (k). From s ∈ Rd (h) we obtain (s) ≤ d.

10.2 The Spherical Case

213

By Lemma 10.1.3, there exist elements i in {1, . . . , (s)} and l in L such that s ∈ Ri (l). Thus, as s ∈ Rd (l), 1 ∈ s∗ s ⊆ Rd (l)∗ Ri (l). Thus, as (s) ≤ d, there exists an integer n such that d − i ≤ n ≤ d + i − 1 and s ∈ Rn (h) ∪ Rn (k); cf. Lemma 10.1.1. From this we conclude that i = d. However, we have i ≤ (s) ≤ d, so that (s) = d. It follows that s ∈ R(s) (h) ∩ R(s) (k). Thus, by Corollary 10.1.4, s∗ ∈ S−1 (L). Recall that L is called spherical if S−1 (L) is not empty. Lemma 10.1.6 Assume L to be spherical, and let s be an element in S−1 (L). Then, for each element l in L, 1 ∈ R2(s) (l). Proof. We have fixed an element s in S−1 (L). Thus, for each element l in L, s in S−1 (l). Thus, for each element l in L, s∗ ∈ R(s) (l); cf. Corollary 10.1.4. Thus, by Corollary 10.1.2(i), 1 ∈ R2(s) (l). From Lemma 10.1.5 together with Lemma 10.1.6 we obtain, in particular, that L is spherical if and only if there exists a positive integer j and an element l in L such that 1 ∈ Rj (l). Lemma 10.1.7 If L is finite, L is spherical. Proof. Let l be an element in L, and let us assume L to be finite. Then there exist positive integers i and j such that i + 1 ≤ j and Ri (l) ∩ Rj (l) is not empty. Thus, the claim follows from Corollary 10.1.2(ii) together with Lemma 10.1.5.

10.2 The Spherical Case Throughout this section, the set L is assumed to be spherical. We define dL to be the smallest element in (S−1 (L)). However, having fixed L for the remainder of this section, we shall write d instead of dL . Note that 2 ≤ d. Let h and k be elements in L such that h = k. Since S−1 (L) contains an element of length d, we obtain from Corollary 10.1.4 that Rd (h) ∩ Rd (k) is not empty. Of course, we also have that R0 (h) ∩ R0 (k) is not empty. The first part of the following lemma is a partial converse of this. Lemma 10.2.1 For any two integers i and j satisfying 0 ≤ i ≤ d and 0 ≤ j ≤ d, we have the following.

214


(i) Let h and k be elements in L such that h = k and ∅ = Ri (h) ∩ Rj (k). Then i = j ∈ {0, d}. (ii) Let l be an element in L with ∅ = Ri (l) ∩ Rj (l). Then i = j. Proof. (i) We are assuming that Ri (h) ∩ Rj (k) is not empty. Thus, by Corollary 10.1.2(i), 1 ∈ Ri+j (h) ∪ Ri+j (k). Thus, we must have i + j = 0 or 2d ≤ i + j; cf. Lemma 10.1.5. In the first case, we are done. In the second case, we obtain from 0 ≤ i ≤ d and 0 ≤ j ≤ d that i = j = d. (ii) The claim is obviously true if i = 0. Thus, we may assume that 1 ≤ i ≤ j. In this case, we obtain from Corollary 10.1.2(ii) an integer n with j − i ≤ n ≤ j + i − 1 and an element k in L with 1 ∈ Rn (k). From i ≤ d, j ≤ d, and n ≤ i + j − 1 we obtain n ≤ 2d − 1. Thus, by Lemma 10.1.5, n = 0. Thus, as 1 ≤ i ≤ j and j − i ≤ n, i = j. Lemma 10.2.2 Let j be an integer such that 1 ≤ j ≤ d, and let l be an element in L such that |Rj (l)| = 1. Then |Rj−1 (l)| = 1. Proof. We are assuming that |Rj (l)| = 1. Thus, there exists an element s in Rj (l) such that {s} = Rj (l). Let us fix elements p and q in Rj−1 (l). We shall see that p = q. Let us first assume j to be odd. In this case, j − 1 is even, so that pl ⊆ Rj−1 (l)l = Rj (l) = {s}. Similarly, ql ⊆ {s}, so that pl = {s} = ql. It follows that q ∈ qll = pll ⊆ {p} ∪ pl = {p, s}. If q = s, Rj−1 (l)∩Rj (l) is not empty, contrary to Lemma 10.2.1(ii). Therefore, p = q. Similarly, the assumption that j is even leads to p = q. Lemma 10.2.3 For any two elements s in L and l in L, there exists a non-negative integer j such that s ∈ Rj (l). Proof. Let s be an element in L. Then, by Lemma 3.1.1(i), there exists a non-negative integer n such that s ∈ Ln . Thus, by Lemma 10.1.3, there exists a non-negative integer i and an element l in L such that s ∈ Ri (l). Let j be a multiple of 2d such that i ≤ j. From Lemma 10.1.6 we know that 1 ∈ R2d (l). Thus, 1 ∈ Rj (l). It follows that s ∈ 1s ⊆ Rj (l)∗ Ri (l).

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Now the claim follows from Lemma 10.1.1(ii). Let s be an element in L, and let l be an element in L. According to Lemma 10.2.3, there exists a non-negative integer j such that s ∈ Rj (l). In the following, we shall denote by l (s) the smallest non-negative integer j satisfying s ∈ Rj (l). Note that, for each element l in L, l (1) = 0 and l (l) = 1. It follows immediately from the definition of that, for any two elements s in L and l in L, (s) ≤ l (s). From Lemma 10.1.3 (together with Lemma 10.2.3) we obtain a little more, namely that (s) = min{l (s) | l ∈ L} for each element s in L. We shall use this little observation frequently without further reference. Lemma 10.2.4 For each element s in L, we have the following. (i) Let l be an element in L such that l (s) is odd. Then l (s∗ ) ≤ l (s). (ii) Let h and k be elements in L such that h = k and h (s) is even. Then k (s∗ ) ≤ h (s). Proof. (i) Set j := l (s). Then s ∈ Rj (l). It follows that s∗ ∈ Rj (l)∗ . On the other hand, we are assuming j to be odd, so that Rj (l)∗ = Rj (l). Thus, s∗ ∈ Rj (l), and it follows that l (s∗ ) ≤ j. (ii) Set j := h (s). Then s ∈ Rj (h). It follows that s∗ ∈ Rj (h)∗ . On the other hand, we are assuming j to be even, so that Rj (h)∗ = Rj (k). Thus, s∗ ∈ Rj (k), and it follows that k (s∗ ) ≤ j. Lemma 10.2.5 Let p and q be elements in L, and let l be an element in L. Then, if q ∈ pL and 1 = p, l (q) ≤ l (p) + 1. Proof. Set j := l (p). Then, by definition, p ∈ Rj (l). Moreover, as 1 = p, 0 = j. From q ∈ pL and p ∈ Rj (l) we obtain q ∈ Rj (l)L ⊆ Rj−1 (l) ∪ Rj (l) ∪ Rj+1 (l). It follows that l (q) ≤ j + 1. Lemma 10.2.6 For each element s in L, we have the following. (i) Let h and k be elements in L such that h = k. Then, if 1 = s, 2d ≤ h (s) + k (s). (ii) Let j be an integer with 0 ≤ j ≤ d, and let l be an element in L such that s ∈ Rj (l). Then (s) = j.

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Proof. (i) Set i := h (s) and j := k (s). Then s ∈ Ri (h) ∩ Rj (k). Thus, by Corollary 10.1.2(i), 1 ∈ Ri+j (h) ∪ Ri+j (k). On the other hand, we are assuming that 1 = s, so that 0 = i and 0 = j. Thus, by Lemma 10.1.5, 2d ≤ i + j. (ii) We are assuming that s ∈ Rj (l). Thus, l (s) ≤ j. Set i := l (s). Then i ≤ j and s ∈ Ri (l). It follows that s ∈ Ri (l) ∩ Rj (l). Thus, as 0 ≤ i ≤ d and 0 ≤ j ≤ d, we obtain from Lemma 10.2.1(ii) that i = j. Thus, l (s) = j. The statement is obviously true if 1 = s. Thus, we may assume that 1 = s. Thus, as l (s) = j and j ≤ d, we obtain from (i) that (s) = l (s). Lemma 10.2.7 Let j be an integer with 1 ≤ j ≤ d, and let l be an element in L. Then we have the following. (i) If j is odd, Rj (l) ⊆ S−1 (l). (ii) Let h and k be elements in L such that h = k. Then, if j is even, Rj (h) ⊆ S−1 (k). Proof. (i) Assume j to be odd, and let s be an element in Rj (l). Then there exists an element r in Rj−1 (l) such that s ∈ rl. From s ∈ Rj (l) and j ≤ d we obtain (s) = j; cf. Lemma 10.2.6(ii). From r ∈ Rj−1 (l) we obtain (r) ≤ j − 1. Thus, as s ∈ rl, s ∈ S−1 (l). (ii) Assume j to be even, and let s be an element in Rj (h). Then there exists an element r in Rj−1 (h) such that s ∈ rk. From s ∈ Rj (h) and j ≤ d we obtain (s) = j; cf. Lemma 10.2.6(ii). From r ∈ Rj−1 (h) we obtain (r) ≤ j − 1. Thus, as s ∈ rk, s ∈ S−1 (k). Let us briefly look at the consequences of our results for S−1 (L) and L. Lemma 10.2.8 We have S−1 (L)∗ = S−1 (L). Proof. Let s be an element in S−1 (L). Then, for each element l in L, s∗ ∈ R(s) (l); cf. Corollary 10.1.4. Thus, for each element l in L, l (s∗ ) ≤ (s) = (s∗ ), and that means that (s∗ ) = l (s∗ ). Thus, by Lemma 10.2.4, for each element l in L, l (s) ≤ (s) = (s∗ ). Thus, by Corollary 10.1.4, s∗ ∈ S−1 (L). Lemma 10.2.9 We have 2d ≤ | L|. Proof. This follows from Lemma 10.2.1.

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10.3 Arithmetic of the Length Function Throughout this section, the set L is assumed to be spherical. Let h and k be elements in L such that h = k. For each element s in L, we define νh (s) := h (s) − k (s). Note that, for any two elements s in L and l in L, (s) = l (s) if and only if νl (s) ≤ 0. Lemma 10.3.1 For each element l in L, we have S−1 (l) = {s ∈ L | νl (s∗ ) ≤ 0} \ {1}. Proof. Let l be an element in L, and let s be an element in L. Then we have s∗ ∈ R(s) (l) if and only if (s∗ ) = l (s∗ ), and that is equivalent to νl (s∗ ) ≤ 0. On the other hand, we know from Corollary 10.1.4 that s ∈ S−1 (l) is equivalent to s∗ ∈ R(s) (l) \ {1}. As a consequence of Lemma 10.3.1 and Lemma 10.2.8 we obtain S−1 (L) = {s ∈ L | νl (s) = 0} \ {1} for each element l in L. Lemma 10.3.2 For any two elements s in L and l in L, we have the following. (i) If (s) is odd, 0 ≤ νl (s∗ )νl (s). (ii) If (s) is even, νl (s∗ )νl (s) ≤ 0. Proof. (i) Assume first that νl (s) ≤ 0. Then (s) = l (s). Thus, assuming (s) to be odd we obtain from Lemma 10.2.4(i) that l (s∗ ) ≤ (s∗ ). It follows that νl (s∗ ) ≤ 0. Similarly, 0 ≤ νl (s) yields 0 ≤ νl (s∗ ). (ii) Assume first that νl (s) ≤ 0. Then (s) = l (s). Thus, by Lemma 10.2.4(ii), 0 ≤ νl (s∗ ). Similarly, 0 ≤ νl (s) yields νl (s∗ ) ≤ 0. Lemma 10.3.3 Let p and q be elements in L such that q ∈ pL. Then, for each element l in L, 0 ≤ νl (p)νl (q). Proof. Assume, by way of contradiction, that νl (p)νl (q) ≤ −1. Then there exist elements h and k in L such that h (p) ≤ k (p) − 1 and k (q) ≤ h (q) − 1.

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From νl (p)νl (q) ≤ −1 we obtain 1 = p and 1 = q. Thus, by Lemma 10.2.5, h (p) ≤ k (p) − 1 ≤ k (q) ≤ h (q) − 1 ≤ h (p). It follows that h (p) = k (q). We set j := h (p). Since the claim is true for p = 1, we may assume that 1 ≤ j. Since, by hypothesis, q ∈ pL, we have q ∈ ph or q ∈ pk. Let us assume first that q ∈ ph. If j is odd, q ∈ ph ⊆ Rj (h)∗ h. Thus, by Lemma 10.1.1(i), q ∈ Rj−1 (h) ∪ Rj (h), contrary to j + 1 ≤ h (q). If j is even, p ∈ qh ⊆ Rj (h)∗ h. Thus, by Lemma 10.1.1(ii), p ∈ Rj−1 (k) ∪ Rj (k), contrary to j + 1 ≤ k (p). Assume now that q ∈ pk. If j is odd, p ∈ qk ⊆ Rj (k)∗ k. Thus, by Lemma 10.1.1(i), p ∈ Rj−1 (k) ∪ Rj (k), contrary to j + 1 ≤ k (p). If j is even, q ∈ pk ⊆ Rj (k)∗ k. Thus, by Lemma 10.1.1(ii), q ∈ Rj−1 (h) ∪ Rj (h), contrary to j + 1 ≤ h (q). Lemma 10.3.4 Let p, q, r, and s be elements in L \ S−1 (L) such that p ∈ Lq ∩ rL and s ∈ Lr ∩ qL. Then (p) = (r) if and only if (q) = (s). Proof. Since p, q, r, s ∈ / S−1 (L), we obtain from Lemma 10.3.3 that 1 ≤ νl (p∗ )νl (q ∗ ), 1 ≤ νl (r∗ )νl (s∗ ), 1 ≤ νl (p)νl (r), 1 ≤ νl (q)νl (s) for each element l in L. Let l be an element in L, and let us assume that (p) = (r). Then, by Lemma 10.3.2, 1 ≤ νl (p∗ )νl (p)νl (r∗ )νl (r). Thus, by the above four observations, 1 ≤ νl (q ∗ )νl (q)νl (s∗ )νl (s). Thus, by Lemma 10.3.2, (q) − (s) is even. Now recall that we are assuming that s ∈ qL. Thus, (q) − (s) ∈ {−1, 0, 1}. It follows that (q) = (s). Lemma 10.3.5 Let p, q, r, and s be elements in L \ S−1 (L), and let l be an element in L such that p ∈ Lq ∩ rl and s ∈ Lr ∩ ql. Then (p) + 1 = (r) if and only if (q) + 1 = (s). Proof. Let us assume that (p) + 1 = (r). Then, by Lemma 10.3.4, (q) = (s). Thus, as we are assuming that s ∈ qL, we conclude that (s) ∈ {(q) − 1, (q) + 1}. Let us assume, by way of contradiction, that (s) = (q) − 1.

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From s ∈ ql we obtain q ∈ sl; cf. Lemma 1.3.3(i). Thus, as we are assuming that (q) = (s) + 1, q ∈ S−1 (l). Thus, by Lemma 10.3.1, νl (q ∗ ) ≤ −1. Similarly, we obtain from p ∈ rl and (r) = (p) + 1 that νl (r∗ ) ≤ −1. Now recall that, by Lemma 10.3.3, 1 ≤ νl (p∗ )νl (q ∗ ). Thus, as νl (q ∗ ) ≤ −1 and νl (r∗ ) ≤ −1, 1 ≤ νl (p∗ )νl (r∗ ). From Lemma 10.3.3 we also obtain 1 ≤ νl (p)νl (r). Thus, 1 ≤ νl (p∗ )νl (p)νl (r∗ )νl (r). But, as (p) + 1 = (r), this contradicts Lemma 10.3.2. Proposition 10.3.6 Let p, q, and r be elements in L such that p ∈ S1 (r) and r ∈ S1 (q). Assume that, for each element s in L with (r) + 1 ≤ (s) ≤ (p) + (r) + (q), s∈ / S−1 (L). Then there exists an element s in prq such that (s) = (p) + (r) + (q). Proof. Let y be an element in X, and let z be an element in yr. We are assuming that r ∈ S1 (q). Thus, there exists an element u in rq such that (u) = (r) + (q). From u ∈ rq we obtain r ∈ uq ∗ ; cf. Lemma 1.3.3(i). Thus, as z ∈ yr, z ∈ yuq ∗ . Thus, there exists an element w in yu such that z ∈ wq ∗ . Set n := (q). Then q ∈ Ln . Thus, as w ∈ zq, there exist elements z0 , . . . , zn in X such that z0 = z, zn = w, and, for each element j in {1, . . . , n}, zj ∈ zj−1 L; cf. Lemma 1.3.9. We are assuming that p ∈ S1 (r). Thus, there exists an element t in pr such that (t) = (p) + (r). From t ∈ pr we obtain r ∈ p∗ t; cf. Lemma 1.3.3(ii). Thus, as z ∈ yr, z ∈ yp∗ t. Thus, there exists an element v in yp∗ such that z ∈ vt. Set m := (p). Then, as before, we find elements y0 , . . . , ym in X such that y0 = y, ym = v, and, for each element i in {1, . . . , m}, yi ∈ Lyi−1 . For any two elements i in {0, . . . , m} and j in {0, . . . , n}, we define sij to be the element in L which satisfies zj ∈ yi sij . Then s00 = r, sm0 = t, and s0n = u. Moreover, for any two elements i in {1, . . . , m} and j in {1, . . . , n}, we have sij ∈ Lsi−1,j ∩ si,j−1 L.

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From s00 = r and s0n = u we obtain (s00 ) + n = (s0n ). Thus, we obtain from Lemma 3.4.1 that (s0j ) = j for each element j in {1, . . . , n}. Similarly, we have (si0 ) = i for each element i in {1, . . . , m}. Our claim is that, for any two elements i in {1, . . . , m} and j in {1, . . . , n}, (sij ) = i + (r) + j. By induction, we may assume that (si−1,j−1 ) = i+(r)+j−2, that (si,j−1 ) = i + (r) + j − 1, and that (si−1,j ) = i + (r) + j − 1. Thus, by Lemma 10.3.5, (sij ) = i + (r) + j. Since we have that, for any two elements i in {1, . . . , m} and j in {1, . . . , n}, (sij ) = i + (r) + j, (smn ) = m + (r) + n = (p) + (r) + (q), and that finishes the proof.

10.4 Two Characteristic Subsets Throughout this section, the set L is assumed to be spherical. For each element l in L, we define S0 (l) := {s ∈ L | min (sl) = (s)}. Note that, for each element l in L, 1 ∈ / S0 (l). Lemma 10.4.1 Let l be an element in L, let q be an element in S0 (l), and let p be an element in L such that q ∈ S−1 (p). Then the following hold. (i) We have p ∈ / S−1 (l). (ii) For each element r in pl \ S−1 (l), (r) = (p). Proof. (i) Assume, by way of contradiction, that p ∈ S−1 (l). Then, by Lemma 3.4.4(iii), q ∈ S−1 (l). (We apply this lemma to p in place of q and l in place of p.) This contradicts our hypothesis that q ∈ S0 (l). (ii) Let r be an element in pl \ S−1 (l), and let us assume, by way of contradiction, that (r) = (p). Then, as r ∈ pl, (r) = (p) − 1 or (r) = (p) + 1. In the first case, we obtain p ∈ S−1 (l), contrary to (i). In the second case, we obtain r ∈ S−1 (l), contrary to our hypothesis.

