Characters of finite groups [2 ed.] 9783110224061, 3110224062, 9783111740461, 3111740463

Table of contents :
List of frequently met concepts and notations
Preface
Contents
I. Basic concepts
II. Characters
III. On arithmetical properties of characters
IV. Products of characters
V. Induced characters and representations
VI. Projective representations
VII. Clifford theory
VIII. Brauer’s induction theorems
IX. Faithful characters
X. Existence of normal subgroups
XI. On sums of degrees of irreducible characters
Bibliography
Author index
Subject index

Citation preview

Yakov G. Berkovich, Lev S. Kazarin, Emmanuel M. Zhmud’ Characters of Finite Groups

De Gruyter Expositions in Mathematics

| Edited by Lev Birbrair, Universidade Federal do Ceará, Fortaleza, Brazil Victor P. Maslov, Russian Academy of Sciences, Moscow, Russia Walter D. Neumann, Columbia University, New York, USA Markus J. Pflaum, University of Colorado, Boulder, USA Dierk Schleicher, Jacobs University, Bremen, Germany

Volume 63

Yakov G. Berkovich, Lev S. Kazarin, Emmanuel M. Zhmud’

Characters of Finite Groups | Volume 1 2nd edition

Mathematics Subject Classification 2010 20C15 Authors Prof. Dr. Yakov G. Berkovich 18251 Afula Israel [email protected] Prof. Dr. Lev S. Kazarin Department of Mathematics Yaroslavl P. Demidov State University Sovetskaya Str. 14 150000 Yaroslav Russia [email protected]

ISBN 978-3-11-022406-1 e-ISBN (PDF) 978-3-11-022407-8 e-ISBN (EPUB) 978-3-11-039054-4 Set-ISBN 978-3-11-174046-1 ISSN 0938-6572 Library of Congress Cataloging-in-Publication Data A CIP catalog record for this book has been applied for at the Library of Congress. Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2018 Walter de Gruyter GmbH, Berlin/Boston Typesetting: Dimler & Albroscheit, Müncheberg Printing and binding: CPI books GmbH, Leck ♾ Printed on acid-free paper Printed in Germany www.degruyter.com

List of frequently met concepts and notations Set theory ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙

|M| is the cardinality of a set M (if G is a group, then |G| is called the order of G). x ∈ M means that x is an element of M. N ⊆ M means that N is a subset of M; if N ≠ M, we write N ⊂ M. 0 is the empty set. N is called a nontrivial subset of M, if N ≠ 0 and N ⊆ M. If N ⊂ M, we say that N is a proper subset of M. M ∩ N is the intersection and M ∪ N is the union of sets M and N. If M, N are sets, then N − M is the difference of N and M. ℂ is the set (field) of complex numbers. ℝ is the set (field) of real numbers. ℚ is the set (field) of rational numbers. ℤ is the set (ring) of integers: ℤ = {0, ±1, ±2, . . . }. ℕ is the set of natural numbers.

Number theory and general algebra ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙

p is always a prime number. m, n are always natural numbers. GCD(m, n) is the greatest common divisor of m and n. m | n should be read as follows: m divides n. π(m) is the set of all prime divisors of m. π is a set of primes (it may be the empty set). π󸀠 is the set of primes not contained in π. m π is the number satisfying the following three conditions: π(m π ) ⊂ π, m π | m, π(m/m π ) ⊂ π󸀠 . We write m p , p󸀠 instead of m{p} , {p󸀠 }, respectively. m is a π-number if π(m) ⊆ π (or m π = m). GF(p m ) is the finite field containing p m elements. 𝔽∗ is the multiplicative group of a field 𝔽. 𝔽n is the n-dimensional vector space over 𝔽. 𝔽n is the set of all n × n matrices over 𝔽. If A is a square matrix, then det A and tr A are the determinant and the trace of A (that is, the sum of elements on its principal diagonal), respectively. I n is the n × n identity matrix. α is the number conjugate to α ∈ ℂ. [x] is the integer part of x ∈ ℝ.

DOI 10.1515/9783110224078-202

VI | Characters of Finite Groups 1

Groups ∙ ∙ ∙

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G is, after §I.1, always a finite group. H ≤ G means that H is a subgroup of G. H < G means that H ⊆ G and H ≠ G (in this case H is called a proper subgroup of G); {1} denotes the group of order 1; H is a nontrivial subgroup of G if {1} < H < G. H is a maximal subgroup of G if H < G and H ≤ M < G imply that H = M. H ⊴ G means that H is a normal subgroup of G; moreover, if H ≠ G, we write H ⊲ G and say that H is a proper normal subgroup of G. H ⊲ G is called a nontrivial normal subgroup of G if |H| > 1. H is a minimal normal subgroup of G if (a) H ⊴ G, (b) H > {1}, (c) N ⊲ G and N < H imply N = {e}. Thus, {1} has no minimal normal subgroups. G is simple if it is a minimal normal subgroup of G (in particular, |G| > 1). H is a maximal normal subgroup of G if G/H is simple. G is a monolith if G = {1} or if G contains only one minimal normal subgroup. The subgroup generated by all minimal normal subgroups of G is called the socle of G and is denoted by Sc(G). One can represent Sc(G) as the direct product of certain minimal normal subgroups of G. We put Sc({1}) = {1}. Obviously, Sc(G) is a characteristic subgroup of G. NG (M) = {x ∈ G | x−1 Mx = M} is the normalizer of a subset M in G. CG (x) is the centralizer of an element x in G: CG (x) = {z ∈ G | zx = xz}. CG (M) = ⋂x∈M CG (x) is the centralizer of a subset M in G. Aut G is the group of all automorphisms of G (the automorphism group of G). Inn G is the group of all inner automorphisms of G. Out G = Aut G/Inn G. [x, y] = x−1 y−1 xy is the commutator of elements x, y of G. If M, N ⊂ G, then [M, N] = ⟨[x, y] | x ∈ M, y ∈ N⟩. (However, we use [M, N] = {[x, y] | x ∈ M, y ∈ N} in Chapter XI.) If M ⊂ G, then ⟨M⟩ is the subgroup of G generated by M. G󸀠 is the subgroup generated by all commutators [x, y], x, y ∈ G (i.e., G󸀠 = [G, G]), G󸀠󸀠 = (G󸀠 )󸀠 , G󸀠󸀠󸀠 = (G󸀠󸀠 )󸀠 and so on. Z(G) = ⋂x∈G C G (x) is the center of G. Φ(G) is the Frattini subgroup of G (the intersection of all maximal subgroups of G). F(G) is the Fitting subgroup of G (the maximal normal nilpotent subgroup of G). S(G) is the solvable radical of G (the maximal solvable normal subgroup of G). exp G is the exponent of G (the least common multiple of the orders of the elements of G). o(x) is the order of an element x of G. k(G) is the number of conjugacy classes of G (= G-classes), the class number of G. If M ⊆ G, then kG (M) is the number of G-classes containing elements of M. π(G) = π(|G|).

List of frequently met concepts and notations |

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VII

Oπ (G) is the maximal normal π-subgroup of G, O(G) = O2󸀠 (G) (obviously, one has Op (G) ∈ Sylp (F(G))). Oπ (G) is the subgroup generated by all π󸀠 -elements of G. Cm is the cyclic group of order m. A × B is the direct product of groups A and B. A ∗ B is a central product of groups A and B. G0 = {1}; G m is the direct product of m copies of G. Ep m = (Cp )m is the elementary abelian group of order p m . A group G is said to be homocyclic if it is a direct product of isomorphic cyclic subgroups (obviously, elementary abelian p-groups are homocyclic ). ES(m, p) is an extraspecial group of order p l+2m (a p-group G is said to be extraspecial if G󸀠 = Φ(G) = Z(G) is of order p). A special p-group is a nonabelian p-group G such that G󸀠 = Φ(G) = Z(G) is elementary abelian. (A, B) is a Frobenius group with kernel B and Frobenius complement A (A and B do not determine (A, B) up to isomorphism). D2m is the dihedral group of order 2m, m > 2. Q2m is the generalized quaternion group of order 2m > 2, m > 3. SD2m is the semidihedral group of order 2m ≥ 24 . cl G is the nilpotency class of a p-group G. CL G is the set of all G-classes. A p-group of maximal class is a nonabelian group G of order p m with cl G = m − 1. m If G is a p-group, then Ω m (G) = ⟨x ∈ G | x p = 1⟩, and m.G = ⟨x m | x ∈ G⟩. Syl(G) is the set of all Sylow subgroups of G. Sylp (G) is the set of all Sylow p-subgroups of G. H is a Hall subgroup of G if (|H|, |G : H|) = 1. H is a π-Hall subgroup of G if |H| = |G|π . Sn is the symmetric group of degree n. An is the alternating group of degree n. GL(n, F) is the set of all nonsingular n × n matrices with entries in a field F, the general linear group over F. SL(n, F) = {A ∈ GL(n, F) | det A = 1 ∈ F}, the special linear group over F. PGL(m, F) = GL(n, F)/Z(GL(n, F)). PSL(n, F) = SL(n, F)/Z(SL(n, F)). AGL(n, F) is the natural extension of F n by GL(n, F), the affine general linear group. Sz(2m ) is the simple Suzuki group, m > 1 being odd. For H < G, H G = ⋂x∈G x−1 Hx is called the core of the subgroup H in G. Obviously, H G ⊴ G. An element x ∈ G is a π-element if π(o(x)) ⊆ π. G is a π-group if π(G) ⊆ π. Obviously, G is a π-group if and only if all its elements are π-elements.

VIII | Characters of Finite Groups 1

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Oπ (G) = ⟨x ∈ G | π(o(x)) ⊆ π󸀠 ⟩. Oπ,σ (G) = Oσ (Oπ (G)). A group G is an extension of N ⊴ G by a group H if G/N ≃ H. A group G splits over N if G = HN with H < G and H ∩ N = 1 (in that case, G is a semidirect product of H and N with kernel N). A group G is p-solvable if all indices of its composition series are equal to p or are p󸀠 -numbers. A group G is π-solvable if it is p-solvable for all p ∈ π. A group G is said to be π-separable if all indices of its composition series are π- or π󸀠 -numbers. If M ⊆ G, x ∈ G, then M x = x−1 Mx = {x−1 ax | a ∈ M}. H is a TI-subgroup of G if H ∩ H x = 1 for all x ∈ G − NG (H); M is a TI-subset of G if M ∩ M x ⊆ {1} for all x ∈ G − NG (M). H # = H − {e H }, where e H is the identity element of the group H. If M ⊆ G, then M # = M − {e G }. A permutation σ of a set M is regular if σ(x) ≠ x for all x ∈ M. An automorphism a of G is regular (= fixed-point free) if it induces a regular permutation on G# . If x, y ∈ G, then the expression “x ∼ y in G” means that x, y are conjugate in G. Similarly, “M ∼ N in G” means that the subsets M, N are conjugate in G. An involution is an element of order 2 in a group. An element x ∈ G is real if x ∼ x−1 in G. An element x is rational if all generators of the subgroup ⟨x⟩ are conjugate in G. An involution is a real and rational element. A section of a group G is an epimorphic image of some subgroup of G. A group G is p-closed if |Sylp (G)| = 1 (i.e., Op (G) ∈ Sylp (G)). A group G is p-nilpotent if it has a normal p-complement, i.e., a normal subgroup H of order |G|p󸀠 . An S(p a , q b , q c )-group is a q-closed minimal nonnilpotent group G of order p a q b+c with |Z(G)| = p a−1 q c (see Chapter XI). If F = GF(p n ), then we write GL(m, p n ), etc., instead of GL(m, F), etc. If M ⊆ G, then M G is the normal closure of M in G.

Characters and representations ∙ ∙ ∙ ∙ ∙ ∙ ∙

F[G] is the set of all functions from G to ℂ. CF[G] is the set of all central (= class) functions from G to ℂ. Char(G) is the set of all complex characters of G. It is convenient to consider the zero function OG→ℂ as an element of the set Char(G). Irr(G) is the set of all irreducible characters of G. A character of degree 1 is said to be linear. Lin(G) is the set of all linear characters of G (obviously, Irr(G) ⊆ Irr(G)). Irr1 (G) = Irr(G) − Lin(G) is the set of all nonlinear irreducible characters of G; n(G) = |Irr1 (G)| is the number of nonlinear irreducible characters of G.

List of frequently met concepts and notations | IX

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A class function θ is said to be a generalized character of G if θ = χ1 − χ2 , where χ1 , χ2 ∈ Char(G). Ch(G) is the set of all generalized characters of G. If θ, λ ∈ F[G], x ∈ G, then (θλ)(x) = θ(x)λ(x). FG is the group algebra of G over the field F. χ(1) is the degree of a character χ of G; deg T is the degree of a representation T of G. If χ ∈ Char(G), ϕ ∈ Char(H), H < G, then χ H is the restriction of χ to H, and ϕ G is the induced character (ϕ G ∈ Char(G)). If ϑ, ψ ∈ CF[G], then ⟨ϑ, ψ⟩ = |G|−1 ∑x∈G ϑ(x)ψ(x) is the scalar (or inner) product of ϑ and ψ. If H ≤ G, ϕ ∈ Irr(H), then I G (ϕ) = {x ∈ G | ϕ x = ϕ} is the inertia group of ϕ in G (where ϕ x (h) = ϕ(xhx−1 ) for h ∈ H). If H ≤ G and ϕ ∈ CF[H], then ϕ̇ is the function in CF[G] that coincides with ϕ on H and vanishes on G − H. 1G is the principal character of G (1G (x) = 1 for all x ∈ G). ρ G is the regular character of G. Irr(χ) is the set of all irreducible constituents of a character χ of G. Furthermore, Irr1 (χ) = Irr(χ) ∩ Irr1 (G). (The expression ψ ∈ Irr(χ) means that the character ψ is a constituent of χ.) X(G) is the character table of G, and X1 (G) is its first column (consisting of the degrees of irreducible characters, counting multiplicities). M(G) is the Schur multiplier of G. If M is a set, the Kronecker symbol δ : M × M → {0, 1} is defined as follows: if a = b, then δ a,b = 1, and if a ≠ b, then δ a,b = 0. cd G = {χ(1) | χ ∈ Irr(G)}. cd1 G = {χ(1) | χ ∈ Irr1 (G)} = cd G − {1}. b(G) = max{n | n ∈ cd G}. ker T is the kernel of a representation T. ker χ is the kernel of a character χ. Z(χ) = {x ∈ G | |χ(x)| = χ(1)} is the quasikernel of χ ∈ Char(G). Tχ = {x ∈ G | χ(x) = 0} is the set of zeros of χ ∈ Ch(G). Uχ = {x ∈ G | |χ(x)| = 1} is the set of χ-unitary elements of G (where χ ∈ Ch(G)). Let N ⊴ G. Then IrrN (G) = {χ ∈ Irr(G) | N < ker χ}. We often identify the sets IrrN (G) and Irr(G/N). Next, Irr(G, N) = Irr(G) − Irr(G/N); LinN (G) = Lin(G) ∩ IrrN (G). Irrϕ (G) = {χ ∈ Irr(G) | ⟨χ N , ϕ⟩ > 0}, where N ⊴ G, ϕ ∈ Irr(N). Let H < G, ϕ ∈ Irr(H), χ ∈ Irr(G). Then χ is an extension of ϕ to G if χ H = ϕ. ν2 (χ) is the Frobenius–Schur indicator of χ ∈ Irr(G) (see Chapter IV). mc(G) = k(G)/|G| is the measure of commutativity of G. T(G) = ∑χ∈Irr(G) χ(1), and f(G) = T(G)/|G|. Let T be a representation, affording the character χ of G. Then the function det χ : G → ℂ∗ is defined by det(χ)(x) = det T(x), x ∈ G. Obviously, det χ ∈ Lin(G).

X | Characters of Finite Groups 1

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If χ ∈ CF(G), then χ : G → ℂ is defined by χ(x) = χ(x), x ∈ G. If X ⊆ Irr(G), then X # = X − 1G . In particular, Irr# (G) is the set of all nonprincipal characters of G. Irr1 (G, p󸀠 ) = {χ ∈ Irr1 (G) | p ∤ χ(1)}. T1 (G, p󸀠 ) = ∑χ∈Irr1 (G,p󸀠 ) χ(1). If P ∈ Sylp (G), then T1 (G, P, p󸀠 ) = ∑χ∈Irr1 (G,p󸀠 )P≤ker ̸ χ χ(1). Kern G = {ker χ | χ ∈ Irr1 (G)}. v(x) = |{χ(x) | x ∈ G}|. A character χ of G is monolithic if χ ∈ Irr(G) and G/ker χ is a monolith. Irrm (G) is the set of all monolithic characters of G, Irr1,m (G) = Irrm (G) ∩ Irr1 (G).

Preface The representation theory of finite groups is now more than 100 years old. Its foundations were laid down by Frobenius, Burnside, Schur and, later, Brauer. It was Frobenius and Burnside who first realized the importance of representation theory for analyzing the structure of finite groups. Their classical papers amaze us even today with their depth and originality, and experts are still pondering the same fundamental problems. It suffices to note that Frobenius, in his very first paper on character theory, constructed the character table of the group PSL(2, p), a result regarded even now as highly nontrivial. The representation theory of finite groups is still developing vigorously, with the active participation of prominent mathematicians. Characters constitute one of the main tools of the representation theory of finite groups over the complex field (we will consider only such representations in Chapters I–XXVI). The goal of this book is to place character theory and its applications to finite groups within the reach of people with a comparatively modest mathematical background, exceeding the usual algebra course only with respect to finite groups. In our opinion, it should not be very difficult for people with a good knowledge of the theory of finite groups to read this book (which is indeed intended primarily for such readers). But even people with a rather superficial knowledge of finite groups will be able to master the basics of representation theory if they read the first sections of Chapters I–VIII. The first edition was published by Y. G. Berkovich and E. M. Zhmud’ in 1997. This book prepared by the authors together with L. S. Kazarin follows the main road of the previous edition. In fact, the titles of the chapters in the first volume are the same as in the earlier version, but the containment is wider. First of all, this is concerned the exercises (the whole number of them in the first volume is now 446, while the previous one has 257 only). The book consists of two volumes. The present Volume 1 contains Chapters I–XI, Volume 2 (currently scheduled to be published in 2018) contains Chapters XII–XXVII. Although a very detailed table of contents is provided, we think it is appropriate to survey the contents of the book. In this survey we will describe the structure of the book, emphasize its main themes and point out connections between chapters. Chapter I. We introduce the main notions, prove such facts of primary importance as Schur’s lemma and Maschke’s theorem, and study the group Lin(G) of all irreducible representations of a finite abelian group over the complex field. §I.10 contains some important corollaries and applications of Maschke’s theorem. On the other hand, we include in Chapter I the results concerning the structure of semisimple finitedimensional algebras and their representations. This chapter ends by two appendices, that are interesting due to their applications. Chapter II. We prove the orthogonality relations and deduce their simplest consequences. Then we begin the study the relations between the character table X(G) of DOI 10.1515/9783110224078-203

XII | Characters of Finite Groups 1 a group G and the properties of the group (this theme recurs through the whole book). For example, it is shown in [Gar2] that the character table of a solvable group enables one to determine its Frattini subgroup (i.e., to describe all G-classes that belong to it); but, as Garrison has shown, the character table does not generally determine the Frattini subgroup. Chapter III. Starting from the fact that the values of characters are algebraic integers, we deduce a series of deep theorems, the most important of which is Burnside’s theorem on the solvability of {p, q}-groups. L. S. Kazarin [Kaz8] in Chapter XXVII obtained a substantial improvement of the p a -lemma that implies Burnside’s theorem. Namely, he showed that any element of G, the index of whose centralizer is a prime power, belongs to a solvable radical of G (his proof uses modular theory). Note that in the 1970s there appeared a proof of Burnside’s {p, q}-theorem that does not use character theory (Goldschmidt, Bender, Matsuyama). In §III.8 we present a short introduction to the theory of rational groups, i.e., groups all of whose characters are rational-valued. Next an extensive survey of the characters of p-groups follows (many of the results that we present about the characters of p-groups are due to A. Mann). The chapter is ended by P. Gallagher’s theorem on commutators. Chapter IV. We begin this chapter with a short introduction to multilinear algebra. We define some operations with representations and characters and prove the important Burnside–Brauer theorem on powers of faithful characters, as well as an analogous result on powers of conjugacy classes (Garrison). These two results inspired a large amount of work on analogies between irreducible characters and conjugacy classes (Arad, Brenner, Mann, Blau, Chillag, Herzog, and others). We introduce the Frobenius–Schur indicator and present a formula for the number of involutions in a group. The irreducible characters of direct products and their kernels and quasikernels are then studied in detail. From these results we deduce an important corollary, due to Schur, stating that the degree of an irreducible character divides the index of the center (the presented proof is due to Tate). The chapter is concluded with Frobenius’ proof of his fundamental theorem on the number of solutions of the equation x n = 1 in a group (in Chapter V we present two more proofs of this theorem, one of them not using characters). Chapters V and VII are the central of the book. Some of the results proved there are among those we refer to most often in the sequel. Chapter V. Our presentation of the theory of induced characters follows an approach outlined in the well-known paper of Brauer and Tate [BrT], according to which the Reciprocity Law is postulated. Nevertheless, the chapter ends with another way to develop the theory of induced representations, due to Mackey. We prove the important Mackey theorems on restrictions of induced characters [Mac1]; these theorems are then used to deduce irreducibility criteria for induced characters, due to Mackey and Shoda. As a nice application of the theory we present a proof of the Brauer–Suzuki–

Preface | XIII

Wall theorem on groups with elementary abelian centralizers of involutions [Bra1] – historically speaking, this was one of the first characterization results. We must mention that this theorem appeared in 1900 in a paper of Burnside, which was forgotten afterwards. Another proof of this theorem, using Bender’s method, is presented in Chapter XVI. Incidentally, the whole direction of characterization results in the theory of finite groups is rather poorly represented in the book. We believe that the theorems on intersections of kernels of certain characters are of some interest (see Chapter XIV for more results of this type). The chapter ends with a short survey of results about the number of elements of a given order and the number of subgroups of given structure in a group. Chapter VI. We develop a theory of projective representations, study Schur multipliers and representation groups, and calculate multipliers of abelian groups (Schur). We prove that an abelian group possesses a faithful projective representation with at most two irreducible components (Zhmud’). From this result we deduce a description of the abelian groups that admit faithful irreducible projective representations (Frucht [Fru]). We also prove a realization theorem (a projective representation of a group G may be realized over the cyclotomic field generated by the |G|-th root of unity; the proof uses Brauer’s theorem on the realization of ordinary representations; see Chapter VIII). We study the p-groups with large multipliers. For example, we show that a group of order p n may have a multiplier of order p n(n−1)/2 (i.e., of maximal possible order) if and only if it is an elementary abelian group; we also describe the p-groups satisfying the same condition with a multiplier of order p n(n−1)/2−1 , cf. [Ber14]. We also prove the Gaschütz–Neubüser–Ti Yen estimate for the order of the multiplier of a p-group; this proof, due to the third author, does not use cohomology theory. The following result deserves attention: if |G/Z(G)| = p n and |G󸀠 | = p n(n−1)/2 , then G/Z(G) is either elementary abelian or nonabelian group of order p3 and exponent p, cf. [Ber14] (the remark about the order is due to A. Mann). Chapter VII. In this chapter we present Clifford’s classical work [Cli]. His result on the ramification index of an irreducible character over a normal subgroup will be especially important further on. This result implies Ito’s Theorem VII.2.3, which states that the degree of an irreducible character divides the index of an abelian normal subgroup [Ito2]. Similarly we prove a more general Reynolds’ theorem, which states that the degree of an irreducible character ϕ of a group G divides |G : H|ϕ(1), where H ⊴ G. Gallagher’s fundamental results (see [Gal4]) on the extension of invariant characters of a normal subgroup are presented. As a corollary, we obtain the inequality k(G) ≤ k(H)k(G/H), where H ⊴ G (see [Gal8]). Among other important results we want to emphasize Tate’s p-nilpotency criterion [Tat]. The chapter ends with a results on CR- and NR-subgroups of finite groups. The definition of a CR-subgroup is due to I. M. Isaacs. Theorems giving criteria of the existence of normal complements for certain subgroups are obtained. The appendix contains results on the squares of characters due to Isaacs and Zisser [IZ].

XIV | Characters of Finite Groups 1 Chapter VIII. We prove the Brauer induction theorems. These theorems, together with their corollary, the Realization Theorem, are among the most important achievements of character theory. The rest of the chapter is devoted to various applications of these theorems. In particular, we present a complete exposition of Brauer’s important paper [Bra8] on quotient groups of finite groups. The induction theorems undoubtedly have a potential far beyond this. Chapter IX. A necessary and sufficient condition for a group to admit a faithful representation with at most k irreducible constituents is presented [Zhm1, Zhm3]. We believe that special attention should be paid to the theorem stating that the number of kernels equals the number of antikernels (by “kernel” we mean the kernel of an irreducible character, and by “antikernel” the subgroup generated by a conjugacy class). For example, if distinct irreducible characters have distinct kernels (such a group is called a CM-group) and H is an antikernel, then the class generating H is uniquely determined. We study the structure of CM-groups and their generalizations. CM-groups are rational. All the results of §IX.1 and §IX.3 were first proved by the third author [Zhm8, Zhm14], but our presentation in §IX.3 differs considerably from the original one (see [BZ1]). Chapter X. The chapter revolves around Frobenius’ famous theorem on transitive groups in which the stabilizer of any two different points is trivial. We also consider other types of groups arising naturally in this connection. We prove several generalizations and converse theorems. Frobenius’ theorem treats an important particular case of the following problem also formulated by Frobenius: If a natural number n divides G and the number of solutions of the equation x n = 1 in G is n, is it true that the solutions constitute a subgroup? This problem was recently solved (in the affirmative) using the classification of finite simple groups [Iiy]. It is shown that the first column, X1 (G) of the character table enables one to decide whether G is a Frobenius group. We give a brief introduction to the theory of exceptional characters and Suzuki’s theorem on the solvability of CA-groups of odd order (a group is called a CA-group if the centralizer of any element ≠ 1 is abelian). We note that X1 (G) determines the complex group algebra ℂG and vice versa. We consider several examples illustrating the influence of X1 (G) on the structure of G. For example, we prove Isaacs’ Theorem which asserts that X1 (G) enables one to decide whether G is p-nilpotent. This approach is generalized in Chapter XI. Chapter XI. We introduce the functions T(G) (= the sum of degrees of irreducible characters of G), f(G) = T(G)/|G|, and mc(G) = k(G)/|G|. Of course, the knowledge of X1 (G) permits to calculate these functions (but the converse is not true). The main theorem classifies those groups G for which f(G) > l/p, where p is the smallest prime divisor of G. Note that T(G) > |{x ∈ G | x2 = 1}| (by the Frobenius–Schur formula; see Chapter IV), and this makes it possible to use the main theorem to obtain a description of groups at least half of whose elements are involutions (see [Wall]; the proof

Preface |

XV

presented is due to K. G. Nekrasov). It is further shown that if H ⊴ G, ϕ ∈ Irr(H) and |Irr(ϕ G )| > |G : H|/4, then G/H is solvable (see [Ber19]). Chapter XII. We study groups all of whose nonlinear irreducible characters take exactly three values, also proving some related results. All the results of this chapter were proved jointly by Chillag and the authors (see [BCZ]). Chapter XIII. The Sylow 2-subgroup A(m, Θ) of a Suzuki simple group Sz(2m ) has interesting extremal properties. Some of them are inherited by its generalization A p (m, Θ). In fact, these groups arising naturally, when some group G ≤ Aut(H) has small number of orbits on a group H. I. A. Sagirov (student of L. S. Kazarin) described the character degrees of a groups A p (m, Θ) in the cases p = 2 and p > 2. His results are presented in this chapter. Chapter XIV. This chapter is devoted to one of the central themes of the book, the connection between the degrees and kernels of irreducible characters. It is shown that the quasikernel (and, therefore, the kernel) of an irreducible character of maximal degree is nilpotent. The same can be also said about the minimum (with respect to inclusion) quasikernels and kernels. We prove Thompson’s theorem [Tho4] on p-nilpotency of a group such that all its nonlinear irreducible characters are of degree divisible by a fixed prime p, and present some related results. Incidentally, Chapter XXV contains a proof of the fact that the groups, arising in Thompson’s theorem, are even solvable (here we use the classification of finite simple groups; see Proposition 25.9 and Remark 1 after it). As an easy consequence of Thompson’s theorem one obtains another theorem, also due to Thompson, which asserts that G has an ordered Sylow tower if cd(G) is a chain with respect to divisibility. A similar situation, in which cd(G)−{χ(1)} is a chain for a certain χ ∈ Irr1 (G), and in addition it is assumed that χ(1) is prime to any element of cd(G) − χ(1), is much more difficult; nevertheless, here too can achieve a good description of G. Note that Tate’s theorem on p-nilpotency plays a considerable role in the proof of the latter result. Thompson’s theorem is also an easy corollary of Tate’s theorem. Tate’s theorem is also used to prove Isaacs’ theorem on the solvability of a group G with |cd(G)| ≤ 3. A criterion for π-closure is formulated in terms of characters. Three appendices to the chapter are of independent interest. Chapter XV. In this chapter we study the structure of a group G satisfying the following condition: If χ ∈ Irr(G) is such that G/ker(χ) is a monolith (such a character is said to be monolithic), then χ assumes at most three distinct values. We show that G is solvable, π(G) ⊆ (2, 3, 7}. In §XV.3 we consider a generalization of CM-groups (see §IX.3). §XV.4 contains generalizations of some known results. In §XV.5 we will generalize Isaacs’ three character degrees theorem (see Theorem XIV.7.1). If the degrees of nonmonomial irreducible characters of the group G are primes, then G is solvable (I. A. Sagirov). In Appendix A.15 we will give a fairly easy proof of the number-theoretic Zsigmondy theorem [Roi1, Roi2], which is important in finite group theory. This chapter is a continuation of Chapter XII.

XVI | Characters of Finite Groups 1 Chapter XVI. We begin with an exposition of the classical Brauer–Fowler paper [Bra4] on groups of even order. This paper, published in 1955, played (and still plays) an important role in the subsequent development of finite group theory, especially now that the solvability of groups of odd order has been proved [FeiT3]. We describe Bender’s method computing the order of a group under some natural condition. At the end we present an approach of bounding the order of a group by L. S. Kazarin and B. Amberg. Chapter XVII. We study the lattice of disconnected groups and double Frobenius groups. Then we prove a theorem of Veitsblit, which gives a classification of groups with two infinitely distant involutions. The chapter ends with the characterization of Zassenhaus groups by the third author. Chapter XVIII. We obtain a description of the irreducible linear groups G of degree p, where p is the least prime divisor of |G|; we also prove the important Jordan theorem on linear groups (the proof presented here is due to Frobenius). Chapter XIX. The first half of the chapter is an introduction to the character theory of multiply transitive groups. In the last section we present the first steps of Young’s approach to representations of symmetric groups. Chapter XX. The material of this chapter is taken from unpublished notes of the third author. Here we study so-called S-characters. A nonprincipal character θ of a group G is said to be an S-character if θ is a generalized character assuming nonnegative values and such that ⟨θ, 1G ⟩ = 1. An obvious example of such a character is the regular character of a group. There are many nontrivial results concerning such characters. Chapter XXI. As far as we know, this is the most complete presentation of the theme “zeros of characters” (Karpilovsky’s book [Kar5] contains a special chapter on this topic). Burnside has showed that any nonlinear irreducible character has a zero (i.e., an element of the group on which the character takes the value zero). Further results on zeros, obtained by Gallagher [Gal6], are complemented and strengthened in this chapter. An important role is played by Veitsblit’s inequality, which gives an estimate for the number of zeros of an irreducible character [Vei]. The chapter contains an extensive survey of the theory of the so-called S-characters, a notion due to the second author (as are all results of the survey). Chapter XXII. The results of this chapter are due to S. Ostrovskaya (former student of the third author). An element u of a group G is said to be totally unitary (= TU-element) provided |χ(u)| = 1 for all χ ∈ Irr(G). The set of TU-elements of G will be denoted by U(G). A group G with U(G) ≠ 0 is said to be a TU-group. The complete characterization of a metabelian TU-group is the main result of the chapter. Chapter XXIII. This is an elementary introduction to the theory of the Schur index, constituting an important part of the theme “arithmetic properties of characters”. This material was moved from Chapter III to Volume 2 for reasons of continuity only.

Preface |

XVII

Chapter XXIV. We study the groups that possess only two nonlinear irreducible characters of the same degree (D1 -groups). Solvable D1 -groups are classified. It follows from the result of Kazarin and the first author [Ber4] that PSL(2, 5) and PSL(2, 7) are the only nonsolvable D1 -groups. An important role is played here the characterizations of Frobenius groups proved from Chapter X. Chapter XXV. This chapter is devoted to Saksonov’s results on isomorphisms of character rings and of units of the character rings. Behavior of the regular character under isomorphisms described by E. M. Zhmud’. Chapter XXVI. The group G is called an SR-group (simply reducible) if every element of G is real and the tensor product of any two irreducible representations have multiplicities in its decomposition into the sum of irreducible representations of G not exceeding 1. These groups were introduced by E. Wigner and are interesting by many reasons. The main result of the chapter is solvability of finite SR-groups, obtained by L. S. Kazarin and his former students V. Yanishevskii and E. Chankov. Chapter XXVII. This chapter is an introduction for modular character theory. The first ten parts are prepared by E. M. Zhmud’ and the remaining three (and almost all exercises) by L. S. Kazarin. It is evident that we concentrate mostly on applications, while purely theoretic questions occupy a relatively modest part of the book. A reader interested in a more detailed study of the theory is referred to the books by Isaacs, Feit, Dornhoff, Huppert, Gorenstein, Suzuki, Collins, Navarro and Curtis–Reiner, and also to the books of Karpilovsky, which can be viewed as an encyclopedia of representation theory. Moreover, we regard our book as a complement to those mentioned above. It is especially useful to read it in parallel with one of them (especially those of Isaacs and Karpilovsky). In the contents and the style of the presentation we were greatly influenced by Isaacs’ text, while the influence of other authors is comparatively small. The material of Chapters IX, XI, XII, XIV, XVI, XVII, XXI, XXIII–XXVI is presented in monographic form for the first time. Other chapters also contain much new material. Our presentation is fairly detailed. In Chapters I–XXVI we restricted ourselves to ordinary representations, so the mathematical prerequisites are rather modest. We hope that readers acquainted with the basics of the theory of finite groups will not find the book difficult. The exercises scattered through the whole book form an important part of the presentation. They are of varying degrees of difficulty – from purely technical ones, used in the main text, to really hard, often unsolved problems. In addition, we provide a long list of open problems at the end of Volume 2, written by the first author (many unsolved questions are formulated in the main text of the book as well). Most of the problems in the list were posed by the first author, but there are also some known problems. We also append a list of frequently met concepts and notations (containing definitions).

XVIII | Characters of Finite Groups 1 M. Roitman (University of Haifa) read some portions of the final text and made numerous useful remarks. His help to the first author with PC was invaluable. The authors of the results are mentioned wherever their names are known to us (unfortunately, the literature sometimes shows discrepancies on this point). The bibliography at the end of the book, though very incomplete, nevertheless contains many important works. Yakov Berkovich (Afula, Israel) Lev Kazarin (Yaroslavl, Russia) Emmanuel Zhmud’ (Kharkov, Ukraine)

Contents List of frequently met concepts and notations | V Preface | XI I 1 2 3 4 5 6 7 8 9 10 11 12 A B

Basic concepts | 1 Permutation representations | 1 Modules, operator representations | 9 Representations of algebras | 10 Group algebras | 13 Matrix representations | 15 Completely reducible modules | 18 Schur’s lemma | 19 Maschke’s theorem | 23 Representations of abelian groups | 25 Homogeneous completely reducible modules | 30 Semisimple algebras | 35 Representations of semisimple algebras | 47 On q󸀠 -automorphisms of q-groups | 54 B. H. Neumann’s lemma on covering | 61

II 1 2 3 4 5 6 7 8 9

Characters | 65 Functions on a group | 65 Schur’s relations | 65 Characters. The First Orthogonality Relation | 67 Irreducible constituents of a character | 71 The number of irreducible characters of a group | 74 The Second Orthogonality Relation | 77 The character table and the Frattini subgroup | 84 Nagao’s theorem | 87 Saksonov’s example | 88

III 1 2 3 4 5 6 7

On arithmetical properties of characters | 91 Algebraic integers | 91 Values of characters are algebraic integers | 92 The Frobenius–Molien theorem on degrees | 92 Algebraically conjugate characters | 96 Kernels and quasikernels | 100 Solvability of {p, q}-groups | 106 Burnside’s theorems on commutators | 109

XX | Characters of Finite Groups 1 8 9 10 11 12

Rational groups | 113 On characters of p-groups | 118 On the class equation | 123 Mann’s theorems | 124 Gallagher’s theorems on commutators | 129

IV 1 2 3 4 5 6 7 8 9 10 11 12

Products of characters | 135 Tensor product | 135 Exterior power | 140 Symmetric powers | 143 Products of representations | 145 On some generalized characters | 152 The Frobenius–Schur indicator | 154 Solutions of systems of two-member equations | 163 Characters of direct products | 165 The vanishing of defect zero characters | 171 On products of irreducible characters | 172 On the number of solutions of x n = 1 in a group | 173 Mann’s theorem on squares of characters | 178

V 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Induced characters and representations | 181 Induced characters | 181 Kernels of induced characters | 187 Mackey’s theorems | 189 Permutation representations | 192 Induced representations | 202 The Brauer–Suzuki–Wall theorem | 206 Small Burnside’s theorem on 2-transitive groups | 209 Gallagher’s theorem on restriction of characters | 210 Monomial representations and characters | 215 M-groups | 219 On the number of solutions of x n = 1 in a group | 220 Further Mackey’s theorems | 228 Generalizations of M-groups | 230 On intersections of kernels of some characters | 232 Varia | 234

VI 1 2 3 4

Projective representations | 237 Basic notions | 237 Twisted group algebra | 240 The Schur multiplier | 241 The exponent of the Schur multiplier | 253

Contents

5 6 7 8 9 10 11 12 13

| XXI

The Schur multipliers of a group and its quotient group | 258 Projective representations of abelian groups | 270 Commutator subgroups of representation groups | 281 Intersection of kernels | 282 The Schur multiplier of a direct product | 284 Realization theorem | 286 p-groups with large multipliers | 288 More about orders of multipliers | 297 Representation groups of nonabelian metacyclic p-groups | 299

VII 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 A

Clifford theory | 301 Clifford’s theorem | 301 Ito’s theorem on degrees | 305 Clifford’s theorem on ramification | 309 Gallagher’s extension theorem | 319 Isaacs’ restriction theorem | 323 Gallagher’s general extension theorem | 324 p-rational characters | 330 The class number of an extension | 333 Dornhoff’s theorem | 339 Tate’s normal p-complement theorem | 342 Further results on the class number | 344 Miscellaneous | 345 Monomiality criterion | 348 Commutativity criterion of a normal section | 349 Dade’s example | 350 Generalizations of M-groups | 352 On the derived length of a p-group | 353 CR- and NR-subgroups | 355 Squares of characters | 362

VIII 1 2 3 4 5 6

Brauer’s induction theorems | 373 Characterization of characters | 373 Realization theorem | 382 Groups with partitions of special type | 384 Characters of p-defect 0 | 385 Brauer’s paper on quotient groups | 387 Existence of a normal complement to a section | 395

IX 1 2

Faithful characters | 397 Introduction | 397 G-groups | 398

XXII | Characters of Finite Groups 1 3 4 5 6 7

Group-theoretical functions of Möbius and Euler | 407 G-kernels and G-antikernels | 419 Application to the group G | 429 CM-groups | 434 Varia | 449

X 1 2 3 4 5 6 7 8 9 10 11 12

Existence of normal subgroups | 455 Introductory remarks | 455 Wielandt’s theorems | 456 Characters of Frobenius groups | 466 Converses to Frobenius’ theorem | 474 Exercises | 483 Exceptional characters. CA-groups | 486 Cossey–Hawkes–Mann’s and Isaacs’ theorems | 494 Characterization of Frobenius complements | 497 Groups with few irreducible characters of small degrees | 499 Frobenius’ theorem for solvable Frobenius complement | 504 On irreducible constituents of induced characters | 507 The Brauer–Suzuki theorem | 510

XI 1 2 3 4 5 6 7 8 9

On sums of degrees of irreducible characters | 521 Introduction | 521 Minimal nonnilpotent groups | 523 Some properties of functions f(∗) and mc(∗) | 527 Nonnilpotent case of Theorem 1.0 | 540 Examples of groups G satisfying f(G/H) < f(G) (H ⊲ G) | 541 Further properties of functions f(∗) and mc(∗) | 545 Groups with many involutions | 547 Irreducible constituents of induced characters | 555 Mann’s theorem | 557

Bibliography | 559 Author index | 587 Subject index | 591

I Basic concepts In this chapter we consider the basic concepts of the representation theory of finite groups and prove such fundamental results as Maschke’s theorem and Schur’s lemma. At first we will not assume that the groups in question are finite. The ground field will be denoted by 𝔽. We denote by n a natural number. Throughout the chapter, V stands for the n-dimensional linear space over 𝔽 for some n (we sometimes write V = V(n, 𝔽)); the elements of V are called vectors. Familiarity with this chapter is necessary for further reading of the book.

1 Permutation representations We recall the definition and principal properties of homomorphisms of groups. Let G and H be groups. A mapping ϕ : G → H is called a homomorphism of G into H if ϕ(xy) = ϕ(x)ϕ(y) for all x, y ∈ G. The mapping that sends every element of G into the identity element of H is a homomorphism (the trivial homomorphism). With every homomorphism ϕ : G → H we associate two sets, the kernel and the image of ϕ, denoted by ker(ϕ) and im(ϕ), respectively, and defined as follows: ker(ϕ) = {x ∈ G | ϕ(x) = 1 ∈ H},

im(ϕ) = {ϕ(x) | x ∈ G}.

For example, if ϕ : G → H is the trivial homomorphism, then we have ker(ϕ) = G and im(ϕ) = {1H }, the identity subgroup of H. Note. Below and throughout the book, to avoid, in many cases, imperatives, most exercises are phrased as assertions to be proved. Exercise 1.1. The kernel of ϕ is a normal subgroup of G (we write ker(ϕ) ⊴ G), and the image of ϕ is a subgroup of H (we write im(ϕ) ≤ G), G/ker(ϕ) ≅ im(ϕ). Hint. Define a mapping τ : G/ker(ϕ) → im(ϕ) as follows: τ(x ker(ϕ)) = ϕ(x) (x ∈ G). It is easy to check that τ is an isomorphism. If ker(ϕ) = {1}, then a homomorphism ϕ is said to be faithful (in this case, ϕ is called a monomorphism). If im(ϕ) = H, then ϕ is a surjection of G onto H (ϕ is also called an epimorphism). If ϕ is both injective and surjective (i.e., bijective), then ϕ is called an isomorphism of G onto H, and we write ϕ : G ≅ H. Exercise 1.2. Let GL(n, 𝔽) be the group of all n × n nonsingular matrices, i.e., matrices with nonzero determinants, over the field 𝔽, and let 𝔽∗ be the multiplicative group of 𝔽. Given a ∈ GL(n, 𝔽), let det(a) denote the determinant of a. Since det(ab) = det(a) det(b) for a, b ∈ GL(n, 𝔽), we obtain a homomorphism det : GL(n, 𝔽) → 𝔽∗ . The kernel of DOI 10.1515/9783110224078-001

2 | Characters of Finite Groups 1 this homomorphism, which is {a ∈ GL(n, 𝔽) | det(a) = 1𝔽 }, is denoted by SL(n, 𝔽). It follows from im(det) = 𝔽∗ that GL(n, 𝔽)/SL(n, 𝔽) ≅ 𝔽∗ . If 𝔽 is a finite field, then |𝔽| = p n for some prime p, the characteristic of 𝔽, and the group F ∗ is cyclic. The group GL(n, 𝔽) is called the general linear group of degree n over 𝔽 and the group SL(n, 𝔽) is called the special linear group of degree n over 𝔽. If |𝔽| = p m (p is a prime), then the finite cardinality of the field determines it up to isomorphism (E. H. Moore). In this case, instead of GL(n, 𝔽), SL(n, 𝔽) we write GL(n, p m ), SL(n, p m ), respectively. Find |GL(n, p m )| and |SL(n, p m )|. Answer. The number of bases of V(n, p m ) is equal to |GL(n, p m )| = (p nm − 1)(p nm − p m ) . . . (p nm − p(n−1)m )

(n factors).

It follows from the above that |SL(n, p m )| = (p m − 1)−1 |GL(n, p m )|. As we have noted, given a prime power p n , there is exactly one field of cardinality p n . Following L. E. Dickson, we denote this field by GF(p n ) (“GF” is the abbreviation of “Galois field”). Exercise 1.3. Let Sn be the set of all bijections of the set {1, 2, . . . , n}. Then Sn is a group with respect to the following operation: (στ)(x) = σ(τ(x)),

σ, τ ∈ Sn , x ∈ {1, 2, . . . , n},

known as the symmetric group of degree n. Elements of the group Sn are called permutations. Every permutation may be presented as a product of transpositions in many ways, however, the number of factors in all such products have the same parity. Given π ∈ Sn , we define sgn(π) = (−1)i π , where i π is the number of factors in certain decomposition of π in a product of transpositions. The number sgn(π) is called the sign of permutation π. A permutation π is even if sgn(π) = 1 and odd if sgn(π) = −1. Thus, { 1 if σ is even, sgn(σ) = { −1 if σ is odd. { If n > 1, then the mapping sgn : Sn → {1, −1} is a surjective homomorphism of Sn onto C2 , the cyclic group of order 2 (if n = 1, then sgn is the trivial homomorphism of Sn into C2 ); in this case, ker(sgn) = An , the set of all even permutations in Sn ; this is a subgroup of Sn , which is called the alternating group of degree n. In this case, if n > 1, then |Sn : An | = 2 (Exercise 1.1) and An ⊲ Sn . We also set A1 = S1 . As we know, |Sn | = n!. By Cayley’s theorem, a finite group G of order n is isomorphic to a subgroup of the symmetric group Sn . Homomorphisms into a group acting on a set present a particular interest. Let S be a multiplicative semigroup with the identity element 1 (= monoid) and let M be a set. Then a mapping δ l : S × M → M (for s, m ∈ M we denote δ l (s, m) by sm) is called the left action of S on M if the following conditions are fulfilled: (g1 g2 )m = g1 (g2 m)

(D1)

I Basic concepts | 3

and 1m = m

(D2)

for 1, g1 , g2 ∈ S, m ∈ M. If an element s ∈ S is invertible, the mapping m 󳨃→ sm is a bijection P(s) of the set M. Indeed, if m, m1 ∈ M and sm = sm1 , then m = s−1 (sm) = (s−1 s)m) = m, i.e., P(s) is injective. Next, s(s−1 m) = m so that P(s) is surjective. The set of all bijections of the set M is denoted by S(M); it is a group with respect to the natural composition, known as the symmetric group on M. Clearly, Sn ≅ S({1, 2, . . . , n}). If S = G is a group, then the left action δ l affords a homomorphism P : G → S(M) given by P(g)(m) = gm (g ∈ G, m ∈ M). Indeed, if g, h ∈ G, m ∈ M, then P(gh)(m) = (gh)m = g(hm) = P(g)(P(h)(m)) 󳨐⇒ P(gh) = P(g)P(h). If the sets M and N have the same cardinality, then S(M) ≅ S(N) (in fact, if λ : N → M is a bijection, then π 󳨃→ λ−1 πλ is an isomorphism (π ∈ S(M)). The group GL(n, 𝔽) can be considered as a subgroup of the symmetric group S(𝔽n ), where 𝔽n is the set of all columns of height n with entries from 𝔽. However, GL(n, 𝔽) ≠ S(𝔽n ), because GL(n, 𝔽) leaves the zero column fixed. Besides, these groups are not isomorphic as permutation groups. Indeed, if 𝔽 is finite, they have different orders; if 𝔽 is infinite, the center of the first group in not trivial (it is isomorphic to 𝔽∗ , the multiplicative group of 𝔽 and consists of all scalar matrices) and the center of the second group has order 1. Now let a group G act on a set M; denote the action by δ : G × M → M. Set ker(δ) = {g ∈ G | gm = m for every m ∈ M}. Since ker(δ) = ker(P), where P is a homomorphism of G into S(M) afforded by δ, it follows that ker(δ) ⊴ G (see Exercise 1.1). We say that an action δ of a group G on a set M is faithful if ker(δ) = {1}. In this case G is isomorphic to a subgroup of S(M). If x, y ∈ M, we write x ∼ y if and only if y = gx for some g ∈ G. The relation ∼ is an equivalence relation on M. The equivalence classes of the relation ∼ are called G-orbits. If G has only one orbit on M, the action of G on M is said to be transitive; otherwise it is intransitive. It is easy to check that the symmetric group Sn acts transitively on the set {1, 2, . . . , n}. We offer to readers to show that, if n > 2, then the alternating group An acts transitively on the set {1, 2, . . . , n}. Suppose that x is an element of M. Let O(x) denote the G-orbit containing x and let G x = {g ∈ G | gx = x}. Obviously, G x is a subgroup of G; it is called the stabilizer of x in G (or G-stabilizer of x). There exists a one-to-one correspondence between the G-orbit O(x) and the set of (left) cosets of G x in G. Then, for h ∈ G, let the point hx corresponds to the coset hG x . (It is easy to check that {g ∈ G | gx = hx} = hG x .) Hence, if |O(x)| < ∞, then |G : G x | = |O(x)|. g If x, y ∈ M, g ∈ G and x = gy, then G y = G x . Indeed, if h ∈ G y , then (hg−1 )x = hy = y = g −1 x 󳨐⇒ (ghg−1 )x = x

4 | Characters of Finite Groups 1 g

g

so that G y ≤ G x . If h ∈ G x , then (ghg −1 )x = x 󳨐⇒ hy = (hg −1 )x = g −1 x = y, g

and we conclude that (G x )g ≤ G y . Thus, G y = G x . Hence, stabilizers of points from the same G-orbit are conjugate in G. In particular, if G is transitive, then the G-stabilizers of points are G-conjugate. If G is a transitive permutation subgroup of Sn , then the intersection of G-stabilizers of all points is the identity subgroup (indeed, any permutation from this intersection fixes all points). The mapping δ l : G × G → G defined by δ l (g)(m) = gm (g, m ∈ G) is an action of G on itself, called the left regular action. This action is faithful and transitive. Indeed, if gm = m, then g = 1. Given m, m1 ∈ G, there is g ∈ G such that gm = m1 (indeed, take g = m1 m−1 ). Similarly, one can define a right action M × G → M. The right regular action δ r : G × G → G, defined by δ r (g)m = mg−1 , is an example of a right action (this action is also transitive). Indeed, δ r (gh)m = m(gh)−1 = mh−1 g −1 = δ r (g)mh−1 = δ r (g)δ r (h)m so that δ r (gh) = δ r (g)δ r (h). Sometimes we write δ r (g)(m) = m g . Another action of the group G on itself is the mapping τ : G × G → G defined by τ(g)m = gmg−1 , called conjugation. The kernel of conjugation is the center Z(G) of G. If |G| > 1, then conjugation is intransitive. The G-orbits of the above action of G on G are called conjugacy classes; the number of conjugacy classes, denoted by k(G), is called the class number of G. If x ∈ G, then G x , the G-stabilizer of x, is said to be the centralizer of x and denoted by CG (x). If K x is the conjugacy class containing x, then |K x | = |G : CG (x)|. Definition 1.1. A homomorphism ϕ of G into the symmetric group S(M), where M is a nonempty set, is called a permutation representation of G of degree |M|. It follows from the above that any left action δ l : G × M → M of the group G on the finite set M affords a representation P : G → S(M) of G by permutations of M. In particular, the left regular action of G (on G) affords so-called left regular permutation representation L of G. If G = {x1 , . . . , x h } and g ∈ G, then L(g) = (

x1 , . . . , x h ). gx1 , . . . , gx h

is the left shift on g. Let ϕ󸀠 : G → S(M 󸀠 ) be another permutation representation of G. Then ϕ and ϕ󸀠 are said to be equivalent (we write ϕ ∼ ϕ󸀠 ) if there is a bijection τ : M 󸀠 → M such that ϕ󸀠 (g) = τ−1 ϕ(g)τ for all g ∈ G. Equivalent permutation representations have the same degree and their kernels coincide. Later, the symbol ∼ may have another meaning (for example, if x, y ∈ G, we write x ∼ y provided x and y are conjugate in G), but it will always be clear from the context what we mean.

I Basic concepts | 5

As we have noted, equivalent permutation representations of G have the same kernel. If M ≠ M 󸀠 , then im(ϕ) ≠ im(ϕ󸀠 ) as subsets of the disjoint sets S(M) and S(M 󸀠 ), but if M = M 󸀠 , then ϕ ∼ ϕ󸀠 implies that im(ϕ) and im(ϕ󸀠 ) are conjugate in S(M). A representation ϕ : G → S(M) defines an action of G on M by gm = ϕ(g)m (g ∈ G, m ∈ M). The kernel of this action coincides with ker(ϕ). Given an action of G on M, the last equality, if it is read backwards, defines a permutation representation of G. More generally, if H is a group acting on a set M and ϕ : G → H is a homomorphism, then the formula gm = ϕ(g)m defines an action of G on M. Indeed, if g1 , g2 ∈ G, then (g1 g2 )m = ϕ(g1 g2 )m = ϕ(g1 )(ϕ(g2 )m) = g1 (g2 m). Exercise 1.4. Let L : G → G be the left shift of the group G = {x1 , . . . , x h } and L(g) = (

x1 , . . . , x h ), x i1 , . . . , x i h

where i1 , . . . , i h is a permutation of numbers 1, . . . , h. Writing Preg (g) = (

1, . . . , h ), i1 , . . . , i h

prove that the mapping g 󳨃→ P req (g) is the permutation representation of G equivalent to L. The representation Preg is said to be the standard permutation regular representation of G. Exercise 1.5. Let H ≤ G and let a mapping δ : H → G be such that δ(h)g = hg (h ∈ H, g ∈ G). Show that δ is a left action of the subgroup H on the set G. Find the orbits of this action. Prove the Lagrange theorem. Exercise 1.6. Let A, B ≤ G and let a mapping δ : A × B → G be such that δ((a, b))g = agb−1

((a, b) ∈ A × B, g ∈ G).

Prove that δ is a left action of the group A × B on G and find the orbits of this action. Next, δ is a product of left and right shifts. Exercise 1.7. Let a mapping δ : G → G be such that δ(g)x = g −1 xg = x g (g, x ∈ G). Show that this is the exponential (= right) action of G on itself. Find the orbits of this action and prove that the size of the orbit containing x is equal to |G : CG (x)|. Exercise 1.8. For elements a, b, g ∈ G, set g (a,b) = a−1 gb, where we consider (a, b) as an element of the group G × G. Show that the so-defined action is exponential. Solution. If also c, d ∈ G, then (g(a,b) )(c,d) = (a−1 gb)(c,d) = c−1 a−1 gbd = g (ac,bd) = g (a,b)(c,d) . Exercise 1.9. Let A and B be subgroups of a finite group G. Prove the following product formula: |A||B| |AB| = . |A ∩ B|

6 | Characters of Finite Groups 1 Solution. Let AB = a1 B ∪ a2 B ∪ ⋅ ⋅ ⋅ ∪ a n B be a partition, where a i ∈ A for all i. Then A acts on the set {a1 B, . . . , a n B} by (a, a i B) 󳨃→ aa i B (a ∈ A, i ∈ {1, . . . , n}. This action is transitive. We may assume that a1 = 1. Then the stabilizer of the point B in A is A ∩ B. It follows that |A : (A ∩ B)| = n. Thus, |A||B| . |AB| = n|B| = |A : (A ∩ B)||B| = |A ∩ B| Exercise 1.10. Let A, B ≤ G be finite and let x ∈ G. Find |AxB|. Solution. We have, by Exercise 1.9, −1

|AxB| = |AB x | =

−1

|A||B x | |A||B| = . −1 x |A ∩ B | |A ∩ xBx−1 |

Exercise 1.11. Let G be the symmetry group of the cube. Then G acts transitively on the set of six faces of the cube. Obviously, the stabilizer of a face in G is the symmetry group of this face, and this is isomorphic to D8 , the dihedral group of order 8. It follows that |G| = 6 ⋅ 8 = 48. Let H be the rotation group of the cube. Prove that |H| ≅ S4 . Since |G : H| = 2 and Aut(H) ≅ H, Z(H) = {1}, we get G = H × Z, where |Z| = 2. Prove that Z = ⟨σ⟩, where σ is the central symmetry. Exercise 1.12. Prove that the symmetry group of the regular tetrahedron is isomorphic to S4 . Exercise 1.13. Find the order of the symmetry group G of the regular icosahedron. Let H be the rotation group of the regular icosahedron. Prove that |G : H| = 2. Is it true that G = H × Z, where |Z| = 2? Exercise 1.14. Using that the group GL(n, q) acts transitively on the set of all nonzero vectors of the linear space V = V(n, q), find |GL(n, q)|. Prove that the number of bases of the space V is equal to |GL(n, q)|. If σ ∈ S(M), then the set supp(σ) = {x ∈ M | σ(x) ≠ x} is said to be the support of σ. The following result is classical. If σ, τ ∈ S(M) and |supp(σ) ∩ supp(τ)| = 1, then the commutator [σ, τ] is a 3-cycle. The following generalization was offered by Moshe Roitman (University of Haifa). Exercise 1.15. Let σ and τ be permutations of a (nonempty) set A, not necessarily finite, and suppose that S = supp(σ) ∩ supp(τ) ≠ 0. We have (1) supp([σ, τ]) ⊆ S∪σ−1 (S)∪τ−1 (S). Thus, if k := |S| is finite, then |supp([σ, τ])| ≤ 3k. (2) The following three conditions are equivalent for k finite: (a) |supp([σ, τ])| = 3k. (b) supp([σ, τ]) = S ∪̇ σ−1 (S) ∪̇ τ−1 (S) (a partition). (c) The commutator [σ, τ] is a product of k pairwise independent 3-cycles. Solution. First we prove: (∗) Let f ∈ {σ, σ−1 } and g ∈ {τ, τ−1 }, or conversely. If a ∈ supp(g) − g −1 (S), then we have f(g(a)) = g(a).

I Basic concepts | 7

Indeed, g(a) ≠ a, thus g(g(a)) ≠ g(a), that is, g(a) ∈ supp(g). Therefore, since g(a) ∈ ̸ S, by the choice of a, and S = supp(f) ∩ supp(g), we conclude that g(a) ∈ ̸ supp(f), that is, that f(g(a)) = g(a). (1) Let a ∈ A − (S ∪ σ−1 (S) ∪ τ−1 (S)). We show that a ∈ ̸ supp([σ, τ]). Indeed, since a ∈ ̸ S, and since supp([σ, τ]) = supp([σ, τ]−1 ) = supp([τ, σ]), one may assume that a ∈ ̸ supp(σ), i.e., σ(a) = a. If τ(a) = a, we have [σ, τ](a) = a, i.e., a ∈ ̸ supp([σ, τ]). If τ(a) ≠ a, then by (∗) we obtain that σ(τ(a)) = τ(a) = τ(σ(a)), thus a ∈ ̸ supp([σ, τ]), and (1) holds. (2) Clearly (b) ⇒ (a) and (b) ⇒ (c). By (1), we have (a) ⇒ (b). Thus we assume (c) and prove (b). To this end, let s ∈ S. Using (∗), we obtain [σ, τ](s) = σ−1 τ−1 στ(s) = σ−1 τ−1 τ(s) = σ−1 (s). Also Finally,

[σ, τ](σ−1 (s)) = σ−1 [τ, σ−1 ](s) = σ−1 τ−1 (s) = τ−1 (s). [σ, τ](τ−1 (s)) = σ−1 τ−1 σ(s) = σ−1 σ(s) = s.

Hence [σ, τ] is the product of the k disjoint 3-cycles (s, σ−1 (s), τ−1 (s)) for s ∈ S, thus (b) holds. In condition (2)(b) (Exercise 1.15), we may not drop the requirement that the union be disjoint. For example, let A = {1, 2, 3, 4}, σ = (123) and τ = (234). Then we have [σ, τ] = (12)(34), S = {2, 3}, and supp([σ, τ]) = A = S ∪ σ−1 (S) ∪ τ−1 (S) is not a partition. Remark (Roitman, Zhmud, independently). Given π ∈ Sn , set ‖π‖ = |supp(π)|. Given σ, τ ∈ Sn , one has ‖στ‖ ≤ ‖σ‖ + ‖τ‖ (triangle inequality). Let ρ(σ, τ) = ‖σ−1 τ‖. Then ρ(σ, τ) ≤ ρ(σ, ω) + ρ(ω, τ) (σ, τ, ω ∈ Sn ), hence ρ : Sn → ℕ ∪ {0} is a metric on Sn . Exercise 1.16. Suppose that a finite group G acts on a finite set Ω. Given x ∈ G, let χ(x) be the number of points in Ω fixed by x. Then χ(x y ) = χ(x) for all y ∈ G and |G|−1 ∑x∈G χ(x) is the number of G-orbits on Ω. (Remark: Here χ is a permutation character of G and |G|−1 ∑x∈G χ(x) = ⟨χ, 1G ⟩. See also Chapter V.) Solution. Consider the |G| × |M| matrix Γ = (γ x,α ) (x ∈ G, α ∈ M), where γ x,α = δ xα,α and δ is the Kronecker delta on the set M. Let O1 , . . . , O k be the set of G-orbits, α i ∈ O i , and G i = G α i , the G-stabilizer of α i . Then the sum of elements in the α-th column of the matrix Γ is equal to |G α |, so the sum of all entries of the matrix Γ is equal to k

k

S = ∑ |G α | = ∑ ∑ |G α | = ∑ |O i ||G α i | = k|G|. α∈M

i=1 α∈O i

i=1

Indeed, if v = xu (u, v ∈ M, x ∈ G), then G u and G v are conjugate in G so have the same order. The sum of elements of x-th row of the matrix Γ is equal to χ(x) so the sum of all entries of the matrix Γ is equal to S1 = ∑x∈G χ(x). Since S = S1 (by double counting), we are done.

8 | Characters of Finite Groups 1 Exercise 1.17. Let V = V(n, 𝔽) be the n-dimensional vector space over the field 𝔽. Then the group GL(V) acts on the set of subspaces of V. Suppose that f ∈ GL(V) fixes all subspaces of V. Prove that f is a scalar transformation, i.e., f(x) = λx, where λ ∈ 𝔽 is the same for all x ∈ V − {0} (obviously, λ depends on f ). Solution. We may assume that n > 1. Then V has two linearly independent vectors u, v. Set f(u) = αu, f(v) = βv(α, β ∈ 𝔽∗ ). Then f(u + v) = αu + βv is a scalar multiple of u + v if and only if α = β. A group G is said to be k-transitive on the set Ω (k ∈ ℕ) if for any two ordered k-subsets {α1 , . . . , α k } and {β1 , . . . , β k } of the set Ω there exists x ∈ G such that x(α i ) = β i for all i. It is easy to check that: (i) The group Sn is n-transitive. (ii) The group An , n > 2, is (n − 2)-transitive. Exercise 1.18. A permutation group G is 2-transitive (= doubly transitive) on the set {1, . . . , n}, n > 1, if and only if it is transitive and, for any i ∈ {1, . . . , n} and some x ∈ G, one has G = G i ∪ G i xG i (here G i is the G-stabilizer of a point i). Solution. Let G be 2-transitive. Then G i is transitive on the set {1, . . . , n} − {i} and |G : G i | = n,

|G i : (G i )j | = |G i : (G i ∩ G j )| = n − 1

for j ≠ i.

Therefore, for x ∈ ̸ G i and j = x−1 (i), we get |G i |2 |G i | = |G i |(1 + x ) |G xi ∩ G i | |G i ∩ G i | |G| |G| (1 + |G i : (G i )j |) = (1 + n − 1) = |G|, = n n and we conclude that G = G i ∪ G i xG i . Now suppose that G = G i ∪ G i xG i . Reading the displayed formula from right to left, we get |G i : (G xi ∩ G i )| = n − 1, so G i is transitive on the set {1, . . . , n} − {i}. It follows that G is 2-transitive. |G i ∪ G i xG i | = |G i | +

Another solution (M. Roitman). Consider the regular left action of G on the left cosets of the stabilizer G i . The group G acts 2-transitively on the set {1, . . . , n} if and only if G i acts transitively on the set {xG i | x ∈ G − G i }. If the action of G is 2-transitive, it follows that G − G i = G i xG i so G = G i ∪ G i xG i . Conversely, assume that G = G i ∪ G i xG i . Let g ∈ G − G i . Then we have g ∈ G i xG i . Hence g ∈ hxG i for some h ∈ G i . It follows that gG i = h(xG i ), so G i acts transitively on the set {xG i | x ∈ G − G i }. The following two exercises are not trivial. Exercise 1.19. Let δ(G) be the minimal degree of a faithful representation of a group G by permutations (in this case, we will consider G as a subgroup of the symmetric group Sδ(G) ). Prove that π(CSδ(G) (G)) ⊆ π(G), where π(G) is the set of prime divisors of |G|.

I Basic concepts | 9

Hint. Prove that δ(A × B) = δ(A) + δ(B) if A and B are nonidentity groups of coprime orders. Exercise 1.20. Let i({1}) = 1 and, if G > {1}, then i(G) = min{|G : H| | H < G}. Prove that G > {1} is simple if and only if δ(G) = i(G). In particular, if G > {1} is solvable and H < G with |G : H| = i(G), then H ⊲ G.) Exercise 1.21. Let ϕ be a permutation representation of a group G and deg(ϕ) = n. Then |G/ker(ϕ)| divides n!.

2 Modules, operator representations If a set M is equipped with some algebraic structure, we might be interested only in actions preserving this structure. An important case is M = V = V(n, 𝔽), the n-dimensional linear 𝔽-space. In this case an action of a semigroup S with identity element 1 = 1S on V must satisfy the following additional conditions: s(αv) = α(sv),

(M1)

s(v + w) = sv + sw,

(M2)

1S v = v

(M3)

for all s ∈ S, w, v ∈ V, α ∈ 𝔽. Such actions are called unital. If S = G is a group and the action of G on V satisfies the above conditions, then V is called a G-module. In what follows G will always denote a finite group, and all considered linear spaces will be of finite dimension over 𝔽. If V and V 󸀠 are linear spaces over the same field 𝔽, then Hom(V, V 󸀠 ) is defined as the set of all linear mappings of V into V 󸀠 . The set Hom(V, V) = End(V) of all linear operators (endomorphisms) of V is a ring (and even, as we shall see later, an 𝔽-algebra). The set of all invertible elements of the ring End(V) is denoted by GL(V) and called the general linear group of V. Definition 2.1. Let V and V 󸀠 be G-modules. A linear mapping τ : V → V 󸀠 is a homomorphism of G-modules (= G-homomorphism) if τ ∈ Hom(V, V 󸀠 )

and

τ(gv) = gτ(v)

(g ∈ G, v ∈ V);

this means that τ commutes with a given action of G. The set of all homomorphisms of a G-module V in a G-module V 󸀠 is a subset of Hom(V, V 󸀠 ) which is denoted by HomG (V, V 󸀠 ). Instead of HomG (V, V) we write EndG (V). Evidently, EndG (V) is a subring of the ring End(V). The identity endomorphism of V is denoted by idV ; then idV is the identity element of the ring EndG (V). Exercise 2.1. If V is a G-module, the mapping ϕ : G → GL(V), defined by ϕ(g)v = gv (v ∈ V, g ∈ G), is a group homomorphism. Conversely, if ϕ : G → GL(V) is a group homomorphism, the above equality, read backward, makes V a G-module.

10 | Characters of Finite Groups 1 Exercise 2.2. Let ϕ : G → H be a homomorphism of groups and let V be an H-module. Then the equality gv = ϕ(g)v (g ∈ G, v ∈ V) converts V into a G-module. Solution. If a, b ∈ G, v ∈ V, then (ab)v = ϕ(ab)v = (ϕ(a)ϕ(b))v = ϕ(a)(ϕ(b)v) = a(bv). Since, if e is the identity element of G, we have ev = ϕ(e)v = v, our claim follows (linearity of the mapping v 󳨃→ gv is obvious). Definition 2.2. A homomorphism ϕ : G → GL(V) is called an operator representation of a group G on a vector 𝔽-space V. The degree deg(ϕ) of ϕ is equal to dim𝔽 (V), the dimension of V. Representations ϕ : G → GL(V) and ϕ󸀠 : G → GL(V 󸀠 ) are called operator-equivalent (this is denoted again by ϕ ∼ ϕ󸀠 ) if ϕ󸀠 (g) = τ−1 ϕ(g)τ

for all g ∈ G and some isomorphism τ : V 󸀠 → V.

An operator representation is a particular case of a permutation representation. The degrees of these two kinds of representations are defined differently. If two operator representations are equivalent as permutation representations, this need not imply that they are equivalent as operator representations. The type of representation will always be clear from the context. When considering operator representations, we shall always speak of operator-equivalence. It follows from Exercise 2.1 that a G-module V defines an operator representation of G on V (in this case we speak on the representation afforded by the G-module V). This relationship is reversible. We state this rigorously as follows. Theorem 2.1. The set of isomorphism classes of G-modules and the set of equivalence classes of operator representations of a group G can be put in one-to-one correspondence by pairing every G-module V with the representation of G afforded by V, and any representation ϕ : G → GL(V) with V, where V is viewed as G-module. Evidently, to prove this, it suffices to show the following statements: (1) Two isomorphic G-modules V, V 󸀠 afford equivalent representations of G. (2) If ϕ, ϕ󸀠 are equivalent representations of G on linear 𝔽-spaces V, V 󸀠 , respectively, then, equipping V and V 󸀠 with the structures of G-modules as in Exercise 2.1, we get the isomorphic G-modules. We leave the proofs to the reader.

3 Representations of algebras In this section we consider homomorphisms of 𝔽-algebras. Definition 3.1. An 𝔽-algebra A is a finite-dimensional linear 𝔽-space, which is at the same time a ring with 1, such that α(xy) = (αx)y = x(αy),

where α ∈ 𝔽, x, y ∈ A.

I Basic concepts | 11

The set 𝔽n of n × n matrices over 𝔽 is an 𝔽-algebra with respect to usual matrix operations (𝔽n is said to be the full matrix algebra over 𝔽). Let A be an 𝔽-algebra. An 𝔽-subspace A󸀠 of the 𝔽-space A is a subalgebra of A if it is a subring of A (in that case, the identity element of A is contained in A󸀠 ). The set of diagonal n × n matrices over 𝔽 is a subalgebra of the algebra 𝔽n . A mapping ϕ : A → A󸀠 is a homomorphism of an 𝔽-algebra A into an 𝔽-algebra A󸀠 if it is a homomorphism of linear spaces and ϕ(xy) = ϕ(x)ϕ(y),

ϕ(1) = 1󸀠

(1, x, y ∈ A, 1󸀠 ∈ A󸀠 ).

The image of ϕ is defined as usual, while the kernel is defined as follows: ker(ϕ) = {a ∈ A | ϕ(a) = 0󸀠 ∈ A󸀠 }. A (left ideal, right ideal) ideal of the algebra A is a respective ideal of the ring A (since A contains the identity element, every ideal of the algebra A is a subspace of the linear 𝔽-space A). It is clear that a proper ideal of A is not a subalgebra of A since it does not contain the identity element of A. A homomorphism ϕ : A → A󸀠 is called an algebra isomorphism if ker(ϕ) = {0} and im(ϕ) = A󸀠 . In that case we write A ≅ A󸀠 . Exercise 3.1. If ϕ : A → A󸀠 is a homomorphism of algebras, then ker(ϕ) is an ideal (= two-sided ideal) of A, im(ϕ) is a subalgebra of A󸀠 and A/ker(ϕ) ≅ im(ϕ). We say that an 𝔽-algebra A acts on an 𝔽-space V if for all 1, a, b ∈ A, v, w ∈ V, α ∈ 𝔽 we have (i) a(bv) = (ab)v, 1v = v, (ii) the mapping A × V → V, (a, v) 󳨃→ av is bilinear, i.e., (1) (a + b)v = av + bv, a(v + w) = av + aw, (2) (αa)v = a(αv) = α(av). Here a left action is defined. Similarly, a right action is defined. Let V be a linear 𝔽-space, let f ∈ End(V) and α ∈ 𝔽. Define αf by (αf)(v) = α ⋅ f(v), v ∈ V. Then one can view End(V) as an 𝔽-algebra. Any action of an 𝔽-algebra A on an 𝔽-space V defines a homomorphism ϕ : A → End(V); conversely, every homomorphism of A → End(V) defines an action of A on V. Definition 3.2. If an 𝔽-algebra A acts on an 𝔽-space V, then V is said to be an A-module. A subspace V1 of V is called a submodule if av1 ∈ V1 for all a ∈ A, v1 ∈ V1 . Submodules of an 𝔽-algebra A considered as A-modules are left ideals of A. Given an A-submodule V1 of V, we put a(v + V1 ) = av + V1 (a ∈ A, v ∈ V). This enables us to consider the quotient space V/V1 as an A-module. A homomorphism from an A-module V into an A-module V 󸀠 is a mapping ϕ ∈ Hom(V, V 󸀠 ) such that ϕ(av) = aϕ(v), where a ∈ A, v ∈ V. It follows that homomorphisms of an A-module V into an A-module V 󸀠 commute with the action of A on V. We denote the set of all homomorphisms satisfying the above condition by HomA (V, V 󸀠 ). It is obvious that EndA (V) = HomA (V, V) is a subalgebra of the algebra End(V). If ϕ ∈ HomA (V, V 󸀠 ),

12 | Characters of Finite Groups 1 then ker(ϕ), im(ϕ) are submodules of V, V 󸀠 , respectively, and V/ker(ϕ) ≅ im(ϕ). If V1 and V2 are submodules of an A-module V, then V1 + V2 = {v1 + v2 | v1 ∈ V1 , v2 ∈ V2 } and V1 ∩ V2 are also A-submodules. If V = V1 ⊕ V2 , then V1 ≅ V/V2 (module isomorphism). The partially ordered set of submodules of V is a lattice. Definition 3.3. Let A be an 𝔽-algebra and V an 𝔽-space. Then a homomorphism of 𝔽-algebras ϕ : A → End(V) is called an operator 𝔽-representation of A. Equivalence of operator representations of algebras is defined in the same way as for groups. Analogs of Theorem 2.1 and Exercise 3.1 are also true for algebras. Exercise 3.2. If A is an 𝔽-algebra and V an A-module, then the set Ann(V) = {a ∈ A | aV = 0} is called the annihilator of V. If ϕ : A → End(V) is a representation afforded by the A-module V, then ker(ϕ) = Ann(V). Next, Ann(V) is a two-sided ideal of the algebra A. Exercise 3.3. Let I be a two-sided ideal of an 𝔽-algebra A. Then the quotient space A/I possesses a natural 𝔽-algebra structure. Suppose that V is an A/I-module. We turn V into an A-module by setting av = (a + I)v (a ∈ A, v ∈ V). It is evident that I ⊆ Ann(V) (where V is considered as an A-module). What is the relationship between Ann(V) and the annihilator of V, considered as an A/I-module? Let us consider an 𝔽-algebra A as a linear 𝔽-space V. For every (a, v) ∈ A × V, we define av as the product of a and v in A. Then V is an A-module (denoted by A A), called the left regular A-module. Its submodules are just the left ideals of the algebra A. So far we have always considered left modules. Right modules can be defined similarly and a corresponding theory can be constructed. However, as a rule, only left modules will be considered in this book. Definition 3.4. Let V ≠ {0} be an A-module. If V has no submodules except {0} and V, then V is called irreducible; otherwise V is reducible. The trivial module {0} is neither reducible nor irreducible. A module V is decomposable if V = V1 ⊕V2 , where V1 and V2 are nonzero submodules (otherwise, V is indecomposable). A module is completely reducible if it is either the zero module or a direct sum of irreducible submodules. We do not give separate definitions for G-modules because, as we shall see in the following section, every G-module can be considered as an A-module, where A is an 𝔽-algebra that is canonically associated with G and the correspondence between A-modules and G-modules is one-to-one. Exercise 3.4. Any nonzero A-module is either indecomposable or a direct sum of indecomposable modules. Definition 3.5. An operator representation ϕ : A → End(V) is reducible, irreducible, completely reducible, etc., if and only if the A-module V is reducible, irreducible, completely reducible, etc., respectively.

I Basic concepts | 13

Given an operator representation of an algebra A on V, there is a natural way of giving V an A-module structure. The representation of A afforded by the regular module A A is called the left regular representation. If we denote it by ρ, then ρ(a)v = av

(v ∈ A A, a ∈ A).

Exercise 3.5 (compare with Exercise 2.2). Let A, A1 be algebras and V an A1 -module. If ϕ : A → A1 is an algebra homomorphism, then the equality av = ϕ(a)v (a ∈ A, v ∈ V) equips V the structure of an A-module.

4 Group algebras Now we define the group algebra 𝔽G of a finite group G over a field 𝔽. This is an object that for all nearest purposes can completely replace the group G itself; any statement about representations of G has an exact equivalent statement about the group algebra 𝔽G and conversely. If x, y, z ∈ G satisfying xy = z, then, in the basis G, the structure constants of the algebra 𝔽G are c zx,y = δ xy,z , where δ is the Kronecker delta on the set G. Recall that all groups in question are finite. Definition 4.1. The group algebra 𝔽G of a group G over a field 𝔽 is the set of all formal linear combinations a = ∑x∈G a(x)x (a(x) ∈ 𝔽 for all x ∈ G) such that a = b ⇐⇒ a(x) = b(x) for each x ∈ G

(A1)

(that is, a : G → 𝔽 is a function and 𝔽G is a subset of the set of all such functions). We define the algebra structure as follows: (a + b)(x) = a(x) + b(x), (γa)(x) = γa(x)

i.e., a + b = ∑ (a(x) + b(x))x,

(A2)

x∈G

(γ ∈ 𝔽, x ∈ G),

(A3) (A4)

(ab)(g) = ∑ a(x)b(y) xy=g

that is, ab = ∑g∈G (∑x,y∈G,xy=g a(x)b(y))g. The dimension of the group algebra 𝔽G as an vector 𝔽-space is equal to |G|. Suppose that V is a G-module, a ∈ 𝔽G is as in Definition 4.1, v ∈ V and put av = ∑x∈G a(x)(xv). This enables us to view V as an 𝔽G-module. Now, identifying an element g of G with the element ∑x∈G δ x,g x of 𝔽G, one may consider G as a subset of 𝔽G. In that case every 𝔽G-module can be considered as a G-module and, vice versa, every G-module can be considered as an 𝔽G-module. In what follows, no distinction will be made between these two types of modules. If τ : G → GL(V) is a representation (here V is a finite-dimensional linear 𝔽-space), we get a representation of the algebra 𝔽G by setting τ(a) = ∑ a(x)τ(x) (a = ∑ a(x)x ∈ 𝔽G), x∈G

x∈G

14 | Characters of Finite Groups 1 which will be denoted by the same letter τ. Let i be the embedding of G into 𝔽G as described in the previous paragraph, and let τ : 𝔽G → End(V) be a representation. Then the composition τ ∘ i is a representation of G. Representations of 𝔽G and G related in this way will be denoted by the same letter. Any submodule of an 𝔽G-module V is a submodule of the G-module V, and every submodule of the G-module V is a submodule of the corresponding 𝔽G-module. Therefore, a representation τ : G → GL(V) is reducible, completely reducible, etc., if and only if the corresponding representation of 𝔽G has the same property. Thus, the study of G-modules is equivalent to the study of 𝔽G-modules (and, similarly, the study of the representations of a group G is equivalent to the study of the representations of the group algebra 𝔽G). In terms of modules, the regular representation of a group G is that afforded by the regular module 𝔽G 𝔽G. Exercise 4.1. If |G| > 1, then the left module 𝔽G 𝔽G is reducible. Solution. Let e = ∑g∈G g ∈ 𝔽G. Then L = 𝔽e is a proper submodule of 𝔽G since we have dim𝔽 (L) = 1 < |G| = dim𝔽 (𝔽G). If two G-modules are isomorphic, so are the corresponding 𝔽G-modules. Consequently, two representations of a group G are equivalent if and only if the corresponding representations of the group algebra 𝔽G are equivalent. A representation ϕ of a group or an algebra is faithful if it is injective. If a representation of a group G is faithful, then the corresponding representation of the algebra 𝔽G need not be faithful. But if a representation of the algebra 𝔽G is faithful, then the corresponding representation of G is always faithful. This incomplete correspondence will almost never cause confusion. Exercise 4.2. The regular representation of an algebra A is faithful. Hint. By definition, A has the identity element. Replacing, in Definition 4.1, the field 𝔽 by any ring R with identity element, we lead to the definition of the group ring RG of a group G over the ring R. We shall consider group rings over commutative rings only. Exercise 4.3. Given X ⊆ G, define X = G − X, [X] = ∑x∈X x ∈ ℤG (in particular, [0] = 0; here ℤ is the ring of rational integers). Given Ai ⊆ G define

(i = 1, . . . , n),

A = (A1 , . . . , A n ), n

A = (A1 , . . . , A n ),

n

d(A) = |G|−1 (∏ |A i | − (−1)n ∏ |A i |), i=1

i=1

and for g ∈ G, NA (g) = |{a1 . . . a n = g | a i ∈ A i , i = 1, . . . , n}|, NA (g) = |{a1 . . . a n = g | a i ∈ A i , i = 1, . . . , n}|.

I Basic concepts | 15

Prove the following assertions: (a) NA (g) − (−1)n NA (g) = d(A). (b) If n is even, then (i) A1 . . . A n = G for d(A) > 0, (ii) A1 . . . A n = G for d(A) < 0, (iii) A1 . . . A n = A1 . . . A n for d(A) = 0. (c) If A, B are nonempty subsets of G, then (i) AB = G if |A| + |B| > |G|, (ii) A B = G if |A| + |B| < |G|, (iii) AB = A B if |A| + |B| = |G|. Hint. Consider

n

n

∏[A i ] = ∏([G] − [A i ]) i=1

i=1

and use the equality [X][G] = [G][X] for X ⊆ G.

5 Matrix representations Let B = {v i }1n be a basis of the linear space V = V(n, 𝔽) and let τ ∈ End(V). If v ∈ V, then n

n

v = ∑ α i v i , τ(v) = ∑ α i τ(v i ) i=1

(α i ∈ 𝔽).

i=1

Thus, τ is defined by the τ-images of the basis vectors. If n

τ(v j ) = ∑ τ ij v i

(τ ij ∈ 𝔽, j ∈ {1, . . . , n}),

i=1

we will associate with τ the matrix (τ ij ) = MB (τ) ∈ 𝔽n , where 𝔽n is the set of all n × n matrices over 𝔽 (as we have noted, 𝔽n is an 𝔽-algebra of dimension n2 ). The coefficients of the vector τ(v j ) are located in j-th column of the matrix MB (τ). It is known that MB is an isomorphism of the algebras End(V) and 𝔽n , and MB (GL(V)) = GL(n, 𝔽). Definition 5.1. A homomorphism T : G → GL(n, 𝔽) is called a matrix representation of a group G. The degree deg(T) of T is equal to n. The representations T : G → GL(n, 𝔽) and T 󸀠 : G → GL(n󸀠 , 𝔽) are said to be equivalent (we write T ∼ T 󸀠 ) if n = n󸀠 and T 󸀠 (g) = S−1 T(g)S for some S ∈ GL(n, 𝔽) and all g ∈ G. Any matrix representation of a group G, extended by linearity, yields a corresponding matrix representation of the group algebra 𝔽G. Everything stated in §I.4 about linear representations of G and 𝔽G related in this way remains valid here. Exercise 5.1. Suppose that T : G → GL(n, 𝔽) is a representation of a group G. Construct a G-module which affords T.

16 | Characters of Finite Groups 1 Hint. Convert the linear space V = 𝔽n of columns of height n with entries from 𝔽 into a G-module using the homomorphism T. See Exercise 2.2. Let τ : G → GL(V) be a representation of a group G, let B = {v i }1n and B󸀠 = {v󸀠i }1n be bases of the 𝔽-space V and let n

v󸀠j = ∑ σ ij v i ∈ 𝔽n

(j ∈ {1, . . . , n}),

i=1

so that S = (σ ij ) is the n × n matrix of transition from the basis B of V to the basis B󸀠 of V 󸀠 . The representation τ affords matrix representations T = MB (τ), T 󸀠 = MB󸀠 (τ) of G such that T 󸀠 (g) = S−1 T(g)S (g ∈ G), i.e., T ∼ T 󸀠 . Since one can view an element of GL(n, 𝔽) as a transition matrix, it follows that τ affords the entire equivalence class of matrix representations of G. Hence any G-module determines the class of matrix representations. Exercise 5.2. Isomorphic G-modules afford equivalent matrix representations and, consequently, equivalent operator representations afford equivalent matrix representations. As follows from Exercise 5.1, a matrix representation can be considered as an operator representation. Exercise 5.3. Let τ and τ󸀠 be equivalent operator representations of a group G afforded by 𝔽G-modules V and V 󸀠 , respectively. Then there exist bases B and B󸀠 of V and V 󸀠 , respectively, such that MB (τ) = MB󸀠 (τ󸀠 ). Definition 5.2. A matrix representation is irreducible, reducible, etc., if and only if the corresponding operator representation possesses this property. Let T : G → GL(n, 𝔽) be a representation afforded by a reducible 𝔽G-module V and suppose that V1 is a nontrivial submodule of V. Let B = {v i }1n be a basis of V whose first m < n elements span V1 . The matrix of any element x ∈ G in this basis has the form T(x) = (

T1 (x) 0

θ(x) ), T2 (x)

where T1 (x) is an m × m matrix (matrices of this form are called block matrices). The representation T1 is afforded by the G-module V1 . It is easy to see that the G-module V/V1 affords a representation T2 of G, and deg(T2 ) = n − m. If a representation τ : G → GL(V) is decomposable with a direct factor V1 , then there exists a basis B of V such that MB (τ)(x) = (

T1 (x) 0

0 ) T2 (x)

We shall now define sums of representations.

(x ∈ G).

I Basic concepts | 17

Let V1 and V2 be G-modules and let V = V1 +̇ V2 = {(v1 , v2 ) ∈ V1 × V2 } be the outer direct sum of V1 and V2 as 𝔽-spaces. The equality g(v1 , v2 ) = (gv1 , gv2 )

(g ∈ G, v1 ∈ V1 , v2 ∈ V2 )

makes V a G-module. Identify v1 ∈ V1 with (v1 , 0) ∈ V and v2 ∈ V2 with (0, v2 ) ∈ V. Then V1 and V2 are embedded in V, so we can consider V as the inner direct sum V1 ⊕ V2 of its submodules V1 and V2 . If τ i ∈ End(V i ), i = 1, 2, then the operator representation of G afforded by V is called the direct sum of τ1 and τ2 and denoted by τ1 + τ2 . Similarly, one can define the sum of any finite number of representations. Let B1 = {v i }1m and B2 = {v i }nm+1 be bases of V1 and V2 , respectively, T i the matrix representation of G afforded by V i , written in terms of the basis B i (i = 1, 2). Then the matrix representation of G afforded by V = V1 ⊕ V2 and written in terms of the basis B = B1 ∪ B2 = {v i }1n is called the sum of the matrix representations T1 and T2 (we write T = T1 + T2 ). We have T(x) = (

T1 (x) 0

0 ) T2 (x)

(x ∈ G).

Exercise 5.4. We have T1 + T2 ∼ T2 + T1 . If T i ∼ T i󸀠 , i = 1, 2, then T1 + T2 ∼ T1󸀠 + T2󸀠 . Hint. Using the previous notation, let B󸀠 = B2 ∪ B1 . Then T2 + T1 = MB󸀠 (τ1 + τ2 ). Exercise 5.5. If T1 , . . . , T k are 𝔽-representations of a group G, then k

ker(T1 + ⋅ ⋅ ⋅ + T k ) = ⋂ ker(T i ). i=1

The above statements remain true in the more general context of associative algebras. Exercise 5.6. Let δ : G × M → M be a left action of the group G = {x1 , . . . , x h } on a finite set M = {m1 , . . . , m n }. For i, j ∈ {1, . . . , n}, set {1 if gm j = m i , α ij (g) = { 0 if gm j ≠ m i . {

(1)

Considering the n × n matrix T(g) = (α ij (g)), prove that the mapping g 󳨃→ T(g) is a matrix representation of the group G afforded by the action δ. The elements α ij (i, j = 1, . . . , n} are said to be matrix elements of the matrix representation T. Let P be a permutation representation of G afforded by the action δ, i.e., P(g)(m) = gm (g ∈ G, m ∈ M). The matrix representation T is said to be the matrix form of the permutation representation P. In particular, the matrix form Treg of the permutation regular representation L of the group G on G (see §I.1) is said to be the matrix regular representation of G. Obviously, deg(Treg ) = h, where h = |G|. The matrix elements α ij (i, j = 1, . . . , h) of the representation Treg are given as in (1).

18 | Characters of Finite Groups 1 Hint. Consider an n-dimensional linear space V over field 𝔽. Let B = {m1 , . . . , m n } be a basis of V. For x = ∑ni=1 ξ i m i , where ξ i ∈ 𝔽 (i = 1, . . . , n), we set gx = ∑ni=1 ξ i (gm i ). Show that V is a left G-module. In the basis B, this module affords the matrix representation whose matrix elements α ij are such as in (1).

6 Completely reducible modules In this section we consider properties of completely reducible A-modules (see Definition 3.3; all algebras are assumed to be associative and all A-modules over 𝔽-algebras have finite dimension over 𝔽). Lemma 6.1. The following assertions are equivalent: (a) A module V is completely reducible, i.e., V is a direct sum of irreducible submodules. (b) V is a sum (not necessarily direct) of a set of irreducible submodules. (c) Each submodule V1 of V is complemented, i.e., we have V = V1 ⊕ V2 , where V2 is a submodule. Proof. We may assume that V is a nonzero A-module. (a) ⇒ (b): Obvious. (b) ⇒ (c): Suppose that V = ∑i∈I L i , where L i are irreducible submodules for all i ∈ I. Let V1 be a proper submodule of V and V2 a submodule of V of maximal dimension such that V1 ∩ V2 = {0}; then V2 ≠ {0} since V1 ∩ L j = {0} for some j ∈ I. In this case, V1 + V2 = V1 ⊕ V2 . If V1 + V2 ≠ V, there is L i ⊈ V1 + V2 for some i ∈ I. Then (V1 ⊕ V2 ) ∩ L i = {0} since L i is irreducible, and so (V1 ⊕ V2 ) + L i = V1 ⊕ V2 ⊕ L i . Therefore, V1 ∩ (V2 ⊕ L i ) = {0}, contrary to the choice of V2 since V2 < V2 ⊕ L i . Thus, V1 ⊕ V2 = V, i.e., the submodule V1 is complemented. (c) ⇒ (a): Suppose that each submodule of V is complemented. Let V1 denote a completely reducible submodule of V of maximal dimension. If V1 ≠ V, then, by hypothesis, V = V1 ⊕ V2 , where V2 is a nonzero submodule of V. Let L be a minimal submodule of V2 . Then V1 ⊕ L is completely reducible and dim(V1 ⊕ L) > dim(V1 ), contrary to the choice of V1 . Hence V = V1 is completely reducible. Lemma 6.2. Let V be an A-module, let L be an irreducible A-module and let ϕ : L → V be a nonzero homomorphism. Then ker(ϕ) = {0} (in particular, im(ϕ) ≅ L). Proof. Since ker(ϕ) is a submodule of the irreducible module L and ϕ is nonzero, it follows that ker(ϕ) = {0}. Corollary 6.3. All nonzero submodules and quotient modules of a completely reducible module V are completely reducible. Proof. By assumption, we have V = L1 ⊕ ⋅ ⋅ ⋅ ⊕ L k , where all L i are irreducible submodules. Let L be a nontrivial submodule of V. We claim that L and V/L are completely reducible.

I Basic concepts | 19

Let ϕ : V → V = V/L be the natural module homomorphism. Then we have that V = L1 + ⋅ ⋅ ⋅ + L k , where L i = ϕ(L i ). By Lemma 6.2, either L i = {0} or L i ≅ L i . Thus, V is completely reducible, by Lemma 6.1. Next, by Lemma 6.1, V = L ⊕ N, where N is a submodule of V. By what has just been proved, L ≅ V/N is completely reducible. If T, T1 , T2 are representations of a group G and T = T1 + T2 , then T1 , T2 are called components of T. Corollary 6.3 implies that components of a completely reducible representation are completely reducible. A linear operator π ∈ End(V) is called a projection of a space V onto its subspace L if im(π) = L and π2 = π. It follows from the definition that π L = idL , where π L is the restriction of π to L. Indeed, if x ∈ L, then there is y ∈ V such that x = π(y). In this case, π(x) = π(π(y)) = π2 (y) = π(y) = x, as required. In particular, if L = V, then π = idV . Lemma 6.4. A submodule L is complemented in an A-module V if and only if there exists a projection π of V onto L which belongs to EndA (V). Proof. Suppose that V = L ⊕ L󸀠 , where L󸀠 is a submodule of V. For v ∈ V such that v = m + m󸀠 , where m ∈ L, m󸀠 ∈ L󸀠 , put π(v) = m. If x ∈ V, then we have x = u + v, where u = π(x) ∈ L, v ∈ L󸀠 . Then ax = au + av for a ∈ A. Since au ∈ L and av ∈ L󸀠 , it follows that au = π(ax), i.e., π(ax) = aπ(x). This means that π ∈ EndA (V). Let π be a projection of V onto L such that π ∈ EndA (V). Setting N = ker(π), we obtain V = L ⊕ N. It remains to check that N is a submodule. Let x ∈ N. We have π(x) = 0, and so π(ax) = aπ(x) = 0. Therefore, ax ∈ N, and so N is a submodule of V. It follows that L is complemented.

7 Schur’s lemma In this and the next section we will prove Schur’s lemma and Maschke’s theorem, which are the most important basic results of representation theory. Some formulations of these results depend upon properties of the field 𝔽. A ring R is said to be a skew field if R∗ = R − {0} is a multiplicative group. Lemma 7.1. Suppose that A is an algebra over a field 𝔽 and V, V 󸀠 are irreducible A-modules. Then {0 if V ≇ V 󸀠 , HomA (V, V 󸀠 ) = { a skew field if V = V 󸀠 . { If, in addition, the field 𝔽 is algebraically closed, then the 𝔽-algebra EndA (V) = 𝔽 ⋅ idV . Proof. Let 0 ≠ ϕ ∈ HomA (V, V 󸀠 ). Since im(ϕ) is a nonzero submodule of the irreducible module V 󸀠 and ker(ϕ) is a proper submodule of the irreducible module V, we get im(ϕ) = V 󸀠 and ker(ϕ) = {0}. Thus, ϕ : V → V 󸀠 is an isomorphism. Suppose that V ≇ V 󸀠 . It follows from V/ker(χ) ≅ im(ϕ) and the previous paragraph that ker(ϕ) = V hence ϕ = 0.

20 | Characters of Finite Groups 1 Suppose that V = V 󸀠 . Then, as we now know, HomA (V, V) = EndA (V) is a skew field. Suppose, in addition, that the field 𝔽 is algebraically closed. Then the characteristic polynomial det(ϕ − λ ⋅idV ) of ϕ has a root α ∈ 𝔽. It follows from det(ϕ − α ⋅idV ) = 0 that the operator ϕ − α ⋅ idV is singular. Since ϕ − α ⋅ idV is an element of the skew field EndA (V), we get ϕ − α ⋅ idV = 0, i.e., ϕ = α ⋅ idV . Thus, EndA (V) = 𝔽 ⋅ idV . It follows that if the A-modules V, V 󸀠 are irreducible and V ≅ V 󸀠 , then all nonzero elements of the set HomA (V, V 󸀠 ) are isomorphisms. The matrix form of Schur’s lemma is a basis of a new approach to character theory in Schur’s paper [Sch2]. Lemma 7.2. Let T and T 󸀠 be irreducible matrix representations of an 𝔽-algebra A with deg(T) = n, deg(T 󸀠 ) = n󸀠 . Suppose that S is an n × n󸀠 matrix over 𝔽 such that (1)

T(a)S = ST 󸀠 (a) for all a ∈ A. Then either S = 0 or n = and S ∈ GL(n, 𝔽), T ∼ algebraically closed, then S is a scalar matrix. n󸀠

T󸀠.

If, in addition, T =

T󸀠

and 𝔽 is

󸀠

Proof. Let V = 𝔽n and V 󸀠 = 𝔽n in Lemma 7.1. Consider V, V 󸀠 as A-modules, i.e., av = T(a)v, av󸀠 = T 󸀠 (a)v󸀠

(a ∈ A, v ∈ V, v󸀠 ∈ V 󸀠 ).

We claim that S ∈ HomA (V 󸀠 , V). Define a linear mapping ϕ : V 󸀠 → V by ϕ(v󸀠 ) = Sv󸀠 for v󸀠 ∈ V 󸀠 . We have, for v󸀠 ∈ V 󸀠 and a ∈ A, aϕ(v󸀠 ) = T(a)Sv󸀠 = ST 󸀠 (a)v󸀠 = Sav󸀠 = ϕ(av󸀠 ), and our claim follows. Now Lemma 7.1 implies the last assertion. A matrix S that satisfies (1) intertwines the representations T and T 󸀠 . Thus, if a representation T : G → GL(n, 𝔽) is irreducible and 𝔽 is algebraically closed, then any n × n matrix commuting with T(g) for all g ∈ G is scalar. Exercise 7.1. The matrices of 𝔽n (the set of n × n matrices over 𝔽) can be considered as linear operators of the space 𝔽n . A set B ⊆ 𝔽n is called irreducible if any B-invariant subspace of 𝔽n is either 0 or 𝔽n . Another version of Schur’s lemma: If the field 𝔽 is algebraically closed, then the centralizer in 𝔽n of an irreducible set of matrices is the field consisting of scalar matrices. Exercise 7.2. We have Z(𝔽n ) = {αIn | α ∈ 𝔽}, where In is the n × n identity matrix. Deduce that Z(GL(n, 𝔽)) = {αIn | α ∈ 𝔽∗ }. Hint. 𝔽n and GL(n, 𝔽) are irreducible sets of matrices. The proof of Theorem 7.3, below, is based on the following: Exercise 7.3. A finite subgroup G of the multiplicative group 𝔽∗ of a field 𝔽 is cyclic. Solution. By hypothesis, G is a finite abelian group. Let p be a prime divisor of |G|. Then the number of solutions of the equation x p = 1 in G is p since a polynomial of

I Basic concepts | 21

degree n over a field has at most n distinct roots. It follows from the basic theorem of abelian p-group theory that the Sylow p-subgroup of G is cyclic for every choice of p. Therefore, G is cyclic as a direct product of cyclic Sylow subgroups. We might have considered the analogs of Lemmas 7.1 and 7.2 for groups. But the context of algebras is more general, owing to the existence of group algebras. In particular, we have the following result. Theorem 7.3. The center of an irreducible finite group G of matrices over the field 𝔽 is cyclic. If, in addition, 𝔽 is algebraically closed, this center consists of scalar matrices. Proof. It follows from the matrix form of Schur’s lemma (Exercise 7.1) that C𝔽n (G) is a skew field. Hence Z(G) is a finite abelian subgroup of this skew field. It is easy to see that the linear hull of Z(G) is a subfield K of the skew field C𝔽n (G). Since Z(G) ≤ K ∗ (the multiplicative group of the field K), the subgroup Z(G) is cyclic, by Exercise 7.3. The last assertion now follows from Schur’s lemma. Exercise 7.4. Let T be a faithful completely reducible representation of a group G. If all irreducible components of T are one-dimensional, then G is abelian. Solution. Indeed, let T1 , . . . , T k be all the irreducible components of T. Then k

{1} = ker(T) = ⋂ ker(T i ), i=1

whence G is isomorphic to a subgroup of the direct product of the abelian groups G/ker(T i ), i = 1, . . . , k. Example. Let B be a nonzero subgroup of the additive group 𝔽+ of a field 𝔽 and let 󵄨 1 b 󵄨󵄨󵄨 G = {( ) 󵄨󵄨󵄨 b ∈ B} . 0 1 󵄨󵄨

Then we can regard G as a group of linear operators of the space 𝔽2 . It is reducible but indecomposable. Note that G ≅ B. Theorem 7.4. The following statements hold: (a) Any irreducible (finite) abelian group of matrices is cyclic. (b) Any irreducible representation of an abelian group over an algebraically closed field is one-dimensional. This is a corollary of Theorem 7.3 and Schur’s lemma. Let i : H ≤ G be an embedding and let T : G → GL(n, 𝔽) be a representation. Then T H = T ∘ i is the restriction of T to H, which maps every x ∈ H into T(x). Clearly, T H : H → GL(n, 𝔽) is a representation of H. Theorem 7.5. Let T be a faithful irreducible matrix representation of a group G over an algebraically closed field 𝔽 and H ≤ G. If the restriction T H is irreducible, then CG (H) = Z(G).

22 | Characters of Finite Groups 1 Proof. Let n = deg(T). Then T(H) = {T(x) | x ∈ H} is an irreducible set of matrices hence, by Exercise 7.1, the set CGL(n,𝔽) (T(H)) consists of scalar matrices. Let g ∈ CG (H). Then the matrix T(g) is scalar, i.e., T(g) ∈ CGL(n,𝔽) (T(H)). Since T is faithful, we get g ∈ Z(G), i.e., CG (H) ≤ Z(G). The reverse inclusion is obvious. Theorem 7.6. Let m ∈ ℕ and 𝔽 = GF(q), where q = p ν , p is a prime, and suppose that G = Cm is a cyclic group of order m, p ∤ m. Then the degree of a faithful irreducible 𝔽-representation T of G is equal to the order of q (mod m). In the proof of this theorem we use the following two lemmas. Lemma 7.7. Let p be a prime and 𝔽 = GF(q), where q = p ν , and let 𝕂 = 𝔽(α) be a finite field with q n elements. If m = o(α) is the order of α in the multiplicative group 𝕂∗ of the field 𝕂, then n is equal to the order of q (mod m). Proof. It follows from α q Then

n

−1

󸀠

= 1 that m | q n − 1. Assume that m | q n − 1, where n󸀠 ≤ n. αq

n󸀠

n󸀠

−1

n ξq

󸀠

=1

so that = α. As 𝕂 = 𝔽(α), we have = ξ for all ξ ∈ 𝕂 (note that ξ is a power n󸀠 n󸀠 of α). In particular, if ⟨ρ⟩ = 𝕂∗ , we get ρ q = ρ. It follows that ρ q −1 = 1 so that q n − 1 󸀠 divides q n − 1, i.e., n ≤ n󸀠 . Thus, n = n󸀠 is equal to the order of q (mod m). αq

Lemma 7.8. Let A be an irreducible n × n matrix over an arbitrary field 𝔽. Then the minimal polynomial m(λ) and characteristic polynomial χ(λ) of A coincide. In particular, deg(m(λ)) = n. Proof. Let V be an n-dimensional 𝔽-space, let B = {e1 , . . . , e n } be a basis of V and let A = M B (ϕ) be the matrix of a linear operator ϕ of V in the basis B. Take x ∈ V − {0}. Suppose that m(λ) = a0 + a1 λ + ⋅ ⋅ ⋅ + a k−1 λ k−1 + λ k , where a i ∈ 𝔽 for all i. Assume that k < n. Then m(ϕ)x = a0 x + a1 ϕ(x) + ⋅ ⋅ ⋅ + a k−1 ϕ k−1 (x) + ϕ k (x) = 0 so the vectors x, ϕ(x), . . . , ϕ k−1 (x), ϕ k (x) are linearly dependent. Therefore, their linear span L is a proper subspace of V. This is impossible since L ≠ {0} is ϕ-invariant and A is irreducible. Thus we must have k = n, i.e., m(λ) = χ(λ) since m(λ) divides χ(λ) and deg(χ(λ)) = n(= deg(m(λ))). Proof of Theorem 7.6. Let 𝕂 be a linear 𝔽-span of matrices T(g), all g ∈ G (here 𝔽 = GF(q)). By Schur’s lemma, 𝕂 is a finite field with q n elements, where n = dim𝔽 (𝕂). Let G = ⟨s⟩. Then A = T(s) is an irreducible matrix over 𝔽 and 𝕂 = 𝔽(A) (i.e., A is a primitive element of the extension 𝕂/𝔽). By Lemma 7.8, the minimal polynomial of A coincides with its characteristic polynomial. Therefore, n = deg(χ(λ)), and this equals the size of A, which is equal to deg(T). Thus deg(T) = n. Since o(A) = o(s) = m, it follows that n equals the order of q (mod m), by Lemma 7.7.

I Basic concepts | 23

8 Maschke’s theorem In this section we prove Maschke’s theorem. Theorem 8.1 (Maschke). Suppose that G is a (finite) group and 𝔽 is a field whose characteristic does not divide |G| (in particular, it is possible that the characteristic of 𝔽 is equal to 0). Then any 𝔽G-module is completely reducible so is any 𝔽-representation of G. Proof. Let L be a submodule of an 𝔽G-module V. By Lemma 6.1, it suffices to show that V = L ⊕ L1 , where L1 is an 𝔽G-submodule of V. By Lemma 6.4, it suffices to construct a projection π of V onto L such that π ∈ End𝔽G (V). Let π0 be any projection of V onto L, i.e., im(π0 ) = L and π02 = π0 (generally speaking, π0 is not an element of End𝔽G (V)). Let τ be the representation of G afforded by the 𝔽G-module V. Put π = (|G|e)−1 ∑ τ(x)−1 π0 τ(x),

(1)

x∈G

where e is the identity element of 𝔽 (since the characteristic of 𝔽 does not divide |G|, we have (|G|e)−1 ∈ 𝔽). Since π is a sum of members of the ring End(V), it follows that π ∈ End(V). We have τ(y)−1 πτ(y) = (|G|e)−1 ∑ τ(y)−1 τ(x)−1 π0 τ(x)τ(y) x∈G

= (|G|e)−1 ∑ τ(xy)−1 π0 τ(xy) x∈G

−1

= (|G|e)

∑ τ(z)−1 π0 τ(z) = π

z∈G

for each y ∈ G (see (1)). Therefore, one has π ∈ End𝔽G (V). If v ∈ V, then xv ∈ L so that π0 (xv) = xv, and we get τ(x)−1 π0 τ(x)v = τ(x)−1 π0 xv = x−1 π0 (xv) = x−1 xv = v ∈ L, whence im(π) = L and π L = idL . Consequently, π2 = π and so π is a projection of V onto L. The averaging method used in the proof of Theorem 8.2 can be applied to prove the following more general result. Theorem 8.2. Let G be a finite group acting on an abelian group A (it is convenient to assume that A is an additive group; if g ∈ G, a, b ∈ A, then g(a + b) = ga + gb). Suppose that for any a ∈ A there exists a unique element |G|−1 a, i.e., the element b ∈ A such that |G|b = a. Let L be a G-invariant subgroup of A such that A = L ⊕ L1 for some subgroup L1 of A. Then A possesses a G-invariant subgroup M such that A = L ⊕ M. Proof. Let a ∈ A, a = x + x1 , x ∈ L, x1 ∈ L1 . Set π0 (a) = x. Obviously, (π0 )L = idL and π02 = π0 . Indeed, using the same notation, we have π02 (a) = π0 (π0 (a)) = π0 (x) = x = π0 (a).

24 | Characters of Finite Groups 1 Let us define π : A → A as follows: π = |G|−1 ∑g∈G g −1 π0 g. As in the proof of Theorem 8.1, π L = idL and π2 = π. Set M = ker(π); then M is a G-invariant subgroup of A. If a ∈ L ∩ M, then a = π(a) = 0, i.e., L ∩ M = {0}. Furthermore, for b ∈ A, one has π(b − π(b)) = π(b) − π2 (b) = π(b) − π(b) = 0 󳨐⇒ b − π(b) ∈ M. Since im(π) = L, it follows that b ∈ L + M and A = L + M = L ⊕ M, as desired. For us, in Chapters I–XXVI, the most interesting case is when 𝔽 = ℂ or ℝ. We present another proof of Maschke’s theorem for this case. It will provide us with valuable information about representations over number fields. Let V be a ℂG-module, dimℂ (V) = n. It is well known that V possesses an inner product, i.e., a function ( ⋅ , ⋅ ) : V × V → ℂ satisfying the following conditions (below α, β ∈ ℂ, u, v, w ∈ V): (1) (u, v) = (v, u) (α is the complex conjugate of α ∈ ℂ), (2) (αu + βv, w) = α(u, w) + β(v, w), (3) (v, v) ≥ 0; if v ≠ 0, then (v, v) > 0. Put ⟨v, w⟩ = ∑ (xv, xw) (v, w ∈ V). x∈G

It is not difficult to check that so defined ⟨ ⋅ , ⋅ ⟩ is a new inner product, which is G-invariant, i.e., ⟨gv, gw⟩ = ⟨v, w⟩ (v, w ∈ V, g ∈ G). Let L be a submodule of V and L⊥ the orthogonal complement of L, i.e., L⊥ = {w ∈ V | ⟨v, w⟩ = 0} for each v ∈ L. As we know from linear algebra, V = L ⊕ L⊥ . If v ∈ L, w ∈ L⊥ , g ∈ G, then ⟨v, gw⟩ = ⟨g(g −1 v), gw⟩ = ⟨g −1 v, w⟩ = 0 since g−1 v ∈ L. Consequently, L⊥ is a submodule of V. Thus, each submodule of V is complemented and so, by Lemma 6.1, V is completely reducible. This proof has the following important consequence. Let τ : G → GL(V) be a complex or real representation of a group G, afforded by a G-module V. Then, since ⟨gv, gw⟩ = ⟨v, w⟩, it follows that ⟨τ(g)v, τ(g)w⟩ = ⟨v, w⟩. This means that the linear operators τ(g) (g ∈ G) are unitary with respect to the metric defined by the inner product ⟨ ⋅ , ⋅ ⟩. Let B = {v i }1n be a basis of V and T = MB (τ) the matrix of τ in the basis B. If B󸀠 = {v󸀠i }1n is a basis of V which is orthonormal with respect to the invariant metric and T1 = MB󸀠 (τ), then each matrix T1 (x) is unitary. Recall that a matrix U ∈ ℂn 󸀠 is unitary if UU = In is the identity n × n matrix (if U = (u ij ), then U 󸀠 = (u󸀠ij ), where u󸀠ij = u ji , and U = (u ij )). Unitary matrices over ℝ are called orthogonal matrices. A matrix representation T of a group G is said to be unitary if every matrix T(g) (g ∈ G), is unitary. So every matrix ℂ-representation is equivalent to some unitary one. Hence every ℝ-representation is equivalent to an orthogonal one. Therefore it follows that every matrix T(g) for any fixed g ∈ G is similar to a diagonal matrix. The example preceding Theorem 7.4 shows that the assumption about the characteristic in Maschke’s theorem is necessary. This is yet another instance in which the properties of the ground field affect our results. Starting from Chapter II, we shall not encounter this phenomenon since we will always have 𝔽 = ℂ in Chapters I–XXVI.

I Basic concepts | 25

9 Representations of abelian groups We now apply the previous results to representations of (finite) abelian groups. Recall that, by the basic theorem on abelian groups, a finite abelian group G contains cyclic subgroups Z1 , . . . , Z d such that G = Z1 × ⋅ ⋅ ⋅ × Z d (Kronecker–Frobenius– Stickelberger). In what follows, unless otherwise stated, the ground field 𝔽 coincides with ℂ, the field of complex numbers. All other fields are subfields of ℂ (such fields are called number fields). Theorems 7.4 (b) and 8.1 imply the following result. Corollary 9.1. Let T be a matrix ℂ-representation of an abelian group G and deg(T) = n. Then there exists a matrix S ∈ GL(n, ℂ) such that the matrix S−1 T(g)S is diagonal for each g ∈ G. Indeed, the representation T, being completely reducible, is a sum of irreducible representations of G which are one-dimensional, by Schur’s lemma. (Note that any two diagonal matrices commute.) Let T be a one-dimensional matrix representation of an arbitrary (finite) group G. Then T(g) = (λ(g)), λ : G → ℂ∗ , where λ(xy) = λ(x)λ(y) for all x, y ∈ G. A function λ : G → ℂ∗ with this property is called a linear character of G. The set of all linear characters of G is denoted by Lin(G). Sometimes we will write G∗ instead of Lin(G). Clearly, linear characters are not more than one-dimensional representations of G, and conversely. If λ ∈ Lin(G), then λ−1 (g) = λ(g)−1 = λ(g −1 ) since λ is a homomorphism. For λ, μ ∈ Lin(G) and g ∈ G, define λ(g) λ . (1) (λμ)(g) = λ(g)μ(g), μ−1 (g) = μ(g)−1 , ( )(g) = μ μ(g) It is easy to check that λμ, λ−1 ∈ Lin(G). Thus, Lin(G) = Hom(G, ℂ∗ ) is a (multiplicative) abelian group with identity element 1G : g → 1 ∈ ℂ for all g ∈ G (1G is said to be the principal character of G). Before we have not assumed that G is abelian. Throughout the remainder of this section, G is abelian of order m (so that g m = 1 for all g ∈ G), unless otherwise stated. Theorem 9.2. We have Lin(G) ≅ G. Proof. Let W m be the group of all m-th roots of 1 in ℂ. Then W m ≅ Cm since every element of W m is a power of a primitive m-th root of 1. If λ ∈ Lin(G), g ∈ G, then λ(g) ∈ W m since λ(g)m = λ(g m ) = λ(1) = 1. Let G = ⟨b1 ⟩ × ⋅ ⋅ ⋅ × ⟨b r ⟩,

o(b i ) = m i

(i = 1, . . . , r).

If χ ∈ Lin(G), then χ(b i ) ∈ W m i since χ(b i )m i = χ(b m i ) = χ(1) = 1. The mapping f : χ 󳨃→ (χ(b1 ), . . . , χ(b r )) is a homomorphism of Lin(G) into W = W m1 × ⋅ ⋅ ⋅ × W m r . Note that W is isomorphic to G since W m i ≅ ⟨b i ⟩ ≅ Cm i for i = 1, . . . , r. If χ ∈ ker(f), then (1, . . . , 1) = f(χ) = (χ(b1 ), . . . , χ(b r )),

26 | Characters of Finite Groups 1 and we get χ(b i ) = 1 ∈ ℂ for all i, whence χ(g) = 1 for all g ∈ G, and so χ = 1G . Thus, ker(f) = {1G }, i.e., the mapping f is faithful. Let (a1 , . . . , a r ) ∈ W = W m1 × ⋅ ⋅ ⋅ × W m r . Define a function χ : G → ℂ∗ by β

β

β

β

χ(b11 . . . b r r ) = a11 . . . a r r β

β

(β1 , . . . , β r ∈ ℤ). γ

γ

Then χ is well defined, because b11 . . . b r r = b11 . . . b r r implies β i ≡ γ i (mod m i ) for each i, whence β

β

β

β

γ

γ

γ

γ

χ(b11 . . . b r r ) = a11 . . . a r r = a11 . . . a r r = χ(b11 . . . b r r ). Clearly, χ ∈ Lin(G). Since χ(b i ) = a i for all i, it follows that f(χ) = (a1 , . . . , a r ), and we conclude that im(f) = W. Thus f is an isomorphism of the group Lin(G) onto W. Since, as has been noted above, W ≅ G, we get Lin(G) ≅ G. Given χ ∈ Lin(G), define ker(χ) = {g ∈ G | χ(g) = 1 ∈ ℂ}. Since χ : G → ℂ∗ is a homomorphism, we have ker(χ) ≤ G. Let H ≤ G and let g 󳨃→ g be the natural homomorphism of G onto G = G/H. Given θ ∈ Lin(G), define the inflation of θ as the function χ : G → ℂ∗ such that χ(g) = θ(g). If, in addition, x ∈ G, then χ(gx) = θ(gx) = θ(g x) = θ(g)θ(x) = χ(g)χ(x).

(2)

It follows that χ ∈ Lin(G), and we write χ = inf(θ). We have inf(θ)(g) = θ(g)

(θ ∈ Lin(G), g ∈ G).

Thus, inf : Lin(G) → Lin(G). If h ∈ H, then χ(h) = inf(θ)(h) = θ(h) = θ(1) = 1 ∈ ℂ 󳨐⇒ H ≤ ker(inf(θ)) = ker(χ). Define H 0 = {χ ∈ Lin(G) | H ≤ ker(χ)}, i.e., H 0 is the set of those linear characters of G whose kernels contain H; clearly, H 0 ≤ Lin(G). It follows from the definition that the mapping inf : Lin(G) → H 0 is a homomorphism. Indeed, if θ1 , θ2 ∈ Lin(G) and g ∈ G, then inf(θ1 θ2 )(g) = (θ1 θ2 )(g) = θ1 (g)θ2 (g) = inf(θ1 )(g) ⋅ inf(θ2 )(g), and we conclude that inf(θ1 θ2 ) = inf(θ1 ) ⋅ inf(θ2 ), proving our claim. Moreover, the following assertion holds. Theorem 9.3. The map inf : Lin(G) → H 0 is an isomorphism. Proof. We have proved that inf is a homomorphism; therefore, it suffices to prove that it is one-to-one and surjective. Let θ ∈ ker(inf). Then, for g ∈ G, we have θ(g) = inf(θ)(g) = 1G (g) = 1 ∈ ℂ 󳨐⇒ ker(inf) = {1G }, i.e., inf is a monomorphism. Given χ ∈ H 0 , define a function θ : G → ℂ∗ by θ(g) = χ(g) (g ∈ g ∈ G). Then θ is well defined since H ≤ ker(χ). It follows from (2) that θ ∈ Lin(G), and so χ = inf(θ). Thus, im(inf) = H 0 .

I Basic concepts | 27

The subgroup H 0 ≤ Lin(G) is called the annihilator of H ≤ G since H 0 consists of all those linear characters of G which “annihilate” H. Theorems 9.2 and 9.3 imply that H 0 ≅ G/H. If χ ∈ Lin(G), then χ(x) = 1 for all x ∈ H if and only if χ ∈ H 0 , and χ(x) = 1 for all χ ∈ H 0 if and only if x ∈ H. Exercise 9.1. If χ1 , χ2 ∈ Lin(G) and χ1 (g) = χ2 (g) for each g ∈ G, then χ1 = χ2 . Compare this obvious fact with Exercise 9.3, below. Exercise 9.2. The following assertions hold: (a) If χ ∈ Lin(G), then ∑x∈G χ(x) = |G|δ χ,1G , where δ is the Kronecker delta on the set Lin(G). (b) If a ∈ G, then ∑χ∈Lin(G) χ(a) = |G|δ a,1 , where δ is the Kronecker delta on the set G. (c) ⋂χ∈Lin(G) ker(χ) = {1}. Solution. (a) This is true for χ = 1G . Now let χ ≠ 1G . Then there is a ∈ G such that χ(a) ≠ 1. In this case, χ(a) ∑ χ(x) = ∑ χ(ax) = ∑ χ(x) 󳨐⇒ ∑ χ(x) = 0 = |G|δ χ,1G . x∈G

x∈G

x∈G

x∈G

(b) This is true for a = 1 since |Lin(G)| = |G|. Now let a ≠ 1. Then there is τ ∈ Lin(G) such that τ(a) ≠ 1. It follows from {τχ | χ ∈ Lin(G)} = Lin(G) that τ(a)

∑ χ∈Lin(G)

χ(a) =

∑ (τχ)(a) = χ∈Lin(G)

∑ χ∈Lin(G)

χ(a) 󳨐⇒

∑

χ(a) = 0 = |G|δ a,1 .

χ∈Lin(G)

(c) This is obvious. Exercise 9.3. If g1 , g2 ∈ G and χ(g1 ) = χ(g2 ) for all χ ∈ Lin(G), then g1 = g2 . Hint. We have g1−1 g2 ∈ ⋂χ∈Lin(G) ker(χ) = {1}, by Exercise 9.2 (c). Exercises 9.1 and 9.3 actually mean that the linear characters of an abelian group G separate its elements and, conversely, the linear characters of G are separated by the elements of G. We shall see in Chapter II that these results are special cases of theorems about irreducible characters of arbitrary finite groups. Recall that if H ≤ G, then H 0 = {χ ∈ Lin(G) | H ≤ ker(χ)}, i.e., H 0 is the set of those linear characters of G whose kernels contain H. It follows from H ≤ K ≤ G that H 0 ≥ K 0 (indeed, if λ ∈ K 0 , then ker(λ) ≥ K ≥ H so λ ∈ H 0 ). If, in addition, H 0 = K 0 , then, by the above, |G/H| = |H 0 | = |K 0 | = |G/K| so that |H| = |K| and hence H = K. Now suppose that subgroups H and K are arbitrary and H 0 = K 0 ; then, as above, |H| = |K|. If χ ∈ H 0 (= K 0 ), then ⟨H, K⟩ ≤ ker(χ), and so H 0 = K 0 ≤ ⟨H, K⟩0 . By what has been proved above, we have H = ⟨H, K⟩ = K. Thus, the mapping H 󳨃→ H 0 (H ∈ L(G), H 0 ∈ L(G∗ )) is injective (here L(G) is the lattice of subgroups of G). Since, by Theorem 9.2, |L(G)| = |L(G∗ )|, this mapping is bijective. Since this mapping inverts inclusion, it is an anti-isomorphism of L(G) onto L(G∗ ). As G∗ ≅ G (Theorem 9.2), it follows that the lattice L(G) is anti-isomorphic to itself, i.e., it is self-dual.

28 | Characters of Finite Groups 1 We claim that abelian groups G and G∗ = Lin(G) are in a certain sense equivalent, and G can be identified with G∗∗ = (G∗ )∗ . Let g ∈ G. The mapping ϕ g : χ 󳨃→ χ(g) (χ ∈ G∗ ) is a linear character of G∗ , since ϕ g (χ1 χ2 ) = (χ1 χ2 )(g) = χ1 (g)χ2 (g) = ϕ g (χ1 )ϕ g (χ2 )

(χ1 , χ2 ∈ G∗ )

and ϕ g (χ) = χ(g) ≠ 0 for all g ∈ G and χ ∈ G∗ . Thus, ϕ g ∈ G∗∗ . Let us show that the mapping δ : g 󳨃→ ϕ g is an isomorphism of G onto Lin(G∗ ) = G∗∗ . One has δ(gh)(χ) = ϕ gh (χ) = χ(gh) = χ(g)χ(h) = ϕ g (χ)ϕ h (χ) = δ(g)(χ)δ(h)(χ), that is, ϕ gh = ϕ g ϕ h and δ(gh) = δ(g)δ(h). Thus, δ is a homomorphism. If δ(g) = δ(h), then δ(g)(χ) = δ(h)(χ) for all χ ∈ G∗ , or χ(g) = ϕ g (χ) = ϕ h (χ) = χ(h) for all χ ∈ G∗ , whence g = h (Exercise 9.3), and δ is a monomorphism. Therefore, ker(δ) = {1}. It follows from Theorem 9.2 that G∗∗ ≅ G∗ ≅ G, so that δ is an isomorphism of G onto G∗∗ . Thus, we can identify ϕ g with g and write g(χ) instead of ϕ g (χ), so that g(χ) = χ(g). In this manner G∗∗ is identified with G, the elements of G being considered as linear characters of G∗ . We have proved the Duality Theorem for finite abelian groups: Theorem 9.4. Any valid assertion about elements and linear characters of a finite abelian group remains valid if the terms “elements” and “linear characters” are interchanged. In this sense Theorem 9.2 is dual to itself, and Exercises 9.1 and 9.3 are dual to each other. Given H ≤ G∗ , we define its annihilator in G as follows: H0 = {g ∈ G | χ(g) = 1 for all χ ∈ H} = ⋂ ker(χ). χ∈H

= ⋂χ∈H ker(χ) is the subgroup of G. Since χ(g) = g(χ), H0 is the annihilator of H in G (considered as the group of linear characters of G∗ ). Therefore, by the above, H0 ≅ G∗ /H. We claim that a mapping H 󳨃→ H0 of the lattice L(G∗ ) onto the lattice L(G) is inverse to a mapping H 󳨃→ H ∗ of L(G) onto L(G∗ ). This is true, as the following remark shows.

Thus, H0

Remark. If H ≤ G and H = H 0 , then H0 = H (or, what is the same, (H 0 )0 = H). Indeed, |H0 | = |G∗ /H| = |G∗ /H 0 | = |G|/|H 0 | = |H| since |G∗ | = |G| and H 0 ≅ G/H. Thus, we have |H0 | = |H|. Since H ≤ H0 , we get H0 = H, as desired. In particular, H ≅ G∗ /H 0 for every H ≤ G. Exercise 9.4. The subgroup lattice L(G) of a finite abelian group G possesses an antiisomorphism α such that H α ≅ G/H for every subgroup H of G.

I Basic concepts | 29

Exercise 9.5. Let G be a finite abelian group of rank r. Then r = min{n | G has a faithful matrix ℂ-representation of degree n}. Solution. Let T be a faithful matrix ℂ-representation of the abelian group G and deg(T) = n. By Corollary 9.1, one can assume that the matrices T(g) (g ∈ G) are diagonal: T(g) = diag(χ1 (g), . . . , χ n (g)), where χ i ∈ Lin(G) for all i. Next, n

n

i=1

i−1

⋂ ker(χ i ) = ker(T) = {1} 󳨐⇒ ∏(ker(χ i ))0 = G∗ . ))0

Noting that (ker(χ i ≅ G/ker(χ i ) is cyclic, we conclude that the (abelian) group G is decomposed in direct product of n cyclic subgroups. It follows that n ≥ rank(G∗ ). Since G ≅ G∗ , we get n ≥ rank(G). It is easy to build a faithful representation of G of degree r. Thus, n = r. In the following exercise particular cases of orthogonality relations for general finite groups are proved. Exercise 9.6. Let G be an abelian group. (a) If χ1 , χ2 ∈ Lin(G), then ∑g∈G χ1 (g)χ2 (g) = |G|δ χ1 ,χ2 . (b) If g1 , g2 ∈ G, then ∑χ∈Lin(G) χ(g1 )χ(g2 ) = |G|δ g1 ,g2 . Hint. (a) Given χ ∈ Lin(G), let us prove that S χ = ∑g∈G χ(g) = |G|δ χ,1G . To this end, show that χ(h)S χ = S χ for all h ∈ H. If χ ≠ 1G , choose h so that χ(h) ≠ 1. As result, we get S χ = 0. The case χ = 1G is trivial. Apply this to χ = χ1 χ2 (χ1 , χ2 ∈ Lin(G)); this completes the proof of (a). Assertion (b) is proved similarly, however, one can make use of Theorem 9.4 and Exercise 9.2. Exercise 9.7. Any linear character of a subgroup H of an abelian group G can be extended in |G : H| ways to a linear character of G. Hint. Prove that the mapping ϕ : χ 󳨃→ χ H (χ ∈ Lin(G)) is a surjective homomorphism of Lin(G) onto Lin(H) with ker(ϕ) = H 0 . (This also follows, by Frobenius reciprocity; see Chapter V.) For a group G (not necessarily abelian) we define Lin(G) as the set of all homomorphisms of G into ℂ∗ , and define multiplication in Lin(G) as in abelian case. This definition makes Lin(G) a group. As in abelian case, we sometimes write G∗ instead of Lin(G). Theorem 9.5. We have Lin(G) ≅ G/G󸀠 for any (finite) group G. Proof. Let G = G/G󸀠 and θ ∈ Lin(G). Set inf(θ)(g) = θ(g), where g ∈ g ∈ G. As for abel∗ ian groups, the mapping inf : G → ℂ∗ is an injective homomorphism. If χ ∈ G∗ , then we have G󸀠 ≤ ker(χ) since G/ker(χ) is isomorphic to a finite abelian subgroup of ℂ∗ . Now, putting θ(g) = χ(g), we get a well-defined element θ of Lin(G). Since χ = inf(θ), we obtain im(inf) = Lin(G) = G∗ .

30 | Characters of Finite Groups 1 In Exercises 9.8–9.10 groups are arbitrary (finite). Exercise 9.8. Let D(G) = {x ∈ G | det(T(x)) = 1 for all irreducible representations T of G}. Then D(G) = G󸀠 . Hint. If x, y ∈ G󸀠 , then det(T[x, y]) = det(T(x))−1 det(T(y))−1 det(T(x)) det(T(y)) = 1 so that G󸀠 ≤ D(G). Let x ∈ D(G) − G󸀠 . There exists λ ∈ Lin(G) such that λ(x) ≠ 1. Then λ is a representation of G of degree 1 with det(λ(x)) = λ(x) ≠ 1, contrary to definition of D(G). Exercise 9.9. Suppose that x ∈ G󸀠 ∩ Z(G) is of prime order p. If T is an irreducible representation of G such that x ∈ ̸ ker(T), then p | deg(T). Hint. By Schur’s lemma, T(x) = ωI, where ω ≠ 1 is a p-th root of unity (o(x) = p) and I is the identity deg(T) × deg(T) matrix. By Exercise 9.8, 1 = det(T(x)) = ω n , where n = deg(T), hence p | n = deg(T). Exercise 9.10. Let x ∈ G󸀠 ∩ Z(G). If T is a faithful irreducible representation of G, then the order o(x) of x divides the degree of T. Hint. Mimic the solution of the previous exercise. Note that G󸀠 ∩ Z(G) is contained in Φ(G), the Frattini subgroup of G (check!).

10 Homogeneous completely reducible modules In this section A is an algebra over a field 𝔽 and M is a finitely generated completely reducible (left) A-module. Let I(M) be the set of all irreducible submodules of M. Then I(M) is partitioned in classes of isomorphic (irreducible) submodules of M. Lemma 10.1. Let M be a completely reducible A-module and let M = L1 ⊕ ⋅ ⋅ ⋅ ⊕ L n

(1)

be a decomposition of M in a direct sum of irreducible submodules L1 , . . . , L n . If L ∈ I(M), then: (a) L ≅ L i for some i ∈ {1, . . . , n}. (b) L is contained in the direct sum of all those summands L i of M that are isomorphic to L. Proof. (a) Let x ∈ L with x = x1 + ⋅ ⋅ ⋅ + x n , where x i ∈ L i for all i. Then the mapping ϕ i : L → L i defined by ϕ i (x) = x i , is a module homomorphism. By Lemma 6.2, either ϕ i = 0 or ϕ i is an isomorphism of L onto L i . Since L ≠ {0}, the homomorphism ϕ i ≠ 0 for some i ∈ {1, . . . , n}, and so L ≅ L i .

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(b) Suppose that ϕ i1 , . . . , ϕ i k are all nonzero homomorphisms among ϕ i for i = 1, . . . , n (see the proof of (a)). Then, for x ∈ L, n

x = ∑ ϕ i (x) = ϕ i1 (x) + ⋅ ⋅ ⋅ + ϕ i k (x) = x i1 + ⋅ ⋅ ⋅ + x i k ∈ L i1 ⊕ ⋅ ⋅ ⋅ ⊕ L i k . i=1

By the proof of (a), L i j ≅ L for j = 1, . . . , k. Definition 10.1. A completely reducible A-module M is said to be homogeneous if all irreducible submodules of M are isomorphic. We say that the homogeneous completely reducible A-modules M and N have the same type if the irreducible submodules of M are isomorphic to irreducible submodules of N. It is easy to prove, using Lemma 6.2, that nonzero submodules and quotient modules of a homogeneous completely reducible module are also homogeneous completely reducible of the same type. It follows from Lemma 10.1 that if M = L1 ⊕ ⋅ ⋅ ⋅ ⊕ L k , where all irreducible submodules L i are isomorphic, then every irreducible submodule of M is isomorphic to L1 , i.e., M is homogeneous. Therefore: Corollary 10.2. A completely reducible A-module M is homogeneous if there exists a direct decomposition M = L1 ⊕ ⋅ ⋅ ⋅ ⊕ L n , where L i are isomorphic irreducible submodules. Definition 10.2. Maximal by inclusion homogeneous submodules of a completely reducible module M are called homogeneous components of M. Let M be a completely reducible A-module. It follows from Lemma 10.1 that I(M) is partitioned in the finite number of classes C1 , . . . , C k of isomorphic irreducible submodules. Let H i be the sum of all submodules of M belonging to the class C i . We retain the notation introduced in this paragraph. Lemma 10.3. Let M be a completely reducible module. (a) The set of homogeneous components of M coincides with {H1 , . . . , H k } (see the paragraph preceding the lemma). In particular, this set is finite. (b) M = H1 ⊕ ⋅ ⋅ ⋅ ⊕ H k . If M = L1 ⊕ ⋅ ⋅ ⋅ ⊕ L n is a decomposition of M in direct sum of irreducible submodules, then H i = ⨁L α ∈C i L α , i = 1, . . . , k. (c) Every homogeneous submodule of M is contained in H i for some i. Proof. It follows from Lemma 10.1 that all homogeneous components H i are homogeneous modules. Since every irreducible submodule L of M is a member of C i for some i, we get L ⊆ H i . Therefore, M = H1 + ⋅ ⋅ ⋅ + H k . Let M = ⨁ni=1 L i be a decomposition of M in a direct sum of irreducible submodules. Set H i󸀠 = ⨁L α ∈C i L α . Since C i has at least one representative among submodules L α (Lemma 10.1 (a)), we get H i󸀠 ≠ {0} for all i, and so M = H1󸀠 ⊕ ⋅ ⋅ ⋅ ⊕ H k󸀠 . It is clear that H i󸀠 ⊆ H i . Next, if L ∈ C i , then, by Lemma 10.1 (b), L ⊆ ⨁L α ≅L L α = ⨁L α ∈C i L α = H i󸀠 , and so H i ⊆ H i󸀠 for all i, whence H i = H i󸀠 . If i ≠ j, then H i ∩ H j = {0} (Lemma 10.1). Therefore, M = H1 + ⋅ ⋅ ⋅ + H k = H1 ⊕ ⋅ ⋅ ⋅ ⊕ H k . It follows from the definition that H i is

32 | Characters of Finite Groups 1 a maximal homogeneous submodule of M, i.e., H i is a homogeneous component of M, i = 1, . . . , k. Let K be a homogeneous component of M. Since all irreducible submodules of K are belong to some class C i , it follows that K ⊆ H i , and so K = H i in view of maximality of K. Thus H i (i = 1, . . . , k) are all homogeneous components of M. Remark 10.1. By agreement, all modules are finite-dimensional vector spaces over the field 𝔽. As above, let M be a completely reducible module. Set h i = dim𝔽 (H i ), i = 1, . . . , k, where H i are homogeneous components of M, and let f i be the dimension of a member of C i . Obviously, h i and f i are invariants of M. Let M = ⨁ni=1 L i with irreducible submodules L1 , . . . , L n , and let n i = |{L α | L α ∈ C i }|. Since H i = ⨁L α ∈C i L α , we get h i = n i f i (i = 1, . . . , k), and so n i is also an invariant of M for all i. Therefore, L1 , . . . , L n are determined by M up to order and isomorphism. (This also follows from the Jordan–Hölder theorem.) We see that the number n of irreducible direct summands of M is an invariant of M. This number n = n1 + ⋅ ⋅ ⋅ + n k is said to be the length of the module M and will be denoted by l(M). Lemma 10.4. Let ϕ be an endomorphism of the completely reducible module M and let H1 , . . . , H k be its homogeneous components. Then ϕ(H i ) ⊆ H i , i = 1, . . . , k. Proof. Take L ∈ C i . By Lemma 6.4, either ϕ(L) = {0} or ϕ(L) ≅ L. so that ϕ(L) ∈ C i , unless ϕ(L) = {0}. It follows from H i = ∑L∈C i L that ϕ(H i ) = ∑L∈C i ϕ(L) ⊆ H i . We want to describe the algebra EndA (M)(= HomA (M, M)), where M is a homogeneous completely reducible A-module. Let L be a fixed irreducible submodule of M. By Schur’s lemma, D = EndA (L) is an algebra with division. It is clear that D is determined by M up to isomorphism. We call D the skew field corresponding to M. Write V M = HomA (L, M), where, as above, L is a fixed irreducible submodule of the homogeneous completely reducible A-module M. If ξ ∈ D(= EndA (L)) and θ ∈ V M , then the product θξ is defined and θξ ∈ V M . Since V M is closed with respect to addition, it is a right linear space over the skew field D. Let us prove that the dimension of this space equals the length n of M, i.e., dimD (V M ) = l(M). To this end, we have to construct a D-basis of the space V M . Let M = L1 ⊕ ⋅ ⋅ ⋅ ⊕ L n be a decomposition of M in a direct sum of (isomorphic) irreducible submodules. Since L ≅ L i (i = 1, . . . , n), it follows that for each i ∈ {1, . . . , n} there exists a monomorphism θ i ∈ V M such that θ i (L) = L i . Assume that a linear combination θ1 ξ1 + ⋅ ⋅ ⋅ + θ n ξ n = 0 for some ξ1 , . . . , ξ n ∈ D(= EndA (L)). Let x ∈ L be arbitrary. Then θ1 ⋅ ξ1 (x) + ⋅ ⋅ ⋅ + θ n ⋅ ξ n (x) = (θ1 ξ1 + ⋅ ⋅ ⋅ + θ n ξ n )(x) = 0. It follows from ξ i (x) ∈ L that θ i ⋅ ξ i (x) ∈ L i for i = 1, . . . , n. Therefore, θ i ⋅ ξ i (x) = 0 for all i since M = L1 ⊕ ⋅ ⋅ ⋅ ⊕ L n is a direct sum, and so ξ i (x) = 0 since θ i is a monomorphism, i = 1, . . . , n. Therefore, ξ1 = ⋅ ⋅ ⋅ = ξ n = 0, and so θ1 , . . . , θ n are linearly independent over D. It remains to show that θ1 , . . . , θ n is a D-basis of V M . It suffices to show that these n elements span the D-space V M . Take θ ∈ V M and x ∈ L. Then

I Basic concepts | 33

θ(x) = y1 + ⋅ ⋅ ⋅ + y n = θ1 (x1 ) + ⋅ ⋅ ⋅ + θ n (x n ), where θ i (x i ) = y i ∈ L i for an appropriate x i ∈ L, i = 1, . . . , n. The mapping ξ i : x 󳨃→ x i , i = 1, . . . , n, is an endomorphism of L, i.e., ξ i ∈ EndA (L) = D for all i. Since θ(x) = θ1 ⋅ ξ1 (x) + ⋅ ⋅ ⋅ + θ n ⋅ ξ n (x) = (θ1 ξ1 + ⋅ ⋅ ⋅ + θ n ξ n )(x), it follows that θ = θ1 ξ1 + ⋅ ⋅ ⋅ + θ n ξ n , and so θ1 , . . . , θ n is a basis of V M over D. In particular, dimD (V M ) = n. Theorem 10.5. Let M be a homogeneous completely reducible A-module and let D be the corresponding skew field (i.e., D = EndA (L), where L is an irreducible submodule of M). Then EndA (M) ≅ D n , where n = l(M), the length of M, and D n is the ring of n × n matrices over D. Proof. Let L0 be an irreducible submodule of M and D = EndA (L0 ). Then we have V M = HomA (L0 , M) is an n-dimensional linear space over D (see the paragraph preceding the theorem). If f ∈ HomA (M, M) and θ ∈ V M , the product fθ is determined and fθ ∈ HomA (L0 , M) = V M . The mapping ϕ(f) : θ 󳨃→ fθ (f ∈ HomA (M, M) and θ ∈ V M ) is a linear operator of the space V M . Indeed, the additivity of ϕ(f) is obvious; next, ϕ(f)(θξ) = f(θξ) = (fθ)ξ = (ϕ(f)(θ))ξ

for any ξ ∈ D.

Let us prove that the mapping ϕ is a ring isomorphism of HomA (M, M)(= EndA (M)) onto Φ = HomD (V M , V M ). First, it is easy to check that ϕ(f1 + f2 ) = ϕ(f1 ) + ϕ(f2 ), ϕ(f1 f2 ) = ϕ(f1 )ϕ(f2 ). Next, if ϕ(f) = 0, then fθ = 0 for all θ ∈ V M . In particular, fθ i = 0, i = 1, . . . , n (here θ1 , . . . , θ n is a D-basis of V M ). Let x ∈ M. Then x = θ1 (x1 ) + ⋅ ⋅ ⋅ + θ n (x n )

(x i ∈ L0 , i = 1, . . . , n).

Therefore, f(x) = (fθ1 )(x1 ) + ⋅ ⋅ ⋅ + (fθ n )(x n ) = 0 so f = 0. This means that ϕ is injective. It remains to prove that ϕ is surjective. Let μ ∈ Φ(= HomD (V M , V M )). We have to show that ϕ(f) = μ for some f ∈ HomA (M, M). Let x ∈ M. Then x = θ1 (x1 ) + ⋅ ⋅ ⋅ + θ n (x n ), where x i ∈ L0 is determined uniquely by x. Consider a mapping f : x 󳨃→ μ(θ1 )(x1 ) + ⋅ ⋅ ⋅ + μ(θ n )(x n ). If a ∈ A (M is an A-module), x ∈ M, then f(ax) = f(θ1 (ax1 ) + ⋅ ⋅ ⋅ + θ n (ax n )) = μ(θ1 )(ax1 ) + ⋅ ⋅ ⋅ + μ(θ n )(ax n ) = aμ(θ1 )(x1 ) + ⋅ ⋅ ⋅ + aμ(θ n )(x n ) = af(x). Thus, f ∈ HomA (M, M). Now let θ ∈ V M . Then θ = θ1 ξ1 + ⋅ ⋅ ⋅ + θ n ξ n ,

34 | Characters of Finite Groups 1 where ξ i ∈ D, i = 1, . . . , n. Since μ is a linear operator of the space V M , it follows that μ(θ) = μ(θ1 )ξ1 + ⋅ ⋅ ⋅ + μ(θ n )ξ n . If x ∈ L0 , then μ(θ)(x) = μ(θ1 )(ξ1 (x)) + ⋅ ⋅ ⋅ + μ(θ n )(ξ n (x)) = f(θ(x)). Thus, μ(θ)(x) = f(θ(x)) = (fθ)(x) for all x ∈ L0 . Therefore, μ(θ) = fθ = ϕ(f)(θ), and so μ = ϕ(f). Hence, ϕ is an isomorphism. Since Φ ≅ D n , it follows that EndA (M) = HomA (M, M) ≅ D n , proving the theorem. Using the space V M , one can obtain an important information on the lattice L(M) of submodules of a homogeneous completely reducible A-module M. Theorem 10.6. Let M be a homogeneous completely reducible A-module. The lattice L(M) is isomorphic to L(V M ), the lattice of subspaces of the D-space V M . Proof. Let N ∈ L(M) and let L0 be an irreducible submodule of M. Put V N = {θ ∈ V M | θ(L0 ) ⊆ N}. It is clear that V N is a subspace of V M . Obviously, V N = j(HomA (L0 , N)), where j is an injective homomorphism of N into M. Thus, we have obtained a mapping of L(M) in L(V M ) given by N 󳨃→ V N . Let us prove that this mapping is a lattice isomorphism. It follows from complete reducibility of N that N = ∑θ∈V N θ(L0 ). In fact, let N = N1 ⊕ ⋅ ⋅ ⋅ ⊕ N k , where N i are irreducible submodules. Since N i ≅ L0 , i = 1, . . . , k, there exists, for every i = 1, . . . , k, a monomorphism ϑ i : L0 → M such that ϑ i (L0 ) = N i . Obviously, ϑ i ∈ V N . It follows from N = ∑ni=1 ϑ i (L0 ) that N = ∑θ∈V N θ(L0 ), as desired. Therefore, if V N 󸀠 = V N , where N, N 󸀠 are submodules of M, then N = N 󸀠 . This means that the mapping N 󳨃→ V N is injective. Note that dim(V N ) = dim(HomA (L0 , N)). Therefore (see the paragraph preceding Theorem 10.5), dim(V N ) = k, where k = l(N) is the length of N. Now we have to prove that the mapping N 󳨃→ V N is surjective. Let H be an arbitrary subspace of V N and let N = ∑θ∈H θ(L0 ). Since H ⊆ V N , we get dim(H) ≤ k, where k is the length of N. Let {θ1 , . . . , θ k󸀠 } be a D-basis of H. If θ ∈ H, then θ = θ1 ξ1 +⋅ ⋅ ⋅+ θ k󸀠 ξ k󸀠 , 󸀠 where ξ i ∈ D for all i. Therefore, θ(L0 ) ⊆ ∑ki=1 θ i (L0 ), and so k󸀠

N = ∑ θ(L0 ) ⊆ ∑ θ i (L0 ). θ∈H

i=1

󸀠 ∑ki=1

It follows from this that N = θ i (L0 ). Since θ i (L0 ) are irreducible, we get k ≤ k󸀠 , and so dim(H) = k, i.e., dim(H) = dim(V N ). Together with what we have proved already, this shows that the mapping N 󳨃→ V N is a bijection. Since it preserves the inclusion, it is a lattice isomorphism, completing the proof. Remark 10.2. It follows from the proof of Theorem 10.6, that the length l(N) of a submodule N of Theorem 10.6 is equal to the dimension of the corresponding subspace V N .

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Remark 10.3. Theorem 10.6 shows that a homogeneous completely reducible module of length n may be considered as (n − 1)-dimensional projective space over skew field D corresponding to this module. Theorem 10.7. If the skew field D is commutative (i.e., it is a field) and M is a homogeneous completely reducible module, then the lattice L(M) is strongly auto-dual. This means that there exists a bijection α : L(M) → L(M) such that (a) α reverses inclusion, (b) N α ≅ M/N (or, what is the same, l(N α ) = l(M) − l(N)), where l(M) is the length of M.¹ Proof. Let l(M) = n. Then n = dim(V M ) (see the paragraph preceding Theorem 10.5). Choosing a basis {θ1 , . . . , θ n } of V M , define in V M a scalar product as follows: n

n

n

if θ, ϑ ∈ V M , θ = ∑ θ1 ξ i , ϑ = ∑ θ1 η i (ξ i , η i ∈ D, then ⟨θ, ϑ⟩ = ∑ ξ i η i . i=1

i=1

i=1

If H is a subspace of V M , then H 0 = {θ ∈ V M | ⟨θ, ϑ⟩ = 0 for all ϑ ∈ H} is the annihilator of H. The subspace H 0 , generally speaking, is not a direct complement of H. Since the form ∑ni=1 ξ i η i is not singular, one has dim(H 0 ) = n − dim(H) = codim(H). Now let N ∈ L(M), H = V N . Then there exists a unique submodule N α such that V N α = H 0 . Since the mapping H 󳨃→ H 0 is a bijection of the set L(V M ), reversing inclusion, the mapping N 󳨃→ N α has the same property. Next, l(N α ) = dim(H 0 ) = n − dim(H) = n − l(N), i.e., l(N α ) + l(N) = l(M). Since l(M/N) = l(M) − l(N), we get l(N α ) = l(M/N). Since N α and M/N are homogeneous modules of the same type (i.e., their irreducible submodules are isomorphic), we get N α ≅ M/N. This means that α is a strong antiautomorphism of the lattice L(M). It follows from the construction of α that α2 = id (i.e., α is an anti-automorphism of order 2).

11 Semisimple algebras In what follows, A is a finite-dimensional associative 𝔽-algebra, where 𝔽 is a field. Recall that the dimension of A is equal to the dimension of the linear 𝔽-space A. Definition 11.1. An algebra A is said to be semisimple if all A-modules are completely reducible. Obviously, submodules of the regular A-module A A are left ideals of A. It follows from the definition that, if A is semisimple, then the regular A-module A A is completely reducible. The reverse assertion also holds as the following theorem shows. 1 The bijection α is called a strong anti-automorphism of L(M).

36 | Characters of Finite Groups 1 Theorem 11.1. If the regular module A A is completely reducible, then the algebra A is semisimple. Proof. By assumption, A = L1 ⊕ ⋅ ⋅ ⋅ ⊕ L n , where L1 , . . . , L n are minimal left ideals of A. Let M be an arbitrary left A-module and let {m1 , . . . , m r } be its 𝔽-basis (as a linear 𝔽-space). Then M = AM = ∑n,r i,j=1 L i m j . Setting M ij = L i m j , we obtain either M ij = {0} or M ij ≅ L i . Indeed, consider a mapping ϕ ij : L i → M ij , defined as follows: ϕ ij (x) = xm j (x ∈ L i ). It is easy to check that ϕ ij is a module homomorphism (recall that all our modules are left). Since L i is irreducible, we have either ϕ ij = 0 or ϕ ij is an isomorphism of L i onto M ij . In the second case M ij is irreducible, and so M, as a sum of irreducible submodules, is completely reducible. By definition, the algebra A is semisimple. Remark. It follows from Maschke’s theorem that the group algebra 𝔽G is semisimple if the characteristic of the field 𝔽 does not divide |G|. Now suppose that the characteristic of 𝔽 is a prime p | |G|. Set e = ∑g∈G g ∈ 𝔽G. Since se = e for all s ∈ G, we get e2 = |G|e = 0. Set I = 𝔽Ge; then I is a left ideal of 𝔽G, {0} ≠ I ≠ 𝔽G (indeed, the identity element e of G is not contained in I). Moreover, I 2 = {0} since e2 = 0. Assume that 𝔽G is semisimple. Then 𝔽G = I ⊕ L, where L is a nonzero left ideal of 𝔽G, and we have 1 = u + v, where 1 is the identity element of 𝔽G, u ∈ I, v ∈ L. Therefore, u = u ⋅ 1 = u(u + v) = u2 + uv = uv since u2 = 0 in view of u ∈ I. Since uv ∈ L, it follows that u = uv ∈ I ∩ L = {0}, and so 1 = v ∈ L. We conclude that 𝔽G = L, I = {0}, which is not the case. This means that the algebra 𝔽G is not semisimple. Definition 11.2. An algebra A is said to be simple if the regular module A A is completely reducible and homogeneous. It follows from Theorem 11.1 that simple algebras are semisimple. An element e ∈ A is said to be an idempotent if e ≠ 0 and e2 = e. Lemma 11.2. Let an algebra A be of the form A = A1 ⊕ ⋅ ⋅ ⋅ ⊕ A k , where A1 , . . . , A k are nonzero two-sided ideals of A. Then the following assertions hold: (a) A i A j = {0} for i ≠ j. (b) If 1 = e1 + ⋅ ⋅ ⋅ + e k is the identity element of A, where e i ∈ A i for all i, then e1 , . . . , e k is a set of pairwise orthogonal idempotents of A and A i is an algebra with the identity element e i , A i = Ae i for i = 1, . . . , k. Proof. (a) Since A i , A j are distinct two-sided ideals of A, we get A i A j ⊆ A i ∩ A j = {0} for i ≠ j. (b) It follows from (a) that e i e j = 0 for i ≠ j. If x ∈ A i , then x = x ⋅ 1 = x(e1 + ⋅ ⋅ ⋅ + e k ) = xe i ,

x = 1 ⋅ x = (e1 + ⋅ ⋅ ⋅ + e k )x = e i x.

Therefore, e i is the identity element of A i whence A i is an algebra (of course, if k > 1, then 1 ∈ ̸ A i so that A i is not an subalgebra of A). Finally, one has Ae i = (A1 ⊕ ⋅ ⋅ ⋅ ⊕ A k )e i = A i e i = A i since, by (a), A j e i ⊆ A j A i = {0} for i ≠ j. Similarly, e i A j = {0} for j ≠ i.

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Theorem 11.3. Let A be a semisimple algebra. Then A = A1 ⊕⋅ ⋅ ⋅⊕A k , where A1 , . . . , A k are two-sided ideals that are simple algebras. Proof. Suppose that the left ideals A1 , . . . , A k are the homogeneous components of the completely reducible regular module A A (see the remark following Definition 11.1). Then A = A1 ⊕ ⋅ ⋅ ⋅ ⊕ A k (see Lemma 10.3). We claim that A1 , . . . , A k are two-sided ideals. Fix a ∈ A. Then the mapping ϕ : x 󳨃→ xa (x ∈ A) is an endomorphism of the regular A-module A A. By Lemma 10.4, ϕ(A i ) ⊆ A i , i.e., A i a ⊆ A i for all i = 1, . . . , k. Therefore, A1 , . . . , A k are right and so two-sided ideals of A. Let the identity element of A is 1 = e1 + ⋅ ⋅ ⋅ + e k , where e i ∈ A i (i = 1, . . . , k). By Lemma 11.2, A i is an algebra with identity element e i . The regular module A i A i is completely reducible and homogeneous (see Lemma 10.3). This means that A i is a simple algebra. Let a ∈ A. A mapping ϕ : x 󳨃→ xa (x ∈ A) is said to be the right multiplication by an element a. Obviously, right multiplications are endomorphisms of the left regular A-module A A. Lemma 11.4. Every endomorphism of the left regular A-module A A is a right multiplication. Proof. Let ϕ be an endomorphism of A A. Then ϕ(x) = ϕ(x ⋅ 1) = xϕ(1) for all x ∈ A. Thus, ϕ is the right multiplication by a = ϕ(1). Lemma 11.5. Let M = L1 ⊕ ⋅ ⋅ ⋅ ⊕ L k = N1 ⊕ ⋅ ⋅ ⋅ ⊕ N k be two direct decompositions of a homogeneous completely reducible module M, where L1 , . . . , L k , N1 , . . . , N k are (left) irreducible submodules. Then there exists ϕ ∈ Aut(M) such that ϕ(L i ) = N i for all i = 1, . . . , k. Proof. All submodules L i , N j are isomorphic. Let ϕ i be an isomorphism of L i onto N i , i = 1, . . . , k. For x = x i + ⋅ ⋅ ⋅ + x k ∈ M, x i ∈ L i (i = 1, . . . , k), set ϕ(x) = ϕ1 (x1 ) + ⋅ ⋅ ⋅ + ϕ k (x k ). It is clear that ϕ is the desired automorphism of M. Corollary 11.6. Let M = L1 ⊕ ⋅ ⋅ ⋅ ⊕ L k be a decomposition of an homogeneous completely reducible module in a direct sum of (left) irreducible submodules. Then for every permutation π ∈ Sk there exists ϕ ∈ Aut(M) such that ϕ(L i ) = L π(i) . Theorem 11.7. A simple algebra has no nontrivial two-sided ideals. Proof. Let I be a nontrivial two-sided ideal of the simple algebra A, i.e., {0} ⊂ I ⊂ A. Since A A is a completely reducible module, there exists a nonzero left ideal N such that A = I ⊕ N. Let I = I1 ⊕ ⋅ ⋅ ⋅ ⊕ I r , N = N1 ⊕ ⋅ ⋅ ⋅ ⊕ N s , where I α , N β are minimal left ideals. There exists an endomorphism ϕ of A A such that ϕ(I1 ) = N1 . On the other hand, ϕ is a right multiplication, by Lemma 11.4, and so ϕ(I1 ) ⊆ I (since I is a two-sided ideal). Therefore, N1 ⊆ I, which is impossible.

38 | Characters of Finite Groups 1 The converse is also true. The proof uses the notion of radical which we define in what follows. The following two theorems lead to a description of simple algebras. Let K be a division algebra over the field 𝔽 and let K n be the ring of n × n matrices over K. Let e ij be a square matrix (a kl ) all of whose entries except a ij = 1 are equal to 0. Theorem 11.8. The ring K n is a simple algebra. Proof. We have K n = ⨁ni,j=1 Ke ij , where e ij e kl = δ jk e il and all elements of K are permutable with e ij for all i, j. Set L i = ⨁nk=1 Ke ki (i = 1, . . . , n). Then L i consists of n × n matrices whose j-th columns are zero columns for all j ≠ i. It is clear that L i are left ideals of the algebra K n for all i. Let us prove that L i is a minimal left ideal of K n . Let L be a left ideal of K n such that {0} ⊂ L ⊆ L i , let 0 ≠ x = ∑nj=1 ξ j e ji ∈ L, where all ξ j ∈ K. Then ξ s ≠ 0 for some s ∈ {1, . . . , n}. It follows from ξ s e si = e ss x ∈ L that e si ∈ L. Then e ji = e js ⋅ e si ∈ L for every j ∈ {1, . . . , n}. Therefore, L = L i , and so L i is a minimal left ideal for all i. We claim that L1 ≅ L i for i ≠ 1 (as K n -modules). We have n

n

L1 e1i = ∑ Ke s1 e1i = ∑ Ke si = L i . s=1

s=1

A mapping ϕ i : x 󳨃→ xe1i (x ∈ L1 ) is a nonzero element of HomK n (L1 , L i ). Since the K n -modules L1 and L i are irreducible (as minimal left ideals of K n ), ϕ i is an isomorphism, by Schur’s lemma. It follows from K n = L1 ⊕ ⋅ ⋅ ⋅ ⊕ L n that K n as a K n -module is completely reducible and homogeneous. Therefore, K n is a simple algebra (see Theorem 11.1). Remark. The algebra Aop is defined as follows. As a set it coincides with A. Addition in Aop coincides with one in A. Multiplication ∗ in Aop is defined as follows: a ∗ b = ba

(a, b ∈ Aop = A).

The following result is basic. Theorem 11.9 (Wedderburn). A simple algebra A is isomorphic to a full matrix algebra over some skew field K. Proof. Since A A is a homogeneous completely reducible module (see Definition 11.2), we get, by Theorem 10.5, EndA (A A) ≅ D n , where D is a skew field (D is the set of endomorphisms of a minimal left ideal of the algebra A). On the other hand, by Lemma 11.4, EndA (A A) is isomorphic to the algebra of right multiplications of the algebra A. Therefore, EndA (A A) ≅ Aop , where Aop is the algebra opposite to A. Thus, Aop ≅ D n , and so A ≅ K n , where K = Dop is the skew field opposite to the skew field D. It follows from the proof of Theorem 11.8 that if A is a simple algebra isomorphic to K n , where K is a skew field, then n is an invariant, i.e., A determines n. Indeed, n is the

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length of the left D-module A A (K is a skew field opposite to D). As to the skew field K, it is determined by the algebra A up to isomorphism. Exercise 11.1. Prove the last assertion. Hint. Show that the skew field K is isomorphic inversely to the skew field of endomorphisms of a minimal left ideal of the algebra A. (If B, C are 𝔽-algebras, then a linear bijection ϕ : B → C is said to be an inverse isomorphism if ϕ(xy) = ϕ(y)ϕ(x) for all x, y ∈ B). Let S1 , S2 be subspaces of an 𝔽-algebra A; then S1 S2 denotes the linear hull of the set {s1 s2 | s1 ∈ S1 , s2 ∈ S2 }. If S1 is a left ideal, so is S1 S2 ; if S2 is a right ideal, so is S1 S2 . Therefore, the product S1 S2 , where S1 is a left ideal and S2 a right ideal, is a two-sided ideal. An ideal L is said to be nilpotent if L n = {0} for some n ∈ N (in such an ideal a product of arbitrary n its elements equals 0). If M is a left A-module and S1 , S2 are subspaces in the 𝔽-algebra A and M, respectively, then S1 S2 is the linear hull of the set {s1 s2 | s1 ∈ S1 , s2 ∈ S2 }. If, in addition, S1 is a left ideal of A, then S1 S2 is a (left) submodule of M. Lemma 11.10. If I, J are nilpotent two-sided ideals of the algebra A, so is I + J. Proof. By assumption, I m = {0} = J n for some m, n ∈ ℕ. Obviously, I + J is a two-sided ideal. We claim that (I + J)m+n = {0}. Indeed, if x i ∈ I, y i ∈ J (i = 1, . . . , m + n), then (x1 + y1 ) . . . (x m+n + y m+n ) is a sum of 2m+n products such that every of them contains either at least m factors from I or at least n factors from J. It follows that every of these m+n = {0}, i.e., I + J products is equal to 0, and so ∏m+n i=1 (x i + y i ) = 0. Therefore, (I + J) is nilpotent. Lemma 11.10 is true if I, J are both left (right) ideals; then I + J is a left (right) ideal. Corollary 11.11. The 𝔽-algebra A possesses a nilpotent two-sided ideal R such that I ⊆ R for every nilpotent left, right or two-sided ideal I of A. Proof. Since dim𝔽 (A) < ∞, it follows that A has a maximal by inclusion nilpotent twosided ideal R, by Lemma 11.10. Let I be an arbitrary nilpotent two-sided ideal. Then R + I is also nilpotent two-sided ideal by Lemma 11.10, and so R + I = R, i.e., I ⊆ R. Thus, R is a unique maximal nilpotent two-sided ideal of A. Now let I be a nilpotent left ideal. Then I m = {0} for some m ∈ ℕ. Obviously, J = IA is a two-sided ideal. Since J m = (I(AI)m−1 A ⊆ I ⋅ I m−1 AA = I m A = {0}, it follows that J is nilpotent, and so, by what we have proved already, J ⊆ R. Since I ⊆ J (indeed, 1 ∈ A), we get I ⊆ R. Similarly, R contains all nilpotent right ideals of the algebra A. We say that a left A-module M is annihilated by a ∈ A if aM = {0}, i.e., am = 0 for all m ∈ M.

40 | Characters of Finite Groups 1 Definition 11.3. The set of all elements of the algebra A annihilating all irreducible A-modules is called the radical of A and denoted by Rad(A). Obviously, Rad(A) annihilates every completely reducible A-module. If b ∈ A− Rad(A), there exists an irreducible (left) A-module M such that bM ≠ {0} and so bM = M. It is easy to check that Rad(A) is a two-sided ideal of A. Indeed, if r ∈ Rad(A), x, y ∈ A and M is an irreducible (left) A-module, then (xry)M = (xr)yM ⊆ (xr)M = x(rM) = {0}. Theorem 11.12. The radical Rad(A) of the algebra A coincides with the maximal nilpotent two-sided ideal of A. Proof. Let R be the maximal nilpotent two-sided ideal of A and M an irreducible A-module. Since RM is a submodule of M, it follows that either RM = M or RM = 0. If RM = M, then R n M = M for all n ∈ ℕ, which is not the case since R n = {0} for some n ∈ ℕ. Thus, RM = {0}, and so R ⊆ Rad(A). To prove that Rad(A) = R, it remains to prove, in view of Lemma 11.10, that Rad(A) is nilpotent. Let A A = A0 ⊃ A1 ⊃ ⋅ ⋅ ⋅ ⊃ A s = {0} be a composition series of the left regular module A A (i.e., A i are left ideals of A and A i /A i+1 is irreducible for i = 0, 1, . . . , s − 1). Since A i /A i+1 is an irreducible A-module for i = 0, . . . , s − 1, it is annihilated by Rad(A), and therefore we get Rad(A)A i ⊆ A i+1 for i = 0, . . . , s − 1, and so Rad(A)s = {0}, i.e., Rad(A) is nilpotent. Thus, Rad(A) = R, as was to be shown. Lemma 11.13. Suppose that L is a left ideal of the semisimple algebra A. Then L = Ae, where e is an idempotent of A (in other words, L is generated by the idempotent e). Proof. Since the regular module A A is completely reducible, L = L ⊕ L󸀠 for some left ideal L󸀠 of A. It follows that 1 = e + e󸀠 , where 1 is the identity element of A, e ∈ L, e󸀠 ∈ L󸀠 . It follows that e = e ⋅ 1 = e2 + ee󸀠 . Since e, e2 ∈ L, ee󸀠 ∈ L󸀠 , we get e2 = e and ee󸀠 = 0. Similarly, (e󸀠 )2 = e󸀠 , e󸀠 e = 0. We have Ae = {ae | a ∈ A} ⊆ L. If x ∈ L, then x = x ⋅ 1 = xe + xe󸀠 , and so, since x, xe ∈ L, xe󸀠 ∈ L󸀠 , we get x = xe ∈ Ae. Thus, L ⊆ Ae, and we conclude that L = Ae. We see that L from Lemma 11.13 is the least left ideal of A containing the idempotent e. Theorem 11.14. The algebra A is semisimple if and only if Rad(A) = {0}. Proof. (i) Suppose that A is semisimple but Rad(A) ≠ {0}. Then, by Lemma 11.13, Rad(A) = Ae for some idempotent e (recall, that e ≠ 0). Since e ∈ Rad(A), e is nilpotent, by Theorem 11.12, i.e., e m = 0 for some m ∈ ℕ. Since e m = e, we get e = 0, which is a contradiction. Thus, Rad(A) = {0}. (ii) Now suppose that Rad(A) = {0} and prove that A is semisimple. Let L1 be a nonzero minimal left ideal of A. Then L21 ≠ {0}, by Theorem 11.12. Therefore, there exists a ∈ L1 such that L1 a ≠ {0}. Since a left ideal L1 a is contained in a minimal left ideal L1 , we get L1 a = L1 . A mapping ϕ : x 󳨃→ xa (x ∈ L1 ) is a surjective linear opera-

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tor of the finite-dimensional linear 𝔽-space L1 . Therefore, ϕ is not singular. Since a ∈ L1 , we have ϕ(e) = a for some e ∈ L1 , i.e., ea = a. It follows that e2 a = ea, i.e., ϕ(e2 ) = ϕ(e). Therefore, e2 = e since ϕ is nonsingular, i.e., e is an idempotent. In view of e ∈ L1 , we get Ae ⊆ L1 since L1 is a left ideal of A. Since Ae is a nonzero left ideal of A and L1 is minimal, we get L1 = Ae, i.e., L1 is generated by the idempotent e. Set L󸀠 = {x − xe | x ∈ A}. Obviously, L󸀠 is a left ideal, L󸀠 = Ae󸀠 , where e󸀠 = 1 − e is an idempotent; indeed, (e󸀠 )2 = (1 − e)2 = 1 − 2e + e2 = 1 − 2e + e = 1 − e = e󸀠 and

e󸀠 e = (1 − e)e = e − e2 = e − e = 0.

Since x = xe + xe󸀠 ∈ L1 + L󸀠 for x ∈ A, we have A = L1 + L󸀠 . Let x ∈ L1 ∩ L󸀠 . Then xe = x since L1 = Ae. It follows from L󸀠 = Ae󸀠 that xe ∈ (Ae󸀠 )e = A(e󸀠 e) = {0}. Therefore x = 0, and so L1 ∩ L󸀠 = {0} and A = L1 ⊕ L󸀠 . If L󸀠 ≠ {0}, let L2 ⊆ L󸀠 be a minimal left ideal of A. Then, by what has just been proved, A = L2 ⊕ L∗ , where L∗ is a left ideal. Since L2 ⊆ L󸀠 , it follows, by the modular law, that L󸀠 = L2 ⊕ L󸀠󸀠 , where L󸀠󸀠 = L󸀠 ∩ L∗ is a left ideal. Continuing so, we obtain A = L1 ⊕ L2 ⊕ ⋅ ⋅ ⋅ ⊕ L n , where L i (i = 1, . . . , n) are minimal left ideals of A. Thus, A A is completely reducible, i.e., A is a semisimple algebra (see Theorem 11.1). Idempotents e, e󸀠 ∈ A satisfying ee󸀠 = e󸀠 e = 0 are called orthogonal. Corollary 11.15. The algebra A/Rad(A) is semisimple. Proof. It is easy to see that the algebra A/Rad(A) has no nontrivial nilpotent two-sided ideals. Therefore, we have Rad(A/Rad(A)) = {0}, and so A/Rad(A) is semisimple, by Theorem 11.14. Lemma 11.16. The intersection D of all maximal submodules of a completely reducible module M is trivial. Proof. By assumption, M = D ⊕ N for some submodule N of M. Assume that D ≠ {0}; then N ≠ M. Therefore, N ⊂ S, where S is a maximal submodule of M. However, D ⊈ S, a contradiction. Thus, D = {0}. Lemma 11.17. Let I be a nilpotent two-sided ideal of A. Then Rad(A/I) = Rad(A)/I. Proof. Since I is nilpotent, I ⊆ Rad(A). Let x 󳨃→ x be the natural homomorphism of A onto A = A/I. Then Rad(A)/I = Rad(A) is a nilpotent two-sided ideal of A. Therefore, Rad(A)/I ⊆ Rad(A/I). Now let J be a nilpotent ideal of A (and J is its preimage in A). n Then J = {0} for some positive integer n. Therefore, J n ⊆ I, and J is nilpotent (since I is nilpotent). Hence J ⊆ Rad(A), and so J ⊆ Rad(A) = Rad(A)/I. Thus, Rad(A)/I is the maximal nilpotent two-sided ideal of A/I, i.e., Rad(A)/I = Rad(A/I). Theorem 11.18. The radical Rad(A) coincides with the intersection of all maximal left ideals of the algebra A.

42 | Characters of Finite Groups 1 Proof. Since the algebra A/Rad(A) is semisimple (Corollary 11.15), the intersection of all its maximal left ideals is trivial (Lemma 11.16). Therefore, the intersection of all maximal left ideals of A, containing Rad(A), coincides with Rad(A). It follows that the intersection D of all maximal left ideals of A is contained in Rad(A). If M is a maximal left ideal of A, then A A/M is an irreducible A-module. Therefore, it is annihilated by Rad(A). This means that Rad(A) ⋅ A ⊆ M, and this implies Rad(A) ⊆ M since A has the identity element. Therefore, Rad(A) ⊆ D, and so Rad(A) = D. Similarly, Rad(A) coincides with the intersection of all maximal right ideals of the algebra A. Now we can prove the converse to Theorem 11.7: If A has no nontrivial two-sided ideal, it is simple. Indeed, since Rad(A) is a two-sided ideal of A, we get Rad(A) = {0}, i.e., A is semisimple. Now the result follows from Theorem 11.3. Definition 11.4. A two-sided ideal I of A is called minimal if it is nonzero and, whenever J ⊆ I is a nonzero two-sided ideal of A, then J = I. Lemma 11.19. If the algebra A = A1 ⊕⋅ ⋅ ⋅⊕ A r , where A1 , . . . , A r are minimal two-sided ideals of A, then the set {A1 , . . . , A r } contains all minimal two-sided ideals of A. In particular, the above decomposition is unique up to order of summands. Proof. Let B be a minimal two-sided ideal of A such that B ∈ ̸ {A1 , . . . , A r }. Then BA i ⊆ B ∩ A i = {0} for all i, and so BA = {0}, which is not the case in view of 1 ∈ A. Corollary 11.20. The following statements hold: (a) The simple algebras A1 , . . . , A r from Theorem 11.3 are all minimal two-sided ideals of the semisimple algebra A. (b) Let A = A1 ⊕⋅ ⋅ ⋅⊕A r , where A i , . . . , A r are minimal two-sided ideals of the algebra A and let 1 = e1 + ⋅ ⋅ ⋅ + e r , where e i ∈ A i for all i. Then e1 , . . . , e r is a system of pairwise orthogonal idempotents of A and A i is a simple algebra with the identity element e i for i = 1, . . . , r. In particular, A is a semisimple algebra. Proof. Assertion (a) follows from Theorem 11.7 and Lemma 11.2 (a). Assertion (b) follows from Lemma 11.2 (a) and converse to Theorem 11.7. Definition 11.5. The simple algebras A1 , . . . , A r of Theorem 11.3 are called the simple components of the semisimple algebra A. In what follows, we need an information on the center Z(A) of the semisimple algebra A with simple components A1 , . . . , A r (see Theorem 11.3). Recall that Z(A) = {z ∈ A | za = az for all a ∈ A}. Obviously, Z(A) is a commutative subalgebra of A. First we consider the general case of a semisimple algebra A decomposed in direct sum of two-sided ideals A i (i = 1, . . . , k). Recall that A i = Ae i is an algebra with the identity element e i (see Lemma 11.2 (b)). It follows from Lemma 11.2 (a) that e i ∈ Z(A) (i = 1, . . . , k).

I Basic concepts | 43

Lemma 11.21. Let A = A1 ⊕ ⋅ ⋅ ⋅ ⊕ A r be a decomposition of the algebra A in a direct sum of two-sided ideals A1 , . . . , A r . Then Z(A) = Z(A1 ) ⊕ ⋅ ⋅ ⋅ ⊕ Z(A r ). Proof. It follows from Lemma 11.2 (a) that Z(A i ) ⊆ Z(A) so that ⨁ri=1 Z(A i ) ⊆ Z(A). Let x ∈ Z(A), x = x1 + ⋅ ⋅ ⋅ + x r , where x i ∈ A i for all i. If y ∈ A i , then xy = yx, and this implies, by Lemma 11.2 (a), that x i y = yx i . Therefore, x i ∈ Z(A i ) and x ∈ ⨁ri=1 Z(A i ), completing the proof. Let I(A) denote a set of all idempotents of the algebra A. This set is not empty since 1 ∈ I(A). Then, if u, v are orthogonal, then (u + v)2 = u2 + v2 + 2uv = u + v so that u + v is also an idempotent. We have u + v ≠ 0 (otherwise, u = −v so that u = u2 = −uv = 0, a contradiction). Exercise 11.2. Let u1 , . . . , u n be a set of mutually orthogonal idempotents of an algebra A. Prove that these idempotents are linearly independent. Solution. If α1 e1 + ⋅ ⋅ ⋅ + α n e n = 0, where α1 , . . . , α n ∈ 𝔽, then 0 = e i (α1 e1 + ⋅ ⋅ ⋅ + α n e n ) = α i e i for any i so that α i = 0. Let us define on the set I(A) a binary relation ≥ as follows: if u, v ∈ I(A), then u ≥ v means that uv = vu = v. If the idempotents u, w are orthogonal, then (u + w)u = u(u + w) = u so that u + w ≥ u. It follows that if u1 , . . . , u n are pairwise orthogonal idempotents, then u1 + ⋅ ⋅ ⋅ + u n ≥ u1 + ⋅ ⋅ ⋅ + u n−1 ≥ ⋅ ⋅ ⋅ ≥ u1 . Exercise 11.3. Prove that the relation ≥ is reflexive, antisymmetric and transitive. Solution. Obviously, the relation ≥ is reflexive (indeed, uu = u) and antisymmetric (if u ≥ v and v ≥ u, then uv = vu = v and vu = uv = u so that u = v). Let u ≥ v ≥ w be idempotents in A. Then uw = u(vw) = (uv)w = vw = w, wu = (wv)u = w(vu) = wv = w 󳨐⇒ u ≥ w, as desired. It follows from Exercise 11.3 that ≥ is a partial order relation on the set I(A) of idempotents of the algebra A. Obviously, 1 ∈ A is the largest element of the poset (= partially ordered set) I(A). Minimal elements of the poset I(A) are called minimal idempotents of the algebra A. As we shall prove below, the set of minimal idempotents of A is nonempty. Given the relation ≥, we can define a relation > of strong order as follows: u > v if and only if u ≥ v and u ≠ v (u, v ∈ I(A)). If the idempotents u, w are orthogonal, then, as we have seen, u + w > w. Obviously, > is a transitive relation.

44 | Characters of Finite Groups 1 Lemma 11.22. Given u, v ∈ I(A), the following conditions are equivalent: (a) u > v. (b) u = v + w, where w ∈ I(A) and v, w are orthogonal. Proof. Let u > v. Then (u − v)v = uv − v = v − v = 0, and

v(u − v) = 0

(u − v)2 = u2 − 2uv + v2 = u − 2v + v = u − v

so that the idempotents v and w = u − v are orthogonal, hence (a) ⇒ (b). The reverse implication has been proved above. Lemma 11.23. Let u1 , . . . , u m be mutually orthogonal idempotents of the algebra A. Then m does not exceed the dimension of A. Proof. By Exercise 11.2, elements u1 , . . . , u m are linearly independent. A system C = {u0 , u1 , . . . , u k } of idempotents of the algebra A is called a chain of an idempotent u ∈ I(A) if u = u0 > u1 > ⋅ ⋅ ⋅ > u k . The number k is called the length of the chain C. Exercise 11.4. Suppose that C = {u0 , u1 , . . . , u k } is a chain of an idempotent u0 ∈ I(A). Then u0 , . . . , u k are pairwise permutable. Hint. Use Exercise 11.3. Lemma 11.24. The length of a chain of u ∈ I(A) does not exceed the dimension of the algebra A. Proof. Let C = {u0 , . . . , u k } be a chain of an idempotent u0 = u ∈ I(A) of length k. Then v i = u i − u i+1 ≠ 0 and v i ∈ I(A) for i = 0, . . . , k − 1, by Lemma 11.22. Besides, if 0 ≤ i < j ≤ k − 1, then v i v j = (u i − u i+1 )(u j − u j+1 ) = u i u j − u i u j+1 − u i+1 u j + u i+1 u j+1 = u j − u j+1 − u j + u j+1 = 0 (we have used transitivity of the relation >). Since the elements u0 , . . . , u k are mutually permutable (Exercise 11.4), it follows that v j v i = 0. Hence v0 , . . . , v k−1 is a set of mutually orthogonal idempotents of the algebra A, and so we have k ≤ dim(A), by Lemma 11.23. Chains of maximal length of u ∈ I(A) are called maximal. This notion is well defined, as follows from Lemma 11.24. Lemma 11.25. For u ∈ I(A), the following conditions are equivalent: (a) u is a minimal idempotent of A. (b) If uv = vu = v ∈ I(A), then v = u.

I Basic concepts | 45

Proof. (a) ⇒ (b): Suppose that u is a minimal idempotent of A and uv = vu = v for v ∈ I(A). Then u ≥ v so u = v, and (b) follows. (b) ⇒ (a): Suppose that u ≥ v for v ∈ I(A). Then uv = vu = v so u = v, by hypothesis, and so (a) is true. Definition 11.6. An idempotent v ∈ I(A) is covered by u ∈ I(A) if u > v and, provided u ≥ w > v with w ∈ I(A), then w = u. In this case, we write u >̇ v. Lemma 11.26. For u, v ∈ I(A), the following conditions are equivalent: (a) u >̇ v. (b) u − v is a minimal idempotent of the algebra A. Proof. (a) ⇒ (b): Suppose that u >̇ v. Then u − v ∈ I(A), by Lemma 11.22. Assume that u − v is not a minimal idempotent. Then u − v > w, where w ∈ I(A). By Lemma 11.22, u − v = w + e, where e ∈ I(A) and we = ew = 0. Next, (u − v)v = uv − v2 = v − v = 0 󳨐⇒ (w + e)v = wv + ev = 0. Similarly, it follows from v(u − v) = 0 that v(w + e) = vw + ve = 0. Therefore, wv = w(wv) = −w(ev) = −(we)v = 0, vw = (vw)w = −(ve)w = −v(ew) = 0. Thus, vw = wv = 0, and so (v + w)2 = v + w, v + w ∈ I(A) since v + w ≠ 0 (otherwise, 0 = vw = (−w)w = −w so w = 0). It follows from (u − v)w = (w + e)w = w = w(u − v) that uw = wu = w since wv = vw = 0. Therefore, u(v + w) = uv + uw = v + w, (v + w)u = vu + wu = v + w. Since v + w = u − e ≠ u, we get u > v + w. Noting that v + w > v, we see that u does not cover v, contrary to the hypothesis. Hence u − v is a minimal idempotent of A. (b) ⇒ (a): Assume that u − v is a minimal idempotent of the algebra A. Since u − v ∈ I(A), it follows that (u − v)2 = u − v whence uv + vu = 2v, i.e., vuv + vu = v(uv + vu) = 2v, vuv + uv = (vu + uv)v = 2v. Therefore, uv = vu and, in view of uv + vu = 2v, we obtain uv = vu = v, i.e., u ≥ v. Since u ≠ v, we have u > v. If u > w > v, where w ∈ I(A), then u − w and w − v are orthogonal idempotents of A (Lemma 11.24). Therefore, u − v = (u − w) + (w − v) > u − w (recall that u − v is a minimal idempotent), which is a contradiction. Thus, u >̇ v. Lemma 11.27. If C = {u0 , . . . , u k } is a maximal chain of u0 = u ∈ I(A), then the elements v0 = u0 − u1 , v1 = u1 − u2 , . . . , v k−1 = u k−1 − u k form a system of mutually orthogonal minimal idempotents of the algebra A.

46 | Characters of Finite Groups 1 Proof. It follows from Lemma 11.26 and the proof of Lemma 11.24, that v0 , . . . , v k−1 are mutually orthogonal minimal idempotents of A. Corollary 11.28. The set of minimal idempotents of the algebra A is nonempty. Lemma 11.29. If u, v are permutable minimal idempotents of A, then either u = v or u and v are orthogonal. Proof. Suppose that uv = e ≠ 0. Since eu = uvu = u2 v = uv = e,

ue = uuv = uv = e,

ev = uvv = uv = e,

ve = vuv = uv = e,

we get u ≥ e, v ≥ e. Since e ∈ I(A), then, taking into account the minimality of u and v, we get u = e = v so u = v, as desired. Corollary 11.30. The set of minimal idempotents of a commutative algebra A is finite. Proof. By Lemma 11.29, distinct minimal idempotent of the commutative algebra form an orthogonal system, so they are linearly independent (Exercise 11.2). Therefore, their number does not exceed dim𝔽 (A), the dimension of A. Exercise 11.5. Let {e1 , . . . , e m } be the set of minimal idempotents of a commutative algebra A. Given a nonempty subset S ⊆ {1, . . . , m} = M, put e S = ∑i∈S e i ; obviously, e S is an idempotent of A. Prove that 1 = e M and I(A) = {e S | 0 ≠ S ⊆ M}. In particular, |I(A)| = 2m − 1. Hint. Use Lemmas 11.27 and 11.29. Definition 11.7. A nonzero two-sided ideal of the algebra A is said to be indecomposable if it is not a sum of two nonzero proper two-sided ideals of A. Minimal two-sided ideals of A are indecomposable. However, the converse is not necessarily true, i.e., generally speaking, not all indecomposable two-sided ideals are minimal. Any finite-dimensional algebra is a direct sum of indecomposable two-sided ideals. The following important result holds. Lemma 11.31. A decomposition of the algebra A into direct sum of indecomposable two-sided ideals is unique up to the order of summands. Proof. Let A = A1 ⊕ ⋅ ⋅ ⋅ ⊕ A r = B1 ⊕ ⋅ ⋅ ⋅ ⊕ B s be decompositions of A into direct sum of indecomposable two-sided ideals. Since A has the identity element, it follows that B i = B i A = B i A1 ⊕ ⋅ ⋅ ⋅ ⊕ B i A r for i ∈ {1, . . . , s}. Since B i A j are two-sided ideals of A, we get B i = B i A j (⊆ B i ∩ A j ) for some j ∈ {1. . . . , r}, taking into account the indecomposability of B i (here we use the modular law). Therefore, B i ⊆ A j . Similarly, A j ⊆ B k for some k ∈ {1, . . . , s}. Thus B i ⊆ A j ⊆ B k so i = k and B i = A j . This shows that the given decompositions are the same up to the order of summands.

I Basic concepts | 47

Theorem 11.32. Let A = A1 ⊕ ⋅ ⋅ ⋅ ⊕ A r be a decomposition of an 𝔽-algebra A into direct sum of indecomposable two-sided ideals. Then the identity elements e1 , . . . , e r of the algebras A1 , . . . , A s , respectively, form the full system of minimal idempotents of the center Z(A) of A. Proof. By Lemma 11.2 (a), e i ∈ Z(A i ) for all i and e1 + ⋅ ⋅ ⋅ + e r = 1. It remains to show that all e1 , . . . , e r are minimal idempotents of Z(A) and there are no other minimal idempotents in Z(A). If some e i is not a minimal idempotent of Z(A), then by Lemma 11.22, e i = v + w, where v and w are orthogonal idempotents of Z(A). Clearly, Av and Aw are nonzero two-sided ideals of A and Av = A(ve i ) ⊆ Ae i = A i , and similarly Aw ⊆ A i . Therefore, we have A i = A(v + w) ⊆ Av + Aw ⊆ A i , and we conclude that A i = Av + Aw. We claim that A i = Av ⊕ Aw. Assume that x ∈ Av ∩ Aw. Then xv = x = xw 󳨐⇒ x = xe i = xv + xw = 2x since x ∈ A i = Ae i . Therefore, x = 0 so Av∩Aw = {0} and the sum A i = Av+Aw is direct, which a contradiction. Thus, e1 , . . . , e r are minimal idempotents of the algebra Z(A). Now let e be an arbitrary minimal idempotent of Z(A). If e ∈ ̸ {e1 , . . . , e r }, then ee i = 0 for all i, by Lemma 11.29. Then e = e ⋅ 1 = e(e1 + ⋅ ⋅ ⋅ + e r ) = 0, which is a contradiction. Corollary 11.33. Let A be a semisimple algebra and let e1 , . . . , e r be the identity elements of its simple components A1 , . . . , A r . Then {e1 , . . . , e r } is the set of minimal idempotents of Z(A).

12 Representations of semisimple algebras The semisimplicity of the algebra A can be expressed in the language of representation theory as follows: Theorem 12.1. The following assertions on the algebra A are equivalent: (a) A is semisimple. (b) All representations of A are completely reducible. Proof. The implication (a) ⇒ (b) follows from the definition of the semisimple algebra (see Definition 11.1). If (b) holds, then the (faithful) regular representation of A is completely reducible so the left regular module A A is completely reducible. Then A is semisimple by Theorem 11.1, proving (a). It follows from Theorem 12.1 that in order to find all representations of a semisimple algebra A, it is enough to know all its irreducible representations. Lemma 12.2. Any irreducible representation of the semisimple algebra A is afforded by some its minimal left ideal considered as an irreducible A-module.

48 | Characters of Finite Groups 1 Proof. It suffices to show that every irreducible A-module M is isomorphic to some minimal left ideal of A considered as an A-module. Let A = L1 ⊕ ⋅ ⋅ ⋅ ⊕ L n be a direct sum of minimal left ideals of A. (Such a decomposition exists by Theorem 11.1.) We have M = AM = ∑ni=1 L i M since A has the identity element. It follows that L i M ≠ {0} for some i ∈ {1, . . . , n}. Therefore, L i m ≠ {0} for some m ∈ M. Hence L i m = M since L i m is a (nonzero) submodule of the irreducible module M. Then a mapping ϕ : x → xm (x ∈ L i ) is a homomorphism L i → M. Since ϕ is surjective, it is an isomorphism, by Lemma 6.2. Theorem 12.3. The number of equivalence classes of irreducible representations of the semisimple algebra A is equal to the number of its simple components so this number is finite. Proof. The simple components A1 , . . . , A r of the algebra A are homogeneous components of the left regular module A A (see the proof of Theorem 11.3). Let L(i) be a minimal left ideal of A contained in A i and let T i be an irreducible matrix representation of A afforded by L(i) in some basis, i = 1, . . . , r. Since L(1) , . . . , L(r) are pairwise nonisomorphic A-modules, the representations T1 , . . . , T r are nonequivalent. Let T be an irreducible representation of A. Then T is afforded by some minimal left ideal L of A, by Lemma 12.2. However, L is contained in some simple component A i of A. Then L ≅ L(i) since A i is a homogeneous completely reducible A-module, and we conclude that T ∼ T i . Therefore, T1 , . . . , T r is a transversal of equivalence classes of irreducible representations of A. It follows from the proof of Theorem 12.3 the following: Theorem 12.4. Let A = A1 ⊕ ⋅ ⋅ ⋅ ⊕ A r be the decomposition of a semisimple algebra A in a direct sum of simple components and let R be the regular representation of A. Then in the notation of the proof of Theorem 12.3, we have R ∼ n1 T1 + ⋅ ⋅ ⋅ + n r T r , where n i , the multiplicity of T i in R, is equal to the length of the simple component A i considered as A-module (i.e., n i is the number of summands in decomposition of A i in a direct sum of minimal left ideals).² Corollary 12.5. A semisimple algebra is simple if and only if it has only one class of irreducible representations. This follows from the definition of simple algebras and Theorem 12.4. Lemma 12.6. Every representation of an arbitrary simple algebra A is faithful. In particular, all irreducible representations of A are faithful (by Theorem 12.3, all these representations are equivalent).

2 Since distinct simple components of A are orthogonal, minimal left ideals of a component A i are minimal left ideals of A.

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Proof. Let T be a representation of the simple algebra A. Since T(1) = I, the identity matrix, ker(T) ≠ A and so ker(T) = {0} as a proper two-sided ideal of the simple algebra A. Corollary 12.7. Any simple algebra A is isomorphic to an irreducible matrix algebra. Proof. If T is an irreducible matrix representation of A, then A ≅ T(A), by Lemma 12.6. It remains to note that T(A) is an irreducible matrix algebra. Exercise 12.1. Let R be the radical of an algebra A and let T be an irreducible representation of A. Then T(R) = {0}. Hint. Use Definition 11.3. Exercise 12.2. All irreducible matrix algebras are simple. Hint. Let A be an irreducible matrix algebra. Since Rad(A) = {0}, the algebra A is semisimple (Theorem 11.14). Use Corollary 11.33 and Schur’s lemma to prove that the identity element of any simple component A i of A coincides with the identity element 1 of A. Since 1 ∈ A i and A i is an ideal of A, we get A = A i . Corollary 12.8. The center of a simple algebra is a field. Proof. By Corollary 12.7, A ≅ A, where A is an irreducible matrix algebra. It follows that Z(A) ≅ Z(A). By Schur’s lemma, Z(A) is a field. Theorem 12.9. The center of a semisimple algebra A is a direct sum of r mutually orthogonal fields, where r is the number of simple components of A. In particular, a semisimple algebra is simple if and only if its center is a field. Proof. Let A = A1 ⊕ ⋅ ⋅ ⋅ ⊕ A r be a decomposition of A in a direct sum of simple components. Then, by Lemma 11.21, Z(A) = Z(A1 ) ⊕ ⋅ ⋅ ⋅ ⊕ Z(A r ), and the claim follows from Corollary 12.8. Corollary 12.10. Let A be a semisimple algebra over an algebraically closed field 𝔽. Then r, the number of classes of irreducible representations of A, is equal to dim𝔽 (Z(A)). Proof. By Theorem 12.3, r is the number of simple components A i of A. Since Z(A) = Z(A1 ) ⊕ ⋅ ⋅ ⋅ ⊕ Z(A r ), it suffices to show that dim𝔽 (Z(A i )) = 1 for all i. This is true since Z(A i ) is a field that is a finite extension of an algebraically closed field 𝔽 so coincides with 𝔽. Wedderburn’s theorem allows us to improve the above results. Let A be a simple algebra. Then A ≅ K n , where K is a division algebra over 𝔽, by Theorem 11.9. It follows from the proof of Theorem 11.9 that A = ⨁ni,j=1 Ke ij , where e ij e kl = δ jk e il and all elements from K (that is assumed embedded in A) are permutable with all e ij . Using the above relation, it is easy to show that e11 + ⋅ ⋅ ⋅ + e nn = 1.

50 | Characters of Finite Groups 1 Lemma 12.11. We have Z(K n ) = Z(K) (here we assume that the skew field K is embedded in the algebra K n ). Proof. It is clear that Z(K) ⊆ Z(K n ). Let α ij ∈ K

(i, j = 1, . . . , n) and

n

z = ∑ α ij e ij ∈ Z(K n ). i,j=1

Then ze kl = e kl z for all k, l. Noting that ze kl = ∑ni=1 α ik e il and e kl z = ∑nj=1 α lj e kj , we get α kk = α ll (since α kk is the coefficient of e kl in ze kl and α ll is the coefficient of e kl in e kl z). Thus, α11 = ⋅ ⋅ ⋅ = α nn . Next, if k ≠ l, then e kl ze kk = ze kl e kk = 0. Noting that n

e kl ze kk = ∑ α lj e kj e kk = α lk e kk , j=1

we obtain α lk = 0. Setting α ii = α (i = 1, . . . , n), we get z = α ∑ni=1 e ii = α ∈ K, and so z ∈ Z(K). This proves that Z(K n ) ⊆ Z(K). Thus Z(K n ) = Z(K). We see that Corollary 12.8 follows from Lemma 12.11 and Theorem 11.9. Now we improve Theorem 12.4 using Wedderburn’s theorem. Let A1 , . . . , A r be the simple components of the semisimple 𝔽-algebra A. By Theorem 11.9, A i ≅ (K (i) )n i , where K (i) is a division 𝔽-algebra, i = 1, . . . , r. Write ni

(i)

A i = ⨁ K (i) e kl , k,l=1 (i) (i)

(i)

n

(i)

i where e kl e pq = δ lq e kq . Then (see the proof of Theorem 11.8) L(i) = ⨁k=1 K (i) e k1 is (i) a minimal left ideal of the algebra A i and so L , in view of mutual orthogonality of A1 , . . . , A r , is a minimal left ideal of the algebra A. It follows from the definition that L(i) is a linear K (i) -space of dimension n i . Let d i be the dimension of the skew field K (i) over the field 𝔽; then dim𝔽 (L(i) ) = n i d i . Let T i be the irreducible representation of the algebra A afforded by a minimal left ideal L(i) (i = 1, . . . , r). Since deg(T i ) = dim𝔽 (L(i) ), we get deg(T i ) = n i d i . Thus, the representations T1 , . . . , T r form a transversal of equivalence classes of irreducible representations of the algebra A (see the proof of Theorem 12.3). Since the length of the A-module A i is n i (see the proof of Theorem 11.8), for the regular representation R of the algebra A we have R ∼ n1 T1 + ⋅ ⋅ ⋅ + n r T r , by Theorem 12.4. Since dim𝔽 (A) = deg(R) = ∑ri=1 n i deg(T i ), it follows by what has been proved already that dim(A) = n21 d1 + ⋅ ⋅ ⋅ + n2r d r . We have proved the following:

Theorem 12.12. Let A = A1 ⊕ ⋅ ⋅ ⋅ ⊕ A r , where A1 , . . . , A r be the simple components of (i) the semisimple 𝔽-algebra A. Let A i = K n i , where K (i) is a skew field, i = 1, . . . , r. Then R ∼ n1 T1 + ⋅ ⋅ ⋅ + n r T r , where R is the regular representation of A, T1 , . . . , T r is a complete system of representatives of equivalence classes of irreducible representations of A and T i is afforded by a minimal left ideal of A that contained in A i . If dim𝔽 (K i ) = d i for all i, then the following equalities hold: deg(T i ) = n i d i , dim(A) = n21 d1 + ⋅ ⋅ ⋅ + n2r d r , d i = dim𝔽 (K (i) ),

i = 1, . . . , r.

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Now suppose that the field 𝔽 is algebraically closed. Then d i = 1 for all i, and the relations in Theorem 12.12 take the form deg(T i ) = n i

for all i,

dim𝔽 (A) = n21 + ⋅ ⋅ ⋅ + n2r .

Thus the following assertion holds: Theorem 12.13. Let A be a semisimple 𝔽-algebra, where the field 𝔽 is algebraically closed. Let T1 , . . . , T r be a complete system of representatives of classes of irreducible representations of A and R the regular representations of A. Then R ∼ deg(T1 )T1 + ⋅ ⋅ ⋅ + deg(T r )T r , dim𝔽 (A) = deg(T1 )2 + ⋅ ⋅ ⋅ + deg(T r )2 . Exercise 12.3. Prove that the mapping ϕ : A → 𝔽n1 +̇ ⋅ ⋅ ⋅ +̇ 𝔽n r defined by ϕ(x) = (T1 (x), . . . , T r (x))

(x ∈ A)

is an isomorphism of the algebra A onto direct sum of matrix algebras 𝔽n i , i = 1, . . . , r. Hint. Since the regular representation of the algebra A is faithful, we have r

⋂ ker(T i ) = {0}. i=1

Now let T be an arbitrary representation of the algebra A and let M be a nonempty subset of A. Set T(M) = {T(x) | x ∈ M}. Lemma 12.14. Let A be a semisimple algebra and let A1 , . . . , A r be all its simple components. Then, in the notation of Theorem 12.12, T i (A) = T i (A i ) (i = 1, . . . , r). Proof. Obviously, we have T i (A i ) ⊆ T i (A). Let x ∈ A. Since A = A1 ⊕ + ⋅ ⋅ ⋅ ⊕ A r , we have x = x1 + ⋅ ⋅ ⋅ + x r with x j ∈ A j , j = 1, . . . , r. Let L(i) be a minimal left ideal of A contained in A i . Then L(i) affords (in some basis) the representation T i . If j ≠ i, then x j L(i) = {0}, by Lemma 11.2. Hence, we have T i (x j ) = 0, and so T i (x) = ∑rj=1 T i (x j ) = T i (x i ) ∈ T i (A i ). Thus, T i (A) ⊆ T i (A i ), completing the proof. Corollary 12.15. We have that (T i )A i , the restriction of T i to A i , is an irreducible representation of the algebra A i .³ Theorem 12.16. Let A be a semisimple algebra over an algebraically closed field 𝔽 and let T be its irreducible matrix representation of degree n. Then T(A) = 𝔽n is the full algebra of n × n matrices over 𝔽. Proof. In view of Lemma 12.14 and Corollary 12.15, one may consider the case when A is simple. By Wedderburn’s Theorem 11.9, A ≅ 𝔽n . Let {e ij } be the standard basis of the matrix algebra 𝔽n . As we know (see the proof of Theorem 11.8), L1 = ∑ni=1 𝔽e i1 is a minimal left ideal of A. Without loss of generality, one may assume 3 As we have noted, if r > 1, then A i is not a subalgebra A since 1 ∉ A i .

52 | Characters of Finite Groups 1 that T is afforded by the ideal L1 in a basis {e11 , . . . , e n1 }. Let a = ∑ni,j=1 α ij e ij ∈ A. Then ae j1 = ∑ni,k=1 α ik e ik e j1 = ∑ni=1 α ij e i1 . Therefore, T(a) = (α ij )1n is an n × n matrix. Since the elements α ij ∈ 𝔽 are arbitrary, this implies that T(A) = 𝔽n . Definition. Let T be a matrix representation of an 𝔽-algebra A. A function χ T : A → 𝔽 such that χ T (a) = tr(T(a))(a ∈ A) is said to be the character of T. The following theorem is obvious. Theorem 12.17. If T and U are matrix representations of an 𝔽-algebra A, then χ T+U = χ T + χ U . Theorem 12.18. Characters of equivalent matrix representations of an 𝔽-algebra A are equal. Proof. Let T and T 󸀠 be equivalent matrix representations of the algebra A. Then deg(T) = n = deg(T 󸀠 ) and, for all a ∈ A, we have T 󸀠 (a) = S−1 T(a)S for some matrix S ∈ GL(n, 𝔽). It follows from the known property of the trace of a matrix that 󸀠

χ T (a) = tr(T 󸀠 (a)) = tr(S−1 T(a)S) = tr(T(a)) = χ T (a), 󸀠

and so χ T = χ T . If the characteristic of 𝔽 is 0, then, as the following theorem shows, the reverse assertion is true for semisimple algebras: Theorem 12.19. If an 𝔽-algebra A is semisimple and the characteristic of 𝔽 is 0, then the representations T and T 󸀠 of A, having equal characters, are equivalent. Proof. Let A1 , . . . , A r be all simple components of A and T1 , . . . , T r corresponding irreducible representations of A. By Theorem 12.1, the representations T and T 󸀠 are completely reducible: T ∼ m1 T1 + ⋅ ⋅ ⋅ + m r T r ,

T 󸀠 ∼ m󸀠1 T1 + ⋅ ⋅ ⋅ + m󸀠r T r ,

where all coefficients m i , m󸀠i are nonnegative integers. It suffices to prove that m i = m󸀠i for all i. Let χ i = χ T i be the character of the representation T i for all i. Then, by Theorems 12.17 and 12.18, r

󸀠

r

χT = ∑ mi χi ,

χ T = ∑ m󸀠i χ i .

i=1

i=1

Let L(i) be a minimal left ideal of A contained in A i , all i. Then the representation T i is afforded (in some basis B i ) by the ideal L(i) considered as a left A-module: if a ∈ A, then T i (a) is the matrix (in the basis B i ) of a linear operator τ i (a) : L(i) → L(i) given by τ i (a)(x) = ax (x ∈ L(i) ). Since different minimal two-sided ideals of A are mutually orthogonal, it follows, for j ≠ i and a ∈ A j , that τ i (a) = 0. Next, if e i is the identity

I Basic concepts | 53

element of A i , then τ i (e i ) is the identity operator, and so T i (e i ) = If i is the f i × f i identity matrix, where f i = dim𝔽 (L(i) ) = deg(T i ) = n i d i in the notation of Theorem 12.12. If j ≠ i, then T i (e j ) = O f i is the f i × f i zero matrix. Therefore, χ i (e i ) = tr(I f i ) = f i . If j ≠ i, then, by the above, χ(e j ) = tr(T(e j )) = tr(O f i ) = 0. Thus, χ i (e j ) = δ ij f i , i, j = 1, . . . , r. 󸀠 󸀠 It follows that χ T (e i ) = m i f i , χ T (e i ) = m󸀠i f i for all i. Since, by assumption, χ T = χ T , we get m i f i = m󸀠i f i , and so m i = m󸀠j , all i, since the characteristic of 𝔽 is 0. Therefore, T ∼ T 󸀠 , as desired. Theorem 12.20. If the field 𝔽 is algebraically closed and the algebra A is semisimple, then the characters χ1 , . . . , χ r are linearly independent (we retain the notation of the previous theorem). Proof. In the notation of Theorem 12.19, deg(T i ) = dim𝔽 (L(i) ) = f i for all i. Then we have T i (A) = T i (A i ) = 𝔽f i , by Lemma 12.14 and Theorem 12.16. It follows that, for each i, there exists an element a i ∈ A i such that χ i (a i ) = tr(T i (a i )) = 1. If j ≠ i, then χ i (a j ) = 0 (see the proof of the previous theorem). Thus, χ i (a j ) = δ ij , all i, j. Suppose that λ1 χ1 + ⋅ ⋅ ⋅ + λ r χ r = 0, where λ i ∈ 𝔽 for all i. Then, for each j, r

r

0 = ∑ λ i χ i (a j ) = ∑ λ i δ ij = λ j . i=1

i=1

Thus, we have λ1 = ⋅ ⋅ ⋅ = λ r = 0, proving the linear independence of the characters χ1 , . . . , χ r . Remark. It follows from the definition of Rad(A) (the radical of algebra A) that there exists the natural one-to-one correspondence between irreducible representations of algebras A and A = A/Rad(A). Therefore, the number of classes of irreducible representations of A is finite. If χ is an irreducible character of A, it may be considered as an irreducible character of A, and conversely. If, in addition, 𝔽 is algebraically closed, the irreducible characters χ1 , . . . , χ r of A are linearly independent even if Rad(A) ≠ {0}. In conclusion of this section we consider homomorphisms of the center Z(A) of a semisimple K-algebra A to the algebraic closed field K of characteristic 0. We retain the above notation. Let e i (i = 1, . . . , r) be the minimal idempotents of the algebra Z(A) and take z ∈ Z(A). Since these idempotents form the basis of Z(A), we get z = ω i (z)e1 + ⋅ ⋅ ⋅ + ω r (z)e r ,

ω i (z) ∈ K (i = 1, . . . , r).

Since the system {e i }1r is orthogonal, it follows that the mappings ω i : Z(A) → K are homomorphisms (these homomorphisms are also called linear characters of Z(A)). Indeed, if also z󸀠 = ∑ri=1 ω i (z󸀠 )e i , then r

r

r

r

∑ ω i (zz󸀠 )e i = ( ∑ ω i (z)e i )( ∑ ω i (z󸀠 )e i ) = ∑ ω i (z)ω i (z󸀠 )e i i=1

so that ω i

(zz󸀠 )

i=1

= ω i (z)ω i

(z󸀠 ),

i=1

i=1

and our claim follows. It follows from the definition of

54 | Characters of Finite Groups 1 the mappings ω i that ω i (e j ) = δ ij

(i, j = 1, . . . , r),

(1)

where δ ij is Kronecker’s delta on the set {1, . . . , r}. Let ω be an arbitrary linear character of the algebra Z(A). Then ω(e i )2 = ω(e2i ) = ω(e i ) 󳨐⇒ ω(e i ) ∈ {0, 1} (i = 1, . . . , r). Next, e1 + ⋅ ⋅ ⋅ + e r = 1 󳨐⇒ ω(e1 ) + ⋅ ⋅ ⋅ + ω(e r ) = ω(1) = 1, and we conclude that ω(e j ) = δ ij for some i ∈ {1, . . . , r}. Next, (1) implies that ω = ω i . Thus, the following important result holds: Lemma 12.21. The set {ω i }1r is the set of all linear characters of the algebra Z(A). Now let, as above, χ i be the character of an irreducible representation T i of our algebra A, afforded by a minimal left ideal contained in the simple component A i of A. As was proved above (see the proof of Theorem 12.19), χ i (e j ) = δ ij f i ,

where f i = deg(T i ) = χ i (1), i, j = 1, . . . , r.

(2)

Take z ∈ Z(A). Since z = ω i (z)e1 + ⋅ ⋅ ⋅ + ω r (z)e r , it follows from (2) that r

r

χ i (z) = ∑ ω j (z)χ i (z j ) = ∑ ω j (z)δ ij f i = ω i (z)f i , i=1

i=1

and we obtain ω i (z) =

χ i (z) fi

(i = 1, . . . , r).

(3)

This establishes a connection between irreducible characters χ i of the algebra A and homomorphisms ω i of its center Z(A) (i = 1, . . . , r).

A On q󸀠 -automorphisms of q-groups We assume that the reader is familiar with an extension of Maschke’s theorem to abelian groups, Theorem 8.2. We will prove one consequence of Maschke’s theorem, Theorem A.1. That theorem is a fairly deep result on finite abelian groups with a coprime operator group. We use semidirect products instead of actions. The proof of Theorem A.1 is based on some nontrivial assertions on abelian groups (which are not new). A p-group is said to be homocyclic if it is a direct product of cyclic groups of equal order. A p-group is homocyclic if and only if it is abelian of type (p n , p n , . . . , p n ) for

I Basic concepts | 55

some n ∈ ℕ. In particular, the elementary abelian group Ep n of order p n is homocyclic. An abelian p-group of exponent p e > p is homocyclic if and only if at least one of the following holds: (i) Φ(G) is homocyclic with d(Φ(G)) = d(G), (ii) ℧e−1 (G) = Ω1 (G). Any nonhomocyclic abelian p-group > {1} is a direct product of homocyclic subgroups of pairwise distinct exponents. In what follows, p and q are distinct primes and b is the least natural number such that q b ≡ 1 (mod p), i.e., b is the order of q modulo p (or, what is the same, p is a Zsigmondy prime for the pair ⟨q, b⟩). Exercise A.1. Suppose that Q = C1 × ⋅ ⋅ ⋅ × C d , where C1 , . . . , C d are cyclic groups of orders q m1 , . . . , q m d , respectively, and let m = m1 = ⋅ ⋅ ⋅ = m s < m s+1 ≤ ⋅ ⋅ ⋅ ≤ m d = μ. Set F = Ω m (Q),

m

M = ℧m (Q) = ⟨x q | x ∈ Q⟩,

Q1 = C1 × . . . C s ,

Q2 = C s+1 × ⋅ ⋅ ⋅ × C d .

Then the following hold: (a) Every cyclic subgroup of order q μ is a direct factor of Q. (b) M = ℧m (Q2 ). (c) F ≅ Q/M is homocyclic of rank d and exponent q m . (d) If Q1 ≅ L ≤ F and L ∩ M = {1}, then Q = L × Q2 . (e) If m = μ (i.e., Q is homocyclic), then every homocyclic subgroup of exponent q m is a direct factor of Q. (f) ℧μ−1 (Q) = Ω1 (Q) if and only if Q is homocyclic. Hint. Statements (a), (b), (c), (f) are well known (for (a), see [Ber31, Introduction, Lemma 4 (b)]; (a) implies (e). Let us prove (d). It follows from Ω1 (L) ∩ Q2 = Ω1 (L) ∩ Ω1 (Q2 ) = Ω1 (L) ∩ Ω1 (M) ≤ L ∩ M = {1} and |Q : Q2 | = |Q1 | = |L| that Q = L × Q2 . The main result of this section is the following theorem extending Maschke’s theorem to abelian prime power groups. Theorem A.1 ([Ber2, Theorem 1]). Suppose that Q is a normal abelian Sylow q-subgroup of a group G and R is a minimal normal q-subgroup of G. Then there exists a normal homocyclic q-subgroup S of G such that the following hold: (a) Ω1 (S) = R. (b) Q = S × S1 , where S1 is normal in G. (c) exp(S) = q e(R) depends only on R. (d) If R ≤ U ≤ Q and Ω1 (U) = R, then exp(U) ≤ q e(R) = exp(S), i.e., U is isomorphic to a subgroup of S. Proof. We retain the notation of Exercise A.1 for Q. It is clear that R ≤ Q.

56 | Characters of Finite Groups 1 If Q is elementary abelian, the result follows from Maschke’s theorem. In what follows, we assume that exp(Q) > q. Suppose that the desired decomposition has been proved for all groups of order < |G|. Let A be a q󸀠 -Hall subgroup of G (Schur–Zassenhaus). Since Q is abelian and G = A ⋅ Q, a semidirect product with kernel Q, we have: Claim (i). Any A-invariant q-subgroup of G is normal in G. Claim (ii). If T is a normal q-subgroup of G such that Ω1 (T) = R, then T is homocyclic. To prove this assertion, one may assume that exp(T) = q r > q. It follows from {1} < ℧r−1 (T) ≤ Ω1 (T) = R, minimality of R and normality of ℧r−1 (T) in G that ℧r−1 (T) = R since R is a minimal normal subgroup of G. Hence T is homocyclic (Exercise A.1 (f)). Claim (iii). Let T be a normal q-subgroup of G of maximal order such that Ω1 (T) = R. Then the rank of Q/T is equal to d − d(R), where d = d(Q) is the rank of Q; next,we have d(R) = d(T). In particular, T ≰ Φ(Q). Indeed, by (ii), T is homocyclic, Ω1 (Q) and Ω1 (Q/T) are G-invariant elementary abelian q-subgroups of G and G/T, respectively; therefore, by Maschke’s theorem, there exists a normal subgroup L of G such that Ω1 (Q) = R × L, Ω1 (Q/T) = (LT/T) × (U/T),

where U ⊲ G.

We have U ∩ L = U ∩ (LT ∩ L) = (U ∩ LT) ∩ L = T ∩ L = Ω1 (T) ∩ L = R ∩ L = {1}. Therefore, |R| ≤ |Ω1 (U)| ≤ |Ω1 (Q) : L| = |R| 󳨐⇒ |Ω1 (U)| = |R|. Since R ≤ U, we get Ω1 (U) = R. By the choice of T, we have U = T. Thus, Ω1 (Q/T) = LT/T = LU/U ≅ L, i.e., the rank of Q/T is d(L) = d − d(R) < d. In particular, T ≰ Φ(Q). Claim (iv). If Q is homocyclic, i.e., Q = F (see Exercise A.1), the theorem is true. Indeed, let T be as in (iii) and exp(Q) = q m (see Exercise A.1). Since Φ(Q) = Ω m−1 (Q), we have exp(T) = q m . By (ii), T is homocyclic, so that Q = T × T1 , by Exercise A.1 (e). By Theorem 8.2, Q = T × T2 , where T2 ⊲ G. All other assertions of the theorem are trivial in the case under consideration. In what follows, we assume that Q is not homocyclic. Then m < μ, Ω m (Q) = F < Q, ℧m (Q) = M > {1} (as in Exercise A.1, m is the minimal invariant of Q and μ is the maximal invariant of Q).

I Basic concepts | 57

Claim (v). We claim that if R ≰ M, assertions (a) and (b) of the theorem are true. Indeed, in this case, R ∩ M = {1} since R is a minimal normal subgroup of G and M is normal in G. By (iv), F = F1 × F2 , where F = Ω m (Q) is G-invariant homocyclic, F1 , F2 of exponent q m are normal in AF (here A is a q󸀠 -Hall subgroup of G) and Ω1 (F1 ) = R. By (i), F1 , F2 ⊲ G. Since Ω1 (F1 ) ∩ M = R ∩ M = {1}, it follows that F1 ∩ M = {1}. Set F3 = F ∩ Q2 = Ω m (Q2 ) (recall, that we retain the notation of Exercise A.1). Then F1 F3 = F1 × F3 is homocyclic of exponent q m . By (iv), F = (F1 F3 ) × F4 = (F1 × F4 ) × F3

for some F4 < F,

and we obtain Q = FQ2 = [(F1 × F4 ) × F3 ]Q2 = (F1 × F4 )Q2 . Since (F1 × F4 ) ∩ Q2 = (F1 × F4 ) ∩ Ω m (Q2 ) = (F1 × F4 ) ∩ F3 = {1}, we get Q = (F1 × F4 ) × Q2 = F1 × (F4 × Q2 ), i.e., F1 (⊲ G) is complemented in Q. By Theorem 8.2, Q = F1 × S1 , where S1 ⊲ G, and assertions (a)–(b) of the theorem hold with S = F1 . In what follows, we assume that R ≤ M. By Maschke’s theorem, Ω1 (Q) = L1 × Ω1 (M), where M = ℧m (Q) and L1 ⊲ G. We have L1 ∩ M = L1 ∩ Ω1 (M) = {1} since exp(L1 ) = p. Let R1 be a minimal normal subgroup of G contained in L1 . Then R1 ∩ M = {1} so that, by Theorem 8.2, Q = T × T1 , where T, T1 are normal in G, T is of maximal order such that Ω1 (T) = R1 . It follows from Ω1 (T) ∩ M = R1 ∩ M = {1} that T ∩ M = {1}. Next, ℧m (T) ≤ ℧m (Q) ∩ T = M ∩ T = {1} 󳨐⇒ exp(T) = q m . By Exercise A.1 (b), R ≤ M < T1 . It follows from AT1 < G and the inductive hypothesis that T1 = S × T2 , where S, T2 are normal in AT1 and Ω1 (S) = R. By (i), S, T2 are normal in G. Next, Q = T × T1 = T × (S × T2 ) = S × (T × T2 ). By what has been proved already, S, T × T2 are normal in G, Ω1 (S) = R, and assertions (a)–(b) of the theorem hold with S1 = T × T2 (it follows from (ii) that S is homocyclic). Let us prove that if Q = S × S1 = T × T1 are decompositions of Q such that S, S1 , T, T1 are normal in G and Ω1 (S) = Ω1 (T) = R, then T and S have the same exponent (and so S ≅ T). Assuming that this is false, one may assume, without loss of generality, that exp(T) > exp(S). Since S, T are homocyclic of the same rank, it follows that |T| > |S|. Since |Q| = |S| ⋅ |S1 |, we have T ∩ S1 > {1}, by the product formula, and hence {1} = R ∩ S1 = Ω1 (T) ∩ S1 > {1}, a contradiction. Thus, exp(T) = exp(S).

58 | Characters of Finite Groups 1 Let R ≤ U ≤ Q and Ω1 (U) = R. Then U ∩ S1 = {1}, and since US1 = U × S1 ≤ Q = S × S1 , it follows that U is isomorphic to a subgroup of Q/S1 ≅ S. The proof is complete. It follows from Theorem A.1 that if Q is not a direct product of two nontrivial normal subgroups of G, it is homocyclic and Ω1 (Q) is a minimal normal subgroup of G. Moreover, the following assertion holds. Corollary A.2. Let Q be a normal abelian Sylow q-subgroup of a group G and R a normal elementary abelian q-subgroup of G. Then Q = S × S1 , where S, S1 are normal in G and Ω1 (S) = R. Furthermore (M. Harris, D. Taunt), Q = Q1 × ⋅ ⋅ ⋅ × Q s , where Q1 , . . . , Q s are normal in G and Ω1 (Q i ) is a minimal normal subgroup of G, i = 1, . . . , s. Proof. We proceed by induction on |Q|. Let A be a q󸀠 -Hall subgroup of the group G. One has R ≤ Q. Let R1 be a minimal normal subgroup of G such that R1 ≤ R. If R1 = R, the result follows from Theorem A.1. Assume that R1 < R. By Theorem A.1, Q = T × T1 , where T and T1 are normal in G and Ω1 (T) = R1 so R is homocyclic. Set R2 = R ∩ T1 , G1 = AT1 . By the inductive hypothesis, T1 = T2 × S1 , where T2 and S1 are normal in G1 so in G and Ω1 (T2 ) = R2 . Set S = TT2 = T × T2 . Then Ω1 (S) = R1 R2 = R, Q = S × S1 and S, S1 are normal in G (see part (i) of the proof of Theorem A.1), completing the proof of the first assertion. Now the second assertion follows. The subgroup S in Corollary A.2 need not be homocyclic. Corollary A.2 is not obvious even in the case G = Q (in that case it asserts that if R < Ω1 (Q), then Q = Q1 × Q2 and Ω1 (Q1 ) = R). Theorem A.3. Let H be a normal abelian π-Hall subgroup of G (where π is a set of primes), let R be a normal π-subgroup of G and let exp(R) be square free. Then there exists a normal π-subgroup S of G such that the following hold: (a) Ω1 (S) = R (here Ω1 (S) is the subgroup of S generated by all elements of S of prime orders). (b) H = S × S1 , where S1 is normal in G. To prove this, it is enough to apply Corollary A.2 to all Sylow subgroups of H and Theorem 8.2. Exercise A.2. Let Q be a normal abelian Sylow q-subgroup of G. Let L be a normal q-subgroup of G such that L ∩ Φ(Q) = {1}. Then Q = L × S, where S is normal in G. Similarly, Q = S × S1 , where S, S1 are normal in G and Ω1 (S) = Ω1 (Φ(Q)). Remark. Let Q be a normal abelian Sylow q-subgroup of a group G with Ω1 (Q) = R1 × ⋅ ⋅ ⋅ × R s , where R1 , . . . , R s are minimal normal subgroups of G. It is not necessarily true that there exist Q1 , . . . , Q s that are normal in G and such that Q = Q1 × ⋅ ⋅ ⋅ × Q s ,

Ω1 (Q i ) = R i , i = 1, . . . , s.

I Basic concepts | 59

Indeed, let G = Q = ⟨a, b | a4 = b2 = 1, ab = ba⟩,

R1 = ⟨b⟩,

R2 = ⟨a2 b⟩.

Then R1 and R2 are maximal cyclic subgroups of Q and R1 × R2 < Q. Corollary A.4 (Fitting). Let Q be a normal abelian Sylow q-subgroup of G. Then Q ∩ Z(G) is a direct factor of G. Proof. By Theorem A.3, we have Q = Q1 × Q2 , where Q1 , Q2 are normal in G and Ω1 (Q1 ) = Ω1 (Q) ∩ Z(G). Therefore, a q󸀠 -Hall subgroup F of G centralizes Q1 (indeed, since FQ1 has no q-closed minimal nonnilpotent subgroups, we have FQ1 = F × Q1 , by Frobenius’ normal complement theorem so that CG (Q1 ) ≥ FQ = G). Since Q is abelian, this implies Z(G) ∩ Q = Q1 , and so G = Q1 × FQ2 = (Q ∩ Z(G)) × FQ2 . Exercise A.3. Suppose that a finite group F acts on a group H = A × B, where A is F-invariant. If exp(Z(A)) exp(B/B󸀠 ) < ∞,

GCD(|F|, exp(Z(A)) exp(B/B󸀠 )) = 1,

then H = A × B1 , where B1 is F-invariant. For a solution, see [BZ3, Theorem 1.10.16”]. Exercise A.4. Let G be an abelian group of order p m and type (α1 ⋅ p e1 , . . . , α r ⋅ p e r ) (i.e., G has α i > 0 invariants p e i , i = 1, . . . , r), where e1 > ⋅ ⋅ ⋅ > e r . Set d = α1 + ⋅ ⋅ ⋅ + α r ; obviously, d = d(G), m = α1 e1 + ⋅ ⋅ ⋅ + α r e r . Find |Aut(G)|. Solution. First we will choose α1 basis elements of maximal order p e1 . All these elements lie in G − Ω e1 −1 (G) (all elements in the last set have order p e1 ). Note that |G : Ω e1 −1 (G)| = p α1 . The first basis element x1 can be chosen in |G − Ω e1 −1 (G)| ways, the second x2 , after the choice of x1 , in |G| − p|Ω e1 −1 (G)| ways since |G − ⟨x1 , Ω e1 −1 (G)⟩| = |G| − p|Ω e1 −1 (G)|, and so on. Then we will choose α2 basis elements of order p e2 . All these elements lie in the set Ω e2 (G) − Ω e2 (Φ(G))Ω e2 −1 (G), and so the first basis element y1 of order p e2 can be chosen, after the choice of x1 , . . . , x α1 , in |Ω e2 (G) − Ω e2 (Φ(G))Ω e2 −1 (G)| ways, the second basis element y2 can be chosen, after the choice of x1 , . . . , x α1 , y1 , in |Ω e2 | − p|Ω e2 (Φ(G))Ω e2 −1 (G)| ways, and so on. At last, we will choose α r basis elements of order p e r . All these elements lie in the set Ω e r (G) − Ω e r (Φ(G))Ω e r −1 (G), and so the first basis element z1 of order p e r can be chosen, after all previous choices, in |Ω e r (G)−Ω e r (Φ(G))Ω e r −1 (G)| ways, the second z2 in |Ω e r (G)|−p|Ω e r (Φ(G))Ω e r −1 (G)| ways, and so on. Note that logp (|Ω e i (G)|) = (α1 + ⋅ ⋅ ⋅ + α i )e i + α i+1 e i+1 + ⋅ ⋅ ⋅ + α r e r , |Ω e i (G) : Ω e i −1 (G)| = p(α1 +⋅⋅⋅+α i )(e i −e i−1 ) .

60 | Characters of Finite Groups 1 The elements x1 , . . . , x α1 , y1 , . . . , y α2 , . . . , z1 , . . . , z α r , chosen in this way, in view of their choice, generate G. Since the product of their orders is equal to |G|, it follows from the product formula that G = ⟨x1 ⟩ × ⋅ ⋅ ⋅ × ⟨x α1 ⟩ × ⟨y1 ⟩ × ⋅ ⋅ ⋅ × ⟨y α2 ⟩ × ⋅ ⋅ ⋅ × ⟨z1 ⟩ × ⋅ ⋅ ⋅ × ⟨z α r ⟩, i.e., the chosen elements, in fact, form a basis of G. Now it is not very difficult to write out |Aut(G)| in terms of e1 , α1 , . . . , e r , α r : this number equals the product of the obtained numbers. For example, let G be abelian of type (2 ⋅ p6 , 1 ⋅ p4 , 3 ⋅ p2 ). Then |G| = p2⋅6+1⋅4+3⋅2 = p22 ,

|Ω5 (G)| = p22−2 = p20 ,

|Ω4 (G)| = p2⋅4+4+3⋅2 = p18 ,

|Ω2 (G)| = p2(2+1+3) = p12 ,

|Ω4 (Φ(G))Ω3 (G)| = p2⋅4+3+3⋅2 = p17 ,

|Ω2 (Φ(G))Ω1 (G)| = p2⋅2+2+3⋅1 = p9 .

Therefore, |Aut(G)| = (p22 − p20 )(p22 − p21 )(p18 − p17 )(p12 − p9 )(p12 − p10 )(p12 − p11 ) = p88 [(p2 − 1)(p − 1)][p − 1][(p3 − 1)(p2 − 1)(p − 1)]. We see that the p󸀠 -part of |Aut(G)|, where G is the abelian group of type given above, ei is equal to ∏ri=1 [∏j=1 (p j − 1)]. Using the proven formula, one can show that the automorphism group of the above abelian 2-group is a 2-group if and only if α i = 1 for all i. We recommend to the reader to classify the abelian p-groups G such that Aut(G) is a solvable group. Suppose that p > 2 is the minimal prime divisor of |G| and P ∈ Sylp (G) is abelian of the same type as above. Then G is p-nilpotent if all α i = 1, by the Burnside normal p-complement theorem. Theorem A.5. Let Q be a normal abelian q-Sylow subgroup of a group G with Ω1 (Q) = R1 × R2 , where R1 and R2 are normal in G. Then the following assertions are equivalent: (a) Q = S1 × S2 for some S1 , S2 ≤ Q with Ω1 (S i ) = R i , i = 1, 2. (b) Q = T1 × T2 for some G-invariant T1 , T2 ≤ G with Ω1 (T i ) = R i , i = 1, 2. Proof. Clearly, (b) ⇒ (a). (b) Suppose that (a) holds. By Corollary A.2, Q has G-invariant direct factors T1 , T2 of Q satisfying Ω1 (T i ) = R i , i = 1, 2. We have T i ≅ S i , i = 1, 2 (it is possible to deduce this from Theorem A.1 (c)–(d)). Therefore, we obatin |T1 × T2 | = |S1 × S2 | = |Q| so Q = T1 × T2 , as required. Lemma A.6. Let α ∈ Aut(G) be an automorphism of a group G satisfying [α, G] ≤ Φ(G). If GCD(o(α), |Φ(G)|) = 1, then α = idG .

I Basic concepts | 61

Proof. Let W = ⟨α⟩ ⋅ G be the natural semidirect product. Since ⟨α, Φ(G)⟩ ⊲ W, by hypothesis, we obtain, using Hall’s theorem on solvable groups and Frattini’s argument, that W = NW (⟨α⟩)⟨α⟩Φ(G) = NW (⟨α⟩)Φ(G). Since G ⊲ W, we have Φ(G) ≤ Φ(W), so that W = NW (⟨α⟩), and we conclude that α = idG .

Problems Problem 1. Find the maximal rank of abelian p-subgroups of the general linear group GL(n, q m ), where p, q are distinct primes. (For m = 1, see [Ber31, Corollary 6.8].) Problem 2. Let G be a group of Exercise A.4. Describe the normal structure of Aut(G). Consider in detail the case when G is homocyclic. Problem 3. Let G be the group of Exercise A.4. Find the number of subgroups of given type in G. Find the number of subgroups N in G such that G/N has a prescribed type. Problem 4. Find a necessary and sufficient condition under which an abelian p-group possesses a characteristic subgroup (a) of given type, (b) of given order, (c) of all possible orders. Problem 5. Given an abelian p-group G, find the class, the derived length, exponent and minimal number of generators of a Sylow p-subgroups of the automorphism group of G and the holomorph of G.

B B. H. Neumann’s lemma on covering Here we prove the following theorem which is true for arbitrary groups. Theorem B.1 (B. H. Neumann [Neu1]). Let G be a group, let H1 , H2 , . . . , H n be subgroups of G and let x1 , x2 . . . , x n elements of G. If G = x1 H1 ∪ x2 H2 ∪ ⋅ ⋅ ⋅ ∪ x n H n ,

(∗)

then |G : H i | ≤ n for some i. This result is best possible, Indeed, let H1 = ⋅ ⋅ ⋅ = H n = H be a subgroup of index n and (∗) is the decomposition of G modulo H. Proof. One may assume that G is not a union of n − 1 cosets x i H i .

62 | Characters of Finite Groups 1 Step 1. We claim that |G : H i | < ∞ for some i ≤ n. Suppose that the number of distinct subgroups H i is k. If k = 1, then (∗) is a decomposition of G modulo H1 so |G : H1 | = n. We prove the assertion using induction by k. Assume that y ∈ ̸ ⋃ni=1 x i H1 . Then yH1 ⊆ ⋃H j =H ̸ 1 x j H j since yH 1 has empty inter−1 section with each coset of H1 from (∗). Thus, H1 ⊆ ⋃H j =H ̸ 1 y x j H j . We can exclude −1 each coset of the form x s H1 from expression (∗) replacing x s H s by ⋃H j =H ̸ 1 xs y xj Hj . After this replacing, G will be still the union of finitely many cosets of k − 1 distinct subgroups. By induction, at least one of these k − 1 subgroups has finite index in G. Step 2. Suppose that among H1 , . . . , H n there is one, say F = H s+1 , of finite index in G and such that all members of the set {H1 , . . . , H n } having finite index in G, coincide with F. Let us prove that then all subgroups H j are equal to F. Let the subgroups H1 = I1 , H2 = I2 , . . . , H s = I s have infinite index in G and let F = H s+1 = H s+2 = H n have finite index in G, and assume that s > 0. Then G = (x1 I1 ∪ ⋅ ⋅ ⋅ ∪ x s I s ) ∪ (x s+1 F ∪ ⋅ ⋅ ⋅ ∪ x n F). Since s > 0, there exists g ∈ ̸ ⋃ni=s+1 x i F such that gF ∩ (⋃ni=s+1 x i F) = 0. Therefore gF ⊆ ⋃sj=1 x j I j and x u F ⊆ ⋃sj=1 x u g−1 x j I j for each u = s + 1, . . . , n. Thus G is a union of a finite number of cosets of I j for j ∈ {1, 2, . . . , s}. By Step 1, at least one of subgroups I j (j ≤ s) has finite index in G, which is a contradiction. Step 3. Let us prove that we can reduce the general situation when there are subgroups of finite and infinite index to the case when only one H r is of finite index in G. Let, as before, H1 = I1 , H2 = I2 , . . . , H s = I s have infinite index, s > 0, and H s+1 = F s+1 , . . . , H n = F n have finite index in G. Denote by D the maximal G-invariant subgroup contained in all subgroups F s+1 , . . . , F n ; then |G : D| < ∞ (Poincaré). Set G = G/D; then G is finite and G = (x1 I 1 ∪ ⋅ ⋅ ⋅ ∪ x s I s ) ∪ (x s+1 F s+1 ∪ ⋅ ⋅ ⋅ ∪ x n F n ). Since |F i : D| = |F i | < ∞, every x i F i is a union of a finite number of cosets of D. It follows that then G is a union of s > 0 cosets x1 I1 , . . . , x s I s and a finite number of cosets of D, which is impossible, by Step 2. Step 4. Now we may assume that all subgroups H i (1 ≤ i ≤ n) have finite index in G. Let D be as in Step 3 and set G = G/D. Suppose that |H 1 | ≤ ⋅ ⋅ ⋅ ≤ |H n |. In that case, we have n

|G| ≤ ∑ |H i | ≤ n|H n | 󳨐⇒ |G : H n | = |G : H n | ≤ n, i=1

completing the proof.

I Basic concepts | 63

Corollary B.2. Let G be a permutation group, acting on a set Ω and let Γ, ∆ be finite subsets of Ω such that g(Γ) ∩ ∆ ≠ 0 for all g ∈ G. Then G has an orbit on Ω of length that do not exceed |Γ||∆|. Proof. Let Γ = {γ1 , γ2 , . . . , γ m },

∆ = {δ1 , δ2 , . . . , δ n }.

Define elements g ij ∈ G, if they exist, as follows: g ij (γ i ) = δ j . Let x ∈ G be arbitrary. As x(Γ) ∩ ∆ ≠ 0, there exist elements γ = γ i ∈ Γ and δ = δ j ∈ ∆ such that x(γ) = δ. Then x(γ) = g ij (γ) = δ and g −1 ij x(γ) = γ. Hence x ∈ g ij G γ , the right coset of the stabilizer of γ. There are at most mn = |Γ||∆| cosets of this form and these cosets cover G since x is an arbitrary element of G, and Theorem B.1 implies the result. Note that the approach from [Came] does not give an estimate of the length of the orbit. Exercise B.1. Let ϕ be a permutation representation of a group G on the set Ω and let Γ be a finite subset of Ω such that every element of G has a fixed point on Γ. Then G has an orbit on Ω of length not exceeding |Γ|. Hint. Stabilizers G γ (γ ∈ Γ) cover G. Apply Theorem B.1. The following application of Corollary B.2 can be found in [KazSt]. Exercise B.2. Let G be a permutation group, acting on sets A, B and C. Suppose that to every pair of elements a ∈ A and b ∈ B corresponds a nonempty subset of elements c = a ∗ b ∈ C satisfying g(a ∗ b) = g(a) ∗ g(b) for all g ∈ G. If there exist a ∈ A and b ∈ B such that the sets S1 = ⋃ (g(a) ∗ b), g∈G

S2 = ⋃ (a ∗ g(b)) g∈G

are finite, then |G : G x | ≤ |S1 ||S2 | for some x ∈ S2 . Solution. Let {a1 , a2 , . . . , a m } = A, {b1 , b2 , . . . , b n } = B and let a1 = a and b1 = b satisfy the condition of the exercise. Let m

S1 = ⋃ a i ∗ b1 , i=1

n

S2 = ⋃ a1 ∗ b j . j=1

We have (x ∈ G) x(a1 ∗ b i ) = x(a1 ) ∗ x(b i ) = a s ∗ b l

for some s = s(x) ≤ m, l = l(i, x) ≤ n.

It is easy to see that x induces a permutation on the set {b1 , . . . , b m } and for some i ≤ n we have b l = b1 . Therefore x(S2 ) ∩ S1 ≠ 0. By Corollary B.2, we are done. We use this for the proof of the analog of Brauer–Fowler’s theorem on involutions.

64 | Characters of Finite Groups 1 Theorem B.3. Suppose that G is a (finite) group containing two nonconjugate involutions a and b such that the element ag−1 bg has finite order for each g ∈ G. If |CG (a) ∩ (a G ∪ b G )| ≤ m,

|CG (b) ∩ (a G ∪ b G )| ≤ n,

then there exists an involution μ ∈ Z(⟨a g , b h ⟩) for some elements g, h ∈ G such that |G : CG (μ)| ≤ mn. Proof. Define, for involutions a1 ∈ a G = A and b1 ∈ b G = B, an involution j = a1 ∗ b1 ∈ Z(⟨a1 , b1 ⟩), which is distinct from a1 and b1 . Clearly, ⟨a1 , b1 ⟩ is a dihedral group and an element a1 b1 is of even order 2q. Hence we have ja1 ∈ C G (a1 ) ∩ a1G , if q is even. If q is odd, then ja1 ∈ C G (a1 ) ∩ b1G . Since C G (a1 ) ∩ (a1G ∪ b1G ) is finite, the set S1 = ⋃g∈G a ∗ g(b) is finite (here g(b) = g−1 bg). Similarly, S2 = ⋃ g(a) ∗ b g∈G

is finite. Clearly, |S1 | ≤ m, |S2 | ≤ n. Now the result follows from Exercise B.2. Corollary B.4. Let G be a finite simple group, having nonconjugate involutions a and b. Then the order of G is bounded in terms of |CG (a)| and |CG (b)|.

II Characters In what follows, the ground field will always be the field ℂ of complex numbers. In this chapter we introduce characters, essentially the main effective tools in the representation theory of finite groups, and prove the first and second orthogonality relations for irreducible characters. These relations play important role in applications of characters to finite group theory. G. Frobenius (1849–1917) is the creator of character theory. W. Burnside and I. Schur also contributed essentially in the first decades after creating this theory. Our exposition in this chapter is based on the seminal paper [Sch2]. The chapter also contains simplest properties of characters. More deep applications are contained in subsequent chapters.

1 Functions on a group Let G be a finite group of order h, F[G] the set of all functions θ : G → ℂ. F[G] is an h-dimensional linear ℂ-space with respect to the natural operations of addition of functions and multiplication of function by scalars: if α ∈ ℂ, θ1 , θ2 ∈ F[G], g ∈ G, then (θ1 + θ2 )(g) = θ1 (g) + θ2 (g),

(αθ1 )(g) = α(θ1 (g)) = αθ1 (g).

The space F[G] has a basis consisting of the h functions f x (x ∈ G) such that {1 ∈ ℂ if g = x, f x (g) = δ g,x = { 0 if g ≠ x. { Here δ is the Kronecker delta on the set G. Indeed, let G = {x1 , . . . , x h } and let ∑hi=1 α i f x i = 0. Computing the left-hand side of that equality at x i , we get α i = 0, and we conclude that functions f x i are linearly independent. Since f = ∑hi=1 f(x i )f x i for all f ∈ F[G], our claim follows. Define the inner product ⟨θ1 , θ2 ⟩ of functions θ1 , θ2 ∈ F[G] by ⟨θ1 , θ2 ⟩ = h−1 ∑ θ1 (x)θ2 (x). x∈G

Clearly, ⟨θ1 , θ2 ⟩ = ⟨θ2 , θ1 ⟩ and ⟨θ1 , θ1 ⟩ ≥ 0 with equality if and only if θ1 is the zerofunction. It is easy to see that ⟨ ⋅ , ⋅ ⟩ satisfies the axioms for inner product. In what follows, F[G] is considered as an unitary linear space equipped with the above defined inner product.

2 Schur’s relations Let T : g 󳨃→ T(g) = (α ij (g)) DOI 10.1515/9783110224078-002

(i, j ∈ {1, . . . , n}, g ∈ G)

66 | Characters of Finite Groups 1 be a matrix representation of a group G of degree n. The functions α ij : G → ℂ are called the matrix elements of T. As the proof of Theorem I.8.1 for ℂ shows, one may assume that the representation T is unitary, that is, T(g)󸀠 = T(g)−1 for all g ∈ G (A󸀠 denotes the transpose of a matrix A; if A = (α ij ), then A = (α ij ), where α is the complex conjugate of α ∈ ℂ). Since T is a homomorphism, we have T(g −1 ) = T(g)−1 , and so the matrix elements of the representation T satisfy the following condition: α ij (g−1 ) = α ji (g)

(i, j ∈ {1, . . . , n}, g ∈ G).

(1)

The following theorems of Schur on the matrix elements of irreducible representations are the basis of our exposition. Theorem 2.1 (Schur [Sch2]). Suppose that T and U are nonequivalent unitary irreducible matrix representations of a group G, and let {α ij } and {β ij } be their matrix elements, respectively. Then ⟨α ij , β kl ⟩ = 0,

i, j = 1, . . . , n = deg(T), k, l = 1, . . . , m = deg(U).

(2)

Proof. Let X ij denote the element at the intersection of the i-th row and j-th column of a matrix X. Let X be an arbitrary n × m matrix. Consider the following function of X: H(X) = ∑ T(g)XU(g)−1 .

(3)

g∈G

If s ∈ G, then, taking into account that T, U are homomorphisms and {sg | g ∈ G} = G, we get T(s)H(X)U(s)−1 = ∑ T(s)T(g)XU(g)−1 U(s)−1 g∈G

= ∑ T(sg)XU(sg)−1 g∈G

= ∑ T(g)XU(g)−1 = H(X). g∈G

Hence T(s)H(X) = H(X)U(s), i.e., H(X) intertwines nonequivalent irreducible representations T and U of G, and we conclude that H(X) = 0, the zero matrix (Schur’s lemma). Since the representation U is unitary, it follows from (3) and (1) that H(X)ik ∑ ∑ T(g)ij X jl U(g−1 )lk = ∑( ∑ α ij (g)β lk (g −1 ))X jl g∈G j,l

j,l

g∈G

= ∑( ∑ α ij (g)β kl (g))X jl , j,l

g∈G

whence we have 0 = H(X)ik = h ∑j,l ⟨α ij , β kl ⟩X jl . Since the matrix X was arbitrary, we get ⟨α ij , β kl ⟩ = 0 for all i, j, k, l. Recall that the trace tr(A) of a square matrix A is equal to the sum of its diagonal elements.

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Theorem 2.2 (Schur). Let T be an n-dimensional irreducible unitary matrix representation of a group G, {α ij } the matrix elements of T. Then ⟨α ij , α kl ⟩ =

δ ik δ jl n

(4)

for all i, j, k, l ∈ {1, . . . , n}. Proof. As in the proof of Theorem 2.1, take the n × n matrix X ∈ ℂn and consider the function H : ℂn → ℂn defined by H(X) = ∑ T(g)XT(g)−1 .

(5)

g∈G

As in the proof of Theorem 2.1, one has T(s)H(X) = H(X)T(s) for any s ∈ G. Therefore, by Schur’s lemma, H(X) is a scalar matrix for any X ∈ ℂn , i.e., H(X) = λ(X)In , where λ is some function from ℂn into ℂ, In is the n × n identity matrix. We have tr(H(X)) = nλ(X). On the other hand, since tr(T(g)XT(g)−1 ) = tr(X), it follows from (5) that tr(H(X)) = h ⋅ tr(X), where h = |G|. Consequently, λ(X) =

h n h h ⋅ tr(X) = ∑ X kk = ∑ δ jl X jl , n n k=1 n j,l

where δ is Kronecker’s delta on the set {1, . . . , n}. This gives H(X)ik = λ(X)δ ik =

h ∑ δ ik δ jl X jl . n j,l

(6)

As in the proof of Theorem 2.1, one can show that H(X)ik = h ∑⟨α ij , α kl ⟩X jl .

(7)

j,l

Now, combining (6) and (7) together and taking into account that X was chosen arbitrarily, we get ⟨α ij , α kl ⟩ = 1n δ ik δ jl . Let T be an irreducible matrix representation of a group G of degree n with matrix elements α ij , i, j = 1, . . . , n (we consider α ij as a function G → ℂ). It follows from Theorem 2.2 that these n2 matrix elements α ij are linearly independent (indeed, these elements form an orthogonal system of nonzero functions).

3 Characters. The First Orthogonality Relation We now introduce the concept of the character of a representation of a group G. If In is the n × n identity matrix, then tr(In ) = n. It is known that tr(A) is the sum of eigenvalues of A (multiplicities are counted). Definition 3.1. Let T be a matrix representation of a group G. The function χ T : G → ℂ defined by χ T (g) = tr(T(g)) (g ∈ G) is called the character of T.

68 | Characters of Finite Groups 1 If {α ij } are the matrix elements of T, then n

χ T = ∑ α ii , i=1

n

χ T (g) = ∑ α ii (g)

(n = deg(T)).

i=1

We say that the character χ T is afforded by the representation T. A character, afforded by an irreducible representation of G, is said to be irreducible. The character of the principal (= trivial) representation of a group G is called the principal character of G and denoted by 1G . The set of all characters of a group G is denoted by Char(G), the set of all its irreducible characters by Irr(G). Since the traces of similar matrices are equal, we have: Theorem 3.1. If T, U are equivalent matrix representations of a group G, then χ T = χ U . Since an operator representation τ of a group G affords a class of equivalent matrix representations, the character of any one of them can be taken as the character of τ. Hence, when speaking of the character of a representation, we need not to specify whether we are dealing with operator or matrix representations. It should now be clear how to define the character of a G-module: this is the character of the (operator or matrix) representation afforded by this module. Corollary 3.2. Every character of a group G is afforded by some unitary representation. This follows from the proof of Maschke’s Theorem I.8.1 for the field ℂ and Theorem 3.1. Definition 3.2. A central function (= a class function) on a group G is a function ψ : G → ℂ which is constant on conjugacy classes of G (i.e., if x, y belong to the same G-class, then ψ(x) = ψ(y)). The set of all central functions on G is denoted by CF[G]. Obviously, CF[G] is a k(G)-dimensional subspace of the space F[G] of all functions G → ℂ, where k(G) is the number of conjugacy classes of G (= the class number of G). Clearly, CF[G] = F[G] if and only if G is abelian. Theorem 3.3. Every character of a group G is a central function. Proof. Indeed, for any two n × n matrices A and B one has tr(AB) = tr(BA). Hence, if T is a representation of G, χ = χ T and g, t ∈ G, then χ(t−1 gt) = tr(T(t−1 gt)) = tr(T(t)−1 T(g)T(t)) = tr(T(g)) = χ(g), as desired. It follows from Theorem 3.3 that |Irr(G)| ≤ dimℂ (CF[G]) = k(G). Theorem 3.4. If T is an n-dimensional representation of a group G, then n = χ T (1). Proof. Indeed, T(1) = In so that χ T (1) = tr(In ) = n. The number χ(1) is the degree of a character χ and as we have seen, it coincides with degree of any representation that affords χ. The principal character 1G of a group G

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has degree 1: 1G (g) = 1 ∈ ℂ for all g ∈ G. A character of degree 1 is said to be linear. The set of all linear characters of G is denoted by Lin(G). Every linear character of G is irreducible since it afforded by a one-dimensional representation which is irreducible so that Lin(G) ⊆ Irr(G). As we know, Lin(G) = Hom(G, ℂ∗ ). If G is abelian, this definition agrees with that of the set Lin(G) from Chapter I. If χ ∈ Irr(G) is nonlinear, i.e., χ(1) > 1, then χ(12 ) = χ(1) < χ(1)2 , and so χ is not a homomorphism from G into ℂ∗ . Lemma 3.5. If χ is a character of a group G, then χ(g −1 ) = χ(g). Proof. Let T be a unitary representation of G which affords the character χ (see Corollary 3.2). It follows from T(g −1 ) = T(g)󸀠 that χ(g −1 ) = tr(T(g −1 )) = tr(T(g)󸀠 ) = tr(T(g)) = χ(g), as desired. If θ : G → ℂ is a central function, then, generally speaking, θ(g −1 ) ≠ θ(g). If elements x, y of a group G are conjugate, we write x ∼ y. An element g ∈ G is called real if g ∼ g−1 . It is known that all elements of symmetric groups are real, however, the alternating group An contains elements which are not real for most n (note that all elements of the alternating group A5 are real). Involutions (= elements of order 2) are real. Elements of order 7 in the group PSL(2, 7) are not real. Corollary 3.6. Any character is real-valued on real elements. Proof. Indeed, by Lemma 3.5, χ(g) = χ(g−1 ) = χ(g) since g −1 ∼ g. Exercise 3.1. If T, U are representations of a group G, then χ T+U = χ T + χ U . Solution. Indeed, (T + U)(g) is similar to a block-diagonal matrix diag(T(g), U(g)) for g ∈ G, and the required equality follows. By Exercise 3.1, the set Char(G) is closed under addition. In what follows we show the following more deep result: Char(G) is also closed under multiplication, i.e., the set Char(G) is a semiring. Theorem 3.7 (First Orthogonality Relation). Let T and U be nonequivalent irreducible representations of a group G. Then ⟨χ T , χ U ⟩ = 0,

⟨χ T , χ T ⟩ = 1.

Proof. One may assume that T, U are unitary representations and let {α ij }, {β ij } be their respective matrix elements. Then χ T = ∑i α ii and χ U = ∑k β kk . Since ⟨α ii , β kk ⟩ = 0, by Theorem 2.1, it follows that ⟨χ T , χ U ⟩ = ∑⟨α ii , β kk ⟩ = 0, i,k

which proves the first equality.

70 | Characters of Finite Groups 1 To prove the second equality, let n = deg(T). Since, by Theorem 2.2, 1 1 ⟨α ii , α jj ⟩ = δ2ij = δ ij , n n one obtains n

n

n

⟨χ T , χ T ⟩ = ⟨ ∑ α ii , ∑ α jj ⟩ = ∑ ⟨α ii , α jj ⟩ = i=1

j=1

i,j=1

1 n ∑ δ ij = 1. n i,j=1

Theorem 3.7 shows that the set of irreducible characters of a group G is orthonormal. It follows that the above set is linearly independent since all its members are nonzero functions (for example, χ(1) ≠ 0 for χ ∈ Irr(G)). Indeed, if we have a linear combination of pairwise distinct irreducible characters χ1 , . . . , χ s , say a1 χ1 +⋅ ⋅ ⋅+a s χ r = 0, then 0 = ⟨a1 χ1 + ⋅ ⋅ ⋅ + a r χ s , χ i ⟩ = a i , and our claim follows. We shall see that the set Irr(G) of irreducible characters forms the basis of the space CF[G] of functions constant on G-classes. The last space is r-dimensional, where r = k(G), the class number of G. To prove our assertion, it suffices to show that |Irr(G)| = r. This will be done later (see Theorem 5.4). Theorems 3.7 and 3.1 imply the following: Corollary 3.8. Irreducible representations T an U of a group G are equivalent if and only if χ T = χ U . Proof. Indeed, χ T = χ U implies ⟨χ T , χ U ⟩ = ⟨χ T , χ T ⟩ = 1 so that T and U are equivalent, by Theorem 3.7. Conversely, if T and U are equivalent, then we have χ T = χ U , by Theorem 3.1. Corollary 3.9. We have |Irr(G)| ≤ k(G). Proof. Theorems 3.1, 3.3 and 3.7 imply that the irreducible characters of a group G form an orthonormal system of functions in the unitary space CF[G], whose dimension is k(G). Corollary 3.8 implies that there exists a one-to-one correspondence between the irreducible characters of a group G and the equivalence classes of irreducible representations of G. If r is the number of the latter, then, by Corollary 3.9, r = |Irr(G)| ≤ k(G). Suppose that {T1 , . . . , T r } is a complete system of distinct representatives of equivalence classes of irreducible matrix representations of a group G; then the characters χ i = χ T i (i ∈ {1, . . . , r}) form the full system of irreducible characters of G, i.e., {χ1 , . . . , χ r } = Irr(G). In the above notation. Theorem 3.7 can now be formulated as a system of r2 equalities ⟨χ i , χ j ⟩ = δ ij (i, j ∈ {1, . . . , r}), or, written out in detail, ∑ χ i (x)χ j (x) = hδ ij ,

x∈G

i, j ∈ {1, . . . , r}, h = |G|.

Here δ is the Kronecker delta on the set {1, . . . , r}. In the above relation i, j are fixed members of the set {1, . . . , r = k(G)} and x runs over the set G.

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We have obtained the first system of orthogonality relations for the irreducible characters of G, which is the most important property of characters. Now, if T is a representation of G, then, in the above notation, T ∼ m1 T1 + ⋅ ⋅ ⋅ + m r T r 󳨐⇒ χ T = m1 χ T1 + ⋅ ⋅ ⋅ + m r χ T r . If, in the above equality, m i > 0, then the character χ T i is said to be an irreducible constituent of the character χ T with multiplicity m i = ⟨χ T , χ T i ⟩. We also have ⟨χ T , χ T ⟩ = m21 + ⋅ ⋅ ⋅ + m2k . In particular, χ T ∈ Irr(G) if and only if ⟨χ T , χ T ⟩ = 1. Exercise 3.2 ([Isa11, Problem 2.9 (a)]). If χ is a character of an abelian group A, then ∑x∈A |χ(x)|2 ≥ |A|χ(1). Solution. Let χ = m1 χ1 + ⋅ ⋅ ⋅ + m k χ k , where χ1 , . . . , χ k ∈ Irr(A) are pairwise distinct. Since χ(1) = m1 + ⋅ ⋅ ⋅ + m k , one obtains ∑ |χ(x)|2 = |A|⟨χ, χ⟩ = |A|(m21 + ⋅ ⋅ ⋅ + m2k ) ≥ |A|(m1 + ⋅ ⋅ ⋅ + m k ) = |A|χ(1).

x∈A

4 Irreducible constituents of a character Let χ ∈ Char(G). Then χ = χ T for some matrix representation T of the group G. Since, by Maschke’s theorem, T is completely reducible, it is equivalent to a sum of irreducible representations of G (and each summand of this sum is equivalent to some element of a system {T1 , . . . , T r } of representatives of equivalence classes of irreducible matrix representations of G). Suppose that exactly m i summands of T are equivalent to T i . Then we write T ∼ m1 T1 + ⋅ ⋅ ⋅ + m r T r (we are not excluding here the possibility that m i = 0 for some i). By Theorem 3.1 and Exercise 3.1, setting χ i = χ T i , one obtains χ = χ T = m1 χ1 + ⋅ ⋅ ⋅ + m r χ r ; the coefficients m1 , . . . , m r in the above linear combination are determined uniquely since the set Irr(G) is linearly independent. Theorem 3.7 now gives r

⟨χ, χ i ⟩ = ∑ m j ⟨χ j , χ i ⟩ = m i . j=1

Thus, we get: Corollary 4.1. Any character χ of a group G can be expressed uniquely as χ = m1 χ1 + ⋅ ⋅ ⋅ + m r χ r , where Irr(G) = {χ1 , . . . , χ r } and m i = ⟨χ, χ i ⟩. If T is a representation of G such that χ = χ T and T ∼ m󸀠1 T1 + ⋅ ⋅ ⋅ + m󸀠r T r , then m i = m󸀠i (i ∈ {1, . . . , r}).

72 | Characters of Finite Groups 1 The number m i = ⟨χ, χ i ⟩ is called the multiplicity of the irreducible character χ i in the character χ, or the multiplicity of the irreducible representation T i in the representation T. An irreducible character (representation) which is of nonzero multiplicity in a character χ (a representation T) is called an irreducible constituent of χ (of T). Let Irr(χ) denote the set of all (distinct) irreducible constituents of a character χ. Note that |Irr(χ)| = 1 ⇐⇒ χ = mλ, where λ ∈ Irr(χ)(⊆ Irr(G)), m ∈ ℕ. Exercise 4.1. If χ = m1 χ1 + ⋅ ⋅ ⋅ + m r χ r , then ⟨χ, χ⟩ = m21 + ⋅ ⋅ ⋅ + m2r . Hint. Use the First Orthogonality Relation. The following important result complements Theorem 3.1. Theorem 4.2. Let T, U be representations of a group G. If χ T = χ U , then T ∼ U. Proof. It follows from the hypothesis that m i = ⟨χ T , χ i ⟩ = ⟨χ U , χ i ⟩

(i ∈ {1, . . . , r}).

By Corollary 4.1, T ∼ m1 T1 + ⋅ ⋅ ⋅ + m r T r ∼ U. Hence T ∼ U. Theorems 3.1 and 4.2 show that characters separate representation classes. Let T be a representation of a group G with matrix elements {α ij }. Then the representation T of G with matrix elements {α ij } is called the complex conjugate of T. Since χ T (g) = χ T (g) for all g ∈ G, the character χ = χ T is completely determined by the character χ = χ T . The character χ is called the complex conjugate of χ. Using Lemma ̂ 3.5, we get χ(g) = χ(g) = χ(g −1 ) (g ∈ G). Further, setting T(g) = (T(g)󸀠 )−1 , we obtain another matrix representation of the group G. Indeed, if x, y ∈ G, then ̂ ̂ T(y), ̂ T(xy) = (T(xy)󸀠 )−1 = (T(y)󸀠 T(x)󸀠 )−1 = (T(x)󸀠 )−1 (T(y)󸀠 )−1 = T(x) ̂ ̂ ∼ T. The representaand our claim follows. Obviously, χ T = χ, so by Theorem 4.2, T ̂ is said to be contragredient to the representation T. tion T Let T be a representation of a group G and χ = χ T . Then ker(χ) = ker(T) is said to be the kernel of χ. A character χ is said to be faithful if ker(χ) = {1}.

Definition 4.1. The character of the regular representation Treg of a group G is called the regular character of G and denoted by ρ G . Thus, ρ G = χ Treg . Suppose that {α ij } are the matrix elements of the regular representation Treg of a group G = {x1 , . . . , x h }. Then α ij (g) = 1 if and only if gx j = x i , α ij = 0 if and only if gx j ≠ x i . In particular, it follows that α ii (g) = 1 if and only if g = 1, α ii (g) = 0 if and only if g ≠ 1 (i ∈ {1, . . . , h}). Therefore, ρ G (g) = |G|δ1,g . where δ is the Kronecker delta on the set G. It follows from this formula that the regular character is faithful and its degree is equal to ρ G (1) = h = |G|. Theorem 4.3. We have ρ G = χ1 (1)χ1 +⋅ ⋅ ⋅+χ r (1)χ r . In particular, χ1 (1)2 +⋅ ⋅ ⋅+χ r (1)2 = |G| and Irr(ρ G ) = Irr(G).

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| 73

Proof. By the First Orthogonality Relation, we have, for all i ∈ {1, . . . , r}, ⟨ρ G , χ i ⟩ = h−1 ∑ ρ G (x)χ i (x) = h−1 ρ G (1)χ i (1) = h−1 hχ i (1) = χ i (1); x∈G

here h = |G|. This proves the first relation. (In particular, Irr(ρ G ) = Irr(G).) Calculating both sides of the relation ρ G = χ1 (1)χ1 + ⋅ ⋅ ⋅ + χ r (1)χ r at x = 1, we get the second relation. (This theorem explains importance of the regular character and regular representation.) In particular, we have Treg ∼ χ1 (1)T1 + ⋅ ⋅ ⋅ + χ r (1)T r . Therefore, each T i , i ∈ {1, . . . , r}, is a constituent of the regular representation Treg . Every T i appears in Treg with multiplicity deg(T i ). Exercise 4.2. Since Treg ∼ χ1 (1)T1 + ⋅ ⋅ ⋅ + χ r (1)T r and Treg is faithful, it follows that {1} = ker(Treg ) = ⋂ri=1 ker(T i ). The last equality implies that, if G has a unique minimal normal subgroup (i.e., G is a monolith), then it has a faithful irreducible representation. Taking Theorem I.7.3 into account, we conclude that a p-group (where p is a prime) has a faithful irreducible representation if and only if its center is cyclic. The reader should now construct a group with cyclic center which does not have a faithful irreducible representation. Show that a group with exactly two minimal normal subgroups also has a faithful irreducible representation. Remark. The main results of this section, Theorems 4.2 and 4.3, can be proved by means of the theory of representations of semisimple algebras (see §I.12). First of all, by Maschke’s theorem, the group algebra ℂG is semisimple. The one-to-one correspondence between ℂ-representations of G and representations of ℂG can be established in the natural way. Let T1 , . . . , T r be a complete system of representatives of classes of equivalent irreducible ℂ-representations of G and let χ i be the character of T i , all i. Theorems 4.2 and 4.3 follow immediately from Theorems I.12.13, I.12.18 and I.12.19. We have to take into account that dimℂ (ℂG) = |G|. In conclusion we will present the elegant proof of the semisimplicity of the complex group algebra ℂG taken from [Hers]. By Theorem I.11.14, it suffices to show that the radical of the algebra ℂG is trivial. The proof uses the following simple fact. Let x = ∑ x g g ∈ ℂG, g∈G

where x g ∈ ℂ (g ∈ G),

x∗ = ∑ x g g −1 . g∈G

It is easy to see that the mapping x 󳨃→ x∗ (x ∈ ℂG) has the following properties (below x, y ∈ ℂG, α, β ∈ ℂ): (a) (x∗ )∗ = x, (b) (αx + βy)∗ = αx∗ + βy∗ , (c) (xy)∗ = y∗ x∗ . If x ∈ ℂG, then (xx∗ )1 = ∑g∈G |x g |2 ((xx∗ )1 is the coefficient of xx∗ at 1 ∈ G). It follows that xx∗ = 0 if and only if x = 0.

74 | Characters of Finite Groups 1 Now suppose that x ∈ Rad(ℂG). Then v = xx∗ ∈ Rad(ℂG), since Rad(ℂG) is a twosided ideal of ℂG. Since the radical is nilpotent, it follows that v m = 0 for some m ∈ ℕ. λ It is clear that v2 = 0 for all sufficiently large λ ∈ ℕ. If v ≠ 0, there exists μ ∈ ℕ such μ μ−1 that v2 = 0 but v2 ≠ 0. Since v∗ = (xx∗ )∗ = (x∗ )∗ x∗ = xx∗ = v, we get 0 = v2 = v2 μ

μ−1

v2

μ−1

= v2

μ−1

(v∗ )2

μ−1

= v2

μ−1

(v2

μ−1

)∗ .

By the previous paragraph, it follows that v2 = 0, contrary to the choice of μ. Thus v = 0 so x = 0. Hence Rad(ℂG) = {0}, i.e., ℂG is semisimple, as was to be shown. μ−1

Exercise 4.3. If T is a nontrivial irreducible matrix representation of a group G, then ∑g∈G T(g) = 0. Solution. Set ∑g∈G T(g) = A. Since T is nontrivial, there is x ∈ G such that T(x) ≠ I, where I the identity matrix of the same size as T. We have T(x)A = A and T(x)A = AT(x) for all x ∈ G. It follows from the first equality that A is singular. On the other hand, the second equality implies that A is scalar (Schur’s lemma). Since A is a singular scalar matrix, we conclude that A is the zero matrix. Exercise 4.4 ([Isa11, Problem 2.1]). Let H ≤ G and g ∈ G be such that all elements of the coset Hg are conjugate in G. Let χ ∈ Char(G) be such that ⟨χ H , 1H ⟩ = 0. Then we have χ(g) = 0. Solution. Let T be a representation of the group G affording the character χ. By hypothesis, T H has no principal constituent so, by Exercise 4.3, we get ∑h∈H T(h) = 0. It follows that ∑ T(hg) = ∑ T(h)T(g) = ( ∑ T(h))T(g) = 0. h∈H

h∈H

h∈H

By hypothesis, since χ is a class function, we get χ(g)|H| = ∑ χ(hg) = tr( ∑ T(hg)) = 0, h∈G

h∈G

and we conclude that χ(g) = 0.

5 The number of irreducible characters of a group The main goal of this section is to prove the following basic result: r = k(G), where r = |Irr(G)| and k(G) is the class number of a group G. In view of Corollary I.12.10, it suffices to prove that dimℂ (Z(ℂG)) = k(G). Arguments that lead to this result are of independent interest. Lemma 5.1. An element z = ∑g∈G ξ(g)g (ξ(g) ∈ ℂ for g ∈ G) of the group algebra ℂG belongs to the center Z(ℂG) of ℂG if and only if the function ξ : G → ℂ is central.

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Proof. Let z ∈ Z(ℂG). Then tzt−1 = z for all t ∈ G. Hence, ∑ ξ(g)tgt−1 = tzt−1 = z = ∑ ξ(tgt−1 )tgt−1 .

g∈G

g∈G

Since G = | g ∈ G} is a basis of the algebra ℂG, we conclude that ξ(tgt−1 ) = ξ(g) for all t, g ∈ G, i.e., ξ ∈ CF[G], where CF[G] is the set of central functions G → ℂ. Conversely, if a function g 󳨃→ ξ(g) is central, then ξ(tgt−1 ) = ξ(g) for all t, g ∈ G, whence {tgt−1

tzt−1 = ∑ ξ(g)tgt−1 = ∑ ξ(tgt−1 )tgt−1 = ∑ ξ(g)g = z, g∈G

g∈G

g∈G

and so z ∈ Z(ℂG). Let K1 , . . . , K r be the set of G-classes (r = k(G), the class number of G). Put k i = ∑x∈K i x (i ∈ {1, . . . , r}), where the sum is considered in the group algebra ℂG. The elements k1 , . . . , k r are called the class sums. Lemma 5.2. The set {k i }1r is a basis of the algebra Z(ℂG). Proof. Writing k i = ∑g∈G ψ i (g)g, we see that ψ i : G → ℂ is a central function for all i = 1, . . . , r (this is easy to check directly). Therefore, by Lemma 5.1, k i ∈ Z(ℂG). If z = ∑g∈G ξ(g)g ∈ Z(ℂG) and α i denotes the value of ξ(g) for g ∈ K i , we get r

s

z = ∑ ∑ ξ(g)g = ∑ α i k i . i=1 g∈K i

i=1

Since k1 , . . . , k r are linearly independent over ℂ (this follows from linear independence of the elements of G), the displayed formula implies the result. Corollary 5.3. We have dimℂ (Z(ℂG)) = k(G). The following theorem is the main result of this section. Theorem 5.4. Let G be a group. Then the number of classes of equivalent irreducible ℂ-representations of G is equal to k(G), the class number of G. Proof. The theorem follows from Corollary 5.3 and I.12.9. Theorem 5.4 has numerous applications in what follows. For example, using this theorem, it is easy to show that S4 has irreducible characters of degrees 1, 1, 2, 3, 3 and the degrees of irreducible characters of the alternating group A5 are 1, 3, 3, 4, 5. We suggest to readers to find the degrees of irreducible characters of the symmetric group S5 (note that k(S5 ) = 7). The character table of a group G is the matrix X(G) = (χ i (g j )),

g j ∈ K j , i, j ∈ {1, . . . , r = k(G)},

where {K1 , . . . , K r } and {χ1 , . . . , χ r } are the sets of conjugacy classes and irreducible characters of G, respectively, g j ∈ K j for all j. Since characters are class functions, we

76 | Characters of Finite Groups 1 obtain that X(G) independent of a choice g j in K j . It follows from the First Orthogonality Relation and Theorem 5.4 that X(G) ∈ GL(k(G), ℂ). We shall always assume that K1 = {1}. Then the entries of the first column X1 (G) of the character table X(G) of G are the degrees of the irreducible characters of G (counting multiplicities). What kind of information about G can be derived from a knowledge of its character table X(G)? Since |G| = ∑χ∈Irr(G) χ(1)2 , it follows that X(G) (even X1 (G)) determines |G|, the order of G. Theorem I.9.5 enables us to compute the index of the derived subgroup G󸀠 of G once X1 (G) (the first column of the matrix X(G)) is known. We shall see later that a considerable amount of information about G can be deduced from X(G). Nevertheless, X(G) does not determine G up to isomorphism. For example, character tables of two nonisomorphic extraspecial 2-groups of the same order are identical (check!). Exercise 5.1. Let χ be a nonlinear irreducible character of an extraspecial 2-group G of order 21+2m . Show that: (a) χ(1) = 2m . (b) If g ∈ G − Z(G), then χ(g) = 0, i.e., χ vanishes on G − Z(G). Solution. (a) We have k(G) = 22m + 1 and |G : G󸀠 | = 22m . It follows that χ is the unique nonlinear irreducible character of G. Therefore 22m+1 = |G| = |G : G󸀠 | + χ(1)2 󳨐⇒ χ(1)2 = 22m+1 − 22m = 22m 󳨐⇒ χ(1) = 2m . Assertion (b) follows from Exercise 5.5 (a), below. (State a similar assertion for the case p > 2.) Exercise 5.2. Construct the character table of an extraspecial group of order 22m+1 . Exercise 5.3. Construct the character table of the symmetric group S3 . Exercise 5.4. Construct the character table of the alternating group A4 . Exercise 5.5. Let A < G be abelian and χ ∈ Irr(G). Then: (a) χ(1) ≤ |G : A| with equality if and only if χ vanishes on G − A. (b) If χ(1) = |G : A| > 1, then A G > {1}. Solution. (a) It follows from Theorem 3.7 and Exercise 3.2 that |G| = ∑ |χ(x)|2 ≥ ∑ |χ(x)|2 ≥ |A|χ(1) 󳨐⇒ |G : A| ≥ χ(1). x∈G

x∈A

The equality in the displayed formula holds if and only if χ vanishes on G − A. (b) Now let χ(1) = |G : A| > 1. Let V χ = {x ∈ G | χ(x) ≠ 0}. By (a), V χ ⊆ A. Clearly, the set V χ is G-invariant and nonempty. It follows that ⟨V χ ⟩ is a nonidentity G-invariant subgroup contained in A. It remains to show that |V χ | > 1. Assume that this is false. Then V χ = {1} so that χ vanishes on G# . In that case, |G| = ∑x∈G |χ(x)|2 = χ(1)2 . It follows that k(G) = 1 so G = {1} which is a contradiction.

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| 77

Exercise 5.6 ([Isa11, Problem 2.10]). Let G = ⋃ni=1 A i be a partition of a nonabelian group with abelian components. Then the following assertions hold: (a) If χ ∈ Irr1 (G) = Irr(G) − Lin(G), then (n − 1)χ(1) ≥ |G|. (b) |A i | ≤ n − 1 (i = 1, . . . , n) and |G| ≤ (n − 1)2 . Solution. (a) We have |G| = ∑ |χ(x)|2 x∈G

n

= χ(1)2 + ∑ ∑ |χ(x)|2 i=1 x∈A# i

n

= χ(1)2 + ∑ ∑ |χ(x)|2 − nχ(1)2 i=1 x∈A i

n

= ∑ ∑ |χ(x)|2 − (n − 1)χ(1)2 . i=1 x∈A i

By Exercise 3.2, ∑x∈A i

|χ(x)|2

≥ |A i |χ(1) for all i. It follows from

n

n

i=1

i=1

∑ |A i | = n + ∑ |A#i | = n + |G# | = n − 1 + |G| that n

|G| ≥ ∑ |A i |χ(1) − (n − 1)χ(1)2 i=1

= (n − 1 + |G|)χ(1) − (n − 1)χ(1)2 = −(n − 1)χ(1)(χ(1) − 1) + |G|χ(1). Then (n − 1)χ(1)(χ(1) − 1) ≥ |G|(χ(1) − 1). Now χ(1) > 1 implies (n − 1)χ(1) ≥ |G|. |G| , by (a), so we get |A i | ≤ n − 1 (b) By Exercise 5.5 (a), χ(1) ≤ |G : A i |, and χ(1) ≥ n−1 for all i. Next (see, in solution of (a), the second displayed formula), n

n − 1 + |G| = ∑ |A i | ≤ (n − 1)n 󳨐⇒ |G| ≤ (n − 1)2 . i=1

6 The Second Orthogonality Relation In this section we will derive the Second Orthogonality Relation, which is in a sense dual to the First Orthogonality Relation. Let CL(G) = {K1 , . . . , K r } be the set of all G-classes, r = k(G),

Irr(G) = {χ1 , . . . , χ r },

h α = |K α |,

g α ∈ K α (α ∈ {1, . . . , r}),

h = |G|.

78 | Characters of Finite Groups 1 Theorem 6.1 (Second Orthogonality Relation). Let G be a group. Then r

∑ χ i (g α )χ i (g β ) = δ αβ |CG (g α )|. i=1

(In this relation, α and β are fixed and χ i runs over the set Irr(G). Note that we have δ αβ |CG (g α )| = δ αβ |CG (g β )|.) Proof. Rewrite the First Orthogonality Relation (Theorem 3.7) in the following form: r

∑

α=1

hα χ i (g α )χ j (g α ) = δ ij , h

i, j ∈ {1, . . . , r}.

(Here i, j are fixed.) Let H = h−1 diag(h1 , . . . , h r ) and let X = X(G) be the character table of G. Putting Y = HX 󸀠 , one may rewrite the First Orthogonality Relation in the following matrix form: XY = Ir . This yields YX = Ir , whence, noting that h α = |G : CG (α)| =

h , |CG (g α )|

we obtain the desired result. The Second Orthogonality Relation shows that different columns of the character table X(G) are orthogonal. Since these columns are nonzero (indeed, all members of the first line of the matrix X(G) are equal to 1 ∈ ℂ), it follows that they are linearly independent so that X(G) ∈ GL(r, ℂ) (this also follows from the First Orthogonality Relation). Another proof of the Second Orthogonality Relation. Let ϵ i be the characteristic function of the class K i , i = 1, . . . , r (this means that ϵ i (x j ) = δ ij , where x j ∈ K j ). If χ ∈ Irr(G) and x i ∈ K i , then, using the above notation, we get since χ is a class function, ⟨ϵ i , χ⟩ = |G|−1 ∑ ϵ i (g)χ(g) = h i |G|−1 χ(x i ), g∈G

i = 1, . . . , r.

Therefore, ϵ i = h i |G|−1 ∑χ∈Irr(G) χ(x i )χ(g i ∈ K i ). It follows that, for any i, j ∈ {1, . . . , r}, one has |G| |G| ϵ i (x j ) = δ ij = δ ij |CG (x i )|. ∑ χ(x i )χ(g j ) = h hi i χ∈Irr(G) The Second Orthogonality Relation can be written in another form. Let s, t ∈ G. Then {|CG (s)| χ(s)χ(t) = { 0 χ∈Irr(G) { ∑

if s ∼ t, if s ≁ t.

(1)

In particular, putting s = t, we obtain the following important relation: ∑ |χ(s)|2 = |CG (s)|.

(2)

χ∈Irr(G)

This relation enables us to determine the length of any G-class, once X(G) is given. Putting s = 1 in (2), we get the already known identity ∑χ∈Irr(G) χ(1)2 = |G|.

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Another important consequence of (1) is the following assertion: (a) Two elements s, t are conjugate in G if and only if χ(s) = χ(t) for every χ ∈ Irr(G). In particular, the columns of the character table X(G) are pairwise distinct. This follows, though, from the fact that the matrix X(G) is nonsingular. In other words, assertion (a) expresses the fact that irreducible characters separate the conjugacy classes of G. The assertion dual to (a) is trivial: (b) Characters χ1 , χ2 ∈ Irr(G) are equal if and only if χ1 (s) = χ2 (s) for all s ∈ G. The meaning of (b) is that the rows of the matrix X(G) are pairwise distinct. Compare with Exercises I.9.2 and I.9.3, which refer to abelian groups. Recall that an element g ∈ G is real if g ∼ g −1 . Exercise 6.1. An element g ∈ G is real if and only if χ(g) ∈ ℝ for all χ ∈ Irr(G). Solution. One has for all χ ∈ Irr(G): χ(g) ∈ ℝ 󳨐⇒ χ(g) = χ(g) = χ(g−1 ) 󳨐⇒ g ∼ g −1 , i.e., the element g is real. On the other hand, if g is real, i.e., g ∼ g −1 , then χ(g) = χ(g−1 ) = χ(g) hence χ(g) ∈ ℝ for all χ ∈ Irr(G). The following lemma is purely group-theoretic. Lemma 6.2. Let H < G, let a subset M be G-invariant and let H G = ⋂x∈G H x be the core of H in G. Then H ∪ M = G if and only if H G ∪ M = G. Proof. Clearly, H G ∪ M = G implies H ∪ M = G. Now assume that H ∪ M = G. Then G − M ⊆ H is G-invariant so G − M ⊆ H G . In that case, H G ∪ M ⊇ (G − M) ∪ M = G. Given χ ∈ Char(G), let Tχ = {x ∈ G | χ(x) = 0} denote the set of zeros of the character χ. If z ∈ Tχ and x ∈ G, then, since χ is a class function, we get χ(z x ) = χ(z) = 0, i.e., the set Tχ is G-invariant. If H ≤ G and χ = χ T is the character of a representation T of G, then the restriction χ H of χ to H is the character of the restriction T H of T to H, i.e., χ H is a character of H. Lemma 6.3. If χ ∈ Char(G) and H < G, then ⟨χ H , χ H ⟩ ≤ |G : H|⟨χ, χ⟩ with equality ⇐⇒ Tχ ∪ H G = G. Proof. We have |H|⟨χ H , χ H ⟩ = ∑ |χ(x)|2 ≤ ∑ |χ(x)|2 = |G|⟨χ, χ⟩, x∈H

x∈G

whence ⟨χ H , χ H ⟩ ≤ |G : H|⟨χ, χ⟩, with equality if and only if G − H ⊆ Tχ . Since the set Tχ is G-invariant, we conclude, using Lemma 6.2, that G = H ∪ T χ implies G = H G ∪ Tχ . The converse assertion follows from the displayed formula.

80 | Characters of Finite Groups 1 Recall that b(G) = max{χ(1) | χ ∈ Irr(G)}, and let b(χ) = max{ψ(1) | ψ ∈ Irr(χ)}(χ ∈ Char(G)). χ(1) Lemma 6.4. Let χ ∈ Char(G). Then ⟨χ, χ⟩ ≥ b(χ) with equality if and only if all the irreducible constituents of χ are of the same degree and multiplicity 1 in χ.

Proof. Let Irr(χ) = {χ1 , . . . , χ s },

χ1 (1) ≤ ⋅ ⋅ ⋅ ≤ χ s (1), χ = a1 χ1 + ⋅ ⋅ ⋅ + a s χ s .

Then, by Exercise 4.1, ⟨χ, χ⟩ = a21 + ⋅ ⋅ ⋅ + a2s . We observe that b(χ) = χ s (1) and χ(1) = a1 χ1 (1) + ⋅ ⋅ ⋅ + a s χ s (1) ≤ (a1 + ⋅ ⋅ ⋅ + a s )b(χ), with equality if and only if χ1 (1) = ⋅ ⋅ ⋅ = χ s (1). Since ⟨χ, χ⟩ = a21 + ⋅ ⋅ ⋅ + a2s ≥ a1 + ⋅ ⋅ ⋅ + a s ≥ the inequality is established. Now suppose that ⟨χ, χ⟩ = above displayed formula that a21 + ⋅ ⋅ ⋅ + a2s = a1 + ⋅ ⋅ ⋅ + a s =

χ(1) , b(χ)

χ(1) b(χ) .

Then it follows from the

χ(1) b(χ)

so that a i ∈ {0, 1}, and the lemma follows. Exercise 6.2. If H ≤ G and χ ∈ Irr(G), then χ(1) ≤ |G : H|b(χ H ), b(H) ≤ b(G) ≤ |G : H|b(H). Hint. Let χ H = a1 ϕ1 + ⋅ ⋅ ⋅ + a s ϕ s , Irr(χ H ) = {ϕ1 , . . . , ϕ s }. Then, by Lemma 6.3, a21 + ⋅ ⋅ ⋅ + a2s = ⟨χ H , χ H ⟩ ≤ |G : H|, Since a1 + ⋅ ⋅ ⋅ + a s ≤

a21

+ ⋅⋅⋅ +

a2s ,

χ(1) ≤ (a1 + ⋅ ⋅ ⋅ + a s )b(χ H ).

it follows, taking χ(1) = b(G), that

b(G) = χ(1) ≤ |G : H|b(χ H ) ≤ |G : H|b(H). We advise the reader to postpone the proof that b(H) ≤ b(G) until after the Frobenius reciprocity law (Chapter V). It is natural to devote closer attention to the situation when b(G) = |G : H|b(H), where H < G. Slightly modifying the above argument and using the same notation, we see that if χ ∈ Irr(G) and χ(1) = b(G), then, in the case under consideration, we have ⟨χ H , χ H ⟩ = |G : H|. Indeed, |G : H|b(H) = b(G) = χ(1) ≤ (a1 + ⋅ ⋅ ⋅ + a s )b(H) ≤ (a21 + ⋅ ⋅ ⋅ + a2s )b(H) ≤ |G : H|b(H) hence we get |G : H| = a21 + ⋅ ⋅ ⋅ + a2s = ⟨χ H , χ H ⟩, and our claim follows. Consequently, by Lemma 6.3, G = Tχ ∪ H G . In that case, H G > {1}. Indeed, otherwise, χ is a multiple of ρ G , the (reducible) regular character of G > {1} (see the following exercise), which is a contradiction.

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Exercise 6.3. (a) Suppose that χ ∈ Char(G) vanishes on G# . Then χ is a multiple of ρ G . In particular, |G| | χ(1). (b) Suppose that χ ∈ Char(G) is not a multiple of 1G and suppose that χ is constant on G# . Then χ = a1G + bρ G , where a ∈ ℤ, b ∈ ℕ, ρ G is a regular character of G, a + b ≥ 0, χ(1) = a + b|G| = a + b + b(|G| − 1) ≥ b(|G| − 1). Solution. (a) We have ⟨χ, τ⟩ = |G|−1 χ(1)τ(1) for all τ ∈ Irr(G). Taking τ = 1G , we get χ(1) = |G|s for some s ∈ ℕ. If χ = ∑τ∈Irr(G) m τ τ, then m τ = ⟨χ, τ⟩ = sτ(1) 󳨐⇒ χ = s

τ(1)τ = sρ G .

∑ τ∈Irr(G)

(b) Let χ(x) = a for all x ∈ G# . In view of (a), one may assume that a ≠ 0. Set θ = χ − a1G ; then θ(x) = 0 for all x ∈ G# . In that case, by (a), θ = bρ G for some integer b = |G|−1 θ(1). We have χ = a1G + bρ G so that χ(1) = a + b|G|, and we conclude that a ∈ ℤ. We have ρ G = 1G + ∑τ∈Irr# (G) τ(1)τ. Therefore, χ = a1G + bρ G = (a + b)1G + b

τ(1)τ.

∑

τ∈Irr# (G)

It follows that a + b = ⟨χ, 1G ⟩ ≥ 0. Next χ(1) = a + b|G| = a + b + b(|G| − 1) ≥ b(|G| − 1). It follows from the first displayed formula that b > 0. Exercise 6.4. Let x ∈ G with o(x) = m and let θ ∈ Char(G). Then the value θ(x) is the sum of m-th roots of unity. Hint. Let H = ⟨x⟩. Then θ H is a sum of λ i ∈ Irr(H) = Lin(H). Exercise 6.5. Let T be a representation of a group G with the character θ. Then ker(T) = {x ∈ G | θ(x) = θ(1)}. Hint. Use Exercise 6.4. Using this exercise, it is possible to determine the lattice of the normal subgroups of a finite group G from the character table X(G). Exercise 6.6 (Gallagher). Let G = AB be a group which is a product of two subgroups A and B. Then a ∈ Z(A), b ∈ Z(B) implies θ(ab)θ(1) = θ(a)θ(b) for all θ ∈ Irr(G). Solution. Let K a , K b and K ab be the G-classes containing a, b and ab, respectively, and set k a = ∑x∈K a x, k b = ∑y∈K b y, k ab = ∑z∈K ab z. We claim that k a k b = ck ab for some c ∈ ℕ. Indeed, if x, y ∈ G, then x = a1 b1 , y = b󸀠1 a󸀠1 for some a1 , a󸀠1 ∈ A, b1 , b󸀠1 ∈ B since AB = BA. Then 󸀠

󸀠

󸀠

󸀠

−1

a x b y = a a1 b1 b b1 a1 = a b1 b a1 = (ab a1 b1 )b1 .

82 | Characters of Finite Groups 1 It follows from AB = BA that a󸀠1 b−1 1 = b 2 a 2 with a 2 ∈ A, b 2 ∈ B. Therefore we have a x b y = (ab)a2 b1 . This means that k a k b = ck ab for some c ∈ ℕ. Since Z(T(G)) consists of scalar matrices for an irreducible representation T of G affording θ, we conclude that θ(k a ) = λ1 θ(1),

θ(k b ) = λ2 θ(1),

θ(k ab ) = λ3 θ(1) for some λ1 , λ2 , λ3 ∈ ℂ.

On the other hand, θ(k a ) = |K a |θ(a),

θ(k b ) = |K b |θ(b),

θ(k ab ) = |K ab |θ(ab).

Hence

θ(b) θ(ab) θ(a) , λ2 = |K b | , λ3 = |K ab | . θ(1) θ(1) θ(1) Since T(k a ) = λ1 I and T(k b ) = λ2 I, where I is the identity θ(1) × θ(1) matrix, we have λ1 = |K a |

θ(k a k b ) = λ1 λ2 θ(1) = |K a |

θ(b) θ(a)θ(b) θ(a) |K b | θ(1) = |K a ||K b | . θ(1) θ(1) θ(1)

It follows that |K a ||K b |

θ(a)θ(b) = θ(k a k b ) = θ(ck ab ) = c|K ab |θ(ab). θ(1)

This is true for each irreducible character θ of G. Setting in the obtained displayed formula θ = 1G , we conclude that |K a ||K b | = c|K ab |. Hence it follows from the same displayed formula that θ(a)θ(b) = θ(1)θ(ab), as was to be shown. Exercise 6.7. Suppose that all nonlinear irreducible characters of a group G have degrees divisible by a fixed prime p. Prove then G has a normal subgroup of index p. (In Chapter XIV we prove that, in the case under consideration, G has a normal p-complement; this is Thompson’s theorem.) Exercise 6.8. Let Irr(n) (G) denote the set of characters of degree n in Irr(G). A 2-group G is of maximal class if and only if the number |Irr(2) (G)| is odd. Solution. Suppose that |Irr(2) (G)| = 2k + 1 is odd. Then (we use the Frobenius–Molien Theorem III.3.2) |G| = |G : G󸀠 | +

∑ χ∈Irr(2) (G)

χ(1)2 +

∑

χ(1)2

χ∈Irr(G), χ(1)>2

≡ |G : G󸀠 | + 4(2k + 1) ≡ |G| ≡ 0 (mod 16). It follows that |G : G󸀠 | = 4 so G is of maximal class (Taussky). If a 2-group of maximal class and order 2n , then |Irr(2) (G)| = 2n−2 − 1 is odd. (In the case under consideration, k = 2n−3 − 1 is odd.) In particular, a 2-group G has exactly one irreducible character of degree 2 if and only if it is nonabelian of order 8. Moreover, if G is a 2-group and |Irr(2) (G)| ≡ 1 (mod 4), then G is nonabelian of order 8. Indeed, in the case under consideration, we have that |Irr(2) (G)| = 2n−2 − 1 ≡ 1 (mod 4) implies n = 3.

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Exercise 6.9. Let G be a p-group, p > 2. If |Irr(p) (G)| ≢ 0 (mod p), then |Irr(p) (G)| ≡ −1 (mod p) and |G : G󸀠 | = p2 . Classify the p-groups G with |Irr(p) (G)| = p − 1. In particular, if |Irr(p) (G)| ≡ 1 (mod p), where G is a p-group, then p = 2 so that G is of maximal class (Exercise 6.8). For solutions of Exercises 6.8 and 6.9, we refer to Exercises VII.2.4 and VII.2.5. However, we suggest to the reader to solve these exercises now. Exercise 6.10. If G = D2n , the dihedral group of order 2n, then the degrees of all nonlinear irreducible characters of G are equal to 2. Solution. Let U < G be cyclic of index 2. First suppose that n is odd. Then |G/G󸀠 | = 2 so |Lin(G)| = 2. If x ∈ G − U, then |CG (x)| = 2, so G − U is a G-class. If x ∈ U # , then n+1 n+1 CG (x) = U, so U is the union of 1+ n−1 2 = 2 conjugacy G-classes. Thus, k(G) = 1+ 2 . 󸀠 Now let n is even. Then |G/G | = 4 so |Lin(G)| = 4. If x ∈ G − U, then |G : CG (x)| = 2n , so the set G − U of cardinality n is a union of two G-classes. As above, U is a union n+2 n+2 of 2 + n−2 2 = 2 conjugacy G-classes. Thus, k(G) = 2 + 2 . Let Irr1 (G) be the set of nonlinear irreducible characters. If n is odd, then |Irr1 (G)| = k(G) − |G/G󸀠 | =

n−1 . 2

|Irr1 (G)| = k(G) − |G/G󸀠 | =

n−2 . 2

If n is even, then In the first case, 2n = |G/G󸀠 | +

∑

χ(1)2 ≥ 2 + 2(n − 1) = 2n

χ∈Irr1 (G)

so that χ(1) = 2 for all χ ∈ Irr1 (G). In the second case 2n = |G/G󸀠 | +

∑

χ(1)2 ≥ 4 + 2(n − 2) = 2n,

χ∈Irr1 (G)

and so again χ(1) = 2 for all χ ∈ Irr1 (G). Exercise 6.11. Let a nonabelian group G = ⟨x⟩ ⋅ A be such that x2 = 1 and a x = a−1 for all a ∈ A. Prove that χ(1) = 2 for all χ ∈ Irr1 (G). Exercise 6.12. Suppose that G is a 2-group of maximal class, i.e., dihedral, semidihedral or generalized quaternion. Prove that χ(1) = 2 for all χ ∈ Irr1 (G). Exercise 6.13. Prove that all nonlinear irreducible characters of minimal nonabelian group have the same prime degree. Later we shall see that Exercises 6.10–6.13 are easy corollaries of Theorem VII.2.3.

84 | Characters of Finite Groups 1

7 The character table and the Frattini subgroup As was pointed out by R. Brauer [Bra7], the Frattini subgroup Φ(G) of a p-group G is uniquely determined by its character table X(G). In general, however, this is not the case. S. Garrison [Gar2] showed that if G1 is a split extension of Ep3 by PSL(2, p) and G2 is a nonsplit extension of the same groups, p ≥ 5 and Z(G1 ) = Z(G2 ) = {1}, then X(G1 ) = X(G2 ), while Φ(G1 ) = {1} and Φ(G2 ) ≅ Ep3 . However, Garrison has shown that knowledge of X(G) is sufficient to determine the Frattini subgroup of a solvable group G. (Note that solvability can be defined in terms of the character table: if X(G1 ) = X(G2 ), there exists an isomorphism of the lattice of normal subgroups of G1 onto that of G2 which preserves indices). Given H ≤ G, the core of H in G is defined as H G = ⋂x∈G H x . Let Γ1 denote the set of maximal subgroups of G. Theorem 7.1 ([Gar2]). Let G > {1} be a solvable group and let M be the set of all cores of maximal subgroups of G. Then: (a) Any M ∈ M can be characterized by the property that the Fitting subgroup F(G/M) of G/M is the unique minimal normal subgroup of G/M. (b) The members of the set M are uniquely determined by the character table X(G) of the group G. (c) The Frattini subgroup Φ(G) is uniquely determined by X(G). Proof. The equality ⋂M∈M M = Φ(G) is valid for any finite group G (this intersection coincides with the intersection of all maximal subgroups of G). Hence, (c) follows from (a) and (b). (a) If M ∈ M, then M = H G for some H ∈ Γ1 and the core of H/H G in G/H G is {1}. We will show that if L ∈ Γ1 with L G = {1}, then G is a monolith, i.e., it possesses only one minimal normal subgroup. Let F(G) be the Fitting subgroup of G. Put D = L ∩ F(G). Since L ≤ NG (D) and D < NF(G) (D), we conclude that NG (D) = G and so D ≤ L G = {1}. It follows that F(G) is a minimal normal subgroup of G. Since all minimal normal subgroups of G are contained in F(G), it follows that F(G) is the unique minimal normal subgroup of G, as was to be shown. Thus, if M ∈ M, then, by the previous paragraph, G/M has only one minimal normal subgroup coinciding with F(G/M). Now let M ⊲ G be such that F(G/M) is the unique minimal normal subgroup of G/M. We show that M = H G for some H ∈ Γ1 . It suffices to prove that G/M possesses a maximal subgroup with trivial core. Without loss of generality, one may assume that M = {1}. It follows from Frattini’s lemma that if G > {1} is solvable, then F(G) > Φ(G). Hence, there is H ∈ Γ1 not containing F(G). If H G > {1}, then H G ≥ F(G) because F(G) is the unique minimal normal subgroup of G, whence H ≥ F(G), contrary to the choice of H. Thus H G = {1}. This shows that, in general, M ∈ M, completing the proof of (a). (b) Since the normal subgroups of G and their orders are uniquely determined by X(G), knowledge of X(G) is sufficient to determine whether or not G possesses

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at most one minimal normal subgroup. Therefore, the set M is uniquely determined by X(G). Let G = S × S, where S is a nonabelian simple group. Then G has a maximal subgroup (for example, the diagonal of G) with identity core but G has two distinct minimal normal subgroups. Exercise 7.1. Let π be a nonempty set of primes. A group G is said to be π-solvable if indices of some composition series of G are either primes from π or π󸀠 -numbers. Let Φ π (G) denote the intersection of the maximal subgroups of G whose indices are π-numbers (if π is empty we put Φ π (G) = G). Given X(G), is it possible to determine Φ π (G) for an arbitrary solvable group? Exercise 7.2. If a group G has a maximal subgroup with identity core, then F(G) is either {1} or a minimal normal subgroup of G. Exercise 7.3. Let M be defined as in Theorem 7.1. Suppose that F(G/M) > {1} for all M ∈ M. Is it true that G is solvable? Exercise 7.4. Let G = ⟨a, b | a4 = b4 = 1, a b = a3 ⟩. Find all H satisfying X(H) = X(G). Solution. One has |H| = |G| = 24 and H/H 󸀠 ≅ G/G󸀠 since Lin(H) ≅ Lin(G) as multiplicative groups. It follows that H is minimal nonabelian. Clearly, all nonlinear irreducible characters of G have the same degree 2, and so |Irr(H)| = |Irr1 (G)| = 2. As Z(G) ≅ E4 is noncyclic, G has no cyclic subgroups of index 2 and all members of Irr(G) are not faithful; then the same is true for H. Let U, V be kernels of nonlinear irreducible characters of G; then, say, G/U ≅ D8 , G/V ≅ Q8 . If H has an epimorphic image ≅ Q8 , it has exactly three involutions. and we conclude that H is metacyclic. Since there is only one nonabelian metacyclic group of order 16 and exponent 4, it follows that H ≅ G. If all proper nonabelian epimorphic images of H are ≅ D8 , then H is nonmetacyclic, H = ⟨a, b | a4 = b2 = 1, [a, b] = c, c2 = 1, [a, c] = [b, c] = 1⟩ has seven involutions. It is easy to show that X(G) = X(H) in this case. Exercise 7.5. Let G be a 2-group of maximal class. Find all H such that X1 (H) = X1 (G). Hint. Find |Irr1 (G)|. Use Exercise 7.8. Exercise 7.6. Let G = ⟨a, b | a2 = b2 = 1, a b = a1+2 ⟩ ≅ M2n , and suppose that H is a group such that X(H) = X(G). Is it true that H ≅ G? n−1

n

Solution. We have |H 󸀠 | = |G󸀠 | = 2 and G/G󸀠 ≅ H/H 󸀠 is abelian of type (2n−2 , 2). Thus, H is minimal nonabelian. Since H has only one normal subgroup of order 2, it follows that H is metacyclic and so has a cyclic subgroup of index 2 hence H ≅ M2n ≅ G. Exercise 7.7. Let G be a Sylow 2-subgroup of the simple Suzuki group Sz(23 ) and let H be a group such that X(H) = X(G). Is it true that H ≅ G?

86 | Characters of Finite Groups 1 Exercise 7.8. Let G and H be extraspecial 2-groups of the same order. Is it true that X(G) = X(H)? Hint. Use the orthogonality relations. Exercise 7.9. (a) If Sylow subgroups of a group of order 24 are not normal, then G is isomorphic to S4 .¹ (b) Let G ≅ S4 . Is it true that every group with the same character table as G is isomorphic to G? Solution. (b) Let H be such that X(H) = X(G). Then |H| = 24, |H/H 󸀠 | = |G/G󸀠 | = 2 so the Sylow subgroups of H are not normal. Now it follows from (a) that H ≅ S4 . Exercise 7.10. Let G be a minimal nonabelian p-group and let H be a group such that X(H) = X(G). Prove that H is also minimal nonabelian. Solution. We have H/H 󸀠 ≅ G/G󸀠 is abelian two-generator, H 󸀠 ≤ Z(H) order p, and the result follows from [BJ1, Lemma 65.2 (a)]. Exercise 7.11. Let A and B be two nonisomorphic nonabelian groups of order p3 , p > 2. Is it true that X(A) = X(B)? Exercise 7.12. Suppose that G ≅ A5 and H is a group such that X(H) = X(G). Prove that H ≅ G. Hint. We have |H| = |G| = 60, H 󸀠 = G󸀠 so H is nonsolvable. Show that if a group X of order 60 has no normal subgroup of order 5, then X ≅ A5 . It follows that H ≅ A5 . Exercise 7.13. (a) Let G ≅ S5 and let H be a group such that X(H) = X(G). Prove that H is isomorphic to G. (b) Let G ≅ Sn , n > 5, and let H be a group such that X(H) = X(G). Prove that H is isomorphic to G. (See [BZ3, Chapter 17].) Solution. (a) We have |H/H 󸀠 | = |G/G󸀠 | = 2, where H 󸀠 is a unique minimal normal subgroup of G. It follows that H 󸀠 ≅ A5 so H ≅ Aut(H 󸀠 ) ≅ S5 . Exercise 7.14. Let G ≅ SL(2, 5) and let H be a group such that X(H) = X(G). Prove that H is isomorphic to G. Hint. Note that the representation group of PSL(2, 5) is isomorphic ro SL(2, 5) (see Chapter VI).

1 It is easy to prove that if Sylow subgroups of a group of order p3 q are not normal, then |G| = 24. Indeed, q ≡ 1 (mod p) and p μ ≡ 1 (mod q) for some integer μ > 1 (Sylow). If μ = 2, then q = p + 1 so p = 2, q = 3, and we are done. Now assume that μ = 3. Then G is 2-nilpotent, by Burnside, a contradiction.

II Characters

| 87

Exercise 7.15. Let G ≅ PSL(2, 7) and let H be a group such that X(H) = X(G). Prove that H ≅ G. Hint. If a subgroup of order 7 is not normal is a group X of order 168, then X is isomorphic to PSL(2, 7). Exercise 7.16. Let G ≅ SL(2, 7) and let H be a group such that X(H) = X(G). Prove that H ≅ G. Hint. The representation group of PSL(2, 7) is isomorphic to SL(2, 7) (see Chapter VII; it follows from the content of Chapter VII that if X = X 󸀠 , then X has only one representation group). Exercise 7.17. Let G ≅ PGL(2, 7) and let H be such that X(H) = X(G). Is it true that H is isomorphic to G? Exercise 7.18. Let G ≅ AGL(1, p n ) and let H be such that X(H) = X(G). Is it true that H is isomorphic to G? Exercise 7.19. Let G be a nonabelian Dedekindian group and let H be such that X(H) = X(G). Find all H ≇ G. (As we have noted, X(Q8 ) = X(D8 ).)

8 Nagao’s theorem As we have seen before, Theorem 2.2 (Schur’s relations for matrix elements of irreducible representations) implies Theorem 3.7 (the Second Orthogonality Relation). Now, following H. Nagao [Nag3], we will prove the reverse implication. Let G be a group of order h, {χ1 , . . . , χ r } = Irr(G), χ i (1) = f i for all i. Suppose that the following relation holds (in fact, it is a consequence of the Second Orthogonality Relation): r

h−1 ⋅ ∑ f i χ i (t) = δ1,t

for all t ∈ G.

(1)

i=1

Putting t = 1 in (1), we get r

∑ f i2 = h.

(2)

i=1

It follows from (2) that there exist exactly h triples (i; m, n) such that i ∈ {1, . . . , r} and m, n ∈ {1, . . . , f i }. Let A be an h × h matrix whose rows are indexed by the elements t ∈ G and columns by the above triples (i; m, n). Let the (t, (i; m, n))-entry of A be a im,n (t), γ where {a α,β } are the matrix elements of a representation of G affording the character χ γ . Let B be an h × h matrix whose rows are indexed by the triples (i; m, n) (i ∈ {1, . . . , r}, m, n ∈ {1, . . . , f i }) and columns by t ∈ G and let the ((i; m, n), t)-entry

88 | Characters of Finite Groups 1 of B be fhi a in,m (t−1 ). Then the (t, s)-entry of the matrix AB (t, s ∈ G) is equal to r fi fi ∑ a im,n (t) a in,m (s−1 ) = ∑ a im,m (ts−1 ) = h−1 ∑ f i χ i (ts−1 ) = δ1,ts−1 = δ t,s , h h i;m i=1 i;m,n

by hypothesis. This shows that AB = Ih , and so we obtain that BA = Ih . However, the ((i; m, n), (j; k, l))-entry of the matrix BA is equal to j

h−1 f i ∑ a im,n (t)a k,l (t−1 ), t∈G

and, on the other hand, since BA = Ih , it is equal to δ(i;m,n),(j;k,l) = δ ij δ mk δ nl . This shows that Theorem 2.2 follows from (1).

9 Saksonov’s example We shall now consider an example, due to A. I. Saksonov [Sak2], showing that a knowledge of the character table X(G) of a group G is not sufficient to determine whether a given normal subgroup N of G is abelian. Let G = ES(m, p) be the extraspecial group of order p1+2m and exponent p > 2 (there is, up to isomorphism, only one such group). A permutation of the rows and columns of the character table X(G) is called an automorphism of X(G) if it preserves X(G). Any automorphism of X(G) induces a permutation of the set of normal subgroups of G. One can show that the automorphism group of X(G) is isomorphic to GL(2m, p) × Cp−1 . Since GL(2m, p) acts transitively on the set of n-dimensional subspaces (1 ≤ n ≤ 2m), it follows that the index-preserving automorphisms of the lattice of normal subgroups of G form a group that acts transitively on the set of normal subgroups of order p1+n . Therefore, any two normal subgroups of the same order greater than p are mapped onto each other by automorphisms of X(G). This means that they are equivalently represented in X(G). If m > 1, however, G possesses both abelian and nonabelian normal subgroups of order p3 , and so the commutativity of a normal subgroup of G cannot be determined from X(G). An analogous example was constructed by E. C. Dade [Dad1]. Theorem 9.1 (L. Solomon). The sum of all entries of any row of the matrix X(G) is a nonnegative rational integer. Proof. With every g ∈ G we associate a permutation π(g) = (g−1xxg) of the set G. This yields a permutation representation π of G. Let θ be the character of π. Obviously, θ(g) = |CG (g)|. Let θ = ∑1r a i χ i , where χ i are the irreducible constituents of θ and a i = ⟨θ, χ i ⟩ = |G|−1 ∑ θ(g)χ i (g) g∈G

II Characters

| 89

is a nonnegative integer, i = 1, . . . , r. Let {K1 , . . . , K r } be the set of G-classes, g j ∈ K j . Then one can rewrite the previous displayed formula as follows: r

r

r

a i = |G|−1 ∑ θ(g j )|K j |χ i (g j ) = |G|−1 ∑ |CG (g j )||K j |χ i (g j ) = ∑ χ i (g j ), j=1

j=1

j=1

and we are done since a i ∈ ℕ ∪ {0} for all i. According to Chapter III, the sum of all entries of any column of X(G) is a rational integer. The example of the Mathieu group M11 shows that number may be negative. Exercise. Let G = ES(m, 3) be the extraspecial group of order 32m+1 . Calculate the degrees of all irreducible characters of G. Hint. It is natural to find k(G), the class number of G, and use the two equalities k(G) = |Irr(G)|, |Lin(G)| = |G : G󸀠 | and the inequality χ(1)2 < 32m+1 for all χ ∈ Irr(G).

III On arithmetical properties of characters 1 Algebraic integers Since many important properties of characters are consequences of the fact that the values of characters are algebraic integers, we recall some relevant definitions and results relating to the latter. An algebraic number is a root of a polynomial X n + a n−1 X n−1 + ⋅ ⋅ ⋅ + a0 over the field ℚ of rational numbers. The algebraic numbers form a field, denoted by 𝔸. Algebraic numbers α, β are said to be conjugate with respect to ℚ if they are roots of the same irreducible polynomial over ℚ. It follows that the number of conjugates with any given algebraic number is finite, A polynomial is said to be monic if its leading coefficient is equal to 1. Let α be an algebraic number. A monic polynomial f ∈ ℚ(X) of minimal degree such that f(α) = 0 is said to be a minimal polynomial of α. Since a field has no zero divisors, it follows that minimal polynomial of α is irreducible over ℚ and uniquely defined. Let ℚ(α) denote the field extension of ℚ generated by α ∈ ℂ, that is, ℚ(α) = {

m(α) n(α)

󵄨󵄨 󵄨󵄨 m(X), n(X) are polynomials over ℚ, n(α) ≠ 0}. 󵄨󵄨 󵄨

In particular, if α is an algebraic number, then ℚ(α) ⊂ 𝔸 and ℚ(α) = {f(α) | f is a polynomial over ℚ} = ℚ ⊕ ℚα ⊕ ⋅ ⋅ ⋅ ⊕ ℚα n−1 , where n is the degree of the minimal polynomial of α over ℚ. If ϵ is a root of 1, the field ℚ(ϵ) is called cyclotomic. An algebraic integer is a complex number which is a root of a monic polynomial with coefficients in ℤ. The algebraic integers form a ring which will be denoted by 𝕀. Clearly, algebraic integers are algebraic numbers. Any root of 1 is an algebraic integer. Lemma 1.1. We have 𝕀 ∩ ℚ = ℤ. Proof. Obviously, ℤ ⊆ 𝕀 ∩ ℚ. Let sr = α ∈ 𝕀 ∩ ℚ, where r, s ∈ ℤ with GCD(r, s) = 1, s > 0. Let a polynomial f(x) = x n +a n−1 x n−1 +⋅ ⋅ ⋅+a1 x+a0 ∈ ℤ[x] be such that f(α) = 0. It follows from rn = −(a n−1 r n−1 + a n−2 r n−2 s + ⋅ ⋅ ⋅ + a1 rs n−2 + a0 s n−1 ) ∈ ℤ s that s = 1 so α ∈ ℤ, completing the proof. Lemma 1.2. Let α ∈ 𝕀 and a, b ∈ ℤ with GCD(a, b) = 1. If

aα b

∈ 𝕀, then

Proof. Let u, v be rational integers such that au + bv = 1. Then α (au + bv)α aα = =u + vα ∈ 𝕀 b b b since 𝕀 is a ring. DOI 10.1515/9783110224078-003

α b

∈ 𝕀.

92 | Characters of Finite Groups 1 Lemma 1.3 (Kronecker). Let α ∈ 𝕀 and let α1 = α, α2 , . . . , α n be all of its conjugates. Let |α i | ≤ 1 for all i. Then either α = 0 or α is a root of 1. Proof. Let k ∈ ℕ and n

f k (X) = ∏(X − α kν ) = X n + a k1 X n−1 + a k2 X n−2 + ⋅ ⋅ ⋅ + a kn . ν=1

By the theory of symmetric polynomials, a ki ∈ ℤ for i = 1, . . . , n. It follows from Vieta’s formulas that |a ki | ≤ (ni) for all i; hence the sequence f1 (X), f2 (X), . . . contains only finitely many pairwise distinct polynomials. Since f k (α k ) = 0, there are only finitely many pairwise distinct numbers among α, α2 , . . . . Therefore, there exist positive integers r, s such that α r+s = α r , whence either α = 0 or α s = 1.

2 Values of characters are algebraic integers Let h = |G| denote the order of a group G. Theorem 2.1. The values of the characters of a group G are algebraic integers in the field ℚ(ϵ), where ϵ is a primitive h-th root of 1. Proof. Let χ ∈ Char(G) and let T be a representation of degree n of G affording the character χ. If g ∈ G, then for some nonsingular n × n matrix S we have S−1 T(g)S = diag(ϵ1 , . . . , ϵ n ) (see the end of §I.8). Since g h = 1, it follows that (diag(ϵ1 , . . . , ϵ n ))h = diag(ϵ1h , . . . , ϵ hn ) = (S−1 T(g)S)h = S−1 T(g h )S = In , whence ϵ hi = 1, i ∈ {1, . . . , n}. Hence, χ(g) = ϵ1 +⋅ ⋅ ⋅+ϵ n . If ϵ is a primitive h-th root of 1, then ϵ i = ϵ ν i for some ν i ∈ ℤ, i ∈ {1, . . . , n}. Therefore, χ(g) = ϵ ν1 + ⋅ ⋅ ⋅ + ϵ ν n ∈ ℚ(ϵ). Since ϵ i ∈ 𝕀, the ring of algebraic integers, we finally conclude that χ(g) ∈ ℚ(ϵ) ∩ 𝕀. Exercise 2.1. Derive from the proof of Theorem 2.1 that χ(g −1 ) = χ(g). Hint. If ϵ is a root of 1, then ϵ = ϵ−1 .

3 The Frobenius–Molien theorem on degrees In this section we prove the Frobenius–Molien theorem, which asserts that the degrees of irreducible characters of a group divide its order. To this end, we will apply the results proved in the end of §I.12 to the semisimple group algebra ℂG. Let Irr(G) = {χ1 , . . . , χ r }, where all χ i are extended to the group algebra ℂG, by linearity: if z = a1 g1 + ⋅ ⋅ ⋅ + a n g n , a1 , . . . , a n ∈ ℂ, g1 , . . . , g n ∈ G, then we have χ i (z) = a1 χ i (g1 ) + ⋅ ⋅ ⋅ + a n χ1 (g n ). Here r = k(G), the class number of G (see Chapter II).

III On arithmetical properties of characters | 93

Let ω i be the linear (= of degree 1) character of Z(ℂG) corresponding to a charac, where f i = χ i (1) (see formula (3) at the ter χ i ∈ Irr(G): if z ∈ Z(ℂG), then ω i (z) = χ if(z) i end of §I.12). As we know, ω i : Z(ℂG) → ℂ is a homomorphism. If T is an irreducible representation of G with character χ i and z ∈ Z(ℂG), then T(z) = ω i (z)Iχ i (1) . The formula for ω i (z) follows from χ i (z) = tr(T(z)) = ω i (z)χ i (1) = ω i (z)f i . If χ = χ i , we write χ also ω χ instead of ω i . Thus, ω χ = χ(1) . By Lemma I.12.21, {ω i }1r is the set of all linear characters of Z(ℂG). Let CL(G) = {K1 , . . . , K r } be the set of G-classes and let k α = ∑x∈K α x (α = 1, . . . , r) be the α-th class sum. Since {k1 , . . . , k r } is a basis of the algebra Z(ℂG) (see Chapter II), we get r

γ

k α k β = ∑ h αβ k γ ,

(1)

γ=1

γ

where h αβ ∈ ℂ (α, β, γ = 1, . . . , r) are the structure constants of the algebra Z(ℂG). If γ g γ ∈ K γ , then h αβ , as the number of solutions (s, t) ∈ K α × K β of the equation st = g γ , is a nonnegative rational integer. We have h α χ i (g α ) ω i (k α ) = , χ i (1) where h α = |K α |, g α ∈ K α . Indeed,

χ(k α ) = ∑ χ(g) = |K α |χ(g α ), g∈K α

since χ is a central function (= constant on G-classes). Lemma 3.1. We have ω i (k α ) ∈ 𝕀 for i = 1, . . . , r (i.e., ω i (k α ) is an algebraic integer). Proof. Since ω i is a representation of the algebra Z(ℂG), it follows from (1) that r

γ

ω i (k α k β ) = ω i (k α )ω i (k β ) = ∑ h αβ ω i (k γ ),

(2)

γ=1

where α, β ∈ {1, . . . , r} and α is arbitrary but fixed. Setting ω i (k α ) = λ α ,

j

H α = (h αi ),

X = (λ1 , . . . , λ r )󸀠 ,

we rewrite (2) in the matrix form (λ α Ir − H α )X = 0r , where 0r is the zero column of height r. If K1 = {1}, say, then λ1 = χ1f(1) = 1 ≠ 0, so 1 that X ≠ 0r . Consequently, the matrix λ α Ir − H α is singular, and so det(λ α Ir − H α ) = 0. Since all entries of the matrix H α are rational integers, we conclude that λ α (= ω i (k α )) is an algebraic integer for all α, i ∈ {1, . . . , r}. Theorem 3.2 (Frobenius–Molien). The degree of any irreducible character of a group G divides its order.

94 | Characters of Finite Groups 1 Proof. Let χ ∈ Irr(G). Then, by (2), ω χ (k α ) =

χ(k α ) χ(1) .

Since

where h α = |K α |, g α ∈ K α ,

χ(k α ) = ∑ χ(g) = h α χ(g α ), g∈K α

we get ω χ (k α ) =

h α χ(g α ) . χ(1)

(3)

Since the set of all algebraic integers 𝕀 is a ring and ω χ (k α ), χ(g α ) ∈ 𝕀 (see Theorem 2.1 and Lemma 3.1), it follows that h α χ(g α )χ(g α ) = ω χ (k α )χ(g α ) ∈ 𝕀. χ(1) Now, summing the last equality over α ∈ {1, . . . , r} and taking into account that h is an ∑rα=1 h α χ(g α )χ(g α ) = |G| = h (the First Orthogonality Relation), we see that χ(1) element of 𝕀 ∩ ℚ = ℤ (Lemma 1.1), i.e., χ(1) | h, completing the proof. Using this result, it is easy to show that the degrees of irreducible characters of the alternating group A5 are 1, 3, 3, 4, 5. Exercise 3.1 (Brauer, see [Isa11, Problem 3.13]). Let k1 , . . . , k r be all class sums in the group algebra ℂG. If there is c ∈ ℂ such that ∑ri=1 k i = c ∏ri−1 k i , then G󸀠 = G. Solution. For any χ ∈ Irr# (G) = Irr(G) − {1G } it suffices to show that χ is nonlinear, i.e., χ(1) > 1. Since ω χ : Z(ℂG) → ℂ is a homomorphism, it follows that r

r

r

ω χ ( ∑ k i ) = ∑ ω χ (k i ) = χ(1)−1 ∑ χ(k i ) i=1

i=1

i=1

= χ(1)−1 ∑ χ(g) = χ(1)−1 |G|⟨χ, 1G ⟩ = 0. g∈G

On the other hand, r

r

r

ω χ (c ∏ k i ) = cω χ (∏ k i ) = c ∏ ω χ (k i ). i=1

i=1

i=1

∏ri=1

By the first displayed formula and hypothesis, ω χ (k i ) = 0 so there is i ≤ r such that ω χ (k i ) = 0, i.e., χ(g i ) = 0. It follows that χ is nonlinear. Thus, every nonprincipal irreducible character of G is nonlinear, and we conclude that Lin(G) = {1G }, hence we have G󸀠 = G. Exercise 3.2 (Brauer, see [Isa11, Problem 3.14]). In the notation of Exercise 3.1, if we have G = G󸀠 > {1}, then ∑ri=1 k i = c ⋅ ∏ri=1 k i for some c ∈ ℚ. Hint. The elements a = ∑ri=1 k i and b = ∏ri=1 k i belong to the ring Z(ℂG). For all χ ∈ Irr(G), one has ω χ (a) =

1 |G| ⟨χ, 1G ⟩ = |G|δ χ,1G ∑ χ(g) = χ(1) g∈G χ(1)

III On arithmetical properties of characters | 95

and

r

r

ω χ (b) = ∏ ω χ (k i ) = ∏ i=1

i=1

h i χ(g i ) , χ(1)

where h i = |K i |. By Burnside’s theorem, every nonlinear irreducible character has a zero (see Chapter XXI), and so ω χ (b) = 0 provided χ is nonlinear. Thus, r

ω χ (b) = mδ χ,1G for m = ∏ h i > 0 󳨐⇒ ω χ ( i=1

It follows that a = cb, where c =

|G| m

|G| b) = |G|δ χ,1G = ω χ (a). m

or, what is the same, ∑ri=1 k i = c ⋅ ∏ri=1 k i .

Proposition 3.3 (Schur). If χ ∈ Irr(G), then χ(1) | |G : Z(G)|.¹ Proof (Glauberman). Since Z(G/ker(χ)) ≥ Z(G) ker(χ)/ker(χ), it is sufficient, in view of Lagrange’s theorem, to consider the case when χ is faithful. In that case, Z(χ) = Z(G), where Z(χ) = {x ∈ G | |χ(x)| = χ(1)} is the quasikernel of χ (it is checked easily that Z(χ) ⊴ G; see §III.5, below). Let us consider the following equivalence relation ≡ on G: x ≡ y if and only if x = yz for some z ∈ Z(G). Let C1 , . . . , Cm be all ≡-classes. If T is a representation of G affording χ, then T(x) = T(y)T(z) = λ(z)T(y), where λ(z) is a root of unity. In that case, χ(x) = λ(z)χ(y) so that |χ(x)| = |χ(y)|, and we conclude that |χ(x)| is a function constant on ≡-classes. Let C1 , . . . , Cs be all ≡-classes on which χ does not vanish. We have, by the First Orthogonality Relation, s

|G| = ∑ |χ(g)|2 = ∑ |Ci ||χ(g i )|2 , g∈G

where g i ∈ Ci , i = 1, . . . , s.

i=1

Let K be the G-class containing g i . If x ∈ Ci , then x ∼ zg i , where z ∈ Z(G), i.e., x = z ⋅ y, where y ∈ K. Conversely, if x = zy, where z ∈ Z(G), y ∈ K, then x ∼ zg i , i.e., x ≡ g i so that x ∈ Ci . Thus, Ci = Z(G)K. If zy = z󸀠 y󸀠 , where z, z󸀠 ∈ Z(G) and y, y󸀠 ∈ K, then χ(y) = ϵχ(y󸀠 ), where ϵχ(1) = χ(z󸀠 z−1 ). Since y ∼ y󸀠 , we get χ(y) = χ(y󸀠 ). Since χ does not vanish on Ci , we have ϵ = 1, i.e., χ(z󸀠 z−1 ) = χ(1) so, since χ is faithful, it follows that z󸀠 z−1 = 1 hence z󸀠 = z. Then y = y󸀠 . Thus, every element of Ci is presented uniquely in the form zy, where z ∈ Z(G), y ∈ K. Therefore, |Ci | = |K i ||Z(G)| so that s

s

s

i=1

i=1

i=1

|G| = ∑ |Ci )|χ(g i )|2 = ∑ |K|χ(g i )χ(g i )|Z(G)| = ∑ χ(1)ω χ (k)χ(g −1 i )|Z(G)|, s where k = ∑x∈K x. It follows that the rational number |G:Z(G)| χ(1) = ∑i=1 ω χ (k)χ(g i ) is an algebraic integer so it belongs to ℤ (Lemma 1.1), and we finally conclude that χ(1) divides |G : Z(G)|.

1 This is a generalization of the Frobenius–Molien theorem. As Ito has shown, χ(1) divides the index of every abelian normal subgroup; see Theorem VII.2.3 whose proof is based on Clifford theory.

96 | Characters of Finite Groups 1

4 Algebraically conjugate characters If χ is a character of a group G (we also write χ ∈ Char(G)), then the function χ : G → ℂ defined by χ(g) = χ(g) is also a character of G (see the example following Theô rem II.4.2). If χ = χ T , then χ = χ T = χ T . In this section the operation of complex conjugation of characters will be generalized in a natural way. Let G be a group of order h and let ϵ be a primitive h-th root of unity. The cyclotomic field ℚ(ϵ) is a normal extension of ℚ of degree φ(h), where φ is the numbertheoretic Euler totient function.² The Galois group G = Gal(ℚ(ϵ)/ℚ) (= the group of automorphisms of ℚ(ϵ)) of this extension is isomorphic to the multiplicative group of residue classes modulo h = |G| hence it is abelian. If σ ∈ G and χ is a character of G, then χ(g) ∈ ℚ(ϵ) for any g ∈ G (see Theorem 2.1). Define the function σχ from G into ℂ as follows: (σχ)(g) = σ(χ(g)) (g ∈ G). It is easy to see that σ(ϵ) = ϵ a , where a is a rational integer such that GCD(a, h) = 1 (indeed, σ(ϵ) is a primitive h-th root of unity since σ is an automorphism of ℚ(ϵ)). Thus, if χ(g) = ϵ1 + ⋅ ⋅ ⋅ + ϵ n (n = χ(1)), where the ϵ i are |G|-th roots of unity, then n

(σχ)(g) = ∑ ϵ ai , i=1

so that the function σχ is defined. Our aim is to prove the following important result. Theorem 4.1. Let χ be a character of a group G and σ ∈ G. Then σχ is a character of G. Moreover, if χ is irreducible, then so is σχ. To prove, it is natural to construct a representation of our group G with the character σχ. However, our proof will be another. We prove that σ ∈ G is a permutation of the set Irr(G), and this is sufficient to finish the proof. The proof of the theorem will follow from a sequence of lemmas. Throughout the section, we denote by ϵ some fixed primitive h-th root of unity, where h = |G|. Lemma 4.2. If σ ∈ G, χ is a character of a group G of order h and σ(ϵ) = ϵ a , where GCD(h, a) = 1, then (σχ)(g) = χ(g a ). Proof. Suppose that χ(1) = n, and let T : G → GL(n, ℂ) be a representation affording the character χ. Then χ(g) = ϵ1 + ⋅ ⋅ ⋅ + ϵ n , where ϵ1 , . . . , ϵ n is the spectrum of the matrix T(g), i.e., the set of eigenvalues of this matrix, counting multiplicities (see 2 Recall that a field extension E/ℚ is said to be normal if it is algebraic and every irreducible polynomial over ℚ which has a zero in E splits completely in E, or, which is the same, if a ∈ E, then all conjugates of a are contained in E.

III On arithmetical properties of characters | 97

the proof of Theorem 2.1). We have ϵ k = ϵ ν k , where ν k ∈ ℤ, k ∈ {1, . . . , n}. Therefore, aν σ(ϵ k ) = σ(ϵ ν k ) = ϵ k k = ϵ ak , whence (see the definition of σχ) one has (σχ)(g) = σ(χ(g)) = σ(ϵ1 ) + ⋅ ⋅ ⋅ + σ(ϵ n ) = ϵ1a + ⋅ ⋅ ⋅ + ϵ an = tr(T(g)a ) = tr(T(g a )) = χ(g a ), since T is a homomorphism. Now we construct an idempotent basis of the algebra Z(ℂG). Let us establish an important connection between an irreducible character χ i of G and the corresponding minimal idempotent e i of the algebra Z(ℂG). Recall that e i is the identity element of the simple component A i of the algebra ℂG; a minimal left ideal of A i affords an irreducible representation T i with character χ i (see §I.12). Denoting ω χ i by ω i , we have ω i (z) = χ i (z)/f i for all z ∈ Z(ℂG), where f i = χ i (1) (see also the first paragraphs of §III.3). Lemma 4.3. For any i ∈ {1, . . . , r}, one has ei =

fi ∑ χ i (g)g. h g∈G

(1)

Next, e i ∈ Z(ℂG), e i e j = δ ij e i , i, j = 1, . . . , r (here δ is the Kronecker delta on the set ℕ). In particular, the set {e i }1r is an orthogonal basis of the algebra Z(ℂG). Proof. Denote the right-hand side of (1) by z i . Since χ i is a class function, we get z i ∈ Z(ℂG), by Lemma II.5.1. By Theorem II.3.7, one has for any j ∈ {1, . . . , r}, ω i (z j ) = ω i (

fj fj fj ∑ χ j (g)g) = ∑ χ j (g)χ i (g) = ⋅ ⟨χ i , χ j ⟩ = δ ij . h g∈G hf i g∈G fi

(2)

Since ω j (e i ) = δ ij (see the end of §I.12), it follows that ω j (z i ) = ω j (e i ) for all j, and we conclude that z i = e i . Suppose that λ1 e1 + ⋅ ⋅ ⋅ + λ r e r = 0, where all λ i ∈ ℂ. Applying ω i to this equality and using the equality ω j (e i ) = δ ij , one obtains λ i = 0. Thus, {e i }1r is a linearly independent system. The previous paragraph implies that {e i }1r is a basis of the algebra Z(ℚ(ϵ)G). Indeed, we have r = dim Z(ℚ(ϵ)G), e i ∈ ℚ(ϵ)G ∩ Z(ℂG) = Z(ℚ(ϵ)G) and the elements e i (i = 1, . . . , r) are linearly independent in Z(ℚ(ϵ)G). We compute using (2) that ω k (e i e j − δ ij e i ) = ω k (e i )ω k (e j ) − δ ij ω k (e i ) = δ ki δ kj − δ ij δ ki = δ ki (δ kj − δ ij ) = 0. Since e i e j − δ ij e i = z ∈ Z(ℂG), we have z = λ1 e1 + ⋅ ⋅ ⋅ + λ r e r for some λ1 , . . . , λ r ∈ ℂ. Then, by (2), 0 = ω k (z) = λ k for all k = 1, . . . , r, and we conclude that z = 0 hence e i e j = δ ij e i . This means that {e i }1r is an orthonormal basis of the algebra Z(ℂG). In particular, e1 + ⋅ ⋅ ⋅ + e r = 1 (see the last three sections of Chapter I). Proof of Theorem 4.1. Let x = ∑g∈G x(g)g ∈ Z(ℚ(ϵ)G), σ ∈ G. Set σ(x) = ∑g∈G σ(x(g))g. Then the mapping x 󳨃→ σ(x) is an automorphism of the algebra Z(ℚ(ϵ)G). Therefore,

98 | Characters of Finite Groups 1 if e󸀠i = σ(e i ), then {e󸀠i }1r is another idempotent basis of Z(ℚ(ϵ)G). Since σ is an automorphism of the field ℚ(ϵ) and {e i }1r is a basis of the algebra Z(ℂG), we get e󸀠i e󸀠j = δ ij e󸀠i ,

e󸀠i + ⋅ ⋅ ⋅ + e󸀠r = 1,

(3)

and r

e󸀠i = ∑ α ij e j ,

(4)

α ij ∈ ℚ(ϵ).

j=1

It follows from (2) that r

ω j (e󸀠i ) = ω j ( ∑ α ik e k ) = α ij e j

(j ∈ {1, . . . , r}).

k=1

Since (e󸀠i )2 = σ(e i )2 = σ(e2i ) = σ(e i ) = e󸀠i , it follows that ω j (e󸀠i )2 = ω j ((e󸀠i )2 ) = ω j (e󸀠i ), i.e., α2ij = α ij . Thus, α ij = 1 or 0. The equalities in (3) and (4) give r

r

r

r

r

∑ ( ∑ α ij e j ) = ∑ ( ∑ α ij )e j = 1 = ∑ e j , i=1 j=1

j=1 i=1

j=1

whence, by Lemma 4.3, ∑ri=1

α ij = 1 (j ∈ {1, . . . , r}). Therefore, all entries in each column of the matrix (α ij ) are equal 0, except one, which is equal to 1. Since (α ij ) is the matrix of transition to another basis in Z(ℚ(ϵ)G), it is nonsingular. As it contains exactly r nonzero entries, each of its rows possesses exactly one nonzero entry, which is equal to 1. Hence (α ij ) is a permutation matrix, i.e., e󸀠1 , . . . , e󸀠r is a permutation of the idempotents e1 , . . . , e r . Let e󸀠i = e k . Since fk fi ∑ σ(χ i (g))g = σ(e i ) = e󸀠i = e k = ∑ χ k (g)g, h g∈G h g∈G it follows from linear independence of the set G in the algebra ℂG that f i σ(χ i (g)) = f k χ k (g)

(g ∈ G).

Since G is abelian, it follows that σ commutes with complex conjugation, and so f i σ(χ i (g)) = f k χ k (g)(g ∈ G). Taking g = 1, we get f i2 = f k2 , i.e., f i = f k (all f i are positive integers), and we obtain σ(χ i (g)) = χ k (g). Thus, the automorphism σ permutes the irreducible characters of the group G. If χ = ∑ri m i χ i is a character of G, then σχ = ∑ri m i σ(χ i ) is also a character. Now the last assertion of the theorem is obvious. This completes the proof. Definition. Let χ ∈ Char(G) and σ ∈ G. Then σχ is called the character algebraically conjugate to χ. In particular, characters χ and χ are algebraically conjugate. Theorem 4.4. Let χ ∈ Char(G), g ∈ G. If |χ(g)| = 1, then χ(g) is a root of 1.

III On arithmetical properties of characters | 99

Proof. Let ϵ and G have the same meaning as above. Denote complex conjugation by σ0 . Since G is abelian, for all σ ∈ G, it follows that σ(α) = σ0 σ(α) = σσ0 (α) = σ(α),

α ∈ ℚ(ϵ).

Let χ(g) = ϵ1 + ⋅ ⋅ ⋅ + ϵ n , where the ϵ i are h-th roots of unity, n = χ(1). Then |χ(g)| = 1 implies 1 = |χ(g)|2 = χ(g)χ(g). Hence, for any σ ∈ G, |σ(χ(g))|2 = σ(χ(g))σ(χ(g)) = σ(χ(g))σ(χ(g)) = σ(χ(g)χ(g)) = σ(|χ(g)|2 ) = σ(1) = 1. Thus, |σ(χ(g))| = 1 for all σ ∈ G hence, by Kronecker’s Lemma 1.3, the algebraic integer χ(g) is a root of 1. Lemma 4.5. An algebraic number coinciding with all its conjugates is rational. Proof. Indeed, let an algebraic number α coincide with all its conjugates. If α is a root of an irreducible monic polynomial f(X) ∈ ℚ(X), then f(X) = (X − α)n , where deg(f) = n (all roots of F are conjugate!). The second coefficient of f is rational and equals −nα. Thus, α is rational. Definition. An element g of a group G is said to be rational if it is conjugate in G to all generators of the cyclic subgroup ⟨g⟩. A character χ of a group G is said to be rational if all its values are rational (and therefore belong to ℤ). Theorem 4.6. An element g of a group G is rational if and only if χ(g) is a rational number for all χ ∈ Irr(G). In particular, all characters of the symmetric group Sn are rational. Proof. Let g ∈ G be rational. If χ ∈ Irr(G) and σ ∈ G = Gal(ℚ(ϵ)/ℚ), then by Lemma 4.2, σ(χ(g)) = χ(g a ) for some a ∈ ℤ such that GCD(a, o(g)) = 1, where o(g) is the order of g ∈ G. Since g a ∼ g in G and χ is a central function, it follows that σ(χ(g)) = χ(g) for all σ ∈ G, whence χ(g) is rational (Lemma 4.5). Now let χ(g) ∈ ℚ for any χ ∈ Irr(G). Let a ∈ ℤ be such that GCD(a, o(g)) = 1. Then one can choose σ ∈ G such that σ(ϵ) = ϵ a . Now, if χ ∈ Irr(G), then χ(g a ) = σ(χ(g)) = χ(g), whence g a is conjugate to g (see Chapter II) and so the element g is rational. It is well known that any element of the symmetric group Sn is rational, and so the last assertion follows. It follows from Theorem 4.6 that the alternating group A5 has a character which is not rational: indeed, any element of A5 of order 5 is not rational. A dihedral group D2n is rational if and only if n ∈ {3, 4, 6} (see Exercise 4.1, below). The ordinary quaternion group Q8 is rational. Involutions are rational. Exercise 4.1. Prove that an element g of a group G is rational if and only if it satisfies |NG (⟨g⟩)/C G (g)| = φ(o(g)). Here φ is the Euler totient function. Hint. We have |Aut(⟨g⟩)| = φ(o(g)), the number of generators of ⟨g⟩ Exercise 4.2 ([Isa11, Problem 3.3]). Prove that a nonabelian simple group has no irreducible character of degree 2.

100 | Characters of Finite Groups 1 Solution. Assume that a nonabelian simple group G has an irreducible character χ of degree 2. Since χ(1) divides |G| (Frobenius–Molien), there is in G an involution g. If Γ is a representation of G affording the character χ, then eigenvalues of matrix Γ(g) are contained in the set {1, −1}. The matrix Γ(g) is diagonalizable, so one may assume that Γ(g) = diag(α, β), where α, β ∈ {1, −1}. If α = β, then g ∈ Z(χ) = Z(G) so that G is not simple, a contradiction. Thus, α = −β. Therefore, det(χ)(g) = −1 so that det(χ) ≠ 1G , and since ‘det’ is a linear character, we conclude that G󸀠 < G, and G is not simple. Thus, χ does not exist. Exercise 4.3. If Irr(G) = {χ1 , . . . , χ r }, then for all m, n ∈ {1, . . . , r} one has ∑ χ m (s)χ n (s−1 g) = δ mn

s∈G

|G| χ m (g) for all g ∈ G, χ m (1)

where δ is the Kronecker delta on the set {1, . . . , r}.³ Solution. We use (1) and the equality e m e n = δ mn e m (recall that the idempotents e1 , . . . , e r form an orthonormal basis of the algebra Z(ℂG)). One can rewrite equality (1) for e m , e n as follows: em =

χ m (1) χ m (1) ∑ χ m (s)s = ∑ χ m (s)s−1 , |G| s∈G |G| s∈G

en =

χ n (1) χ n (1) ∑ χ n ((sg−1 )−1 )sg −1 = ∑ χ n (gs−1 )sg−1 . |G| g∈G |G| g∈G

Comparing now the coefficients for g−1 on the left and right sides of the equation e m e n = δ mn e m , we obtain χ m (1) χ n (1) χ m (1) χ m (g). ∑ χ m (s)χ n (gs−1 ) = δ mn |G| |G| s∈G |G| As χ(gs−1 ) = χ(s−1 (gs−1 )s) = χ(s−1 g) for each χ ∈ Irr(G), we obtain the required.

5 Kernels and quasikernels Let T be a representation of a group G affording the character χ = χ T . Then, by definition, the kernel of χ is the normal subgroup ker(χ) = ker(T). We call χ faithful provided ker(χ) = {1}. Thus, T is faithful if and only if χ T is faithful. The regular character ρ G of G is faithful since it vanishes on G# . Exercise 5.1. If χ, ϑ ∈ Char(G), then ker(χ + ϑ) = ker(χ) ∩ ker(ϑ). Derive from this and the previous paragraph that ⋂χ∈Irr(G) ker(χ) = {1}. Hint. We have Irr(G) = Irr(ρ G ). 3 For g = 1, this coincides with the First Orthogonality Relation.

III On arithmetical properties of characters | 101

The following theorem shows that the kernel of a character is uniquely determined by the character table X(G) of G. Theorem 5.1. If χ ∈ Char(G), then ker(χ) = {g ∈ G | χ(g) = χ(1)}. Proof. Let χ be the character of a representation T : G → GL(n, ℂ) of a group G. If g ∈ ker(χ), then g ∈ ker(T), by definition, and T(g) = In , the identity n × n matrix. Hence, χ(g) = tr(T(g)) = tr(In ) = n = χ(1). Now suppose that χ(g) = χ(1). If {ϵ1 , . . . , ϵ n } is the spectrum of the matrix T(g), then n = χ(g) = ϵ1 + ⋅ ⋅ ⋅ + ϵ n , all ϵ i are roots of 1. Since |ϵ i | = 1, it follows that ϵ i = 1 for all i. Since T(g) is conjugate in GL(n, ℂ) to diag(ϵ1 , . . . , ϵ n ) = In , we get T(g) = In , i.e., g ∈ ker(T) = ker(χ). It follows from the proof that for all χ ∈ Char(G) and all g ∈ G one has |χ(g)| ≤ χ(1) with equality if and only if g ∈ ker(χ). Definition. If χ ∈ Char(G), then the quasikernel of χ is the set Z(χ) = {g ∈ G | |χ(g)| = χ(1)}. By Theorem 5.1, Z(G) ⊆ Z(χ) and ker(χ) ⊆ Z(χ). If χ ∈ Lin(G), then Z(χ) = G. As the following theorem shows, Z(χ) ⊴ G. Theorem 5.2. Suppose that χ ∈ Char(G) is afforded by a representation T of a group G. Then Z(χ) = {g ∈ G | T(g) is a scalar matrix}. In particular, Z(χ) ⊴ G. If, in addition, one has χ ∈ Irr(G), then Z(G/ker(χ)) = Z(χ)/ker(χ) and Z(G) ≤ Z(χ) with equality provided the character χ is faithful. Proof. If T(g) = ϵIχ(1) is a scalar matrix, where ϵ is a o(g)-th root of 1, then χ(g) = tr(T(g)) = χ(1)ϵ, |χ(g)| = χ(1)|ϵ| = χ(1) 󳨐⇒ g ∈ Z(χ). Now let g ∈ Z(χ). In that case, for some o(g)-roots ϵ1 , . . . , ϵ n of 1, we have χ(g) = ϵ1 + ⋅ ⋅ ⋅ + ϵ n ,

n = χ(1) and |χ(g)| = n.

This shows that ϵ1 = ⋅ ⋅ ⋅ = ϵ n = ϵ. Therefore, T(g) = ϵIn , i.e., T(g) is a scalar matrix. If g1 , g2 ∈ Z(χ), then T(g i ) = ϵ i In , i = 1, 2. Thus, T(g1 g2 ) = T(g1 )T(g2 ) = ϵ1 ϵ2 In is a scalar matrix, and so g1 g2 ∈ Z(χ). Thus, Z(χ) ≤ G. If t ∈ G, then T(t−1 g1 t) = T(t)−1 ϵ1 In T(t) = ϵ1 In 󳨐⇒ Z(χ) ⊴ G. If χ is irreducible, so is T and, by Schur’s lemma, T(Z(G)) consists of scalar matrices. Consequently, Z(G) ≤ Z(χ). Now let, in addition, χ be faithful. Then Z(χ) ≅ T(Z(χ)) ≤ Z(T(G)) ≅ Z(G) which shows that Z(G) = Z(χ). In the general case, χ can be considered as a faithful character of G/ker(χ); then we have Z(G/ker(χ)) = Z(χ)/ker(χ), by what has been proved above.