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Let l be an element in L. Similarly to S−1 (l), the set S0 (l) can be described in terms of νl . Lemma 10.4.2 For each element l in L, we have S0 (l) = {s ∈ L | νl (s∗ ) = 1}. Proof. Let us first prove that, for each element s in S0 (l), νl (s∗ ) = 1. In order / S−1 (l). to do so we pick an element s in S0 (l). From s ∈ S0 (l) we obtain s ∈ Thus, by Lemma 10.3.1, 1 ≤ νl (s∗ ). Set j := (s). Then, as s ∈ S0 (l) by definition, there exists an element r in sl such that (r) = j. From r ∈ sl we obtain s∗ ∈ lr∗ ; cf. Lemma 1.3.3(iii). From (r) = j we obtain (r∗ ) = j. Let k be the element in L \ {l}. Suppose that l (r∗ ) = j. Then r∗ ∈ Rj (l), whence s∗ ∈ lRj (l). Thus, as 1 ≤ j, s∗ ∈ Rj−1 (k) ∪ Rj (l). If s∗ ∈ Rj−1 (k), (s∗ ) ≤ j − 1, contrary to (s) = j. If s∗ ∈ Rj (l), l (s∗ ) ≤ j, so that νl (s∗ ) ≤ 0, contrary to 1 ≤ νl (s∗ ). Thus, we must have k (r∗ ) = j. It follows that r∗ ∈ Rj (k), whence s∗ ∈ lRj (k) = Rj+1 (l). Thus, l (s∗ ) ≤ j + 1. It follows that νl (s∗ ) ≤ 1. Let us now prove that only elements s in S0 (l) may satisfy νl (s∗ ) = 1. Let s be an element in L such that νl (s∗ ) = 1. Set j := (s∗ ). Then, l (s∗ ) = j + 1. In particular, s∗ ∈ Rj+1 (l). Thus, there exists an element r in Lj such that s∗ ∈ lr. From s∗ ∈ lr we obtain s ∈ r∗ l; cf. Lemma 1.3.2(iii). From r ∈ Lj we obtain r∗ ∈ Lj , so that (r∗ ) ≤ j = (s). Thus, s ∈ S−1 (l) or s ∈ S0 (l). / S−1 (l). Therefore, we Since νl (s∗ ) = 1, Lemma 10.3.1 tells us also that s ∈ must have s ∈ S0 (l). From Lemma 3.4.5(ii) (together with Lemma 10.2.8) we know that, for each element s in S−1 (L), {s} = S−1 (s). The following lemma shows that, for any two elements l in L and s in S0 (l), there exists an element q in S−1 (L) such that s is ‘almost’ in S−1 (q). Proposition 10.4.3 Let l be an element in L, and let s be an element in S0 (l). Then there exist elements p in L and q in S−1 (L) such that s ∈ pq and (s) + 1 = (p) + (q). Proof. We set j := l (s∗ ). Then s∗ ∈ Rj (l). Note also that 2 ≤ j. Thus, there exist elements l1 , . . . , lj in L such that s∗ ∈ l1 · · · lj and, for each element i in {1, . . . , j}, li = l if and only if i is odd.

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Since s∗ ∈ l1 · · · lj , Lemma 1.3.5 yields elements s0 , . . . , sj in L such that s0 = 1, sj = s, and, for each element i in {1, . . . , j}, si ∈ li si−1 . Since s1 = l, l (s∗1 ) = 1. Therefore, by induction, Lemma 10.2.5 yields that, for each element i in {1, . . . , j}, l (s∗i ) ≤ i. Similarly, as sj = s, l (s∗j ) = j. Thus, by induction, Lemma 10.2.5 yields that, for each element i in {1, . . . , j}, i ≤ l (s∗i ). It follows that, for each element i in {1, . . . , j}, l (s∗i ) = i. From s1 = l we obtain νl (s∗1 ) ≤ −1. On the other hand, as sj = s, sj ∈ S0 (l), so that, according to Lemma 10.4.2, νl (s∗j ) = 1. We set

I := {i ∈ {1, . . . , j} | 0 ≤ νl (s∗i )}.

Since νl (s∗j ) = 1, j ∈ I. Thus, I is not empty. We define i := min I. Then 0 ≤ νl (s∗i ). / I. Thus, 2 ≤ i and νl (s∗i−1 ) ≤ −1. Since νl (s∗1 ) ≤ −1, 1 ∈ However, we have that s∗i ∈ s∗i−1 L. Thus, by Lemma 10.3.3, 0 ≤ νl (s∗i )νl (s∗i−1 ). Thus, as 0 ≤ νl (s∗i ) and νl (s∗i−1 ) ≤ −1, νl (s∗i ) = 0. This means that s∗i ∈ S−1 (L). We set q := s∗i . Then l (q) = i. Thus, as q ∈ S−1 (L), (q) = i. Since νl (s∗i ) = 0 and νl (s∗j ) = 1, i ≤ j − 1. Since νl (s∗ ) = 1 and l (s∗ ) = j, (s∗ ) = j − 1. Thus, (s) = j − 1. Note, finally, that s ∈ lj · · · li+1 q. Thus, there exists an element p in lj · · · li+1 such that s ∈ pq. From p ∈ lj · · · li+1 we obtain (p) ≤ j − i. From s ∈ pq and (s) = j − 1 we obtain j − i − 1 ≤ (p). Thus, j − i − 1 ≤ (p) ≤ j − i. Assume that (p) = j − i − 1. Then s ∈ S−1 (q). Thus, as q ∈ S−1 (L), q = s; cf. Lemma 3.4.5(ii). This contradicts q ∈ S−1 (L) and s ∈ S0 (l). Thus, we have (p) = j − i. Now we obtain from (q) = i and (s) = j − 1 that (s) + 1 = (p) + (q). Recall that dL is our notation for the smallest element in (S−1 (L)). For the remainder of this section, we shall write d instead of dL .


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Corollary 10.4.4 Let s be an element in L such that (s) ≤ d − 1. Then the following hold. (i) For each element l in L, s ∈ / S0 (l). (ii) There exists an element l in L such that s ∈ / S−1 (l). Proof. (i) This is an immediate consequence of Proposition 10.4.3. (ii) Since (s) ≤ d − 1, s ∈ / S−1 (L). Thus, there exists an element l in L with s∈ / S−1 (l). Let l be an element in L. We define S∞ (l) := {s ∈ L | (sl) = {(s)}}. Note that L is the disjoint union of the sets S−1 (l), S∞ (l), and S1 (l). Note also that S∞ (l) ⊆ S0 (l). We shall use this latter observation frequently without further mentioning. Lemma 10.4.5 For any two elements l in L and s in S∞ (l), s l ⊆ S∞ (l). Proof. Let r be an element in s l. Then, by Lemma 2.1.4, r l = s l. Thus, |(r l)| = |(s l)| = 1. It follows that r ∈ S∞ (l). Lemma 10.4.6 Let s be an element in L such that, for each element r in

L, (r) ≤ (s). Then s ∈ S−1 (L) or there exists an element l in L such that s ∈ S∞ (l). Proof. Let l be an element in L. Then our hypothesis on (s) implies that, for each element r in s l, (r) ∈ {(s), (s) − 1}. It follows that s ∈ L−1 (l) or that s ∈ S∞ (l). The following lemma is similar to Lemma 10.4.1(ii). Lemma 10.4.7 Let l be an element in L, and let q be an element in S∞ (l). Let p be an element in L such that q ∈ Lp and (q) = 1 + (p). Then, for each element r in pl, we have r ∈ S−1 (L) or (r) = (p). Proof. Let r be an element in pl. Then, by Lemma 1.3.3(i), p ∈ rl. Similarly, as q ∈ Lp, p ∈ Lq; cf. Lemma 1.3.3(ii). Thus, p ∈ Lq ∩ rl. Thus, by Lemma 1.3.4, Lr ∩ ql is not empty. Let s be an element in Lr ∩ ql. From q ∈ Lp and (q) = 1 + (p) we obtain q ∈ S−1 (p). Thus, by Lemma / S−1 (L). 10.4.1(i), p ∈ / S−1 (l). In particular, p ∈ / S−1 (l). In particular, q ∈ / S−1 (L). Since q ∈ S∞ (l), q ∈

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Since q ∈ S∞ (l) and s ∈ ql, s ∈ S∞ (l); cf. Lemma 10.4.5, Thus, s ∈ / S−1 (l). In particular, s ∈ / S−1 (L). Since q ∈ S∞ (l) and s ∈ ql, (q) = (s). Thus, by Lemma 10.3.4, r ∈ S−1 (L) or (p) = (r). For each element l in L, we define d S∞ (l) := {s ∈ S∞ (l) | (s) = d}. d It may happen that there exists an element l in L such that S∞ (l) is empty. Indeed, it will be an immediate consequence of Proposition 10.4.9(i) that, if d (l) with l ∈ L both elements in L have finite valency, one of the two sets S∞ must be empty.

Later, in Section 10.6, we shall see that, if L has finite valency and S−1 (L) d (l) (with l ∈ L) decide on the structure exactly one element, then the sets S∞ of L. d Lemma 10.4.8 Let l be an element in L, and let us assume that S∞ (l) is not empty. Then there exists an element s in S−1 (L) such that (s) = d and, for each element r in s l \ S−1 (L), (r) = d − 1. d Proof. We are assuming that S∞ (l) is not empty. Let q be an element in d S∞ (l). Then (q) = d. Thus, as 2 ≤ d, there exists an element p in L such that q ∈ Lp and (q) = 1 + (p); cf. Lemma 3.1.2. Thus, by Lemma 10.4.7, pl ∩ S−1 (L) is not empty or p ∈ S∞ (l).

Since (q) = d and (q) = 1 + (p), (p) = d − 1. Thus, by Corollary 10.4.4(i), / S∞ (l). Thus, by the above dichotomy, the set p ∈ / S0 (l). In particular, p ∈ pl ∩ S−1 (L) is not empty. Let s be an element in pl ∩ S−1 (L). From s ∈ pl we obtain (s) ≤ (p) + 1, from s ∈ S−1 (L) we obtain d ≤ (s). Thus, as (p) = d − 1, (s) = d. From p ∈ sl we obtain s l = p l; cf. Lemma 2.1.4. Since q ∈ Lp, (q) = 1 + (p), and q ∈ S∞ (l), we may apply Lemma 10.4.7. Thus, for each element r in pl \ S−1 (L), (r) = (p) = d − 1. Proposition 10.4.9 Let h and k be elements in L such that h = k. Assume d that nh and nk are finite and that S∞ (h) is not empty. Then the following hold. (i) We have nh + 1 ≤ nk . (ii) The integer d is odd.


225

d Proof. Let s be an element in S∞ (h), let y be an element in X, and let z be an element in ys. Let l be an element in L such that (s) = l (s). Then νl (s) ≤ 0.

For each element x in z h, we set Wx := yl ∩ xLd−1 . We shall first prove that, for each element x in z h, Wx is not empty. Let x be an element in z h, and let us denote by r the element in L which satisfies x ∈ yr. From x ∈ z h and z ∈ ys we obtain x ∈ ys h. Thus, as x ∈ yr, r ∈ s h. Thus, as νl (s) ≤ 0, νl (r) ≤ 0; cf. Lemma 10.3.3. Thus, (r) = l (r). d (h) and r ∈ s h, (r) = d. Thus, r ∈ lLd−1 . On the other hand, as s ∈ S∞ d−1 Thus, as x ∈ yr, x ∈ ylL , and that implies that Wx is not empty.

We now shall prove that, for any two elements x and x in z h, ∅ = Wx ∩ Wx implies x = x . Let x and x be elements in z h such that Wx ∩ Wx is not empty, and let w be an element in Wx ∩ Wx . Since w ∈ xLd−1 , x ∈ wLd−1 . Thus, there exists an element q in Ld−1 such that x ∈ wq. Thus, as w ∈ yl, x ∈ ylq. Thus, there exists an element r in lq such that x ∈ yr. Then r ∈ S−1 (q). From z ∈ x h and x ∈ yr we obtain z ∈ yr h. Thus, as z ∈ ys, s ∈ r h. / S−1 (h); cf. Lemma Thus, as s ∈ S∞ (h), r ∈ S∞ (h). Thus, as r ∈ S−1 (q), q ∈ 10.4.1(i). Since (q) ≤ d − 1, q ∈ / S0 (h); cf. Corollary 10.4.4(i). / S−1 (h) Let q be the element in Ld−1 such that x ∈ wq . Then, similarly q ∈ / S0 (h). However, as x ∈ x z, q ∈ q h. Thus, x = x. and q ∈ Now we conclude from

Wx ⊆ yl

x∈zh

that nh + 1 = |z h| ≤ |yl| = nl . It follows that h = l. (Recall that we are assuming that h and k have finite valency.) This means that l = k, whence (s) = k (s). This means that 0 ≤ νh (s). On the other hand, we obtain from s ∈ S∞ (h) that s ∈ S0 (h), and this implies νh (s∗ ) = 1; cf. Lemma 10.4.2. Thus, we have 0 ≤ νh (s∗ )νh (s). Now we obtain from Lemma 10.3.2(ii) that d / S−1 (L).) is odd. (Recall that s ∈ S∞ (h), so that s ∈

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Assume that each of the two elements in L has finite valency. Then we obtain d (l) with l in L must be from Proposition 10.4.9(i) that one of the two sets S∞ empty.

10.5 The Constrained Spherical Case Throughout this section, the set L is assumed to be spherical. We investigate the case where L is constrained. Instead of dL we shall write d. Recall that L is called constrained if, for any two elements q in L and p in S1 (q), 1 = |pq|. Lemma 10.5.1 Assume L to be constrained, let j be an integer with 0 ≤ j ≤ d, and let l be an element in L. Then |Rj (l)| = 1. Proof. By definition, R0 (l) = {1}. Let us assume that |Rj−1 (l)| = 1. We shall see that |Rj (l)| = 1. Assuming that |Rj−1 (l)| = 1 we obtain an element p in Rj−1 such that {p} = Rj−1 (l). From p ∈ Rj−1 (l) and j − 1 ≤ d we obtain (p) = j − 1; cf. Lemma 10.2.6(ii). Let k be an element in L such that Rj (l) = Rj−1 (l)k. Then, as {p} = Rj−1 (l), Rj (l) = pk. Let q be an element in pk. Then q ∈ Rj (l). Thus, as we are assuming that j ≤ d, (q) = j; cf. Lemma 10.2.6(ii). Thus, as (p) = j − 1 and q ∈ pk, p ∈ S1 (k). Thus, as L is assumed to be constrained, |pk| = 1. Thus, as Rj (l) = pk, |Rj (l)| = 1. Corollary 10.5.2 Assume L to be constrained. Then the following hold. (i) For each element l in L, Rd (l) = S−1 (L). (ii) We have |S−1 (L)| = 1. Proof. (i) From Corollary 10.1.4 we know that, for each element l in L, S−1 (L) ⊆ Rd (l). However, Lemma 10.5.1 tells us that, for each element l in L, |Rj (l)| = 1. Thus, the claim follows from the hypothesis that S−1 (L) is not empty. (ii) Considering Lemma 10.5.1 this follows from (i). Lemma 10.5.3 Assume L to be constrained. Then, for each element s in

L, (s) ≤ d. Proof. It is enough to show that there is no element in L of length d + 1. By way of contradiction, we assume that L contains an element s such that (s) = d + 1. Then, by Lemma 3.1.2, there exist elements r in L and l in

10.5 The Constrained Spherical Case

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L such that s ∈ rl and (s) = (r) + 1. From s ∈ rl and (s) = (r) + 1 we obtain r ∈ S1 (l). From d + 1 = (s) and (s) = (r) + 1 we obtain (r) = d. Thus, there exists an element l in L such that s ∈ Rd (l); cf. Lemma 10.1.3. Thus, by Corollary 10.5.2(i), r ∈ S−1 (L), contradiction. Let j be an element in {0, . . . , d}, and let l be an element in L. From Lemma 10.5.1 we know that |Rj (l)| = 1. In the following, we shall denote by rj (l) the uniquely determined element in Rj (l). Theorem 10.5.4 The following statements are equivalent. (a) We have 2d = | L|. (b) The set L is constrained. (c) We have {rj (l) | l ∈ L, j ∈ {0, . . . , d}} = L. Proof. (a) ⇒ (b) Let q be an element in L, and let p be an element in S1 (q). We have to show that 1 = |pq|. Since p ∈ S1 (q), there exists an element r in pq such that (r) = (p) + (q). Since we are assuming that 2d = | L|, (r) ≤ d; cf. Lemma 10.2.1. Thus, 1 = |pq|. (b) ⇒ (c) Assume L to be constrained. Then, for each element s in L, (s) ≤ d; cf. Lemma 10.5.3. Thus, the desired equation follows from Lemma 10.1.3. (c) ⇒ (a) By definition, we have that, for any two elements h and k in L, r0 (h) = r0 (k). From Corollary 10.5.2 we also know that, for any two elements h and k in L, rd (h) = rd (k). Thus, the equation in (c) yields | L| ≤ 2d, so that we obtain the desired equation in (a) from Lemma 10.2.1. Theorem 10.5.5 If L is constrained, L satisfies the exchange condition. Proof. Let us assume L to be constrained, let us fix elements h, k in L and s in S1 (k), and let us assume that h ∈ S1 (s). We have to show that hs = sk or that hs ⊆ S1 (k). From h ∈ S1 (s) we obtain an element t in hs such that (t) = 1 + (s). Similarly, as s ∈ S1 (k), there exists an element u in sk such that (u) = (s) + 1. From t ∈ hs, h ∈ S1 (s), and the hypothesis that L is constrained we obtain {t} = hs. Similarly, we obtain {u} = sk. Assume first that (s) = d − 1. Then (t) = d = (u). Thus, by Lemma 10.5.1, t = u. Thus, as {t} = hs and {u} = sk, hs = sk.

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Assume now that (s) ≤ d − 2, and let us fix an element r in tk. Then r ∈ hsk = hu. Thus, as s ∈ ht ∩ uk and (s) + 1 = (u), (t) + 1 = (r); cf. Lemma 10.3.5. It follows that hs = {t} ⊆ S1 (k).