102 | Characters of Finite Groups 1 Let N ⊴ G. Our next goal is to study how the characters of the quotient group G/N are related to characters of the group G (for an abelian group G, see §I.9). Write CharN (G) = {χ ∈ Char(G) | N ≤ ker(χ)},

IrrN (G) = CharN (G) ∩ Irr(G)

(thus, IrrN (G) is the set of those irreducible characters of G whose kernels contain N). Obviously, Lin(G) = IrrG󸀠 (G). Exercise 5.2. We have χ ∈ CharN (G) if and only if χ = m1 χ1 + ⋅ ⋅ ⋅ + m k χ k , where the m i are nonnegative integers and χ i ∈ IrrN (G) if m i > 0. Hint. Let χ = m1 χ1 + ⋅ ⋅ ⋅ + m k χ k ∈ CharN (G), where all χ i are irreducible. Then N ≤ ker(χ) = ⋂ ker(χ i ) ≤ ker(χ j ) for all j ∈ {1, . . . , k} with m j > 0. m i >0

Let N ⊴ G, let Γ be a representation of the quotient group G = G/N and let ν : g 󳨃→ g̃ be the natural homomorphism of G onto G. One can see that the composition T Γ = Γ ∘ ν is a representation of G such that ker(T Γ ) ≥ N. Since ν is a surjection, the mapping Γ → T Γ is an injection of the set of all representations of G onto the set of all representations of G whose kernels contain N. Let T be a representation of a group G whose kernel contains N ⊴ G. We define a representation Γ of G = G/N by Γ(g) = T(g), where g = gN = ν(g) (ν : G → G is a natural epimorphism). The direct checking shows that this definition is legitimate. Since T = Γ ∘ ν = T Γ (see the previous paragraph), the mapping Γ 󳨃→ T Γ is surjective, and so the above mapping is bijective. Now let θ be the character of G afforded by the representation Γ, T = T Γ and χ = χ T . Since θ uniquely determines χ (indeed, χ(g) = θ(g)), we write χ = χ θ . Note that ker(χ θ ) ≥ N, because ker(χ θ ) = ker(T). As we have seen, the mapping θ 󳨃→ χ θ is a bijection of Char(G) onto CharN (G). It is easy to see that this mapping induces a bijection of Irr(G) onto IrrN (G). We call the character χ θ the inflation of θ and denote it by inf(θ). Thus, if θ ∈ Char(G), then inf(θ) ∈ Char(G). Next, {inf(θ) | θ ∈ Irr(G)} ⊆ Irr(G). Exercise 5.3. In the above notation, if ker(θ) = M/N, then ker(inf(θ)) = M. Show that the mapping inf : Char(G/N) → CharN (G) is an isometry, that is, if θ i ∈ Char(G/N), i = 1, 2, then ⟨θ1 , θ2 ⟩ = ⟨inf(θ1 ), inf(θ2 )⟩. Exercise 5.4. If N ⊴ G, then N = ⋂χ∈IrrN (G) ker(χ). In particular, ⋂χ∈Lin(G) ker(χ) = G󸀠 since Lin(G) = IrrG󸀠 (G). Theorem 5.3. Let N ⊴ G and let g 󳨃→ g be the natural homomorphism of the group G onto G = G/N. Then |CG (g)| ≤ |CG (g)|. Proof. Using the Second Orthogonality Relation, we get |CG (g)| =

∑ |θ(g)|2 = θ∈Irr(G)

=

∑ χ∈IrrN (G)

∑ |(inf(θ))(g)|2 θ∈Irr(G)

2

|χ(g)| ≤

∑ |χ(g)|2 = |CG (g)|. χ∈Irr(G)

III On arithmetical properties of characters | 103

It is possible to produce a character-free proof of Theorem 5.3. Exercise 5.5. Let N, G, G have the same meaning as in Theorem 5.3 and g ∈ G − N. (a) If |CG (g)| = |CG (g)|, then χ(g) = 0 for any χ ∈ Irr(G) whose kernel does not contain N. In particular (Thompson), this is true when CN (g) = {1}. (b) If |CG (g)| = |CG (g)| for g ∈ G − N, then N ≤ G󸀠 . Hint to (b). By (a), χ(g) = 0 for any χ ∈ Irr(G | N)(= Irr(G) − IrrN (G)) and g ∈ G − N. Therefore, Lin(G) ⊆ IrrN (G) (linear characters of G have no zeros!), whence we get N ≤ ⋂χ∈Lin(G) ker(χ) = G󸀠 . Lemma 5.4. If χ ∈ Irr(G), then χ(1)2 ≤ |G : Z(χ)| with equality if and only if χ vanishes on G − Z(χ). Proof. Applying Lemma II.6.2, we get ⟨χZ(χ) , χZ(χ) ⟩ ≤ |G : Z(χ)| with equality if and only if χ vanishes on G − Z(χ). Since χZ(χ) = χ(1)λ, where λ ∈ Lin(Z(χ)), it follows that ⟨χZ(χ) , χZ(χ) ⟩ = χ(1)2 ⟨λ, λ⟩ = χ(1)2 , and the first assertion is proved. Now the last assertion is obvious. A group G is said to be of central type if it is nonabelian and χ(1)2 = |G : Z(G)| for some χ ∈ Irr(G). It follows from Lemma 5.4 that then Z(χ) = Z(G). Using the classification of finite simple groups (= CFSG), Howlett and Isaacs [HowI] have shown that the groups of central type are solvable. Let M i (i = 1, . . . , n) be subsets of a nonempty set M. Given S ⊆ {1, . . . , n}, put MS = ⋂i∈S M i . Let A be an additive abelian group. Given a function f : M → A, set f S = ∑x∈MS f(x). Then the following inclusion-exclusion principle holds (see any textbook on combinatorial theory): ∑

f(x) = ∑ f{i} − ∑ f{i,j} + ⋅ ⋅ ⋅ + (−1)n f{1,...,n} .

x∈M{1,...,n}

i

i {1}. Then ∑χ∈Irr(θ) χ(1)2 ≤ |G/M| ≤ |G/N|, a contradiction. Exercise 5.15. Let G be a nonabelian p-group such that the number of characters of degree p in Irr(G) is not a multiple of p. Then |G : G󸀠 | = p2 so, if p = 2, then G is of maximal class, by Taussky’s theorem. If, in addition, G has an abelian subgroup of index p and p > 2, then G is of maximal class. Solution. Let the number of characters of degree p in Irr(G) is k ≢ 0 (mod p). Assume that |G : G󸀠 | > p2 . Then |G| =

χ(1) +

∑ χ∈Lin(G)

χ(1)2 +

∑ χ∈Irr1 (G),χ(1)=p

2

∑

χ(1)2

χ∈Irr1 (G),χ(1)>p

3

≡ |G : G | + kp ≢ 0 (mod p ), 󸀠

since |G| ≥ p3 , and this is a contradiction. Thus, |G : G󸀠 | = p2 . The second assertion is a consequence of the following fact: If G is a nonabelian p-group with abelian subgroup of index p and |G/G󸀠 | = p2 , then G is of maximal class. To prove this, we proceed by induction on |G|. In that case, |Z(G)| = 1p |G : G󸀠 | = p. If |G| = p3 , then G is of maximal class. If |G| > p3 , then G/Z(G) is of maximal class, by induction. Then G is of maximal class since |Z(G)| = p. Exercise 5.16 (= Lemma 9.2, see [Isa11, Theorem 2.31]). If χ ∈ Irr(G) and the quotient group G/Z(χ) is abelian, then χ(1)2 = |G : Z(G)|. Exercise 5.17. If G is such that X(G) = X(A5 ), then G ≅ A5 . Solution. The character table determines |G| so that |G| = 60. It follows from the character table that all nonprincipal characters of G are faithful so G is simple. Next, A5 is the only simple group of order 60. Exercise 5.18. If G is such that X(G) = X(SL(2, 5)), then G ≅ SL(2, 5).

106 | Characters of Finite Groups 1 Solution. We have |G| = 120 and Z(G) of order 2 is the unique nontrivial normal subgroup of G. Now result follows from Exercise 5.17 and results of §VI.4 on Schur multipliers. Exercise 5.19. Let H be a nonabelian group of order p3 . Suppose that G is such that X(G) = X(H). Is it true that G ≅ H? Exercise 5.20 (G. Seitz [Sei1]). Classify the groups with exactly one nonlinear irreducible character. Exercise 5.21 (Burnside, Miller, independently). Classify the groups G with k(G) ≤ 5. Hint. Use the equality k(G) = |Irr(G)| and the theorem of Frobenius–Molien. Note that the class number of the following groups is equal to 5: Frobenius group of orders 18, 20, 21, groups C5 , D8 , Q8 , S4 and A5 . Is there another group G with k(G) = 5? Exercise 5.22. Show that X(D16 ) ≠ X(Q16 ). Hint. Construct the character tables of groups D16 and Q16 . It is also possible to use Frobenius–Schur formula for the number of involutions (see §IV.6). Exercise 5.23. Find the number of nonlinear irreducible characters and their degrees m−1 m−2 of the group Mp m = ⟨a, b | a p = b p = 1, a b = a1+p ⟩, where m > 2 and m > 3 provided p = 2. Answer. All nonlinear irreducible characters of G have degree p. Therefore, |Irr1 (G)| =

|G| − |G : G󸀠 | = p n−2 − p n−3 . p2

Exercise 5.24. Prove that all nonlinear irreducible characters of a minimal nonabelian p-group G have the same degree p. Find the number of such characters provided |G| = p m .

6 Solvability of {p, q}-groups We shall now prove celebrated Burnside’s p a q b -theorem, one of the earliest nontrivial applications of character theory. Note that the most theorems concerning applications of characters to finite groups have conditions involving characters. The hypothesis of Burnside’s theorem does not involve characters. The proof of this theorem is one of the most brilliant triumphs of character theory. We begin with several lemmas, which are of independent interest. χ(g) Lemma 6.1 (Burnside). Let G be a group, g ∈ G and χ ∈ Char(G). Suppose that χ(1) is χ(g) an algebraic integer (i.e., χ(1) ∈ 𝕀). Then g ∈ Z(χ) ∪ Tχ , where Tχ = {x ∈ G | χ(x) = 0} is the set of zeros of the character χ and Z(χ) = {x ∈ G | |χ(x) = χ(1)} is the quasikernel of χ.

III On arithmetical properties of characters | 107

Proof. Let ϵ be a primitive |G|-th root of unity and let σ ∈ G = Gal(ℚ(ϵ)/ℚ). Since χ(g) = ϵ1 + ⋅ ⋅ ⋅ + ϵ n , where ϵ i and σ(ϵ i ) are |G|-th roots of 1, n = χ(1), we see that χ(g) )| ≤ 1. Then, |σ(χ(g))| ≤ χ(1) (indeed, σχ is a character of degree χ(1)) so that |σ( χ(1) by Lemma 1.3, χ(g)/χ(1) is either 0 or a root of 1. In the first case g ∈ Tχ . In the second case, |χ(g)| = χ(1) and so g ∈ Z(χ). The proof of Lemma 6.1 shows that if χ(g)/χ(1) is not an algebraic integer, then 0 < |χ(g)/χ(1)| < 1. Lemma 6.2 (Burnside). Let G be a group, let K be a G-class, let g ∈ K and let χ ∈ Irr(G). If GCD(χ(1), |K|) = 1, then g ∈ Tχ ∪ Z(χ). Proof. Let k = ∑g∈K g be the class sum in ℂG and let ω χ be the linear character of the algebra Z(ℂG) corresponding to χ. By Lemma 3.1, ω χ (k) is an algebraic integer. Since ω χ (k) = |K|χ(g)/χ(1), it follows by the hypothesis and Lemma 1.2 that χ(g)/χ(1) is an algebraic integer. The result now follows from Lemma 6.1. If χ(1) is a power of p and g ∈ Z(P)# , where P is a Sylow subgroup of a nonabelian simple group, then χ(g) = 0. Indeed, if K is a G-class containing g, then we have GCD(|K|, χ(1)) = 1, and the assertion follows from Lemma 6.2 since Z(G) = {1}. Lemma 6.3 (Burnside’s non-simplicity criterion = Burnside’s p n -lemma). If a group G contains a G-class K of size p n > 1, where p is a prime, then G is not simple. Proof. Since |K| > 1, G is not abelian. Assume, by way of contradiction, that G is simple. Then Lin(G) = 1G . Therefore, for the regular character ρ = ρ G of G, we have ρ = 1G +

∑

χ(1)χ,

χ∈Irr# (G)

where Irr# (G) = Irr(G) − {1G },

(1)

and so ρ(1) = |G| = h = 1 +

χ(1)2 .

∑

(2)

#

χ∈Irr (G)

Since ρ vanishes on G# , it follows from (1) that ∑ #

χ(1)χ(g) = −1 for all g ∈ G# .

(3)

χ∈Irr (G)

Put

Irr(1) (G) = {χ ∈ Irr# (G) | p ∤ χ(1)},

Irr(2) (G) = {χ ∈ Irr(G) | p | χ(1)}

and ψ(i) (g) =

∑

i = 1, 2.

χ(1)χ(g),

χ∈Irr(i) (G)

If g ∈ K, then ψ(1) (g) + ψ(2) (g) = −1, by (3), whence ψ(1) (g) ≡ −1 (mod p)

(in 𝕀)

(4)

108 | Characters of Finite Groups 1 so that Irr(1) (G) ≠ 0. If χ ∈ Irr(1) (G), then χ(1) and |K| are coprime. Therefore, by Lemma 6.2, g ∈ Tχ ∪ Z(χ). Since G is simple and nonabelian, it follows that Z(χ) = {1}. Since g ≠ 1, we have g ∈ Tχ , that is, any character in Irr(1) (G) vanishes on K. Consequently, ψ(1) (g) = 0 for g ∈ K, contrary to (4). Kazarin [Kaz8] has proved that, under hypothesis of Lemma 6.3, one has K ⊆ S(G), where S(G) is the solvable radical of G (see §XXVII.13). This is very strong and useful generalization of Lemma 6.3. Theorem 6.4 (Burnside’s p a q b -theorem = Burnside’s two-prime theorem). Let G satisfy |G| = p a q b , where p, q are distinct primes. Then G is solvable. Proof. Let G be a minimal counterexample. Then ab > 0 since prime power groups are solvable. By induction, G is nonabelian simple. Let P ∈ Sylp (G), g ∈ Z(P)# . If K is the G-class of g, then |K| = |G : CG (g)| = q c for some positive integer c. This, however, contradicts Lemma 6.3. Exercise 6.1 (P. Hall, S. A. Chunikhin). A group G that possesses a p󸀠 -Hall subgroup for every prime divisor p of |G| is solvable. Hint. Every normal subgroup and every epimorphic image of G possesses the property. It will therefore suffice to prove that G is not simple. By the product formula, for any distinct prime divisors p, q of |G|, there exists a {p, q}-Hall subgroup M of G. By Theorem 6.4, M contains a normal subgroup of prime power order, say a q-subgroup Q ≠ {1}. If H is a p󸀠 -Hall subgroup of G, then {1} < Q ≤ H G , and now the result follows by induction. Exercise 6.2 (P. Hall). Let π be a set of primes. Then any two maximal π-subgroups of a solvable group G are conjugate in G. See [BJ2, Theorem A.28.4] for a stronger result which also due to Hall. Exercise 6.3 (H. Wielandt). If G = AB, where A, B are nilpotent Hall subgroups of G, then G is solvable. See [BJ2, Theorem A.28.14] for more general result (that result uses the Odd Order Theorem). If H ≤ G and χ is a character of a group G that vanishes on H # , then |H| | χ(1). Indeed, if μ ∈ Irr(H), then the rational integer ⟨μ, χ H ⟩ = |H|−1 ∑ μ(x)χ(x) = |H|−1 μ(1)χ(1) > 0. x∈H

Taking μ ∈ Lin(G), we conclude that |H| | χ(1). Theorem 6.5. Let χ ∈ Irr(G) be faithful, χ(1) = p s > 1, P ∈ Sylp (G). Suppose that, for any g ∈ P# , one has p ∤ |G : CG (g)|. If p ∤ |Z(G)|, then |P| = p s . If, in addition, s = 1, then |P| = p.

III On arithmetical properties of characters | 109

Proof. Let g ∈ P# . Since GCD(χ(1), |G : CG (g)|) = 1, it follows from Lemma 6.2 that g ∈ Z(χ) ∪ Tχ . We have Z(χ) = Z(G), because χ is faithful. Consequently, as o(g) ∤ |Z(G)|, it follows that g ∈ Tχ . Thus, χ vanishes on P# so |P| | χ(1) = p s , by the paragraph preceding the theorem. Since χ(1) | |G| (Frobenius–Molien), it follows that |P| = p s . Now let χ(1) = p. By the above, it suffices to show that P is abelian. Suppose that P is nonabelian. Since χ is faithful of degree p, we get χ P ∈ Irr(P) and {1} < Z(P) = Z(χ P ) ≤ Z(χ) = Z(G), whence p | |Z(G)|, contrary to the hypothesis. Exercise 6.4 ([Isa11, Theorem 3.13]). If P ∈ Sylp (G) is abelian and χ ∈ Irr(G) is faithful of degree p a , then |PZ(G)/Z(G)| = p a . Solution (Zhmud). One may assume that a > 0 (otherwise, G is cyclic and then the desired equality holds). By Proposition 3.3, p a | |G : Z(G)|. Take x ∈ P# ; then P ≤ CG (x) since P is abelian so that GCD(χ(1), |G : CG (x)|) = 1. Then, by Lemma 6.2, either χ(x) = 0 or x ∈ Z(χ) = Z(G). It follows that if ∆ = P ∩ Z(G), then χ vanishes on P − ∆. Then, as it is easy to show, we get |P : ∆| | χ(1) (see also [Isa11, Problem 2.16]. Since P/∆ ≅ PZ(G)/Z(G) ∈ Sylp (G/Z(G)), the result follows since χ(1) | |G/Z(G)|p = |P/∆|. Exercise 6.5 ([Isa11, Problem 3.7]). If P ∈ Sylp (G), χ ∈ Irr1 (G) is faithful of degree p a and CG (P) ≰ P, then G󸀠 < G. Solution (Zhmud). Assume that G󸀠 = G. Since CG (P) ≰ P, it follows that CG (P) contains p󸀠 -elements (apply Sylow’s theorem to PCG (P)). Let {1} < Q < CG (P) be a p󸀠 -subgroup. Assume that x ∈ Q ∩ Z(G) is of order m; then χ(x) = ϵχ(1), where ϵ m = 1. Let Γ be a matrix representation of G affording the character χ. Then Γ(x) = ϵI, where I is the identity matrix. Therefore, det(Γ(x)) = ϵ χ(1) = 1 since G󸀠 = G and det(Γ) ∈ Lin(G). Since χ(1) = p a and p ∤ m, we conclude that ϵ = 1. Therefore, we have χ(x) = χ(1), i.e., x ∈ ker(χ) = {1}. Thus, we obtain that Q ∩ Z(G) = {1}. Let g ∈ Q# , Then P ⊂ CG (g), and so h g = |G : CG (g)| is not divisible by p. Thus, GCD(χ(1), h g ) = 1 so, by Lemma 6.2, either χ(g) = 0 or g ∈ Z(χ) = Z(G). Since g ∈ Q# and Q ∩ Z(G) = {1}, we must have χ(g) = 0, i.e., Q# ⊆ Tχ (the set of zeros of χ). Let o(g) = q be prime; then q ≠ p. Let ϵ = e2πi/q and λ = 1 − ϵ. Then λ is a prime element of the ring J of algebraic integers of the field K = ℚ(ϵ), and λ divides NK/ℚ (λ) = q, where NK/ℚ (λ) is the norm of λ from K to ℚ(λ). It is easily seen that χ(g) ≡ χ(1) (mod λ). Since χ(g) = 0, it follows that p α = χ(1) ≡ 0 (mod λ), and this implies that λ | p in 𝕀, which is impossible since GCD(p, q) = 1. Thus, G󸀠 < G, as asserted.

7 Burnside’s theorems on commutators Let {K1 , . . . , K r } be the set of G-classes (r = k(G)), g i ∈ K i , k1 , . . . , k r the corresponding class sums and h νij the structure constants with respect to the basis {k i }1r of the

110 | Characters of Finite Groups 1 algebra Z(ℂG), i.e., r

(i, j = 1, . . . , r).

k i k j = ∑ h νij k ν ν=1

(1)

Since ω χ ∈ Hom(Z(ℂG), ℂ) for all χ ∈ Irr(G) (see §III.3), it follows from (1) that r

(2)

ω χ (k i k j ) = ω χ (k i )ω χ (k j ) = ∑ h νij ω χ (k ν ). ν=1

As we know, r

χ i (1) = max{|χ i (g α )| | i = 1, . . . , r},

|G| = h = ∑ χ i (1)2 . i=1

It follows from the Second Orthogonality Relation that r h = ∑ |χ i (g α )|2 , h α i=1

where h α = |K α |. Recalling that ω χ (k i ) =

h i χ(g i ) χ(1)

and using (2), one obtains r

h i h j χ(g i )χ(g j ) = χ(1) ∑ h νij χ(g ν )h ν .

(3)

ν=1

α) Multiplying both sides of (3) by χ(g χ(1) and summing over all χ ∈ Irr(G), we use the Second Orthogonality Relation to get

h αij |G| = h i h j

∑ χ∈Irr(G)

χ(g i )χ(g j )χ(g α ) . χ(1)

Thus we have proved the following: Lemma 7.1. We have h αij =

hi hj |G|

∑ χ∈Irr(G)

χ(g i )χ(g j )χ(g α ) . χ(1)

Consequently, given only X(G), one can construct the entire multiplication table of G-classes: The converse is also true. The multiplication table of G-classes provides sufficient information to construct the character table X(G). Let us prove the last assertion. γ Given {h αβ }, we know the algebra Z(ℂG) and all its homomorphisms ω1 , . . . , ω r in ℂ. Thus, the matrix (ω α (k β )) is known. Since h1αβ = h α δ αβ󸀠 , where K β󸀠 = K −1 β (this obtained from Lemma 7.1 and the Second Orthogonality Relation), one can determine i (g α ) allow us to the class sizes h α , and so |G| = ∑rα=1 h α . The equalities ω i (k α ) = h αχχi (1) χ i (g α ) determine the numbers χ i (1) . Since r r |G| 1 χ i (g α ) = ⋅ ∑ h α |χ i (g α )|2 = ∑ ω i (k α ), 2 2 χ (1) χ i (1) χ i (1) α=1 α=1 i

one can determine χ i (1) and so χ i (g α ) (the right-hand side of the displayed formula is known). We assume that K1 = {1}. Then h1ii󸀠 = h i .

III On arithmetical properties of characters | 111

The following four remarkable theorems are due to Burnside. Theorem 7.2. If g, x ∈ G, then g is conjugate to a commutator [x, y] for some y ∈ G if and only if χ(x)2 χ(g) > 0. ∑ χ(1) χ∈Irr(G) Theorem 7.3. We have ∑ χ∈Irr(G)

χ(g) ≠ 0 χ(1)

if and only if g ∈ G is a commutator. The following result is useful for general group theory. It will be interesting to produce its character free proof. Theorem 7.4. If g ∈ G is a commutator, then any generator of the cyclic group ⟨g⟩ is a commutator. Theorem 7.5. If χ ∈ Irr(G), x, y ∈ G, then χ(x)χ(y) =

χ(1) ∑ χ(xy t ). |G| t∈G

Proof of Theorem 7.2. Applying Lemma 7.1 and taking into account that h j󸀠 = h j and χ(g j󸀠 ) = χ(g j ), we get h kj󸀠 j = h2j |G|−1

|χ(g j )|2 χ(g k ) . χ(1) χ∈Irr(G) ∑

(4)

Let x ∈ K j , g ∈ K k . Then it follows from (4) that h kj󸀠 j = h2j |G|−1

|χ(x)|2 χ(g) . χ(1) χ∈Irr(G) ∑

(5)

Suppose that h kj󸀠 j > 0. Then there exist u ∈ K j󸀠 and v ∈ K j such that uv = g. Next, u ∼ x−1 , v ∼ x, g k ∼ g 󳨐⇒ g ∼ x−1 v󸀠 , where v󸀠 = y−1 xy for some y ∈ G. Thus g ∼ x−1 y−1 xy = [x, y] is a commutator. Conversely, if g ∼ [x, y] for some y ∈ G, then g k ∼ x−1 v, where v = y−1 xy ∈ K j , so that g k ∈ K j󸀠 K j . Hence h kj󸀠 j > 0, and the result follows, by (5). Proof of Theorem 7.3. If g ∈ K s , then by (4) and the Second Orthogonality Relation, r

∑ j=1

h sj󸀠 j hj

r

=∑

h j |G|−1

∑

j=1 χ∈Irr(G) r

=

∑

∑ hj

χ∈Irr(G) j=1

|χ(g j )|2 χ(g) χ(1)

|χ(g j )|2 χ(g) χ(g) χ(g) = ∑ ⟨χ, χ⟩ = ∑ . |G|χ(1) χ(1) χ∈Irr(G) χ(1) χ∈Irr(G)

112 | Characters of Finite Groups 1 Therefore, ∑ χ∈Irr(G)

χ(g) ≠ 0 ⇐⇒ h sj󸀠 j ≠ 0 for some j ∈ {1, . . . , r}. χ(1)

It follows from the proof of Theorem 7.2 that the inequality h sj󸀠 j ≠ 0 is equivalent to g = [x, y], where x ∈ K j and y ∈ G. O. Ore has shown that every element of the alternating group An , n > 4 (then An is simple), is a commutator. Proof of Theorem 7.4. Suppose that g ∈ G is a commutator. Then, by Theorem 7.3, we have χ(g) α= ∑ ≠ 0. χ(1) χ∈Irr(G) Let ϵ be a primitive o(g)-th root of 1 and σ ∈ G = Gal(ℚ(ϵ)/ℚ). Then σ(α) ≠ 0. Let g m be a generator of ⟨g⟩. Since GCD(m, o(g)) = 1, there exists an automorphism σ ∈ G such that σ(ϵ) = ϵ m . In that case, by Lemma 4.2, σ(χ(g)) = χ(g m ). Hence ∑ χ∈Irr(G)

χ(g m ) = σ(α) ≠ 0 χ(1)

so g m is a commutator, by Theorem 7.3. We suggest to the reader to produce a character free proof of Theorem 7.4. Proof of Theorem 7.5. We use the notation that was introduced in the first paragraph of this section. Let ω χ be a linear character of the algebra Z(ℂG) associated with χ; χ(z) , z ∈ Z(ℂG). In that case, then ω χ (z) = χ(1) χ(k i k j ) χ(g i )χ(g j ) = ω χ (k i k j ) = ω χ (k i )ω χ (k j ) = h i h j χ(1) χ(1)2 implies χ(g i )χ(g j ) =

χ(1) ⋅ χ(k i k j ), hi hj

and so χ(k i k j ) = χ( =

∑

u∈K i , v∈K j

uv) =

1 ∑ χ(g is g jt ) |CG (g i )||CG (g j )| s,t∈G

−1 |G| |G| ∑ χ(g i g jts ) = ∑ χ(g i g jt ). |CG (g i )||CG (g j )| s,t∈G |CG (g i )||CG (g j )| t∈G

Therefore, χ(g i )χ(g j ) =

χ(1) |G|χ(1) ∑ χ(g i g jt ) = ∑ χ(g i g jt ). h i |CG (g i )||h j CG (g j )| t∈G |G| t∈G

Setting g i = x, g j = y, one obtains χ(x)χ(y) =

χ(1) |G|

∑t∈G χ(xy t ), as required.

III On arithmetical properties of characters | 113

Thus, a knowledge of the character table X(G) will enable us to find all G-classes whose representatives are commutators. Note that X(G) does not enable us to determine the orders of the elements (e.g., the ordinary quaternion group and the dihedral group of order 8 have the same character table but different numbers of involutions). However, it enables us to find all prime divisors of the order of an element (G. Higman, A. Saksonov).

8 Rational groups Recall that a character χ is rational if χ(g) ∈ ℚ for all g ∈ G. An element g ∈ G is rational if all generators of the cyclic subgroup ⟨g⟩ are conjugate in G. Definition. A group G is said to be a ℚ-group (= rational group) if all its characters are rational. As shows the following exercise, a group G is a ℚ-group if and only if all its elements are rational. Exercise 8.1. A group G is a ℚ-group if and only if NG (Z)/CG (Z) ≅ Aut(Z) for any cyclic subgroup Z ≤ G. Thus, the order of a nonidentity ℚ-group is even. Solution. Clearly, NG (Z)/CG (Z) ≅ H ≤ Aut(Z). Now suppose that G is a Q-group. If ν is an integer such that GCD(ν, |G|) = 1, then g ν ∼ g, i.e., g ν = g t , where Z = ⟨g⟩ and t ∈ NG (Z). Therefore, H ≅ Aut(Z). The reverse implication is proved so easy. Nonabelian groups of order 8 are ℚ-groups. Moreover, extraspecial 2-groups are ℚ-groups. A 2-group G of maximal class and order > 8 is not a ℚ-group (indeed, if Z < G is cyclic of index 2, then NG (Z)/CG (Z) is of order 2 < |Aut(Z)|). All symmetric groups are ℚ-groups. Exercise 8.2. The following statements hold: (a) All epimorphic images of a ℚ-group are ℚ-groups. (b) A group G > {1} is an abelian ℚ-group if and only if exp(G) = 2. (c) If G is a ℚ-group, then exp(G/G󸀠 ) ≤ 2. In particular, all nonidentity nilpotent ℚ-groups G are 2-groups (in that case G󸀠 = Φ(G)). Exercise 8.3. The following statements hold: (a) If G > {1} is a ℚ-group, then its order is even. (b) If a ℚ-group G possesses a normal subgroup of prime order p, then p ≤ 3. (c) If a ℚ-group G is supersolvable, then π(G) ∈ {{2}, {2, 3}}. Solution of (b). Assume that L ⊲ G is cyclic of order p > 3. Then G/CG (L) ≅ Cp−1 and p − 1 > 2. Then exp(G/G󸀠 ) is divisible by p − 1, contrary to Exercise 8.2 (c). Note that (a) follows easily from (b). (Hint to (c). The group G contains a normal subgroup of order q, where q is the maximal prime divisor of |G|. We have q ≤ 3, by (b)).

114 | Characters of Finite Groups 1 Exercise 8.4. If G = P ⋅ Q, where P ≅ Ep n , Q ∈ Sylq (G), q ≡ 1 (mod p), Q ⊲ G, then G is supersolvable. Solution. One may assume that G is nonnilpotent. It suffices to show that there is in Z(Q) a P-invariant subgroup N of order q (indeed, then N ⊲ G and G/N will be supersolvable, by induction). Therefore, without loss of generality, one may assume that G = PΩ1 (Z(Q)), i.e., Q = Ω1 (Z(Q)). First let n = 1 and let H ≤ PZ(Q) be minimal nonnilpotent. Since q ≡ 1 (mod p), we get |H|q = q (see Lemma XI.2.1 (d)). Then H ∩ Q is a normal subgroup of G of order q. Now suppose that n > 1. In that case, by Lemma X.2.3, there is x ∈ P of order p that centralizes a cyclic subgroup Y of order q in Q (since, in view of n > 1, G is not a Frobenius group). Then Y ⊲ G, completing the solution. Second solution. As above, it suffices to show that G has a normal subgroup, say N, of order q. One may assume that G is not nilpotent. Let F ≅ Eq k be minimal normal subgroup of G, where k ∈ ℕ; then F ≤ Z(Q) since G is solvable and Q ≤ F(G). Let us prove that k = 1. We consider F as an 𝔽q P-module. This module is irreducible. The module F affords a faithful irreducible 𝔽q -representation ∆ of the abelian group A = P/CP (F). By Schur’s lemma, A is cyclic so |A| = p since A is elementary abelian. Since deg(∆) is the order of q modulo p, it follows that deg(∆) = 1 and so k = 1 (here we make use of Theorem I.7.6). Exercise 8.5. The following statements hold: (a) If G is a ℚ-group having an abelian Sylow 2-subgroup P, then P is elementary abelian. (b) A ℚ-group G with an abelian Sylow 2-subgroup is 2-nilpotent, its normal 2-complement coincides with G󸀠 . (c) Let Z be a normal cyclic subgroup of a ℚ-group G. Then |Z| ∈ {1, 2, 3, 4, 6, 8, 12, 24}. Solution. (a) Assume that L < P is cyclic of order 4. Then NG (L) = CG (L), contrary to Exercise 8.1. (b) By statement (a), P is elementary abelian. Assume that G is nonsolvable. By Exercise 8.2 (c) and Walter’s theorem [Walt], G = M × S, where M is an elementary abelian 2-group and S is a direct product of nonabelian simple groups of the following types: L2 (q), q ≡ 3, 5 (mod 8), or q = 2n , J1 or of Ree type. Then G/M ≅ S and all simple direct factors of S are not ℚ-groups, and this is a contradiction. Thus, G is solvable. We claim that G is 2-nilpotent. Indeed, by Hall-Higman, the 2-length of G is equal to 1. Write G = G/O2󸀠 (G). Assume that G is not a 2-group. Without loss of generality, one may assume that O2󸀠 (G) = {1}. Then P ∈ Syl2 (G) is normal in G, contrary to Exercise 8.2 (c). By Exercise 8.2 (a), G󸀠 = O2󸀠 (G). (By Lemma 8.1, G󸀠 is a 3-group.)

III On arithmetical properties of characters | 115

(c) One has G/CG (Z) = NG (Z)/CG (Z) ≅ Aut(Z). Since Aut(Z) is abelian, it follows that G󸀠 ≤ CG (Z), and since G/G󸀠 is elementary abelian, Aut(Z) must be elementary abelian. By small modification of Exercise 8.3 (b), |Z|2󸀠 ∈ {1, 3}. It is known that the automorphism group of a cyclic 2-group of order 2e > 22 has exponent 2e−2 so that e ≤ 3 and |Z|2 ∈ {1, 2, 4, 8}. Lemma 8.1. If a Sylow 2-subgroup P of a ℚ-group G is abelian, then G is a supersolvable {2, 3}-group. Proof. By Exercise 8.5 (a)–(b), P is elementary abelian and G is 2-nilpotent. Suppose that NG (P) > P; then P is a direct factor of NG (P). Let Z ≠ {1} be a cyclic subgroup of NG (P) of odd order. Since |G : CG (Z)| is odd, we get NG (Z)/CG (Z) ≇ Aut(Z) (the order of Aut(Z) is even), contrary to Exercise 8.1. Thus, NG (P) = P and so G = P⋅Q, a semidirect product with kernel Q, by Burnside’s normal p-complement theorem. It now remains to show that Q is a 3-group (see Exercise 8.4). We use induction on |G|. One may assume that Q is not a 3-group and that G possesses only one minimal normal subgroup R, |R| = p n , p > 3. As G/R is a ℚ-group, we have that G/R is a {2, 3}-group, by induction. By Exercise 8.4, PR is supersolvable, so there exists a normal subgroup T of order p in PR. By Exercise 8.1, NG (T)/CG (T) ≅ Cp−1 , whence NG (T)󸀠 ≤ CG (T). Since P ≤ NG (T), we have PCG (T) ⊴ NG (T), and NG (P) = P implies that PCG (T) = NG (T), by Frattini’s lemma. Therefore, NG (T)/CG (T) ≅ P/(P ∩ CG (T)) is an elementary abelian 2-group. Since the first of these groups is isomorphic to Cp−1 , we see that p − 1 = 2, p = 3, contrary to the assumption. Lemma 8.2. Let P be a nonabelian metacyclic Sylow 2-subgroup of a ℚ-group G. Then |P| = 8. Proof. We proceed by induction on |G|. Let Z be a normal cyclic subgroup of P such that P/Z is cyclic. Since P is nonabelian, |Z| > 2. Since NG (Z)/CG (Z) ≅ Aut(Z) is a 2-group and NG (Z)/CG (Z) has a cyclic Sylow 2-subgroup, it follows that |Z| = 4 (indeed, if |Z| > 4, then Aut(Z) is noncyclic). We have P = ⟨x, Z⟩. Suppose that |P| > 23 . Then P has no cyclic subgroup of index 2 (otherwise, G is not a ℚ-group), so that ⟨x⟩ ∩ Z = {1}. Since P is nonabelian, ⟨x⟩ is nonnormal in P. By Exercise 8.1, |⟨x⟩| ≤ 8, because |P : ⟨x⟩| = 4 ≥ |Aut(⟨x⟩)|. Suppose that |⟨x⟩| = 8. Again by Exercise 8.1, we get NG (⟨x⟩)/CG (⟨x⟩) ≅ Aut(⟨x⟩) ≅ E4 . Let P1 ∈ Syl2 (NG (⟨x⟩)). It follows from |P| = |P1 | that P1 ∈ Syl2 (G). Since P1 is generated by two elements and P1 /⟨x⟩ ≅ E22 , we have ⟨x⟩ = Φ(P1 ) = ⟨y2 | y ∈ P1 ⟩, so that P1 contains a cyclic subgroup of index 2. Since P1 ≅ P (Sylow’s theorem), this is a contradiction. Thus |⟨x⟩| < 8, |P| ≤ 16. Assume that |P| = 16. Then P = ⟨a, b | a4 = b4 = 1, a b = a3 ⟩ is the unique nonabelian metacyclic group of order 16 and exponent 4. Suppose now that G is solvable. By induction, O2󸀠 (G) = {1}. Then R = O2 (G) > {1} and CG (R) ≤ R (Hall–Higman). As P is minimal nonabelian, we get |R| > 4. If R = P, then, since an abelian group of type (4, 2) does not admit a nontrivial automorphism of odd order and P/P󸀠 is of type (4, 2), it follows that Aut(P) is a 2-group, and we

116 | Characters of Finite Groups 1 get G = P, a contradiction since P is not a ℚ-group. Thus R < P so that R is abelian of type (4, 2). Since G/CG (R) = G/R is not a 2-group, we get a contradiction since Aut(R) ≅ D8 . It remains to show that if G is not solvable, then |P| = 8. Suppose that this is false and |P| = 16. We know that |P󸀠 | = 2, Z(P) ≅ E4 and Aut(P) is a 2-group. Suppose that G contains an element x of order 4q, where q is an odd prime. Since Aut(⟨x⟩) contains a subgroup of type (2, 2), a Sylow 2-subgroup of NG (⟨x⟩)/CG (⟨x⟩) is of type (2, 2). Hence we may assume that P ∈ Syl2 (NG (⟨x⟩)). Obviously, x q ∈ P. Then P/⟨x q ⟩ is of type (2, 2). Since P is metacyclic, we have ⟨x q ⟩ = Φ(P) = ℧1 (P) = ⟨y2 | y ∈ P⟩ and P possesses a cyclic subgroup of index 2, a contradiction. Thus x does not exist. This shows that NG (P) = P. The group G is not 2-nilpotent (otherwise, we must have exp(G/G󸀠 ) > 2, contrary to Exercise 8.2 (b)). Therefore, by Frobenius’ normal p-complement theorem, G contains a minimal nonnilpotent subgroup H of even order with an invariant Sylow 2-subgroup R. Without loss of generality one may assume that R < P. Then R = Z(P) ≅ E4 , P < NG (R) (note that H < NG (R); clearly, H is a {2, 3}-subgroup; see Lemma XI.2.1 (d)). One may assume that R < P; then we have R = Z(P). Since P/R is self-normalizing in NG (R)/R, the latter has the normal 2-complement L/R, by Burnside’s normal p-complement theorem. Let Q1 /R be a Sylow 3-subgroup of L/R (we have Q1 /R > {1} because H < NG (R)). Set K = PQ = QP, where Q ∈ Syl3 (Q1 ). Since QR/R = Q1 /R is normal in K/R and P/R is abelian of type (2, 2), it follows that P/R contains an involution zR which commutes with an element yR of order 3 of Q1 /R (note that o(z) = 4 since R = Ω1 (P)). One may assume that o(y) = 3. Consider the subgroup F = ⟨z, R, y⟩ = ⟨y⟩ ⋅ ⟨z, R⟩, a semidirect product with kernel ⟨z, R⟩. Since ⟨z, R⟩ is abelian of type (4, 2), it follows that F is abelian (indeed, Aut(⟨z, R⟩) ≅ D8 ). Therefore G contains the element yz = zy of order 12 = 4 ⋅ 3, contrary to what has been proved already. Remark. Making use of the Brauer–Suzuki theorem on groups with generalized quaternion Sylow 2-subgroup (Chapter X) and the classification of nonsolvable groups with a dihedral Sylow 2-subgroup (Gorenstein–Walter), one can now easily classify the nonsolvable ℚ-groups with a nonabelian metacyclic Sylow 2-subgroup. We leave this to the reader. Lemma 8.3. Let P = D8 be a Sylow 2-subgroup of a nonnilpotent solvable ℚ-group G. If q is an odd prime divisor of |G|, then q ≡ 3 (mod 4). Proof. Suppose that the lemma is true for all groups of order less than |G|. Applying Exercise 8.1 to a cyclic subgroup Z of odd order centralizing P, we see that Z = {1}. Therefore, since Aut(P) ≅ P, we get NG (P) = P. Let R be a minimal normal subgroup of G, |R| = p n ; R ≠ P by what has just been said. If R < P, then P/R is self normalizing abelian of order 2 so G/R is a {2, 3}-group (Lemma 8.1), and we are done.

III On arithmetical properties of characters | 117

Now suppose that p > 2. By induction, it suffices to prove that p ≡ 3 (mod 4). One may assume that R is the unique minimal normal subgroup of G. Since PR is not a Frobenius group (Lemma X.2.3), it contains a cyclic subgroup Z of order 2p. By Exercise 8.1, NG (Z)/CG (Z) ≅ Cp−1 . If 4 | p − 1, then a Sylow 2-subgroup of NG (Z), being an extension of a group of order 2 by a cyclic group of order > 2, is abelian of order 8, contrary to the Sylow theorem. Thus, 4 ∤ p − 1, whence p ≡ 3 (mod 4). The symmetric group S4 satisfies the hypothesis of Lemma 8.3, so the groups of that lemma are not necessarily 2-nilpotent. If G of Lemma 8.3 is not 2-nilpotent, then |G/G󸀠 | = 2. Lemma 8.4 (R. Gow). If G is a solvable ℚ-group then π(G) ⊆ {2, 3, 5}. The only known proof of this deep result involves modular theory (see [Gow2] and [Kle]). As an immediate consequence, we get: Corollary 8.5. If G is as in Lemma 8.3, then π(G) ⊆ {2, 3}. We suggest to the reader to produce a proof of Corollary 8.5 independent of Lemma 8.4. Below,(H, N) denotes a Frobenius group with kernel N and complement H (see Chapter X). Theorem 8.6. Let G be a solvable nonnilpotent ℚ-group and let P ≅ Q8 be a Sylow 2-subgroup of G. Then G = (Q8 , E3n ). Proof. Suppose that the theorem is valid for groups of order < |G|. We claim that G is 2-nilpotent. Indeed, 2 divides |G/G󸀠 | (Exercise 8.2 (c)) and so G󸀠 has a cyclic Sylow 2-subgroup hence it has a normal 2-complement H; then H is a normal 2-complement of G. Since P ≅ Q8 , it follows from Exercise 8.1 that G possesses no element of order 4q for all primes q > 2. Let x be an involution of P and suppose that CG (x) > P. By the above, CG (x) = PQ, where Q > {1} is q-subgroup, a prime q > 2. By Exercise 8.4, the quotient PQ/⟨x⟩ is supersolvable. It follows that CG (x) contains a subgroup PQ1 , where |Q1 | = q. Then CPQ1 (Q1 ) is cyclic of order 4q, contrary to what that has just been said. Thus, CG (x) = P. It follows that G = (P, Q), a Frobenius group with kernel Q, and involution x inverts Q. By Burnside, Q is abelian. If Z < Q is cyclic, then NG (Z)/Z is a group of exponent dividing exp(P/⟨x⟩) = 2, and we conclude that exp(Q) = 3. It follows from [FeiS] that the groups Sp6 (2) and O∗8 (2)󸀠 are the only nonabelian simple ℚ-groups. Exercise 8.6. Direct products of ℚ-groups are ℚ-groups. Hint. Use Theorem IV.8.1. Exercise 8.7. The groups PSL(2, 7) and PGL(2, 7) are not ℚ-groups. Hint. The group PGL(2, 7) has a cyclic subgroup of order 8. Apply Exercise 8.1 to subgroups of order 7 in PSL(2, 7).

118 | Characters of Finite Groups 1 Exercise 8.8. Classify the Frobenius groups (see Chapter X) that are ℚ-groups. Hint. Let G = (H, F) be a Frobenius ℚ-group with kernel F and complement H (see Chapter X), P ∈ Syl2 (H); then H ≅ G/F is a ℚ-group. The subgroup of order 2 from P is contained in Z(H) so H = P ∈ {C2 , Q8 }. If H ≅ C2 , then F is elementary abelian so, by Exercise 8.4, F ≅ E3n . If H ≅ Q8 , then again F ≅ E3n (Theorem 8.6). Exercise 8.9. Is it true that all PSL(2, p n ), p n > 3, are not ℚ-groups? A p-group G is said to be extraspecial if G󸀠 = Z(G) = Φ(G) is of order p. Exercise 8.10. Study the solvable ℚ-groups G with extraspecial Sylow 2-subgroups P of order > 8. Describe the structure of P provided G is not 2-nilpotent. Exercise 8.11. Let G = Q ⋅ E, where Q ≅ Q8 and let E ≅ E3n be minimal G-invariant, n > 1. Prove that n = 2. Solution. Clearly, Q is maximal in G. If the involution u ∈ Q centralizes a nonidentity element of E, then CG (u) = G. By Exercise 8.4, G/⟨u⟩ so G is supersolvable, a contradiction since n > 1. Thus, G = (Q, E) is a Frobenius group. In that case, u inverts E. Let a, b be generators of two distinct cyclic subgroups of order 4 in Q; then ⟨a, b⟩ = Q and a2 = b2 = u. Take x ∈ E# . Then ⟨x⟩a = ⟨y⟩ and ⟨x⟩ ≠ ⟨y⟩ (otherwise, u centralizes x since Aut(⟨x⟩) ≅ C2 ). But 2

⟨x⟩ = ⟨x⟩u = ⟨x⟩a = (⟨x⟩a )a = ⟨y⟩a . It follows that the subgroup ⟨x, y⟩ of order 32 is ⟨a⟩-invariant. Now let ⟨x⟩b = ⟨z⟩, where ⟨x⟩ ≠ ⟨z⟩. As above, we have ⟨x⟩ = ⟨z⟩b so the subgroup ⟨x, z⟩ of order 32 is ⟨b⟩-invariant. Assume that ⟨x, y⟩ ≠ ⟨x, z⟩. Then the subgroup ⟨x, y, z⟩ of order 33 is Q-invariant. By Maschke’s theorem, ⟨x, y, z⟩ = ⟨x, y⟩ × ⟨w⟩, where ⟨w⟩ is a-invariant. Then a2 = u centralizes w ∈ E# , which is s contradiction. Thus, ⟨x, y⟩ ⊲ G. Exercise 8.12. Is it true that Exercise 8.11 holds in the case where E ≅ Ep n with p ≡ 3 (mod 4)? Consider the case where p ≡ 1 (mod 4).

9 On characters of p-groups We shall write δ(G) = (a0 ⋅ d0 , a1 ⋅ d1 , . . . , a t ⋅ d t ) to mean that cd(G) = {χ(1) | χ ∈ Irr(G)} = {d0 = 1 < d1 < ⋅ ⋅ ⋅ < d t } and Irr(G) contains exactly a i characters of degree d i , i ∈ {0, 1, . . . , t}. Obviously, a0 = |G : G󸀠 |,

|G| = a0 + a1 d21 + ⋅ ⋅ ⋅ + a t d2t .

In this section |G| = p n , a0 = p k , d i = p c i , i ∈ {1, . . . , t}, c1 < ⋅ ⋅ ⋅ < c t . Then p n = p k + a1 p2c1 + ⋅ ⋅ ⋅ + a t p2c t .

(1)

III On arithmetical properties of characters | 119

Assume that G is nonabelian so that t > 0. Since (1) implies k ≥ 2c1 , one has p n−2c1 = p k−2c1 + a1 + a2 p2(c2 −c1 ) + ⋅ ⋅ ⋅ + a t p2(c t −c1 ) .

(2)

Lemma 9.1. If G is a nonabelian p-group such that a1 < p2 − p, then it is extraspecial. Proof. Suppose that t > 1. It follows from (2) that p k−2c1 + a1 ≡ 0 (mod p2 ), whence a1 ≥ p2 − p, a contradiction. Thus, t = 1 and p n−2c1 = p k−2c1 + a1 . Since a1 < p2 − p and k < n, it follows that a1 = p k−2c1 (p n−k − 1) 󳨐⇒ k = 2c1 , a1 = p − 1, n = 2c1 + 1. If χ ∈ Irr1 (G), then, by Lemma 5.4, p n−1 = p2c1 = χ(1)2 ≤ |G : Z(G)| 󳨐⇒ |Z(G)| = p = |G󸀠 |, Z(G) = G󸀠 . If x, y ∈ G, then 1 = [x, y]p = [x, y p ], whence y p ∈ Z(G) for all y ∈ G. Therefore, we have ⟨y p | y ∈ G⟩G󸀠 = Φ(G) ≤ Z(G) = G󸀠 ≤ Φ(G), hence Φ(G) = Z(G) = G󸀠 is of order p. It follows that G is extraspecial. The following lemma is of independent interest. Lemma 9.2. If χ ∈ Irr(G) and G/Z(χ) is abelian, then |G/Z(χ)| = χ(1)2 . Proof. One may assume that the group G is nonabelian and χ is faithful (indeed, Z(χ)/ker(χ) = Z(G/ker(χ)). Then Z(χ) = Z(G). Let x ∈ G − Z(G). Then xh ≠ hx for some h ∈ G. Since G/Z(G) is abelian, it follows that 1 ≠ [x, h] = z ∈ Z(G). We have χ(z) = ϵχ(1), where ϵ ≠ 1 is a root of 1. Note that χ(xz) = χ(x) because xz = h−1 xh. If D is a representation affording χ, then D(xz) = D(x)D(z) = ϵD(x) 󳨐⇒ χ(x) = χ(xz) = ϵχ(x), whence, using the fact that ϵ ≠ 1, we get χ(x) = 0, i.e., χ vanishes on G − Z(G). Therefore, ⟨χZ(G) , χZ(G) ⟩ = |G : Z(G)|. On the other hand, χZ(G) = χ(1)λ, where λ ∈ Lin(Z(G)). This implies that ⟨χZ(G) , χZ(G) ⟩ = χ(1)2 ⟨λ, λ⟩ = χ(1)2 . Thus, |G : Z(G)| = χ(1)2 . Lemma 9.3. If |G| = p n = p k + a1 p2c1 , where |G : G󸀠 | = p k and 0 < a1 < p2 , then one of the following holds: (a) G is extraspecial. (b) n = 2c1 + 2, k = 2c1 + 1, a1 = p2 − p, |Z(G)| = p2 , Φ(G) ≤ Z(G). (c) n = 2c1 + 2, k = 2c1 , |G󸀠 | = p2 , a1 = p2 − 1, |Z(G)| ≤ p2 . Proof. By hypothesis, G is nonabelian so k < n. Since k ≥ 2c1 , we get p n−2c1 = p k−2c1 + a1 󳨐⇒ p k−2c1 (p n−k − 1) = a1 < p2 . It follows that k − 2c1 ≤ 1, n − k ≤ 2.

(3)

120 | Characters of Finite Groups 1 Case (i): Let k − 2c1 = 0, i.e., k = 2c1 . If n − k = 1, then a1 = p − 1 and G is extraspecial (Lemma 9.1). Let n − k = 2. Then |G󸀠 | = p2 , a1 = p2 − 1 and |G : Z(G)| ≥ p2c1 = p k = p n−2 󳨐⇒ |Z(G)| ≤ p2 , so G is as in (c). Case (ii): Let k − 2c1 = 1. Then n − k = 1,

|G󸀠 | = p,

p2 = p + a1 ,

a1 = p2 − p.

Since G is not extraspecial (Lemma 9.1), we get |Z(G)| > p. Next, |G : Z(G)| ≥ p2c1 = p n−2 󳨐⇒ |Z(G)| = p2 . If x, y ∈ G, then 1 = [x, y]p = [x, y p ] hence ⟨x p | x ∈ G⟩ ≤ Z(G), and so Φ(G) ≤ Z(G). Thus, G is as in (b). We will now present without proof several results about characters of finite p-groups, related to those already proved. The following results (from Theorem 9.4 to Proposition 9.15) are taken from [Man4]. Analysis of (1) leads to the following (we use above introduced notation): Theorem 9.4. If G is a nonabelian p-group, then one of the following holds: (a) a1 = p − 1, G is extraspecial. (b) k = 2c1 , a1 ≡ −1 (mod p2 ). (c) k = 2c1 + 1, a1 ≡ −p (mod p2 ). (d) k ≥ 2c2 + 2, a1 ≡ 0 (mod p2 ). Of course, Theorem 9.4 (a) is just Lemma 9.1. The proofs of most of the other results require familiarity with many deep results of theory of p-groups. Definition. A group G is of Frobenius type with respect to N ⊲ G if any two elements of an arbitrary coset xN ≠ N are conjugate in G. If a p-group G is of Frobenius type with respect to G󸀠 , it is simply said to be of Frobenius type. We say that a group G is of type (b) if it meets the description in Theorem 9.4 (b). Similarly, we define groups of types (c) and (d). Given n, put Kn (G) = [G, G, . . . , G] (n times). The characteristic subgroup Kn (G) is the n-th member of the lower central series of G. Proposition 9.5. A group G is of type (b) if and only if G/K3 (G) is of Frobenius type. Proposition 9.6. If G is of type (b), then |G󸀠 : K3 (G)| ≤ p2c1 . If K3 (G) > {1}, then one has |G󸀠 : K3 (G)| = p c1 and |K3 (G) : K4 (G)| ≤ p2c1 . In addition, a1 ≥ p c1 − 1. Proposition 9.7. If G is of Frobenius type and of class 2 or 3, then it is extraspecial or of type (b). Theorem 9.8. If p = 2, then G is of type (b) if and only if G is of Frobenius type and either of class 2 or of maximal class.

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Theorem 9.9. One has a1 = p2 − p if and only if one of the following holds: (a) |G󸀠 | = p and |Z(G)| = p2 . (b) |G : G󸀠 | = p3 , |G󸀠 : K3 (G)| = p, p c1 = d1 = p. If H is a maximal subgroup of G, then either H 󸀠 = G󸀠 or H 󸀠 = K3 (G). Maximal subgroups H such that H 󸀠 = G󸀠 exist if and only if d(G) = 3; there always exist maximal subgroups H satisfying H 󸀠 = K3 (G). Corollary 9.10. There exist exactly seven 3-groups that have exactly 8 = 32 − 1 irreducible characters of degree 3. Of these, four are of order 34 and class 3, and three of order 35 and class 4 (i.e., all these groups are of maximal class). Corollary 9.11. The following conditions on a p-group G are equivalent: (a) G has exactly p2 − 1 irreducible characters of degree p. (b) |G : K4 (G)| = p4 and C󸀠 = K4 (G), where C = CG (G󸀠 /K4 (G)). (c) Either G is of order p4 and class 3, or |G : K5 (G)| = p5 and G/K5 (G) has no abelian maximal subgroups. Proposition 9.12. Let G be a metabelian p-group with exactly p2 − 1 irreducible characters of degree p. Then G is of maximal class. For any given p, there exist only finitely many groups with this property. If G is as in the conclusion Proposition 9.12 and R be a G-invariant subgroup of index p2 in G󸀠 = Φ(G), then all nonlinear irreducible characters of G = G/R have 󸀠 degree p. Then |Irr(G)| = p12 (|G| − |G : G |) = p2 − 1. It follows that the intersection of kernels of all irreducible characters of G of degree p is equal to R. Theorem 9.13. The number p c1 = d1 is the least index of a subgroup H of a p-group G such that G󸀠 > H 󸀠 . The irreducible characters of G of degree d1 are exactly the irreducible characters induced from linear characters of such subgroups of minimal index, and they are not induced from linear characters of other subgroups. If H is such a subgroup of minimal index, then H ≥ G󸀠 , and H has linear characters that induce irreducible characters of G. The intersection of the kernels of all irreducible characters of degree d1 of G is the intersection of all subgroups H 󸀠 , where H runs over the set of all subgroups of minimal index such that H ≥ G󸀠 > H 󸀠 . Given such a subgroup H and an irreducible character χ of degree d1 , χ is induced from a linear character of H if and only if ker(χ) ≥ H 󸀠 . Proposition 9.14. Let G be a p-group having exactly p2 − 1 irreducible characters of each of the degrees p, p2 , . . . , p r−1 . Then r ≤ 12 (p + 1) and G/K2r+1 (G) is of maximal class. If K2r (G) > {1}, then G has characters of degree p r , and the number a r of such characters is p − 1 if |G| = p2r+1 , and ≡ p2 − 1 (mod p2 (p − 1)) otherwise. If r = 21 (p + 1), then G is of maximal class. Proposition 9.15. If c1 = 2 and a1 = p2 − 1, then G/K4 (G) is of Frobenius type and |G : G󸀠 | = p4 . Remark (Mann). We claim that a i ≡ 0 (mod p − 1) for all i ≥ 1. Let χ ∈ Irr1 (G) and let an element x ⋅ ker(χ) be central of order p in G/ker(χ). Then in the representation

122 | Characters of Finite Groups 1 affording χ, the element x is represented by a matrix ϵ ⋅ Iχ(1) , where ϵ is a primitive p-th root of 1. In this case, the number of algebraic conjugates of χ is divisible by the number of algebraic conjugates of ϵ, which is p − 1 (indeed, ϵ is a root of the irreducible polynomial 1 + X + ⋅ ⋅ ⋅ + X p−1 ∈ ℚ(X)). Summing over all characters of degree d i , we get our claim. The following problems are still open: Problem 1. Classify all 2-groups for which a1 = 4. Problem 2. What is the structure of a p-group such that the intersection of kernels of all its nonlinear irreducible characters of minimal degree is trivial? In the following theorem we present a classification of the p-groups G for which the number n(G) = a1 + ⋅ ⋅ ⋅ + a t of nonlinear irreducible characters does not exceed p2 (if p > 2, then we have n(G) ≡ 0 (mod p − 1), so that, in the case under consideration, a1 + ⋅ ⋅ ⋅ + a t < p2 ). Theorem 9.16. Let G be a nonabelian group of order p n such that δ(G) = (p k ⋅ 1, a1 ⋅ p c1 , . . . , a t ⋅ p c t ), n(G) = a1 + ⋅ ⋅ ⋅ + a t ≤ p2 . Then one of the following holds: (a) t = 1, G is extraspecial, a1 = p − 1. (b) t = 1, n = 2c1 + 2, a1 = p2 − 1, |G󸀠 | = p2 , |Z(G)| = p. (c) t = 1, n = 2c1 + 2, a1 = p2 − p, |G󸀠 | = p, |Z(G)| = p2 . (d) t = 1, n = 2c1 + 3, a1 = 4, p = 2, |G󸀠 | = 2, |Z(G)| = 8. (e) t = 1, n = 2c1 + 2, a1 = p2 − 1, G is special, |Z(G)| = p2 . (f) t = 2, n = 2c1 + 3, a1 = p2 − p, a2 = p − 1, |G󸀠 | = p2 , |Z(G)| = p. (g) t = 2, p = 2, n = 2c1 + 3, |G󸀠 | = 8, |Z(G)| = 2, a1 = 3, a2 = 1. The proof of Theorem 9.16 is quite elementary, and is therefore left to the reader as an exercise. Exercise 9.1. Classify the p-groups G with n(G) ≤ p2 + 1. Is it true that t ≤ 2? Exercise 9.2. Is it true that if G is a p-group with a1 + a2 ≤ p2 , then t ≤ 2? Exercise 9.3. Classify the p-groups G with a1 = p. (By Lemma 9.1, p = 2. See also the remark following Proposition 9.15.) Exercise 9.4. Classify the p-groups G with k(G) = |G : G󸀠 | + kG (G󸀠 ), where kG (G󸀠 ) is the number of G-classes contained in (G󸀠 )# . Exercise 9.5. Classify the p-groups G with k(G) = |Z(G)| + k(G/Z(G)) − 1. The groups G in Exercise 9.4 are groups of Frobenius type, while those in Exercise 9.5 are groups of Frobenius type with respect to Z(G). Since p n − p k −n(G) ≡ 0 (mod p −1),

III On arithmetical properties of characters | 123

it follows that n(G) ≡ 0 (mod p − 1). (This is a special case of the remark following Proposition 9.15.) Lemma 9.17 (= Lemma VII.1.1 (Kazarin)). Let G be a p-group with |G󸀠 | = p k . If k is even, then n(G) ≥ 21 (p2 − 1)k. If k is odd, then n(G) ≥ 12 (p2 − 1)(k − 1) + (p − 1). We now state some results on 2-groups G with n(G) < 6. We can extract the description of 2-groups with n(G) ≤ 3 from Theorem 9.16 in more convenient form. Lemma 9.18. Let G be a nonabelian 2-group with n(G) ≤ 3. Then one of the following holds: (a) n(G) = 1 and G is an extraspecial 2-group of order 22n+1 . (b) n(G) = 2, |G| = 22n+2 , |G󸀠 | = 2, |Z(G)| = 4 and cd(G) = {1, 2n }. (c) n(G) = 3, G is either a special group of order 22n+4 with |Z(G)| = 4, cd(G) = {1, 2n }, or a 2-group of maximal class and order 16, or |G| = 25 , |Z(G)| = 2, |G󸀠 | = 4 and cd(G) = {1, 2, 4}. Using this result, E. I. Chankov (Kazarin’s student) has described finite 2-groups with n(G) ∈ {4, 5, 6}. Independently, similar results were obtained by A. Mann. Theorem 9.19. Let G be a 2-group. If n(G) = 4, then |G| = 22n+3 , |G󸀠 | = 2, |Z(G)| = 8, cd(G) = {1, 2n }. Iff n(G) = 5, then |G| = 26 , |G󸀠 | = 8, cd(G) = {1, 2, 4}. The example of a group G with n(G) = 5 is as follows: G = ⟨a, b, c | a4 = b4 = c4 = 1, [a, b] = 1, a c = ab, b c = a2 b⟩.

10 On the class equation In this section we state some results concerning the sizes (or lengths) of conjugacy classes of p-groups. The results are similar to those of §III.9. Let G be a group, let CL(G) = {K0 , K1 , . . . , K r } be the set of conjugacy classes of G (= G-classes), let |K i | = h i , and let h0 = 1, h1 , . . . , h s be the distinct numbers among {h i }0r , 1 = h0 < h1 < h2 < ⋅ ⋅ ⋅ < h s . We write h(G) = {b0 ⋅ 1, b1 ⋅ h1 , . . . , b s ⋅ h s } to express the fact that there are exactly b i classes of size h i (i ∈ {0, 1, . . . , s}). Obviously, b0 = |Z(G)|. The equality b0 + b1 h1 + ⋅ ⋅ ⋅ + b s h s = |G|

(1)

is called the class equation. We have h i | |G| for all i. Let G be a group of order p n , b0 = p z , h i = p d i (i > 0). Then h(G) = {p z ⋅ 1, b1 ⋅ p d1 , . . . , b s ⋅ p d s }, where p n = p z + b 1 p d1 + ⋅ ⋅ ⋅ + b s p d s ,

d1 < ⋅ ⋅ ⋅ < d s .

(2)

The group G is abelian if and only if s = 0. Suppose that s > 0. Since d1 ≤ z, by (2),

124 | Characters of Finite Groups 1 one obtains p n−d1 = p z−d1 + b1 + b2 p d2 −d1 + ⋅ ⋅ ⋅ + b s p d s −d1 ,

(3)

p z−d1 + b1 ≡ 0 (mod p).

(4)

whence Suppose that b1 < p. Then z = d1 , b1 = p − 1 and n = d1 + 1. If s = 1, then we have p n−d1 = p z−d1 + b1 . It follows that b1 = (p − 1)p z−d−1 , n = z + 1. Exercise 10.1. Classify the p-groups G with b1 , . . . , b s ≤ p. Exercise 10.2. Study the p-groups G with b1 + b2 ≤ 2p − 2. Exercise 10.3. Let {1} < A ⊲ G, and let kG (G − A) be the number of G-classes contained in G − A. Study the pairs A < G such that kG (G − A) = k(G/A). Exercise 10.4. Study the p-groups of order p n satisfying h(G) = {p z ⋅ 1, b1 ⋅ p d1 , . . . , b s ⋅ p d s }, where b1 < p, b2 , . . . , b s−1 ≤ p, 1 < d1 < ⋅ ⋅ ⋅ < d s . Exercise 10.5. Prove that p − 1 divides GCD(b1 , . . . , b s ). It would be interesting to obtain a sharp upper bound for the order of a p-group having at most r conjugacy classes. This is part of a more general question concerning arbitrary finite groups. More information about the class number of p-groups can be found in [BZ3, Chapter 31].

11 Mann’s theorems This section (besides of exercises) is taken from a paper of A. Mann. Theorem 11.1 (Mann). Let the two greatest sizes of the conjugacy classes of a group G be two consecutive integers n and n + 1, where n > 1. Then G/Z(G) = (Cp a −1 , Ep a ) is a Frobenius group with kernel Ep a and complement Cp a −1 , n + 1 = p a and the preimage in G of the kernel of G/Z(G) is abelian. Any conjugacy class of G is of length 1, n or n + 1. Exercise 11.1. Let A and B be two G-classes such that GCD(|A|, |B|) = 1. Then G = CG (a)CG (b), where a ∈ A, b ∈ B. Solution. By hypothesis, GCD(|G : CG (a)|, |G : CG (b)|) = GCD(|A|, |B|) = 1 so, by the product formula, |G|p | |CG (a)CG (b)| for all p ∈ π(G). Exercise 11.2 (Burnside). If A and B are G-classes such that GCD(|A|, |B|) = 1, then AB is also a G-class.

III On arithmetical properties of characters | 125

Solution. If a ∈ A, b ∈ B, z ∈ AB, then, by Exercise 11.1, z = a x b y for some x ∈ CG (b), y ∈ CG (a). In that case, z = a x b y ∼ ab yx

−1

= ab uv

(u ∈ CG (b), v ∈ CG (a)),

and finally, z ∼ ab v (v ∈ CG (a)). We see that z ∼ ab v = (ab)v ∼ ab. Proof of Theorem 11.1 (the original proof was partially reworked by the third author). Let A, B be G-classes of sizes n, n + 1, respectively, n > 1 (by hypothesis, G has no class of size > n + 1). By Exercise 11.2, AB is a G-class since GCD(n, n + 1) = 1. Since |AB| ≥ |B| = n + 1 and n + 1 is the greatest size, it follows that |AB| = n + 1. Then, by the same argument, A−1 AB is a G-class. And since B ⊆ A−1 AB (indeed, 1 ∈ AA−1 ), it follows that A−1 AB = B. Given S ⊂ G, put ker(S) = {x ∈ G | xS = S}. Then ker(S) ≤ G and it follows from ker(S)S = S that S is the union of the right cosets of ker(S), and we conclude that |ker(S)| | |S|. It is easily checked that if S is a normal subset of G, then ker(S) ⊴ G. By the result of the first paragraph, A−1 A ⊆ N = ker(B). Since |N| ≥ |A| = n and |N| | |B| = n + 1, we conclude that |N| = n + 1. Clearly, N = ⟨A−1 A⟩ implies N ⊲ G since A is a G-invariant set. Therefore, we have B = yN = Ny for all y ∈ B (indeed, by the previous paragraph, B is a union of cosets of N and |N| = |B|; note that 1 ∈ ̸ B), and A ⊂ xN for all x ∈ A (indeed, if x ∈ A, then x−1 A ⊂ A−1 A ⊂ N hence A ⊂ xN). It follows that xN − A = {z}, z ∈ G. Since, for every v ∈ G, one has A = A v ⊂ x v N, it follows that x v N = xN, i.e., the set xN is G-invariant. Then {z} = xN − A is also G-invariant since xN and A are G-invariant, whence z ∈ Z(G). Therefore, we have xN = zN and z ∈ ̸ A. Consequently, A = zN # . Multiplying this by z−1 , we see that N # is a G-class (this follows from Exercise 11.2). Therefore, without loss of generality, one may assume that A = N # . This may be the case only if N is an elementary abelian p-group for some prime p, n + 1 = |N| = p a . Since N # is a G-class, it follows that N is a minimal normal subgroup of G. Let x ∈ A, C = CG (x). Then |G : C| = |A| = n > 1, and therefore there exists an element y ∈ G − ⋃t∈G C t (indeed, proper conjugate subgroups do not cover G). It follows that y acts on A in a fixed-point-free manner. Therefore, putting D = CG (y), we get D ∩ N = {1} since A = N # , whence, by the product formula, |G : D| ≥ |ND : D| = |N| = n + 1. Since |G : D| = |G : CG (y)| is the size of a G-class containing y, we get |G : D| = n + 1. Let Z/N = Z(G/N) (by the above, we have Z > N). Set G = G/N. It follows from |G : D| = n + 1 = |N| and D ∩ N = {1} that ND = G. Therefore, G = D = CG (y) ≤ CG (y), where y = yN. It follows that CG (y) = G, and hence y ∈ Z(G) so that y ∈ Z. Thus, we have G − ⋃t∈G C t ⊆ Z, where C = CG (x), and we conclude that G = Z ∪ (⋃ C t ) ⊆ ⋃ (ZC)t , t∈G

(1)

t∈G

whence ZC = G (the conjugates of a proper subgroup do not cover a finite group) and

126 | Characters of Finite Groups 1 so ZC = G. Here N ≤ Z ∩ C and, since Z = Z/N = Z(G/N), it follows that C ⊲ G. By (1), G = Z ∪ C, so G = Z since C < G, i.e., G/N = Z/N = Z(G/N). In particular, the quotient group G/N is abelian so G󸀠 = N (since G󸀠 > {1} and N is a minimal normal subgroup of G). Since C ⊲ G and A = N # , it follows that C = CG (N).

(2)

Indeed, C = CG (x), where x ∈ A, so C = C t = CG (x t ) for all t ∈ G, and (2) follows since A = N # is a G-class. Let G/C act on N via conjugation. Since CG (x) = C for all x ∈ N # (= A) and G/C is an abelian group all of whose nonidentity elements act fixed-point-freely on N. Therefore, G/C is cyclic of order n = p a − 1 = |A| (see Lemma X.2.3). Let H ⊲ G with H ∩ N = {1}. If g ∈ G, x ∈ H, then [x, g] ∈ G󸀠 ∩ H = N ∩ H = {1} implies [x, g] = 1, and we conclude that H ≤ Z(G). Obviously, we have N ∩ Z(G) = {1} (otherwise, a minimal G-invariant subgroup) N ≤ Z(G) and so C = CG (N) = G, a contradiction). Thus, Z(G) is the largest normal subgroup of G that has trivial intersection with N. Therefore, G = G/Z(G) is a monolith with minimal normal subgroup N = NZ(G)/Z(G) ≅ N. Next, C󸀠 ≤ G󸀠 = N. Since C = CG (N), we have C󸀠 ≤ N ≤ Z(C) hence cl(C) ≤ 2. Let P ∈ Sylp (C) (recall that N is a p-subgroup). As p ∤ p a − 1 = |G : C|, it follows that P ∈ Sylp (G). Since |N| = p a and N ⊲ G, we get G󸀠 = N ≤ P. Therefore, P ⊲ G and G/P is abelian as an epimorphic image of G/N. Then G = H ⋅ P is a semidirect product (Schur–Zassenhaus) and H ≅ G/P is abelian. Since C is nilpotent, it follows that C = Q × P, Q󸀠 = {1}.

(3)

Since a p󸀠 -subgroup Q ⊲ G, we get Q ≤ H so that CG (Q) ≥ HP = G and Q ∩ N = {1}. Therefore, Q ≤ Z(G). Set ∆ = G − C. The set ∆ is G-invariant since C is. Let y ∈ ∆, D = CG (y). Then D ∩ N = {1} and G = D ⋅ N, so, by the modular law, P = D1 ⋅ N,

where D1 = D ∩ P = CP (y).

(4)

Since P󸀠 ≤ G󸀠 = N and N is minimal normal, we get P󸀠 ∈ {{1}, N}. Assume that P󸀠 = N. Then N ≤ Φ(P), and the equality P = D1 ⋅ N implies D1 = P. In that case, N ≤ D1 , which is impossible since N ∩D1 = {1}. Thus P󸀠 = {1}, i.e., P is abelian so that C = Q×P is abelian (recall that Q ≤ Z(G)). Since G = CG (y) ⋅ N (semidirect product), we get P = N × CP (y).

(5)

CG (CP (y)) ≥ ⟨C, y⟩.

(6)

Next, By the above, G = G/C is cyclic. We have G = ⟨y⟩, where y = yC. Then y ∈ G − C and ⟨C, y⟩ = ⟨y⟩ = G, and we conclude that ⟨C, y⟩ = G. Therefore, CG (CP (y)) = G 󳨐⇒ CP (y) ≤ Z(G).

III On arithmetical properties of characters | 127

It follows from (4) and (5) that C = P × Q = NCP (y)Q ≤ N ⋅ Z(G) = N × Z(G). On the other hand, N ⋅ Z(G) ≤ C since C = CG (N). Thus, C = N × Z(G). Set G = G/Z(G), N = NZ(G)/Z(G) = C/Z(G) ≅ N. Then N is a minimal normal subgroup of G. Since N # = A is a G-class, we conclude that N# is also a G-class. If x ∈ N# , x ∈ N # , then CG (x) ≥ N, since N is abelian. Next, since N# is a G-class, we get |G : CG (x)| = |N# | = |N # | = p a − 1. Now |G : N| = |G : C| = p a − 1 󳨐⇒ CG (N) = N. Thus, G is a Frobenius group with kernel N ≅ Ep a and complement G/N ≅ G/C ≅ Cp a −1 . Let G = G/Z(G). Let K be a G-class. Then K is a G-class. The size of the G-class N# is p a − 1; the sizes of G-classes contained in G − N are p a . Thus, the set of sizes of G-classes is {1, p a , p a − 1} = {1, n, n + 1}. It is easy to see that |K| divides |K|. If |K| = n + 1, then n + 1 divides |K| so |K| = n + 1. If |K| = n, then n divides |K| so |K| = n. If |K| = 1, then K ⊆ Z(G) so |K| = 1. Thus, the sizes of G-classes are 1, n, n + 1. The preimage of the kernel N of the Frobenius group G in G is the abelian group N × Z(G) = C. The proof is complete. The above arguments also yield the following result. Proposition 11.2 (Mann). Let k be the greatest size of a conjugacy class of a group G, and let N be the subgroup generated by the G-classes whose sizes are coprime to k. Then N is nilpotent of class ≤ 2. If A is a G-class with GCD(|A|, k) = 1, then the subgroup ⟨A⟩ is abelian. Proof. Let A and B be G-classes with |A| = n, |B| = k and GCD(n, k) = 1. Put K = ker(B)(= {x ∈ G | xB = B}); then (see the proof of Theorem 11.1) K ⊲ G of order dividing k. It follows from the proof of Theorem 11.1 that ⟨A−1 A⟩ ≤ ker(B) = K. Let x ∈ A, C = CG (x); then we have |G : C| = |A| = n. Since |K| | k, we conclude that |K : (K ∩ C)| = |KC : C| is a divisor of GCD(k, n) = 1, whence K ≤ C. Furthermore, if g ∈ G, then [x, g] = x−1 x g ∈ A−1 A ⊆ K implies x g ∈ xK, i.e., ⟨A⟩ ≤ ⟨x, K⟩, and we get AK = ⟨x, K⟩. As we now know, K centralizes ⟨A⟩. In particular, ⟨A⟩ ∩ K ≤ Z(⟨A⟩). Since ⟨A⟩/(⟨A⟩ ∩ K) ≅ ⟨A⟩K/K = ⟨x, K⟩/K is cyclic, it follows that ⟨A⟩ is abelian. We have thus shown that K ∩ N ≤ Z(N) (N is defined in the statement of the theorem) and that, if A is a G-class such that GCD(|A|, k) = 1, then ⟨A⟩K/K ≤ Z(G/K). Therefore, NK/K ≤ Z(G/K), so that N/(N ∩ K) is abelian, and therefore cl(N) ≤ 2.

128 | Characters of Finite Groups 1 The following proposition, which also follows from Theorem 11.1, uses, among other things, the classification of the finite simple groups (see also [Kor]). Proposition 11.3. Assume that the size of any G-class has at most two prime divisors. Then either G is solvable or G/S(G) is isomorphic to PSL(2, 5) or PSL(2, 8) (here S(G) is the solvable radical of G). Exercise 11.3. If the sizes of all G-classes are powers of a fixed prime p, then G has a normal p-complement (=p-nilpotent). Solution. We proceed by induction on |G|. Let P ∈ Sylp (G) and z ∈ Z(P)# . Then z ∈ Z(G). By induction, G/Z(P) has a normal p-complement F/Z(P). By Burnside, F = H × Z(P); then H is a normal p-complement in G. Exercise 11.4 (Ferguson, Casolo). Under the assumptions of Proposition 11.3, one has |π(G/Z(G))| ≤ 4. Exercise 11.5. If, in Theorem 11.1, n = 1, then G/Z(G) is an elementary abelian 2-group. The proof of the following proposition is similar to that of Theorem 11.1. Proposition 11.4 (Mann). There exists no finite group for which the two maximal sizes of G-classes are consecutive odd numbers greater than 1. Proof. Let n and n + 2 be the maximal sizes of G-classes, where n > 1 is odd. Let A, B be G-classes such that |A| = n, |B| = n + 2, and let N = ker(B); then GCD(n, n + 2) = 1. As we have seen in Theorem 11.1, N = ⟨A−1 A⟩,

|N| = n + 2,

B = xN,

A ⊂ zN

(z ∈ A).

Then zN − A is a normal subset of size 2 and, therefore, either zN − A ⊆ Z(G) or zN − A is a G-class of size 2. Let zN − A ⊆ Z(G). Then one may assume that z ∈ Z(G). In that case, N = z−1 zN = z−1 A ∪ T, where z−1 A is a G-class (Exercise 11.2) and T ⊆ Z(G). In that case T = N ∩ Z(G) is a subgroup of order 2, a contradiction because T ≤ N and |N| = n + 2 is odd. Now let zN − A = C be a G-class. Then C−1 A is a G-class since GCD(|C|, |A|) = 1 (Exercise 11.2) and C−1 A ⊆ N. It follows from the hypothesis that |C−1 A| = n. Then N − C−1 A is a normal subset of G of size 2 and 1 ∈ N − C−1 A. We again have N − C−1 A ⊆ Z(G),

N ∩ Z(G) = N − C−1 A,

which leads to a contradiction. Exercise 11.6 (Baer). If any G-class of a group G is of prime power size, then G is solvable.

III On arithmetical properties of characters | 129

Hint. One may assume that there are two non-one-element G-classes A and B such that GCD(|A|, |B|) = 1 (see Exercise 11.3). Use Proposition 11.2. Exercise 11.7. Suppose that m is the largest size of conjugacy classes in G. If B is a conjugacy class of G with |B| = m and A is a G-class with GCD(|A|, m) = 1, then A is contained in those quasikernels of all irreducible characters of G which are not vanished on B. Solution. By Exercise 11.2, AB is a G-class. As a consequence of Exercise 11.1, we have G = CG (a)CG (b) (a ∈ A, b ∈ B). By Exercise II.6.6, χ(ab)χ(1) = χ(a)χ(b) for any χ ∈ Irr(G).

(7)

It follows from the Second Orthogonality Relation, formula (7) and the maximality of |B| that |CG (b)| = |CG (ab)| =

∑ |χ(ab)|2 = χ∈Irr(G)

Since |χ(ab)| =

|χ(a)|2 |χ(b)|2 . χ(1)2 χ∈Irr(G) ∑

(8)

|χ(a)||χ(b)| ≤ |χ(b)| χ(1)

for any χ ∈ Irr(G) such that χ(b) ≠ 0, it follows that |χ(ab)| = |χ(b)| for each character χ which does not vanish on B (this is a consequence of (8)). The equality holds if and only if |χ(a)| = χ(1), i.e., when a ∈ Z(χ), the quasikernel of χ.

12 Gallagher’s theorems on commutators The results of this section are due to Gallagher [Gal1]. Let χ ∈ Irr(G), Tχ = {g ∈ G | χ(g) = 0}, f χ = χ(1). Next, Irr1 (G) = Irr(G) − Lin(G) is the set of nonlinear irreducible characters of a group G. Below, three theorems of Gallagher on commutators are stated. Theorem 12.1. Let G be a group and let n ∈ ℕ be such that ∑χ∈Irr1 (G) f χ2−2n < |G : G󸀠 |. Then any element of G󸀠 can be represented as a product of n commutators. In particular, if |Irr1 (G)| < |G : G󸀠 |, then any element of G󸀠 is a commutator. It follows from Theorem 12.1 (take there n = 1) that if G is the holomorph of cyclic subgroup of order 8, then all elements of G󸀠 are commutators. Indeed, in that case, |Irr1 (G)|
∑χ∈Irr1 (G) f χ2−2n , then every element of G󸀠 is a product of n commutators. (Taking n = 1, we see that if |G : G󸀠 | > |Irr1 (G)|, then all elements of G󸀠 are commutators.) Remark. If g ∈ ̸ G󸀠 , then ρ(g[t1 , v1 ] . . . [t n , v n ]) = 0

(t i , v i ∈ G),

∑

χ(g) = ρ G/G󸀠 (g) = 0.

χ∈Lin(G)

It therefore follows from (6) that χ(g)

∑ χ∈Irr1 (G)

f χ2n−1

= 0 (g ∈ G − G󸀠 ).

(8)

Putting n = 1, 2, . . . in (8) and summing up, we get f χ χ(g)

∑ χ∈Irr1 (G)

f χ2 − 1

=0

(g ∈ G − G󸀠 ).

(9)

We now deduce from the first equality in (6) that, if g ∈ G󸀠 , then χ(g)

∑ χ∈Irr(G)

f χ2n−1

=

f n (g) , |G|2n−1

(10)

where f n (g) is the number of representations of g as a product of n commutators. In particular, χ(g) f1 (g) = |G| ∑ . fχ χ∈Irr(G) Exercise 12.1. If g ∈ ̸ G󸀠 , then there is χ ∈ Irr1 (G) such that χ(g) ≠ 0. Solution. Indeed, in the case under consideration, the right-hand side of (7) is equal to |G|2n |G : G󸀠 |. On the other hand, since every one of the |G|2n summands on the left-hand side of (7) is equal to 0, then g[t1 , v1 ] . . . [t n , v n ] ≠ 1 for all n ∈ ℕ, and we get |G|2n |G : G󸀠 | = 0, which is a contradiction. Proof of Theorem 12.2. Let n ∈ ℕ be such that 4n ≥ |G󸀠 |. One has ∑

f χ2 = |G| − |G/G󸀠 | = |G/G󸀠 |(|G󸀠 | − 1).

χ∈Irr1 (G)

Next, ∑ χ∈Irr1 (G)

f χ2−2n =

∑

f χ2

2n χ∈Irr1 (G) f χ

.

134 | Characters of Finite Groups 1 But f χ ≥ 2 implies f χ2n ≥ 4n , f χ−2n ≤ 4−n . Therefore, we get since 4−n |G󸀠 | ≤ 1, f χ2−2n ≤ 4−n |G/G󸀠 |(|G󸀠 | − 1) < 4−n |G/G󸀠 ||G󸀠 | ≤ |G/G󸀠 |.

∑ χ∈Irr1 (G)

The result now follows from Theorem 12.1. Remark. It is easy to see from this proof that if a = min{f χ | χ ∈ Irr1 (G)} and a2n ≥ |G󸀠 |, then any element of G󸀠 is a product of n commutators. In particular, if G is a p-group and p2n ≥ |G󸀠 |, then any element of G󸀠 is a product of n commutators. Proof of Theorem 12.3. Multiplying both sides of (3) by f χ , summing over all χ ∈ Irr(G) and replacing g by g−1 , we get ∑

t1 ,...,t n ∈G

ρ(g −1 [t1 , v1 ] . . . [t n , v n ]) = |G|n

∑

1

n−1 χ∈Irr(G) f χ

χ(g −1 v1 . . . v n )χ(v1 ) . . . χ(v n ).

In particular, if {v1 , . . . , v n } is a complete system of representatives of the sets Tχ (χ ∈ Irr1 (G)), then χ(v1 ) . . . χ(v n ) = 0 for any χ ∈ Irr1 (G). In that case 1

∑

n−1 χ∈Irr(G) f χ

=

χ(g −1 v1 . . . v n )χ(v1 ) . . . χ(v n ) ∑

1

n−1 χ∈Lin(G) f χ

=

∑

−1 χ(g −1 v1 . . . v n )χ(v−1 1 ) . . . χ(v n )

χ(g−1 )

χ∈Lin(G)

(we use the fact that χ ∈ Lin(G) is a homomorphism of G into ℂ∗ , the multiplicative group of ℂ). Now, if g ∈ G󸀠 , we obtain ∑ χ∈Irr(G)

1 f χn−1

χ(g −1 v1 . . . v n )χ(v1 ) . . . χ(v n ) = |Lin(G)| = |G/G󸀠 |.

Therefore, if g ∈ G󸀠 , then ∑

t1 ,...,t n ∈G

ρ(g −1 [t1 , v1 ] . . . [t n , v n ]) = |G/G󸀠 | ≠ 0.

Hence we can choose t1 , . . . , t n so that g −1 [t1 , v1 ] . . . [t n , v n ] = 1 ⇐⇒ g = [t1 , v1 ] . . . [t n , v n ], as was to be shown. Remark. Actually, Gallagher has proved Theorem 12.3 only under the condition that {v1 , . . . , v n } = ⋃χ∈Irr1 (G) Tχ , so Theorem 12.3 is somewhat stronger than his original result. Exercise 12.2. If G is a p-group such that |G󸀠 | ≤ p2 , then all elements of G󸀠 are commutators.

IV Products of characters In the first half of the chapter we present some facts from multilinear algebra that are necessary in what follows.

1 Tensor product Let V i (i ∈ {1, . . . , m}) be an n i -dimensional linear ℂ-space (in fact, nearly all the results of §§IV.1–IV.3 are valid for any ground field), n i ∈ ℕ and N i = {1, . . . , n i }, N = N1 × ⋅ ⋅ ⋅ × N m (the cartesian product) so that |N| = n1 . . . n m . Any element of the set N is a sequence (i) = (i1 , . . . , i m ), where i k ∈ N k (k ∈ {1, . . . , m}). Let V be the cartesian product V1 × ⋅ ⋅ ⋅ × V m . The vector space operations are defined on V elementwise as usually. We retain the introduced notation throughout the section. A function F : V → ℂ is called an m-linear form if it is linear in each of its m arguments. For example, if (v1 , . . . , v m ) ∈ V and v ik is the k-th coordinate of vector v i , then the form (v1 , v2 , . . . , v m ) 󳨃→ v11 v21 . . . v m1 is m-linear. Next, if A is an m × m matrix, then the function det : A 󳨃→ det(A) is linear in each of m columns of A so it is an m-linear form. The set of all m-linear forms is a linear space over ℂ with respect to the natural operations of addition and multiplication by scalars. If F1 , F2 are m-linear forms on the space V, (v1 , . . . , v m ) ∈ V and α ∈ ℂ, then (F1 + F2 )(v1 , . . . , v m ) = F1 (v1 , . . . , v m ) + F2 (v1 , . . . , v m ) and (αF1 )(v1 , . . . , v m ) = αF1 (v1 , . . . , v m ). We denote the space of m-linear forms on V = V1 × ⋅ ⋅ ⋅ × V m by P(V1 , . . . , V m ). The two displayed formulas above show that F1 + F2 , αF1 ∈ P(v1 , . . . , v m ) provided we have F1 , F2 ∈ P(V1 , . . . , V m ) and α ∈ ℂ. Choosing in each linear space V i a basis B i = {e ij | j ∈ {1, . . . , n i }}

(i ∈ {1, . . . , m}),

we obtain for any (v1 , . . . , v m ) ∈ V, F(v1 , . . . , v m ) = ∑ α(j) v1j1 . . . v mj m ,

(∗)

(j)∈N

where (j) = (j1 , . . . , j m ) ∈ N,

α(j) = F(e1j1 , . . . , e mj m ),

and v ij is the j-th coordinate of the vector v i in the basis B i ; clearly, F(v1 , . . . , v m ) ∈ ℂ. The number of summands in (∗) is equal to |B1 | . . . |B m | = n1 . . . n m . Let F(j) be the m-linear form defined by F(j) (v1 , . . . , v m ) = v1j1 . . . v mj m . DOI 10.1515/9783110224078-004

136 | Characters of Finite Groups 1 The number of forms F(j) ((j) ∈ N) is equal to |N| = n1 . . . n m . The form F = ∑ α(j) F(j) (j)∈N

(see (∗)) is a linear combination of the forms F(j) ((j) ∈ N). A form F is said to be the zero-form if F(v1 , . . . , v m ) = 0 for all (v1 , . . . , v m ) ∈ V (in that case we write F = 0). It follows that the forms F(j) span the space P(v1 , . . . , v m ). Next, 0 = F = ∑ α(j) F(j) 󳨐⇒ α(j) = F(e1j1 , . . . , e mj m ) = 0 (j)∈N

so that the |N| forms F(j) ((j) ∈ N) are linearly independent. It follows that the |N| forms F(j) ((j) ∈ N) constitute a basis of the space P(V1 , . . . , V m ), and therefore dim(P(V1 , . . . , V m )) = n1 . . . n m = |N|. Definition 1.1. The tensor product m

V1 ⊗ ⋅ ⋅ ⋅ ⊗ V m = ⨂ V i i=1

of the spaces V i (i ∈ {1, . . . , m}) is the space P(V1 , . . . , V m )∗ = Hom(P(V1 , . . . , V m ), ℂ) dual to the space P(V1 , . . . , V m ). Let (v1 , . . . , v m ) ∈ V and F ∈ P(V1 , . . . , V m ). The mapping F 󳨃→ F(v1 , . . . , v m ) is a linear form on the space P(V1 , . . . , V m ), i.e., belongs to the dual space m

P(V1 , . . . , V s )∗ = Hom(P(V1 , . . . , V m ), ℂ) = ⨂ V i . i=1

If we denote this form by v1 ⊗ ⋅ ⋅ ⋅ ⊗ v m , then we obtain the following equality: (v1 ⊗ ⋅ ⋅ ⋅ ⊗ v m )(F) = F(v1 , . . . , v m ).

(1)

Thus, v1 ⊗ ⋅ ⋅ ⋅ ⊗ v m depends of (v1 , . . . , v m ) ∈ V1 × ⋅ ⋅ ⋅ × V m . Exercise 1.1. The mapping ⊗ : (v1 , . . . , v m ) 󳨃→ v1 ⊗ ⋅ ⋅ ⋅ ⊗ v m

((v1 , . . . , v m ) ∈ V)

∗ of V in ⨂m i=1 V i = P(V 1 , . . . , V m ) is m-linear.

Solution. If c1 , d1 ∈ ℂ, v󸀠1 ∈ V1 and F ∈ P(V1 , . . . , V m ), then, since the form F is linear at the first argument, we get ((c1 v1 + d1 v󸀠1 ) ⊗ v2 ⊗ ⋅ ⋅ ⋅ ⊗ v m )(F) = F(c1 v1 + d1 v󸀠1 , v2 , . . . , v m ) = c1 F(v1 , v2 , . . . , v m ) + d1 F(v󸀠1 , v2 , . . . , v m ) = c1 (v1 ⊗ v2 ⊗ ⋅ ⋅ ⋅ ⊗ v m ) + d1 (v󸀠1 ⊗ v2 ⊗ ⋅ ⋅ ⋅ ⊗ v m ), i.e., our mapping is linear at the first argument. The same calculation shows that it is linear at m − 1 remaining arguments.

IV Products of characters | 137

Lemma 1.1 (Universality of the tensor product). The space ⨂m i=1 V i possesses the following universality property. Let U be a linear ℂ-space and let f : V = V1 × ⋅ ⋅ ⋅ × V m → U be an m-linear mapping. Then there exists a unique homomorphism of linear spaces ψ : ⨂m i=1 V i → U such that ψ ∘ ⊗ = f , i.e., (ψ ∘ ⊗)(v1 , . . . , v m ) = ψ(v1 ⊗ ⋅ ⋅ ⋅ ⊗ v m ) = f(v1 , . . . , v m )

((v1 , . . . , v m ) ∈ V).

Proof. Let B i = {e ij | j = 1, 2, . . . , n i } be a basis of the space V i (i = 1, 2, . . . , m)). Put e(j) = e1j1 ⊗ ⋅ ⋅ ⋅ ⊗ e mj m

((j) = (j1 , . . . , j m ) ∈ N).

Here, for example, e1j1 ∈ V1 is a vector whose j1 -th coordinate is 1 and all other coordinates are equal to 0. Suppose that ∑ λ(j) e(j) = 0 (λ(j) ∈ ℂ). (j)∈N

Then for any form F ∈ P(V1 , . . . , V m ) we have ∑ λ(j) e(j) (F) = 0 󳨐⇒ ∑ λ(j) F(e1j1 , . . . , e mj m ) = 0. (j)∈N

(j)∈N

Note that F(i) (e1j1 , . . . , e mj m ) = δ(i),(j) , where δ is the Kronecker delta on the set N. Therefore, putting F = F(i) in the above equality, we get 0 = ∑ λ(j) δ(i),(j) = λ(i) (j)∈N

so that the elements e(j) ((j) ∈ N) are linearly independent. Recall that if W is a finitedimensional ℂ-linear space, then dim(W ∗ ) = dim(W). Since m

|N| = dim(P(V1 , . . . , V m )) = dim(P(V1 , . . . , V m )∗ ) = dim(⨂ V i ) i=1

(see the definition of tensor product), it follows that {e(j) | (j) ∈ N} is a basis of the m space ⨂m i=1 V i . Thus, the dimensions of ⨂i=1 V i and V are equal so these spaces are isomorphic as linear spaces. Now we are ready to complete the proof. By the previous arguments, there exists a homomorphism of linear spaces ψ : ⨂m i=1 V i → U such that ψ(e(j) ) = f(e1j1 , . . . , e mj m ) for any (j) ∈ N. Now, ψ(e(j) ) = (ψ ∘ ⊗)(e1j1 , . . . , e mj m ) 󳨐⇒ ψ ∘ ⊗ = f. ⨂m i=1

V i → U is another homomorphism of the same linear spaces such that If : ψ󸀠 ∘ ⊗ = f , then, evaluating both sides of that equality at e(j) = (e1j1 , . . . , e mj m ) for all (j) ∈ N and taking into account the result of the previous paragraph, we get ψ󸀠

(ψ󸀠 ∘ ⊗)(e1j1 , . . . , e mj m ) = ψ󸀠 (e1j1 ⊗ ⋅ ⋅ ⋅ ⊗ e mj m ) = ψ󸀠 (e(j) ) = f(e1j1 , . . . , e mj m ) = ψ(e(j) ), so that ψ󸀠 = ψ since {e(j) | (j) ∈ N} is a basis of the space ⨂m i=1 V i . Thus, ψ is uniquely determined.

138 | Characters of Finite Groups 1 Let ϕ i ∈ End(V i ) (i ∈ {1, . . . , m}). Since (v1 , . . . , v m ) 󳨃→ ϕ1 (v1 ) ⊗ ⋅ ⋅ ⋅ ⊗ ϕ m (v m ) is an m-linear mapping of V = V1 × ⋅ ⋅ ⋅ × V m into ⨂1m V i , it follows by Lemma 1.1 that there exists a unique linear operator ω ∈ End(⨂m i=1 V i ) such that ω(v1 ⊗ ⋅ ⋅ ⋅ ⊗ v m ) = ϕ1 (v1 ) ⊗ ⋅ ⋅ ⋅ ⊗ ϕ m (v m )

(2)

for each (v1 , . . . , v m ) ∈ V. Definition 1.2. The operator ω that satisfies (2) is called the tensor product of the operators ϕ i (i ∈ {1, . . . , m}); it is denoted by ϕ1 ⊗ ⋅ ⋅ ⋅ ⊗ ϕ m = ⨂m i=1 ϕ i . Now we can rewrite (2) as follows: (ϕ1 ⊗ ⋅ ⋅ ⋅ ⊗ ϕ m )(v1 ⊗ ⋅ ⋅ ⋅ ⊗ v m ) = ϕ1 (v1 ) ⊗ ⋅ ⋅ ⋅ ⊗ ϕ m (v m ).

(3)

Exercise 1.2. Prove that m

m

m

(⨂ ϕ i )(⨂ ψ i ) = ⨂ ϕ i ψ i i=1

i=1

(ϕ i , ψ i ∈ End(V i ), i ∈ {1, . . . , m})).

i=1

Solution for m = 2. Let v i ∈ V i , i = 1, 2. We have (ϕ1 ⊗ ϕ2 )(ψ1 ⊗ ψ2 )(v1 ⊗ v2 ) = (ϕ1 ⊗ ϕ2 )(ψ1 (v1 ) ⊗ ψ2 (v2 )) = (ϕ1 ψ1 )(v1 ) ⊗ (ϕ2 ψ2 )(v2 ) = ((ϕ1 ψ1 ) ⊗ (ϕ2 ψ2 ))(v1 ⊗ v2 ), proving the required equality since V1 ⊗ V2 = {v1 ⊗ v2 | v1 ∈ V1 , v2 ∈ V2 }. m Exercise 1.3. Prove that ⨂m i=1 idV i = id⨂i=1 V i . It follows from this and Exercises 1.1 m and 1.2 that the operator ⨂i=1 ϕ i is nonsingular if and only if each ϕ i is nonsingular, −1 = ⨂m ϕ −1 . and in that case we have (⨂m i=1 ϕ i ) i=1 i

Let A i = (α ikj ) = MB i (ϕ i ) be the matrix of the linear mapping ϕ i in the basis B i = {e ij | j ∈ {1, . . . , n i }}. Since

ni

ϕ i (e ij ) = ∑ α ikj e ik

(i ∈ {1, . . . , m}),

k=1

it follows that (recall that e(j) = e1j1 ⊗ ⋅ ⋅ ⋅ ⊗ e mj m ) m

(⨂ ϕ i )(e(j) ) = ϕ1 (e1j1 ) ⊗ ⋅ ⋅ ⋅ ⊗ ϕ m (e mj m ) i=1 ni

nm

= ∑ α1k1 j1 e1k1 ⊗ ⋅ ⋅ ⋅ ⊗ ∑ α m k m j m e mk m k1 =1

= ∑ (k)∈N

k m =1

α1k1 j1

...

αm k m j m e (k) .

IV Products of characters | 139

For all (i), (j) ∈ N, we set α(i),(j) = α1i1 j1 . . . α m i m j m . Then we can rewrite the above displayed equality as follows: m

(⨂ ϕ i )(e(j) ) = ∑ α(k),(j) e(k) . i=1

(k)∈N

Thus, we determined the matrix of the operator ⨂m i=1 ϕ i in the basis B = {e (i) | (i) ∈ N} m of the space ⨂i=1 V i : it is the |N | ×|N| matrix MB (⨂m i=1 ϕ i ) = (α (i),(j) ) called the tensor or Kronecker product of the matrices A i (i ∈ {1, . . . , m}) and denoted by m

A1 ⊗ ⋅ ⋅ ⋅ ⊗ A m = ⨂ A i . i=1

It should be noted that the matrix ⨂m i=1 A i depends on the ordering of m-tuples (i) ∈ N. We shall use the most natural order, the one induced by the lexicographical order. The corresponding tensor product is called standard. Any other product is similar to the standard product. Any permutation of factors in the tensor product also yields a similar matrix. When m = 2, the standard tensor product of square matrices A1 = (a1ij ) and A2 = (a2ij ) of sizes n1 × n1 and n2 × n2 , respectively, may be written in block form: α111 A2 . . . α11n1 A2 .. .. A1 ⊗ A2 = ( ... ). . . 1 1 α n1 1 A 2 . . . α n1 n1 A 2 It follows that tr(A1 ⊗ A2 ) = (a111 + ⋅ ⋅ ⋅ + a1n1 n1 )tr(A2 ) = tr(A1 ) ⋅ tr(A2 ),

(4)

and an analogous formula is valid for more than two factors. Exercise 1.4. We have (A1 ⊗ ⋅ ⋅ ⋅ ⊗ A m )(B1 ⊗ ⋅ ⋅ ⋅ ⊗ B m ) = A1 B1 ⊗ ⋅ ⋅ ⋅ ⊗ A m B m , and In1 ⊗ ⋅ ⋅ ⋅ ⊗ In m = In1 ...n m . The matrix ⨂m i=1 A i is nonsingular if and only if all matrices A i are nonsingular, and −1 = ⨂m A −1 . A ) then (⨂m i=1 i i=1 i Hint. See Exercise 1.2. Definition 1.3. Let V be a linear space. The space V ⊗ ⋅ ⋅ ⋅ ⊗ V (m factors) is denoted by ⨂m V and called the m-th tensor power of the space V. If ϕ ∈ End(V), then the operator ϕ ⊗ ⋅ ⋅ ⋅ ⊗ ϕ (m factors) is denoted by ⨂m ϕ and called the m-th tensor power of the operator ϕ. Clearly, ⨂m ϕ ∈ End(⨂m V). We define ⨂m A, the m-th tensor power of a matrix A, similarly.

140 | Characters of Finite Groups 1 Exercise 1.5. For ϕ, ψ ∈ End(V), A, B ∈ ℂn , we have m

m

m

⨂ (ϕψ) = ⨂ ϕ ⋅ ⨂ ψ, m

⨂ idV = id⨂m V , m

m

m

⨂ (AB) = ⨂ A ⋅ ⨂ B, m

⨂ In = In m , m

−1

(⨂ A)

m

= ⨂ A−1

(A is nonsingular).

2 Exterior power We now define the exterior power of an n-dimensional linear space V. Let m ∈ ℕ and let V m = V × ⋅ ⋅ ⋅ × V (m factors) be the m-th cartesian power of the space V. The form F ∈ P(V, . . . , V) = P m (V) (m arguments) is called antisymmetric if, for any permutation σ ∈ Sm and any (v1 , . . . , v m ) ∈ V m , we have F(v σ(1) , . . . , v σ(m) ) = sgn(σ) ⋅ F(v1 , . . . , v m ), where sgn(σ) is the sign of the permutation σ, i.e., sgn(σ) = 1 if σ is even and −1 if σ is odd. The set of all antisymmetric m-linear forms F ∈ P m (V) is the proper subspace P Am (V) of P m (V) for m > 1. If the vectors v1 , . . . , v m are linearly dependent, then F(v1 , . . . , v m ) = 0. If v1 , . . . , v m ∈ ℂm (here ℂm is the set of columns of height m with complex entries), then det : (v1 , . . . , v m ) → ℂ is an antisymmetric m-linear form (here (v1 , . . . , v m ) is the m × m matrix with columns v1 , . . . , v m ). n! Let n, m be nonnegative integers. If m ≤ n, then (mn ) = m!(n−m)! denotes the binon mial coefficient; if m > n, then (m) = 0 (by agreement, 0! = 1! = 1). It follows from what we have noted above that, if m > 1, then dim(P Am (V)) < dim(P m (V)). However, more precise assertion holds. Lemma 2.1. We have dim(P Am (V)) = (mn ), where n = dim(V). Proof. Put N = {1, . . . , n}m (m-th cartesian power consisting of all sequences of length m with entries from the set {1, . . . , n}). If B V = {e i }1n is a basis of the space V, F ∈ P Am (V) and (v1 , . . . , v m ) ∈ V m , then F(v1 , . . . , v m ) = ∑ α(i) v(i) , (i)

where α(i) = α i1 . . . α i m = F(e i1 , . . . , e i m ), v(i) = v1i1 . . . v mi m , v kj is the j-th coordinate of the vector v k ∈ V relative to the basis B V and (i) ∈ N (here we use the fact that our form F is m-linear).

IV Products of characters | 141

If a sequence (i) has repetitions, then α(i) = F(e i1 , . . . , e i m ) = 0. It follows, therefore, that if m > n, then P Am (V) = 0 = (mn ) (in that case, every sequence (i) ∈ N has repetitions), and this proves lemma in the considered case. Now let m ≤ n. If a sequence (i) has no repetitions (so that i1 , . . . , i m are pairwise distinct) and j1 , . . . , j m are the same numbers in increasing order, then α(i) = sgn(

j1 . . . j m ) ⋅ α(j) . i1 . . . i m

Let N< denote the lexicographically ordered set of all m-tuples (j) ∈ N such that j1 < ⋅ ⋅ ⋅ < j m . We associate with each m-tuple (j) ∈ N< an antisymmetric m-linear form D(j) , defined by 󵄨󵄨󵄨 v1j1 . . . v1j m 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 .. 󵄨󵄨󵄨 . .. D(j) (v1 , . . . , v m ) = 󵄨󵄨󵄨 ... . . 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨v mj1 . . . v mj m 󵄨󵄨󵄨 It is clear that F = ∑ α(j) D(j) . (j)∈N
χ(1). Exercise 4.1. Let N be a normal subgroup of a group G such that the quotient group G/N is nonabelian. Then ⋂χ∈Irr1 (G), N≤ker(χ) ker(χ) = N. In particular, the intersection of kernels of all nonlinear irreducible characters of a nonabelian group G is the identity subgroup. Solution. Let χ ∈ Irr1 (G), λ ∈ Lin(G), g ∈ D = ⋂τ∈Irr1 (G) ker(τ). Then λχ ∈ Irr1 (G) so that χ(1) = (λχ)(g) = λ(g)χ(g) = λ(g)χ(1) 󳨐⇒ λ(g) = 1 󳨐⇒ g ∈

⋂ ker(τ) = ker(ρ G ) = {1} τ∈Irr(G)

󳨐⇒ D = {1} (here ρ G is the regular character of G; as we know, Irr(ρ G ) = Irr(G)). Now, if N > {1}, then application of the obtained result to G/N completes the solution. Definition 4.1. The difference (χ󸀠 − χ󸀠󸀠 )(g) = χ󸀠 (g) − χ󸀠󸀠 (g) (g ∈ G) of two characters χ󸀠 and χ󸀠󸀠 of a group G is called a generalized character of G. Ordinary characters are also considered as generalized characters. If, in that definition, χ󸀠 = χ󸀠󸀠 , then χ󸀠 − χ󸀠󸀠 is the zero function which is considered as a generalized character of G. Exercise 4.2. A central function θ is a generalized character of G if and only if ⟨θ, χ⟩ ∈ ℤ for all χ ∈ Irr(G). Thus, a generalized character of G is a ℤ-linear combination of irreducible characters of G. It is a consequence of Theorem 4.1 that the set ℤ[Irr(G)] = Ch(G) of all generalized characters of G is a commutative ring with identity 1G , the principal character of G (we have 1G (x) = 1 for all x ∈ G). A generalized character τ is an ordinary character if and only if it is a nonzero function and ⟨τ, χ⟩ ≥ 0 for all χ ∈ Irr(G). If, in addition, ⟨τ, τ⟩ = 1 and τ(1) > 0, then τ ∈ Irr(G).

148 | Characters of Finite Groups 1 γ

γ

If χ α , χ β ∈ Irr(G), then χ α χ β = ∑χ γ ∈Irr(G) c αβ χ γ , where c αβ = ⟨χ α χ β , χ γ ⟩ are nonnegative integers – the structure constants of the ring ℤ[Irr(G)] = Ch(G) of generalized γ characters of G. We claim that the knowledge of the numbers c αβ is sufficient to reconstruct the character table X(G) of G. The following argument is due to D. Chillag [Chi5]. Let θ be a character of a group G, and V = CF[G],

CL(G) = {K1 , . . . , K r },

=K ,

Irr(G) = {χ1 , . . . , χ r },

−1

(K i )

i󸀠

xi ∈ Ki , r = k(G).

The linearly independent subset Irr(G) is a basis of the space V of class functions (indeed, we have dim(V) = k(G) = |Irr(G)|; see Chapter II). Define a linear operator T(θ) : V → V by (T(θ)f)(g) = θ(g)f(g) (f ∈ V, g ∈ G). Put m ij (θ) = ⟨T(θ)χ j , χ i ⟩

(i, j ∈ {1, . . . , r}).

∑ri=1

m ij (θ)χ i , that is, M(θ) = (m ij (θ)) is the matrix of the operator T(θ) Then T(θ)χ j = in the basis Irr(G) of the linear space V = CF[G]. Define the elements f s ∈ V by f s (x j ) = δ js |G|/|K s | (s ∈ {1, . . . , r}). Then B = {f s󸀠 | s ∈ {1, . . . , r}} is a basis of V. We have (T(θ)f j󸀠 )(x i ) = θ(x i )f j󸀠 (x i ) = θ(x i )δ ij󸀠 = θ(x j )δ ij󸀠

{0, |G| ={ |K j󸀠 | θ(x ) |G| , { j |K j󸀠 |

i ≠ j 󸀠 , i = j󸀠 ,

|G| = θ(x j )f j󸀠 (x i ). |K j󸀠 |

Therefore T(θ)f j󸀠 = θ(x j )f j󸀠 , so that the matrix of operator T(θ) in basis B is a diagonal matrix MB (T(θ)) = diag(θ(x1 ), . . . , θ(x r )). The Second Orthogonality Relation can be rewritten as f j󸀠 = ∑ri=1 χ i (x j )χ i . Therefore, the character table X(G) of G is the transition matrix from the basis Irr(G) to the basis B of V, and we have X(G)−1 M(θ) X(G) = diag(θ(x1 ), . . . , θ(x r )) (as we know, the matrix X(G) is not singular; see Chapter II). Therefore, given the matrix M(θ), we can evaluate θ(x1 ), . . . , θ(x r ). Putting consecutively θ = χ1 , . . . , χ r , we see that knowledge of the structure constants of the ring ℤ[Irr(G)] = Ch(G) enables us to construct the character table X(G) of G. Again, let θ ∈ Char(G) with m pairwise distinct values θ(1) = α1 , α2 , . . . , α m (these numbers are algebraic integers; see Chapter III). The polynomial m

m−1

f(x) = ∏(x − α i ) = ∑ a i x i ∈ ℤ[x] i=2

i=0

IV Products of characters | 149

since f as a function on α1 , α2 , . . . α m is symmetric. On one hand, m

m

∑ f(θ(g)) = ∑ ∏(θ(g) − α i ) = |ker(θ)| ⋅ ∏(α1 − α i ).

g∈G

g∈G i=2

i=2

On the other hand, m−1

m−1

∑ f(θ(g)) = ∑ ∑ a i θ(g)i = |G| ∑ a i ⟨θ i , 1G ⟩ ≡ 0 (mod |G|).

g∈G

g∈G i=0

Thus

i=0

m

|ker(θ)| ⋅ ∑ (α1 − α i ) ≡ 0 (mod |G|). i=2

A particular case of this result was obtained by Blichfeldt. We now turn to the question of the decomposition of powers of a faithful character of a group G into irreducible constituents. Lemma 4.2 (Zhmud). Let M be a nonempty set, 𝔽 a field, N ∈ ℕ and f : M → 𝔽 a function taking exactly N distinct values α1 , . . . , α N . Let C i = {m ∈ M | f(m) = α i } and let V f be the linear 𝔽-space of functions constant on classes C i . Then {1M , f, . . . , f N−1 } is a basis of V f , where f k (m) = f(m)k and 1M (m) = 1 ∈ 𝔽 for all m ∈ M. Proof. For each i ∈ {1, . . . , N} define a function e i : M → 𝔽 by e i (m) = δ ij (m ∈ C j ). Then {e i }1N is a basis of the space V f and therefore dim(V f ) = N. Therefore, it suffices to show that the system of functions described in the statement of the lemma is linearly independent. Suppose that λ0 + λ1 f + ⋅ ⋅ ⋅ + λ N−1 f N−1 = 0

(λ i ∈ 𝔽).

Then, for any m ∈ M, we have λ0 + λ1 f(m) + ⋅ ⋅ ⋅ + λ N−1 f(m)N−1 = 0. Putting consecutively m = m1 , . . . , m N (m i ∈ C i , i ∈ {1, . . . , N}), we obtain the following system of linear equations in λ0 , λ1 , . . . , λ N−1 : λ0 + λ1 α1 + ⋅ ⋅ ⋅ + λ N−1 α1N−1 = 0, λ0 + λ1 α2 + ⋅ ⋅ ⋅ + λ N−1 α2N−1 = 0, .. . λ0 + λ1 α N + ⋅ ⋅ ⋅ + λ N−1 α N−1 = 0. N As the determinant of this system is the Vandermonde determinant, it is different from zero. This proves that λ0 = λ1 = ⋅ ⋅ ⋅ = λ N−1 = 0, and hence the functions 1G , f, . . . , f N−1 are linearly independent.

150 | Characters of Finite Groups 1 Theorem 4.3 (Burnside–Brauer). Let χ be a faithful character of a group G that takes exactly N different values. Then the sum 1G + χ + ⋅ ⋅ ⋅ + χ N−1 contains all irreducible characters of G, that is, Irr(1G + χ + ⋅ ⋅ ⋅ + χ N−1 ) = Irr(G). Proof. Let C1 , . . . , C N be the subsets of G where χ takes constant values (see the statement of Lemma 4.2). Applying Lemma 4.2 with M = G, 𝔽 = ℂ and f = χ, we see that {1G , χ, . . . , χ N−1 } is a basis of the space of functions constant on each C i , i ∈ {1, . . . , N}. Since χ is faithful, one of the sets C i is {1}, while the regular character ρ = ρ G of G vanishes on G# . Hence, ρ ∈ V χ (see the statement of Lemma 4.2 again). Therefore, ρ = λ0 1G + λ1 χ + ⋅ ⋅ ⋅ + λ N−1 χ N−1 (λ i ∈ ℂ), which implies Irr(1G + χ + ⋅ ⋅ ⋅ + χ N−1 ) = Irr(ρ) = Irr(G). In particular, if a faithful nonlinear irreducible character χ of a nonabelian group G assumes exactly three values (“three” is minimal possible), then Irr(1+χ+χ2 ) = Irr(G). We use this observation in one of subsequent chapters. Second proof of Theorem 4.3. Let α1 , . . . , α r be all (pairwise distinct) values of the character χ and let ϵ = e2πi/|G| , G = Gal(ℚ(ϵ)/ℚ), σ ∈ G. If σ(ϵ) = ϵ ν , where GCD(ν, |G|) = 1, then σ(χ(g)) = χ(g ν ). Therefore, {σ(α1 ), . . . , σ(α r )} = {α1 , . . . , α r } (as sets), i.e., the set {α1 , . . . , α r } is G-invariant. It follows that the symmetric in α1 , . . . , α r polynomial F χ (t) = ∏ri=1 (t − α i ) is contained in ℤ[t]. Therefore, the polynomial r

F 󸀠χ (t) = ∑ i=1

F χ (t) t − αi

is also contained in ℤ[t]. Let α1 = f χ , where f χ = χ(1). Then F 󸀠χ (f χ ) = (f χ − α2 ) . . . (f χ − α r ) ∈ ℤ[x]. Consider the generalized character r

θ(g) = ∏(χ(g) − α i )

(g ∈ G).

(7)

i=2

We have θ(1) = (f χ − α2 ) . . . (f χ − α r ) ≠ 0. Since χ is faithful, we get θ(g) = 0 if and only if g ≠ 1. Let ψ ∈ Irr(G). Then ⟨θ, ψ⟩ = |G|−1 ∑ θ(g)ψ(g) = |G|−1 θ(1)ψ(1) ≠ 0. g∈G

On the other hand (see (7)), r

θ = ∏(χ − α i 1G ) = F 󸀠χ (χ) = χ r−1 + σ1 χ r−2 + ⋅ ⋅ ⋅ + σ r−1 1G , i=2

(8)

IV Products of characters | 151

where σ1 , . . . , σ r−1 ∈ ℂ. Therefore (see (8)), ⟨χ r−1 , ψ⟩ + σ2 ⟨χ r−2 , ψ⟩ + ⋅ ⋅ ⋅ + σ r−1 ⟨1G , ψ⟩ ≠ 0 It follows that ⟨χ i , ψ⟩ > 0 for some i ∈ {0, 1, . . . , r − 1}. Exercise 4.3 (Frobenius). Let χ ∈ Irr(G) be faithful. Prove that if α1 , . . . , α r are all different values of χ on the group G > {1}, then the following congruence in the ring of algebraic integers of the field ℚ(ϵ) holds (ϵ = e2πi/|G| , f χ = χ(1) = α1 ): Φ χ = (f χ − α2 ) . . . (f χ − α r ) ≡ 0 (mod |G|). Solution. Let

r

θ = ∏(χ − α i 1G ) ∈ ℤ[Irr(G)] = Ch(G). i=2

Then θ(g) = δ g,1 Φ χ , where δ is Kronecker’s delta on the set G. Therefore, ∑ θ(g) = θ(1) = Φ χ .

g∈G

Since ∑g∈G θ(g) = |G|⟨θ, 1G ⟩ ≡ 0 (mod |G|), we get Φ χ ≡ 0 (mod |G|). Exercise 4.4 ([Isa11, Problem 4.12]). If χ, ψ, θ ∈ Irr(G), then ⟨χψ, θ⟩ ≤ θ(1). Solution. Given a character τ ∈ Char(G), let ‖τ‖ = √⟨τ, τ⟩ be the norm of τ. By the First Orthogonality Relation, the norm of any irreducible character is equal to 1. The Cauchy–Schwarz inequality yields ⟨χψ, θ⟩ = |⟨χψ, θ⟩| = |⟨χ, ψθ⟩| ≤ ‖χ‖ ⋅ ‖ψθ‖ = ‖ψθ‖, and ‖ψθ‖2 = ⟨ψθ, ψθ⟩ = |G|−1 ∑ |(ψθ)(g)|2 g∈G

−1

= |G|

2

∑ |ψ(g)| |θ(g)|2 ≤ θ(1)2 |G|−1 ∑ |ψ(g)|2 = θ(1)2 ‖ψ‖2 = θ(1)2 . g∈G

g∈G

Therefore, ‖ψθ‖ ≤ θ(1) so that ⟨χψ, θ⟩ ≤ θ(1), by the first displayed formula. It follows that if ρ G is the regular character of a group G > {1} and χ, ψ ∈ Irr(G), then ρ G − χψ ∈ Char(G). Indeed, ρG =

∑ χ∈Irr(G)

θ(1)θ,

χψ =

∑ ⟨χψ, θ⟩θ. θ∈Irr(G)

We conclude that in the decomposition of the function ρ G − χψ all coefficients are nonnegative integers, by Exercise 4.4. Since (χψ)(1) < |G| = ρ G (1), it follows that ρ G − χψ is nonzero function so it is a character of G. This fact is equivalent to Exercise 4.4. The following result is in a certain sense dual to Theorem 4.3.

152 | Characters of Finite Groups 1 Theorem 4.4 (Garrison). Let K be a G-class, ⟨K⟩ = G, k = ∑g∈K g ∈ ℂG, m the number of distinct values of the function χ 󳨃→ ω χ (k) (χ ∈ Irr(G)), where ω χ (k) = |K|χ(g) χ(1) (g ∈ K). Then each element of G can be written as a product of less than m elements of K. Proof. Let M = Irr(G) and 𝔽 = ℂ (we use the same notation as in Lemma 4.2 and Theorem 4.3). Define a function f : M → ℂ by f(χ) = ω χ (k) =

|K|χ(g) χ(k) = χ(1) χ(1)

(χ ∈ M)

(we denoted by χ the extension of χ ∈ Irr(G) to ℂG by linearity; note that k ∈ Z(ℂG)). Let C1 , . . . , C m , V f be defined as in Lemma 4.2. Suppose that 1G ∈ C1 . Then f(1G ) =

|K|1G (g) |K|χ(g) = |K| 󳨐⇒ = f(χ) = |K| for each χ ∈ C1 , 1G (1) χ(1)

so that χ(g) = χ(1) for all g ∈ K. Since ⟨K⟩ = G, this gives χ = 1G . Thus C1 = {1G }. By Lemma 4.2, {1M , f, . . . , f m−1 } is a basis of the space V f . Define a function ψ : M → ℂ by ψ(χ) = ⟨χ, 1G ⟩ = δ χ,1G so that ψ(χ) = 0 for all nonprincipal χ ∈ Irr(G). Since C1 = {1G }, it follows that ψ ∈ V f . Consequently, by Lemma 4.2, ψ = λ0 1M + λ1 f + ⋅ ⋅ ⋅ + λ m−1 f m−1 for some λ0 , λ1 , . . . , λ m−1 ∈ ℂ. It follows from the definition of function ψ that, for each g ∈ G, one has m−1

1=

χ(1)ψ(χ)χ(g) = ∑ λ i

∑

i=0

χ∈Irr(G)

∑

χ(1)f(χ)i χ(g).

χ∈Irr(G)

Hence there exists i ≤ m − 1 such that ∑χ∈Irr(G) χ(1)f(χ)i χ(g) ≠ 0. Since f(χ)i = ω χ (k)i = ω χ (k i ) =

χ(k i ) , χ(1)

we have ∑χ∈Irr(G) χ(k i )χ(g) ≠ 0. Noting that k i = ∑x1 ,...,x i ∈K x1 . . . x i , we obtain ∑

∑

x1 ,...,x i ∈K χ∈Irr(G)

χ(x1 . . . x i )χ(g) ≠ 0.

Therefore, we have ∑χ∈Irr(G) χ(x1 . . . x i )χ(g) ≠ 0 for some x1 , . . . , x i ∈ K. According to the Second Orthogonality Relation, g ∼ x1 . . . x i ∈ K i . Since the set K i is G-invariant, it follows that g ∈ K i .

5 On some generalized characters In this section we consider characters of exterior and symmetric powers of a representation of a group G.

IV Products of characters | 153

Let X g (t) = t n − σ1 (g)t n−1 + ⋅ ⋅ ⋅ + (−1)n σ n (g) be the characteristic polynomial of the matrix T(g), where T : G → GL(n, ℂ) is a representation of G and g ∈ G. Then, m m by Lemma 2.3, χ⋀ T = σ m . The character χ⋀ T is called the m-th exterior power m of the character χ and denoted by ⋀ χ. Thus ⋀m χ = σ m . Evidently, σ1 = χ and σ n = ⋀n χ = det(χ) ∈ Lin(G) is defined by (det(χ))(g) = det(T(g)) (g ∈ G), Let m ∈ ℕ and let T : G → GL(n, ℂ) be a representation of a group G. Then m χ S T = S m χ is called the m-th symmetric power of the character χ = χ T . Let T : G → GL(n, ℂ) be a representation, g ∈ G and let {ϵ1 (g), . . . , ϵ n (g)} be the spectrum of the matrix T(g), m ∈ {1, . . . , n}. Then σ m (g) denotes the m-th elementary symmetric function on ϵ1 (g), . . . , ϵ n (g). Let f(X1 , . . . , X n ) be a symmetric polynomial over ℤ. According to the theory of symmetric polynomials, f(ϵ1 (g), . . . , ϵ n (g)) is a polynomial in σ1 (g), . . . , σ n (g) with coefficients in ℤ. Since σ i (g) are characters of G, it follows that f(ϵ1 (g), . . . , ϵ n (g)) is a generalized character of G. Thus: Theorem 5.1. Let f(X1 , . . . , X n ) be a symmetric polynomial over ℤ, let χ be the character of a representation T : G → GL(n, ℂ), let g ∈ G and {ϵ1 (g), . . . , ϵ n (g)} the spectrum of the matrix T(g). Then the mapping λ : g 󳨃→ f(ϵ1 (g), . . . , ϵ n (g)) is a generalized character of G. Let m ∈ ℕ and let θ : G → ℂ be a central function. Define a function θ(m) : G → ℂ by θ(m) (g) = θ(g m ), g ∈ G. It is obvious that θ(m) ∈ CF[G]. Theorem 5.2. If χ ∈ Char(G), then χ(m) ∈ Ch(G) for all m ∈ ℕ. Proof. Let χ = χ T , where T : G → GL(n, ℂ) is a representation. Let {ϵ1 (g), . . . , ϵ n (g)} denote the spectrum of the matrix T(g). Then χ(m) (g) = χ(g m ) = tr (T(g m )) = tr (T(g)m ) = ϵ1 (g)m + ⋅ ⋅ ⋅ + ϵ n (g)m is a symmetric function on ϵ1 (g), . . . ϵ n (g), and we are done, by Theorem 5.1. Exercise 5.1. If m ∈ ℕ and χ ∈ Char(G), then |G|−1 ∑g∈G χ(g m ) ∈ ℤ. Hint. The above sum is equal to ⟨χ(g m ), 1G ⟩. Use Theorem 5.2. The generalized characters χ(m) are closely related to binomial equations in a group G. Let ϑ m (g) = |{x ∈ G | x m = g}| (g ∈ G). If X is the set of solutions of the equation x m = g in G, then z−1 Xz is the set of solutions of the equation x m = z−1 gz. It follows that ϑ m (z−1 gz) = ϑ m (g) so that ϑ m ∈ CF[G] (the set of class functions on G). If GCD(m, |G|) = 1, then the equation x m = g has exactly one solution. We conclude that, in the case under consideration, ϑ m = 1G . Put ϑm =

∑

ν m (χ)χ,

(1)

χ∈Irr(G)

where ν m (χ) ∈ ℂ (in fact, as Lemma 5.3 shows, ν m (χ) ∈ ℤ). We have ν m (χ) = ⟨ϑ m , χ⟩. The function ν2 : Irr(G) → ℂ plays important role in what follows.

154 | Characters of Finite Groups 1 Lemma 5.3. If χ ∈ Irr(G), then ν m (χ) = ⟨χ(m) , 1G ⟩ ∈ ℤ. Proof. Since ν m (χ) = ⟨ϑ m , χ⟩ = |G|−1 ∑ ϑ m (g)χ(g), g∈G

ϑ m (g)χ(g) = ∑ χ(x m ), x∈G x m =g

it follows that ν m (χ) = |G|−1 ∑ ∑ χ(x m ) = |G|−1 ∑ χ(x m ) = ⟨χ(m) , 1G ⟩, g∈G x∈G x m =g

x∈G

and now the result follows from Theorem 5.2. Equality (1) implies the following: Corollary 5.4. We have ϑ m ∈ ℤ[Irr(G)] = Ch(G).

6 The Frobenius–Schur indicator The following series of statements concerns the important case m = 2. By Lemma 5.3, for every χ ∈ Irr(G), one has ν2 (χ) = |G|−1 ∑ χ(g 2 ) ∈ ℤ.

(1)

g∈G

The rational integer ν2 (χ) is said to be the Frobenius–Schur indicator of χ ∈ Irr(G). Theorem 6.1 (Frobenius–Schur). Let t(G) be the number of involutions in a group G. Then 1 + t(G) =

∑

ν2 (χ)χ(1).

(2)

χ∈Irr(G)

Proof. By definition, ϑ2 (1) = |{x ∈ G | x2 = 1}| = 1 + t(G) (recall that ϑ m (1) is the number of solutions of x m = 1 in G). On the other hand, it follows from equality (5.1) that ϑ2 (1) = ∑χ∈Irr(G) ν2 (χ)χ(1). We say that a matrix 𝔽-representation T of a group G is realized over a subfield 𝔽0 of the field 𝔽 if there exists T 󸀠 ∼ T such that, for every g ∈ G, all entries of the matrix T 󸀠 (g) belong to the subfield 𝔽0 . Maschke conjectured that every ℂ-representation of a group G is realized over the cyclotomic field ℚ(ϵ), where ϵ is a primitive |G|-th root of 1. As Brauer showed, this is true (see Chapter VIII). The proof of this deep result is based on Brauer’s characterization of generalized characters. The following fundamental theorem is the main result of this section.

IV Products of characters | 155

Theorem 6.2 (Frobenius–Schur). Let χ be the character of an irreducible representation T of G. Then 1 if χ = χ and T is realized over ℝ, { { { ν2 (χ) = {−1 if χ = χ and T is not realized over ℝ, { { if χ ≠ χ. {0

Clearly, in the third case, T is not realized over field ℝ of real numbers. All values of an irreducible character χ are real if and only if |ν2 (χ)| = 1, but the converse is not true. It follows from Theorem 6.2 that the representation T is realized over ℝ if and only if ν2 (χ T ) = 1. Theorem 6.2 is a corollary of several lemmas, some of them being of independent interest. It follows from Theorems 6.1, 6.2 and Lemma 2.3 that 1 + t(G) ≤ ∑χ∈Irr(G) χ(1) with equality if and only if all irreducible representations of G are realized over ℝ (in particular, in that case, exp(G/G󸀠 ) ≤ 2). Lemma 6.3. Let U be a linear space which is the direct sum of two subspaces U1 and U2 . Let f ∈ U ∗ = Hom(U, ℂ) and let f i = f U i denote the restriction of f to U i , i = 1, 2. Then the mapping f 󳨃→ (f1 , f2 ) is an isomorphism of U ∗ onto U1∗ ⊕ U2∗ . Proof. Obviously, f i ∈ U i∗ (i = 1, 2) and f → 󳨃 (f1 , f2 ) is a homomorphism of U ∗ into ∗ ∗ U1 ⊕ U2 . If u ∈ U, u = u1 + u2 , where u i ∈ U i , i = 1, 2, then f(u) = f(u1 ) + f(u2 ) = f1 (u1 ) + f2 (u2 ). Hence, the mapping f 󳨃→ (f1 , f2 ) is injective. Conversely, let f i ∈ U i∗ , i = 1, 2. Given u ∈ U and u = u1 + u2 , where u i ∈ U i , set f(u) = f1 (u1 ) + f2 (u2 ). Obviously, f ∈

U∗

and f i = f U i (i = 1, 2), so that our mapping f 󳨃→ (f1 , f2 ) is surjective.

Lemma 6.4. If χ is an irreducible character of a group G, then 2

χ2 = ⋀ χ + S2 χ.

(3)

Proof. Let V be the module of a representation τ affording the character χ. Then the ℂG-modules ⨂2 V = W, ⋀2 V = W A , S2 V = W S afford representations ⨂2 τ, ⋀2 τ, S2 τ, respectively. The characters of these representations are χ2 , ⋀2 χ, S2 χ, respectively. Recall (see §§IV.1–IV.3) that W = P2 (V)∗ ,

W A = P2A (V)∗ ,

W S = P2S (V)∗ ,

where P2 (V) is the space of bilinear forms on V, P2A (V) is the space of antisymmetric bilinear forms on V, and P2S (V) is the space of symmetric bilinear forms on V. As we know, P2A (V) and P2S (V) are subspaces of the space P2 (V), and, moreover, the following equality holds: P2 (V) = P2A (V) ⊕ P2S (V).

(4)

156 | Characters of Finite Groups 1 Indeed, every form F ∈ P2 (V) can be represented in a unique way as a sum F A + F S , where F A ∈ P2A (V), F S ∈ P2S (V), namely (v1 , v2 ∈ V): F A (v1 , v2 ) =

1 [F(v1 , v2 ) − F(v2 , v1 )], 2

F S (v1 , v2 ) =

1 [F(v1 , v2 ) + F(v2 , v1 )]. 2

Let w ∈ W, and let w A = w P2A (V) and w S = w P2S (V) be restrictions of w to P2A (V) and P2S (V), respectively. Then, by Lemma 6.3, the mapping w 󳨃→ (w A , w S ) (w ∈ W) is an isomorphism of W onto the space W A ⊕ W S . We claim that this is an isomorphism of ℂG-modules. Indeed, let w = v1 ⊗ v2 (v i ∈ V i , i = 1, 2). If F ∈ P2A (V), then, by formulas (1.1) and (2.1), w A (F) = w(F) = (v1 ⊗ v2 )(F) = F(v1 , v2 ) = (v1 ∧ v2 )(F). Therefore, w A = v1 ∧ v2 , that is, (v1 ⊗ v2 )A = v1 ∧ v2 .

(5)

Also, if F ∈ P2S (V), then, by formulas (1.1) and (3.1), w S (F) = w(F) = (v1 ⊗ v2 )(F) = F(v1 , v2 ) = (v1 Sv2 )(F). Therefore, w S = v1 Sv2 , that is, (v1 ⊗ v2 )S = v1 Sv2 .

(6)

Let g ∈ G. Then, by formulas (4.1), (4.3), (4.4), (5) and (6), [g(v1 ⊗ v2 )]A = (gv1 ⊗ gv2 )A = gv1 ∧ gv2 = g(v1 ∧ v2 ) and [g(v1 ⊗ v2 )]S = (gv1 ⊗ gv2 )S = gv1 Sgv2 = g(v1 Sv2 ). Consequently, [g(v1 ⊗ v2 )]A = g(v1 ⊗ v2 )A ,

[g(v1 ⊗ v2 )]S = g(v1 ⊗ v2 )S .

Since W is the linear span of all elements v1 ⊗ v2 , it follows that (gw)A = gw A and (gw)S = gw S for every w ∈ W. Since g(w A , w S ) = (gw A , gw S ), the mapping w 󳨃→ (w A , w S )

(w ∈ W)

is an isomorphism of the ℂG-module W onto the direct sum W A ⊕ W S . Thus, we have W ≅ W A ⊕ W S (as ℂG-modules), that is, ⨂2 V ≅ ⋀2 V ∔ S2 V (as ℂG-modules), whence ⨂2 τ ∼ ⋀2 τ ∔ S2 τ. Consequently, taking traces of both parts of the displayed equivalence, we get 2 χ2 = ⋀ χ + S2 χ, and the proof is complete.

IV Products of characters | 157

Lemma 6.5. If χ ∈ Irr(G), then {1 if χ = χ, 2 ⟨⋀ χ, 1G ⟩ + ⟨S2 χ, 1G ⟩ = { 0 if χ ≠ χ. {

In particular, if χ ≠ χ, then ⟨⋀2 χ, 1G ⟩ = ⟨S2 χ, 1G ⟩ = 0. If χ = χ, then either 2

⟨S2 χ, 1G ⟩ = 0

2

⟨S2 χ, 1G ⟩ = 1.

⟨⋀ χ, 1G ⟩ = 1, or

⟨⋀ χ, 1G ⟩ = 0,

Proof. By Lemma 6.4, ⟨⋀2 χ, 1G ⟩ + ⟨S2 χ, 1G ⟩ = ⟨χ2 , 1G ⟩ = ⟨χ, χ⟩ whence the lemma follows since all inner products are nonnegative (indeed, ⋀2 χ, S2 χ ∈ Char(G)). Lemma 6.6. If χ ∈ Irr(G), then 1 if χ = χ and ⟨S2 χ, 1G ⟩ = 1, { { { ν2 (χ) = {−1 if χ = χ and ⟨⋀2 χ, 1G ⟩ = 1, { { if χ ≠ χ. {0

(7)

In the latter case, ⟨⋀2 χ, 1G ⟩ = ⟨S2 χ, 1G ⟩ = 0. Proof. By Lemma 5.3, ν2 (χ) = ⟨χ(2) , 1G ⟩. Now (see the proof of Theorem 5.2), if g ∈ G, then χ(2) (g) = ϵ1 (g)2 + ⋅ ⋅ ⋅ + ϵ n (g)2 = σ1 (g)2 − 2σ2 (g). Consequently, χ(2) = σ21 − 2σ2 , whence it follows (see the beginning of §IV.5) that 2

χ(2) = χ2 − 2 ⋀ χ.

(8)

Using equality (3), we obtain χ(2) = S2 χ − ⋀2 χ from (8). Therefore, 2

ν2 (χ) = ⟨χ(2) , 1G ⟩ = ⟨S2 χ, 1G ⟩ − ⟨⋀ χ, 1G ⟩. The result now follows from Lemma 6.5. Let χ ∈ Irr(G). We claim that Irr(χ(2) ) ⊆ Irr(χ2 ). Indeed, we have χ(2) = χ2 − 2 ⋀2 χ (see formula (8) in the proof of Lemma 6.6) so it suffices to show that Irr(⋀2 χ) ⊆ Irr(χ2 ). By Lemma 6.4, χ2 = ⋀2 χ + S2 χ, and we conclude that Irr(⋀2 χ) ⊆ Irr(χ2 ). (Note that χ(2) is a generalized character of G.) We will now determine when ⟨S2 χ, 1G ⟩ = 1. Let V be a ℂG-module. An element v ∈ V is said to be an invariant of the group G if gv = v for all g ∈ G. The set J(V) = {v ∈ V | v is an invariant of G} is a submodule of V. Lemma 6.7. Let V be the module of a representation of a group G with a character χ. Then ⟨χ, 1G ⟩ = dim(J(V)).

158 | Characters of Finite Groups 1 Proof. By Maschke’s theorem, J(V) is a direct summand of the module V, that is, V = J(V) ⊕ N, where N is a submodule of V. Let dim(J(V)) = m and let {v1 , . . . , v m } be a basis of J(V). Then the one-dimensional subspaces ℂv i (i = 1, . . . , m), are submodules affording the principal representation of G. Let N = N1 ⊕ ⋅ ⋅ ⋅ ⊕ N l be a decomposition of N into a direct sum of irreducible submodules. Then none of the N i affords the principal representation of G. Consequently, if θ i is the character of submodule N i , then ⟨θ i , 1G ⟩ = 0. Since V = ℂv1 ⊕ ⋅ ⋅ ⋅ ⊕ ℂv m ⊕ N1 ⊕ ⋅ ⋅ ⋅ ⊕ N l , it follows that χ = m1G + θ1 + ⋅ ⋅ ⋅ + θ l 󳨐⇒ ⟨χ, 1G ⟩ = m = dim(J(V)). Corollary 6.8. If χ ∈ Irr(G) and V is a G-module affording χ, then 2

⟨S2 χ, 1G ⟩ = dim(J(S2 V)),

2

⟨⋀ χ, 1G ⟩ = dim(J(⋀ V)).

In particular, if χ = χ, then dim(J(S2 V)) + dim(J(⋀2 V)) = 1 (Lemma 6.5). If χ ≠ χ, then J(S2 V) = J(⋀2 V) = 0. Corollary 6.9. If χ and V are as in Corollary 6.8, then ν2 (χ) = 1 if and only if J(S2 V) ≠ 0. Retaining the notation introduced in Corollary 6.8, we will explain the meaning of the condition J(S2 V) ≠ 0. Choose a basis B V = {e1 , . . . , e n } of the space V, where n = deg(T) = χ(1), T being a representation of G affording the character χ. The elements e i Se j , 1 ≤ i ≤ j ≤ n, form a basis of the space S2 V (see §IV.3). Since the symmetric product v1 Sv2 (v i ∈ V) is commutative, it follows that for each w ∈ S2 V there exists a unique decomposition n

(9)

w = ∑ ω ij (e i Se j ), i,j=1

where (ω ij ) = Ω w is a symmetric n × n matrix. Lemma 6.10. Let w ∈ S2 V and g ∈ G. Then Ω gw = T(g)Ω w T(g)󸀠 . Proof. Let {α ij } be the matrix elements of the representation T. Then n

ge k = ∑ α lk (g)e l

(k = 1, . . . , n).

l=1

It follows from formulas (9) and (4.4) that n

n

gw = ∑ ω ij g(e i Se j ) = ∑ ω ij (ge i Sge j ) i,j=1

i,j=1

n

n

n

= ∑ ω ij [( ∑ α pi (g)e p )S( ∑ α qj (g)e q )] p=1

i,j=1

q=1

n

= ∑ ω ij ∑ α pi (g)α qj (g)(e p Se q ). i,j=1

p,q

(10)

IV Products of characters | 159

Therefore, n

n

gw = ∑ γ pq (g)(e p Se q ), p,q=1

where γ pq (g) = ∑ α pi (g)ω ij α qj (g). p,q=1

Consequently, Ω gw = (γ pq (g)) = T(g)Ω w T(g)󸀠 . Corollary 6.11. If w ∈ S2 V, then w ∈ J(S2 V) ⇐⇒ g ∈ J(S2 V) ⇐⇒ Ω w = T(g)Ω w T(g)󸀠 for all g ∈ G.

(11)

Lemma 6.12. Suppose that Ω ≠ 0 is an n × n matrix such that Ω = T(g)ΩT(g)󸀠

for all g ∈ G.

(12)

Then the following statements hold: (a) Ω is nonsingular. (b) For every n × n matrix Ω1 satisfying (12) there exists λ ∈ ℂ such that Ω1 = λΩ. (c) The matrix Ω is either symmetric or skew-symmetric. Proof. Equality (12) can be rewritten in the form ̂ T(g)Ω = Ω T(G)

(g ∈ G),

̂ where T̂ is contragredient to T: T(g) = T(g−1 )󸀠 . Since both representations T and T̂ are irreducible and Ω ≠ 0, it follows, by Schur’s lemma, that Ω is nonsingular, and (a) is proved. If Ω1 = T(g)Ω1 T(g)󸀠 (g ∈ G), then Ω1 Ω−1 = T(g)Ω1 T(g)󸀠 (T(g)ΩT(g)󸀠 )−1 = T(g)Ω1 Ω−1 T(g)−1 so that Ω1 Ω−1 , commuting with all T(g) (g ∈ G), is a scalar operator, by Schur’s lemma. Consequently, Ω1 = λΩ, where λ ∈ ℂ∗ . This proves (b). Transposing both sides of (12) for each g ∈ G, we obtain that Ω󸀠 = T(g)Ω󸀠 T(g)󸀠 , whence, by (b), Ω󸀠 = λΩ, where λ ∈ ℂ. Therefore, Ω = (Ω󸀠 )󸀠 = λΩ󸀠 = λ2 Ω. Since Ω ≠ 0, it follows that λ2 = 1. Hence, either Ω󸀠 = Ω or Ω󸀠 = −Ω, i.e., the matrix Ω is either symmetric or skew-symmetric. Corollary 6.13. If 0 ≠ w ∈ J(S2 V), then the matrix Ω w is nonsingular. Indeed, in that case, Ω w ≠ 0 so it is nonsingular (Lemma 6.12 (a)). Lemma 6.14. We have J(S2 V) ≠ 0 if and only if there exists a nonzero symmetric n × n matrix Ω satisfying (12). Proof. Suppose that J(S2 V) ≠ 0 and 0 ≠ w ∈ J(S2 V). It follows from (9) that Ω w ≠ 0. By Corollary 6.11, the symmetric matrix Ω = Ω w satisfies (12). Conversely, let Ω = (ω ij ) be a nonzero symmetric matrix satisfying (12). Write w = ∑ni,j=1 ω ij (e i Se j ). Then w ≠ 0 and Ω = Ω w . Since w ∈ J(S2 V), it follows from Corollary 6.11 that J(S2 V) ≠ 0.

160 | Characters of Finite Groups 1 Recall that a (complex) n × n matrix A is said to be orthogonal if it is nonsingular and A󸀠 = A−1 . Definition 6.1. A matrix ℂ-representation T of a group G is said to be orthogonal if the matrix T(g) is orthogonal for every g ∈ G. Lemma 6.15. If χ is the character of an irreducible ℂ-representation T of a group G, then ν2 (χ) = 1 if and only if T is equivalent to some orthogonal representation. Proof. By Corollary 6.9 and Lemma 6.14, ν2 (χ) = 1 if and only if there exists a symmetric n × n matrix Ω ≠ 0 satisfying (12). Let ν2 (χ) = 1 and let Ω = (ω ij ) be such matrix. Let deg(T) = n. Since, by Lemma 6.12, det(Ω) ≠ 0, we have Ω = CC󸀠 , where C is a nonsingular n × n matrix. The existence of such C follows from the main theorem of the theory of quadratic forms which claims that every nonsingular complex quadratic form can be transformed into a form with the n × n identity matrix In . Since Ω = T(g)ΩT(g)󸀠 = T(g)CC󸀠 T(g)󸀠 = CC󸀠 , we see, setting T1 (g) = C−1 T(g)C, that T ∼ T1 and T1 (g)T1 (g)󸀠 = C−1 (T(g)C ⋅ C󸀠 T(g)󸀠 )(C󸀠 )−1 = C−1 (CC󸀠 )(C󸀠 )−1 = In . Hence, the matrix T1 (g) is orthogonal for every g ∈ G, that is, T1 is an orthogonal representation of G. Conversely, let T ∼ T1 , where T1 is an orthogonal representation of G. Then, for every g ∈ G, we have T1 (g)T1 (g)󸀠 = In , where n = deg(T), and there exists a nonsingular n × n matrix C such that T1 (g) = C−1 T(g)C for all g ∈ G. We claim that the nonzero symmetric matrix Ω = CC󸀠 satisfies (12). Indeed, T(g)ΩT(g)󸀠 = CT1 (g)C−1 ⋅ CC󸀠 ⋅ (C󸀠 )−1 T1 (g)󸀠 C󸀠 = CT1 (g)T1 (g)󸀠 C󸀠 = CIn C󸀠 = Ω. Hence ν2 (χ) = 1 (this follows from the above lemmas). Lemma 6.16. Every irreducible orthogonal matrix representation of G is equivalent to some real orthogonal representation. In particular, it is realizable over ℝ. Proof. Let T be an irreducible orthogonal matrix representation of G of degree n. By the remark in the end of §I.8, T is equivalent to some unitary representation U: T(g) = S−1 U(g)S (Recall that a matrix A is unitary if A =

(g ∈ G, S ∈ GL(n, ℂ)). (A󸀠 )−1 .)

(13)

Therefore,

T(g) = (T(g)󸀠 )−1 = (S󸀠 U(g)󸀠 (S󸀠 )−1 )−1 = S󸀠 U(g)(S󸀠 )−1 , so that S−1 U(g)S = S󸀠 U(g)(S󸀠 )−1 (see (13)). Setting C = SS󸀠 , we get C−1 U(g)C = U(g). Consequently, for all g ∈ G, we obtain −1

(CC)−1 U(g)CC = C (C−1 U(g)C)C

−1

= C−1 U(g)C = U(g).

(14)

IV Products of characters | 161

Since U is irreducible, CC = λIn is a scalar matrix, by Schur’s lemma. Since the matrix C = SS󸀠 is symmetric, it follows that C = C󸀠 = C∗ , where C∗ is conjugate to C. Consequently, CC = CC∗ is a Hermitian positive matrix, whence λ > 0. Multiplying C by λ−1/2 , we renormalize C so that CC∗ = In . Then C = W −1 DW,

(15)

where the matrix W is unitary, D = diag(α1 In1 , . . . , α k In k ), α1 , . . . , α k are pairwise distinct and |α i | = 1. Transposing both sides of (15), we obtain (recall that the matrix C is symmetric) C = W 󸀠 D(W 󸀠 )−1 .

(16)

Therefore, W −1 DW = W 󸀠 D(W 󸀠 )−1 , whence (WW 󸀠 )−1 D(WW 󸀠 ) = D.

(17)

It follows from (16) and (17) that WW 󸀠 = diag(B1 , B2 , . . . , B k ),

(18)

where B i is a square block of order n i . Put D0 = √D = diag(√α1 In1 , . . . , √α k In k ) (we take √α i arbitrarily). Since WW 󸀠 commutes with D0 , we get W −1 D0 W = W 󸀠 D0 (W 󸀠 )−1 .

(19)

Φ = W −1 D0 W.

(20)

Put

Then, by (19), Φ = Φ󸀠 . Since D0 and W are unitary, so is Φ. Taking into account that D = D20 , we deduce from (15) and (20) that C = Φ2 ,

(21)

that is, some square root of C is symmetric and unitary. Now, by (21), we can rewrite (14) in the form Φ−2 U(g)Φ2 = U(g). Therefore, Φ−1 U(g)Φ = ΦU(g)Φ−1 . Since Φ = (Φ󸀠 )∗ = Φ∗ = Φ−1 , the previous equality becomes Φ−1 U(g)Φ = Φ−1 U(g)Φ. This means that Φ−1 U(g)Φ is real, unitary and, therefore, orthogonal. Thus, the mapping g 󳨃→ Φ−1 U(g)Φ is a real orthogonal representation of the group G equivalent to T.

162 | Characters of Finite Groups 1 Lemma 6.17. If χ is the character of an irreducible representation T of a group G, then ν2 (χ) = 1 if and only if T can be realized over ℝ. Proof. If ν2 (χ) = 1, then the irreducible representation T is equivalent to some orthogonal representation (Lemma 6.15), which is in turn equivalent to some orthogonal real representation (Lemma 6.16). Suppose that the representation T can be realized over ℝ. Then T ∼ T1 , where all entries of matrices T1 (g) (g ∈ G) are real. By the remark following Theorem I.8.2, T1 ∼ T2 , where T2 is a real orthogonal representation. Thus, T ∼ T2 , whence ν2 (χ) = 1 (Lemma 6.15). Theorem 6.2 now follows from Lemmas 6.6 and 6.17. Exercise 6.1 (Burnside). If |G| is odd and χ ∈ Irr(G) − {1G }, then χ ≠ χ. Solution. Assume that χ = χ. If s ∈ G# , then s ≠ s−1 , and we have χ(s) + χ(s−1 ) = χ(s) + χ(s) = 2χ(s). Therefore, in view of ⟨χ, 1G ⟩ = 0, the integer χ(1) = − ∑g∈G# χ(g) is even so does not divide |G|, contrary to the Frobenius–Molien theorem. (We use this exercise in the proof of the Brauer–Suzuki–Wall Theorem V.6.2.) The second solution. Since |G| is odd, the equation x2 = g has only one solution in G. Therefore, ν2 (G) = ∑ χ(g2 ) = ∑ χ(g) = ⟨χ, 1G ⟩ = 0. g∈G

g∈G

It follows from Theorem 6.2 that χ ≠ χ. The third solution. By Brauer’s permutation Lemma X.3.1 (c), the number of realvalued nonprincipal irreducible characters of G equals the number of real G-classes ≠ {1}. The last number equals 0 since |G| is odd. Therefore, all nonprincipal irreducible characters of G are not real-valued. Exercise 6.2. Let χ be an irreducible character of the group G = S4 of degree 3. Compute the characters S2 χ and Λ2 χ, using the character table of G. Solution. The character table of G looks as follows. χ1 χ2 χ3 χ4 χ5

e 1 1 2 3 3

a 1 1 2 −1 −1

b 1 −1 0 1 −1

c 1 1 −1 0 0

d 1 −1 0 −1 1

Here e is the identity element, a = (1, 2)(3, 4), b = (1, 2), c = (1, 2, 3), d = (1, 2, 3, 4). Choose χ = χ4 . Then, computing the inner products ⟨χ24 , χ i ⟩, it is easy to see that (2) χ24 = χ1 + χ3 + χ4 + χ5 . Recall that χ(2) (x) = χ(x2 ). The values of χ4 are (we obtain

IV Products of characters | 163

these values from the above character table) (2)

χ4 (e) = 3,

(2)

(2)

χ4 (a) = 3,

χ4 (b) = 3,

(2)

χ4 (c) = 0,

(2)

χ4 (d) = −1.

Similar calculations show that χ4 = χ1 + χ3 + χ4 − χ5 . Since χ24 − χ4 = 2Λ2 χ4 , we have Λ2 χ4 = χ5 . On the other hand, χ24 = Λ2 χ4 + S2 χ4 . Hence S2 χ4 = χ1 + χ3 + χ4 . (2)

(2)

Exercise 6.3. Let G, G1 be groups such that X(G) = X(G1 ). Let χ ∈ Irr(G), χ1 ∈ Irr(G1 ) be corresponding characters. Show that, generally speaking, ν2 (χ) ≠ ν2 (χ1 ). Solution. Let G = D8 and G1 = Q8 , Irr1 (G) = {χ} and Irr1 (G1 ) = {χ1 }. We have 6 = ϑ2 (G) = 4 + 2ν2 (χ) 󳨐⇒ ν2 (χ) = 1 and 2 = ϑ2 (G1 ) = 4 + 2ν2 (χ1 ) 󳨐⇒ ν2 (χ1 ) = −1. Thus, the representation of G affording χ, is realized over ℝ, but the representation affording χ1 , is not realized over ℝ (here we use Theorem 6.2).

7 Solutions of systems of two-member equations We continue the study of generalized characters ϑ m = |{x ∈ G | x m = g}|. Let kℝ (G) denote the number of real classes of a group G (a G-class is real if it contains a real element; see §II.3). Next, let Irrℝ (G) = {χ ∈ Irr(G) | χ = χ} be the set of real irreducible characters of G. (By Theorem 6.2, χ ∈ Irr(G) is real if and only if ν2 (χ) ∈ {−1, 1}.) Using Brauer’s permutation Lemma X.3.1 (c), it is easy to show that |Irrℝ (G)| = kℝ (G). Theorem 7.1. Suppose that G is a group, m, n ∈ ℕ and s, t ∈ ℕ ∪ {0}. Then the following statements hold: (i) ⟨ϑ2 , ϑ2 ⟩ = kℝ (G). (ii) If exp(G) > 2 and t(G) is the number of involutions in G (obviously, we have t(G) = ϑ2 (G) − 1), then [1 + t(G)]2 < |G|kℝ (G). (iii) |G| | ∑g∈G ϑ m (g)s ϑ n (g)t . (iv) |G| | ∑g∈G ϑ m (g)s . (v) The number of solutions (x, y) ∈ G × G of the equation x2 = y2 in G is equal to |G|kℝ (G) so is divisible by |G|. (vi) |G| divides the number of solutions (x1 , . . . , x s , y1 . . . , y t ) of the system of equations n n x1m = ⋅ ⋅ ⋅ = x m s = y1 = ⋅ ⋅ ⋅ = y t in G. (vii) |G| divides the number of solutions (x1 , . . . , x s ) of the system of equations x1m = ⋅ ⋅ ⋅ = x m s in G.

164 | Characters of Finite Groups 1 Proof. (i) Assume that |G| is even. As ϑ2 = ∑χ∈Irr(G) ν2 (χ)χ, the result follows from Theorem 6.2. (ii) We have ϑ2 (1) = 1 + t(G). Next, if ϑ2 (g) = 0 for all g ∈ G# , then exp(G) = 2. Therefore, since exp(G) > 2, we have, by (i), |G|kℝ (G) = |G|⟨ϑ2 , ϑ2 ⟩ = ∑ ϑ2 (g)2 > ϑ2 (1)2 = (1 + t(G))2 . g∈G

(iii)–(iv) Since ϑ m , ϑ n ∈ Ch(G), so are ϑ sm and ϑ tn . Therefore, ⟨ϑ sm , ϑ tn ⟩ ∈ ℤ. Consequently, |G| = |G|⟨ϑ m (g)s , ϑ n (g)t ⟩ = ∑ ϑ m (g)s ϑ n (g)t g∈G

so (iii) holds. Setting t = 0, we obtain (iv). (v)–(vii) As ϑ2 (g) is the number of solutions of the equation x2 = g, it follows that ϑ2 (g)2 is the number of solutions of the system x2 = y2 = g. It follows that the number of solutions of the equation x2 = y2 in G is equal to ∑g∈G ϑ2 (g)2 = |G|⟨ϑ2 , ϑ2 ⟩, and (v) follows. Assertions (vi) and (vii) follow from (iii) and (iv), respectively. The following theorem was proved by A. MacWilliams [MacW] in case p = 2. The below-presented proof is almost the same as in [MacW]. Theorem 7.2. If G is a p-group such that |G/Φ(G)| ≥ p2k+1 , then ϑ p (1) ≡ 0 (mod p k+1 ). In particular, if k > 0, then p2 | ϑ p (1). In particular, if d(G) > 2, then c1 (G) > 1. Proof. Note that ϑ p (1) = |{x ∈ G | x p = 1}|. By formula (5.1), ϑ p (1) = ∑χ∈Irr(G) ν p (χ)χ(1) and ν p (χ) ∈ ℤ. Let A = {λ ∈ Lin(G) | λ p = 1G }. Since A ≅ G/Φ(G), we get |A| ≥ p2k+1 , by hypothesis. Let O χ be the A-orbit of χ ∈ Irr(G) (A acts on Irr(G) via multiplication) and λ ∈ A; then λ(g p ) = λ p (g) = 1 for g ∈ G and ν p (λχ) = |G|−1 ∑ (λχ)(g p ) = |G|−1 ∑ λ(g p )χ(g p ) g∈G

g∈G

−1

= |G|

p

∑ χ(g ) = |G|

−1

g∈G

∑ χ(p) (g) = ⟨χ(p) , 1G ⟩ = ν p (χ).

g∈G

Therefore, if T is a transversal of the set of all A-orbits, then ∑ ν p (χ)|O χ |χ(1) = ϑ p (1).

χ∈T

It suffices to show that for χ ∈ T one has |O χ |χ(1) ≡ 0 (mod p k+1 ). We need to consider only those χ ∈ T for which χ(1) ≤ p k . Let χ(1) = p k−e (e ≥ 0). If C χ is the A-stabilizer of χ, then |O χ | = |A : Cχ |. Let N = ⋂λ∈C χ ker(λ). Then, by §I.1, |G : N| = |C χ |. If g ∈ ̸ N, there exists λ ∈ C χ such that λ(g) ≠ 1. Since λχ = χ, it follows that λ(g)χ(g) = χ(g), whence χ(g) = 0, that is, χ vanishes on G − N. Since |G| = |G|⟨χ, χ⟩ = ∑ |χ(g)|2 = ∑ |χ(g)|2 ≤ |N|χ(1)2 , g∈G

g∈N

IV Products of characters | 165

it follows that χ(1)2 ≥ |G : N| = |C χ |. Consequently, |C χ | ≤ |χ(1)|2 = (p k−e )2 = p2k−2e . This gives

|O χ |χ(1) = |A : C χ |p k−e ≥ p2k+1−(2k−2e)+k−e = p k+e+1 .

Therefore, ϑ p (1) = ∑ ν p (χ)|O χ |χ(1) ≡ 0 (mod p k+1 ). χ∈T

There are known deeper results about function ϑ p (1) on p-groups (see [Ber31, Theorem 13.2 (a)]). Let G be an extraspecial group of order 22k+1 ; then |G/Φ(G)| = 22k > 22(k−1)+1 . The group G has only one nonlinear irreducible character and its degree is equal to 2k . Therefore, by Theorem 6.1, ϑ2 (1) = 22k ± 2k is divisible by 2k = 2(k−1)+1 but not divisible by 2k+1 = 2(k−1)+2 . Thus, the result of Theorem 7.2 is best possible. For example, if k = 2, then ϑ2 (1) ∈ {12, 20} is not divisible by 23 = 2k+1 .

8 Characters of direct products We will now consider the characters of direct products. Let G i be groups, let G = G1 × ⋅ ⋅ ⋅ × G n , and let T i be a representation of degree n i of G i , i = 1, . . . , n. If g = (g1 , . . . , g n ) ∈ G, we put T(g) = T1 (g1 ) ⊗ ⋅ ⋅ ⋅ ⊗ T n (g n ). It follows from Exercise 1.5 that the mapping g 󳨃→ T(g) is a representation of G. This representation is called the exterior tensor product of representations T1 , . . . , T n and denoted by T1 × ⋅ ⋅ ⋅ × T n . Let χ i = χ T i (i = 1, . . . , n). The character of the representation T is called the exterior tensor product of characters χ1 , . . . , χ n and denoted by χ1 × ⋅ ⋅ ⋅ × χ n . We have (χ1 × ⋅ ⋅ ⋅ × χ n )(g) = χ1 (g) . . . χ n (g), where g = (g1 , . . . , g n ) ∈ G. Theorem 8.1. If G = G1 × ⋅ ⋅ ⋅ × G n , then Irr(G) = {χ1 × ⋅ ⋅ ⋅ × χ n | χ i ∈ Irr(G i ), i = 1, . . . , n}. For every χ ∈ Irr(G) there exists a unique decomposition χ = χ1 × ⋅ ⋅ ⋅ × χ n , where χ i ∈ Irr(G i ), i = 1, . . . , n. Proof. Let χ i , θ i ∈ Irr(G i ),

i = 1, . . . , n,

χ = χ1 × ⋅ ⋅ ⋅ × χ n ,

It is easy to check that ⟨χ, θ⟩ = ⟨χ1 , θ1 ⟩ × ⋅ ⋅ ⋅ × ⟨χ n , θ n ⟩.

θ = θ1 × ⋅ ⋅ ⋅ × θ n .

(1)

166 | Characters of Finite Groups 1 In particular, if all the χ i , θ i are irreducible, then the last equality and the First Orthogonality Relation yield ⟨χ, θ⟩ = δ χ,θ , where δ is the Kronecker delta on the set Irr(G). Therefore, we have k(G1 ) . . . k(G n ) pairwise distinct irreducible characters χ1 × ⋅ ⋅ ⋅ × χ n of G. The result now follows from the equality |Irr(G)| = k(G) = k(G1 ) . . . k(G n ). Recall that if χ is an irreducible character of a group G, then Z(χ) = {x ∈ G | |χ(x)| = χ(1)} (as we know, the subgroup Z(χ) which is called the quasikernel of χ, is G-invariant). Below we use a partial case of Clifford’s Theorem VII.1.3: if χ ∈ Irr(G), then the restriction χZ(χ) = χ(1)λ, where λ ∈ Lin(Z(χ)) (this is clear since Z(χ)/ker(χ) is abelian). Theorem 8.2. If G = G1 × ⋅ ⋅ ⋅ × G n and θ = χ1 × ⋅ ⋅ ⋅ × χ n , where χ i ∈ Irr(G i ) (i = 1, . . . , n), then the following statements hold true: (a) Z(θ) = Z(χ1 ) × ⋅ ⋅ ⋅ × Z(χ n ). (b) We have |Z(θ) : ker(θ)| = m, where m is the least common multiple of the numbers |Z(χ i ) : ker(χ i )| (i = 1, . . . , n). Proof. Statement (a) can be verified immediately. Let us prove (b). By the paragraph preceding the theorem, the restriction (χ i )Z(χ i ) = χ i (1)λ i (λ i ∈ Lin(Z(χ i )), i = 1, . . . , n. If g = (g1 , . . . , g n ) ∈ Z(θ), it follows from Theorem 8.1 that g i ∈ Z(χ i ) (i = 1, . . . , n), so that, since θ(1) = χ1 (1) . . . χ n (1), we get θ(g) = χ1 (1)λ1 (g1 ) . . . χ n (1)λ n (g n ) = θ(1)λ1 (g1 ) . . . λ n (g n ). Since ker(θ) ≤ Z(θ), it follows that ker(θ) = {(g1 , . . . , g n ) ∈ Z(χ1 ) × ⋅ ⋅ ⋅ × Z(χ n ) | λ1 (g1 ) . . . λ n (g n ) = 1}. Note that λ i (Z(χ i )) = {λ i (z) | z ∈ Z(χ i )} coincides with the group W m i of all m i -th roots of 1, where m i = |Z(χ i ) : ker(λ i )| = |Z(χ i ) : ker(χ i )| (see the last assertion of Theorem III.5.2). Since the character λ i takes each value ϵ i ∈ W m i exactly |ker(λ i )| = |ker(χ i )| times, we obtain n

|ker(θ)| = ∏ |ker(χ i )| ⋅ |{(ϵ1 , . . . , ϵ n ) ∈ W m1 × ⋅ ⋅ ⋅ × W m n | ϵ1 . . . ϵ n = 1}|. i=1

If (ϵ1 , . . . , ϵ n ) ∈ W m1 × ⋅ ⋅ ⋅ × W m n , then (ϵ1 . . . ϵ n )m = 1, where m is the least common multiple of the numbers m i , i = 1, . . . , n. Consider the homomorphism ψ : (ϵ1 , . . . , ϵ n ) 󳨃→ ϵ1 . . . ϵ n of the group W m1 × ⋅ ⋅ ⋅ × W m n into the group W m of all m-th roots of unity. Let ϵ ∈ W m .

IV Products of characters | 167

Since GCD( mm1 , . . . , mmn ) = 1, there exist integers u i such that mm1 u1 + ⋅ ⋅ ⋅ + mmn u n = 1. Taking ϵ i = ϵ mu i /m i (i = 1, . . . , n), we obtain ϵ i ∈ W m i and ϵ1 . . . ϵ n = ϵ. Therefore ψ is a surjection. Since ker ψ = {(ϵ1 , . . . , ϵ n ) ∈ W m1 × ⋅ ⋅ ⋅ × W m n | ϵ1 . . . ϵ n = 1}, it follows that |{(ϵ1 , . . . , ϵ n ∈ W m1 × ⋅ ⋅ ⋅ × W m n | ϵ1 . . . ϵ n = 1}| = |W m |−1 |W m1 × ⋅ ⋅ ⋅ × W m n | m1 . . . m n = . m Consequently, |ker θ| = |ker(χ1 )| . . . |ker(χ n )| ⋅

m1 . . . m n |Z(χ1 )| . . . |Z(χ n )| |Z(θ)| = = , m m m

and the result follows. Let G = Z1 × Z2 , where Z1 ≅ Cp ≅ Z2 and let χ ∈ Lin(G) be nonprincipal such that ker(χ) = Z3 , where Z3 ≠ Z i , i = 1, 2. Set χ i = χ Z i , i = 1, 2. Then χ = χ1 ×χ2 , ker(χ i ) = {1}, i = 1, 2. Therefore, ker(χ) > ker(χ1 ) × ker(χ2 ). Corollary 8.3. Let G[n] be the direct product of n copies of a group G, let χ be a character of G, and let θ = χ × ⋅ ⋅ ⋅ × χ (n times). Then |ker(θ)| = |Z(χ)|n−1 |ker(χ)|. In particular, if χ is faithful then |ker(θ)| = |Z(G)|n−1 . Theorem 8.4 (I. Schur). If χ ∈ Irr(G), then χ(1) divides |G : Z(χ)|. Proof (J. Tate). For a positive integer n, let G[n] = G × ⋅ ⋅ ⋅ × G (n times),

θ n = χ × ⋅ ⋅ ⋅ × χ (n times) .

As θ n ∈ Irr(G[n] ), it follows from Proposition III.3.3 that θ n (1) divides |G[n] : ker(θ n )|. Using Corollary 8.3, we obtain χ(1)n | |G : Z(χ)|n ⋅ m, where m = |Z(χ) : ker(χ)| is independent of n. Let |G : Z(χ)| a = χ(1) b be a reduced fraction. Then we conclude that b = 1.

an bn m

∈ ℤ, so that b n | m. Since n was chosen arbitrarily,

Theorem 8.4 is a generalization of the Frobenius–Molien theorem, but it is not most strong result (see Ito’s theorem on degrees, Theorem VII.2.3). Remark. Let us present another proof of the Second Orthogonality Relation assuming the first one (this proof is due to Mann). Let the direct product G × G act on G as a permutation group with the action defined as x(a,b) = a−1 xb, x ∈ G, (a, b) ∈ G × G. Let π be the character of this permutation representation of the group G × G. Then π((a, b)) = |{x ∈ G | x(a,b) = x}| = |{x ∈ G | x−1 ax = b}

168 | Characters of Finite Groups 1 so that π((a, b)) = |CG (a)|δ K a ,K b , where K a , K b are G-classes containing a, b, respectively. Let Irr(G) = {χ1 , . . . , χ r }. Then Irr(G × G) = {χ i × χ j | | i, j = 1, . . . , r}. Suppose that π = ∑ri,j=1 m ij (χ i × χ j ). Let us find the multiplicities m ij . We have m ij = |G × G|−1 π((a, b))(χ i χ j )((a, b)) = |G|−2 ∑ |CG (a)|χ i (a)χ j (b) a∼b

= |G|−2 ∑ |K a |CG (a)|χ i (a)χ j (a) a∈G

−1

= |G|

∑ χ i (a)χ j (a) = ⟨χ i , χ j ⟩ = δ ij .

a∈G

It follows that π = ∑ri=1 (χ i × χ i ), and we conclude that ∑ri=1 χ i (a)χ i (b) = |CG (a)|δ K a ,K b . It follows from equation (6.8) that Irr(χ(2) ) ⊆ Irr(χ2 ). We use this fact to prove the following: Theorem 8.5 (Isaacs–Zisser [IZ, Theorem 1]). For a group G the following assertions are equivalent: (a) There exists a faithful irreducible character χ of G satisfying χ2 = aψ, where a ∈ ℕ and ψ ∈ Irr(G). (b) G is a direct product of a cyclic 2-group and a group of central type¹ and odd order. Proof. (a) ⇒ (b): By the paragraph preceding the theorem, we have Irr(χ(2) ) ⊆ Irr(χ2 ) so χ(2) = bψ for some b ∈ ℤ (recall that χ(2) ∈ Ch(G)). We have χ(2) (1) = χ(12 ) = χ(1) so b ∈ ℕ. By the above, χ2 = cχ(2) , where c = ba . Since χ(1)2 = cχ(2) (1) = cχ(1), we get c = χ(1), and we obtain χ(g)2 = aψ(g) = χ(1)χ(g 2 ) for all g ∈ G.

(1)

Replacing g by g2 in (1), we get χ(g2 )2 = χ(1)χ(g 4 ), g ∈ G. Therefore, squaring equality (1), we obtain χ(g)4 = χ(1)2 χ(g2 )2 = χ(1)2 χ(1)χ(g 4 ) = χ(1)3 χ(g 4 ). By appropriate induction, χ(g)2 = χ(1)2 n

n

−1

χ(g 2 ) for all g ∈ G. n

(2)

We claim now that every 2-element z is in the center of G. Indeed, if o(z) = 2n , n n then by (2), χ(z)2 = χ(1)2 so |χ(z)| = χ(1), and our claim follows since the character χ is faithful. By the Schur–Zassenhaus theorem, we get G = P × H, where P ∈ Syl2 (G). By the above, P ≤ Z(G) so P is cyclic since P is abelian and χ is faithful. It remains to show that H is of central type. Note that the restriction χ H is irreducible and faithful (see Theorem 8.1). It suffices to show that for each g ∈ G of odd order, either χ(g) = 0 1 Recall that a group G is of central type there is χ ∈ Irr(G) such that χ(1)2 = |G : Z(G).

IV Products of characters | 169

or else g ∈ Z(G). Let g be such an element and o(g) = m. Since m is odd, we have 2n ≡ 1 (mod m), where n = φ(m) (here φ(∗) is Euler’s totient function). Therefore, n g2 = g, hence (2) implies χ(g)2 = χ(1)2 n

n

−1

χ(g) ⇒ χ(g)(χ(g)2

n

−1

− χ(1)2

n

−1

) = 0.

If χ(g) ≠ 0, then |χ(g)| = χ(1), i.e., in view of faithfulness of χ, one has g ∈ Z(χ) = Z(G), as required. (b) ⇒ (a): Suppose that G = Z × H, where Z is a cyclic Sylow 2-subgroup of G and H is of central type. Let ϕ be a faithful irreducible character of H vanishing on H − Z(H). Then the relation ϕ(h)2 = ϕ(1)ϕ(2) (h) = ϕ(1)ϕ(h2 )

(∗)

holds for all h ∈ H. Indeed, if h ∈ Z(H), then ϕ(h) = ϕ(1)λ(h) for some λ ∈ Lin(H); then ϕ(h2 ) = ϕ(1)λ(h2 ). In the case, ϕ(h)2 = ϕ(1)2 λ(h)2 = ϕ(1) ⋅ ϕ(1)λ(h)2 = ϕ(1) ⋅ ϕ(1)λ(h2 ) = ϕ(1)ϕ(h2 ). If h ∈ H − Z(H), then both sides of (∗) are equal to 0, by hypothesis (see the definition of characters of central type). Since G = Z × H, we have χ = τ × ϕ, where τ ∈ Lin(Z) is faithful and ϕ ∈ Irr(H) (see Theorem 8.1). By Theorem 8.2, ϕ is also faithful. Now, χ2 = τ2 × ϕ2 = τ2 × ϕ(1)ϕ(2) = ϕ(1)(τ2 × ϕ(2) ) = χ(1)(τ2 × ϕ(2) ). It follows that χ2 = χ(1)ψ, where ψ = τ2 × ϕ(2) . Since τ2 ∈ Lin(Z) and ϕ(2) ∈ Irr(H), we get ψ ∈ Irr(G), completing the proof. Theorem 8.6 ([IZ]). Suppose that a faithful χ ∈ Irr1 (G) satisfies χ2 = aχ + bχ,

(3)

where a, b ∈ ℕ ∪ {0}. Then the following holds: (a) G has an odd order and χ(1) = a + b. (b) If g ∈ G# and χ(g) ≠ 0, then χ(g) + χ(g) = a − b and χ(g)χ(g) = b2 − ab. Proof. By hypothesis, the group G is nonabelian. Evaluating formula (3) at 1 ∈ G, we get χ(1)2 = (a + b)χ(1) so that χ(1) = a + b since χ(1) ≠ 0. We claim the following: (i) The only real value that χ can have on an element from G# is zero. Indeed, if χ(x) ∈ ℝ for x ∈ G# , then, evaluating (3) at x, we get χ(x)2 = (a + b)χ(x) so, if χ(x) ≠ 0, we get χ(x) = a + b = χ(1), by (i), and we conclude that x ∈ ker(χ) = {1}, a contradiction. In particular, χ ≠ χ so that ν2 (χ) = 0 (Theorem 6.2). (ii) Now we complete the proof of (a). Using the generalized character χ(2) , we get χ(2) = uχ + vχ for u, v ∈ ℤ (see the paragraph preceding Theorem 8.5). If |G| is even, pick an involution t ∈ G. Since t is a real element, it follows from (i) that χ(t) = 0 so that χ(1) = χ(t2 ) = χ(2) (t) = (u + v)χ(t) = 0, a contradiction. Thus, |G| is odd.

170 | Characters of Finite Groups 1 (iii) Let us prove (b). It follows from (3) that χ2 = aχ + bχ. Subtract this equality from (3) to obtain χ(g)2 − χ(g)2 = (a − b)(χ(g) − χ(g)),

g ∈ G# .

Since χ(g) − χ(g) ≠ 0, by (i), the displayed equality implies χ(g) + χ(g) = a − b. Now, if we add equation (3) to its complex conjugate and use the above displayed formula, we obtain χ(g)2 + χ(g)2 = (a + b)(χ(g) + χ(g)) = (a + b)(a − b) = a2 − b2 , and therefore (a − b)2 = (χ(g) + χ(g))2 = (a2 − b2 ) + 2χ(g)χ(g) 󳨐⇒ χ(g)χ(g) = b2 − ab. Until the end of this section, we assume that χ is the character from Theorem 8.6. Observe that if χ ∈ Lin(G), then a + b = χ(1) = 1 and there are two possibilities. If a = 1 and b = 0, then χ2 = χ, and this yields |G| = 1. If a = 0 and b = 1, then χ2 = χ. In that case, χ is linear so χ3 = χχ = 1G , and we conclude that |G| ∈ {1, 3}. We assume now that |G| > 3, and so χ is nonlinear. Consider the character χχ. Since ⟨χχ, χ⟩ = ⟨χχ, χ⟩ = ⟨χ2 , χ⟩ = a

and

⟨χχ, 1G ⟩ = ⟨χ, χ⟩ = 1,

we get χχ = 1G + aχ + aχ + ψ,

(4)

where ψ is either a character of G or the zero function. Using (4), we compute ψ(1) = χ(1)2 − 2aχ(1) − 1 = (a + b)2 − 2a(a + b) − 1 = b2 − a2 − 1. This quantity can be zero only if b = 1 and a = 0; then χ is linear, which is not the case. It follows that ψ is actually character, and we define the subgroup N = ker(ψ).

(5)

We deduce also that b > a, since ψ(1) > 0. Theorem 8.7 ([IZ]). If G, χ, N, ψ are as above, then the elements of N are precisely the elements of G on which χ does not vanish (i.e., Tχ , the set of zeros of χ, coincides with G − N). Proof. If χ(g) = 0, we see from (4) that ψ(g) = −1 so g ∈ ̸ ker(ψ) = N 󳨐⇒ Tχ ∩ N = 0. Conversely, if g ∈ G# and χ(g) ≠ 0, then by Theorem 8.6 (b), χ(g) + χ(g) = a − b and χ(g)χ(g) = b2 − ab. Using this and (4), we obtain b2 − ab = χ(g)χ(g) = 1G (g) + a(χ(g) + χ(g)) + ψ(g) = 1 + a(a − b) + ψ(g), and thus ψ(g) = b2 − a2 − 1. Since this number is also equal to ψ(1), we conclude that g ∈ N. It follows from what has been proved already that N > {1}. For further information on the considered situation, see the Appendix to Chapter VII.

IV Products of characters | 171

9 The vanishing of defect zero characters This section, written by Isaacs, presents an argument of M. Leitz [Leit] giving an elementary proof that p-defect zero characters vanish on p-singular elements. Definition. Let p be a prime. A character χ ∈ Irr(G) is of p-defect zero if p ∤ element x is said to be p-singular if p | o(x).

|G| χ(1) .

An

We begin with a bit of notation. Given m ∈ ℕ, we consider m-tuples [x] ∈ G(m) . In other words, we have [x] = (x1 , . . . , x m ) with x1 , . . . , x m ∈ G. In this situation we write χ󸀠 ([x]) = χ(x1 )χ(x2 ) . . . χ(x m ) for a character χ of G (i.e., χ󸀠 is a character of the group G(m) , the direct product of m copies of G). Given g ∈ G, we write T m (g) to denote the set of m-tuples [x] for which x1 x2 . . . x m = g. Lemma 9.1. If χ ∈ Irr(G) and g ∈ G, then χ󸀠 ([x]) = (

∑ [x]∈T m (g)

|G| m−1 χ(g). ) χ(1)

Proof. Let e be the central idempotent corresponding to the character χ (see the proof of Lemma III.4.3). The coefficient of x ∈ G in e is exactly (χ(1)/|G|)χ(x), and so comparison of the coefficients of g in both sides of the equation e m = e yields the equality χ(1) m χ(1) ( ) )χ(g), ∑ χ(x1 ) . . . χ(x m ) = ( |G| |G| [x]∈T (g) m

where the sum runs over all m-tuples in T m (g). The result now follows.

Now let S m (g) be the union of the disjoint sets T m (x) as x runs over the conjugacy class of g ∈ G. If σ is the cyclic “shift” operator defined by [x]σ = (x m , x1 , x x , . . . , x m−1 ), we see that the cyclic group ⟨σ⟩ permutes the set S m (g). If σ r fixes [x] ∈ S m (g), where r | m, then [x] must be the concatenation of mr identical r-tuples, and thus g = h m/r for some element h ∈ G. Theorem 9.2 (= Theorem VIII.4.1). Suppose that χ ∈ Irr(G) is such that a prime p does |G| not divide χ(1) (i.e., χ is of p-defect 0). If g ∈ G is p-singular, then χ(g) = 0. In other words, an irreducible character of defect 0 vanishes on all elements of G whose orders are divisible by p. Proof. Let u = p a be the p-part of |G| and fix an integer m = uv, where v is a power of p. Let ⟨σ⟩ be the cyclic shift group of order m acting on the set S m (g), and note that since g has order divisible by p, we cannot have g = h t for any element h ∈ G, where t > 1 is a power of p (otherwise, o(g) will not divide |G|). Then σ v fixes no member of S m (g), and it follows that each ⟨σ⟩-orbit on S m (g) has size divisible by pv. The quantity χ󸀠 ([x]) is constant on ⟨σ⟩-orbits, and thus ∑[x]∈S m (g) χ󸀠 ([x]) is divisible by pv in the ring of algebraic integers. By the lemma, however, ∑ [x]∈S m (g)

χ󸀠 ([x]) = |K g |(

|G| m−1 χ(g), ) χ(1)

172 | Characters of Finite Groups 1 where K g is the conjugacy class of g in G. Since |G|/χ(1) is an integer not divisible by p, we see that χ(g) is divisible by arbitrarily large powers of p in the ring of algebraic integers. This implies that χ(g) = 0, as desired. (One may see this if to observe that the norm of χ(g) is a rational integer divisible by arbitrarily large power of p.)

10 On products of irreducible characters We begin with the following: Exercise. Let χ be a character of a group G > {1} and Z(χ) = {1}. Then, by Theorem 4.3, there exists a integer n > 0 such that Irr(χ n ) = Irr(G) (the smallest n with this property will be denoted by s(χ)). If G is nonabelian and s(G) = max {s(χ) | χ ∈ Irr1 (G)} < ∞, then G󸀠 is the only minimal normal subgroup in G. Furthermore, G is a nonabelian simple group if and only if s(G) = max {s(χ) | χ ∈ Irr# (G)} < ∞. Hint. For μ ∈ Irr(G), use the relation ⟨χ n , μ⟩ μ(1) 1 χ(g) n = + ] μ(g). ∑[ n χ(1) |G| |G| χ(1) # g∈G

χ(g) Since Z(χ) = {1}, it follows that | χ(1) | < 1 for g ∈ G# . Therefore,

lim

n→∞

⟨χ n , μ⟩ μ(1) = . χ(1)n |G|

Hence ⟨χ n , μ⟩ > 0 for a sufficiently large n. M. Herzog and D. Chillag have proved the following result: Theorem. If G is a nonabelian simple group, then s(G) ≤ k(G)2 − 3k(G) + 4. In this section we present some results similar to those considered in this chapter, without proofs. They are due to Arad, Blau, Chillag, Fisman, Herzog and Mann. Let χ be a character of G with Z(χ) = {1}. Then (see the previous exercise) there exists a least natural number n = s(χ) such that Irr(χ n ) = Irr(G). Theorem 10.1. Let G be a nonabelian simple group, k = k(G). (a) If k is even, then s(G) ≤ k2 − 4k + 6. (b) If G is real (that is, x ∼ x−1 for all x ∈ G), then s(G) ≤ 2k − 2. (c) If Irr(G) = {1G = χ1 , χ2 , . . . , χ k }, then Irr# (G) ⊆ Irr(χ1 . . . χ k ) and

Irr(G) = Irr((χ1 χ2 . . . χ k )2 ).

(d) If s(G) = 2, then G ≅ J1 , the first Janko simple group. (e) If χ ∈ Irr# (G), then s(χ) ≤ (k − 1)2 ; if, additionally, χ is real, then s(G) ≤ 2k − 2.

IV Products of characters | 173

I. Zisser has proved that for every nonabelian simple group G, one has the equality Irr(G) = Irr(χ1 χ2 . . . χ k ). It is conjectured that s(G) ≤ k for a nonabelian simple group G. A group G is called a CF-group if there is F ⊲ G such that, for every a ∈ G − F, all elements of the coset aF are conjugate in G. In that case {G, F} is called a CF-pair. Theorem 10.2 (Arad, Mann, Fisman). A group G has two distinct irreducible characters χ and ψ such that χψ = mχ + nψ, m, n ∈ ℕ, if and only if G possesses two normal subgroups M < N such that {G/M, N/M} is a CF-pair and N/M is an abelian minimal normal subgroup of G/M having exactly two nontrivial G/M-classes, both of which are real. Theorem 10.3 (Blau, Chillag). Let χ ≠ 1G , ψ be irreducible characters of a group G such that χ n = kψ for certain natural numbers n ≥ 2 and k. Then χ vanishes on G − Z(χ), ψ = χ(n) and |G|π(n) divides |Z(χ)|. (See Theorems 8.6 and 8.7 for case n = 2.) Conjecture. If G is a nonabelian simple group and χ is an irreducible character of G of maximal degree, then |Irr(χ i )| ≥ i for each i ≤ k(G). Results concerning products of characters are in fact strikingly similar to those concerning products of conjugacy classes. Earlier in this chapter we noted some results of this kind (see the theorems of Brauer–Burnside and Garrison). This should have been expected, in view of the fact that knowledge of either the character table or the class multiplication table enables us to determine the other. In recent work of Arad, Blau and Fisman, this similarity is elucidated clear in the theory of table algebras.

11 On the number of solutions of x n = 1 in a group This section is a reworked version of the classical paper [Fro8] by G. Frobenius. Recall that, given g ∈ G and n ∈ ℕ, ϑ n (g) is the number of solutions of x n = g in G (see §IV.6). If x is such a solution, then ⟨g⟩ ≤ ⟨x⟩. The case g = 1 is more important. We present a proof of the following fundamental result. Theorem 11.1 (Frobenius). Let G be a group of order h, n | h and ϑ(g) =

{h 1 (n) ρ (g) = { n n 0 {

if g n = 1, if g n ≠ 1.

Then ϑ = 1n ρ(n) is a generalized character of G.

Recall that μ(∗) is the number-theoretic Möbius function defined by 1 { { { μ(d) = {(−1)r { { {0

if d = 1, if d is a product of r > 0 pairwise distinct primes, otherwise.

We need the following two lemmas.

174 | Characters of Finite Groups 1 Lemma 11.2 (Möbius inversion formula). Let K be a commutative ring, let f : ℕ → K and g : ℕ → K be functions satisfying g(n) = ∑d|n f(d) for all n ∈ ℕ. Then n f(n) = ∑ g(d)μ( ) d d|n for all n ∈ ℕ. Lemma 11.3. Let d ∈ ℕ and let P d be the set of all primitive d-th roots of 1. Then ∑ ϵ = μ(d).

ϵ∈P d

Proof. Let W m be the group of m-th roots of 1. Then W m = ⋃d|m P d is a partition. Setting ∑ϵ∈P d ϵ = ψ(d), we obtain ∑ ψ(d) = ∑ ϵ = 1 + ξ + ⋅ ⋅ ⋅ + ξ m−1 . d|m

ϵ∈W m

Therefore, ∑d|m ψ(d) = δ(m), where {1 if m = 1, δ(m) = { 0 if m > 1. { Applying the Möbius inversion formula (see Lemma 11.2), we obtain n ψ(m) = ∑ δ(d)μ( ) = μ(m) d d|m

for all m ∈ ℕ.

Therefore, ψ = μ. Thus, we have ∑ μ(d) = δ(m).

(∗)

d|m

Proof of Theorem 11.1. Let R be the standard regular matrix representation of the x group G: if g ∈ G and L(g) = (gx ) is the corresponding left shift, then R(g) is the matrix of the permutation L(g) (the group G is assumed to be ordered in some fixed way). If o(g) = d, then L(g) is decomposed in a product of dh independent cycles of length d (one of such cycles is (x, gx, . . . , g d−1 x)). Therefore, there exists some enumeration of the elements of G such that the matrix R(g) consists of dh identical blocks B d of size d × d: 00 ... 01 10 ... 00 ( 0 1 . . . 0 0) Bd = .. .. . . (0 0

...

1 0)

(the matrix of a d-cycle); these blocks lie on the leading diagonal.

IV Products of characters | 175

Let X g (t) be the characteristic polynomial of the matrix R(g): X g (t) = det(t Ih −R(g)), where Ih is the h × h identity matrix. As det(tId − B d ) = t d − 1, we get X g (t) = (det(tId − B d ))h/d = (t d − 1)h/d

(d = o(g)).

(1)

Let X g (t) = t h − σ1 (g)t h−1 + ⋅ ⋅ ⋅ + (−1)h σ h (g). It follows that σ k = ⋀k ρ is the exterior k-th power of the regular character ρ = ρ G (k ∈ {1, . . . , h}) (see §IV.2). In particular, we have σ h (g) = det(R(g)) = det(ρ)(g), that is, σ h = det(ρ) is a real linear character of our group G. Note that σ h (g) = sgn(L(g)) is the sign of the permutation L(g): since σ h (g) = (−1)h X g (0), it follows from (1) that σ h (g) = (−1)h−h/d , where h − dh is the decrement of the permutation L(g). It follows from (1) that (1 − t d )h/d = (−1)h/d X g (t) = (−1)h σ h (g)X g (t). Since ((−1)h σ h (g))2 = 1, we get (1 − t d )h/d = 1 − ψ1 (g)t + ⋅ ⋅ ⋅ + (−1)h ψ h (g)t h = F g (t), where ψ k = σ h σ h−k (k = 1, . . . , h, ψ h = σ h ) are real characters of G. Given χ ∈ Irr(G), we obtain ∑ F g (t)χ(g) = hΦ χ (t), g∈G

where Φ χ (t) = ⟨1G , χ⟩ − ⟨ψ1 , χ⟩t + ⋅ ⋅ ⋅ + (−1)h ⟨ψ h , χ⟩t h ∈ ℤ[t].

(2)

As all the characters ψ k are real, we have ⟨ψ k , χ⟩ = ⟨ψ k , χ⟩. Therefore, ∑ F g (t)χ(g) = hΦ χ (t),

that is,

g∈G

∑ χ(g)(1 − t o(g) )h/o(g) = hΦ χ (t).

g∈G

Given r | h, we put f χ (r) = ∑ χ(g).

(3)

o(g)=r

Then hΦ χ (t) = ∑ f χ (r)(1 − t r )h/r .

(4)

r|h

Given a polynomial P(t) ∈ ℂ[t], let C t n (P) denote the coefficient of t n in P. Since f χ (1) = χ(1), it follows from (4) that C t (Φ χ ) = −χ(1).

(5)

176 | Characters of Finite Groups 1 Now we consider a special case where G = ⟨s⟩ ≅ Cm is the cyclic group of order m and χ a faithful linear character of G. If d | m, we have f χ (d) = ∑o(g)=d χ(g), the sum of all primitive d-th roots of 1. Thus, by Lemma 11.3, f χ (d) = μ(d). Relation (4) now becomes mΦ χ (t) = ∑ μ(r)(1 − t r )m/r . r|m

This shows that, in the case under consideration, Φ χ (t) is independent of the choice of the faithful linear character χ of G. We may therefore use the notation Φ(m) (t) instead of Φ χ (t). Thus, for every m ∈ ℕ, we have, using (∗), that mΦ(m) (t) = ∑ μ(r)(1 − t r )m/r r|m

or mΦ(m) (t) = ∑ μ(r)(1 − t r )s .

(6)

rs=m

Therefore, ∑ dΦ(d) (t m/d ) = ∑ ∑ μ(r)(1 − t mr/d )s = ∑ μ(r)(1 − t m/s )s d|m

rs|m

d|m rs=d

= ∑ (1 − t m/s )s ∑ μ(r) = ∑ (1 − t m/s )s δ( r| ms

s|m

s|m

m ) = (1 − t)m , s

since ∑ μ(r) = δ(

r| ms

Thus,

{0 if s ≠ m, m )={ s 1 if s = m. {

(1 − t)m = ∑ dΦ(d) (t m/d ) d|m

(the same result can easily be derived from Lemma 11.2), whence we obtain (1 − t r )h/r = ∑ dΦ(d) (t h/d ).

(7)

d| hr

Returning to the general case and using formulas (4) and (7), we see that, for every χ ∈ Irr(G), hΦ χ (t) = ∑ f χ (t) ∑ dΦ(d) (t h/d ) = ∑ f χ (t)dΦ(d) (t h/d ). r|h

d| hr

(8)

rd|h

If rd | h, we put h = rds. Then (8) can be rewritten as Φ χ (t) = ∑ f χ (r) rs|h

1 (h/rs) rs Φ (t ). rs

(9)

IV Products of characters | 177

If d | h, we put (10)

v χ (d) = d−1 ∑ f χ (r). r|d

Then one can rewrite (9) as follows: Φ χ (t) = ∑ v χ (d)Φ(h/d) (t d ).

(11)

d|h

If n | h and d | h, put c n,d = C t n (Φ(h/d) (t d )) (see the paragraph following (4)). Then ∈ ℤ[t] implies c n,d ∈ ℤ. Taking into account that the degrees of the nonconstant terms of the polynomial Φ(h/d) (t d ) are ≥ d, we conclude that Φ(h/d) (t d )

c n,d = 0

for n < d.

(12)

Applying (5) to the case in which G is cyclic of order m and χ ∈ Lin(G) is faithful (then Φ χ = Φ(m) ), we obtain C t (Φ(m) ) = −1. Therefore, c n,n = C t n (Φ(h/n) (t n )) = −1.

(13)

It now follows from (11) that ∑ v χ (d)C t n (Φ(h/d) (t d )) = C t n (Φ χ ),

i.e.,

∑ c n,d v χ (d) = C t n (Φ χ ).

d|h

d|h

In view of (12) and (13), we obtain v χ (n) = a χ (n) +

∑

c n,d v χ (d),

(14)

d|h, d 1 since G is nonabelian. If λ ∈ Irr(χ2 ) ∩ Lin(G), then ⟨χ, λχ⟩ = ⟨χ2 , λ⟩ > 0 ⇐⇒ χ = λχ, ⟨χ2 , λ⟩ = 1. Therefore: Claim (i). We have χ2 = λ1 + ⋅ ⋅ ⋅ + λ k , where λ1 , . . . , λ k are pairwise distinct linear characters of G such that λ i χ = χ for each i. In particular, k = χ(1)2 . Claim (ii). The character χ vanishes on G − Z(G). Indeed, let x ∈ G. Applying (i), we obtain k

χ2 ⋅ χ = ∑ λ i χ = kχ, i=1

whence χ(χχ − k ⋅ 1G ) = 0. Hence, if χ(x) ≠ 0, then |χ(x)|2 = k. Since k = χ(1)2 , it follows that |χ(x)| = χ(1), and so x ∈ Z(χ) = Z(G) since χ is faithful. Thus, χ vanishes on G − Z(G). Claim (iii). We have G󸀠 ≤ Z(G). Indeed, if x ∈ G󸀠 , then λ i (x) = 1 for all i, and therefore χ(x)2 = λ1 (x) + ⋅ ⋅ ⋅ + λ k (x) = k ≠ 0. It follows from (ii) that x ∈ Z(G). Claim (iv). We claim that |G󸀠 | = 2. Indeed, by Clifford’s theorem (see Chapter VII), 1G󸀠 ∈ ̸ Irr(χ G󸀠 ) (otherwise, since G acts transitively on the set Irr(χ G󸀠 ), we conclude that G󸀠 ≤ ker(χ) = {1] so G is abelian). Hence, 0 = |G󸀠 |⟨χ G󸀠 , 1G󸀠 ⟩ = ∑ χ(x) = χ(1) + ∑ χ(x). x∈G󸀠

(∗)

x∈(G󸀠 )#

Next, χ(x) = −χ(1) for each x ∈ (G󸀠 )# . Indeed, x ∈ G󸀠 ≤ Z(G) and so |χ(x)| = χ(1) which implies χ(x) = −χ(1) since χ is faithful. Therefore, by (∗), we have χ(1) − χ(1)(|G󸀠 | − 1) = 0 󳨐⇒ |G󸀠 | = 2. Claim (v). The group Z(G) is cyclic since χ is faithful. Claim (vi). We have |G : Z(G)| = χ(1)2 (= k). Indeed, by (ii), we obtain |G| = |G|⟨χ, χ⟩ = ∑ |χ(x)|2 = ∑ |χ(x)|2 = |Z(G)|χ(1)2 . x∈G

x∈Z(G)

180 | Characters of Finite Groups 1 (b) Now let G be a nonabelian group with cyclic Z(G) and |G󸀠 | = 2. Then G is nilpotent. There is a faithful χ ∈ Irr(G). Then |G : Z(G)| = χ(1)2 since Z(χ) = Z(G) and G/Z(G) is abelian (Lemma III.9.2), and so χ vanishes on G − Z(G) (see Lemma II.6.3). Let λ ∈ Lin(G) be such that Z(G) ≤ ker(λ) so that λ(z) = 1 for z ∈ Z(G). Then, since χ vanishes off Z(G), we obtain ⟨χ2 , λ⟩ = |G|−1 ∑ χ(x)2 λ(x) = |G|−1 ∑ χ(x)2 = |G|−1 |Z(G)|⟨χZ(G) , χZ(G) ⟩. x∈Z(G)

x∈Z(G)

If χZ(G) = χ(1)μ, where μ ∈ Lin(Z(G)), then ⟨χ2 , λ⟩ = |G|−1 |Z(G)|χ(1)2 ⟨μ, μ⟩ = ⟨μ, μ⟩. Since G/Z(G) is an elementary abelian 2-group, the linear characters of G whose kernels contain Z(G) can take values −1 and 1 only so all linear characters are realvalued. Therefore ⟨χ2 , λ⟩ = ⟨μ, μ⟩ = ⟨μ, μ⟩ = 1, and since |G/Z(G)| = χ(1)2 , it follows that Irr(χ2 ) ⊆ Lin(G). Exercise 12.1 (A. Mann). Classify the nonabelian groups G that possess a faithful irreducible character χ such that Irr(χχ) ⊆ Lin(G). Exercise 12.2. Let χ be a faithful (not necessarily irreducible) character of a nonabelian group G and let m, n be nonnegative integers such that m + n ≥ 2. Describe the structure of G provided Irr(χ m χ n ) ⊆ Lin(G).

V Induced characters and representations 1 Induced characters In this chapter we consider the relationship between the characters of a group G and the characters of a subgroup H of G, a question that arises naturally in the theory of characters. Let Char(G) be the set of characters of G and χ ∈ Char(G). Then we have χ H ∈ Char(H), where χ H is the restriction of χ to H ≤ G. It is natural to consider the mapping Char(G) → Char(H), which associates to each character of G its restriction to H. This mapping is a homomorphism of additive semigroups: indeed, (χ + τ)H = χ H + τ H

(χ, τ ∈ Char(G)).

The restriction mapping is generally noninvertible. Nevertheless, as Frobenius could show, there exists a natural mapping Char(H) → Char(G) which is a homomorphism of additive semigroups and which is in a sense the inverse of the restriction mapping. The image of a character ψ of the subgroup H is the induced character ψ G of G. Frobenius also proposed a construction which enables one to find a representation of G affording the character ψ G , given a representation ∆ of H such that ψ is the character of ∆. The accepted notation for this induced representation of the group G is ∆ G . Frobenius reciprocity law connecting characters of a group and its subgroups is very important in constructing character tables. Let CF[G] be the unitary space of central (= class) functions on G (as we know, all characters of G are members of the set CF[G] and Irr(G) is its orthonormal basis). Take θ ∈ CF[G]; then, obviously, θ H ∈ CF[H]. The mapping R G→H : CF[G] → CF[H] that sends a function θ ∈ CF[G] to its restriction θ H to H (i.e., R G→H θ = θ H ) is a homomorphism of linear spaces. Let I H→G : CF[H] → CF[G] be the homomorphism dual to R G→H : if ψ ∈ CF[H] and θ ∈ CF[G], then ⟨I H→G ψ, θ⟩ = ⟨ψ, R G→H θ⟩.

(1)

Equality (1) is called the Frobenius reciprocity law. Instead of the expression “by Frobenius reciprocity law” we write, in most cases, “by reciprocity”. It is well known from linear algebra that the “inducing homomorphism” I H→G exists and is uniquely determined by the restriction homomorphism R G→H . One usually writes, within the same argument, R instead of R G→H and I instead of I H→G . If ψ ∈ CF[H], then the accepted notation for Iψ is ψ G , so one can rewrite formula (1) as follows: ⟨ψ G , θ⟩ = ⟨ψ, θ H ⟩

(ψ ∈ CF[H], θ ∈ CF[G]).

(2)

It follows from the definition of I that (φ + ψ)G = φ G + ψ G , DOI 10.1515/9783110224078-005

(αφ)G = αφ G

(φ, ψ ∈ CF[H], α ∈ ℂ).

(3)

182 | Characters of Finite Groups 1 Let us find an explicit form of the function ψ G (ψ ∈ CF[H]). To this end, we define a function ψ̇ : G → ℂ as follows: {ψ(g) if g ∈ H, ̇ ψ(g) ={ 0 if g ∈ G − H. {

(4)

We do not assert that ψ̇ (ψ ∈ CF[G]) is a class function on G even in the case when H is normal in G. Given a function θ : G → ℂ and t ∈ G, we define a function θ t : G → ℂ as follows: θ t (g) = θ(tgt−1 )

(5)

(g ∈ G).

The so-defined function θ t is said to be G-conjugate to θ. If x, y ∈ G, then θ xy = (θ x )y −1 (therefore, we are writing tgt−1 = g t but not t−1 gt = g t in formula (5)). Since θ1 = θ (1 ∈ G), one can speak of the action of the group G on the set CF[G] of function G → ℂ. It follows from (5) that θ ∈ CF[G] if and only if θ t = θ for all t ∈ G. It is easily checked that, for a fixed t ∈ G, the mapping θ 󳨃→ θ t (θ ∈ F[G] (the set of functions G → ℂ) is an isometry of F[G]: ⟨θ1t , θ2t ⟩ = ⟨θ1 , θ2 ⟩

(θ1 , θ2 ∈ F[G]).

Theorem 1.1. If ψ ∈ CF[H], then ψ G = |H|−1 ∑ ψ̇ t .

(6)

t∈G

Proof. Take θ ∈ CF[G]. Then ⟨|H|−1 ∑ ψ̇ t , θ⟩ = |H|−1 ∑ ⟨ψ̇ t , θ t ⟩ = |H|−1 ∑ ⟨ψ,̇ θ⟩ t∈G

t∈G

t∈G

̇ = |G||H| ⟨ψ,̇ θ⟩ = |G||H|−1 |G|−1 ∑ ψ(g)θ(g) −1

g∈G

−1

= |H|

∑ ψ(g)θ(g) = ⟨ψ, θ H ⟩.

g∈H

Thus, it follows from the above and (2) that ⟨|H|−1 ∑ ψ̇ t , θ⟩ = ⟨ϕ, θ H ⟩ = ⟨ϕ G , θ⟩ t∈G

so that ⟨|H|−1 ∑ ψ̇ t − ψ G , θ⟩ = 0.

(7)

t∈G

Since |H|−1 ∑t∈G ψ̇ t − ψ G ∈ CF[G], it follows, putting θ = |H|−1 ∑t ψ̇ t − ψ G in (7), that ⟨θ, θ⟩ = 0. Therefore θ = 0 is the zero function, and (6) is proved.

V Induced characters and representations | 183

Let g ∈ G. Equality (6) implies −1 ̇ ψ G (g) = |H|−1 ∑ ψ(tgt ) = |H|−1 t∈G

∑

t∈G,tgt−1 ∈H

ψ(tgt−1 ).

(8)

Let |G : H| = n and let G = ⋃ni=1 Ht i be a decomposition of G into right cosets of H. Since the function ψ̇ is H-invariant, it follows from (8) that n

̇ i gt−1 ) ψ G (g) = ∑ ψ(t i

(g ∈ G).

(9)

i=1

There is a formula similar to (9) for any left transversal {u i }1n of G over H. Let K g be the G-class containing g ∈ G. Relation (8) implies that ψ G (g) = |CG (g)| ⋅ |H|−1

∑

ψ(s).

(10)

s∈H∩K g

Indeed, for a fixed s ∈ H ∩ K g , the equation tgt−1 = s has exactly |CG (g)| solutions −1 G (if, in addition, t1 gt−1 1 = s, then t t 1 ∈ CG (g)), and our formula for ψ (g) holds). Exercise 1.1. If H < G and ϕ ∈ CF(H), then ψ G vanishes on G − ⋃t∈G H t . −1 ) = 0 for all t ∈ G. (It follows that if H < G, ̇ Solution. If x ∈ G − ⋃t∈G H t , then ψ(txt G then the set of zeros of character ψ is nonempty since conjugates of a proper subgroup H do not cover G).

Exercise 1.2. If H ≤ G and ψ ∈ CF(G), then ψ G (1) = |G : H|ψ(1).

(11)

This follows from formula (9) immediately. Theorem 1.2. The set im(I H→G ) = (CF[H])G is the set of all θ ∈ CF[G] that vanish on the set G − ⋃t∈G H t . Proof. The central functions that vanish on G − ⋃t∈G H t form a subspace V of CF[G], and moreover im(I H→G ) ⊆ V, by Theorem 1.1 (see also Exercises 1.1 and 1.3). Assume that im(I H→G ) ≠ V. Then there exists a nonzero function θ ∈ V which is orthogonal to all functions belonging to im(I H→G ), i.e., ⟨ψ G , θ⟩ = 0 for all ψ ∈ CF[H]. By reciprocity (see (2)), it follows from that equality that ⟨ψ, θ H ⟩ = 0 for all ψ ∈ CF[H]. Putting ψ = θ H , we obtain ⟨θ H , θ H ⟩ = 0, hence we get θ H = 0. Therefore θ vanishes on H. Since θ ∈ CF[G], it follows that θ vanishes on H t for all t ∈ G. But then θ vanishes on G (since it vanishes on G − ⋃t∈G H t ), i.e., θ = 0, a contradiction. Thus im(I H→G ) = V. Theorem 1.2 implies that im(I H→G ) = (CF[H])G is an ideal of the algebra CF[G]. This is also consequence of the following important property of inducing homomorphisms (this property will be used in what follows). Lemma 1.3. Let H ≤ G, ψ ∈ CF[H], θ ∈ CF[G]. Then θψ G = (θ H ψ)G .

184 | Characters of Finite Groups 1 Proof. By Theorem 1.1, since θ ψ̇ = θ H ψ,̇ one obtains ̇ t θψ G = |H|−1 ∑ θ ψ̇ t = |H|−1 ∑ θ t ψ̇ t = |H|−1 ∑ (θ ψ) t∈G

−1

= |H|

t∈G

t∈G

̇ t = (θ H ψ)G . ∑ (θ H ψ)

t∈G

Exercise 1.3. Let H ≤ G. The set I(H) = {θ ∈ CF[G] | θ = ψ G , where ψ ∈ CF[H]} is an ideal of the algebra CF[G] (see Lemma 1.3). Prove that I(H) = V(H), where 󵄨󵄨 V(H) = {θ ∈ C[G] 󵄨󵄨󵄨󵄨 θ vanishes on G − ⋃ H t }. 󵄨 t∈H Hint. The set V(H) is an ideal in CF[G] and I(H) ⊆ V(H) (Exercise 1.1). If I(H) ≠ V(H), there is a nonzero function θ ∈ V(H) which is orthogonal to all functions from I(H). Use (2) (reciprocity) to get a contradiction. Exercise 1.4. Let H i < G, i = 1, . . . , n, and G = ⋃ni=1 H i . Prove that CF[G] = ∑ni=1 I(H i ) (see Exercise 1.3). Deduce from this the following “induction theorem”: Every class function θ ∈ CF[G] can be written as a sum of class functions induced from cyclic subgroups of the group G. Hint. Mimic the solution of Exercise 1.3. Theorem 1.4 (Transitivity of induction). If H ≤ K ≤ G and ψ ∈ CF[H], then ψ G = (ψ K )G . Proof. It is obvious that R G→H = R K→H ⋅ R G→K (this is equivalent to the obvious equality θ H = (θ K )H for θ ∈ CF[G]). Therefore, in view of well-known properties of dual homomorphisms of linear spaces, we have (R G→H )∗ = (R G→K )∗ (R K→H )∗ , hence I H→G = I K→G ⋅ I H→K . It follows that ψ G = I H→G ψ = (I K→G ⋅ I H→K )ψ = I K→G (I H→K ψ) = I K→G (ψ K ) = (ψ K )G . Observe one more property of the inducing homomorphism (see Exercise 1.5). Let H ≤ G, g ∈ G. If ψ ∈ CF[H] and we define a mapping ψ g : H g → ℂ by ψ g (x) = ψ(gxg −1 )

(x ∈ H g = g −1 Hg),

then ψ g ∈ CF[H g ]. Exercise 1.5. If ψ ∈ CF[H], H ≤ G, g ∈ G, then (ψ g )G = ψ G . Hint. We have gG = G for all g ∈ G. Now the result follows from (6). Exercise 1.6. Suppose that G is a nonabelian group and f = min{χ(1) | χ ∈ Irr1 (G)}. Then, if H < G and |G : H| ≤ f , then G󸀠 ≤ H. In particular, H ⊲ G. Solution. Let χ ∈ Irr(1GH ). Since ⟨1GH , 1G ⟩ = ⟨1H , 1H ⟩ = 1, by reciprocity, it follows that the induced character (1H )G is reducible, and so χ(1) ≤ |G : H| < f . We conclude that χ

V Induced characters and representations | 185

is linear. Thus, all irreducible constituents of the character (1H )G are linear. By reciprocity, we obtain (1GH )H = |G : H|1H so that H = ker(1GH ). By the above, G󸀠 ≤ ker(1GH ) whence G󸀠 ≤ H. Exercise 1.7. If N ⊲ G, g ∈ N and K g is the G-class containing g, then |K g | ≤ |[G, N]| with equality if and only if K g = g[G, N]. Solution. For t ∈ G, one has g t = g[g, t] ∈ g[G, N] so that K g ⊆ g[G, N]. In particular, |K g | ≤ |G󸀠 | with equality if and only if K g = gG󸀠 . Exercise 1.8. If N ⊲ G, then |[G, N]| ≤ f = min{χ(1) | χ ∈ Irr1 (G)} 󳨐⇒ N ≤ CG (G󸀠 ). Solution. Every G-class, contained in N, has size ≤ |[G, N]| (Exercise 1.7) so its size, by hypothesis, is ≤ f . Therefore, if g ∈ N, then |G : CG (g)| ≤ f with strong inequality if N > {1} (this is clear if g = 1, and this is true if g ≠ 1 since then 1 ∈ ̸ K g ). Therefore, by Exercise 1.6, G󸀠 ≤ CG (g). Since g ∈ N is arbitrary, it follows that G󸀠 ≤ CG (N). Lemma 1.5 is a reducibility criterion of induced characters. Lemma 1.5 ([Isa14, Lemma 4.3]). Let H < G and suppose that ψ ∈ Irr(H) is such that ψ(h) = ψ(k) whenever h, k ∈ H are conjugate in G. Then ψ G is reducible. Proof (Zhmud). Since the regular character ρ G of G is reducible (it is easy to check, using reciprocity, that if H = {1}, then 1GH = ρ G ), one may assume that H > {1}. If g ∈ H and K g is the G-class containing g, then, by (10), ψ G (g) = |CG (g)||H|−1 |H ∩ K g |ψ(g), and, by reciprocity, ⟨ψ G , ψ G ⟩ = ⟨(ψ G )H , ψ⟩ = |H|−1 ∑ (ψ G )H (g)ψ(g) g∈H

−1

= |H|

∑ |CG (g)||H| |H ∩ K g ||ψ(g)|2 . −1

g∈H

Since |H|−1 |H ∩ K g | ≥ |H|−1 |H ∩ g H | = |H|−1 |g H | = |CH (g)|−1 (here g H is the H-class containing g), we get ⟨ψ G , ψ G ⟩ ≥ |H|−1 ∑

g∈H

It follows from

|CG (1)| |CH (1)|

|CG (g)| |ψ(g)|2 . |CH (g)|

= |G : H| > 1 that ⟨ψ G , ψ G ⟩ > |H|−1 ∑ |ψ(g)|2 = ⟨ψ, ψ⟩ = 1, g∈H

and hence

ψG

is reducible.

186 | Characters of Finite Groups 1 It follows from Lemma 1.5 that if ϕ ∈ Irr(H) is the restriction of a class function of G to H < G, then ϕ G is reducible. Indeed, if ϕ = θ H for some θ ∈ CF[G] and h, k ∈ H are conjugate in G, then ϕ(h) = θ(h) = θ(k) = ϕ(k) so that ϕ satisfies the hypothesis of Lemma 1.5. Lemma 1.6 ([Isa14, Lemma 4.2]). Let H < G and suppose that χ ∈ Irr(G) vanishes on the set g ∈ G − ⋃t∈G H t . Then χ H is reducible. Proof. By reciprocity and Lemma 1.3, ⟨χ H , χ H ⟩ = ⟨χ, (χ H )G ⟩ = ⟨χ, (1H )G χ⟩ = |G|−1 ∑ (1H )G (g)|χ(g)|2 > |G|−1 ∑ |χ(g)|2 = ⟨χ, χ⟩ = 1 g∈G

g∈G

since (1H

)G (g)

≥ 1 whenever χ(g) ≠ 0, and (1H

)G (1)

= |G : H| > 1.

It follows from Lemma 1.6 that if χ ∈ Irr(G) is induced from a class function of H < G, then χ H is reducible. Indeed, in that case, χ vanishes on G − ⋃t∈G H t (Exercise 1.1). Exercise 1.9. Let H < G. The set I(H) = {θ | θ = ψ G , where ψ ∈ CF[H]}, as follows from Lemma 1.3, is an ideal of the algebra CF[G]. Prove that I(H) is equal to V(H), where V(H) = {θ ∈ CF[G] | θ vanishes on G − ⋃t∈G H t }. Hint. The set V(H) is an ideal of CF(G) containing I(H) (see Exercise 1.1). If one has I(H) ≠ V(H), there is a nonzero function θ ∈ V(H) orthogonal to all functions contained in I(H). Use reciprocity to obtain a contradiction. Exercise 1.10. Let H i < G (i = 1, . . . , n) cover a group G, i.e., G = ⋃ni=1 H i . Prove that CF[G] = ∑ni=1 I(H i ) (see Exercise 1.9). Deduce from this the “weakened induction theorem”: Every class function θ ∈ CF[G] can be presented as a linear combination of class functions induced from cyclic subgroups of G (indeed, G is covered by cyclic subgroups). Hint. Apply the method used in the Hint to Exercise 1.9. Exercise 1.11. Let H ⊲ G and θ ∈ Irr(G). Then (θ H )G = ρ G/H θ, where ρ G/H is the regular character of the quotient group G/H. If, in addition, G/H is abelian, then all irreducible constituents of the induced character (θ H )G have the same degree θ(1). Hint. Use Lemma 1.3 with ψ = 1H . Take into account that (1H )G = ρ G/H . Exercise 1.12. Let G󸀠 ≤ H < G, μ ∈ Irr(H). Prove that all irreducible constituents of μ G have the same degree. Solution. Let θ ∈ Irr(μ G ). Then, by Lemma 1.3, (θ H )G = (θ H 1H )G = θ(1H )G = θρ G/H . Since all irreducible constituents of the regular character ρ G/H of G/H are linear, it follows that all irreducible constituents of θρ G/H have degree θ(1). Since μ ∈ Irr(θ H ),

V Induced characters and representations | 187

we get Irr(μ G ) ⊆ Irr((θ H )G ) = Irr(θρ G/H ), and we are done. In particular, if μ ∈ Irr(G󸀠 ), then all irreducible constituents of the character μ G have the same degree and the same multiplicity. Indeed, if χ ∈ Irr(μ G ) and χ G󸀠 = e(μ1 + ⋅ ⋅ ⋅ + μ t ) (μ1 = μ) is the Clifford decomposition (see Chapter VII), χ(1) then e = ⟨χ, μ G ⟩ = tμ(1) .

2 Kernels of induced characters In this section we find kernels of induced characters. The introduction of this notion is justified by the following: Exercise 2.1. If ψ ∈ Char(H), then ψ G ∈ Char(G). Solution. If θ ∈ Irr(G), then, by reciprocity, ⟨ψ G , θ⟩ = ⟨ψ, θ H ⟩ ∈ ℕ ∪ {0}. As we have noted, the relation ⟨ψ G , θ⟩ = ⟨ψ, θ H ⟩, where ψ ∈ Char(H), θ ∈ Char(G), is usually called the Frobenius reciprocity law. Incidentally, Frobenius defined induced characters by formula (1.6); he then proved equality (1.2) and the last relation as theorems. The Frobenius reciprocity law is equivalent to the following assertion: If ψ, χ are irreducible characters of H and G, respectively, H ≤ G, then the multiplicity of χ in ψ G is equal to the multiplicity of ψ in χ H . Exercise 2.2. Let H < G. For any ψ ∈ Irr(H), there exists χ ∈ Irr(G) such that ⟨χ H , ψ⟩ > 0. In particular, if G is abelian, then ψ admits an extension χ to G such that χ ∈ Irr(G) (more precisely, there are exactly |G : H| such extensions). Hint. Let χ be an irreducible constituent of ψ G (in that case, we write χ ∈ Irr(ψ G )). Use the Frobenius reciprocity law. If G is abelian, then ψ G is the sum of |G : H| pairwise distinct constituents, by reciprocity, so their restrictions to H coincide with ψ. Exercise 2.3. Let H ≤ G. If one has ψ ∈ Char(H) and ψ G ∈ Irr(G), then ψ ∈ Irr(H). But if ψ ∈ CF[H] and ψ G ∈ Char(G), then the inclusion ψ ∈ Char(H) may be false (give an example). Solution. Let ϕ = a1 μ1 + ⋅ ⋅ ⋅ + a n μ n , where {μ1 , . . . , μ n } ⊆ Irr(H) and a1 + ⋅ ⋅ ⋅ + a n > 1. Then, by hypothesis, we have ϕ G = a1 μ1G + ⋅ ⋅ ⋅ + a n μ Gn , and the first assertion follows. Example: Let ψ ∈ Irr(H) be such that ψ G = eχ, where e > 1 and χ ∈ Irr(G). Then one has 1e ψ ∈ CF(G) and ( 1e ψ)G = χ. Exercise 2.4. If ψ ∈ Irr(H) and χ ∈ Irr(ψ G ) (this notation is used from now on for the set of the irreducible constituents of the character ψ G ), then ψ(1) ≤ χ(1) ≤ |G : H|ψ(1).

188 | Characters of Finite Groups 1 Solution. By reciprocity, ψ ∈ Irr(χ H ) so ψ(1) ≤ χ(1). The second inequality follows from |G : H|ψ(1) = ψ G (1) ≥ χ(1). Given H ≤ G, the subgroup H G = ⋂t∈G H t is said to be the core of H in G. Obviously, H G is the product of all G-invariant subgroups of H. Lemma 2.1. If ψ ∈ Char(H), then ker(ψ G ) = ker(ψ)G . In particular, ker((1H )G ) = H G and if ψ is faithful, then ψ G is also faithful. Proof. Let g ∈ ker(ψ G ). Then, by (1.6), −1 ̇ |G : H|ψ(1) = ψ G (1) = ψ G (g) = |H|−1 ∑ ψ(tgt ). t∈G

−1 )| ≤ ψ(1), the last equality implies that ψ(tgt −1 ) = ψ(1) ≠ 0 for all t ∈ G. ̇ ̇ Since |ψ(tgt −1 −1 ̇ Therefore, tgt ∈ H for all t ∈ G, so that ψ(tgt ) = ψ(tgt−1 ) = ψ(1). Consequently, we get tgt−1 ∈ ker(ψ) for all t ∈ G, hence we have g ∈ ker(ψ)G , and we conclude that ker(ψ G ) ≤ ker(ψ)G . Conversely,

g ∈ ker(ψ)G 󳨐⇒ g t ∈ ker(ψ) for all t ∈ G 󳨐⇒ ψ(g t ) = ψ(1). By the displayed formula, we get ψ G (g) = ψ G (1), and this implies that g ∈ ker(ψ G ). Thus, ker(ψ G ) = ker(ψ)G . In particular, ker(ψ) = {1} implies ker(ψ G ) = {1}G = {1}. Exercise 2.5. If H ⊲ G, χ ∈ Irr(G) and H ≰ ker(χ), then ⟨χ H , 1H ⟩ = 0. Solution. If ⟨χ H , 1H ⟩ > 0, then χ ∈ Irr((1H )G ), by reciprocity, whence it follows that ker(χ) ≥ ker((1H )G ) = H G = H (see Lemma 2.1), contrary to the hypothesis. (Another proof follows from Clifford’s theorem; see Chapter VII.) Exercise 2.6. Let b(G) = max{χ(1) | χ ∈ Irr(G)}. Then H ≤ G implies b(H) ≤ b(G) ≤ |G : H|b(H). Hint. Use Exercises 2.2 and 2.4. Exercise 2.7. If H ≤ G is abelian, then b(G) ≤ |G : H|. Hint. This follows from Exercise 2.6. Exercise 2.8. If G is a nonabelian group, then D = ⋂χ∈Irr1 (G) ker(χ) = {1}. Solution. Assume that D > {1}. If χ ∈ Irr1 (G), then we have χ D = χ(1)1D . Therefore, if ϕ ∈ Irr# (D), then Irr(ϕ G ) ⊂ Lin(G) and Irr1 (G) ⊂ Irr(1GD ), by reciprocity. It follows that D ≤ G󸀠 = ⋂χ∈Lin(G) ker(χ). We conclude that D ≤ ⋂χ∈Irr(G) ker(χ) = ker(ρ G ) = {1}, contrary to the assumption. Second solution. Assume that, as above, |D| = d > 1 and let |G| = g, |G󸀠 | = g󸀠 . Then g g g g ≥ ∑ χ(1)2 = g − 󸀠 󳨐⇒ g ≤ + 󸀠 󳨐⇒ d = g 󸀠 = 2. d χ∈Irr (G) g d g 1

V Induced characters and representations | 189

If ϕ ∈ Lin# (D), then Irr(ϕ G ) ⊆ Lin(G) so that G󸀠 ≤ ker(ϕ G ) = ker(ϕ)G = {1}, since ker(ϕ)G < D and |D| = 2. Thus, G󸀠 = {1} hence G is abelian, contrary to the assumption.

3 Mackey’s theorems It is natural to ask what the behavior of an induced character will be when restricted to a given subgroup of G. Important Mackey’s theorems, proved below, answer this question. We introduce some additional notation that is used only in this section. Let H, K ≤ G,

ψ ∈ Char(H),

ψ s = (ψ s )D s .

D s = H s ∩ K,

s ∈ G,

Recall that ψ s is the character of the subgroup H s = s−1 Hs defined by ψ s (x) = ψ(sxs−1 )

(x ∈ H s ).

Since sxs−1 ∈ H, the value ψ s (x) is determined. The expression ψ Ks is also determined since ψ s ∈ Char(D s ) and D s ≤ K. In what follows we retain the notation introduced in this paragraph. Theorem 3.1 (Mackey). Let H, K ≤ G, ψ ∈ Char(H), and let G = ⋃s∈I HsK be a decomposition of G into double cosets of H, K. Then (ψ G )K = ∑s∈I ψ Ks . Proof. Put (ψ G )K = θ. Then, for x ∈ K, one has θ(x) =

∑

t∈J,

txt−1 ∈H

(1)

ψ(txt−1 ),

where J is some right transversal of G over H. Choose J as follows. Let s ∈ I (see the statement of the theorem) and K = ⋃v∈V s D s v, where V s is a right transversal of K over D s . Then HsK = sH s K = ⋃ sH s D s v = ⋃ sH s (H s ∩ K)v = ⋃ sH s v = ⋃ Hsv v∈V s

v∈V s

v∈V s

v∈V s

is a partition. Take J to be the set ⋃s∈I J s , where J s = sV s = {sv | v ∈ V s }. For this choice of J, equality (1) takes the following form: θ(x) = ∑

s∈I t∈J s ,

∑

txt−1 ∈H

=∑

ψ(txt−1 ) = ∑ s∈I

s∈I v∈V s ,

v∈V s

∑

ψ s (vxv−1 ) = ∑ ∑ (ψ s )D s (vxv−1 )

∑

ψ s (vxv ) = ∑ ψ Ks (x).

s∈I vxv−1 ∈H s , v∈V s

=∑

ψ(svxv−1 s−1 )

∑

svxv−1 s−1 ∈H,

vxv−1 ∈D

s∈I v∈V s

−1

s

s∈I

190 | Characters of Finite Groups 1 The character (θ =)(ψ G )K of Theorem 3.1 is decomposed into the sum of |I| summands. However, if G is a doubly transitive permutation group and H = K is the stabilizer of a point, then |I| = 2. Another interesting case is obtained when G = HK and H ∩ K = {1} (see Exercise 3.3). Theorem 3.1 implies the following useful irreducibility criterion for induced characters, also due to Mackey: Theorem 3.2. If H ≤ G, ψ ∈ Irr(H), then ψ G ∈ Irr(G) if and only if the characters ψ s and ψ D s (where D s = H s ∩ H) have no common irreducible constituents for all s ∈ G − H. Proof. By reciprocity, we have ⟨ψ G , ψ G ⟩ = ⟨ψ, (ψ G )H ⟩. By Mackey’s Theorem 3.1, we s get (ψ G )H = ∑s∈I ψ H s , where I is a transversal of G over (H, H), D s = H ∩ H (put K = H in Theorem 3.1). Therefore, ⟨ψ G , ψ G ⟩ = ∑ ⟨ψ H s , ψ⟩. s∈I

Since, by reciprocity,

⟨ψ H s ,

ψ⟩ = ⟨ψ s , ψ D s ⟩ (recall that ψ s ∈ Char(D s )), it follows that ⟨ψ G , ψ G ⟩ = ∑ ⟨ψ s , ψ D s ⟩. s∈I

Choose the system I so that it contains the identity element s = 1. Then D1 = H 1 ∩ H = H,

ψ1 = ψ,

ψ D1 = ψ H = ψ,

and therefore ⟨ψ G , ψ G ⟩ = ⟨ψ, ψ⟩ + ∑ ⟨ψ s , ψ D s ⟩ = 1 + ∑ ⟨ψ s , ψ D s ⟩. s∈I #

(2)

s∈I #

If ψ G ∈ Irr(G), then ⟨ψ G , ψ G ⟩ = 1, and (2) implies that ∑s∈I # ⟨ψ s , ψ D s ⟩ = 0, so that Irr(ψ s ) ∩ Irr(ψ D s ) = 0 for all s ∈ I # = I − {1}. Since I may be chosen so that it contains any given element s ∈ G − H, it follows that ψ s and ψ D s have no common irreducible constituents. Conversely, if Irr(ψ s ) ∩ Irr(ψ D s ) = 0 for all s ∈ G − H, then formula (2) implies that G ⟨ψ , ψ G ⟩ = 1, hence ψ G ∈ Irr(G). Exercise 3.1. Use Theorem 3.2 to prove Lemma 1.5. Corollary 3.3 (Shoda’s irreducibility criterion). Let H ≤ G and ψ ∈ Lin(H). Then one has ψ G ∈ Irr(G) if and only if for every s ∈ G − H, there exists x ∈ D s = H ∩ H s such that ψ(sxs−1 ) ≠ ψ(x) (i.e., ψ s ≠ ψ D s for every s ∈ G − H). To prove this, observe that if ψ ∈ Lin(H), then ψ s ≠ ψ D s if and only if ⟨ψ s , ψ D s ⟩ = 0. Exercise 3.2. If H ⊲ G and ψ ∈ Irr(H), then ψ G ∈ Irr(G) if and only if ψ s ≠ ψ for all s ∈ G − H. (In the terminology of Chapter VII this means that ψ G ∈ Irr(G) if and only if the inertia group IG (ψ) of the character ψ in G coincides with H.)

V Induced characters and representations | 191

Exercise 3.3. Let H, K ≤ G with HK = G and H ∩ K = {1}. Then ((1H )G )K = ρ K , the regular character of K. Hint. In the notation of Theorem 3.1, we have I = {1} hence D1 = H ∩ K = {1}. Although it is not true in general that the induced character ψ G can be explicitly decomposed into irreducible constituents, this may be successfully done in some particular cases. Theorem 3.4. Let G be a group and (G, G) the diagonal subgroup of the group G × G: (G, G) = {(g, g) | g ∈ G}. For ϑ ∈ CF[G], define ϑ0 ∈ CF[(G, G)] by putting ϑ0 ((g, g)) = ϑ(g)

for all g ∈ G.

Then ϑ0G×G =

⟨χψ, ϑ⟩(χ × ψ).

∑

(3)

χ,ψ∈Irr(G)

Proof. Indeed, for χ, ψ ∈ Irr(G) we have (note that |(G, G)| = |G|) ⟨ϑ0G×G , χ × ψ⟩ = ⟨ϑ0 , (χ × ψ)(G,G) ⟩ = |(G, G)|−1 ∑ ϑ0 ((g, g))(χ × ψ)((g, g)) g∈G

−1

= |G|

∑ ϑ(g)χ(g)ψ(g) = ⟨ϑ, χψ⟩.

g∈G

Here we used the equality Irr(G × G) = {χ × ψ | χ, ψ ∈ Irr(G) (see §IV.8). In particular, if ϑ ∈ Lin(G), then ⟨χψ, ϑ⟩ ≠ 0 󳨐⇒ ⟨χ, ϑψ⟩ ≠ 0 󳨐⇒ ⟨χ, ϑψ⟩ = 1 since ϑψ is irreducible; hence, χ = ϑψ, i.e., ψ = ϑχ, and (3) may be rewritten as follows: ϑ0G×G = ∑ χ × ϑχ. χ∈Irr(G)

Finally, we get (1(G,G) )G×G =

∑

χ × χ.

χ∈Irr(G) G×G (1) = |G|, we obtain from the above formula the following known equality: Since 1(G,G)

|G| =

∑

χ(1)2 .

χ∈Irr(G)

We use induced characters to prove the following useful result: Proposition 3.5. Let p be a prime, G a group, P ∈ Sylp (G) and D = G󸀠 ∩ Z(G) ∩ P > {1}. If λ ∈ Lin# (D), then λ has no extension to P, i.e., there are no μ ∈ Irr(P) such that μ D = λ.

192 | Characters of Finite Groups 1 Proof. Assume, by way of contradiction, that μ ∈ Irr(P) is such that μ D = λ. Take χ ∈ Irr(μ G ) such that p ∤ χ(1) (such a character χ exists, since p ∤ |G : P| = μ G (1)). By reciprocity, μ ∈ Irr(χ P ); consequently, λ = μ D ∈ Irr(χ D ). Next, it follows from D ≤ Z(G) that χ D = χ(1)λ (here we use Clifford’s theorem). Let T be a representation of G affording the character χ. If x ∈ D, then the χ(1) × χ(1) matrix λ(x) . T(x) = ( .. 0

... .. . ...

0 .. . ) λ(x)

is scalar. Consequently, det(T(x)) = λ(x)χ(1) . Next, φ = det(T) ∈ Lin(G). The inclusions D ≤ G󸀠 ≤ ker(φ) imply that λ χ(1) = 1D . Since, by the choice, p ∤ χ(1) and Lin(D) ≅ D is a nonidentity p-group, we have arrived at a contradiction. Thus, μ does not exist. It should be noticed that, if a subgroup P ∈ Sylp (G) is abelian, then D = P ∩ Z(G) ∩ G󸀠 = {1}. Indeed, assume that D > {1} and let λ ∈ Irr(D) − {1D }. Then, by Exercise 2.2, there is χ ∈ Irr(G) such that λ ∈ Irr(χ D ). One has χ D = (χ P )D . Therefore, there is μ ∈ Irr(χ P ) such that λ ∈ Irr(μ D ). Since P is abelian, μ(1) = 1 so that μ D = λ, contrary to Proposition 3.5. Thus, D = {1}, as asserted.

4 Permutation representations We repeat some results of §I.1. Let M be a finite set. A homomorphism of a group G into the symmetric group S(M) on a set M is called a permutation representation of G. The number |M| is called the degree of this representation. It is known that a representation π of G by permutations of a set M determines an action of G on M. Indeed, if α ∈ M, then the equality gα = π(g)(α) defines the following action of G on M: (gh)α = π(gh)(α) = (π(g)π(h))(α) = π(g)(π(h)(α)) = g(hα),

1G (α) = α

Conversely, provided an action of G on M is given, reading the displayed formula from right to left, we define a representation of G by permutations of M. If α, β ∈ M and β = gα for g ∈ G, then α and β are said to be G-equivalent (we write G α ∼ β or simply α ∼ β if G is clear from the context); G-equivalence classes are called G-orbits of M. If M itself is a G-orbit, the action of G on M is said to be transitive. Exercise 4.1. Let O α be the G-orbit containing a point α ∈ M, and let G α = {g ∈ G | gα = α} be the stabilizer of α in G (obviously, G α ≤ G). Then |O α | = |G : G α |. (See §I.1.)

V Induced characters and representations | 193

Since the result of Exercise 4.1 is basic, we present its solution. Let α ∈ M and let G = x1 G α ∪x2 G α ∪⋅ ⋅ ⋅∪x n G α be a decomposition of G modulo G α . If g = x i h with h ∈ G α , then gα = (x i h)α = x i α, i.e., |O α | ≤ n. Now let x i h1 , x j h2 ∈ G, i ≠ j and h1 , h2 ∈ G α . Then x i h1 α = x i α ≠ x j α, we get |O α | ≥ n. Exercise 4.2. If α, β ∈ M, β = gα (g ∈ G), then G β = gG α g −1 . (In particular, if G acts transitively on M, then stabilizers of all points are conjugate. In that case, if |M| > 1, there exists an element g ∈ G such that gα ≠ α for all α ∈ M. Solution. Let f ∈ G β , i.e., fβ = β or fgα = gα. It follows that g −1 fgα = α hence we obtain f ∈ gG α g−1 . Thus, G β ≤ gG α g−1 . The reverse containment is proved similarly. Now the first assertion follows. The last assertion is true since, if H < G, then ⋃x∈G H x ≠ G. Exercise 4.2 has an interesting interpretation, due to Frobenius. Let G be a permutation group of degree n having k orbits, and let σ(G) be the sum of numbers of points moved by permutations of G. Then σ(G) |G| = n − k. Indeed, let χ be the natural character of G (i.e., χ(g) is the number of points fixed by g ∈ G). Then σ(G) = ∑ (n − χ(g)) = n|G| − ∑ χ(g) g∈G

g∈G

= n|G| − ⟨χ, 1G ⟩ ⋅ |G| = n|G| − k|G| = (n − k)|G|. In particular, the average number of points moved by permutations of G is n − k, where k is the number of G-orbits; σ(G) |G| = n − 1 if and only if G is transitive (in that case, ⟨χ, 1G ⟩ = 1, i.e., the sum of numbers of points fixed by all permutations of a transitive group G is equal to |G|). If G acts on G by conjugation, then G-orbits of this action are the conjugacy classes of G (= G-classes). It follows from Exercise 4.1 that the class sizes divide |G|. If G acts on the set of its subgroups by conjugation, then the stabilizer of a point H ≤ G is equal to NG (H), the normalizer of H is G, so |G : NG (H)| is the number of subgroups conjugate with H in G, and that number is a divisor of |G|. Remark. The following application of the results of Exercises 4.1 and 4.2 is due to Burnside. Let ϕ be an automorphism of G fixing all G-classes as sets, ϕ ≠ idG . We will show that the prime divisors of o(ϕ) divide |G|. Assume the contrary. Without loss of generality, we may assume that o(ϕ) = p, a prime. Assume that p ∤ |G|. Let K be a conjugacy class of G. It follows from K ϕ = K that ⟨ϕ⟩ acts on K. Since p = |⟨ϕ⟩| ∤ |K|, there is g ∈ K such that g ϕ = g. Let g1 , . . . , g r be representatives of all of the G-classes ϕ such that g i = g i for all i. Put D = ⟨g1 , . . . , g r ⟩. Then ⋃x∈G D x = G since all G-classes are contained in this union, and so D = G. But then ϕ acts on G trivially, which is a contradiction. Thus, π(ϕ(1)) ⊆ π(G). If a group G acts on the set M and g ∈ G, let I(g) denote the set of all α ∈ M such that gα = α. The function χ(g) = |I(g)| (g ∈ G) is called the character of the permutation representation (or of the corresponding action) of G on M (above we have called this character natural). Such characters are called permutation characters of G. The per-

194 | Characters of Finite Groups 1 mutation characters are indeed characters of G, i.e., elements of Char(G) (just replace the permutations by the corresponding permutation matrices). Let P be a permutation representation of a group G on a set M, and let f be the action of G on M induced by P. Then P (or f ) induces a transitive representation P α (a transitive action f α ) on every G-orbit O α of M. The transitive representation P α (α ∈ M) is called a transitive constituent of P. Let χ α be the character of P α . If χ is the character of P and O α1 , . . . , O α k is the complete set of G-orbits, then χ = χ1 + ⋅ ⋅ ⋅ + χ k , where we put χ i = χ α i for simplicity. By Frobenius’ interpretation (see Exercise 4.2), if G is transitive on the set {1, . . . , n} and χ is the natural character of that action, then σ(G) ⟨χ, 1G ⟩ = n − = n − (n − 1) = 1. |G| It follows from this and the above formula for χ that, in the general case, k

⟨χ, 1G ⟩ = ∑ ⟨χ i , 1G ⟩ = k. i=1

Let V be an n-dimensional ℂ-space with a basis B = {v α }α∈M (we assume that the set M has been ordered in some fixed way). The action f : G × M → M uniquely determines an operator representation τ : G → GL(V); namely, if g ∈ G, then τ(g)v α = v gα . If we consider V as a left ℂG-module, then gv α = v gα (in other words, the action of G on M corresponds to the action of G on the basis B). The ℂG-module V (more precisely, the pair (V, B)) is called a permutation module. To an operator representation τ in the basis B there corresponds a matrix representation T = T P . Let g ∈ G and T P (g) = (p αβ (g))1n , where p αβ (g) ∈ ℂ. Then p αβ (g) = δ α,gβ , where δ is the Kronecker delta on the set M. In particular, p αα (g) = 1 if and only if gα = α if and only if α ∈ I(g), where, as above, I(g) = {α ∈ M | gα = α}. This implies that the character of the matrix representation T P is equal to the character χ of the action P, i.e., the character of a permutation representation of G is a character of G (as remarked above). The same result is contained in the following: Theorem 4.1. Let χ be the character of a transitive permutation representation P of a group G and let H be the G-stabilizer of an arbitrary point. Then χ = (1H )G . Proof. Let G act on a set M = {1, . . . , n}, where n is the degree of P. Since stabilizers of points are conjugate in G, one can take H to be G1 , the stabilizer of the point 1. Let G = ⋃nα=1 t α H be a decomposition of G into left cosets of H, with t α 1 = α (i.e., α is the t α -image of 1 ∈ M). Then n

(1H )G (g) = ∑ 1̇ H (t−1 α gt α ) = α=1

∑

t−1 α gt α ∈H

1=

∑

(t−1 α gt α )1=1

1=

∑

(gt α )1=t α 1

1= ∑ 1 gα=α

= |I(g)| = χ(g), which implies that χ = (1H )G . Exercise 4.3. If χ is the character of a transitive permutation representation of a group G > {1}, then ⟨χ, 1G ⟩ = 1. In that case, χ = 1G + θ, where θ ∈ Char(G) and ⟨1G , θ⟩ = 0.

V Induced characters and representations | 195

Solution. This was solved above, and that solution was character-free. Below we offer the proof based on the above theorem. We use the notation of Theorem 4.1. By Theorem 4.1 and reciprocity, ⟨χ, 1G ⟩ = ⟨(1H )G , 1G ⟩ = ⟨1H , 1H ⟩ = 1 since χ = (1H )G for some H ≤ G. Since the character θ = χ − 1G has no constituent equal 1G , the last assertion follows. Exercise 4.4 (Cauchy–Frobenius–Burnside). Let P be a permutation representation of+G having k orbits and χ the character of P. Then ⟨χ, 1G ⟩ = k. Solution. This follows easily from the previous exercise. Now we offer another solution. Let G act on the set Ω and let Ω1 , . . . , Ω k be the set of G-orbits. Set S = {(g, ω) | g ∈ G, ω ∈ Ω, g ⋅ ω = ω}. Then |S| = ∑g∈G χ(g) = |G|⟨χ, 1G ⟩, and, on the other hand, by Exercises 4.1 and 4.2 (we take ω i ∈ Ω i , i = 1, . . . , k), one obtains k

k

k

k

|S| = ∑ |G ω | = ∑ ∑ |G ω | = ∑ ∑ |G ω i | = ∑ |Ω i ||G ω i | = ∑ |G| = k|G|. ω∈Ω

i=1 ω∈Ω i

i=1 ω i ∈Ω i

i=1

i=1

Therefore, ⟨χ, 1G ⟩ = k. Definition. If G acts transitively on the set of ordered k-subsets of M, then the action of G on M is said to be k-transitive. Exercise 4.5 (Burnside). Let χ be the character of a transitive permutation representation P of a group G and t the number of P H -orbits, where P H is the restriction of P to the stabilizer H of one of the points. Then ⟨χ, χ⟩ = t. Solution. By Theorem 4.1, we get χ = 1GH . Therefore, by reciprocity and Exercise 4.4, we obtain ⟨χ, χ⟩ = ⟨χ, (1H )G ⟩ = ⟨χ H , 1H ⟩ = t. One can rewrite the result of Exercise 4.5 as follows: |G|−1 ∑g∈G |χ(g)|2 = t. We offer to present a character free proof of that relation. Exercise 4.6. The following statements hold: (a) Suppose that a group G acts transitively on the set Ω of cardinality n and ω ∈ Ω. If G ω acts transitively on Ω − {ω}, then G is 2-transitive. (b) Suppose that P is a 2-transitive permutation representation of G on the set Ω with character χ. Then χ = 1G + θ, where θ ∈ Irr(G). (c) Suppose that the group G is 2-transitive on the set Ω, ω ∈ Ω and H = G ω . Then G = H ∪ HxH for every x ∈ G − H. Solution. (a) The number of ordered pairs (α, β), where α, β ∈ Ω, α ≠ β, equals n(n−1). If β ∈ Ω − {α}, then |G : (G α )β | = |G : G α ||G α : (G α )β | = n(n − 1), and we conclude that G is transitive on the set of ordered pairs hence G is 2-transitive.

196 | Characters of Finite Groups 1 (b) We have χ = 1G + θ, where θ ∈ Char(G) and ⟨θ, 1G ⟩ = 0 (Exercise 4.3). Since G is doubly transitive on Ω, then H = G α (α ∈ Ω) has exactly two orbits. Therefore, in the notation of Exercise 4.5, t = 2, i.e., ⟨χ, χ⟩ = 2. It follows that ⟨θ, θ⟩ = ⟨χ − 1G , χ − 1G ⟩ = ⟨χ, χ⟩ − 2⟨χ, 1G ⟩ + ⟨1G , 1G ⟩ = 2 − 2 + 1 = 1 so that θ ∈ Irr(G). (c) We have |G : H| = n and |H : (H x ∩ H)| = n − 1 since H is transitive on the set Ω − {ω} of cardinality n − 1 and H ∩ H x is the stabilizer of a point in H. Therefore, by the product formula applied to HxH = xH x H, we obtain |H ∪ HxH| = |H| +

|H|2 = |H|(1 + n − 1) = |H|n = |G|, |H x ∩ H|

and we conclude that H ∪ HxH = G. Exercise 4.7. Let G = Sn (n ≥ 2) or An (n ≥ 4). Then G possesses an irreducible integervalued character of degree n − 1. Hint. Let T be a 2-transitive permutation representation of Sn of degree n. If θ is the character of T, then, by Exercise 4.6 (b), we have θ = 1Sn + χ, where χ ∈ Irr(Sn ). Since χ(1) = θ(1) − 1 = n − 1, we are done. The same argument also holds for the groups G = An , n > 3. Exercise 4.8. The symmetric group Sn (n ≥ 4) possesses at least two irreducible characters of degree n − 1 whose restrictions to An coincide. Solution. Multiplying the character χ from Exercise 4.7 by a nonprincipal linear character λ of Sn , we get the irreducible character of Sn of degree n − 1 which is ≠ χ. Then (χλ)An = χAn λAn = χAn 1An = χAn . A character χ of a group G is said to be rational if it takes only rational values (see §III.8). Let Charℚ (G) denote the set of all rational characters of G. It is easy to show that χ is rational if and only if χ(x) = χ(y) whenever elements x, y generate the same (cyclic) subgroup; see Lemma 4.6, below, and also §III.8). If χ1 , χ2 ∈ Charℚ (G), then the function χ1 − χ2 : G → ℂ is said to be a generalized rational character of G. We denote by Chℚ (G) the set of all generalized rational characters of G. It is obvious that Chℚ (G) is a subring of the ring Ch(G) of all generalized characters of G (see the following paragraph). Let Irr(G) = {χ1 , . . . , χ r } (r = k(G), the class number of G). Then Ch(G) = {m1 χ1 + ⋅ ⋅ ⋅ + m r χ r | m i ∈ ℤ, i ∈ {1, . . . , r}} is the set of generalized characters of G. Therefore the additive group of the ring Ch(G) is the free abelian group of rank r = k(G) with (the orthonormal) basis Irr(G) (indeed, the set Irr(G) is linearly independent over ℂ and every class function on G is a linear

V Induced characters and representations | 197

combination of members of the set Irr(G)). The additive group of the subring Chℚ (G) is a subgroup of the free abelian group Ch(G) of rank r so it is itself a free abelian group, of rank ≤ r. It follows from the formula for induced characters that λ G is rational if λ ∈ Irr(H) is a rational character of H ≤ G. Therefore, the characters of permutation representations of G are rational. Any generalized character of the form ∑H≤G a H (1H )G belongs to Chℚ (G)(a H ∈ ℤ). But the ordinary quaternion group has a rational irreducible character of degree 2, which cannot be written in this form. Denote by Λ(G) the set of all ℤ-linear combinations of the characters (1H )G (H ≤ G). Lemma 4.2. The set Λ(G) is a subring of the ring Chℚ (G). Proof. It suffices to prove that, for all H, K ≤ G, we have (1H )G (1K )G ∈ Λ(G). To this end, let Γ : G → S(M) and ∆ : G → S(N) be transitive permutation representations of the group G affording the characters (1H )G , (1K )G , respectively, and let M and N be the sets on which Γ(G) and ∆(G) act, respectively, |M| = |G : H|, |N| = |G : K|. Define an action of G on the set M × N as follows: g(μ, ν) = (gμ, gν)

(g ∈ G, (μ, ν) ∈ M × N)

(here gμ = Γ(g)μ and gν = ∆(g)ν). Let P be the permutation representation of G, afforded by this action. Since g(μ, ν) = (μ, ν) if and only if gμ = μ, gν = ν, it follows that the number of fixed points of the permutation P(g) is the product of the number (1H )G (g) of fixed points of the permutation Γ(g) and the number (1K )G (g) of fixed points of the permutation ∆(g), i.e., (1H )G (1K )G = χ, where χ = χ P is the character of P. But χ = χ1 + ⋅ ⋅ ⋅ + χ k , where χ i is the character of a transitive constituent P i of P, i.e., χ i = (1H i )G for some H i ≤ G. Therefore, k

(1H )G (1K )G = ∑ (1H i )G ∈ Λ(G). i=1

Exercise 4.9. By Lemma 1.3, we have (1H )G (1K )G = [((1H )G )K ⋅ 1K ]G = [((1H )G )K ]G . Give another proof of Lemma 4.2, using this equality and Mackey’s theorem. The additive group Λ(G)+ of the ring Λ(G) (see Lemma 4.2) is a subgroup of the additive group Chℚ (G)+ of the ring Chℚ (G); Chℚ (G)+ is a free abelian group of rank not exceeding r = k(G). Artin’s theorem, stated and proved below, shows, in particular, that |Chℚ (G)+ : Λ(G)+ | < ∞, exp(Chℚ (G)+ /Λ(G)+ ) | |G|. Let Λ C (G) be the set of all ℤ-linear combinations of characters (1H )G , where H ≤ G is cyclic; then Λ C (G)+ ≤ Λ(G)+ . Observe that the groups Λ C (G)+ , Λ(G)+ , considered as subgroups of a free abelian group of rank not exceeding r = k(G), are themselves free abelian of ranks not exceeding r. Theorem 4.3 (E. Artin). For every finite group G, one has |G|χ ∈ Λ(G) for all χ ∈ Chℚ (G). The proof of Theorem 4.3, presented below, is due to Lam [Lam].

198 | Characters of Finite Groups 1 Let G1 = {1}, G2 , . . . , G m be representatives of all the conjugacy classes of cyclic subgroups of the group G, G i = ⟨x i ⟩, |G1 | ≤ |G2 | ≤ ⋅ ⋅ ⋅ ≤ |G m |. Put μ i = (1G i )G . Then μ1 = (1G1 )G = ρ G is the regular character of G. Put a ij = μ j (x i ) (i, j = 1, . . . , m); then a ij ∈ ℤ for all i, j (Theorem 1.1). Lemma 4.4. With the above notation, the following statements hold: (a) a ij ≠ 0 if and only if G j contains a subgroup conjugate to G i . (b) If i ≠ j and a ij ≠ 0, then i < j and a ij = |NG|G(Gj | i )| . (c) a ii = |NG (G i ) : G i | (i ∈ {1, . . . , m}). Proof. (a) For all i, j ∈ {1, . . . , m}, one obtains, using formula (1.2), a ij = μ j (x i ) = |G j |−1 ∑ (1̇ G j )(x ti ) = |G j |−1 t∈G

∑ t∈G,

x ti ∈G j

1 = |G j |−1

∑

1.

t∈G, G ti ≤G j

Therefore, a ij ≠ 0 if and only if G ti ≤ G j for some t ∈ G. (b) Let i ≠ j and a ij ≠ 0. Then, by (a), the cyclic subgroup G j contains a unique subgroup L conjugate to G i in G, and therefore G ti = L for some t ∈ G. Next, L < G j , since G i , G j are not conjugate in G. Therefore, |G i | < |G j |, and so i < j. Finally, a ij = |G j |−1

∑

t∈G, G ti =L

1=

|NG (G i )| . |G j |

(c) It follows from (a) that a ii ≠ 0 for all i. We have a ii = μ i (x i ) = |G i |−1

∑

t∈G, G ti =G i

1 = |NG (G i ) : G i |.

It follows from Lemma 4.4 that A = (a ij ) is a nonsingular upper triangular matrix and all entries on its leading diagonal are positive integers dividing |G|. Proof of Theorem 4.3. Consider the following system of equations: m

∑ a ij u j = |G|χ(x i )

(i = 1, . . . , m).

(1)

j=1

Since (a ij ) is a nonsingular upper triangular ℤ-matrix (see the previous paragraph) and χ(x i ) ∈ ℤ (because of the rationality of χ; recall that χ(x i ) is an algebraic integer), system (1) has a unique solution (u1 , . . . , u m ), where u1 , . . . , u m ∈ ℚ. We assert that |G j |−1 u j ∈ ℤ (j ∈ {1, . . . , m}). For j = m this is obvious: the last equation of system (1) is of the form a mm u m = |G|χ(x m ), and as a mm = |NG (G m ) : G m |, this gives u m |G m |−1 = |G : NG (G m )|χ(x m ) ∈ ℤ (indeed, χ(x) is a rational algebraic integer for all x ∈ G; see Lemma III.1.1). Suppose that we have already proved that u j |G j |−1 ∈ ℤ for j = i + 1, . . . , m. We prove that then u i |G i |−1 ∈ ℤ. To do this, consider the i-th equation of system (1): a ii u i + a i,i+1 u i+1 + ⋅ ⋅ ⋅ + a im u m = |G|χ(x i ).

(2)

V Induced characters and representations | 199

Let a ij1 , . . . , a ij s be all nonzero numbers among a i,i+1 , . . . , a im . By Lemma 4.4, |NG (G i )| |G j ν |

a ij ν =

(ν = 1, . . . , s),

a ii = |NG (G i ) : G i |.

Dividing both sides of equation (2) by |NG (G i )|, one obtains s

u i |G i |−1 + ∑ u j ν |G j ν |−1 = |G : NG (G i )|χ(x i ) ∈ ℤ, ν=1

hence, by induction, u i |G i |−1 ∈ ℤ. In particular, u i ∈ ℤ for all i. Let x ∈ G. Then there exists i ∈ {1, . . . , m} such that the subgroup ⟨x⟩ is conjugate to G i = ⟨x i ⟩. Since χ(x) is a rational algebraic integer, we get χ(x) = χ(x i ) ∈ ℤ. Similarly, μ j (x) = μ j (x i ) = a ij (recall that μ i = (1G i )G ). Therefore, m

m

|G|χ(x) = |G|χ(x i ) = ∑ a ij u j = ∑ μ j (x)u j . j=1

j=1

Since x ∈ G is arbitrary, we get m

|G|χ = ∑ u j μ j ∈ Λ C (G). j=1

A different proof of Theorem 4.3, due to Brauer, may be outlined as follows: Let L C (G) denote the set of all cyclic subgroups of the group G. Given H ∈ L C (G), define a function θ(H) : H → ℂ by {|H| if ⟨x⟩ = H, θ(H) (x) = { 0 if ⟨x⟩ ≠ H. { Brauer has proved the following lemmas. Lemma 4.5. The following statements hold: G . (a) |G| ⋅ 1G = ∑H∈L C (G) θ(H) (b) If H ∈ L C (G), then θ(H) ∈ Λ C (H). Proof. (a) Given x ∈ G and H ∈ L C (G), we get G θ(H) (x) = |H|−1

∑

t∈G, x t ∈H

θ(H) (x t ) = |H|−1

∑

t∈G, ⟨x t ⟩=H

|H| =

∑

1.

t∈G, ⟨x t ⟩=H

Therefore, ∑ H∈L C (G)

G θ(H) (x) =

∑

∑

H∈L C (G) t∈G, ⟨x t ⟩=H

1 = ∑ 1 = |G|, t∈G

whence = |G| ⋅ 1G , completing the proof of (a). (b) We proceed by induction on |H|. Obviously, one may assume that |H| > 1 (otherwise, it is nothing to prove). Then, by (a), G ∑H∈L C (G) θ(H)

H H |H| ⋅ 1H = ∑ θ(D) = θ(H) + ∑ θ(D) . D≤H

D 0. This means that ∑x∈S χ(x) ≥ 2|S| = 2q. Since χ(x) = χ(1) − q for x ∈ S# , it follows that ∑ χ(x) = χ(1) + (q − 1)[χ(1) − q] = q[χ(1) − (q − 1)].

x∈S

Thus, q[χ(1) − (q − 1)] ≥ 2q implies χ(1) ≥ 1 + q. Consequently, if a χ > 0, then we have χ(1) − q ≥ 1. Therefore, λ(1) =

∑ χ∈Irr(G)

a χ (χ(1) − q) ≥

∑ χ∈Irr(G)

aχ .

V Induced characters and representations | 209

One can rewrite (4) as follows: (r − 1)λ(1) = q

∑

aχ .

χ∈Irr(G)

Next, λ(1) ≥

∑

a χ 󳨐⇒ r − 1 ≤ q 󳨐⇒ r = 1 + q,

χ∈Irr(G)

and we conclude that |G| = (q − 1)qr = (q + 1)q(q − 1). Let γ be the action of G by conjugation on the r-element set M = Syl2 (G) = {S = S0 , S1 , . . . , S q }. Let N i = NG (S i ) (i = 0, 1, . . . , q). Then N i ≅ N = NG (S0 ) is a Frobenius group with kernel S i of order q and Frobenius complement of order q − 1 (see Chapter X for details on Frobenius groups). It is obvious that N i is the γ-stabilizer of the “point” S i in G. Let γ0 be the action of N on M0 = M − {S0 }, the restriction of γ. Let t be the length of the γ0 -orbit of the point S1 . We have t = |N : (N ∩ N1 )| = |N ∩ N1 |−1 ⋅ q(q − 1). Since S ∩ S1 = {1}, it follows that |N ∩ N1 | is an odd number. Since q = 2n and |N ∩ N1 | divides q(q − 1), we get |N ∩ N1 | | q − 1. Therefore t ≥ q. But t ≤ |M0 | = q. Therefore t = q, and the action γ0 is transitive. But then γ is a 2-transitive action. Since N is a Frobenius group of order q(q − 1) with kernel S of order q, it follows that γ0 is a strictly 2-transitive action of N on M0 . But then γ is a strictly 3-transitive action. Since |M| = 1 + q = 1 + 2n , Lemma 6.1 now gives G ≅ PSL(2, q).

7 Small Burnside’s theorem on 2-transitive groups The following theorem of Burnside shows that a subgroup having index < n in a 2-transitive permutation group of degree n, must be transitive. Theorem 7.1. Let Γ be a 2-transitive representation of a group G by permutations of a set M of cardinality n. If H < G and |G : H| < n, then the restriction Γ H is transitive. Proof. By hypothesis, n > 2. Let θ be the character of Γ. Since Γ is a 2-transitive representation, we get θ = 1G + χ, where χ ∈ Irr# (G) (see Exercise 4.4) so that χ ∈ Irr1 (G) since n > 2. To prove the transitivity of H, it suffices to show that there is exactly one H-orbit on the set M. Therefore, in view of Exercise 4.4, we have to show that ⟨θ H , 1H ⟩ = 1. It suffices to show that ⟨χ H , 1H ⟩ = 0. Assume that ⟨χ H , 1H ⟩ > 0. Then m = ⟨(1H )G , χ⟩ = ⟨1H , χ H ⟩ > 0. Next, ⟨(1H )G , 1G ⟩ = ⟨1H , 1H ⟩ = 1 󳨐⇒ (1H )G = mχ + 1G + ϑ, where ϑ is either a character of G or the zero function. Note that χ(1) = θ(1) − 1 = n − 1.

210 | Characters of Finite Groups 1 Consequently, n > |G : H| = (1H )G (1) = mχ(1) + 1 + ϑ(1) ≥ χ(1) + 1 = (n − 1) + 1 = n, which is a contradiction. Thus, ⟨χ H , 1H ⟩ = 0 and, therefore, ⟨θ H , 1H ⟩ = ⟨1H + χ H , 1H ⟩ = 1 implying that the representation Γ H is indeed transitive, by the result of Cauchy– Frobenius–Burnside (see Exercise 4.4). Note that G = Sn (n > 4) does not contain subgroups H such that 2 < |G : H| < n (Bertrand’s theorem). Indeed, suppose that such a subgroup H exists. Then it follows from the normal structure of Sn that H G = {1}. But then G = Sn is isomorphic to a subgroup of the symmetric group S|G:H| of degree |G : H| < n, which is impossible since |Sn | = n! > |G : H|! = |S|G:H| |.

8 Gallagher’s theorem on restriction of characters Let us consider the following two properties that may or may not satisfy a finite group G: (i) If x1 , . . . , x t ∈ G such that GCD(o(x i ), o(x j )) = 1 for i ≠ j and x1 . . . x t = 1, then x1 = ⋅ ⋅ ⋅ = x t = 1. (ii) If χ ∈ Irr(G) and ⟨χ S , 1S ⟩ > 0 for all S ∈ Syl(G), then χ = 1G . Exercises 8.1 and 8.2, below, show that solvable groups satisfy both properties (i) and (ii). The main result of this section is Theorem 8.1, due to P. Gallagher [Gal3], asserting that, for a finite group G, properties (i) and (ii) are equivalent. The starting point of that theorem is the following Exercise 8.1 (P. Hall). A solvable group G possesses property (i). Solution. We use induction on |G|. The statement is trivial if G is abelian. Now let G be nonabelian. Therefore, if G = G/G󸀠 , then x1 . . . x t = 1, and so x1 = ⋅ ⋅ ⋅ = x t = 1, i.e., x1 , . . . , x t ∈ G󸀠 . Since G󸀠 < G, it follows by induction that x1 = ⋅ ⋅ ⋅ = x t = 1. P. Hall conjectured that (i) is true if and only if G is solvable. This is indeed the case, but the only proof is based on Thompson’s classification of minimal simple groups (a nonabelian simple group is said to be minimal if all its proper subgroups are solvable). In what follows we do not use this deep result. Exercise 8.2. A solvable group G satisfies property (ii). Hint. Without loss of generality, one may assume that ker(χ) = {1} and G > {1}; then F(G) > {1} and χ ≠ 1G . Take a nontrivial P ∈ Sylp (F(G)), and let P ≤ S ∈ Sylp (G). Now ⟨χ S , 1S ⟩ > 0 󳨐⇒ ⟨χ P , 1P ⟩ > 0.

V Induced characters and representations | 211

Then, by Clifford’s theorem (Chapter VII), we obtain P ≤ ker(χ) = {1}, contrary to the choice of P. Theorem 8.1. For each group G, (i) if and only if (ii). Set π(G) = {p1 , . . . , p s }. Consider the following condition: (i󸀠 ) If x i ∈ G is a p i -element (i = 1, . . . , s), then x1 . . . x s = 1 implies x1 = ⋅ ⋅ ⋅ = x s = 1. Proof of Theorem 8.1. Let S i ∈ Sylp i (G), i = 1, . . . , s. Put φ i = (1S i )G

(i = 1, . . . , s),

σ = |G|1−s

∑

x1 ...x s =1

φ1 (x1 ) . . . φ s (x s ).

By formula (1.10) with H = S i and ψ = 1S i , one obtains φ i (g) = |S i |−1 ∑ 1̇ S i (t−1 gt) = |S i |−1 t∈G

1 = |CG (g)| ⋅ |S i |−1 |K g ∩ S i |,

∑

t−1 gt∈S

i

where K g is the G-class of g ∈ G. Therefore, φ i (g) > 0 if and only if K g ∩ S i ≠ 0 (this is true if and only if o(g) is a power of p i ), and so {= 0 φ i (g) { >0 {

if g is not a p i -element, if g is a p i -element.

In that case, σ = |G|1−s

∑

x i is a p i -element x1 ...x s =1

φ1 (x1 ) . . . φ s (x s ) ≥ |G|1−s φ1 (1) . . . φ s (1) = 1,

since φ i (1) = |G : S i |, i = 1, . . . , s, and so |S1 | . . . |S s | = |G| 󳨐⇒ φ1 (1) . . . φ s (1) = |G|s−1 . Thus, σ ≥ 1, and moreover, the following equivalence relations hold: σ = 1 ⇐⇒ {x1 . . . x s = 1, x i is a p i -element for all i if and only if x1 = ⋅ ⋅ ⋅ = x s = 1} ⇐⇒ (i󸀠 ) holds for some rearrangement of the numbers p i . Let Irr(G) = {χ0 = 1G , χ1 , . . . , χ r−1 }. Then, if c iα = ⟨(χ α )S i , 1S i ⟩(= ⟨χ α , φ i ⟩), we get r−1

φ i = ∑ c iα χ α , α=0

i = 1, . . . , s.

Consider the following condition: (ii󸀠 ) If α > 0, then ∏si=1 c iα = 0, i.e., c iα = 0 for some i ≤ s. We claim that (ii󸀠 ) if and only if (ii). Indeed, if (ii) holds, then r−1

∏ c iα ≠ 0 󳨐⇒ c iα > 0 for i = 1, . . . , s 󳨐⇒ ⟨φ i , χ α ⟩ > 0 for all i. i=1

(1)

212 | Characters of Finite Groups 1 Let S ∈ Syl(G). Then S is conjugate to S i for some i; hence φ i = (1S )G , i.e., ⟨1S , (χ α )S ⟩ = ⟨(1S )G , χ α ⟩ = ⟨φ i , χ α ⟩ = c iα > 0, so that, by (ii), χ α = 1G , i.e., α = 0. Thus, (ii) implies (ii󸀠 ). Now suppose that (ii󸀠 ) holds. Let χ ∈ Irr(G) and ⟨χ S , 1S ⟩ > 0 for all S ∈ Syl(G). 󸀠 Putting χ = χ α , S ∼ S i , one obtains c iα > 0 for all i, i.e., ∏r−1 i=1 c iα > 0. It follows from (ii ) 󸀠 that α = 0, i.e., χ = 1G , and so (ii ) implies (ii). Let us continue the calculation of σ. Let x i be a p i -element as above, i = 1, . . . , s, and set g1 = x1 , g2 = x1 x2 , . . . , g s = x1 . . . x s . Then x1 = g1 , x2 = g1−1 g2 , . . . , x s = g −1 s−1 g s .

(2)

It is obvious that x1 . . . x s = 1 if and only if (2) holds and g s = 1. Therefore, σ = |G|1−s

∑

g1 ,...,g s ∈G, g s =1

φ1 (g1 )φ2 (g1−1 g2 ) . . . φ s (g −1 s−1 g s ).

Hence, by (1), σ = |G|1−s

s r−1

∏ ∑ c iα χ α (g −1 i−1 g i ),

∑

g1 ,...,g s−1 ∈G i=1 α=0

where g0 = g s = 1. Thus, σ = |G|1−s = |G|1−s

r−1

∑

∑

g1 ,...,g s−1 ∈G α1 ,...,α s =0

c1α1 . . . c sα s χ α1 (g1 )χ α2 (g1−1 g2 ) . . . χ α s (g −1 s−1 g s )

r−1

∑

α1 ,...,α s =0

c1α1 . . . c sα s

∑

g1 ,...,g s−1 ∈G

χ α1 (g1 )χ α2 (g1−1 g2 ) . . . χ α s (g −1 s−1 g s ).

Note that g s = 1. But (see Chapter III, Exercise III.4.1) ∑

g1 ,...,g s−1 ∈G

=

χ α1 (g1 )χ α2 (g1−1 g2 ) . . . χ α s (g−1 s−1 g s ) ∑

g2 ,...,g s−1 ∈G

−1 χ α3 (g2−1 g3 ) . . . χ α s (g −1 s−1 g s ) ∑ χ α1 (g 1 )χ α2 (g 1 g 2 ) g1 ∈G

= δ α1 ,α2 |G|χ α1 (1)−1 2

∑

g2 ,...,g s−1 ∈G

χ α2 (g2 )χ α3 (g2−1 g3 ) . . . χ α s (g −1 s−1 g s )

−1

= δ α1 ,α2 |G| χ α1 (1) δ α2 ,α3 χ α2 (1)−1

∑

g3 ,...,g s−1 ∈G

χ α3 (g3 )χ α4 (g3−1 g4 ) . . . χ α s (g −1 s−1 g s ).

Continuing in this way, we finally obtain ∑

g1 ,...,g s−1 ∈G

χ α1 (g1 )χ α2 (g1−1 g2 ) . . . χ α s (g −1 s−1 g s )

= δ α1 ,α2 δ α2 ,α3 . . . δ α s−1 ,α s |G|s−1 [χ α1 (1) . . . χ α s−1 (1)]−1 χ α s (1). We have used the relation ∑ χ α (x)χ β (x−1 y) = δ α,β |G|χ β (y)χ α (1)−1 .

x∈G

V Induced characters and representations | 213

Therefore, σ = |G|1−s

r−1

∑

α1 ,...,α s =0

c1α1 . . . sα s δ α1 ,α2 δ α2 ,α3 . . . δ α s−1 ,α s |G|s−1 [χ α1 (1) . . . χ α s−1 (1)]−1 χ α s (1)

r−1

r−1

s

α=0

α=0

i=1

= ∑ c1α . . . c sα χ α (1)2−s = ∑ (∏ c iα )χ α (1)2−s . Since

s

s

i=1

i=1

(∏ c i0 )χ0 (1)2−s = ∏⟨1S i , 1S i ⟩ = 1, it follows that σ ≥ 1 and, moreover, σ = 1 if and only if ∏si=1 c iα = 0 for all α > 0, i.e., σ = 1 ⇔ (ii󸀠 ). Since, on the other hand, σ = 1 ⇔ (i󸀠 ), it follows that (i󸀠 ) ⇔ (ii󸀠 ), and since (ii󸀠 ) ⇔ (ii), one obtains (ii) ⇔ (i󸀠 ). Now suppose that (i) holds. Then, for any rearrangement of the primes p i , condition (i󸀠 ) holds; hence (ii) holds as well. Thus, (i) ⇔ (ii), and the theorem is proved. Corollary 8.2. For some arrangement of the primes p i , (i󸀠 ) implies (i). For a more compact formulation of the following exercises, we introduce some notation. Given χ ∈ Char(G), we set w(χ) =

∑ ⟨χ, τ⟩ τ∈Irr(G)

(if, for example, χ = ∑τ∈Irr(χ) c τ τ, then w(χ) = ∑τ∈Irr(G) c τ ). Let H ≤ G, g = |G|, h = |H|, φ ∈ Irr(H), χ ∈ Irr(G). Set χ H = ∑i∈I b i φ i , where φ i runs over the set Irr(χ H ) (and so b i > 0 for all i) and φ1 (1) ≤ φ2 (1) ≤ ⋅ ⋅ ⋅ ; in that case, w(χ H ) = ∑i∈I b i . Next, let φ G = ∑j∈J a j χ j , where χ j runs over the set Irr(φ G ) (so that a j > 0 for all j ∈ J); in that case, w(φ G ) = ∑j∈J a j . For better acquainting as Frobenius reciprocity law works, we suggest to read solutions of the following exercises. Exercise 8.3. The following statements hold: (a) w(φ G ) ≤ hg . (b) w(χ H ) ≤ hg . (c) If w(φ G ) = hg , then a τ = 1 for all τ ∈ Irr(φ G ). For solution, see Lemmas VII.8.1 and VII.8.2. Exercise 8.4. If w(χ H ) = hg , then: (a) b i = 1 for all i. (b) φ α (1) = φ β (1) for all α, β ∈ I. (c) φ Gα = χ for all α ∈ I. (d) ⟨χ H , τ H ⟩ = 0 for all τ ∈ Irr(G) − {χ}.

214 | Characters of Finite Groups 1 Solution. Write φ = φ1 . By the above agreement, χ(1) ≥ φ(1)w(χ H ) = hg φ(1) = φ G (1). It follows from χ ∈ Irr(φ G ) that φ G = χ, and so χ(1) = hg φ(1). In that case, by reciprocity, b1 = ⟨χ H , φ⟩ = ⟨χ, φ G ⟩ = ⟨χ, χ⟩ = 1. On the other hand, if (b) is not true, then hg φ(1) = χ(1) = ∑i∈I b i φ i (1) > hg φ(1), a contradiction. Thus, all irreducible constituents of χ H have the same degree, proving (b), and now (a) follows from what has just has been proved. Now (c) follows, by reciprocity, and (d) follows from (c) and reciprocity. Exercise 8.5. If w(χ H ) >

g 2h ,

then min{b i | i ∈ I} = 1.

Hint. Assume that b i > 1 for all i. Then χ(1) = ∑i∈I b i φ i (1) ≥ w(χ H )φ1 (1) > It follows from ⟨φ1G , χ⟩ = b1 ≥ 2 that g g g φ1 (1) = φ1G (1) ≥ b1 χ(1) > 2 ⋅ φ1 (1) = φ1 (1), h 2h h

g 2h φ 1 (1).

a contradiction. Exercise 8.6. If w(χ H ) =

g h

− 1, then b i = 1 for all i.

Exercise 8.7. If w(φ G ) = hg , then: (a) a j = 1 for all j. j

(b) χ H = φ for all j.

(c) If λ ∈ Irr(H) − {φ}, then ⟨φ G , λ G ⟩ = 0.

(d) Is it true that, provided H ⊲ G, then G/H is abelian? Solution. Part (a) coincides with Exercise 8.3 (c). Statement (b) follows by hypothesis since φ G (1) = hg φ(1) and all hg irreducible constituents of φ G have degrees ≥ φ(1). Statement (c) follows from (b). To solve (d), apply Clifford theory (Chapter VII). g Exercise 8.8. If w(φ G ) = − 1, then a i = 1 for all i. h Solution. We have χ j (1) ≥ a j φ(1) for all j ∈ J, by reciprocity. Therefore, g φ(1) = φ G (1) = ∑ a j χ j (1) ≥ ∑ a2j ϕ(1). h j∈J j∈J Assume that a i > 1 for some i; then a2i ≥ 2 + a i , and we obtain g g ∑ a2j ≥ 2 + ∑ a j = 2 + w(φ G ) = 2 + − 1 = + 1. h h j∈J j∈J It follows that hg φ(1) ≥ ( hg + 1)φ(1), a contradiction. Exercise 8.9. If H is normal in G and w(φ G ) >

g 2h ,

then G/H is solvable.

Hint. This is a difficult exercise, whose solution involves results from Chapters VII and XI. Exercise 8.10. Let n ∈ ℕ, G[n] the direct product of n copies of G and D n the diagonal subgroup of G[n] . For χ ∈ Irr(G), define χ0 ∈ Irr(D n ) by χ0 ((g, . . . , g)) = χ(g) for

V Induced characters and representations | 215

all (g, . . . , g) ∈ G[n] . Then, for τ ∈ Irr(G), one has τ0G

[n]

=

(χ1 × ⋅ ⋅ ⋅ × χ n )⟨χ1 . . . χ n , τ⟩.

∑

χ1 ,...,χ n ∈Irr(G)

Solution. By reciprocity, for χ1 , . . . , χ n ∈ Irr(G), one obtains [n]

⟨τ0G , χ1 × ⋅ ⋅ ⋅ × χ n ⟩ = ⟨τ0 , (χ1 × ⋅ ⋅ ⋅ × χ n )D n ⟩ = |G|−1 ∑ χ1 (g) . . . χ n (g)τ(g) = ⟨χ1 . . . χ n , τ⟩, g∈G

since D n ≅ G, and we are done. (Compare with Theorem 3.4.)

9 Monomial representations and characters Let V be a ℂG-module, and let V = L1 ⊕ ⋅ ⋅ ⋅ ⊕ L n , where the L i are one-dimensional ℂ-subspaces. The module V is said to be monomial if the subspaces L i are permuted by the elements of G: if g ∈ G, then gL j = L j󸀠 , where {1󸀠 , . . . , n󸀠 } is a permutation (depending on g) of the ordered set {1, . . . , n}. The mapping g 󳨃→ (

1 1󸀠

... ...

n ) = p(g) n󸀠

is a permutation representation of the group G. A matrix representation of G afforded by a monomial ℂG-module V is said to be monomial. Characters of monomial representations of G are called monomial characters. Let v i be a basis vector of the subspace L i , i ∈ {1, . . . , n}; then B = {v i }ni is a basis of V. The matrix representation T afforded by the module V and written in terms of the basis B is called a standard monomial representation of G. In particular, permutation representations are monomial. If g ∈ G and gL j = L j󸀠 , then we have gv j = λ j (g)v j󸀠 , where λ j (g) ∈ ℂ∗ (the functions λ j : G → ℂ∗ are called the multipliers of the monomial representation T). It follows that T(g) = (τ ij (g))ni , where {λ j (g) if j󸀠 = i (i.e., gv j = λ j (g)v i ), τ ij (g) = { 0 if j󸀠 ≠ i. { Let P(g) be the matrix of the permutation p(g) associated with matrix T(g). Then we have T(g) = P(g)D(g), where D(g) = diag(λ1 (g), . . . , λ n (g)). Note that matrices of the form PD, where P is a permutation matrix and D is a nonsingular diagonal matrix, are called monomial matrices (in partial case, when D is the identity matrix, we obtain a permutation matrix). These matrices form a group which is a semidirect product Sn ⋅ D(n, ℂ) of the matrix symmetric group Sn and the diagonal group D(n, ℂ) (the kernel of this semidirect product is D(n, ℂ)). This group is called the complex monomial group of degree n and denoted by GM(n, ℂ). A monomial representation of the group G can be defined as a homomorphism of G into the group GM(n, ℂ).

216 | Characters of Finite Groups 1 The concept of monomial representation of a group G as defined above is too general for our purposes. In what follows we will consider only transitive monomial representations. A monomial ℂG-module V and the corresponding monomial representation T are said to be transitive if G acts transitively on the set (of one-dimensional ℂ-subspaces) {L i , . . . , L n }. Observe that transitive monomial modules are identical to imprimitive modules, as defined above, with one-dimensional subspaces of imprimitivity. Let H = {g ∈ G | gL = L} be the stabilizer of the ℂ-subspace L = L1 . Then L is a ℂH-module, and, if V is transitive, we have V = L G and T ∼ ∆ G , where ∆ is the one-dimensional representation of the subgroup H afforded by the ℂH-module L (in particular, this implies that one can choose the basis B = {v i }1n so that the multipliers of the representation T are the |G|-th roots of unity). Let χ = χ T be the character of the monomial representation T. Then χ is called a monomial character of G, and χ = ψ G , where ψ is the linear character of H afforded by one-dimensional representation ∆. Conversely, if H ≤ G and ∆ is a one-dimensional representation of H, then ∆ G is the transitive standard monomial representation of G and χ = ψ G is the corresponding monomial character (where ψ is the character of ∆). Therefore, in the narrower sense just adopted, the monomial characters of G are the characters induced by linear characters of subgroups of G. In conclusion we will establish a connection between the transfer and monomial representations of groups that is important in many questions of finite group theory. Let H < G, |G : H| = n and let Σ = {t1 , . . . , t n } be a transversal of the set of left cosets of H. If g ∈ G and i ∈ {1, . . . , n}, then gt i = t i󸀠 h g,i , where i󸀠 ∈ {1, . . . , n}, h g,i ∈ H (this corresponds to the equality gt i H = t i󸀠 H); then h g,i = t−1 i󸀠 gt i . The elements h g,i are called the multipliers of the element g with respect to the transversal Σ. Let us form the product of all multipliers of g: n

v Σ (g) = ∏ h g,i . i=1

The element v Σ (g) ∈ H depends on Σ and the ordering of factors. However, as we shall see below, the element v Σ (g)H 󸀠 of the quotient group H/H 󸀠 is independent of Σ and ordering of factors (the second assertion is obvious since the quotient group H/H 󸀠 is abelian), and this allows us to introduce the notation V(g) = v Σ (g)H 󸀠 . The mapping V : g 󳨃→ V(g), as we shall prove, is a homomorphism of G into H/H 󸀠 . This homomorphism is called the transfer of G into H. Let ψ ∈ Lin(H) and let ∆ be the corresponding one-dimensional representation of H < G. Consider the induced representation ∆ G of G with respect to the transversal Σ (in our case this representation is monomial). By (5.1), for every g ∈ G, one has ̇ −1 gt j ))i,j=1,...,n , ∆ G (g) = (ψ(t i where the function ψ̇ : G → ℂ vanishes on G − H and coincides with ψ on H, i.e., ψ̇ H = ψ.

V Induced characters and representations | 217

It is easy to see that the only nonzero element of i-th column of the matrix ∆ G (g) is equal to ψ(h g,i ). Denoting by σ(g) the signature of the permutation P(g) = (

t1 H gt1 H

... ...

tn H ) gt n H

of left cosets of G by H, we get det(ψ G )(g) = σ(g)ψ(v Σ (g))

(g ∈ G),

where det(ψ G ) is a linear character g 󳨃→ det(∆ G (g)) of G (depending on ψ). Since σ(g) ∈ {−1, 1}, we get ψ(v Σ (g)) = σ(g) det(ψ G (g)). It is important that the right-hand side of this equality independent of Σ. Therefore, if Σ󸀠 is another transversal, then ψ(v Σ (g)) = ψ(v Σ󸀠 (g)) for all g ∈ G and ψ ∈ Lin(G). It follows that v Σ (g)v Σ󸀠 (g)−1 ∈

⋂

ker(ψ) = H 󸀠 ,

i.e.,

v Σ (g)H 󸀠 = v Σ󸀠 (g)H 󸀠 .

ψ∈Lin(G)

Thus, the element v Σ (g)H 󸀠 of the quotient group H/H 󸀠 is independent of Σ, and this allows us to write V(g) = v Σ (g)H 󸀠 . Let us prove that the mapping g 󳨃→ V(g) is a homomorphism of G into the quotient group H/H 󸀠 . Let g1 , g2 ∈ G. Since the mapping g 󳨃→ ψ(v Σ (g)) is a linear character of G coinciding with σ ⋅ det(ϕ G ), we get ψ(v Σ (g1 g2 )) = ψ(v Σ (g1 )) ⋅ ψ(v Σ (g2 )). Since this is true for all ψ ∈ Lin(H), one obtains v Σ (g1 g2 )H 󸀠 = v Σ (g1 )H 󸀠 ⋅ v Σ (g2 )H 󸀠 󳨐⇒ V(g1 g2 ) = V(g1 )V(g2 ), hence V is a homomorphism. Exercise 9.1. Prove that G󸀠 ≤ ker(V), where V is the transfer of G in H < G. Hint. The group im(V) is abelian. We will obtain a formula for V(g) that is important for applications. Let C1 , . . . , C t be the cycles of the permutation P(g) = (

s1 H gs1 H

... ...

sn H ) gs n H

defined above. To obtain the desired formula, we use independence of V(g) on the choice of transversal Σ. Denoting by l i the length of the cycle C i , one can write this cycle in the following form: C i = (x i H, gx i H, . . . , g l i −1 x i H),

i = 1, . . . , t, where x i ∈ G.

One has l1 + ⋅ ⋅ ⋅ + l t = n.

218 | Characters of Finite Groups 1 As a transversal Σ we take the union of the sets {x i , gx i , . . . , g l i −1 x i }, i = 1, . . . , t. Now, li g l i x i H = x i H 󳨐⇒ x−1 i g xi = hi ∈ H

for all i.

It is clear that the ≠ 1 multipliers of g with respect to Σ are equal to h1 = h g,l1 , h2 = h g,l1 +l2 , . . . , h t = h g,n . Therefore the product v Σ (g) of all multipliers of g ∈ G is equal to h1 . . . h t , i.e., t li v Σ (g) = ∏ x−1 i g xi . i=1

It follows that t li V(g) = H 󸀠 ∏ x−1 i g xi .

(1)

i=1 li It is important to recall that x−1 i g x i = h i ∈ H for all i. A simplest application of equality (1) is the following:

Lemma 9.1. If g ∈ Z(G), then g n ∈ H and V(g) = g n H 󸀠 , where n = |G : H|. li li li Proof. In the case under consideration, one has h i = x−1 i g x i = g so that g ∈ H for t t n l all i ∈ {1, . . . , t}. Noting that ∑i=1 l i = n = |G : H|, one obtains g = ∏i=1 g i ∈ H and, by (1), t

t

li 󸀠 li n 󸀠 V(g) = H 󸀠 ∏ x−1 i g xi = H ∏ g = g H . i=1

i=1

We shall use Lemma 9.1 in the proof of Lemma VI.4.5 (c). Burnside applied transfer in the proof of his celebrated normal p-complement theorem. This is the earliest significant application of transfer. In that proof the following lemma of independent interest is essential. Lemma 9.2 (Burnside). Let p be a prime divisor of |G| and P ∈ Sylp (G). If x, y ∈ Z(P) are conjugate in G, then they are also conjugate in NG (P). Proof. By hypothesis, we have y = x u for some u ∈ G. It follows from x, y ∈ Z(P) that P ≤ CG (x) ∩ CG (y). Next, CG (y) = CG (x u ) = CG (x)u ≥ P u . Since P, P u ∈ Sylp (CG (y)), we get, by Sylow’s theorem, that P = (P u )v = P uv for some v ∈ CG (y) so that uv ∈ NG (P). Therefore, x uv = (x u )v = y v = y so that x, y are conjugate in NG (P). Remark. We use the notation of Lemma 9.2. The same argument shows that if two P-invariant subsets X, Y ⊆ P are conjugate in G, then they are also conjugate in NG (P). If, in particular, X = {x} and Y = {y}, then x, y ∈ Z(P), and we get Lemma 9.2.

V Induced characters and representations | 219

Theorem 9.3 (Burnside). Let P ∈ Sylp (G). If P ∈ Z(NG (P)), then G is p-nilpotent, i.e., G has a normal p󸀠 -subgroup N such that G = P ⋅ N (N is said to be the normal p-complement of G). Proof. Let V be the transfer of G in P. Since P is abelian, V is a homomorphism of G −1 l i li li in P = P/P󸀠 . Let g ∈ P. Then, by (1), V(g) = ∏ti=1 x−1 i g x i . Since g and x i g x i are G-conjugate, it follows that they are also conjugate in NG (P) (Lemma 9.2). In view t li li li n of P ≤ Z(NG (P)), by hypothesis, we get x−1 i g x i = g . Therefore, V(g) = ∏i=1 g = g , n where n = |G : P|. Since GCD(n, |P|) = 1, the map g 󳨃→ g is a bijection of P. Therefore, im(V) = P. Since im(V) ≅ G/N, where N = ker(V), the order of N is equal to |G : P| = n. It follows that P ∩ N = {1} and G = P ⋅ N (semidirect product with kernel N), by the product formula. Thus, N is the normal p-complement of G. Wielandt generalized Theorem 9.3 as follows. If P ∈ Sylp (G) is regular in P. Hall’s sense and NG (P) is p-nilpotent, then G is also p-nilpotent. (A p-group P is regular if for all x, y ∈ G we have (xy)p = x p y p z p for some z ∈ ⟨x, y⟩󸀠 .)

10 M-groups We begin with the following: Definition. A group G is said to be an M-group if all of its irreducible characters are monomial. Note that M-groups are of great interest since they present a fairly wide class of groups. For example, supersolvable groups, in particular, nilpotent groups, are M-groups (Blichfeldt was the first who observed the last fact). Next, the symmetric group S4 is an M-group. Moreover, as Dade has shown, every solvable group is isomorphic to a subgroup of an appropriate M-group. Therefore, it is impossible to classify the M-groups. The special linear group SL(2, 3) is not monomial since it has an irreducible character of degree 2 but has no subgroup of index 2 (it follows from definition that if an M-group has an irreducible character of degree n, then it possesses a subgroup of index n). In this section we present a proof of the classical Taketa’s theorem asserting that all M-groups are solvable. Let cd(G) denote the set of degrees of the irreducible characters of a group G (for example, cd(S4 ) = {1, 2, 3}); let G(i) be the i-th derived subgroup of G. Theorem 10.1 (K. Taketa [Tak]). Suppose that G is an M-group such that cd(G) is equal to {f1 , . . . , f k }, where 1 = f1 < ⋅ ⋅ ⋅ < f k . If χ ∈ Irr(G), χ(1) ≤ f i , then G(i) ≤ ker(χ). Proof. We use induction on i. If i = 1, then the result is obvious since χ (of degree f1 = 1) is linear so its kernel contains G(1) = G󸀠 . Now let i > 1 and suppose that the statement has already been proved for irreducible characters of degrees less than f i .

220 | Characters of Finite Groups 1 Since G is an M-group, we obtain that χ = λ G , where λ ∈ Lin(H) for some H < G; then χ(1) = |G : H|. Since ⟨(1H )G , 1G ⟩ = ⟨1H , 1H ⟩ = 1,

1GH (1) = |G : H| > 1,

it follows that the character (1H )G is reducible. If ψ ∈ Irr((1H )G ), then ψ(1) < (1H )G (1) = |G : H| = λ G (1) = χ(1) ≤ f i so that ψ(1) < f i−1 , and therefore G(i−1) ≤ ker(ψ), by induction. Since ker((1H )G ) =

⋂

ker(ψ),

ψ∈Irr((1H )G )

it follows from Lemma 2.1 that G(i−1) ≤ ker((1H )G ) = H G ≤ H, and therefore we obtain G(i) = (G(i−1) )󸀠 ≤ H 󸀠 . Thus, G(i) ≤ H 󸀠 ≤ ker(λ). Since G(i) ⊲ G, we get G(i) ≤ (ker(λ))G = ker(λ G ) = ker(χ), by Lemma 2.1. It follows from Theorem 10.1 that an M-group G is solvable and its derived length does not exceed k = |cd(G)| (in that case, G(k) ≤ ker(ρ G ) = {1}, where ρ G is the regular character of G). There is a conjecture that the last relation is true for all solvable groups which are need not M-groups. T. Berger [Berg] has proved this conjecture for groups of odd order. A normal subgroup of an M-group does not necessarily possess this property (S. D. Berman, E. Dade, R. van der Waall). Below some generalizations and variants of Taketa’s theorem will be presented. For further information, we refer to the numerous papers on M-groups mentioned in the bibliography.

11 On the number of solutions of x n = 1 in a group In this section, character theory will be applied to prove an important theorem of Frobenius about the number of solutions of the equation x n = 1 in G: if n | |G|, then that number is divisible by n (see also Theorem IV.11.1). In fact, Frobenius proved the following, more general assertion (the previously mentioned result follows from it when χ = 1G ): Theorem 11.1. Suppose that n ∈ ℕ divides the order h = |G| of a group G and χ ∈ Irr(G). Then the number a n (χ) = 1n ∑g n =1 χ(g) ∈ ℤ. The proof, due to Frobenius, is based on consideration of the n-th exterior power of the regular representation of G (see Theorem IV.11.1). There is a proof, due to R. Brauer, using his Induction Theorem. Here we present a comparatively simple proof of Theorem 11.1, due to L. Solomon, which uses some ideas of Wielandt relating to permutation representations.

V Induced characters and representations | 221

Let G n be the set of all n-element subsets of G. Consider the action of G by left multiplication on G n , and let θ n denote the character of the corresponding permutation representation of G (thus, if g ∈ G, then θ n (g) is the set of all n-element subsets M of G such that gM = M). We have θ n (1) = |G n | = (nh), where h = |G|. The remaining values of the character θ n are given by the following lemma. Lemma 11.2. If g ∈ G, d = o(g), the order of g ∈ G, then {0 θ n (g) = { h/d ( ) { n/d

if d ∤ n, if d | n.

Proof. Since θ n (g) is the number of g-invariant subsets X ∈ G n , it follows that, if θ n (g) ≠ 0, there exists X ∈ G n such that gX = X. This gives an action of the cyclic subgroup ⟨g⟩ by left multiplication on the set X. It follows from g i a = a (a ∈ G) that g i = 1 so that o(g) | i. Therefore, the above defined action is semi-regular, that is all of ⟨g⟩-orbits on X have the same length d = o(g), and hence d | |X| = n. Thus, if d ∤ n, then θ n (g) = 0. Now let d | n, and let the set X be the union of any dn distinct right cosets of the subgroup ⟨g⟩ ≤ G (g ∈ G, o(g) = d). It is obvious that X is a g-invariant n-element subset in G. Conversely, any g-invariant n-element subset X of G can be obtained in a similar manner in view of gX = X. Therefore, θ n (g) equals the number of ordered dn -selections from all the dh right cosets of the subgroup ⟨g⟩ ≤ G. Conseh/d quently, θ n (g) = (n/d ). In the following lemmas, h ∈ ℕ (here we do not assume that h is the order of G). Let T h = {d ∈ ℕ | d | h} be the set of all divisors of h, and let F = Fh be the set of all functions α : T h → ℝ. Lemma 11.3. Let α, β ∈ F, and let μ be the number-theoretic Möbius function. Then for any n ∈ T h , n n ∑ α( )β(d) = ∑ ∑ α( )μ(r) ∑ β(d). d rk d|n k|n r| n d|k

(1)

k

Proof. We define the convolution ∗ on the set F as follows: given α, β ∈ F, we put n (α ∗ β)(n) = ∑ α( )β(d) d d|n

(n ∈ T h ),

so that the left-hand side of formula (1) coincides with (α ∗ β)(n). Obviously, one has α ∗ β = β ∗ α and so the defined operation ∗ is commutative. Let ε ∈ F be the function n 󳨃→ δ1,n (n ∈ T h ). One has (ε ∗ α)(n) = ∑ δ1, dn α(d) = α(n), d|n

n (α ∗ ε)(n) = ∑ α( )δ1,d = α(n), d d|n

222 | Characters of Finite Groups 1 so that ε ∗ α = α = α ∗ ε. It follows that F is a commutative semigroup with respect to convolution with identity element ε (= commutative monoid). If u ∈ F is the function identically equal to 1, then n u ∗ μ = ∑ u( )μ(d) = ∑ μ(d) = δ1,n = ε, d d|n d|n n (α ∗ u)(n) = (u ∗ α)(n) = ∑ u( )α(d) = ∑ α(d), d d|n d|n and hence α ∗ β = (α ∗ β) ∗ ε = (α ∗ β) ∗ (u ∗ μ) = (α ∗ μ) ∗ (β ∗ u), which is equivalent to (1). Lemma 11.4. If n ∈ T h and a prime p | n, p a | n, then (

h −1 h−1 ) (mod p a ). ) ≡ ( pn n−1 − 1 p

Proof. For z ∈ ℤ, define [z] = (z − 1)(z − 2) . . . (z − p + 1). If p | z, then p ∤ [z]. If z, z1 , z2 ∈ ℤ and m ∈ ℕ, then z1 ≡ z2 (mod m) 󳨐⇒ [z1 ] ≡ [z2 ] (mod m)

(2)

[−z] = (−1)p−1 [z + p].

(3)

and

An obvious grouping of factors in the numerator and denominator of the formula (

h−1 (h − 1)(h − 2) . . . (h − n + 1) )= n−1 (n − 1)(n − 2) . . . 1

gives (

h−1 [h](h − p)[h − p] . . . (h − n + p)[h − n + p] )= n−1 [n](n − p)[n − p] . . . p[p] h

= ( pn p

− 1 [h][h − p] . . . [h − n + p] . ) [n][n − p] . . . [p] −1

Hence (

h h −1 − 1 [h][h − p] . . . [h − n + p] h−1 − 1}. ) − ( pn ) = ( pn ){ n−1 [n][n − p] . . . [p] p −1 p −1

Since h ≡ n ≡ 0 (mod p a ), it follows by (3) and (4) that, for any z ∈ ℤ, [h − zp] ≡ [−zp] ≡ (−1)p−1 [zp + p] ≡ [n − zp] (mod p a ).

(4)

V Induced characters and representations | 223

Putting z = 0, 1, . . . , pn − 1 in the first and last terms of the above displayed formula and multiplying, we obtain [h][h − p] . . . [h − n + p] ≡ [n][n − p] . . . [p] (mod p a ). Since GCD([n][n − p] . . . [p], p) = 1 (see the second sentence of the proof), the righthand side of (4) is divisible by p a , and we obtain the desired congruence. Lemma 11.5. If n | h, then h 1 ∑ ( nr )μ(r) ∈ ℤ. h r|n r

Proof. Since

h

( nr ) = r

(5)

h hr − 1 ( ), n nr − 1

formula (5) is equivalent to h

∑ ( nr r

r|n

−1 )μ(r) ≡ 0 (mod n). −1

(6)

Let Ṫ n be the set of all square-free divisors of n. Since μ(r) = 0 if r ∈ T n − Ṫ n , we can rewrite (6) as h

∑ ( nr r

r∈Ṫ n

−1 )μ(r) ≡ 0 (mod n). −1

(7)

Let p | n,

Ṫ n󸀠 = {r ∈ Ṫ n | p ∤ r},

Ṫ n󸀠󸀠 = {r ∈ Ṫ n | p | r}.

Then Ṫ n󸀠󸀠 = p Ṫ n󸀠 , and hence h

∑ ( nr r∈Ṫ n

r

h h −1 −1 −1 )μ(r) = ∑ ( nr )μ(r) + ∑ ( nr )μ(r) −1 ̇󸀠 ̇ 󸀠󸀠 r − 1 r −1 r∈T n

r∈T n

h −1 −1 )μ(r) + ∑ ( pr )μ(pr) n −1 pr − 1 r∈Ṫ 󸀠

=

h ∑ ( nr r r∈Ṫ n󸀠

=

h ∑ {( nr r r∈Ṫ n󸀠

n

−1 )− −1

h ( pr n pr

−1

)}μ(r)

−1

since μ(pr) = −μ(r) (note that pr is square free). By Lemma 11.4, for r ∈ Ṫ n󸀠 , one has h

( nr r

and so

h −1 −1 ) ≡ ( pr ) (mod p a ), n −1 − 1 pr h

∑ ( nr r∈Ṫ n

r

−1 )μ(r) ≡ 0 (mod p a ). −1

As p | n is arbitrary, this implies (7) and (5).

224 | Characters of Finite Groups 1 Corollary 11.6. If k | n | h, then h k ∑ ( kr n )μ(r) ∈ ℤ. n r| n kr k

Proof. Replace h and n in (5) by

h k

and nk , respectively.

Proof of Theorem 11.1. Let θ n = ∑χ∈Irr(G) c n (χ)χ be the decomposition of the character θ n into irreducible constituents. Then c n (χ) = ⟨θ n , χ⟩ ∈ ℕ ∪ {0} for all χ ∈ Irr(G). Since θ n = θ n , it follows that c n (χ) = ⟨θ n , χ⟩, i.e., 1 (8) c n (χ) = ∑ θ n (g)χ(g). h g∈G Writing (8) as c n (χ) =

1 h

∑d|h ∑g∈G,o(g)=d θ n (g)χ(g), we obtain, by Lemma 11.2, c n (χ) =

h 1 ∑ ( dn ) ∑ χ(g). h d|n d g∈G

(9)

o(g)=d

For any j ∈ ℕ, put jn α j (n) = ( ) n

(n ∈ T h = {d ∈ ℕ | d divides h}).

Define a function β ∈ F by β(n) = ∑ χ(g) g∈G o(g)=n

(n ∈ T h ).

Applying Lemma 11.3 to the functions α = α j and β, one obtains jn

jn

∑ ( dn ) ∑ χ(g) = ∑ ∑ ( rk n )μ(r) ∑ ∑ χ(g). g∈G o(g)=d

d

d|n

k|n r| nk

d|k g∈G o(g)=d

rk

Observing that ∑ ∑ χ(g) = ∑ χ(g) = ka k (χ), d|k g∈G o(g)=d

we find that

g∈G g k =1

jn

h

∑ ( dn ) ∑ χ(g) = ∑ ∑ ( rk n )μ(r) ⋅ ka k (χ). d|n

Putting j =

h n

d

g∈G o(g)=d

k|n r| dn

rk

and using (9), we obtain

c n (χ) = ∑ ∑ k|n r| nk

h k rk h h ( n )μ(r)a k (χ) = a n (χ) + ∑ ∑ ( rk n )μ(r)a k (χ). n rk n rk k|n r| n k n p . It follows that P1 is cyclic. Since |P| ≥ pn p > p, we conclude that P is either cyclic or p = 2, |P1 | = 4 and P is a generalized quaternion group. In the second case, clearly, n2 = 2. Note that all indices of a principal series of G are elementary abelian. It follows that all p-indices of a principal series of G are equal to p in the first case and 2 or 4 in the second case. Suppose that P ∈ Sylp (G) is cyclic. By the last sentence of the previous paragraph and Lemmas 11.8 and 11.10, G has a maximal subgroup H of index p. Then n | |H|, and the proof is complete in this case. Now suppose that p = 2 and a Sylow 2-subgroup P of G is generalized quaternion; then n2 = 2. In that case, by Lemma 11.8 and the first paragraph of the proof and Lemma 11.10, G has a maximal subgroup H of index dividing 4. Since |P| ≥ 8, the order of H is even so n2 = 2 | |H|. The proof is complete. Now we use Lemma 11.8 to prove the following classical theorem. Theorem 11.11 (P. Hall). If m is a Hall divisor of the order of a solvable group G, then all largest π(m)-subgroups of G have order m and conjugate in G. Proof. We use induction on |G| and m. Let R be a minimal normal subgroup of G and let |R| = p α . (i) First we prove that G contains a subgroup of order m. Let q be a prime divisor of |G| m ; then there is H < G such that |G : H| is a power of q (Lemma 11.8). In that case, m | |H| and GCD(m, |H| m ) = 1 so, by induction, H contains a subgroup of order m. (ii) We claim that all subgroups of order m are conjugate in G. Let F, H < G be of order m. Then the π(m)-Hall subgroups FR/R and HR/R of G/R are conjugate, by induction, so (FR)x = HR for some x ∈ G whence F x ≤ HR. If p | m, then R ≤ F ∩ H so FR = F, HR = H and F x = H. Now let p ∤ m. If HR < G, there is y ∈ HR such that (F x )y = H, by induction. Now assume that HR = G for any choice of R; then also FR = G and R ∈ Sylp (G) is the

228 | Characters of Finite Groups 1 unique minimal normal subgroup of G so H and F are maximal in G. Let Q/R be a minimal normal subgroup of G/R; then Q/R is a q-subgroup for some prime q ≠ p. In that case, H ∩ Q and F ∩ Q are Sylow q-subgroups of Q and they are not normal in G. Therefore, NH (H ∩ Q) = H and NF (F ∩ Q) = F. By Sylow’s theorem, F ∩ Q = (H ∩ Q)y for some y ∈ Q. Then F = NG (F ∩ Q) = NG ((H ∩ Q)y ) = NG (H ∩ Q)y = H y . (iii) It remains to show that if K < G is the greatest π(m)-subgroup, then |K| = m. By induction, KR/R ≤ H/R, where H/R is a π(m)-Hall subgroup of G/R (see (ii)); then m | |H|. If p | m, then |H| = m, R ≤ K so K = H, by maximality of K. Now let p ∤ m. We have KR ≤ H. If H < G, then, by induction, K is contained in a subgroup of order m in H so |K| = m, by maximality of K. Next suppose that H = G. Then G = FR, where F < G is of order m (F exists, by (i)). Set K1 = KR ∩ F; then |K1 | = |K|, by the product formula. By (ii), K = K1z for some z ∈ KR so K ≤ F z , and we are done since |F z | = |F| = m. Suppose that L n is such that L n is a subgroup of G of order n (here we do not assume that G is solvable). Let p ∈ π(|G : L n |) ∩ π(n); then P ∈ Sylp (G) is not contained in L n . Then L n ∩ P is the unique subgroup of its order in P since all subgroups of P of exponent ≤ n p are contained in L n and P ∩ L n ∈ Sylp (L n ). In that case, either P is cyclic or p = 2 and P is a generalized quaternion group ([Ber31, Theorem 1.2] or Exercise VI.4.2). In the first case, L n is p-nilpotent, by Tate’s theorem (see Theorem VII.10.1). In the second case, 4 ∤ |L n |, so L n is again 2-nilpotent. Exercise. Find the number of involutions in Sn , An and in Sylow 2-subgroups of the above groups.

12 Further Mackey’s theorems The results of this section are due to Mackey. Theorem 12.1. Suppose that G1 and G2 are groups, and G = G1 ×G2 , H i ≤ G i , i ∈ {1, 2}, H = H1 × H2 . If ψ i ∈ Char(H i ), i ∈ {1, 2}, then ψ = ψ1 × ψ2 ∈ Char(H),

G

G

ψ G = ψ1 1 × ψ2 2 .

Proof. As usual, if H < G and ψ ∈ Char(H), then ψ̇ : G → ℂ coincides with ψ on H and vanishes on G − H. Then ψ̇ = ψ̇ 1 × ψ̇ 2 (here ψ̇ i coincides with ψ i on H i and vanishes

V Induced characters and representations | 229

on H − H i , i = 1, 2). Therefore, by the formula for induced characters, |H|ψ G = ∑ ψ̇ g = g∈G

=

∑

(g1 ,g2 )∈G1 ×G2

∑

g1 ∈G1 ,g2 ∈G2

= (|H1 | ∑

(ψ̇ 1 × ψ̇ 2 )(g1 ,g2 )

g g g g (ψ̇ 11 × ψ̇ 22 ) = ( ∑ ψ̇ 11 ) × ( ∑ ψ̇ 22 )

g1 ∈G1

g2 ∈G2

g1 ∈G1

g ψ̇ 11 )

× (|H2 | ∑

g2 ∈G2

g ψ̇ 22 )

G

G

= |H|(ψ1 1 × ψ2 2 ),

completing the proof. Let H, K ≤ G. We say that pairs (x, y), (x1 , y1 ) ∈ G × G are equivalent if and only if Hxy−1 K = Hx1 y−1 1 K. It is easy to check that this is indeed an equivalence relation. Let I denote a complete system of representatives of classes of equivalent pairs (x, y) ∈ G × G. Given x, y ∈ G, put D x,y = H x ∩ K y , and given characters φ, ψ of H, K, respectively, denote φ x,y = (φ x )D x,y , ψ x,y = (ψ y )D x,y . Recall that φ x (h) = φ(xhx−1 ), ψ y (k) = ψ(yky−1 ) Then

φ x (h x )

= φ(h) so that

φx

∈

Char(H x ).

(h ∈ H, k ∈ K).

Similarly, ψ y ∈ Char(K y ).

Theorem 12.2. In the notation of the previous paragraph, φ G ψ G = ∑ (φ x,y ⋅ ψ x,y )G . (x,y)∈I

Proof. Put Γ = G × G, H0 = H × K, and let ∆ = {g = (g, g) | g ∈ G} be the diagonal of Γ. If χ1 , χ2 ∈ Char(G), then χ = χ1 × χ2 ∈ Char(Γ) (see §IV.8), and moreover χ(g) = χ1 (g)χ2 (g) = (χ1 χ2 )(g)

(g ∈ G).

In particular, if χ1 = φ G , χ2 = ψ G , then (φ G × ψ G )(g) = (φ G ψ G )(g).

(1)

But by Theorem 12.1, φ G × ψ G = (φ × ψ)G×G . Therefore we can rewrite (1) as (φ G ψ G )(g) = (φ × ψ)Γ (g).

(2)

Denoting φ × ψ = θ, we obtain (φ G ψ G )(g) = (θ Γ )∆ (g)

(g ∈ G).

(3)

In order to apply the first Mackey’s theorem (Theorem 3.1) to (θ Γ )∆ , let us decompose the group Γ into double cosets H0 (x, y)∆. We claim that H0 (x, y)∆ = H0 (x1 , y1 )∆ if and only if the pairs (x, y), (x1 , y1 ) are equivalent. Indeed, H0 (x, y)∆ = H0 (x1 , y1 )∆ ⇐⇒ (x1 , y1 ) = (h, k)(x, y)(g, g) (h ∈ H, k ∈ K, g ∈ G) −1 −1 ⇐⇒ x1 = hxg, y1 = kyg ⇐⇒ x1 y−1 1 = hxy k

⇐⇒ the pairs (x, y) and (x1 , y1 ) are equivalent.

230 | Characters of Finite Groups 1 Therefore the necessary decomposition is Γ = ⋃(x,y)∈I H0 (x, y)∆. By Theorem 3.1, ∆ (θ Γ )∆ = ∑ θ(x,y) ,

(4)

(x,y)∈I

where θ(x,y) = (θ(x,y) )D(x,y) ,

(x,y)

D(x,y) = H0

∩ ∆.

Note that (x,y)

H0

= (H × K)(x,y) = H x × K y .

Therefore, (x,y)

D(x,y) = H0

∩ ∆ = {(h x , k y ) | h ∈ H, k ∈ K, h x = k y }

= {(h x , h x ) | h x ∈ K y } = {(g, g) | g ∈ H x ∩ K y } = {g | g ∈ D x,y }, so that D(x,y) is the diagonal of the direct product D x,y × D x,y = D x,y . Since θ(x,y) = (φ × ψ)(x,y) = φ x × ψ y (recall that θ = φ × ψ), it follows that θ(x,y) = (θ(x,y) )D(x,y) = (φ x × ψ y )D x,y = φ x,y × ψ x,y . Therefore, for any g ∈ G, ∆ θ(x,y) (g) =

∑

θ(x,y) (g t ) =

t

t∈∆, g ∈D(x,y)

∑

t∈G, g t ∈D x,y

φ x,y (g t )ψ x,y (g t ) = (φ x,y ψ x,y )G (g).

Thus, ∆ θ(x,y) (g) = (φ x,y ψ x,y )G (g)

(5)

It follows from (4) and (5) that (θ Γ )∆ (g) = ∑ (φ x,y ψ x,y )G (g). (x,y)∈I

13 Generalizations of M-groups In this section we will generalize the solvability criterion of Taketa’s Theorem 10.1. As far as we know, Theorem 13.1 was first proved by L. Dornhoff (see also [Isa16]). Let S be some class of simple groups (not necessarily all of them are nonabelian) such that, whenever G ∈ S and G1 ≅ G, then G1 ∈ S. We assume that {1} ∈ S. A group all of whose composition factors belong to S is called an S-group. Obviously, the property of being an S-group carries over to subnormal subgroups, epimorphic images and extensions.

V Induced characters and representations | 231

Theorem 13.1. Suppose that for every χ ∈ Irr(G) there exist H ≤ G, λ ∈ Irr(H) such that χ = λ G and H/ker(λ) is an S-group. Then G is also an S-group. Proof. Suppose that the theorem has been proved for groups of orders < |G|. All epimorphic images of G satisfy the assumption (indeed, if M ⊲ G and χ ∈ Irr(G/M), then M ≤ ker(λ) for the character λ defined in the statement of the theorem; see Lemma 2.1). Let A, B be distinct minimal normal subgroups of G. Then, by induction, G/A is an S-group. Since BA/A ≅ B is a normal subgroup of the S-group G/A, it follows that B is an S-group. But then G is an S-group, as an extension of the S-group B by the S-group G/B. From now on, therefore, one may assume that G contains only one minimal normal subgroup R (i.e., G is a monolith). Then G has a faithful irreducible character, namely, any irreducible character of G whose kernel does not contain R (note that ⋂χ∈Irr(G) ker(χ) = ker(ρ G ) = {1}). Let χ ∈ Irr(G) be faithful of minimal degree. By assumption, we have χ = λ G , where λ ∈ Irr(H), H ≤ G, and H/ker(λ) is an S-group. Let ϑ = (1H )G = ∑ τ i ,

τ i ∈ Irr(G).

i

We may assume that H < G (otherwise, it is nothing to prove). By reciprocity, ⟨ϑ, 1G ⟩ = ⟨1H , 1H ⟩ = 1, and so ϑ is reducible since ϑ(1) = |G : H| > 1. Therefore, τ i (1) < ϑ(1) = |G : H| ≤ χ(1) for all i. Then, by assumption, R ≤ ker(τ i ) for all i, since G is a monolith. Consequently, R ≤ ker(ϑ) = H G = ⋂ x−1 Hx. x∈G

Since {1} = ker(χ) = ker(λ)G , it follows that R ≰ ker(λ). Let R = R1 × ⋅ ⋅ ⋅ × R n , where R1 , . . . , R n are isomorphic simple groups. Since R ∩ ker(λ) is normal in R (as the intersection of two normal subgroups of H), there exists i ∈ {1, . . . , n} such that R i ∩ ker(λ) = {1}. Since H/ker(λ) is an S-group and R i ker(λ)/ker(λ) ≅ R i is subnormal in an S-group H/ker(λ), it follows that R i is an S-group. Since R1 , . . . , R n are isomorphic, their direct product R is also an S-group. By induction, G/R is an S-group so is G. Problem. Investigate the structure of a group G all of whose nonlinear irreducible characters χ equal λ G , where λ ∈ Irr(H) for some H < G. Clearly, Theorem 13.1 implies that M-groups are solvable (Taketa’s Theorem 10.1). Moreover, G is solvable if, for every χ ∈ Irr(G), there exist H ≤ G and λ ∈ Irr(H) such that χ = λ G and H/ker(λ) is solvable. Let π be a fixed set of primes. If, for every χ ∈ Irr(G), there exist H ≤ G and λ ∈ Irr(H) such that χ = λ G and H/ker(λ) is a π-group, then G is also a π-group.

232 | Characters of Finite Groups 1

14 On intersections of kernels of some characters In this section we will consider intersections of the kernels of some irreducible characters of a group G. As we know, if G is nonabelian, then (see Exercise IV.4.1) ⋂

ker(χ) = {1}.

χ∈Irr1 (G)

We introduce the following notation: b(G) = max{χ(1) | χ ∈ Irr(G)}, B(G) = {χ ∈ Irr(G) | χ(1) = b(G)}, Ψ(G) is the intersection of all maximal normal subgroups of G, n is a maximal element of the set cd(G), partially ordered by divisibility (there may be several such elements), ∙ Irr(n) (G) = {χ ∈ Irr(G) | χ(1) = n}, ∙ m is a minimal element of the set {|ker(χ)| | χ ∈ Irr(G)}, partially ordered by divisibility, ∙ X m (G) = {χ ∈ Irr(G) | |ker(χ)| = m}. Part (b) of Theorem 14.1 was proved by A. Mann, parts (a) and (c) are due to the first author.

∙ ∙ ∙ ∙

Theorem 14.1. The following statements hold: (a) The subgroup D1 = ⋂χ∈B(G) ker(χ) ≤ Φ(G). (b) The subgroup ⋂χ∈Irr(n) (G) ker(χ) ≤ Ψ(G). (c) The subgroup ⋂χ∈X m (G) ker(χ) ≤ Φ(G). The following lemma is well known. Lemma 14.2. The following statements hold: (a) Let A ≤ G and let N ⊲ G. If N ≤ Φ(A), then N ≤ Φ(G). (b) Let N ⊲ G. If A ≤ G is minimal subjecting AN = G, then N ∩ A ≤ Φ(A). Proof. (a) Assume that N ≰ Φ(G). Then HN = G for some maximal H < G. By the modular law, we have A = N(A ∩ H) and so A = A ∩ H since N ≤ Φ(A). In that case, we obtain N ≤ Φ(A) ≤ A ≤ H and H = NH = G, a contradiction. (b) Assume that N ∩ A ≰ Φ(A). Then (N ∩ A)B = A for some maximal B < A. Then G = NA = N(N ∩ A)B = NB, B < A, contrary to the choice of A. Proof of Theorem 14.1. (a) Let χ ∈ B(G) and suppose that K = ker(χ) ≰ Φ(G). Then there exists A < G such that KA = G and A is as small as possible. In that case, we have K ∩ A ≤ Φ(A) (Lemma 14.2 (b)), χ A = ϑ ∈ Irr(A) (Lemma XIV.1.1 (f)) and ker(ϑ G ) = ker(ϑ)G ≤ ker(ϑ) = ker(χ A ) = K ∩ A ≤ Φ(A).

(1)

It follows from (1) and Lemma 14.2 (a) that ker(ϑ G ) ≤ Φ(G) since ker(ϑ G ) ⊲ G. Note that ϑ(1) = χ(1) and χ ∈ B(G). Therefore, if ψ ∈ Irr(ϑ G ), then ψ ∈ B(G), by reciprocity. Consequently, D1 ≤ ker(ϑ G ) ≤ Φ(G), and (a) is proved.

V Induced characters and representations | 233

(b) Assume that χ ∈ Irr(n) (G) is such that ker(χ) ≰ Ψ(G). Then there exists a maximal normal subgroup H of G such that K = ker(χ) ≰ H. In that case, KH = G; therefore, χ H = ϑ ∈ Irr(H) (Lemma XIV.1.1 (f)). By assumption and by Clifford’s theorem (see Chapter VII), Irr(ϑ G ) ⊆ Irr(n) (G). Therefore, D0 =

ker(τ) ≤ ker(ϑ G ) ≤ ker(ϑ) = ker(χ H ) = H ∩ K ≤ H.

⋂ τ∈Irr(n) (G)

Assume that D0 ≰ Ψ(G). Then D0 ≰ H for some maximal normal subgroup H of G. But then ker(χ) ≰ H for some χ ∈ Irr(n) (G) and, by what we have proved above, D0 ≤ H, a contradiction. This proves part (b). (c) Assume that there is χ ∈ X m (G) such that K = ker(χ) ≰ Φ(G). Then there is A < G such that AK = G and A is as small as possible; then we get A ∩ K ≤ Φ(A), by Lemma 14.2 (b). It follows that ϑ = χ A ∈ Irr(A). Take ψ ∈ Irr(ϑ G ). Then ϑ ∈ Irr(ψ A ), by reciprocity, so A ∩ ker(ψ) = ker(ψ A ) ≤ ker(ϑ) = ker(χ A ) = A ∩ K. Next, |G : A ker(ψ)| =

|K| |A ∩ ker(ψ)| |A| ⋅ |K| ⋅ |A ∩ ker(ψ)| = ⋅ |A ∩ K||A||ker(ψ)| |ker(ψ)| |A ∩ K|

is an integer. By hypothesis, |K| ≤1 |ker(ψ)| and, by the displayed line in the previous paragraph, |A ∩ ker(ψ)| ≤ 1, |A ∩ K| and we conclude that m = |K| = |ker(ψ)|, ψ ∈ X m (G),

|A ∩ K| = |A ∩ ker(ψ)|,

A ∩ ker(ψ) = A ∩ K,

G = A ker(ψ).

But then N=

⋂

ker(τ) ≤ ker(ϑ G ) ≤ ker(ϑ) = ker(χ A ) = A ∩ K ≤ Φ(A).

τ∈X m (G)

Since N ⊲ G, it follows that N ≤ Φ(G) (Lemma 14.2 (a)), completing the proof. Remark (Mann). Let χ ∈ B(G) and let A < G be such that AK = G, where K = ker(χ). Then χ A = ϑ ∈ Irr(A). If ψ ∈ Irr(ϑ G ), then, as above, A ∩ ker(ψ) = A ∩ K, and therefore we get ker(ϑ G ) = ⋂ ker(ψ) = A ∩ K 󳨐⇒ A ∩ K ⊲ G. ψ∈Irr(ϑ G )

234 | Characters of Finite Groups 1

15 Varia Most results of this section are taken from [Kar5]. 1°. In the following theorem a generalization of the Second Orthogonality Relation is presented. In what follows x ∼ y denotes that x, y ∈ G are conjugate. Theorem 15.1 (Osima). Let H ≤ G and suppose that Irr(H) = {χ1 , . . . , χ r }. Then for all g ∈ G and h ∈ H, r {|CG (g)|, g ∼ h. ∑ χ Gk (g)χ k (h−1 ) = { 0, g ≁ h. k=1 {

Setting H = G, we get the Second Orthogonality Relation. Also the proof uses that relation. Theorem 15.2 (Nakayama (1938) and Osima (1952)). Let H ≤ G and let Irr(H) be given by {ϕ1 , . . . , ϕ r }. Then the number of linearly independent characters of G among the induced characters ϕ1G , . . . , ϕ Gr is equal to the number t of G-conjugate classes having nonempty intersections with H. If H < G, then there is a G-class whose intersection with H is empty (otherwise, ⋃x∈G H x = G, which is impossible), and so t < k(G), the class number of G. If H = G, we get k(G) = r, the known consequence of the Second Orthogonality Relation. 2°. Let Irr(G) = {χ1 , . . . , χ r }. Recall that ρ G = ∑ri=1 χ i (1)χ i is the regular character of G. Exercise 15.1. If x = ∑g∈G x g g ∈ ℂG, then x g = |G|−1 ∑ri=1 χ i (1)χ i (xg−1 ). Solution. As x = x g g + ∑t∈G−{g} x t t, multiplying this equality by g −1 , one obtains xg −1 = x g 1 +

∑ x t tg −1 . t∈G−{g}

Taking the values of the regular character of both sides, we get r

∑ χ i (1)χ i (xg−1 ) = x g |G| i=1

since tg−1 ∈ G# for t ∈ G − {g} so that ρ G ( ∑ x t tg−1 ) = 0. t∈G−{g}

We have

r

x g |G| = ρ G (x g 1) = ρ G (xg −1 ) = ∑ χ i (1)χ i (xg−1 ). 1=1

Here the characters χ i are extended on the group algebra ℂG by linearity. Exercise 15.2. Let ℂG = ℂGe1 ⊕ ⋅ ⋅ ⋅ ⊕ ℂGe r be the decomposition of the group algebra ℂG in the direct sum of its minimal two-sided ideals ({e1 , . . . , e r } is the idempotent basis of the algebra Z(ℂG)) and let χ i be the irreducible character of G afforded by an

V Induced characters and representations | 235

irreducible module belonging to ℂGe i , 1 ≤ i ≤ r (see §III.4). Then e i = |G|−1 χ i (1) ∑ χ i (g−1 )g. g∈G

Solution. Write e i = ∑g∈G e gi g, where i = 1, . . . , r and e gi ∈ ℂ. Then, by Exercise 15.1, e gi = |G|−1 ∑rk=1 χ k (1)χ k (e i g−1 ). Note that e i is the identity element of ℂGe i and χ k vanishes on ℂGe i provided i ≠ k so χ k (e i g −1 ) = δ ik χ k (g−1 ). It follows that r

e gi = |G|−1 ∑ δ ki χ k (1)χ k (g −1 ) = |G|−1 χ i (1)χ i (g −1 ). k=1

Exercise 15.3 (Generalized Orthogonality Relation). Let x ∈ G. In the above notation, ⋅ χ i (x). ∑g∈G χ i (gx)χ j (g−1 ) = δ ij χ|G| i (1) Solution. Let the idempotents e1 , . . . , e r be as in Exercise 15.2; then e i e j = δ ij e i , and we compare the coefficients of the group elements on both sides of that equality. The coefficient of x ∈ G on the right-hand side equals |G|−1 δ ij χ i (1)χ i (x−1 ) and the coefficient on the left-hand side equals |G|−2 χ i (1)χ j (1) ∑g∈G χ((xg−1 )−1 )χ j (g −1 ). Equating these expressions and substituting x for x−1 , we get the desired result. Substituting x = 1 in the equality in the statement of the exercise, we get ⟨χ i , χ j ⟩ = δ ij , the First Orthogonality Relation. Proposition 15.3 ([Isa11, Exercise 2.17]). If A < G is nonidentity abelian, χ ∈ Irr(G) and χ(1) = |G : A|, then A G > {1}. Proof. One may assume that G is nonabelian; then we have χ(1) > 1. It follows from χ(1) = |G : A| that CG (A) = A, i.e., A < G is maximal abelian. Using Exercise II.3.2, one obtains |G| = ∑ |χ(x)|2 + ∑ |χ(x)|2 ≥ |A|χ(1) + ∑ |χ(x)|2 x∈A

x∈G−A

x∈G−A

2

= |A||G : A| + ∑ |χ(x)| = |G| + ∑ |χ(x)|2 , x∈G−A

x∈G−A

and we conclude that ∑x∈G−A |χ(x)|2 = 0, i.e., χ vanishes on G − A. Thus, we have G − A ⊆ Tχ , where Tχ is the set of zeros of χ. Set D = A G = ⋂t∈G A t . Then, by the formula of de Morgan, G − D = G − ⋂ A t = ⋃ (G − A)t ⊆ ⋃ Ttχ = T χ t∈G

t∈G

i∈G

since the set Tχ is G-invariant. Thus, D ⊲ G and G − D ⊆ Tχ . If D = {1}, then G# ⊆ Tχ and so χ = ρ G is the regular character of G. Since |G| > 1, ρ is reducible, contrary to the hypothesis. Thus, D = A G > {1}.

VI Projective representations The theory of projective representations was created by Issai Schur (January 10, 1875, Mohilev, Belarus–January 10, 1941, Petakh Tikvah, Israel). Our exposition is based on Schur’s papers [Sch1, Sch3] and on papers of the third author [Zhm4, Zhm11, Zhm25, Zhm12]. Many group theorists also contributed in this theory essentially (see Karpilovsky’s monographs in the bibliography).

1 Basic notions Let G be a finite group, ℂ∗ the multiplicative group of the field ℂ of complex numbers, n ∈ ℕ. Definition 1.1. A mapping P : G → GL(n, ℂ) is called a projective π-representation of a group G of degree n if there exists a function π : G × G → ℂ∗ such that P(s)P(t) = π(s, t)P(st)

(s, t ∈ G).

(1)

The function π is called the factor set of P. We say that the representation P belongs to the factor set π, or that P is a π-representation of G. Everything presented below remains valid for an algebraically closed field of characteristic not dividing |G|. In this chapter we will consider only matrix projective representations, therefore we usually omit the adjective “matrix”. A factor set π is said to be trivial if π(s, t) = 1 for all s, t ∈ G; such a factor set is often denoted by 1 so that 1(x, y) = 1 for all x, y ∈ G. A projective representation with a trivial factor set is an ordinary linear representation (such representations were considered in the previous chapters). Let F[G, ℂ∗ ] be the set of all functions λ : G → ℂ∗ . If P is a π-representation of a group G and λ ∈ F[G, ℂ∗ ], then P󸀠 = λP (i.e., P󸀠 (g) = λ(g)P(g) for all g ∈ G) is also a projective π󸀠 -representation, where π󸀠 (s, t) =

λ(s)λ(t) π(s, t) λ(st)

(s, t ∈ G).

(2)

Indeed, P󸀠 (s)P󸀠 (t) = λ(s)λ(t)P(s)P(t) = λ(s)λ(t)π(s, t)P(st) = λ(s)λ(t)π(s, t)λ(st)−1 P󸀠 (st), and (2) follows since P󸀠 (s), P󸀠 (t), P󸀠 (st) ∈ GL(deg(P), ℂ) are non-degenerate. Definition 1.2. Two factor sets π, π󸀠 : G × G → ℂ∗ are said to be associated (we write π ∼ π󸀠 ) if relation (2) holds for some λ ∈ F[G, ℂ∗ ].¹ 1 Thus, the representations P󸀠 , λP have the associated factor sets π, π 󸀠 , respectively. DOI 10.1515/9783110224078-006

238 | Characters of Finite Groups 1 This is an equivalence relation on the set of factor sets. Let [π] denote the class of factor sets associated with π. It follows from (2) that π ∼ 1 (or, what is the same, [π] = [1]) if and only if there exists λ ∈ F[G, ℂ∗ ] such that, for all s, t ∈ G, one has π(s, t) =

λ(s)λ(t) . λ(st)

(2’)

Definition 1.3. Two projective representations P, P󸀠 of a group G are said to be linLIN early equivalent (we write (P ∼ P󸀠 ) if they have the same degree, say n, and P󸀠 (g) = S−1 P(g)S

for some S ∈ GL(n, ℂ) and all g ∈ G. PR

These representations P, P󸀠 are said to be projectively equivalent (we write P ∼ P󸀠 ) LIN provided there exists λ ∈ F[G, ℂ∗ ] such that P󸀠 ∼ λP, i.e., P󸀠 (g) = S−1 λ(g)P(g)S

for some S ∈ GL(n, ℂ) and all g ∈ G.

Clearly, the so-defined relations are equivalence relations. Linearly equivalent linear representations are equivalent in the usual sense (see Chapter I). LIN Let π P denote the factor set of a projective representation P of a group G. If P ∼ P󸀠 , 󸀠 then π P = π P (see (2’)). PR

󸀠

Exercise 1.1. If P ∼ P󸀠 , then [π P ] = [π P ]. Hint. See the previous paragraph, Definition 1.1 and the paragraph following it. Definition 1.4. Given a projective representation P of a group G, write KER(P) = {g ∈ G | P(g) is a scalar matrix}. The set KER(P) is said to be the projective kernel of P. It follows from (1) that if s, t ∈ KER(P), then P(st) is a scalar matrix hence st ∈ KER(P), and therefore KER(P) ≤ G. If L is a linear representation of G, then its projective kernel KER(L) coincides with Z(L), the quasikernel of L (by definition, Z(L) coincides with Z(χ) = {s ∈ G | |χ(s)| = χ(1)}, where χ is the character of L; thus, all members of the set Z(L) are scalar matrices). Exercise 1.2. Let P be a π-representation of a group G. Deduce from the equalities π(s, t)−1 P(s)P(t) = P(st) = P(ts t ) = π(t, s t )−1 P(t)P(s t ) that P(t)−1 P(s)P(t) = ω π (s, t)P(s t ),

where ω π (s, t) =

π(s, t) π(t, s t )

(s, t ∈ G).

(3)

Therefore, if s ∈ KER(P), then P(s) is a scalar matrix so is P(s t ), i.e., s t ∈ KER(P), and we conclude that KER(P) ⊴ G.

VI Projective representations | 239

Exercise 1.3. If P and P󸀠 are projectively equivalent projective representations of PR a group G, i.e., P ∼ P󸀠 , then KER(P) = KER(P󸀠 ). Solution. We have P󸀠 (g) = S−1 λ(g)P(g)S for some (non-degenerate) square matrix S and λ ∈ F[G, ℂ∗ ]. If g ∈ KER(P), then P󸀠 (g) = λ(g)P(g) is a scalar matrix since P(g) is so that g ∈ KER(P󸀠 ) and hence KER(P) ≤ KER(P󸀠 ). As P(g) = λ(g)−1 P󸀠 (g), the reverse inclusion holds. Let LIN(G) denote the set of all linear representations of a group G and let PR(G) be the set of all projective representations of G. Obviously, LIN(G) ⊂ PR(G). Projective representations will usually be denoted by P and linear representations by L. If L ∈ LIN(G), then π L = 1 is the trivial factor set. The identity element of G is denoted by e. The relations P(g) = P(eg) = π(e, g)−1 P(e)P(g) (g ∈ G) 󳨐⇒ P(e) = π(e, g)In (n = deg(P)), where In is the n × n identity matrix. Similarly, P(e) = π(g, e)In (g ∈ G). Therefore π(e, g) = π(g, e) does not depend on g ∈ G, so that π(e, g) = π(e, e) = π(g, e)

(g ∈ G).

(4)

A factor set π is said to be normalized if π(e, e) = 1. If π = π P , then π is normalized if and only if P(e) = In (n = deg(P)). Indeed, P(e) = P(ee) = π(e, e)−1 P(e)P(e) = P(e)P(e) 󳨐⇒ P(e) = In . The converse assertion is also true. In that case, π(e, g) = π(g, e) = 1 (g ∈ G). If π = π P and the function λ ∈ F[G, ℂ∗ ] is such that λ(e) = π(e, e)−1 , then π λP is normalized and π ∼ π λP , i.e., every factor set is associated with a normalized factor set. Indeed, if π󸀠 is a factor set for the representation λP, then, by (2), π󸀠 (e, e) =

λ(e)λ(e) ⋅ π(e, e) = 1, λ(e)

and our claim follows. A projective representation P ∈ PR(G) is said to be reducible if the system of matrices {P(g)}g∈G is reducible. Similarly, one can define irreducibility, decomposability and complete reducibility of projective representations. Note that the sum M + N of two square matrices equals diag(M, N). It is easy to 󸀠 check that if P, P󸀠 ∈ PR(G), then P + P󸀠 ∈ PR(G) if and only if π P = π P . Let PGL(n, ℂ) = GL(n, ℂ)/ℂ∗ In be the projective general linear group.² Further, let p : G → PGL(n, ℂ) be a homomorphism and let ν : GL(n, ℂ) → PGL(n, ℂ) be the natural homomorphism. Choosing a matrix P(g) ∈ ν−1 (p(g)) for every g ∈ G, one obtains ν(P(s)P(t)) = ν(P(s))ν(P(t)) = p(s)p(t) = p(st) = ν(P(st)) 2 It is known that Z(GL(n, ℂ)) = ℂ∗ I n .

240 | Characters of Finite Groups 1 for all s, t ∈ G, so that P(s)P(t)P(st)−1 = π(s, t)In ∈ ℂ∗ In

(n = deg(P)).

Therefore, P(s)P(t) = π(s, t)P(st) (s, t ∈ G) and P ∈ PR(G) is a transversal of the homomorphism p. Conversely, if P ∈ PR(G), then, putting p(s) = ν(P(s)) (s ∈ G), one obtains a homomorphism p : G → PGL(n, ℂ), and the representation P is a transversal of p. All transversals of the same homomorphism p : G → GL(n, ℂ) are projectively equivalent.

2 Twisted group algebra Let P ∈ PR(G). It follows from (P(x)P(y))P(z) = P(x)(P(y)P(z)) for all x, y, z ∈ G that π(x, y)π(xy, z) = π(x, yz)π(y, z).

(1)

In homological algebra a function π : G × G → ℂ∗ satisfying (1) is called a 2-cocycle of G over ℂ∗ (considered as a trivial G-module). The set of all such 2-cocycles is an abelian group with respect to pointwise multiplication, and this group is denoted by Z2 (G, ℂ∗ ). Thus, the factor set of a projective representation of G is an element of the group Z2 (G, ℂ∗ ). To prove the converse we need the construction of a twisted group algebra. Exercise 2.1. Using relation (1), prove that π(x, x−1 ) = π(x−1 , x)(x ∈ G). Hint. Put, in (1), y = x−1 and z = x and use equalities (1.4). Thus, if π ∈ Z2 (G, ℂ∗ ), then it follows from (1) that π(g, e) = π(e, g) = π(e, e), π(g, g −1 ) = π(g −1 , g) for all g ∈ G.

(∗)

The support of the twisted group algebra ℂπ G is a ℂ-space of dimension |G| whose basis elements u g are indexed by the elements g ∈ G and satisfy the relations u s u t = π(s, t)u st

(2)

(s, t ∈ G).

The basis {u g }g∈G is called a standard basis. Relations (2) and (1) imply that the algebra ℂπ G is associative and its identity element is π(e, e)−1 u e . Indeed, by (∗), (π(e, e)−1 u e )u g = π(e, e)−1 π(e, g)u g = u g

for all g ∈ G.

The elements u g are invertible: −1 −1 −1 u−1 g = π(g, g ) π(e, e) u g −1

(g ∈ G).

VI Projective representations | 241

Indeed, using (2) and (∗), we see that (π(g, g−1 )−1 π(e, e)−1 u g−1 )u g = π(g, g−1 )−1 π(e, e)−1 π(g −1 , g)u e = π(e, e)−1 u e is the identity element of the algebra ℂπ G. Of course, if the factor ser π is normalized, then the identity element of ℂπ G is equal to u e . Let L be a linear representation of the twisted group algebra ℂπ G. Put P(g) = L(u g ) (g ∈ G). Then it follows from (2) that, for s, t ∈ G, one has P(s)P(t) = L(u s )L(u t ) = L(u s u t ) = π(s, t)L(st) = π(s, t)P(st), so that P ∈ PR(G) and π P = π, i.e., the 2-cocycle π is a factor set of the projective representation P of G. Therefore, the set of all factor sets of projective representations of G is precisely Z 2 (G, ℂ∗ ). It is easy to see that every (projective) π-representation P of G is afforded by some linear representation L of the algebra ℂπ G. Thus, we arrive at a one-to-one correspondence between the set of linear representations of the twisted group algebra ℂπ G and the set of π-representations of the group G such that if L i corresponds to P i (i = 1, 2), then LIN

L1 ∼ L2 ⇐⇒ P1 ∼ P2 ). A projective representation P of a group G is reducible, irreducible, decomposable and completely reducible if and only if the corresponding linear representation L of its twisted group algebra ℂπ G is. Thus, the description of π-representations of a group G is reduced to the description of linear representations of the twisted group algebra ℂπ G. Schur based his approach to the description of the set PR(G) upon the construction of central extensions. He went on to introduce the notions of multiplier and representation (or covering) group. In §§VI.3–VI.4 we will present the basic principles of Schur theory, at the same time combining his approach with the approach based on twisted group algebras.

3 The Schur multiplier We have seen that the set Z2 (G, ℂ∗ ) of factor sets of projective representations of a group G is an abelian group with respect to pointwise multiplication. Recall that a factor set π is said to be trivial if π(x, y) =

λ(x)λ(y) λ(xy)

for some λ ∈ F[G, ℂ∗ ].

The class [1] of the trivial factor set is closed with respect to multiplication so is a subgroup B2 (G, ℂ∗ ) of the (abelian) group Z2 (G, ℂ∗ ). The group H2 (G, ℂ∗ ) = Z2 (G, ℂ∗ )/B2 (G, ℂ∗ )

242 | Characters of Finite Groups 1 of classes of factor sets of G is called the second cohomology group of G over ℂ∗ (considered as a trivial G-module). It is also called the Schur multiplier of G, denoted by M(G). Thus, M(G) = H2 (G, ℂ∗ ). Exercise 3.1. One has M(G) = {1} if and only if every projective representation of G is projectively equivalent to a linear representation. Prove that if G is cyclic, then M(G) = {1}. Hint. See Definition 1.3 and Exercise 1.1. Lemma 3.1. One has exp(M(G)) | |G|. Proof. Let π ∈ Z2 (G, ℂ∗ ). Multiplying equalities (2.1) over all z ∈ G and denoting λ(g) = ∏z∈G π(g, z), one obtains π(x, y)|G| λ(xy) = λ(x)λ(y) 󳨐⇒ π|G| ∼ 1 󳨐⇒ [π]|G| = [1] (see the paragraph between Definitions 1.2 and 1.3). The result follows since M(G) = {[π] | π ∈ Z2 (G, ℂ∗ )}. Let Z2N (G, ℂ∗ ) denote the set of all normalized factor sets of G (recall that a factor set π is normalized if π(e, e) = 1). Write B2N (G, ℂ∗ ) = B2 (G, ℂ∗ ) ∩ Z2N (G, ℂ∗ ), where B2 (G, ℂ∗ ) = {[1]}. Since [π] ∩ Z2N (G, ℂ∗ ) ≠ 0 for all π ∈ Z2 (G, ℂ∗ ) (see the paragraph following Definition 1.2), it follows that M(G) ≅ Z2N (G, ℂ∗ )/B2N (G, ℂ∗ ). Lemma 3.2. Let Π ∈ M(G) and o(Π) = m (the order of a group element). Then Π = [π], where π ∈ Z2N (G, ℂ∗ ) and o(π) = m. In particular, |M(G)| < ∞. Proof. Since Π = [π0 ], where π0 ∈ Z2N (G, ℂ∗ ) and π0m ∼ 1, it follows that π0 (s, t)m =

λ(s)λ(t) λ(st)

(s, t ∈ G)

for a suitable function λ : G → ℂ∗ . Next, π0 (e, e) = 1 implies λ(e) = 1 (indeed, we have π0 (e, e) = λ(e)). Define a function μ ∈ F[G, ℂ∗ ] by μ m = λ−1 , μ(e) = 1. Putting π(s, t) = one obtains

μ(s)μ(t) π0 (s, t) μ(st)

π ∈ Z2N (G, ℂ∗ ),

Indeed,

(s, t ∈ G),

[π] = [π0 ] = Π,

o(π) = m.

μ(e)μ(e) π0 (e, e) = 1. μ(ee) Since the set of all functions ϕ : G × G → ℂ∗ satisfying the condition ϕ|G| = 1 ∈ ℂ is finite, the second statement is also proved. π(e, e) =

VI Projective representations | 243

Definition 3.1. A triple (Γ, f, A) is called a central extension of a group G if Γ is a group, A ≤ Z(Γ) and f is a homomorphism of Γ onto G such that ker(f) = A. (In particular, one has Γ/A ≅ G.) For every s ∈ G, choose an element g s in the preimage f −1 (s) ∈ Γ (thus f(g s ) = s; see Definition 3.1). Then for s, t ∈ G, one has f(g s g t ) = f(g s )f(g t ) = st = f(g st ), and therefore g s g t = p(s, t)g st ,

p(s, t) ∈ ker(f) = A.