10.6 Dihedral Closed Subsets of Finite Valency Throughout this section, L is assumed to have finite valency. Assuming L to have finite valency we obtain that L is finite, so that, according to Lemma 10.1.7, L is spherical. Instead of dL we shall write d. Lemma 10.6.1 Let h and k be elements in L such that h = k, and assume

L to have finite valency. Let R be the set of all elements s of S such that (s) ≤ d. Then the following hold. (i) Assume d to be odd, and let c be the integer satisfying d = 2c + 1. Then (nh + 1)((nk + 1)

c−1

(nh nk )j + (nh nk )c ) ≤ nR

j=0

with equality if and only if L is constrained. (ii) Assume d to be even, and let c be the integer satisfying d = 2c. Then (nh + 1)(nk + 1)

c−1

(nh nk )j ≤ nR

j=0

with equality if and only if L is constrained. Proof. For any two elements j in {0, . . . , d} and l in L, we have Rj (l) ⊆ R. Moreover, the sets R1 (h), . . . , Rd (h), R0 (k), . . . , Rd−1 (k) are pairwise disjoint; cf. Lemma 10.2.1. Thus, d j=1

nRj (h) +

d−1

nRj (k) ≤ nR .

j=0

Let us now fix an integer j with 0 ≤ j ≤ d − 1 and compute nRj (h) . If j is odd and i the integer satisfying j = 2i + 1, we have nRj (h) = (nh nk )i nh . If j is even and i the integer satisfying j = 2i, we have

10.6 Dihedral Closed Subsets of Finite Valency

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nRj (h) = (nh nk )i . The values nRj (k) are computed similarly. This proves the lemma. We are assuming L to have finite valency. As a consequence, L is a finite set. In particular, ( L) has a maximal element. In the following, we shall denote this element by max ( L). For the remainder of this section, we shall focus on the case where |S−1 (L)| = 1. In this case, we shall obtain precise information about the structure of L. Let l be an element in L. Recall that S∞ (l) is our notation for the set of all d (l) denotes the elements s in L such that (sl) = {(s)}. Recall also that S∞ set of all elements s in S∞ (h) with (s) = d. Proposition 10.6.2 If |S−1 (L)| = 1, d = max ( L). Proof. Let us assume that |S−1 (L)| = 1, and let us denote by m the element in S−1 (L). Since {m} = S−1 (L), (m) = d. We set e := max ( L). Then d ≤ e, and our claim is that d = e. In order to prove d = e we assume, by way of contradiction, that d + 1 ≤ e. Let s be an element in L such that (s) = e. Then, as (m) = d ≤ e − 1, / S−1 (L). Thus, by Lemma 10.4.6, there m = s. Thus, as {m} = S−1 (L), s ∈ exists an element h in L such that s ∈ S∞ (h). From s ∈ S∞ (h) we obtain s ∈ S0 (h). Thus, by Proposition 10.4.3, there exists an element p in L such that s ∈ pm and e + 1 = (p) + d. Since (p) = e − d + 1, (p∗ ) = e − d + 1. Thus, there exist elements l1 , . . . , le−d+1 in L such that p∗ ∈ l1 · · · le−d+1 and, for each element i in {2, . . . , e − d + 1}, li−1 = li . Since p∗ ∈ l1 · · · le−d+1 and y ∈ xp∗ , there exist elements y0 , . . . , ye−d+1 in X such that ye−d+1 ∈ y0 p∗ and, for each element i in {1, . . . , e − d + 1},

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yi ∈ yi−1 li ; cf. Lemma 1.3.9. From {m} = S−1 (L) we conclude that m∗ = m; cf. Lemma 10.2.8. Thus, as s ∈ pm, p ∈ sm; cf. Lemma 1.3.3(i). Thus, as y0 ∈ ye−d+1 p, y0 ∈ ye−d+1 sm. Thus, there exists an element z0 in ye−d+1 s such that y0 ∈ z0 m. For each element i in {0, . . . , e − d + 1}, we denote by si0 the element in L which satisfies z0 ∈ yi si0 . (This implies s00 = m and se−d+1,0 = s.) Let i be an element in {1, . . . , e − d + 1}. Then, as z0 ∈ yi si0 and yi ∈ yi−1 li , z0 ∈ yi−1 li si0 . Thus, as z0 ∈ yi−1 si−1,0 , si−1,0 ∈ li si0 . For each element i in {1, . . . , e − d + 1}, we set s0i := s∗i0 . Our first claim is now that there exist elements z1 , . . . , ze−d+1 in X such that, for each element i in {1, . . . , e − d + 1}, zi ∈ y0 s0i ∩ zi−1 li . Recall that m∗ = m. Thus, as m = s00 , s∗00 = s00 . Thus, as s00 ∈ l1 s10 , s00 ∈ s01 l1 ; cf. Lemma 1.3.2(iii). Thus, as z0 ∈ y0 s00 , z0 ∈ y0 s01 l1 . Thus, there exists an element z1 in y0 s01 ∩ z0 l1 . This proves our claim for i = 1. Let us now fix an element i in {2, . . . , e−d+1}, and let us assume, by induction hypothesis, that there exists an element zi−1 in X with zi−1 ∈ y0 s0,i−1 ∩ zi−2 li−1 . Since si−1,0 ∈ li si0 , s0,i−1 ∈ s0i li . Thus, as zi−1 ∈ y0 s0,i−1 , zi−1 ∈ y0 s0i li . Thus, there exists an element zi in y0 s0i ∩ zi−1 li , so that we have shown our first claim. From s00 ∈ l1 s10 and m = s00 we obtain m ∈ l1 s10 . Thus, as m∗ = m, s01 ∈ ml1 ; cf. Lemma 1.3.3(iii). Thus, as m ∈ S−1 (L), (s01 ) ≤ d. Thus, for each element i in {1, . . . , e − d + 1}, (s0i ) = d − 1 + i; cf. Lemma 3.4.1. It follows that, for each element i ∈ {1, . . . , e − d + 1},


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(si0 ) = d − 1 + i. (Recall that, for each element i in {1, . . . , e − d + 1}, si0 := s∗0i .) d (h); Recall that se−d+1,0 = s ∈ S∞ (h). Thus, as se−d+1,0 ∈ S−1 (s10 ), s10 ∈ S∞ d cf. Lemma 10.4.1(ii). In particular, S∞ (h) is not empty, so that, according to Proposition 10.4.9(ii), d is odd. Moreover, Lemma 10.4.8 yields

(m h \ {m}) = {d − 1}. d Since s10 ∈ S∞ (h), s10 = m. Thus, as s01 ∈ s00 l1 ⊆ m l1 and (s01 ) = d, the last equation implies h = l1 .

For any two elements i and j in {1, . . . , e − d + 1}, we define sij to be the element in L which satisfies zj ∈ yi sij . Suppose that s11 = m. Then, as h = l1 , s12 ∈ m h. Thus, by the above equation, s12 = m or (s12 ) = d − 1. If s12 = m, s21 ∈ S−1 (L), s20 ∈ s21 L, (s21 ) = d, and (s20 ) = d + 1, contradiction. If (s12 ) = d − 1, (s02 ) ≤ d, contradiction. Therefore, we must have s11 = m. Assume that (s11 ) = d − 1. Since d is odd, d − 1 is even. Thus, s11 ∈ Rd−1 (l1 ) or s∗11 ∈ Rd−1 (l1 ). In the first case, s01 ∈ l1 s11 ⊆ l1 Rd−1 (l1 ) ⊆ Rd−2 (h) ∪ Rd−1 (l1 ), which yields (s01 ) ≤ d − 1, contradiction. In the second case, we obtain, similarly, the contradiction (s10 ) ≤ d − 1. Thus, we have (s11 ) = d − 1. From (s10 ) = d and s11 ∈ s10 L we now conclude that (s11 ) ∈ {d, d + 1}. Assume that (s11 ) = d + 1. Then {s∗11 , s11 } ⊆ S−1 (l1 ). Therefore, by Lemma 10.3.1, νl1 (s∗11 ), νl1 (s11 ) ≤ 0. It follows that 0 ≤ νl1 (s∗11 )νl1 (s11 ). On the other hand, as d + 1 is even, Lemma 10.3.2(ii) yields νl1 (s∗11 )νl1 (s11 ) ≤ 0. Thus, νl1 (s∗11 )νl1 (s11 ) = 0, which means that s11 ∈ S−1 (L). It follows that s11 = m. This contradiction yields (s11 ) = d. We claim that, more generally, for each element i in {1, . . . , e − d + 1}, (si1 ) = d − 1 + i.

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The case i = 1 has just been proven. By induction, we may assume that (si−1,1 ) = d − 1 + i − 1. We have that (si−1,0 ) = d − 1 + i − 1 and that (si0 ) = d − 1 + i. Thus, as si−1,0 ∈ Lsi0 ∩ si−1,1 L and si1 ∈ Lsi−1,1 ∩ si0 L, (si1 ) = d − 1 + i; cf. Lemma 10.3.4. It follows that (se−d+1,1 ) = e. Let p be the element in S satisfying y1 ∈ ye−d+1 p . Then (p ) = e − d and p ∈ S1 (s11 ). Similarly, let q be the element in S satisfying ze−d+1 ∈ y1 q . Then (q ) = e − d and s11 ∈ S1 (q ). From p ∈ S1 (s11 ) and s11 ∈ S1 (q ) we obtain an element t in p s11 q such that (t) = (p ) + (s11 ) + (q ); cf. Proposition 10.3.6. Thus, as (p ) = e − d, (s11 ) = d, and (q ) = e − d, we obtain (t) = 2e − d. It follows that (s) + 1 = e + 1 ≤ 2e − d = (t), contrary to the (maximal) choice of s. Corollary 10.6.3 Assume that |S−1 (L)| = 1. Then, for each element h in d (k). L, there exists an element k in L such that Rd (h) ⊆ S−1 (L) ∪ S∞ Proof. Let h be an element in L, and let s be an element in Rd (h). Then, by Lemma 10.2.6(ii), (s) = d. Thus, by Proposition 10.6.2, (s) = max ( L), so that the claim follows from Lemma 10.4.6. Proposition 10.6.4 Assume that |S−1 (L)| = 1 and that L is not constrained. Then there exist elements h and k in L such that h = k, Rd (h) = S−1 (L), d d (h), and S∞ (h) ⊆ S−1 (L)k. Rd (k) = S−1 (L) ∪ S∞ d Proof. Let us assume that, for each element l in L, S∞ (l) is empty. Then, for each element l in L, S−1 (L) = Rd (l); cf. Corollary 10.6.3. Thus, we obtain from Lemma 10.2.2 and Lemma 10.2.1 that 2d = | L|. From 2d = | L| we obtain that L is constrained; cf. Theorem 10.5.4. d (h) is Thus, we may assume that there exists an element h in L such that S∞ not empty. Then, by Proposition 10.4.9(ii), d is odd. Thus, for each element l in L, Rd (l) ⊆ S−1 (l); cf. Lemma 10.2.7(i).

Let us denote by k the element in L \ {h}.


233

From Rd (h) ⊆ S−1 (h) we obtain that Rd (h) ∩ S∞ (h) is empty, from Rd (k) ⊆ S−1 (k) that Rd (k) ∩ S∞ (k) is empty. Thus, by Corollary 10.6.3, Rd (h) ⊆ d (k) and S−1 (L) ∪ S∞ d (h). Rd (k) ⊆ S−1 (L) ∪ S∞ d From Proposition 10.4.9(i) we obtain that S∞ (k) is empty. Thus, as Rd (h) ⊆ d S−1 (L) ∪ S∞ (k) and S−1 (L) ⊆ Rd (h),

Rd (h) = S−1 (L). d d In order to prove that S∞ (h) ⊆ S−1 (L)k we fix an element in S∞ (h) and call it s. d (h) we obtain s∗ ∈ Rd (h) ∪ Rd (k). Thus, as Rd (k) ⊆ S−1 (L) ∪ From s ∈ S∞ d d d (h). From s∗ ∈ S∞ (h) we obtain s∗ ∈ S∞ (h) and Rd (h) ⊆ S−1 (L), s∗ ∈ S∞ S0 (h). Thus, by Proposition 10.4.3, there exists an element l in L such that s∗ ∈ lS−1 (L). Thus, by Lemma 1.3.2(iii), s ∈ S−1 (L)l. d (h) we obtain s ∈ / S−1 (L)h. Thus, s ∈ S−1 (L)k. From s ∈ S∞

Proposition 10.6.5 Assume that |S−1 (L)| = 1. Then we have the following. (i) For each element s in L, there exist elements j in {0, . . . , d} and l in L such that s ∈ Rj (l). (ii) Let j be an element in {0, . . . , d − 1}, and let l be an element in L. Then |Rj (l)| = 1. Proof. (i) Let s be an element in L. Set j := (s), and let l be an element in L such that (s) = l (s). Then s ∈ Rj (l) and, by Proposition 10.6.2, j ≤ d. (ii) According to Lemma 10.5.1, we may assume that L is not constrained. By Lemma 10.2.2, it suffices to show that |Rd−1 (l)| = 1. Let r be an element in Rd−1 (l). Then, by Lemma 10.2.6(ii), (r) = d − 1. Thus, by Corollary 10.4.4, there exists an element s in rL such that (s) = d. We are assuming that |S−1 (L)| = 1. Thus, there exists an element m in S−1 (L) such that {m} = S−1 (L). We wish to prove that r ∈ mL. If s = m, this follows from s ∈ rL. Therefore, we assume that s = m. If s = m, there exist elements h and k in L such that h = k, s ∈ S∞ (h), and s ∈ mk; cf. Proposition 10.6.4. (Recall that (s) = d and that L is not constrained.) From s ∈ S∞ (h), r ∈ sL, and (r) + 1 = (s) we conclude that r ∈ sk. Thus, as s ∈ mk, r ∈ mk; cf. Lemma 2.1.4. In particular, r ∈ mL. Since r has been chosen arbitrarily in Rd−1 (l) , we have shown that Rd−1 (l) ⊆ mL. Now Corollary 10.4.4(i) yields |Rd−1 (l)| = 1.

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Recall that, according to Theorem 10.5.4, L is constrained if and only if 2d = | L|. In this case, L satisfies the exchange condition; cf. Theorem 10.5.5. Assume that |S−1 (L)| = 1. We call L a Moore set if there exist elements h d (h), and and k in L such that h = k, Rd (h) = S−1 (L), Rd (k) = S−1 (L) ∪ S∞ d S∞ (h) ⊆ S−1 (L)k. Assume that L is a Coxeter set. Then, by Corollary 10.5.2(ii), |S−1 (L)| = 1. By definition, this equation also holds if L is a Moore set. The following theorem is the main result of this chapter. (Recall that L is assumed to have finite valency.) Theorem 10.6.6 Assume that |S−1 (L)| = 1. Then L is a Coxeter set or a Moore set. Proof. If L is constrained, L is a Coxeter set; cf. Theorem 10.5.5. If L is not constrained, we obtain from Proposition 10.6.4 elements h and k in L such that d d h = k, Rd (h) = S−1 (L), Rd (k) = S−1 (L) ∪ S∞ (h), and S∞ (h) ⊆ S−1 (L)k. Much more can be said in either of the two cases of Theorem 10.6.6. The first case of Theorem 10.6.6 will be investigated in more detail in Section 12.4. In Theorem 12.4.6, for instance, we shall see that, if L is a Coxeter set,

L is thin or d ∈ {2, 3, 4, 6, 8, 12}. More details about the valencies of the elements in L will be given in each of the six cases in the subsequent six theorems of Section 12.4. Similar investigations have been made about the second case of Theorem 10.6.6. In fact, in this case, one can show that d ∈ {3, 5}. However, the proof of this latter fact is beyond the scope of this monograph. It is based on modular representation theory of finite schemes and follows from [13; Theorem 3], [6; Theorem 2], and [14; Theorem 1]. In the introduction to this chapter, we mentioned a representation theoretic condition which implies |S−1 (L)| = 1. Let us now, at the end of this section, look at this condition. Theorem 10.6.7 Assume that there exists an algebraically closed field C of characteristic 0 such that C[L] = C L. Then |S−1 (L)| = 1. Proof. We may assume that L = S, so that C[L] = CS. Let h and k be elements in L such that h = k. We first shall prove that |S| ≤ |S/ h| + |S/ k|


235

and | |S/ h| − |S/ k| | ≤ 1. For each element i in {1, 2}, we define Mi to be the sum of the i-dimensional submodules of the CS-module CS. Then, by Lemma 9.4.5, CS = M1 ⊕ M2 . Let l be an element in L. Set σl := σlC , and define Cl := {σ ∈ CS | σσl = nl σ}. Let σ be an element in Cl . Then there exist elements σ1 in M1 and σ2 in M2 such that σ = σ1 + σ2 . Since σ ∈ Cl , σσl = nl σ = nl (σ1 + σ2 ) = nl σ1 + nl σ2 . On the other hand, σσl = (σ1 + σ2 )σl = σ1 σl + σ2 σl . Thus, as CS = M1 ⊕ M2 , we obtain σ1 σl = nl σ1 and σ2 σl = nl σ2 . Since σ ∈ CS has been chosen arbitrarily, we have shown that Cl = (M1 ∩ Cl ) ⊕ (M2 ∩ Cl ). Let i be an element in {1, 2}. We set bi := dimC (Mi ), and, for each element l in L, we define ali := dimC (Mi ∩ Cl ). Since CS = M1 ⊕ M2 , this notation yields |S| = b1 + b2 . Moreover, for each element l in L, |S/ l| = al1 + al2 . (Recall that Cl = (M1 ∩ Cl ) ⊕ (M2 ∩ Cl ), and use Theorem 9.4.7.) From Corollary 9.4.4(ii) we obtain 1 b2 = al2 2 for each element l in L. From Corollary 9.4.4(ii) we also deduce that

236


b1 ≤

al1 .

l∈L

(Recall that CS possesses a 1-dimensional right ideal which affords the character 1CS .) Thus, we have (al1 + al2 ) = |S/ l|. |S| ≤ l∈L

l∈L

From Corollary 9.4.4(ii) we finally deduce that, for each element l in L, al1 ∈ {1, 2}. Thus, | |S/ h| − |S/ k| | = |ah1 − ak1 | ≤ 1. Let us now define C :=

S/ l,

l∈L

r := {(s h, s k) | s ∈ S}, and Γ := (C, r). Then, by Theorem 9.4.1, the graph Γ is connected. Note that Γ is not a tree. Thus, |C| ≤ |r|. Moreover, the definition of r yields |r| ≤ |S|. Finally, as |S| ≤ |S/ h| + |S/ k|, we have |S| ≤ |C|. It follows that |C| = |r| = |S|. In particular, for each element s in S, s h∩s k = {s}. Therefore, we conclude that |S−1 (L)| = 1. Theorem 10.6.6 (together with Corollary 10.5.2(ii)) says that Coxeter sets as well as Moore sets are characterized by the equation |S−1 (L)| = 1. The following corollary is a representation theoretic characterization of Coxeter sets and Moore sets. It was proved first in [42; Theorem B]. Here, we obtain it as a corollary of Theorem 10.6.6 and Theorem 10.6.7. Corollary 10.6.8 Assume that there exists an algebraically closed field C of characteristic 0 with C[L] = C L. Then L is a Coxeter set or a Moore set.