(1)

The mapping p : G × G → A from (1) is called the factor set of the central extension (Γ, f, A). In what follows we assume that g e = 1 ∈ Γ. In that case, the factor set p is said to be normalized. Equality (1), in view of g e = 1, implies p(e, s) = p(e, e) = p(s, e) = 1

for all s ∈ G

(compare with (1.4)). The inclusion A ≤ Z(Γ) and the associativity of multiplication in Γ imply p(s, t)p(st, r) = p(s, tr)p(t, r)

(2)

(s, t, r ∈ G)

(see the similar formula (2.1)). Given ψ ∈ Lin(A) (the set of linear characters of the abelian group A), put π ψ (s, t) = ψ(p(s, t))

(3)

(s, t ∈ G).

Since the factor set p is normalized, it follows from (2) that π ψ is also normalized, i.e., π ψ ∈ Z2N (G, ℂ∗ ). Indeed, π ψ (e, t) = ψ(p(e, t)) = ψ(1) = 1 = ψ(p(s, e)) = π ψ (s, e)

(s, t ∈ G).

Since the mapping ψ 󳨃→ π ψ is a group homomorphism of Lin(A) into Z2N (G, ℂ∗ ), it follows that the mapping ψ 󳨃→ [π ψ ] = τ(ψ) is a homomorphism τ of Lin(A) into M(G). A central extension (Γ, f, A) is said to be faithful if the homomorphism τ is injective. We have im(τ) = τ(Lin(A)) = H ≤ M(G). The homomorphism τ (hence also the subgroup H) is independent of the choice of the transversal {g s }s∈G of the group Γ over A. Indeed, if {g 󸀠s }s∈G is another transversal of Γ over A, then g󸀠s = l(s)g s , with l(s) ∈ A (s ∈ G). Therefore, g󸀠s g󸀠t = p󸀠 (s, t)g󸀠st ,

where p󸀠 (s, t) =

l(s)l(t) p(s, t) l(st)

Let ψ ∈ Lin(A),

λ(s) = ψ(l(s))

(s ∈ G).

(s, t ∈ G).

(∗∗)

244 | Characters of Finite Groups 1 Applying ψ to the second equality in (∗∗), we get π󸀠ψ (s, t) = ψ(p󸀠 (s, t)) =

λ(s)λ(t) π ψ (s, t), λ(st)

so that [π󸀠ψ ] = [π ψ ] (see the paragraph following Definition 1.2). We say that the subgroup H = im(τ) = τ(Lin(A)) ≤ M(G) is produced by the central extension (Γ, f, A). Note that H = {[π ψ ] | ψ ∈ Lin(A)}. Since H ≅ Lin(A)/ker(τ) and Lin(A) ≅ A, it follows that |H| ⋅ |ker(τ)| = |A|, i.e., |H| | |A|. Definition 3.2. A central extension (Γ, f, A) of a group G is said to be a covering group for G if H = im(τ) = M(G). If (Γ, f, A) is a covering group of G, then |M(G)| | |A|. In particular, |Γ| ≥ |G| ⋅ |M(G)|. Definition 3.3. A covering group (Γ, f, A) of G is called a representation group (Darstellungsgruppe) of G if |Γ| = |G| ⋅ |M(G)|, i.e., |M(G)| = |A|. If (Γ1 , f, A) is another representation group of a group G, then it is not necessarily true that Γ1 ≅ Γ (examples of this will be presented later), however, |Γ1 | = |Γ| since M(G) is defined up to isomorphism. By considering projectively equivalent representations, Schur reduced the theory of projective representations of finite groups to the theory of linear representations of their representation groups. Lemma 3.3. A central extension (Γ, f, A) of a group G is a representation group of G if and only if τ is an isomorphism of Lin(A) onto M(G) if and only if (Γ, f, A) is a faithful covering central extension of G. Proof. The result follows from the equivalence of the following pairs of assertions: (a) im(τ) = M(G), |A| = |M(G)|, (b) Lin(A)/ker(τ) ≅ M(G), |Lin(A)| = |M(G)|, (c) ker(τ) = {1}, im(τ) = M(G), (d) τ : Lin(A) → M(G) is an isomorphism. Let (Γ, f, A) be a central extension of a group G, L ∈ LIN(Γ) (the set of linear representations of Γ), P ∈ PR(G) (the set of projective representations of G). Assume that there is a function λ ∈ F[Γ, ℂ∗ ] such that L(g) = λ(g)P(f(g)) (g ∈ Γ). Then we say that the linear representation L of G is obtained by lift the (projective) representation P to Γ, and P is obtained by descent of L to G. The equality L(g) = λ(g)P(f(g)) (g ∈ Γ) determines a correspondence on LIN(Γ) × PR(G), which is called the Schur correspondence and S denoted by ↔. Lemma 3.4. A representation L ∈ LIN(Γ) admits descent to G if and only if Z(L) ≥ A.³

3 The quasikernel Z(L) of L is the set of g ∈ Γ such that L(g) is a scalar matrix.

VI Projective representations | 245 S

Proof. If L ↔ P, then L(g) = λ(g)P(f(g)) (g ∈ Γ) 󳨐⇒ L(a) = λ(a)P(e)(a ∈ A) is a scalar matrix since A = ker(f). Therefore A ≤ Z(L), the quasikernel of L. Let A ≤ Z(L). Then, for ψ ∈ Lin(A), one has L(a) = ψ(a)In (a ∈ A, n = deg(L)) since P(f(a)) = P(e) = I n . Put P(s) = L(g s ) for s ∈ G (where {g s }s∈G is a transversal of Γ over A). If s, t ∈ G, then P(s)P(t) = L(g s )L(g t ) = L(g s g t ) = L(p(s, t)g st ) = L(p(s, t))L(g st ) = ψ(p(s, t))In P(st) = π ψ (s, t)P(st) (recall that p(s, t) ∈ A so L(p(s, t)) = ψ(p(s, t))In ), i.e., P is a projective π ψ -representation of G. Put ag s = g ∈ Γ (a ∈ A, s ∈ G); then L(g) = L(a)L(g s ) = ψ(a)P(s) = ψ(a)P(f(g)) = λ(g)P(f(g)), S

where λ : Γ → ℂ∗ is defined by λ(g) = ψ(a). Thus, L ↔ P, i.e., the linear representation L of Γ admits descent to G. If one has L ∈ LIN(Γ),

Z(L) ≥ A,

ψ ∈ Lin(A),

L(a) = ψ(a)In

(a ∈ A, n = deg(L)),

we say that the linear character ψ is afforded by the linear representation L (writing ψ = ψ L ). Lemma 3.5. Suppose that L, L1 ∈ LIN(Γ), Then

P ∈ PR(G),

S

L ↔ P.

S

L1 ↔ P, ψ L1 = ψ L ⇐⇒ L1 = χL, where χ ∈ Lin(Γ), ker(χ) ≥ A. Proof. Suppose that L1 = χL and χ ∈ LinA (Γ) (i.e., χ is a linear character of Γ with A ≤ ker(χ); we do not assume that A ≤ Γ󸀠 ). Then, for any g ∈ Γ, L1 (g) = χ(g)L(g) = χ(g)λ(g)P(f(g)) = λ1 (g)P(f(g)), S

where λ1 = χλ, g ∈ Γ. Therefore, L1 ↔ P and, moreover, ψ L1 = ψ L . S Now suppose that L1 ↔ P, ψ L1 = ψ L . Then L1 (g) = λ1 (g)P(f(g)),

λ1 ∈ F[Γ, ℂ∗ ], g ∈ Γ.

Therefore, L1 (g) = χ(g)L(g), where χ = λ1 λ−1 ∈ F[Γ, ℂ∗ ] (recall that L(g) = λ(g)P(f(g)). It follows from the relations L, L1 ∈ LIN(Γ) and ψ L1 = ψ L that χ ∈ LinA (Γ). S

Exercise 3.2. If L ∈ LIN(Γ), P ∈ PR(G) and L ↔ P, then L is irreducible if and only if P is irreducible.

246 | Characters of Finite Groups 1 Lemma 3.6. Suppose that L ∈ LIN(Γ),

P ∈ PR(G),

S

L ↔ P.

(a) If L = L1 + L2 , L1 , L2 ∈ LIN(Γ), then P = P1 + P2 ,

P1 , P2 ∈ PR(G),

Li ↔ Pi ,

S

i = 1, 2.

S

i = 1, 2.

(b) If P = P1 + P2 (where P1 , P2 ∈ PR(G)), then L = L1 + L2 ,

L1 , L2 ∈ LIN(Γ),

Li ↔ Pi ,

S

Proof. (a) Since L ↔ P, there exists a function λ : Γ → ℂ∗ such that L(g) = λ(g)P(f(g)) (g ∈ Γ). In particular, if s ∈ G, then P(s) = P(f(g s )) = λ(g s )−1 L(g s ) = λ(g s )−1 diag(L1 (g s ), L2 (g s )) = diag(P1 (s), P2 (s)), where P i (s) = λ(g s )−1 L(g s ) (i = 1, 2). This implies that P1 , P2 ∈ PR(G) and P = P1 +P2 . If g ∈ Γ, then diag(L1 (g), L2 (g)) = L(g) = λ(g)P(f(g)) = diag(λ(g)P1 (f(g)), λ(g)P2 (f(g))), and hence

S

L i (g) = λ(g)P i (f(g)) 󳨐⇒ L i ↔ P i

(i = 1, 2).

Part (b) can be proved similarly. Lemma 3.7. Let (Γ, f, A) be a central extension of a group G, producing a subgroup H ≤ M(G). (a) Every L in LIN(Γ) satisfying Z(L) ≥ A is the lifting of some π-representation of G to Γ, where [π] ∈ H. (b) Every π-representation with [π] ∈ H can be lifted to Γ. Proof. (a) Suppose that ψ = ψ L . By Lemma 3.4, there exists a projective π ψ -represenS tation P of G such that L ↔ P. It remains to note that [π ψ ] ∈ H by the definition of H. (b) It follows from the definition of H that π ∼ π ψ for some ψ ∈ Lin(A). It was shown above that there exists a function μ : G → ℂ∗ such that P󸀠 = μP is a projective π ψ -representation of G. Let g = ag s ∈ Γ, a ∈ A, s ∈ G. Putting L(g) = ψ(a)P󸀠 (s), we prove that L ∈ LIN(Γ). Let g i = a i g s i ∈ Γ,

a i ∈ A,

si ∈ G

(i = 1, 2).

Then, since a1 , a2 ∈ Z(Γ), the center of Γ, we obtain L(g1 g2 ) = L(a1 a2 g s1 g s2 ) = L(a1 a2 p(s1 , s2 )g s1 s2 )ψ(a1 a2 p(s1 , s2 ))P󸀠 (s1 s2 ) = ψ(a1 )ψ(a2 )π ψ (s1 , s2 )P󸀠 (s1 s2 ) = ψ(a1 )ψ(a2 )P󸀠 (s1 )P󸀠 (s2 ) = L(g1 )L(g2 ),

VI Projective representations | 247

i.e., L ∈ LIN(Γ). Since L(g) = ψ(a)P󸀠 (s) = ψ(a)μ(s)P(s) = λ(g)P(f(g)), where (g = ag s ∈ Γ, a ∈ A, s ∈),

λ(g) = ψ(a)μ(s) S

we get L ↔ P. Corollary 3.8. If (Γ, f, A) is a covering of a group G, then every P ∈ PR(G) can be obtained by descent of some L ∈ LIN(Γ) to G. Proof. If π ∈ Z2 (G, ℂ∗ ) and P is a π-representation of G, then, by Lemma 3.7, the inclusion [π] ∈ M(G) and the fact that (Γ, f, A) produces M(G) imply that P can be lifted to L ∈ LIN(Γ). Then P is obtained by descent of L to G. Consequently, if (Γ, f, A) is a covering of G (in particular, its representation group) then, by Lemma 3.7, the knowledge of LIN(Γ) implies the knowledge of PR(G). The following theorem contains the statement about the existence of a representation group. Theorem 3.9 (Schur). Each subgroup H ≤ M(G) is produced by some faithful central extension of a group G. In particular, every finite group possesses a representation group. Proof. Let {Π i }1r be a basis of the abelian group H, m i = o(Π i ) and let A i be the group of m i -th roots of unity in ℂ (i ∈ {1, . . . , r}). Then H ≅ A1 × ⋅ ⋅ ⋅ × A r . Using Lemma 3.2, choose a normalized factor set π i in each class Π i such that o(π i ) = m i (i ∈ {1, . . . , r}). (i) Let R i = ℂπ i G be the twisted group algebra corresponding to π i and let {u s }s∈G be its standard basis. Then (i) (i)

(i)

u s u t = π i (s, t)u st

(s, t ∈ G, i ∈ {1, . . . , r}).

(4)

(i)

Since π i is normalized, e i = u e (e ∈ G) is the identity element of the algebra R i . Consider the direct sum R = R1 ⊕ ⋅ ⋅ ⋅ ⊕ R r of the algebras R i . The elements of R are (1) (r) the r-tuples x = (x1 , . . . , x r ), x1 ∈ R1 , . . . , x r ∈ R r . Put g s = (u s , . . . , u s ) (s ∈ G). Then g e = (e1 , . . . , e s ) is the identity element of R. For any system of elements α i ∈ A i (i ∈ {1, . . . , r}) define a = (α1 e1 , . . . , α r e r ) ∈ R. These elements form a multiplicative group A ≅ A1 ×⋅ ⋅ ⋅× A r ≅ H. Obviously, A ⊆ Z(R), the center of the algebra R. Since π i (s, t) ∈ A i (s, t ∈ G) (i ∈ {1, . . . , r}), A contains the elements p(s, t) = (π1 (s, t)e1 , . . . , π r (s, t)e r ) for all s, t ∈ G. It follows from (4) that g s g t = p(s, t)g st

(s, t ∈ G).

The elements g s are invertible in R: (i)

(i)

−1 (u s )−1 = π i (s, s−1 )−1 u s−1 󳨐⇒ g −1 s = p(s, s )g s−1 .

(5)

248 | Characters of Finite Groups 1 Therefore, the set Γ = {ag s | a ∈ A, s ∈ G} is a finite group. Since A ⊆ Z(R), we have A ≤ Z(Γ). Let a = (α1 e1 , . . . , α r e r ),

b = (β1 e1 , . . . , β r e r ), (i)

(i)

α i , β i ∈ A (i ∈ {1, . . . , r}).

(i)

If ag s = bg t (s, t ∈ G), then α j u s = β j u t . Since {u s }s∈G is a basis of the algebra R i , it follows that s = t and α i = β i (i ∈ {1, . . . , r}), i.e., a = b. This proves that |Γ| = |H| ⋅ |G|. Let g = ag s ∈ Γ, a ∈ A, s ∈ G. The mapping g 󳨃→ s = f(g) is a homomorphism of Γ onto G. Indeed, if g i = a i g s i , a i ∈ A, s i ∈ G (i = 1, 2), then, by (5), f(g1 g2 ) = f(a1 a2 g s1 g s2 ) = f(a1 a2 p(s1 , s2 )g s1 s2 ) = s1 s2 = f(g1 )f(g2 ) since a1 a2 p(s1 , s2 ) ∈ A. Next, as g e is the identity element of the group Γ, we have f(a) = f(ag e ) = e ∈ G. Conversely, if g = ag s (a ∈ A, s ∈ G) and f(g) = e, then s = e so g = ag e = a ∈ A. Thus, we have constructed a central extension (Γ, f, A) of G with factor set p(s, t) (s, t ∈ G). Let ν ν Π = Π11 . . . Π r r , where the integer ν i is uniquely determined (mod m i ) for i ∈ {1, . . . , r}. Then ν

ν

π = π11 . . . π r r ∈ Π, ν

π i ∈ Π i (i ∈ {1, . . . , r}).

ν

Defining ψ(a) = α11 . . . α r r for an element a = (α1 e1 , . . . , α r e r ) ∈ A, we obtain a welldefined element ψ ∈ Lin(A). Since ψ(p(s, t)) = π1 (s, t)ν1 . . . π r (s, t)ν r = π(s, t), it follows that π = π ψ , [π ψ ] = Π. Therefore, the mapping ψ 󳨃→ [π ψ ] = τ(ψ) (ψ ∈ Lin(A)) is a surjective homomorphism of Lin(A) onto H, and H is produced by the central extension (Γ, f, A). Since |Lin(A)| = |A| = |H|, it follows that τ is an isomorphism of Lin(A) onto H, i.e., (Γ, f, A) is a faithful central extension of G producing H. The last statement is a particular case of the first. Exercise 3.3. All projective representations of the group G are completely reducible. Hint. Theorem 3.9, Maschke’s theorem (our ground field is ℂ), Lemma 3.6. Corollary 3.10. The degree of an irreducible projective representation of a group G divides its order. Proof. Let P ∈ PR(G) be irreducible and (Γ, f, A) a representation group of G. By CorolS lary 3.8, there exists an irreducible L ∈ LIN(Γ) such that L ↔ P. Since deg(P) = deg(L), it follows that deg(P) = deg(L) | |Γ : Z(Γ)|, by Theorem IV.8.4, and therefore deg(P) divides |Γ : A| = |G| since A ≤ Z(Γ).

VI Projective representations | 249

Corollary 3.11. Given π ∈ Z2 (G, ℂ∗ ), the set of classes of linearly equivalent irreducible projective π-representations of a group G is finite. If r π is the cardinality of that set, (Γ, f, A) a representation group of G and π ∼ π ψ , where ψ ∈ Lin(A), then r π is the number of classes of irreducible linear representations of Γ with the character ψ of A. Proof. Let P be an irreducible projective π ψ -representation of G. Putting L(ag s ) = ψ(a)P(s)

(a ∈ A, s ∈ G),

we lift P to an irreducible linear representation L of Γ with the character ψ of A (see the proof of Lemma 3.7). We call this lift of P to L the standard lift. In particular, L(g s ) = ψ(1)P(s) = P(s) (s ∈ G). Conversely, let L be an irreducible linear representation of Γ with a linear character ψ of A (the existence of L follows from reciprocity). Putting P(s) = L(g s ), we obtain an irreducible projective π ψ -representation P of G. We call this particular descent of L to G the standard descent. As the standard descent and the standard lift are mutually inverse, we obtain a one-to-one correspondence between classes of linearly equivalent irreducible projective π ψ -representations of G and classes of irreducible linear representations of Γ corresponding to ψ ∈ Lin(A). It remains to note that the classes of linearly equivalent π-representations of G are in one-to-one correspondence with the classes of linearly equivalent π ψ -representations of G. Lemma 3.12. Suppose that (Γ, f, A) is a central extension of a group G, and ψ ∈ Lin(A),

Irrψ (Γ) = {θ ∈ Irr(Γ) | θ A = θ(1)ψ}.

Then ∑θ∈Irrψ (Γ) θ(1)2 = |G|. Proof. By reciprocity, Irr(ψ Γ ) = Irrψ (Γ),

ψΓ =

θ(1)θ

∑ θ∈Irrψ (Γ)

since ψ ∈ Lin(A) and A ≤ Z(Γ). Therefore, |G| = |Γ : A| = ψ Γ (1) =

∑

θ(1)2 .

θ∈Irrψ (Γ)

Theorem 3.13. Suppose that P i (i ∈ {1, . . . , r π }) is a complete system of representatives of classes of linearly equivalent irreducible projective π-representations of a group rπ G and f i = deg(P i ). Then |G| = ∑i=1 f i2 . Proof. Let (Γ, f, A) be a representation group of G. There exists a unique ψ ∈ Lin(A) such that π ∼ π ψ . Since equivalent projective representations have the same degrees, we may assume that {P i } is a complete system of representatives of classes of linearly equivalent projective π ψ -representations of G. If L i ∈ LIN(Γ) is obtained by the stanr dard lift of P i to Γ (i ∈ {1, . . . , r π }), then {L i }1π is a complete system of representatives

250 | Characters of Finite Groups 1 of classes of irreducible linear representations of Γ associated with ψ ∈ Lin(A). Let θ i r rπ be the character of L i ; then {θ i }1π = Irrψ (Γ), and, by Lemma 3.12, ∑i=1 θ i (1)2 = |G|. It remains to note that θ i (1) = deg(L i ) = deg(P i ) = f i . Of course, this is a generalization of known equality |G| = ∑χ∈Irr(G) χ(1)2 . Exercise 3.4. All linear representations of the twisted group algebra ℂπ G, where π ∈ Z 2 (G, ℂ∗ ), are completely reducible. Hint. Use Exercise 3.3. It follows from Exercise 3.4 that the twisted group algebra ℂπ G is semisimple (see §I.12). Below will be offered a direct proof of this fact. Note that Theorem 3.13 is the immediate consequence of semisimplicity of the algebra ℂπ G and some results from the last three paragraphs of Chapter I. Exercise 3.5. The number r π , defined in Theorem 3.13, equals the number of classes of irreducible linear representations of the twisted group algebra ℂπ G. Hint. Determine a one-to-one correspondence between the classes in the exercise and the classes of linearly equivalent irreducible projective π-representations of the group G. Lemma 3.14. One has r π = dimℂ (Z(ℂπ G)). Proof. This follows from the results of §I.12 since ℂπ G is semisimple (see the paragraph after Exercise 3.4). We now introduce the notion of a π-element of a group G. Let {u s }s∈G be the standard basis of the twisted group algebra ℂπ G. It follows easily from (2.2) that u−1 t u s u t = ω π (s, t)u s t ,

where ω π (s, t) =

π(s, t) π(t, s t )

(6)

(see formula (1.3)). For s, t1 , t2 ∈ G, it follows from the equalities −1 −1 t1 u−1 t2 (u t1 u s u t1 )u t2 = ω π (s, t 1 )u t2 u s t1 u t2 = ω π (s, t 1 )ω π (s , t 2 )u s t1 t2 ,

u−1 t1 t2 u s u t1 t2 = ω π (s, t 1 t 2 )u s t1 t2 that ω π (s, t1 t2 ) = ω π (s, t1 )ω π (s t1 , t2 ).

(7)

Next, since −1 −1 −1 −1 −1 u−1 g (u t u s u t )u g = (u g u t u g ) (u g u s u g )(u g u t u g ),

it follows that ω π (s g , t g ) = ω π (s, t)ω π (s t , g)ω π (s, g)−1 . By (7), for a fixed s ∈ G, the function (π)

χ s (t) = ω π (s, t),

where t ∈ CG (s),

(8)

VI Projective representations | 251

is a linear character of the group CG (s). It is evident from the equality (π)

χ s (t) = π(s, t)π(t, s)−1

(s ∈ G, t ∈ CG (s))

(9)

(π)

(see (1.3)) that the function χ s depends only on the class [π] of the factor set π (for a fixed s ∈ G). (π)

Definition 3.4. An element s ∈ G is called a π-element if χ s = 1CG (s) , i.e., 1 = ω π (s, t) =

π(s, t) π(s, t) = π(t, s t ) π(t, s)

for all t ∈ CG (s);

see the expression for ω π (s, t) in (1.3). (π)

(π)

Thus, χ s is a linear character of CG (s) but, if s ∈ G is a π-element, then χ s is the principal character of CG (s). Let G π be the set of π-elements of G. It is evident from (6) and (7) that the following statements are equivalent: (a) s ∈ G π . (b) st = ts implies u s u t = u t u s . (c) st = ts implies π(s, t) = π(t, s). Exercise 3.6. Let s ∈ G π . (a) If P is a projective π-representation of the group G, then st = ts(t ∈ G) 󳨐⇒ P(s)P(t) = P(t)P(s). (b) If s󸀠 ∼ s in G, then s󸀠 ∈ G π , i.e., the set G π is G-invariant. (c) If s t1 = s t2 , then ω π (s, t1 ) = ω π (s, t2 ). Solution. (a) Since 1 = ω π (s, t) =

π(s,t) π(t,s) ,

we get π(s, t) = π(t, s) so that

P(s)P(t) = π(s, t)P(st) = π(t, s)P(ts) = P(t)P(s). (b) Let s󸀠 = s g (g ∈ G). If st = ts, then s g t g = (st)g = (ts)g = t g s g , CG (s g ) = {t g | t ∈ CG (s)} so, by (8), since ω π (s, t) = 1 and s t = s, we get ω π (s󸀠 , t g ) = ω π (s g , t g ) = ω π (s, t)ω π (s t , g)ω π (s, g)−1 = 1. Exercise 3.6 (b) justifies the following: Definition 3.5. A G-class consisting of π-elements is called a π-class of a group G. Let s π denote the number of all π-classes of G. Thus, G π is the union of all π-classes of a group G. The class {e} is a π-class for any factor set π. Definition 3.6. A function α : G → ℂ is said to be π-central (or a π-class function) provided α(s t ) = ω π (s, t)α(s) (s, t ∈ G).

252 | Characters of Finite Groups 1 Exercise 3.7. A π-central function vanishes on G − G π . Solution. Let α : G → ℂ be a π-central function. Then there exist s ∈ G − G π and t ∈ CG (s) such that ω π (s, t) ≠ 1. In that case, α(s) = α(s t ) = ω π (s, t)α(s) 󳨐⇒ α(s) = 0 since 0 ≠ ω π (s, t) ≠ 1. Exercise 3.8. One has x = ∑s∈G α(s)u s ∈ Z(ℂπ G) if and only if the function s 󳨃→ α(s) is π-central. Hint. Here {u s }s∈G is the standard basis of the twisted group algebra ℂπ G. For t ∈ G, we obtain, using formula (8), t ∑ α(s)u s = x = u−1 t xu t = ∑ α(s)ω π (s, t)u s t = ∑ α(s )u s t .

Let

s∈G

s∈G

s∈G (π) Ki

be π-classes of G, g i ∈

(π) Ki ;

(π)

ki

then

= |C G (g i )|−1 ∑ u−1 t u gi u t t∈G

are called π-class sums (i = 1, . . . , s π ). Obviously, these sums depend on the choice of the representatives g i of the π-classes. (π)

Exercise 3.9. One has k i

∈ Z(ℂπ G). Moreover, (π)

ki

= |C G (g i )|−1 ∑ ω π (g i , t)u g it . t∈G

(π)

We prove that the elements k i (i ∈ {1, . . . , s π }) form a basis of the algebra Z(ℂπ G). (π) To this end, note that, by Exercises 3.9 and 3.8, one has k i = ∑g∈G α i (g)u g , where α i (π) (π) is a π-central function, that does not vanish on K i but vanishes on G − K i . There(π) fore, the elements k i (i ∈ {1, . . . , s π }) are linearly independent in ℂπ G. Let x = ∑ α(g)u g ∈ ℂπ G. g∈G

If g ∈ that

(π) Ki ,

then g = g it , where t ∈ G. Since α and α i are π-central functions, it follows α(g) = α(g it ) = ω π (g i , t)α(g i ),

α i (g) = α i (g it ) = ω π (g i , t)α i (g i ).

(π)

Therefore, α(g) = β i α i (g) for g ∈ K i , where β i = Exercise 3.7, sπ

α(g i ) α i (g i )

is independent of g. Hence, by sπ

(π)

x = ∑ α(g)u g = ∑ ∑ α(g)u g = ∑ β i k i . g∈G

(π) s

i=1 g∈K (π) i

i=1

Consequently, {k i }1π is a basis in Z(ℂπ G). Taking Lemma 3.14 into account, we arrive at the following result.

VI Projective representations | 253

Lemma 3.15. The following numbers are equal: (i) dimℂ (Z(ℂπ G)). (ii) The number s π of π-classes of G. (iii) The number r π of classes of linearly equivalent irreducible π-representations of G. It is clear that if a factor set π ∼ 1, Lemma 3.15 implies some of the results proved previously (for example, the class number s1 = k(G) of a group G equals the number of classes of equivalent irreducible linear representations of G; see §II.6). Let us present, in conclusion, another proof of the semisimplicity of the twisted group algebra ℂπ G. Let R be the regular (matrix) representation of the twisted group algebra ℂπ G: R(x) = (ρ s,t (x)) (x ∈ ℂπ G, s, t ∈ G). If x = ∑g∈G α(g)u g , where {u g }g∈G is the standard basis of the algebra ℂπ G, then ρ s,t (x) = π(t, t−1 s)α(t−1 s). In particular, one has ρ s,s = π(t, 1)α(1). Assuming that the factor set π is normalized and denoting by tr(x) the trace of the matrix R(x), we get tr(x) = |G|α(1). In particular, tr(u g ) = |G|δ1,g

(g ∈ G),

(∗)

where δ is the Kronecker delta on the set G. To prove that the algebra ℂπ G is semisimple, it suffices to check that ℂπ G has no nontrivial nilpotent ideals. Assume, by way of contradiction, that J is such an ideal −1 and x = ∑g∈G α(g)u g ∈ J. Since u−1 g x ∈ J (g ∈ G), the matrix R(u g x) is nilpotent, which −1 implies that tr(u g x) = 0 (g ∈ G). Noting that −1 −1 −1 −1 u−1 ∑ π(g −1 , s)α(s)u g−1 s , g x = π(g, g ) u g−1 x = π(g, g ) s∈G

we get using formula (∗) and taking into account that π(g, g −1 ) = π(g −1 , g) (this follows from the equalities P(g)P(g−1 ) = π(g, g −1 )P(1) and P(g −1 )P(g) = π(g −1 , g)P(1); see also Exercise 2.1), we get tr(u g−1 x) = |G|α(g). It follows that α(g) = 0 for all g ∈ G. Hence x = 0. Thus, J = {0}, i.e., the algebra ℂπ G is semisimple.

4 The exponent of the Schur multiplier Continuing our study of the Schur multiplier M(G) of a group G, let us prove a stronger version of Lemma 3.1. We retain the notation introduced in the previous section. Lemma 4.1. If P is a projective π-representation of a group G, then o([π]) | deg(P). Proof. Let n = deg(P). Taking determinants in formula (1.1), one obtains π(s, t)n = det(P(s)) det(P(t)) det(P(st))−1 (see (1.2’)). Putting det(P(s)) = λ(s)(s ∈ G), we get [π]n = 1.

254 | Characters of Finite Groups 1 In particular, exp(M(G)) divides the least common multiple of the degrees of the irreducible projective representations of G. Theorem 4.2 (Schur). One has exp(M(G))2 | |G|. Proof. Let π ∈ M(G) = Z2 (G, ℂ∗ ) and let P i (i ∈ {1, . . . , r π }) be a complete system of representatives of classes of linearly equivalent irreducible projective π-representations of G, and let deg(P i ) = f i ; then o([π]) | f i for all i (Lemma 4.1). Therefore, o([π])2 rπ divides ∑i=1 f i2 = |G| (Theorem 3.13). Since LCM{o([π]) | [π] ∈ M(G)} = exp(M(G)), the theorem follows. It follows from Theorem 4.2 that π(M(G)) ⊆ π(G) (π(X) is the set of all prime divisors of a group X). Next, if a p-group G is of order ≤ p3 , then M(G) is elementary abelian. Let (Γ, f, A) be a central extension of a group G. Retaining the previous notation, we prove the following: Lemma 4.3. If ψ ∈ Lin(A), then π ψ ∼ 1 if and only if there is χ ∈ Lin(Γ) such that ψ = χ A (the restriction χ to A). Proof. See formulas (3.1) and (3.3). (a) Let ψ = χ A , where χ ∈ Lin(Γ). Since g s g t = p(s, t)g st ,

χ(p(s, t)) = ψ(p(s, t)) = π ψ (s, t)

(s, t ∈ G),

it follows that χ(g s )χ(g t ) = χ(g s g t ) = χ(p(s, t)g st ) = χ(p(s, t))χ(g st ) = π ψ (s, t)χ(g st ). Putting λ(s) = χ(g s ), we obtain π ψ (s, t) = λ(s)λ(t)λ(st)−1 (s, t ∈ G),

i.e.,

πψ ∼ 1

(see the paragraph preceding Definition 1.3). (b) Now let π ψ ∼ 1. Then π ψ (s, t) = λ(s)λ(t)λ(st)−1

(s, t ∈ G)

for some λ ∈ F[G, ℂ∗ ]. Define χ : Γ → ℂ∗ by χ(ag s ) = ψ(a)λ(s)

(a ∈ A, s ∈ G).

If a, b ∈ A and s, t ∈ G, then A ≤ Z(Γ) 󳨐⇒ p(s, t) ∈ A, ψ(p(s, t)) = π ψ (s, t) 󳨐⇒ ψ(p(s, t))λ(st) = λ(s)λ(t), and hence χ(ag s ⋅ bg t ) = χ(abp(s, t)g st ) = ψ(abp(s, t))λ(st) = ψ(a)ψ(b)ψ(p(s, t))λ(st) = ψ(a)ψ(b)λ(s)λ(t) = χ(ag s )χ(bg t ),

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i.e., χ ∈ Lin(Γ) since χ : Γ → ℂ∗ is a homomorphism. Since, as stipulated in §VI.3, π ψ ∈ Z2N (G, ℂ∗ ), it follows that λ(e) = 1 ∈ ℂ, g e is the identity element of the group Γ. Therefore, χ(a) = χ(ag e ) = ψ(a)λ(e) = ψ(a), i.e., χ A = ψ. The mapping σ : χ 󳨃→ χ A (χ ∈ Lin(Γ)) is a homomorphism of Lin(Γ) into Lin(A). Let L = {χ A | χ ∈ Lin(Γ)} = im(σ). Exercise 4.1. One has L = ker(τ), where τ is the homomorphism of Lin(A) into M(G) introduced in §VI.3 (see the paragraph following formula (3.3)). In other words, the sequence σ τ Lin(Γ) → Lin(A) → M(G) is exact. In particular, if a central extension (Γ, f, A) is faithful, then L = {1A }. If H ≤ M(G) is produced by a central extension (Γ, f, A) of G, then H ≅ Lin(A)/L. Hint. By Lemma 4.3, L = {ψ ∈ Lin(A) | [π ψ ] = [1]}. Lemma 4.4. The following statements hold: (a) If a central extension (Γ, f, A) of a group G produces a subgroup H ≤ M(G), then H ≅ D = A ∩ Γ󸀠 . (b) ker(τ) = {ψ ∈ Lin(A) | ker(ψ) ≥ D} = LinD (A). (c) If (Γ, f, A) is a covering of G, then M(G) ≅ A ∩ Γ󸀠 . Proof. (a) Let us find Ann(L), the annihilator of L ≤ Lin(A) in A. Since Ann(L) = {a ∈ A | ψ(a) = 1 for all ψ ∈ L}, it follows from the equalities ⋂

ker(χ) = Γ󸀠 ,

L = {χ A | χ ∈ Lin(Γ)}

χ∈Lin(Γ)

that Ann(L) = {a ∈ A | χ(a) = 1 for all χ ∈ Lin(Γ)} = A ∩ Γ󸀠 . Since Ann(L) ≅ Lin(A)/L (Chapter I), it follows that H ≅ D = A ∩ Γ󸀠 . (b) One has Ann(L) = D, whence it follows that ker(τ) = L = Ann(D) = LinD (A). (c) The central extension (Γ, f, A) produces M(G), and the result follows from statement (a). Lemma 4.5. Let (Γ, f, A) be a central extension of a group G. (a) (Γ, f, A) is faithful if and only if A ≤ Γ󸀠 . (b) (Γ, f, A) is a representation group of G if and only if A ≤ Γ󸀠 and |A| = |M(G)|. (c) Let (Γ, f, A) be faithful and let S be a subgroup of index n in G, Σ = f −1 (S) and D S = Σ󸀠 ∩ A. Then exp(A/D S ) | n. (d) If a subgroup S from (c) is cyclic, then exp(A) | n. In particular, if all Sylow subgroups of G are cyclic, then M(G) = {1}.

256 | Characters of Finite Groups 1 Proof. (a) If (Γ, f, A) is faithful, then H = im(τ) ≅ Lin(A) ≅ A 󳨐⇒ |H| = |A|. It follows from |H| = |A ∩ Γ󸀠 | that A ≤ Γ󸀠 . Conversely, if A ≤ Γ󸀠 , then D = A ∩ Γ󸀠 = A, and therefore |H| = |A| (Lemma 4.4 (a)). Since H ≅ Lin(A)/ker(τ) and |Lin(A)| = |A|, we have ker(τ) = {1A }, i.e., the central extension (Γ, f, A) of G is faithful. (b) The implication (⇒) follows from (a) and Definition 3.3. Conversely, if A ≤ Γ󸀠 and |A| = |M(G)|, then (Γ, f, A) produces M(G) and is faithful (Lemma 4.4 (a)), i.e., (Γ, f, A) is a representation group of G, and this proves the implication (⇐). (c) Let V be the transfer of the group Γ in Σ(= f −1 (S)). If a ∈ A, then V(a) = a n Σ󸀠 (Lemma V.9.1). On the other hand, since A ≤ Γ󸀠 and Γ󸀠 ≤ ker(V), we get V(a) = Σ󸀠 (here Σ󸀠 is considered as the identity element of the quotient group Σ/Σ󸀠 ). Therefore, an ∈

⋂

ker(φ) = Σ󸀠 󳨐⇒ a n ∈ A ∩ Σ󸀠 = D S .

φ∈Lin(Σ)

(d) Since ≅ S is cyclic and A ≤ Z(f −1 (S)), the group f −1 (S) = Σ is abelian. 󸀠 Therefore, D S = A ∩ Σ = {1}, and the result follows from (c). f −1 (S)/A

For S ≤ G we will establish the relationship between M(S) and M(G). Theorem 4.6 (Schur). The following statements hold: (a) If (Γ, f, A) is a faithful central extension of a group G and S < G is of index n, then A(n) = {a ∈ A | GCD(o(a), n) = 1} is isomorphic to a subgroup of M(S). (b) Let S1 , . . . , S k be representatives of all the conjugate classes of Sylow subgroups of G. Then M(G) is isomorphic to a subgroup of the direct product M(S1 )×⋅ ⋅ ⋅×M(S k ). Proof. (a) Note that A(n) is a π(n)󸀠 -Hall subgroup of A. Recall that S < G is of index n, D S = Σ󸀠 ∩ A, where Σ = f −1 (S). Let a ∈ A(n) . By Lemma 4.5 (c) (retaining the above notation), a n ∈ Σ󸀠 ∩ A = D S 󳨐⇒ A(n) ≤ D S . Define a homomorphism f S : Σ → S by f S (g) = f(g). Since ker(f S ) = A ≤ Z(Σ), we obtain a central extension (Σ, f S , A) of the group S. Applying Lemma 4.4 (a) to this extension, we infer that D S is isomorphic to a subgroup of M(S); this proves (a), since A(n) ≤ D S . (b) Let (Γ, f, A) be a representation group of G; then A ≅ M(G) (Lemma 4.5 (a)). Let p be a prime dividing |M(G)| = |A|. Then p | |G| (Lemma 3.1). Let S ∈ Sylp (G). Then n = |G : S| is the p󸀠 -part of |G|. It is clear that A(n) = A p ∈ Sylp (A). Therefore, A p is isomorphic to a subgroup of M(S), by (a). If π(A) = {p1 , . . . , p k }, where A = A1 × ⋅ ⋅ ⋅ × A k , A i ∈ Sylp i (A), S i ∈ Sylp i (G), i = 1, . . . , k, then, by what has just been said, A = A1 × ⋅ ⋅ ⋅ × A k is isomorphic to a subgroup of M(S1 ) × ⋅ ⋅ ⋅ × M(S k ), proving (b).

VI Projective representations | 257

The following series of exercises enables one to find the multipliers of certain groups. Recall that a nonabelian group of order p n > p2 is said to be a p-group of maximal class if its nilpotency class is n − 1. Exercise 4.2. The following statements hold: (a) (L. Redei) If G is a minimal nonabelian p-group, then |G󸀠 | = p. (b) (O. Taussky) If G is a nonabelian 2-group of order 2n such that |G : G󸀠 | = 4, then G has a cyclic subgroup of index 2 and G ∈ {Q2n , D2n , SD2n }, i.e., G is a 2-group of maximal class. Solution. (a) If a, b ∈ G are noncommuting elements, then ⟨a, b⟩ = G, so that |G : Φ(G)| = Obviously, Φ(G) = Z(G), and so |G󸀠 | = 1p |G : Z(G)| = p. (b) One may assume that n > 3. We proceed by induction on n. Let R ≤ G󸀠 ∩ Z(G) be of order 2. By induction G/R has a cyclic subgroup T/R of index 2. Assume that T is noncyclic; then T is abelian of type (2n−2 , 2). Let R1 ≤ Φ(T) be of order 2. It is easy to see that R1 < G󸀠 ∩ Z(G). By induction, G/R1 must have a cyclic subgroup T1 /R1 of index 2. Clearly, T1 ≠ T are abelian so T ∩ T1 = Z(G). In that case, G is minimal nonabelian so |G󸀠 | = 2, by (a). Then |G| = |G󸀠 ||G : G󸀠 | = 8, contrary to the assumption. Thus, G has a cyclic subgroup of index 2, and now the result follows from the classification of 2-groups with cyclic subgroup of index 2, see [Ber31, Theorem 1.2]. p2 .

Exercise 4.3. If a p-group G contains a cyclic subgroup of index p and M(G) ≠ {1}, then either G is a dihedral 2-group or an abelian group of type (p, p), |M(G)| = p. Hint. If G contains two distinct cyclic subgroups of index p, then any representation group Γ of G is minimal nonabelian since it has two distinct abelian maximal subgroups and two generators, and therefore |Γ󸀠 | = p, by Exercise 4.2 (a); in that case, G is abelian of type (p, p). If G contains only one cyclic subgroup of index p, then p = 2 and |G : G󸀠 | = 4. But then |Γ : Γ󸀠 | = 4, and so Γ is of maximal class, by Exercise 4.2. Then M(G) = Z(Γ) and G ≅ G/M(G) is dihedral. Suppose that any Sylow subgroup of a group G contains only one subgroup of prime order. In that case (see [Ber31, Proposition 1.3]), M(S) = {1} (Exercises 3.1 and 4.3) so M(G) = {1}, by Theorem 4.6 (b). Exercise 4.4. If the Sylow subgroups of a group G are cyclic or 2-groups of maximal class, then |M(G)| ≤ 2. If, in addition, a 2-Sylow subgroup of G has exactly one involution, then M(G) = {1}. In particular, if p > 2, then |M(PSL(2, p))| = 2. Moreover, by Schur’s theorem (see [Sch1, Sch3]), a representation group of the group PSL(2, p), p > 2, is isomorphic to SL(2, p). Hint. Use Exercise 4.3 and Theorem 4.6 (a). Moreover, the group PSL(2, p), p > 2, has only one representation group, and it is isomorphic to the group SL(2, p) (Schur); see §IV.7.

258 | Characters of Finite Groups 1 Example 4.1. Let a nonnilpotent group G = ⟨a⟩ ⋅ G󸀠 , where o(a) = 7 and G󸀠 is elementary abelian of order 23 . Then G is a minimal nonabelian group. Let Γ be a representation group of G. Then |Γ| = 2s |G| (Theorem 4.6 (b)). Since Γ󸀠 ∈ Syl2 (Γ), it follows that Γ is a minimal nonnilpotent group, so s = 0 (Lemma XI.2.1). Therefore we have M(G) = {1}, though M(G󸀠 ) > {1}. Thus, in Theorem 4.6 (b), M(G) may be a proper subgroup of M(S1 ) × ⋅ ⋅ ⋅ × M(S k ). Example 4.2. Let n > 1 be odd, let p be a prime and let q be a prime divisor of p n − 1 such that q ∤ p i − 1 for i ∈ {1, . . . , n − 1} (such a prime divisor q exists by Zsigmondy’s theorem; see [BZ3, Chapter 30, Appendix B]). Let G be a subgroup of order qp n in the group AGL(1, p n ). By Theorem 4.6 (b), M(G) is a p-group. Let Γ be a representation group of G. Take H, a minimal nonnilpotent subgroup of Γ (see Lemma XI.2.1). Since Z(H) = {1} (Lemma XI.2.1), we get |H| = |G| and so H ∩ M(G) = {1}, whence Γ = H × M(G). As M(G) < Γ󸀠 , this implies M(G) = {1}. If Γ is a representation of a group G, then one has M(G) ≤ Φ(G) (if H ≤ Z(G) ∩ G󸀠 , then H ≤ Φ(G)). Exercise 4.5. Let Θ be a group-theoretic property inherited by subgroups, epimorphic images and central extensions. A minimal non-Θ-group is called a Θ1 -group. Prove that a representation group of a Θ1 -group is a Θ1 -group. Solution. Let G be a Θ1 -group and Γ its representation group. One may assume that M(G) > {1}. Assume that a maximal subgroup H of Γ does not contain M(G). Then Γ = HM(G). In that case, M(G) ≤ Γ󸀠 ≤ H since Γ/H is abelian, a contradiction. It follows that M(G) ≤ Φ(Γ). Let F < Γ be maximal; then M(G) < F so F/M(G) as a proper subgroup of the Θ1 -group Γ/M(G) ≅ G is a Θ-group. In that case, F as a central extension of a Θ-group is a Θ-group, by hypothesis, and we conclude that Γ is a Θ1 -group. In particular, a representation group of a minimal nonnilpotent group is minimal nonnilpotent.

5 The Schur multipliers of a group and its quotient group Given H ⊲ G, let us study the relationship between Schur multipliers M(G) and M(G/H) of G and G/H, respectively. The results of this section are due to the third author. We repeat the following: Definition 5.1. Given π ∈ Z2 (G, ℂ∗ ), a normal subgroup H of a group G is said to be a π-kernel if H is the kernel of some projective π-representation P of G (write H ∈ KER(P)). Recall that (see Exercise 1.2) ω π (s, t) =

π(s, t) π(t, s t )

(s, t ∈ G).

VI Projective representations | 259

Definition 5.2. Let H ⊲ G and let π ∈ Z2 (G, ℂ∗ ). A function λ : H → ℂ∗ is said to be a π-character of H if the following conditions are satisfied: λ(s)λ(t) = π(s, t)λ(st) t

−1

λ(s ) = ω π (s, t) λ(s)

(s, t ∈ H),

(1)

(s ∈ H, t ∈ G).

(2)

Of course, any ordinary G-invariant linear character of H is a π-character provided π is trivial. Let X π (H) be the set of all π-characters of H. If a factor set π is trivial, then 1H ∈ X π (H). However, if π is nontrivial, the set X π (H) may be empty. Let X π (H) ≠ 0. If λ, λ󸀠 ∈ X π (H), it is evident from (1) and (2) that λ/λ󸀠 ∈ Lininv (H), where Lininv (H) = {ψ ∈ Lin(H) | ψ(s t ) = ψ(s) (s ∈ H, t ∈ G)} is the group of G-invariant linear characters of H. Therefore, X π (H) ≠ 0 󳨐⇒ X π (H) ≅ Lininv (H) ≅ H/[H, G]. Let us prove this. If h ∈ H and g ∈ G, then ψ([h, g]) = ψ(h−1 )ψ(h g ) = 1 if and only if ψ is G-invariant. It follows that ψ ∈ Lininv (H) if and only if ψ[H,G] = 1[H,G] ; therefore (see Chapter I) Lininv (H) ≅ Lin(H/[H, G]) ≅ H/[H, G]. Lemma 5.1. The following statements hold: (a) H ⊴ G is a π-kernel of the group G if and only if X π (H) ≠ 0. (b) If H is a π-kernel of the group G, then H ⊆ G π , where G π is the set of π-elements of G (see Definition 3.4). (c) Let H be a π-kernel of G. If N ⊴ G and N ≤ H, then N is also a π-kernel. (d) If H is a π-kernel of G and π󸀠 ∼ π, then H is a π󸀠 -kernel. Proof. (a) Let H = KER(P), where P is a projective π-representation of G (see Definition 1.4). Then P(g) = λ(g)In

(g ∈ H), where n = deg(P), λ ∈ F[H, ℂ∗ ].

This, together with (1.1), implies (1). Applying (1.3), one obtains (2). Thus, X π (H) ≠ 0. Conversely, suppose that X π (H) ≠ 0 and let λ ∈ X π (H). Put (λ)

j H = |H|−1 ∑ λ(g)−1 u g , g∈H

where {u g }g∈G is the standard basis of the twisted group algebra ℂπ G. We extend λ−1 to a function μ ∈ F[G, ℂ∗ ] that vanishes on G − H and coincides with λ−1 on H. As a result, one obtains (λ)

j H = |H|−1 ∑ μ(g)u g . g∈G

260 | Characters of Finite Groups 1 Since μ(s t ) = ω π (s, t)μ(s) for s, t ∈ G, it follows that μ is a π-central function, and (λ) hence j H ∈ Z(ℂπ G), by Exercise 3.8. Let t ∈ H. Then, by (1), (λ)

(λ)

u t j H = j H u t = |H|−1 ∑ λ(s)−1 π(s, t)u st s∈H

−1

= |H|

(λ)

∑ λ(t)λ(st)−1 u st = λ(t)j H .

s∈H

Thus, (λ)

(λ)

u t j H = λ(t)j H

(3)

(t ∈ H).

(λ)

It follows from (3) that j H is an idempotent. (λ) (λ) Consider the subspace J = (ℂπ G)j H of the algebra ℂπ G. Since j H ∈ Z(ℂπ G), it follows that J is a two-sided ideal of ℂπ G. Let L be the linear representation of ℂπ G afforded by the module J relative to some basis. Then, putting P(g) = L(u g ) (g ∈ G), one obtains a projective π-representation P of the group G (see §VI.2). Let g ∈ H, x ∈ ℂπ G. By (3), (λ)

(λ)

(λ)

u g (xj H ) = (u g j H )x = λ(g)(xj H ), and so P(g) = λ(g)In , n = deg(P) = dim(J) 󳨐⇒ H ≤ KER(P). Let t ∈ KER(P). Then L(u t ) = P(t) = μIn (μ ∈ ℂ∗ ). It follows from the definition of L (λ) (λ) that u t j H = μj H , i.e., ∑ λ(s)−1 u t u s = μ ∑ λ(s)−1 u s . s∈H

s∈H

This gives ∑ λ(s)−1 π(t, s)u ts = μ ∑ λ(s)−1 u s .

s∈H

s∈H

Hence λ(e)−1 π(t, e)u te = u t is one of the terms of the sum μ ∑s∈H λ(s)−1 u s , i.e., t ∈ H. Thus KER(P) ≤ H, i.e., H = KER(P) and H is a π-kernel. (b) By (a), there exists a function λ ∈ X π (H). If s ∈ H and t ∈ CG (s), then, by (2), λ(s) = ω π (s, t)λ(s), hence ω π (s, t) = 1 and s ∈ G π . (c) Since the restriction λ N belongs to X π (N), (c) follows from (b). (d) This follows from Exercise 1.3. It follows from Lemma 5.1 (c) that the set Lπ (G) of all π-kernels of a group G is a semilattice. See Exercise 5.9, below, for a sufficient condition for Lπ (G) to be a lattice. Definition 5.3. A factor set π ∈ Z2 (G, ℂ∗ ) is said to be compatible with a normal subgroup H of a group G provided Hs = Hs󸀠 , Ht = Ht󸀠 󳨐⇒ π(s, t) = π(s󸀠 , t󸀠 )

(s, s󸀠 , t, t󸀠 ∈ G).

If π ∈ Z2 (G, ℂ∗ ), Π = [π] and H is a π-kernel, we shall also call H a Π-kernel (this is legitimate, by Lemma 5.1 (d)).

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Lemma 5.2. The following statements hold: (a) If Π ∈ M(G), then H ⊴ G is a Π-kernel if and only if Π = [π], where π is compatible with H. (b) If Π1 , Π2 ∈ M(G) and H ⊴ G is a Π i -kernel (i = 1, 2), then H is a Π1 Π2 -kernel. Proof. (a) Let ν be the natural homomorphism of the group G onto G = G/H. For every α ∈ G, choose an element g α ∈ ν−1 (α)(∈ G), g ε = 1, where ε is the identity element of G. Then G = ⋃ g α H, g α g β = g αβ h(α, β)

(α, β ∈ G, h(α, β) ∈ H).

α∈G

Let H be a Π-kernel, Π = [π]̇ for some π̇ ∈ Z2 (G, ℂ∗ ) and let P be a projective ̇ π-representation of G such that KER(P) = H. If s ∈ H, then one has P(s) = λ(s)In , ∗ where λ ∈ F[H, ℂ ] and n = deg(P). If Hs = Hg α , we put P󸀠 (s) = P(g α ). Moreover, let Ht = Hg β (s, t ∈ G). Then ̇ α , g β )P(g α g β ) = π(g ̇ α , g β )P(h(g α , g β )g αβ ) P󸀠 (s)P󸀠 (t) = P(g α )P(g β ) = π(g ̇ α , g β )π(h(α, ̇ = π(g β), g αβ )−1 P(h(α, β))P(g αβ ) ̇ α , g β )π(h(α, ̇ = π(g β), g αβ )−1 λ(h(α, β))P󸀠 (st), since h(α, β) ∈ H = KER(P). Consequently, P󸀠 (s)P󸀠 (t) = π(s, t)P󸀠 (st), where ̇ α , g β )π(h(α, ̇ π(s, t) = π(g β), g αβ )−1 λ(h(α, β)). This expression for π(s, t) shows that π is compatible with H. Since g = hg α , h ∈ H and α ∈ G, it follows that ̇ ̇ P(g) = P(hg α ) = π(h, g α )−1 P(h)P(g α ) = π(h, g α )−1 λ(h)In P(g α ) = μ(g)P󸀠 (g), ̇ where μ(g) = π(h, g α )−1 λ(h), and hence π ∼ π,̇ i.e., [π] = Π. Conversely, let Π = [π], where π is compatible with H. If s, t ∈ H, then one has π(s, t) = π(e, e); if s ∈ H, t ∈ G, then ω π (s, t) = π(s, t)π(t, s t )−1 = π(e, t)π(t, e)−1 = 1 ∈ ℂ (see the text after Exercise 1.3). Therefore, we can rewrite (1) and (2) as follows: λ(s)λ(t) = π(e, e)λ(st) t

λ(s ) = λ(s)

(s, t ∈ H), (s ∈ H, t ∈ G).

To satisfy these relations, we take λ(s) = π(e, e) (s ∈ H). Therefore H is a π-kernel (Lemma 5.1 (a)). (b) By what we have already proved, Π i = [π i ], where π i is a factor set compatible with H (i = 1, 2). Since Π1 Π2 = [π1 π2 ] and the factor set π1 π2 is compatible with H, it follows that H is a Π1 Π2 -kernel, by (a).

262 | Characters of Finite Groups 1 It follows from Lemma 5.2 (b) that the classes Π of factor sets for which H is a Π-kernel form a subgroup MH (G) of M(G), called the H-multiplier of G. Suppose that H ⊴ G. Let us establish the relationship between MH (G) and M(G), where G = G/H. We define the following homomorphisms: φ : Lin(G) → Lininv (H),

τ : Lininv (H) → M(G),

σ : M(G) → MH (G)

and prove that im(φ) = ker(τ),

im(τ) = ker(σ),

im(σ) = MH (G).

In other words, we prove that the sequence φ

τ

σ

Lin(G) → Lininv (H) → M(G) → MH (G) → {1} is exact. Let χ ∈ Lin(G). Then the mapping χ 󳨃→ χ H is a homomorphism of Lin(G) into Lininv (H). Denote this homomorphism by φ : φ(χ) = χ H (χ ∈ Lin(G); here χ H is the restriction of χ to H). Let ψ ∈ Lininv (H). Retaining the notation of the proof of Lemma 5.2, put η(ψ) (α, β) = ψ(h(α, β))

(α, β ∈ G).

Using the invariance of ψ and the relation h(α, βγ)h(β, γ) = h(αβ, γ)h(α, β)g γ

(α, β, γ ∈ G)

(which follows from the equality g α g β = g αβ h(α, β)), we obtain, applying ψ to the displayed relation, η(ψ) (α, βγ)η(ψ) (β, γ) = η(ψ) (αβ, γ)η(ψ) (α, β). Since h(ε, ε) = 1 (as implied by g ε = 1), we have η(ψ) ∈ Z2 (G, ℂ∗ ). The mapping ψ 󳨃→ [η(ψ) ] is a homomorphism of Lininv (H) into M(G). Denoting it by τ, we get τ(ψ) = [η(ψ) ] (ψ ∈ Lininv (H)). It is clear that τ is independent of the transversal {g α } of G over H. Let ρ ∈ Z2N (G, ℂ∗ ). Define a function π ρ : G × G → ℂ∗ by π ρ (s, t) = ρ(ν(s), ν(t))

(s, t ∈ G),

(4)

where ν : G → G = G/H is the natural homomorphism. It is clear that π ρ ∈ Z2N (G, ℂ∗ ) (the set of normalized factor sets). Equality (4) implies that π ρ is compatible with H, and so [π ρ ] ∈ MH (G). If ρ, ρ󸀠 ∈ Z2N (G, ℂ∗ ), [ρ] = [ρ󸀠 ], then it follows from (4) that [π ρ ] = [π ρ󸀠 ]. Hence the mapping σ : [ρ] 󳨃→ [π ρ ] from M(G) into MH (G) given by σ([ρ]) = [π ρ ]

(ρ ∈ Z 2N (G, ℂ∗ ))

is well defined. Thus, we have defined three mappings φ, τ and σ.

(5)

VI Projective representations | 263

Lemma 5.3. Let H ⊴ G,

ψ ∈ Lininv (H),

G = G/H,

λ ∈ F[G, ℂ∗ ]

and define a function χ : G → ℂ∗ by χ(g α h) = λ(α)ψ(h)

(α ∈ G, h ∈ H).

Then η(ψ) (α, β) = λ(α)λ(β)λ(αβ)−1 χ(st)χ(s)−1 χ(t)−1

(s, t ∈ G, α = ν(s), β = ν(t)),

where ν : G → G = G/H is the natural homomorphism. Proof. Putting s = g α h1 , t = g β h2 (h1 , h2 ∈ H), we obtain χ(s) = λ(α)ψ(h1 ), χ(t) = λ(β)ψ(h2 ) 󳨐⇒ χ(s)χ(t) = λ(α)λ(β)ψ(h1 )ψ(h2 ). As

gβ

st = g α h1 g β h2 = g αβ h(α, β)h1 h2 , we get, in view of the invariance of ψ, that gβ

χ(st) = λ(αβ)ψ(h(α, β)h1 h2 ) = λ(αβ)η(ψ) (α, β)ψ(h1 )ψ(h2 ). Comparing the expressions obtained for χ(st) and χ(s)χ(t), we obtain the required equality. Lemma 5.4. One has im(φ) = ker(τ). Proof. Let ψ ∈ im(φ). Then ψ = χ H , where χ ∈ Lin(G). In that case, g α g β = g αβ h(α, β) 󳨐⇒ χ(g α )χ(g β ) = χ(g αβ )ψ(h(α, β)) = χ(g αβ )η(ψ) (α, β). Hence η(ψ) (α, β) = λ(α)λ(β)λ(αβ)−1 ,

where λ(γ) = χ(g γ ),

γ ∈ G.

Therefore, [η(ψ) ] = [1], i.e., τ(ψ) = [1] and ψ ∈ ker(τ). Thus, im(φ) ≤ ker(τ). Now let ψ ∈ ker(τ). Then [η(ψ) ] = [1], i.e., η(ψ) (α, β) = λ(α)λ(β)λ(αβ)−1

(α, β ∈ G, λ ∈ F[G, ℂ∗ ]).

Define a function χ : G → ℂ∗ by χ(g) = λ(α)ψ(h)

(g = g α h, α ∈ G, h ∈ H).

Then χ(st) = χ(s)χ(t) for s, t ∈ G (Lemma 5.3) so that χ ∈ Lin(G). If h ∈ H, then χ(h) = χ(g ε h) = λ(ε)ψ(h) = ψ(h) (since η(ψ) (ε, ε) = 1, we have λ(ε) = 1). Therefore, χ H = ψ, i.e., ψ ∈ im(φ). Thus, we get ker(τ) ≤ im(φ), so im(φ) = ker(τ).

264 | Characters of Finite Groups 1 Lemma 5.5. One has im(τ) = ker(σ). Proof. Let ρ ∈ Z2N (G, ℂ∗ ), [ρ] ∈ im(τ). Then [ρ] = τ(ψ) = [η(ψ) ]

(ψ ∈ Lininv (H).

Define a function χ : G → ℂ∗ , putting χ(g) = ψ(h)

(g = g α h, α ∈ G, h ∈ H).

Writing this equality as χ(g) = λ(α)ψ(h)

(λ(α) = 1 for all α ∈ G),

and applying Lemma 5.3, we obtain η(ψ) (α, β) = χ(st)χ(s)−1 χ(t)−1

(s, t ∈ G, α = ν(s), β = ν(t)).

This implies π η(ψ) (s, t) = η(ψ) (ν(s), ν(t)) = μ(s)μ(t)μ(st)−1 , where μ(g) = χ(g)−1 (g ∈ G). Consequently, [π η(ψ) ] = [1]. Since [π η(ψ) ] = σ([η(ψ) ]) and [η(ψ) ] = [ρ], it follows that σ([ρ]) = [1], i.e., [ρ] ∈ ker(σ). Thus, im(τ) ≤ ker(σ). Let ρ ∈ Z2 (G, ℂ∗ ) be such that [ρ] ∈ ker(σ). Then [π ρ ] = [1], i.e., π ρ (s, t) = χ(s)χ(t)χ(st)−1

(s, t ∈ G),

where χ ∈ F[G, ℂ∗ ]. Put ψ = χ H , λ(α) = χ(g α ) (α ∈ G). Since π ρ is compatible with H, we have for h ∈ H and g ∈ G, χ(gh) = χ(g)χ(h)π ρ (g, h)−1 = χ(g)χ(h) = χ(g)ψ(h) = χ(hg). Hence χ(gh) = χ(hg) = χ(g)ψ(h)

(g ∈ G, h ∈ H).

(6)

Putting g = h1 , h = h2 (h1 , h2 ∈ H) in (6), we get ψ(h1 h2 ) = χ(h1 h2 ) = χ(h1 )ψ(h2 ) = ψ(h1 )ψ(h2 ). Therefore ψ ∈ Lin(H). Next, for h ∈ H and g ∈ G we have ψ(h)χ(g) = χ(hg) = χ(g ⋅ h g ) = χ(g)ψ(h g ). Consequently, ψ(h g ) = ψ(h), i.e., ψ ∈ Lininv (H). Putting g = g α (α ∈ G) in (6), we obtain χ(g α h) = χ(g α )ψ(h) = λ(α)ψ(h). Therefore, Lemma 5.3 gives η(ψ) (α, β) = λ(α)λ(β)λ(αβ)−1 χ(s)−1 χ(t)−1 χ(st) = λ(α)λ(β)λ(αβ)−1 π ρ (s, t),

VI Projective representations | 265

where s, t ∈ G, α = ν(s), β = ν(t). Observe that π ρ (s, t) = ρ(ν(s), ν(t)) = ρ(α, β) 󳨐⇒ η(ψ) (α, β) = λ(α)λ(β)λ(αβ)−1 ρ(α, β)−1 , and this means that [η(ψ) ] = [ρ−1 ]. Therefore, [ρ] = [η(ψ) ]−1 = τ(ψ)−1 = τ(ψ−1 ) ∈ im(τ). Thus, ker(σ) ≤ im(τ). Lemma 5.6. One has im(σ) = MH (G). Proof. Let Π ∈ MH (G). Then, by Lemma 5.2, Π = [π], where π ∈ Z2 (G, ℂ∗ ) is compatible with H. Define a function ρ : G × G → ℂ∗ by ρ(α, β) = π(g α , g β )

(α, β ∈ G).

Using the compatibility of π with H, it is easy to check that ρ(α, β)ρ(αβ, γ) = ρ(β, γ)ρ(α, βγ) i.e., ρ ∈

Z2 (G, ℂ∗ ).

(α, β, γ ∈ G),

Since ρ(ε, ε) = π(g ε , g ε ) = π(e, e) = 1, we get ρ ∈ Z2N (G, ℂ∗ ). Let

s, t ∈ G,

s = g α h1 ,

t = g β h2

(α, β ∈ G, h1 , h2 ∈ H).

Then π(s, t) = π(g α , g β ) = ρ(ν(s), ν(t)) = π ρ (s, t), where ν : G → G = G/H is the natural homomorphism. Therefore, we have π = π ρ , i.e., Π = [π] = [π ρ ] = σ([ρ]). Consequently, im(σ) = MH (G). Lemmas 5.4–5.6 imply the following: Theorem 5.7. The following sequence is exact: φ

τ

σ

Lin(G) → Lininv (H) → M(G) → M H (G) → {1}. Lemma 5.8. If X is a finite group and X 󸀠 ≤ Y ≤ Z ≤ X, then LinY (X)/LinZ (X) ≅ Z/Y. Proof. Define a homomorphism f : LinY (X) → LinY (Z), putting f(ψ) = ψ Z

(ψ ∈ LinY (X)).

It follows from ker(f) = LinZ (X) that im(f) ≅ LinY (X)/LinZ (X), and hence |im(f)| =

|LinY (X)| |Lin(X/Y)| |X/Y| = = = |Z/Y| = |LinY (Z)|. |LinZ (X)| |Lin(X/Z)| |X/Z|

Therefore, im(f) = LinY (Z), and by (7) and the commutativity of Z/Y, we have LinY (X)/LinY (Z) ≅ LinY (Z) ≅ Lin(Z/Y) ≅ Z/Y.

(7)

266 | Characters of Finite Groups 1 Corollary 5.9. In the notation of Lemma 5.8, Lin(X)/LinZ (X) ≅ Z/X 󸀠 . Proof. Put, in Lemma 5.8, Y = X 󸀠 and observe that LinX󸀠 (X) = Lin(X). Theorem 5.10. Let H ⊴ G, G = G/H. Then there exists a subgroup N ≤ M(G) = M(G/H) such that the following hold: (a) One has MH (G) ≅ M(G)/N (see the paragraph following Lemma 5.2, for the definition of MH (G)). (b) Let Linprol (H) be the group of linear characters of H extendible to G. Then N ≅ Lininv (H)/Linprol (H) ≅ (H ∩ G󸀠 )/[H, G]. Proof. By Theorem 5.7, MH (G) = M(G)/N, where N = ker(σ) = im(τ) ≅ Lininv (H)/ker(τ) = Lininv (H)/im(φ). Since im(φ) ≅ Lin(G)/ker(φ) = Lin(G)/LinH (G) = Lin(G)/LinHG󸀠 (G) and G󸀠 ≤ HG󸀠 ≤ G, it follows, by Corollary 5.9, that im(φ) ≅ HG󸀠 /G󸀠 ≅ H/(H ∩ G󸀠 ) ≅ LinH∩G󸀠 (H). Next, im(φ) ≤ LinH∩G󸀠 (H) 󳨐⇒ im(φ) = LinH∩G󸀠 (H). Thus, N ≅ Lininv (H)/LinH∩G󸀠 (H) = Lin[H,G] (H)/LinH∩G󸀠 (H), and hence, by Lemma 5.8, N ≅ (H ∩ G󸀠 )/[H, G]. This gives (a) and the first part of (b). It remains to observe that N ≅ Lininv (H)/im(φ) and im(φ) = Linprol (H). Corollary 5.11. If M(G) = {1} and H ⊴ G, then M(G/H) ≅ (H ∩ G󸀠 )/[H, G]. If M(G) > {1}, then |(H ∩ G󸀠 ) : [H, G]| divides |M(G/H)|. In particular, if M(G/H) = {1}, then all G-invariant linear characters of H are extendible to G. Corollary 5.12. If H ⊴ G, then the following conditions are equivalent: (a) MH (G) ≅ M(G/H). (b) H ∩ G󸀠 = [H, G]. (c) Lininv (H) = Linprol (H). Corollary 5.13. Let G = K ⋅ H, a semidirect product with kernel H. Then the following statements hold: (a) M(K) = MH (G). (b) If |H| = n and H is a Hall subgroup of G, then M(K) = M(G)(n) is the π(n)󸀠 -part of the (abelian) group M(G). In particular, if M(G) = {1}, then M(K) = {1}.

VI Projective representations | 267

Proof. (a) Let λ ∈ Lininv (H), μ ∈ Lin(K). Define a mapping χ : G → ℂ∗ by χ(hk) = λ(h)μ(k)

(h ∈ H, k ∈ K). k−1

If g i = h i k i (h i ∈ H, k i ∈ K, i = 1, 2), then g1 g2 = h1 h21 k1 k2 . Hence k−1

χ(g1 g2 ) = λ(h1 h21 )μ(k1 k2 ) = λ(h1 )λ(h2 )μ(k1 )μ(k2 ) = χ(g1 )χ(g2 ), i.e., χ ∈ Lin(G), χ H = λ, so that Lininv (H) = Linprol (H). By Corollary 5.12 (a), MH (G) ≅ M(G/H) ≅ M(K). (b) By (a), M(K) is isomorphic to a subgroup of M(G). Since exp(M(K)) divides |K| (Theorem 4.2), it follows that GCD(n, |M(K)|) = 1, i.e., M(K) is isomorphic to a subgroup of M(G)(n) , which is isomorphic to a subgroup of M(K), by Theorem 4.6. Thus, M(K) = M(G)(n) . Now the last assertion follows. Exercise 5.1. If H 󸀠 = H ⊴ G, then one has MH (G) ≅ M(G/H). If, besides, M(G) = {1}, then M(G/H) = {1}. Exercise 5.2. Find the multiplier of the group G = ⟨a, b | a4 = b4 = 1, a b = a3 ⟩. Hint. By [Ber31, Remark 148.53], |M(G)| = 2. Exercise 5.3. Show that |M(D2n )| = 1 if n is odd, and |M(D2n )| = 2 if n is even. Hint. The result also follows from Taussky’s theorem if n is a power 2. See also Theorem 4.6 (b). Exercise 5.4. The multiplier of an elementary abelian p-group is an elementary abelian p-group (this also follows from Lemma 3.1). Exercise 5.5. The multiplier of the special linear group SL(2, p) (p > 2) is trivial. Actually, if |M(SL(2, p n ))| > {1}, then p n ∈ {4, 9} (Alperin–Gorenstein [AG]). Exercise 5.6. If p > 2, then |M(PSL(2, p))| = 2. Recall that X π (H) = H π is the set of π-elements of a group H, where π ∈ Z2 (H, ℂ∗ ). Lemma 5.14. Suppose that π ∈ Z2 (G, ℂ∗ ), K π = G π ∩ Z(G) and let P be an irreducible projective π-representation of the group G. Then the following statements hold: (a) KER(P) ∩ Z(G) = K π . (b) K π is the greatest π-kernel of G contained in Z(G). Proof. If s ∈ G and t ∈ CG (s), then, by (1.3), (π)

[P(s), P(t)] = ω π (s, t)In = χ s (t)In ,

n = deg(P),

ω π (s, t) = (π)

π(s, t) (π) = χ s (t). π(t, s)

In particular, if s ∈ G π , then [P(s), P(t)] = In since then χ s = 1 (see Definition 3.4). Thus, s ∈ G π , t ∈ CG (s) 󳨐⇒ P(s)P(t) = P(t)P(s).

268 | Characters of Finite Groups 1 Therefore, we obtain [P(s), P(t)] = In for every t ∈ G provided s ∈ G π ∩ Z(G) = K π . Since P is irreducible, it follows, by Schur’s lemma, that P(s) is a scalar matrix, i.e., s ∈ KER(P) ∩ Z(G). Thus, K π ≤ KER(P) ∩ Z(G) so that K π is a π-kernel (Lemma 5.1 (c)). Since KER(P) ∩ Z(G) ≤ K π (Lemma 5.1 (b)), we get KER(P) ∩ Z(G) = K π . It follows that K π is the greatest π-kernel of G contained in Z(G). In conclusion we formulate some additional assertions on π-kernels (see [Zhm25]). Exercise 5.7. Let H be a normal subgroup that is a π-kernel of a group G, Further, let (λ) π ∈ Z2 (G, ℂ∗ ), λ ∈ X π (H), and let j H = |H|−1 ∑g∈H λ(g)−1 u g be the idempotent of the algebra Z(ℂπ G) introduced in the proof of Lemma 5.1. Put (λ)

{∑ j j H = { λ∈X π (H) H 0 {

if H is a π-kernel, otherwise.

Then the following statements hold: (a) Let e = ∑g∈G ξ(g)u g ∈ Z(ℂπ G) satisfies e2 = e. Then e = 0 if and only if ξ(1) = 0. (b) j H is an idempotent of Z(ℂπ G), j H ≠ 0 if and only if H is a π-kernel. (c) If H1 , H2 ⊴ G, then j H1 j H2 = j H1 H2 . (d) If H1 , H2 are π-kernels, then H1 H2 is a π-kernel if and only if j H1 ⋅ j H2 ≠ 0. (e) If H is a π-kernel and λ ∈ X π (H), then λ[H,G] is independent of λ. (f) If H1 , H2 are π-kernels of G and λ i ∈ X π (H i ), i = 1, 2, then H1 H2 is a π-kernel if and only if (λ1 )D = (λ2 )D , where D = [H1 , G]∩[H2 , G]. In particular, if H1 ∩H2 = {1}, then H1 × H2 is a π-kernel. (g) If H is a π-kernel of G, then H ≤ Z(G) if and only if j H = {1}. Solutions and hints. (a) Let R be a matrix regular representation of the twisted group algebra ℂπ G in the standard basis {u g }. Then where α s,t (g) = δ s,gt π(g, s),

R(u g ) = (α s,t (g)),

(s, t ∈ G)

(δ is the Kronecker delta on G). Next, tr(R(u g )) = ∑ α s,s (g) = ∑ δ s,gs π(g, s) = δ1,g π(1, 1) s∈G

s∈G

and this implies tr(R(e)) = ∑ ξ(g) tr(R(u g )) = |G|ξ(1)π(1, 1). g∈G

Since the matrix R(e) satisfies R(e)2 = R(e) and R is faithful, then e = 0 ⇐⇒ R(e) = 0 ⇐⇒ tr(R(e)) = 0 ⇐⇒ ξ(1) = 0. (λ)

(b) Let H be a π-kernel of G. Prove that the idempotents j H (λ ∈ X π (H)) form an (λ) (μ) (λ) orthogonal system, i.e., j H j H = δ λ,μ j H (λ, μ ∈ X π (H)). Use the definition of j H .

VI Projective representations | 269

(c) Let ∆ π be the set of all minimal idempotents of the algebra Z(ℂπ G). Given e ∈ ∆ π , put N e = {g ∈ G | u g e ∈ ℂ∗ e}. Prove that N e is a π-kernel of G and u g e = λ e (g)e (g ∈ N e ), where λ e ∈ X π (N e ). Given H ⊴ G, put ∆ π (H) = {e ∈ ∆ π | H ≤ N e }. It follows from Lemma 5.1 (b) that ∆ π (H) ≠ 0 if and only if H is a π-kernel of G. Given a π-kernel H of G and λ ∈ X π (H), put ∆ π,λ (H) = {e ∈ ∆ π (H) | (λ e )H = λ}. It is clear that ∆ π (H) = ⋃λ∈X π (H) ∆ π,λ (H). Let H ⊴ G. Then j H = ∑e∈∆ π (H) e. Indeed, if H is not a π-kernel, then ∆ π (H) = 0 and, by (b), j H = 0 = ∑e∈∆ π (H) e. If H is a π-kernel and (λ) λ ∈ X π (H), it is easy to prove that j H = ∑e∈∆ π,λ (H) e. So jH =

(λ)

jH =

∑

∑

e

e∈∆ π (H)

λ∈X π (H)

since ∆ π (H) = ⋃λ∈X π (H) ∆ π,λ (H), and ∆ π,λ ∩∆ π,μ = 0 provided λ, μ ∈ X π (H) are distinct. Now, for H1 , H2 ⊴ G, j H1 j H2 = (

e)(

∑ e∈∆ π (H1 )

e) =

∑ e∈∆ π (H2 )

e=

∑ e∈∆ π (H1 )∩∆ π (H2 )

∑

e = j H1 H2 .

e∈∆ π (H1 H2 )

Statement (d) follows from (b) and (c). (e) If λ, μ ∈ X π (H), then μ = ψλ, where ψ ∈ Lininv (H). Therefore, λ[H,G] = μ[H,G] . (f) Let H be a π-kernel of G. Prove that j H = |[H, G]|−1

∑

for all λ ∈ X π (H).

λ−1 (g)u g

g∈[H,G]

Now, if H1 , H2 ⊴ G, then, by (d), H1 H2 is a π-kernel if and only if j H1 ⋅ j H2 ≠ 0. Let λ i ∈ X π (H i ), i = 1, 2. Then, by the above, j H1 j H2 = |[H1 , G]|−1 |[H2 , G]|−1 = |[H1 H2 , G]|−1

∑

s∈[H1 ,G],t∈[H2 ,G]

∑

s∈[H1 ,G],t∈[H2 ,G]

λ−1 (s)λ−1 (t)u s u t

−1 λ−1 1 (s)λ 2 (t)π(s, t)u st .

Let j H1 j H2 = ∑ ξ(g)u g , g∈G

ξ(g) ∈ ℂ, g ∈ G.

Then −1 −1 −1 ξ(1) = |[H1 H2 , G]|−1 ∑ λ−1 1 (s)λ 2 (s )π(s, s ), s∈D

where D = [H1 , G] ∩ [H2 , G]. Let (λ i )D = μ i , i = 1, 2. Then one has μ2 = μ1 ψ, where ψ ∈ Lininv (D). Therefore, −1 −1 −1 ξ(1) = |[H1 H2 , G]|−1 ∑ μ−1 1 (s)μ 2 (s )π(s, s ) s∈D

= |[H1 H2 , G]|−1 ∑ (μ1 (s)μ1 (s−1 ))−1 ψ(s)π(s, s−1 ). s∈D

270 | Characters of Finite Groups 1 Since μ1 (s)μ1 (s−1 ) = π(s, s−1 )μ1 (1), it follows that −1 −1 ξ(1) = |[H1 H2 , G]|−1 μ−1 1 (1)|D|⟨ψ, 1D ⟩ = |[H 1 H 2 , G]| μ 1 (1)δ ψ,1D

= δ μ1 ,μ2 |[H1 H2 , G]|−1 μ−1 1 (1)|D|. Therefore, ξ(1) ≠ 0 if and only if μ1 = μ2 . Since j H1 ⋅ j H2 is an idempotent of ℂπ G, we obtain from (a) that H1 H2 is a π-kernel if and only if μ1 = μ2 . If H1 ∩ H2 = {1}, then D = {1}, whence it follows that (λ1 )D = (λ s )D = π(1, 1). By the above, H1 × H2 is a π-kernel. (g) Recall that j H = |[H, G]|−1

∑

λ−1 (g)u g ,

where λ ∈ X π (H).

g∈[H,G]

If H ≤ Z(G), then one has j H = λ−1 (1)u1 = π(1, 1)−1 u1 = 1. Conversely, if j H = 1, then [H, G] = {1}, i.e., H ≤ Z(G), since the elements u g (g ∈ G) are linearly independent. Exercise 5.8. Let H1 , . . . , H m be maximal π-kernels of a group G. Then the following statements hold: (a) H i are kernels of irreducible π-representations of G, i ∈ {1, . . . , m}. −1 ≤ 1, with equality if and only if {H }m is a complete system of ker(b) ∑m i 1 i=1 |[H i , G]| nels of irreducible π-representations of G. (c) Let Scπ (G) be the product of all the minimal normal subgroups of G which are π-kernels. Prove that Scπ (G) is a π-kernel contained in ⋂m i=1 H i . In particular, if m > 1, then ⋂m H > {1}. i=1 i Exercise 5.9. The set Lπ (G) of π-kernels of G is a lattice if one of maximal π-kernels of G is contained in Z(G). Hint. Use Exercise 5.8 (b) and Lemma 5.14. Exercise 5.10. One has H1 , H2 ⊴ G 󳨐⇒ MH1 H2 (G) ≤ MH1 (G) ∩ MH2 (G) and H1 ∩ H2 = {1} 󳨐⇒ MH1 ×H2 (G) = MH1 (G) ∩ MH2 (G).

6 Projective representations of abelian groups In this section we assume that a group G is abelian. Definition 6.1. A function ω : G × G → ℂ∗ is said to be a bicharacter of a (multiplicative) abelian group G if ω(s, t) (s, t ∈ G) is a linear character when either of its arguments is held fixed. A bicharacter ω is said to be antisymmetric if ω(s, s) = 1 for all s ∈ G. We write Ba (G) to denote the set of all antisymmetric bicharacters of G.

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Bicharacters of a group can be multiplied pointwise. A bicharacter ω is said to be nontrivial if ω(s, t) ≠ 1 for some s, t ∈ G and trivial otherwise. The set Ba (G) is an abelian group with respect to this multiplication, its identity element is the trivial bicharacter. Henceforth we will consider only antisymmetric bicharacters. It follows from Definition 5.1 that values of (antisymmetric) bicharacters are roots of unity. Indeed, if ω ∈ Ba (G), then ω(s, t)o(s)o(t) = ω(s o(s)o(t) , t o(s)o(t) ) = ω(1, 1) = 1. If ω ∈ Ba (G), then, for all s, t ∈ G, 1 = ω(st, st) = ω(s, s)ω(t, t)ω(s, t)ω(t, s) = ω(s, t)ω(t, s) implies ω(s, t) = ω(t, s) since the values of ω are roots of unity. Two elements s, t ∈ G are said to be orthogonal with respect to ω ∈ Ba (G) if ω(s, t) = 1 (we write s ⊥ t or, what is the same, t ⊥ s since ω is antisymmetric). By definition, s ⊥ s for all s ∈ G. A group G (which is abelian always) with a given nontrivial ω ∈ Ba (G) is said to be symplectic. Definition 6.2. Let ω be a bicharacter of a group G. (a) Let H ≤ G. The set H 0 = {t ∈ G | ω(s, t) = 1 for all s ∈ H} is said to be the annihilator of H. We have H 0 ≤ G. Indeed, if s ∈ G, t1 , t2 ∈ H, then ω(s, t1 t2 ) = ω(s, t1 )ω(s, t2 ) = 1 ⋅ 1 = 1. (b) H1 , H2 ≤ G are said to be (mutually) orthogonal (we write H1 ⊥ H2 ), if H2 ≤ H10 (or H1 ≤ H20 , what is the same). (c) The set ker(ω) = G0 = {s ∈ G | ω(s, t) = 1 for all t ∈ G} is called the kernel of ω ∈ Ba (G). If ker(ω) = {1} (by the above, ker(ω) ⊥ G), then ω is said to be non-degenerate, otherwise it is degenerate. If ω ∈ B a (G) is non-degenerate, H < G and HH 0 = G, then H ∩ H 0 = {1}. Indeed, let x ∈ (H ∩ H 0 )# and y ∈ G; then y = hh0 , where h ∈ H and h0 ∈ H 0 . Using the linearity of ω by the second argument, one obtains ω(x, y) = ω(x, hh0 ) = ω(x, h)ω(x, h0 ) = 1 ⋅ 1 = 1 󳨐⇒ x ∈ ker(ω) = 1. Let ω ∈ B a (G) be non-degenerate. Put φ s (t) = ω(s, t) (s, t ∈ G); then φ s ∈ Lin(G) since ω is linear by the second argument. If φ s = 1G , the principal character of G, then s ∈ ker(ω) = {1} so that the mapping s 󳨃→ φ s is an isomorphism of G in Lin(G). We use freely this fact in the sequel.

272 | Characters of Finite Groups 1 Lemma 6.1. If ω ∈ Ba (G) is non-degenerate and H ≤ G, then the following statements hold: (a) H 0 ≅ G/H. (b) If the restriction ω H of ω to H is non-degenerate, then G = H × H 0 . Proof. (a) Let s ∈ G. Put φ s (t) = ω(s, t) (t ∈ G). Then, by the above, s 󳨃→ φ s is an isomorphism of G onto Lin(G). Set K = {φ s | s ∈ H 0 }. If χ ∈ Lin(G), χ = φ s (s ∈ G), then χ ∈ K ⇐⇒ s ∈ H 0 ⇐⇒ χ(h) = φ s (h) = ω(s, h) = 1 for all h ∈ H ⇐⇒ χ ∈ Ann(H). Conversely, if χ ∈ Ann(H), then χ ∈ K. Therefore, K = Ann(H) and hence (see Chapter I) K ≅ G/H. Since K ≅ H o , it follows that H o ≅ G/H. (b) Now let ω H be non-degenerate. Since |G| = |H| ⋅ |H 0 |, by (a), and H ∩ H 0 = {s ∈ H | ω(s, t) = 1 for all t ∈ H} = ker(ω H ) = {1}, part (b) follows immediately. Exercise 6.1. Let G = H × F, let ω ∈ B a (G) be non-degenerate and H ⊥ F. Then ker(ω) = ker(ω H ) × ker(ω F ). In particular, ω is non-degenerate if and only if ω H and ω F are non-degenerate. Exercise 6.2. If s, t ∈ G and GCD(o(s), o(t)) = 1, then ω(s, t) = 1. Solution. Let o(s) = m, o(t) = n; then GCD(m, n) = 1. In that case, o(st) = mn so that t = (st)km with GCD(k, n) = 1, and we have ω(st, t) = ω(st, (st)km ) = ω(st, st)km = 1km = 1. In that case, 1 = ω(st, t) = ω(s, t)ω(t, t) = ω(s, t) 󳨐⇒ s ⊥ t. By Exercise 6.1, ω ∈ Ba (G) is non-degenerate if and only if ω Q is non-degenerate for all Q ∈ Syl(G). Definition 6.3. An abelian group G is called a group of symmetric type if G = G1 × G2 with G1 ≅ G2 . If, in addition, G1 , G2 are cyclic p-groups of equal order, then G is said to be hyperbolic. It follows from the basic theorem on abelian groups that G is of symmetric type if and only if it is a direct product of hyperbolic subgroups. All Sylow subgroups of such G are of symmetric type as well. Lemma 6.2. A non-degenerate bicharacter ω ∈ Ba (G) exists if and only if G is of symmetric type.

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Proof. (a) (Yamasaki and Zhmud, independently) Let ω ∈ Ba (G) be non-degenerate. We have to prove that G is of symmetric type. Then ω Q is non-degenerate for all Q ∈ Syl(G), by Exercises 6.1 and 6.2. Therefore, one may assume that exp(G) = p σ . Putting φ s (t) = ω(s, t) (s, t ∈ G), we obtain an isomorphism s 󳨃→ φ s of G onto Lin(G) (see the proof of Lemma 6.1 (a)). Consequently, if o(s) = p σ , then o(φ s ) = p σ . Therefore, there exists t ∈ G such that ω(s, t) = φ s (t) = ε is a primitive p σ -th root of unity. Let s i = t j . Then ε i = ω(s, t)i = ω(s i , t) = ω(t j , t) = ω(t, t)j = 1j = 1, so that i ≡ 0 (mod p σ ), and so s i = 1. Thus, ⟨s⟩ ∩ ⟨t⟩ = {1}. On the other hand, since φ s (t) = ε is a primitive p σ -th root of unity, it follows that o(t) = p σ . Therefore, the subgroup H = ⟨s⟩ × ⟨t⟩ is a hyperbolic subgroup. We claim that ω H is non-degenerate. Assume that s i t j ∈ ker(ω H ). Then 1 = ω(s i t j , s) = ω(s i t j , t) 󳨐⇒ ω(t, s)j = 1 = ω(s, t)i 󳨐⇒ i ≡ j ≡ 0 (mod p σ ), s i t j = 1. Hence ω H is non-degenerate. Then, by Lemma 6.1 (b), G = H × H 0 . By Exercise 6.1, ω H 0 is non-degenerate. Then, by induction, H 0 is of symmetric type. It follows that G is also of symmetric type since the hyperbolic group H is of symmetric type (see Definition 6.3). (b) Let the (abelian) group G be of symmetric type. We prove that there exists a non-degenerate bicharacter in Ba (G). First, let G = ⟨s⟩ × ⟨t⟩ be hyperbolic, o(s) = p σ = o(t), and let ε be a primitive p σ -th root of unity. Given x = s i t j , y = s k t l , put ω(x, y) = ε il−jk . One has ω(y, x) = ϵ jk−il = ω(x, y) so that this definition is legitimate and ω ∈ Ba (G). Let x ∈ ker(ω). In that case, ϵ−j = ω(x, s) = 1 = ω(x, t) = ϵ i , and hence ε−j = ε i = 1, i ≡ j ≡ 0 (mod p σ ) 󳨐⇒ x = 1, i.e., so constructed bicharacter ω is non-degenerate. Now let G = H1 × ⋅ ⋅ ⋅ × H k be a direct product of hyperbolic subgroups. By what has just been proved, there exists a non-degenerate ω i ∈ Ba (H i ) for each i. Let s, t ∈ G,

s = s1 . . . s k ,

t = t1 . . . t k ,

si , ti ∈ Hi

(i ∈ {1, . . . , k}).

Putting ω(s, t) = ω1 (s1 , t1 ) . . . ω k (s k , t k ), we obtain a non-degenerate bicharacter ω ∈ Ba (G). Let us establish the relationship between the projective representations of an abelian group G and its antisymmetric bicharacters.

274 | Characters of Finite Groups 1 Let π ∈ Z2 (G, ℂ∗ ) and let ω π : G × G → ℂ∗ be the function given by (1.3). As G is abelian, we have π(s, t) . ω π (s, t) = π(t, s) It follows from (3.7) and from the equality ω π (s, s) = 1 (s ∈ G) that ω π ∈ Ba (G). The relation π ∼ π󸀠 implies ω π = ω π󸀠 . We may therefore write ω[π] or ω Π instead of ω π if Π = [π]. It is evident from the formula for ω π that the mapping π 󳨃→ ω π is a homomorphism from Z2 (G, ℂ∗ ) into Ba (G). In fact, we shall prove the following result. Theorem 6.3. The mapping Π 󳨃→ ω Π is an isomorphism of the Schur multiplier M(G) onto Ba (G), M(G) ≅ Ba (G). We need the following lemma. Lemma 6.4. Let π ∈ Z2 (G, ℂ∗ ). Then the following statements hold: (a) ker(ω π ) = G π , the set of π-elements of the group G. (b) ker(ω π ) is the greatest π-kernel of G. (c) All irreducible π-representations of G have the same kernel, namely, ker(ω π ) = K π . Proof. Statement (a) follows from Definition 3.4, (b) and (c) follow from (a) and Lemma 5.14. Proof of Theorem 6.3. Let Π ∈ M(G), Π = [π], where π ∈ Z2 (G, ℂ∗ ). If ω Π = 1, then ω π = 1, i.e., ker(ω π ) = G. By Lemma 6.4, G is a π-kernel of G. Therefore, in view of (1.1), a function λ : G → ℂ∗ satisfying λ(s)λ(t) = π(s, t)λ(st) for any s, t ∈ G does exist. It follows that π ∼ 1, Π = [1]. Hence, the homomorphism Π 󳨃→ ω Π is injective. We will now prove that it is surjective. Choosing a basis {b i }1r of G, put ω ij = ω(b i , b j ), where ω ∈ Ba (G). Let s, t ∈ G,

α

α

s = b 11 . . . b r r ,

β

β

t = b11 . . . b r r .

Since α i , β i are uniquely determined by the elements s, t modulo o(b i ), and since o(b i )

ω ij

o(b j )

= ω ij

= 1,

αi βj it follows that the number ∏1≤j≤r ω ij is also uniquely can define a function π ∈ Z2 (G, ℂ∗ ) by putting αi βj

π(s, t) = ∏ ω ij . 1≤i ε r . If W is a cyclic subgroup of G such that G/W is of symmetric type, then |W| = p σ ,

where σ = ε1 − ε2 + ⋅ ⋅ ⋅ + (−1)r−1 ε r .

Proof. One may assume that r > 1. Assume that the lemma has been proved for groups of order < |G|. Write G = G/W. Claim (i). Suppose that ε r > 1. Then ℧1 (G)(= Φ(G)) is of type (p ε1 −1 , . . . , p ε r −1 ). Assume that r is even. Then W ≤ ℧1 (G) (otherwise, d(G) = r−1 is odd, by Exercise 6.5, so G/W is not of symmetric type). In that case, ℧1 (G)/W = ℧1 (G) = ℧1 (G). Since G is of symmetric type, by hypothesis, the group ℧1 (G) = ℧1 (G)/W is also of symmet󸀠 ric type. Since |℧1 (G)| < |G|, it follows, by induction applied to ℧1 (G), that |W| = p σ , where, in view of evenness of r, one obtains σ󸀠 = (ε1 − 1) − (ε2 − 1) + ⋅ ⋅ ⋅ + (−1)r−1 (ε r − 1) = σ. Now let r be odd. Then W ≰ ℧1 (G) (Exercise 6.5), W ∩ ℧1 (G) = ℧1 (W) and ℧1 (G) = ℧1 (G) = ℧1 (G)W/W ≅ ℧1 (G)/(℧1 (G) ∩ W) ≅ ℧1 (G)/℧1 (W). Therefore, since ℧1 (G) is of symmetric type, ℧1 (G)/℧1 (W) is also of symmetric type. 󸀠 Since |℧1 (G)| < |G|, we have, by induction, that |℧1 (W)| = p σ , where, in view of oddness of r, one obtains σ󸀠 = (ε1 − 1) − (ε2 − 1) + ⋅ ⋅ ⋅ + (ε r − 1) = σ − 1, 󸀠

and hence |W| = p|℧1 (W)| = p σ +1 = p σ . Claim (ii). Suppose that ε r = 1. In that case ℧1 (G) is of type (p ε1 −1 , . . . , p ε r−1 −1 ).

278 | Characters of Finite Groups 1 If r is even, then W ≤ ℧1 (G) (Exercise 6.5) and, as in (i), we get, by induction, logp |W| = (ε1 − 1) − (ε2 − 1) + ⋅ ⋅ ⋅ + (ε r−1 − 1) = ε1 − ε2 + ⋅ ⋅ ⋅ + ε r−1 − 1 = ε1 − ε2 + ⋅ ⋅ ⋅ + ε r−1 − ε r = σ since ε r = 1. If r is odd, then W ∩ ℧1 (G) = ℧1 (W) (since W ≰ Φ(G), by Exercise 6.5) and, as in (i), we get, by induction, logp |℧1 (W)| = (ε1 − 1) − (ε2 − 1) + ⋅ ⋅ ⋅ − (ε r−1 − 1) = ε1 − ε2 + ⋅ ⋅ ⋅ − ε r−1 = σ − 1, so that logp (|W|) = 1 + logp (|℧1 (W)|) = σ. Lemma 6.10. Any abelian group G possesses a cyclic subgroup W such that quotient group G/W is of symmetric type. Proof. Let G = S1 × ⋅ ⋅ ⋅ × S k , where S i ∈ Syl(G). If S i contains a cyclic subgroup W i such that S i /W i is of symmetric type (i ∈ {1, . . . , k}), then W = W1 × ⋅ ⋅ ⋅ × W k is cyclic and G/W is of symmetric type. It is therefore sufficient to consider the case in which G is a p-group. Then G = G1 × G2 , where G1 is of symmetric type and G2 has pairwise distinct invariants. If there exists a cyclic subgroup W of G2 such that G2 /W is of symmetric type, then G/W ≅ G1 × (G2 /W) is of symmetric type. It is therefore sufficient to consider the case in which G is of type (p ε1 , . . . , p ε r ), where ε1 > ε2 > ⋅ ⋅ ⋅ > ε r . Let {b i }1r be a basis of G, o(b i ) = p ε i (i ∈ {1, . . . , r}). Since the lemma is obvious for r = 1, we may assume that r ≥ 2. It is easy to check the theorem for r = 2 (this case we consider later). Let r ≥ 3. Put B = ⟨b1 , b2 ⟩, H = ⟨b3 , . . . , b r ⟩ 󳨐⇒ G = B × H. By induction, H contains a cyclic subgroup U = ⟨u⟩ such that H/U is of symmetric type and logp (|U|) = ε3 − ε4 + ⋅ ⋅ ⋅ + (−1)r−1 ε r (by Lemma 6.9; note that (−1)r−3 = (−1)r−1 ). Set W = ⟨w⟩,

r

where w = b1τ u, τ = p ε2 −ε3 +⋅⋅⋅+(−1) ε r .

We shall prove that G = G/W is of symmetric type. Let x 󳨃→ x be the natural homomorphism of G onto G. Let ρ be the projection of G onto H corresponding to the direct decomposition G = B × H. Now G/B = (G/W)/(BW/W) ≅ G/BW and BW = ρ−1 (U) 󳨐⇒ G/B ≅ H/U, so that G/B is of symmetric type. Therefore, to complete the proof, it is sufficient to verify the following statements: (a) B is of symmetric type. (b) B is a direct factor of G. (c) o(b1 ) = o(b2 ) = p ε2 = exp(G). (d) B = ⟨b1 ⟩ × ⟨b2 ⟩.

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The equality o(b2 ) = p ε2 follows from ⟨b2 ⟩ ∩ W = {1}. Let us find o(b1 ). We have wp

ε3 −ε4 +⋅⋅⋅+(−1)r−1 ε r

p ε2

= b1

p ε2

󳨐⇒ b1

∈W

(the element w is defined in the second paragraph of the proof). Let λ be the minimal positive integer such that b1λ ∈ W. Then μp ε2 −ε3 +⋅⋅⋅+(−1)

b1λ = w μ (μ ∈ ℤ) 󳨐⇒ b1λ = b1

rε

r

⋅ uμ .

Therefore, u μ ∈ B ∩ H = {1} 󳨐⇒ u μ = 1. This gives r

μ ≡ 0 (mod p ε3 −ε4 +⋅⋅⋅+(−1) ε r ),

i.e.,

r

μ = cp ε3 −ε4 +⋅⋅⋅+(−1) ε r ,

c ∈ ℤ.

We conclude that cp ε2

b1λ = b1

󳨐⇒ λ ≡ cp ε2 (mod p ε1 ) 󳨐⇒ λ ≡ 0 (mod p ε2 ).

Thus, o(b1 ) = p ε2 . α β β Assume that b1 = b2 (α, β ∈ ℤ). Then b1α = b2 w γ with γ ∈ ℤ. Therefore, γp ε2 −ε3 +⋅⋅⋅+(−1)

b1α = b1

r ]ε

r

β

β

β

u γ b2 󳨐⇒ b2 = 1, b2 = 1 󳨐⇒ ⟨b1 ⟩ ∩ ⟨b2 ⟩ = {1}.

It follows that B = ⟨b1 ⟩ × ⟨b2 ⟩, and (d) is proved. But then (c) is also true, as well as (a) and (b) (for example, (b) follows from Exercise 6.4). Since G/B is of symmetric type, so is G. If r = 2, we set W = ⟨w⟩, where w = b1τ , τ = p ϵ2 . As above, we prove that G = G/W is of symmetric type: G = ⟨b1 ⟩ × ⟨b2 ⟩, o(b1 ) = o(b2 ) = p ϵ2 = exp(G)⟩. Exercise 6.6 (Zhmud). If W1 and W2 are two cyclic subgroups of an abelian group G such that G/W1 and G/W2 are of symmetric type, then W2 = W1α for some α ∈ Aut(G). Now we will establish some correspondences between the irreducible π-representations of an abelian group G and the linear characters of the π-kernel K π = G π (see Lemma 6.4 (c)). Let π ∈ Z2N (G, ℂ∗ ) (the set of normalized 2-cocycles). As K π = ker(ω π ) is a π-kernel, it follows from Lemma 5.2 that we can take π compatible with K π . Then, in particular, the restriction π K π = 1. If P is an irreducible π-representation of G, then KER(P) = K π , by Lemma 6.4 (c). Therefore, if deg(P) = n, then n2 = |G : K π |, P(s) = λ(P) (s)In (s ∈ K π ),

where λ(P) ∈ F[K π , ℂ∗ ].

Since π(s, t) = 1 for s, t ∈ K π , it follows that λ(P) (s)λ(P) (t) = λ(P) (st)

(s, t ∈ K π ),

i.e.,

λ(P) ∈ Lin(K π ).

Lemma 6.11. Irreducible π-representations P1 , P2 of a group G are linearly equivalent if and only if λ(P1 ) = λ(P2 ) .

280 | Characters of Finite Groups 1 LIN

Proof. If P1 ∼ P2 , then, obviously, λ(P1 ) = λ(P2 ) . Conversely, let λ(P1 ) = λ(P2 ) = λ. Put G = G/K π and let G = ⋃ Kπ gα α∈G

be the decomposition of G modulo K π such that g α g β = p(α, β)g αβ

p(α, β) ∈ K π .

(α, β ∈ G),

Put ∆ i (α) = P i (g α ) (i = 1, 2, α ∈ G). Then ∆ i (α)∆ i (β) = ρ(α, β)∆ i (α, β)

(α, β ∈ G),

where ρ(α, β) = π(g α , g β )λ(p(α, β)). Thus, ∆1 and ∆2 are projective ρ-representations of the quotient group G, where ρ ∈ Z2 (G, ℂ∗ ) is independent of i = 1, 2. Next, ∆ i is irreducible, since P i is irreducible. Assume that α ∈ KER(∆ i ). Then the matrix ∆ i (α) is scalar, and hence g α ∈ ker(P i ) = K π 󳨐⇒ α = 1 󳨐⇒ KER(∆ i ) = {1} (i = 1, 2). LIN

LIN

By Theorem 6.6 (a), ∆1 ∼ ∆2 , and hence P1 ∼ P2 . By Exercise 6.3 (a), the number r π = |K π | = |Lin(K π )|. Hence, Lemma 6.11 establishes a one-to-one correspondence between the classes of linearly equivalent π-representations of G and the linear characters of the subgroup K π , namely λ(P) ∈ Lin(K π ) corresponds to the class of P. Proof of Theorem 6.8. If G is of symmetric type, the result follows from Frucht’s Theorem 6.7. Assume that G is not of symmetric type and let W be a cyclic subgroup of G such that G = G/W is of symmetric type (Lemma 6.10). By Lemma 6.2, the quotient group G possesses a non-degenerate antisymmetric bicharacter θ. Putting ω(s, t) = θ(s, t) (s, t ∈ G), we get ω ∈ Ba (G) and ker(ω) = W. By Theorem 6.3, ω = ω Π , where Π ∈ M(G). Then ω = ω π , where [π] = Π and the factor set π is compatible with K π = ker(ω π ) = W. Let λ be a faithful linear character of the cyclic group W = K π . Putting λ1 = λ and λ2 = 1K π , we find two irreducible π-representations P1 , P2 such that λ(P i ) = λ i (i = 1, 2), and form the π-representation P = P1 + P2 . It is obvious that KER(P) ≤ KER(P1 ) ∩ KER(P2 ) = K π . If s ∈ KER(P), then P(s) = diag(P1 (s), P2 (s)) = diag(λ1 (s)If , λ2 (s)If ), where

1

f = deg(P i ) = |G : K π | 2

(i = 1, 2).

VI Projective representations | 281

Therefore, one has s ∈ KER(P) if and only if λ1 (s) = λ2 (s). The last condition means that s ∈ ker(λ). Hence s = 1 because λ is faithful. Thus, KER(P) = {1}. Since G does not possess a faithful irreducible projective representation (by Theorem 6.7), it follows that P has the minimal number (namely, two) of irreducible constituents. Exercise 6.7. The minimal number of irreducible constituents of a faithful π-representation of an abelian group G is 1 + rk(K π ).

7 Commutator subgroups of representation groups In this section we will show that the commutator subgroup of a representation group of a group G is uniquely determined by G. In particular, a group G = G󸀠 has only one representation group (hence, a nonabelian simple group has only one representation group). Note that the dihedral group D2n has exactly three representation groups. The abelian group of type (2n , 2) has exactly two representation groups, one of which has a cyclic subgroup of index p and another is nonmetacyclic. Theorem 7.1. Let (Γ i , f i , A i ) be representation groups of a group G, where i = 1, 2. Then Γ󸀠1 ≅ Γ󸀠2 . Proof. Define the group Γ ≤ Γ1 × Γ2 as follows: Γ = {(x1 , x2 ) ∈ Γ1 × Γ2 | f1 (x1 ) = f2 (x2 )}. The mapping f : (x1 , x2 ) 󳨃→ f1 (x1 ) is an epimorphism of Γ onto G. Indeed, if x ∈ G, then, since both f1 and f2 are surjective, there exist x1 ∈ Γ1 and x2 ∈ Γ2 such that f1 (x1 ) = x = f2 (x2 ). Therefore, f(x1 , x2 ) = x. Next, ker(f) = {(x1 , x2 ) ∈ Γ1 × Γ2 | f1 (x1 ) = f2 (x2 ) = 1} = ker(f1 ) × ker(f2 ) = A1 × A2 = A. As A i ≤ Z(Γ i ) (i = 1, 2), it follows that A ≤ Z(Γ1 × Γ2 ) so that A ≤ Z(Γ) (this follows from A ≤ Γ < Γ1 × Γ2 ). We have thus obtained a central extension (Γ, f, A) of G. We claim (i) that (Γ, f, A) is a covering of G. Let Γ i = ⋃x∈G A i g x be the decomposition of Γ i modulo A i and let (i) (i) (i) g s g t = p i (s, t)g st (i = 1, 2, s, t ∈ G), where p i : G × G → A i is the factor set of the central extension (Γ i , f i , A i ). Since (i) (1) (2) f i (g x ) = x (i = 1, 2), it follows that g x = (g x , g x ) ∈ Γ for any x ∈ G. If Ag s = Ag t , then g t = ag s (a ∈ A). Putting a = (a1 , a2 ), where a1 ∈ A1 , a2 ∈ A2 , we obtain (1)

(2)

(1)

(2)

(1)

(2)

(g t , g t ) = (a1 , a2 )(g s , g s ) = (a1 g s , a2 g s ); (i)

(i)

(i)

(i)

hence g t = a i g s , i.e., A i g t = A i g s (i = 1, 2) and so s = t. Thus, the |G| cosets Ag x (x ∈ G) are pairwise distinct. Since Γ/A ≅ G (because f : Γ → G is an epimorphism with kernel A), we have |Γ| = |A| ⋅ |G|. Thus, Γ = ⋃x∈G Ag x

282 | Characters of Finite Groups 1 is a decomposition of Γ modulo A. Next, putting p(s, t) = (p1 (s, t), p2 (s, t)), we obtain p(s, t) ∈ A, g s g t = p(s, t)g st , s, t ∈ G. The mapping ψ 󳨃→ [π ψ ] = [ψ(p(s, t))]

(ψ ∈ Lin(A), s, t ∈ G)

is a homomorphism τ : Lin(A) → M(G). Γ󸀠 ,

Denoting D = A ∩ we have im(τ) ≅ D (see Lemma 4.4 (a)). Let τ i : Lin(A i ) → M(G) be an isomorphism (it exists, since (Γ i , f i , A i ) is a representation group of G, i = 1, 2). Recall that π ψ i (s, t) = ψ i (p i (s, t)) (i = 1, 2, s, t ∈ G). There exists a canonical isomorphism Lin(A1 × Lin(A2 ) → Lin(A): if (ψ1 , ψ2 ) ∈ Lin(A1 ) × Lin(A2 ), then, putting ψ((a1 , a2 )) = ψ(a1 )ψ(a2 )

((a1 , a2 ) ∈ A1 × A2 = A),

one obtains ψ ∈ Lin(A), where (ψ1 , ψ2 ) 󳨃→ ψ is an isomorphism. If (ψ1 , ψ2 ) ∈ Lin(A1 ) × Lin(A2 ) and (ψ1 , ψ2 ) 󳨃→ ψ, then π ψ (s, t) = ψ(p(s, t)) = ψ(p1 (s, t), p2 (s, t)) = ψ1 (p1 (s, t))ψ2 (p2 (s, t)) = π ψ1 (s, t)π ψ2 (s, t), i.e., π ψ = π ψ1 ⋅π ψ2 . Hence [π ψ ] = [π ψ1 ]⋅[π ψ2 ] or (what is the same) τ(ψ) = τ(ψ1 )τ(ψ2 ). Since ψ i can be chosen arbitrarily in Lin(A i ), it follows that im(τ) = M(G) (because im(τ i ) = M(G), i = 1, 2), and this means that (Γ, f, A) is a covering of G. We have seen that |M(G)| = |D|. Thus, we get |A ∩ Γ󸀠 | = |M(G)| = |A i | (i = 1, 2). Since f : Γ → G is an epimorphism, f(Γ󸀠 ) = G󸀠 . Similarly, f i (Γ i ) = G󸀠 (i = 1, 2). Since ker(f) = A, it follows that Γ󸀠 /(A ∩ Γ󸀠 ) ≅ G󸀠 . Since A i ≤ Γ󸀠i , we obtain Γ󸀠i /A i ≅ G󸀠 , and hence Γ󸀠 /(A ∩ Γ󸀠 ) ≅ Γ󸀠i /A i , |Γ󸀠 | ⋅ |A ∩ Γ󸀠 |−1 = |Γ󸀠i /A i | (i = 1, 2). It follows from |A i | = |A ∩ Γ󸀠 | that |Γ󸀠 | = |Γ󸀠i | (i = 1, 2). The mapping (x1 , x2 ) 󳨃→ x1 = φ(x1 , x2 ) is an epimorphism of Γ onto Γ1 (if x1 ∈ Γ1 , then, putting x = f1 (x1 ), find an element x2 ∈ Γ2 such that f2 (x2 ) = x; then (x1 , x2 ) ∈ Γ and φ(x1 , x2 ) = x1 ). Therefore, φ(Γ󸀠 ) = Γ󸀠1 . Since |Γ󸀠 | = |Γ󸀠1 |, it follows that φ Γ󸀠 is an isomorphism of Γ󸀠 onto Γ󸀠1 . Thus, Γ󸀠 ≅ Γ󸀠1 . Similarly Γ󸀠 ≅ Γ󸀠2 and so Γ󸀠1 ≅ Γ󸀠2 .

8 Intersection of kernels It is known that the intersection of kernels of irreducible characters of a group G is the identity subgroup. Indeed, that intersection is the kernel of the regular character of G which is faithful. Now let D = ⋂χ∈Irr(G) Z(χ), where Z(χ) = {x ∈ G | |χ(x)| = χ(1)} is the quasikernel of χ. If x ∈ D, then, by the Second Orthogonality Relation, |CG (x)| =

∑ |χ(x)|2 = χ∈Irr(G)

∑

χ(1)2 = |G|

χ∈Irr(G)

so that x ∈ Z(G). Thus, D ≤ Z(G). Since the reverse inclusion holds, we get D = Z(G).

VI Projective representations | 283

In this section we will determine the intersection of kernels of all irreducible projective representations of a group G. Let IPR(G) be the set of all irreducible projective representations of G and K = ⋂P∈IPR(G) KER(P). Let ILIN(G) be the set of all irreducible linear representations of G. If L ∈ ILIN(G), then KER(L) = Z(L) is a quasikernel of L, which coincides with quasikernel of the character χ L afforded by L (here KER(L) is the projective kernel of a linear representation L of G so that it consists of all x ∈ G such that L(x) is a scalar matrix). Therefore, ILIN(G) ⊆ IPR(G) 󳨐⇒ K ≤

KER(L) = Z(G).

⋂ L∈ILIN(G)

Usually, K < Z(G) (the simplest example is a noncyclic abelian group). Let (Γ, f, A) be a representation group of G. Consider the Schur correspondence S between IPR(G) and ILIN(Γ): P ↔ L (P ∈ IPR(G), L ∈ ILIN(Γ)). We have Z(L) = f −1 (KER(P)),

KER(P) = f(Z(L)).

(1)

It follows from (1) that K = f(Z(Γ)). Indeed, if L ∈ ILIN(Γ), then L can be lowered to G S and, hence, there exists P ∈ IPR(G) such that L ↔ P. We have f(Z(L)) = KER(P) ≥ K, implying, since Z(L) ≥ A = ker(f), that f(Z(Γ)) = f (

⋂ L∈ILIN(Γ)

f(Z(L)) =

⋂

f(Z(L)) ≥ K.

L∈ILIN(Γ) S

Now let P ∈ IPR(G) be given. In that case, there exists L ∈ ILIN(Γ) such that L ↔ P. Next, KER(P) = f(Z(L)) ≥ f(Z(Γ)) 󳨐⇒ K =

⋂

KER(P) ≥ f(Z(Γ)).

P∈IPR(G)

Thus, K = f(Z(Γ)), and we have proved the following: Theorem 8.1. Let K = ⋂P∈IPR(G) KER(P). If (Γ, f, A) is a representation group of G, then K = f(Z(Γ)) ≅ Z(Γ)/A. Corollary 8.2. If G, (Γ, f, A) and K are such as in Theorem 8.1, then the following statements hold: (a) Γ/Z(Γ) ≅ G/K. (b) M(G) ≅ M(G/K)/N, where N = K ∩ G󸀠 . Proof. (a) This follows from Theorem 8.1. (b) Let π ∈ Π ∈ M(G) and let P be an irreducible π-representation of G. Then one has KER(P) ≥ K, so that K is a π-kernel and, hence, a Π-kernel (see Lemma 5.1 (c)). Therefore, Π ∈ MK (G) and so M(G) ≤ MK (G). Since the reverse inclusion is obvious, we get MK (G) = M(G). On the other hand, by Theorem 5.10 (a)–(b), MK (G) ≅ M(G/K)/N, where N ≅ (K ∩ G󸀠 )/[K, G]. Since K ≤ Z(G), we obtain [K, G] = {1}, so that N ≅ K ∩ G󸀠 . Thus, M(G) ≅ M(G/K)/(K ∩ G󸀠 ).

284 | Characters of Finite Groups 1

9 The Schur multiplier of a direct product Let us find the multiplier of the direct product of two groups A and B. Definition. Let G = A × B. A function ω : A × B → ℂ∗ is called a pairing of the groups A and B if ω(a, b) is a linear character of A when b ∈ B is fixed and a linear character of B when a ∈ A is fixed. The set P(A, B) of all pairings of A and B is an abelian group with respect to pointwise multiplication. The group Lin(P(A, B)) = Hom(P(A, B), ℂ∗ ) is called the tensor product of groups A and B, denoted by A ⊗ B (compare with Definition IV.1.1). It is known that A ⊗ B ≅ P(A, B), so that the group P(A, B) may also be called the tensor product of groups A and B. Our aim is to prove the following: Theorem 9.1. We have M(A × B) = M(A) × M(B) × P(A, B). Proof. Let G = A × B, π ∈ Z2 (G, ℂ∗ ) and let π A , π B be the restrictions of π to A, B, respectively. Define a function β π : A × B → ℂ∗ by β π (a, b) = ω π (a, b) where ω π (g1 , g2 ) =

(a ∈ A, b ∈ B),

π(g1 , g2 ) g

π(g2 , g12 )

(g1 , g2 ∈ G)

(see (1.3)). As ab = ba, we get β π (a, b) =

π(a, b) . π(b, a)

Clearly, β π ∈ P(A, B). The classes [π A ], [π B ] and the pairing β π are uniquely determined by the class [π]. Consider the mapping [π] 󳨃→ ([π A ], [π B ], β π ) = f([π]) of the group M(G) into the group M(A) × M(B) × P(A, B). We claim that f is an isomorphism. Let ℂπ G be the twisted group algebra corresponding to π, and let {u g }g∈G be its standard basis. Then for g1 , g2 ∈ G we have (see (2.2) and (3.6) u g1 u g2 = π(g1 , g2 )u g1 g2 ,

g u−1 g2 u g1 u g2 = ω π (g 1 , g 2 )u g 2 . 1

Putting g i = a i b i (i = 1, 2, a i ∈ A, b i ∈ B), we obtain u g1 g2 = π(g1 , g2 )−1 u g1 u g2 = π(g1 , g2 )−1 π(a1 , b1 )−1 π(a2 , b2 )−1 ⋅ u a1 u b1 u a2 u b2 . Since b1 a2 = a2 b1 , it follows that u b−1 u a2 u b1 = β π (a2 , b1 )u a2 . Therefore, 1 u g1 g2 = π(g1 , g2 )−1 π(a1 , b1 )−1 π(a2 , b2 )−1 β π (a2 , b1 )−1 v,

VI Projective representations | 285

where v = u a1 u a2 u b1 u b2 is an invertible element in the twisted group algebra ℂπ G. On the other hand, u g1 g2 = u a1 a2 b1 b2 = π(a1 a2 , b1 b2 )−1 u a1 a2 u b1 b2 = π(a1 a2 , b1 b2 )−1 π(a1 , a2 )−1 π(b1 , b2 )−1 v. It follows from the above two expressions obtained for u g1 g2 that π(g1 , g2 )−1 π(a1 , b1 )−1 π(a2 , b2 )−1 β π (a2 , b1 )−1 = π(a1 a2 , b1 b2 )−1 π(a1 , a2 )−1 π(b1 , b2 )−1 .

(1)

Define a function λ : G → ℂ∗ by λ(g) = π(a, b)−1

(g = ab, a ∈ A, b ∈ B).

Using this definition, one can rewrite (1) as follows: π(g1 , g2 ) = λ(g1 )λ(g2 )λ(g1 g2 )−1 π A (a1 , a2 )π B (b1 , b2 )β π (a2 , b1 ).

(2)

Suppose that [π] ∈ ker(f). Then π A ∼ 1, π B ∼ 1, β π = 1. Put π A (a1 , a2 ) = σ(a1 )σ(a2 )σ(a1 a2 )−1 ,

π B (b1 , b2 ) = τ(b1 )τ(b2 )τ(b1 b2 )−1 ,

and define a function ν : G → ℂ∗ by ν(g) = σ(a)τ(b)

(g = ab, a ∈ A, b ∈ B).

Then (2) gives π(g1 , g2 ) =

(λν)(g1 )(λν)(g2 ) 󳨐⇒ π ∼ 1 and [π] = [1], (λν)(g1 g2 )

i.e., f is a monomorphism. Now let ξ ∈ Z2 (A, ℂ∗ ),

η ∈ Z2 (B, ℂ∗ ),

be given. Define a function π : G × G →

ℂ∗

β ∈ P(A, B)

by

π(g1 , g2 ) = ξ(a1 , a2 )η(b1 , b2 )β(a2 , b1 )−1 ,

(3)

where, as before, g i = a i b i (i = 1, 2, a i ∈ A, b i ∈ B). We claim that π ∈ Z2 (G, ℂ∗ ),

π A = ξ,

π B = η, quadβ π = β.

To show this, we put π1 (g1 , g2 ) = ξ(a1 , a2 ),

π2 (g1 , g2 ) = η(b1 , b2 ),

π3 (g1 , g2 ) = β(a2 , b1 )−1 .

One has π1 (g1 , g2 )π1 (g1 g2 , g3 ) = ξ(a1 , a2 )ξ(a1 a2 , a3 ) = ξ(a2 , a3 )ξ(a1 , a2 a3 ) = π1 (g2 , g3 )π1 (g1 , g2 g3 ).

(4)

286 | Characters of Finite Groups 1 Thus, π1 ∈ Z2 (G, ℂ∗ ) and, similarly, π2 ∈ Z2 (G, ℂ∗ ). Finally, π3 (g1 , g2 )π3 (g1 g2 , g3 ) = β(a2 , b1 )−1 β(a3 , b1 b2 )−1 = β(a2 , b1 )−1 β(a3 , b1 )−1 β(a3 , b2 )−1 , π3 (g2 , g3 )π3 (g1 , g2 g3 ) = β(a3 , b2 )−1 β(a2 a3 , b1 )−1 = β(a3 , b2 )−1 β(a2 , b1 )−1 β(a3 , b1 )−1 . Consequently, π3 ∈ Z2 (G, ℂ∗ ). Since π1 , π2 , π3 ∈ Z2 (G, ℂ∗ ) and π = π1 π2 π3 (by (4)), it follows that π ∈ Z2 (G, ℂ∗ ) since Z2 (G, ℂ∗ ) is a group. Next, it follows from (4) that π(a1 , a2 ) = ξ(a1 , a2 )η(1, 1)β(a2 , 1)−1 = ξ(a1 , a2 ) and π(b1 , b2 ) = ξ(1, 1)η(b1 , b2 )β(a2 , 1)−1 = η(b1 , b2 ). Therefore π A = ξ, π B = η. Finally, β π (a, b) = π(a, b)π(b, a)−1 = β(a, b)

(a ∈ A, b ∈ B),

i.e., β π = β. Thus, [π A ] = [ξ], [π B ] = [η], β π = β. This shows that f is an epimorphism and, hence, an isomorphism as well.

10 Realization theorem In this section we will prove Reynolds’ realization theorem for projective representations. We begin with the following useful result. Lemma 10.1 (Alperin and Kuo Tzee-Nan [AT]). For every finite group G, one has that exp(G) divides |G : (G󸀠 ∩ Z(G))|. Proof. (Note that G󸀠 ∩ Z(G) ≤ Φ(G).) We divide the proof into two steps. First step. Let G = P be a p-group, where p is a prime. Put A = Z(P) ∩ P󸀠 . Then the lemma asserts that exp(P) divides |P : A|. Let P = P1 > P2 = P󸀠 > P3 > ⋅ ⋅ ⋅ be the lower central series of P. Denote A i = A ∩ P i and Q i = A i P i+1 for all i. Then A1 = A = A2 and Q1 = P󸀠 . Now A i ∩ P i+1 = (A ∩ P i ) ∩ P i+1 = A ∩ P i+1 = A i+1 implies Q i /P i+1 = A i P i+1 /P i+1 ≅ A i /(A i ∩ P i+1 ) = A i /A i+1 . Therefore, |A| = ∏ |A i : A i+1 | = ∏ |Q i : P i+1 |, i≥1

i≥1

VI Projective representations | 287

and so |P : A| = ∏ |P i : P i+1 | ⋅ |Q i : P i+1 |−1 = ∏ |P i : Q i |. i≥1

i≥1

Put |P i : Q i | = p h i , exp(P i /P i+1 ) = p e i . Since exp(P) | p e1 +e2 +⋅⋅⋅ , it is sufficient to prove that e1 + e2 + ⋅ ⋅ ⋅ ≤ h1 + h2 + ⋅ ⋅ ⋅ . We shall prove the following two statements: (a) e1 + e2 ≤ h1 . (b) e i ≤ h i−1 for i ≥ 3. It is clear that (a) and (b) imply the result. If P is cyclic, then (a) and (b) hold. Let P be noncyclic. Then P/P󸀠 is noncyclic, so that there exist elements x1 , . . . , x m in P such that the cosets x i P󸀠 (i ∈ {1, . . . , m}) form a basis of the abelian group P/P󸀠 . Let o(x i P󸀠 ) = p a i , a1 ≥ ⋅ ⋅ ⋅ ≥ a m . Thus, e1 = a1 , h1 = a1 + a2 + ⋅ ⋅ ⋅ = log(|P/P󸀠 |). The group P2 /P3 is generated by the elements [x i , x j ]P3 , 1 ≤ i < j ≤ m. Next, cl(P/P3 ) = 2 󳨐⇒ o([x i , x j ]P3 ) ≤ p a j 󳨐⇒ e2 ≤ a2 . Hence, e1 + e2 ≤ a1 + a2 ≤ h1 , and (a) is proved. It remains to prove (b). The group P i /P i+1 is generated by all the elements of the form P i+1 [x, g], where x ∈ P i−1 and g ∈ P. Since |P i−1 /Q i−1 | = p h i−1 , it follows that h h x p i−1 = zy, where z ∈ A, y ∈ P i (since x p i−1 ∈ Q i−1 = A i−1 P i ≤ AP i ). But z ∈ A ≤ Z(P), and h i−1 h i−1 (P i+1 [x, g])p = P i+1 [x p , g] = P i+1 [zy, g] = P i+1 [y, g] = P i+1 . Thus, o(P i+1 [x, g]) ≤ p h i−1 . Since P i /P i+1 is abelian and is generated by elements of order at most p h i−1 , it follows that p e i = exp(P i /P i+1 ) ≤ p h i−1 , e i ≤ h i−1 , hence (b) is proved. The lemma for p-groups follows from (a) and (b). Second step. Consider the general case. Since exp(G) is the product of the exponents of representatives of the classes of Sylow subgroups of G, it is sufficient to show that Z(G) ∩ G󸀠 ∩ S ≤ Z(G) ∩ S󸀠

(S ∈ Syl(G)).

Indeed, if T is a transversal of classes of Sylow subgroups of G, then G󸀠 ∩ Z(G) = A = ∏ (A ∩ S), S∈T

and, as we have proved, exp(G) = ∏ exp(S) | ∏ |S/(S󸀠 ∩ Z(G))| | |G|/ ∏ |A ∩ S| = |G : A|. S∈T

S∈T

Z(G)∩ G󸀠 ∩ S (so that x

S∈T

Let x ∈ ∈ Z(S)) and let V G→S be the transfer homomorphism of G into S/S󸀠 . If x ∈ ̸ S󸀠 , then x n ∈ ̸ S󸀠 , because GCD(n, |S|) = 1 (where n = |G : S|), i.e., x n ∈ ̸ ker(V G→S ), and hence, since ker(V G→S ) ≥ G󸀠 , it follows that x ∈ ̸ G󸀠 , a contradiction. Thus, Z(G) ∩ G󸀠 ∩ S ≤ Z(G) ∩ S󸀠 (in fact, this is an equality, since the reverse inclusion is obvious).

288 | Characters of Finite Groups 1 As the nonabelian p-groups with cyclic subgroup of index p show, the equality exp(G) = |G : (G󸀠 ∩ Z(G))| is possible. Now the proof of the main result of this section follows. Theorem 10.2 (Reynolds). Any projective representation of a group G can be realized in the field of |G|-th roots of unity. Proof. Let P be a projective representation of G, deg(P) = n, and let (Γ, f, A) be a representation group of G. Let L be a linear representation of Γ obtained by lifting P (see Lemma 3.7 and Corollary 3.8). Then one has L(g) = λ(g)P(f(g)) (g ∈ Γ), where λ : Γ → ℂ∗ . By Brauer’s realization theorem (see Chapter VIII), one can write L in the field F of e-th roots of unity, where e = exp(Γ), i.e., there exists T ∈ GL(n, ℂ) such that T −1 L(g)T ∈ GL(n, F) for all g ∈ Γ. If s ∈ G, g s = f −1 (s), then L(g s ) = λ(g s )P(s) = μ(s)P(s)

(μ(s) = λ(g s )).

Put P󸀠 = μP. It follows that P󸀠 ∼ P and P󸀠 (s) = L(g s ). Replacing P󸀠 by a linearly equivalent projective representation P󸀠󸀠 such that P󸀠󸀠 = T −1 P󸀠 (s)T, we obtain P ∼ P󸀠󸀠 , where P󸀠󸀠 (s) = T −1 L(g s )T ∈ GL(n, F), so that P󸀠󸀠 is written in the field F. By Lemma 10.1, e | |Γ/(Γ󸀠 ∩ Z(Γ))|. Next, Z(Γ) ∩ Γ󸀠 ≥ A implies |Γ/(Z(Γ) ∩ Γ󸀠 )| | |Γ/A| = |G| (here we used Lemma 4.5). Thus, e | |G|, and this implies that F is contained in the field of |G|-th roots of unity.

11 p-groups with large multipliers In this section we will derive an upper estimate for |M(G)|, where G is a group of order p n , where p is a prime. Put |M(G)| = p m(G) . The estimate m(G) ≤ n(n−1) is due 2 to J. A. Green [Gre2]. Lemma 11.1. Let G be a p-group such that |G/Z(G)| = p n . Then (see [Wie1]) 1

|G󸀠 | = p 2 n(n−1)−s ,

s ∈ ℕ ∪ {0},

and |(G/Z(G))󸀠 | ≤ p1+s . If |(G/Z(G))󸀠 | = p1+s , then exp(Z2 (G)/Z(G)) = p, where Z2 (G) is the second term of the upper central series of G. Proof. One may assume that G is nonabelian. Then G/Z(G) is noncyclic so that n > 1. Let z0 ∈ Z2 (G) − Z(G). Consider the mapping φ : G → [G, z0 ](≤ Z(G)) defined by φ(x) = [x, z0 ] = x−1 z−1 0 xz 0

(x ∈ G).

It is well known that φ is a homomorphism. Write im(φ) = N and |N| = p ν . Clearly, ker(φ) = CG (z0 ) ≥ ⟨z0 , Z(G)⟩ > Z(G)

VI Projective representations | 289 y

(note that all elements z0 = z0 [z0 , y] ∈ z0 Z(G) have the same centralizer in G). Therefore, |G : CG (z0 )| ≤ p n−1 . Now, N ≅ G/ker(φ) = G/CG (z0 ) 󳨐⇒ ν = logp (|N|) ≤ n − 1. Put p b = |(G/N) : Z(G/N)|. Now N ≠ z0 N ∈ Z(G/N) − Z(G)/N 󳨐⇒ b ≤ n − 1. Next, N = [z0 , G] ≤ G󸀠 , so that |G󸀠 | = |N| ⋅ |G󸀠 : N|. By induction, 1

|G󸀠 /N| = |(G/N)󸀠 | ≤ p 2 b(b−1) . Therefore, 1

|G󸀠 | = |G󸀠 /N||N| ≤ p 2 b(b−1)+ν .

(1)

Since b ≤ n − 1, ν ≤ n − 1, we have 1 1 1 b(b − 1) + ν ≤ (n − 1)(n − 2) + (n − 1) = n(n − 1) 2 2 2 so that 1

|G󸀠 | ≤ p 2 n(n−1) ,

1

(2)

|G󸀠 | = p 2 n(n−1)−s

with s ∈ ℕ ∪ {0}. By (1) and (2), ν≥

1 1 1 1 n(n − 1) − s − b(b − 1) ≥ n(n − 1) − s − (n − 1)(n − 2) = n − 1 − s. 2 2 2 2

Consequently, p n−1−s ≤ p ν = |N| = |G : CG (z0 )|. Since G/CG (z0 ) ≅ N and N ≤ Z(G) is abelian, it follows that G󸀠 Z(G) ≤ CG (z0 ). Therefore, |G/Z(G) : (G/Z(G))󸀠 | = |G/Z(G) : G󸀠 Z(G)/Z(G)| = |G : G󸀠 Z(G)| ≥ |G : CG (z0 )| = |N| = p ν ≥ p n−1−s and |(G/Z(G))󸀠 | ≤

|G/Z(G)| pn = n−1−s = p1+s . n−1−s p p p

Suppose that |(G/Z(G))󸀠 | = p1+s . Assume that z0 ∈ ̸ Z(G) for some z0 ∈ Z2 (G) − Z(G) (in other words, exp(Z2 (G)/Z(G)) > p). It follows from the inclusion z0 N ∈ Z(G/N) (since N = [z0 , G]) that |G/N : Z(G/N)| ≤ p n−2 . Therefore, by induction (or by what was proved previously), 1 |G󸀠 /N| ≤ p 2 (n−2)(n−3) and therefore 1

|G󸀠 | = |N| ⋅ |G󸀠 /N| = |G : CG (z0 )| ⋅ |G󸀠 /N| ≤ p n−1−s+ 2 (n−2)(n−3) ,

290 | Characters of Finite Groups 1 or 1 1 n(n − 1) − s ≤ n − 1 − s + (n − 2)(n − 3) and (n − 2)(n − 1) ≤ (n − 2)(n − 3), 2 2 whence n = 2. Then we obtain that G/Z(G) is abelian and G = Z2 (G). But it follows from exp(G/Z(G)) > p that G/Z(G) is cyclic of order p2 , which is impossible. Thus, in the case under consideration, Z2 (G)/Z(G) is elementary abelian. Exercise 11.1. Let G be an abelian group of order p n with invariants {p e i }1d , where e1 ≤ ⋅ ⋅ ⋅ ≤ e d . Then, by Theorem 6.5 (below p m(G) is the order of the Schur multiplier of G), m(G) = (d − 1)e1 + (d − 2)e2 + ⋅ ⋅ ⋅ + e d−1 = (n − e d ) + (n − e d − e d−1 ) + (n − e d − ⋅ ⋅ ⋅ − e2 ). If e d > 1, d > 1, then 1 (n − 1)(n − 2). 2 Let m(G) = 21 n(n − 1) − s, s ≤ 2. Then the following statements hold: (a) If s = 0, then d = 1 and G = Ep n . (b) If s = 1, then d = 1, G = Cp2 . (c) If s = 2, then G = Cp2 × Cp . m(G) ≤ (n − 2) + ⋅ ⋅ ⋅ + (n − d) < (n − 2) + ⋅ ⋅ ⋅ + 1 =

Hint. Use Theorem 6.5. Theorem 11.2. The following statements hold: 1 (i) (Green [Gre2]) If G is a group of order p n , then |M(G)| ≤ p 2 n(n−1) . 1 (ii) (Berkovich [Ber14]) If |M(G)| = p 2 n(n−1) , then G = Ep n . 1

Proof. First we observe that, by Lemma 11.1, |M(G)| ≤ p 2 n(n−1) . Indeed, let Γ be a representation group of the group G. Then there exists a subgroup M of Γ󸀠 ∩ Z(Γ) such that M ≅ M(G) and Γ/M ≅ G. Now, 1

M ≤ Z(Γ) 󳨐⇒ |Γ/Z(Γ)| ≤ |G| = p n 󳨐⇒ |Γ󸀠 | ≤ p 2 n(n−1) , where the last implication is a consequence of Lemma 11.1. It follows from M ≤ Γ󸀠 1 that |M| ≤ p 2 n(n−1) . 1 1 1 Let |M(G)| = p 2 n(n−1) . Then the relation p 2 n(n−1) = |M| ≤ |Γ󸀠 | ≤ p 2 n(n−1) implies that M = Γ󸀠 , so that Γ/M ≅ G is abelian. The isomorphism G ≅ Ep n follows from Exercise 11.1. As a representation group of the elementary abelian group of order p n shows, estimate of Theorem 11.2 is best possible (see Theorem 6.5). Remark 11.1. Let G be a p-group such that G/Z(G) = (U1 /Z(G)(U2 /Z(G)), where both U1 /Z(G) and U2 /Z(G) are cyclic. Then U1 ∩ U2 = Z(G). Indeed, since U1 and U2 are abelian, CG (U1 ∩ U2 ) ≥ ⟨U1 , U2 ⟩ = G so that U1 ∩ U2 ≤ Z(G). As Z(G) ≤ U1 ∩ U2 , it follows that U1 ∩ U2 = Z(G).

VI Projective representations | 291

Remark 11.2 (reported by Mann). Let G be a p-group such that G/Z(G) is extraspecial. We claim that then G/Z(G) is either of order p3 and exponent p > 2 or isomorphic to D8 . Write Z = Z(G). Assume that |G/Z| = p2n+1 , n > 1. Then there exist elements x1 , y1 , . . . , x n , y n ∈ G − Z such that G = ⟨x1 , y1 , . . . , x n , y n , z, Z⟩, where ⟨zZ(G)⟩ = Z(G/Z), G/Z = (⟨x1 , y1 , Z⟩/Z) ∗ ⋅ ⋅ ⋅ ∗ (⟨x n , y n , Z⟩/Z),

|⟨x i , y i , Z⟩/Z| = p3 ,

[x i , y i ] ∈ ⟨zZ⟩.

One has Z2 (G) = ⟨z, Z⟩. We can select elements x1 , y1 , . . . , x n , y n in such a way that [x i , y−1 i ] = zu i , where u i ∈ Z, all i. Let i ≠ j. By the Hall–Witt commutator identity, yi −1 xj −1 xi [x i , y−1 i , x j ] [y i , x j , x i ] [x j , x i , y i ] = 1. −1 The second and third factors are equal to 1 since [y i , x−1 j ], [x j , x i ] ∈ Z, so the first 󸀠 󸀠 factor is equal to 1 = [z, x j ]y i since [x i , y−1 i ] = zu i with u i ∈ Z. Thus, z ∈ CG (x j ) for all j. Similarly, [z, y j ] = 1 for all j, and so z ∈ Z, which is a contradiction. Thus, n = 1, i.e., |G/Z| = p3 , and the result follows from Remark 11.1. 1

Theorem 11.3 ([Ber14]). Let G be a p-group such that |G/Z(G)| = p n . If |G󸀠 | = p 2 n(n−1) , then G/Z(G) is either elementary abelian or nonabelian of order p3 and exponent p. Proof. We have s = 0 in the notation of Lemma 11.1. Therefore, by Lemma 11.1, |(G/Z(G))󸀠 | ≤ p. p Assume that G/Z(G) is abelian. Take z0 ∈ G − Z(G). If z0 ∈ ̸ Z(G) (in that case, exp(G/Z(G)) > p), then b ≤ n − 2 (in the notation of Lemma 11.1), so that 1 1 n(n − 1) = logp (|G󸀠 |) ≤ (n − 2)(n − 3) + logp (|N|). 2 2 It follow from CG (z0 ) ≥ ⟨z0 , Z(G)⟩ that |N| = |G : CG (z0 )| ≤ p n−2 , and therefore 1 1 n(n − 1) ≤ (n − 2)(n − 1), 2 2 which is impossible. Thus, exp(G/Z(G)) = p and so G/Z(G) = Ep n . Now let |(G/Z(G))󸀠 | = p. Then exp(Z2 (G)/Z(G)) = p (Lemma 11.1). Suppose that Z2 (G)/Z(G) contains two distinct subgroups A/Z(G) and B/Z(G) of order p, and let A = ⟨z0 , Z(G)⟩, B = ⟨y0 , Z(G)⟩. In that case, one has, by (1), ν = n − 1 = b for any choice of z0 ∈ Z2 (G) − Z(G). But then A = CG (z0 ), B = CG (y0 ) 󳨐⇒ G󸀠 ≤ A ∩ B = Z(G), a contradiction, since G/Z(G) is nonabelian, by assumption. Therefore the abelian group Z2 (G)/Z(G) of exponent p has only one subgroup of order p. Since by assumption, |(G/Z(G))󸀠 | = p, (G/Z(G))󸀠 ≤ Z2 (G)/Z(G), we obtain (G/Z(G))󸀠 = Z2 (G)/Z(G) = Z(G/Z(G)),

292 | Characters of Finite Groups 1 and this group is of order p. It follows that G/Z(G) is extraspecial of order p n , i.e., we have G/Z(G) ≅ ES( 21 (n − 1), p). Let us prove that, provided our quotient group G/Z(G) is nonabelian, it is of order p3 and exponent p. We have |G/Z(G)| = p3 (cf. Remark 11.2). Assume now that exp(G/Z(G)) = p2 . In that case, G/Z(G) has two distinct cyclic subgroups A/Z(G) and B/Z(G) of index p; then A and B are abelian, and so A ∩ B = Z(G) has index p2 in G, a contradiction. Thus, G/Z(G) is either of exponent p > 2 or dihedral (of order 8). Assume that G/Z(G) ≅ D8 . In that case, by hypothesis, |G󸀠 | = 23 . However, we have 8 = |G : Z(G)| = 2|G󸀠 | since G has an abelian subgroup of index 2 ([Ber31, Lemma 1.1]), and so |G󸀠 | = 22 , a final contradiction. We will need the following theorem to prove subsequent results (another proof may be found in books by G. Karpilovsky; see the bibliography). Theorem 11.4 (Gaschütz–Neibüser–Ti Yen [GasNT]). If G is a p-group and G satisfies |(G/Z(G)) : Φ(G/Z(G))| = p d , then |M(G)| ≤ |M(G/G󸀠 )| ⋅ |G󸀠 |d−1 . For example, if G is a minimal nonabelian p-group and G/G󸀠 is abelian of type (p a , p b ), a ≤ b, then |M(G)| ≤ p a ⋅ p = p a+1 , by Theorems 6.5 and 11.4. For abelian p-groups, Theorem 11.4 is trivial. We introduce the following notation. Let H ⊴ G. Then D(H) = H ∩ G󸀠 ,

LinH (G) = {λ ∈ Lin(G) | ker(λ) ≥ H}.

It is clear that LinH (G) ≅ Lin(G/H) ≅ G/HG󸀠 . We proceed to the proof of Theorem 11.4 (this proof is due to the third author). Lemma 11.5. Let N1 , N2 ⊲ G, N1 ≤ N2 . Then |N1 |−1 |N2 ∩ N1 G󸀠 | ⋅ |D(N1 )| = |D(N2 )|. Proof. Using the product formula and the modular law (noting that N2 N1 = N2 ), one obtains |N2 ∩ N1 G󸀠 | =

|N2 ||N1 G󸀠 | |N2 ||N1 ||G󸀠 ||N2 ∩ G󸀠 | |N1 ||D(N2 )| = = . |N2 N1 G󸀠 | |N1 ∩ G󸀠 ||N2 ||G󸀠 | |D(N1 )|

Let H ⊴ G. Then (see Theorem 5.10) |M(G)| = u H (G)v H (G) (u H (G) =

|M(G/H)| |M(G)| , v H (G) = ⋅ |[H, G]|). |D(H)| |MH (G)|

(3)

Lemma 11.6. Let N1 , N2 ⊴ G, N1 ≤ N2 . Then v N2 (G) = v N2 /N1 (G/N1 ). v N1 (G)

(4)

Proof. Put G = G/N1 . Then, by (3), |M(G)| = u N1 (G)v N1 (G) = |M(G)| ⋅ |D(N1 )|−1 v N1 (G).

(5)

VI Projective representations | 293

It follows from G/N 2 ≅ G/N2 that |M(G)| = u N 2 (G)v N 2 (G) = |M(G/N 2 )| ⋅ |D(N 2 )|−1 v N 2 (G) = |M(G/N2 )| ⋅ |D(N 2 )|−1 v N 2 (G). By Lemma 11.5, 󸀠

|D(N 2 )| = |N 2 ∩ G | = |(N2 ∩ N1 G󸀠 )/N1 | = |N1 |−1 |N2 ∩ N1 G󸀠 |, so we obtain |M(G)| = |M(G/N2 )| ⋅ |N1 | ⋅ |N2 ∩ N1 G󸀠 |−1 v N 2 (G).

(6)

Relations (5) and (6) imply the following equality: |M(G)| = |M(G/N2 )| ⋅ |N1 | ⋅ |N2 ∩ N1 G󸀠 |−1 ⋅ |D(N1 )|−1 v N 2 (G)v N1 (G). Hence, in view of Lemma 11.5, we obtain |M(G)| = |M(G/N2 )| ⋅ |D(N2 )|−1 v N 2 (G)v N1 (G) = u N2 (G)v N 2 (G)v N1 (G). Since, on the other hand, |M(G)| = u N2 (G)v N2 (G), it follows that v N2 (G) = v N 2 (G)v N1 (G), proving (4). (π)

Let s ∈ G and let χ s , where (π)

χ s (t) = π(s, t)π(t, s)−1

(s ∈ G, t ∈ CG (s)), (π)

be a linear character of CG (s) defined in (3.9). It follows from this formula that χ s (Π) depends only on the class [π] of the factor set π. Therefore we shall write χ s instead (π) (Π) of χ s , where Π = [π]. If s ∈ Z(G), then, obviously, χ s ∈ Lin(G). Since the set G π depends only on the class Π = [π], we shall write G Π instead of G π . Lemma 11.7. If Z = ⟨z⟩ < Z(G) is of order f , then the following hold: (Π) (a) The mapping ψ z : Π 󳨃→ χ z is a homomorphism of M(G) into Lin(G). (b) ker(ψ z ) = MZ (G). (Π) (Π) (c) (χ z )f = 1G for any Π ∈ M(G) (i.e., o(χ s ) | f ). (d) If Z ≤ G󸀠 (i.e., Z ≤ D(Z(G)), then im(ψ z ) ≤ LinZ(G) (G). Proof. (a) This follows from the definitions. (b) We have Π ∈ ker(ψ z ) ⇐⇒ χ Π z = 1G ⇐⇒ z ∈ G Π ∩ Z(G) = K Π (here we used Definition 3.4). Thus, Π ∈ ker(ψ z ) if and only if z ∈ K Π . By Lemma 5.14, K Π ≤ G so that Π ∈ ker(ψ z ) if and only if Z ≤ K Π . Hence, by Lemma 5.14, Π ∈ ker(ψ z ) if and only if Z is a Π-kernel. Therefore, ker(ψ z ) = MZ (G).

294 | Characters of Finite Groups 1 (Π)

(c) Since, by (3.7), s 󳨃→ χ s (s ∈ Z(G)) is a homomorphism of Z(G) into Lin(G), it follows, in view of f = |Z|, that (Π)

(Π)

(Π)

(χ z )f = χ z f = χ1 = 1G . (Π)

(d) Assume that Z ≤ G󸀠 . If s ∈ Z(G), then χ s (Π) z ∈ G󸀠 and χ s is a linear character). Next, (Π)

(Π)

(Π)

∈ Lin(G), whence χ s (z) = 1 (since

(Π)

(Π)

χ z (s)χ s (z) = 1 󳨐⇒ χ z (s) = 1 󳨐⇒ s ∈ ker(χ z ). Thus, (Π)

Z(G) ≤ ker(χ z ),

i.e.,

(Π)

χz

∈ LinZ(G) (G) = Lin(G/Z(G)).

(Π)

Observing that χ z = ψ z (Π), we obtain ψ z (Π) ∈ LinZ(G) (G) for all Π ∈ M(G). In other words, im(ψ z ) ≤ LinZ(G) (G). In what follows Sc(G) will denote the socle of G. It is known that if G is a p-group, then |G : Φ(G)| = |Sc(G/G󸀠 )|. Lemma 11.8. Let G be a p-group, let Z < D(Z(G)) = Z(G) ∩ G󸀠 be of order p and let |G/Z(G) : Φ(G/Z(G))| = p d . Then v Z (G) | p d (v Z (G) is defined in (3)). Proof. Let Z = ⟨z⟩. Then [Z, G] = {1} 󳨐⇒ v Z (G) = |M(G)/MZ (G)|. But, by Lemma 11.7 (b), M(G)/MZ (G) ≅ im(ψ z ) so that v Z (G) = |im(ψ z )|. Next, Z ≤ D(Z(G))(= Z(G) ∩ G󸀠 ) 󳨐⇒ |im(ψ z )| | |Sc(LinZ(G) (G))| (Π)

(Lemma 11.7 (c)–(d)). Indeed, in view of |Z| = p, Lemma 11.7 (c) gives (χ z )p = 1G , (Π) i.e., ψ z (Π) = χ z ∈ Sc(LinZ(G) (G)) for all Π ∈ M(G). Since LinZ(G) (G) ≅ G/G󸀠 Z(G), the remark before the lemma implies that |Sc(LinZ(G) (G))| = |Sc(G/G󸀠 Z(G))| = |Sc(G/Z(G)/G󸀠 Z(G)/Z(G))| = |Sc(G/Z(G)/(G/Z(G))󸀠 )| = |G/Z(G)/Φ(G/Z(G))| = p d . Hence v Z (G) | p d . Theorem 11.9. Let G be a p-group and let |G/Z(G) : Φ(G/Z(G))| = p d . If H ≤ G󸀠 is G-invariant, then |M(G)| ≤ |M(G/H)| ⋅ |H|d−1 .

(7)

Proof. Let {1} = Z0 < Z1 < ⋅ ⋅ ⋅ < Z l = H be a segment of the principal series of G. Since all indices of this series are p, we have |H| = p l . Since v Z0 (G) = v1 (G) = 1, it follows from Lemma 11.6 that l

v H (G) = ∏ i=1

l v Z i (G) = ∏ v Z i /Z i−1 (G/Z i−1 ). v Z i−1 (G) i=1

(8)

VI Projective representations | 295

For a fixed i ∈ {1, . . . , l}, put G = G/Z i−1 . Then |Z i | = p, Z i ≤ D(Z(G)) (recall that 󸀠 H ≤ G󸀠 ). We have Z i ≤ G and Z i ≤ Z(G). Let |G/Z(G) : Φ(G/Z(G))| = p d . By Lemma 11.8 (applied to G and Z i ), we have v Z i (G) | p d . It is easy to see, that d ≤ d. Therefore, v Z i /Z i−1 (G/Z i−1 ) divides p d . Hence, in view of (8), we obtain v H (G) | (p d )l = (p l )d = |H|d . Finally, the equalities |M(G)| = u H (G)v H (G) = |M(G/H)| ⋅ |D(H)|−1 v H (G) = |M(G/H)| ⋅ |H|−1 v H (G) imply that |M(G)| ≤ |M(G/H)| ⋅ |H|d−1 . Putting H = G󸀠 in Theorem 11.9, we obtain Theorem 11.4. The proof of Lemma 11.10 may be found in [Kar6, Theorem 11.8.23]. Lemma 11.10. If G = ES(m, p) is an extraspecial p-group of order p1+2m with m > 1, then logp (|M(G)|) = m(G) = 2m2 − m − 1. If m = 1 and M(G) > {1}, then M(G) = C2 for G = D8 , and M(G) = Ep2 for G nonabelian of order p3 and exponent p > 2. It follows from §VI.3 that if G is nonabelian of order p3 and exponent p2 , p > 2, or G ≅ Q8 , then M(G) = {1}. Theorem 11.11 ([Ber14]). Let G be a p-group of order p n and m(G) = 12 n(n − 1) − 1, where p m(G) = |M(G)|. Then G = Cp2 or G is nonabelian of order p3 and exponent p > 2. Proof. Obviously, the groups mentioned in the conclusion of the theorem satisfy the condition (see Lemma 11.10). Let m(G) = 12 n(n − 1) − 1 and let Γ be a representation group of G. The group 󸀠 Γ ∩ Z(Γ) contains a subgroup M ≅ M(G) with Γ/M ≅ G. Then we have G󸀠 ≅ Γ󸀠 /M and |Γ : Z(Γ)| ≤ |Γ : M| = |G| = p n , so that, for some s ∈ ℕ ∪ {0}, 1 1 n(n − 1) − s ≥ m(G) = n(n − 1) − 1 2 2 (see Lemma 11.1), and therefore s ≤ 1. Assume that M < Z(Γ). Then |Γ : Z(Γ)| ≤ p n−1 and, by Lemma 11.1, logp (|Γ󸀠 |) =

1 1 (n − 1)(n − 2) ≥ logp (|Γ󸀠 |) ≥ logp (|M|) = m(G) = n(n − 1) − 1 2 2 and so n ≤ 2. In this case G ≅ Cp2 . Now let M = Z(Γ) and assume that G is noncyclic. By Theorem 11.2, G is not elementary abelian so that n > 2. If G is abelian then, as in Exercise 11.1, d > 1, e d > 1 imply m(G) = (n − e d ) + (n − e d − e d−1 ) + ⋅ ⋅ ⋅ + (n − e d − ⋅ ⋅ ⋅ − e2 ) 1 1 ≤ (n − 2) + (n − 3) + ⋅ ⋅ ⋅ + 1 = (n − 1)(n − 2) < n(n − 1) − 1 2 2 (since, by assumption, n ≥ 3), and this is a contradiction. Now assume that G is nonabelian; then M < Γ󸀠 . Therefore, logp (|Γ󸀠 |) =

1 1 n(n − 1) − s > m(G) = n(n − 1) − 1, 2 2

296 | Characters of Finite Groups 1 so that s = 0. By Theorem 11.3, G ≅ Γ/Z(Γ) = Γ/M ≅ ES( 12 (n − 1), p). If n > 3, then (see Lemma 11.10) 2 1 1 1 m(G) = 2[ (n − 1)] − (n − 1) − 1 = (n − 1)(n − 2) − 1 2 2 2 1 < n(n − 1) − 1 = m(G), 2

a contradiction. Thus, n = 3 and, again by Theorem 11.4 (b), G is a nonabelian group of order p3 and exponent p > 2. 1

Exercise 11.2. Let |G| = p n , |M(G)| = p 2 n(n−1)−s , |G󸀠 | = p s . Then s ≤ 1. Hint. Let Γ be a representation group of G and M ≅ M(G) a subgroup of Γ󸀠 ∩ Z(Γ) such 1 that Γ/M ≅ G. Since Γ󸀠 /M ≅ G󸀠 , it follows that |Γ󸀠 | = |M| ⋅ |G󸀠 | = p 2 n(n−1) . Obviously, |Γ/Z(Γ)| ≤ |G| = p n . Therefore, by Theorem 11.3, either Γ/Z(Γ) is elementary abelian or nonabelian of order p3 and exponent p. Hence s ≤ 1. Exercise 11.3. Let |G| = p n , |G󸀠 | = p k and m(G) ≥ 12 (n − 1)(n − 2). Then one of the following holds: (a) G = Ep n . (b) G = Ep n−2 × Cp2 . (c) G󸀠 = Φ(G), k2 + k + 2 ≤ 2n. Hint. If G is abelian, the result follows from Exercise 11.1. If G is nonabelian, then, by Theorem 11.4 (see also Lemma 11.1), 1 1 (n − 1)(n − 2) ≤ m(G) ≤ (n − k)(n − k − 1) + k(n − k − 1) 2 2 1 = (n + k)(n − k − 1), 2 and hence k2 + k + 2 ≤ 2n. Theorem 11.4 shows that case (c) holds. Exercise 11.4. Let |G| = p n , |G󸀠 | = p k , m(G) = 21 n(n − 1) − s. Then k2 + k ≤ 2s. In particular, if s ≤ 5, then k ≤ 2. Hint. See the previous hint. 1

Exercise 11.5. Study the p-groups G such that |G/Z(G)| = p n and |G󸀠 | ≤ p 2 n(n−1)−2 . 1

Exercise 11.6. Let G be a p-group, |G/Z(G)| = p n and |G󸀠 | > p 2 (n−1)(n−2) . Is it true that exp(Z2 (G)/Z(G)) = p? Below we state, without proofs, some results due to the third author (unpublished). Theorem 11.12. Let G be a nonabelian group of order p n and |G󸀠 | = p n1 . Then one has 1 |M(G)| = p 2 n(n−1)−s , where s ≥ 12 n1 (n1 + 1). Theorem 11.12 follows from Theorem 11.4, but its proof is independent of the last theorem.

VI Projective representations | 297

Theorem 11.13. Let G be a nonabelian group of order p n and let G > G󸀠 > ⋅ ⋅ ⋅ > G(k−1) > G(k) = {1} be its derived series, |G(i) | = p n i (i = 1, . . . , k). Then 1

M(G)||M(G󸀠 )| . . . |M(G(k) )| = p 2 n(n−1)−t , where t ∈ ℕ, t ≥ ∑ki=1 n i ≥ l − 1.

12 More about orders of multipliers The results of this section are due to the third author. In this section we first consider arbitrary finite groups. Recall that the character (Π) χ s (s ∈ G, Π = [π], π ∈ Z2 (G, ℂ)) was defined in formula (3.9) (see also paragraph preceding Lemma 11.7). Next, v H = v H (G) =

|M(G)| ⋅ |[H, G]| |MH (G)|

(see the paragraph following Lemma 11.5). Theorem 12.1. Let Z = ⟨z⟩ ≤ Z(G). Then v Z (G) | |G : G󸀠 |. Proof. It follows from [Z, G] = {1} that v Z (G) = |M(G)/M Z (G)|. By Lemma 11.7 (b), we have MZ (G) = ker(ψ z ). Therefore, v Z (G) = |im(ψ z )|, and this implies the assertion, since im(ψ z ) ≤ Lin(G) and |Lin(G)| = |G : G󸀠 |. A chain {1} = Z0 < Z1 < Z2 < ⋅ ⋅ ⋅ of normal subgroups of G will be called a C-chain if it is central and all the quotients Z i /Z i−1 are cyclic. Theorem 12.2. Let H be a term of some C-chain, say, {1} = Z0 < Z1 < ⋅ ⋅ ⋅ < Z l = H. Then |M(G)| =

|M(G/H)| k, |D(H)|

where D(H) = H ∩ G󸀠 , k ∈ ℕ, k | |G : G󸀠 |l . Proof. In view of (11.3), it suffices to prove that v H (G) | |G : G󸀠 |l . By Lemma 11.6 (see there (11.4)), one has l

v H (G) = ∏ v Z i /Z i−1 (G/Z i−1 ).

(1)

i=1

Since Z i /Z i−1 ≤ Z(G/Z i−1 ) and Z i /Z i−1 is cyclic, it follows that v Z i /Z i−1 (G/Z i−1 ) | |G/Z i−1 : (G/Z i−1 )󸀠 | = |G : G󸀠 Z i−1 |, by Theorem 12.1. Therefore, v Z i /Z i−1 (G/Z i−1 ) | |G : G󸀠 | and so the assertion follows from (1).

(i ∈ {1, . . . , l}),

(2)

298 | Characters of Finite Groups 1 Corollary 12.3. Let G be a p-group, H ⊲ G and |G : G󸀠 | = p m . Then |M(G)| ≤

|M(G/H)| m |H| . |H ∩ G󸀠 |

(3)

In particular, if H ≤ G󸀠 , then |M(G)| ≤ |M(G/H)| ⋅ |H|m−1 . Proof. Obviously, H is a member of a C-chain. Putting |H| = p l , we obtain, by Theorem 12.2, that |M(G/H)| |M(G)| = k, |H ∩ G󸀠 | where k ∈ ℕ, k | p ml = |H|m . Corollary 12.4. Let |G| = p n , |G󸀠 | = p n1 . Then logp (|M(G)|) = m(G) =

1 n(n − 1) − s, 2

s≥

1 n1 (n1 + 1). 2

Proof. Putting n − n1 = m, and taking H = G󸀠 in (3), we obtain |M(G)| ≤ |M(G/G󸀠 )| ⋅ |G󸀠 |m−1 = |M(G/G󸀠 )|p(m−1)n1 . 1

Since G/G󸀠 is abelian of order p m , it follows that |M(G/G󸀠 )| ≤ p 2 m(m−1) (see Theorem 6.5). Therefore, by Corollary 12.3, 1 1 m(m − 1) + n1 (m − 1) = (n − n1 )(n − n1 − 1) + n1 (n − n1 − 1) 2 2 1 1 1 = (n + n1 )(n − n1 − 1) = n(n − 1) − n1 (n1 + 1). 2 2 2

m(G) ≤

Corollary 12.5. Let |G| = p n and let G = G0 > G󸀠 > ⋅ ⋅ ⋅ > G(σ) = {1} be the derived series of G, |G(i) | = p n i (i = 1, . . . , σ). Then σ

1

∏ |M(G(i) )| = p 2 n(n−1)−t ,

t ∈ ℕ, t ≥ (σ − 1)2 .

i=0

Proof. By Corollary 12.4, logp (|M(G(i) )|) ≤

1 1 n i (n i−1 ) − n i+1 (n i+1 − 1) − n i+1 2 2

(i = 0, 1, . . . , σ − 1).

It follows from n σ = 0 and n0 = n that σ−1

∑ logp (|M(G(i) )|) ≤ i=0

σ−1 1 n(n − 1) − ∑ n i+1 . 2 i=0

It remains to observe that n σ−1 ≥ 1, n i − n i−1 ≥ 2 (i = 1, . . . , σ − 2). Therefore, n1 + n2 + ⋅ ⋅ ⋅ + n σ−1 ≥ (2(σ − 1) − 1) + ⋅ ⋅ ⋅ + 3 + 1 = (σ − 1)2 , hence

σ−1

∑ logp (|M(G(i) )|) ≤ i=0

1 n(n − 1) − (σ − 1)2 . 2

VI Projective representations | 299

Note that the first assertion of Theorem 11.2 is an immediate consequence of Corollary 12.5. Given a cyclic Z ≤ Z(G), put Σ Z = {F ≤ G | Z ≤ F, |M(F)| = 1}. Since Z ∈ Σ Z , we obtain Σ Z ≠ 0. Set H Z = ⟨⋃F∈Σ Z F⟩; then H Z ⊴ G. The H-multiplier (H ≤ G) of G is defined after the proof of Lemma 5.2. Theorem 12.6. We use the above introduced notation. If Z ≤ Z(G) is cyclic, then |M(G)| =

|M(G/Z)| ⋅ m Z , where m Z ∈ ℕ, m Z | |G : G󸀠 H Z |. |Z ∩ G󸀠 | (Π)

Proof. Let Z = ⟨z⟩ and let ψ z : Π 󳨃→ χ z be a homomorphism of M(G) in Lin(G) (see Lemma 11.7). By Lemma 11.7 (b), we have ker(ψ z ) = MZ (G) so that |M(G)| = |MZ (G)| ⋅ m Z , where m Z = |im(ψ z )|. If Π = [π], where π ∈ Z2 (G, ℂ∗ ) and H ∈ Σ Z , then π H ∼ 1, i.e., there exists a function λ : H → ℂ∗ such that λ(s)λ(t) π(s, t) = (s, t ∈ H). λ(st) Therefore, π(s, t) = π(t, s)(s, t ∈ H) if st = ts. In particular, it follows from z ∈ H that ω π (z, t) =

π(z, t) =1 π(t, z)

for all t ∈ H.

(Π)

This means that the restriction (χ z )H = 1H . Next, (Π)

χz

(Π)

∈ Lin(G) 󳨐⇒ G󸀠 H Z ≤ ker(χ z ),

i.e.,

(Π)

χz

∈ LinG󸀠 H Z (G).

It follows that im(ψ z ) ≤ LinG󸀠 H Z (G), i.e., m Z divides |LinG󸀠 H Z (G)|. It remains to note that |LinG󸀠 H Z (G)| = |G : G󸀠 H Z | and, by Theorem 5.10, |MZ (G)| = |M(G/Z)| . |Z∩G󸀠 |

13 Representation groups of nonabelian metacyclic p-groups In this section we prove the following: Theorem 13.1 ([BJ1, Theorem 47.2]). If G is a nonabelian metacyclic p-group, then all representation groups of G are also metacyclic so that M(G), the Schur multiplier of G, is cyclic. Our proof is based on the following theorem which is due essentially to Blackburn (see [Ber31, Theorem 36.1]). Theorem 13.2. If G is a nonabelian p-group, the following conditions are equivalent: (a) G is metacyclic. (b) G/R is metacyclic for some G-invariant subgroup R of index p in G󸀠 .

300 | Characters of Finite Groups 1 Lemma 13.3. The following statements hold: (a) If a p-group G is two-generator of class 2, then G󸀠 is cyclic. (b) (Blackburn) If G is a nonabelian two-generator p-group, then G󸀠 /K3 (G) is cyclic. Proof. (a) Since cl(G) = 2, we have [xy, uv] = [x, u][x, v][y, u][y, v]

(x, y, u, v ∈ G).

Let G = ⟨a, b⟩ and w, z ∈ G. Expressing w, z in terms of a and b and using the above identity, we see that the commutator [w, z] is a power of [a, b] so G󸀠 = ⟨[a, b]⟩. (b) This follows from (a): G/K3 (G) is two-generator of class 2. Proof of Theorem 13.2. It suffices to show that (b) implies (a). There is a cyclic subgroup U/R ⊲ G/R such that G/U is cyclic. Assume that U is noncyclic; then U is abelian with a cyclic subgroup of index p. In that case, U has a G-invariant subgroup T such that U/T is abelian of type (p, p) (the subgroup T = Φ(U) is cyclic). Set G = G/T. Then R ≰ T since (the noncyclic group) U = U/T cannot be an epimorphic image of 󸀠 the cyclic group U/R; in that case, G󸀠 ≰ T so G is nonabelian. Next, G/G is noncyclic 󸀠 󸀠 󸀠 so G < U and |G | = p since |U| = p2 . Since G = G󸀠 T/T ≅ G󸀠 /(G󸀠 ∩ T) is of order p, we get G󸀠 ∩ T = R (Lemma 13.3 (b)). Thus, R = G󸀠 ∩ T < T, contrary to what has just been established. Proof of Theorem 13.1. Let Γ be a representation group of G. By definition, there is in Z(Γ) ∩ Γ󸀠 a subgroup M ≅ M(G) such that Γ/M ≅ G. Since G is nonabelian, we get M < Γ󸀠 . Since Γ/M ≅ G is metacyclic, Γ is also metacyclic, by Theorem 13.2. The cyclicity of the Schur multiplier of a metacyclic p-group is known, its order is computed in [Kar4, Theorem 10.1.25]. Exercise 13.1. A p-group is said absolutely regular if |G/℧1 (G)| < p p . Show that all representation groups of any absolutely regular p-group are regular. Solution. Let Γ be a representation group of G. Since M(G) ≤ Z(Γ) and Γ/M(G) ≅ G is absolutely regular, Γ is regular [Ber31, Remark 7.2].

VII Clifford theory 1 Clifford’s theorem In general, one cannot say much about the restriction χ N of a character χ ∈ Irr(G) to N ≤ G. The situation turns to be much simpler if N ⊴ G. From now on, we assume that N ⊴ G. Let us recall some notation. Let F[G] be the unitary space of functions θ : G → ℂ. If N is normal in G and θ ∈ F[N], g ∈ G, then the function θ g ∈ F[N] is defined by θ g (x) = θ(gxg−1 ) (x ∈ N) (the function θ(gxg−1 ) make sense since gxg−1 ∈ N). The functions θ and θ g are said to be conjugate under G or G-conjugate. A function θ ∈ F[G] is said to be central if θ g = θ for all g ∈ G, i.e., θ coincides with all its G-conjugates. It is obvious that θ ∈ F[G] is central if and only if it is constant on G-classes (= conjugacy classes of G). For example, characters of G are central functions. Let CF[G] denote the set of all central functions from G to ℂ. If θ ∈ CF[N] and g ∈ G, then θ g ∈ CF[N] −1 since N ⊴ G, by assumption. Indeed, if ν ∈ N, then (θ g )ν = (θ gνg )g = θ g . Therefore, one may speak about the action of G on the sets F[N] and CF[N]. Given ψ ∈ F[N], put IG (ψ) = {g ∈ G | ψ g = ψ}. One has N ≤ IG (ψ) ≤ G. The group IG (ψ) is called the inertia group of ψ in G. Next, ψ ∈ CF[N] if and only if N ≤ IG (ψ). If IG (ψ) = G, then ψ is said to be G-invariant. Denoting the G-orbit of ψ ∈ F[N] by O G (ψ), we get |O G (ψ)| = |G : IG (ψ)|

(1)

(see §I.1). Let O G (ψ) = {ψ1 = ψ, ψ2 , . . . , ψ t },

σ ψ = ψ1 + ⋅ ⋅ ⋅ + ψ t = |IG (ψ)|−1 ∑ ψ g

(2)

g∈G

(note that every summand is appeared exactly |IG (ψ)| times in the sum on the righthand side). Obviously, σ ψ = σ ψ i for all i ∈ {1, . . . , t}. Let ψ G be the function in CF[G] induced by a function ψ ∈ CF[N], i.e., ψ G = |N|−1 ∑ ψ̇ g , g∈G

where ψ̇ : G → ℂ is a function equal to ψ on N and vanishing on G − N. Then (2) implies ψ G = |IG (ψ) : N|σ̇ ψ

(3)

(ψ G )N = |IG (ψ) : N|σ ψ .

(4)

and

In this chapter N ⊴ G throughout. DOI 10.1515/9783110224078-007

302 | Characters of Finite Groups 1 Lemma 1.1. Given g ∈ G, the mapping ψ 󳨃→ ψ g (ψ ∈ F[N]) of the space F[N] onto itself is isometric: g g ψ1 , ψ2 ∈ F[N] 󳨐⇒ ⟨ψ1 , ψ2 ⟩ = ⟨ψ1 , ψ2 ⟩. Recall that Ch(X) is the set of generalized characters of a group X. Lemma 1.2. If ψ ∈ ℤ[Irr(N)](= Ch(N)) and g ∈ G, then ψ g ∈ ℤ[Irr(N)]. If ψ ∈ Irr(N), then ψ g ∈ Irr(N). Exercise 1.1. If ψ ∈ Char(N) and g ∈ G, then ψ g ∈ Char(N). Indeed, if T is a representation of N that affords ψ, then T g : x 󳨃→ T(gxg−1 ) (x ∈ N) is a representation of N: if, in addition, y ∈ N, then x g , y g ∈ N so that −1

−1

−1

T g (xy) = T((xy)g ) = T(x g y g ) = T(x g )T(y g ) = T g (x)T g (y). Theorem 1.3 (Clifford [Cli]). Let χ ∈ Irr(G), let ψ ∈ Irr(N) with ⟨χ N , ψ⟩ = e > 0 and let O G (ψ) = {ψ1 = ψ, ψ2 , . . . , ψ t }. Setting σ ψ = ψ1 + ⋅ ⋅ ⋅ + ψ t , one obtains χ N = e(ψ1 + ⋅ ⋅ ⋅ + ψ t ) = eσ ψ .