11 Coxeter Sets

In this chapter, we start to look more thoroughly at Coxeter sets. In order to do so we first recall the definition of a Coxeter set. We fix a set of involutions of S and call it L. Instead of L we shall write . Recall that, for each element q in L, S1 (q, L) is our notation for the set of all elements p in L such that pq contains an element r satisfying (r) = (p) + (q). In accordance with Section 3.4 we shall write, for each element s in L, S1 (s) instead of S1 (s, L). Recall that L is called constrained if, for any two elements q in L and p in S1 (q), 1 = |pq|. Recall also that L is said to satisfy the exchange condition if, for any three elements h, k in L and s in S1 (k), h ∈ S1 (s) implies hs ⊆ sk ∪ S1 (k). Recall, finally, that L is called a Coxeter set if L is constrained and satisfies the exchange condition. For the remainder of this chapter, we assume L to be a Coxeter set. It is the goal of this chapter to provide global information about the structure of L. In the first section, we compile a few general facts about closed subsets of

L generated by subsets of L. Such closed subsets are called parabolic. The section can be viewed as a continuation of Section 3.6. In the second section, we show that, if L is a finite set and does not contain thin elements, the closed subset generated by L is a direct product of simple closed subsets each of which is generated by the elements of L which it contains. In the third section of this chapter, we investigate faithful maps defined on subsets of x L with x ∈ X. Recall that a map χ from a subset Y of X to X is called faithful if, for any three elements v, w in Y and s in S, w ∈ vs implies wχ ∈ vχs. The goal of the last of the four sections of this chapter is the proof of a specific extension theorem (Theorem 11.4.6) for Coxeter sets.

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By an extension theorem we mean a theorem which guarantees that, given subsets Y and Z of X, a faithful map χ from Y of X extends faithfully to a map from Z of X. In Theorem 11.4.6, we deal with a case where the set Z contains only one element which is not in Y . This theorem will be crucial in the proof of Theorem 12.3.4.

11.1 Parabolic Subsets Throughout this section, the letter K stands for a subset of L. Lemma 11.1.1 Let p be an element in S1 (K), and let q be an element in p K. Then q ∗ ∈ S−1 (p∗ ). Proof. We are assuming that q ∈ p K. Thus, there exists an element s in

K such that q ∈ ps. We are assuming that p ∈ S1 (K). Thus, by Theorem 3.6.4, p ∈ S1 ( K). Thus, as s ∈ K, p ∈ S1 (s). Thus, as q ∈ ps and L is assumed to be constrained, (q) = (p) + (s). From q ∈ ps, we obtain q ∗ ∈ s∗ p∗ . From (q) = (p) + (s) we obtain (q ∗ ) = (s∗ ) + (p∗ ). Thus, q ∗ ∈ S−1 (p∗ ). Lemma 11.1.2 We have L \ K ⊆ S1 ( K). Proof. Let l be an element in L \ K. Then, as L is assumed to satisfy the exchange condition, l ∈ S1 (K). Thus, by Theorem 3.6.4, l ∈ S1 ( K). Thus, by Lemma 3.4.4(i), K ⊆ S1 (l). Since l has been chosen arbitrarily in L \ K, we have shown that K ⊆ S1 (L \ K). Thus, by Theorem 3.6.4, K ⊆ S1 ( L \ K). Thus, by Lemma 3.4.4(i), L \ K ⊆ S1 ( K). The following lemma generalizes Lemma 3.4.8 for Coxeter sets. Lemma 11.1.3 For each subset H of L, ( H ∩ S1 (K)) H ∩ K = H. Proof. From Lemma 11.1.2 we know that H ⊆ S1 (K \ H). Thus,

H ∩ S1 (H ∩ K) = H ∩ S1 (K). On the other hand, we obtain from Lemma 3.4.8 that

H ∩ S1 (H ∩ K) H ∩ K = H, and according to Lemma 2.2.1(ii), the left hand side of this equation is equal to ( H ∩ S1 (H ∩ K)) H ∩ K.

11.1 Parabolic Subsets

239

Lemma 11.1.4 Let p and q be elements in L, and let l be an element in L ∩ S1 (p) ∩ S1 (q). Then, if lp ⊆ lq K, p ∈ q K. Proof. Let us assume, by way of contradiction, that lp ⊆ lq K and that p∈ / q K. Assuming lp ⊆ lq K we must have p ∈ q K or p ∈ lq K. Thus, as p ∈ / q K, p ∈ lq K. From Lemma 3.4.8 we know that S1 (K) K = L. Thus, as q ∈ L, there exists an element u in S1 (K) such that q ∈ u K. From u ∈ S1 (K) and q ∈ u K we obtain q ∗ ∈ S−1 (u∗ ); cf. Lemma 11.1.1. On the other hand, we are assuming that l ∈ S1 (q), and that means that q ∗ ∈ S1 (l); cf. Lemma 3.4.4(i). Thus, by Lemma 3.4.4(ii), u∗ ∈ S1 (l), and this implies l ∈ S1 (u); cf. Lemma 3.4.4(i). Now recall that L is assumed to satisfy the exchange condition. Thus, we obtain from l ∈ S1 (u) and u ∈ S1 (K) that lu ⊆ uK or lu ⊆ S1 (K). However, as p ∈ lq K, q K = u K, and p ∈ / q K, we cannot have lu ⊆ uK. Thus, lu ⊆ S1 (K). Similarly, we obtain an element t in S1 (K) such that p ∈ t K, l ∈ S1 (t), and lt ⊆ S1 (K). Thus, referring to Theorem 3.6.4 we now obtain lt = lu. (Note that lt K = lu K.) Thus, by Lemma 3.5.1, t = u. Thus, as p ∈ t K and q ∈ u K, p ∈ q K, contradiction. Lemma 11.1.5 Let l be an element in L, and let s be an element in S1 (K). ∗ Assume there exists an element r in s K such that l ∈ S1 (r) \ Kr . Then ls ⊆ S1 (K). Proof. From r ∈ s K we obtain r K = s K. According to Lemma ∗ ∗ 1.3.2(iii), this implies Kr∗ = Ks∗ . Thus, by Lemma 1.3.7, Kr = Ks . ∗ ∗ / Ks . Thus, by definition, Thus, as we are assuming that l ∈ / Kr , l ∈ ∗ ∗ s l ⊆ Ks . Thus, by Lemma 1.3.2(iii), ls ⊆ s K. From r ∈ s K and s ∈ S1 (K) we obtain r∗ ∈ S−1 (s∗ ); cf. Lemma 11.1.1. Thus, by Lemma 3.4.4(iv), S1 (r) ⊆ S1 (s). Thus, as we are assuming that l ∈ S1 (r), l ∈ S1 (s). Thus, as L is assumed to satisfy the exchange condition, we have ls ⊆ S1 (K). Lemma 11.1.6 For any two elements y and z in X, |yS1 (K) ∩ z K| = 1. Proof. Let v and w be elements in yS1 (K) ∩ z K. Since v, w ∈ z K, w ∈ v K. Thus, there exists an element s in K such that w ∈ vs. Since v ∈ yS1 (K), there exists an element p in S1 (K) such that v ∈ yp. From w ∈ vs and v ∈ yp we obtain w ∈ yps. Thus, there exists an element q in ps such that w ∈ yq. Since p ∈ S1 (K), s ∈ K, and q ∈ ps, (q) = (p) + (s); cf. Theorem 3.6.4.

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From w ∈ yS1 (K) we, similarly, obtain (p) = (q) + (s∗ ). Thus, (s) = 0. It follows that 1 = s. Thus, as w ∈ vs, v = w. Lemma 11.1.7 Let x be an element in X, let l be an element in L, and let y and z be elements in x l such that y = z. Then, for each subset H of L, y H ∩ z K ⊆ x( H ∪ K). Proof. Let w be an element in y H ∩ z K. Since w ∈ z K, z ∈ w K. Thus, as w ∈ y H, z ∈ y H K. Thus, as z ∈ yl, l ∈ H K ⊆ H ∪ K. Thus, as l ∈ L, l ∈ H ∪ K; cf. Lemma 3.6.2(ii). Since w ∈ y H and y ∈ x l, w ∈ x l H. Similarly, w ∈ x l K. Thus, as l ∈ H ∪ K, we must have w ∈ x H or w ∈ x K. Thus, w ∈ x( H ∪ K).

11.2 Direct Products In this section, we shall see that L is the direct product of simple closed subsets each of which is generated by the elements of L which it contains. As a consequence, we shall also look at the case where L is simple. Theorem 11.2.1 Let H and K be subsets of L. Then the following hold. (i) We have H ∩ K = H ∩ K. (ii) If H = K, then H = K. Proof. (i) It is clear that H ∩ K ⊆ H ∩ K. Thus, we just have to show that H ∩ K ⊆ H ∩ K. Let us assume, by way of contradiction, that H ∩ K ⊆ H ∩ K. Among the elements in H ∩ K which are not in H ∩ K we fix an element s such that (s) is as small as possible. From s ∈ / H ∩ K we obtain 1 = s. From 1 = s we obtain elements r in L and l in L such that s ∈ rl and (s) = (r) + 1; cf. Lemma 3.1.2. From s ∈ rl and (s) = (r) + 1 we obtain s ∈ S−1 (l). Assume that l ∈ / H. Then, as s ∈ H, s ∈ S1 (l); cf. Lemma 11.1.2. This contradiction yields l ∈ H. Similarly, we obtain from s ∈ K that l ∈ K. It follows that l ∈ H ∩ K. From s ∈ rl we obtain r ∈ sl; cf. Lemma 1.3.3(i). Thus, as s ∈ H ∩ K and l ∈ H ∩ K, r ∈ H ∩ K. Thus, the (minimal) choice of s yields r ∈ H ∩ K. Thus, as s ∈ rl and l ∈ H ∩ K, s ∈ H ∩ K, contradiction. (ii) This follows from Lemma 3.6.2(ii). Theorem 11.2.2 Assume that L does not contain thin elements. Then K →

K is a bijective map from the power set of L to the set of all closed subsets of L.

11.2 Direct Products

241

Proof. From Corollary 3.5.4(ii) we know that, for each closed subset T of

L, there exists a subset K of L such that K = T . This shows that the map in question is surjective. Injectivity follows from Theorem 11.2.1(ii). Recall that, by Lemma 3.5.5(iii), L does not contain thin elements if and only if {1} = Oϑ ( L). Let P and Q be subsets of S, and let us assume Q to be not empty. Recall that CP (Q) is our notation for the set of all elements p in P such that, for each element q in Q, qp = pq. Theorem 11.2.3 For each subset K of L, the following conditions are equivalent. (a) For any two elements k in K and l in L \ K, kl = lk. (b) We have L \ K ⊆ CL ( K). (c) The closed subset K is normal in L. Proof. (a) ⇒ (b) Let p be an element in K, and let q be an element in

L \ K. We have to show that pq = qp. If 1 = p or 1 = q, the claim is obvious. Thus, we may assume that 1 = p and that 1 = q. From 1 = p we obtain elements t in K and h in K such that p ∈ th and (p) = (t) + 1; cf. Lemma 3.1.2. Similarly, we obtain elements k in L \ K and u in L \ K such that q ∈ ku and (q) = 1 + (u). From p ∈ th and (p) = (t) + 1 we obtain {p} = th. From q ∈ ku and (q) = 1 + (u) we obtain {q} = ku. We are assuming that hk = kh, and, by induction, we may assume that tk = kt, that hu = uh, and that tu = ut. Thus, pq = thku = tkhu = ktuh = kuth = qp. (b) ⇒ (c) Let s be an element in L. We have to show that Ks ⊆ s K. There is nothing to show if 1 = s. Therefore, we assume that 1 = s. From 1 = s we obtain elements r in L and l in L such that s ∈ rl and (s) = (r) + 1; cf. Lemma 3.1.2. From s ∈ rl and (s) = (r) + 1 we obtain r ∈ S1 (l). Thus, as L is assumed to be constrained, we conclude that {s} = rl. By induction, we have Kr ⊆ r K, and from (b) we obtain Kl ⊆ l K. Thus, as {s} = rl,

Ks = Krl ⊆ r Kl ⊆ rl K = s K. (c) ⇒ (a) Let k be an element in K and let l be an element in L \ K. We have to show that kl = lk.

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Since k ∈ K and l ∈ L \ K, k ∈ S1 (l); cf. Lemma 11.1.2. Thus, there exists an element s in kl such that (s) = (k) + (l) = 2. Since L is assumed to be constrained, we obtain from k ∈ S1 (l) and s ∈ kl that {s} = kl. Since K is assumed to be normal in L, we have kl ⊆ l K. Thus, s ∈ l K. Thus, there exists an element r in K such that s ∈ lr. From l ∈ L \ K and r ∈ K we obtain l ∈ S1 (r); cf. Lemma 11.1.2. From l ∈ S1 (r) and s ∈ lr we obtain (s) = 1 + (r) and {s} = lr. From (s) = 2 and (s) = 1 + (r) we obtain (r) = 1. Thus, r ∈ L. From s ∈ lr we obtain r ∈ ls; cf. Lemma 1.3.3(ii). Thus, as s ∈ kl, r ∈ lkl. In particular, r ∈ k, l. Thus, as r ∈ L, r ∈ {k, l}; cf. Lemma 3.6.2(ii). From kl = {s} = lr we obtain r = l. Thus, r = k. Thus, as kl = {s} = lr, kl = lk. Recall that a closed subset T of S different from {1} is called simple if {1} and T are the only normal closed subsets of T . Theorem 11.2.4 Assume that L is finite and does not contain thin elements. Then L contains subsets L1 , . . . , Ln such that {L1 , . . . , Ln } is a partition of L,

L1 × . . . × Ln = L1 , and, for each element i in {1, . . . , n}, Li is simple. Proof. Assume that L is not simple. Then, by definition, L possesses a normal closed subset T with {1} = T = L. Since we are assuming that L has no thin element, we obtain from Theorem 11.2.2 a subset K of L such that K = T . Since {1} = T , K is not empty, and that means that L\K = L. Since T = L, K = L, cf. Theorem 11.2.1(ii). Since T is normal in L and K = T , K is normal in L. Thus, by Theorem 11.2.3, L \ K is normal in L. From Theorem 11.2.1(i) we obtain {1} = K ∩ L \ K. Thus, by definition,

K × L \ K = L. Now the claim follows by induction. Theorem 11.2.4 tells us that, in order to investigate Coxeter sets without thin elements, it is enough to look at Coxeter sets which generate a simple closed subset. The following lemma is a result about simple closed subsets generated by a Coxeter set. It will be needed in the proof of Lemma 12.3.1. Lemma 11.2.5 Assume that 3 ≤ |L| and that L is simple. Let l be an element in L. Then there exists an element s in S1 (l) such that, for each subset K of L with |K| = 2, Ks ⊆ S1 (l).

11.3 Faithful Maps

243

Proof. Since {l} = L and L is assumed to be simple, we find an element k in L \ CS (l); cf. Theorem 11.2.3. Since L is assumed to have at least three elements and L is assumed to be simple, we find an element h in L \ {k, l} such that h ∈ / CS ( k, l); cf. Theorem 11.2.3. Let s be the element in hk, let K be a subset of L with |K| = 2, and let q be an element in L such that K ⊆ S1 (q) and Ks = Kq; cf. Lemma 3.4.8. From s ∈ Kq and K ⊆ S1 (q) we obtain s ∈ S−1 (q). From s ∈ hk and l∈ / {h, k} we obtain s ∈ S1 (l). Thus, S−1 (q) ∩ S1 (l) is not empty. Thus, by Lemma 3.4.4(ii), q ∈ S1 (l). Note also that hkl ⊆ Ks. Thus, by Lemma 11.1.5, K ∈ S1 (ql). Thus, by Lemma 3.4.6(ii), Kq ⊆ S1 (l). Thus, as Kq = Ks, Ks ⊆ S1 (l).

11.3 Faithful Maps Throughout this section, the letter K stands for a subset of L. Lemma 11.3.1 Let y be an element in X, let z be an element in yS1 (K), and let χ be a map from {y} ∪ z K to X. Then, if χ|{y,z} and χ|zK are faithful, so is χ. Proof. Let x be an element in z K. We shall be done if we succeed in showing that χ|{y,x} is faithful. Since x ∈ z K, there exists an element q in K such that x ∈ zq. Since we are assuming that z ∈ yS1 (K), there exists an element p in S1 (K) such that z ∈ yp. Since p ∈ S1 (K), p ∈ S1 ( K); cf. Theorem 3.6.4. Thus, as q ∈ K, p ∈ S1 (q). Thus, as χ|{y,z} and χ|{z,x} are faithful, χ|{y,x} is faithful; cf. Lemma 6.6.1. Lemma 11.3.2 Let x be an element in X, let H be a subset of L, and let χ be a map from x( H ∪ K) to X. Then, if χ|xH and χ|xK are faithful, so is χ. Proof. Let y be an element in x H. Then, x ∈ y H. Thus, by Lemma 11.1.3, x ∈ y( H ∩ S1 (K)) K. Thus, there exists an element z in y( H ∩ S1 (K)) such that x ∈ z K. Since z ∈ y H, χ|{y,z} is faithful. Since x ∈ z K and χ|xK is faithful, χ|zK is faithful. Thus, as z ∈ yS1 (K), χ|{y}∪xK must be faithful; cf. Lemma 11.3.1. Lemma 11.3.3 Let s be an element in L, and let l be an element in L ∗ with l ∈ S1 (s) \ Ks . Let x be an element in X, y an element in xl, and z an element in ys. Finally, let χ be a map from {x, y} ∪ z K to X. Then, if χ|{x,y} and χ|{y}∪zK are faithful, so is χ.

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Proof. By Lemma 3.4.8, there exists an element r in S1 (K) such that s ∈ ∗ r K. Thus, as we are assuming that l ∈ S1 (s)\ Ks , lr ⊆ S1 (K); cf. Lemma 11.1.5. Since z ∈ ys and s ∈ r K, z ∈ yr K. Thus, there exists an element w in yr such that z ∈ w K. Since w ∈ yr and y ∈ xl, w ∈ xlr. Thus, as lr ⊆ S1 (K), w ∈ xS1 (K). Since r in S1 (K) and s ∈ r K, s∗ ∈ S−1 (r∗ ). Thus, by Lemma 3.4.4(iv), S1 (s) ⊆ S1 (r). Thus, as as l ∈ S1 (s), l ∈ S1 (r). By hypothesis, χ|{x,y} and χ|{y,w} are faithful. (Note that w ∈ z K.) Thus, as y ∈ xl, w ∈ yr, and l ∈ S1 (r), we obtain that χ|{x,w} is faithful; cf. Lemma 6.6.1. On the other hand, as χ|zK is assumed to be faithful, we obtain from z K = w K that χ|wK is faithful. Thus, as w ∈ xS1 (K), χ|{x}∪wK is faithful; cf. Lemma 11.3.1. Thus, as w K = z K, χ|{x}∪zK is faithful. Lemma 11.3.4 Let H be a subset of L, and let s be an element in S1 (H) such that H ⊆ Ks . Let y be an element in X, let z be an element in ys H, and let z be an element in z H. Then, for each faithful map χ from y K ∪ {z} to X, there exists at most one faithful map χ from y K ∪ {z } to X such that χ |yK = χ|yK and z χ ∈ zχ H. Proof. Since we are assuming that z ∈ z H and z ∈ ys H, z ∈ ys H. On the other hand, we are assuming that H ⊆ Ks , and that means that s H ⊆ Ks. Thus, as z ∈ ys H, z ∈ y Ks. Thus, there exists an element y in y K with z ∈ y s. Let us fix a faithful map χ from y K ∪ {z } to X which satisfies χ |yK = χ|yK and z χ ∈ zχ H. Since y ∈ y K, χ is defined on y . Moreover, we have z ∈ y s. Thus, as χ is assumed to be faithful, we have z χ ∈ y χ s. On the other hand, as we are assuming that χ |yK = χ|yK , we have y χ = y χ. Thus, z χ ∈ y χs. Thus, as s ∈ S1 (H), z χ ∈ y χS1 (H). Thus, as z χ ∈ zχ H, the claim follows from Lemma 11.1.6.