(5)

In other words, the irreducible constituents of χ N are all of the same multiplicity and form the G-orbit of ψ. Note that t = |G : IG (ψ)|, by (1). Proof. Let g ∈ G. Taking into account that (χ N )g = (χ g )N = χ N , Lemma 1.1 implies ⟨χ N , ψ g ⟩ = ⟨(χ N )g , ψ g ⟩ = ⟨χ N , ψ⟩ = e. Therefore, χ N = eσ ψ + θ, where θ is either a character of N or the zero function. In that case, if θ ∈ Char(N), one has ⟨θ, ψ i ⟩ = 0 for i ∈ {1, . . . , t}. By reciprocity, ⟨ψ G , χ⟩ = ⟨ψ, χ N ⟩ = ⟨ψ, eσ ψ ⟩ + ⟨ψ, θ⟩ = e. Therefore, ψ G = eχ + λ, where λ is either a character of G or the zero function and ⟨λ, χ⟩ = 0. Then, by (4), |IG (ψ) : N|σ ψ = (ψ G )N = eχ N + λ N = e2 σ ψ + eθ + λ N , i.e., eθ + λ N = (|IG (ψ) : N| − e2 )σ ψ . Taking the inner product of both sides of this equality with θ and recalling that ⟨θ, σ ψ ⟩ = 0, we obtain e⟨θ, θ⟩ + ⟨λ N , θ⟩ = 0. Since both terms on the left are nonnegative, it follows that ⟨θ, θ⟩ = 0, i.e., θ = 0, and hence χ N = eσ ψ . This theorem will be referred to as Clifford’s theorem and the equality χ N = e(ψ1 + ⋅ ⋅ ⋅ + ψ t ) the Clifford decomposition. As we know, [Cli] is the unique paper that Clifford devoted to representation theory of finite groups. The multiplicity e of the irreducible constituents of χ N in Theorem 1.3 is called the ramification of χ over N and denoted sometimes by e χ . If e χ = 1, then χ is said to be unramified over N. If χ N = e χ ψ and ψ ∈ Irr(N), then χ is said to be completely ramified

VII Clifford theory | 303

over N, and χ N is said to be homogeneous. It follows from Clifford’s theorem that χ is completely ramified over N if and only if IG (ψ) = G, i.e., ψ is G-invariant (in that case, σ ψ = ψ). We can rewrite (5) in the new notation as follows: χN = eχ σψ ,

(6)

where ψ is any irreducible constituent of χ N . Since σ ψ (1) = tψ(1), where t is the number of irreducible constituents of χ N , it follows from (6) and Theorem 1.3 that χ(1) = te χ ψ(1) = e χ |G : IG (ψ)|ψ(1).

(7)

Theorem 1.3 is a particular case of the following remarkable general result, also due to Clifford: Theorem 1.4 (Clifford’s general theorem). Let G be an arbitrary group, K a field and T an irreducible finite-dimensional representation of G over K. Then T N , the restriction of T to N ⊴ G, is completely reducible. The irreducible constituents of T N are conjugate with respect to G and occur in T N with the same multiplicity. Let us sketch the proof. Let V be an (irreducible) G-module affording the representation T, and let V N be the restriction of V to N (i.e., now V is considered as an N-module; then V N may be reducible). Let L be any irreducible N-submodule of V N affording a representation Ψ of N. Since N ⊴ G, it follows that gL = {gv | v ∈ L} is an irreducible submodule of V N for all g ∈ G. Indeed, if x ∈ N, then xgL = g(x g L) = gL since x g ∈ N. Since V 󸀠 = ∑g∈G gL is a nonzero submodule of the irreducible KG-module V, it follows that V 󸀠 = V. But then, as we know, we have V N = g1 L ⊕ ⋅ ⋅ ⋅ ⊕ g m L, where g i are elements of G. The partition of the set {g i L | i ∈ {1, . . . , m}} into classes of isomorphic KN-submodules corresponds to this decomposition. Summing up the submodules g i L belonging to the same class, we obtain the homogeneous constituents M j (j ∈ {1, . . . , k}) of V N . One can show that these constituents M j do not depend on the choice of the initial direct decomposition V N = ⨁m i=1 g i L and, hence, they are uniquely determined by the module V N (see §I.10). The group G acts transitively on the set {M1 , . . . , M k } via multiplication. To fix ideas, let L ⊆ M1 . The stabilizer of M1 is the subgroup H = {g ∈ G | gL ≅ L as KN-modules} = {g ∈ G | Ψ g ∼ Ψ}. Clearly, M1 is a KN-module. Let G = ⋃kj=1 t j H be the decomposition of G into left cosets of H, t1 = 1 ∈ G. We can arrange the N-submodules M j in such a way that M j = t j M1 (j ∈ {1, . . . , k}). Since V = t1 M1 ⊕ ⋅ ⋅ ⋅ ⊕ t k M1 , we have V = M1G and, consequently, ∆ G ∼ T, where ∆ is the representation of H afforded by the KH-module M1 . The irreducibility of T implies the irreducibility of ∆. Since (M1 )N is decomposable into the sum of e irreducible submodules isomorphic to L, it follows that ∆ N ∼ eΨ, where Ψ is the representation of N afforded by one of these terms. In particular, if K = ℂ, χ is a character of T, θ is a character of ∆ and ψ is a character of Ψ, then H = IG (ψ) is the

304 | Characters of Finite Groups 1 inertia group of ψ in G, θ ∈ Irr(H),

χ = θG ,

θ N = eψ.

These results in case of finite groups will be reproved later by methods of character theory. It should note that the field K of Theorem 1.4 is arbitrary, and, in contrast to Maschke’s theorem, the characteristic of K may divide the order of G. Exercise 1.2. Let H be a subnormal subgroup of a group G and χ ∈ Irr(G). Consider decomposition of χ H in irreducible constituents. Is it true that all irreducible constituents of χ H have equal degrees? Hint. Let N = H G , the normal closure of H in G. If H < G, then N ⊲ G. Then χ H = (χ N )H . Apply Clifford’s theorem to H < N and use induction. Let n(G) denote the number of nonlinear irreducible characters of a group G: n(G) = |Irr1 (G)| = |Irr(G)| − |Lin(G)|. Proposition 1.5 (Kazarin). Let G be a p-group with |G󸀠 | = p k . If k is even, then one has n(G) ≥ 21 (p2 − 1)k. If k is odd, then n(G) ≥ 12 (p2 − 1)(k − 1) + (p − 1). In other words, if k = 2s + e, e ∈ {0, 1}, then n(G) ≥ (p2 − 1)s + (p − 1)e. Proof. Let |G| = p m and N ≤ G󸀠 is G-invariant. Clearly, Irr(G | N) = {χ ∈ Irr(G) | N ≰ ker(χ} ⊆ Irr1 (G). (i) Suppose that |N| = p; then N ≤ Z(G). Let Lin(N) = {1N , μ1 , . . . , μ p−1 } (all μ are G-invariant). Then Irr(μ Gi ) ⊆ Irr1 (G) and, by reciprocity and the Clifford theorem, Irr(μ Gi ) ∩ Irr(μ Gj ) = 0 for i ≠ j. It follows that |Irr1 (G)| ≥ |Lin(N)| − 1 = p − 1. (ii) Suppose that |N| = p2 . Then p m−2 (p2 − 1) = |G| − |G/N| =

∑

χ(1)2 ≡ |Irr1 (G | N)| mod (p2 − 1)

χ∈Irr1 (G|N)

and we conclude that p2 − 1 | |Irr1 (G | N)|. Since |Irr(G | N)| > 0, we get |Irr(G | N)| ≥ p2 − 1. In view of (i) and (ii), one may assume in what follows that |G󸀠 | = p k > p2 . We proceed by induction on k. Set k = 2s + e, where e ∈ {0, 1}. We have a partition Irr1 (G) = Irr1 (G/N) ∪ Irr(G | N) or, what is the same, n(G) = n(G/N) + |Irr(G | N)|. Recall that Irr(G | N) = {χ ∈ Irr(G) | N ≰ ker(χ)}.

(8)

VII Clifford theory | 305

(iii) Let e = 1. Then, by induction, n(G/N) ≥ (p2 − 1)s = 12 (p2 − 1)(k − 1). Since |Irr(G | N)| ≥ p − 1, by (i), we get, by (4), n(G) ≥

1 2 (p − 1)(k − 1) + (p − 1). 2

(iv) Let e = 0 and let N be of order p2 as in (ii). Then, by (4), (ii) and induction, we get 1 n(G) ≥ (p2 − 1)(s − 1) + (p2 − 1) = (p2 − 1)s = (p2 − 1)k. 2

2 Ito’s theorem on degrees We retain the previous notation and assumptions. In particular, as in the previous section, N ⊴ G. Throughout remaining part of this chapter, K = ℂ. Given ψ ∈ Irr(N), define the set Irrψ (G) = {χ ∈ Irr(G) | ⟨χ N , ψ⟩ > 0} = Irr(ψ G ). Theorem 1.3 implies that χ ∈ Irrψ (G) if and only if χ N = eσ ψ , where e ∈ ℕ. By reciprocity, Irrψ (G) = {χ ∈ Irr(G) | ⟨ψ G , χ⟩ > 0} 󳨐⇒ Irrψ (G) = Irr(ψ G ), where Irr(ψ G ) is the set of irreducible constituents of the induced character ψ G . Therefore, Irrψ (G) ≠ 0 for any ψ ∈ Irr(N). Then χ ∈ Irrψ (G) 󳨐⇒ ⟨ψ G , χ⟩ = ⟨ψ, χ N ⟩ = e χ ; hence ψG =

(1)

e χ χ,

∑ χ∈Irrψ (G)

and therefore, by the First Orthogonality Relation, e2χ .

(2)

⟨ψ G , ψ G ⟩ = |IG (ψ)/N|.

(3)

⟨ψ G , ψ G ⟩ =

∑ χ∈Irrψ (G)

Theorem 2.1. If ψ ∈ Irr(N), then

Proof. By (1.4) and reciprocity, ⟨ψ G , ψ G ⟩ = ⟨ψ, (ψ G )N ⟩ = ⟨ψ, |N|−1 ∑ ψ g ⟩ = |N|−1 ∑ ⟨ψ, ψ g ⟩. g∈G

g∈G

Since ⟨ψ, ψ g ⟩ = 1 if g ∈ IG (ψ), and ⟨ψ, ψ g ⟩ = 0 otherwise, it follows that ⟨ψ G , ψ G ⟩ = |IG (ψ)/N|, completing the proof.

306 | Characters of Finite Groups 1 Exercise 2.1. If ψ ∈ Irr(N), then ∑

e2χ = |IG (ψ)/N|.

χ∈Irrψ (G)

Exercise 2.2. If ψ ∈ Irr(N), then ψ G ∈ Irr(G) if and only if IG (ψ) = N. Hint. Compare ψ G (1) and χ(1), where χ = ψ G ∈ Irr(ψ G ). Now we are ready to prove the following important theorem (recall that, by agreement, N ⊴ G): Theorem 2.2. If ψ ∈ Irr(N) and H = IG (ψ) (the inertia group of ψ in G), then the following statements hold: (a) θ G ∈ Irrψ (G) for any θ ∈ Irrψ (H)(= Irr(ψ H )). (b) If χ ∈ Irrψ (G), θ ∈ Irrψ (H) and θ G = χ, then θ N = e χ ψ (i.e., e θ = e χ ), and ⟨χ H , θ⟩ = 1. (c) The mapping θ 󳨃→ θ G (θ ∈ Irrψ (H)) is a bijection of Irrψ (H) onto Irrψ (G); in particular, |Irrψ (H)| = |Irrψ (G)|. Proof. (a) Let θ ∈ Irrψ (H), χ ∈ Irr(θ G ) so that ⟨θ G , χ⟩ > 0. Then, by reciprocity, one has ⟨χ H , θ⟩ = ⟨χ, θ G ⟩ > 0 hence χ H = θ + λ,

where λ ∈ Char(H) ∪ {0H→ℂ };

here 0H→ℂ is the zero function. Since θ ∈ Irrψ (H) and ψ is invariant with respect to H(= IG (ψ) ≥ N), it follows by Clifford’s Theorem 1.3 that θ N = e θ ψ. Therefore, we have χ N = θ N + λ N = e θ ψ + λ N , and consequently, e χ = ⟨χ N , ψ⟩ = ⟨θ N , ψ⟩ + ⟨λ N , ψ⟩ = e θ + ⟨λ N , ψ⟩ ≥ e θ > 0, that is, χ ∈ Irrψ (G) and e χ ≥ e θ . On the other hand, by reciprocity, ⟨θ G , χ⟩ = ⟨θ, χ H ⟩ = ⟨θ, θ + λ⟩ = ⟨θ, θ⟩ + ⟨θ, λ⟩ ≥ ⟨θ, θ⟩ = 1 > 0. Therefore, θ G = χ + ϑ, where ϑ is either the zero function or a character of G. Hence, we obtain θ G (1) = χ(1) + ϑ(1) ≥ χ(1). Since χ(1) = e χ |G : H|ψ(1) (in fact, χ N has exactly |G : IG (ψ)| = |G : H| irreducible constituents which are G-conjugates of ψ) and θ G (1) = |G : H|θ(1) = |G : H|e θ ψ(1) ≥ χ(1) = e χ |G : H|ψ(1), it follows that e θ ≥ e χ . Thus, e θ = e χ and θ G (1) = χ(1). Therefore, θ G = χ, completing the proof of (a). (b) Let χ ∈ Irrψ (G), χ = θ G , θ ∈ Irrψ (H). In this situation, as we have seen (see (a)), e χ = e θ , θ N = e χ ψ,

θ G = χ,

⟨χ H , θ⟩ = ⟨χ, θ G ⟩ = ⟨χ, χ⟩ = 1,

and (b) is proved. (c) Let χ ∈ Irrψ (G). Then χH =

∑ ⟨χ H , θ⟩θ 󳨐⇒ χ N = θ∈Irr(H)

∑ ⟨χ H , θ⟩θ N , θ∈Irr(H)

VII Clifford theory | 307

so that ∑ ⟨χ H , θ⟩ ⋅ ⟨θ N , ψ⟩ = ⟨χ N , ψ⟩ = e χ > 0. θ∈Irr(H)

Therefore, ⟨χ H , θ⟩ ⋅ ⟨θ N , ψ⟩ > 0 for some θ ∈ Irr(H). This means that ⟨χ H , θ⟩ > 0 and θ ∈ Irrψ (H). But θ G ∈ Irrψ (G), by (a). It follows from ⟨χ, θ G ⟩ = ⟨χ H , θ⟩ > 0 that θ G = χ. Therefore, the mapping θ 󳨃→ θ G (θ ∈ Irrψ (H)) is a surjection of Irrψ (H) onto Irrψ (G). Assume that θ, θ1 ∈ Irrψ (H) and θ G = θ1G = χ. Then, by (a) and (b), χ ∈ Irrψ (G) and θ N = (θ1 )N = e χ ψ. Next, by (b), ⟨χ H , θ⟩ = 1, and, similarly, ⟨χ H , θ1 ⟩ = 1. Therefore, supposing that θ ≠ θ1 , we obtain χ H = θ + θ1 + ω, where ω is either the zero function or a character of H. By what was proved above, χ N = θ N + (θ1 )N + ω N = 2e χ ψ + ω N . But χ N = e χ ψ. This contradiction proves that the mapping θ 󳨃→ θ G is a bijection of Irrψ (H) onto Irrψ (G). The following important theorem, due to N. Ito, allows us to control the degrees of irreducible characters of a group G containing nontrivial abelian normal subgroup. Theorem 2.3 (Ito [Ito2]). If N ⊲ G is abelian and χ ∈ Irr(G), then χ(1) divides the index of N in a group G. Proof. Let ψ ∈ Irr(χ N ); then ψ ∈ Lin(N) since N is abelian. In that case, χ ∈ Irrψ (G), by reciprocity, and therefore χ = θ G for θ ∈ Irrψ (H), where H = IG (ψ), the inertia group of ψ in G (Theorem 2.2). Again, by Theorem 2.2, θ N = e χ ψ so that N ≤ Z(θ) (the quasikernel of θ) since ψ(1) = 1. Therefore, by Theorem IV.8.4, θ(1) | |H : N| hence χ(1) = θ G (1) = |G : H|θ(1) | |G : H| ⋅ |H : N| = |G : N|. In particular, if χ ∈ Irr(G), then χ(1) | |G : Z(χ)| (Theorem IV.8.4). It is evident from the proof of Theorem 2.3 that e χ | |IG (ψ) : N|. In §VII.3 we shall show that more general fact is true even if N is nonabelian. Exercise 2.3 ([Isa11, Problem 2.19]). Let E = ⟨x1 , x2 , x3 , x4 ⟩ be an elementary abelian group of order 16. Let P = ⟨y⟩ be cyclic of order 3. Let P act on E by y

x1 = x2 ,

y

x2 = x1 x2 ,

y

x3 = x4 ,

y

x4 = x3 x4

and let G = P ⋅ E be the natural semidirect product. Show that Z(G) = {1} but that G has no a faithful irreducible character. Solution. Suppose that (x1a x2b x3c x4d )y = x1a x2b x3c x4d . One may assume that a, b, c, d ∈ {0, 1}. Since x1a x2b x3c x4d = (x1a x2b x3c x4d )y = x1b x2a+b x3d x4c+d , it follows that a = b = c = d = 0 so, since Z(G) ≤ E, we get Z(G) = {1}. Now assume that χ ∈ Irr(G) is faithful. Then χ(1) = 3, by Theorem 2.3 so that χ E is a faithful character

308 | Characters of Finite Groups 1 of degree 3 of elementary abelian 2-group E of rank 4. Hence, χ E has at most three irreducible constituents. Intersection of their kernels, which are maximal in E, is ≠ {1}, and this is a contradiction. Let Irr(n) (G) = {χ ∈ Irr(G) | χ(1) = n}. Exercise 2.4. A nonabelian 2-group G has an odd number of irreducible characters of degree 2 if and only if G is of maximal class. Solution. If G is of maximal class and order 2n , then cd(G) = {1, 2} (Theorem 2.3) and |G : G󸀠 | = 4 so the number |G| − |G : G󸀠 | |Irr(2) (G)| = |Irr1 (G)| = = 2n−2 − 1 22 is odd. Now suppose that G is a nonabelian 2-group, n > 3 and |Irr(2) (G)| = 2k + 1. Then 2n = |G| ≡ |G : G󸀠 | +

∑

χ(1)2 ≡ |G : G󸀠 | + 4(2k + 1) (mod 16).

χ∈Irr(2) (G)

Since |G : G󸀠 | is a power of 2, we conclude that |G : G󸀠 | = 4. In that case, by Taussky’s theorem [Ber31, Proposition 1.6], G is of maximal class. Exercise 2.5. Let G be a p-group, p > 2. If |Irr(p) (G)| ≢ 0 (mod p), then |Irr(p) (G)| ≡ −1 (mod p) and

|G : G󸀠 | = p2 .

Classify the p-groups G such that |Irr(p) (G)| = p − 1. Solution of the last part. Let |Irr(p) (G)| = sp − 1. Then |G| = |G : G󸀠 | + (sp − 1)p2 + kp4 ≡ |G : G󸀠 | − p2 ≡ 0 (mod p3 ). It follows that |G : G󸀠 | = p2 . Now assume, in addition, that s = 1. If |G| = p n ≥ p4 and N < G󸀠 is G-invariant of index p2 , then |Irr1 (G/N)| = |Irr(p) (G/N)| =

p4 − p2 = p2 − 1 > p − 1, p2

a contradiction. Thus N does not exist so |G󸀠 | = p and, |G| = p3 . Theorem 2.4 ([Isa11, Problem 6.3]). Let N ⊲ G, ψ ∈ Irr(N) and χ ∈ Irr(ψ G )(= Irrψ (G)). Then the following conditions are equivalent: (a) χ N = eψ and e2 = |G : N|. (b) ψ is G-invariant and χ vanishes on G − N. (c) |Irr(ψ G )| = 1 and ψ is G-invariant. Proof. If ψ G = ∑θ∈Irr(ψ G ) e θ θ, then (see (2) and (3)) |IG (ψ) : N| = ⟨ψ G , ψ G ⟩ =

∑

e2θ .

(4)

θ∈∈Irr(ψ G )

If θ ∈ Irr(ψ G ), then θ N = e θ ∑ti=1 ψ i is the Clifford decomposition, where ψ1 = ψ and t = |G : IG (ψ)|.

VII Clifford theory | 309

Suppose that (a) holds. It follows from χ N = eψ that t = 1 and ψ is G-invariant. Next, e2 = |G : N| and (4) imply that Irr(ψ G ) = {χ}. In that case, ψ G = e χ χ vanishes on G − N so the same is true for χ. Thus, (a) implies (b) and (c). Now assume that χ vanishes of G − N and ψ is G-invariant so that χ N = e χ ψ, by Clifford’s theorem, i.e., (b) holds. In that case, t = 1 and hence, by (2), ∑

e2θ = |G : N|.

θ∈∈Irr(ψ G )

One has

|G| = ∑ |χ(g)|2 = ∑ |χ(g)|2 = e2χ ∑ |ψ(g)|2 = e2χ |N| g∈G

g∈N

g∈N

(the last equality follows from the First Orthogonality Relation), therefore, e2χ = |G : N|. Thus, (b) implies (a). Now let (c) hold. Then χ vanishes on G − N so (b) holds. By the previous paragraph, (a) holds. Thus, (c) implies (a) and (b). We shall show, in Exercise 3.14, that Theorem 2.3 remains to be true under assumption that N is subnormal in G.

3 Clifford’s theorem on ramification As before, let ψ ∈ Irr(N), where N ⊴ G. Theorem 2.2 reduces the description of the set Irrψ (G) = Irr(ψ G ) to the description of Irrψ (H), where H = IG (ψ) = {g ∈ G | ψ g = ψ} is the inertia subgroup of ψ in G. The character ψ is H-invariant. The solution of this problem requires projective representation theory. We retain the notation introduced in this paragraph. Since the case N = G is trivial, we assume in what follows that N < G. Let Ψ be an irreducible linear representation of N ⊲ G affording the character ψ. Since ψ is H-invariant, Ψ h ∼ Ψ for any h ∈ H (indeed, Ψ h and Ψ have equal characters; see Chapter II). Consequently, for x ∈ N, Ψ h (x) = P(h)Ψ(x)P(h)−1 ,

(1)

where P(h) ∈ GL(f, ℂ), f = deg(Ψ) = ψ(1). Lemma 3.1. We retain the above notation. (a) P is a projective representation of H = IG (ψ). (b) P is determined by the character ψ up to projective equivalence; if the representation Ψ (with character ψ) is given, then P is uniquely determined up to a factor μ ∈ F[H, ℂ∗ ]. (c) The class of the factor set of P is uniquely determined by ψ. (d) The restriction P N = λ−1 Ψ, where λ ∈ F[N, ℂ∗ ]. (e) P is irreducible.

310 | Characters of Finite Groups 1 Proof. (a) Replacing h by st (s, t ∈ H) in (1) and observing that Ψ st = (Ψ s )t , one obtains P(st)Ψ(x)P(st)−1 = (Ψ s )t (x) = Ψ s (txt−1 ) = P(s)Ψ(txt−1 )P(s)−1 = P(s)Ψ t (x)P(s)−1 = (P(s)P(t))Ψ(x)(P(s)P(t))−1

(s, t ∈ H).

Since Ψ is irreducible, it follows from Schur’s lemma that P(st)−1 P(s)P(t) = π(s, t)−1 If ,

where π(s, t) ∈ ℂ∗ ,

whence P(s)P(t) = π(s, t)−1 P(st).

(2)

It follows from (2) that P is a projective representation of H belonging to the factor set π−1 . (b, c) Since Ψ is irreducible, it follows from Schur’s lemma that the matrix P(h) in (1) is determined up to a factor μ(h) ∈ ℂ∗ , μ ∈ F[H, ℂ∗ ]. For given ψ, the representation Ψ is determined up to equivalence; hence, if Ψ is not fixed, P will be uniquely determined up to projective equivalence. Therefore, the equivalence class of the factor set π is uniquely determined by ψ. (d) If s, x ∈ N, then Ψ s (x) = Ψ(sxs−1 ) = Ψ(s)Ψ(x)Ψ(s)−1 since Ψ ∈ ILIN(N) (the set of irreducible linear representations of N). On the other hand, Ψ s (x) = P(s)Ψ(x)P(s)−1 , by (1). Hence, the matrix Ψ(s)−1 P(s) commutes with matrix Ψ(x) for all x ∈ N. Therefore, by Schur’s lemma, Ψ(s)−1 P(s) is a scalar matrix, i.e., P(s) = λ(s)−1 Ψ(s), where λ ∈ F[N, ℂ∗ ]. Hence P N = λ−1 Ψ. (e) By (d), since Ψ is irreducible, it follows that P N is irreducible so is P. Lemma 3.2. Let A = (α ij ) and B = (β ij ) be m × m matrices and C an arbitrary nonzero square matrix. Then A ⊗ C = B ⊗ C implies A = B. Proof. This follows from the form of the tensor product of two matrices. If P1 and P2 are projective representations of G, then (as in the case of linear representations) their tensor product is defined by (P1 ⊗ P2 )(g) = P1 (g) ⊗ P2 (g)

(g ∈ G).

If π1 , π2 , π are the factor sets of P1 , P2 , P1 ⊗ P2 , respectively, and s, t ∈ G, then (P1 (s) ⊗ P2 (s))(P1 (t) ⊗ P2 (t)) = P1 (s)P1 (t) ⊗ P2 (s)P2 (t) = π1 (s, t)P1 (st) ⊗ π2 (s, t)(P2 (st) = π1 (s, t)π2 (s, t)(P1 (st) ⊗ P2 (st)) = π1 (s, t)π2 (s, t)(P1 ⊗ P2 )(st),

VII Clifford theory | 311

i.e., π = π1 π2 and P1 ⊗ P2 is a projective π1 π2 -representation. Thus, if P1 is a π-representation and P2 a π−1 -representation, then P1 ⊗ P2 is a linear representation. Lemma 3.3. Let G be a group, N ⊲ G, ψ ∈ Irr(N), H = IG (ψ), θ ∈ Irrψ (H). Then the following statements hold: (a) There is an irreducible linear representation ∆ of H affording θ such that ∆ N = eΨ, where e = e θ is the ramification of θ over N and Ψ ∈ ILIN(N) with the character ψ. (b) There is a π-representation Ω of H of degree e such that ∆ = Ω ⊗ P (for the definition of P see (1)). (c) Ω is uniquely determined by ∆. (d) Ω is irreducible. (e) Ω satisfies the condition Ω(x) = λ(x)Ie (x ∈ N, λ ∈ F[H, ℂ∗ ]). Proof. (a) Let ∆0 be a linear representation of H affording the character θ. Since ψ is H-invariant, we get θ N = e θ ψ, by Clifford’s theorem. Consequently, (∆0 )N ∼ e θ Ψ, where Ψ ∈ ILIN(N) affords the character ψ. Therefore, there exists an irreducible linear representation ∆ of H such that ∆ ∼ ∆0 and ∆ N = e θ Ψ. It is obvious that θ is the character of ∆. (b) If x ∈ N, then, by (a), ∆(x) is an ef × ef block-diagonal matrix, where e = e θ , f = deg(Ψ) = ψ(1), with the block Ψ(x) on the principal diagonal. This matrix can be expressed in the more compact form ∆(x) = Ie ⊗ Ψ(x). Therefore, ∆ h (x) = Ie ⊗ Ψ h (x) for any h ∈ H. Since ∆ ∈ ILIN(H), one can rewrite the last equality as follows (see (1)): ∆(h)∆(x)∆(h)−1 = ∆ h (x) = Ie ⊗ (P(h)Ψ(x)P(h)−1 ).

(3)

Since the right-hand side of (3) is equal to (Ie ⊗ P(h))(Ie ⊗ Ψ(x))(Ie ⊗ P(h))−1 = Φ(h)∆(x)Φ(h)−1 , where Φ(h) = Ie ⊗ P(h), it follows that the matrix Φ(h)−1 ∆(h) commutes with ∆(x) for all x ∈ N. Writing Φ(h)−1 ∆(h) in the block form (Ω ij (h))1≤i,j≤e , where Ω ij (h) are f × f blocks, we easily see that the blocks Ω ij (h) commute with Ψ(x). Since Ψ is irreducible, Schur’s lemma implies Ω ij (h) = ω ij (h)If ,

where ω ij (h) ∈ ℂ

(i, j ∈ {1, . . . , e}).

Consequently, Φ(h)−1 ∆(h) = (ω ij (h)If ) = Ω(h) ⊗ If ,

where Ω(h) = (ω ij (h))1≤i,j≤e .

Therefore, ∆(h) = Φ(h)(Ω(h) ⊗ If ) = (Ie ⊗ P(h))(Ω(h) ⊗ If ) = Ω(h) ⊗ P(h). If s, t ∈ H, then, by (4), Ω(s)Ω(t) ⊗ P(s)P(t) = (Ω(s) ⊗ P(s)) ⋅ (Ω(t) ⊗ P(t)) = ∆(s)∆(t) = ∆(st) = Ω(st) ⊗ P(st).

(4)

312 | Characters of Finite Groups 1 Hence, by (2), since P(s)P(t) = π(s, t)−1 P(st), we obtain Ω(s)Ω(t) ⊗ P(st) = π(s, t)Ω(st) ⊗ P(st). As P(st) ≠ 0 (zero matrix), Lemma 3.2 implies that Ω(s)Ω(t) = π(s, t)Ω(st). By (4), the matrix Ω(h) (h ∈ H) is nonsingular. Indeed, det(∆(h)) = det((Ie ⊗ P(h))) det((Ω(h) ⊗ If )) = (det(P(h)))e (det(Ω(h)))f , hence, taking into account that det(∆(h)) ≠ 0, we obtain det(Ω(h)) ≠ 0. We have thus proved that Ω is a π-representation of H, and moreover, deg(Ω) = e. Finally, it follows from (4) that ∆ = Ω ⊗ P, and (b) is proved. (c) Suppose that, together with the decomposition (b) in the statement of the lemma, we also have ∆ = Ω󸀠 ⊗ P, where Ω󸀠 is a π-representation of H. Then we get Ω󸀠 (h) ⊗ P(h) = Ω(h) ⊗ P(h)

for all h ∈ H.

Hence, Ω󸀠 (h) = Ω(h), by Lemma 3.2, so that Ω󸀠 = Ω, and (c) is proved. (d) Suppose that Ω is reducible. Then Ω is linearly equivalent to Ω1 + Ω2 , where Ω1 and Ω2 are π-representations of H. Let us form the linear representations ∆ i = Ω i ⊗ P (i = 1, 2) of H. Let γ, θ i , ω, ω i denote the characters (traces) of the representations P, ∆ i , Ω, Ω i , respectively (i = 1, 2). Then ω = ω1 + ω2 and θ i = ω i γ (i = 1, 2). Since ∆ = Ω ⊗ P and the character of ∆ is equal to θ, we have θ = ωγ = ω1 γ + ω2 γ = θ1 + θ2 , but this contradicts the irreducibility of θ. Thus, Ω is irreducible. (e) Let x ∈ N. Then ∆(x) = Ie ⊗ Ψ(x). On the other hand, by (4) and Lemma 3.1 (d), ∆(x) = Ω(x) ⊗ P(x) = Ω(x) ⊗ λ(x)−1 Ψ(x) = λ−1 (x)Ω(x) ⊗ Ψ(x)

(λ ∈ F[H, ℂ∗ ]).

Comparing these two expressions for ∆(x), we get, by Lemma 3.2, λ(x)−1 Ω(x) = Ie

(λ ∈ F[H, ℂ∗ ]),

i.e.,

Ω(x) = λ(x)Ie

((x ∈ N),

and (e) is proved. Exercise 3.1. In the notation of Lemma 3.3, the subgroup N is a π-kernel. Hint. One has N ≤ KER(Ω), where Ω is a π-representation (Lemma 3.3 (b)) i.e., N is a π-kernel (see Chapter VI). Therefore, π from Exercise 3.1 is associated with a certain factor set compatible with N (see Chapter VI). Multiplying P by a suitable function μ ∈ F[H, ℂ∗ ], if necessary, one may assume that π compatible with N, i.e., π(s󸀠 , t󸀠 ) = π(s, t) if Ns󸀠 = Ns, Nt󸀠 = Nt,

(5)

and π(s, t) = 1

if s ∈ N or t ∈ N.

From now on, we assume that (5) and (6) are true.

(6)

VII Clifford theory | 313

Let Iπ,λ (H) be the set of all irreducible π-representations Ω of the group H = IG (Ψ) such that λ−1 Ω N is the identity representation of N for some λ ∈ F[N, ℂ∗ ]. Then Iπ,λ (H) ≠ 0 (Lemma 3.3). Exercise 3.2. If Ω ∈ Iπ,λ (H), then λ(x)Ω(h) = Ω(xh) (x ∈ N, h ∈ H). Solution. Indeed, if deg(Ω) = e, then Ω(xh) = Ω(x)Ω(h) = λ(x)Ie Ω(h) = λ(x)Ω(h). Lemma 3.4. The function λ of Exercise 3.2 belongs to Lin(N) and is H-invariant. Proof. If x ∈ N, then Ω(x) = λ(x)Ie , e = deg(Ω). If x1 , x2 ∈ N, then λ(x1 )λ(x2 )Ie = Ω(x1 )Ω(x2 ) = π(x1 , x2 )Ω(x1 x2 ) = Ω(x1 x2 ) = λ(x1 x2 )Ie (since π(x1 , x2 ) = 1). This implies λ(x1 )λ(x2 ) = λ(x1 x2 ), i.e., λ ∈ Lin(N). Next, if x ∈ N, h ∈ H, then Ω(h)Ω(x h ) = π(x, x h )Ω(hx h ) = π(x, x h )Ω(xh) = λ(x)Ω(h), so that Ω(x h ) = λ(x)Ie . Hence λ(x h ) = λ(x), i.e., λ is an H-invariant linear character of N. Now we establish a one-to-one correspondence between the set Iπ,λ (H) and the set of irreducible projective representations of the quotient group H = H/N belonging to some factor set π̃ ∈ Z 2 (H, ℂ∗ ). Let H = ⋃α∈H Nh α be the decomposition of H into cosets of N, where, by assumption, h1 = 1,

h α h β = f(α, β)h αβ ,

f(α, β) ∈ N

(α, β ∈ H).

(7)

Let Ω ∈ Iπ,λ (H), deg(Ω) = e. Define a mapping Ω̃ : H → GL(e, ℂ) by ̃ Ω(α) = Ω(h α )

(α ∈ H).

Lemma 3.5. The above defined Ω̃ is an irreducible projective representation of degree e of H. The factor set π̃ of this representation is defined by ̃ π(α, β) = π(h α , h β )λ(f(α, β))

(α, β ∈ H).

(8)

Proof. If α, β ∈ H, then (see Exercise 3.2) ̃ Ω(β) ̃ Ω(α) = Ω(h α )Ω(h β ) = π(h α , h β )Ω(h α h β ) = π(h α , h β )Ω(f(α, β)h αβ ) = π(h α , h β )π−1 (f(α, β), h αβ )Ω(f(α, β))Ω(h αβ ) = π(h α , h β )λ(f(α, β)Ω(h αβ ) ̃ Ω(β) ̃ ̃ ̃ ̃ since π(f(α, β), h αβ ) = 1, by (6). Thus, Ω(α) = π(α, β)Ω(αβ), where π(α, β) is ̃ defined in (8) and Ω(h αβ ) = Ω(αβ); see the display preceding the lemma. To prove that Ω̃ is irreducible, we note that any element of H admits a unique presentation of the form xh α , where x ∈ N and α ∈ H(= H/N). Since, by Exercise 3.2, ̃ Ω(xh α ) = λ(x)Ω(h α ) = λ(x)Ω(α),

314 | Characters of Finite Groups 1 the irreducibility of Ω̃ follows from the irreducibility of Ω. The definition of Ω̃ implies ̃ = deg(Ω) = e. that deg(Ω) ̃ Let Iπ̃ (H) be the set of all irreducible projective π-representations of the quotient group H. By Lemma 3.5, Ω ∈ Iπ,λ (H) implies Ω̃ ∈ Iπ̃ (H). We thus have a mapping Ω 󳨃→ Ω̃ from Iπ,λ (H) to Iπ̃ (H). Lemma 3.6. The mapping Ω 󳨃→ Ω̃ is a bijection of Iπ,λ (H) onto Iπ̃ (H). Proof. (i) Let Γ ∈ Iπ̃ (H). Define a mapping Ω : H → GL(deg(Γ), ℂ) by Ω(xh α ) = λ(x)Γ(α)

(x ∈ N, α ∈ H(= H/N)).

(9)

Given x, y ∈ N and α, β ∈ H, we obtain ̃ Ω(xh α )Ω(yh β ) = λ(x)Γ(α)λ(y)Γ(β) = λ(x)λ(y)π(α, β)Γ(αβ). Putting y α = h α yh−1 α , we get, in view of (6) and (9), π(xh α , yh β )Ω(xh α ⋅ yh β ) = π(h α , h β )Ω(xy α f(α, β)h αβ ) = π(h α , h β )λ(xy α f(α, β))Γ(αβ). Since the linear character λ is H-invariant, it follows that λ(xy α f(α, β)) = λ(x)λ(y)λ(f(α, β)). Therefore, π(xh α , yh β )Ω(xh α yh β ) = π(h α , h β )λ(x)λ(y)λ(f(α, β))Γ(αβ) ̃ = λ(x)λ(y)π(α, β)Γ(αβ). Thus, Ω(xh α )Ω(yh β ) = π(xh α , yh β )Ω(xh α ⋅ yh β ), i.e., Ω is a π-representation of H. The fact that Ω is irreducible follows from the irreducibility of Γ and (9). Next, if x ∈ N, then, by (9), we obtain Ω(x) = Ω(xh1 ) = λ(x)Γ(1) = λ(x)Ie (π is normalized because h1 = 1; hence Γ(1) = Ie , where e = deg(Γ)). Thus, Ω ∈ Iπ,λ (H). But ̃ Ω(α) = Ω(h α ) = Ω(1 ⋅ h α ) = λ(1)Γ(α) = Γ(α), and therefore Γ = Ω.̃ Thus, the mapping Ω 󳨃→ Ω̃ is a surjection of Iπ,λ (H) onto Iπ̃ (H). (ii) Let Γ ∈ Iπ̃ (H) and Γ = Ω,̃ where Ω ∈ Iπ,λ (H). If x ∈ N and α ∈ H, then, using ̃ Exercise 3.2 and the equality Ω(α) = Ω(h α ), where α ∈ H (see the paragraph follow-

ing (7)), we obtain

̃ Ω(xh α ) = λ(x)Ω(h α ) = λ(x)Ω(α) = λ(x)Γ(α). Therefore, the representation Ω ∈ Iπ,λ (H) satisfying the condition Γ = Ω,̃ is uniquely determined by Γ. Consequently, the mapping Ω 󳨃→ Ω̃ is an injection of the set Iπ,λ (H) into the set Iπ̃ (H).

VII Clifford theory | 315

The following result is basic and very important. Theorem 3.7 (Clifford [Cli]). Let ψ ∈ Irr(N) and χ ∈ Irrψ (G). Then the number e χ divides |IG (ψ)/N|. Proof. Let H = IG (ψ); then one has χ = θ G for some θ ∈ Irrψ (H) (see Theorem 2.2). Since, by the same theorem, e χ = e θ , it is sufficient to prove that e θ divides |H/N|. By ̃ where Ω̃ is an irreducible projective representation Lemmas 3.3 and 3.5, e θ = deg(Ω), of H/N. Therefore e θ | |H/N| (see Lemma VI.3.1).

Ito’s Theorem 2.3 follows easily from Theorem 3.7. Indeed, let A ⊲ G be abelian and χ ∈ Irr(G). Let χ A = e(μ1 + ⋅ ⋅ ⋅ + μ t ) be Clifford’s decomposition and H = IG (μ1 ). Then t = |G : H|, χ(1) = etμ1 (1) = et. By Theorem 3.7, e divides |H : A|. It follows that χ(1) = et | |H : A|t = |H : A||G : H| = |G : A|. Exercise 3.3 (Burnside). If G/N is cyclic, then e χ = 1 for χ ∈ Irr(G). In fact, in Exercise 3.3 one needs only the fact that IG (ψ)/N is cyclic. Since all irreducible projective representations of a cyclic group have degree 1, the equality e χ = 1 follows from Theorem 3.7. If, under hypothesis of Exercise 3.3, ψ ∈ Irr(N), then either ψ G ∈ Irr(G) or ψ G is the sum of |G : N| pairwise distinct irreducible characters of G of the same degree ψ(1). Next, if χ ∈ Irr(G), then either χ N ∈ Irr(N) or χ N = ψ1 + ⋅ ⋅ ⋅ + ψ|G:N| is the Clifford decomposition. This fact will be used in solution of Exercise 3.9. We will apply Exercise 3.3 to show that if the degrees of irreducible characters of G are 1, 1, 1, 2, 2, 2, 3, then G ≅ SL(2, 3). Let χ ∈ Irr(G) be of degree 2. Since |G : G󸀠 | = |Lin(G)| = 3, it follows from Exercise 3.3 that χ G󸀠 ∈ Irr(G󸀠 ) so G󸀠 is nonabelian. As |G| = 3 ⋅ 12 + 3 ⋅ 22 + 33 = 24, it follows that G󸀠 is nonabelian of order 8. Next, it follows from |G : G󸀠 | = 3 that G is not nilpotent. Hence, 3 | |Aut(G󸀠 )| so G󸀠 ≅ Q8 . It follows easily that then G ≅ SL(2, 3). Similarly, it is easy to show that if the degrees of irreducible characters of G are 1, 1, 2, 3, 3, then G ≅ S4 . Indeed, |G| = 24. If χ ∈ Irr(G) is of degree 3, then one has χ G󸀠 ∈ Irr(G󸀠 ) (Exercise 3.3) so that G󸀠 of order 12 has an irreducible character of degree 3, and we conclude that G󸀠 ≅ A4 . Since CG (G󸀠 ) = {1}, we get G ≅ S4 , as was to be shown. We suggest to the reader to show that if |G| = 24 and |G : G󸀠 | = 2, then G ≅ S4 . (Hint. If G󸀠 is nilpotent, then N of order 3 is G-invariant; then |G/N| = 8 and the derived group of G/N has index ≥ 4 > 2, which is a contradiction. Thus, G󸀠 is nonnilpotent so all Sylow subgroups are not normal in G. In this case, G ≅ S4 ; see Exercise XVIII.1.1.) We now assume that G/N is abelian. Theorem 3.8 ([Isa11, Problem 6.2]). Let G/N be abelian and ψ ∈ Irr(N). Then Irrψ (G) (= Irr(ψ G )) is an orbit with respect to the action of the group LinN (G)(≅ G/N) on the set Irr(G) by multiplication. The characters χ ∈ Irrψ (G) have the same ramification over N.

316 | Characters of Finite Groups 1 Proof. Let χ ∈ Irrψ (G). By Theorem V.1.2 (b), (χ N )G vanishes on G − N. If x ∈ N, then (χ N )G (x) = |N|−1 ∑ χ(t−1 xt) = |G : N|χ(x) t∈G

since χ N is G-invariant. Let ρ(N) be an inflation of the regular character ρ G/N of G/N to G: ρ(N) vanishes on G − N and ρ(N) (x) = |G : N| for x ∈ N (clearly, ρ(N) = (1N )G ). Then (see Lemma V.1.3) (χ N )G = (χ N ⋅ 1N )G = χ(1N )G = χρ(N) . On the other hand, since G/N is abelian, ρ(N) = ∑λ∈LinN (G) λ, and, consequently, one has (χ N )G = ∑λ∈LinN (G) λχ. But, by Clifford’s theorem, χ N = e χ (ψ1 + ⋅ ⋅ ⋅ + ψ t ), where {ψ1 , . . . , ψ t } is the G-orbit of the character ψ1 = ψ and t = |G : IG (ψ)|. Since the characters ψ i are G-conjugate, ψ Gi = ψ G (see Exercise V.1.3). Consequently, (χ N )G = e χ tψ G 󳨐⇒ e χ tψ G =

∑

λχ.

λ∈LinN (G)

Let {χ = χ1 , . . . , χ s } be the LinN (G)-orbit of χ and m χ = |StLinN (G) (χ)|, where StLinN (G) (χ) is the LinN (G)-stabilizer of χ. Then s

∑ λ∈LinN (G)

s

λχ = m χ ∑ χ i 󳨐⇒ e χ tψ G = m χ ∑ χ i 󳨐⇒ ψ G = i=1

i=1

mχ s ∑ χi . te χ i=1

mχ te χ

But = ∑ϑ∈Irrψ (G) e ϑ ϑ. Therefore, e ϑ = for all ϑ ∈ Irrψ (G). Thus, e ϑ = e χ for all χ, ϑ ∈ Irrψ (G), i.e., e ϑ independent of ϑ ∈ Irrψ (G). Moreover, Irrψ (G) = {χ1 = χ, . . . , χ s } is the LinN (G)-orbit of χ. As IG (ψ) = G and e χ = 1 for all χ ∈ Irrψ (G), it follows that |Irrψ (G)| = |G/N| (here we used Exercise 2.1). ψG

The problem of extending irreducible characters from N ⊴ G to G is of great interest. Exercise 3.4. If N ⊴ G and ψ ∈ Irr(N) admits an extension onto G, then the character ψ is G-invariant. Hint. Use Clifford’s theorem. Theorem 3.9. If G/N is cyclic, then every G-invariant irreducible character of N admits exactly |G/N| extensions to G. Proof. Let Irrinv (N) be the set of all G-invariant characters in Irr(N). Let ψ ∈ Irrinv (N) and χ ∈ Irrψ (G)(= Irr(ψ G )). Since IG (ψ) = G and e χ = 1 (Exercise 3.3), it follows by Clifford’s theorem, that χ N = ψ, i.e., χ is an extension of ψ to G. So Irrψ (G) is the set of all extensions of ψ to G. Since ψ G (1) = |G/N|ψ(1), the character ψ G is the sum of |G/N| pairwise distinct irreducible characters of G, extensions of ψ to G. Let N ⊲ G be of prime index p and ψ ∈ Irr(N). Then ψ G is either irreducible or one has ψ G = χ1 + ⋅ ⋅ ⋅ + χ p , where χ1 , . . . , χ p are pairwise distinct irreducible characters of G, by Theorem 3.9.

VII Clifford theory | 317

Again, let N ⊲ G, ψ ∈ Irr(N) and let H = IG (ψ). We will reduce the problem of describing the set Irrψ (H)(= Irr(ψ H )) to the determination of a certain one-to-one correspondence between Irrψ (H) and the equivalence classes of linear irreducible ̃ π-representations of the quotient group H = H/N. Let [T]L denote the class of linearly equivalent projective representations of some group G containing T. ̃ L is determined by Exercise 3.5. Let Ω ∈ Iπ,λ (H). Then each of the classes [Ω]L and [Ω] other. (For definition of the set Iπ,λ (H) see the paragraph preceding Exercise 3.2.) ̃ Hint. One has Ω(xh α ) = λ(x)Ω(α) (x ∈ N, α ∈ H). Exercise 3.6. There exists a one-to-one correspondence between the sets [Iπ,λ (H)]L = {[Ω]L | Ω ∈ Iπ,λ (H)}

̃ L | Ω̃ ∈ Iπ̃ (H)}. and [Iπ̃ (H)]L = {[Ω]

Hint. Lemma 3.6 and Exercise 3.5 enable us to define the required bijection. Let θ ∈ Irrψ (H) (here ψ ∈ Irr(N)). By Lemma 3.3, there exists an irreducible linear representation ∆ of H satisfying condition (a) of that lemma. Choosing one of such representations, ∆ θ say, we have (∆ θ )N = e θ Ψ. Again, by Lemma 3.3, there exists a unique projective representation Ω = Ω θ ∈ Iπ,λ (H) of degree e θ such that ∆ θ = Ω θ ⊗ P. Let ω θ denote the character of Ω θ ; then θ = ω θ ⋅ γ, where γ is the character of P defined previously. ̃ Let Irr(π)̃ (H) denote the set of all characters of irreducible π-representations of H = H/N. Given θ ∈ Irrψ (H), ω̃ θ denotes the character of the representation Ω̃ θ . If x ∈ N, α ∈ H and xh α ∈ Nh α = α, then (see the proof of Lemma 3.5) λ(x)ω̃ θ (α) = ω θ (xh α ).

(10)

Theorem 3.10. The mapping θ 󳨃→ ω̃ θ is a bijection of Irrψ (H) onto Irr(π)̃ (H). Proof. If θ, θ󸀠 ∈ Irrψ (H) and ω̃ θ = ω̃ θ󸀠 , then ω θ = ω θ󸀠 by (10), so θ = ω θ γ = ω θ󸀠 γ = θ󸀠 . Thus, the mapping θ 󳨃→ ω̃ θ is an injection. To prove that it is also surjective, we use Exercise 2.2, which implies that e2θ = |H/N| = |H|.

∑

(11)

θ∈Irrψ (H)

But e θ = deg(Ω̃ θ ); hence one can rewrite (11) in the following form: (deg(Ω̃ θ ))2 = |H|.

∑

(12)

θ∈Irrψ (H)

Since the characters ω̃ θ (θ ∈ Irrψ (H)) are pairwise distinct, the classes [Ω̃ θ ]L are also pairwise distinct. It follows from Theorem 2.2 and (11) that Ω̃ θ (θ ∈ Irrψ (H)) form ̃ a complete system of classes of irreducible π-representations of H. Therefore, {ω̃ θ | θ ∈ Irrψ (H)} = Irr(π)̃ (H), i.e., the mapping θ 󳨃→ ω̃ θ is surjective.

318 | Characters of Finite Groups 1 Exercise 3.7. The following statements hold: (a) The mapping θ 󳨃→ [Ω̃ θ ]L (θ ∈ Irrψ (H)) is a bijection of the set Irrψ (H) onto the set [Iπ̃ (H)]L . (b) The mapping θ 󳨃→ [Ω θ ]L is a bijection of the set Irrψ (H) onto the set [Iπ,λ (H)]L . Hint. For (a), see the end of the proof of Theorem 3.10, and (b) follows from (a) and Exercise 3.7. Let Irr(π,λ) (H) denote the set of all characters of representations Ω ∈ Iπ,λ (H). Exercise 3.8. The mapping θ 󳨃→ ω θ is a bijection of Irrψ (H) onto Irr(π,λ) (H). Hint. For injectivity, see the beginning of the proof of Theorem 3.10. Let ω ∈ Irr(π,λ) (H). Then ω is the character of a representation Ω ∈ Iπ,λ (H). Since Ω̃ ∈ Iπ̃ (H), it follows ̃ L = [Ω̃ θ ]L for some θ ∈ Irrψ (H) and hence [Ω]L = [Ω θ ]L from Exercise 3.7 (a) that [Ω] (see Exercise 3.6). Therefore, ω = ω θ and so our mapping is surjective. Corollary 3.11. The mapping ω 󳨃→ ωγ is a bijection of Irr(π,λ) (H) onto Irrψ (H). This follows from Exercise 3.8 and the equality θ = ω θ γ. Exercise 3.9. Let N ⊲ G be of index p, a prime. If s is the number of G-invariant members of the set Irr(N), then pk(G) = k(N) + (p2 − 1)s. Solution. We use the equality k(X) = |Irr(X)| for any group X. Take ϕ ∈ Irr(N). If ϕ is G-invariant, then |Irr(ϕ G )| = p. Indeed, if χ ∈ Irr(ϕ G ), then χ N = ϕ, by Clifford and Exercise 3.3. If ϕ is not G-invariant, then ϕ G ∈ Irr(G) since IG (ϕ) = N. Thus, inducpairwise distinct irreducible characters of G. ing from N produces exactly ps + k(N)−s p Since, for every χ ∈ Irr(G), there is ϕ ∈ Irr(N) ∩ Irr(χ N ), we get k(G) = ps +

k(N) − s ⇐⇒ pk(G) = k(N) + (p2 − 1)s. p

Remark. Suppose that, as in Exercise 3.9, N ⊲ G is of index p. Let t be the number of N-classes which are G-classes. By Burnside’s result (see [Gro, Proposition 3.1.4]), pk(G) = k(N) + (p2 − 1)t. It follows from the previous displayed formulas that s = t. This is a partial case of Brauer permutation Lemma X.3.1 (c). Exercise 3.10. If N ⊲ G and G/N is cyclic of order p n , then k(G) = p n k(N) − (p2 − 1)p n−1 h1 − (p4 − 1)p n−2 h2 − (p6 − 1)p n−3 h3 − ⋅ ⋅ ⋅ − (p2n−2 − 1)ph n−1 − (p2n − 1)h n , where h1 , . . . , h n ∈ ℕ ∪ {0}. Exercise 3.11. Let N ⊲ G, ψ ∈ Irr(N), χ ∈ Irr(ψ G ). Then

χ(1) ψ(1)

| |G : N|.

Solution. Let χ N = e(ψ1 + ⋅ ⋅ ⋅ + ψ t ) be the Clifford decomposition, ψ1 = ψ. Then χ(1) = etψ(1) = eψ(1)|G : IG (ψ)|.

VII Clifford theory | 319

By Theorem 3.7, e | |IG (ψ) : N|. Therefore, χ(1) = e|G : IG (ψ)| | |IG (ψ) : N||G : IG (ψ)| = |G : N|. ψ(1) It follows from Exercise 3.11 that if N is abelian, then χ(1) | |G : N| (this coincides with Ito’s Theorem 2.3). Exercise 3.12. If N ⊲ G, χ ∈ Irr(G) and GCD(χ(1), |G/N|) = 1, then χ N is irreducible. Solution. Write t = |Irr(χ N )|. Then both t and the ramification e χ of χ over N divide GCD(|G : N|, χ(1)) = 1 (see Theorems 1.3 and 3.7). Exercise 3.13. Let G/N be cyclic. All character ψ ∈ Irr(N) are G-invariant if and only if k(G) = |G/N|k(N). Solution. If χ ∈ Irr(G), then χ N ∈ Irr(N) (Exercise 3.3 and the hypothesis). Therefore, if ψ ∈ Irr(N), then |Irr(ψ G )| = |G/N| since all irreducible constituents of the induced character ψ G have the same degree ψ(1) (Theorem 3.8). If ν ∈ Irr(N) − {ψ}, then ⟨ψ G , ν G ⟩ = 0, and the result follows. Exercise 3.14 (Reynolds). Let S be a subnormal subgroup of a group G and χ ∈ Irr(G). Then χ(1) | |G : S|λ(1) for an appropriate λ ∈ Irr(S). Solution. In view of Exercise 3.11, one may assume that S ⋬ G; then G > N = S G > S, where S G is the normal closure of S in G. By Exercise 3.11, χ(1) | |G : N|ψ(1) for some ψ ∈ Irr(N). By induction, ψ(1) | |N : S|λ(1) for some λ ∈ Irr(S). It follows that χ(1) | |G : N|ψ(1) | |G : N||N : S|λ(1) = |G : S|λ(1).

4 Gallagher’s extension theorem We now consider the problem of extension of characters. As above, N ⊲ G, ψ ∈ Irr(N) and H = IG (ψ). We also retain the notation from §VII.3. Exercise 4.1. The numbers e θ (θ ∈ Irrψ (H)) form a complete system of degrees of irrẽ ducible projective π-representations of the quotient group H = H/N. Hint. See the proof of Theorem 3.10. Theorem 4.1. A character ψ ∈ Irr(N) is extendible to H = IG (ψ) if and only if π̃ ∼ 1. Proof. (i) Let θ ∈ Irr(H) be an extension of ψ to H. Then θ N = ψ, so that θ ∈ Irrψ (H) and e θ = 1. In that case, deg(Ω̃ θ ) = 1, i.e., we may identify Ω̃ θ with a function μ ∈ F[H, ℂ∗ ] satisfying the condition ̃ μ(α)μ(β) = π(α, β)μ(αβ)

(α, β ∈ H).

Hence π̃ ∼ 1 (see the paragraph following Definition VI.1.2).

320 | Characters of Finite Groups 1 (ii) Assume that π̃ ∼ 1. Then there exists a function μ ∈ F[H, ℂ∗ ] such that ̃ π(α, β) =

μ(α)μ(β) μ(αβ)

(α, β ∈ H).

Since the representations μ−1 Ω̃ θ (θ ∈ Irrψ (H)) are linear, it follows by Exercise 3.7 that they form a complete system of representatives of classes of irreducible linear representations of H. Since some of these representations are one-dimensional, there exists θ ∈ Irrψ (H) such that e θ = 1. One has θ N = ψ for this θ and hence θ is an extension of ψ to H. Corollary 4.2. If the Schur multiplier M(H/N) is equal to {1}, then the character ψ can be extended to H. In particular, if all Sylow subgroups of H/N are either cyclic or generalized quaternion, then ψ can be extended to H (see Theorem VI.4.6). Assume that ψ ∈ Irr(N) admits an extension of γ ∈ Irr(H). Let P be an irreducible linear representation of H affording the character γ. Then Ψ = P N is an irreducible linear representation of N with the character ψ. If h ∈ H, then Ψ h (x) = P(h)Ψ(x)P(h)−1 (having equal characters, representations Ψ and Ψ h are equivalent), i.e., condition (3.1) is fulfilled. Since P is linear and P N = Ψ, it follows that π = 1 and λ = 1. The set Iπ,λ (H) = I1,1 (H) is now the set of all irreducible linear representations Ω of H such that Ω N is a multiple of the identity representation of N. In other words, I1,1 (H) is the set of all irreducible linear representations of H whose kernels contain N. Hence (see the paragraph preceding Exercise 3.8), Irr(1,1) (H) = {ω ∈ Irr(H) | ker(ω) ≥ N} = IrrN (H). Hence Corollary 3.11 leads to the following result: Lemma 4.3. Suppose that ψ ∈ Irr(N) has an extension γ to H = IG (ψ). Then the mapping ω 󳨃→ ωγ is a bijection of the set IrrN (H) = {χ ∈ Irr(H) | N ≤ ker(χ)} onto the set Irrψ (H)(= Irr(ψ H )). Let ProlH (ψ) be the set of all extensions of ψ ∈ Irr(N) to H = IG (ψ) and let LinN (H) = {ω ∈ Lin(H) | ker(ω) ≥ N}. Corollary 4.4. Let γ ∈ Irr(H) be an extension of ψ ∈ Irr(N) to H = IG (ψ). Then the mapping ω 󳨃→ ωγ is a bijection of LinN (H) onto ProlH (ψ). In particular, we have |ProlH (ψ)| = |H/H 󸀠 N|. Proof. Since ProlH (ψ) ⊆ Irrψ (H), it follows from Lemma 4.3 that θ ∈ ProlH (ψ) ⇐⇒ θ = ωγ, where ω ∈ IrrN (H), θ(1) = ψ(1). Since ψ(1) = γ(1) and θ(1) = ω(1)γ(1), the character ω is linear. Thus, θ ∈ ProlH (ψ) ⇐⇒ θ = ωγ, where ω ∈ LinN (H).

VII Clifford theory | 321

That our mapping is bijective follows from Lemma 4.3. The required equalities follow from bijectivity and from the isomorphisms LinN (H) ≅ Lin(H/N) ≅ H/NH 󸀠 (see Chapter I). Theorem 4.5. Suppose that ψ ∈ Irr(N) has an extension χ ∈ Irr(G). Then ω 󳨃→ ωψ is a bijection of IrrN (G) onto Irrψ (G). Indeed, in the case under consideration, H = G. Exercise 4.2. Suppose that χ ∈ Irr(G) is an extension of ψ ∈ Irr(N). Then ω 󳨃→ ωχ is a bijection of LinN (G) onto ProlG (ψ). In particular, |ProlG (ψ)| = |LinN (G)| = |G/NG󸀠 |. Theorem 4.5 is a reformulation of the two previous statements. Because of the importance of Theorem 4.5, we now present another proof of it, not using projective representations. We will prove a more general assertion, due to P. Gallagher. Recall that Irrinv (N) is the set of G-invariant irreducible characters of N ⊴ G. Theorem 4.6. Let φ, ψ, φψ ∈ Irrinv (N), and moreover ψ = χ N , where χ ∈ Irr(G). Then α 󳨃→ αχ is a bijection of the set Irrφ (G) onto Irrφψ (G). Proof. By Theorem 2.1, ⟨φ G , φ G ⟩ = |G/N| = ⟨(φψ)G , (φψ)G ⟩. Next, (φψ)G = (φχ N )G = φ G χ (Lemma V.1.3) 󳨐⇒ ⟨φ G , φ G ⟩ = ⟨φ G χ, φ G χ⟩. Since, by formula (2.1), φG =

∑

e α α,

α∈Irrφ (G)

it follows that ∑ e2α = ∑ e α e β ⟨αχ, βχ⟩ (α, β run over the set Irrφ (G) both here and in the following equality), or ∑ e2α [⟨αχ, αχ⟩ − 1] + ∑ e α e β ⟨αχ, βχ⟩ = 0. α=β ̸

Since all terms on the left are nonnegative, we obtain {1 if α = β ∈ Irrφ (G), ⟨αχ, βχ⟩ = { 0 if α, β ∈ Irrφ (G), α ≠ β. { Therefore, αχ ∈ Irr(G)

(α ∈ Irrφ (G)),

αχ ≠ βχ if α, β ∈ Irrφ (G), α ≠ β.

(1)

322 | Characters of Finite Groups 1 Next, (αχ)N = α N χ N = e α φψ 󳨐⇒ αχ ∈ Irrφψ (G),

e αχ = e α

(α ∈ Irrφ (G)).

Consequently, α 󳨃→ αχ is an injection of Irrφ (G) into Irrφψ (G). Since (φψ)G = φ G χ =

e α (αχ)

∑ α∈Irrφ (G)

and, on the other hand, (φψ)G =

∑

e β β,

β∈Irrφψ (G)

it follows that for any β ∈ Irrφψ (G), there is α ∈ Irrφ (G) such that β = αχ. Therefore, α 󳨃→ αχ is a surjection and, hence, a bijection of Irrφ (G) onto Irrφψ (G). Putting φ = 1N in Theorem 4.6, we obtain Theorem 4.5 since, in view N ⊴ G, one has Irr1N (G) = IrrN (G). Lemma 4.7. If GCD(|N|, |H/N|) = 1, then ProlH (ψ) ≠ 0. Proof. By the Schur–Zassenhaus theorem, H = K ⋅ N, where K ≤ H and K ∩ N = {1}. We may therefore assume that the elements h α of a transversal of H over N belong to K. Then f(α, β) = 1 (α, β ∈ H). Let π K be the restriction of the factor set π ∈ Z2 (G, ℂ∗ ) to K |K| (see the proof of Lemma 3.1 (a)). We claim that π K ∼ 1. Indeed, π K ∼ 1. On the other hand, taking determinants in formula (3.2), one obtains π(s, t)deg(P) =

det(P(s)) ⋅ det(P(t)) det(P(st))

(s, t ∈ H),

deg(P)

which implies that πdeg(P) ∼ 1. Hence π K ∼ 1. But deg(P) | |N| and we conclude that π K ∼ 1 since GCD(|N|, |K|) = 1. As f(α, β) = 1 (α, β ∈ H) and λ ∈ Lin(N), we can rewrite (3.8) (see Lemma 3.5) as follows: ̃ π(α, β) = π(h α , h β ) = π K (h α , h β ) (α, β ∈ H). It follows from π K ∼ 1 that π K (k1 , k2 ) =

μ(k1 )μ(k2 ) μ(k1 k2 )

Therefore, ̃ π(α, β) =

(k1 , k2 ∈ K, μ ∈ F[K, ℂ∗ ]).

μ(h α )μ(h β ) μ(h α )μ(h β ) = . μ(h α h β ) μ(h αβ )

Consequently, π̃ ∼ 1. Then, by Theorem 4.1, ProlH (ψ) ≠ 0. Lemma 4.7 implies the following result. Corollary 4.8. One has GCD(|N|, |G/N|) = 1 if and only if ProlG (ψ) ≠ 0 for all characters ψ ∈ Irrinv (N).

VII Clifford theory | 323

5 Isaacs’ restriction theorem In this section we prove the following theorem: Theorem 5.1 (Isaacs). Let K/N be an abelian chief factor of a group G (i.e., N, K ⊴ G and K/N is an abelian minimal normal subgroup of G/N) and ϑ ∈ Irrinv (K). Then one of the following holds: (a) ϑ N ∈ Irr(N). (b) ϑ N = eψ for some ψ ∈ Irr(N) and e2 = |K/N|. (c) ϑ N = ψ1 + ⋅ ⋅ ⋅ + ψ l , where l = |K/N|, the characters ψ i ∈ Irr(N) are pairwise distinct and they form the K-orbit (and the G-orbit) of the character ψ1 . Proof. Let ψ ∈ Irr(N),

⟨ϑ N , ψ⟩ = e > 0,

H = IG (ψ).

Then ϑ N = e(ψ1 + ⋅ ⋅ ⋅ + ψ l ) (the Clifford decomposition) and {ψ1 , . . . , ψ l } = O K (ψ), the K-orbit of the character ψ = ψ1 . Let χ ∈ Irrϑ (G)(= Irr(ϑ G )). By Clifford’s theorem, χ K = e󸀠 ϑ (e󸀠 ∈ ℕ) since ϑ is G-invariant. Hence, χ N = (χ K )N = e󸀠 ϑ N = ee󸀠 (ψ1 + ψ2 + ⋅ ⋅ ⋅ + ψ l ) 󳨐⇒ O G (ψ) = {ψ1 , . . . , ψ l }, by Clifford’s theorem. It follows that |K : IK (ψ)| = l = |G : IG (ψ)|. Next, IK (ψ) = K ∩ H 󳨐⇒ |G : H| = |K : (K ∩ H)| = |KH : H| 󳨐⇒ G = KH. Since K ∩ H ≥ N and K/N is abelian, we get K ∩ H ⊴ K. But K ⊴ G implies K ∩ H ⊴ H. Therefore, NG (K ∩ H) ≥ ⟨K, H⟩ = G and so K ∩ H ⊴ G. Since N ≤ K ∩ H and K/N is a minimal normal subgroup of G/N, it follows that either K ∩ H = N or K ∩ H = K. Case 1. Suppose that K ∩ H = N. Then l = |K/N|,

ϑ(1) = elψ(1) = e|K/N|ψ(1) = eψ K (1).

On the other hand, ⟨ψ K , ϑ⟩ = ⟨ϑ N , ψ⟩ = e > 0 󳨐⇒ ψ K (1) ≥ ϑ(1). Therefore, ϑ(1) ≥ eϑ((1) 󳨐⇒ e = 1 󳨐⇒ ϑ N = ψ1 + ⋅ ⋅ ⋅ + ψ l , and (c) holds. Case 2. Suppose that K ∩ H = K, i.e., K ≤ H. In this case ψ is K-invariant, so we have ϑ N = eψ. Let λ ∈ LinN (K) (in the case under consideration, this means that λ ∈ Lin(K) and ker(λ) ≥ N). Then λϑ ∈ Irr(K), (λϑ)N = λ N ϑ N = ϑ N = eψ.

324 | Characters of Finite Groups 1 Assume that the |K/N| characters λϑ (λ ∈ LinN (K)) are pairwise distinct. Since all of them have the ramification e over N, and since by Exercise 2.1 the sum of squares of ramifications over N of the characters in IrrN (K) is equal to |K/N|, it follows that |K/N|e2 = |K/N| 󳨐⇒ e = 1,

i.e.,

ϑ N = ψ ∈ Irr(N),

and (a) holds. Suppose that there exist distinct characters λ󸀠 , λ󸀠󸀠 ∈ IrrN (K) such that λ󸀠 ϑ = λ󸀠󸀠 ϑ. Then one has λϑ = ϑ, where λ = λ󸀠 (λ󸀠󸀠 )−1 ≠ 1K . Let g ∈ K − ker(λ). Then λ(g) ≠ 1 and λ(g)ϑ(g) = ϑ(g), hence ϑ(g) = 0, i.e., ϑ vanishes on the set K − ker(λ). Therefore, ϑ vanishes on the set K − ⋂g∈G ker(λ)g . Since ⋂g∈G ker(λ)g ⊲ G, we have N ≤ ⋂ ker(λ)g < K 󳨐⇒ N = ⋂ ker(λ)g g∈G

g∈G

(recall that K/N is a minimal normal subgroup of G/N) and ϑ vanishes on K − N. But then |K/N| = |K/N|⟨ϑ, ϑ⟩ = ⟨ϑ N , ϑ N ⟩ = e2 . The particular case |G : N| = p was considered by Burnside before the advent of the theory of projective representations: If ϑ ∈ Irr(G) and N is a normal subgroup of G of prime index p, then either ϑ N ∈ Irr(N) or |Irr(ϑ N )| = p (see Exercise 2.3). We see that possibility (b) does not hold, unless |K/N| is a square.

6 Gallagher’s general extension theorem In this section we will prove several remarkable results, due to Gallagher [Gal4], on the extension of invariant irreducible characters from N ⊴ G to G. These results have many applications. Our exposition is close to one in [Isa11]. Let T be a matrix linear representation of a group G with character χ. Define a function det(χ) : G → ℂ∗ by (det(χ))(g) = det(T(g))

(g ∈ G).

Then det(χ) ∈ Lin(G). Indeed, (det(χ))(gh) = det(T(gh)) = det(T(g) det(T(h)) = (det(χ))(g)(det(χ))(h). The order o(det(χ)) of the linear character det(χ) in the group Lin(G) is called the order of χ and we denote it by o(χ). Since Lin(G) ≅ G/G󸀠 , we see that if G = G󸀠 , then det(χ) = 1G for all χ ∈ Irr(G). If λ ∈ Lin(G), then, assuming λ to be a faithful character of the cyclic group G/ker(λ), we get o(λ) = |G/ker(λ)|. Lemma 6.1. If χ ∈ Char(G) and λ ∈ Lin(G), then det(λχ) = λ χ(1) det(χ). Proof. Let T be a linear representation of the group G with character χ. Then λT is a representation with character λχ, so that for all g ∈ G, (det(λχ))(g) = det(λT)(g)) = det(λ(g)T(g)) = λ(g)χ(1) det(T(g)) = λ(g)χ(1) (det(χ))(g) = (λ χ(1) det(χ))(g).

VII Clifford theory | 325

Exercise 6.1. If one has N ⊲ G, ψ ∈ Irr(N) and GCD(ψ(1), |G/N|) = 1, then the mapping λ 󳨃→ λ ψ(1) (λ ∈ LinN (G)) is an automorphism of the group LinN (G). Hint. If A is an abelian group and GCD(n, |A|) = 1, then a 󳨃→ a n is an automorphism of A. Apply this to the abelian group LinN (G) ≅ G/G󸀠 N. Let N ⊲ G and ψ ∈ Irr(N). Recall that ProlG (ψ) is the set of all extensions of ψ to G. This set may be empty. Our aim in this section is to give sufficient conditions for ProlG (ψ) ≠ 0. Lemma 6.2. Let N ⊲ G, ψ ∈ Lin(N), GCD(ψ(1), |G/N|) = 1 and ProlG (ψ) ≠ 0. Then one has ProlG (det(ψ)) ≠ 0 and θ 󳨃→ det(θ) is a bijection of ProlG (ψ) onto ProlG (det(ψ)). Proof. If χ ∈ ProlG (ψ), then det(χ) ∈ ProlG (det(ψ)), hence the last set is nonempty. By Exercise 4.2, ProlG (det(ψ)) = LinN (G) det(χ) and therefore |ProlG (ψ)| = |ProlG (det(ψ))| = |LinN (G)|. Let η ∈ ProlG (det(ψ)). On the one hand there exists λ ∈ LinN (G) such that η = λ det(χ). On the other hand, by Exercise 6.1, there exists μ ∈ LinN (G) such that λ = μ ψ(1) . Therefore one has η = μ ψ(1) det(χ). Consequently, by Lemma 6.1, η = det(θ), where θ = μχ ∈ ProlG (ψ) and our mapping is a surjection. Since |ProlG (ψ)| = |ProlG (det(ψ))|, it is also a bijection. Exercise 6.2. Let N ⊴ G, θ ∈ Char(N), g ∈ G. Then det(θ g ) = det(θ)g . Solution. Let T be a representation of N affording the character θ and x ∈ N. Then det(θ g )(x) = det(T g (x)) = det(T(gxg−1 )), det(θ)g (x) = det(θ)(gxg−1 ) = det(T(gxg−1 )), and we are done. Lemma 6.3. Let N ⊲ G, N ≤ K ⊲ G, |G/K| = p, ψ ∈ Irrinv (N) and GCD(ψ(1), |G/N|) = 1. If ProlG (det(ψ)) ≠ 0, then ProlG (ψ) = 0 if and only if ProlK (ψ) = 0. Proof. Let μ ∈ ProlG (det(ψ)). Then μ K ∈ ProlK (det(ψ)), μ K ∈ Lininv (K).

(1)

Assume that ProlG (ψ) = 0, but ProlK (ψ) ≠ 0. Since GCD(ψ(1), |K/N|) = 1, it follows from (1) and Lemma 6.2 that there exists a unique character θ ∈ ProlK (ψ) such that det(θ) = μ K . Let us show that θ ∈ Irrinv (K). Take g ∈ G. Then ψ ∈ Irrinv (N) implies θ g ∈ ProlK (ψ). In addition, by Exercise 6.2 and (1), we get g

det(θ g ) = det(θ)g = μ K = μ K 󳨐⇒ θ, θ g ∈ ProlK (ψ), det(θ g ) = det(θ). Since G/K is cyclic, θ can be extended to G (Corollary 4.2). Let χ ∈ ProlG (θ). Since θ ∈ ProlK (ψ), this contradicts the assumption. Thus, if ProlK (ψ) = 0, then we obtain ProlG (ψ) = 0.

326 | Characters of Finite Groups 1 Remark. By Corollary 4.2, the condition |G/K| = p in Lemma 6.3 can be replaced by the weaker condition M(G/K) = {1}. It is evident from the proof that, instead of GCD(ψ(1), |G/N|) = 1, it is sufficient to assume that GCD(ψ(1), |K/N|) = 1. Lemma 6.4. Let G be a solvable group, N ⊴ G, ψ ∈ Irrinv (N), GCD(ψ(1), |G/N|) = 1. Then ProlG (ψ) ≠ 0 if and only if ProlG (det(ψ)) ≠ 0. Proof. If ProlG (ψ) ≠ 0, then ProlG (det(ψ)) ≠ 0 as well (Lemma 6.2). Now let ProlG (det(ψ)) ≠ 0 but ProlG (ψ) = 0. Let M ≤ G be minimal such that N ≤ M and ProlM (ψ) = 0; then N < M. Let K be a maximal normal subgroup of M containing N. Then |M/K| is a prime since G is solvable. Since GCD(ψ(1), |M/N|) = 1 and ProlM (ψ) = 0, it follows from Lemma 6.3 that ProlK (ψ) = 0, contrary to the choice of M. As we will prove later, the condition of solvability in Lemma 6.4 is too strong. The proof makes use of Brauer’s first induction theorem (see Chapter VIII), with Lemma 6.4 and the following lemma (due to Gallagher) playing essential roles. Lemma 6.5. Let N ⊲ G, ψ ∈ Irrinv (N), χ ∈ ProlG (ψ) and ϑ ∈ Irr(G). Define a function η : G → ℂ as follows: η(g) = |G|−1 ∑ ϑ(x)χ(x) x∈Ng

(g ∈ G).

(2)

Then the following statements hold: (a) η ∈ IrrN (G) if ⟨ϑ N , ψ⟩ > 0, i.e., ϑ ∈ Irrψ (G)(= Irr(ψ G ). (b) η = 0 if ⟨ϑ N , ψ⟩ = 0 (i.e., ϑ ∈ ̸ Irrψ (G)). (c) η = 1G if ϑ = χ. Proof. It is easy to check that η is a central function which is constant on cosets Ng. Indeed, if z ∈ G, then, taking into account that ϑ and χ are central functions, we get −1

η(g z ) = |G|−1

∑

x∈Ng z−1

ϑ(x)χ(x) = |G|−1 ∑ ϑ(x z )χ(x z ) x z ∈Ng

= |G|−1 ∑ ϑ(x)χ(x) = η(g), x∈Ng

proving the first assertion. The second assertion is obvious since Ntg = Ng for t ∈ N. Define a function η̃ : G = G/N → ℂ by ̃ η(g) = η(g)

(g ∈ G), where g = Ng.

By the previous remark, η̃ is well defined. Since η is central, so is η.̃ Therefore η̃ may be represented as a linear combination of irreducible characters of the group G. Since ̃ Irr(G) = {θ̃ | θ ∈ IrrN (G)}, where θ̃ is defined by θ(g) = θ(g) (g ∈ G), and since the map̃ ping θ 󳨃→ θ is a bijection of IrrN (G) onto Irr(G), it follows that η̃ =

∑ θ∈IrrN (G)

c θ θ̃

(c θ ∈ ℂ).

VII Clifford theory | 327

Consequently, ̃ = η(g) = η(g)

∑

̃ c θ θ(g) =

θ∈IrrN (G)

∑

c θ θ(g),

θ∈IrrN (G)

i.e., η = ∑θ∈IrrN (G) c θ θ. This implies c θ = ⟨η, θ⟩ = |G|−1 ∑ η(g)θ(g) g∈G

−1

= |G|

−1

∑ |N|

g∈G

∑ ϑ(x)χ(x)θ(g) = |G|−1 ∑ ϑ(x)χ(x)θ(x) = ⟨ϑ, χθ⟩.

x∈Ng

(3)

x∈G

By Theorem 4.5, χθ ∈ Irrψ (G) for any θ ∈ IrrN (G). Hence we have, by (3), c θ = δ ϑ,χθ (Kronecker’s delta). If ϑ ∈ ̸ Irrψ (G), then c θ = 0 for all θ ∈ IrrN (G), since otherwise ϑ = χθ for some θ ∈ IrrN (G), and hence ϑ ∈ Irrψ (G), a contradiction. In that case, then η = 0. If ϑ ∈ Irrψ (G), then, by Theorem 4.5, there exists a unique character θ0 ∈ IrrN (G) such that ϑ = χθ0 . Therefore, if θ ∈ IrrN (G), then c θ = δ θ,θ0 , whence we obtain that η = c θ0 θ0 = θ0 ∈ IrrN (G). In particular, if ϑ = χ, then θ0 = 1G , and so η = 1G . Corollary 6.6. If N is a normal subgroup of a group G, ψ ∈ Irrinv (N) and χ ∈ ProlG (ψ), then ∑x∈Ng |χ(x)|2 = |N| for any g ∈ G. Proof. Put ϑ = χ in the definition of η (see Lemma 6.5) and take into account that then η = 1G , by Lemma 6.5 (c). We have now reached the main result of the section: Theorem 6.7 (Gallagher [Gal4]). Let N ⊴ G, ψ ∈ Irrinv (N), GCD(ψ(1), |G/N|) = 1. Then ProlG (ψ) ≠ 0 if and only if ProlG (det(ψ)) ≠ 0. Proof. We have already observed that ProlG (ψ) ≠ 0 implies ProlG (det(ψ)) ≠ 0 (see Lemma 6.2). Now assume, conversely, that ProlG (det(ψ)) ≠ 0 and let μ ∈ ProlG (det(ψ)). Assume that N ≤ H ≤ G and H/N is solvable. It follows from ProlH (det(ψ)) ≠ 0 that ProlH (ψ) ≠ 0, by Lemma 6.4. Since μ H ∈ ProlH (det(ψ)), Lemma 6.2 implies the existence of a unique character θ(H) ∈ ProlH (ψ) such that μ H = det(θ(H) ). x The character θ(H) has the following property: if x ∈ G, then θ(H x ) = θ(H) . Indeed, x x it is obvious, first, that θ(H) ∈ Irr(H ). Next, if y ∈ N, then x θ(H) (y) = θ(H) (xyx−1 ) = ψ(xyx−1 ) = ψ(y) x x since ψ is G-invariant. Consequently, one has (θ(H) )N = ψ, i.e., θ(H) ∈ ProlH (ψ). Next, by Exercise 6.2, x det(θ(H) ) = det(θ(H) )x = μ xH = μ H x . x x x Thus, θ(H) is an extension of ψ to H x such that det(θ(H) ) = μ H x . Hence, θ(H) = θ(H x ) , as claimed. Define a function χ : G → ℂ as follows: if g ∈ G and H = ⟨N, g⟩, then

χ(g) = θ(H) (g).

(4)

328 | Characters of Finite Groups 1 Using Brauer’s induction Theorem VIII.1.2, we will prove that χ ∈ Irr(G). First, we show that χ is a central function). Let g, x ∈ G and H = ⟨N, g⟩. Since H x = ⟨N, g x ⟩ (recall that N ⊲ G), it follows from (4) that χ(g x ) = θ(H x ) (g x ), which implies x χ(g x ) = θ(H) (g x ) = θ(H) (xg x x−1 ) = θ(H) (g) = χ(g),

and our claim follows. Let E ≤ G be a Brauer elementary subgroup (see definition in §VIII.1) and H = NE. Since H/N ≅ E/(N ∩ E) is nilpotent, the character θ(H) is well defined. We claim that χ H = θ(H) . Let g ∈ H. Then ⟨N, g⟩ = K ≤ H. It is obvious that (θ(H) )K ∈ ProlK (ψ), and det((θ(H) )K ) = (det(θ(H) ))K = (μ H )K = μ K . Therefore, (θ(H) )K = θ(K) . Consequently, χ H (g) = χ(g) = θ(K) (g) = (θ(H) )K (g) = θ(H) (g), so that χ H = θ(H) . Hence, we get χ E = (θ(H) )E ∈ Char(E). But then, by Brauer’s Induction Theorem VIII.1.2, χ is a generalized character of G. To prove that χ is irreducible, let us compute ⟨χ, χ⟩. For every coset Ng (g ∈ G), we have ∑ |χ(x)|2 = ∑ |θ(H) (x)|2 ,

x∈Ng

x∈Ng

where H = ⟨N, g⟩ = ⟨N, x⟩

(x ∈ Ng).

Since θ(H) ∈ ProlH (ψ), it follows by Corollary 6.6 (replacing G by H and θ by θ(H) ) that ∑x∈Ng |θ(H) (x)|2 = |N| and hence ⟨χ, χ⟩ = |G|−1 ∑ |χ(x)|2 = |G|−1 |G : N| ⋅ |N| = 1. x∈G

Next, χ(1) = θ(N) (1) = ψ(1) > 0 󳨐⇒ χ ∈ Irr(G). If x ∈ N, then H = ⟨N, x⟩ = N 󳨐⇒ χ(x) = θ(N) (x) = ψ(x) 󳨐⇒ χ N = ψ. Consequently, χ belongs to the set ProlG (ψ), which, therefore, is nonempty. Theorem 6.7 reduces the problem of extending a character ψ ∈ Irrinv (N) to G to the case in which ψ is linear. We will now present some results on extension of linear characters (obviously, a character ψ ∈ ̸ Irrinv (N) has no extension to G). Theorem 6.8. Let N ⊴ G and λ ∈ Lininv (N). For every prime p | o(λ), choose a subgroup H p ≤ G such that H p ≥ N and H p /N ∈ Sylp (G/N). If ProlH p (λ) ≠ 0 for all p | o(λ), then ProlG (λ) ≠ 0. The converse is also true. Proof. Let m = o(λ) and let λ = ∏p|m λ p be a decomposition of λ into primary factors: λ p is a power of λ and the number o(λ p ) is a power of p for every p | m such that p ∤ m/o(λ). Since λ p is a power of λ, we have ProlH q (λ p ) ≠ 0 for all q ∈ π(m); in par-

VII Clifford theory | 329

ticular, ProlH p (λ p ) ≠ 0. If we can prove that the set ProlG (λ p ) ≠ 0 for all p | m, this will imply that ProlG (λ) ≠ 0. Indeed, μ p ∈ ProlG (λ p ) 󳨐⇒ μ = ∏ μ p ∈ ProlG (λ). p∈π(m)

pr .

Hence it suffices to consider the case m = Let ν ∈ ProlH p (λ). Since ν G (1) = |G : H p | ≢ 0 (mod p) (we have ν(1) = λ(1) = 1), there exists χ ∈ Irr(G) such that ⟨ν G , χ⟩ > 0 and p ∤ χ(1). Since ⟨ν, χ H p ⟩ = ⟨ν G , χ⟩ > 0,

⟨χ N , λ⟩ = ⟨χ N , ν N ⟩ > 0

(λ ∈ Irrinv (N)),

it follows by Clifford’s theorem that χ N = eλ. Now, χ(1) = eλ(1) = e 󳨐⇒ χ N = χ(1)λ 󳨐⇒ det(χ)N = det(χ N ) = λ χ(1) . The equality GCD(χ(1), p) = 1 implies that χ(1)c ≡ 1 (mod m) for some integer c (m = p r ). Putting μ = det(χ)c , we obtain λ = λ χ(1)c = (det(χ N ))c = μ N . Thus, μ ∈ ProlG (λ). Corollary 6.9. Let N ⊴ G, λ ∈ Lininv (N), GCD(o(λ), |G/N|) = 1. Then λ has a unique extension μ to G such that GCD(|G/N|, o(μ)) = 1. Moreover, o(μ) = o(λ). Proof. In the case under consideration, H p = N for any prime divisor p of o(λ) (we use the notation introduced in the statement of Theorem 6.8). Therefore the fact that λ is extendible to G follows from Theorem 6.8. It remains to prove that there is only one such extension. Let ν ∈ ProlG (λ). We claim that there exists μ ∈ ProlG (λ) such that o(μ) = o(λ). We look for μ in the form ν k , where k ∈ ℤ. Since necessarily μ N = λ, we have (ν k )N = λ. This gives k ≡ 1 (mod o(λ)), which guarantees that μ N = λ. As GCD(o(λ), |G/N|) = 1, one may additionally require that k ≡ 0 (mod |G/N|). It follows from μ N = λ that λ o(μ) = (μ o(μ) )N = (1G )N = 1N . Therefore, o(λ) | o(μ). Next, (ν o(λ) )N = λ o(λ) = 1N 󳨐⇒ ν o(λ) ∈ LinN (G). It follows that ν o(λ)|G/N| = 1G 󳨐⇒ ν o(λ)k = 1G , i.e., μ o(λ) = 1G . Therefore, o(μ) = o(λ). Let ω be an extension of the character λ to G such that GCD(o(ω), |G/N|) = 1. Then o(μω) | o(μ)o(ω) 󳨐⇒ GCD(o(μω), |G/N|) = 1. Now μω ∈ LinN (G), |LinN (G)| | |G/N| 󳨐⇒ μω = 1G 󳨐⇒ ω = μ.

330 | Characters of Finite Groups 1 Corollary 6.10. Let N ⊴ G and ψ ∈ Irrinv (N), GCD(|G/N|, o(ψ)ψ(1)) = 1. Then ψ has a unique extension χ to G such that GCD(|G/N|, o(χ)) = 1. Moreover, o(χ) = o(ψ). Proof. Let λ = det(ψ). It follows from ψ ∈ Lininv (N) that λ ∈ Lininv (N) (see Exercise 6.2). Moreover, o(λ) = o(ψ) implies GCD(|G/N|, o(λ)) = 1. Therefore, by Corollary 6.9, the character λ has a unique extension μ to G such that GCD(|G/N|, o(μ)) = 1, and then o(μ) = o(λ). By Theorem 6.7, ψ admits an extension χ to G. By Lemma 6.2, there exists a unique χ ∈ ProlG (ψ) such that det(χ) = μ, and in that case o(χ) = o(μ) = o(ψ). If χ0 is another extension of ψ to G such that GCD(|G/N|, o(χ0 )) = 1, then, putting μ0 = det(χ0 ), one obtains GCD(|G/N|, o(μ0 )) = 1. Since χ0 ∈ ProlG (ψ), it follows by Lemma 6.2 that μ0 ∈ ProlG (λ). Therefore, μ0 = μ, i.e., det(χ0 ) = det(χ) = μ. Hence, χ0 = χ, since χ is unique. Exercise 6.3 ([Isa11, Problem 7.17]). If N ⊲ G, G/N is cyclic and ψ ∈ Irr(N) is G-invariant, then ψ is extendible to G. Solution. Let χ ∈ Irr(ψ G ). Since ψ is G-invariant, we have χ N = eψ. Since e is a degree of irreducible projective representation of the cyclic group IG (ψ)/N = G/N, we get e = 1, i.e., χ N = ψ. This means that χ is an extension of ψ to G. Moreover, ψ has exactly |G/N| extensions to G.

7 p-rational characters The proof of the main theorem of this section is based on the results of §VII.6. In what follows ℚr denotes the cyclotomic field ℚ(ε) generated by ϵ, a primitive r-th root of unity. If r | s, then ℚr ⊆ ℚs with strong inclusion if r < s. Definition. Let χ ∈ Char(G) and p a prime. A character χ is said to be p-rational if there exists a positive integer r such that p ∤ r and all values of χ belong to the field ℚr . Let ℚ(χ) be the field of the character χ, i.e., the field obtained by adjoining all the values of χ to ℚ. It follows from Realization Theorem VIII.2.1 that ℚ(χ) ⊆ ℚ|G| . By the definition, χ is p-rational if and only if ℚ(χ) ⊆ ℚr , where p ∤ r. If p ∤ |G|, then all the characters of G are p-rational, by Theorem VIII.2.1. From now on, therefore, we shall assume that p | |G|. We need the following well-known lemma from number theory. Lemma 7.1. Let m1 , m2 ∈ ℕ and d = GCD(m1 , m2 ). Then ℚm1 ∩ ℚm2 = ℚd . Let |G| = p a m, where p ∤ m. The character χ of G p-rational if and only if ℚ(χ) ⊆ ℚm . Put Gp (G) = Gal(ℚ|G| /ℚm ) (this is the set of all ℚm -automorphisms of the field ℚ|G| ).¹ 1 If K is a subfield of the field F, then α ∈ AutK (F), the set of K-automorphisms of F if and only if x α = x for all x ∈ K.

VII Clifford theory | 331

It follows from the previous remark that χ is p-rational if and only if σ(χ) = χ for all σ ∈ Gp (G). The following lemma from number theory is also well known. Lemma 7.2. In the above notation, one has Gp (G) ≅ Gal(ℚp a /ℚ). Since Gal(ℚp a /ℚ) is isomorphic to the multiplicative group of residue classes modulo p a , this group is cyclic if p > 2. Therefore, by Lemma 7.2, the group Gp (G) is cyclic if p > 2. Lemma 7.3. Let p > 2, λ ∈ Lin(G) and p | o(λ). Then λ is not p-rational. Proof. If λ is p-rational, then one has ℚ(λ) ⊆ ℚm , by definition. Consequently, putting μ = λ o(λ)/p , we obtain ℚ(μ) ⊆ ℚm ∩ ℚp = ℚGCD(m,p) = ℚ1 = ℚ. Therefore, μ(g) = ±1 for all g ∈ G, hence μ2 = 1G , i.e., p = 2, a contradiction. Hence λ is not p-rational. Lemma 7.4. Let H ≤ G and ψ ∈ Char(H). Then ψ is p-rational if and only if σ(ψ) = ψ for all σ ∈ Gp (G). Proof. Let |H| = p b m󸀠 , where p ∤ m󸀠 . If ψ is p-rational, then ℚ(ψ) ⊆ ℚm󸀠 ⊆ ℚm (since m󸀠 | m, by Lagrange’s theorem). Thus, ψ(h) ∈ ℚm (h ∈ H), whence σ(ψ(h)) = ψ(h) for all σ ∈ Gp (G), and so σ(ψ) = ψ (see the paragraph following Lemma 7.1). Conversely, let σ(ψ) = ψ for all σ ∈ Gp (G). Then ℚ(ψ) ⊆ ℚm , so that ℚ(ψ) ⊆ ℚm ∩ ℚ|H| = ℚGCD(m,|H|) = ℚm󸀠 , and so ψ is p-rational since p ∤ m󸀠 . Theorem 7.5. Let N ⊴ G, let G/N be a p-group, p > 2, ψ ∈ Irrinv (N), and let ψ be p-rational. Then the induced character ψ G contains a unique p-rational irreducible constituent χ, and moreover χ N = ψ (in particular, ProlG (ψ) ≠ 0). Proof. We argue by induction on |G/N|. If |G/N| = 1, the theorem is obvious. Let N < K ⊴ G and |K/N| = p. Then, by Theorem 3.9, ψ has an extension η to K. Since p > 2, the group Gp (K) is cyclic, Gp (K) = ⟨σ⟩ (see the paragraph following Lemma 7.2). Then σ(η) ∈ Irr(K) and, since ψ is p-rational, (σ(η))N = σ(ψ) = ψ. Thus, η, σ(η) ∈ ProlK (ψ) 󳨐⇒ σ(η) = λη (λ ∈ LinN (K)), by Corollary 4.4. If λ = 1K , then σ(η) = η, i.e., η is p-rational. Let λ ≠ 1K . Then |LinN (K)| = p = |Lin(K/N)| 󳨐⇒ o(λ) = p. Therefore, by Lemma 7.3, λ is not p-rational, i.e., σ(λ) ≠ λ. Since LinN (K) is cyclic of order p and σ(λ) ∈ LinN (K), it follows that σ(λ) = λ m , where p ∤ m, and m ≢ 1 (mod p).

332 | Characters of Finite Groups 1 Choose b ∈ ℤ such that (1 − m)b ≡ 1 (mod p) and put ω = λ b η. Next, λ N = 1N 󳨐⇒ ω N = η N = ψ 󳨐⇒ ω ∈ ProlK (ψ). Moreover, in view of mb + 1 ≡ b (mod p), we get σ(ω) = σ(λ)b σ(η) = λ mb λη = λ b η = ω, i.e., ω is p-rational. Thus, ω is a p-rational extension of ψ to K. Let ω󸀠 be an arbitrary p-rational extension of ψ to K. Then ω󸀠 = μω, where μ ∈ LinN (K), and μ is uniquely determined (see Corollary 4.4). Since μ is unique, ω󸀠 = σ(ω󸀠 ) = σ(μ)σ(ω) = σ(μ)ω 󳨐⇒ σ(μ) = μ. If μ ≠ 1K , then o(μ) = p and μ is not p-rational by Lemma 7.3, i.e., σ(μ) ≠ μ. This contradiction proves that μ = 1K , i.e., ω󸀠 = ω. Thus, ω is the unique p-rational extension of ψ to K. If g ∈ G, then σ(ω g ) = σ(ω)g = ω g ,

(ω g )N = (ω N )g = ψ g = ψ.

(1)

Therefore, ω g is a p-rational extension of ψ to K. Consequently, we have ω g = ω, i.e., ω ∈ Irrinv (K). Since |G : K| < |G : N|, it follows by induction that ω G possesses a unique p-rational irreducible constituent χ, and χ K = ω; hence we get χ N = ω N = ψ. It is clear that ⟨ψ G , χ⟩ = ⟨ψ, χ N ⟩ = ⟨ψ, ψ⟩ = 1. Thus, χ is a p-rational irreducible constituent of the character ψ G and χ N = ψ. Suppose that χ0 ∈ Irr(ψ G ) is p-rational. We claim that χ0 = χ. Let φ ∈ Irr((χ0 )K ). Since (χ0 )N = eψ and K/N is cyclic, it follows from Exercise 3.3 that φ N = ψ. On the other hand, we have ω N = ψ. Then, by Corollary 4.4, φ = ων, where ν ∈ LinN (K). Since K/N ≤ Z(G/N) (K/N is a normal subgroup of order p in a p-group G/N), it follows that ν g = ν for any g ∈ G. Therefore, in view of the G-invariance of ω, we obtain φ g = ω g ν g = ων = φ 󳨐⇒ φ ∈ Irrinv (K). Hence, by Clifford’s theorem, we must have (χ0 )K = e0 φ. If k ∈ K, then, as χ0 is p-rational, φ(k) = e10 χ0 (1) ∈ ℚr , where p ∤ r. Consequently, φ is p-rational. Since ω is the unique p-rational extension of ψ to K, it follows that φ = ω, i.e., ⟨ω, (χ0 )K ⟩ > 0, and so ⟨ω G , χ0 ⟩ > 0. Since χ is the unique p-rational constituent of ω G , we have χ0 = χ, i.e., χ is the unique p-rational constituent of ψ G . Corollary 7.6. Let N ⊴ G, ψ ∈ Irrinv (N), GCD(ψ(1), |G/N|) = 1, and let ψ be p-rational for all p | o(ψ). Then ProlG (ψ) ≠ 0. Proof. For every p | o(ψ), choose H p ≤ G such that H p ≥ N and H p /N ∈ Sylp (G/N). Since ψ is p-rational, it follows from Theorem 7.5, that ProlH p (ψ) ≠ 0. Then, putting λ = det(ψ), we obtain ProlH p (λ) ≠ 0, by Theorem 6.7. Since o(λ) = o(ψ), it follows that ProlH p (λ) ≠ 0 for every p | o(λ). Hence, by Theorem 6.8, we have ProlG (λ) ≠ 0. Since GCD(ψ(1), |G/N|) = 1, it follows that ProlG (ψ) ≠ 0, by Lemma 6.4.

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8 The class number of an extension In this section, Gallagher’s results [Gal8] will be strengthened slightly. Let H ≤ G, h = |H|, g = |G|, g󸀠 = |G󸀠 | and let k(G) be the class number of G. Given a character ψ, let w(ψ) = |Irr(ψ)|, the number of pairwise distinct irreducible constituents of ψ. We put n(G) = |Irr1 (G)|, where Irr1 (G) = Irr(G) − Lin(G). For ψ = a1 τ1 + ⋅ ⋅ ⋅ + a k τ k ,

where Irr(ψ) = {τ1 , . . . , τ k },

we set σ(ψ) = a1 + ⋅ ⋅ ⋅ + a k . Obviously, w(ψ) ≤ σ(ψ) with equality if and only if a1 = ⋅ ⋅ ⋅ = a k = 1. In what follows we use repeatedly the following two lemmas. Lemma 8.1. Let H < G and χ ∈ Irr(G). Suppose that χ H = a1 ψ1 + ⋅ ⋅ ⋅ + a k ψ k , where Irr(χ H ) = {ψ1 , . . . , ψ k }. Then the following statements hold: (a) σ(χ H ) ≤ hg . (b) σ(χ H ) = hg implies ψ1 (1) = ⋅ ⋅ ⋅ = ψ k (1) = hg χ(1). In particular, w(χ H ) = hg implies a1 = ⋅ ⋅ ⋅ = a k = 1 so that k = hg and G − H ⊆ Tχ , where Tχ = {x ∈ G | χ(x) = 0} is the set of zeros of the character χ. Proof. (a) Let ψ1 (1) ≤ ⋅ ⋅ ⋅ ≤ ψ k (1). Then g ψ1 (1) = ψ1G (1) ≥ χ(1) = a1 ψ1 (1) + ⋅ ⋅ ⋅ + a k ψ k (1) h ≥ (a1 + ⋅ ⋅ ⋅ + a k )ψ1 (1) = σ(χ H )ψ1 (1) so that σ(χ H ) ≤

g , h

proving (a). (b) Now suppose that σ(χ H ) = hg . It follows from the first displayed formula in (a) that ψ1 (1) = ⋅ ⋅ ⋅ = ψ k (1); indeed, if ψ1 (1) < ψ k (1), then g g ψ1 (1) = ψ1G (1) > σ(χ H )ψ1 (1) 󳨐⇒ > σ(χ H ), h h whence ψ1 (1) =

h χ(1). g

334 | Characters of Finite Groups 1 Assume, in addition, that k = w(χ H ) = hg . Then a1 = ⋅ ⋅ ⋅ = a k = 1. We have k

⟨χ H , χ H ⟩ = ∑ a2i = k = w(χ H ) = i=1

g . h

(1)

On the other hand, since ⟨χ H , χ H ⟩ = |H|−1 ∑ |χ(x)|2 ≤ |H|−1 ∑ |χ(x)|2 = |G : H|⟨χ, χ⟩ = x∈H

x∈G

g , h

we conclude that χ vanishes on the set G − H. Lemma 8.2 is some sense is dual to Lemma 8.1. Lemma 8.2. Let H < G and ψ ∈ Irr(H). Then the following statements hold: (a) σ(ψ G ) ≤ hg with equality if and only if all irreducible constituents of ψ G have the same degree ψ(1). (b) If w(ψ G ) = hg , then ⟨χ, ψ G ⟩ = 1 for all χ ∈ Irr(ψ G ). Proof. If χ ∈ Irr(ψ G ), then χ(1) ≥ ψ(1), by reciprocity, and, since g ψ(1) = ψ G (1) = ∑ a χ χ(1) ≥ σ(ψ G )ψ(1), h χ∈Irr(ψ G ) the first assertion of (a) follows. If, in addition, σ(ψ G ) = hg , then all irreducible constituents of ψ G have the same degree ψ(1), and the second assertion of (a) follows. Now (b) is obvious. Given a nonabelian group G and H < G, put τ=

∑

χ,

S = {λ ∈ Irr(H) | ⟨τ H , λ⟩ = 0},

|S| = s.

χ∈Irr1 (G)

By definition, if λ ∈ S, then Irr(λ G ) ⊆ Lin(G) so the linear character λ is extendible to G and S ⊆ Lin(H). By reciprocity, Irr1 (H) ⊆ Irr(τ H ). Let LinProl (H) denote the set of all linear characters of H that are extendible to G. The above defined set S, if it is nonempty, has the following properties: Property (i). One has S ⊆ LinProl (H) (this was noted above). Property (ii). Suppose that λ ∈ S. Then λ G is a multiplicity-free character and the set ProlG (λ) = Irr(λ G ) has cardinality hg . Indeed, S ⊆ Lin(H) and for λ ∈ S, we have Irr(λ G ) ⊆ Lin(G). In particular, if μ ∈ S, then |Irr(μ G )| = hg . Property (iii). If S ≠ 0, then G󸀠 ≤ H so that H ⊲ G. Indeed, let λ ∈ S. Then, by (i) and (ii), Irr(λ G ) ⊆ Lin(G), so that G󸀠 ≤ ker(λ G ) = ker(λ)G ≤ H.

VII Clifford theory | 335

Property (iv). One has |S| = s ≤ (by (iii), gh󸀠 ∈ ℕ).

h g󸀠

with equality if and only if ⋃λ∈S Irr(λ G ) = Lin(G)

Indeed, λ1 , λ2 ∈ S, λ1 ≠ λ2 implies Irr(λ1G ) ∩ Irr(λ2G ) = 0, by (i) and (ii) and reciprocity. Therefore, by (ii), s⋅ Now let s =

h . g󸀠

󵄨󵄨 g 󵄨󵄨󵄨 h g ≤ 󵄨󵄨󵄨 ⋃ Irr(λ G )󵄨󵄨󵄨󵄨 ≤ |Lin(G)| = 󸀠 󳨐⇒ s ≤ 󸀠 . h 󵄨λ∈S g g 󵄨

Then S ≠ 0 and G󸀠 ≤ H, by (iii). Moreover, 󵄨󵄨 󵄨 󵄨󵄨 ⋃ Irr(λ G )󵄨󵄨󵄨 = s ⋅ g = h ⋅ g = g = |Lin(G)|, 󵄨󵄨 󵄨󵄨 h g󸀠 h g󸀠 󵄨λ∈S 󵄨

and, hence, Lin(G) = ⋃λ∈S Irr(λ G ) is a partition. Conversely, if the last equality is true, then s = gh󸀠 . Supposing that S ≠ 0, we claim that the following inequality holds: k(H) ≤

h g + n(G). g󸀠 h

(2)

Indeed (see the paragraph following the proof of Lemma 8.2), put T = {η ∈ Irr(H) | ⟨τ H , η⟩ > 0}. Then Irr(H) = S ∪ T is a partition. Therefore, k(H) = |Irr(H)| = |T| + |S| = |T| + s ≤

h + |T|, g󸀠

by (iv). It follows from the definition that T = Irr(τ H ). Therefore, by Lemma 8.1 (a), |T| ≤

∑

|Irr(χ H )| =

χ∈Irr1 (G)

∑

w(χ H ) ≤

χ∈Irr1 (G)

so that k(H) ≤ s + |T| ≤

g g |Irr1 (G)| = n(G), h h

g h n(G) + 󸀠 , h g

proving (2). 3 2 Let G = Mp4 = ⟨x, y | x p = y p = 1, x y = x1+p ⟩ and let H < G be maximal. Then k(H) = |H| = p3 , so that

g󸀠 = p,

h = p2 , g󸀠

n(G) =

p4 − p3 = p2 − p p2

h g + n(G) = p2 + p ⋅ (p2 − p) = p3 = k(H), g󸀠 h

i.e., equality in (2) is possible for nonabelian groups. If G is extraspecial of order p2m+1 and H < G is maximal, then (2) holds again.

336 | Characters of Finite Groups 1 Lemma 8.3. If k(H) = gh󸀠 + hg n(G), then the following statements hold: (a) G󸀠 ≤ H. (b) S = Irrinv (H) = Lininv (H). (c) If χ ∈ Irr1 (G), then χ H = η1 + ⋅ ⋅ ⋅ + η m , (d) (e) (f) (g)

m=

g , Irr(χ H ) = {η1 , . . . , η m }, h

χ = η Gi , i = 1 . . . , m.

G − H ⊆ Tχ = {x ∈ G | χ(x) = 0} for any χ ∈ Irr1 (G). G󸀠 = [H, G]. If g ∈ G − H, then all elements of gG󸀠 are conjugate in G. G has a normal π( hg )-complement, say K, and G/K is nilpotent.

Proof. It follows from the proof of (2) that |T| =

∑ χ∈Irr1 (G)

|Irr(χ H )| =

g n(G), h

s=

h ∈ ℕ 󳨐⇒ S ≠ 0. g󸀠

(∗)

By properties (iii) and (iv) of the (nonempty) set S, the second equality in (∗) implies that G󸀠 ≤ H 󳨐⇒ H ⊲ G, ⋃ Irr(λ G ) = Lin(G). λ∈S

The first equality in (∗) implies that the subsets Irr(χ H ) (χ ∈ Irr1 (G)) are pairwise disjoint and w(χ H )(= |Irr(χ H )|) = hg for all χ ∈ Irr1 (G). Therefore, χ H is a character without multiplicities and G − H ⊆ Tχ for all χ ∈ Irr1 (G) (here Tχ is the set of zeros of the character χ). Since H ⊲ G, it follows by Clifford’s theorem that χ H = η1 + ⋅ ⋅ ⋅ + η m , where m = hg and {η1 , . . . , η m } is the G-orbit of η1 ∈ Irr(χ H ). Next, g m = |G : IG (η1 )| = |G : H| = 󳨐⇒ IG (η1 ) = H. h Therefore χ = η1G . This completes the proofs of (a), (c) and (d). Since all the characters in S are extendible to G, we have S ⊆ Lininv (H). Let η ∈ Irrinv (H) and χ ∈ Irr(η G ). Then ⟨χ H , η⟩ = ⟨χ, η G ⟩ = e > 0. If χ ∈ Irr1 (G), then we have χ H = η1 + ⋅ ⋅ ⋅ + η m , where η = η1 and {η1 , . . . , η m } is the G-orbit of η, m = hg , by (c). Therefore, η ∈ ̸ Irrinv (H). This contradiction shows that χ ∈ Lin(G). Since Lin(G) = ⋃λ∈S Irr(λ G ), there exists λ ∈ S such that χ ∈ Irr(λ G ), i.e., χ ∈ ProlG (λ), χ H = λ. Next, ⟨χ H , η⟩ > 0 󳨐⇒ η = χ H 󳨐⇒ η = λ ∈ S. Thus, Irrinv (H) ⊆ S, implying that S = Lininv (H) = Irrinv (H) and completing the proof of (b). It follows from Lininv (H) = Lin[H,G] (H) = {λ ∈ Irr(H) | ker(λ) ≥ [H, G]}

VII Clifford theory | 337

that s = |Lininv (H)| = |Lin[H,G] (H)| = |Lin(H/[H, G])| = |H/[H, G]| =

h . |[H, G]|

Since, on the other hand, s = gh󸀠 , it follows that |[H, G]| = g 󸀠 , implying that [H, G] = G󸀠 and proving (e). If x ∈ G − H, then, taking into account that all members of the set Irr1 (G) vanish on the set G − H, we get g |CG (x)| = ∑ |χ(x)|2 = |Lin(G)| + ∑ |χ(x)|2 = |Lin(G)| = 󸀠 . g χ∈Irr(G) χ∈Irr (G) 1

This implies that |G : CG (x)| = |G󸀠 |. Let K x denote the G-class of x. Since x−1 K x = {x−1 (t−1 xt) | t ∈ G} = {[x, t] | t ∈ G} ⊆ G󸀠 and |x−1 K x | = |K x | = |G󸀠 |, it follows that x−1 K x = G󸀠 , i.e., K x = xG󸀠 , and (f) is proved. It remains to prove (g). If χ ∈ Irr1 (G), then, by (c), χ H = η1 + ⋅ ⋅ ⋅ + η m is the Clifford decomposition with m = hg . Hence χ(1) = hg η1 (1). Then, by Theorem XIV.3.3 (a), G possesses an invariant p-complement G p for any p ∈ π( hg ). But then K = ⋂p∈π( g ) G p is h an invariant π( hg )-complement in G. It is clear that K ≤ H and G/K is nilpotent. This proves (g). Under the assumptions of Lemma 8.3 (g), if G/K is nonabelian, then hg is a prime power. In fact, G/K has a character χ ∈ Irr1 (G/K) such that χ(1) = p k , p ∈ π( hg ). Since g g h | χ(1), we obtain π( h ) = {p}. If N ⊴ G, then k0 = k0 (N) denotes the number of G-classes contained in N. Let G be a permutation group and let χ be its natural character. Then ⟨χ, 1G ⟩ is the number of G-orbits (Cauchy–Frobenius–Burnside). This fact and Brauer’s permutation Lemma X.3.1 show that k0 (N) is the number of G-orbits of the set Irr(N). Theorem 8.4. Let N ⊴ G, let C1 , . . . , Ck0 be all the G-orbits of the set Irr(N) (here k0 = k0 (N)), and let ψi ∈ Ci ,

T i = IG (ψ i ),

t i = |G : T i |.

Then k0

k(G) ≤ k(N)n(G/N) + |G : G󸀠 N| ∑ t−1 i .

(3)

i=1 k

0 Proof. Note that |C i | = t i , so that ∑i=1 t i = k(N). By the above, one has for all i ≤ k0 (see Theorem 2.2 (b)–(c)):

|Irrψ i (G)| = |Irrψ i (T i )| ≤ k(T i /N).

(4)

Next, by (2) (applied to T i /N and G/N), k(T i /N) ≤

|T i /N| |G/N| |T i | + n(G/N) = 󸀠 + t i n(G/N). |(G/N)󸀠 | |T i /N| |G N|

(5)

338 | Characters of Finite Groups 1 Note that Irrψ i (G) ∩ Irrψ j (G) = 0 for i ≠ j since characters ψ i and ψ j belong to different G-orbits (see Clifford’s theorem). Therefore, summing (5) over i ∈ {1, . . . , k0 } and taking into account (4), we obtain k0

k0

i=1

i=1

k(G) = |Irr(G)| = ∑ |Irrψ i (G)| ≤ ∑ k(T i /N) k0

k0 |T i | + n(G/N) ti ∑ |G󸀠 N| i=1 i=1

≤∑

= k(N)n(G/N) + Indeed, completing the proof.

|G| k0 −1 ∑t . |G󸀠 N| i=1 i

|T i | |G| |G| |T i | = ⋅ 󸀠 = 󸀠 ⋅ t−1 , 󸀠 |G N| |G| |G N| |G N| i

Corollary 8.5 ([Gal8]). If N ⊴ G, then k(G) ≤ k(N)k(G/N). If k(G) = k(N)k(G/N), then all members of the set Irr(N) are G-invariant. Proof. In the notation of Theorem 8.4, we have k0

k0

i=1

i=1

∑ t−1 i ≤ ∑ 1 = k 0 (N) with equality if and only if all members of the set Irr(N) are G-invariant. Therefore, k(G) ≤ k(N)n(G/N) + |G/G󸀠 N|k0 (N).

(6)

Since k0 (N) ≤ k(N), it follows from (6) that k(G) ≤ k(N)n(G/N) + |G/G󸀠 N|k(N) = k(N)[n(G/N) + |G/N : (G/N)󸀠 |] = k(N)k(G/N). Exercise 8.1. Let G = (F, H) be a Frobenius group with cyclic Frobenius complement F (see Chapter X). Then k(H) = 1 + hg n(G), i.e., for the pair H < G, we have in (2) equality. Prove that if G = (F, H) is a Frobenius group with equality in (2), then F is cyclic. Solution (formula for k(H)). By Theorem X.3.2 (a), if ϕ ∈ Irr(H) − {1H }, then one has ϕ G ∈ Irr(G). By Theorem X.3.2 (b), if χ ∈ Irr(G | H), then χ = ϕ G for some character ϕ ∈ Irr(H) − {1H }, so that |Irr(χ H )| = hg . It follows that n(G) =

k(H) − 1 , g/h

and this proves the formula for k(H). Exercise 8.2. If N ⊴ G, then k(G) ≤ k(N)n(G/N) + |G/G󸀠 N|k0 (N). The inequality of Corollary 8.5 cannot be substantially improved without taking the action of G on N into account.

VII Clifford theory | 339

9 Dornhoff’s theorem Recall that χ ∈ Irr(G) is monomial if there exist H ≤ G and λ ∈ Lin(H) such that χ = λ G . A group G is an M-group if all its irreducible characters are monomial. Let G be an M-group. If N ⊲ G, then N is not necessarily an M-group (counterexamples have been constructed by S. D. Berman, E. Dade and R. van der Waall). Below we will slightly improve a result due to L. Dornhoff (Theorem 9.2). Lemma 9.1. Suppose that χ ∈ Irr(G) is monomial and K ≤ G. If χ K ∈ Irr(K), then χ K is monomial. Proof. By assumption, χ = λ G , where λ ∈ Lin(H), H ≤ G. Let G = ⋃s∈J HsK be the decomposition of G into double cosets of H and K. Define the function λ s : H s → ℂ by λ s (x) = λ(sxs−1 ) (x ∈ H s ). Obviously, λ s ∈ Lin(H s ). Put D s = H s ∩ K, λ s = (λ s )D s ; then λ s ∈ Lin(D s ). By Mackey’s first theorem (see Theorem V.3.1), we get χ K = (λ G )K = ∑ λ Ks .

(1)

s∈J

Since χ K ∈ Irr(K), by hypothesis, it follows that |J| = 1 and we may assume that J = {1} (at the same time, we have proved that G = HK). By (1), χ K = λ1K = (λ D1 )K ,

where D1 = H ∩ K, λ D1 ∈ Lin(D1 ).

Therefore χ K is a monomial character of K. Exercise 9.1 (Reynolds). If N ⊴ G, χ ∈ Irr(N), ψ ∈ Irrχ (G), then

ψ(1) χ(1)

divides |G : N|.

Solution. Let T = IG (χ) the inertia group of χ and let ψ N = e(χ1 + ⋅ ⋅ ⋅ + χ l ) be the Clifford decomposition, where l = |G : T| and e ∈ ℕ is the degree of an irreducible projective representation of the group T/N. Therefore e | |T/N| (see Corollary VI.3.10). Then ψ(1) = el | |T/N| ⋅ |G/T| = |G/N|. χ(1) (In particular, if N is abelian, we obtain Ito’s Theorem 2.3.) Theorem 9.2. Let N ⊴ G, GCD(|N|, |G/N|, χ(1)) = 1, ψ ∈ Irr(N). If χ ∈ Irrψ (G) is monomial, then ψ is a monomial character of N. Proof. By assumption, χ = λ G , where λ ∈ Lin(H) and H ≤ G. Write φ = λ NH ; then φ G = (λ NH )G = λ G = χ. Since χ is irreducible, so is φ. Let a function φ̇ coincide with φ on NH and vanish outside NH. Then −1 ̇ χ(x) = φ G (x) = |NH|−1 ∑ φ(txt ). t∈G

340 | Characters of Finite Groups 1 In particular, if x ∈ N, then χ(x) = |NH|−1 ∑ φ(txt−1 ) = |NH|−1 ∑ φ t (x) 󳨐⇒ χ N = |NH|−1 ∑ (φ t )N . t∈G

t∈G

t∈G

We have 0 < ⟨χ N , ψ⟩ = |NH|−1 ∑ ⟨(φ t )N , ψ⟩; t∈G

therefore,

⟨(φ t )

N,

ψ⟩ > 0 for some t ∈ G. Next, φ ∈ Irr(NH) 󳨐⇒ φ t ∈ Irr((NH)t ) = Irr(NH t ).

Applying Exercise 9.1 to φ t ∈ Irr(NH t ) and ψ ∈ Irr(N), we see that φ(1) φ t (1) = | |NH t : N| ψ(1) ψ(1) hence

φ(1) | |G : N| ψ(1)

since NH t ≤ G. Next, φ(1) = λ NH (1) = |NH : H| = Therefore,

φ(1) ψ(1)

|N| | |N|. |N ∩ H|

| |N|. Next,

χ(1) = φ G (1) = |G : NH|φ(1) 󳨐⇒ φ(1) | χ(1) 󳨐⇒ Then

φ(1) | χ(1). ψ(1)

φ(1) | GCD(|G/N|, |N|, χ(1)) = 1 󳨐⇒ φ(1) = ψ(1), (φ t )N = ψ. ψ(1)

But φ = λ NH is a monomial character of NH, and φ t is a monomial character of (NH)t = NH t . Therefore, by Lemma 9.1, ψ = (φ t )N is a monomial character of N. In particulary, if N is a normal Hall subgroup of G, then ψ is a monomial character of N (L. Dornhoff). Thus: Corollary 9.3. If G is an M-group and N is a normal Hall subgroup of G, then N is also an M-group. Note that in Theorem 9.2, N is not assumed to be a Hall subgroup; besides, as we are speaking of only one character of G, G is also not assumed to be an M-group. Let N ⊲ G, ψ ∈ Irr(N) and suppose that all χ ∈ Irrψ (G) are monomial. This does not imply that ψ is monomial, even if one assumes additionally that G is solvable of odd order, as the following example due to T. Berger shows. Example 9.1. Let F be a nonabelian group of order 53 and exponent 5, and T a nonabelian group of order 33 and exponent 3. Suppose that T acts nontrivially on F,

VII Clifford theory | 341

centralizes Z(F) and acts irreducibly on F/Z(F)(≅ E52 ). Let K denote the kernel of this action. Since |Aut(F/Z(F))| = (52 − 1)(52 − 5) is not a multiple of 9, one has |T/K| = 3. Let G = T ⋅ F be the semidirect product corresponding to this action. Let L ⊲ T and |T : L| = 3, where L ≠ K. Put N = LF ⊲ G; then |G : N| = 3 and N 󸀠 = F. Since F has exactly 5 − 1 = 4 nonlinear irreducible characters, one of them, say φ, is invariant with respect to N. Note that φ(1) = 5. By Corollary 6.10, there exists an extension θ of φ to N. If Z(T) ≤ ker(θ), then, multiplying θ by a suitable linear character of N/F, we can ensure that the kernel of the character does not contain Z(T); it is also an extension of φ (consider the restriction of the obtained character to F). We may thus assume that Z(T) ≰ ker(θ). One has θ(1) = φ(1) = 5. Since LK = T (in view of L and K are distinct subgroups of index 3 in T), it follows that L acts irreducibly on F/Z(F) (indeed, K acts trivially on that quotient group). Therefore, there is no subgroup of index 5 in N = LF. It follows that the defined in the previous paragraph character θ (of degree 5) is not monomial. Let χ ∈ Irrθ (G). Then χ(1) ∈ {5, 15} (note that |G : N| = 3 so θ G (1) = 15). Assume that χ(1) = 5. Then all characters in Irrθ (G) = Irr(θ G ) are of degree 5 since, in the case under consideration, θ G (1) = 15, θ G is reducible and degrees of all irreducible constituents of θ G are multiples of θ(1) = 5 and divide |G|. Since χ N = θ, it follows from Theorem 4.5 that χβ ∈ Irrθ (G) for every β ∈ Irr(G/F), and so Irrθ (G) contains a character of degree 15, a contradiction. Thus, χ(1) = 15, so that χ = θ G . If U < F is of index 5, then A = U × K is an abelian subgroup of index 15 in G. The equality χ(1) = |G : A| implies that χ = λ G for some λ ∈ Lin(A), that is χ is monomial. Recall that χ = θ G . But χ N has an irreducible constituent θ, where θ is not a monomial character of N. Example 9.2. Take F to be Q8 , and take T as in Example 9.1, and suppose that T act nontrivially on F. Then we obtain a group G = T ⋅ F, a semidirect product with kernel F, of order 216 with a monomial character χ of degree 6 and a normal subgroup N of index 3 such that θ, a faithful character of N of degree 2, is an irreducible constituent of χ N and θ is not monomial. We now present an additional information about normal subgroups of M-groups (the results are due to Isaacs). A character χ ∈ Irr(G) is said to be primitive if χ = λ G and λ ∈ Irr(H) with H ≤ G imply H = G; otherwise, it is imprimitive. All nonlinear irreducible characters of an M-group are imprimitive (the converse is most probably false). Theorem 9.4 (Isaacs [Isa12]). Let N be a normal subgroup of an M-group G and let θ ∈ Irr(N) be primitive. Then: (a) θ(1) is a power of 2. (b) If |G/N| is odd, then θ ∈ Lin(N). Theorem 9.5 (Isaacs [Isa12]). Let G be a p-solvable group, p > 2, and let χ ∈ Irr(G) be primitive of degree p s . If N ⊲ G and G/N is a p-group, then χ N is primitive and irreducible.

342 | Characters of Finite Groups 1 So far, it is not known if normal subgroups of odd order and nonnormal Hall subgroups of M-groups are M-groups.

10 Tate’s normal p-complement theorem We are now going to use Corollary 6.10 to prove an important theorem due to J. Tate; the original proof was cohomological. The following proof of Tate’s result is due essentially to J. Thompson [Tho4] (this proof was reworked by the third author). Let p be a prime divisor of |G|. Let Op (G) denote the intersection of all N ⊴ G such that G/N is a p-group (obviously, Op (G) is a subgroup generated by all p󸀠 -elements p of G). Let Ap (G) denote the complete preimage of (G/Op (G))󸀠 in G and let A1 (G) denote the complete preimage of Φ(G/Ap (G)) in G. Then G/Ap (G) is an abelian p-group and p G/A1 (G) is an elementary abelian p-group (both of maximal possible orders). Thus, p if U is a proper G-invariant subgroup of Ap (G) (of A1 (G)), then G/U is not an abelian p-group (is not an elementary abelian p-group). Theorem 10.1 (Tate [Tat]). If P ∈ Sylp (G), P ≤ N ≤ G, then p

p

N ∩ A1 (G) = A1 (N) 󳨐⇒ N ∩ Op (G) = Op (N). Proof. We have NOp (G) ≥ POp (G) = G. Put U = N ∩ Op (G). Then U ⊲ G and G/Op (G) ≅ N/U is a p-group, so that U ≥ Op (N). We have to prove that U = Op (N). Assume, by way of contradiction, that U > Op (N). Let V ⊲ N be such that V < U and |U/V| = p; then U/V ≤ Z(N/V). Write |G| = p a m, where p ∤ m; let ε be a primitive |G|-th root of unity, and δ = εp

a−1

m

,

G = Gal(ℚ(ε)/ℚ(δ)).

Then |G| = p a−1 . Note that G fixes all p-th roots of unity. Let λ be a faithful character of U/V, considered as a character of U. Since there are exactly p − 1 such characters and the values of λ are p-th roots of unity, it follows that λ is invariant with respect to the p-group P and with respect to G. p Put ϑ = λO (G) . Since λ is invariant with respect to P and G, so is ϑ. The number p ϑ(1) = |O (G) : U| is prime to p. Let us prove that ϑ possesses an irreducible constituent θ invariant with respect to P and G and such that p ∤ ⟨ϑ, θ⟩θ(1). To this end, consider the direct product Π = P × G. By the above, Π is a p-group. This group acts on characters of normal subgroups of G: if H ⊲ G, χ is a character of H and (t, σ) ∈ Π (t ∈ P, σ ∈ G), then, by definition, χ(t,σ) = (χ t )σ = (χ σ )t . Since the character ϑ of Op (G) is Π-invariant (see above), then the action of Π on the set Irr(ϑ) is defined. Let ϑ = ∑θ∈Irr(ϑ) a θ θ, where all a θ ∈ ℕ. It is clear that the coefficient a θ depends on the Π-orbit of θ only. Therefore, ϑ = ∑η c η η, where the η are orbit sums and c η ∈ ℕ.

VII Clifford theory | 343

Obviously, all the η are Π-invariant. Note (see above) that ϑ(1) = λO

p

(G)

(1) = |Op (G) : U|

(U = N ∩ Op (G)).

We get NOp (G) = G since P ≤ N. Therefore, p ∤ |Op (G) : U| = |G : N|. Since p ∤ ϑ(1) and ϑ = ∑η c η η, we get p ∤ c η η(1) for some η. Let η = ∑θ∈O θ, where O is the Π-orbit containing θ. We have η(1) = |O|θ(1), and we conclude that p ∤ |O| and p ∤ θ(1). On the other hand, since Π is a p-group, |O| is a power of p. It follows that |O| = 1 hence η = θ and so θ is a Π-invariant irreducible character of Op (G), p ∤ θ(1) and p ∤ c θ = ⟨ϑ, θ⟩. Since G = POp (G) and θ is P-invariant, it follows that θ is G-invariant. We claim that there exists an extension of θ to G. Since |G : Op (G)| is a power of p, it suffices to show that p ∤ o(θ)θ(1) (see Corollary 6.10). By the above, it suffices to show that p ∤ o(θ). This follows from Op (Op (G)) = Op (G). Therefore, by Corollary 6.10, there exists a unique extension χ of θ to G such that o(χ) = o(θ). Since such extension is unique, χ is Π-invariant. Let χ N = ∑φ∈Irr(N) b φ φ, where b φ ≥ 0. Since χ ∈ ProlG (θ), it follows that ∑

b φ ⟨φ U , λ⟩ = ⟨χ U , λ⟩ = ⟨θ U , λ⟩ = ⟨θ, λO

p

(G)

⟩ = ⟨θ, ϑ⟩ ≢ 0 (mod p).

φ∈Irr(N)

Therefore, there is φ ∈ Irr(N) such that b φ ⟨φ U , λ⟩ ≢ 0 (mod p). Using the same arguments as above for a choice of θ, one can choose φ so that it is Π-invariant. Since ⟨φ U , λ⟩ ≠ 0, it follows from Clifford’s theorem that φ U = eλ,

φ(1) = e = ⟨φ U , λ⟩ ≢ 0 (mod p).

Since V = ker(λ) and φ U = eλ (recall that λ(1) = 1), we obtain that V ≤ ker(φ) and so φ ∈ Irr(N/V). Since N/V is a p-group and, by the Frobenius–Molien theorem, φ(1) | |N/V|, it follows that φ(1) = 1, and so φ U = λ. We conclude that N/ker(φ) is a cyclic p-group of order, say p s . In that case, there is x ∈ N such that φ(x) = α is a primitive p s -th root of 1. Since the element φ(x) of the field ℚ(ϵ) is G-invariant in view of G-invariance of the character φ, we have α ∈ ℚ(δ) (see the second paragraph of the proof), so that p s−1 (p − 1) = |ℚ(α) : ℚ| | |ℚ(δ) : ℚ| = p − 1. p

Therefore, s = 1 so that |N/ker(φ)| = p and hence A1 (N) ≤ ker(φ). By condition, we p p have A1 (N) = N ∩ A1 (G). Therefore, p

ker(φ) ≥ N ∩ A1 (G) ≥ N ∩ Op (G) = U which implies that λ = φ U = 1U , contrary to the choice of λ. Thus, the desired equality N ∩ Op (G) = Op (N) is proved. The condition of Theorem 10.1 is often formulated as follows: If the subgroup N of G p controls A1 (G), then N controls Op (G).

344 | Characters of Finite Groups 1 Exercise 10.1. If P ∈ Sylp (G), M ⊲ G, M ∩ P ≤ Φ(P), then M is p-nilpotent (i.e., it has a normal p-complement). Solution. Without loss of generality, one may assume that PM = G. Then we have G/M ≅ P/(P ∩ M). Set T = Φ(P)M. Then p

T = A1 (G),

p

p

P ∩ A1 (G) = P ∩ T = Φ(P) = A1 (P).

By Theorem 10.1, we get P ∩ Op (G) = Op (P) = {1} so Op (G) is a normal p-complement of G. It follows that M, as a subgroup of G, has a normal p-complement as well (we use Theorem 10.1 with N = P), as desired. Exercise 10.2 (Wielandt). Let P ∈ Sylp (G) be cyclic and suppose that at least two composition indices of G are multiples of p. Then G is p-solvable. Solution. One may assume that Op󸀠 (G) = {1}. By hypothesis, there is N ⊲ G such that p | |N| and p | |G : N|. Then N ∩ P ≤ Φ(P) since P is cyclic; then NΦ(P) ∩ P = Φ(P). It follows from Exercise 10.1 that N has a normal p-complement N1 . By assumption, N1 = {1}; then N is a normal cyclic p-subgroup. It follows that G/CG (N) is abelian as a subgroup of Aut(N). Next, CG (N) has a normal p-complement T since it has no minimal nonnilpotent subgroup with derived subgroup of order divisible by p (here we use Frobenius’ normal p-complement theorem). By assumption, T = {1}. Since G is the extension of a p-group CG (N) by the abelian group G/CG (N), it is solvable. Theorem 10.2 (P. Roquette [Roq2]; see [Hup1, Satz IV.4.6]). Assume that H is a π-Hall subgroup of G and N ⊲ G. If N ∩ H ≤ Φ(H), then N has a normal π-complement.

11 Further results on the class number In this section we present results related with inequality (8.2) and Lemma 8.3. Exercise 11.1. If H < G, then k(H) ≤ hg n(G) +

g g󸀠

− 1.

Recall that w(τ) = |Irr(τ)| for τ ∈ Char(G). Theorem 11.1. Let H be a proper subgroup of a nonabelian group G. Then one has k(H) = hg n(G) + gg󸀠 − 1 if and only if G is a Frobenius group with abelian kernel H of index 2. Proof. Let k(H) = hg n(G) + gg󸀠 − 1. By reciprocity, we have w(χ H ) ≤ hg for χ ∈ Irr1 (G). Since (1H )G = 1G + θ, where θ ∈ Char(G) has degree hg − 1, and there is only one member of Irr(G) − {1G }, say χ, such that 1H ∈ Irr(χ H ), it follows that θ ∈ Irr(G) . As for any τ ∈ Irr1 (G) we have w(τ h ) = hg , it follows that θ (of degree hg − 1) is linear. We conclude that hg = 2. Thus, the degree of every nonlinear irreducible character of G is even so G is 2-nilpotent, by Theorem XIV.3.3. Besides, ⋃χ∈Irr# (G) Irr(χ H ) is a partition. It follows that μ G ∈ Irr(G) for any μ ∈ Irr∗ (H). By Corollary X.4.2, G is a Frobenius group with

VII Clifford theory | 345

Frobenius kernel H is index hg = 2; then H is abelian. Conversely, if G is a Frobenius group with the (abelian) Frobenius kernel of index 2, then k(H) = 2 + 2 ⋅

g g h−1 = 2h + 1 = n(G) + 󸀠 − 1, 2 h g

by Theorem X.3.2. In particular, if H < G and k(H) = 2n(G) + 1, then G is a Frobenius group with kernel H of index 2. Let G = D2n and C2n−1 ≅ H < G. Then g = 2n , We have

h = 2n−1 ,

g󸀠 = 2n−2 ,

n(G) = 2n−2 − 1,

k(H) = 2n−1 .

g g󸀠 n(G) + − 1 = 2 ⋅ (2n−2 − 1) + 4 − 1 = 2n−1 + 1 > k(H). h g

Exercise 11.2. Classify the pairs of groups H < G satisfying k(H) =

g g n(G) + 󸀠 − 2. h g

Let G be a Frobenius group with kernel H of index 3. Then k(H) = 3n(G) + 1 and g g k(H) − 1 n(G) + 󸀠 − 2 = 3n(G) + 1 = 3 ⋅ + 3 − 2 = k(H). h g 3

12 Miscellaneous This section contains some additional information about the subject of this chapter. Exercise 12.1. Let H < G be subnormal, χ ∈ Irr(G) and ψ ∈ Irr(χ H ). Then ψ(1) | χ(1) χ(1) | |G : H|. and ψ(1) Solution. By hypothesis, M = H G ⊲ G. We have χ H = (χ M )H . There exists a character θ ∈ Irr(χ M ) such that ψ ∈ Irr(θ H ). By induction applied to M, we have ψ(1) | θ(1). By Clifford’s theorem, θ(1) | χ(1) so that ψ(1) | χ(1), proving the first assertion. Next, by χ(1) θ(1) induction, ψ(1) divides |M : H|. By Clifford theory, θ(1) divides |G : M|. Therefore χ(1) χ(1) θ(1) = ⋅ | |G : M||M : H| = |G : H|. ψ(1) θ(1) ψ(1) Exercise 12.2 (Gallagher [Gal8]). Let H ≤ G. Prove the following statements without using character theory): (a) k(H) ≤ |G : H|k(G) with equality if and only if H = G. (b) k(G) ≤ |G : H|k(H) with equality if and only if G = CG (s)H for all s ∈ G. (c) If H ⊴ G, then k(G) ≤ k(H)k(G/H) with equality if and only if for all s ∈ G one has {t ∈ G | s t H = sH}H = G.

346 | Characters of Finite Groups 1 Exercise 12.3. One can rewrite inequality (8.2) in the following form: gk(G) − hk(H) ≥

g2 − h2 . g󸀠

(1)

If in (1) we have equality, then H ≥ G󸀠 (see Lemma 8.3 (a)). Hint. Replacing n(G) by k(G) −

g , g󸀠

we rewrite (8.2) in the form

hk(H) ≤ g(k(G) −

g h2 )+ 󸀠 , 󸀠 g g

which, after simplifications, yields (1). Exercise 12.4. If a nonabelian group G contains a normal abelian subgroup H of index p, then in (1) we have equality, but G is not necessarily a Frobenius group. Solution. In our case, g = ph so, rewriting (1) and dividing by h, we get pk(G) − k(H) ≥

(p2 − 1)h . g󸀠

(2)

Using Exercise 3.9, we can replace pk(G) by k(H) + (p2 − 1)s, where s is the number of G-invariant characters in Irr(H), and obtain k(H) + (p2 − 1)s − k(H) = (p2 − 1)s. Since s = gh󸀠 , we get finally equality in (2) so in (1). Exercise 12.5 (Isaacs). Let {1} < P ∈ Sylp (G), N = NG (P) and R = Op󸀠 (N). Then the following conditions are equivalent: (a) The p-length of G is equal to 1. (b) If ψ ∈ Irr(N/R), then ψ = χ N for some χ ∈ Irr(G). (c) If ψ ∈ Irr(N/R) and p ∤ ψ(1), then ψ = χ N for some χ ∈ Irr(G). Exercise 12.6. Let G = ES(m, p), an extraspecial group of order p2m+1 . Then cd(G) = {1, p m },

k(G) = p2m + p − 1,

n(G) = p − 1.

If χ ∈ Irr1 (G), then χZ(G) = p m λ, where λ is a nonprincipal linear character of Z(G). The character χ is uniquely determined by the character λ in the last equality. Solution. If x ∈ G − Z(G), then |G : CG (x)| = p. It follows that k(G) = |Z(G)| + Next,

1 |G − Z(G)| = p2m + p − 1. p

|Lin(G)| = |G/G󸀠 | = p2m 󳨐⇒ n(G) = p − 1.

Let Irr1 (G) = {χ1 , . . . , χ p−1 }, By Theorem VII.2.3, c p−1 ≤ p−1

pm .

χ1 (1) = p c i ,

c1 ≤ ⋅ ⋅ ⋅ ≤ c p−1 .

We have

∑ p2c i = p2m+1 − p2m = p2m (p − 1).

i=1

VII Clifford theory | 347

It follows that

c1 = ⋅ ⋅ ⋅ = c p−1 = p m .

By Clifford’ theorem, (χ i )Z(G) = p m λ i , where λ i ∈ Lin(Z(G)) is faithful. Since χ i vanishes on G − Z(G), we conclude that λ1 , . . . , λ p−1 are pairwise distinct. It follows that two extraspecial p-groups of equal order have coinciding character tables. Hence, character tables do not determine orders of group elements. The following theorem is useful in many applications. Theorem 12.1. Let G = H ⋅ P be a semidirect product with kernel P ∈ Sylp (G) and let Z(P) be cyclic of order p n . Then there are at least p n−1 (p − 1) faithful characters in Irr(P). If, in addition, Z(P) ≤ Z(G) and Irr(P) contains exactly p n−1 (p − 1) faithful characters, then all faithful characters from Irr(P) are extendible to G. We leave the easy proof of this theorem to the reader. The last statement follows from Corollary 4.8. Exercise 12.6 shows that the subgroup P of Theorem 12.1 may be taken as ES(m, p). Exercise 12.7. Using Theorem 12.1, construct a group G such that |cd(G)| = 4 = dl(G), where dl(G) is the derived length of G. Exercise 12.8 (Tuan). Let H be an abelian normal subgroup of a nonabelian group G and |G : H| = p, a prime. Then |G| = p|G󸀠 ||Z(G)|. Solution. First we find the class number k(G) of G. Let |G| = g,

|H| = h,

|G󸀠 | = g󸀠 ,

|Z(G)| = z.

If x ∈ H − Z(G), then CG (x) = H so H is the union of z + 1p (h − z) = z + pg2 − pz conjugate classes of G. If x ∈ G − H, then we have |CG (x)| = pz, so the set G − H is the union of pz g g (g − p ) = pz − z conjugacy classes of G. It follows that g z g z k(G) = z + 2 − + pz − z = 2 − + pz. p p p p By Ito’s Theorem 2.3 on degrees, we have cd(G) = {1, p}. Therefore, g g g g = 󸀠 + p2 (k(G) − 󸀠 ) = p2 k(G) − (p2 − 1) 󸀠 . g g g After simplifications, we obtain (p2 − 1) gg󸀠 = pz(p2 − 1) so g = pg󸀠 z, as required. Below we formulate a deep theorem, due to Dade, on the extension of characters. Let G, H, E and K be groups satisfying the following conditions: (a) G = H⋅E is a semidirect product with kernel E which is an extraspecial p-subgroup. (b) If K ⊴ H and p ∤ |K|, then [E, K] = E. (c) Z(E) ≤ Z(G). Theorem 12.2 (Dade [Dad6]). Let G, H, E be as above and φ ∈ Irr1 (E). Then φ admits an extension χ to G. If p > 2, then χ is uniquely determined (this is false for p = 2). It is important in Theorem 12.2 that |H| and |E| are not necessarily coprime.

348 | Characters of Finite Groups 1 Definition. Let L, K ⊲ G, L ≤ K and let K/L be abelian. We say that K/L is a strict section of G if there exists N ⊴ G such that the group S ≅ N/CN (K/L) of automorphisms of K/L induced by N satisfies the following conditions: (a) CK/L (S) = {1}. (b) GCD(|S|, |K/L|) = 1. (c) S is solvable. A section K/L is strict in G if and only if it is strict in G/L. If G is solvable, then any of its noncentral chief factors is a strict section. Theorem 12.3 (Dade [Dad6]). Let K/L be a strict section of G. Let θ ∈ Irr(K) be G-invariant and θ L = eφ, where e2 = |K : L| (see Theorem 5.1) and φ ∈ Irr(L). Then there exist H ≤ G and a bijection π : Irrφ (H) → Irrθ (G) such that (a) KH = G and K ∩ H = L, (b) ψ π (1) = eψ(1) for ψ ∈ Irrφ (H). If |K/L| is odd, we need not assume in the definition of a strict section that S is solvable. Therefore, as groups of odd order are solvable, Theorem 12.3 also holds when K/L is an abelian noncentral chief factor of G. The following two results (due to Isaacs) may be derived from Theorem 12.3. Theorem 12.4. Let A act on a solvable group G, GCD(|A|, |G|) = 1. Then the number of A-invariant irreducible characters of G is equal to |Irr(CG (A))|. Theorem 12.5. Let P be a π-Hall subgroup of a π-solvable group G, N = NG (P). Then G and N have the same numbers of irreducible characters whose degrees are π󸀠 -numbers.

13 Monomiality criterion Below we will prove a criterion which enables us to determine when an extension is an M-group. Lemma 13.1 (Huppert [Hup1]). Let Θ be a class of groups with the following properties: (a) Θ is inherited by subgroups and epimorphic images. (b) If G ∈ Θ is nonabelian, then G contains a normal abelian subgroup A ≰ Z(G). Then all Θ-groups are M-groups. Proof. We use induction on |G|. By induction, G is solvable (here we use (b)). Let χ ∈ Irr(G). We have to prove that χ is monomial. If χ is not faithful, the result follows by induction applied to G/ker(χ). Now let χ be faithful. One may assume that G is nonabelian. By (b), it contains a normal abelian subgroup A ≰ Z(G). Let χ A = e(λ1 +⋅ ⋅ ⋅+λ t ) be the Clifford decomposition. By the assumption on A, we have t > 1.² Therefore, by 2 Indeed, if t = 1, then |χ(a)| = χ(1) for all a ∈ A so A ≤ Z(χ) = Z(G) since χ is faithful.

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Theorem 2.2 (a), χ = ψ G , where ψ ∈ Irr(IG (λ1 )). Since IG (λ1 ) < G, it follows, by induction, that ψ = μIG (λ1 ) , where μ ∈ Lin(H) and H ≤ IG (λ1 ). Therefore, χ = ψ G = (μIG (λ1 ) )G = μ G so that χ is monomial. Theorem 13.2 (Huppert [Hup1]). Let A ⊴ G and let the quotient group G/A be supersolvable. If A is solvable and all Sylow subgroups of A are abelian, then G is an M-group. Proof. We may assume that, if A is abelian, it is a maximal abelian normal subgroup of G. On this assumption, if B/A is a normal subgroup of prime order in G/A, then B is not abelian, so that A ≰ Z(G). Now let A be nonabelian. Let B be a G-invariant abelian subgroup of maximal order in A. If D/B is a minimal normal subgroup in G/B contained in A/B, then D󸀠 ≤ B. If B ≤ Z(G), then D is nilpotent and so abelian as a subgroup of A (recall that all Sylow subgroups of A are abelian), a contradiction. Thus, G satisfies condition (b) of Lemma 13.1. Condition (a) of that lemma is obviously satisfied. Therefore, by Lemma 13.1, G is an M-group. It follows from Theorem 13.2 that an extension of abelian group by supersolvable group is an M-group. Corollary 13.3 (Blichfeldt). Supersolvable groups are M-groups. In particular, nilpotent groups are M-groups.

14 Commutativity criterion of a normal section The following useful fact follows from Clifford’s theorem: Theorem 14.1. If H ≤ G, λ ∈ Lin(H), χ = λ G , then H G /ker(χ) is abelian. Proof. Since ker(χ) = ker(λ)G ≤ H G , we get ker(χ H G ) = H G ∩ ker(χ) = H G ∩ ker(λ)G = ker(λ)G = ker(χ). We have to prove that H G /ker(χ H G ) is abelian. Let Irr(χ) = {χ1 , . . . , χ k }. Then, by reciprocity, λ ∈ Irr(χ iH ) hence λ H G ∈ Irr(χ iH G ), so that, by Clifford’s theorem, all irreducible constituents of χ iH G are linear, i ∈ {1, . . . , k}. Therefore, all irreducible constituents of χ H G are linear, so that (H G )󸀠 ≤ ker(χ H G )(= ker(χ)). It follows that H G /ker(χ) as an epimorphic image of H G /(H G )󸀠 , is abelian. Exercise 14.1. Under the assumptions of Theorem 14.1, if ker(χ) = {1}, then the degree of any irreducible character of G divides n!, where n = |G : H|. Hint. The subgroup H G is abelian and G/H G is isomorphic to a subgroup of the symmetric group Sn . Apply Theorem 2.3.

350 | Characters of Finite Groups 1 Exercise 14.2. Let G be a p-group and let χ ∈ Irr(G) be faithful with χ(1) = p n . Let b(G) = max{ψ(1) | ψ ∈ Irr(G)} = p j(n) , where j(n) = 1 + p + ⋅ ⋅ ⋅ + p n−1 . Is it true that G contains a normal abelian subgroup A such that G/A is isomorphic to a Sylow p-subgroup of Sp n ? It is always true that b(G) ≤ p j(n) (D. Passman).

15 Dade’s example By Theorem 9.2, a normal Hall subgroup of an M-group is an M-group. But, as one can see from the examples constructed by S. D. Berman, R. van der Waall and E. Dade, a normal subgroup of an M-group is not necessarily an M-group. The following example is due to Dade. Let K be a finite extension of degree n of the field F. Then K is a finite-dimensional F-algebra, and the regular representation gives an inclusion ψ : K → F n (the ring of n × n matrices over F). Set tr(α) = trK/F (α) = tr(ψ(α))

(α ∈ K).

Obviously, tr(α) is an element of F (henceforth we shall omit the index K/F). Let us use the map tr : GF(8) → GF(2) to convert the cartesian product GF(8) × GF(8) × GF(2) into ES(3, 2), the extraspecial group E of order 27 with the multiplication (x, y, z)(x󸀠 , y󸀠 , z󸀠 ) = (x + x󸀠 , y + y󸀠 , z + z󸀠 + tr(xy󸀠 )) for all x, x󸀠 , y, y󸀠 ∈ GF(8), z, z󸀠 ∈ GF(2). The triple (0, 0, 0) is the identity element of the group E. It is easy to check that A1 = GF(8)×0×0 and A2 = 0×GF(8)×0 are elementary abelian subgroups of order 8 in E, and 0×0×GF(2) is Z(E), the center of E. Let ⟨ξ⟩ = GF(8)∗ be the multiplicative group of GF(8). Define a mapping σ : E → E as follows: (x, y, z)σ = (ξx, ξ −1 y, z),

x, y ∈ GF(8), z ∈ GF(2).

It follows from (x, y, z)σ (x1 , y1 , z1 )σ = (ξ(x + x1 ), ξ −1 (y + y1 ), z + z1 + tr(xy1 )) = ((x, y, z)(x1 , y1 , z1 ))σ that σ is a homomorphism. If (x, y, z) ∈ ker(σ), then we have x = y = 0 ∈ GF(8) and z = 0 ∈ GF(2) so (x, y, z) = (0, 0, 0) is the identity element of E, and we conclude that σ ∈ Aut(E). We have o(σ) = o(ξ) = 7. It is obvious that A1 , A2 , Z(E) as just defined are irreducible σ-subgroups which are not pairwise σ-isomorphic. We have A1 ∩ A2 = A1 ∩ Z(E) = A2 ∩ Z(E) = {1}.

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As we have noted, the group E is extraspecial of order 27 . Thus, |⟨A1 , A2 ⟩| ≥ 26 implies Z(E) < ⟨A1 , A2 ⟩. Next, (A1 × Z(E)) ∩ A2 = {1} 󳨐⇒ ⟨A1 , A2 ⟩ = ⟨A1 , A2 , Z(E)⟩ = E (here we used the product formula). Let (x, y, z) ∈ E, where x ≠ 0 ≠ y. It is easy to prove that the normal closure of (x, y, z) in ⟨E, σ⟩ is equal to E. Moreover, for an element (x, y, z) as above we have i ⟨(x, y, z)σ | i = 1, . . . , 7⟩ = E. Therefore: Claim (i). The subgroups {1}, Z(E), A1 , A1 × Z(E), A2 , A2 × Z(E), E are the only σ-invariant subgroups of E. Claim (ii). The semidirect product ⟨σ⟩ ⋅ E with kernel E admits an automorphism α of order 2 such that σ α = σ−1 , (x, y, z)α = (y, x, z + tr(yx)) ((x, y, z) ∈ E). Claim (iii). Put N = ⟨α⟩ ⋅ (⟨σ⟩ ⋅ E), a semidirect product with kernel ⟨σ⟩ ⋅ E. The group N admits an automorphism β of order 2 such that ⟨[α, β]⟩ = Z(E) and β centralizes ⟨σ⟩ ⋅ E. It is clear that ⟨α, β⟩ ≅ D8 . Put G = ⟨β⟩ ⋅ N, a semidirect product with kernel N. Obviously, |G| = 7 ⋅ 29 . By Theorem 13.2, G/Z(E) is an M-group since it is an extension of elementary abelian group E/Z(E) of order 26 by supersolvable group ⟨α, β, σ⟩. In fact, the following statement is true: Claim (iv). We claim that G is an M-group. Assume, by way of contradiction, that χ ∈ Irr(G) is not monomial. Then Z(E) ≰ ker(χ) (as we have noted, G/Z(E) is an M-group). Therefore, χ E = eη, where η ∈ Irr(E) is faithful and is therefore of degree 8 (note that Irr(E) contains only one faithful member and the kernel of any linear character of E contains Z(E) = E󸀠 ). Since Irr1 (E) = {η}, it follows that η is therefore ⟨σ⟩ ⋅ E-invariant. Therefore, by Corollary 4.8 and reciprocity, η has exactly |⟨σ, E⟩ : E| = 7 extensions to ⟨σ, E⟩, and these extensions are of the form μi φ

(i = 0, 1, . . . , 6),

μ ∈ Lin(⟨σ, E⟩/E),

φ ∈ Irr(⟨σ, E⟩),

φ(1) = 8.

Since cd(⟨σ⟩A1 Z(E)) = {1, 7}, by Theorem 2.3, and = 8, it follows that the i restriction of μ φ to ⟨σ⟩A1 Z(E) contains a linear constituent λ i . Next, (μ i φ)(1)

⟨σ,E⟩

|⟨σ, E⟩ : ⟨σ⟩A1 Z(E)| = 8 = (μ i φ)(1) 󳨐⇒ λ i

= μ i φ,

i = 0, 1, . . . , 6.

Note that the G-orbits of the set {μ i φ}60 coincide with its α-orbits (see the definition of β in (iii)); therefore, |G : IG (μ i φ)| = 2. Since o(α) = 2 ∤ 7 = |{μ i φ}60 |, one of these α-orbits consists of a single element (say, {φ}), and the others consist of two elements each (since μ α = μ−1 ): {μφ, μ−1 φ}, {μ2 φ, μ−2 φ}, {μ3 φ, μ−3 φ}. It follows from ⟨σ, E⟩ ⊲ G that Irr(χ⟨σ,E⟩ ) is one of these four orbits. The equality |G : ⟨σ, E⟩| = 4 implies that χ(1) | 16 (we are using the Clifford theory). If μ i φ ∈ Irr(χ⟨σ,E⟩ ) for some i ∈ {1, 2, 3}, then χ(1) = 16. Next, ⟨β⟩⟨σ, E⟩ = ⟨β⟩ × ⟨σ, E⟩ = IG (μ i φ) 󳨐⇒ χ = τ G

(τ ∈ Irr(⟨β⟩ × ⟨σ, E⟩).

352 | Characters of Finite Groups 1 Since ⟨β⟩ × ⟨σ, E⟩ is an M-group, by Theorem 13.2, we have τ = λ⟨β⟩×⟨σ,E⟩

(λ ∈ Lin(H)),

where H ≤ ⟨β⟩ × ⟨σ, E⟩.

Therefore, χ = λ G , by transitivity of inducing. Now let φ ∈ Irr(χ⟨σ,E⟩ ) and ν ∈ Lin# (⟨β⟩). Then φ × 1⟨β⟩ and φ × ν are conjugate with respect to α and therefore, again, χ(1) = 16. Since (φ × μ)G (1) = 16 and μ ∈ {1⟨β⟩ , ν}, it follows that (φ × μ)G = χ. Next, φ × β = λ⟨β⟩×⟨σ,E⟩

(λ ∈ Lin(H)),

where H ≤ ⟨β⟩ × ⟨σ, E⟩,

and therefore χ = λ G . This proves (iv). It remains to prove that: Claim (v). N = ⟨α, σ, E⟩ is not an M-group. Indeed, as before, let φ be a faithful irreducible character of the group ⟨σ, E⟩ = R which is invariant with respect to G, φ(1) = 8. Since N/R (of order 2) is cyclic, φ admits an extension ψ to N = ⟨α⟩ ⋅ (⟨σ⟩ ⋅ E), by Theorem 3.9. To prove that ψ is not monomial, it is sufficient to show that N has no subgroup of index ψ(1) = 8. Assume that such an H of index 8 in N exists. Put H N = ⋂x∈N H x . Since |N : E| = 14, it follows that E ≰ H. Therefore, A i ≰ H N (i = 1, 2), i.e., |H N | ≤ 2. But then N/H N is isomorphic to a subgroup of S|N:H| = S8 . But any subgroup of a Sylow 2-subgroup of S8 is generated by four elements. Since Sylow 2-subgroup of N/H N is not generated by four elements (its rank is 6), we have arrived at a contradiction. Thus N is not an M-group. This proves (v). Up to now, there is no analog of Dade’s example of odd order. It is impossible to obtain a complete group-theoretic classification of M-groups. In fact, according to a result proved by Dade, any solvable group is isomorphic to a subgroup of a suitable M-group.

16 Generalizations of M-groups In this section we will generalize the notion of an M-group. Definition. Let N ⊴ G and χ ∈ Irr(G). Then χ is an M-N-character if there exist H ≤ G with N ≤ H, and φ ∈ Irr(H) such that χ = φ G and φ N ∈ Irr(N). If all χ ∈ Irr(G) are M-N-characters for a fixed N, then G is called an M-N-group. Note that M-{1}-groups are exactly the M-groups since, in that case, φ must be linear. In that case, if χ ∈ Irr(G), N ⊴ G and χ N ∈ Irr(N), then χ is an M-N-character (take H = G and φ = χ in the Definition); however, this extreme case is not interesting. If G is an M-group and N ⊲ G, then G is not necessarily an M-N-group (but, as we shall soon see the last assertion is true for supersolvable groups G). Theorem 16.1 (D. Price [Pri]). Suppose that N ⊲ G and G/N is solvable. Assume that the order of every chief factor of any nonidentity subgroup of G/N is not a square. Then G is an M-N-group.

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Proof. Let χ ∈ Irr(G). We have to show that χ is an M-N-character. If χ N ∈ Irr(N), then χ is an M-N-character (see the paragraph preceding the theorem). We therefore assume in what follows that χ N is reducible. Let K ≤ G be a minimal G-invariant subgroup with the properties N ≤ K and χ K ∈ Irr(K) (such K exists since χ G = χ ∈ Irr(G)). Then N < K, by assumption. Let L ⊲ G and let N ≤ L < K be such that K/L is a chief factor of G. Since χ K ∈ Irr(K) and χ L = (χ K )L is reducible, by assumption, it follows, as |K/L| is not a square, that the following Clifford decomposition holds (see Theorem 5.1): t

χL = ∑ φi ,

t = |Irr(χ L )| = |K : L| > 1.

i=1

Obviously, T = IG (φ1 ) ≥ L ≥ N, |G : T| = t > 1. Then χ = ψ G for some ψ ∈ Irr(T) (Theorem 2.2 (a)). Since T/N < G/N, it follows that T is an M-N-group, by induction, and therefore ψ = ϑ T for some ϑ ∈ Irr(H), where N ≤ H ≤ T and ϑ N ∈ Irr(N). Now, χ = ψ G = (ϑ T )G = ϑ G , by Theorem V.1.4, and hence χ is an M-N-character. Since χ is arbitrary, we are done. If G is supersolvable and N ⊲ G, then G is an M-N-group, by Theorem 16.1. In particular, a supersolvable group G is an M-G󸀠 -group. This means that if χ ∈ Irr(G) and G is supersolvable, then χ = ϑ G , where ϑ ∈ Irr(H), G󸀠 ≤ H ≤ G, ϑ G󸀠 ∈ Irr(G󸀠 ).

17 On the derived length of a p-group By dl(G) we denote the derived length of a solvable group G. Theorem 17.1. If a p-group G has a faithful irreducible character χ of degree p n , then dl(G) ≤ n + 1, and this estimate is best possible for all n. Proof. We use induction on n. If G has no normal abelian subgroup of type (p, p), then p = 2 and G is of maximal class hence n = 1 and dl(G) = 2 = n + 1. Next we assume that Ep2 ≅ R ⊲ G; then |G : CG (R)| ≤ p and so there exists a maximal subgroup M of G such that R ≤ M ≤ CG (R). Since Z(M) is noncyclic, it follows that χ M is reducible. Let χ M = μ1 + ⋅ ⋅ ⋅ + μ p be the Clifford decomposition; we have p μ i (1) = p n−1 for i = 1, . . . , p. Since χ is faithful, we get ⋂i=1 ker(μ i ) = {1}. Then, by induction, dl(M/ker(μ i )) ≤ (n − 1) + 1 = n. Since M is isomorphic to a subgroup of (M/ker(μ1 )) × ⋅ ⋅ ⋅ × (M/ker(μ p )), it follows that dl(M) ≤ n. Since G/M is abelian (of order p), we conclude that dl(G) ≤ n + 1.³ It remains to show that G = Σ p n+1 ∈ Sylp (Sp n+1 ) has derived length n + 1 and a faithful irreducible character of degree p n . We have G = H wr Cp , the standard wreath product, with “passive” factor H ≅ Σ p n and “active” factor Cp of order p. Then the 3 According to the letter of N. Ito, he also proved the result of this paragraph.

354 | Characters of Finite Groups 1 base B = H1 × ⋅ ⋅ ⋅ × H p of G has index p in G and H i ≅ H. It is known that dl(G) = n + 1. Let F = H2 × ⋅ ⋅ ⋅ × H p and let ϕ be a faithful irreducible character of degree p n−1 of B/F ≅ H, existing by induction. Set χ = ϕ G , and prove that χ ∈ Irr(G) is faithful (of degree p n ), thereby completing the proof. Since ker(χ) = ker(ϕ)G = F G = {1}, the character χ is faithful. Assume, however, that χ is reducible. Then, by Clifford theory, we have χ = τ1 + ⋅ ⋅ ⋅ + τ p , where τ i (1) = p n−1 for i = 1, . . . , p. In that case, by reciprocity, χ B = p ⋅ ϕ so {1} < F = ker(ϕ) ≤ ker(χ) = {1}, which is a contradiction. Definition. A group G is said to be an M∗ -group if it satisfies the following condition. Whenever H is a subnormal subgroup of G and χ is a nonlinear irreducible character of H, there exists in H a normal subgroup A of prime index such that χ A is reducible. We consider abelian groups as M∗ -groups. Obviously, subnormal subgroups and epimorphic images of M∗-groups are M∗-groups. It follows, by induction, that M∗ -groups are solvable. The symmetric group S4 is an M-group but not an M∗ -group (let H = G, χ ∈ Irr(G) be of degree 3 and A = A4 ). If m ∈ ℕ, then λ(m) denotes the number of prime factors of m (multiplicities counted). For example, λ(32) = 5, λ(96) = 6. The following proposition was inspired by Isaacs’ letter at January 29, 2005. Theorem 17.2. Suppose that an M∗ -group G has a faithful irreducible character χ. Then dl(G) ≤ λ(χ(1)) + 1. Proof (Isaacs). One may assume that G is nonabelian; then λ(χ(1)) = n > 0. We use induction on n. Let T ⊲ G be of prime index, say p, such that χ T is reducible and let χ T = μ1 + ⋅ ⋅ ⋅ + μ p be the Clifford decomposition. We have μ i (1) = χ(1) p and so we obtain λ(μ i (1)) = n − 1 and dl(T/ker(μ i ) ≤ (n − 1) + 1 = n, by induction. It follows that (below T (n) is the n-th member of the derived series of T) p

T (n) ≤ ⋂ ker(μ i ) = T ∩ ker(χ) = {1} 󳨐⇒ dl(T) ≤ n. i=1

Since G/T is abelian, we get dl(G) ≤ dl(T) + 1 ≤ n + 1. Given n ∈ ℕ and a prime p, set j p (n) = 1 + p + ⋅ ⋅ ⋅ + p n−1 =

pn − 1 . p−1

Given a group G, set b(G) = max{χ(1) | χ ∈ Irr(G)}. If f : ℕ → ℕ be such that f(1) = 1 and f(n + 1) = pf(n) + 1, then, by induction, f(n) = j p (n). Suppose that a p-group G has a faithful irreducible character χ of degree p. Since G is an M-group, there exist A < G and a linear character μ1 of A such that χ = μ1G . By Clifford’s theorem, χ A = μ1 + ⋅ ⋅ ⋅ + μ p , where all μ i are distinct G-conjugate linear. Then {1} = ker(χ) ≥ (A󸀠 )G = A󸀠 so A is abelian. Theorem 17.3 (Passman). Suppose that a p-group G has a faithful irreducible character χ of degree p n . Then b(G) ≤ p j p (n) .

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Proof. In view of the paragraph preceding the theorem, one may assume that n > 1. Then, as in the proof of Theorem 17.2, G has a maximal subgroup T of index p such that χ T = μ1 + ⋅ ⋅ ⋅ + μ p is the Clifford decomposition. By induction, b(T/ker(μ i )) ≤ p j p (n−1) and T is a subgroup of the direct product D = (T/ker(μ1 )) × ⋅ ⋅ ⋅ × (T/ker(μ p )). We have b(T) ≤ b(D) ≤ (p j p (n−1) )p = p pj p (n−1) . Take τ ∈ Irr(G); then τ T (1) ≤ pb(T) ≤ p pj p (n−1)+1 = p j p (n) 󳨐⇒ τ(1) = τ T (1) ≤ p j p (n) .

18 CR- and NR-subgroups We begin with the following definition. Definition 18.1 (Isaacs). Let G be a group, H ≤ G. We say that H has the character restriction property in G if each irreducible character of H is the restriction of some irreducible character of G. We also call H a CR-subgroup of G. In the case under consideration, cd(H) ⊆ cd(G). Exercise 18.1. Let H be a CR-subgroup of G. If H < T < G, then H is a CR-subgroup of T. Solution. Let ϕ ∈ Irr(H) and let χ ∈ Irr(G) be such that ϕ = χ H . Let θ ∈ Irr(T) be such that θ ∈ Irr(χ T ). One has ϕ = χ H = (χ T )H = θ H . The result of Exercise 18.1 allows us to use in some proofs induction on |G|. If H ≤ G󸀠 and H is a CR-subgroup of G, then H 󸀠 = H. Indeed, if μ ∈ Lin# (H) and χ is an extension of μ to G, then H ≤ G󸀠 ≤ ker(χ) so χ H = 1H ≠ μ, a contradiction. In this argument we have used extendability of all linear characters of H to G only. Isaacs [Isa20, Theorem B] has proved that if N is the normalizer of P ∈ Sylp (G) in a group G and N is a CR-subgroup of G, then N has a normal complement in G. We show that this assertion follows from more general results of pure group theory (see Theorem 18.5) and give a simplified proof of another related Isaacs’ result, Theorem 18.2. Proposition 18.6 is a weak form of a character free version of Theorem 18.2: if a maximal solvable π-subgroup H is a CR-subgroup of G, then H is a π-Hall subgroup of G and G contains a normal π-complement [Isa20]. Definition 18.2. A subgroup H is called an NR-subgroup of a group G if, whenever K ⊴ H, then K G ∩ H = K. Below we state the above mentioned Isaacs’ results. Theorem 18.1 ([Isa20]). Let P ∈ Sylp (G), where p is a prime with p | |G|. If N = NG (P) is a CR-subgroup of G, then N has a normal complement in G.

356 | Characters of Finite Groups 1 In fact, as the proof of Theorem 18.5 shows, a normal complement in Theorem 18.1 is solvable (this follows from CFSG (the classification of finite simple groups); see Lemma 18.3 (c). Theorem 18.2 ([Isa20]). Let π be a set of primes and let H be a maximal solvable π-subgroup of G. If H is a CR-subgroup of G, then H is a π-Hall subgroup of G possessing a normal complement in G. In this section we generalize Theorem 18.1 and present alternate proof of Theorem 18.2. A number of related results are proved as well. Some known results are gathered in the following lemma. Lemma 18.3. The following statements hold: (a) [Kar5, Lemma 27.5.3] Let S be a maximal solvable π-subgroup of G. If L ≤ G is such that S ≤ NG (L) and L ∩ S = {1}, then GCD(|L|, |S|) = 1. (b) (Theorem 10.1, Exercise 10.1) If H ⊴ G, P ∈ Sylp (G), then H ∩ P ≤ Φ(P) implies that H is p-nilpotent. (c) [Cle] Suppose that a p-group P acts on a p󸀠 -group G in such a way that CG (P) = {1}. Then G is solvable. (d) [Kar5, Lemma 27.5.5] Suppose that H is a CR-subgroup of G. If K ⊴ H, then one has K G ∩ H = K and HK G /K G is a CR-subgroup of G/K G . (e) ([Sah] or [Kar5, Theorem 26.2.2]) Let H be a solvable Hall subgroup of G. If H is a CR-subgroup of G, then H has a normal complement in G. Let us prove the first part of Lemma 18.3 (d). Let ϕ1 , . . . , ϕ n be all members of the set Irr(H) containing K in their kernels; then ⋂ni=1 ker(ϕ i ) = K. Let χ i be an extension of ϕ i to G, i = 1, . . . , n, and set D = ⋂ni=1 ker(χ i ). Then D ∩ H = K so K G ≤ D since D ⊲ G. It follows that K ≤ K G ∩ H ≤ D ∩ H = K, and we are done. This result shows that CR-subgroup is an NR-subgroup. Recall that a group is p-closed if its Sylow p-subgroup is normal. Definition 18.3. A subgroup H is an NRp -subgroup of G if it is p-closed and, whenever K is a normal p-subgroup of H, then H ∩ K G = K. Lemma 18.4. Let H < G. (a) Let H be an NR-subgroup of a group G. (1) If H < T < G, then H is an NR-subgroup of T. (2) If K ⊲ H and L/K G ⊴ HK G /K G , then (L/K G )G ∩ (HK G /K G ) = L/K G , i.e., HK G /K G is an NR-subgroup of G/K G . (b) Let H be an NRp -subgroup of G. (1) If H < T < G, then H is an NRp -subgroup of T. (2) Let K be a normal p-subgroup of H. Then HK G /K G is an NRp -subgroup of G/K G . Proof. (a) Let K ⊲ H. Then K ≤ H ∩ K T ≤ H ∩ K G = K 󳨐⇒ H ∩ K T = K,

VII Clifford theory | 357

and this proves (a1). To prove (a2), it suffices to show that (L G /K G ) ∩ (HK G /K G ) = (L ∩ HK G )/K G (= L/K G ).

(1)

Indeed, by the modular law, L = K G (L ∩ H) since K G ≤ L ≤ K G H. Therefore, L G = K G (L ∩ H)G = [K(L ∩ H)]G = (L ∩ H)G , and so, by the modular law, L G ∩ K G H = (L ∩ H)G ∩ HK G = K G [(L ∩ H)G ∩ H] = K G (L ∩ H) = L, since K G ≤ L and L ∩ H ⊴ H. This implies (1). (b) For part (b1), see the proof of part (a1). Let us prove part (b2). Set G = G/K G and let L be a normal p-subgroup of H. It suffices to prove that G

L ∩ H = L. Let P ∈ Sylp (L) be such that K ≤ P. Then L = PK G since K G ≤ L and |L : K G | is a power of p, so that L G = (PK G )G = P G K G = (PK)G = P G 󳨐⇒ L = P. G

G

G

Therefore, L = P . It suffices, therefore, to show that P ∩ H = P, what is equivalent to the equality P G ∩ HK G = PK G (= L). By the modular law, P G ∩ HK G = K G (P G ∩ H) = K G P, completing the proof. The following theorem is a character free version of Theorem 18.1. Theorem 18.5. Suppose that P ∈ Sylp (G), NG (P) = N, P G ∩ N = P, Φ(P)G ∩ N = Φ(P). Then N has a normal solvable complement in G. Proof. By hypothesis, NP G (P) = N ∩ P G = P. Suppose that H is a normal complement to N in G. Then NPH (P) = PH ∩ N = P, and so H is solvable, by Lemma 18.3 (c). Therefore, it suffices to prove that N has a normal complement in G. Let L = P G . By Frattini’s lemma, G = LN. By hypothesis, P ≤ NL (P) = L ∩ N = P G ∩ N = P. Therefore, if P is abelian, L has the normal p-complement R, by Burnside’s normal p-complement theorem, so that L = PR and R ⊲ G since R is characteristic in L ⊲ G. In that case, by Frattini’s lemma, G = NG (P)P G = N(PR) = (NP) ⋅ R = N ⋅ R so that R is a normal complement to N in G since N ∩ R = {1}. In what follows we assume that P is nonabelian. Then Φ(P) > {1}. Set T = Φ(P)G . By hypothesis, T ∩ N = Φ(P) so T ∩ P = Φ(P). In that case, PT/T ∈ Sylp (G/T) is abelian. We claim that NG/T (PT/T) = NT/T (if this holds, then, by the above, NT/T has

358 | Characters of Finite Groups 1 a normal p-complement). It suffices to show that NG (PT) = NT. Set D = NG (PT). By Frattini’s lemma, D = ND (P)PT = ND (P)T since P ∈ Sylp (PT). Since N ≤ D, it follows that N = ND (P). Thus, D = NT, as desired. Now, PT/T ≅ P/(P ∩ T) = P/Φ(P) is abelian and (PT)G = [PΦ(P)G ]G = P G Φ(P)G = [PΦ(P)]G = P G . Therefore (PT/T)G/T ∩ (NT/T) = (P G ∩ NT)/T = PT/T. By the above, G/T = (NT/T) ⋅ (S/T), where S ⊲ G and NT ∩ S = T. Then N ∩ S = (N ∩ NT) ∩ S = N ∩ (NT ∩ S) = N ∩ T = Φ(P), and so P ∩ S = Φ(P). By Lemma 18.3 (b), S has the normal p-complement H. Since H ⊲ G, H ∩ N = {1} and G = N ⋅ H, the proof is complete. Theorem 18.5 implies Theorem 18.1. Supplement to Theorem 18.5. Let P, N, G be as in Theorem 18.1 and Φ(P) ⊲ G. If every irreducible character of N of p󸀠 -degree is the restriction of an appropriate irreducible character of G, then N has a solvable normal complement in G. Proof. All irreducible characters of N/Φ(P) have p󸀠 -degrees, by Ito’s Theorem 2.3. By Theorem 18.1, N/Φ(P) has a solvable normal complement R1 /Φ(P) in G/Φ(P). We have R1 ∩ P = Φ(P) so, by Lemma 18.3 (b), R1 has a normal p-complement R. Obviously, R is a normal complement of N in G. Proof of Theorem 18.2. Assume that the theorem has proved for all groups of order < |G|. Therefore, if H < T < G, then H is a π-Hall subgroup of T = H ⋅ D, where D is a normal π󸀠 -Hall subgroup of T. Let R be a minimal normal subgroup of H. Then R is an elementary abelian p-group, p ∈ π. Set K = R G . By hypothesis, K ∩ H = R and HK/K is a CR-subgroup of G/K (Lemma 18.3 (d)). Suppose that HK < G. Then, by induction, HK = H ⋅ L, where L ∩ H = {1} and L is a normal π󸀠 -Hall subgroup of HK. Since HK/K ≅ H/(H ∩ K) is a π-group, it follows that L ≤ K, and so L ⊲ G since L is characteristic in K ⊲ G. Suppose that HK/K ≤ F/K, where F/K is a maximal solvable π-subgroup of G/K. Obviously, L is a normal π󸀠 -Hall subgroup of F. Therefore, by the theorem of Schur–Zassenhaus, F contains a π-Hall subgroup H1 such that H ≤ H1 . Since H1 is solvable, it follows that H1 = H, by maximality of H, and HK/K = F/K. Thus, H/K is a maximal solvable π-subgroup of G/K. Therefore, by induction, G/K = (HK/K) ⋅ (S/K) is a semidirect product with kernel S/K so S ∩ HK = K. We have HS = G and S ∩ H = S ∩ (HK ∩ H) = (S ∩ HK) ∩ H = K ∩ H = R. If HK = G, then H ∩ K = R, by the choice of K (recall that K = R G ). Thus, in any case, G contains a normal subgroup S such that G = HS and H ∩ S = R.

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(i) Suppose that R ⊲ G (in that case, K = R). Then H/R is a maximal solvable π-subgroup of G/R, and hence, by Lemma 18.3 (a), S/R is a π󸀠 -group. Then H is a π-Hall subgroup of G, and it follows from Lemma 18.3 (e) that G has a normal π-complement. (ii) Let R ⋬ G. Then N = NG (R) < G and H ≤ N. By induction, N = H ⋅ T is a semidirect product with kernel T and T is a π󸀠 -Hall subgroup of N. Therefore, [R, T] = {1}. We have T ≤ S since G/S ≅ H/(H ∩ S) is a π-group. By the modular law, N = H(N ∩ S). Now, H ∩ (N ∩ S) = (H ∩ N) ∩ S = H ∩ S = R, so that |N| =

|H| ⋅ |N ∩ S| , |R|

by the product formula. Thus, |N ∩ S| =

|N| ⋅ |R| = |T| ⋅ |R| 󳨐⇒ R × T = RT = N ∩ S = NS (R). |H|

In particular, R is an abelian Sylow p-subgroup of S. By Burnside’s normal p-complement theorem, S contains the normal p-complement U (in that case, U is characteristic in S ⊲ G). It follows that U ⊲ G, U ∩ H = {1} and G = H ⋅ U. By Lemma 18.3 (a), H is a π-Hall subgroup of G. If an NR-subgroup H is a maximal solvable π-subgroup of G, then H is not necessarily a Hall subgroup of G (for example, G = PSL(2, 5) × C3 , π = {3, 5} and H is cyclic of order 15). But the following weaker result holds: Proposition 18.6. Let H be a maximal solvable subgroup of a group G. If H is an NR-subgroup of G, then H = G. Proof. Suppose that G is a counterexample of minimal order; then we have H < G. By Lemma 18.4 (a1) and induction, H is maximal in G. Since G is not solvable, H > {1}. Let K be a nontrivial normal subgroup of H. Then K G ∩ H = K. If K G = K, then H/K is a maximal solvable subgroup of G/K and H/K is an NR-subgroup of G/K (Lemma 18.4 (a2)). By induction, we have H/K = G/K, contrary to the assumption. Thus, H G = {1}. Let K be a minimal normal subgroup of H; then K is an elementary abelian p-subgroup for some prime p. By the previous paragraph, K G > K and K G ≰ H so that HK G = G (recall that H is maximal in G). It follows that NG (K) = H, and so NK G (K) = H ∩ K G = K. Therefore, K ∈ Sylp (K G ). By Burnside, K G has a normal p-complement R since K is abelian. Then R ⊲ G and G = HK G = HKR = HR, R ∩ H = {1}. By Lemma 18.3 (a), H is a Hall subgroup of G, and we conclude that G is π(H)-solvable. Let q be a prime divisor of |R|, Q ∈ Sylq (R) and σ = π(H) ∪ {q}. Then G = RNG (Q) (Frattini) and NG (Q) is σ-solvable. It follows that there is in NG (Q) a σ-Hall subgroup F; then F is a solvable σ-Hall subgroup of G. Then H is conjugate with some π-Hall

360 | Characters of Finite Groups 1 subgroup of F, and this is a contradiction since |F| > |H|, F is solvable and H is a maximal solvable subgroup of G. Thus, H = G. Definition 18.4. A subgroup H is said to be a Carter subgroup of a group G if H is nilpotent and NG (L) = L for all L such that H ≤ L ≤ G. Here we do not assume that G is solvable; a Carter subgroup need not exist in an arbitrary finite group (for example, A5 has no Carter subgroup). If a Carter subgroup H is a Hall subgroup of G and a CR-subgroup of G, it has a normal complement in G (Theorem 18.2). Proposition 18.7. Let H be a Carter subgroup of a group G. If H is an NR-subgroup of G, then H has a normal complement in G. Proof. Let H = P1 ×⋅ ⋅ ⋅× P t , where P i ∈ Sylp i (H), i = 1, . . . , t. Then one has H = P i × K i , i = 1, . . . , t. By assumption, NG/K G (L/K iG ) = L/K iG i

for all L ≤ G containing of G/K iG . In particular,

HK iG

=

P i K iG .

By definition, HK iG /K iG is a Carter subgroup

HK iG /K iG ∈ Sylp i (G/K iG ).

In that case, by Lemma 18.4 (a2) and Theorem 18.5, G/K iG = (HK iG /K iG ) ⋅ (S i /K iG ) is a semidirect product with kernel S i /K iG , i = 1, . . . , t. Therefore we have H ∩ S i = (H ∩ HK iG ) ∩ S i = H ∩ (HK iG ∩ S i ) = H ∩ K iG = K i ,

i = 1, . . . , t.

In particular, G = HS i = P i K i S i = P i ⋅ S i is a semidirect product with kernel S i for all i. Set S = S1 ∩ ⋅ ⋅ ⋅ ∩ S t . Then H ∩ S = {1} and |G : S| = |G : S1 | ⋅ ⋅ ⋅ |G : S t | = |P1 | ⋅ ⋅ ⋅ |P t | = |H|, and so G = H ⋅ S is a semidirect product with kernel S. Proposition 18.8. If all Sylow subgroups of G are NR-subgroups, then G is supersolvable. Proof. Let p be the least prime divisor of |G|, P ∈ Sylp (G). We prove by induction on |G| that G is p-nilpotent. Let R ≤ Z(P) be of order p. Then P ∩ RG = R and PR G /R G (∈ Sylp (G/R G )) is an NR-subgroup of G/R G (Lemma 18.4 (a2)). By induction, G/R G = (PR G /R G ) ⋅ (S/R G ) is a semidirect product with kernel S/R G . Then G = PS and P ∩ S = (P ∩ PR G ) ∩ S = P ∩ (PR G ∩ S) = P ∩ R G = R, and so R ∈ Sylp (S). Since p = |R| is the least prime divisor of |S|, it follows that S has the normal p-complement H; then G = P ⋅ H, the semidirect product with kernel H.

VII Clifford theory | 361

It follows from the previous paragraph, by induction, that G has an ordered Sylow tower such as a supersolvable group. Let q be the greatest prime divisor of G, Q ∈ Sylq (G); then Q ⊲ G. If Q0 ⊴ Q, then G Q0 = Q0 since Q0G ≤ Q. This implies the q-supersolvability of G. Let H be a q󸀠 -Hall subgroup of G. Since all Sylow subgroups of H are NR-subgroups, it follows that H is supersolvable, by induction. Since G/Q ≅ H, G is supersolvable. If all Sylow subgroups of G are cyclic, it satisfies the assumption of Proposition 18.8 but G is not necessarily nilpotent. By Lemma 18.3 (e), G is nilpotent if all its Sylow subgroups are CR-subgroups. On minimal nonnilpotent groups see [BZ3, §11.2]. Definition. A group G is said to be a B-group if G is a nonnilpotent q-closed group for some prime q and |G/Φ(G)| = pq b , where p ≠ q is a prime number and b is the least natural number such that p | q b − 1 (in that case, G/Φ(G) is minimal nonabelian). By [BZ3, Lemma 11.2.2], minimal nonnilpotent groups are B-groups. Proposition 18.9. Let q be a prime divisor of |G|. If every B-subgroup H ≤ G is such that, whenever q | |H| and Q = H 󸀠 ∈ Sylq (H) implies Q G ∩ H = Q, then G is q-solvable. Proof. Suppose that G is a counterexample of minimal order. Then all proper subgroups of G are q-solvable. It follows easily that G/Φ(G) is nonabelian simple of order divisible by q. By Frobenius’ normal p-complement theorem (see [Hup3, Satz IV.5.8]), G/Φ(G) contains a minimal nonnilpotent subgroup F/Φ(G) such that q | |F/Φ(G)| and a Sylow q-subgroup is normal in F/Φ(G). Let H be a minimal subgroup such that F = HΦ(G). Then H∩Φ(G) ≤ Φ(H), by Lemma XIV.1.1 (b), and F/Φ(G) ≅ H/(Φ(G)∩H) (in particular, H is not nilpotent). By Schur–Zassenhaus, one has π(H) = π(F) and the Sylow q-subgroup Q is normal in H. It follows that H is a B-group. By assumption, H ∩ Q G = Q. Since G/Φ(G) is nonabelian simple and Q G Φ(G)/Φ(G) is a q-group, we get Q ≤ Φ(G), which is impossible since H/H ∩ Φ(G) ≅ HΦ(G)/Φ(G) = F/Φ(G) is not nilpotent.

Problems Problem 1. Let H be a Hall subgroup of G and let N = NG (H) be a CR-subgroup of G. Is it true that N has a normal complement in G? Problem 2. Is it true that if all maximal subgroups are NR-subgroups in G, then G is solvable. Problem 3. Let G󸀠 < G. Study the structure of G if G󸀠 is a CR-subgroup of G. Problem 4. Suppose that H is the solvable residual of G. Study the structure of G if H is a CR-subgroup of G.

362 | Characters of Finite Groups 1

A Squares of characters In this appendix written by the third author, we continue considerations of Theorem IV.8.7. Let G be a group and χ ∈ Irr(G). Recall that χ(2) (x) = χ(x2 ). Then, as we know, Irr(χ(2) ) ⊆ Irr(χ2 ).

(S1)

If Irr(χ2 ) = {ψ1 , . . . , ψ n ) and n

χ2 = ∑ a j ψ j

(a j ∈ ℕ),

(S2)

(b j ∈ ℤ).

(S3)

j=1

then, by (S1), n

χ(2) = ∑ b j ψ j j=1

Write n

n

a = ∑ aj ,

b = ∑ bj .

j−1

j=1

(S4)

Let ϵ be a |G|-th primitive root of 1 and G = Gal(ℚ(ϵ)/ℚ). The group G acts on the set Irr(G) (see Chapter III). Characters ψ, ψ󸀠 ∈ Irr(G) are said to be algebraic conjugate if ψ󸀠 = σψ for some σ ∈ G. If ψ, ψ󸀠 are algebraic conjugate, then ψ(1) = ψ󸀠 (1), ker(ψ) = ker(ψ󸀠 ), Z(ψ) = Z(ψ󸀠 ). In what follows we use the notation introduced above. As usual, Inv(G) is the set of involutions of G. Theorem A.1 ([IZ]). Suppose that N ⊲ G and ϕ ∈ Irrinv (N) (i.e., we have IG (ϕ) = G). If |Irr(ϕ G )| = 1, then π(|G : N|) ⊆ π(N). Proof. Let p ∈ π(|G : N|). By assumption, ϕ G = eχ for some e ∈ ℕ and χ ∈ Irr(G); then, by Clifford’s theorem, χ N = eϕ, and so |G : N|ϕ(1) = ϕ G (1) = eχ(1) = e2 ϕ(1). It follows that e2 = |G : N| so that π(|G : N|) = π(e) ⊆ π(eϕ(1)) = π(χ(1)) 󳨐⇒ p | χ(1). Assume that p ∈ π(|G : N|) does not divide |N|. If P/N ∈ Sylp (G/N), then N is a p󸀠 -Hall subgroup of P. By Corollary 4.8, ϕ can be extended to a character ϕ̃ of P. ̃ Note that ϕ(1) = ϕ(1) and ϕ̃ G is a constituent of ϕ G . Indeed, since ̃ = ⟨ϕ, (ϕ) ̃ N ⟩ = ⟨ϕ, ϕ⟩ = 1, ⟨ϕ P , ϕ⟩

VII Clifford theory | 363

̃ we get ϕ P = ϕ+ϑ, where ϑ is a character of P or zero function. Therefore, ϕ G = ϕ̃ G +ϑ G . G Since ϕ = eχ, it follows that ϕ̃ G = mχ for some m ∈ ℕ, m ≤ e. Equating degrees, we see that |G : P|ϕ(1) = mχ(1). We conclude that p ∤ χ(1), contrary to the result of the previous paragraph. Now we are ready to prove the following: Theorem A.2 ([IZ]). Suppose that χ is a faithful irreducible character of a nonabelian group G such that χ2 = aχ + bχ

(a, b ∈ ℕ ∪ {0}).

(1)

Then the following statements hold: (a) One has χχ = 1G + aχ + aχ + ψ, where ψ is a character of degree b2 − a2 − 1. Write N = ker(ψ). (b) The elements of N are precisely those elements of G on which χ does not vanish. In particular, N > {1}. (c) If γ ∈ Irr# (N), then γ is a constituent of χ N or χ N , and |Irr(γ G )| = 1. (d) N is the unique minimal normal subgroup of G. It is an elementary abelian p-group for some prime p. (e) |N| = (3b + a)/(b − a) and |G| = (3b + a)(a + b). (f) Let T = IG (λ) for some λ ∈ Lin# (N). Then |T/N| = (b − a)2 is a power of p (see (d)). Proof. Parts (a) and (b) have been proved in Theorem IV.8.6 (see in the proof of that theorem formula (8.4)) and Theorem IV.8.7 and the text between them. (c) Define a function ϕ : G → ℂ by ϕ = χ + χ + (b − a)1G , and observe that this is a character of G since b > a (see Theorem IV.8.6). If g ∈ N # , then by Theorem IV.8.7 and Theorem IV.8.6 (b), we have χ(g) + χ(g) = a − b, and thus ϕ(g) = 0. It follows that ϕ N is a multiple of ρ N , the regular character of N, and, in particular, any nonprincipal irreducible character of N is a constituent of ϕ N , and hence of χ N or χ N . Therefore, one may assume that γ ∈ Irr(χ N ). Let Γ denote the sum of all G-conjugates of γ. Then γ G vanishes on G − N and agrees with a multiple of Γ on N. We also know that χ vanishes of G − N and agrees with a multiple of Γ on N, and from this we deduce that γ G is a multiple of χ. (d) Assume that M is a normal subgroup of G not containing N. Then H = MN > M. Choose γ ∈ Irr# (H) with M ≤ ker(γ); then N ≰ ker(γ). Let λ ∈ Irr# (γ N ); then λ ≠ 1N . It follows from (c) that λ ∈ Irr(χ N ) ∪ Irr(χ N ). If λ ∈ Irr(χ N ), then, by (c), λ G is a multiple of χ. Since ⟨γ, λ H ⟩ = ⟨γ N , λ⟩ > 0, we get λ H = γ + ϑ, where ϑ is either character of H or the zero function. Hence, λ G = γ G + ϑ G so γ G is a multiple of χ; then ⟨γ, χ H ⟩ = ⟨γ G , χ⟩ > 0. Therefore, χ H = γ + θ, where either θ is a character of H or the zero function. Hence, χ M = γ M + θ M . Since γ M is a multiple of 1M , we have ⟨χ M , 1M ⟩ > 0 so that M ≤ ker(χ), by Clifford’s theorem. As χ is faithful, we get M = {1}, a contradiction. If λ ∈ Irr(χ N ), we get a contradiction as well. It follows that N is the unique minimal normal subgroup of G. Next, N is solvable since |N| is odd (Theorem IV.8.5). It follows that N is an elementary abelian p-group for some odd prime p.

364 | Characters of Finite Groups 1 (e) Consider the character χ + χ and recall that it vanishes on G − N and has the constant value a − b on N # . Since (χ + χ)(1) = 2χ(1) = 2(a + b) (Theorem IV.8.6), we compute that 0 = ⟨χ + χ, 1G ⟩|G| = 2(a + b) + (|N| − 1)(a − b). It follows that |N| = Similarly,

2(a + b) 3b + a +1= . b−a b−a

2|G| = ⟨χ + χ, χ + χ⟩|G| = 4(a + b)2 + (|N| − 1)(a − b)2 2(a + b) ⋅ (a − b)2 b−a = 2(3b + a)(a + b),

= 4(a + b)2 +

which implies |G| = (3b + a)(a + b). (f) By (c) we know that Irr(χ N ) ∪ Irr(χ N ) = Lin# (N), and |N| − 1 elements of the set # Lin (N) are partitioned in exactly two orbits under the action of G, with each orbit containing exactly 12 (|N| − 1) characters (recall that N is an elementary abelian p-subgroup). Therefore, if λ ∈ Lin# (N) and T = IG (λ), then |G : T| = 21 (|N| − 1) = a+b b−a , by (e). Since |G : N| = (a + b)(b − a), by (e), we deduce that |T : N| = (b − a)2 . The last assertion follows from Theorem A.1. We know that

3b + a = p m , |T : N| = (b − a)2 = p2n , b−a where m, n are nonnegative integers. Solving these equations, we get |N| =

a = pn

pm − 3 , 4

b = pn

pm + 1 . 4

(2)

Since these numbers are integers and p > 2 (see Theorem IV.8.6), we deduce that p ≡ 3 (mod 4), and so m is odd. From |G : T| = 21 (p m − 1), it follows that T ∈ Sylp (G) and |G : T| = a+b b−a . Since χ(1) = p n

pn − 1 = p n |G : T| > |G : T|, 2

the subgroup T is nonabelian. It is interesting to construct all the groups satisfying the hypothesis of Theorem A.2. Theorem A.3 ([IZ]). Let P be a p-group of odd order with center Z = Z(P) and assume that, for each μ ∈ Lin# (Z), we have μ P = eτ with τ ∈ Irr(P) (then |P : Z| = e2 , by Clifford). Suppose that H is an odd order group of automorphisms of P and that its action on Z is Frobenius and decomposes Z # into exactly two orbits. Let G = H ⋅ P be the semidirect product. Then G has exactly two faithful irreducible characters χ and χ, and χ2 = aχ+bχ, where a and b are determined as in (2), taking p m = |Z| and p2n = |P : Z|.

VII Clifford theory | 365

Proof. (i) Note that Φ = H ⋅ Z is a Frobenius group with the kernel Z and a complement H. By hypothesis, the set Z # is the union of two H-orbits. If z ∈ Z # , then these orbits are the sets z H = {z h | h ∈ H} and (z−1 )H . These orbits are different since |H| is odd. Thus, Z # = z H ∪ (z−1 )H is a partition. Let F be a minimal normal subgroup of the group G = H ⋅ P. We claim that F = Z. Assume that F ≰ Z; then F ∩ Z = {1} so that FZ = F × Z. Now, F ∩ H = {1}, since H ⋅ Z is a Frobenius group. It follows that F ≤ P. Then F ∩ Z > {1} (recall that Z = Z(P)) so F ≤ Z and then F = Z since one of the two orbits z H , (z−1 )H , so both, are contained in F, and our claim is proved. Thus, G is a monolith. Clearly, Z is elementary abelian. Set |Z| = p m . (ii) Let μ ∈ Lin# (Z). By hypothesis, μ P = eτ,

τ ∈ Irr(P), e ∈ ℕ.

(3)

Let us show that τ G ∈ Irr(G). To this end it suffices to show that (see Theorem 2.1) IG (τ) = P.

(4)

It follows that τZ =

|P : Z| μ. e

(5)

Indeed, if z ∈ Z, then by (3), in view of Z = Z(P), one obtains τ(z) =

1 |P : Z| 1 P 1 μ (z) = μ(z), ∑ μ(z t ) = ∑ μ(z) = e e|Z| t∈P e|Z| t∈P e

and (5) follows. Now let g ∈ IG (τ); then g = uh, where u ∈ P, h ∈ H. Since τ u = τ, it follows that τ = τ g = τ uh = τ h . By (5), this implies μ h = μ, i.e., h ∈ IΦ (μ). It follows that h = 1 since Φ = H ⋅ Z is a Frobenius group with kernel Z so that IΦ (μ) = Z (see Chapter X). Thus, τ G ∈ Irr(G). (iii) If μ ∈ Lin# (Z) and h ∈ H, z ∈ Z, then the equality μ h (z) = μ(hzh−1 ) defines an action of H on the set Lin# (Z). By Brauer’s permutation lemma, Lin# (Z) is the join of two H-orbits since the set Z # is. If μ1 , μ2 ∈ Lin# (Z) are in the same H-orbit, then μ1G = μ2G since μ1 , μ2 are H-conjugate so G-conjugate. If μ ∈ Lin# (Z), then, by (3), μ G = eχ, where χ = τ G ∈ Irr(G) (see (ii)). Note that μ and μ lie in different H-orbits (otherwise, μ G = μ G , i.e., χ = χ, which is impossible since |G| is odd (recall that χ ≠ 1G ). It follows that Lin# (Z)G = {λ G | λ ∈ Lin# (Z)} = {eχ, eχ}.

(6)

(iv) The characters χ and χ (see (iii)) are faithful. Indeed, if this is false, then Z ≤ ker(χ) (by (i), Z is the unique minimal normal subgroup of G). Since χ = e−1 μ G vanishes on G − Z, we get ⟨χ, 1G ⟩ > 0, which is not the case since χ ≠ 1G . It follows from (6) that a faithful irreducible character of G equals either χ or χ.

366 | Characters of Finite Groups 1 (v) Let us show that χ2 = aχ + bχ,

where a, b ∈ ℕ ∪ {0}.

(7)

Irr(χ2 )

By (iv), it suffices to show that every θ ∈ is faithful. Assume that this is false. Then Z ≤ ker(θ) implies ⟨θ Z , 1Z ⟩ > 0. In that case, ⟨χ2Z , 1Z ⟩ > 0. Since χ2 vanishes on G − Z, we get 0 < ⟨χ2 , 1G ⟩ = ⟨χ, χ⟩, a contradiction since χ ≠ χ. Thus, θ is faithful. By (iv), we get Irr(χ2 ) ⊆ {χ, χ}, proving (7). (vi) In the notation of Theorem A.2, we have Z = N. If λ ∈ Lin# (Z), then by the same p m −1 Indeed, theorem, IG (λ) ∈ Sylp (G) (in fact, by (i), |H| = |Z|−1 2 = 2 ). Also P ∈ Sylp (G). p m −1 |Z|−1 Φ = H ⋅ Z is a Frobenius group so p ∤ |H| (in fact, by (i), |H| = 2 = 2 ). Now, G = H ⋅ P implies P ∈ Sylp (G). Therefore, |P : Z| = p2n . Now the expressions for a and b follow from (2). By assumption, there exist exactly |Z| − 1 characters in Irr(P | Z) = Irr(P) − Irr(P/Z), and these are permuted into just two orbits under the action of H. If τ ∈ Irr(P | Z), then Irr(τ Z ) = {μ}, and μ is moved by all nonidentity elements of H since H ⋅ Z is a Frobenius group. It follows that P is the inertia group of τ in G, and thus τ G is irreducible. Theorem A.4 ([IZ]). Let G > {1} be a group and suppose that there is a faithful χ ∈ Irr(G) such that χ2 = aψ + bψ,

(8)

where a, b ∈ ℕ and ψ ∈ Irr(G). Then G is a direct product of a cyclic 2-group of order not exceeding 4 and a group of odd order. The proof, which is due to Zhmud, follows after the proof of Theorem A.5. First we will to prove the Zhmud theorem. Let χ ∈ Irr(G). Recall that χ(2) (g) = χ(g 2 ) (g ∈ G). Then Irr(χ(2) ) ⊆ Irr(χ2 ).

(9)

Indeed, χ(2) = χ2 − 2 ⋀2 χ (see (6.8) in §IV.6) so it suffices to show Irr(⋀2 χ) ⊆ Irr(χ2 ). Next, χ2 = ⋀2 χ + θ, where θ is the exterior square of χ (see Lemma IV.6.4) so we have Irr(⋀2 χ) ⊆ Irr(χ2 ), as desired. Therefore, if Irr(χ2 ) = {ψ1 , . . . , ψ n } and n

χ2 = ∑ a j ψ j ,

(10)

j=1

where all a j ∈ ℕ, then, by (9), n

χ(2) = ∑ b j ψ j ,

(11)

j=1

where all b j ∈ ℤ. Set a = a1 + ⋅ ⋅ ⋅ + a n ,

b = b1 + ⋅ ⋅ ⋅ + b n .

(12)

Let ϵ be a |G|-th primitive root of 1, G = Gal(ℚ(ϵ)/ℚ). Then the group G acts in the natural way on the set Irr(G) (see Chapter III). Characters ψ, ψ󸀠 ∈ Irr(G) are said to be

VII Clifford theory | 367

algebraically conjugate if ψ󸀠 = σψ for some σ ∈ G. In that case, ψ(1) = ψ󸀠 (1),

ker(ψ) = ker(ψ󸀠 ),

Z(ψ) = Z(ψ󸀠 ).

In what follows we will retain this notation. Definition. A group G possesses a property A if it satisfies the following conditions: (A1) |G| is even. (A2) χ ∈ Irr(G) is faithful. (A3) χ2 = ∑nj=1 a j ψ j , n ≥ 2, all a j ∈ ℕ and ψ j ∈ Irr(G) (j = 1, . . . , n) are algebraically conjugate with ψ = ψ1 , i.e., ψ j = σ j ψ for some σ j ∈ G (j = 1, . . . , n). It follows from (A3) that ψ j (1) = ψ(1) and Z(ψ j ) = Z(ψ) for all j ∈ {1, . . . , n}. Theorem A.5 (Zhmud). If the group G satisfies the condition A, then the following statements hold: (a) G is nonabelian and χ(1) > 1. (b) G has only one involution u. (c) ker(ψ) = ⟨u⟩. (d) Z(ψ) is abelian. (e) Sylow 2-subgroups of G are cyclic. (f) G = P ⋅ N, a semidirect product, where P ∈ Syl2 (G) and {1} < N ⊲ G. (g) χ(1) = ψ(1) and b = 1. (h) χ N ∈ Irr(N). (i) All (ψ j )N are irreducible and nonreal for j = 1, . . . , n. (j) If w ∈ G with o(w) = 8, then ψ(w) = m√−1, where m ∈ ℤ, m | χ(1). Proof. (a) As ψ j is algebraically conjugate with ψ1 = ψ, it follows that ψ j (1) = ψ(1) for j = 1, . . . , n. It follows from (10) and (12) that χ(1)2 = aψ(1) so χ(1) > 1 since a ≥ n > 1. Since χ is faithful, the group G is nonabelian. (b) Let u ∈ G be an involution (see (A1)). Since ψ(u) ∈ ℤ, we have ψ j (u) = ψ(u) since ψ j is algebraically conjugate with ψ. Taking into account (12), one obtains χ(2) (u) = bψ(u). Since χ(2) (u) = χ(u2 ) = χ(1), we get χ(1) = bψ(u).

(13)

∑nj=1

b j ψ j (1) = bψ(1) so that ψ(u) = ψ(1), by (13), On the other hand, χ(1) = = i.e., u ∈ ker(ψ). By (10), in view of ker(ψ j ) = ker(ψ), one obtains χ(2) (1)

n

ker(χ2 ) = ⋂ ker(ψ j ) = ker(ψ) 󳨐⇒ u ∈ ker(χ2 ) 󳨐⇒ χ(u)2 = χ(1)2 . j=1

Hence, χ(u) = ±χ(1), and we conclude that χ(u) = −χ(1)

(14)

since the character χ is faithful and u ≠ 1. By formula (14), we have u ∈ Z(χ) = Z(G), i.e., |Inv(G)| = |Inv(Z(G)|. Since Z(G) is cyclic (χ is faithful), it follows that |Inv(G)| = 1, i.e., Inv(G) = {u}, and (b) is proved.

368 | Characters of Finite Groups 1 (c) By the sentence after (13), u ∈ ker(ψ) so that |ker(ψ)| ≥ 2. Let x ∈ ker(ψ)# . Then, for all j (see (10)), n

ψ j (x) = ψ(1) 󳨐⇒ χ(x)2 = ∑ a j ψ j (x) = aψ(1) = χ(1)2 󳨐⇒ χ(x) = −χ(1) j=1

since χ is faithful. In particular, x ∈ Z(χ) so that χ(x2 ) = χ(1) 󳨐⇒ x2 ∈ ker(χ) = {1}. It follows from this and (b) that x = u. Thus ker(ψ) = {1, u} = ⟨u⟩, and (c) is proved. (d) By (c), Z(ψ) is abelian since Z(ψ)/ker(ψ) is cyclic, in view of irreducibility of ψ and |ker(ψ)| = 2, by (c). (e) Let P ∈ Syl2 (G). By (b), the subgroup P is either cyclic or generalized quaternion. Assume that P is generalized quaternion. Take in P an element v of order 4. Then, by (b), v2 = u.

(15)

Let j ∈ {1, . . . , n}. Then, by (A3), ψ j = σ j ψ so that ψ j (v) = σ j ψ(v). We have σ j ϵ = ϵ ν j with ν j ∈ ℕ and GCD(ν j , |G|) = 1 (here ϵ is a primitive |G|-th root of 1); then ν j is odd. Setting ν j = 2λ j + 1 and taking into account that ψ(v) is a sum of powers of ϵ, one obtains σ j ψ(v) = ψ(v ν j ) = ψ(v2λ j +1 ) = ψ(u λ j v) = ψ(v) since u ∈ ker(ψ), by (c). Thus ψ j (v) = ψ(v)

(j = 1, . . . , n).

(16)

Then, by (14) and (16), n

−χ(1) = χ(u) = χ(v2 ) = χ(2) (v) = ∑ b j ψ j (v) = bψ(v). j=1

On the other hand, in view of the above displayed line, χ(1) = χ(2) (1) = bψ(1) 󳨐⇒ ψ(v) = −ψ(1). It follows from this and (16), that ψ j (v) = −ψ(1)

(j = 1, . . . , n).

(17)

Therefore, χ(v)2 = ∑nj=1 a j ψ j (v) = −aψ(1) = −χ(1)2 , and so χ(v) = ciχ(1),

(18)

where c = ±1 and i = √−1. It follows from |χ(v)| = χ(1) that v ∈ Z(χ) = Z(G), a contradiction, since P, a generalized quaternion group, has no central element of order 4. Thus, the subgroup P ∈ Syl2 (G) is cyclic. (f) By (e) and Burnside’s normal complement theorem, G has a normal complement N; note that N > {1} since G is nonabelian.

VII Clifford theory | 369

(g) Since G/N ≅ P is cyclic, it follows that χ is not ramified over N, by Exercise 3.3, and so the following Clifford decomposition holds (see Exercise 3.3): l

χN = ∑ ϕk ,

(19)

k=1

where {ϕ1 , . . . , ϕ l } is the G-orbit of ϕ = ϕ1 ∈ Irr(N). It follows from (19) that l

(2)

(χ(2) )N = ∑ ϕ k .

(20)

k=1 (2)

Since |N| is odd, it follows that ϕ k are pairwise distinct irreducible characters of N, (2) (2) and we conclude that (χ(2) )N ∈ Char(N).⁴ Thus, {ϕ1 , . . . , ϕ l } is the G-orbit of the (2) n (2) (2) character ϕ = ϕ1 . By (11), (χ )N = ∑j=1 b j (ψ j )N . Since for ϕ1 = ϕ, n

1 = ⟨(χ(2) )N , ϕ(2) ⟩ = ∑ b j ⟨(ψ j )N , ϕ(2) ⟩, j=1

we get ⟨(ψ s )N , ϕ(2) ⟩ ≠ 0 for some s ∈ {1, . . . , n}. This means that ϕ(2) ∈ Irr((ψ s )N ). By Clifford’s theorem, the set Irr((ψ s )N ) is a G-orbit of the character ϕ(2) , i.e., Irr((ψ s )N ) is (2) (2) equal to {ϕ1 , . . . , ϕ l }, where ϕ1 = ϕ. Since ψ s is unramified over N (Exercise 3.3), (2) we conclude that (ψ s )N = ∑lk=1 ϕ k so, by (20), (χ(2) )N = (ψ s )N . (2)

(∗) (2)

Since |N| is odd, we have (χ(2) )N , ϕ k ∈ Char(N) (k = 1, . . . , l) and the characters ϕ k are pairwise distinct. Next, by (20), χ(1) = χ(2) (1) = bψ(1) implies b = 1. (h) It follows from (10) and (11) that χ2 − χ(2) = ∑nj=1 (a j − b j )ψ j . Since (see §IV.6) 2 1 2 (2) 2 (χ − χ ) = ⋀ χ ∈ Char(G), we get a j ≡ b j (mod 2) for all j. Summing up over all j this congruence, one obtains a ≡ b (mod 2) so a is odd since b = 1, by (g). As we have noted in the proof of (a), χ(1)2 = aψ(1). Therefore, since χ(1) = ψ(1), by (g), we get χ(1) = a, and hence χ(1) is odd. It follows from (19) that l = |Irr(χ N )| = |G : IG (ϕ)| and from χ(1) = lϕ(1) that l is odd. Since IG (ϕ) ≥ N and |G : N| = |P| is a power of 2, we get l = 1. Therefore, χ N = ϕ ∈ Irr(N), proving (h). (i) Since l = 1, there exists s ∈ {1, . . . , n} such that (ψ s )N = ϕ(2) ∈ Irr(N) (see the displayed formula preceding (∗)). Since all ψ j are algebraically conjugate, we get (ψ j )N ∈ Irr(N) for all j. If (ψ j )N is real for some j, it is the principal character 1N of N since |N| is odd. It follows that ψ(1) = ψ j (1) = 1. Then, by (g), χ(1) = 1, a contradiction. Thus, all (ψ j )N are not real. (2)

4 Indeed, assume that ψ i

(2)

= ψ j . Then, for x ∈ N, we have ψ i (x2 ) = ψ i (x) = ψ j (x) = ψ j (x2 ) (2)

(2)

and so ψ i = ψ j since {x2 | x ∈ N} = N. This proves the first assertion. Now the second assertion is obvious.

370 | Characters of Finite Groups 1 (j) Let w ∈ G be of order 8. Set v = w2 ; then o(v) = 4. By (b), v2 = u. It follows from (18) that ciχ(1) = χ(v) = χ(w2 ) = χ(2) (w) so, in view of (17), n

(21)

∑ b j ψ j (w) = ciχ(1) j=1

(here i = √−1). Recall that ψ j (w) = ψ(w ν j ), where ν j ∈ ℤ is odd (see the proof of (e)). It follows from o(w) = 8 that ν j ∈ {1, 3, 5, 7}. As u ∈ ker(ψ) and ψ(v) = −ψ(1) (see (17)), we have ψ(w3 ) = ψ(vw) = −ψ(w), ψ(w5 ) = ψ(uw) = ψ(w), ψ(w7 ) = ψ(uvw) = −ψ(w). Thus, ψ j (w) ∈ {ψ(w), −ψ(w)}. Setting ψ j (w) = c j ψ(w), where c j = ±1, rewrite (21) in the form n

ciχ(1) = dψ(w),

where d = ∑ b j c j ∈ ℤ.

(22)

j=1

Note that

ψ(w) = ψ(w−1 ) = ψ(w7 ) = −ψ(w).

Therefore, we have ψ(w) = im, where m ∈ ℝ. Now, (22) yields χ(1) = cdm. It follows that m = c ⋅ χ(1) d ∈ ℚ. Since m = −iψ(w) is an algebraic integer, we get m ∈ ℤ hence m | χ(1). This completes the proof. Proof of Theorem A.4 (Zhmud). We use the same notation as in the statement. Since ψ1 = ψ and ψ2 = ψ1 are algebraically conjugate, one can apply Theorem A.5 to our group G. It remains to prove that |P| ≤ 4 and P ⊲ G. We claim that |Irr(χ(2) )| = 1 (recall that χ(2) : x 󳨃→ χ(x2 )). Otherwise, χ(2) = b1 ψ1 + b2 ψ2 ,

where b1 , b2 ∈ ℤ − {0}.

It follows that (χ(2) )N = b1 (ψ1 )N + b2 (ψ2 )N (recall that (χ(2) )N , (ψ1 )N , (ψ2 )N are irreducible, by Theorem A.5 (h)–(i)). We have (χ(2) )N = (ψ s )N for some s ∈ {1, 2} (see (∗)). Using this and the equality b1 + b2 = 1, we get, by the above, that (ψ1 )N = (ψ2 )N . It follows that (ψ1 )N = (ψ1 )N , i.e., the character (ψ1 )N is real, contrary to Theorem A.5 (i). Thus, one of the numbers b1 , b2 equals 0 and our claim is proved, i.e., χ(2) ∈ {ψ1 , ψ2 }. Assume, for definiteness, that χ(2) = ψ1 . Thus, χ2 = a1 ψ1 + a2 ψ2 and χ(2) = ψ1 . As above, set ψ = ψ1 . Assume that |P| > 4. Take in the cyclic P an element w of order 8. Set v = w2 . Then, by (18) (recall that χ(2) = ψ), ψ(w) = χ(2) (w) = χ(v) = ciχ(1),

where c = ±1.

(23)

It follows from χ(1) = ψ(1) that ψ(w) = ciψ(1) = ±iψ(1) 󳨐⇒ w ∈ Z(ψ).

(24)

VII Clifford theory | 371

Let H ∈ Syl2 (Z(ψ)); then we have H ⊲ G, by Theorem A.5 (d), so HN = H × N and w centralizes N, the normal 2-complement of G (Theorem A.5 (d)). Since w is an element in Z(ψ), it follows that w ∈ H, and we conclude that w centralizes N. Since w centralizes P ∈ Syl2 (G), which is a consequence of Theorem A.5 (e), we see that w ∈ Z(G) since G = P ⋅ N, and so |χ(w)| = χ(1). Then, in view of (24) and (23), ψ2 (w) = ψ1 (w) = −ψ1 (w) = −ψ(w) and |ψ(w)| = χ(1), by (23), we have χ(1)2 = |χ(w)|2 = |χ2 (w)| = |a1 ψ(w) + a2 ψ2 (w)| = |(a1 − a2 )ψ(w)| = |a1 − a2 ||ψ(w)| = |a1 − a2 |χ(1). Hence χ(1) = |a1 − a2 |. On the other hand, by (17), we have χ(1) = a = a1 + a2 . Thus a1 + a2 = |a1 − a2 |, a contradiction since a1 , a2 > 0. Therefore, |P| ≤ 4. Let us prove that P ⊲ G. In view of Theorem A.5 (b), G contains only one involution, so one may assume that |P| = 4 and P = ⟨v⟩. Then, by (17), ψ(v) = −ψ(1) 󳨐⇒ v ∈ Z(ψ) 󳨐⇒ P ≤ Z(ψ). Since P is characteristic in Z(ψ) (indeed, P is a Sylow 2-subgroup of the abelian group Z(ψ) ⊲ G; see Theorem A.5 (d)), it follows that P ⊲ G, and hence G = P × N, completing the proof.

VIII Brauer’s induction theorems In this chapter we prove the fundamental Brauer theorems, which make it possible to solve such important problems of representation theory as the characterization of group characters and Maschke’s problem of the realization of representations in cyclotomic fields. We also present several applications of these theorems.

1 Characterization of characters Recall that a function τ is said to be a generalized character provided τ = χ󸀠 − χ󸀠󸀠 , where χ󸀠 , χ󸀠󸀠 are ordinary characters or zero functions. Let Ch(G) be the ring of generalized characters of a group G (formally speaking, it is possible to denote this ring by ℤ[Irr(G)]). By the definition, the zero function G → ℂ is a member of Ch(G). Since the values of generalized characters of a group are algebraic integers, Ch(G) is a proper subring of the ring CF[G] of central functions on G. However not every element of CF[G] whose values are algebraic integers belongs to Ch(G). Definition. A group H is said to be a Brauer elementary group with respect to a prime p (or, more simply, p-elementary group) if H = P × C, where C is cyclic and P is a p-group. Let Ep = Ep (G) denote the set of all p-elementary subgroups of a group G. A group H is said to be (Brauer) elementary if it is p-elementary for some prime p. Let E = E(G) denote the set of all (Brauer) elementary subgroups of a group G. Obviously, the property of being p-elementary is inherited by subgroups and epimorphic images. A nonprimary group H is elementary with respect to all primes p if and only if it is cyclic. A group H is elementary if and only if H is nilpotent and at most one of its Sylow subgroups is noncyclic. Let J(G) denote the set of generalized characters χ ∈ Ch(G) which are ℤ-linear combinations of characters φ G (φ ∈ Irr(H), H ∈ E = E(G)). Then J(G) ⊆ Ch(G). The following important theorem of Brauer asserts that the converse inclusion is also true. Theorem 1.1 (Brauer’s induction theorem). One has CH(G) = J(G) for every group G. In other words, every generalized character of a group G is a ℤ-linear combination of characters induced from its elementary subgroups. Since elementary groups are nilpotent, they are M-groups (see Corollary VII.13.3); therefore, since induction of class functions is transitive, we may confine our attention in the definition of J(G) to linear characters. Thus, in fact, Theorem 1.1 asserts that every generalized character of a group G is a ℤ-linear combination of characters of the form λ G , where λ are linear characters of elementary subgroups of G. DOI 10.1515/9783110224078-008

374 | Characters of Finite Groups 1 Theorem 1.2 (Characterization of characters). A function θ ∈ CF[G] is a generalized character of a group G (i.e., belongs to Ch(G)) if and only if the restriction θ H belongs to Ch(H) for all H ∈ E. Let us show how to deduce Theorem 1.2 from Theorem 1.1. Write B(G) = {θ ∈ CF[G] | θ H ∈ Ch(H) for all H ∈ E}. It is clear that B(G) is a ring and Ch(G) is a subring of B(G). Let us prove, using Theorem 1.1, the converse inclusion. Since the principal character 1G ∈ Ch(G), it suffices to prove that Ch(G) is an ideal in the ring B(G). Let φ ∈ Ch(G) and θ ∈ B(G). We have to show that φθ ∈ Ch(G). Write M = ⋃H∈E Irr(H). By Theorem 1.1, one has φ = ∑λ∈M c φ,λ λ G , (c φ,λ ∈ ℤ). Then (see Lemma V.1.3) φθ = ∑ c φ,λ (λ G θ) = ∑ c φ,λ (λθ H )G λ∈M

λ∈M

(here λθ H ∈ Ch(H), since θ H ∈ Ch(H), by definition of B(G), and Ch(H) is a ring). This equality shows that φθ ∈ Ch(G), so that Ch(G) is an ideal in B(G) and, since 1G ∈ Ch(G), we conclude that Ch(G) = B(G), and Theorem 1.2 is proved. It remains to prove Theorem 1.1. Let R be a ring such that ℤ ⊆ R ⊆ ℂ. If we replace ℤ by R in the definitions of the rings Ch(G), J(G) and B(G), we obtain the rings ChR (G), J R (G) and B R (G), respectively (for example, the inclusion θ H ∈ Ch(H) in the definition of B R (G) must be replaced by θ H ∈ ChR (H) = R[Irr(H)]). Obviously, J R (G) ⊆ ChR (G) ⊆ B R (G). As in the deduction of Theorem 1.2 from Theorem 1.1, one can show that J R (G) is an ideal of the ring B R (G). Therefore, to prove that J R (G) = ChR (G), it suffices to show that 1G ∈ J R (G). Then, for any ring R with ℤ ⊆ R ⊆ ℂ, we have J R (G) = B R (G); hence J R (G) = ChR (G). Letting R = ℤ, we complete the proof of Theorem 1.1. Let ω ∈ ℂ be a primitive |G|-th root of unity, R = ℤ[ω], and let f(X) ∈ ℚ[X] be an irreducible polynomial with the root ω (= minimal polynomial of ω), deg(f) = n. It is easy to see that R has a ℤ-basis 1, ω, . . . , ω n−1 . Lemma 1.3. If 1G ∈ J R (G), then Ch(G) = J(G). Proof. Let Irr(G) = {χ1 , . . . , χ r }. Suppose that 1, ω, . . . , ω n−1 are linearly dependent over Ch(G), i.e., n−1

r

∑ ( ∑ c ij χ i )ω j = 0, j=0

i=1

c ij ∈ ℤ, ∑ |c ij | > 0. i,j

Since the set Irr(G) is linearly independent over ℤ, it follows by setting n−1

d i = ∑ c ij ω j j=0

(i = 1, . . . , r)

VIII Brauer’s induction theorems | 375

that

r

∑ d i χ i = 0 󳨐⇒ d i = 0. i=1

Since 1, ω, . . . , is a ℤ-basis of the ring R, we get c ij = 0 for all i, j, contrary to the assumption. This implies immediately that ω n−1

n−1

J R (G) = ⨁ J(G)ω j .

(1)

j=0

Indeed, as follows from the previous remark, the right-hand side of (1) is a direct sum. But if θ = ∑ r φ φ G ∈ J R (G), φ

where r φ ∈ R, then

n−1

r φ = ∑ r φj ω j ,

r φj ∈ ℤ,

j=0

and it is now clear that in (1) we have equality. If 1G ∈ J R (G), then it follows from (1) that 1G ∈ J(G), and J(G) = Ch(G) since J(G) is an ideal in Ch(G). It remains to show that 1G ∈ J R (G). If x, y ∈ G are conjugate, we write x ∼ y. Let x ∈ G and let p be a prime; then x = x p x p󸀠 = x p󸀠 x p , where the order of x p is a power of p and the order of x p󸀠 a p󸀠 -number. This representation is unique. The element x p is called the p-part of x, and x p󸀠 , the p󸀠 -part of x. Two elements x, y ∈ G are said to p p be p-conjugate in G (we write x ∼ y) if x p󸀠 ∼ y p󸀠 . Clearly, x ∼ x p󸀠 for all x ∈ G. Since p-conjugacy is an equivalence relation, one can consider the p-classes of G (sets of p-conjugate elements). Each such p-class, being G-invariant, consists of several G-classes. In a p󸀠 -group G, p-conjugacy is the same as ordinary conjugacy. Note that all p-elements of G are p-conjugate with 1. If p does not divide |G|, i.e., G is a p󸀠 -group, then p-conjugacy is the same as usual conjugacy. p

Lemma 1.4. If χ ∈ ChR (G) takes values in ℤ, then x ∼ y implies χ(x) ≡ χ(y) (mod p). Proof. As χ is a class function, it follows, by definition, that χ(x p󸀠 ) = χ(y p󸀠 ), and so it suffices to prove that χ(z) ≡ χ(z p󸀠 ) (mod p) for all z ∈ G, since in that case we have χ(x) ≡ χ(x p󸀠 ) = χ(y p󸀠 ) ≡ χ(y) (mod p). If Y = ⟨y⟩, then χY = ∑ ai φi ,

φ i ∈ Lin(Y), a i ∈ R.

i

Since the characters φ i are linear, we get φ i (y)o(y p ) = φ i (y p y p󸀠 )o(y p ) = φ i (y p )o(y p ) φ i (y p󸀠 )o(y p) = φ i (y p󸀠 )o(y p ) .

(2)

376 | Characters of Finite Groups 1 Raising (2) to the power p m = o(y p ) and using the binomial theorem and the equalities in the previous display, we obtain m

m

(3)

χ(y)p ≡ χ(y p󸀠 )p (mod pR), where pR is the principal ideal in R generated by the prime p. It follows from n−1

R = ⨁ ℤω j j=0

that ℤ ∩ pR = pℤ. Since χ(y), χ(y ) ∈ ℤ, by hypothesis, it follows from (3) that p󸀠 m

m

(4)

χ(y)p ≡ χ(y p󸀠 )p (mod pℤ).

By Fermat’s theorem, the result now follows from (4), since congruence (mod pℤ) is equivalent to congruence (mod p). The set E appears for the first time in the following lemma. Lemma 1.5. Let U = ⟨u⟩ ≤ G, where u is a p󸀠 -element, let P ∈ Sylp (CG (u)) and U×P = H (so that H ∈ Ep ). Then there exists ψ ∈ ChR (H) such that the following hold: (a) One has ψ G (g) ∈ ℤ for all g ∈ G. p (b) If g ≁ u, then ψ G (g) = 0. (c) One has ψ G (u) = |CG (u) : P|(≢ 0 (mod p)). Proof. Let t = |U| and let {ψ1 , . . . , ψ t } = Irr(H/P)(= Lin(H/P) since H/P ≅ U is cyclic). Then ψ i ∈ Lin(H) and P ≤ ker(ψ i ) for all i. Put t

ψ = ∑ ψ i (u)ψ i .

(5)

i=1

If v ∈ P, then t

t

t

i=1

i=1

i=1

ψ(uv) = ∑ ψ i (u)ψ i (uv) = ∑ (ψ i (u)ψ i (u))ψ i (v) = ∑ |ψ i (u)|2 = |U|. Indeed, ψ i (uv) = ψ i (u)ψ i (v) by the linearity of ψ i , and ψ i (v) = 1 since v ∈ P ≤ ker(ψ i ) (i ∈ {1, . . . , t}). If u j ≠ u, then, by the Second Orthogonality Relation, t

t

ψ(u j v) = ∑ ψ i (u)ψ i (u j )ψ i (v) = ∑ ψ i (u)ψ i (u j ) = 0 i=1

i=1

for all v ∈ P. The last two displayed formulas show that {0 for x ∈ H − uP, ψ(x) = { |U| for x ∈ uP. { Since ψ i (u) (i = 1, . . . , t) are |H|-th roots of unity, we have ψ ∈ ChR (H) and ψ G ∈ J R (G) (by (5)). By the induced character formula (V.1.6), we have (see there definition of the function ψ)̇ −1 ̇ ψ G (g) = |H|−1 ∑ ψ(zgz ) z∈G

(g ∈ G).

(6)

VIII Brauer’s induction theorems | 377

Given g ∈ G, put S(g) = {z ∈ G | zgz−1 ∈ uP},

σ(g) = |S(g)|.

Then we can rewrite (6) as follows (recall that ψ vanishes on H − uP): ψ G (g) = |H|−1 |U|σ(g) = |P|−1 σ(g).

(7)

If z ∈ S(g), v ∈ P, then (vz)g(vz)−1 ∈ vuv−1 P = uP,

i.e.,

vz ∈ S(g).

Therefore, z ∈ S(g) implies Pz ⊆ S(g), whence it follows that |P| | σ(g). Thus (see (7)), ψ G (g) ∈ ℤ for all g ∈ G. This proves (a). p If g ≁ u, then ψ G (g) = 0, by (7), and (b) is proved. Let us calculate σ(u). If zuz−1 = uv, where v ∈ P, then v = 1 and so z ∈ CG (u). This means that σ(u) = |CG (u)| and, by (7), we obtain ψ G (u) = |P|−1 σ(u) = |CG (u) : P| ≢ 0 (mod p), completing the proof of (c) and thereby the lemma. Lemma 1.6. If θ ∈ CF[G] takes values in ℤ and |G| | θ(g) for all g ∈ G, then θ ∈ J R (G). Proof. Let {K1 , . . . , K t } be the set of p-classes in G. By Lemma 1.5, for a prime p ∤ |G|, for every j ≤ t there exists a ℤ-valued χ j ∈ J R (G) that vanishes on G − K j and takes a value d j on K j such that d j | |G|. If θ(g j ) = m j |G| for g j ∈ K j , then r

θ=∑ j=1

It follows from

m j |G| dj

m j |G| ⋅ χj . dj

∈ ℤ that θ ∈ J R (G).

Lemma 1.7. For any prime p ∈ π(G), there exists a ℤ-valued χ ∈ J R (G) that satisfies χ(g) ≡ 1 (mod p) for all g ∈ G. Proof. Let K1 , . . . , K t be all p-classes in G. By Lemma 1.5, there exists a ℤ-valued χ j ∈ J R (G) that vanishes on G − K j but its value on K j is not divisible by p (Lemma 1.4). Then a j χ j (g j ) ≡ 1 (mod p) for suitable a j ∈ ℤ, where g j ∈ K j (j ∈ {1, . . . , t}). It is obvious that χ = ∑tj=1 a j χ j possesses the necessary property. Lemma 1.8. Let p be a prime, |G| = p a m, p ∤ m. Then m ⋅ 1G ∈ J(G). Proof. In view of (1), we need only prove that m⋅1G ∈ J R (G). Choose χ as in Lemma 1.7 a and put θ = χ p . Since χ(g)p − 1 = (χ(g) − 1)(1 + χ(g) + ⋅ ⋅ ⋅ + χ(g)p−1 ) is divisible by p2 (both factors on the right are divisible by p since χ(g) ≡ 1 (mod p)), it follows that χ(g)p ≡ 1 (mod p2 ). Next, using induction on a, one can prove that a χ(g)p ≡ 1 (mod p a+1 ). In particular, θ(g) ≡ 1 (mod p a ). Further, m ⋅ 1G = m(1G − θ) + mθ.

(8)

378 | Characters of Finite Groups 1 Since 1G (g) = 1 for all g ∈ G, it follows that m(1G − θ)(g) is divisible by mp a = |G| (since θ(g) ≡ 1 (mod p a )). Therefore, m(1G − θ) ∈ J R (G), by Lemma 1.6. As J R (G) is an ideal of ChR (G), we get θ ∈ J R (G), and so m ⋅ 1G ∈ J R (G), by (8). a

a

Proof of Theorem 1.1. Let |G| = p11 ⋅ ⋅ ⋅ p k k be the canonical decomposition. In case a that |G| = p i i m i , then m i ⋅ 1G ∈ J(G) (Lemma 1.8). Since GCD(m1 , . . . , m k ) = 1, there exist c1 , . . . , c k ∈ ℤ such that c1 m1 + ⋅ ⋅ ⋅ + c k m k = 1. Therefore, 1G = (c1 m1 + ⋅ ⋅ ⋅ + c k m k )1G ∈ J(G). As we have already seen, this completes the proof of Theorem 1.1. Theorem 1.9 (Frobenius). Let |G| = mn, GCD(m, n) = 1. Define the function ϑ : G → ℂ by {m if x n = 1, ϑ(x) = { 0 if x n ≠ 1. { Then ϑ ∈ Ch(G). Proof. Let E ∈ E; then E = N × M, where |N|, |M| divide n, m, respectively. If x ∈ N, then x n = 1 implies ϑ(x) = m. If x ∈ E − N, then x n ≠ 1 implies ϑ(x) = 0. Therefore, {m ϑ E (x) = { 0 {

if x ∈ N, if x ∈ E − N.

Let ψ be a regular character of E/N considered as a character of E. Then {|E/N| = |M| if x ∈ N, ψ(x) = { 0 if x ∈ E − N. { We have m = |M|k for some k ∈ ℕ since |M| | m. Then ϑ E = kψ ∈ Char(E). By Theorem 1.2, ϑ ∈ Ch(G), as was to be shown. All solutions and proofs up to the end of the section are due to the third author. All of this material is taken from [Isa11, section “Problems” in Chapter 8]. Exercise 1.1 ([Isa11, Problem 8.4]). Let χ, ψ ∈ Char(G), |G| = mn, GCD(m, n) = 1 and α = ∑g∈G,o(g)|n χ(g)ψ(g). Then αn ∈ ℤ. Solution. Let θ = χψ; then θ ∈ Char(G) and α = ∑g n =1 θ(g). Let ϑ ∈ Ch(G) be as in Theorem 1.9. Then mα = ∑ θ(g)ϑ(g) = ∑ θ(g)ϑ(g) = |G|⟨θ, ϑ⟩ = mn⟨θ, ϑ⟩, g n =1

and we conclude that

α n

g∈G

= ⟨θ, ϑ⟩ ∈ ℤ.

Lemma 1.10. Let χ ∈ Irr(G) and let the subset I ⊆ G be G-invariant. Then for any ψ ∈ Char(G) one has χ(1) | ∑g∈I χ(g)ψ(g) in the ring R of algebraic integers.

VIII Brauer’s induction theorems | 379

Proof. Let I = ⋃si=1 K i be the union of the G-classes, g i ∈ K i . Then s

s

∑ χ(g)ψ(g) = ∑ ∑ χ(g)ψ(g) = ∑ |K i |χ(g i )ψ(g i ).

g∈I

i=1 g∈K i

i=1

Let ω χ : Z(ℂG) → ℂ be a homomorphism (see Chapter III) such that ω χ (z) = k i = ∑g∈K i g, then |K i |χ(g i ) ω χ (k i ) = . χ(1)

χ(z) χ(1) :

if

The number ω χ (k i ) ∈ R, the ring of algebraic integers. Therefore, χ(1) | |K i |χ(g i ) in the ring R. It follows from s

s

∑ χ(g)ψ(g) = ∑ |K i |χ(g i )ψ(g i ) = ∑ χ(1)ω χ (k i )ψ(g i )

g∈I

i=1

i=1

that χ(1) | ∑g∈I χ(g)ψ(g) in R. Corollary 1.11. If I ⊆ G is G-invariant. then χ(1) | ∑g∈I χ(g) and χ(1) | ∑g∈I |χ(g)|2 in the ring R. Proof. Set, in Lemma 1.10, first ψ = 1G and then ψ = χ. Exercise 1.2 ([Isa11, Problem 8.5]). Let |G| = mn, GCD(m, n) = 1, g ∈ G# , χ ∈ Irr(G). (a) If n | χ(1), then g m ≠ 1 implies χ(g) = 0. In other words, if GCD(o(g), n) > 1, then χ(g) = 0. (b) Assume that g n = 1 implies χ(g) = 0. Then g m ≠ 1 implies χ(g) = 0. Solution. (a) Write I = {g ∈ G | g m = 1}. By Exercise 1.1, m | ∑g∈I |χ(g)|2 in ℤ. Since I is G-invariant, we get χ(1) | ∑g∈I |χ(g)|2 (Corollary 1.11). Since n | χ(1), it follows that n | ∑g∈I |χ(g)|2 , and we conclude that |G| = mn | ∑g∈I |χ(g)|2 in view of GCD(m, n) = 1. Thus, ∑ |χ(g)|2 ≥ |G| = ∑ |χ(1)|2 . g∈I

g∈G

Therefore, ∑g m =1̸ |χ(g)|2 = 0 so that χ(g) = 0 if g m ≠ 1. (b) By hypothesis, ∑g n =1 χ(g) = χ(1). By Exercise 1.1, n | χ(1), and so, by (a), g n ≠ 1 implies χ(g) = 0.

χ(1) n

=

1 n

∑g n =1 χ(g) ∈ ℤ, i.e.,

As a consequence, we get the following two results. Exercise 1.3 (= Theorem VIII.4.1 (R. Brauer)). If χ ∈ Irr(G) and |G|p | χ(1), then χ vanishes on all g ∈ G with p | o(g). Exercise 1.4. Let p ∈ π(G), χ ∈ Irr(G) and χ vanishes on all p-elements of G. Then χ(g) = 0 for any g ∈ G with p | o(g), and then |G|p | χ(1). Proposition 1.12 ([Isa11, Problem 8.6]). Let G and H be groups with conjugacy classes K i and L i , respectively, and irreducible characters χ i and ψ i , respectively. Assume that,

380 | Characters of Finite Groups 1 whenever g i ∈ K i and h i ∈ L i , we have χ j (g i ) = ϕ j (h i ) for all i, j. (In short, G and H have identical character tables.) Let x ∈ Z(P), where P ∈ Sylp (G). If x ∈ K i and y ∈ L i , then o(x) = o(y). Proof. By hypothesis, |G| = |H| (the character tables of our groups have identical first columns). By G. Higman’s theorem (see [Isa11, Theorem 8.21]), the table of characters of G determines the sets π i = π(g i ), where g i ∈ K i . It follows that h i ∈ L i implies π(g i ) = π(h i ). By hypothesis, π(x) = {p} so that π(y) = {p}, i.e., x, y are p-elements. Let o(x) = p a ,

o(y) = p b .

(9)

Then (∗)

|K i | = |L i | 󳨐⇒ |CG (x)| = |CH (y)|.

By hypothesis, P ≤ CG (x) so that, by (∗), |P| | |CH (y)|. Therefore, CG (y) ≥ Q, where Q ∈ Sylp (H). Since y ∈ Z(CH (y)) and y is a p-element, then y ∈ Q. Therefore, y ∈ Z(Q). Thus, x ∈ Ki ,

x ∈ Z(P),

P ∈ Sylp (G),

y ∈ Li ,

y ∈ Z(Q),

Q ∈ Sylp (H).

Assume that b > a (see (9)). Denoting by ℚm the cyclotomic field generated by m-th root of 1, we get ψ j (y) ∈ ℚp b , χ j (x) ∈ ℚp a for all j. By hypothesis, ψ j (y) = χ j (x) so that ψ j (y) ∈ ℚp a for all j. Therefore, writing G0 = ℚp b /ℚp a , we get σ(ψ j (y)) = ψ j (y)

for all σ ∈ G0 .

(10)

b−a

Let ϵ be a primitive p b -th root of 1. Then ϵ p is a primitive p a -th root of 1 and b−a so ℚp a = ℚ(ϵ p ). Let σ ∈ G = ℚp b /ℚ. Then σ(ϵ) = ϵ ν , where p ∤ ν and ν is uniquely determined b−a b−a modulo p b . Obviously, G0 = {σ ∈ G | σ(ϵ p ) = ϵ p }. If σ ∈ G and σ(ϵ) = ϵ ν , then b−a b−a σ(ϵ p ) = σ(ϵ νp ), and we conclude that σ ∈ G0 󳨐⇒ ϵ νp

b−a

= ϵp

b−a

󳨐⇒ ν ≡ 1 (mod p a ).

Thus, σ ∈ G0 implies σ(ϵ) = ϵ ν , where ν ≡ 1 (mod p a ). It follows that |G0 | = p b−a . Next, if σ ∈ G0 and σ(ϵ) = ϵ ν , then, since o(y) = p b , we get σ(ψ j (y)) = ψ j (y ν ) for all j. But then, by (10), we have ψ j (y ν ) = ψ j (y) for all j. Therefore, ν ≡ 1 (mod p a ) 󳨐⇒ y ν ∼ y 󳨐⇒ y ν ∈ L i . It follows from this the following partition of L i onto classes. Elements y, z ∈ L i belong to the same class if and only if z = y ν , where ν ≡ 1 (mod p a ). The cardinality of the class (y) is equal to the number of all noncongruent modulo p b solutions of the congruence ν ≡ 1 (mod p a ). Setting ν = 1 + p a ξ , we find (since for given ν (mod p b ) the number ξ is uniquely determined modulo p b−a ) that |(y)| = p b−a . Hence, L i is partitioned in classes of cardinality p b−a , i.e., p b−a | |L i | implies p | |L i |. This is impossible since |L i | = |H : CH (y)| ≢ 0 (mod p), in view of CG (y) ≥ Q, where Q ∈ Sylp (H). Thus, it is not true that b > a. Similarly, it is not true that a > b. Thus, we obtain that a = b, i.e., o(x) = o(y).

VIII Brauer’s induction theorems | 381

Exercise 1.5 ([Isa11, Problem 8.1]). Let ℤ ⊆ R ⊆ ℂ, where R is a ring and R+ = ℤ+ ⊕S+ , where R+ , ℤ+ are additive groups of R, ℤ, respectively, and S+ < R+ . Let H be the family of subgroups of a group G, let IR (G, H) be the set of R-linear combinations of characters ϕ G (where ϕ ∈ Char(H), H ∈ H) and let I(G, H) = Iℤ (G, H). Prove that IR (G, H) ∩ Ch(G) = I(G, H). Solution. Obviously, I(G, H) ⊆ IR (G, H) ∩ Ch(G). Let ϑ ∈ IR (G, H) ∩ Ch(G); then ϑ=

∑

aψ ψG ,

a ψ ∈ R.

ψ∈Char(H),H∈H S S ℤ Setting a ψ = aℤ ψ + a ψ , where a ψ ∈ ℤ, a ψ ∈ S, we get G S G ϑ = ∑ aℤ ψ ψ + ∑ aψ ψ . ψ

ψ

G Setting θ = ∑ψ a Sψ ψ G , we get θ = ϑ − ∑ψ aℤ ψ ψ ∈ Ch(G). Therefore. ⟨θ, χ⟩ ∈ ℤ for all χ ∈ Irr(G). On the other hand,

⟨ψ G , χ⟩ ∈ ℤ 󳨐⇒ ⟨θ, χ⟩ = ∑ a Sψ ⟨ψ G , χ⟩ ∈ S. ψ

Thus, ⟨θ, χ⟩ ∈ ℤ ∩ S = {0} hence ⟨θ, χ⟩ = 0 for all χ ∈ Irr(G), and we conclude that θ = 0. It follows that G ϑ = ∑ aℤ ψ ψ ∈ I(G, H) 󳨐⇒ IR (G, H) ∩ Ch(G) ⊆ I(G, H). ψ

As the reverse inclusion is true, we are done. Theorem 1.13 ([Isa11, Problem 8.3]; compare with Theorem 1.9). Let χ ∈ Char(G). For a prime p, define the class function χ p by χ p (g) = χ(g p󸀠 ). where g p󸀠 is a p󸀠 -part of g ∈ G. Then the following statements hold: (a) (χ p )N ∈ Char(N) for every nilpotent N ≤ G and so, by Theorem 1.2, χ p ∈ Ch(G). (b) If G is nonnilpotent, then χ p ∈ ̸ Char(G) for some χ ∈ Irr(G) and some prime p | |G|. Proof. (a) Let N ≤ G be nilpotent. (a1) Let p ∤ |N|. Then g ∈ N 󳨐⇒ g p = 1,

g p󸀠 = g 󳨐⇒ χ p (g) = χ(g) 󳨐⇒ (χ p )N = χ N ∈ Char(N).

(a2) Let p | |N|. Then N = N p × N p󸀠 , where N p ∈ Sylp (N). If g ∈ N, then g = g p g p󸀠 , where g p ∈ N p , g p󸀠 ∈ N p󸀠 . Let ϑ = χ N p󸀠 ; then ϑ ∈ Char(N p󸀠 ), χ p (g) = χ(g p󸀠 ) = ϑ(g p󸀠 )

(g ∈ N).

(11)

Let Irr(N p󸀠 ) = {ϑ1 , . . . , ϑ s }. Then s

ϑ = ∑ mi ϑi , i=1

m i ∈ ℕ ∪ {0}, i = 1, . . . , s.

(12)

382 | Characters of Finite Groups 1 Set ϑ̂ j = 1N p × ϑ j ∈ Irr(N)

(j = 1, . . . , s),

(13)

and ϑ̂ j (g) = 1N p (g p )ϑ j (g p󸀠 ) = ϑ j (g p󸀠 ).

(14)

Then, by the definition of χ p and (11), (12), for all g ∈ N one has χ p (g) = ∑sj=1 m j ϑ j (g p󸀠 ) so, by (14), χ p (g) = ∑sj=1 m j ϑ̂ j (g p󸀠 ) for all g ∈ N. Thus, s

(χ p )N = ∑ m j ϑ̂ j . j=1

Since ϑ̂ j ∈ Char(N), we get, by Theorem 1.2, (χ p )N ∈ Char(N) 󳨐⇒ χ p ∈ Ch(G). (b) Assume that for a given p and every χ ∈ Irr(G), one has χ p ∈ Char(G). Let us prove that then P ∈ Sylp (G) is G-invariant. If g ∈ P, then g p󸀠 = 1, and so we obtain χ p (g) = χ(1) = χ p (1) for every χ ∈ Irr(G). As χ p ∈ Char(G), it follows that g ∈ ker(χ p ) hence P ≤ ker(χ p ).

(15)

Let g ∈ ker(χ p ). Then χ(g p󸀠 ) = χ p (g) = χ p (1) = χ(1) and hence g p󸀠 ∈ ker(χ) for all g ∈ ker(χ p ).

(16)

Now let g p󸀠 ∈ D = ⋂χ∈Irr(G) ker(χ p ). Then, by (16), one has g p󸀠 ∈ ⋂χ∈Irr(G) ker(χ) = {1}, i.e., g = g p is a p-element. Thus, D consists of p-elements, i.e., D is a p-subgroup. Then, by (15), P = D ⊲ G. If now χ p ∈ Char(G) for all χ ∈ Irr(G) and all p ∈ π(G), then all Sylow subgroups are normal in G, i.e., G is nilpotent. Thus, if G is nonnilpotent, then there are p ∈ π(G) and χ ∈ Irr(G) such that χ p ∈ ̸ Char(G). Note that χ p ∈ Ch(G) since the restriction of χ p to any Brauer elementary subgroup is a character, by Theorem 1.13 (b), and the assertion follows from Theorem 1.2.

2 Realization theorem One of the most important applications of Brauer’s induction theorems is the solution of Maschke’s problem about the realization of ℂ-representations of a group in a cyclotomic field. Recall that a ℂ-representation T of a group G is realizable in a subfield K ⊆ ℂ if T is equivalent to a representation T 󸀠 such that T 󸀠 (x) is a matrix over K for all x ∈ G. The characters of K-representations of G are called the K-characters of G. It is obvious that a ℂ-representation T of G can be realizable in K if and only if χ T , the character of T, is a K-character. However, if χ(g) ∈ K for all g ∈ G, then χ is not necessarily a K-character.

VIII Brauer’s induction theorems | 383

Let e = exp(G) and let ε be a primitive e-th root of unity. Theorem 2.1 (Brauer’s realization theorem). Every ℂ-representation of a group G is realizable in the cyclotomic field ℚ(ε). According to the previous remark, it suffices to prove that all complex characters of G are ℚ(ε)-characters. We need a generalization of the orthogonality relations for the matrix elements of irreducible representations of G (see Theorems II.2.1 and II.2.2). Using Schur’s lemma and literally repeating the proof of the Schur’s first theorem on matrix elements (see Theorem II.2.1), one obtains the following result. Lemma 2.2. Let K be a subfield of ℂ, let T and U be nonequivalent irreducible K-representations of G, and let {α ij }1n , {β kl }1m be the matrix elements of T and U, respectively. Then ∑ α ij (g)β kl (g−1 ) = 0 (i, j ∈ {1, . . . , n}, k, l ∈ {1, . . . , m}) g∈G

Putting j = i, l = k and summing over i, k, we get ∑g∈G χ T (g)χ U (g −1 ) = 0 (here χ T is a character of T). Since χ U (g−1 ) = χ U (g), it follows that ⟨χ T , χ U ⟩ = 0. We thus obtain the following corollary. Corollary 2.3. The irreducible K-characters of a group G form a finite orthogonal (but, generally speaking, not orthonormal) system (but, generally speaking, not a basis) in the unitary space CF[G] of class functions on G. (If K ≠ ℂ, then an irreducible K-character may be reducible over ℂ and this explain the assertion in the previous parentheses.) Let {θ i }1t be a complete system of irreducible K-characters of a group G. It follows from Maschke’s theorem that every K-character of G can be expressed uniquely as ∑ti=1 m i θ i , where all m i ∈ ℕ ∪ {0}. Proof of Theorem 2.1. Let χ ∈ Char(G). Write L = ⋃ Lin(H), H∈E

where E is the set of all Brauer elementary subgroups of G. Since exp(H) | e = exp(G) (H ∈ E), the values of a character ψ ∈ L are e-th roots of unity. Formulas (V.5.1) and (V.5.2) imply that a representation Ψ of G affording the character ψ G is realizable in ℚ(ε). Hence, ψ G is a ℚ(ε)-character. By Theorem 1.1, χ = ∑ aψ ψG ,

a ψ ∈ ℤ.

ψ∈L

Since ψ G (ψ ∈ L) is a ℚ(ε)-character, it follows that t

(i)

ψG = ∑ cψ θi , i=1

(i)

c ψ ∈ ℕ ∪ {0},

384 | Characters of Finite Groups 1 where {θ i }1t is a complete system of irreducible ℚ(ε)-characters of G. By the above formulas for χ and ψ G , one obtains t

χ = ∑ mi θi , i=1

(i)

m i = ∑ a ψ c ψ ∈ ℤ. ψ∈L

Since the characters θ i (i ∈ {1, . . . , t}) are pairwise orthogonal, we have ⟨χ, θ i ⟩ = m i ⟨θ i , θ i ⟩, so that m i ≥ 0 (i ∈ {1, . . . , t}). Hence, χ is a ℚ(ε)-character.

3 Groups with partitions of special type Let π and π󸀠 be complementary sets of primes. A partition G# = A ∪ B is called a (π, π󸀠 )-partition if all elements of A are π-elements and those of B are π󸀠 -elements. If G# = A ∪ B is a (π, π󸀠 )-partition and E ∈ E, then either E# ⊆ A or E# ⊆ B (the same is true for any nilpotent subgroup of G). It follows that if x ∈ A, then CG (x) ⊆ A. For example, A5 admit {p, p󸀠 }-partitions for p ∈ {2, 3, 5}, PSL(2, 7) and PSL(2, 8) admit {p, p󸀠 }-partitions for p ∈ {2, 3, 7}, Theorem 3.1 ([Zhm18]). A partition G# = A ∪ B of G# into two nonempty subsets A and B is a (π, π󸀠 )-partition if and only if there exists θ ∈ Ch(G) that equals 1 on A and 0 on B. Proof. (i) Suppose that there exists θ ∈ Ch(G) taking value 1 on A and 0 on B. We have to prove that G# = A ∪ B is a (π, π󸀠 )-partition for an appropriate set π of primes. It suffices to show that if x ∈ A and y ∈ B, then GCD(o(x), o(y)) = 1. If x ∈ A and p is a prime, then, by Lemma 1.4, the entire p-class of x is contained a a in A. Let o(x) = p11 ⋅ ⋅ ⋅ p r r be the canonical decomposition, and let x = x1 ⋅ ⋅ ⋅ x r be the corresponding decomposition of x in pairwise permutable primary parts of x, a where o(x i ) = p i i , i ∈ {1, . . . , r}. We prove, by induction on r, that all x i ∈ A. Since p1 x ∼ x2 ⋅ ⋅ ⋅ x r (indeed, x p󸀠1 = x2 ⋅ ⋅ ⋅ x r ), it follows that x2 ⋅ ⋅ ⋅ x r ∈ A, so that, by induction, x i ∈ A for i > 1. Similarly, it follows from x = x2 ⋅ ⋅ ⋅ x r x1 , that x1 ∈ A. Let x ∈ A, p y ∈ B. If p | GCD(o(x), o(y)), then, by the above, x p ∈ A, and y p ∈ B. Since x p ∼ y p (indeed, p󸀠 -parts of x p and y p are equal to 1), we obtain a contradiction. Thus, GCD(o(x), o(y)) = 1, i.e., G# = A ∪ B is a (π, π󸀠 )-partition, where π = ⋃x∈A π(o(x)). (ii) Now suppose that G# = A ∪ B is a (π, π󸀠 )-partition. Using the Chinese remainder theorem, choose m ∈ ℕ such that m ≡ 1 (mod |G|π ) and m ≡ 0 (mod |G|π󸀠 ). Define θ ∈ CF[G] by m if x = 1, { { { θ(x) = {1 if x ∈ A, { { {0 if x ∈ B.

VIII Brauer’s induction theorems | 385

To show that θ ∈ Ch(G), we use Theorem 1.2. Let E ∈ E (the set of Brauer elementary subgroups of G). Then, as we have noted, either E# ⊆ A or E# ⊆ B. If E# ⊆ B, then θE =

m ρE , |E|

where ρ E is the regular character of E. If E# ⊆ A, then (θ − 1G )E = and hence θ E = 1E +

m−1 ρE , |E| m−1 ρE . |E|

In either case, θ E ∈ Ch(E). By Theorem 1.2, θ ∈ Ch(G).

4 Characters of p-defect 0 In order to demonstrate yet another application of Theorem 1.2, we will prove a deep theorem about the zeros of irreducible characters. Theorem 4.1 (Brauer). Let χ ∈ Irr(G) and let a prime p | |G| be such that χ(1)p = |G|p (in that case, χ is said to be a character of defect 0). If p | o(g) (g ∈ G), then χ(g) = 0. Proof. Define a function θ : G → ℂ as follows: {χ(g) θ(g) = { 0 {

if p ∤ o(g), if p | o(g).

We claim that θ ∈ Ch(G). By Theorem 1.2, it suffices to prove that θ E ∈ Ch(E) for any E ∈ E. Since E is a nilpotent, we have E = P × Q, where P ∈ Sylp (E) and Q is cyclic. By definition, θ vanishes on E − Q (of course, provided that the set E − Q is not empty) and θ Q = χ Q . If ψ ∈ Irr(E), then |P|⟨θ E , ψ⟩ = ⟨χ Q , ψ Q ⟩.

(1)

Indeed, since |Q| ⋅ |P| = |E| and |E|⟨θ E , ψ⟩ = ∑ θ(x)ψ(x) = ∑ θ(x)ψ(x) = |Q|⟨χ Q , ψ Q ⟩, x∈Q

x∈E

we obtain equality (1). By (1), |P|⟨θ E , ψ⟩ ∈ ℕ ∪ {0}. Let ω χ ∈ Hom(Z(ℂG), ℂ) (see §III.3). Let K g denote the G-class containing g ∈ G and k g the corresponding class sum. We have ω χ (k g ) =

|K g |χ(g) . χ(1)

386 | Characters of Finite Groups 1 Then, using (1), one obtains |G| |G| |G| |Q|⟨θ E , ψ⟩ = |Q| ⋅ |P|−1 ⟨χ Q , ψ Q ⟩ = ∑ χ(g)ψ(g) χ(1) χ(1) |P|χ(1) g∈Q =

χ(1)ω χ (k g ) |G| ψ(g) = |P|−1 ∑ |CG (g)|ω χ (k g )ψ(g). ∑ |P|χ(1) g∈Q |K g | g∈Q

Note that CG (g) ≥ P for g ∈ Q, and therefore we get |G| |Q|⟨θ E , ψ⟩ = ∑ |CG (g) : P|ω χ (k g )ψ(g). χ(1) g∈Q |G| |Q|⟨θ E , ψ⟩ ∈ ℤ. Since ω χ (k g ) is an algebraic integer for any g ∈ G, it follows that χ(1) |G||Q| As |P|⟨θ E , ψ⟩ ∈ ℤ and χ(1) is an integer coprime with p (in view of the character χ is of defect 0), it follows that 1 ∈ ℕ is a ℤ-linear combination of |P| and |G||Q| χ(1) , and we conclude that ⟨θ E , ψ⟩ ∈ ℤ. Thus, θ E ∈ Ch(E) for all E ∈ E and therefore, by Theorem 1.2, θ ∈ Ch(G). In particular, ⟨θ, χ⟩ ∈ ℤ. We have

⟨θ, χ⟩ = |G|−1 ∑ θ(g)χ(g) = |G|−1 ∑ |χ(g)|2 ≤ |G|−1 ∑ |χ(g)|2 = 1. g∈G

p∤o(g)

g∈G

Thus, 0 < ⟨θ, χ⟩ ≤ 1. This inequality and the inclusion ⟨θ, χ⟩ ∈ ℤ imply that ⟨θ, χ⟩ = 1, i.e., ∑ |χ(g)|2 = ∑ |χ(g)|2 . g∈G

p∤o(g)

Thus, χ(g) = 0 if p | o(g). For more elementary proof of this theorem, due to M. Leitz, see Theorem IV.9.2. Exercise 4.1. Let p ∈ π(G), χ ∈ Irr(G) and χ vanishes on all p-elements of G. Then χ(g) = 0 for all g ∈ G with p | o(g), and |G|p | χ(1). Hint. If P ∈ Sylp (G), then χ vanishes on P# so χ P is a multiple of ρ P , the regular character of P. Therefore, |G|p = |P| | χ(1), and now the result follows from Theorem 4.1. Exercise 4.2. Let |G| = p a m, where p ∤ m and D = ⋂χ∈Irr(G),p a ∤χ(1) ker(χ). Prove that (i) p ∤ |D|, (ii) D is nilpotent. Solution. Let μ ∈ Irr# (D). Then kernels of all members of the set Irr(μ G ) do not contain D so that p a | χ(1) | μ G (1) = |G : D|μ(1) for all χ ∈ Irr(μ G ). If p ∤ μ(1), then p a | |G : D| implies p ∤ |D|. Assume that p a ∤ |G : D|; then p | |D|. In that case, all nonprincipal irreducible characters of D have degrees divisible by p so that D󸀠 = D. Now, |D| = |D : D󸀠 | +

∑

μ(1)2 󳨐⇒ p | |D : D󸀠 |,

μ∈Irr1 (D)

contrary to the previous sentence. Thus, assertion (i) is proved. We suggest to return to (ii) after reading Chapter X.

VIII Brauer’s induction theorems | 387

5 Brauer’s paper on quotient groups In this section we will prove numerous consequences (all due to Brauer) of induction Theorem 1.2 in the structure theory of finite groups. Throughout this section, H0 ⊲ H ≤ G. A normal subgroup G0 of a group G is called a normal complement to H over H0 if G = HG0 and H0 = H ∩ G0 (then G/G0 ≅ H/H0 ). Let us form the conjugacy classes in H/H0 and consider their full preimages in H: H0 , H1 , . . . , H n (each H-invariant subset H i is the union of certain cosets of H0 in H, and the union of certain H-classes as well); then H = ⋃ni=0 H i is a partition, and therefore |H| = ∑ni=0 |H i |. Clearly, H i H0 = H i for all i. Let π = π(|H : H0 |) be the set of prime divisors of |H : H0 |. Then H0 ⊲ H contains all π󸀠 -elements of H. Therefore each set H i , i > 0, contains a π-element. Indeed, if x ∈ H i , i > 0, then x = x π x π󸀠 = x π󸀠 x π ; since x π󸀠 ∈ H0 , we get x π H0 = xH0 ⊆ H i , and hence x π ∈ H i . Moreover, the number of π-elements in H i is at least k if H i is the union of k conjugacy classes of H. Let the set G i (i > 0) consist of those elements g ∈ G whose π-part g π is conjugate in G to an element of H i . Put G0 = G − ⋃ni=1 G i . Generally speaking, the G-invariant subsets G i are not necessarily disjoint. Let us consider the following four conditions: (A1) GCD(|G : H|, |H : H0 |) = 1. (A2) If h1 , h2 ∈ H and h1 ∼ h2 in G, then h1 H0 ∼ h2 H0 in H/H0 (or, equivalently, in H). (A3) If h is a π-element in H − H0 and p ∈ π does not divide o(h), then p does not divide |CG (h) : CH (h)|. (A4) If G0 , G1 , . . . , G n are defined as above, there are no indices i, j ∈ {0, 1, . . . , n}, i ≠ j, such that |G i | > |G : H| ⋅ |H i |,

|G j | < |G : H| ⋅ |H j |.

The following lemma shows that condition (A2) is fairly restrictive. Lemma 5.1. Assume that condition (A2) holds. Then the following statements are true: (a) Each set G j is G-invariant. (b) G = ⋃nj=0 G j is a partition. (c) H i = G i ∩ H (i ∈ {0, 1, . . . , n}). Proof. (a) This follows from the definition of subsets G j . (b) The definition of G0 implies that G0 ∩ G i = 0, i > 0. Let g ∈ G i , i > 0, and let K be the G-class containing g π . Then K ∩ H i ≠ 0 since g π is conjugate to an element, say h i , of H i , by the definition of G i . If also g ∈ G j , j > 0, then, as above, K ∩ H j ≠ 0 since g π is conjugate to an element, say h j , of H j . Therefore h i ∈ H i and h j ∈ H j are conjugate to each other in G since they are conjugate to g π . Then (A2) implies that h i H0 ∼ h j H0 in H; hence i = j. The proof of (b) is complete.

388 | Characters of Finite Groups 1 (c) Let i > 0. The definition of G i implies that H i ⊆ G i . Indeed, let g = g π g π󸀠 ∈ H i . Then H i = gH0 = g π g π󸀠 H0 = g π H0 󳨐⇒ g π ∈ H i 󳨐⇒ H i ⊆ G i , and we conclude that (c) is true for i > 0 (all π󸀠 -elements of H are in H0 ). The relation n

n

n

G0 ∩ H = (G − ⋃ G i ) ∩ H = H − ⋃(G i ∩ H) = H − ⋃ H i = H0 i=1

i=1

i=1

shows that (c) is true for i = 0, too. Assume that (A1) and (A2) hold. Let p ∈ π(= π(|H : H0 |)) and let x ∈ G be a p-element. Then, by (A1) and Sylow’s theorem, x ∼ y ∈ P ∈ Sylp (H) ⊆ Sylp (G). If this y ∈ H − H0 , then x is called an element of the first type. If y ∈ H0 , then x is called an element of the second type.¹ By (A2), the type of a p-element (p ∈ π) is well defined. Take g ∈ G. Let α(g) be the product of all primary parts of g π that are of the first type, and let β(g) be the product of the remaining primary parts of g π . If g π has no primary parts of the first (second) type, we put α(g) = 1 (respectively, β(g) = 1). We have α(g)β(g) = g π = β(g)α(g),

α(g) = α(g π ),

β(g) = β(g π ).

Conjugate elements have the same type, and x ∼ y implies α(x) ∼ α(y), β(x) ∼ β(y). Obviously, x, y ∈ G,

xy = yx 󳨐⇒ α(xy) = α(x)α(y),

β(xy) = β(x)β(y).

Lemma 5.2. Assume that conditions (A1)–(A3) hold and let g ∈ G. If α(g) ≠ 1, then g π is conjugate to an element of the set H − H0 , i.e., g ∈ G i for some i > 0. Proof. If g π is primary, then α(g) = g π , and the result follows from the definition of α(g). We now proceed by induction on the number of primary parts of g π different from 1. Assuming that the element g π is nonprimary, we have g π = xg p = g p x,

x = g π−{p} .

Suppose that a primary part of the first type occurs in x. Then α(x) ≠ 1 and, by induction, x is conjugate to an element of the set H − H0 . Replacing g (if necessary) by a suitable conjugate element, one may assume that x ∈ H − H0 . By (A3), CH (x) contains P0 ∈ Sylp (CG (x)). Since g p ∈ CG (x), g p is conjugate in CG (x) to an element of P0 ≤ H, and then g π = xg p is conjugate in G to an appropriate h ∈ H. Suppose that h ∈ H0 . Since the primary parts of g π are conjugate to suitable powers of h ∈ H0 , it follows that all primary parts of g π are of the second type whence α(g) = 1, contrary to the assumption. Thus, h ∈ H − H0 .

1 Note that H0 , which contains all π 󸀠 -elements of H, is not necessarily a π 󸀠 -subgroup.

VIII Brauer’s induction theorems | 389

Lemma 5.3. Assume that conditions (A1)–(A3) hold. Then the following statements hold: (a) g ∈ G0 if and only if α(g) = 1. (b) For every g ∈ G, the elements g, g π and α(g) are in the same set G i . (c) If g ∈ G0 , then all powers of g are in the set G0 . (d) All π󸀠 -elements of the group G are in the set G0 . Proof. (a) If α(g) ≠ 1, then g ∈ G i for some i > 0 (Lemma 5.1 (b)). Conversely, if g ∈ G i , i > 0, then g π ∼ h ∈ H i . Hence α(g) ∼ α(h). By (A2), β(h) ∈ H0 . Next, h = h π = α(h)β(h) 󳨐⇒ α(h) = hβ(h)−1 ∈ H i H0 = H i . In particular, α(h) ≠ 1 (indeed, if i > 0, then 1 ∈ ̸ H i ); therefore α(g) ≠ 1. (b) If g ∈ G i (i > 0), then α(g) ∼ α(h) ∈ H i , and since α(h) is a π-element, we have α(g) ∈ G i . By (a), this is also true for i = 0. By the definition of G i , g π and g are in the same G i . Since α(g) ∈ G i , this proves (b). (c) If g ∈ G0 , then α(g) = 1 (by (a)) and g k = β(g)k g πk 󸀠 (k ∈ ℕ) since g = α(g)β(g)g π󸀠 . Since all primary parts of β(g)k are of the second type, it follows that α(g k ) = 1 and so g k ∈ G0 (by (a)), and the proof of (c) is complete. (d) If g is a π󸀠 -element in G, then α(g) = 1 and so g ∈ G0 , by (a). Lemma 5.4. Assume that conditions (A1)–(A3) hold. Let ⟨u⟩ × P = E ∈ E, P ∈ Sylp (E), and suppose that E is a π-subgroup. If α(u) ≠ 1, then E x ≤ H for some x ∈ G. Proof. The element u is conjugate to an element of H − H0 (see Lemma 5.2); we may therefore assume that u ∈ H − H0 . By (A3), CH (u) ≥ P∗ ∈ Sylp (CG (u)). Since P ≤ CG (u), P is conjugate in CG (u) to a subgroup of P∗ , by Sylow’s theorem, and therefore E x = (⟨u⟩ × P)x ≤ ⟨u⟩ × P∗ ≤ H, where x ∈ CG (u), P x ≤ P∗ . Lemma 5.5. Assume that condition (A2) holds. Then (A4) is equivalent to the following condition: (A4’) |G i | = |G : H| ⋅ |H i | for all i ∈ {0, 1, . . . , n}. Proof. By Lemma 5.1, one has n

n

|G| = ∑ |G i |,

|H| = ∑ |H i |.

i=0

i=0

Assume, say, that |G i | ≥ |G : H| ⋅ |H i | for all i. Then n

n

|G| = ∑ |G i | ≥ |G : H| ∑ |H i | = |G : H| ⋅ |H| = |G|, i=0

i=0

and so |G i | = |G : H| ⋅ |H i | for all i. The same is true provided |G i | ≤ |G : H| ⋅ |H i | for all i.

390 | Characters of Finite Groups 1 We are ready to prove the following fundamental result. Theorem 5.6 (Brauer [Bra8]). Let H0 ⊲ H < G and π = π(|H : H0 |). Assume that conditions (A1)–(A4) hold. Then H has a normal complement G0 in G over H0 . Proof. Let ϕ ∈ Ch(H/H0 ) (we consider ϕ as an element of Ch(H)). Then (the generalized character) ϕ is constant on the sets H i , i ∈ {0, 1, . . . , n}. Let {h0 , h1 , . . . , h n }, where h i ∈ H i (i = 0, 1, . . . , n), be a fixed complete system of representatives of the sets H i . Define a function ϑ ϕ : G → ℂ by ϑ ϕ (g) = ϕ(h i )

(g ∈ G i ).

The legitimacy of this definition follows from Lemma 5.1 (b) (in fact, G = ∑ni=0 G i is a partition). Since the sets G i are G-invariant (Lemma 5.1 (a)), we get ϑ ϕ ∈ CF[G]. We claim that (ϑ ϕ )H = ϕ. Take h ∈ H. If h ∈ H i , then h ∈ G i (since H i ⊆ G i ). Therefore, ϑ ϕ (h) = ϕ(h i ) = ϕ(h) 󳨐⇒ (ϑ ϕ )H = ϕ. We claim that ϑ ϕ ∈ Ch(G). To do this, we use Theorem 1.2. Let E ∈ E. It suffices to show that (ϑ ϕ )E ∈ Ch(E). We have E = E0 × E1 , where E0 is a π-subgroup and E1 a π󸀠 -subgroup. Let ρ be the projection operator of E onto E0 . If x ∈ E, then x π = ρ(x). By Lemma 5.3 (b), x and x π are in the same set G i . Therefore, ϑ ϕ (x) = ϑ ϕ (x π ) = ϑ ϕ (ρ(x)). Thus, (ϑ ϕ )E = (ϑ ϕ )E0 ∘ ρ. Therefore, it suffice to prove that (ϑ ϕ )E0 ∈ Ch(E0 ). Indeed, let (ϑ ϕ )E0 = ψ1 − ψ2 , where ψ1 , ψ2 ∈ Char(E0 ). Then, for x ∈ E, we have ϑ ϕ (x) = ϑ ϕ (ρ(x)) = ψ1 (ρ(x)) − ψ2 (ρ(x)). Let Γ i be a representation of E0 affording the character ψ i (i = 1, 2). Then, denoting T i (x) = Γ i (ρ(x)) (i = 1, 2), we obtain a representation T i of E whose character is ψ i ∘ ρ. Thus, ψ i ∘ ρ ∈ Char(E). Since (ϑ ϕ )E = ψ1 ∘ ρ − ψ2 ∘ ρ, it follows that (ϑ ϕ )E ∈ Ch(E), and so ϑ ϕ ∈ Ch(G), by Theorem 1.2. Write E0 = U×P, where U = ⟨u⟩ is a p󸀠 -subgroup and P is a p-subgroup. If α(u) ≠ 1, then, in view of Lemma 5.4, we may assume that E0 ≤ H. Then (ϑ ϕ )H = ϕ ∈ Ch(H) 󳨐⇒ (ϑ ϕ )E0 ∈ Ch(E0 ). Now let α(u) = 1. Then, by Lemma 5.3 (a), u ∈ G0 , and therefore, by Lemma 5.3 (c), u m ∈ G0 for any m ∈ ℕ. Hence, α(u m ) = 1 for any m ∈ ℕ, by Lemma 5.3 (a). Let σ be the projection operator of E0 = U × P onto P. Then α(x) = α(σ(x)) for x ∈ E0 . Indeed, putting x = u m v, where v ∈ P, we obtain α(x) = α(u m )α(v) = α(v) = α(σ(x)). Let x ∈ E0 and x ∈ G i . Then α(x) ∈ G i (see Lemma 5.3 (b)) and therefore also α(σ(x)) ∈ G i (since, as we have proved, α(σ(x)) = α(x)). If, in addition, σ(x) ∈ G j , then α(σ(x)) ∈ G j (Lemma 5.3 (b)), and this implies that i = j since, for i ≠ j, the sets G i

VIII Brauer’s induction theorems | 391

and G j are disjoint. Then ϑ ϕ (σ(x)) = ϕ(h i ) = ϑ ϕ (x). Hence, we get (ϑ ϕ )E0 = (ϑ ϕ )P ∘ σ. By Sylow’s theorem and (A1), we may assume without loss of generality that P ≤ H. Since (ϑ ϕ )H = ϕ ∈ Ch(H), it follows that (ϑ ϕ )P ∈ Ch(P) 󳨐⇒ (ϑ ϕ )E0 = (ϑ ϕ )P ∘ σ ∈ Ch(E0 ). As noted previously, this implies that (ϑ ϕ )E ∈ Ch(E). As E was an arbitrary element of the set E, we have ϑ ϕ ∈ Ch(G), by Theorem 1.2. Next, since G = ⋃ni=0 G i is a partition, we have n

⟨ϑ ϕ , ϑ ϕ ⟩ = |G|−1 ∑ |ϑ ϕ (g)|2 = |G|−1 ∑ ∑ |ϑ ϕ (g)|2 g∈G

j=0 g∈G j

n

n

= |G|−1 ∑ ∑ |ϕ(h j )|2 = |G|−1 ∑ |G j ||ϕ(h j )|2 , j=0 g∈G j

j=0

and hence, by Lemma 5.5, n

⟨ϑ ϕ , ϑ ϕ ⟩ = |G|−1 ∑ |G : H| ⋅ |H j | ⋅ |ϕ(h j )|2 j=0 n

= |H|−1 ∑ |H j | ⋅ |ϕ(h j )|2 = |H|−1 ∑ |ϕ(h)|2 = ⟨ϕ, ϕ⟩. j=0

h∈H

Thus, we obtain ⟨ϑ ϕ , ϑ ϕ ⟩ = ⟨ϕ, ϕ⟩. In particular, if ϕ ∈ Irr(H), then ⟨ϑ ϕ , ϑ ϕ ⟩ = 1 and, since ϑ ϕ (1) = ϕ(1) > 0, one obtains ϑ ϕ ∈ Irr(G). We conclude that any irreducible character of H/H0 (considered as a character of H) can be extended to an irreducible character of G. Let ϕ = ρ H/H0 be the regular character of H/H0 . Then ϕ(h0 ) = ϕ(1) = |H/H0 |, ϕ(h i ) = 0

(i > 0)

and therefore (ϑ ϕ )(g i ) = ϕ(h i ) (g i ∈ G i , i ∈ {0, 1, . . . , n}) 󳨐⇒ (ϑ ϕ )G0 = ϕ(1) ⋅ 1G0 = ϑ ϕ (1) ⋅ 1G0 , and ϑ ϕ vanishes on G i for i > 0. Hence we have G0 = ker(ϑ ϕ ) ⊲ G. By Lemma 5.3 (d), all π󸀠 -elements of G are in G0 , so that G/G0 is a π-group. Since GCD(|G : H|, |H : H0 |) = 1 (by condition (A1)), it follows that |G : H|π = 1, so that |G : HG0 |π = 1 too. Since G/G0 is a π-group, we get G = HG0 . Since G0 ∩ H = H0 (Lemma 5.1), this completes the proof. The following remarks simplify the verification of condition (A4). Remark 5.1. Assume that condition (A2) holds. Let |G i | = |G : H||H i | for all i except one; then condition (A4) is true. This follows from Lemma 5.5 and the equality |G| = ∑ni=0 |G i |.

392 | Characters of Finite Groups 1 Remark 5.2. Let G i = ⋃λ M λ be a disjoint union of subsets M λ for some index i with i ∈ {0, 1, . . . , n}. If one can prove that for every λ we have |M λ | ∗ |G : H| ⋅ |M λ ∩ H| (where ∗ stands for one of the symbols ≥ or ≤), then |G i | ∗ |G : H| ∑ |M λ ∩ H| = |G : H| ⋅ |G i ∩ H| = |G : H| ⋅ |H i |. λ

Then condition (A4) is true if such a partition exists for every i ∈ {0, 1, . . . , n}. Remark 5.3. If x ∈ G is a π-element, then the π-section S(x) = {g ∈ G | g π ∼ x}. Let us calculate |S(x)|. Let r π (X) be the number of π󸀠 -elements in the group X. For g ∈ S(x), there are |G : CG (x)| possibilities for g π . If x󸀠 is one of these possibilities, then g π = x󸀠 if and only if g = x󸀠 y, where y is a π󸀠 -element of CG (x󸀠 ) ≅ CG (x); hence there are r π (CG (x)) possibilities for y. Thus, |S(x)| = |G : CG (x)|r π (CG (x)). The following theorem is a partial converse to Theorem 5.6. Theorem 5.7. Let H0 ⊲ H ≤ G, and assume that H has a normal complement G0 in G over H0 . Then condition (A3) is true. If one assumes additionally that conditions (A1) and (A2) hold, where π is the set of all prime divisors of |H : H0 |, then condition (A4) is also true. Moreover, G0 is uniquely determined by the triple (G, H, H0 ). We leave the proof to the reader. Note that, instead of condition (A2), it is better to prove the equivalent condition (A4’) (see Lemma 5.5). Now we show how to deduce the well-known theorem of Wielandt [Wiel2] from Theorem 5.6 (for another proof, see Theorem X.2.2 (b)). Definition. Let Y ⊂ G. If Y ∩ Y x ⊆ {1} for all x ∈ G − NG (Y), then Y is called a TI-subset of a group G. If, in addition, Y < G, then Y is called a TI-subgroup of G. Lemma 5.8. Let Y be a TI-subset of G. Thenthe following statements hold: (a) Y ⊆ NG (Y). (b) y ∈ Y # implies CG (y) ≤ NG (Y). (c) If y1 , y2 ∈ Y # and y1 ∼ y2 in G, then y1 ∼ y2 in NG (Y). Proof. Let y ∈ Y # and x ∈ CG (y). Then y ∈ Y ∩ Y x implies x ∈ NG (Y) and so we have CG (y) ≤ NG (Y), proving (b). Since y ∈ CG (y) for all y ∈ Y, statement (a) follows from (b). Now let y2 = y1x for y1 , y2 ∈ Y # and some x ∈ G. Then y2 ∈ Y ∩ Y x , and hence x ∈ NG (Y) so that y1 ∼ y2 in NG (Y). Theorem 5.9 (= Theorem X.2.2 (b); see [Wiel2]). Let H0 ⊲ H < G and H ∩ H x ≤ H0 for all x ∈ G − H (in particular, NG (H) = H). Then G0 = G − ⋃x∈G (H − H0 )x is a normal complement of H in G over H0 . Proof. Let π = π(|H : H0 |), p ∈ π and P ∈ Sylp (H). Suppose that P ∈ ̸ Sylp (G). Then P < P1 ∈ Sylp (G) and P1 − P contains an element x such that P x = P. Then P ≤ H ∩ H x .

VIII Brauer’s induction theorems | 393

Since P ≰ H0 (indeed, p | |H : H0 |), it follows that x ∈ NG (H) = H, which is a contradiction, since P < ⟨x⟩P, ⟨x⟩P is a p-subgroup of H and P ∈ Sylp (H). Thus, P ∈ Sylp (G) and so GCD(|G : H|, |H : H0 |) = 1, since p ∈ π is arbitrary. In particular, |G : H| is a π 󸀠 -number. This proves that condition (A1) holds. Let h1 , h2 ∈ H and h1 ∼ h2 in G; then h2 = h1x for some x ∈ G so that h2 ∈ H ∩ H x . −1 Assume that x ∈ G − H. Then h2 ∈ H0 . Similarly, h1 = h2x ∈ H0 , since x−1 ∈ G − H. Therefore h1 H0 = H0 = h2 H0 so condition (A2) holds for this case. But if x ∈ H, then h2 H0 = (h1 H0 )x , and again condition (A2) holds. By hypothesis, H − H0 is a TI-subset of G. Therefore, if h ∈ H − H0 , then we have CG (h) ≤ NG (H) = H (Lemma 5.8 (b)), so that CG (h) = CH (h), and condition (A3) holds. Let K be a G-class, K ⊆ G i , where i > 0. Then H ∩ K ≠ 0. Indeed, if g ∈ K, then, by the definition of G i , there exists t ∈ G such that t−1 g π t = u ∈ H i . It follows from H i ∩ H0 = 0 that u ∈ H − H0 , and so CG (u) = CH (u) ≤ H (Lemma 5.8 (b)). Since t−1 g π t is a power of g t , we have g t ∈ CG (u), i.e., g t ∈ H ∩ K, so that H ∩ K ≠ 0. Since, by Lemma 5.1 (c), H ∩ K ⊆ H ∩ G i = H i , it follows that K = K h , the G-class of h, where h ∈ H i . But then (see Lemma 5.8 (b)) |K| = |G : CG (h)| = |G : CH (h)| = |G : H| ⋅ |H : CH (h)| = |G : H| ⋅ |H ∩ K|, since K hH = H ∩ K, where K hH is the H-class of h. By Remarks 5.1 and 5.2, this means that (A4) is true. Therefore, by Theorem 5.6, the set G0 is a normal complement of H in G over H0 . It remains to prove that G0 = D, where D = G − ⋃x∈G (H − H0 )x . Since H − H0 is a TI-subset of G and NG (H − H0 ) = H (clearly, H − H0 is an H-invariant subset) and |H : H0 | = |G : G0 |, it follows that 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 |D| = 󵄨󵄨󵄨󵄨G − ⋃ (H − H0 )x 󵄨󵄨󵄨󵄨 = |G| − 󵄨󵄨󵄨󵄨 ⋃ (H − H0 )x 󵄨󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 x∈G

x∈G

= |G| − |G : H| ⋅ |H − H0 | = |G : H|[|H| − |H − H0 |)] = |G : H| ⋅ |H0 | = |G0 |. Next, G0 ∩ (H − H0 )x = 0 for any x ∈ G (as G0 ⊲ G and G0 ∩ (H − H0 ) = 0). Therefore, G0 ⊆ D. As the cardinalities of the sets on both sides of the sign ⊆ are equal, we finally obtain G0 = D. Let us show that, under hypothesis of Theorem 5.9, G0 is the unique normal complement of H over H0 . Assume that G1 is another normal complement of H over H0 . Take x ∈ G−NG (H). Then (H −H0 )x ∩G1 = 0 since G1 ⊲ G. It follows that ⋃x∈G−NG (H) (H −H0 )x has empty intersection with G1 so that G1 ≤ G0 (see the proof of Theorem 5.9). Since G/G1 ≅ H/H0 ≅ G/G0 , we get |G1 | = |G0 |, and we conclude that G1 = G0 , as claimed. Letting H0 = {1}, we obtain the following one of most important results of finite group theory. Theorem 5.10 (Frobenius). Let H be a self-normalizing TI-subgroup of G. Then G contains a normal subgroup F such that HF = G and H ∩ F = {1}.

394 | Characters of Finite Groups 1 Theorem 5.11 (Brauer). Let π be an arbitrary set of primes. The following conditions are necessary and sufficient for the existence of a normal π-complement in G: (B1) G contains a π-Hall subgroup H. (B2) If two elements of H are conjugate in G, they are also conjugate in H. (B3) Any elementary π-subgroup of G is conjugate to a certain subgroup of H. This follows also from Theorem 6.1, below. The following weaker version of Theorem 5.11 was proved by Suzuki [Suz7]. Corollary 5.12. Let H be a nilpotent π-Hall subgroup of a group G. If the pair H < G satisfies (B2) of Theorem 5.10, then G has a normal π-complement. If H is a nilpotent π-Hall subgroup of G, then all maximal π-subgroups of G are conjugate (Wielandt) so Corollary 5.12 follows from Theorem 5.11. Corollary 5.13 (Burnside). Let H be a Hall subgroup of G. If CG (H) = NG (H), then H has a normal complement in G. Proof. Let x, y ∈ H be conjugate in G. Since H is abelian, it is sufficient to prove that x = y (see Corollary 5.12). We have for some g ∈ G, x = yg ,

H ≤ CG (x),

H g ≤ CG (y g ) = CG (x).

Therefore, by Wielandt’s D-Theorem, the subgroups H and H g are conjugate in CG (x), and so we have H gz = H for some z ∈ CG (x). We conclude that gz ∈ NG (H) = CG (H). Next, x = y g = (y g )z = y gz . Since y ∈ H, it follows that x = y gz = y. The following lemma is used in the proof of Theorem 5.15 (which we will leave to the reader). Lemma 5.14. If r p (G) is the number of p󸀠 -elements contained in a group G, then p does not divide r p (G). Proof. Indeed, P ∈ Sylp (G) acts via conjugation on Σ p , the set of p󸀠 -elements in G. The length of each P-orbit on Σ p is a p-power. All orbits of length 1 constitute the p󸀠 -Hall subgroup of CG (P) = Z(P) × Op󸀠 (CG (P)). Since |Σ p | ≡ |Op󸀠 (CG (P))| ≢ 0 (mod p) (Lagrange), the proof is complete. Theorem 5.15. Let H0 ⊲ H < G and let π be the set of all prime divisors of |H : H0 |. The following condition is sufficient for the existence of a normal complement to H in G over H0 : (B) If U is an abelian π-subgroup, then for every conjugacy class K in CG (U) that consist of π-elements, its intersection with H is a conjugacy class of CH (U). Exercise. Let χ be an irreducible character of a group G. For g ∈ G and the set π of primes, write g = g π g π󸀠 = g π󸀠 g π . Define the class function χ π as follows: χ π (g) = χ(g π󸀠 ). Is it true that χ π is a generalized character of G and there exists a group G such that χ π ∈ ̸ Char(G)?

VIII Brauer’s induction theorems | 395

6 Existence of a normal complement to a section Let M be a subset of a group G. If a function φ : G → ℂ takes a constant value t on elements of M, we write t = φ(M). Definition. Let H0 ⊲ H < G and H = H/H0 . If the triple (G, H, H0 ) satisfies condition (A2) of §VIII.5, it is called a B-triple. Assume that H = H/H0 is a π-group. We say that the B-triple (G, H, H0 ) is a B π -triple if, in addition, the set ⋃x∈G H x contains all the π-elements of the group G. Let the sets H0 , H1 , . . . , H n and G0 , G1 , . . . , G n be defined as in §VIII.5. Let us assume from now on that (G, H, H0 ) is a B π -triple. Then H = ⋃ni=0 H i and G = ⋃ni=0 G i are partitions (Lemma 5.1 (b)). For any φ ∈ Irr(H) (considered as a character of H), we define a class function φ∗ : G → ℂ by φ∗ (G i ) = φ(H i ),

i ∈ {0, 1, . . . , n}.

(1)

This definition makes sense since (G, H, H0 ) is a B π -triple. As G i ∩ H = H i , the restriction of φ∗ to H coincides with φ: φ∗H = φ. Theorem 6.1. Let (G, H, H0 ) be a B π -triple. If the function φ∗ defined in (1) is a generalized character of G for any φ ∈ Irr(H), then φ∗ ∈ Irr(G),

G0 ⊲ G,

G = HG0 ,

H ∩ G0 = H0 .

Briefly, the section H is normally complementable. Proof. Since φ∗H = φ, φ∗ is a nonzero function. Assume that we have already proved that φ∗ ∈ Irr(G) for all φ ∈ Irr(H). Put N = ⋂φ∈Irr(H) ker(φ∗ ). Next, φ∗ (G0 ) = φ(H0 ) = φ(1) 󳨐⇒ G0 ⊆ N, and since NH ≤ G and the conjugates of NH exhaust G, we get n

⋃ (NH)x ⊇ ⋃ G i = G 󳨐⇒ NH = G.

x∈G

i=1

H0 and φ∗H

= φ for all φ ∈ Irr(H)), we have N ∩ H = H0 . ThereSince ⋂φ∈Irr(H) ker(φ) = fore, |N| = |H0 | ⋅ |G : H|, by the product formula. Let ρ H be the regular character of H, considered as a character of H. Then ρ H (H0 ) = |H| and ρ H (H i ) = 0 for i > 0. The mapping ∗ can be defined in a natural way on all the characters of H. Then ρ∗ (G0 ) = |H| H and ρ∗ (G i ) = 0 for i > 0. Let us calculate the inner product ⟨ρ∗ , 1G ⟩: H

H

n

⟨ρ∗ , 1G ⟩ = |G|−1 ∑ |G i |ρ∗ (G i ) = |G|−1 ⋅ |G0 |ρ∗ (G0 ) = H

i=0

H

H

|G0 | . |H0 | ⋅ |G : H|

As the algebraic integer ⟨ρ∗ , 1G ⟩ is rational and positive, hence a natural number, we H get |G0 | ≥ |H0 | ⋅ |G : H| = |N|, and thus G0 = N, and the theorem is proved.

396 | Characters of Finite Groups 1 It remains to prove that φ∗ ∈ Irr(G) for all characters φ ∈ Irr(H). To this end, assume that |G i | ≥ |H i | ⋅ |G : H| for all i (see Lemma 5.5 and Remark 5.2). Then, by Lemma 5.5, |G i | = |H i | ⋅ |G : H| for all i. By the First Orthogonality Relation, n

⟨φ∗ , φ∗ ⟩ = |G|−1 ∑ |G i | ⋅ |φ∗ (G i )|2 = i=0 n

|G : H| n ∑ |H i | ⋅ |φ(H i )|2 |G| i=0

= |H|−1 ∑ |H i | ⋅ |φ(H i )|2 = ⟨φ, φ⟩ = 1. i=0

Since = φ(1) > 0, it follows that φ∗ ∈ Irr(G). It remains to prove that |G i | ≥ |H i | ⋅ |G : H| for all i. Let us generalize our reasoning regarding ρ H . Define the function ψ i : H → ℂ as follows: φ∗ (1)

ψi =

φ(H i )φ∗ ,

∑

i = 0, 1, . . . , n.

φ∈Irr(H)

It follows from (1) and the Second Orthogonality Relation for H that ψ i (G j ) =

φ(H i )φ∗ (G j ) =

∑ φ∈Irr(H)

so ψ i (G j ) = 0 if j ≠ i and ψ i (G i ) =

φ(H i )φ∗ (H j )

φ∈Irr(H) |H| |H i |

=

|H| |H i | .

Therefore,

n

⟨ψ i , 1G ⟩ = |G|−1 ∑ |G j |ψ i (G j ) = j=0

∑

|G i | |G i | ψ i (G i ) = |G| |H i | ⋅ |G : H|

Thus, the algebraic integer ⟨ψ i , 1G ⟩ is positive and rational, and therefore we obtain that |G i | ≥ |H i | ⋅ |G : H| for all i. Exercise 6.1 (Dade [Dad5]). Let (G, H, H0 ) be a B π -triple. If any Brauer elementary π-subgroup of a group G is conjugate to a subgroup of H, then the section H = H/H0 is normally complementable. By Theorem 6.1, it is sufficient to prove that φ∗ ∈ Irr(G) for all φ ∈ Irr(H). But this follows from Theorem 1.2. Exercise 6.2. If (G, H, {1}) is a B π(H) -triple and φ∗ is a generalized character of G for all φ ∈ Irr(H), then H has a normal complement in G. Note that here H is not necessarily a π-Hall subgroup. Moreover, in Exercise 6.1 we have GCD(|G : H|, |H : H0 |) = 1, but this is not necessarily true in Theorem 6.1.

IX Faithful characters The first six sections are written by the third author, §IX.7 is written by the first author. We use in §IX.7 some results of Chapters X and XIV.

1 Introduction The goal of this chapter is to give a group-theoretic description of kernels of irreducible ℂ-characters of a finite group G. A subgroup N ⊴ G is the kernel of an irreducible character of G if and only if G/N has a faithful irreducible character. Therefore, the above problem is equivalent to one of finding criteria for the existence of faithful irreducible characters of a given group. The existence of a faithful character in Irr(G) restricts the structure of G. In that case, Z(G) is cyclic. First suppose that G is abelian. Then G has a faithful irreducible character if and only if it is cyclic. In that case, the number of faithful irreducible characters of G is φ(|G|), where φ is the Euler totient function (if G is noncyclic, that number is equal to 0). It is natural to consider, on the class of all finite abelian groups, the grouptheoretic Euler function φ(∗), which possesses the following properties: (i) The function φ is constant on classes of isomorphic abelian groups, (ii) φ(G) is the number of faithful irreducible characters of a group G. We continue to suppose that G is an abelian group. It follows that a subgroup N of G is a kernel of an irreducible character of G if and only if φ(G/N) ≠ 0 (in that case, G/N is cyclic). Note that G is cyclic if and only if Sc(G), the socle of G, is cyclic. Recall that the socle of an arbitrary group G is the product of its minimal normal subgroups. It is known that the lattice L(G) of subgroups of an abelian group G is autodual (see Chapter I): there exists an anti-automorphism α of the lattice L(G) such that H α ≅ G/H and G/H α ≅ H for each H ≤ G. The anti-automorphism α of G sends cyclic subgroups to kernels of irreducible characters, i.e., those subgroups H ≤ G for which G/H is cyclic (the latter will simply be called kernels; then it is natural the cyclic subgroups that are dual to kernels, to call antikernels). It follows from the above that, in any abelian group, the number of kernels is equal to the number of antikernels. Then the abelian group possesses a faithful irreducible character if and only if its socle is an antikernel if and only if the socle is cyclic. It is easily seen that the socle of an abelian group G is cyclic if and only if G itself is cyclic. Note also that H ≤ G is an antikernel if and only if φ(|H|) > 0. (It follows that antikernels of an abelian group G are cyclic and, conversely, all cyclic subgroups of G are antikernels.) It is possible to extend the above notions (kernels, antikernels, group theoretic Euler’s function) to arbitrary finite groups. As above, the kernels of irreducible characters are called kernels of G. The subgroup generated by a G-class is called an antikernel DOI 10.1515/9783110224078-009

398 | Characters of Finite Groups 1 of G (i.e., a subgroup K is an antikernel of G if and only if K = C G , the normal closure of a cyclic subgroup C of G). Denote by LN (G) the lattice of all normal subgroups of G. An anti-automorphism α of the lattice LN (G) is said to be strong provided |N α | = |G/N| if and only if |G/N α | = |N| for each N ∈ LN (G). It appears that, as in the case of abelian groups, the anti-automorphism α sends kernels to antikernels and vice versa (however, in the general case, this is not obvious). Moreover, the results obtained above for abelian groups, remain to be true in general case. In particular, in arbitrary finite group, the number of kernels is equal to the number of antikernels. Example. Let G = S4 . Then there are exactly four kernels, their orders are 1, 4, 12, 24 (thus, all normal subgroups of G are kernels). In the case under consideration, these four subgroups are also antikernels since any of these subgroups is generated by a G-class. Exercise. Describe the kernels and antikernels of the nonabelian metacyclic p-group of order p4 and exponent p2 . In what follows, the role of certain generalizations of the Euler function is essential. For every group G we define two (in some sense, dual) group-theoretic Euler functions (1) (2) φ G (of the first kind) and φ G (of the second kind). These functions are defined on certain class of finite groups connected with G (the groups of that class are called in what follows G-groups).

2 G-groups 1°. Let G be a (finite) group with the identity element 1 = 1G and H a finite group. Suppose that the left action γ : G × H → H of G on H is defined (we write γ(g, h) = g ∘ h for g ∈ G and h ∈ H) and satisfies the following conditions: (g1 g2 ) ∘ h = g1 ∘ (g2 ∘ h),

1G ∘ h = h

(g1 , g2 ∈ G, h ∈ H).

(1)

For a fixed g ∈ G, the mapping h 󳨃→ g ∘ h (h ∈ H) is a permutation π(g) of the set H; the mapping g 󳨃→ π(g) is a homomorphism of G into the symmetric group SH of all permutations of H. Let us check, that the mapping h 󳨃→ g ∘ h is a permutation of the set H. Assume that for h1 ∈ H, we have g ∘ h = g ∘ h1 . Then h = 1G ∘ h = (g−1 g) ∘ h = g −1 ∘ (g ∘ h) = g −1 ∘ (g ∘ h1 ) = 1G ∘ h1 = h1 , proving our claim since |H| < ∞. Next, if g, g1 ∈ G and h ∈ H, then π(gg1 )(h) = (gg1 ) ∘ h = g ∘ (g1 ∘ h) = (π(g)π(g1 ))(h) and we conclude that π(gg1 ) = π(g)π(g1 ), i.e., π : G → SH is a homomorphism.

IX Faithful characters | 399

Definition 2.1. A group H is said to be a G-group if the action γ of G on the set H satisfies the following conditions: (i) The permutation π(g) is an automorphism of the group H for all g ∈ G. (ii) Given ψ ∈ Inn(H), there exists g ∈ G such that π(g) = ψ. Thus, a G-group H in a wide sense is a pair (H, γ), where γ is an action of G on H satisfying conditions (i) and (ii) of Definition 2.1. Later we consider a partial case of this notion. A subgroup H1 of the G-group H is said to be admissible if g ∘ H1 = {g ∘ h1 | h1 ∈ H1 } = H1 for all g ∈ G. G

In that case, we write H1 ≤ H. Obviously, every admissible subgroup H1 of the G-group (H, γ) can be converted in a G-group (H1 , γ1 ), where γ1 is the restriction of γ to H1 . We call (H1 , γ1 ) a G-subgroup of the G-group (H, γ). By Definition 2.1 (ii), all G-subgroups of H are H-invariant. Note also that the set LG (H) of all G-subgroups of a G-group H is a lattice. Obviously, LG (H) is a sublattice of the lattice L(H) of all subgroups of H and the lattice LN (H) of all normal subgroups of H (it is possible that LG (H) ≠ LN (H) since, generally speaking, not all normal subgroups of H are G-admissible). G Let H1 ≤ H. Set H = H/H1 . Let us define the action γ of G on H setting g∘h=g∘h

(g ∈ G, h ∈ H).

It is clear that γ is well-defined and the pair (H, γ) is a G-group. Thus, the quotient group H/H1 becomes a G-group. The class of G-groups is not empty since, for example, we can consider G itself as a G-group, defining an action of G on G by conjugation. In what follows we consider G as a G-group in such sense. In the case under consideration, all normal subgroups and all quotient groups of G are G-groups. Suppose that H and K are G-groups. A homomorphism f : H → K is said to be a G-homomorphism provided f(g ∘ h) = g ∘ f(h)

(g ∈ G, h ∈ H),

i.e., f and the action γ are permutable. It is clear how to define G-isomorphisms, G-automorphisms and G-endomorphisms of G-groups. For G-groups hold all basic homomorphism, isomorphism theorems and also the Jordan-Hölder theorem. Let I = IG denote the class of G-groups of order 1. A G-group H ∈ ̸ I is said to be irreducible if LG (H) = {E H , H}, where E H = {1H } (in other words, H has no nontrivial G-subgroups). The irreducible G-groups are characteristically simple and so are direct products of isomorphic simple groups. A G-subgroup F ≠ {1H } of a G-group H is said to be minimal G-subgroup of H ∈ ̸ IG if LG (F) = {{1H }, F}; in that case F itself is an irreducible G-group. Let MH denote the set of all minimal G-subgroups of a G-group H; if H ∈ IG , then the set MH is empty. A G-group H is said to be completely reducible if either H ∈ IG or H is a direct product of irreducible G-subgroups. A G-subgroup N

400 | Characters of Finite Groups 1 of a G-group H is said to be complemented (in H) if H = N × K for some G-subgroup K of H (in particular, {1H } and H are complemented). It is easy to show that if all G-subgroups of a G-group H are complemented, then H is completely reducible (see Proposition 2.1 (c)). If U, V are distinct irreducible subgroups of a G-group H, then U ∩ V = {1}. If U < N < H, where N is admissible in H and U is admissible in N, then U is admissible in H, i.e., admissibility is a transitive relation (it should note that the normality is not a transitive relation). Proposition 2.1. For a G-group H, the following assertions are equivalent: (a) H is completely reducible. (b) H is a product (not necessarily direct) of a family of irreducible G-subgroups. (c) All G-subgroups of H are complemented. (d) All minimal G-subgroups of H are complemented. The proof is analogous of the proof of Lemma I.6.1. Let us show, for example, that (d) ⇒ (a). We use induction on |H|. Let U1 < H be minimal admissible. Then, by hypoG thesis, we have H = U1 × V1 for some V1 ≤ G. If U2 < V1 is minimal admissible, then G H = U2 × V2 for some V2 ≤ G so, by the modular law, V1 = U2 × (V1 ∩ V2 ). Thus, all minimal admissible subgroups of V1 are complemented so, by induction, V1 is completely reducible. Then H = U1 × V1 is completely reducible too, and we are done. Remark. Let a G-group H = F1 ⋅ ⋅ ⋅ F n , where F i ∈ MH for all i (here MH is the set of minimal G-subgroups of a G-group H). Then H is completely reducible, by Proposition 2.1. Moreover, H = F i1 × ⋅ ⋅ ⋅ × F i k for some {i1 , . . . , i k } ⊆ {1, . . . , n} (note that, generally, the subset {i1 , . . . , i k } may be chosen in more than one way). However, if all F1 , . . . , F n are nonsolvable, then {i1 , . . . , i k } = {1, . . . , n}. In that case, if F ∈ MH , then F ∈ {F1 , . . . , F n }. Indeed, if this is false, then we have F ∩ F i = {1} for all i so that CG (F) ≥ F1 ⋅ ⋅ ⋅ F n = G and F is abelian, a contradiction. G

Proposition 2.2. If a G-group H is completely reducible and N ≤ H, then N and H/N are also completely reducible. Proof (compare with Corollary I.6.3). Let U < N be a G-subgroup. By assertion (c) of Proposition 2.1, H = U × V for some G-subgroup V of H. By the modular law, we have N = U × (N ∩ V), where N ∩ V is a G-subgroup since LG (H) is a lattice. By Proposition 2.1, N is completely reducible. It remains to show that H/N is completely reducible. One may assume that N < H. Let K/N < H/N be admissible; then K is admissible. By Proposition 2.1, H = K × L for some admissible L < H. Then H/N = (K/N) × (LN/N), and we are done. Let H = F1 × ⋅ ⋅ ⋅ × F n be a direct product of irreducible G-groups. It follows from the Jordan–Hölder theorem that the factors F i are determined uniquely, up to their ordering and G-isomorphisms. In particular, the number n is also determined uniquely. We call this number the G-length of H and write n = lG (H). If H ∈ IG , then we have, by definition, lG (H) = 0.

IX Faithful characters | 401

Let H be an abelian G-group and EndG (H) the set of its G-endomorphisms. Let us define in EndG (H) the addition and multiplication as follows. If ϕ, ψ ∈ EndG (H), x ∈ H, then (ϕ + ψ)(x) = ϕ(x) ⋅ ψ(x),

(ϕψ)(x) = ϕ(ψ(x)).

It is easy to see that EndG (H) is an associative ring with the identity element. Indeed, the neutral element under addition is the mapping x → 1 = 1H , where 1H is the identity element of H; the neutral element under multiplication is the mapping idH : x → x (x ∈ H). Let H = F be an irreducible abelian G-group. Since F is elementary abelian p-group for some prime p, it follows that |F| = p d , where d = d(F), the rank of F. We may consider F as an irreducible 𝔽p G-module of dimension d (here 𝔽p = GF(p) is the prime field of characteristic p, |𝔽p | = p). By Schur’s lemma, the ring Ω = EndG (F) is a skew field. By the Wedderburn theorem on finite skew fields, Ω = GF(q), the finite field with q = p w elements (since Ω is an extension of 𝔽p ), where w is the degree of Ω over 𝔽p (= dim𝔽p (Ω)). We claim that w | d. Indeed, one can consider F as an ΩG-module. Let j = dimΩ (F) be the Ω-dimension of F. If {f1 , . . . , f j } is an Ω-basis of F and let {ω1 , . . . , ω w } be an 𝔽p -basis of the 𝔽p -vector space Ω, then the products f r ω s form an 𝔽p -basis of the 𝔽p -vector space F, implying jw = d, as claimed. Note that |F| = p d = q j . Definition 2.2. The product of all irreducible G-subgroups of a G-group H ∈ ̸ IG is said to be the G-socle of H (we denote it by ScG (H); if H ∈ IG , then, by definition, ScG (H) = H). In particular, ScG (G) coincides with usual socle of G. A proper G-subgroup M of a G-group H is said to be maximal if the quotient group H/M is irreducible (in that case, H/M is characteristically simple). Definition 2.3. Let H ∈ ̸ IG be a G-group. Denote by H∗ the intersection of all maximal G-subgroups of H. Then AsG (H) = H/H∗ is called the G-antisocle of H. If H ∈ IG , then, by definition, AsG (H) = H. Both G-groups ScG (H) and AsG (H) are completely reducible. For the first of them this follows from Proposition 2.1. For the G-group AsG (H) this follows from Lemma 2.3. A G-group H is completely reducible if and only if H∗ = E H = {1H }. Proof. (i) Suppose that H = F1 ×⋅ ⋅ ⋅× F n is a decomposition of the completely reducible G-group H in direct product of irreducible G-subgroups. The subgroup M i = ∏j=i̸ F j is n a maximal G-subgroup of H since H/M i ≅ F i is irreducible. Since ⋂1=1 M i = E H , we conclude that H∗ = E H . (ii) Supposing that H∗ = E H , we have to show that H is completely reducible. In view of Proposition 2.1 (d), it suffices to show that every minimal admissible subgroup U of H is complemented. By hypothesis, there is in H a maximal admissible subgroup M not containing U so, since U ∩ M = {1}, we get H = U × M. Thus, U is complemented in H so H is completely reducible (Proposition 2.1).

402 | Characters of Finite Groups 1 Corollary 2.4. The G-antisocle of an arbitrary G-group H is completely reducible. Proof. The intersection of all maximal G-subgroups of the G-group AsG (H) = H/H∗ is trivial. Therefore, by Lemma 2.3, AsG (H) is completely reducible. Lemma 2.5. Let D be a G-subgroup of a G-group H. Then H/D is completely reducible if and only if H∗ ≤ D. Proof. (i) If H∗ ≤ D, then H/D is completely reducible as an epimorphic image of the completely reducible G-group H/H∗ , by Proposition 2.2. (ii) Let H/D be completely reducible. Then (H/D)∗ is the identity group, and so D is the intersection of some maximal G-subgroups of H so H∗ ≤ D. 2°. Let F be an irreducible G-group. Definition 2.4. A completely reducible G-group H > {1} is said to be homogeneous of type [F] if all irreducible G-subgroups of H are G-isomorphic to the irreducible G-group F. By definition, H ∈ IG is a homogeneous G-group. Obviously, homogeneous G-groups are direct products of isomorphic simple groups so characteristically simple. The following two assertions are similar to corresponding assertions on completely reducible modules from Chapter I. Proposition 2.6. Let a completely reducible G-group H = F1 ×⋅ ⋅ ⋅×F n , where F1 , . . . , F n are irreducible G-groups. If F is an irreducible G-subgroup of H, then F is G-isomorphic to F i for some i ∈ {1, . . . , n}. Moreover, F ≤ ∏ F i , where F i runs over all direct factors of H that are G-isomorphic to F. We leave the easy proof of this proposition to the reader. Corollary 2.7. If all direct factors F i of Proposition 2.6 are G-isomorphic, then H is homogeneous. It follows that a completely reducible G-group H is homogeneous if and only if all members of the set MH of all irreducible G-subgroups of H are G-isomorphic. Definition 2.5. Let H be a completely reducible G-group. Maximal homogeneous G-subgroups of H are called homogeneous components of H. Definition 2.6. Let a completely reducible G-group H = L1 ×⋅ ⋅ ⋅× L k , where L1 , . . . , L k are G-subgroups. If MH = ⋃ki=1 ML i , we write H = L1 ∘ ⋅ ⋅ ⋅ ∘ L k . Proposition 2.8. If {H1 , . . . , H s } is the set of homogeneous components of a completely reducible G-group H, then H = H1 ∘ ⋅ ⋅ ⋅ ∘ H s . Proof. Repeating word for word the proof of Lemma I.10.3, we prove that H is equal to H1 × ⋅ ⋅ ⋅ × H s . Let H = F1 × ⋅ ⋅ ⋅ × F n , where F1 , . . . , F n are irreducible G-subgroups. If F ∈ MH , then, by Proposition 2.6, F ≤ T = ∏ F α , where F α runs through all those

IX Faithful characters | 403

irreducible G-subgroups of H that are G-isomorphic to F. By Corollary 2.7, T ≤ H j for some j ∈ {1, . . . , s}. Then F ≤ MH j , i.e., F ∈ ⋃si=1 MH i . Therefore, MH = ⋃ MH i , and this is the same as to say that H = H1 ∘ ⋅ ⋅ ⋅ ∘ H s . Exercise 2.1. Suppose that H is a homogeneous completely reducible group. Prove that H ≠ L1 ∘ L2 , where L1 , L2 are nontrivial G-subgroups of H. Hint. Let F i ∈ ML i , where i = 1, 2. Then there exists a G-isomorphism ϕ : F1 → F2 since F1 ≅ F2 . Consider the set F = {xϕ(x) | x ∈ F1 }. Show that F ∈ MH , however F ∈ ̸ ML1 ∪ ML2 . Proposition 2.9. Let H be a homogeneous completely reducible G-group of G-length lG (H) = n > 1. Then H is an elementary abelian p-group for some prime number p. Proof. Since H is completely reducible, it suffices to show that every irreducible G-subgroup F of H is abelian. Since n > 1, there exists an irreducible G-subgroup F1 of H, F1 ≠ F. Since H is homogeneous, there exists a G-isomorphism ϕ : F → F1 . Let y ∈ F. By Definition 2.1 (ii), there exists g ∈ G such that yhy−1 = g ∘ h for all h ∈ H. If, in particular, h ∈ F, then, using the elementwise permutability of irreducible G-subgroups F and F1 , we get ϕ(yhy−1 ) = ϕ(g ∘ h) = g ∘ ϕ(h) = yϕ(h)y−1 = ϕ(h) for all h ∈ F since y ∈ F and ϕ(h) ∈ F1 . Therefore, yhy−1 = h, i.e., the subgroup F is abelian since y, h ∈ F are arbitrary. In that case, H is elementary abelian p-group for some prime p (since H is homogeneous). Corollary 2.10. A nonabelian homogeneous completely reducible G-group H is irreducible. Definition 2.7. Let H be an abelian homogeneous completely reducible G-group of type [F0 ], where F0 is an abelian irreducible G-group. Let Ω = EndG (F0 ) be the field of G-endomorphisms of F0 (Schur’s lemma and Wedderburn’s theorem). The numbers n = lG (H), q = |Ω| and j = dimΩ (F0 ) are called the first, second, and third G-invariants of H, respectively. Recall that n, the G-length of a completely reducible G-group H (Definition 2.7), is equal to the number of factors in decomposition of H in direct product of irreducible G-subgroups; F0 is the elementary abelian group of order q = p w , where w is the degree of Ω over its prime subfield 𝔽p . Since |F0 | = q j , we get |H| = |F0 |n = q nj = p nwj . Let, as above, LG (H) be the lattice of G-subgroups of a G-group H. Recall that a mapping α : LG (H) → LG (H) is called an anti-automorphism if α is bijective and reverses inclusions, i.e., L1 , L2 ∈ LG (H) and L1 ⊆ L2 󳨐⇒ L1α ⊇ L2α . If, in addition, α2 = idLG (H) , then α is said to be involutory anti-automorphism.

404 | Characters of Finite Groups 1 Proposition 2.11. Let H be a completely reducible G-group. Then the lattice LG (H) possesses an involutory anti-automorphism α such that D α is G-isomorphic to H/D for any G-subgroup D of H. Since α2 = idLG (H) , it follows, together with the condition D α ≅ H/D, that also the condition H/D α ≅ D is fulfilled. To prove Proposition 2.11, we have to clear up the structure of the lattice LG (H). Note that if H = H1 ∘ ⋅ ⋅ ⋅ ∘ H s , where H1 , . . . , H s are homogeneous components of the G-group H, then LG (H) = LG (H1 ) × ⋅ ⋅ ⋅ × LG (H s ): every G-subgroup K ≤ H is presented in the form K = K1 ∘ ⋅ ⋅ ⋅ ∘ K s uniquely, where K i = K ∩ H i for all i. Therefore, at first we have to consider the case where H is homogeneous, i.e., is not decomposed in nontrivial ∘-product. In view of Corollary 2.10, if H is nonabelian, it is irreducible, i.e., LG (H) = {E H , H}, and the assertion is trivial. Now suppose that H is abelian homogeneous completely reducible G-group of type [F0 ], where F0 is an irreducible abelian G-group (of prime power order). Let n = lG (H) > 1 (then H is a direct product of n irreducible G-groups that are G-isomorphic with F0 ) and Ω = EndG (F0 ) (as we have noted, Ω is a finite field). We will make use of the construction that we have applied in the proof of Wedderburn’s theorem in Chapter I. Letting V H = HomG (F0 , H), we define the addition in V H as follows: θ1 , θ2 ∈ V H and x ∈ F0 󳨐⇒ (θ1 + θ2 )(x) = θ1 (x) ⋅ θ2 (x). It is easy to see that V H becomes a finite additive abelian group with respect to this addition. Its neutral element coincides with the “zero homomorphism” 0 which is defined by 0(x) = 1H ∈ H for all x ∈ F0 . Since F0 is irreducible, every nonzero homomorphism θ ∈ V H is a monomorphism (indeed, the kernel of this homomorphism is a G-subgroup of the irreducible group F0 ), and so θ(F0 ), the image of θ, is G-isomorphic to F0 . Let θ ∈ V H (= HomG (F0 , H)) and ξ ∈ Ω(= EndG (F0 )). Then θξ , which is determined by (θξ)(x) = θ(ξ(x)) (x ∈ F0 ), is an element of the abelian group V H . Thus, multiplication of elements θ of the abelian group V H by elements ξ of the field Ω is defined, and so V H is a right linear Ω-space. Let H = F1 × ⋅ ⋅ ⋅ × F n , where F1 , . . . , F n are irreducible (G-isomorphic) G-groups. Let us prove that dimΩ (V H ) = n.

(2)

To this end, we construct an Ω-basis of the space V H . Since H is homogeneous of type [F0 ], there exists, for every i ∈ {1, . . . , n}, a G-monomorphism θ i ∈ V H such that θ i (F0 ) = F i . We claim that the set {θ1 , . . . , θ n } is a basis of the linear Ω-space V H . Assume that θ1 ξ1 + ⋅ ⋅ ⋅ + θ n ξ n = 0 for some ξ1 , . . . , ξ n ∈ Ω. It follows that θ i (ξ1 (x)) ⋅ ⋅ ⋅ θ n (ξ n (x)) = 1H

(x ∈ F0 )

IX Faithful characters | 405

(see the definition of the neutral element 0 of V H ). Since θ i (ξ i (x)) ∈ F i , we get θ i (ξ i (x)) = 1H

for all i.

In that case, for all i, we have ξ i (x) = 1F0 since θ i is a monomorphism. This implies that ξ1 = ⋅ ⋅ ⋅ = ξ n = 0, since x ∈ F0 is arbitrary. Thus, θ1 , . . . , θ n are linearly independent over Ω. It remains to show that the elements θ1 , . . . , θ n span the Ω-space V H . We have to show that an arbitrary element θ ∈ V H is a linear combination of elements θ1 , . . . , θ n . If x ∈ F0 , then θ(x) = y1 . . . y n , where y i ∈ F i for all i. Since θ i : F0 → F i is a G-isomorphism, we can write y i = θ i (x i ), where x i ∈ F0 , i = 1, . . . , n. It follows that x i is determined uniquely by y i and so by x. As it is easy to see, the mapping ξ i : x → x i is a G-endomorphism of F0 , i.e., ξ i ∈ Ω. Therefore, θ(x) = θ1 (x1 ) ⋅ ⋅ ⋅ θ n (x n ) = θ1 (ξ1 (x)) ⋅ ⋅ ⋅ θ n (ξ n (x)) = (θ1 ξ1 + ⋅ ⋅ ⋅ + θ n ξ n )(x), and hence θ = θ1 ξ1 + ⋅ ⋅ ⋅ + θ n ξ n is a linear combination of the elements θ i . This completes the proof of (2). Denote by L(V) the lattice of subspaces of a linear space V. Lemma 2.12. Let H be a homogeneous completely reducible abelian G-group. Then there exists an isomorphism σ : LG (H) → L(V H ) such that dimΩ (D σ ) = lG (D) for all D ∈ LG (H). Proof. Letting (recall that V H = HomG (F0 , H)) V D = {θ ∈ V H | θ(F0 ) ⊆ D}

(D ∈ LG (H)),

we obtain a mapping σ : D 󳨃→ V D of LG (H) into L(V H ). Since V D = iHomG (F0 , D), where i : D → H is an inclusion, we have, by the above, dimΩ (V D ) = lG (D). Since D is completely reducible, we have D = ∏ θ(F0 ). θ∈V D

Therefore, D1 , D2 ∈ LG (H), V D1 = V D2 implies D1 = D2 . This means that σ is injective. Let us prove that the mapping σ is also surjective. To this end, let L ∈ L(V H ). Setting D = ∏θ∈L θ(F0 ), we obtain D ∈ LG (H). Let lG (D) = d and dimΩ (L) = l. If θ ∈ L, then θ(F0 ) ⊆ D, i.e., θ ∈ V D . Hence, L ⊆ V D , and we get l = dimΩ (L) ≤ dimΩ (V D ) = d. Let {θ1 , . . . , θ l } be a basis of the space L. If θ ∈ L, then l

θ = ∑ θi ξi ,

where ξ1 , . . . , ξ l ∈ Ω(= EndG (F0 )).

i=1

Therefore, l

l

θ(F0 ) ⊆ ∏ θ i (F0 ) 󳨐⇒ D = ∏ θ(F0 ) ⊆ ∏ θ i (F0 ). i=1

θ∈L

i=1

406 | Characters of Finite Groups 1 As, on the other hand, ∏li=1 θ i (F0 ) ⊆ ∏θ∈L θ(F0 ) = D, we get D = ∏li=1 θ i (F0 ). Since θ i (F0 ) are irreducible G-subgroups of the G-group D, it follows that l ≥ lG (D) = d. Hence, l = d, i.e., dimΩ (L) = dimΩ (V D ) 󳨐⇒ L = V D = D σ . This means that σ is surjective, and so bijective. If D1 , D2 ∈ LG (H) and, D1 ⊂ D2 , then V D1 ⊂ V D2 . Therefore, σ : LG (H) → L(V H ) is a lattice isomorphism. Lemma 2.13. Let V be an n-dimensional linear Ω-space. Then there exists an involutory anti-automorphism α V of the lattice L(V) of subspaces of V such that dimΩ (L α V ) = n − dimΩ (L)(= codimΩ (L)) for any subspace L of the space V. Proof. It is possible to define the anti-automorphism α V by means of a non-degenerate symmetric bilinear form ⟨θ1 , θ2 ⟩ : V × V → Ω. If L ∈ L(V), then L α V = {θ2 ∈ V | ⟨θ1 , θ2 ⟩ = 0 for all θ1 ∈ L}. To construct the form ⟨θ1 , θ2 ⟩, one can choose a basis B = {ϑ1 , . . . , ϑ n } and set n

(1) (2)

⟨ϑ1 , ϑ2 ⟩ = ∑ ξ i ξ i , i=1

where the

(j) ξi

(j = 1, 2) are coordinates of θ j in the basis B.

Now we are ready to prove Proposition 2.11. Proof of Proposition 2.11. (i) Suppose that the G-group H is homogeneous and nonabelian. Then H is irreducible (Corollary 2.10) so that LG (H) = {E H , H}. Letting E αH = H and H α = E H , we get the desired anti-automorphism. (ii) Now suppose that H is homogeneous and abelian. Let V = V H = HomG (F0 , H) be the linear space constructed above, let σ : LG (H) → L(V) be a lattice isomorphism given in Lemma 2.12 and let α V be an anti-automorphism of the lattice L(V) satisfying Lemma 2.13. Then α = σα V σ−1 is an anti-automorphism of the lattice LG (H) possessing desired properties. Indeed, α is involutory since α V is. Take D ∈ LG (H). Then D σ is a subspace of the space V, and dimΩ (D σ ) = lG (D), by Lemma 2.12. Next, lG (D α ) = dimΩ ((D α )σ ) = dimΩ ((D σ )α V ) = n − dimΩ (D σ ) = n − lG (D). Therefore, we get lG (D α ) = lG (H/D) so, since D α and H/D are homogeneous completely reducible G-groups of the same type, they are G-isomorphic. (iii) Let H be an arbitrary completely reducible G-group and H = H1 ∘ ⋅ ⋅ ⋅ ∘ H s , where H1 , . . . , H s are homogeneous components of H. By (i) and (ii), the lattice LG (H i ) possesses an involutory anti-automorphism α i such that, for every D i ∈ LG (H i ), α D i i is G-isomorphic to H i /D i , i = 1, . . . , s, for each D i ∈ LG (H i ).

IX Faithful characters | 407

Let D ∈ LG (H). Then

D = D1 ∘ ⋅ ⋅ ⋅ ∘ D s , α

α

where D i = D ∩ H i ∈ LG (H i ). Setting D α = D i 1 ∘ ⋅ ⋅ ⋅ ∘ D s s , we obtain the desired antiautomorphism α of the lattice LG (H). 3°. Let H be a homogeneous completely reducible G-group, lG (H) = n. Let N ν (H) be the number of G-subgroups of H of a given G-length ν ≤ n. (If H is of type [F0 ], then all such subgroups have the same order |F0 |ν .) If H is nonabelian, then N0 (H) = N1 (H) = 1. Next suppose that H is reducible so abelian (Corollary 2.10). Proposition 2.14. Let H be an abelian homogeneous completely reducible G-group of type [F0 ] with the first two G-invariants n and q. Then, for each integer ν ∈ {1, . . . , n}, the number of G-subgroups of H of length ν is equal to ν−1

N ν (H) = ∏ λ=0

qn − qλ . qν − qλ

(3)

Proof. Note that H is a direct product of n admissible elementary abelian p-groups which are G-isomorphic to F0 , |F0 | = q. By Lemma 2.13, N ν (H) is equal to the number of ν-dimensional subspaces of the linear Ω-space V H = HomG (F0 , H). Every such subspace is determined by any its basis {θ1 , . . . , θ ν }. The number of linearly inden λ pendent systems {θ1 , . . . , θ ν } in V H is equal to A ν = ∏ν−1 λ=0 (q − q ). A ν-dimensional ν−1 ν λ Ω-space has exactly B ν = ∏λ=0 (q − q ) bases. Therefore, N ν (H)B ν = A ν , completing the proof. We make use of formula (3) in the following section.

3 Group-theoretical functions of Möbius and Euler 1°. The class of G-groups introduced in §IX.2 is too wide for our aims. Let CG be the class of all G-groups that are G-isomorphic to sections N1 /N2 , (N1 ≥ N2 are normal in G) of the group G (recall that G is a G-group under action of G via conjugation and all its normal subgroups are G-subgroups). If H ∈ CG and N is a G-subgroup of H, then N, H/N ∈ CG . In what follows, only elements of the class CG will be considered as G-groups. It follows that the set T(G) of the classes of G-isomorphic G-groups has finite cardinality, say t(G). Thus, t(G) = |T(G)|. Let us consider the complex-valued functions f : CG → ℂ constant on classes of G-isomorphic G-groups: G

H1 , H2 ∈ CG , H1 ≅ H2 󳨐⇒ f(H1 ) = f(H2 ). The set of all such functions is denoted by D(G). It is clear that D(G) is a linear t(G)-dimensional ℂ-space with respect to usual pointwise addition of functions and

408 | Characters of Finite Groups 1 their multiplication by complex numbers. Thus, if f, f1 , f2 ∈ D(G), α ∈ ℂ and H ∈ CG , then (f1 + f2 )(H) = f1 (H) + f2 (H), (αf)(H) = αf(H). Let f(A) be the value of function f ∈ D(G) on a class A ∈ T(G). If we define functions eA ∈ D(G) (A ∈ T(G)), setting eA (B) = eA (H) = δA,B

(H ∈ B ∈ T(G)),

(1)

(where δ is the Kronecker delta on the set T(G)), we obtain a basis B = {eA | A ∈ T(G)} of the ℂ-space D(G) (we consider the set T(G) as a linearly ordered, in some way). If f ∈ D(G), then f =

∑

f(A)eA .

(2)

A∈T(G)

Indeed, if B ∈ T(G), then ∑

A∈T(G)

f(A)eA (B) =

∑

A∈T(G)

f(A)δA,B = f(B),

and (2) is proved. Given f, g ∈ D(G), define the function f ∗ g : CG → ℂ by (f ∗ g)(H) = ∑ f(D)g(H/D)

(H ∈ CG ),

(3)

G

D≤H

where D runs over all G-subgroups of H (recall that f(U) = f(B) provided U ∈ B). Clearly, f ∗ g ∈ D(G). It is easy to check that the set D(G) is an associative algebra with respect to the above defined operations (+), ( ⋅ ), (∗); its identity element is e = eI , where I = IG is the class of G-groups of order 1. We have e(B) = eI (B) = δI,B which is not 0 if and only if B = I and e(I) = 1. It follows that (f ∗ e)(H) = ∑ f(D)e(H/D) = f(H) G

D≤H

since the sum on the right-hand side contains only one nonzero summand, namely, f(H)e(H/H) = f(H). If f1 , f2 ∈ D(G), then (f1 ∗ f2 )(I) = f1 (I)f2 (I). Given f ∈ D(G), write |f| = f(I). Then |f1 + f2 | = (f1 + f2 )(I) = f1 (I) + f2 (I) = |f1 | + |f2 |, |f1 ∗ f2 | = (f1 ∗ f2 )(I) = f1 (I)f2 (I) = |f1 ||f2 |, |γ ⋅ f| = γf(I) = γ|f|

(γ ∈ ℂ)

(the second formula follows from (3)) so the mapping f 󳨃→ |f| is a homomorphism of the algebra D(G) onto the field ℂ of complex numbers. Lemma 3.1. An element f ∈ D(G) is nilpotent if and only if |f| = 0. Proof. (i) If f is nilpotent, i.e., f m = 0 for some m ∈ ℕ (here f m = f ∗ f ∗ ⋅ ⋅ ⋅ ∗ f , m times, m ∈ ℕ), then, by the paragraph preceding the lemma, |f|m = |f m | = 0, and so |f| = 0.

IX Faithful characters | 409

(ii) A mapping τ : g 󳨃→ f ∗ g (g ∈ D(G)) is a linear operator of the space D(G). We obtain a homomorphism f 󳨃→ τ(f) (f ∈ D(G)) of the algebra D(G) into the algebra of all linear operators of the space D(G) (the homomorphism τ is the regular representation of D(G)). Denoting by T(f) the t(G) × t(G) matrix of the operator τ(f) in the basis B of the space D(G) (recall that t(G) = |T(G)|), we obtain an injective homomorphism f 󳨃→ T(f) of the algebra D(G) into the matrix algebra ℂt(G) (ℂn denotes the algebra of all n × n matrices over ℂ). Let us compute the trace tr(T(f)) of the matrix T(f). Let A ∈ T(G) and hA = f ∗ eA . By (2), the following decomposition holds: f ∗ eA = hA =

∑

B∈T(G)

hA (B)eB ,

(4)

so that T(f) = (hA (B))A,B∈T(G) (the entries of the matrix T(G) are indexed by the elements of the set T(G)). Therefore, tr(T(f)) =

∑

A∈T(G)

hA (A).

(5)

Let H ∈ A ∈ T(G). Noting that eA (H) = 1 and eB (H) = 0 for B ≠ A, we obtain, making use of (4), hA (A) = (f ∗ eA )(H) = ∑ f(D)eA (H/D). G

D≤H

Since H/D ∈ A if and only if D = E H = {1H }, we get hA (A) = f(E H )eA (H) = |f|δA,A = |f|. Hence, by (5), tr(T(f)) = t(G)|f|.

(6)

It follows from (6) that, for every k ∈ ℕ, we have tr(T(f)k ) = t(G)|f|k . Therefore, if |f| = 0, then tr(T(f)k ) = 0 for all k ∈ ℕ, and so, as it is known, the matrix T(f) is nilpotent. Since the matrix regular representation T of the algebra D(G) is faithful, it follows that f is nilpotent. Lemma 3.2. An element f ∈ D(G) is invertible if and only if |f| ≠ 0. Proof. (i) If f 󸀠 ∈ D(G) and f ∗ f 󸀠 = e = eI , then, since the mapping f 󳨃→ |f| is a homomorphism, |f| ⋅ |f 󸀠 | = |f ∗ f 󸀠 | = |e| = e(I) = δI,I = 1 󳨐⇒ |f| ≠ 0. (ii) Supposing that |f| ≠ 0, we have to prove that f is invertible. Set w = e − then f = (e − w)|f|. Since |w| = |e −

1 1 f| = |e| − |f| = 1 − 1 = 0, |f| |f|

1 |f| f ;

410 | Characters of Finite Groups 1 the element w is nilpotent, by Lemma 3.1. Then w m = 0 for an appropriate m ∈ ℕ. 1 Setting g = |f| (e + w + ⋅ ⋅ ⋅ + w m−1 ), we get f ∗ g = g ∗ f = (e − w)|f| ∗

1 (e + w + ⋅ ⋅ ⋅ + w m−1 ) = e − w m = e − 0 = e, |f|

proving the invertibility of f . Thus, an element of the algebra D(G) is either nilpotent or invertible. Proposition 3.3. If H is a completely reducible G-group, then (f, g ∈ D(G).

(f ∗ g)(H) = (g ∗ f)(H)

Proof. Let α be an anti-automorphism of the lattice LG (H) satisfying the conditions of G G G Proposition 2.11. Since D α ≅ H/D and H/D α ≅ D for each D ≤ H, we get (f ∗ g)(H) = ∑ f(D α )g(H/D α ) = ∑ f(H/D)g(D) = (g ∗ f)(H). G

G

D≤H

D≤H

Indeed, D α runs over the set of all G-subgroups of H if D runs. 2°. Let u G denote a function CG → ℂ that is equal to 1 on CG (i.e., u G (L) = 1 provided L = N1 /N2 , where N1 , N2 ⊴ G); then u G ∈ D(G). Since |u G | = 1, u G is an invertible element of the algebra D(G) (Lemma 3.2). Definition 3.1. A function μ G = u−1 G is said to be the group-theoretic Möbius function associated with the group G. In what follows we set e = eI , where I = IG . It follows from Definition 3.1 and Proposition 3.3 that μ G ∗ u G = u G ∗ μ G = e. If H ∈ I = IG , then (μ G ∗u G )(H) = e(H) = 1. If H ∈ ̸ I, then (μ G ∗u G )(H) = e(H) = 0. This means that, for every G-group H, {1 if H ∈ I, ∑ μ G (D) = ∑ μ G (H/D) = { 0 if H ∈ ̸ I. G G { D≤H D≤H

(7)

It follows from (7) that μ G (H) = 1 if H ∈ I. If f ∈ D(G) and g = f ∗ u G , then f = f ∗ (u G ∗ μ G ) = (f ∗ u G ) ∗ μ G = g ∗ μ G . This leads to a group-theoretic analog of arithmetical Möbius–Dedekind inversion formula: if f ∈ D(G) and g(H) = ∑ f(D)(= (f ∗ u G )(H)),

(8)

G

D≤H

then f(H) = ∑ g(D)μ G (H/D)(= (g ∗ μ G )(H)) G

D≤H

for every G-group H.

(9)

IX Faithful characters | 411

Next, if f ∈ D(G) and g = u G ∗ f , then f = μ G ∗ g and this leads to a dual inversion formula: if f ∈ D and g(H) = ∑ f(H/D)(= (u G ∗ f)(H)),

(10)

G

D≤H

then f(H) = ∑ μ G (D)g(H/D)(= (μ G ∗ g)(H))

(11)

G

D≤H

for every G-group H (recall that the algebra D(G) is commutative). Let us find the explicit form of the Möbius function μ G . To this end, define a function ψ : CG → ℂ, setting for a homogeneous completely reducible G-group H: 1 { { { ψ(H) = {−1 { { n (n) {(−1) q 2

if H ∈ I, if H is irreducible,

(12)

if H is abelian with invariants n and q.

If H is nonhomogeneous completely reducible group and H = L1 ∘ L2 , we demand that ψ(H) = ψ(L1 )ψ(L2 ).

(13)

If H is not completely reducible G-group, we set ψ(H) = 0.

(14)

Let us prove that ψ = μ G . It follows from the definition of ψ that ψ ∈ D(G). It remains to check that (15)

ψ ∗ u G = e.

Indeed, if (15) is true, then, in view of associativity of the algebra D(G), we obtain μ G = e ∗ μ G = (ψ ∗ u G ) ∗ μ G = ψ ∗ (u G ∗ μ G ) = ψ. Setting θ = ψ ∗ u G , we rewrite (15) in the form {1 if H ∈ I, θ(H) = { 0 if H ∈ ̸ I. {

(16)

θ(H) = (ψ ∗ u G )(H) = ∑ ψ(D),

(17)

Noting that G

D≤H

we see that θ(H) = ψ(H) = 1 if H ∈ I. Thus, to prove (15), it remains to show that θ(H) = 0 if H ∈ ̸ I. The general case is reduced to the one when H is completely reducible. Indeed, it follows from (17) and (14) that θ(H) = θ(ScG (H)), where ScG (H)

412 | Characters of Finite Groups 1 is the G-socle of H which is a completely reducible group, by §IX.2. The condition H ∈ ̸ I is equivalent to the condition ScG (H) ∈ ̸ I. Therefore, it is enough to prove the equality θ(H) = 0 for the case where H ∈ ̸ I is completely reducible (then H = ScG (H)). First assume that H ∈ ̸ I is a homogeneous completely reducible G-group. If H is not abelian, it is irreducible, by Corollary 2.10. By (17) and (12), we get θ(H) = ψ(E H ) + ψ(H) = 1 + (−1) = 0 (see the definition of ψ). Now let H be abelian homogeneous completely reducible G-group with the first two invariants n and q. Then we obtain, using (12) and (17), n

ν

θ(H) = ∑ (−1)ν q(2) N ν (H),

(18)

ν=0

where N ν (H) is the number of G-subgroups of H having G-length ν. Let us consider the following rational function F n (x, t) of two variables x and t: n

ν

F n (x, t) = ∑ (−1)ν x(2) G n,ν (x)t ν ,

(19)

ν=0

where {∏ν−1 G n,ν (x) = { λ=0 1 {

x n −x λ x ν −x λ

if 1 ≤ ν ≤ n,

(20)

if ν = 0.

Extend the function G n,ν (x), setting G n,ν (x) = 0 for ν < 0 or for ν > n. Then (19) takes the following form: ∞

ν

F n (x, t) = ∑ (−1)ν x(2) G n,ν (x)t ν .

(21)

ν=−∞

We will prove by induction on n that n−1

F n (x, t) = ∏ (1 − x ν t).

(22)

F n+1 (x, t) = (1 − x n t)F n (x, t).

(23)

ν=0

It is enough to prove that

By (21), ∞

∞

ν

ν

(1 − x n t)F n (x, t) = ∑ (−1)ν x(2) G n,ν (x)t ν − ∑ (−1)ν x n+(2) G n,ν (x)t ν+1 . ν=−∞

ν=−∞

Replacing in the second sum ν + 1 by ν, one obtains ∞

ν

(1 − x n t)F n (x, t) = ∑ (−1)ν x(2) [x n−ν+1 G n,ν−1 (x) + G n,ν (x)]t ν . ν=−∞

(24)

IX Faithful characters | 413

Let us use the following properties of the function G n,ν (x): (25)

G n+1,ν (x) = x ν G n,ν (x) + G n,ν−1 (x),

(26)

G n,ν (x) = G n,n−ν (x). Using (25) and (26), we rewrite (24) as follows: ∞

ν

ν=−∞ ∞

ν

(1 − x n t)F n (x, t) = ∑ (−1)ν x(2) [x n−ν+1 G n,n−ν+1 (x) + G n,n−ν (x)]t ν ∞

ν

= ∑ (−1)ν x(2) G n+1,n+1−ν (x)t ν = ∑ (−1)ν x(2) G n+1,ν (x)t ν . ν=−∞

ν=−∞

In view of (19), this implies (23) and so (22). By (16), (7) and (22), n−1

θ(H) = F n (q, 1) = ∏ (1 − q ν ) = 0. ν=0

Thus, if H ∈ ̸ I is a homogeneous completely reducible G-group, then θ(H) = 0. Now assume that H is an arbitrary completely reducible G-group and H ∈ ̸ I. Let H = H1 ∘ ⋅ ⋅ ⋅ ∘ H s be a decomposition of H in homogeneous components. Since H i ∈ ̸ I for all i, it follows by what has just been proved that θ(H i ) = 0 for all i. By (13) and (17), θ(H) = θ(H1 ) ⋅ ⋅ ⋅ θ(H s ) = 0. This completes the proof of (16) and so (15). Therefore, we have ψ = μ G . Remembering the definition of function ψ(= μ G ) and its properties (see (12), (13) and (14)), we get the following: Proposition 3.4. Let H be a G-group. Then the following statements hold: (a) μ G (H) = 0 if H is not completely reducible. (b) If H is homogeneous and completely reducible, then {1 { { μ G (H) = {−1 { { n (n) {(−1) q 2

if H ∈ I, if H is irreducible,

(27)

if H is abelian with invariants n and q.

(c) If H is completely reducible and H = H1 ∘ ⋅ ⋅ ⋅ ∘ H s is a decomposition of H in homogeneous components, then the following “multiplicative” property of the function μ G holds: μ G (H) = μ G (H1 ) ⋅ ⋅ ⋅ μ G (H s ).

(28)

Proposition 3.4 (c) is equivalent to the following property of “multiplicativity” of the function μ G : if H is a completely reducible G-group and H = L1 ∘ L2 , then μ G (H) = μ G (L1 ∘ L2 ) = μ G (L1 )μ G (L2 ).

(29)

414 | Characters of Finite Groups 1 2°. Let us define a function v G : CG → ℂ by (H ∈ CG ).

v G (H) = |H| Obviously, v G ∈ D(G). (1)

(2)

Definition 3.2. The functions φ G = μ G ∗ v G and φ G = v G ∗ μ G are called grouptheoretical Euler functions of the first and the second kind, respectively, associated with G. Generally speaking, φ(1) ≠ φ(2) since functions μ G and v G do not commute on some G-groups. Of course, φ(1) , φ(2) ∈ D(G). If H is a G-group, it follows from Definition 3.2 that (1)

φ G (H) = ∑ μ G (D)|H/D|,

(30)

G

D≤H (2) φ G (H)

= ∑ μ G (H/D)|D|.

(31)

G

D≤H

We obtain from the inversion formulas (8)–(11) the following relations analogous Gauss’ formula for arithmetical Euler’s φ-function: (1)

(2)

∑ φ G (H/D) = ∑ φ G (D) = |H|, G

(32)

G

D≤H

D≤H

where H is an arbitrary G-group. Now we use formulas (30) and (31). If H is an irreducible G-group, then (1)

φ G (H) = |H| − 1 = φ(2) (H). Next, if H = L1 × L2 , where L1 and L2 are nonabelian simple G-groups, then (1)

φ G (H) = (|L1 | − 1)(|L2 | − 1) = φ(2) (H). Thus, in the case under consideration, φ(1) (H) = φ(2) (H). This is a partial case of the following proposition. (1)

(2)

Proposition 3.5. The functions φ G , φ G coincide on completely reducible G-groups. Proof. By Proposition 3.3, the restriction of the operation ∗ to the set of completely reducible G-groups is commutative. Therefore, if H is a completely reducible group, then (1) (2) φ G (H) = (μ G ∗ v G )(H) = (v G ∗ μ G )(H) = φ G (G). ̂G be the class of all completely reducible G-groups (C ̂G is a proper part of the Let C class CG ; see the beginning of this section). Let φ G denote the (coinciding, by Propo(2) ̂G (see Proposition 3.5): sition 3.5) restrictions of φ1G and φ G to the class C (1)

(2)

φ G = (φ G )C = (φ G )C . ̂ ̂ G G

(33)

IX Faithful characters | 415

(1)

(2)

The following proposition reduces the calculation of φ G (H) and φ G (H) to the case where H is a completely reducible G-group. For the definition of AsG (H) see Definition 2.3. Proposition 3.6. For every G-group H, we have (1)

(2)

φ G (H) φ G (ScG (H)) = , |H| |ScG (H)|

φ G (H) φ G (AsG (H)) = . |H| |AsG (H)|

G

(34)

Proof. Since μ G (D) ≠ 0 if and only if D ≤ ScG (H) (see Proposition 3.4), the first relation in (34) follows from (30): (1)

(1)

φ G (H) μ G (D) = ∑ = |H| |D| G D≤H

∑ G

D≤ScG (H)

μ G (D) φ G (ScG (H)) φ G (ScG (H)) = = . |D| |ScG (H)| |ScG (H)|

Recall that H∗ is the intersection of all maximal G-subgroups of H. In that case, the quotient group H/H∗ is completely reducible. Set H = H/H∗ ; as we know, H = AsG (H), the antisocle of H. To prove the second relation in (34), rewrite (30) in the following form: (2)

φ G (H) = |H|

∑ G

̂ D≤H, H/D∈C G

μ G (H/D) . |H/D|

In our case, D runs through all the G-subgroups of H containing H∗ . Using the previous displayed formula and Lemma 2.5 and complete reducibility of the antisocle AsG (H) of a G-group H, we get (2)

φ G (H) = |H|

∑ G

H∗ ≤D≤H

=

∑ G

H∗ ≤D≤H

=

μ G (H/D) |H/D| μ G ((H/H∗ )/(D/H∗ )) |(H/H∗ )/(D/H∗ )|

∑

μ G (AsG (H)/D)

G

D≤AsG (H)

|AsG (H)/D|

(2)

=

φ G (AsG (H)) φ G (AsG (H)) = . |AsG (H)| |AsG (H)|

Thus, for a G-group H, we have the following formulas that allow us to compute num(1) (2) bers φ G (H) and φ G (H): (1)

φ G (H) = |H| ⋅

φ G (ScG (H)) , |ScG (H)|

(2)

φ G (H) = |H|

φ G (AsG (H)) . |AsG (H)|

It follows from these formulas that φ(1) (H) = φ(2) (H) ⇐⇒ ScG (H)) = AsG (H) ⇐⇒ H is completely reducible.

416 | Characters of Finite Groups 1 Proposition 3.7. If H is a completely reducible G-group, then the following statements hold: (a) φ G (H) = |H| − 1 if H is irreducible (and, in particular, if H is homogeneous and nonabelian; see Proposition 2.9). (b) If H is homogeneous and abelian with G-invariants n, q, j, then n−1

φ(H) = ∏ (q j − q ν ).

(35)

ν=0

(c) If H = H1 ∘ ⋅ ⋅ ⋅ ∘ H s is the decomposition of H in homogeneous components, then φ G (H) = φ G (H1 ) ⋅ ⋅ ⋅ φ G (H s ).