11.4 The Extension Theorem The goal of this section is the proof of Theorem 11.4.6. In this theorem, we focus on faithful maps from certain subsets Y of X to X which extend faithfully to a subset of X containing Y and one additional element. The main idea of the proof of this theorem is the use of Corollary 11.4.3 in the proof of Proposition 11.4.5. Theorem 11.4.6 plays a crucial role in the proof of Lemma 12.3.1. In this section, we fix a nonempty set of subsets of L and call it K.

11.4 The Extension Theorem

245

Lemma 11.4.1 We have K∈K K = K∈K K. Proof. Let us denote by H the intersection of the elements in K and by T the intersection of the sets K where K ∈ K. We have to show that H = T . By way of contradiction, we assume that H = T . Then, as H ⊆ T , T ⊆

H. We pick an element t in T \ H, and we do this in such a way that (t) is as small as possible. Since 1 ∈ H and t ∈ / H, 1 = t. Thus, by Lemma 3.1.2, there exist elements s in L and l in L such that t ∈ sl and (t) = (s) + 1. It follows that t ∈ S−1 (l). Thus, as t ∈ T , l ∈ H; cf. Lemma 11.1.2. Thus, as t ∈ sl and t ∈ T , s ∈ T . Thus, as (t) = (s) + 1, the minimal choice of t forces s ∈ H. Thus, as t ∈ sl and l ∈ H, t ∈ H, contradiction. Here is a generalization of Lemma 11.4.1. Lemma 11.4.2 For each element s in L, s K∈K K = K∈K s K. Proof. Let us denote by R the set of all elements in L which do not satisfy the equation in question. By way of contradiction, we assume that R is not empty. We pick an element r in R, and we do this in such a way that (r) is as small as possible. By Lemma 11.4.1, 1 ∈ / R. Thus, as r ∈ R, 1 = r. Thus, by Lemma 3.1.2, there exist elements l in L and q in L such that r ∈ lq and (r) = 1 + (q). Let us denote by H the intersection of the elements in K. By Q we shall denote the intersection of the sets r K with K ∈ K. Then, as r ∈ R, r H = Q. Thus, as r H ⊆ Q, Q ⊆ r H. Thus, we find an element s in Q such that s∈ / r H. Let us first assume that s∗ ∈ S−1 (l). Then, there exists an element p in L such that s ∈ lp and (s) = 1 + (p). Thus, l ∈ S1 (p). Moreover, since r ∈ lq and (r) = 1 +(q), l ∈ S1 (q). On the other hand, for each element K in K, we have s ∈ r K. Thus, for each element K in K, p ∈ q K; cf. Lemma 11.1.4. Thus, as (r) = 1 + (q), the minimal choice of r forces p ∈ q H. It follows that s ∈ lp ⊆ lq H = r H, contrary to the choice of s. / S−1 (l). Then, by Lemma 3.4.7, s∗ ∈ S1 (l). Thus, Let us now assume that s∗ ∈ by Lemma 3.4.4(i), l ∈ S1 (s). Thus, there exists an element t in L such that {t} = ls. Since s ∈ Q, we have that, for each element K in K, s ∈ r K. Thus, for each element K in K, q ∈ lr ⊆ ls K = t K, and this is equivalent to t ∈ q K. Thus, as (r) = 1+(q), the minimal choice of r forces t ∈ q H. Thus, s ∈ lt ⊆ lq H = r H, and this contradicts our choice of s.

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Corollary 11.4.3 For each element s in L, K∈K Ks = K∈K Ks . Proof. Let us denote by H the intersection of the elements in K. Then, for each element K in K, Hs ⊆ Ks . Conversely, let r be an element in L such that, for each element K in K, r ∈ Ks . Then, for each element K in K, sr ⊆ Ks. Thus, by Lemma 11.4.2, sr ⊆ Hs, and this means that r ∈ Hs . Let y and z be elements in X, and let n be the smallest non-negative integer n with z ∈ yLn . Recall from Section 6.6 that D(y, z) is our notation for the union of the sets yLi ∩ zLj which satisfy i + j = n. For the remainder of this section, we denote by V the union of the sets K with K ∈ K. We also fix two elements y in X and z in y L. Proposition 11.4.4 Let x be an element in D(y, z). Then, each faithful map χ from {y} ∪ zV to X extends faithfully to a bijective map from {x, y} ∪ zV to {xχ, yχ} ∪ zχV . Proof. Let p (respectively q) denote the uniquely determined element in L which satisfies x ∈ yp (respectively z ∈ xq). Then z ∈ ypq. Thus, there exists an element s in pq such that z ∈ ys. Let χ be a faithful map from {y} ∪ zV to X. Then, as z ∈ ys, zχ ∈ yχs. Thus, as s ∈ pq, zχ ∈ yχpq. Therefore, there exists an element v in yχp such that zχ ∈ vq. We set xχ := v. Then, χ|{y,x,z} is faithful. Now we pick an element K in K. Then, by Lemma 3.4.8, z ∈ xS1 (K) K. Thus, there exists an element w in xS1 (K) such that z ∈ w K. Thus, by Theorem 3.6.4, w ∈ D(x, z). On the other hand, as w ∈ z K ⊆ zV , χ|{y,w,z} is faithful. Thus, as x ∈ D(y, z) and χ|{y,x,z} is faithful, χ|{x,w} is faithful; cf. Lemma 6.6.2. Thus, as w ∈ xS1 (K) and z ∈ w K, χ|{x}∪zK is faithful; cf. Lemma 11.3.1. Now the claim follows from the fact that K has been chosen arbitrarily in K. Proposition 11.4.5 Let x be an element in X such that y ∈ D(x, z). Then, if y ∈ xL, each faithful map from {y} ∪ zV to X extends faithfully to a bijective map from {x, y} ∪ zV to {xχ, yχ} ∪ zχV . Proof. Let us denote by s the uniquely determined element in L which satisfies z ∈ ys and by l the uniquely determined element in L which satisfies y ∈ xl. Then, as y ∈ D(x, z), l ∈ S1 (s). Let χ be a faithful map from {y} ∪ zV to X, and let us denote by H the set ∗ of all elements K of K which satisfy l ∈ Ks .

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Assume first that H is empty. In this case, we pick an arbitrary element in yχl, and we denote this element by xχ. By Lemma 11.3.3, the extension of χ defined in this way, must be faithful. Let us now assume that H is not empty, and let us denote by J the intersection ∗ of the elements in H. Then, by Corollary 11.4.3, l ∈ Js , and that means that ls ⊆ s J. On the other hand, we have z ∈ ys and y ∈ xl, whence z ∈ xls. Thus, as ls ⊆ s J, z ∈ xs J. Thus, there exists an element w in xs such that z ∈ w J. From x ∈ yl, w ∈ xs, and l ∈ S1 (s) we obtain x ∈ D(y, w). Let us denote by V the union of the sets H with H ∈ H. Then, as w ∈ z J, wV = zV . Thus, as x ∈ D(y, w), χ|{y}∪zV extends faithfully to a bijective map from {x, y} ∪ zV to {xχ, yχ} ∪ zχV ; cf. Proposition 11.4.4. Thus, by Lemma 11.3.3, χ is faithful also on {x, y} ∪ zV . Here is the Extension Theorem. Theorem 11.4.6 Let x be an element in X. Then, each faithful map χ from yV ∪ {z} to X extends faithfully to a bijective map from yV ∪ {x, z} to yχV ∪ {xχ, zχ}. Proof. Let us denote by s the uniquely determined element in L such that x ∈ zs. Clearly, we may assume that 1 = s. Thus, by Lemma 3.1.2, there exist elements r in L and l in L such that s ∈ rl and (s) = (r) + 1. From x ∈ zs and s ∈ rl we obtain x ∈ zrl. Thus, there exists an element w in zr such that x ∈ wl. Let χ be a faithful map from yV ∪ {z} to X. Then, by induction, χ extends faithfully to yV ∪ {w, z}. From Proposition 11.4.4 and Proposition 11.4.5 we obtain that χ|yV ∪{w} extends faithfully to yV ∪ {w, x}. However, as (s) = (r) + 1 and x ∈ zs, xχ ∈ zχs. For the remaining three results of this section, we shall now fix the letter s to denote the uniquely determined element in L which satisfies z ∈ ys. For each element K in K, we fix a subset K of L such that, if r denotes the uniquely determined element in s K ∩ S1 (K ), K ⊆ Kr . We denote by V the union of the sets K with K ∈ K. Corollary 11.4.7 Each faithful map χ from yV ∪ {z} to X extends faithfully to a bijective map from yV ∪ zV to yχV ∪ zχV . Proof. Let χ be a faithful map from yV ∪ {z} to X. Then, by Theorem 11.4.6, χ extends to yV ∪ zV in such a way that, for each element x in zV , χ|yV ∪{x,z} is faithful.

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Let us now fix an element K in K, and let us pick two elements v and w in z K . By Theorem 11.4.6, χ|yK∪{v} extends faithfully to a bijective map from y K ∪ {v, w} to yχ K ∪ {vχ, wχ}. By Lemma 11.3.4, this extension coincides with χ|yK∪{v,w} . Thus, χ|{v,w} is faithful. Since v and w have been chosen arbitrarily in z K , we have shown that χ|zK is faithful. Thus, the claim follows from Lemma 11.3.2. Corollary 11.4.8 Each faithful map from yV to X extends faithfully to a bijective map from yV ∪ zV to yχV ∪ zχV . Proof. Let χ be a faithful map from yV to X. Then, by Theorem 11.4.6, χ extends faithfully to yV ∪ {z}. Thus, the claim follows from Corollary 11.4.7. Corollary 11.4.9 Let z be an element in ys ∩ zV . Then, for each faithful map χ from yV ∪ zV to X, there exists a faithful map χ from yV ∪ z V to X which coincides with χ on yV and on zV ∩ z V . Proof. We are assuming that z ∈ ys. Thus, there exists a faithful map χ from yV ∪ z V to X which coincides with χ on yV and on z ; cf. Corollary 11.4.7. We shall show that χ and χ coincide on zV ∩ z V . In order to do so, we pick an element x in zV ∩ z V . Since x ∈ z V , there exists an element K in K such that x ∈ z K . Since K ⊆ V , the (faithful) maps χ|yK∪{x} and χ |yK∪{x} coincide on y K. Moreover, as x ∈ z K , xχ ∈ z χ K = z χ K . Finally, as x ∈ z K , xχ ∈ z χ K . Thus, by Lemma 11.3.4, xχ = xχ .

12 Spherical Coxeter Sets

This final chapter is the second part of our investigation of Coxeter sets. It deals with spherical Coxeter sets. Let L be a set of involutions of S. Recall that L is called a Coxeter set if L is constrained and satisfies the exchange condition. Recall also that L is called spherical if S−1 (L) is not empty. Recall, finally, that a closed subset T of S is called faithfully embedded in S if, for any two elements y in X and z in yT , each faithful map χ from {y, z} to X extends to a bijective map from yT to yχT . The first goal of this chapter is to show that L is faithfully embedded in S if L is a spherical Coxeter set having at least three elements none of them thin. The corresponding Schur groups turn out to have a Tits system. The situation will be completely described in the corresponding recognition theorem (Theorem 12.3.4). In the first section of this chapter, we focus on specific characteristics of spherical Coxeter sets such as maximal elements and conjugation. The second section is devoted to an extension theorem for spherical Coxeter sets. Our approach to this theorem (which follows the line of [46]) is partially inspired by a geometrical reasoning provided by Jacques Tits in [37]. In the third section of this chapter, we apply results from the two previous sections in order to prove the above-mentioned recognition theorem for spherical Coxeter sets of cardinality at least 3. In Section 12.4, we shall look closer at the case where L is a spherical Coxeter set consisting of two elements.1 Assuming L = S we shall see that |S| ∈ {4, 6, 8, 12, 16, 24} if S is not thin and has finite valency. This is the scheme theoretic version of a well-known theorem on generalized polygons due to Walter Feit and 1

We promised towards the end of Section 10.6 to look at this case.

250


Graham Higman. Our proof is the scheme-theoretic version of the proof given by Robert Kilmoyer and Louis Solomon in [29]. In Section 12.5, we shall look closer at the individual cases which we obtained in Section 12.4. We finish this chapter by looking at the case where L is a Coxeter set consisting of two elements and L has finite valency and is the kernel of a semidirect product. This is a case which, geometrically, has been investigated by Udo Ott and Stanley Payne.

12.1 Elements of Maximal Length In this section, the letter L stands for a spherical Coxeter set. Instead of L we shall write . Assuming L to be a spherical Coxeter set we obtain from Lemma 3.6.8 that S−1 (L) contains exactly one element. In the following, we shall denote this element by mL . Recall from Section 10.2 that we denote by dL the smallest element in (S−1 (L)). Thus, as {mL } = S−1 (L), (mL ) = dL . From this equation we obtain (s) ≤ dL − 1 for each element s in L \ {mL }; cf. Lemma 3.6.7. In particular, as (mL )∗ ∈ L and ((mL )∗ ) = (mL ), (mL )∗ = mL . Let s be an element in L. Then, as mL ∈ S−1 (L), mL ∈ S−1 (s); cf. Theorem 3.6.6. Thus, by definition, L contains an element r with mL ∈ rs and (mL ) = (r) + (s). Moreover, Lemma 3.5.1 says that L contains at most one such element. Thus, there exists exactly one such element. For the remainder of this section, we shall denote this element by s(L) . Lemma 12.1.1 For any two elements p and q in L, we have the following. (i) If p = q, p(L) = q (L) . (ii) Let r be an element in pq, and assume that (r) = (p) + (q). Then q (L) ∈ r(L) p and (q (L) ) = (r(L) ) + (p). Proof. (i) Let p and q be elements in L such that p(L) = q (L) . Then, by definition, mL ∈ p(L) p, (mL ) = (p(L) ) + (p), mL ∈ p(L) q, and (mL ) = (p(L) ) + (q). Thus, by Lemma 3.5.1, p = q.

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(ii) By definition, we have mL ∈ r(L) r and (mL ) = (r(L) ) + (r). By hypothesis, we have r ∈ pq and (r) = (p) + (q). Thus, by Lemma 3.4.2, there exists an element s in r(L) p such that mL ∈ sq, (s) = (r(L) ) + (p), and (mL ) = (s) + (q). From mL ∈ sq and (mL ) = (s) + (q) we obtain s = q (L) . Thus, the claim follows from s ∈ r(L) p and (s) = (r(L) ) + (p). For each element s in L, we shall write s[L] instead of s(L)(L) . Lemma 12.1.2 For any two elements p and q in L, we have the following. (i) If p = q, p[L] = q [L] . (ii) Let r be an element in pq, and assume that (r) = (p) + (q). Then r[L] ∈ p[L] q [L] and (r[L] ) = (p[L] ) + (q [L] ). Proof. (i) This follows from Lemma 12.1.1(i). (Apply this lemma twice.) (ii) This follows from Lemma 12.1.1(ii). (Apply this lemma three times.) Lemma 12.1.3 For each element s in L, the following hold. (i) We have (s[L] ) = (s). (ii) We have s[L][L] = s. Proof. (i) This follows immediately from the definition of s[L] . (ii) Let s be an element in L. Then, by definition, mL ∈ s[L](L) s[L] . Thus, as (mL )∗ = mL , mL ∈ (s[L] )∗ (s[L](L) )∗ ; cf. Lemma 1.3.2(iii). Thus, as mL ∈ s(L) s, the set (s(L) )∗ (s[L] )∗ ∩ ss[L](L) is not empty; cf. Lemma 1.3.4. Thus, as {mL } = (s(L) )∗ (s[L] )∗ , mL ∈ ss[L](L) . However, by definition, we have mL ∈ s[L][L] s[L](L) . Moreover, by (i), we have (s[L][L] ) = (s). Thus, the claim follows from Lemma 3.5.1. Recall that, for each subset R of L, S1 (R) is our notation for the intersection of the sets S1 (r) with r ∈ {1} ∪ R. Lemma 12.1.4 For each subset K of L, we have (mK )(L) ∈ S1 (K) and (mK )(L) ∈ mL K.

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Proof. Let k be an element in K. Then mK ∈ S−1 (k). Thus, by Lemma 3.4.4(iv), S1 (mK ) ⊆ S1 (k). Thus, as (mK )(L) ∈ S1 (mK ), (mK )(L) ∈ S1 (k). Since k has been chosen arbitrarily in K, we have shown that (mK )(L) ∈ S1 (K). By definition, mL ∈ (mK )(L) mK . Thus, as mK ∈ K, mL ∈ (mK )(L) K, so that the second claim follows from Lemma 2.1.4. For each nonempty subset R of L, we define R[L] to be the set of all elements r[L] with r ∈ R. Lemma 12.1.5 For each nonempty subset K of L, the following hold. (i) We have mK [L] = (mK )[L] . (ii) We have K ⊆ K [L] ((mK )

(L)

)

.

Proof. (i) From Lemma 12.1.2 we obtain by induction that s → s[L] is a bijective map from K to K [L] . Thus, the claim follows from Lemma 12.1.3(i). (ii) From Lemma 1.3.6(iii) together with Lemma 3.1.1(i) we know that it is (L) enough to show that K ⊆ K [L] ((mK ) ) . In order to show this we pick an element k in K. From the first statement of Lemma 12.1.4 we know that (mK )(L) ∈ S1 (k). Thus, there exists an element p in (mK )(L) k such that (p) = ((mK )(L) ) + 1. Assume that K [L] ⊆ S1 (p). Then, by Theorem 3.6.4, K [L] ⊆ S1 (p). Thus, as mK [L] ∈ K [L] , mK [L] ∈ S1 (p). Thus, by (i), (mK )[L] ∈ S1 (p). Thus, there exists an element q in (mK )[L] p such that (q) = ((mK )[L] )+(p). Thus, as (p) = ((mK )(L) ) + 1, (q) = ((mK )[L] ) + ((mK )(L) ) + 1 = (mL ) + 1, contrary to q ∈ L and (mL ) = dL . Thus, as L is assumed to satisfy the exchange condition, there exists an element h in K [L] such that h(mK )(L) = (mK )(L) k. Thus, as h ∈ K [L] , (mK )(L) k ⊆ K [L] (mK )(L) , and that means that k ∈ (L)

K [L] ((mK ) ) .

12.2 Faithful Maps

253

12.2 Faithful Maps In this section, the letter L stands for a spherical Coxeter set. Instead of mL we shall write m. In the previous section, we did not refer to the set X. We shall do this now. In fact, all results of this section deal with faithful maps. Lemma 12.2.1 Let y be an element in X, let z be an element in ym, let l be an element in L, and let χ be a faithful map from yl[L] ∪ {y, z} to X. Then there exists at most one faithful map from y l[L] ∪ z l to X which coincides with χ on yl[L] ∪ {y, z}. Proof. By the first statement of Lemma 12.1.4, l(L) ∈ S1 (l), and, by Lemma (L) 12.1.5(ii), l ⊆ l[L] l . Thus, the claim follows from Lemma 11.3.4 (applied [L] (L) instead of H, K, and s). to {l}, {l }, and l Lemma 12.2.2 Let y be an element in X, let z be an element in ym, and let K be a subset of L. Assume that K does not contain thin elements. Let χ be a faithful map from yK [L] ∪{y, z} to X. Then there exists at most one faithful map from y K [L] ∪z K to X which coincides with χ on yK [L] ∪{y, z}. Proof. Let χ be a faithful map from y K [L] ∪ z K to X. From Lemma 12.2.1 we know that the values of χ on zK are uniquely determined by the values on y K[L] ∪ {y, z}. Let us now prove that the values of χ on zK 2 are determined by the values of χ on yK [L] ∪ {y, z}. In order to show this we pick an element w in zK 2 . Since w ∈ zK 2 , there exists an element v in zK such that w ∈ vK. Since v ∈ zK, there exists an element h in K such that v ∈ zh. Since w ∈ vK, there exists an element l in K such that w ∈ vl. We are assuming that K does not contain thin elements. Thus, h[L] is not thin. Thus, zm∩vm∩y h[L] is not empty. Let x be an element in zm∩vm∩y h[L] . By hypothesis, χ is uniquely defined on yh[L] . In particular, χ is uniquely defined on x. Thus, as x ∈ zm, χ is uniquely defined on x l[L] ; cf. Lemma 12.2.1. Thus, as v ∈ xm, χ is uniquely defined on v l; cf. Lemma 12.2.1 once again. Induction now finishes the proof. For the remainder of this section, we assume that L does not contain thin elements. We fix a positive integer n such that n ≤ |L|, and we denote by K the set of all subsets K of L with 1 ≤ |K| ≤ n. The union of the sets K with K ∈ K will be denoted by V .

254


From Lemma 12.1.3(i) we know that, for each element K of K, K [L] ⊆ L. Moreover, Lemma 12.1.2(i) tells us that, for each element K of K, |K [L] | = |K|. Thus, for each element K of K, K [L] ∈ K. In particular, V [L] = V . The fact that V [L] = V together with Lemma 12.1.4 and Lemma 12.1.5(ii) allows us to apply Corollary 11.4.7, Corollary 11.4.8, and Corollary 11.4.9 to V instead of V . For any two elements y in X and z in ym, we define Ξ{y,z} to be the set of all faithful maps from yV ∪ zV to X. As a consequence of Lemma 12.2.2 we obtain the following. Lemma 12.2.3 Let y be an element in X, and let z be an element in ym. Then each map in Ξ{y,z} is uniquely determined by its action on yL ∪ {y, z}. Let x be an element in X, let y be an element in xm, let z be an element in xm ∩ yV , let φ be an element in Ξ{x,y} , and let ψ be an element in Ξ{x,z} . The maps φ and ψ will be called close to each other if they coincide on xV and on yV ∩ zV . Let K be an element in K, let y be an element in X, let y be an element in y m, let z be an element in y K [L] , and let z be an element in z m ∩ y K. Let φ be an element in Ξ{y ,y} , and let ψ be an element in Ξ{z ,z} . The faithful maps φ and ψ will be called K-compatible if, for each element x in ym ∩ zm ∩ y K [L] , there exist maps η in Ξ{x,y} close to φ and ζ in Ξ{x,z} close to η and ψ. We define Ξ to be the union of the sets Ξ{y,z} with y ∈ X and z ∈ ym. For the remainder of this section, we assume that 2 ≤ n. Lemma 12.2.4 K-compatibility is an equivalence relation on Ξ. Proof. By Corollary 11.4.9, K-compatibility is reflexive. That K-compatibility is symmetric follows immediately from the definition of K-compatibility. Let us prove transitivity. Let (x , x), (y , y), (z , z) be elements in m such that y, z ∈ x K and y , z ∈ x K [L] . Let χx be an element in Ξ{x ,x} , let χz be an element in Ξ{z ,z} , and let χy be an element in Ξ{y ,y} which is K-compatible with both χx and χz . We have to show that χx and χz are K-compatible. By induction, we may assume that z ∈ yK. Thus, there exists an element k in K such that z ∈ yk. In order to show that χx and χz are K-compatible, we pick an element u in xm ∩ zm ∩ x K [L] . Then, as we are assuming that L does not contain thin elements, there exist elements v in xm ∩ ym ∩ u k [L] and w in ym ∩ zm ∩ u k [L] . (Recall that z ∈ yk.)

12.2 Faithful Maps

255

Since χx and χy are assumed to be compatible, there exist maps ρ in Ξ{v ,x} close to χx and σ in Ξ{v ,y} close to ρ and χy . Similarly, as χy and χz are assumed to be compatible, there exist maps ζ in Ξ{w ,y} close to χy and η in Ξ{w ,z} close to ζ and χz . By Corollary 11.4.9, there exists an element φ in Ξ{u ,x} close to χx . Similarly, we find an element ψ in Ξ{u ,z} close to φ. We shall be done if we succeed in showing that ψ is close to χz . From Corollary 11.4.9 we obtain an element η˜ in Ξ{w ,z} close to ψ. We claim that η˜ = η, and in order to see this we pick an element w in w L. Then, as we are assuming that 2 ≤ n, w˜ η = wψ = wφ = wρ = wσ = wζ = wη. Since w has been chosen arbitrarily in w L, we have that η˜|w L = η|w L . Thus, as z η˜ = zψ = zφ = zρ = zσ = zζ = zη, η˜ = η; cf. Lemma 12.2.3. The maps ψ and η˜ coincide on z and on zV . The same is true for η and χz . Thus, ψ and χz coincide on z and on zV . Thus, they coincide on u V ∩ z V and on zV . Thus, ψ is close to χz . Let (y, z) be an element in m, and let K be an element in K. From Corollary 11.4.9 we know that each K-equivalence class intersects Ξ{y,z} in at most one element. The following lemma shows that, if Ξ is not empty, each Kequivalence class intersects Ξ{y,z} in at least one element. Lemma 12.2.5 Let (y , y) and (z , z) be elements in m, let K be an element in K such that z ∈ y K [L] and z ∈ y K, and let χ be an element in Ξ{y ,y} . Then there exists an element in Ξ{z ,z} which is K-compatible with χ. Proof. By Lemma 12.2.4, we may assume that z ∈ yK. In this case, there exists an element k in K such that z ∈ yk. Since z ∈ yk, we may assume that z ∈ y k [L] . Since we are assuming that L does not contain thin elements, there exists an element x in ym∩zm∩y k [L] . By Corollary 11.4.9, there exists a map φ in Ξ{x,v} close to χ. Similarly, we find a map ψ in Ξ{x,z} close to φ. Finally, referring to Corollary 11.4.9 a third time we obtain a map χ in Ξ{z ,z} close to ψ. The maps χ and φ coincide on y V ∩ xV , φ and ψ coincide on xV , and ψ and χ coincide on xV ∩ z V . On the other hand, we know from Lemma 11.1.7 that y V ∩ z V ⊆ xV . Thus, χ and χ coincide on y V ∩ z V . The maps χ and φ coincide on yV , φ and ψ coincide on yV ∩ zV , and ψ and χ coincide on zV . Thus, χ and χ coincide on yV ∩ zV .

256


In order to show that χ and χ are K-compatible, we pick an element t in z K [L] ∩ ym ∩ zm. By Corollary 11.4.9, there exists a map φ in Ξ{t,y} close to χ. Similarly, there exists a map ψ in Ξ{t,z} close to φ. We claim that ψ and χ coincide on zL. In order to see this we pick an element x in zL. Then, as z ∈ zK, xψ = xφ = xχ = xχ . Thus, as z ψ = z φ = z χ = z χw , we conclude that ψ and χ are close to each other. We call K-compatibility the smallest equivalence relation on Ξ which contains K-compatibility for each K in K. Let (y , y) and (z , z) be elements in m, let φ be an element in Ξ{y ,y} , and let ψ be an element in Ξ{z ,z} . Then, if φ and ψ are K-compatible, z ∈ y L and z ∈ y L. In the following lemma, we shall write instead of L . Lemma 12.2.6 Let (y, z) be an element in m. Then, for each element χ in Ξ there exists at most one element in Ξ{y,z} which is K-compatible with χ. Proof. Let us denote by R the set of all elements s in L such that zs contains an element contradicting our claim. By way of contradiction, we assume that R is not empty. We pick an element r in R such that (r) is as small as possible. Since r ∈ R, 1 = r. Thus, by Lemma 3.1.2, there exist elements q in L and l in L such that r ∈ ql and (r) = (q) + 1. Since r ∈ R, there exists an element x in zr such that Ξ{x} contains at least two (different) elements χx and χx which are K-compatible with χ. From x ∈ zr and r ∈ ql we obtain x ∈ zql. Thus, there exists an element u in zq such that x ∈ ul. By Lemma 12.2.5, there exists an element χu in Ξ{u} which is K-compatible with χx . Similarly, we find an element χu in Ξ{u} which is K-compatible with χx . Both, χu and χu are K-compatible with χ. Thus, the choice of r forces χu = χu . Thus, by Lemma 12.2.4, χx = χx . Proposition 12.2.7 Let y be an element in X, let z be an element in ym. Then each faithful map χ from yV ∪ {z} to X extends to a faithful map from y L to yχ L. Proof. Let χ be a faithful map from yV ∪ {z} to X. Then, by Corollary 11.4.7, χ extends faithfully to a map χz from yV ∪ zV to X. Thus, by Lemma 12.2.6, there exists, for each element w in x L, a uniquely determined element χw in Ξ{w} which is K-compatible with χz . For each element w in x L, we set wχ := wχw . Then we obtain wlχ ⊆ wχl for any two elements w in x L and l in L. Thus, by Lemma 6.6.3, χ is a faithful map from x L to X.

12.3 The Main Theorem

257

Let v be an element in xV . Then, by definition, there exists an element K in K such that v ∈ x K. Thus, by induction, χv |xK = χ|xK . This shows that χ extends χx .

12.3 The Main Theorem In this section, we look at spherical Coxeter sets containing at least three elements none of them thin. We shall apply Proposition 12.2.7 in order to prove that closed subsets of S generated by such Coxeter sets are faithfully embedded in S. We also establish the corresponding recognition theorem. Let y and z be elements in X, and let n be the smallest non-negative integer n with z ∈ yLn . Recall from Section 6.6 that we D(y, z) is our notation for the union of the sets yLi ∩ zLj which satisfy i + j = n. Lemma 12.3.1 Let L be a spherical Coxeter set. Assume that L has at least three elements none of them thin. Let l be an element in L, let s be an element in S1 (l), let y be an element in X, and let z be an element in ys. Let us denote by G the Schur group of L with respect to y. Then Gyz acts transitively on zl. Proof. Let v and w be elements in zl. We have to find an element in Gyz with vg = w. Without loss of generality we may assume that s = l(L) . Let us denote by V the union of the sets K with K ⊆ L and |K| ≤ 2. Then, by Lemma 11.2.5, there exists an element x in D(y, z) such that |K| = 2, xV ∩ D(y, z) = xV ∩ D(y, v) and xV ∩ D(y, z) = xV ∩ D(y, v). Let us denote by χ the map on xV ∪{v} which is the identity on xV and maps v to w. Then χ is faithful. Thus, as L is assumed to have no thin element, we obtain from Proposition 12.2.7 an element g in G which is the identity on xV and maps v to w. Since y, z ∈ D(y, z), we have g ∈ Gyz and vg = w. Proposition 12.3.2 Let L be a spherical Coxeter set. Assume that L has at least three elements none of them thin. Let x be an element in X, and let us denote by G the Schur group of L with respect to x. Then we have the following. (i) Let x be an element in X, and let s be an element in L. Then Gx acts transitively on xs. (ii) The group G acts transitively on x L.

258


Proof. (i) Let us denote by R the set of all elements r in L such that Gx does not act transitively on xr. By way of contradiction, we assume that R is not empty. We pick an element r in R such that (r) is as small as possible. Since r ∈ R, 1 = r. Thus, by Lemma 3.1.2, there exist elements q in L and l in L such that r ∈ ql and (r) = (q) + 1. Let y and z be elements in xr. From y ∈ xr and r ∈ ql we obtain y ∈ xql. Thus, there exists an element v in xq such that y ∈ vl. Similarly, as z ∈ xr and r ∈ ql, there exists an element w in xq such that z ∈ wl. Since q ∈ L and (r) = (q) + 1, the minimal choice of r yields q ∈ / R. Thus, as v, w ∈ xq, there exists an element e in Gx such that ve = w. Since e is faithful, we obtain from y ∈ vl and w = ve that ye ∈ vle ⊆ vel = wl; cf. Lemma 6.1.1(i). Thus, as z ∈ wl, there exists an element f in Gxw such that yef = z; cf. Lemma 12.3.1. Setting g := ef we, therefore, obtain g ∈ Gx and yg = z. This contradiction finishes the proof of (i). (ii) Let y be an element in X, and let z be an element in yL. Then there exists an element l in L such that z ∈ yl. Since L is assumed to have no thin element, there exists an element x in X with y ∈ xl and z ∈ xl. Thus, by (i), there exists an element g in Gx with yg = z. (We apply (i) to l in place of s.) What we have seen so far is that, for any two elements y and z in X with z ∈ yL, there exists an element g in G with yg = z. Thus, the claim follows from Lemma 6.3.4. Theorem 12.3.3 Let L be a spherical Coxeter set. Assume that L has at least three elements none of them thin. Then L is schurian. Proof. Referring to Theorem 6.3.1 this follows immediately from Proposition 12.3.2. A scheme which is generated by a Coxeter set L is called a Coxeter scheme with respect to L. If S is a Coxeter scheme with respect to L, |L| is called the rank of S. If {1} = S, S is a Coxeter scheme with respect to the empty set. If S consists of two elements and s denotes the non-identity element of S, s is an involution and S is a Coxeter scheme (of rank 1) with respect to {s}. We shall now see how to obtain Coxeter schemes from thin schemes. Assume S to be thin, let T be a closed subset of S, and let J be a subset of S such that T ∪ J = S. Assume that T ∩ J is normal in J. Assume that, for each element j in J, j 2 ∈ T and

12.4 Coxeter Schemes of Finite Valency and Rank 2

259

T = T jT jT. Assume that, for any two elements j ∈ J and s in J, T sT jT ⊆ T sjT ∪ T sT. Then (T, J) is called a Tits system for S. Theorem 12.3.4 Let L be a set of involutions of S such that L = S. Assume that L is a spherical Coxeter set consisting of at least three elements none of them thin. Then there exists a finite thin scheme S¯ with a Tits system ¯ such that S ∼ ¯ T¯. (T¯, J) = S// Proof. This is a consequence of Theorem 12.3.3. The following theorem is the converse of Theorem 6.5.4. It says that Tits systems give rise to Coxeter sets. It makes Theorem 6.5.4 to be one of our recognition theorems. Theorem 12.3.5 Assume that S is thin and possesses a Tits system (T, J). Then S//T is a Coxeter scheme with respect to J//T . Proof. We are assuming that, for each element j in J, T = T jT jT . Thus, for each element j in J, j T is an involution in S//T . From T ∪ J = S we obtain that S//T is generated by J//T ; cf. Lemma 4.2.2(i). The facts that J//T is constrained and satisfies the exchange condition follow from the hypothesis that T sT jT ⊆ T sjT ∪ T sT for any two elements j in J and s in J.

12.4 Coxeter Schemes of Finite Valency and Rank 2 Throughout this section, the letter L stands for a set of involutions. We assume that S is a Coxeter scheme with respect to L, that |L| = 2, and that S has finite valency. Since S is assumed to have finite valency, S is finite. Thus, by Lemma 10.1.7, L is spherical, and that means that S−1 (L) is not empty. Keeping the notation introduced in Section 10.2 we write dL to denote the smallest element in (S−1 (L)). However, having fixed L for the remainder of this section, we shall write d instead of dL . Recall that, by Theorem 10.5.4, 2d = |S|. The main result of this section is Theorem 12.4.6, a result due to Walter Feit and Graham Higman. This theorem says that

260


d ∈ {2, 3, 4, 6, 8, 12} if S is not thin. After that, we shall give a few arithmetical details in each of the individual cases. The main idea of the proof of Theorem 12.4.6 is to apply Theorem 9.1.7(ii) to an algebraically closed field C of characteristic 0. According to Lemma 9.2.5, the left hand side of the equation in Theorem 9.1.7(ii) is algebraic over the smallest unitary subring Z of C and the right hand side of that equation is in the smallest subfield of C. Thus, according to Lemma 8.2.5, both sides must be in Z, and one obtains an integral divisibility condition. Group theoretic proofs based on this type of reasoning have been given repeatedly by William Burnside. (An example is his proof of the solvability of finite groups of order divisible by at most two primes.) Assuming Burnside would have known and would have been interested in the notion of a finite scheme he might have given a proof of Theorem 12.4.6 long time before Walter Feit and Graham Higman gave their proof. Lemma 12.4.1 Let C be an algebraically closed field of characteristic 0. Then the following hold. (i) Each irreducible character of CS is linear or has degree 2. (ii) If d is odd, CS has two linear characters and of degree 2.

d−1 2

(iii) If d is even, CS has four linear characters and acters of degree 2.

d 2

irreducible characters − 1 irreducible char-

Proof. (i) Since S is assumed to be a Coxeter scheme with respect to L, L is constrained and L. Thus, by Corollary 9.6.3, C[L] = CS, so that the claim follows from Lemma 9.4.5. (ii), (iii) Let us denote by λ1 the number of linear characters of CS and by λ2 the number of irreducible characters of degree 2. Since 2d = |S|, we obtain from (i) together with Corollary 8.6.5 that 2d = λ1 + 4λ2 . From Corollary 9.6.3 we know that C[L] = CS. Thus, each linear character of CS is determined by its values on {σl | l ∈ L}. Thus, by Corollary 9.4.4(ii), λ1 ≤ 4. Since 1CS is linear, we also obtain 1 ≤ λ1 . Our claims are obvious consequences of 2d = λ1 + 4λ2 and 1 ≤ λ1 ≤ 4. For the remainder of this section, we set √ n := nl . l∈L

Let l be an element in L, and let j be a positive integer. Recall from Section 10.1 that Rj (l) is our notation for the complex product l1 · · · lj of elements in


261

L which satisfy l1 = l and, for each integer i with 1 ≤ i ≤ j, li = l if and only if i is odd. We also write R0 (l) instead of {1}. Recall from Lemma 10.5.1 that, for any two elements l in L and j in {0, . . . , d}, |Rj (l)| = 1. Recall also (from Section 10.5) that, for any two elements l in L and j in {0, . . . , d}, rj (l) is our notation for the element in Rj (l). From Theorem 10.5.4 we know that {rj (l) | l ∈ L, j ∈ {0, . . . , d}} = S. From Corollary 10.5.2 we also know that, for any two elements h and k in L, rd (h) = rd (k). Clearly, by definition, we also have that, for any two elements h and k in L, r0 (h) = r0 (k). Let us now denote by Cd the set of all elements c in C \ {−1, 1} satisfying cd = 1. In the following proposition, we completely compute the values of all nonlinear irreducible characters of CS, provided that C is an algebraically closed field of characteristic 0. Proposition 12.4.2 Let C be an algebraically closed field of characteristic 0, and let χ be an irreducible character of CS of degree 2. Then there exists an element c in Cd such that the following hold. (i) Let h and k be elements in L with h = k. Then, for each non-negative integer i with 2i + 1 ≤ d, χ(σr2i+1 (h) ) =

ni nh ((nh − 1)(ci+1 − c−(i+1) ) + (nk − 1) (ci − c−i )). −1 c−c n

(ii) Let l be an element in L. Then, for each non-negative integer i with 2i ≤ d, χ(σr2i (l) ) = ni (ci + c−i ). Proof. Let M be an irreducible CS-module such that χM = χ, let h and k be elements in L such that h = k, and let us denote by a and b the characteristic roots of dM (σh σk ). (Recall from Section 8.1 that dM (σh σk ) is our notation for the endomorphism of M induced by σh σk .) First of all, we have det(dM (σh σk )) = ab. On the other hand, for each element l in L, −1 and nl are the characteristic roots of dM (σl ); cf. Corollary 9.4.4(ii). Thus, det(dM (σh )) = −nh and det(dM (σk )) = −nk . It follows that ab = n2 . Set c :=

a . n

262


Let us first prove that c2 = 1. In order to do this, we assume, by way of contradiction, that c2 = 1. Then a = b. Thus, σh σk induces a scalar multiplication on M , contrary to the irreducibility of M . Thus, c ∈ / {−1, 1}. Set e := d if d is odd and e :=

d 2

if d is even. Define σ := (σh σk )e .

From Corollary 10.5.2 we know that rd (h) = rd (k). Thus, σσh = σh σ and σσk = σk σ. On the other hand, we know from Corollary 9.6.3 that C[L] = CS. Thus, dM (σ) ∈ EndCS (M ). Therefore, by Lemma 8.4.3 and Lemma 8.2.6, σ induces a scalar multiplication on M . It follows that ae = be . Therefore, as / {−1, 1}, c ∈ Cd . c2 = ab , we obtain cd = 1. Thus, as c ∈ From ab = n2 and the definition of c we obtain cb = n. Thus, a + b = n(c + c−1 ). It follows that dM (σh σk )2 = (a + b)dM (σh σk ) − abσ1 = n(c + c−1 )dM (σh σk ) − n2 σ1 . Thus, for each element j in {4, . . . , d}, dM (σrj (h) ) = n(c + c−1 )dM (σrj−2 (h) ) − n2 dM (σrj−4 (h) ); see Lemma 9.6.1(ii). If i ∈ {0, 1}, the equations can be computed explicitly. Let us, therefore, pick an integer i such that 2 ≤ i ≤ d2 . We assume that the equations in question hold for i − 1 and i − 2. Then the last equation yields ni−1 (nh − 1)(ci − c−i ) c − c−1 ni−1 nh (nk − 1) (ci−1 − c−(i−1) ) +n(c + c−1 ) c − c−1 n i−2 n (nh − 1)(ci−1 − c−(i−1) ) −n2 c − c−1 ni−2 nh (nk − 1) (ci−2 − c−(i−2) ) −n2 c − c−1 n i nh n ((nh − 1)(ci+1 − c−(i+1) ) + (nk − 1) (ci − c−i )) = c − c−1 n

χ(σr2i+1 (h) ) = n(c + c−1 )

and χ(σr2i (l) ) = ni (c + c−1 )(ci−1 + c−(i−1) ) − ni (ci−2 + c−(i−2) ) = ni (ci + c−i ). This finishes the proof of the proposition. Let C be an algebraically closed field of characteristic 0, and let χ be an irreducible character of CS of degree 2. We define Cχ to be the set of all elements in Cd satisfying conditions (i) and (ii) of Proposition 12.4.2.


263

According to Proposition 12.4.2, the set Cχ is not empty. In Proposition 12.4.2, we have computed all values of all irreducible characters of CS of degree 2. This allows us to compute, for each of these characters, the left hand side of the equation given in Theorem 9.1.7(ii). We shall do this in the following proposition. Proposition 12.4.3 Let C be an algebraically closed field of characteristic 0, let χ be an irreducible character of CS of degree 2, and let c be an element in Cχ . Then we have the following. (i) If d is odd, 1 (n − 1)2 ). χ(σs∗ )χ(σs ) = 2d(1 − ns∗ n(c + c−1 − 2)

s∈S

(ii) Assume that d is even, and let h and k be elements in L such that h = k. Then 1 χ(σs∗ )χ(σs ) = ns∗ s∈S

2d(1 −

1 (nh − 1)2 (nk − 1)2 (nh − 1)(nk − 1)(c + c−1 ) ( + ) − ). (c − c−1 )2 nh nk n(c − c−1 )2

Proof. (i) Let us assume that d is odd. Then there exists an integer e such that 2e+1 = d. Moreover, for each element l in L, nl = n; cf. Lemma 10.6.1(i). It follows that, for each element l in L, e 1 χ(σr2i+1 (l) )2 χ(σr2i (l) )2 χ(σrd (l) )2 χ(σs∗ )χ(σs ) = (2 + 2 ) − − 4. ns∗ n2i n2i+1 nd i=0

s∈S

Let us now compute the fractions on the right hand side of this equation. In order to do so we fix an element l in L. From Proposition 12.4.2(ii) together with the first equation of Lemma 8.7.2 we obtain e e χ(σr2i (l) )2 2 = 2(ci + c−i )2 = 2(d + 2). 2i n i=0 i=0 From Proposition 12.4.2(i) we obtain χ(σr2i+1 (l) ) =

ni (n − 1)(ci+1 − c−(i+1) + ci − c−i ) c − c−1

for any two elements l in L and i in {0, . . . , d−1 2 }. From this equation together with the second equation of Lemma 8.7.2 we obtain

264

12 Spherical Coxeter Sets e e χ(σr2i+1 (l) )2 2(n − 1)2 i+1 2 = (c − c−(i+1) + ci − c−i )2 2i+1 −1 )2 n n(c − c i=0 i=0

=−

2(n − 1)2 d(c + c−1 + 2). n(c − c−1 )2

Since ce+1 = c−e and c−(e+1) = ce , the above equation also yields χ(σrd (l) )2 (n − 1)2 = (ce+1 − c−(e+1) + ce − c−e )2 = 0. d n n(c − c−1 )2 Thus, 1 2(n − 1)2 χ(σs∗ )χ(σs ) = 2(d + 2) − d(c + c−1 + 2) − 4 ns∗ n(c − c−1 )2

s∈S

= 2d(1 −

(n − 1)2 ). n(c + c−1 − 2)

(ii) Assume that d is even, and let h and k be elements in L with h = k. Since d is assumed to be even, there exists an integer e such that 2e = d. Thus, 1 χ(σs∗ )χ(σs ) ns∗

s∈S

=

e−1 i=0

(2

χ(σr2i+1 (h) )2 χ(σr2i+1 (k) )2 χ(σr2i (h) )2 χ(σrd (h) )2 + + ) + −4 (nh nk )i (nh nk )i nh (nk nh )i nk (nh nk )e

Let us now compute the fractions on the right hand side of this equation. Referring to Proposition 12.4.2(ii) and Lemma 8.7.1 we obtain e−1 e−1 χ(σr2i (h) )2 2 = 2(ci + c−i )2 i (n n ) h k i=0 i=0

=

e−1

2(c2i + 2 + c−2i )

i=0

= 4e + 2

e−1

(c2i + c−2i ) = 2d.

i=0

From Proposition 12.4.2(ii) we also obtain χ(σrd (h) )2 = (ce + c−e )2 = c2e + 2 + c−2e = cd + 2 + c−d = 4. (nh nk )e Finally, referring to Proposition 12.4.2(i) and Lemma 8.7.3 we obtain


265

e−1 χ(σr2i+1 (h) )2 χ(σr2i+1 (k) )2 ( + ) i (nh nk ) nh (nk nh )i nk i=0

=

e−1 i=0

+

1 nh ((nh − 1)(ci+1 − c−(i+1) ) + (nk − 1) (ci − c−i ))2 −1 2 nh (c − c ) n

e−1 i=0

=

e−1 i=0

+

1 nk ((nk − 1)(ci+1 − c−(i+1) ) + (nh − 1) (ci − c−i ))2 nk (c − c−1 )2 n

1 (nh − 1)2 (nk − 1)2 ( + )((ci+1 − c−(i+1) )2 + (ci − c−i )2 ) −1 2 (c − c ) nh nk

e−1 4(nh − 1)(nk − 1) i=0

= −2d(

n(c − c−1 )2

(ci+1 − c−(i+1) )(ci − c−i )

1 (nh − 1)2 (nk − 1)2 (nh − 1)(nk − 1)(c + c−1 ) ( + ) + ), (c − c−1 )2 nh nk n(c − c−1 )2

and that proves the desired equation. Lemma 12.4.4 Let C be an algebraically closed field of characteristic 0, and let c be an element in Cd . Then there exists an irreducible character χ of CS of degree 2 such that c ∈ Cχ . Proof. If d is odd, |Cd | = d − 1, and if d is even, |Cd | = d − 2. Thus, by Lemma 12.4.1(ii), (iii), 12 |Cd | is the number of irreducible characters of CS of degree 2. Let χ be an irreducible character of CS of degree 2, and let c be an element in Cχ . Then, looking at the equations of Proposition 12.4.2, we see that c−1 ∈ Cχ . Thus, for each of the 12 |Cd | irreducible characters χ of CS of degree 2, we have that {c−1 , c} ⊆ Cχ ⊆ Cd . This proves the lemma. Lemma 12.4.5 Let C be an algebraically closed field of characteristic 0, let χ be an irreducible character of CS of degree 2, and let c be an element in Cχ . Let Q be the smallest subfield of C. Assume that S is not thin, that d is even, and that 4 ≤ d. Then (c−c−1 )2 ∈ Q. Proof. By Lemma 12.4.4, CS possesses an irreducible character χ of degree 2 with −c ∈ Cχ . Adding the values computed for χ and χ in Proposition 12.4.3(ii) we obtain from Theorem 9.1.7(ii) that 2d(2 −

2 (nh − 1)2 (nk − 1)2 ( + )) ∈ Q. (c − c−1 )2 nh nk

266


Thus, the claim follows from the fact that S is assumed to be thin. The following two theorems are due to Walter Feit and Graham Higman; cf. [10; Theorem 1]. Theorem 12.4.6 If S is not thin, d ∈ {2, 3, 4, 6, 8, 12}. Proof. Let C be an algebraically closed field of characteristic 0. Let us write Q to denote the smallest subfield of C, and let Z denote the smallest unitary subring of C. Suppose first that d is odd. Then 3 ≤ d. Thus, by Lemma 12.4.1(ii), CS possesses at least one irreducible character χ of degree 2. Let c be an element in Cχ . Then we obtain from Proposition 12.4.3(i) and Theorem 9.1.7(ii) that 2d(1 −

(n − 1)2 ) ∈ Q. n(c + c−1 − 2)

Now recall that S is assumed not to be thin. Thus, by Lemma 3.1.5(i), 1 = n, and this implies c + c−1 ∈ Q. On the other hand, by Lemma 9.2.5 together with Theorem 8.2.4, c + c−1 ∈ IC (Z). Thus, by Lemma 8.2.5, c + c−1 ∈ Z. Now Lemma 8.7.4 yields d = 3. Suppose now that d is even and that 4 ≤ d. Then, according to Lemma 12.4.1(iii), CS possesses at least one irreducible character χ of degree 2. Let c be an element in Cχ . Then, by Lemma 12.4.5, (c − c−1 )2 ∈ Q. It follows that c2 + c−2 ∈ Q. On the other hand, by Lemma 9.2.5 together with Theorem 8.2.4, c2 + c−2 ∈ IC (Z). Thus, by Lemma 8.2.5, c2 + c−2 ∈ Z. Now Lemma 8.7.4 yields d ∈ {4, 6, 8, 12}. Theorem 12.4.7 If {1} = Oϑ (S), we have the following. (i) If d = 6, n is an integer. √ (ii) If d = 8, 2n is an integer. (iii) We have d = 12. Proof. Let C be an algebraically closed field of characteristic 0. Let us write Q to denote the smallest subfield of C, and let Z denote the smallest unitary subring of C. Let χ be an irreducible character of CS of degree 2, and let c be an element in Cχ . Then, by Lemma 12.4.5, (c − c−1 )2 ∈ Q. Now recall that S \ {1} is assumed to have no thin element. Thus, 1 = nh and 1 = nk . Thus, as (c − c−1 )2 ∈ Q, we obtain from Proposition 12.4.3(ii) and Theorem 9.1.7(ii) that

12.5 Valencies and Multiplicities

267

c + c−1 ∈ Q. n (i) Assume that d = 6. Then (c + c−1 )2 = 1. Thus, we must have that n ∈ Q. Therefore, by Lemma 8.2.5, n ∈ Z. −1 2 ) = 2. (ii) Assume that d = 8. By√Lemma 12.4.4, we may assume that (c+c √ Thus, we must have that 2n ∈ Q. Therefore, by Lemma 8.2.5, 2n ∈ Z.

(iii) Assume finally that d = 12. In this case, we may assume that (c+c−1 )2 = 3 or that (c + c−1 )2 = 1. √ √ If (c + c−1 )2 = 3, 3n ∈ Q. Then, by Lemma 8.2.5, 3n ∈ Z. Suppose now that (c + c−1 )2 = 1. Thus, √ we must have that n ∈ Q. Therefore, by Lemma 8.2.5, n ∈ Z, contrary to 3n ∈ Z. As a consequence of Theorem 12.4.7(ii) one obtains that the valencies of the two elements in L must be different if d = 8.

12.5 Valencies and Multiplicities In this section, we continue our investigation on Coxeter schemes of finite valency and rank 2 which we started in the previous section. The letter L stands for a set of involutions. We assume that S is a Coxeter scheme with respect to L, that |L| = 2, and that S has finite valency. Since S is assumed to have finite valency, S is finite. Thus, by Lemma 10.1.7, L is spherical, and that means that S−1 (L) is not empty. Keeping the notation of the previous section we write dL to denote the smallest element in (S−1 (L)). However, having fixed L for the remainder of this section, we shall write d instead of dL . In Theorem 12.4.6, we saw that d ∈ {2, 3, 4, 6, 8, 12}. If d = 2, S is the direct product of two closed subsets each of them generated by an involution. We shall now look separately at each of the individual cases d = 3, d = 4, d = 6, d = 8, and d = 12. In order to do this, we fix an algebraically closed field of characteristic 0 and call it C. From Lemma 12.4.1(ii), (iii) and Corollary 9.4.4(ii) we know that there exists a linear character st such that, for each element l in L, st(σl ) = −1. The linear character st is called the Steinberg character of CS. Theorem 12.5.1 If d = 3, the following hold. (i) We have nS = (n + 1)(n2 + n + 1). (ii) We have mst = n3 . (iii) The multiplicity of the only non-linear irreducible character of CS is n(n + 1).

268


Proof. (i) Applying Theorem 9.1.7(ii) to 1CS in place of φ and ψ we obtain 1 nS = 1CS (σs∗ )1CS (σs ) = 1 + 2n + 2n2 + n3 . m1CS ns∗ s∈S

Thus, the claim follows from Lemma 9.1.8(ii). (ii) Applying Theorem 9.1.7(ii) to st in place of φ and ψ we obtain from (i) that 1 nS 2 2 1 nS = st(σs∗ )st(σs ) = 1 + + 2 + 3 = 3 . mst ns∗ n n n n s∈S

(iii) Let us denote by χ the only non-linear irreducible character of CS. Then, referring to Lemma 9.1.8(ii), Lemma 12.4.1(i), and (ii) we obtain χCX (σ1 ) = m1CS 1CS (σ1 ) + mst st(σ1 ) + mχ χ(σ1 ) = 1 + n3 + mχ χ(σ1 ). On the other hand, by (i), χRX (σ1 ) = nS = 1 + 2n + 2n2 + n3 . Therefore, as χ(σ1 ) = 2, mχ = n(n + 1). It is a well-known open question, whether or not d = 3 implies that n is a prime power. For the remainder of this section, d will be even, and we assume that 4 ≤ d. Since C is assumed to be algebraically closed, Cd contains an element c such that, for each integer i with 1 ≤ i ≤ d − 1, ci = 1. We fix one of these elements. For each integer i with 1 ≤ i ≤ d2 − 1, we define χi to be the non-linear irreducible character of CS which satisfies ci ∈ Cχi ; cf. Lemma 12.4.4. Let us fix elements h and k in L such that h = k. Then, by Lemma 12.4.1(iii) and Corollary 9.4.4(ii), there exists a linear character λh such that λh (σh ) = nh , λh (σk ) = −1 and there exists a linear character λk such that λk (σh ) = −1, λk (σk ) = nk . Without loss of generality, we assume that nh ≤ nk . Theorem 12.5.2 Assume that d = 4. Then the following hold. (i) We have nS = (nh + 1)(nk + 1)(nh nk + 1).

12.5 Valencies and Multiplicities

269

(ii) We have mst = n2h n2k and n2k (nh nk + 1) , nh + nk n2 (nh nk + 1) = h , nh + nk nh nk (nh + 1)(nk + 1) = . nh + nk

mλh = mλk mχ1

Proof. (i) Applying Theorem 9.1.7(ii) to 1CS in place of φ and ψ we obtain 1 nS = 1CS (σs∗ )1CS (σs ) m1CS ns∗ s∈S

= 1 + nh + nk + 2nh nk + nh n2k + n2h nk + n2h n2k . Thus, the claim follows from Lemma 9.1.8(ii). (ii) Applying Theorem 9.1.7(ii) to st in place of φ and ψ we obtain from (i) that 1 nS = st(σs∗ )st(σs ) mst ns∗ s∈S

=

1 1 1 1 1 1 nS 1 + + +2 + + 2 + 2 2 = 2 2. 2 1 nh nk nh nk nh nk nh nk nh nk nh nk

The values of mλh , mλk , and mχ1 are obtained from (i) and Proposition 12.4.3(ii) by applying Theorem 9.1.7(ii) to mλh , mλk , and mχ1 in place of φ and ψ. From Theorem 12.5.2(ii) we obtain, in particular, that nh + nk divides nh nk (nh + 1)(nk + 1). Assume that d = 4 and that {1} = Oϑ (S). Then, if S is isomorphic to a known example, we have (nh , nk ) ∈ {(q − 1, q + 1), (q, q), (q, q 2 ), (q 2 , q 3 )}, where q is a prime power. Donald Higman showed in [21; Theorem 3.2] that, if d = 4 and 2 ≤ nh , nk ≤ n2h . For a proof, see also [43; Theorem 5.3.7(iv)]. Theorem 12.5.3 Assume that d = 6. Then the following hold. (i) We have nS = (nh + 1)(nk + 1)(n2h n2k + nh nk + 1).

270


(ii) We have mst = n3h n3k and mλh =

n3k (n2h n2k + nh nk + 1) , n2h + nh nk + n2k

mλk =

n3h (n2h n2k + nh nk + 1) , n2h + nh nk + n2k

mχ1 =

2((n2h nk

nh nk (nh + 1)(nk + 1)(n2h n2k + nh nk + 1) , + nh − nh nk + nk + nh n2k ) + (nh − 1)(nk − 1)n)

mχ2 =

2((n2h nk

nh nk (nh + 1)(nk + 1)(n2h n2k + nh nk + 1) . + nh − nh nk + nk + nh n2k ) − (nh − 1)(nk − 1)n)

Proof. (i) This equation follows from Theorem 9.1.7(ii) in the same way as the one in Theorem 12.5.2(i). (ii) All five equations are obtained from (i), Proposition 12.4.3(ii) and Theorem 9.1.7(ii) in the same way as the ones in Theorem 12.5.2(ii). Assume that d = 6 and that {1} = Oϑ (S). Then, if S is isomorphic to a known example, we have (nh , nk ) ∈ {(q, q), (q, q 3 )}, where q is a prime power. Willem Haemers and Cornelis Roos showed in [16] that, if d = 6 and 2 ≤ nh , nk ≤ n3h . For a proof, see also [43; Theorem 5.3.8(iv)]. Theorem 12.5.4 Assume that d = 8. Then the following hold. (i) We have nS = (nh + 1)(nk + 1)(nh nk + 1)(n2h n2k + 1). (ii) We have mst = n4h n4k and mλh =

n4k (nh nk + 1)(n2h n2k + 1) , (nh + nk )(n2h + n2k )

mλk =

n4h (nh nk + 1)(n2h n2k + 1) , (nh + nk )(n2h + n2k )

mχ1 =

nh nk (nh + 1)(nk + 1)(nh nk + 1)(n2k n2k + 1) √ , 4((n2h nk + nh − 2nh nk + nk + nh n2k ) + (nh − 1)(nk − 1) 2n)

nh nk (nh + 1)(nk + 1)(n2k n2k + 1) , 2(nh + nk ) nh nk (nh + 1)(nk + 1)(nh nk + 1)(n2k n2k + 1) √ . = 4((n2h nk + nh − 2nh nk + nk + nh n2k ) − (nh − 1)(nk − 1) 2n)

mχ2 = mχ3

12.6 Polarities

271

Proof. (i) This equation follows from Theorem 9.1.7(ii) in the same way as the one in Theorem 12.5.2(i). (ii) All six equations are obtained from (i), Proposition 12.4.3(ii) and Theorem 9.1.7(ii) in the same way as the ones in Theorem 12.5.2(ii). Assume that d = 8 and that {1} = Oϑ (S). Then, if S is isomorphic to a known example, we have (nh , nk ) = (q, q 2 ), where q = 2e for some odd positive integer. Donald Higman showed in [21; Theorem 3.2] that, if d = 8 and 2 ≤ nh , nk ≤ n2h . For a proof, see also [43; Theorem 5.3.9(iv)]. Theorem 12.5.5 Assume that d = 12. Then nh = 1, and the following hold. (i) We have nS = 2(nk + 1)2 (n4k + n2k + 1). (ii) We have mst = n6k , mλh = n6k , mλk = 1, and mχ1 = mχ2 = mχ3 = mχ4 = mχ5 =

nk (nk + 1)2 (n2k + nk + 1), 6 nk (nk + 1)2 (n2k − nk + 1), 2 2nk 4 (nk + n2k + 1), 3 nk (nk + 1)2 (n2k − nk + 1), 2 nk (nk + 1)2 (n2k + nk + 1). 6

Proof. (i) This equation is follows from Theorem 9.1.7(ii) in the same way as the one in Theorem 12.5.2(i). (ii) All eight equations are obtained from (i), Proposition 12.4.3(ii) and Theorem 9.1.7(ii) in the same way as the ones in Theorem 12.5.2(ii).

12.6 Polarities Throughout this section, the letter L stands for a set of two involutions. We assume that S is a Coxeter scheme with respect to L and that S has finite valency. Instead of dL we just write d. For any two elements l in L and j in {0, . . . , d}, rj (l) has the same meaning as in the previous section. An automorphism a of S is called a polarity if 1X = (aX )2 and 1S = aS . Let a be a polarity of S. Then, for any two elements h and k in L with h = k, ha = k and ka = h.

272


Lemma 12.6.1 Let l be an element in L, and let a be a polarity of S. Then we have the following. (i) For each element j in {0, . . . , d}, rj (l)a = rj (la). (ii) Let j be an element in {1, . . . , d}. Assume that X contains an element x with xa ∈ xrj (l). Then j is even or j = d. Proof. (i) This follows immediately from the definition of rj (l). (ii) Let x be an element in X such that xa ∈ xrj (l). Then x ∈ xarj (l)∗ and x = xa2 ∈ xarj (l) = xarj (l)a = xarj (la); see (i). Therefore, rj (l)∗ = rj (la). But, by definition, la = l. Therefore j is even or j = d; cf. Lemma 10.2.1(i). Let a be a polarity of S. Note that, for each element l in L, nla = nl . Thus, as la = l, we have that, for each element l in L, n = nl . In the following, we shall write n instead of nl . Recall that FixX (a) is our notation for the set of all elements x in X such that xa = x. Lemma 12.6.2 Let a be a polarity of S, and assume d to be even. Then |FixX (a)| = nd/2 + 1. Proof. We set Y := FixX (a). By hypothesis, d is even. Therefore, there exists an integer e such that 2e = d. Our first claim is that, for each element x in X, there exist elements y in Y , l in L, and i in {0, . . . , e} such that x ∈ yri (l). Let x be an element in X. Then there exist elements j in {0, . . . , d} and l in L such that xa ∈ xrj (l). Thus, by Lemma 12.6.1(ii), j is even. Let i be an integer such that 2i = j. We are assuming that a is as polarity of S. Thus, {rj (l)} = ri (l)ri (la)∗ . Thus, as xa ∈ xrj (l), xa ∈ xri (l)ri (la)∗ . Thus, there exists an element y in xri (l) such that xa ∈ yri (la)∗ . Thus, by Lemma 12.6.1(i), ya ∈ xari (la) ∩ xri (l). Let us denote by s the element in S which satisfies ya ∈ ys. Then s ∈ ri (l)∗ ri (l) ∩ ri (la)∗ ri (la). Thus, referring to Lemma 10.1.1 and Lemma 10.2.1(i) we obtain s = 1. It follows that y ∈ Y . Thus, x ∈ yri (l)∗ ⊆ Y {ri (l), ri (la)}.

12.6 Polarities

273

Since x has been chosen arbitrarily in X, we have proved our first claim. Let y and z be elements in Y with y = z. Then there exist elements j in {1, . . . , d} and l in L such that z ∈ yrj (l). Thus, by Lemma 10.2.1(i), j = d. So far, we have seen that, for any four elements h and k in L with h = k and i and j in {1, . . . , e}, we have either i = e = j or ∅ = Y {ri (h)} ∩ Y {rj (k)}. Conversely, let y be an element in Y , and let x be an element in yre (l). Then there exists a uniquely determined element z in Y such that x ∈ zes (la). Recall that, for any two elements l in L and i in {1, . . . , s}, nri (l) = ni . Therefore, nS = |Y |(1 + 2n + . . . + 2ne−1 + ne ) = |Y |(n + 1)

e−1

ni

i=0

and nS = (n + 1)

d−1

ni = (n + 1)(ne + 1)

i=0

e−1

ni .

i=0

Thus, we conclude that |Y | = n + 1. e

Let us assume that S possesses a polarity a. We set 1ζ := 1S , aζ := aS , and A := {1, a}. From Lemma 5.2.2(iii) we know that ζ is a group homomorphism from A to Stc(S). Thus, we may consider the semidirect product Sζ A of S and A with respect to ζ; cf. Section 7.3. For any two elements l in L and j in {1, . . . , d}, we set rj+ (l) := rj (l)ζ 1 and

rj− (l) := rj (l)ζ a.

Note that {rj− (l) | l ∈ L, 0 ≤ j ≤ d} ∪ {rj+ (l) | l ∈ L, 0 ≤ j ≤ d} = Sζ A. Let C be an algebraically closed field of characteristic 0. From Theorem 9.4.8(iii) we know that CX is a C(Sζ A)-module. In the following proposition, we assume d to be even and completely compute the character of C(Sζ A) afforded by this module. Proposition 12.6.3 Assume that S possesses a polarity a, and set A := {1, a}. Let C be an algebraically closed field of characteristic 0, and let us denote by φ the character of C(Sζ A) afforded by CX. Let l be an element in L, and let j be an element in {1, . . . , d}. Then we have the following. (i) If j = 0, φ(σr+ (l) ) = 0. j

274


(ii) If d is even and j is odd, φ(σr− (l) ) = 0. j

(iii) If both d and j are even, φ(σr− (l) ) = (nd/2 + 1)nj/2 . j

Proof. (i) For each element x in X, xσr+ (l) = xσrj (l) ; j

see Theorem 9.4.8(i). Therefore, if j = 0, φ(σr+ (l) ) = χCX (σrj (l) ) = 0. j

(ii) Assume that d is even and that j is odd. Then, by Lemma 12.6.1(ii), X has no element x with xa ∈ xrj (l). On the other hand, as j is assumed to be odd, we obtain from Theorem 9.4.8(iv) that φ(σr− (l) ) = |{x ∈ X | xa ∈ xrj (l)}|. j

Thus, φ(σr− (l) ) = 0. j

(iii) Assume that d and j are even. Let i be the integer with 2i = j, and let x be an element in X. Note that xri (l) contains a fixed point of a if and only if xa ∈ xrj (l). Thus, referring to Theorem 9.4.8(iv) we obtain from Lemma 12.6.2 that φ(σr− (l) ) = (nd/2 + 1)ni . j

This proves (iii). The first part of the following theorem was proved by Stanley Payne; cf. [34; Theorem 2]. Its second part is due to Udo Ott [33; Satz 1]. The proof of Theorem 12.6.4 is an application of Theorem 9.1.7(ii) to one of the non-linear irreducible characters of Sζ A and to the character of C(Sζ A) afforded by CX in place of φ and ψ. The left hand side of the equation in Theorem 9.1.7(ii) is easy to compute since we know from Proposition 12.6.3(i), (ii) that φ vanishes on most of the values of Sζ A. Theorem 12.6.4 Assume that S possesses a polarity and that 2 ≤ n. Then we have the following. √ (i) If d = 4, 2n is an integer. √ (ii) If d = 6, 3n is an integer. Proof. Let h and k be elements in L such that {h, k} = L. Let a be a polarity of S, set A := {1, a}, and set R := Sζ A. From the definition of R we obtain |R| = 4d, and from Corollary 7.3.4(ii) we obtain that r1+ (h) and r0− (h) are involutions of R.

12.6 Polarities

275

From Corollary 7.3.4(ii) we also obtain that R is a Coxeter scheme with respect to {r1+ (h), r0− (h)} satisfying {r2− (h)} = r1+ (h)r0− (h)r1+ (h) and

{r2− (k)} = (r0− (h)r1+ (h))2 r0− (h).

Moreover, if 4 ≤ d, we have {r4− (h)} = (r1+ (h)r0− (h))3 r1+ (h) and

{r4− (k)} = (r0− (h)r1+ (h))4 r0− (h).

Finally, if 6 ≤ d, we have {r6− (h)} = (r1+ (h)r0− (h))5 r1+ (h). Let C be an algebraically closed field of characteristic 0, let Q denote the smallest subfield of C, and let Z denote the smallest unitary subring of C. We set I := IC (Z). Recall that C2d is our notation for the set of all elements c in C \ {−1, 1} satisfying c2d = 1. Since C is assumed to be algebraically closed, C2d contains an element c such that, for each integer i with 1 ≤ i ≤ 2d − 1, ci = 1. We fix one of these elements and call it c. We define χ to be the non-linear irreducible character of C(R) which satisfies c ∈ Cχ ; cf. Lemma 12.4.4. Let us denote by φ the character of C(R) afforded by CX. (i) Assume that d = 4. From Proposition 12.4.2(i) (applied to R in the role of S) we compute that √ χ(σr− (h) ) = (n − 1) 2n = χ(σr− (k) ) 2

2

and that χ(σr− (h) ) = 0. Moreover, by Proposition 12.6.3(iii), 4

φ(σr− (h) ) = (n2 + 1)n = φ(σr− (k) ). 2

2

Therefore, referring to Proposition 12.6.3(i), (ii) we obtain 1 1 1 χ(σr∗ )φ(σr ) = 2nS + 2 χ(σr− (h) )φ(σr− (h) ) + 2 χ(σr− (k) )φ(σr− (k) ) 2 2 2 2 ∗ nr n n

r∈R

√ 2 (n − 1) 2nφ(σr− (h) ) 2 n2 √ 2 2 = 2nS + (n − 1)(n + 1) 2n. n = 2nS +

276


Since we assume that 1 = n, we√conclude from Theorem 9.1.7(ii) that Q. Therefore, by Lemma 8.2.5, 2n ∈ Z.

√

2n ∈

(ii) Let us now assume that d = 6. From Proposition 12.4.2(i) (applied to R in the role of S) we compute that √ χ(σr− (h) ) = (n − 1) 3n = χ(σr− (k) ), 2

that

2

√ χ(σr− (h) ) = n(n − 1) 3n = χ(σr− (k) ), 4

4

and that χ(σr− (h) ) = 0. Moreover, by Proposition 12.6.3(iii), 6

φ(σr− (h) ) = (n3 + 1)n = φ(σr− (k) ), 2

that

2

φ(σr− (h) ) = (n3 + 1)n2 = φ(σr− (k) ). 4

4

Therefore, referring to Proposition 12.6.3(i), (ii) we obtain 1 1 1 χ(σr∗ )φ(σr ) = 2nS + 2 χ(σr− (h) )φ(σr− (h) ) + 2 χ(σr− (k) )φ(σr− (k) ) 2 2 2 2 nr∗ n n

r∈R

1 1 χ(σr− (h) )φ(σr− (h) ) + 4 χ(σr− (k) )φ(σr− (k) ) 4 4 4 4 4 n n √ 2 = 2nS + 2 (n − 1) 3nφ(σr− (h) ) 2 n √ 2 + 3 (n − 1) 3nφ(σr− (h) ) 4 n √ 2 = 2nS + 2 (n − 1) 3n(n3 + 1)n n √ 2 + 3 (n − 1) 3n(n3 + 1)n2 n √ 4 = 2nS + (n − 1)(n3 + 1) 3n. n √ Thus, by Lemma 8.2.5, 3n ∈ Z. +

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Index

π-number 75 π-subset 77 π-valenced nonempty subset p-subset 28 p-valenced element 28 p-valenced nonempty subset

completely reducible 165 complex addition 157 complex multiplication 156 complex multiplication in a scheme complex product V, 156 complex sum 157 componentwise addition 159 composition factor 95 composition series 94 conjugate 35 constrained 55 cover 111 Coxeter scheme 258 Coxeter set 58

75

28

act transitively 112 additively written group 96 algebraic ring element 164 algebraically closed field 164 algebraically closed subfield 164 artinian module 167 artinian ring 171 associative operation 95 associative ring 155 automorphism 83 base field

185

center of a ring 155 centralizer of a nonempty subset centralizer of an element 32 character 176 characteristic 161 close faithful maps 254 closed subset VI commutative closed subset 7 commutative group 154 commutative ring 155 commutator 43 commutator subset 43 compatible 254, 256 complement 148

32

decomposable 135 degree 176 dihedral 209 direct product of closed subsets direct product of schemes 140 direct sum 166 division ring 156 Double Centralizer 171 endomorphism 159 exchange condition 54 extend faithfully 103 Extension Theorem 247 factor module 157 faithful 103 faithfully T -embedded 112 faithfully embedded 103 field 156

133

V

282

Index

finite over a unitary subring finitely generated 161 First Isomorphism Theorem Frattini subset 43

161

module homomorphism 158 Moore set 234 morphism 83 multiplication 155 multiplicative inverse 156 multiplicatively written group multiplicity 189

91

generate 39 generators of a module 161 group 95 group correspondence VI group endomorphism 159 group homomorphism 97 group over a commutative group

154

Hall π-subset of a closed subset 77 homogeneous 167 homomorphism 83 Homomorphism Theorem 90, 158 ideal 156 identity element 96 indecomposable 135 integral domain 179 integral ring element 161 inverse element 95 involution VII irreducible character 176 irreducible module 164 isomorphic D-modules 158 isomorphic composition series isomorphic groups 97 isomorphic rings 172 isomorphic schemes 91 isomorphism 83 Isomorphism Theorem 159 Jacobson Density Theorem Jacobson radical 167

left coset 20 left transversal 20 length 40 linear character 176 maximal closed subset 28 maximal submodule 164 minimal ideal 173 module 155

operation 95, 154 orthogonality relations

90

190

parabolic 237 permutation 89 polarity 271 principal character 186 principal module 186 quasidirect product 140 quotient scheme 65

94

170

kernel of a morphism 86 kernel of a semidirect product

natural homomorphism naturally valenced 63 neutral element 95 normal 31 normalize 31 normalizer 31

95

148

rank 258 recognition theorem X regular module 186 regularity condition XI residually thin 98 right transversal 20 ring 155 ring homomorphism 171 ring with 1 155 scheme V scheme ring 185 Schur group 103 Schur relations 194 Schur’s Lemma 169 schurian 103 Second Isomorphism Theorem semidirect product 148 semisimple 172 simple closed subset 95 simple ring 157 spherical 61 stabilizer 113

91

Index standard character 185 standard module 185 Steinberg character 267 strong normalizer 33 strongly normal 34 structure constant VI subfield 156 submodule 156 subnormal series 94 subring 155 subscheme 22 sum of submodules 165 supplement 43 Sylow p-subset 78 symmetric element 5 symmetric subset 7

283

thin element V thin nonempty subset V thin radical 35 thin residue 45 Tits system 259 transversal of a closed subset 20 transversal of two closed subsets 20 unit 156 unital 154 unitary subring

161

valency of a nonempty subset valency of an element 1 vector space 156

2

Theory of Association Schemes (Springer Monographs in Mathematics)

Theory of Association Schemes (Springer Monographs in Mathematics)

Theory of Association Schemes