(36)

Proof. (a) If H is irreducible, then φ G (H) = μ G (E H ) ⋅ |H| + μ G (H)|H/H| = |H| − 1. (b) If H is of type [F], where F is an irreducible abelian G-group, then |F| = q j (see G the remark following Definition 2.7). Thus, it follows from D ≤ H and lG (D) = ν that |D| = |F|ν = q jν . Since the first two invariants of the G-group D are equal to ν and q, we have, by Proposition 3.4 (see there formula (27)), ν

μ G (D) = (−1)ν q(2) . Noting that |H/D| = |H|q−jν , we get, in view of (30) and taking into account that (1) φ G (H) = φ G (H), n

ν

φ G (H) = |H| ∑ (−1)ν q(2) N ν (H)q−jν .

(37)

ν=0

It follows from Proposition 2.14 and the definition of G n,ν (x) (see formula (20)) that N ν (H) = G n,ν (q) (0 ≤ ν ≤ n). Therefore, by (19) and (37), we obtain φ G (H) = |H| ⋅ F n (q, q−j ). Taking into account (22) and equality |H| = |F|n = q nj , we get n−1

n−1

φ G (H) = q nj ∏ (1 − q ν−j ) = ∏ (q j − q ν ). ν=0

ν=0

(c) From (27) and the multiplicativity of the function (μ G )C (see equality (29)) fol̂ G lows that the function φ G is multiplicative: if H is completely reducible and H = L1 ∘L2 , then φ G (H) = φ(L1 ∘ L2 ) = φ G (L1 )φ(L2 ).

(38)

Relation (36) now follows from (38). Corollary 3.8. If H is a completely reducible G-group, then φ G (H) ≠ 0 if and only if the invariants n = lG (K) and j = j G (K) of any abelian homogeneous component K of the G-group H satisfy the inequality n ≤ j.

IX Faithful characters | 417

Proof. It follows from Proposition 3.7 (c) that φ G (H) ≠ 0 if and only if φ G (K) ≠ 0 for all homogeneous components K of the G-group H. If K is nonabelian, then φ G (K) = |K| − 1 ≠ 0, by Proposition 3.7 (a). If K is abelian, then, by Proposition 3.7 (b), n−1

(n = lG (K), j = j G (K)).

φ G (K) = ∏ (q j − q ν ) ν=0

Therefore, φ G (K) ≠ 0 if and only if n ≤ j . Corollary 3.8 and Proposition 3.6 imply the following: (1)

(2)

Proposition 3.9. If H is a G-group, then φ G (H) ≠ 0 (φ G (H) ≠ 0) if and only if the G-invariants n and j of every abelian homogeneous component of the G-socle (correspondingly, of the G-antisocle) of the G-group H satisfy n ≤ j. (1)

(2)

4°. Let us establish a connection between Euler functions φ G and φ G . G Let H be a G-group and N ≤ H (i.e., N is a G-subgroup of H; in particular, N ⊴ H). Recall that N∗ is the intersection of all maximal G-subgroups of N if N ≠ {1}; if N = {1}, then N∗ = {1}. Next, let N ∗ be the product of all G-subgroups of H covering N (a G-subgroup L of H covers N if N < L and L/N is an irreducible G-group) if N ≠ H; if N = H, then N ∗ = H). The quotient groups N/N∗ and N ∗ /N are completely reducible G G-groups. Given a G-group H, we have (H∗ )∗ = H and (E H )∗ = ScG (H). If U < H, then (U ∗ )∗ ≤ U ≤ (U∗ )∗ . G

Definition 3.3. Let U, V ≤ H. A subgroup V is said to be an upper complement of U provided UV = U ∗ . A subgroup U is said to be a lower complement of V provided U ∩ V = V∗ . G

Minimal by inclusion upper complements of U ≤ H are called its upper direct complements; maximal by inclusion lower complements of V are called its lower direct comG plements. The set of all upper direct complements of U ≤ H is denoted by comH (U); G the set of all lower direct complements of V ≤ H is denoted by comH (V). Obviously, both of these sets are nonempty. We have comH (E H ) = {ScG (H)}, G

comH (H) = {H∗ }.

If H is completely reducible and U ≤ H, then comH (U) = comH (U) is the set of all G-subgroups that are usual direct complements of U in H. In that case, U ∗ = H, U∗ = {1}. It follows that if, in addition, H 󸀠 = H, then |comH (U)| = 1 since, in the case under consideration, an upper complement of U is H coincides with CH (U). If F is an irreducible G-subgroup of a G-group H, then F∗ = {1} and comH (F) is the set of all maximal by inclusion G-subgroups N of H such that F ∩ N = {1}. G

Proposition 3.10. Let H > {1} be a G-group and U, V ≤ H. Then the following assertions are equivalent: (a) UV = U ∗ and U ∩ V = V∗ . (b) V ∈ comH (U). (c) U ∈ comH (V).

418 | Characters of Finite Groups 1 Proof. (a) ⇒ (b): If (b) is not true, then U < H. In that case, V ≠ E H ; then U ∗ > U so V > E H since UV = U ∗ > U. Therefore the set M(V) of maximal G-subgroups of the G-group V is nonempty. Denoting by X̃ the image of X ⊆ H under the natural homomorphism of H onto H̃ = H/V∗ , we obtain by (a) that Ũ ∗ = Ũ Ṽ , Ũ ∩ Ṽ = 1̃ 󳨐⇒ Ũ ∗ = Ũ × Ṽ . Since V ∈ ̸ comH (U), there exists V1 ∈ M(V) such that UV1 = U ∗ . Since V1 < V and V∗ ≤ V1 , we get Ṽ 1 < V,̃ and so Ũ ∩ Ṽ 1 = 1.̃ Therefore, ̃1 = Ũ × Ṽ 1 < Ũ × Ṽ = Ũ ∗ , Ũ ∗ = UV a contradiction. Thus, (b) is fulfilled. (b) ⇒ (a): In that case, G

UV = U ∗ 󳨐⇒ V/(U ∩ V) ≅ UV/U = U ∗ /U. Since U ∗ /U = ScG (H/U), it follows that U/(U ∩ V) is a completely reducible G-group, and so, by Lemma 2.5, V∗ ≤ U ∩ V ≤ U. If U ∩ V = D > V∗ , then D̃ = D/V∗ ≠ E H̃ . Since G Ṽ = V/V∗ = AsG (V) is a completely reducible G-group and D̃ ≤ V,̃ we have, by ProposiG G tion 2.1, Ṽ = D̃ × Ṽ 1 , where V∗ ≤ V1 ≤ H. Since D̃ ≠ E H̃ , we have V1 < V. On the other hand, ̃ D̃ × Ṽ 1 ) = Ũ Ṽ 1 = UV ̃1 󳨐⇒ UV1 = U ∗ . Ũ ∗ = ̃ UV = Ũ Ṽ = U( In view of V1 < V, this contradicts (b). Thus U ∩ V = V∗ , and (a) holds. (a) ⇒ (c): If (c) is false, then U ≠ H. Indeed, if U = H, then, by (a), V = H ∩ V = U ∩ V = V∗ 󳨐⇒ V = {1}. In that case, however, (U =)H ∈ comH (V), a contradiction. Therefore, the set M(U) of G-subgroups of H, covering U, is nonempty. Since U ∈ ̸ comH (V), there exists U1 ∈ M(U) such that U1 ∩ V = V∗ . We claim that U1 V = U ∗ . Indeed, U1 V ≤ U ∗ since U1 ∈ M(U) and V ≤ U ∗ (by (a)). On the other hand, U1 V ≥ UV = U ∗ since U1 ≥ U. Therefore, U1 V = U ∗ . Setting, as above, H̃ = H/V ∗ , we get Ũ ∗ = Ũ 1 Ṽ ,

Ũ 1 ∩ Ṽ = E H̃ 󳨐⇒ Ũ ∗ = Ũ 1 × Ṽ .

Similarly, it follows from (a) that Ũ ∗ = Ũ × V.̃ Thus, Ũ 1 × Ṽ = Ũ × Ṽ which implies that Ũ 1 = Ũ since Ũ 1 ≥ U.̃ Since V∗ ≤ U ∩ U1 , we obtain U1 = U, which is impossible since U1 ∈ M(U). This proves (c). (c) ⇒ (a): Since U ∩ V = V∗ , we get G

UV/U ≅ V/(U ∩ V) = V/V∗ = AsG (V) so that the G-group UV/U is completely reducible. Therefore, UV/U ≤ ScG (H/U), and so UV ≤ U ∗ . Assume, by way of contradiction, that UV < U ∗ . Set H = H/U and let X

IX Faithful characters | 419

be the image of X ⊆ H under natural homomorphism H → H. Since UV < U ∗ , we ∗ ∗ get V < U ∗ . Since U = U ∗ /V is completely reducible, we obtain U = U 1 × V, where G U1 ≤ H and U1 > U. It follows that U1 ∩ UV = U 1 ∩ UV = U 1 ∩ V = E H . Therefore, U1 ∩ UV = U, and so, by (a), U1 ∩ V = U1 ∩ (UV ∩ V) = (U1 ∩ UV) ∩ V = U ∩ V = V∗ , contrary to (c) since U1 > U. Thus, UV = U ∗ , and (a) is true. In what follows, we say that U and V are mutually complementary G-subgroups of a G-group H if UV = U ∗ and U ∩ V = V∗ (i.e., condition (a) of Proposition 3.10 is fulfilled). Theorem 3.11. Let U and V be mutually complementary G-subgroups of a G-group H. Then (1)

(2)

φ G (H/U) φ G (V) = . |H/U| |V|

(39)

Proof. Since, by the assumption, UV = U ∗ and U ∩ V = V∗ , then G

ScG (H/U) = U ∗ /U = UV/U ≅ V/(U ∩ V) = V/V∗ = AsG (V), so we get φ G (ScG (H/U)) = φ G (AsG (V)). By Proposition 3.6 (see there equality (28)), we get (1) (2) φ G (H/U) φ G (ScG (H/U)) φ G (AsG (V)) φ G (V) = = = . |H/U| |ScG (H/U)| |AsG (V))| |V|

4 G-kernels and G-antikernels 1°. Let H ∈ CG , where CG is the class of all G-groups that are G-isomorphic to sections N1 /N2 , N1 ≥ N2 and N1 , N2 ⊴ G. Let ⟨⟨x⟩⟩G denote the least G-subgroup of H containing x ∈ H (G-closure of ⟨x⟩ (x ∈ H): ⟨⟨x⟩⟩G is the intersection of all G-subgroups of H containing x). Usually, we will omit the subscript G. Definition 4.1. A subgroup D = ⟨⟨x⟩⟩G of a G-group H, where x ∈ H, is said to be a G-antikernel of H. In particular, H is said to be a G-antikernel provided H = ⟨⟨x⟩⟩G for some x ∈ H. Recall that a certain action γ H of a group G is defined on H. Obviously, ⟨⟨x⟩⟩G = ⟨O x ⟩, where O x is the G-orbit of x ∈ H. Thus, G-antikernels of the G-group H are subgroups generated by G-orbits of elements of H.

420 | Characters of Finite Groups 1 Definition 4.2. Given a G-group H and x, y ∈ H, we will say that x, y belong to the same G-layer of H provided ⟨⟨x⟩⟩G = ⟨⟨y⟩⟩G . There exists a one-to-one correspondence between G-layers and G-antikernels of a G-group H so the number of G-layers in H equals the number of G-antikernels in H. Given a G-group H, set W(H) = {x ∈ H | ⟨⟨x⟩⟩G = H}.

(1)

G

Then W(H) ≠ 0 if and only if H is a G-antikernel. If D ≤ H, then W(D) = {x ∈ H | ⟨⟨x⟩⟩G = D} since the G-orbit of x ∈ D in D coincides with the G-orbit of x in H. Therefore, W(D) ≠ 0 if and only if D is a G-antikernel in D. In that case, W(D) is a G-layer of a G-group H. We say that this G-layer generates the G-subgroup D. Define a function f : CG → ℂ, setting, for a G-group H, f(H) = |W(H)|.

(2)

(If a G-group H is cyclic, then f(H) = φ(|H|), where φ(∗) is the Euler totient function of number theory.) G If D ≤ H, then f(D) ≠ 0 if and only if D is a G-antikernel of H. In that case, f(D) is the cardinality of the G-layer generating D. Obviously, f ∈ D(G), i.e., f is constant on classes of G-isomorphic subgroups of H. It follows that distinct sets W(D) are disjoint, and so the following partition holds: H = ⋃ W(D) 󳨐⇒ ∑ f(D) = |H| G

D≤H

G

D≤H

since x ∈ W(D) (this is similar to a known identity of number theory). Applying to f inversion formulas (3.8) and (3.9), we obtain, in view of (3.31), (2) that f = φ G . Thus, the following assertion holds. G

(2)

Proposition 4.1. If H ∈ CG and D ≤ H, then φ G (D) ≠ 0 if and only if D is a G-antikernel (2) of H. In that case, φ G (D) = |W(D)|. Corollary 4.2. A G-group H is a G-antikernel if and only if G-invariants n and j of every abelian homogeneous component of its antisocle AsG (H) satisfy n ≤ j. Proof. This follows from Proposition 4.1 (for D = H) and Corollary 3.8. Definition 4.3. The G-core S G of a subgroup S of a G-group H is the greatest G-subG group contained in S. Obviously, S G = ⋂g∈G g ∘ S, and S G = S if and only if S ≤ G. Definition 4.4. Given ψ ∈ Char(H), where H is a G-group, the G-core kerG (ψ) = ker(ψ)G is called the G-kernel of ψ. If kerG (ψ) = {1}, then ψ is said to be G-faithful. It is possible that kerG (ψ) < ker(ψ). In what follows, we will call the G-kernels of irreducible characters of a G-group H the G-kernels of H.

IX Faithful characters | 421

Given a G-group H, define a function h : CG → ℂ by h(H) =

ψ(1)2 .

∑

(3)

ψ∈Irr(H), kerG (ψ)={1}

Thus, h(H) is the sum of squares of degrees of G-faithful irreducible characters of the G-group H. G G If H1 , H2 ∈ CG and H1 ≅ H2 , then h(H1 ) = h(H2 ). Therefore, h ∈ D(G). If D ≤ H, we obtain, by (3), h(H/D) =

∑

ψ(1)2 .

(4)

ψ∈Irr(H), kerG (ψ)=D G

Since every irreducible character ψ of a G-group H satisfies ker(ψ)G ≤ H, we obtain from (4) that ∑ h(H/D) = |H|. G

D≤H

Applying to h inversion formulas (3.10) and (3.11), we obtain, in view of (3.30), (1) that h = φ G . Thus, for a G-group H, we obtain (1)

φ G (H) =

∑

ψ(1)2 .

(5)

ψ∈Irr(H), kerG (ψ)={1}

From (4) one obtains the following more general relation for a G-group H and its arbitrary G-subgroup D of H: (1)

φ G (H/D) =

∑

ψ(1)2 .

(6)

ψ∈Irr(H), kerG (ψ)=D

The following assertion follows from (6): G

(1)

Proposition 4.3. If N ≤ H, then one has φ G (H/N) ≠ 0 if and only if N is a G-kernel (1) of a G-group H. In particular, φ G (H) ≠ 0 if and only if H has a G-faithful irreducible character. The following theorem follows from Propositions 4.3, 4.1 and Theorem 3.11. Theorem 4.4. Suppose that U, V is a pair of mutually complementary G-subgroups of a G-group H (see the paragraph preceding Theorem 3.11). Then the following assertions are equivalent: (a) U is a G-kernel of H. (b) V is a G-antikernel of H. Noting that {1} and ScG (H) are mutually complementary subgroups of a G-group H, we get: Theorem 4.5. A G-group H has a G-faithful irreducible character if and only if its G-socle ScG (H) is a G-antikernel, i.e., ScG (H) = ⟨⟨x⟩⟩G for some x ∈ H.

422 | Characters of Finite Groups 1 In particular (Gachütz, see [Gas3]), a usual group H has a faithful irreducible character if and only if its socle is generated by a class of conjugate elements (it suffices to put G = H in Theorem 4.5). It follows that a nilpotent group H has a faithful irreducible character if and only if its center is cyclic. We suggest to the reader to produce independent proof of the last result. (Hint. Using §IV.8, reduce the proof to prime power groups. If Z(G) is noncyclic so is Sc(G). Then Sc(G) ≠ ⟨⟨z⟩⟩G for any z ∈ Z(G).) Let us present a proof of Theorem 4.5 independent of Theorem 4.4. Noting that (1) ScG (H) is a completely reducible G-group, we get φ G (ScG (H)) = φ G (ScG (H)), what allows us to rewrite the first relation (3.34) (in Proposition 3.6) in the following form: (2)

(1)

φ G (H) φ G (ScG (H)) = . |H| |ScG (H)| Now it remains to apply Propositions 4.1 and 4.3. Another existence criterion of G-faithful irreducible characters of a G-group H follows from Proposition 4.3 and Corollary 3.8. Theorem 4.6. A G-group H possesses a G-faithful irreducible character if and only if the G-invariants n and j of every abelian homogeneous component of its G-socle satisfy the inequality n ≤ j. Let the subgroup A (the subgroup B) be the product of all abelian (nonabelian) irreducible G-subgroups of a G-group H; then A is said to be the abelian component and B the nonabelian component of the G-socle ScG (H) of a G-group H. We have ScG (H) = A ∘ B. It follows from Corollary 2.10 that B = F1 ∘ ⋅ ⋅ ⋅ ∘ F k , where F1 , . . . , F k are all nonabelian irreducible subgroups of H. Now let A = P × Q × ⋅ ⋅ ⋅ be the decomposition of A into a direct product of its Sylow subgroups. The latters, as characteristic G subgroups of A ≤ H, are G-subgroups and, obviously, A = P ∘ Q ∘ ⋅ ⋅ ⋅ . Lemma 4.7. A G-group H possesses a G-faithful irreducible character if and only if all the Sylow subgroups P, Q, . . . of the abelian component A of ScG (H) = A ∘ B (see the previous paragraph) possess the same property. Proof. Since ScG (H) = A ∘ B, it follows that φ G (ScG (H)) = φ G (A)φ G (B) since the function φ G is multiplicative. Let B = F1 ∘ ⋅ ⋅ ⋅ ∘ F k , where F1 , . . . , F k are irreducible, and let A = P × Q × ⋅ ⋅ ⋅ , where P, Q, . . . are Sylow subgroups of A. Since k

k

φ G (B) = ∏ φ(F i ) = ∏(|F i | − 1) ≠ 0, i=1

i=1 (1)

(1)

φ G (A) = φ G (P)φ G (Q)⋅ ⋅ ⋅ = φ G (P)φ G (Q)⋅ ⋅ ⋅, by Proposition 3.7, we see that (1)

(1)

(1)

φ G (Sc(H)) ≠ 0 ⇐⇒ φ G (P) ≠ 0, φ G (Q) ≠ 0, . . . ⇐⇒ φ G (H) ≠ 0 (Proposition 3.6). Now the result follows from Proposition 4.3.

IX Faithful characters | 423

Theorem 4.8. A G-group H possesses a G-faithful irreducible character if and only if every Sylow subgroup of the abelian component A of its G-socle ScG (H) has a maximal subgroup not containing nonidentity G-subgroups of H. Proof. Let S ∈ Syl(A) and let λ be a G-faithful irreducible character of S. Since λ ≠ 1S is linear, ker(λ) is maximal in S since S is elementary abelian. By assumption, ker(λ)G = kerG (λ) = {1S } has no nonidentity G-subgroup of H. Conversely, if M is a maximal subgroup of S with identity G-core, then M = ker(λ), where λ ∈ Irr(S), and λ is G-faithful by the choice of M. Thus, S possesses a G-faithful irreducible character if and only if if it contains a maximal subgroup with identity G-core. Now the result follows from Lemma 4.7. Taking G = H, we obtain the following Weisner [Weis2] criterion. A group G possesses a faithful irreducible character if and only if every Sylow subgroup of the abelian part A of the socle Sc(G) contains a maximal subgroup with the identity core in G. 2°. In what follows, let H be a G-group. In this subsection ψ denotes an irreducible G character of H. Given N ≤ H, we set (N)

σG =

ψ(1)ψ.

∑

(7)

ψ∈Irr(H), kerG (ψ)=N

(N)

It is clear that σ G ≠ 0 if and only if N is a G-kernel of H. (N)

Definition 4.5. A function σ G , where N is a G-kernel of a G-group H, is said to be a G-layer character of H. The following lemma justifies this name. (N)

Lemma 4.9. The function σ G is constant on G-layers of H. Proof. Given N ⊴ H, write ρ(N) =

ψ(1)ψ.

∑

ψ∈Irr(H), ker(ψ)≥N

Note that ρ(N) is the inflation to H of the regular character of the group H/N:

G

{|H/N| ρ(N) (x) = { 0 {

if x ∈ N, if x ∈ H − N.

If N ≤ H, then ker(ψ) ≥ N if and only if kerG (ψ) ≥ N. In that case, ρ(N) =

∑

ψ(1)ψ.

{|H/N| ρ(N) (x) = { 0 {

if ⟨⟨x⟩⟩G ≤ N,

ψ∈Irr(H), kerG (ψ)≥N

Then (8) takes the following form:

if ⟨⟨x⟩⟩G ≰ N.

(8)

424 | Characters of Finite Groups 1 G

This shows that, given N ≤ H, the function ρ(N) is constant on G-layers of H. To complete the proof, it is enough to prove the following equality: (N)

∑ μ G (D/N)ρ(D) .

σG =

(9)

G

N≤D≤H

Denoting the right-hand side of (9) by S and using the first of relations in (3.7), we get (below ψ ∈ Irr(H) satisfying a certain condition) S=

∑ μ G (D/N) N≤D≤H

=

ψ(1)ψ =

∑ D≤kerG (ψ)

G

ψ(1)ψ

∑ N≤kerG (ψ)

ψ(1)ψ

∑ N≤kerG (ψ)

μ G (D/N) =

∑ G

D/N ≤kerG (ψ)/N

μ G (D/N)

∑ G

N≤D≤kerG (ψ) (N)

∑

kerG (ψ)=N

ψ(1)ψ = σ G .

Now we are ready to prove the main result of this section. Theorem 4.10. The number of G-kernels of a G-group H is equal to the number of its G-antikernels. Taking G = {1} (or G = H) in Theorem 4.10, we obtain: The numbers of kernels and antikernels of a group H are equal (see [BZ3, Corollary 9.1.8]). G Recall that the set W(D) is defined after formula (1) as follows: If D ≤ H, then W(D) = {x ∈ D | ⟨⟨x⟩⟩G = D}. If D is not an antikernel, then W(D) = 0. The set W(D) is said to be G-layer of the G-group D. First we will prove the following combinatorial identity. Lemma 4.11. Let f : H → A be a mapping of a G-group H into an additive abelian G group A. Then, for every N ≤ H, ∑ f(x) = ∑ μ G (N/D) ∑ f(x). x∈W(N)

x∈D

G

D≤N

Proof. Denoting by T the right-hand side of (10), one obtains T = ∑ f(x) x∈N

μ G (N/D).

∑ G

⟨⟨x⟩⟩G ≤D≤H

G

Given D ≤ N and x ∈ D, put H = H/⟨⟨x⟩⟩G , D = D/⟨⟨x⟩⟩G . We obtain, using the second relation in (3.7), {1 μ G (N/D) = ∑ μ G (N/D) = { 0 G G { ⟨⟨x⟩⟩G ≤D≤N D≤N ∑

Therefore, T=

∑ ⟨⟨x⟩⟩G =N

f(x) =

∑ f(x). x∈W(N)

if ⟨⟨x⟩⟩G = N, otherwise.

(10)

IX Faithful characters | 425

Proof of Theorem 4.10. The functions θ : H → ℂ that are constant on G-layers of the G-group H form a linear ℂ-space FG [H]. The dimension of this space is equal to the number m G = m G (H) of G-layers of H. As we have noted, m G is the number of G-antikernels of H. Define an inner product on θ1 , θ2 ∈ FG [H] by ⟨θ1 , θ2 ⟩ = |H|−1 ∑ θ1 (x)θ2 (x). x∈|H

Let m∗G = m∗G (H) be the number of G-kernels of H and let {N1 , . . . , N m∗G } be the set of (N ) these G-kernels. The functions σ G i (i = 1, . . . , m∗G ), as it is easy to see from (7) and Lemma 4.9, form an orthogonal system in the space FG [H]. Therefore, m∗G ≤ dim(FG [H]) = m G . If m∗G < m G , then there is in the space FG [H] a nonzero function θ orthogonal to all G (N ) (D) functions σ G i , and therefore to all functions σ G (D ≤ H). Thus, (D)

⟨θ, σ G ⟩ = 0

G

(11)

(D ≤ H).

G

Noting that, for N ≤ H, we have (here ψ ∈ Irr(H)) (D)

∑ σG = G

N≤D≤H

∑ G

N≤D≤H

∑

kerG (ψ)=D

ψ(1)ψ =

∑

ψ∈Irr(H), kerG (ψ)≥N

ψ(1)ψ = ρ(N)

G

and summing up (11) over all D such that N ≤ D ≤ H, we get ⟨θ, ρ(N) ⟩ = 0. In view G of (8), this gives ∑x∈N θ(x) = 0 for every N ≤ H. By Lemma 4.11 with A = ℂ, we obtain ∑ θ(x) = ∑ μ G (N/D) ∑ θ(x) = 0. x∈W(N)

x∈D

G

D≤N

Now let S be an arbitrary G-layer of H and N = ⟨⟨x⟩⟩G for some x ∈ S. Then W(N) = S. If x0 ∈ S, then θ(x0 ) = θ(x) for all x ∈ S. Therefore, we get |S|θ(x0 ) = 0, i.e., θ vanishes on S. Since H is the union of G-layers, it follows that θ = 0, contrary to the choice of θ. Thus, m∗G = m G , and the proof is complete. (N i )

Let, as above, {N1 , . . . , N m G } be the set of all G-kernels of H, σ i = σ G

(i = 1, . . . , m G ).

Corollary 4.12. The G-layer characters σ1 , . . . , σ m G of the G-group H form an orthogonal basis of the space FG [H]. Exercise 4.1. Prove that the G-layer characters σ1 , . . . , σ m G of the G-group H satisfy the following orthogonality relations: (1)

⟨σ i , σ j ⟩ = δ ij φ G (H/N i ), where δ is the Kronecker delta on the set {1, . . . , m G }. Hint. Use relations (7) and (6).

(12)

426 | Characters of Finite Groups 1 Exercise 4.2. Let H be a G-group. Prove that, for any x, y ∈ H, one has mG

∑ i=1

Here

{ |H| 1 σ i (x)σ i (y) = { l x fi 0 { (1)

f i = φ G (H/N i ) =

if ⟨⟨x⟩⟩G = ⟨⟨y⟩⟩G , if ⟨⟨x⟩⟩G = ⟨⟨y⟩⟩G . ∑

(13)

ψ(1)2 ,

ψ∈Irr(H), kerG (ψ)=N i

and l x is the cardinality of the G-layer L x of H containing an element x. Hint. Let L1 , . . . , L m G be all G-layers of H, l α = |L α |, x α ∈ L α for α = 1, . . . , m G . Set a iα =

lα σ i (x α ), |H|f i

b αj = σ j (x α ).

Show that m G × m G matrices A = (a iα ) and B = (b αj ) satisfy AB = Im G = BA. Exercise 4.3. Let H be a G-group and x, y ∈ H. Prove that ⟨⟨x⟩⟩G = ⟨⟨y⟩⟩G ⇐⇒ σ i (x) = σ i (y) (i = 1, . . . , m G ). 3°. In this subsection we consider two commutative algebras: the algebra RG [H] of layers and the algebra R∗G [H] of layer characters associated with the G-group H. The support of R∗G [H] is the space FG [H] of functions θ : H → ℂ constant on G-layers with pointwise multiplication of functions. By Corollary 4.12, G-layer characters σ1 , . . . , σ m G form a basis of the algebra R∗G [H]. Let c∗ijk = c∗ijk (H) (i, j, k = 1, . . . , m G ) be the structure constants of this algebra in the basis {σ1 , . . . , σ m G }: mG

σ i σ j = ∑ c∗ijk σ k

(i, j = 1, . . . , m G ).

(14)

k=1

We call R∗G [H] the algebra of G-layer characters of the G-group H. Exercise 4.4. Prove that the structure constants c∗ijk of the algebra R∗G [H] are nonnegative integers. Hint. Since σ i σ j and σ k are ordinary characters of H, it follows that c∗ijk =

⟨σ i σ j , σ k ⟩ ≥ 0. ⟨σ k , σ k ⟩

To prove that c∗ijk are (rational) integers, use relation (9) and the following property of characters ρ(N) : D1 , D2 ⊴ H 󳨐⇒ ρ(D1 ) ρ(D2 ) = |H/D1 D2 |ρ(D1 ∩D2 ) . Definition 4.6. Let {L1 , . . . , L m G } be the set of G-layers of a G-group H. The element s i = ∑x∈L i x of the group algebra ℂH is said to be a G-layer sum, corresponding to the G-layer L i . Since L i is a union of some G-classes, it follows that s i ∈ Z(ℂH), the center of the group algebra ℂH (see Chapter II).

IX Faithful characters | 427

Exercise 4.5. Let A i be a G-antikernel of a G-group H generated by the G-layer L i . Prove that s i = ∑ μ G (A i /D)[D],

where [D] = ∑ x.

(15)

x∈D

G

D≤A i

Hint. Taking into account that L i = W(A i ), make use of Lemma 4.11 with N = A i , A = ℂH, f = id. Exercise 4.6. Prove that for the G-layer sums s i (i = 1, . . . , m G ) the following multiplication table takes place: mG

s i s j = ∑ c ijk s k

(i, j = 1, . . . , m G ),

(16)

k=1

where c ijk = c ijk (H) are nonnegative integers. Hint. Use relation (15), previously proving that [D1 ][D2 ] = |D1 ∩ D2 |[D1 D2 ],

G

D1 , D2 ≤ H

(see Exercise 4.5). Relations (16) show that the ℂ-linear hull RG [H] of the G-layer sums s i , . . . , s m G is a subalgebra of the algebra Z(ℂH). We call it the algebra of the G-layers of H. In the following exercises we find the minimal idempotents of the algebra RG [H]. This allows us to produce a new proof of Theorem 4.10. It is easy to see that RG [H] is a semisimple algebra. Therefore it has exactly m G = dimℂ (RG [H]) minimal idempotents, which is the number of G-antikernels of H. Let u1 , . . . , u k (where k = k(H) is the class number of H) be the complete system of minimal idempotents of the algebra Z(ℂH). As we know, u i u j = δ ij u i (see Chapter I). Let ∆ α be a linear representation of the group H, afforded by the left ideal ℂHu α of the group algebra ℂH. It is known that ∆ α is a multiple of some irreducible representation Γ α of H: Γ α is afforded by every minimal left ideal of the simple algebra ℂHu α . Obviously, ker(Γ α ) = ker(∆ α ) (α = 1, . . . , k). Let ψ α be the character of the representation Γ α (α = 1, . . . , k). Then Irr(H) = {ψ1 , . . . , ψ k }. Exercise 4.7. Prove that ker(ψ α ) = {x ∈ H | xu α = u α }. Hint. Show that {x ∈ H | xu α = u α } coincides with ker(∆ α ) = ker(Γ α ) = ker(ψ α ). Let us correspond to N ≤ H the element j N = |N|−1 [N] = |N|−1 ∑ x ∈ ℂH. x∈N

Exercise 4.8. Prove that j N is an idempotent of the group algebra ℂH. In particular, N ⊴ H 󳨐⇒ j N ∈ Z(ℂH), j N =

∑ ψ α ∈Irr(H), N≤ker(ψ α )

uα .

428 | Characters of Finite Groups 1 Next, G

N ≤ H 󳨐⇒ j N = ∑ u α ,

(17)

I α ≥N

where I α = kerG (ψ α ) is the G-kernel of ψ α . Exercise 4.9. If N1 , N2 ⊴ H, then j N1 j N2 = j N1 N2 . Hint. Use Exercise 4.8. (One may also use the relation [N1 ][N2 ] = [N1 ∩ N2 ][N1 N2 ] for which one can give a group-theoretic proof.) G

Let N ≤ H. Set wN = ∑ uα ,

where I α = kerG (ψ α ).

I α =N

(18)

It follows from (18) that w N ≠ 0 if and only if N is a G-kernel of the G-group H. G

Exercise 4.10. Let N ≤ H. Prove that ∑ μ G (D/N)j D .

wN =

(19)

G

N≤D≤H

Hint. Use (17) and (3.7). G

Exercise 4.11. Let N ≤ H. Prove that w N = ∏D≤M G (N) (j N − j D ), where M G (N) is the set of G-subgroups of H covering N (see §IX.3, Section 4°). G

Exercise 4.12. Prove that j D ∈ RG [H] for all D ≤ H. Hint. Use the equality [D] = ∑A α ⊆D s α . (Recall that A α is the G-antikernel generated by G-layer L α ; see Definition 4.6.) G

Exercise 4.13. Prove that w N ∈ RG [H] for N ≤ H. Hint. Use equality (19) and Exercise 4.12. Exercise 4.14. Let {N1 , . . . , N m∗G } be the set of all G-kernels of the G-group H. Prove that {w N1 , . . . , w N m∗ } is the full system of minimal idempotents of the algebra RG [H]. G Deduce from this that m G = m∗G (Theorem 4.10). G

Hint. Show that j D = ∑N i ≥D w N i for any D ≤ H. It follows, in view of (15), that s α = ∑ μ G (A α /D)|D|j D G

D≤H

is contained in the ℂ-linear hull W of idempotents w N i (i = 1, . . . , m∗G ). Since the elements s α form a basis of the algebra RG [H], we have, by Exercise 4.13, that RG [H] = W. Since the idempotents w N i form an orthogonal system and ≠ 0, they are linear independent, and it follows that m G = dimℂ (RG [H]) = dimℂ (W) = m∗G .

IX Faithful characters | 429

5 Application to the group G 1°. In this subsection we consider a special case when the group H = G. In that case, G-subgroups of G are exactly its normal subgroups (= G-invariant subgroups); the G-socle ScG (G) of G coincides with its usual socle Sc(G) (product of all minimal normal subgroups of G); the G-kernel kerG (χ) = ker(χ) for a character χ of G; G-kernels of G are kernels of irreducible characters of G and will be called kernels of G. If x ∈ G, then ⟨⟨x⟩⟩ coincides with normal closure of x in G. Next, G-antikernels of G are normal closures of elements of G; we will call them antikernels of G. Note that an antikernel is a subgroup generated by conjugacy class of G. We will call G-layers of G layers of G: two elements x, y ∈ G belong to the same layer if and only if ⟨⟨x⟩⟩ = ⟨⟨y⟩⟩. The G-layer characters of G will be called layer characters of G. (i) Note some formulas for Euler functions φ G (i = 1, 2) which are important in what follows. In the sequel χ ∈ Irr(G), N ⊴ G. It follows from (4.5) that (1)

φ G (G/N) =

χ(1)2 .

∑

(1)

χ∈Irr(G), ker(χ)=N

In particular, taking N = {1}, we get (1)

φ G (G) =

∑

χ(1)2 .

(2)

χ∈Irr(G) is faithful

Next, in view of (4.7), (N)

σG =

∑

(3)

χ(1)χ.

χ∈Irr(G), ker(χ)=N (N)

In particular, if N is a kernel of G, then σ G is the layer character of G corresponding (N) to N; otherwise, σ G = 0. It follows from (1) and (2): (1)

Proposition 5.1. A subgroup N ⊴ G is a kernel of a group G if and only if φ G (G/N) ≠ 0. ({1}) In particular, the group G has a faithful irreducible character if and only if φ G (G) ≠ 0. As a consequence of Proposition 4.1 we obtain: (2)

Proposition 5.2. If N ⊴ G, then φ G (N) ≠ 0 if and only if N is an antikernel of G. Under (2) the last condition, φ G (N) is equal to the cardinality of the layer generating the (normal) subgroup N. Theorems 4.4, 4.5, 4.6 and 4.8 in the case H = G lead to the following assertions. Theorem 5.3. If U, V are mutually complementary normal subgroups of a group G (i.e., UV = U ∗ and U ∩ V = V∗ ), then the following assertions are equivalent: (a) U is a kernel of G. (b) V is an antikernel of G.

430 | Characters of Finite Groups 1 Theorem 5.4 (Gaschütz [Gas3]). A group has a faithful irreducible character if and only if its socle is an antikernel. In particular, if Sc(G)󸀠 = Sc(G), then G has a faithful irreducible character. Indeed, in that case, Sc(G) = S1 × ⋅ ⋅ ⋅ × S k , where S1 , . . . , S k are nonabelian simple. If x i ∈ S#i for all i and x = x1 ⋅ ⋅ ⋅ x k , then T = ⟨⟨x⟩⟩G = Sc(G). Assume that this is false. Let, for definiteness, T ≤ U = S1 × ⋅ ⋅ ⋅ × S k−1 . Then x1 ⋅ ⋅ ⋅ x k−1 ∈ U but x = x1 ⋅ ⋅ ⋅ x k−1 x k ∈ ̸ U, which is a contradiction. Theorem 5.5 (Shoda [Sho]). A group G has a faithful irreducible character if and only if the G-invariants n and j of the abelian component of its socle satisfy n ≤ j. Theorem 5.6 (Weisner [Weis2]). A group has a faithful irreducible character if and only if every Sylow subgroup of the abelian component of its socle has a maximal subgroup with identity core in G. In particular, if the abelian component of the socle has square-free order, then a G-group G has a faithful irreducible character. This also follows from Theorem 5.4 since then the abelian component of Sc(G) is cyclic. In particular, the groups all of whose Sylow subgroups are either cyclic or generalized quaternion have a faithful irreducible character. Theorem 5.7 (Zhmud [Zhm1, Zhm3]). In every group, the number of kernels is equal to the number of antikernels. 2°. The obtained results show that there is a duality between antikernels and kernels of a finite group. This duality became especially noticeable if we present the relations (1) (2) for φ G and φ G proved above in another form. Let T be an arbitrary finite set. Set μ(T) = (−1)|T| . (In particular, μ(0) = 1.) Given a nonempty set T of G-subgroups of a G-group H, let C T denote the product of all members of the set T provided T ≠ 0; besides, we put C0 = E H . Next, let D T be the intersection of all members of the set T ≠ 0; besides, we put D0 = H. Let MH and MH be the sets of minimal and maximal G-subgroups of a G-group H, respectively. Then CMH = ScG (H),

DMH = AsG (H).

Exercise 5.1. Let H be a completely reducible G-group. Prove the following two relations: ∑

T⊆MH , D T =E H

μ(T) = μ G (H) =

∑

μ(T).

(4)

T⊆MH , C T =H

Hint. Let f denote a function vanishing on all G-groups that are not completely reducible and coinciding with ∑T⊆MH , C T =H μ(T) provided H is a completely reducible G G-group. Check that ∑D≤H f(D) = 0 if H ≠ {1} and 1, otherwise. This yields the first relation in (4). Next, let h is a function vanishing on all G-groups that are not completely reducible and coinciding with ∑T⊆M(H) , D T ={1} μ(T) if H is a completely

IX Faithful characters | 431

G reducible G-group. Check that ∑D≤H h(H/D) = 0 if H ≠ {1} and 1, otherwise. This yields the second relation in (4).

Exercise 5.2. If H is a G-group, then the following relations hold: (1)

(5)

φ G (H) = ∑ μ(T)|H/C T |, T⊆MH (2)

(6)

φ G (H) = ∑ μ(T)|D T |. T⊆MH

Hint. Use Exercise 5.1. Now let N ⊴ G, let M(N) be the set of all normal subgroups of G that cover N, and let M(N) be the set of all normal subgroups of G that are covered by N. Applying formula (5) to the G-group G/N, we obtain (1)

φ G (G/N) =

∑

μ(T)|G/C T |.

(7)

μ(T)|D T |.

(8)

T⊆M(N), C0 =N

It follows from (6) the following relation: (2)

φ G (N) =

∑ T∈M(N), D0 =N

The lattice LG (G) of G-subgroups of G coincides with the lattice L(N) (G) of normal subgroups of G. In the sequel, G and G denote two finite groups. Definition 5.1. A map α : L(N) (G) → L(N) (G) is said to be a strong isomorphism if it satisfies the following conditions: (a) α is a lattice isomorphism, i.e., a bijection keeping containment of normal subgroups. (b) |M α | = |M| for all M ∈ L(N) (G). Exercise 5.3. Prove that if α is a strong isomorphism of the lattice L(N) (G) onto L(N) (G) and M ∈ L(N) (G) is a kernel (antikernel), then M α is a kernel (antikernel) of G. (1)

(1)

(2)

(2)

Hint. Using (7) and (8), prove that φ G (G/N) = φ (G/N α ), φ G (N) = φ (N α ). Make G G use of Propositions 5.1 and 5.2. Note that there exist groups G and G such that α : L(N) (G) → L(N) (G) is an isomorphism that do not sends kernels (antikernels) into kernels (antikernels); of course, in the case under consideration, α is not a strong isomorphism. Definition 5.2. A map α : L(N) (G) → L(N) (G) is said to be a strong anti-automorphism if it satisfies the following two conditions: (a) α is a lattice anti-automorphism, i.e., a bijection reversing containment of normal subgroups. (b) |M α | = |G/M| for all M ∈ L(N) (G).

432 | Characters of Finite Groups 1 Exercise 5.4. Prove that if there is α a strong anti-automorphism of the lattice L(N) (G) onto the lattice L(N) (G) and M ∈ L(N) (G) is an antikernel (kernel), then M α is a kernel (antikernel). (2)

(1)

(1)

(2)

Hint. Using (7) and (8), prove that φ G (M) = φ (G/M α ), φ G (G/M) = φ (M α ) and G G use Propositions 5.1 and 5.2. 3°. In this subsection we apply to a group G the results of §IX.4 obtained for algebras RG [H] and R∗G [H]. Let S be a layer of a group G; then the corresponding layer sum s = ∑x∈S x is an element of the algebra Z(ℂG), the center of the group algebra ℂG (recall that S is a union of G-classes). Let {S1 , . . . , Sm } be the set of layers of G and s1 , . . . , s m the corresponding layer sums. Here m is the number of antikernels (kernels) of G. The set {s1 , . . . , s m } is a basis of the algebra R[G] of layers of G. These elements satisfy the following relations: m

s i s j = ∑ c ijk s k

(i, j = 1, . . . , m),

(9)

k=1

where c ijk = c ijk (G) (1 ≤ i, j, k ≤ m) are nonnegative rational integers. The explicit form of the structure constants c ijk of the algebra R[G] can be established if we use (4.15) in the case H = G. Let σ(N) be the layer character of G corresponding to its kernel N (see (4.7) where (N) it was denoted by σ G ). Exercise 5.5. Prove that σ(N) =

μ(T)ρ(C T ) ,

∑

(10)

T∈M(N)

where C0 = N and ρ(C T ) is the inflation onto G of the regular character of the quotient group G/C T . Hint. Use relation (4.9) for H = G and Exercise 5.1. Let {N1 , . . . , N m } be the set of kernels of the group G. Set σ i = σ(N i ) (i = 1, . . . , m). The layer characters σ i are linearly independent and form a basis of the algebra R∗ [G] of layer characters of G. They satisfy the following relations: m

σ i σ j = ∑ c∗ijk σ k

(i, j = 1, . . . , m),

k=1

where the structure constants c∗ijk = c∗ijk (G) ∈ ℕ ∪ {0}. The explicit form of these structure constants can be obtained if we use (4.9) (for H = G) or (5.10). Definition 5.3. Given two groups G and G, the lattices of normal subgroups L(N) (G) and L(N) (G) are said to be strongly isomorphic (strongly anti-isomorphic) if there exists a strong isomorphism (strong anti-isomorphism) of these lattices.

IX Faithful characters | 433

Exercise 5.6. Prove that the lattices L(N) (G) and L(N) (G) are strongly isomorphic if and only if there exists a one-to-one correspondence between layer characters of the groups G and G such that the multiplication tables of their layer sums (layer characters) coincide. Exercise 5.7. Prove that the lattices L(N) (G) and L(N) (G) are strongly anti-isomorphic if and only if the multiplication table of the layer sums of one of these groups coincides (under appropriate numbering) with the multiplication table of the layer characters of another group. Hints to Exercises 5.6 and 5.7. Find the explicit form of the structure constants c ijk (G) and c∗ijk (G). For further generalizations, see [BZ3, §9.1]. 4°. Almost all of the above results allow a natural generalization. Let k ∈ ℕ and let X = (x1 , . . . , x k ) be an element of G k , the direct product of k copies of a group G. Let ⟨⟨X⟩⟩ be the intersection of all normal subgroups of G containing all elements x i , i = 1, . . . , k. A subgroup N of G is said to be a k-antikernel of G provided N = ⟨⟨X⟩⟩ for some X ∈ G k . It is clear that k-antikernels of G are subgroups generated by k (not necessarily distinct) G-classes. The elements X, Y ∈ G k are called belonging to the same k-layer if and only if ⟨⟨X⟩⟩ = ⟨⟨Y⟩⟩. There is a natural one-to-one-correspondence between k-layers and k-antikernels of the group G. A normal subgroup N of G is said to be a k-kernel if N = ker(χ1 + ⋅ ⋅ ⋅ + χ k ) for some χ i ∈ Irr(G) not necessarily distinct (i = 1, . . . , k). Since k

N = ⋂ ker(χ i ), i=1

it follows that k-kernels of G are subgroups presentable as intersections of k (not necessarily distinct) kernels. In particular, kernels of G are its 1-kernels. If k > 1, then (k − 1)-kernels are k-kernels. The following generalizations of Theorems 5.3, 5.4, 5.5 and 5.7 hold (for details, see [BZ3, §9.1]). Theorem 5.8 (Zhmud). Let {U, V} be a pair of mutually complemented normal subgroups of a group G (for definition, see the paragraph preceding Theorem 3.11). Then, for every k ∈ ℕ, the following conditions are equivalent: (a) U is a k-kernel of G. (b) V is a k-antikernel of G. Theorem 5.9 (Zhmud). A group G possesses a faithful character with k irreducible constituents if and only if its socle is a k-antikernel. (Note that a character χ ∈ Char(G) of the form χ = a1 μ1 + ⋅ ⋅ ⋅ + a s μ s , where all μ i are irreducible, has a1 + ⋅ ⋅ ⋅ + a s irreducible constituents.)

434 | Characters of Finite Groups 1 Theorem 5.10 (Tazawa [Taz]). A group G has a faithful character with k irreducible constituents if and only if the invariants n and j of every abelian homogeneous component of the socle Sc(G) satisfy n ≥ kj. Theorem 5.11 (Zhmud). Given k ∈ ℕ, the number of k-kernels of G is equal to the number of its k-antikernels. It is also possible to generalize the algebras of layers and layer characters and all the above proved results relative them. The proofs of the above results are based on consideration of generalized group(1) (2) theoretical Euler functions φ G,k and φ G,k . These more general functions are defined on the class CG of G-groups in narrow sense, as follows: (1)

φ G,k = ∑ μ G (H/D)|H/D|k , G

D≤H

(2)

φ G,k = ∑ μ G (H)|D|k , G

D≤H

All proofs (however, more complicated technically) completely coincide with the presented ones in §§IX.1–IX.5. For proofs that do not use group-theoretical Euler functions, we refer to [BZ3, §9.1].

6 CM-groups In this section we apply the apparat of group-theoretical Euler functions to a class of groups which arises in natural way. 1°. Let G be a group and K a field of characteristic not dividing |G|. Characters of K-representations of G are K-characters.¹ It is possible to prove that the set of kernels of irreducible K-characters of G is independent of K and so coincides with the set of kernels of G (see §IX.5). It follows that the number m(G) of kernels of G does not exceed the number r K (G) of its irreducible K-characters. A group G is called a KM-group if m(G) = r K (G) (i.e., distinct irreducible K-characters of G have distinct kernels). In what follows we suppose that K = ℂ. A group G is called a ℂM-group (or, as in [BZ3, §9.3], CM-group) if m(G) = |Irr(G)| = k(G), the class number of G. This means that distinct irreducible characters of a CM-group G have distinct kernels. Epimorphic images of CM-groups are CM-groups. It is easy to check that if an abelian p-group G > {1} is a CM-group, then exp(G) = 2. It follows that if a p-group G > {1} is a CM-group, then p = 2 and Φ(G) = G󸀠 . Theorem 5.7 allows us to give a dual definition of CM-groups (see Proposition 6.1). Let x ∼ y mean that the elements x, y belong to the same G-class. If x ∼ y, then ⟨⟨x⟩⟩ = ⟨⟨y⟩⟩. 1 In that case, χ(x) ∈ K for all x, but it is not true, generally speaking, that if χ(x) ∈ K for all x ∈ G, then χ is a K-character; see Chapter VIII.

IX Faithful characters | 435

From Theorem 5.7 we obtain the following: Proposition 6.1. For a CM-group G and x, y ∈ G we have ⟨⟨x⟩⟩ = ⟨⟨y⟩⟩ if and only if x ∼ y. Examples. The following groups are CM-groups: (a) E2n , an elementary abelian 2-group. (b) ES(m, 2), an extraspecial 2-group of order 21+2m . (c) A Frobenius groups Φ2⋅3n of order 2 ⋅ 3n with the kernel E3n . (d) The Frobenius group Φ72 with a complement Q8 . In the sequel we will show that every nonidentity CM-group has order 2a 3b , where a > 0. We also classify the CM-groups with trivial center. 2°. In this subsection we establish some simple properties of CM-groups. Proposition 6.2. Epimorphic images of CM-groups are CM-groups. Lemma 6.3. Let G = G1 × G2 , χ = χ1 × χ2 , where χ i ∈ Irr(G i ), i = 1, 2 (see §IV.8). Then ker(χ i ) = G i ∩ ker(χ), i = 1, 2. Proposition 6.4. Let G = G1 × G2 . Then G is a CM-group if and only if G1 and G2 are CM-groups. Proof. In view of Proposition 6.2, it is enough to prove that if G1 and G2 are CM-groups so is G = G1 × G2 . If χ, ϑ ∈ Irr(G), then χ = χ1 × χ2 , ϑ = ϑ1 × ϑ2 , χ i , ϑ i ∈ Irr(G i ), i = 1, 2. If ker(χ) = ker(ϑ), then, by Lemma 6.3, ker(χ i ) = G i ∩ ker(χ) = G i ∩ ker(ϑ) = ker(ϑ i ), and so χ i = ϑ i for i = 1, 2, since G i is a CM-group (i = 1, 2). It follows that χ = χ1 × χ2 = ϑ1 × ϑ2 = ϑ. Thus, G is a CM-group. Proposition 6.5. Let F be a minimal normal subgroup of a CM-group G. Then the set F # is a G-class. Proof. If x, y ∈ F # , then ⟨⟨x⟩⟩ = F = ⟨⟨y⟩⟩, and so x ∼ y in G, by Proposition 6.1. Corollary 6.6. A minimal normal subgroup of a CM-group G is elementary abelian. Proof. By the proof of Proposition 6.5, all the elements of the set F # have the same order so, by the lemma of Cauchy, this order is a prime number, say p. Proposition 6.7. CM-groups are solvable. Proof. Use induction on |G| and Proposition 6.1 and Corollary 6.6. Definition 6.1. A group G is said to be rational or ℚ-group, if all its characters are rational valued (see §III.8).

436 | Characters of Finite Groups 1 It is easy to see that G is a ℚ-group if and only if x ν ∼ x for every x ∈ G and every integer ν coprime with |G| (see §III.8). This condition is equivalent to the following one: if x, y ∈ G such that ⟨x⟩ = ⟨y⟩, then x ∼ y. Lemma 6.8. If Z is a cyclic subgroup of a ℚ-group G, then |NG (Z)/CG (Z)| = φ(|Z|), where φ(∗) is the Euler totient function. Proof. The action of NG (Z) on Z via conjugations affords a homomorphism f of this group into Aut(Z). Let σ ∈ Aut(Z), Z = ⟨z⟩. Then ⟨σ(z)⟩ = ⟨z⟩. Since G is a ℚ-group, it follows that σ(z) ∼ z, i.e., σ(z) = tzt−1 , where t ∈ G. However, in the case under consideration, the element t normalizes ⟨z⟩ so that t ∈ NG (Z). This shows that σ = f(t), i.e., f is onto. Since ker(f) = CG (Z), we get NG (Z)/CG (Z) ≅ Aut(Z). As |Aut(Z)| = φ(|Z|), we are done. Corollary 6.9. If Z is a cyclic subgroup of a ℚ-group G, then the following statements hold: (a) φ(|Z|) | |NG (Z) : Z|. (b) If exp(NG (Z) : Z) | 2, then |Z| ∈ {1, 2, 3, 4, 6, 8, 12, 24}. Indeed, (a) follows from Lemma 6.8. Now elementary number-theoretical considerations prove (b). Corollary 6.10. If a ℚ-group G is a 2-group of maximal class, then |G| = 8. This follows from Corollary 6.9 (b) and description of the 2-groups of maximal class (see [Ber31, Proposition 1.6, Theorem 1.2]). In particular, if G is a generalized quaternion group, then G ≅ Q8 . Proposition 6.11. CM-groups are ℚ-groups. Proof. Let x ∈ G and let ν ∈ ℤ be coprime with |G|. Since x ν and x have the same normal closure in G, they are conjugate in G. The claim now follows from Proposition 6.1. Corollary 6.12. A nonidentity CM-group is of even order. Proof. Since Aut(Cp ) is cyclic of order p − 1, the corollary follows from Lemma 6.8. As we have noted, an abelian CM-group has exponent ≤ 2. This is a partial case of the following proposition. Proposition 6.13. Let G be a CM-group. Then the number exp(Z(G)) divides 2. In particular, exp(G/G󸀠 ) divides 2. Proof. Let x ∈ Z(G). We have ⟨x⟩ = ⟨⟨x⟩⟩ = ⟨⟨x−1 ⟩⟩ 󳨐⇒ x ∼ x−1 󳨐⇒ x = x−1 󳨐⇒ x2 = 1. Thus exp(Z(G)) ≤ 2. Now the second assertion follows from Proposition 6.2 and the first assertion.

IX Faithful characters | 437

It follows from Proposition 6.13 that if a group G > {1} is nilpotent, it is a 2-group. It is conjectured by the third author that nilpotent CM-groups have derived length at most 2. Proposition 6.14. Let G be a CM-group, let χ ∈ Irr(G) and N = ker(χ). Then one has (1) χ(1)2 = φ G (G/N). Proof. As χ is the unique irreducible character of the group G with kernel N, the result follows from formula (5.1). (1)

Corollary 6.15. If N is a normal subgroup of a CM-group G such that φ G (G/N) ≠ 0, (1) then φ G (G/N) is a square of a natural number. Proof. This follows from Propositions 5.1 and 6.14. Exercise 6.1. Let G, χ and N be as in Proposition 6.14. Prove that χ(1)χ =

∑ μ G (D/N)ρ(D) .

N≤D⊴G

Hint. Use relations (5.3) and (4.9). 3°. Recall the following: Definition 6.2. A group G is called a monolith if it has a unique minimal normal subgroup. A monolith that is simultaneously a CM-group, is called a ℂM-monolith. A nonidentity p-group is a monolith if and only if its center is cyclic. Below we show how to reduce a study of an arbitrary CM-group to the case when it is a monolith. It is obvious that a monolith possesses a faithful irreducible character (this follows from the equality ⋂χ∈Irr(G) ker(χ) = {1}, which is valid for an arbitrary group G; indeed, the above intersection is the kernel of ρ G , the regular character of G, which is faithful). Therefore, ℂM-monolith has only one faithful irreducible character. Let N ⊴ G. Recall that N ∗ = ∏D∈M(N) D if N < G and N ∗ = N if N = G. (Here M(N) is the set of G-subgroups of H, covering N.) Next, N∗ = ⋂D∈M(G) D if N > {1} and N∗ = N if N = {1}. (Here M(N) is the set of G-subgroups of H, covered by N.) (See §IX.3 and §IX.5.) Lemma 6.16. Let F be a minimal normal subgroup of G. If N is the largest (by inclusion) normal subgroup of G such that F ∩ N = {1}, then G/N is a monolith with the unique G minimal normal subgroup FN/N ≅ F (in that case, N ∗ = FN). Proof. If R/N is a minimal normal subgroup of G/N different from FN/N, then R ∩ FN = N so R ∩ F = {1}. This is a contradiction since R > N. Corollary 6.17. Let G be a CM-group and let F, N be as in Lemma 6.16. Then G/N is a ℂM-monolith. This follows from Proposition 6.2 and Lemma 6.16.

438 | Characters of Finite Groups 1 Lemma 6.18. Let {F1 , . . . , F k } be the set of all minimal normal subgroups of a group G, and let N i be the largest normal subgroup of G such that F i ∩ N i = {1}, i = 1, . . . , k. Then ⋂ki=1 N i = {1}. Proof. Assume that ⋂ki=1 N i = D > {1}. Then F j ≤ D for some j. In that case, F j ≤ N j since D ≤ N j , which is a contradiction. Definition 6.3. A group G is said to be a ℂ-monolith if (i) G is a monolith, (ii) G has a unique faithful irreducible character. Obviously, ℂM-monoliths are ℂ-monoliths, but the converse is not true. However, the following result holds. Proposition 6.19. Let F be (the unique) minimal normal subgroup of a ℂ-monolith G. If G/F is a CM-group, so is G. Proof. Recall that G is a CM-group if and only if its distinct irreducible characters have distinct kernels. Assume that χ, ψ ∈ Irr(G) are such that ker(χ) = K = ker(ψ). We have to prove that χ = ψ. If F ≤ K, then χ and ψ as characters of G/F, have the same kernel K/F so χ = ψ since G/F is a CM-group. Now let F ≰ K. Then K = {1} since F is the unique minimal normal subgroup of G. In that case, χ = ψ since the ℂ-monolith G has only one faithful irreducible character. We will now prove some properties of a ℂ-monolith G with a minimal normal subgroup F and (the unique) faithful irreducible character χ. The following two propositions were proved by Zhmud; however, D. Chillag proved many parts of them, independently. The presented proof is due to Zhmud. Proposition 6.20. A monolith G with (unique) minimal normal subgroup F is a ℂ-monolith if and only if it satisfies the following conditions: (a) F # is a G-class. (b) If f ∈ F # , x ∈ G − F, then fx ∼ x. Proof. (i) Let G be a ℂ-monolith, and let Irr(G) = {χ1 , χ2 , . . . , χ r } (let ker(χ1 ) = {1}; then ker(χ i ) ≥ F for i > 1). Let ρ = ρ G be the regular character of G. Then r

r

ρ = ∑ χ i (1)χ i = χ1 (1)χ1 + ρ(F) ,

where ρ(F) = ∑ χ i (1)χ i .

i=1

i=2

It follows from properties of characters ρ and ρ(F) that χ1 (1)χ1 (x) = ρ − ρ(F)

|G| − |G/F| if x = 1, { { { = {−|G/F| if x ∈ F # , { { if x ∈ G − F. {0

(1)

(Note that G = {1} ∪ F # ∪ (G − F) is a partition.) If x, y ∈ F # , then, by (1), χ1 (x) = χ1 (y) = −

|G/F| . χ1 (1)

(2)

IX Faithful characters | 439

If i > 1, then x, y ∈ ker(χ i ) = F. Therefore, (3)

χ i (x) = χ i (y) = χ i (1).

It follows from (2) and (3) and the Second Orthogonality Relation that x ∼ y. Thus, F # is a G-class so assertion (a) is true. Let f ∈ F and x ∈ G − F. Then fx ∈ G − F, and so, in view of (1), χ1 (fx) = χ1 (x) = 0. If i > 1, then f ∈ ker(χ i ) so that χ i (fx) = χ i (x). Thus, χ i (fx) = χ i (x) for all i = 1, . . . , r. Therefore, fx ∼ x (see the previous paragraph), and so assertion (b) is true. (ii) Now suppose that a monolith G satisfies conditions (a) and (b). It remains to prove that G possesses only one faithful irreducible character. Let r0 be the number of faithful irreducible characters of G. Then (4)

r0 = r − r,

where r is the class number of G = G/F. Let C1 , . . . , C r be the set of G-classes, where C1 = {1} and C2 = F # . The images C i of the G-classes C i under the natural homomorphism x → x of G onto G exhaust all r conjugacy classes of G. Suppose that i, j > 2 and C i = C j . Then x i = x j if x i ∈ C i and x j ∈ C j . Therefore, x j ∼ fx i , where f ∈ F. Since fx i ∼ x i , by (i), we get x i ∼ x j , and so i = j. Thus, the classes C i (i = 3, . . . , r) are pairwise distinct. Since C1 = C2 (= {1}), it follows that {{1}, C3 , . . . , C r } is the set of all G-classes. Therefore, r = r − 1 so that r0 = 1, by (4). Thus, G is a ℂ-monolith. Proposition 6.21. Let G be a monolith with the unique minimal normal subgroup F and the unique faithful irreducible character χ (i.e., G is a ℂ-monolith). Then the following assertions hold: (a) F is an elementary abelian p-group for some prime p. (b) If f ∈ F # , then CG (f) ∈ Sylp (G). (c) If |F| = p λ and p μ = |G|p , then |G| = p μ (p λ − 1) and

μ ≡ λ (mod 2).

(5)

(d) If x ∈ G, then p(μ−λ)/2 (p λ − 1) { { { χ(x) = {−p(μ−λ)/2 { { {0

if x = 1, if x ∈ F # ,

(6)

if x ∈ G − F.

(e) If Z(G) > {1}, then p = 2,

λ = 1,

Z(G) = F,

|F| = 2,

|G| = 22ν+1

so that the group G is of central type. (f) If Z(G) = {1} and Q ∈ Sylq (G), q ≠ p, then Q is either cyclic or generalized quaternion.

440 | Characters of Finite Groups 1 Proof. (a) This follows from Proposition 6.20 (a) (see the proof of Corollary 6.6). Set |F| = p λ . (b) If f ∈ F # , then F ≤ CG (f), by (a). Since F # is a G-class, we get |G : CG (f)| = |F # | = p λ − 1. Let x ∈ CG (f)# . If x ∈ F # , then o(x) = p, by (a). If x ∈ ̸ F, then fx ∼ x (Proposition 6.20 (b)) which implies that o(fx) = o(x). Next, fx = xf 󳨐⇒ f o(x) = f o(x) x o(x) = (fx)o(x) = 1. Since o(f) = p, it follows that p divides o(x). Thus, the orders of all elements of the set CG (f)# are multiples of p. Therefore, CG (f) is a p-subgroup of G (Cauchy). Since |G : CG (f)| = p λ − 1 is not divisible by p, we conclude that CG (f) ∈ Sylp (G), and this completes the proof of (b). (c, d) If f ∈ F # , then, by (b), CG (f) = P ∈ Sylp (G),

|G : P| = |F # | = p λ − 1.

Setting |P| = p μ , we have |G| = p μ (p λ − 1). Next, by (1) with χ1 = χ, we get χ(1)2 = |G| − |G/F| = p μ (p λ − 1) − p μ−λ (p λ − 1) = p μ−λ (p λ − 1)2 , so that the number μ − λ is even and χ(1) = p(μ−λ)/2 (p λ − 1). If x ∈ F # , then χ(x) = −

|G : F| = −p(μ−λ)/2 , χ(1)

by (6). If x ∈ G − F, then χ(x) = 0, by (1). (e) If Z(G) > {1}, then F ≤ Z(G) since G is a monolith, and we conclude that |F| = 2 since F # is a G-class; then p = 2 and λ = 1. Therefore, 2μ = |G|, i.e., G is a 2-group. Since χ vanishes on G − F, we conclude that Z(G) = F. It follows from μ ≡ λ (mod 2) and λ = 1 that μ = 2ν + 1 is odd. Thus, |G| = 22ν+1 . We have χ F = χ(1)τ, where τ is the nonprincipal linear character of F. Then τ G (1) = |G : F| = 22ν . Since χ is the unique faithful irreducible character of G, we get τ G = χ(1)χ, by reciprocity. It follows that |G : Z(G)| = |G : F| = χ(1)2 so that the group G is of central type. (f) Now let Z(G) = {1}. Then G is not a prime power group. If S ∈ Sylq (G), q ≠ p, then, by (b), Φ = S ⋅ F is a Frobenius group with kernel F and complement S. Therefore, S is either cyclic or generalized quaternion (Theorem X.3.3). Proposition 6.22. The following statements hold: (a) Let G be a ℂ-monolith with minimal normal subgroup F of order p λ . If F ∈ Sylp (G), then G is isomorphic to a doubly transitive Frobenius group with kernel F. (b) Let G be a doubly transitive Frobenius group of degree q and let F be its Frobenius kernel. Then G is a ℂ-monolith with the minimal normal subgroup F. Proof. (a) By the theorem of Schur–Zassenhaus and Proposition 6.21 (a)–(b), G = H⋅F is a Frobenius group with kernel F and complement H of order |F| − 1 = p μ − 1. Clearly, G is doubly transitive.

IX Faithful characters | 441

(b) One has |F| = q. If G = (H, F), then |H| = q − 1. If f ∈ F ∗ , then |f H | = |H| = |F # |, so that F # = F H is a G-class. Next, G is a monolith and G has a faithful irreducible character χ of degree q − 1. One has χ(1)2 = |G| − |G/F| = q(q − 1) − (q − 1) = (q − 1)2 . On the other hand, ∑θ∈Irr(G),ker(θ)={1} θ(1)2 . Therefore, χ is the unique faithful irreducible character of G. Thus, G is a ℂ-monolith. Exercise 6.2. A group is a ℂ-monolith if and only if it possesses a faithful irreducible character that takes exactly two nonzero values (see also [BZ3, Lemma 27.2]). Every ℂ-monolith has a faithful irreducible character that takes exactly two nonzero values, by Proposition 6.21 (d). Exercise 6.3. If the socle S = Sc(G) of a CM-group G is a π-subgroup, then CG (S) is also a π-group (π is a set of primes). In particular, if all minimal normal subgroups of a CM-group G are contained in the center, then G is a 2-group. Solution. Due to Proposition 6.21, one may assume that |π(Sc(G))| > 1. Let F1 , . . . , F t be all minimal normal subgroups of G and let N j be a maximal G-invariant subgroup of G such that F j ∩ N j = {1}, j = 1, . . . , t. Then N1 ∩ ⋅ ⋅ ⋅ ∩ N t = {1}, by Lemma 6.18. Set λ |F i | = p i i , i = 1, . . . , t. Obviously, N j ≤ CG (F j ) for all j. By Proposition 6.21 (b), γj

|CG (F j )/N j | = p j

for all j.

γj

If g ∈ CG (F j ), then g p j ∈ N j . If, in addition, g ∈ CG (Sc(G)), then g ∈ CG (F j ) for all j, and therefore (Lemma 6.18) γ1

g p1

γ

...p t t

∈ N1 ∩ ⋅ ⋅ ⋅ ∩ N t = {1} 󳨐⇒ π(o(g)) ⊆ {p1 , . . . , p t } = π.

Thus, CG (Sc(G)) is a π-subgroup. The last assertion follows from Proposition 6.13. Exercise 6.4. If all even indices of a chief series of a group G are equal to 2, then G is 2-nilpotent. Solution. We proceed by induction on |G|. Let N be a minimal normal subgroup of G. By induction, G/N is 2-nilpotent. Therefore, we are done if |N| is odd. Now let |N| is even; then |N| = 2. Let H/N be a normal 2-complement of G/N; then H = N × F (Burnside). Since F is characteristic in H so normal in G and |G : F| is a power of 2, we are done. Similarly, if all indices of a chief series of a group G are equal to p or some p󸀠 -number and p is the least prime divisor of |G|, then G is p-nilpotent. Exercise 6.5. The Diophantine equation 2x − 3y = −1 has only two solutions in the nonnegative integers: {x, y} = {1, 1} and {3, 2}. Solution. If y is odd, then 2x = 3y − 1 ≡ 2 (mod 8) so x = 1; then y = 1. Now let y = 2z be even. Then we have 2x = (3z − 1)(3z + 1) so 3z − 1 and 3z + 1 are powers of 2. As (3z + 1) − (3z − 1) = 2, we get z = 1 so y = 2 and x = 3.

442 | Characters of Finite Groups 1 Exercise 6.6. The Diophantine equation 2x − 3y = 1 has only two solutions in the nonnegative integers: {x, y} = {1, 0} and {2, 1}. Solution. If y is even, then 2x = 3y + 1 ≡ 2 (mod 4) so x = 1; then y = 0. Now let y = 2z + 1 be odd. Then 2x = 32z+1 + 1 ≡ 4 (mod 8) so x = 2; then y = 1. Next we use some basic properties of Frobenius groups (see Chapter X). Theorem 6.23. If G > {1} is a CM-group, then π(G) ⊆ {2, 3}. Proof. Set π = {2, 3}. By Proposition 6.7, G is solvable. Let G be a counterexample of minimal order. Then all proper epimorphic images of G are π-groups. Therefore, G has only one minimal normal subgroup F = Oπ (G), |F| = p λ , p > 3; in that case, G is a ℂM-monolith. Therefore, |G| = 2α ⋅ 3β ⋅ p λ , where α > 0 (see Corollary 6.12). Since F ∈ Sylp (G) and G is a ℂM-monolith, it follows that G is a Frobenius group with kernel F (Proposition 6.21 (b)), G = (H, F), where H is a Frobenius complement. In that case, H ≅ G/F is a CM-group, and, by Proposition 6.21 (f), |H| = 2α 3β = p λ − 1. (i) Assume that β = 0; then |G| = 2α p λ . In this case, H is either cyclic or generalized quaternion (see Chapter X) so, by Proposition 6.13 and Corollary 6.10, H ∈ {C2 , Q8 }. The equality p λ − 1 = |H| implies that p = 3, contrary to the assumption. (ii) Assume that β > 0, i.e., π = π(H) = {2, 3}. Let Q ∈ Syl2 (H), P ∈ Syl3 (H). Then P is cyclic and Q is cyclic or a generalized quaternion group since H is a Frobenius complement. Next, 2 | |Z(H)| (Theorem X.3.3) and, moreover, |Z(H)| = 2 (see Proposition 6.13). By Exercise 6.3, H has a minimal normal subgroup V not contained in Z(H), and its order is 3. Since P and Q have only one subgroup of orders 3 and 2, respectively, it follows that U = Z(H) and V are the only minimal normal subgroups of H. (1) Let us compute φ H . The subgroup ScH (H) = Sc(H) = U ∘ V has order 6. By formula (3.34), multiplicativity of function φ H (see formula (3.38) with G = H) and Proposition 3.7, we get (1)

|H| 1 φ H (ScH (H)) = |H|φ H (U)φ H (V) |ScH (H)| 6 1 1 = |H|(|U| − 1)(|V| − 1) = |H| 6 3 = 2α 3β−1 .

φ H (H) =

(1)

(1)

Since H is a CM-group and φ H (H) ≠ 0, it follows from Corollary 6.15 that φ H (H) must be a square of a natural number so that α is even. Let us prove that this is impossible. Let N be a lower direct complement of U in H (i.e., N ⊲ H is as large as possible such that U ∩ N = {1}). Since U is a unique subgroup of order 2 in H, it follows that |N| is odd. Therefore, N < H is a 3-subgroup. On the other hand, by Lemma 6.16, H/N is a CM-monolith with a minimal normal subgroup ≅ U. Since |U| = 2, we conclude that H/N is a 2-group (Exercise 6.3). Therefore, N ∈ Syl3 (H) so that |H/N| = 2α . Since H/N is a CM-monolith, it follows that α is odd (Proposition 6.21 (e)), which is a contradiction. Thus, case (ii) is also impossible.

IX Faithful characters | 443

4°. In what follows, let the CM-group G be of the form G = PQ, where Q ∈ Syl2 (G) and P ∈ Syl3 (G) (see Theorem 6.23). Lemma 6.24. If f is an index of a principal series of a CM-group G, then f ∈ {2, 3, 9}. In particular, P ∈ Syl3 (G) is normal in G (Exercise 4). Thus, G = Q ⋅ P, a semidirect product with kernel P. Proof. One may assume that π(G) = {2, 3} (otherwise, G is a 2-group, by Corollary 6.12; see Theorem 6.23). We proceed by induction on |G|. Without loss of generality, we may assume that G has only one minimal normal subgroup F of order f . Put |F| = p λ (p ∈ {2, 3}). Then G is a ℂM-monolith and its order is equal to |G| = p μ (p λ − 1) (see Proposition 6.21 (c)). (i) Assume that p = 2. Then, by Proposition 6.21 (c), 2λ − 1 is a power of 3. This implies that λ ≤ 2 (Exercise 6.6) and so f ∈ {2, 4}. If f = 2, then G is a 2-group (Proposition 6.21 (e)), contrary to the assumption. If f = 4, then CG (F) is a 2-group (Exercise 6.3). Since Aut(F) ≅ S3 , we get |P| = 3. Since all even indices of a chief series of G/F, by induction, are equal to 2, a Sylow 3-subgroup PF/F is normal in G/F, by Exercise 6.4. Since G is a monolith, P is not normal in FP. Then G = NF, where N = NG (P) < G (Frattini’s lemma), NG = {1} and N ∩ F = {1}. It follows that G is isomorphic to a subgroup of S|G:N| = S4 . However, S4 and A4 are not CM-groups (for example, S4 has two faithful characters). Thus, f ≠ 4. (ii) Now assume that p = 3. Then it follows from the equalities π(G) = π = {2, 3} and |G| = 3μ (3λ − 1) that 3λ − 1 is a power of 2. Hence f ∈ {3, 9}, by Exercise 6.5. Since |F| ∈ {2, 3, 9}, and, by induction, G/F has chief factors of orders 2, 3, 9 only, a Sylow 3-subgroup P ⊲ G, by Exercise 6.4. Remark. Let us prove that if G is a CM-group such that G = Q ⋅ P, where Q ≅ Q8 and {1} < P ∈ Syl3 (G) is normal in G, then the following statements hold: (i) The involution in Q inverts P so P is abelian. (ii) P is elementary abelian. (iii) G = (Q, P) is a Frobenius group with kernel P. We are working by induction on |P|. By Proposition 6.11, G is a ℚ-group. Let R be a minimal normal 3-subgroup of G, |R| = 3β , where β ≤ 2, by Lemma 6.24. Since G is solvable, we have R ≤ Z(P). Assume that QR is not a Frobenius group. Then |R| = 3 (Exercise III.8.5) and QR has a cyclic subgroup A of order 12 (let us consider CQR (R)). Since |Aut(A)| = 4 does not divide |G : A|, G is not a ℚ-group (Corollary 6.9), contrary to Proposition 6.11. Thus, QR is a Frobenius group so that |R| = 9. Considering G/R, we conclude, using induction, that if R < P, then G/R is a Frobenius group. Thus, the involution in Q inverts P so P is abelian, proving assertions (i) and (iii). Assume that B < G is cyclic of order 9. Since Aut(B) ≅ C6 and 6 ∤ |G : CG (B)|, we get a contradiction. Thus, P is elementary abelian, proving (ii). Remark. Let {1} < N ⊲ G. If GCD(o(x), o(y)) = 1 for any x ∈ N, g ∈ G − N, then G is a Frobenius group with kernel N.

444 | Characters of Finite Groups 1 Exercise 6.7. Let G be a CM-monolith with minimal normal subgroup F. (a) If |F| = 2, then G is a 2-group. (b) If |F| = 3, then G ≅ S3 , the symmetric group of degree 3. (c) If |F| = 9, then G ≅ Φ72 , the Frobenius group of order 72 with complement ≅ Q8 and kernel ≅ E9 . Solution. We have Sc(G) = F. (a) If |F| = 2, then G is a 2-group, by Exercise 6.3. (b) If |F| = 3, then 4 ∤ |G| (Proposition 6.21 (c). Let g ∈ G be of even order. Then o(g) = 2, i.e., all elements of G of even order are involutions. Let P ∈ Syl3 (G), x ∈ P# and y ∈ CG (x). Then 2 ∤ o(y) so that CG (x) = P. Therefore, G is a Frobenius group with kernel P of index 2. As g inverts the 3-subgroup P and G is a monolith, so P is cyclic. Since G is an CM-group, we get |P| = 3 so that G ≅ S3 . (c) We omit the fairly routine but not difficult solution of this part. (In that case, Aut(F) ≅ GL(2, 3) so one can analyze the groups H ⋅ F, where H ≤ GL(2, 3).) Corollary 6.25. If G is a CM-group, then P ∈ Syl3 (G) is elementary abelian. Proof. Let G be a counterexample of minimal order. In that case, G contains a unique minimal normal subgroup R; then R ≤ P ∈ Syl3 (G) so that |R| ∈ {3, 9} (Lemma 6.24). Since k(G) = k(G/R) + 1 (Proposition 6.21), we get QR = (Q, R) (a Frobenius group with kernel R and complement Q so that Q is isomorphic to a 2-subgroup of GL(2, 3)) whence Q ∈ {C2 , Q8 }. Let ⟨u⟩ = Q ≅ C2 . Since R# is a G-class (Proposition 6.21), we have |R| = 3. Write D = ⟨⟨u⟩⟩; then D = G so that G󸀠 = P. Since u inverts P, the subgroup P is abelian. As above, exp(P) = 3. Now let Q ≅ Q8 . Since G is a ℚ-group (Proposition 6.11), it follows from Theorem III.8.6 that G = (Q8 , E3n ). Lemma 6.26. The group G = (Q8 , E34 ) has a faithful irreducible character. Proof. Let P ∈ Syl3 (G). The assertion is true if P is a minimal normal subgroup of G. Now assume that P is not a minimal normal subgroup of G. Then all minimal normal subgroups of G have the same order 32 . By Maschke’s theorem, P = P1 × P2 , where P1 , P2 are minimal normal subgroups of G. Since Q8 has only one 2-dimensional irreducible GF(3)-representation, P1 and P2 are isomorphic GF(3)Q8 -modules. Let P0 be a minimal normal subgroup of G and D = EndGF(3)Q8 (P0 ),

V = HomGF(3)Q8 (P0 , P).

Then D is a finite skew field (Schur’s lemma), so that D is a field (Wedderburn’s theorem). If θ ∈ V, ξ ∈ D, then θξ ∈ V, so that V is a right D-space. Since P0 is an irreducible GF(3)Q8 -module, it affords the irreducible GF(3)-representation Γ of Q8 of degree 2. By Maschke’s theorem, Γ is absolutely irreducible (note that Γ is faithful) since Q8 is not commutative. Hence CGL(2,3) (Γ(Q8 )) is a subgroup of scalar matrices. Thus D = GF(3).

IX Faithful characters | 445

Set P i = θ i (P0 ), where θ i ∈ V, i = 1, 2, is a monomorphism. Let π i : P = P1 + P2 → P i be the projection of P onto P i , i = 1, 2. If θ ∈ V, then 1V = π1 + π2 󳨐⇒ θ = π1 θ + π2 θ, π1 θ, π2 θ ∈ V. The mapping x 󳨃→ (π i θ)(x) (x ∈ P0 ) is a homomorphism of P0 into P i , and the mapping x 󳨃→ θ i (x) is an isomorphism of P0 onto P i , i = 1, 2. Thus, for any x ∈ P0 there exists a unique element x󸀠 ∈ P0 such that (π i θ)(x) = θ i (x󸀠 ), i = 1, 2. The mapping x 󳨃→ x󸀠 is an endomorphism of the module P0 . Denoting it by ξ i , one has ξ i ∈ D (≅ GF(3)), π i θ = θ i ξ i (i = 1, 2). Thus, θ = θ1 ξ1 + θ2 ξ2 . Since π i θ j = δ ij θ j , it follows that θ1 , θ2 is a D-basis of the space V, and dimD V = 2. Since P0 is irreducible, it follows that any nonzero element of V is a monomorphism. If θ ∈ V − {0}, then P0 = θ(P0 ) is an irreducible submodule of P, and for any submodule P1 of P there exists a monomorphism θ ∈ V − {0} such that P1 = θ(P0 ). The equality θ1 (P0 ) = θ(P0 ) (θ1 , θ ∈ V − {0}) holds if and only if θ1 = θξ , where ξ ∈ D − {0}. Thus, the monomorphisms θ ∈ V such that θ(P0 ) = P0 , together with 0 ∈ V, constitute a one-dimensional subspace. Therefore, we obtain a one-to-one correspondence between the one-dimensional subspaces of V and the minimal normal subgroups of G. But the number of one-dimensional subspaces of V is (32 − 1)/(3 − 1) = 4, so that G contains exactly four minimal normal subgroups. Now Irr(G) contains exactly (|G| − |Q8 |)/82 = 10 characters of degree 8. If P0 is a minimal normal subgroup of G, then Irr(G/P0 ) contains exactly one character of degree 8. Hence the number of faithful characters in Irr(G) is equal to 10 − 4 = 6. Exercise 6.8. Let a 2-group G > {1} be a CM-group. Then |G/ker(χ)| = 2χ(1)2 for every nonprincipal χ ∈ Irr(G). Hint. One may assume that ker(χ) = {1}. Then G is a monolith. Use statements (e) and (d) of Proposition 6.21. Conjecture (Zhmud). The derived length of a group G of Exercise 6.8 is at most 2. We omit the easy proofs of the following lemmas. Lemma 6.27. For every n ∈ ℕ, there exists only one Frobenius group Φ2⋅3n of order 2⋅3n with elementary abelian Sylow 3-subgroup. Moreover, Φ2⋅3n is a CM-group with trivial center. For example, let G = Φ2⋅3n ; then G = (⟨a⟩, E), where E ≅ E3n . In that case, a inverts E. This shows that G is defined uniquely. Lemma 6.28. There exists only one Frobenius group Φ72 of order 72 with quaternion Sylow 2-subgroup. Moreover, Φ72 is a CM-group with trivial center.

446 | Characters of Finite Groups 1 Note that the group Aut(E9 ) ≅ GL(2, 3) has a Sylow 2-subgroup isomorphic to SD16 . If Q8 ≅ Q < SD16 , then G = Q ⋅ P, where P ≅ E9 , is a Frobenius group. Since there is only one choice of Q, G is defined uniquely up to isomorphism. Lemma 6.29. Let a Frobenius group G = (H, P) be a CM-group. Then G is isomorphic to one of the groups Φ2⋅3n or Φ72 . Proof. Since H ∈ Syl2 (G), it is cyclic or generalized quaternion. Taking into account that H ≅ G/P is a CM-group, we see that H ∈ {C2 , Q8 }. Since P is elementary abelian, we get G = Φ2⋅3n if H = C2 . Now let H = Q8 . As P = Sc(G), we have P = F1 × ⋅ ⋅ ⋅ × F n , where the F i are minimal normal subgroups of G. Since H # induces on F i regular automorphisms, it follows that an irreducible 𝔽i -representation T i , afforded by F i as an 𝔽3 H-module, is faithful. As H possesses a unique faithful irreducible 𝔽3 -representation, we have T1 ∼ ⋅ ⋅ ⋅ ∼ T n . Thus, P is a homogeneous completely reducible G-group. As deg(T i ) = 2, we get F i = E9 for all i. Let us prove that if n ≥ 2, then G is not a CM-group. Passing to an appropriate quotient group, one may assume that n = 2, i.e., P = F1 × F2 . It follows from the proof of Lemma 6.26 that G has exactly six faithful irreducible characters, i.e., G is not CM-group, a contradiction. Exercise 6.9. Let F be a minimal normal subgroup of a CM-group G and f ∈ F # . Then CG (f) = CG (F) and |G : CG (F)| = |F # |. Now we are ready to prove the main result of this section. Theorem 6.30. A CM-group G has the trivial center if and only if it is a direct product of groups Φ2⋅3n , n ∈ ℕ and Φ72 . Every such product is a CM-group. Proof. (i) By Lemmas 6.27 and 6.28, Φ2⋅3n and Φ72 are CM-groups with trivial centers. Each direct product of such groups is a CM-group (Proposition 6.4). (ii) Let G > {1} be a CM-group with Z(G) = {1}; then, by the above, G = H ⋅ P, where P ∈ Syl3 (G) is G-invariant and elementary abelian and H ∈ Syl2 (G). In that case, by Maschke’s theorem, P = Sc(G) since O2 (G) = {1}. Let P = K1 × ⋅ ⋅ ⋅ × K m , where K1 , . . . , K m are homogeneous components of the (completely reducible) G-group P since P is a completely reducible G-group (see §I.10). Then m

⋂ CG (K i ) = CG (P) = P.

(7)

i=1

Set N i = CG (K i ) ∩ H for i = 1, . . . , m. Then N i ⊲ H and ⋂m i=1 N i = {1}, by (7). Setting D i = ⋂ CG (K j ),

Hi = ⋂ Nj ,

j=i̸

we get

Di ∩ H = Hi ,

j=i̸

H i ∩ N i = {1},

i = 1, . . . , m.

We claim that H = H1 × ⋅ ⋅ ⋅ × H m .

(∗)

IX Faithful characters | 447

To prove this, choose in K i , for every i ∈ {1, . . . , m}, a minimal G-invariant subgroup F i (K i is a direct product of G-isomorphic with F i normal subgroups of G) and set A = F1 × ⋅ ⋅ ⋅ × F m , A i = F1 × ⋅ ⋅ ⋅ × F i−1 × F i+1 × ⋅ ⋅ ⋅ × F m . It is clear that A = F1 ∘ ⋅ ⋅ ⋅ ∘ F m ,

A i = F1 ∘ ⋅ ⋅ ⋅ ∘ F i−1 ∘ F i+1 ∘ ⋅ ⋅ ⋅ ∘ F m .

(8)

Since F i is an antikernel of G, it follows that A and A i (for all i) are also antikernels, by Proposition 4.1, in view of multiplicativity of the function φ G (see formula (3.38)). Since G is a CM-group, the layers C and C(i) , generating the antikernels A and A i , respectively, are G-classes. Let C i be a G-class generating F i ; then C i = F #i ,

(2)

|C| = φ G (A),

(2)

|C(i) | = φ G (A i ),

(2)

|C i | = φ G (F i ),

by Proposition 4.1. Since A, A i and F i are completely reducible G-groups, we have |C| = φ G (A),

|C(i) | = φ G (A i ),

|C i | = φ G (F i ).

Using the multiplicativity of the function φ G , we deduce from (8) that m

i = 1, . . . , m.

|C(i) | = ∏ |C j |,

|C| = ∏ |C j |, j=1

j=i̸

Since C j = F #j , j = 1, . . . , m, we get m

|C| = ∏ |F #j |,

|C(i) | = ∏ |F #j |.

j=1

(9)

j=i̸

On the other hand, if g ∈ C, then |C| = |G : CG (g)|. Noting that g = g1 ⋅ ⋅ ⋅ g m , where g j ∈ F #j (j = 1, . . . , m), we get m

m

CG (g) = ⋂ CG (g j ) = ⋂ CG (F j ), j=1

j=1

by Exercise 6.8. Next, since K i is homogeneous, we have CG (F j ) = CG (K j ) for all j. m Therefore, CG (g) = ⋂m j=1 CG (K j ) so since ⋂j=1 CG (K j ) = P, we get CG (g) = P, in view of (7). Therefore, |C| = |G : P| = |H|, and this implies, in view of (9), that m

|H| = ∏ |F #j |.

(10)

j=1

Similarly, choosing g ∈ C(i) , we obtain |C(i) | = |G : CG (g)|. Since g = g1 ⋅ ⋅ ⋅ g i−1 g i+1 ⋅ ⋅ ⋅ g m , where g j ∈ F #j (j ≠ i), we obtain, as before, CG (g) = ⋂ CG (g j ) = ⋂ CG (F j ) = ⋂ CG (K j ) = D i . j=i̸

Therefore,

|C(i) |

j=i̸

j=i̸

= |G : D i | for all i, so that |D i | = |P||F #i |,

i = 1, . . . , m.

(11)

448 | Characters of Finite Groups 1 Since P ≤ D i , we get D i = P(D i ∩ H) = H i ⋅ P for all i, by the modular law. Therefore, H i ≅ D i /P hence, by (11), we get |H i | = |D i : P| = |F #i |,

i = 1, . . . , m.

(12)

Therefore, we can rewrite (10) as follows: m

(13)

|H| = ∏ |H i |. i=1

Noting that if j ≠ i, then H j = ⋂k=j̸ N k ≤ N i , we get, for all i, H i ∩(∏j=i̸ H j ) ≤ H i ∩N i = {1}, completing the proof of (∗). Next, H i = ⋂ N j = ⋂(CG (K j ) ∩ H) 󳨐⇒ H i ≤ CG (K j ) (j ≠ i). j=i̸

j=i̸

Taking this into account, we obtain from (∗) that G = (K1 H1 ) ⋅ ⋅ ⋅ (K m H m ). If j ≠ i, then H j ≤ CG (K j ) so [K j H j , K i H i ] = {1}. Assume that g1 ⋅ ⋅ ⋅ g m = 1, where g i ∈ K i H i for all i. Setting g i = u i v i , u i ∈ K i , v i ∈ H i for all i, we obtain (u1 ⋅ ⋅ ⋅ u m )(v1 . . . v m ) = 1 󳨐⇒ u1 ⋅ ⋅ ⋅ u m = 1 = v i ⋅ ⋅ ⋅ v m , and hence u i = v i = 1 (i = 1, . . . , m). Since P = K1 × ⋅ ⋅ ⋅ × K m , we get, in view of (∗), g i = 1 for all i, and so G = (K1 H1 ) × ⋅ ⋅ ⋅ × (K m H m ).

(14)

It follows from Proposition 6.2 and (14) that G i = K i H i is a CM-group for all i. We claim that G i is a Frobenius group with kernel K i . Let x ∈ K #i and y ∈ H i . If xy = yx, then y ∈ H ∩ CG (x). Let K i = F (1) × ⋅ ⋅ ⋅ × F (t) be a decomposition in a direct product of minimal normal subgroups of G. Then x = x1 ⋅ ⋅ ⋅ x t , where x j ∈ F (j) for all j, and x j ≠ 1 for some j. Since, by Exercise 6.8, CG (x j ) = CG (F (j) ) if x j ≠ 1, we get t

CG (x) = ⋂ CG (x j ) = ⋂ CG (x j ) = ⋂ CG (F (j) ). j=1

x j =1 ̸

x j =1 ̸

Noting that CG = CG (K i ) for all j since K i is homogeneous, we get CG (x) = CG (K i ). Thus y ∈ H i ∩ CG (K i ) = H i ∩ (CG (K i ) ∩ H) = H i ∩ N i = {1} 󳨐⇒ y = 1 (F (j) )

Since K i is abelian, we get CG i (x) = K i , and so G i is a Frobenius group, as required. By Lemma 6.29, G i is isomorphic to one of the groups Φ2⋅3n (n ∈ ℕ) or Φ72 . The obtained decomposition of G is unique, by the Remak–Schmidt theorem since Z(G) = {1}. This completes the proof. Corollary 6.31. If G is a CM-group, then either G is a 2-group or the center of G/O2 (G) is trivial (so the structure of this group is described by Theorem 6.30). Indeed, being a 2-group (Proposition 6.13), the center of the nonnilpotent CM-group G/O2 (G) must be trivial.

IX Faithful characters | 449

7 Varia Let n ∈ ℕ. A group G is said to be a CMn -group if every of its epimorphic images has at most n faithful irreducible characters. The minimal integer n such that G is a CMn -group is denoted by γ(G). In that case, any normal subgroup of G is the kernel of at most γ(G) distinct irreducible characters of G and there is N ⊲ G such that N is the kernel of exactly γ(G) distinct irreducible characters of G. Thus, the CMn -property is inherited by quotient groups (however, if N ⊲ G, then it may be that γ(G/N) < γ(G)) Exercise 7.1. CM2 -groups are solvable. Solution. Let G be a counterexample of minimal order. Since all proper epimorphic images of G are CM2 -groups, they are solvable, by induction. It follows that G has a unique minimal normal subgroup, say R, and R is nonsolvable. By [BZ3, Theorem 14.19 or Appendix A to Chapter 30], there are in Irr1 (R) three nonlinear characters ψ1 , ψ2 , ψ3 of pairwise distinct degrees. Let χ i ∈ Irr(ψ Gi ), i = 1, 2, 3. Then, by reciprocity and Clifford’s theorem, χ i are pairwise distinct and faithful, i = 1, 2, 3. This is a contradiction. It is interesting to describe the structure of CM2 -groups in more detail. Note that if a p-group G is a CM2 -group, then p ≤ 3. Any 3-group G with |Φ(G)| = 3 is a CM2 -group. In any case, if G is a CM2 -group, then exp(G/G󸀠 ) ≤ 4. The symmetric group S4 is a CM2 -group. If p is the least prime divisor of the order of a CMp -group G, then G is solvable. This follows from Exercise 7.1 and the Odd Order Theorem. We suggest the reader should produce the proof without appealing to the last theorem. Exercise 7.2. If G is an abelian CMn -group, then φ(exp(G)) ≤ n. In particular, if G is a CMn -group, then φ(exp(G/G󸀠 )) ≤ n. Solution. One may assume that G is noncyclic. Let A < G be such that G/A is cyclic. Since A is the kernel of exactly φ(G/A) linear characters of G, we get φ(|G/A|) ≤ n. By the basic theorem on finite abelian groups, there is A < G such that G/A is cyclic of order exp(G). Exercise 7.3. If n > 1, then a direct product of CMn -groups need not be a CMn -group. Exercise 7.4. The groups G whose irreducible characters with equal kernel have equal degree are solvable. Solution. Let G be a counterexample of minimal order. Then G is not simple: indeed, this follows from |G| = 1 +

∑ #

χ(1)2 = 1 + (k(G) − 1)χ(1)2

χ∈Irr (G)

and the Frobenius–Molien theorem. In that case, G is a monolith with minimal normal subgroup R; clearly, R ≤ G󸀠 and G/R is solvable, by induction. If χ ∈ Irr(G | R),

450 | Characters of Finite Groups 1 then χ is faithful. Let p ∈ π(R) and P ∈ Sylp (G). Take a nonprincipal μ ∈ Lin(P | P ∩ R). Then μ G (1) = |G : P| ≢ 0 (mod p) so there is χ ∈ Irr(μ G ) such that p ∤ χ(1). Thus, p does not divide the degrees of all faithful irreducible characters of G, by hypothesis. Since p ∈ π(R) is arbitrary, we get GCD(|R|, χ(1)) = 1 for all faithful irreducible characters of G. However, if μ ∈ Irr(χ R ), then μ(1) > 1 (indeed, R is nonabelian characteristic simple so R󸀠 = R) and μ(1) | |R| (Frobenius–Molien) and μ(1) | χ(1) (Clifford), a final contradiction. Theorem 7.1. If p ∈ π(G), then the following conditions are equivalent: (a) |G/ker(χ)| = pχ(1)2 for all χ ∈ Irr# (G). (b) G is a CMp−1 -group and |G| is a power of p. Proof. For every nonidentity p-group X we have γ(X) ≥ p − 1. Indeed, if K < X is of index p, then X/K has exactly p − 1 faithful linear characters. (a) ⇒ (b): First we prove that G is a p-group. Assume that G is a counterexample of minimal order. By (a), p divides the order of every nonidentity epimorphic image of G. If χ ∈ Irr# (G), then, by assumption, all faithful irreducible characters of G/ker(χ) have the same degree √ 1p |G/ker(χ)|. Therefore, the group G cannot be nonabelian simple. Indeed, otherwise, all members of Irr# (G) have the same degree. It follows that |G| = 1 + χ(1)2 (|Irr(G)| − 1) so GCD(|G|, χ(1)) = 1 for all χ ∈ Irr# (G), contrary to the Frobenius–Molien theorem. Property (a) is inherited by nontrivial epimorphic images of G. Therefore, by induction, all proper epimorphic images of G are p-groups so that G contains only one minimal normal subgroup R and G/R is a nontrivial p-group. All characters in Irr(G) whose kernels do not contain R are faithful, and therefore they have the same degree (|G|/p)1/2 , by hypothesis. Suppose that τ ∈ Irr# (G), p ∤ τ(1), R ≰ ker(τ) (in that case, τ must be faithful). Then, by Clifford theory, τ R = ϕ ∈ Irr(R). Therefore, we have τ(1)2 = ϕ(1)2 < |R| and pτ(1)2 < p|R| ≤ |G| = |G/ker(τ)|, contrary to the hypothesis. Thus, kernels of all characters in Irr(G), whose degrees are not divisible by p, contain R, i.e., R ≤ G(p󸀠 ) =

⋂

ker(χ).

χ∈Irr(G), p∤χ(1)

By Theorem XIV.9.1, G(p󸀠 ) is p-nilpotent. Since, by assumption, G is not a p-group, p ∤ |R|. Take ϕ ∈ Irr# (R) and χ ∈ Irr(ϕ G ). Then, by reciprocity and the uniqueness of R, χ is faithful. Let χ R = e(ϕ1 + ⋅ ⋅ ⋅ + ϕ t ) be the Clifford decomposition and ϕ1 = ϕ. We have χ(1) = etϕ(1), where e and t are powers of p, by Clifford theory. Hence |G| = pχ(1)2 = pe2 t2 ϕ(1)2 (see the previous paragraph). Since |R| is the p󸀠 -part of |G|, and the same is true for ϕ(1)2 , we have ϕ(1)2 = |R|. But this is impossible because R > {1}. Thus, G is a p-group. We claim that γ(G) = p − 1. Taking an arbitrary χ ∈ Irr# (G), it will suffices to show that ker(χ) is the kernel of exactly p − 1 irreducible characters of G. Without loss of generality, one may assume that ker(χ) = {1}. In that case, Z(G) is cyclic so G contains only one minimal normal subgroup N, |N| = p. Let {χ i }1s be the set of faithful irredu-

IX Faithful characters | 451

cible characters of G, and χ i (1) = p a , i = 1, . . . , s. Then, setting |G| = p n , we obtain χ1 (1)2 + ⋅ ⋅ ⋅ + χ s (1)2 = |G| − |G/N| = (p − 1)p n−1 , or

s ⋅ p2a = (p − 1)p n−1 .

By assumption, we have p n = |G| = pχ(1)2 = p2a+1 so that n − 1 = 2a, hence s = p − 1 and γ(G) = p − 1 (b) ⇒ (a): In the case under consideration, G is a p-group with γ(G) = p − 1. Taking χ ∈ Irr# (G), we have to show that |G/ker(χ)| = pχ(1)2 . Without loss of generality, we may assume that ker(χ) = {1}. Let {χ1 , . . . , χ s } be the set of faithful irreducible characters of G. By assumption, s ≤ p − 1. Since ker(χ) = {1}, G contains only one minimal normal subgroup N (indeed, Z(G) is cyclic). If |G| = p n and χ i (1) = p a i for all i, a1 ≤ ⋅ ⋅ ⋅ ≤ a s , then we have, as above, p2a1 + ⋅ ⋅ ⋅ + p2a s = (p − 1)p n−1 . Thus, 1 + p2(a2 −a1 ) + ⋅ ⋅ ⋅ + p2(a s −a1 ) = (p − 1)p(n−1)−2a1 ;

(1)

note that p2a1 = χ1 (1)2 ≤ |G/Z(G)| ≤ p n−1 . It follows from (1) and s < p that n−1 = 2a1 . Since p2a i = χ i (1)2 ≤ |G/Z(G)| ≤ p n−1 for all i, we get a1 = ⋅ ⋅ ⋅ = a s = 12 (n − 1), and therefore |G| = pχ(1)2 , completing the proof. If G is a group of Theorem 7.1, then G󸀠 = Φ(G) (Exercise 7.2). If G is an extraspecial p-group, then γ(G) = p − 1. Let G be a p-group. If χ ∈ Irr1 (G), then χ(1)2 | |G/ker(χ)|. Therefore, every p-group G is a Θ-group, where property Θ is defined as follows: Property Θ. For every χ ∈ Irr1 (G) the number d(χ) =

|G : ker(χ)| χ(1)2

is an odd power of a prime. M. Lewis and S. Gagola [GagL2] have proved that if χ(1)2 | |G/ker(χ)| for all χ ∈ Irr(G), then G is nilpotent. A group G is said to be a θodd -group if, for every χ ∈ Irr1 (G), the number d(χ) is an odd power of a prime. Theorem 7.2. A θodd -group G is a p-group for some prime p.² Proof. Let G be a counterexample of minimal order. It is easy to see that nilpotent θodd -groups are prime power groups. Hence, our counterexample G is nonnilpotent. Therefore, since all epimorphic images of G are θodd -groups, it follows that G is a monolith. Let N be a (unique) minimal normal subgroup of G. Then G/N is a p-group for some prime p, by induction. 2 This follows from the main theorem in [GagL2], which depends on CFSG. The presented proof is elementary.

452 | Characters of Finite Groups 1 (i) Suppose that G is solvable. Let |N| = q s for some prime q. Since G is nonnilpotent, we have N ∈ Sylq (G) so that q ≠ p. By Schur–Zassenhaus, G = AN, where A is maximal in G since N is a minimal normal subgroup of G. In this case, A ≅ G/N is a p-group. Since G is a monolith, we have CG (N) = N. (i1) Suppose that G/N is abelian. If x ∈ A# , then CG (x) = A (otherwise, x ∈ Z(G) so G is not a monolith). Hence G = (A, N), a Frobenius group with kernel N and complement A. In that case G has a faithful irreducible character χ of degree |A| (Theorem X.3.2) and d(χ) is not an integer, a contradiction. (i2) Let G/N be a nonabelian p-group. Then Z(G) = {1}. By Ito’s Theorem VII.2.3, χ(1) is a power of p for all χ ∈ Irr1 (G). Take χ ∈ Irr(G | N). Then χ is faithful and, by assumption, we have |G| = q2e+1 χ(1)2 , where e ∈ ℕ ∪ {0}. In that case, A ∈ Sylp (G) is a θodd -group of order χ(1)2 . Now A acts on N faithfully (since G is a monolith) and irreducibly (since N is a minimal normal subgroup of G). If x ∈ Z(A) is of order p, then CG (x) = A since G is a monolith. It follows that Z(A)N is a Frobenius group so Z(A) is cyclic. Then there is a faithful τ ∈ Irr(A). Since A is nonabelian, the character τ is nonlinear. Since A ≅ G/N so A is a θodd -group, we have, by hypothesis, |A| = p2a+1 τ(1)2 , where a ∈ ℕ ∪ {0}. This means that |A| is an odd power of p. But then the equality |G| = q2e+1 χ(1)2 is not a square contrary to |A| = χ(1)2 . Thus, if G is solvable, it is a prime power group. (ii) In the sequel we assume that G is nonsolvable. Then N (the unique minimal normal subgroup of G) is nonsolvable and G/N is a p-group for some prime p. Let χ ∈ Irr1 (G) be faithful of minimal degree. Then |G| = q2a+1 χ(1)2 for some prime q ∈ π(N) (Clifford). By Thompson’s theorem (see Chapter XIV) there is τ ∈ Irr1 (G) such that q ∤ τ(1). Then q2a+1 χ(1)2 = |G| = q2b+1 τ(1)2 .

(2)

It follows from (2) that a ≥ b. As χ(1) ≤ τ(1) and τ(1) | χ(1), we get χ(1) = τ(1). It follows that q ∤ χ(1), i.e., |G|q = q2a+1 . As G is not q-nilpotent, again, by Clifford’s theory, there is a μ ∈ Irr1 (G) such that q | μ(1). Since μ is faithful, one has for some prime s, |G| = s2c+1 μ(1)2 . Again, by Thompson’s theorem, we may assume that s ≠ q. We get q2a+1 χ(1)2 = |G| = s2c+1 μ(1)2 . q2a+1

It follows from (3) that is the exact power of q dividing Thus, N cannot be nonsolvable.

(3) μ(1)2 ,

a contradiction.

Exercise 7.5. Let P be the nonabelian invariant Sylow p-subgroup of a minimal nonnilpotent group (see Lemma XI.2.1). Show that P is a CMp−1 -group. Hint. If L < Z(P) is of index p, then P/L is extraspecial. It is not known whether there is an upper bound on the derived length of a p-group if it is a CMp−1 -group (for p = 2, see Zhmud’s problem #1). Let G be a p-group which is a CMp−1 -group. Then, by Theorem 7.1, |G/Z(χ)| = χ(1)2 ,

χ ∈ Irr# (G).

IX Faithful characters | 453

The following two conditions are equivalent (check!): (a) |G/Z(χ)| = χ(1)2 for each χ ∈ Irr# (G). (b) G = P × A, where A󸀠 = {1} and P ∈ Sylp (G) satisfies |P/Z(μ)| = μ(1)2 for each character μ ∈ Irr# (P). Note that there are nonsolvable CM4 -groups (for example, SL(2, 5)). However, the following result holds: Theorem 7.3. All CM3 -groups are solvable. Proof. Let G be a counterexample of minimal order. Since every proper epimorphic image of G is a CMn -group with n ≤ 3, it follows from the previous results and by induction that G has only one minimal normal subgroup R; moreover G/R is solvable, by induction, and R is not solvable. Let cd(R) = {1, d1 , . . . , d k },

1 < d1 < ⋅ ⋅ ⋅ < d k ,

and ϕ i ∈ Irr(R),

ϕ i (1) = d i ,

i = 1, . . . , k.

Then, by reciprocity and Clifford’s theorem, ⟨ϕ Gi , ϕ Gj ⟩ = 0

(4)

(i ≠ j).

Take χ i ∈ Irr(ϕ Gi ), i = 1, . . . , k. Again by reciprocity and Clifford’s theorem, all characters χ i are faithful. Since, by the previous arguments, the characters χ1 , . . . , χ k are pairwise distinct, we have k ≤ 3. On the other hand, k ≥ 3 (see Theorem XIV.7.1). Therefore k = 3, so that R is simple (see Theorem IV.8.1). It follows from (4) that ϕ Gi = e i χ i , χ i ∈ Irr(G) are pairwise different, i = 1, 2, 3. Note that, by assumption, the kernel of every character in Irr(G) − {χ1 , χ2 , χ3 } contains R so that |Irr(G)| = |Irr(G/R)| + 3.

(5)

Let |G : IG (ϕ i )| = t i (i = 1, 2, 3), where IG (ϕ i ) is the inertia subgroup of ϕ i in G. Then χ i (1) = e i t i d i , i = 1, 2, 3. On the other hand, |G : R|d i = ϕ Gi (1) = e i χ i (1) = e2i t i d i 󳨐⇒ |G/R| = e2i t i , and π = π(|G : R|) = π(e i t i ) ⊆ π(χ i (1)),

i = 1, 2, 3.

Take p ∈ π. Then χ ∈ Irr(G), p ∤ χ(1) implies R ≤ ker(χ). Therefore, R ≤ G(p󸀠 ), where G(p󸀠 ) = ⋂χ∈Irr1 (G),p∤χ(1) ker(χ). By Theorem XIV.9.1, the subgroup G(p󸀠 ) is p-nilpotent; then R is also p-nilpotent for all p ∈ π. We have R󸀠 = R so that p ∤ |R|, and therefore R is a π󸀠 -Hall subgroup of G.

454 | Characters of Finite Groups 1 By (5), we have k(G) = k(G/R) + 3.

(6)

Since G/R is solvable and R is a π󸀠 -Hall subgroup of G, there is in G a π-Hall subgroup H (Schur–Zassenhaus). Let x be an element of prime order in H (if H = {1}, then G = R and so k(G) = 4; then G is solvable; see [BZ3, Chapter 31]). As, by Thompson’s theorem, R does not admit a fixed-point-free automorphism of prime order, there exists y ∈ R# such that xy = yx. The element xy is not conjugate to elements of the set R ∪ H. Furthermore, R# is the union of at least three G-classes. Therefore we have k(G) ≥ k(G/R) + 3 + 1, contradicting equality (6).

X Existence of normal subgroups In this chapter we prove several theorems on the existence of normal subgroups. To some extent all of them revolve around the classic Frobenius’ theorem (we became acquainted with this theorem in Chapter VIII; here it will be obtained as a corollary of Wielandt’s theorem which also was proved in that chapter). The groups that occur in Frobenius’ theorem are called Frobenius groups, and they play an important part in the theory of finite groups. Throughout in this chapter, H0 ⊲ H < G,

|G| = g,

|H| = h,

|H0 | = h0 .

Given N ⊴ G, we set, following Isaacs, Irr(G | N) = {χ ∈ Irr(G) | N ≰ ker(χ)}; then Irr(G) = Irr(G | N) ∪ Irr(G/N) is a partition. Definition. A triple of groups (G, H, H0 ) is called a W-triple if H0 ⊲ H,

H < G,

H ∩ H x ≤ H0

for all x ∈ G − H.

It is obvious that in the case under consideration NG (H) = H. Note that W-triples were considered in Theorem VIII.5.9.

1 Introductory remarks Let S = (G, H, H0 ) be a triple of groups such that H0 ⊲ H < G. Denote |G| = g,

|H| = h,

|H0 | = h0 ,

∆ = H − H0 ,

G0 = G − ⋃ ∆ t , t∈G

|G0 | = g0 .

As H0 ⊲ H, it follows that H ≤ NG (∆), i.e., the set ∆ is H-invariant. If n = |G : H| and G = ⋃ni=1 Ht i is a partition, then ⋃t∈G ∆ t = ⋃ni=1 ∆ t i (see the first sentence of this paragraph). Therefore, 󵄨󵄨 n 󵄨󵄨 󵄨 󵄨󵄨 n 󵄨󵄨 ⋃ ∆ t 󵄨󵄨󵄨 = 󵄨󵄨󵄨⋃ ∆ t i 󵄨󵄨󵄨 ≤ ∑ |∆ t i | = n|∆| = n(|H| − |H |) = n(h − h ) = g − nh , 0 0 0 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨t∈G 󵄨 󵄨󵄨i=1 󵄨󵄨󵄨 i=1 and we conclude that

󵄨󵄨 󵄨󵄨 g g0 = g − 󵄨󵄨󵄨󵄨 ⋃ ∆ t 󵄨󵄨󵄨󵄨 ≥ nh0 = h0 . h 󵄨t∈G 󵄨

Now assume that g0 = hg h0 . Then one can conclude from the above reasoning that |⋃ni=1 ∆ t i | = ∑ni=1 |∆ t i |, whence the subsets ∆ t i (i ∈ {1, . . . , n}) are pairwise disjoint and NG (∆) = H. Therefore, ∆ t ∩ ∆ = 0 for all t ∈ G − H. In particular, ∆ is a TI-subset. DOI 10.1515/9783110224078-010

456 | Characters of Finite Groups 1 Conversely, if the subsets ∆ t i (i ∈ {1, . . . , n}) are pairwise disjoint (in that case, obviously, NG (∆) = H), then 󵄨󵄨 󵄨 󵄨󵄨 n 󵄨󵄨 󵄨󵄨 ⋃ ∆ t 󵄨󵄨󵄨 = 󵄨󵄨󵄨⋃ ∆ t i 󵄨󵄨󵄨 = n|∆| = g − nh . 0 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨t∈G 󵄨 󵄨󵄨i=1 󵄨󵄨󵄨 Therefore,

󵄨󵄨 󵄨󵄨 g g0 = g − 󵄨󵄨󵄨󵄨 ⋃ ∆ t 󵄨󵄨󵄨󵄨 = nh0 = h0 . h 󵄨t∈G 󵄨

In particular, if (G, H, H0 ) is a W-triple, then g0 = hg h0 .

2 Wielandt’s theorems In this section we prove three of Wielandt’s theorems [Wiel2] and present a converse to the first Wielandt theorem. We retain the notation introduced in §X.1. Given N ⊴ G, set IrrN (G) = {χ ∈ Irr(G) | N ≤ ker(χ)} (as we know, the intersection of kernels of all members of the set IrrN (G) is equal to N; for example, IrrG󸀠 (G) = Lin(G)). One has Irr(G) = IrrN (G) ∪ Irr(G | N). Recall that CF[G] is the set of class functions on G. We have Irr(G) ⊆ CF[G]. Lemma 2.1. Let (G, H, H0 ) be a W-triple. Then the following statements hold: (a) If μ ∈ CF[H] vanishes on H0 , then (μ G )H = μ. (b) Let ϕ ∈ IrrH0 (H) and μ ϕ = ϕ − ϕ(1) ⋅ 1H (∈ Ch(H)). Then (μ Gϕ )H = μ ϕ . ̇ −1 xt) (here μ̇ coincides with μ on H Proof. (a) Let x ∈ H. Since μ G (x) = h−1 ∑t∈G μ(t and vanishes on G − H), it follows that μ G (x) = h−1

∑ t∈G,

x t ∈H

μ(x t ) = h−1

∑ t∈G,

x t ∈∆

μ(x t ),

where ∆ = H − H0 . Next, x t ∈ ∆ implies x t ∈ ∆ ∩ H t . If t ∈ ̸ H, then ∆ ∩ H t ⊆ (∆ ∩ H) ∩ H t = ∆ ∩ (H ∩ H t ) ⊆ ∆ ∩ H0 = 0. Therefore, x t ∈ ∆ implies t ∈ H. Since μ ∈ CF[H] and ∆ is H-invariant, we have x t ∈ ∆ 󳨐⇒ μ(x t ) = μ(x) 󳨐⇒ μ G (x) = h−1 ⋅ ∑ μ(x) = μ(x). t∈H

If x ∈ H0 , then x t ∈ H0 for t ∈ H; hence μ G (x) = 0 = μ(x). Therefore, (μ G )H = μ since H = ∆ ∪ H0 is a partition. (b) Since μ ϕ ∈ CF(H) vanishes on H0 , the result follows from (a).

X Existence of normal subgroups | 457

We use in what follows the notation introduced in Lemma 2.1. Theorem 2.2. Let G be a group, H0 ⊲ H ≤ N = NG (H) < G, ∆ = H − H0 . (a) [Ber12] Assume that for any nonprincipal ϕ ∈ IrrH0 (H) we have (μ Gϕ )H = μ ϕ , where μ ϕ = ϕ − ϕ(1)1H . Then G0 = G − ⋃t∈G ∆ t is a normal subgroup of G, G0 ∩ H = H0 , G0 H = G and (G, H, H0 ) is a W-triple. (b) (First Wielandt’s theorem [Wiel2]) If (G, H, H0 ) is a W-triple, then G0 = G − ⋃ ∆ t ⊲ G, t∈G

G0 ∩ H = H0 ,

G0 H = G.

Conversely, if M ⊲ G, M ∩ H = H0 and MH = G, then M = G0 (uniqueness), where G0 = G − ⋃t∈G (H − H0 )t . (c) (Second Wielandt’s theorem [Wiel2]) Let H ∩ H x ≤ H0 for all x ∈ G − N, and let N = N0 H, where N0 ⊲ N such that N0 ∩ H = H0 . If GCD(|N : H|, |H : H0 |) = 1, then (G, N, N0 ) is a W-triple. In particular, G1 = G − ⋃ (N − N0 )t 󳨐⇒ G1 H = G, G1 ∩ H = H0 . t∈G

(d) (Third Wielandt’s theorem [Wiel2]) Let H ∩ H x ≤ H0 for all x ∈ G − N. Denote |H : H0 | = j, |G| = mj and assume that GCD(m, j) = 1. If there are exactly m solutions of the equation x m = 1 in G, they form a (characteristic) subgroup of G. Proof. (a) Given ϕ ∈ IrrH0 (H) − {1H }, put ϕ0 = μ Gϕ + ϕ(1) ⋅ 1G . Then μ ϕ (1) = 0 󳨐⇒ μ Gϕ (1) = 0 󳨐⇒ ϕ0 (1) = μ Gϕ (1) + ϕ(1) = ϕ(1). Clearly, one has ϕ0 ∈ Ch(G), where Ch(G) is the set of generalized characters of G. Since ⟨ϕ, 1H ⟩ = 0, we have, by reciprocity, Lemma 2.1 (b) and hypothesis, ⟨μ Gϕ , μ Gϕ ⟩ = ⟨μ ϕ , (μ Gϕ )H ⟩ = ⟨μ ϕ , μ ϕ ⟩ = ⟨ϕ − ϕ(1)1H , ϕ − ϕ(1)1H ⟩ = ⟨ϕ, ϕ⟩ + ϕ(1)2 = 1 + ϕ(1)2 . Again, by reciprocity, ⟨μ Gϕ , 1G ⟩ = ⟨μ ϕ , 1H ⟩ = ⟨ϕ − ϕ(1)1H , 1H ⟩ = −ϕ(1). Therefore, ⟨ϕ0 , ϕ0 ⟩ = ⟨μ Gϕ + ϕ(1) ⋅ 1G , μ Gϕ + ϕ(1) ⋅ 1G ⟩ = ⟨μ Gϕ , μ Gϕ ⟩ + 2ϕ(1)⟨μ Gϕ , 1G ⟩ + ϕ(1)2 = 1 + ϕ(1)2 − 2ϕ(1)2 + ϕ(1)2 = 1. Since ϕ0 ∈ Ch(G) = ℤ[Irr(G)] and ϕ0 (1) = μ Gϕ (1) + ϕ(1) = |G : H|μ ϕ (1) + ϕ(1) = ϕ(1) > 0,

458 | Characters of Finite Groups 1 it follows that ϕ0 ∈ Irr(G). Since (ϕ0 )H = (μ Gϕ )H + ϕ(1) ⋅ 1H = μ ϕ + ϕ(1) ⋅ 1H = ϕ, the character ϕ0 is an extension of the character ϕ to G. Note that if ϕ = 1H , then μ1H is the zero function, hence ϕ0 = 1G so that, in this case, ϕ0 ∈ Irr(G) as well. Write K = ⋂ϕ∈IrrH (H) ker(ϕ0 ); then K ⊲ G. If x ∈ K ∩ H, then 0

ϕ(x) = ϕ0 (x) = ϕ0 (1) = ϕ(1) for any ϕ ∈ IrrH0 (H). Therefore K ∩ H ≤ ⋂ϕ∈IrrH (H) ker(ϕ) = H0 . On the other hand, if 0 x ∈ H0 , then ϕ0 (x) = ϕ(x) = ϕ(1) = ϕ0 (1), i.e., H0 ≤ ker(ϕ0 ) for any ϕ ∈ IrrH0 (H), so that H0 ≤ K ∩ H. Thus, K ∩ H = H0 . Given x ∈ G, one has μ Gϕ (x) = h−1 ⋅

∑

t∈G, txt−1 ∈H

μ ϕ (txt−1 ) = h−1 ⋅

∑

t∈G, txt−1 ∈∆

μ ϕ (txt−1 ),

since μ ϕ vanishes on H0 = H − ∆. Therefore, μ Gϕ vanishes on the set G0 = G − ⋃t∈G ∆ t . Hence, for x ∈ G0 , one has ϕ0 (x) = (μ Gϕ + ϕ(1) ⋅ 1G )(x) = ϕ(1) = ϕ0 (1), and so G0 ⊆ ker(ϕ0 ) for any ϕ ∈ IrrH0 (H). Thus, G0 ⊆ K. Since K ⊲ G, it follows, by §X.1 and what has been proved already, that KH/K ≅ H/(H ∩ K) = H/H0 󳨐⇒ |KH| =

h |K|. h0

As G0 ⊆ K, we also have, denoting g0 = |G0 |, that (by §X.1, one has g0 ≥ hg h0 ) |KH| ≥

h h g g0 ≥ ⋅ h0 = g, h0 h0 h

and hence KH = G. It follows that g0 = hg h0 , and so G0 = K. We have thus proved that G0 ⊲ G, G0 H = G, G0 ∩ H = H0 . It remains to prove that (G, H, H0 ) is a W-triple. If M ⊆ G, put M = G − M. Then ∆ = H ∩ H 0 and, by what we have proved, G0 ∩ ∆ = G0 ∩ (H 0 ∩ H) = (G0 ∩ H) ∩ H 0 = H0 ∩ H 0 = 0. As G0 ⊲ G, we have G0 ∩ ∆ t = (G0 ∩ ∆)t = 0 for any t ∈ G. Since H0 < G0 , it follows that H0 ∩ ∆ t ≤ G0 ∩ ∆ t = 0 for any t ∈ G. In particular, if t ∈ G − H, then H t ∩ H = (H0 ∪ ∆)t ∩ (H0 ∪ ∆) = (H0t ∩ H0 ) ∪ (H0t ∩ ∆) ∪ (∆ t ∩ ∆) ∪ (∆ t ∩ H0 ) = H0t ∩ H0 ≤ H0 (since the argument in §X.1 imply that ∆ t ∩ ∆ = 0). Hence, (G, H, H0 ) is a W-triple, and the proof of (a) is complete. (b) In view of (a) and Lemma 2.1 (b), we need only prove the uniqueness of G0 . Let M ⊲ G, M ∩ H = H0 and MH = G. We have to prove that M = G0 . Since ∆ ⊂ H, it

X Existence of normal subgroups | 459

follows that M ∩ ∆ = M ∩ (H ∩ ∆) = (M ∩ H) ∩ ∆ = H0 ∩ ∆ = 0 󳨐⇒ ∆ ⊆ G − M 󳨐⇒ ⋃ ∆ t ⊆ G − M t∈G

since M ⊲ G. One has G0 = G − ⋃t∈G ∆ t ≥ M. As G/M = MH/M ≅ H/(M ∩ H) = H/H0 ,

G/G0 = G0 H/G0 ≅ H/H0 ,

we get |M| = |G0 |, and the inclusion M ≤ G0 implies that M = G0 , proving (b). The subgroup G0 is called the kernel of the W-triple (G, H, H0 ). (c) Let π = π(|H : H0 |), where π(n) is the set of prime divisors of n ∈ ℕ. Let N π 󸀠 denote the subgroup generated by all π-elements of N = NG (H) (i.e., N π = Oπ (N)); then N/N π is a π󸀠 -group. Next, GCD(|N : H|, |H : H0 |) = 1 󳨐⇒ π(|N : H|) ⊆ π󸀠 󳨐⇒ N π ≤ H. Assume that there is x ∈ NG (N) − N. Since N π is characteristic in N, it follows that N x = N 󳨐⇒ N π = (N π )x ≤ H x 󳨐⇒ N π ≤ H ∩ H x ≤ H0 󳨐⇒ |N : H0 | | |N : N π |. This is a contradiction, since |N : N π | is a π󸀠 -number but |N : H0 | = |N : H| ⋅ |H : H0 | is not a π󸀠 -number (because |H : H0 | > 1 is a π-number). We conclude that NG (N) = N. Let a = a π a π󸀠 = a π󸀠 a π ∈ N ∩ N x and x ∈ G − N. Then a π ∈ N π ≤ H, a π ∈ (N π )x ≤ H x 󳨐⇒ a π ∈ H ∩ H x ≤ H0 ≤ N0 . Note that N0 H = N and N0 ∩ H = H0 󳨐⇒ N/N0 ≅ H/(N0 ∩ H) = H/H0 󳨐⇒ |N : N0 | is a π-number, 󸀠

i.e., N π ≤ N0 since N0 ⊲ N. Therefore, we have a π󸀠 ∈ N0 . Thus, a = a π a π󸀠 ∈ N0 , and so N ∩ N x ≤ N0 for all x ∈ G − N. Then (G, N, N0 ) is a W-triple. Let G1 = G − ⋃t∈G (N − N0 )t be the kernel of this W-triple. Then G = G1 N, N ∩ G1 = N0 (see (a)). Therefore, G = G1 N = G1 (N0 H) = (G1 N0 )H = G1 H, G1 ∩ H = G1 ∩ (N ∩ H) = (G1 ∩ N) ∩ H = N0 ∩ H = H0 , and the proof of (c) is complete. (d) Recall that |G| = mj,

j = |H : H0 |,

and so mj = |G| = |G : H||H : H0 ||H0 | = |G : H|j|H0 | 󳨐⇒ m = |G : H||H0 |;

460 | Characters of Finite Groups 1 then H0 is a π-subgroup, where π = π(m). Assume that x ∈ G is not a π-element. Let p ∈ π󸀠 ∩ π(o(x)); then p ∤ m. Let us represent x as x = x p x p󸀠 = x p󸀠 x p . Next, p ∤ m and |G| = mj implies p | j. Let P ∈ Sylp (H); then, by assumption, P ∈ Sylp (G). Without loss of generality, one may assume that x p ∈ P ≤ H. Then x p = x xp ∈ H ∩ H x and x p ∈ ̸ H0 (since H0 is a p󸀠 -subgroup and, by the choice of x, x p ≠ 1). Therefore, by assumption, H x = H, and so x ∈ NG (H) = N. Thus, all non-π-elements of G are contained in the set ⋃y∈G N y . Let N π be the set of all π-elements of N and write D = G − ⋃t∈G (N − N π )t . If y ∈ D is not a π-element, then by what has just been said, y ∈ ⋃ N t = ⋃ (N π ∪ (N − N π ))t = (⋃ (N π )t ) ∪ (⋃ (N − N π )t ). t∈G

t∈G

t∈G

t∈G

Since y ∈ D, it follows that y ∈ ̸ ⋃t∈G (N − N π )t and, therefore, y ∈ ⋃t∈G (N π )t , i.e., y is a π-element, contrary to its choice. Thus, all elements of D are π-elements. Therefore, |D| ≤ m, by hypothesis. On the other hand, since N coincides with its normalizer in G, one has 󵄨󵄨 󵄨󵄨 |D| = |G| − 󵄨󵄨󵄨󵄨 ⋃ (N − N π )t 󵄨󵄨󵄨󵄨 ≥ |G| − |G : NG (N)||N − N π | 󵄨 󵄨 t∈G

= |G| − |G : N| ⋅ |N − N π | = |G : N| ⋅ |N π |.

(∗)

In follows from GCD(m, j) = 1 and |H0 | | m = |G : H||H0 | that H0 is a π-Hall subgroup of H since |H| = j|H0 |, and so H0 is characteristic in H. Thus, H0 ⊲ N(= NG (H)). Next, |N : H0 | = |N : H||H : H0 | = |N : H|j 󳨐⇒ |N : H| | m. Then, since GCD(m, j) = 1, H/H0 is a normal π󸀠 -Hall subgroup of N/H0 and therefore, by the Schur–Zassenhaus theorem, N/H0 = (N0 /H0 )⋅(H/H0 ) (semidirect product with kernel H/H0 ) for some N0 /H0 < N/H0 . Next, |N/H0 | = |H/H0 ||N0 /H0 | = j|N0 /H0 | 󳨐⇒ |N : N0 | = j, whence |G : N||N0 | =

mj |N0 | = m. |N0 |j

In particular, N0 is a π-Hall subgroup of N so N0 ⊆ N π , and hence |N0 | ≤ |N π |. Therefore, by (∗), |D| ≥ |G : N| ⋅ |N π | ≥ |G : N| ⋅ |N0 | = m. Thus, |D| = m, i.e., D is the set of all solutions of the equation x m = 1 in G, and then |N π | = |N0 |. Hence N0 = N π so that N0 ⊲ N. Since |D| = m, it follows from (∗) that NG (N) = N and N − N π = N − N0 is a TI-subset on G, and hence (G, N, N0 ) is a W-triple with kernel D (see (a)). The proof of (d) is complete. Exercise 2.1. If G0 is the kernel of the W-triple (G, H, H0 ), then CG (x) ≤ G0 for all x ∈ G0 − ⋃s∈G H0s .

X Existence of normal subgroups | 461

Solution. Assume that CG (x) ≰ G0 = G − ∑t∈G (H − H0 )t . Then CG (x) ∩ (H − H0 )t ≠ 0 for some t ∈ G. Let y be an element in that intersection. Then y ∈ (H − H0 )t ∩ (H − H0 )tx ,

i.e., (H − H0 ) ∩ (H − H0 )txt

−1

≠ 0.

It follows that txt−1 ∈ H so that x ∈ H t ∩ G0 . Next, H t ∩ G0 = (H ∩ G0 )t = H0t 󳨐⇒ x ∈ H0t . This is impossible since, by condition, x ∈ ̸ ⋃s∈G H0s . Hence CG (x) ≤ G0 . Definition 2.1. If Φ = (G, H, {1}) is a W-triple, then G is called a Frobenius group with Frobenius complement H. The kernel G(Φ) = F of this W-triple is called the Frobenius kernel of G. A Frobenius group with Frobenius complement H and kernel F will be denoted by G = (H, F). It is clear that G = (H, F) implies G = (H t , F) for any t ∈ G. According to the above, Frobenius groups may be defined as follows: Definition 2.2. Let {1} < H < G. Then G is called a Frobenius group with Frobenius complement H if x ∈ G − H implies H ∩ H x = {1}. The first Wielandt theorem implies the following result of fundamental importance: Theorem 2.3. Let G be a Frobenius group with Frobenius complement H. Then: (a) The set F = G − ⋃t∈G (H # )t is a normal subgroup of G. (b) G = H ⋅ F, a semidirect product. Frobenius groups arise naturally in the theory of permutation groups. Let G be a transitive nonregular permutation group acting on the set {1, . . . , n} each of whose nonidentical permutations has at most one fixed point. Let H i be the stabilizer of a point i, H = H1 . Then |G : H1 | = n and the subgroups H = H1 , . . . , H n are conjugate in G. By assumption, H i ∩ H j = {1} for i ≠ j. As n = |G : NG (H), it follows that NG (H i ) = H i . Therefore, G is a Frobenius group with Frobenius complement H. The subgroup F of all permutations without fixed points, together with the identical permutation, is the Frobenius kernel of G. Conversely, let G be a Frobenius group with Frobenius complement H = H1 , and let H1 , . . . , H n be the set of G-conjugates of H. For every element x ∈ G, define the permutation H1 . . . H n x=( x ). H1 . . . H nx It is easy to see that the permutation group G = {x | x ∈ G} thus constructed possesses the same properties as the permutation group G in the previous paragraph. Exercise 2.2. Let (G, H, H0 ) be a W-triple with kernel G0 . Then H G (= the normal closure of H in G) coincides with G. If H < F ≤ G, then (F, H, H0 ) is a W-triple with kernel F ∩ G0 .

462 | Characters of Finite Groups 1 Hint. As G0 < G, it follows from G = G0 ∪H G that H G = G. Then G = HG0 , and if H < F, then F = H(F ∩ G0 ), by the modular law. Now it is easy to check, that (F, H, H0 ) is a W-triple with kernel F ∩ G0 . An automorphism α of a group G > {1} is fixed-point free if x α = x (x ∈ G) implies x = 1. In this case, α induces on the set G# = G − {1} a permutation without fixed points. Exercise 2.3. Let G = (H, F) be a Frobenius group with complement H and kernel F. Then: (a) If a ∈ H # , then the mapping f 󳨃→ afa−1 is a fixed-point-free automorphism of F. (b) |H| | |F # | so that GCD(|H|, |F|) = 1, i.e., H and F are Hall subgroups of G. (c) If P ∈ Sylp (F), then |H| | |Z(P)# |. (d) If F < M ≤ G, then M = (H ∩ M, F). (e) If x ∈ F # , y ∈ H # , then CG (x) ≤ F, CG (y) ≤ H. (f) If H < M ≤ G, then M = (H, M ∩ F) is a Frobenius group with Frobenius complement H and kernel M ∩ F. (g) (Burnside) If |H| is even, then F is abelian. In that case one can prove Frobenius’ theorem without character theory. Solution. (a) If afa−1 = f and f ∈ F # , then a f = a ∈ H ∩ H f = {1}, a contradiction. Thus, f commutes with the identity element of H only; hence conjugation of F with a is a fixed-point-free automorphism of F. (b) By (a), all H-orbits on F # have length |H| so |H| divides |F # | = |F| − 1 whence GCD(|F|, |H|) = 1. (c) Let P ∈ Sylp (F). By Frattini’s lemma, G = FNG (P). It follows from (a) that NG (P) is a Frobenius group with kernel NG (P) ∩ F = NF (P) so that a Frobenius complement of that group is H1 of the same order as H. We have H1 P ≤ G. Next, H1 normalizes Z(P) since Z(P) is characteristic in P so H1 ⋅ Z(P) is a Frobenius group. It follows from (b) that |H1 | = |H| divides |Z(P)| − 1 = |Z(P)# |. (d) By the modular law, M = (H ∩ M)F. Now the claim is obvious. (e) If z ∈ CG (x) ∩ H0# , where H0 is a complement to F, then 1 ≠ z ∈ H0 ∩ H0z , which is a contradiction. If CG (y) ≰ F, then CG (y) ∩ H1 > {1} for some G-conjugate H1 of H. Then H1 -orbit of y has size < |H1 | = |H|, contrary to (a). (f) By the modular law, M = H(M ∩ F), and our claim is obvious. (g) Let H = H1 , H2 , . . . , H n be all the G-conjugates of H in G and let x1 be an invoa a lution in H = H1 . If H i = H1 i (i ∈ {1, . . . , n}, a i ∈ G), denote x i = x1 i ; then x i is an involution in H i . For every i > 1, consider the subgroup D i = ⟨x1 , x i ⟩. Clearly, Z(D i ) = {1} (as we know, D i is dihedral) so that D i is a Frobenius group, by (e). Therefore, the element x1 x i is of odd order for all i = 2, 3, . . . , n. Assume that, for i > 1, x1 x i ∈ H j x for some j. Now (x1 x i )x1 = (x1 x i )−1 ∈ H j implies x1 x i ∈ H j ∩ H j 1 = {1}, which is a contradiction. Then n

(G − ⋃ H i ) ∪ {1} = G0 = {x1 x2 , . . . , x1 x n , 1} i−1

X Existence of normal subgroups | 463

in the notation of §X.1. Indeed, by what has just been proved already, {x1 x2 , . . . , x1 x n , 1} ⊆ G0 . Since |G0 | = n = |{x1 x2 , . . . , x1 x n , 1}|, our claim follows. Let us prove that any two elements of the set G0 commute. We have proved that x1 inverts the set G0 . Since also G0 = {x i x1 , . . . , x i x n , 1}, i > 1, it follows that x i inverts G0 . We will show, for example, that x1 x2 and x1 x3 commute. Indeed, x1 x2 ⋅ x1 x3 = x1 x3 x3 ⋅ x2 x1 x3 = x1 x3 ⋅ (x2 x1 )x3 = x1 x3 ⋅ (x2 x1 )−1 = x1 x3 ⋅ x1 x2 , as was to be shown. We claim that G0 < G. Since G0 = {x i x1 , . . . , x i x n , 1}, i > 1, it follows for i, j ≤ n that for some k ∈ {1, . . . , n}, x1 x i ⋅ x1 x j = x1 x i ⋅ x i x k = x1 x k ∈ G0 . Therefore, G0 < G.¹ We state the result of Exercise 2.3 (e) as follows: Supplement to Frobenius’ theorem. If F is a nontrivial normal subgroup of a group G such that CG (f) ≤ F for all f ∈ F # , then G is a Frobenius group with kernel F. The normalizer NG (P) of a Sylow 2-subgroup P of the simple Suzuki group G = Sz(q) (q = 22m+1 ) is a Frobenius group with nonabelian Frobenius kernel P.² A Frobenius complement of a Frobenius group may be nonsolvable (there exists a Frobenius group (SL(2, 5), E112 )); moreover, SL(2, 5) is an essentially unique nonsolvable Frobenius complement in the sense that if the Frobenius complement coincides with its derived subgroup, it is isomorphic to SL(2, 5). Exercise 2.4. Let α be a fixed-point-free automorphism of prime order of a group F and ⟨α, F⟩ the α-holomorph of F. Then G = (⟨α⟩, F) is a Frobenius group with kernel F and complement ⟨α⟩. (It is easy to construct a group F with a fixed-point-free automorphism α of composite order such that G = (⟨α⟩, F) is not a Frobenius group.) We now digress briefly for a discussion of fixed-point-free automorphisms. To this end, let α ∈ Aut(G) be a fixed-point-free automorphism of a group G. Then the mapping x 󳨃→ x−1 x α is a bijection of G onto itself. Indeed, if x−1 x α = y−1 y α , then xy−1 = x α (y α )−1 = (xy−1 )α 󳨐⇒ xy−1 = 1 󳨐⇒ x = y. Thus, every element a ∈ G may be expressed uniquely in the form x−1 x α (x ∈ G) since G is finite. Let o(α) = n; then α n = idG , and, setting a = x−1 x α , we obtain aa α ⋅ ⋅ ⋅ a α

n−1

2

= x−1 x α (x α )−1 x α ⋅ ⋅ ⋅ (x α

n−1

n

)−1 x α = 1.

1 All trials to produce a character-free proof of the Frobenius theorem in the general case were unsuccessful. 2 The first example of a Frobenius group with nonabelian kernel was constructed by O. Y. Schmidt and has order 3 ⋅ 73 .

464 | Characters of Finite Groups 1 Therefore, if n = 2, we have a α = a−1 for all a ∈ G, so that, if a, b ∈ G, then a−1 b−1 = a α b α = (ab)α = (ab)−1 = b−1 a−1 󳨐⇒ ab = ba and G is abelian (Burnside). Moreover, as easily follows from Sylow’s theorem, G is of odd order. Conversely, if G is an abelian group of odd order, then a 󳨃→ a−1 is a fixedpoint-free automorphism of G of order 2. If G = (H, F), then all the elements of H # induce fixed-point-free automorphisms on F (Exercise 2.3 (a)). J. G. Thompson [Tho1] has proved that a group admitting a fixed-point-free automorphism of prime order is nilpotent. In particular, the Frobenius kernel F of a group G = (H, F) is nilpotent. This fundamental result gives a positive answer to a conjecture of Frobenius. It is easy to deduce from this that if G = (H, F) = (H1 , F1 ), then F = F1 . However, that result can be proved in a simpler way if one uses the fact that a Frobenius complement of a Frobenius group has nontrivial center (we shall prove this assertion later). Let α ∈ Aut(G) be fixed-point free. Suppose that there is u ∈ G such that u α = u x for some x ∈ G. As we know, x = y−1 y α for some y ∈ G so that u α = x−1 ux = y−α yuy−1 y α 󳨐⇒ y α u α y−α = (yuy−1 )α = yuy−1 , and we conclude that yuy−1 = 1, i.e., u = 1. It follows that α induces a regular permutation of G-classes ≠ {1}; in particular, α ∈ ̸ Inn(G). Thus, if G = (H, F) is a Frobenius group, then H is isomorphic to a subgroup of the symmetric group Sk(F)−1 , where k(F) is the class number of F. Let us again α ∈ Aut(G) be fixed-point free and P ∈ Sylp (G). By Sylow, there exists x ∈ G such that P α = x−1 Px. By the above, x = y−1 y α for some y ∈ G. Therefore, P α = y−α yPy−1 y α

or

y α P α y−α = yPy−1 .

It follows that (yPy−1 )α = yPy−1 , i.e., yPy−1 ∈ Sylp (G) is α-invariant. We conclude, that if α and G as above, then for every p ∈ π(G), there is P ∈ Sylp (G) such that P α = P. Moreover, if all proper subgroups of given order n are conjugate in G and α is as above, there is H < G of order n such that H α = H. Lemma 2.4. Let G = (H, F). Then every Sylow subgroup of H is either cyclic or a generalized quaternion group. Proof. Let p ∈ π(F) and P ∈ Sylp (F). By Frattini’s lemma, G = NG (P)F. It is obvious that NG (P) is a Frobenius group with kernel NF (P). Without loss of generality, we may assume that H ∈ Sylq (G). Some subgroup conjugate to H is contained in NG (P). We may therefore assume that G = (H, P) (taking our aim into account). Let A be an abelian subgroup of H, A > {1}. Let M be a minimal normal subgroup of (a Frobenius group) AP = (A, P). Then AM = (A, M), and the abelian group A acts faithfully and irreducibly on M. By Schur’s lemma, A is cyclic. Thus, any abelian subgroup of the q-group H is cyclic. It follows that H is either cyclic or a generalized quaternion group [Ber31, Lemma 1.4, Theorem 1.2].

X Existence of normal subgroups | 465

Let G = (H, F); then π(H)-Hall subgroups are conjugate in G. We claim that one of the subgroups H, F is solvable. If H is of odd order, it is solvable since all its Sylow subgroups are cyclic (Lemma 2.4). If H is even, then F is abelian, by Exercise 2.3 (g). Exercise 2.5. Let α be a fixed-point-free automorphism of a group G > {1}, p ∈ π(G). Then G has only one α-invariant Sylow p-subgroup. Solution. By the above discussion, there is an α-invariant Sylow p-subgroup of G. Let −1 P, Q ∈ Sylp (G) with P α = P, Q α = Q. Let Q = P z , where z ∈ G (Sylow). Then Pz

−1 α

z

α

= Q z = (Q z )α = P α = P 󳨐⇒ z−1 z α ∈ NG (P).

Since NG (P)α = NG (P α ) = NG (P), we get z−1 z α = u−1 u α for some u ∈ NG (P). By the −1 above, z = u ∈ NG (P) so Q = P z = P, and our claim follows.³ Exercise 2.6. Let α and β be commuting fixed-point-free automorphisms of a group G. If P ∈ Sylp (G) is α-invariant, then it is also β-invariant. Solution. It follows from P β = (P α )β = P αβ = P βα = (P β )α that P β is α-invariant. Then, by Exercise 2.5, P β = P, as claimed. Exercise 2.7. Suppose that A ≤ Aut(G) is abelian. If all elements of the set A# are fixed-point-free automorphisms of G, then A is cyclic. Hint. By hypothesis, A ⋅ G is a Frobenius group with complement A and kernel G. Use Lemma 2.4.⁴ W. Burnside, H. Zassenhaus and M. Suzuki have classified the groups all of whose Sylow subgroups are either cyclic or generalized quaternion groups (the result for the more difficult nonsolvable case is due to Suzuki). Now we are ready to prove the following characterization (due to Isaacs) of Frobenius groups. Theorem 2.5. Suppose that a π-group H > {1} acts on a nonidentity π󸀠 -group Q in such a way that CH (x) ≤ H 󸀠 ∩ Z(H) for all x ∈ Q# . Then the natural semidirect group of H and Q is a Frobenius group (H, Q). Proof (compare with [Isa22, Lemma 3.1]). We proceed by induction on |H|. Assume that the action is not Frobenius. Then there exists x ∈ Q# such that Z = CH (x) > {1}. By assumption, H is nonabelian (indeed, {1} ≠ Z ≤ H 󸀠 , by hypothesis) and Z ⊲ H. Write C = CQ (Z) and note that C > {1}. Let Z1 be a subgroup of prime order, say p, in Z.

3 Suppose that H < G is such that all subgroups of G that are isomorphic to H are conjugate. If α is a fixed-point-free automorphism of G, then α fixes exactly one subgroup of G that is isomorphic to H. The proof is the same. 4 It is possible to give the proof of this exercise independent of Lemma 2.4. This allows one to deduce Lemma 2.4 from Exercises 2.5–2.7.

466 | Characters of Finite Groups 1 Then Z1 ⊲ H and H/Z1 acts on C (indeed, CHQ (Z) contains H and has intersection C with Q ⊲ HQ). Without loss of generality, one may assume that H is a p-group. If y ∈ C# , then since GCD(|H|, |Q|) = 1, we get CH/Z1 (y) = CH (y)/Z1 ≤ (H 󸀠 ∩ Z(H))/Z1 ≤ (H/Z1 )󸀠 ∩ Z(H/Z1 ), and so the action of H/Z1 on C satisfies the hypotheses of the theorem. Therefore, by induction, the action of H/Z1 on C is Frobenius. It follows that any Sylow subgroup of H/Z1 is cyclic or generalized quaternion. However, by Theorem VI.4.6 (b), the group M(H/Z1 ), the Schur multiplier of H/Z1 , is trivial. Since Z1 ≤ H 󸀠 ∩ Z(H) and H 󸀠 ∩ Z(H) is isomorphic to a subgroup of M(H/Z1 ) = {1}, we get a contradiction. Exercise 2.8. The following statements hold: (a) (I. N. Herstein) If A, a maximal subgroup of G, is abelian, then G is solvable. (b) (N. Ito) If G = AB, where A, B are abelian subgroups of G, then G is solvable. Solution. Let G be a counterexample of minimal order. Then it is semisimple, i.e., F(G) = {1}; in particular, Z(G) = {1}. (a) Since A ∩ A x ≤ Z(G) = {1} for x ∈ G − A, we see that G is a Frobenius group with complement A and kernel, say M. Since GCD(|A|, |M|) = 1, it follows that G is π(A)-solvable so it has, for any prime divisor q of |M|, a π(A) ∪ {q}-Hall subgroup H such that A < H. Therefore, G = H is solvable. (b) If A is maximal in G, the result follows from (a). If A < M < G, where M is maximal in G, then M = A(M ∩ B) is solvable, by induction, and M G = ⋂x∈B M x > {1}. As M G is solvable, this contradicts the semisimplicity of G.

3 Characters of Frobenius groups In this section we consider characters of Frobenius groups. The following important lemma is useful in many applications. Lemma 3.1. The following statements hold: (a) (A. L. Vishnevetsky, unpublished) Let P, Q be permutations belonging to the symmetric group Sn , considered as the subgroup of permutation matrices of GL(n, 𝔽), where 𝔽 is an arbitrary field. If P, Q are conjugate in GL(n, 𝔽), then they are also conjugate in Sn . (b) (R. Brauer) Let A be a nonsingular matrix. Assume that a matrix B can be obtained from A in two different ways: (i) by rearrangement of rows; (ii) by rearrangement of columns. Then the number of fixed rows in the first transformation is equal to the number of fixed columns in the second. (c) (Brauer permutation lemma) Let A be a group acting on Irr(G) and on the set of G-classes. Assume that χ a (g a ) = χ(g) for all χ ∈ Irr(G), a ∈ A, g ∈ G (where g a is an element of the G-class K ga ). Then, for every a ∈ A, the number of fixed irreducible characters of G is just the number of fixed G-classes.

X Existence of normal subgroups | 467

Proof. (a) (Zhmud) Let χ be the natural character of Sn , i.e., χ(Γ) is equal to the number of fixed points of the permutation Γ ∈ Sn . Let ψ Γ (m) be the number of cycles of length m in Γ (factored into a product of independent cycles). Since χ(Γ m ) is equal to the sum of lengths of those cycles whose lengths divide m, it follows that (1)

χ(Γ m ) = ∑ ψ Γ (d)d. d|m

Putting ϕ Γ (m) = χ(Γ m ), we rewrite (1) in the following form: (2)

ϕ Γ (m) = ∑ ψ Γ (d)d. d|m

By the Möbius inversion formula, we have ψ Γ (m) = m−1 ∑ ϕ Γ (d)μ( d|m

m ), d

(3)

where μ( ⋅ ) is the number-theoretic Möbius function. Let P, Q ∈ Sn be conjugate in GL(n, F). Then P m , Q m are also conjugate in GL(n, F) for all m ∈ ℕ. Therefore, χ(P m ) = tr(P m ) = tr(Q m ) = χ(Q m )

(m ∈ ℕ),

i.e., ϕ P (m) = ϕ Q (m). Therefore, it follows from (3) that ψ P (m) = ψ Q (m) for all m. As it is well known, this implies that P, Q are conjugate in Sn . (b) We have PA = B = AQ, where P, Q are permutation matrices. Then Q = A−1 PA, i.e., P, Q are conjugate in GL(n, ℂ), where A is an n × n matrix. By (a), ψ P (m) = ψ Q (m) for all m ∈ ℕ. In particular, ψ P (1) = ψ Q (1), and (b) is proved. (c) Let A = X(G) be the character table of G. The result follows from (b). Theorem 3.2. Let G = (H, F) be a Frobenius group. (a) If ϕ ∈ Irr# (F) = Irr(F) − {1F }, then ϕ G ∈ Irr(G). (b) If χ ∈ Irr(G) and F ≰ ker(χ), then χ = ϕ G for some ϕ ∈ Irr# (F). Proof. (a) Let x ∈ G − F and K x = K for some F-class K ≠ {1}. Then, for y ∈ K, there exists f ∈ F such that y x = y f , and hence xf −1 ∈ CG (y) ≤ F implies x ∈ F, a contradiction. Thus, the elements of G − F act regularly on the set of F-classes contained in F # (this we have noted above). By Brauer’s Permutation Lemma 3.1 (c), the elements of G − F act regularly on Irr(F) − {1F }. Therefore, if ϕ ∈ Irr# (F), then IG (ϕ), the inertia group of ϕ in G, is equal to F, so that ϕ G ∈ Irr(G) (see Chapter VII). (b) Let χ ∈ Irr(G | F). Then, by Clifford’s theorem, ⟨χ F , 1F ⟩ = 0. Let ϕ ∈ Irr(χ F ). Then ϕ G ∈ Irr(G) (by (a)). Since, by reciprocity, χ ∈ Irr(ϕ G ), it follows that ϕ G = χ by (a).

468 | Characters of Finite Groups 1 Thus, if G = (H, F) and χ ∈ Irr(G | F), then |G : F| | χ(1). Later we prove, in particular, that if F ⊲ G and |G : F| | χ(1), then G is a Frobenius group with Frobenius kernel F. Next, it follows from Theorem 3.2 that Irr(G) = Irr(1GF ) ∪ {ϕ G | ϕ ∈ Irr# (F)}. Exercise 3.1. Let G = (H, F). Then the following statements hold: (a) ∑χ∈Irr(G), |H|∤χ(1) χ(1)2 = |H|. (b) k(G) = k(H) +

k(F)−1 |H| .

Solution. (a) Let χ ∈ Irr(G) with |H| ∤ χ(1). Then F ≤ ker χ hence χ F = χ(1)1F (Theorem 3.2 (a)). It follows that (1F )G = ∑χ∈Irr(G/F) χ(1)χ. Since (1F )G (1) = |H|, the result follows. (b) Use the equality k(G) = |Irr(G)| and properties of the action of H on F. As we shall see later, condition (a) in Exercise 3.1 characterizes the Frobenius groups among groups G = H ⋅ F (semidirect product with kernel F). Exercise 3.2. Let G = (H, F), χ ∈ Irr(G) and F ≰ ker(χ). Then χ H = mρ H , where m ∈ ℕ and ρ H is the regular character of H. Solution. Since χ = ϕ G for some ϕ ∈ Irr(F) (Theorem 3.2 (b)), the character χ vanishes on G − F ⊃ H # . It follows that χ H = mρ H for some m ∈ ℕ. Since χ(1) = |G : F|ϕ(1) = |H|ϕ(1) = ϕ(1)ρ H (1), we get m = ϕ(1). The following statement is more general: Exercise 3.3. Let G = HF, F ⊲ G, H ∩ F = {1}. If ϕ ∈ Irr# (F) is such that ϕ G = χ ∈ Irr(G), then χ H = ϕ(1)ρ H . Hint. The induced character χ = ϕ G vanishes on G − F ⊃ H # so χ H is a multiple of ρ H . Take into account that χ(1) = |H|ϕ(1). Exercise 3.4. Let G = (H, F), ϕ ∈ Irr# (F). Then the number of G-conjugates with ϕ is equal to |H|. Hint. By Theorem 3.2 (a), the character ϕ G ∈ Irr(G) has degree |G : F|ϕ(1) = |H|ϕ(1). Consider (ϕ G )F , make use of Clifford’s theorem and take into account that the inertia subgroup IG (ϕ) = H. It follows from Theorem 3.2, Thompson’s theorem on nilpotency of the kernel F of a Frobenius group G = (H, F), that G is an M-group if and only if H is an M-group. Indeed, let χ ∈ Irr(G). We claim that χ is a monomial character. This is clear if F ≤ ker(χ) since G/F ≅ H is an M-group. If F ≰ ker(χ), then χ = ϕ G for some ϕ ∈ Irr# (F). Since F is an M-group, we obtain that ϕ = μ F for some μ ∈ Lin(U) and U ≤ H. In that case, we have χ = ϕ G = (μ F )G = μ G , as was to be shown.

X Existence of normal subgroups | 469

By Lemma 2.4 and the previous paragraph, if |H| is odd, then G = (H, F) is an M-group since H, all of whose Sylow subgroups are cyclic, is supersolvable and so it is an M-group, by Blichfeldt. The solvable Frobenius group G = (SL(2, 3), E(52 )) is not an M-group since SL(2, 3) is not an M-group (indeed, 2 ∈ cd(G) but G has no subgroup of index 2). Definition 3.1. A matrix representation D of a group G is said to be unit-free if, for any g ∈ G# , 1 is not an eigenvalue of the matrix D(g). If D : G → GL(V), where V = V(n, ℂ), is a unit-free representation and x ∈ V, then D(g)(x) = x implies either g = 1 or x = 0. Let a ℂG-module M afford a representation D. It is clear that D is unit-free if gm ≠ m for g ∈ G# and m ∈ M − {0}. Therefore D is also called a representation without fixed points. A group admitting a unit-free representation is called a group without fixed points. Exercise 3.5. Let D be a unit-free representation of a group G. Then the following statements hold: (a) ker(D) = {1}. (b) If {1} < H < G, then D H is unit-free. Hint. If ker(D) > {1} and g ∈ ker(D)# , then D(g) = Ideg(D) has fixed points. Exercise 3.6. A Frobenius group has no unit-free ℂ-representations. Hint. Use Exercises 3.2 and 3.5. One can show that the result of Exercise 3.6 is also valid for 𝕂-representations, where 𝕂 is a field whose characteristic does not divide the order of the Frobenius group. Exercise 3.7. A group H is a Frobenius complement of a Frobenius group if and only if there is a unit-free representation of H over GF(p) for some prime p that does not divide |H|. Hint. Let a representation D : H → GL(V), where V = V(n, GF(p)), be a representation whose existence is asserted in the exercise. Then the semidirect product H ⋅ V is a Frobenius group with complement H. We claim that a nonabelian group G of order pq, where p and q are distinct primes, say p > q, has no unit-free representation. We have G = (H, F), where |H| = q and |F| = p. Let T be a unit-free matrix representation of G of minimal degree, let 𝕂 be the field of T. As the irreducible blocks of T are also unit-free, T must be irreducible. Since T remains unit-free when the ground field is extended, one may assume that T is absolutely irreducible. We may therefore assume that 𝕂 is algebraically closed. It is not difficult to prove that the characteristic of 𝕂 does not divide |G|, but we shall not use this. Since F ⊲ G, it follows from the general Clifford theorem that the restriction T F of T to F is completely reducible.

470 | Characters of Finite Groups 1 Let M be the module affording T. Restriction of the domain of operators to the subgroup F gives a completely reducible 𝕂F-module M F , which affords the representation T F of F. Let M F = M1 ⊕ ⋅ ⋅ ⋅ ⊕ M r be the decomposition of M F into a direct sum of homogeneous constituents (see §§I.10–I.12). Correspondingly, we have the following decomposition of M as a direct sum of subspaces M i : M = M1 ⊕ ⋅ ⋅ ⋅ ⊕ M r . The F-invariant subspaces M i are transitively permuted by G. Let J = {g ∈ G | gM1 = M1 } be the stabilizer of M1 , and let G = ⋃rj=1 t j J be a decomposition of G into left cosets of J. Then r = |G : J|, and for some enumeration of the cosets we may have M i = t i M1 (i ∈ {1, . . . , r}). We claim that r > 1, i.e., the module M F is not homogeneous. Indeed, otherwise, M F = L1 ⊕ ⋅ ⋅ ⋅ ⊕ L m , where L i are pairwise 𝕂F-isomorphic irreducible submodules. Then T F ∼ m∆, where ∆ is the irreducible representation of the group F, afforded by the 𝕂F-submodule L1 . Since F is abelian and 𝕂 is algebraically closed, it follows that deg(∆) = 1. Hence the matrices T(f) (f ∈ F) are scalar and so commute with all the matrices T(g) (g ∈ G). As T is faithful, it follows that F ≤ Z(G) and G is abelian, a contradiction. Thus |G : J| = r > 1. Since F ≤ J, it follows that J = F, i.e., r = |G : F| = q, q M = M1 ⊕ ⋅ ⋅ ⋅ ⊕ M q . Let H = ⟨s⟩ = {1, s, . . . , s q−1 }. Then G = ⋃i=1 s i−1 F is a decomposition of G over F = J, and we can enumerate the subspaces M i in such a way that M i = s i−1 M1 for i = 1, . . . , q. Taking v ∈ M1 − {0}, put u = v + sv + ⋅ ⋅ ⋅ + s q−1 v. Since v ≠ 0 and s i−1 v ∈ M i , we have u ≠ 0. Noting that su = u, we conclude that 1 is an eigenvalue of the matrix T(s), which is impossible. Thus, the group G has no unit-free representations. We claim that a Frobenius complement H of a Frobenius group G = (H, F) has no noncyclic subgroup of order pq, where p, q are distinct primes. Indeed, let r ∈ π(F) and R ∈ Sylr (F) be such that H ≤ NG (R) (see the proof of Lemma 2.4; here we do not use Thompson’s theorem on nilpotency of F). Let V be a minimal H-invariant subgroup in Z(R); then V is an elementary abelian r-group. One can consider V as an 𝔽r H-module, where 𝔽r is the field with r elements. Then V affords a unit-free representation T of H since H ⋅ V = (H, V) is a Frobenius group. Assume, by way of contradiction, that the assertion is not true. Then we may assume that H itself is nonabelian of order pq, and this contradicts the result of the previous paragraph. It follows from the above result that a Frobenius complement has no subgroups which are Frobenius groups. To prove this, one may assume that G = (H, F), where H is a Frobenius group of minimal possible order; then π(H) = {p, q}, where p, q are distinct primes. By Lemma 2.4, all Sylow subgroups of H are either cyclic or generalized quaternion. Then Frobenius complement of H has prime order, by minimality of H. Therefore, if S ≤ H is minimal nonnilpotent, we get |S| = pq (see Lemma XI.2.1), contrary to what has been proved above. Remark 3.1. Below we will present another proof of the following result. Claim. Let G = (H, F), |H| = pq, where p, q are not necessarily distinct primes. We claim that H is cyclic.

X Existence of normal subgroups | 471

Suppose that H is not cyclic. As above, one may assume that F is a minimal normal elementary abelian subgroup of G. Then |F| = r n , where the prime r ∈ ̸ {p, q}. Let {S1 , . . . , S k } be the set of all subgroups of prime order in H. To fix ideas. we assume that q ≥ p. Then k = q + 1. If v ∈ F # , h i ∈ S#i are fixed, then, considering F as a GF(r)H-module, we have h i ⋅ ∑ hv = ∑ h i hv = ∑ hv 󳨐⇒ ∑ hv = 0. h∈S i

h∈S i

h∈S i

h∈S i

Similarly, ∑h∈H hv = 0. Therefore, q+1

0 = ∑ ∑ hv = qv + ∑ hv = qv + 0 = qv. i=1 h∈S i

h∈H

Since r ≠ q, we have v = 0, contradicting the choice of v. Thus, H is cyclic, as claimed. From this and Exercise 7 one can easily deduce Lemma 2.4. In fact, a similar method enables one to prove that if a group H of order pq (p and q are not necessarily distinct primes) admits a representation without fixed points over an arbitrary field F, then H is cyclic. Burnside and Zassenhaus completely classified the groups which are Frobenius complements of Frobenius groups. In particular, if all Sylow subgroups of a Frobenius complements are cyclic then its derived subgroup is a cyclic Hall subgroup. (See also paragraph following Exercise 3.7.) Theorem 3.3. Let p be the minimal prime divisor of the order of a Frobenius complement H of the Frobenius group G = (H, F). Then p | |Z(H)|. Proof. (a) Suppose that p = 2. Let x, y be distinct involutions in H. Since x, y invert F, it follows that xy ≠ 1 centralizes F, contrary to Exercise 2.3 (d). Thus, H has a unique involution x and therefore x ∈ Z(H), so that 2 | |Z(H)|. (b) Now let p > 2. Then H is p-nilpotent, by Burnside’s normal p-complement theorem since a p-Sylow subgroup of H is cyclic. Let x ∈ H be of order p. By Frattini, x normalizes some Q ∈ Sylq (H) for q ∈ π(H) − {p}. Let K = ⟨x, Q⟩. If Q0 < Q is of order q, then Q0 is x-invariant and, as we know, ⟨x, Q0 ⟩ is cyclic of order pq (recall that Q is cyclic). But then x centralizes Q since ⟨x, Q⟩ has no minimal nonnilpotent subgroups (see Lemma XI.2.1). Thus, x centralizes the p󸀠 -Hall subgroup of H. Since P ∈ Sylp (G) is cyclic, we get x ∈ Z(H), and so p | |Z(H)|. Exercise 3.8. The following statements hold: (a) A Frobenius complement is not a Frobenius group. (b) A Frobenius complement admits a faithful irreducible representation. Hint. (a) Use Theorem 3.3. (b) The socle of a Frobenius complement is cyclic, by Lemma 2.4. Now the claim follows from Gaschütz’ Theorem IX.5.4. The following remark will be useful in the proof of Witt’s Theorem 3.4.

472 | Characters of Finite Groups 1 Remark 3.2. Let α be a regular (= fixed-point free) automorphism of a group G and H an α-invariant normal subgroup of G. We will show that α induces a fixed-pointfree automorphism of the quotient group G/H. Assume that, for x ∈ G − H, we have (xH)α = xH, i.e., x α H = xH. Then x−1 x α ∈ H. Since α H is a fixed-point-free automorphism of H, it follows that x−1 x α = h−1 h α for some h ∈ H. This implies that x = h ∈ H and xH = H. Thus, α induces a fixed-point-free automorphism on G/H. Theorem 3.4 (E. Witt). Let α be a fixed-point-free automorphism of prime order p of a solvable group G. Then G is nilpotent. Proof. It follows from assumption that p ∤ |G|. In fact, the lengths of ⟨α⟩-orbits on G# are equal to p, and therefore |G| ≡ 1 (mod p). (See also Exercise 2.3 (b).) Let G be a counterexample of minimal order. Since G is solvable, Oq (G) = Q > {1} for some q ∈ π(G). Then Q α = Q (Q is characteristic in G) so, by Remark 3.2 and induction, the group G/Q is nilpotent. Therefore, Q ∈ Sylq (G). Let r ∈ π(G) − {q} and R ∈ Sylr (G). Then (RQ)α = RQ since RQ is characteristic in G. If RQ < G for some r ∈ π(G) − {q}, then RQ < G for all r ∈ π(G) − {q}, and (by induction) RQ is nilpotent; then Q is a direct factor of G. In that case, G is nilpotent. Therefore, since G is a minimal counterexample, we get G = RQ, π(G) = {r, q}. Assume that Q0 , Q1 are distinct minimal α-invariant normal subgroups of G. Then G is a subgroup of the direct product of groups G/Q0 and G/Q1 which are nilpotent, by induction. In this case, G is nilpotent, contrary to the assumption. Thus, G has a unique minimal ⟨α⟩-invariant subgroup which we denote Q0 . Then Q0 ≤ Q. If Q0 < Q, then Q0 R/Q0 is the characteristic Sylow r-subgroup in the nilpotent, by induction, group G/Q0 so that (RQ0 )α = RQ0 and, by induction, RQ0 is nilpotent. Then R ≠ Q is an α-invariant and normal subgroup of G, contrary to the result of the previous paragraph. Thus, Q = Q0 is a minimal α-invariant normal (Sylow) subgroup of G so it is elementary abelian. If CG (Q) > Q, then the Sylow r-subgroup of CG (Q) is α-invariant and normal in G, a contradiction. It follows that CG (Q) = Q, and we conclude that R acts on Q faithfully. Therefore, by induction, G/Q has no nontrivial α-invariant subgroups. Hence, R ≅ G/Q is an elementary abelian r-subgroup. Assume that R acts irreducibly on Q. Since R acts faithfully on Q, it follows from Schur’s lemma that R is cyclic. Therefore |R| = r and G = (R, Q) is a Frobenius group with kernel Q and complement R. Considering the action of ⟨α⟩ on the set Sylr (G) and taking into account that p ∤ |Sylr (G)| = |Q| (a power of q), one may assume without loss of generality that R α = R. Then ⟨⟨α, R⟩, Q⟩ is a Frobenius group with kernel Q and complement ⟨α, R⟩. But ⟨α, R⟩ is a Frobenius group, contrary to Theorem 3.3. Thus, R acts reducibly on Q. But Q is an irreducible GF(q)⟨α, R⟩-module. By the general Clifford’s theorem, Q is the direct sum of its homogeneous R-components (see §§I.10–I.12), which are conjugate with respect to ⟨α⟩: Q = Q0 ⊕ Q1 ⊕ ⋅ ⋅ ⋅ ⊕ Q k , i Q i = Q1α , k < p. i p−1 First let k = p − 1, Q i = Q0α , i = 0, . . . , p − 1. If v0 ∈ Q#0 , v = v0 + v0α + ⋅ ⋅ ⋅ + v0α , then v α = v. But v0 ≠ 0 implies v ≠ 0, contrary to the fact that α is fixed-point free.

X Existence of normal subgroups | 473

Now let k < p − 1. Then k = 0. Indeed, α of order p permutes k + 1 < p subgroups k Q0 , Q1α , . . . , Q k = Q0α so it fixes one of them say Q0 , and our claim follows. It follows that Q0 is a normal α-invariant q-subgroup of G so, by what has been proved above, Q0 = Q. Let L0 be a minimal normal subgroup of G. Since CG (Q) = Q, we have L0 ≤ Q. As Q is a homogeneous completely reducible GF(q)G-module, it follows that Q = L0 ⊕ ⋅ ⋅ ⋅ ⊕ L s , where L0 , . . . , L s are G-isomorphic. Since R acts faithfully on Q, R acts faithfully and irreducibly on L0 , . . . , L s . Therefore, by Schur’s lemma, R is cyclic, i.e., |R| = r. But then, as before, we may assume that R α = R, i.e., ⟨α, R⟩ is a Frobenius group. In that case, ⟨α, G⟩ = ⟨α, RQ⟩ = ((⟨α⟩, R), Q) is a Frobenius group with a Frobenius complement (⟨α⟩, R) which is a Frobenius group and so centerless. This contradicts Theorem 3.3 (see also Exercise 3.8 (a)). We use Theorem 3.4 to prove again that, provided G = (H, F) and p ≠ q are distinct primes, then H has no noncyclic subgroup of order pq, If this is false, one may assume that |H| = pq. Next, using Frattini’s lemma, one may assume that F is a Sylow subgroup of G. Let p < q; then H = (P, Q), where |P| = p and |Q| = q. Then a generator of P induces a fixed-point-free automorphism of order p on the solvable nonnilpotent group QF = (Q, F), contrary to Theorem 3.4. Frobenius conjectured that the Frobenius kernel of a Frobenius group is nilpotent. To prove this, it suffices to show, in view of Theorem 3.4, that the Frobenius kernel is solvable. This was done by J. Thompson [Tho1]. However, Thompson’s proof is based on quite different ideas, and we shall not present it here. It may be found, e.g., in [Hup1, Kapitel 4]. Exercise 3.9 (O. Grün). Frobenius’ theorem can be proved without character theory if the Frobenius complement is solvable. (For a solution, see §X.10.) Exercise 3.10 (Zassenhaus; see [Pas2, Theorem 18.6]). If G = (H, F) = G󸀠 , then one has H ≅ SL(2, 5). Exercise 3.11. Assume that a π-Hall subgroup H of a π-solvable group G is not normal in G and is a TI-subset. Then H is isomorphic to a Frobenius complement of some Frobenius group. Solution. We have H G = {1}. Let p be a prime divisor of |G : H| such that p | |G : NG (H)|. Then G possesses a π(H) ∪ {p}-Hall subgroup HP = PH, where P ∈ Sylp (G). Since H is a nonnormal TI-subgroup of HP, one may assume that G = HP. We have P G > {1}. Therefore, there is in P a minimal H-invariant subgroup L > {1}. Consider the semidirect product M = H ⋅ L; in that case, H is maximal in M. Since NM (H) = H and H is a TI-subgroup of M, it follows that M is a Frobenius group with Frobenius complement H. Remark 3.3 (communicated by A. I. Saksonov). A group all of whose Sylow subgroups are cyclic is isomorphic to a Frobenius complement if and only if any two of its elements of distinct prime orders commute.

474 | Characters of Finite Groups 1 It follows from the classification of finite simple groups that a group admitting a fixedpoint-free automorphism is solvable [Cle]. Lemma 3.5. Let G be a Frobenius group with kernel F and N ⊴ G. Then either N ≤ F or F < N. Proof. Assume that N ≰ F and t ∈ N − F. Then t induces a fixed-point-free automorphism of F. Let x, y ∈ F be such that x−1 x t = y−1 y t . Then xy−1 = (xy−1 )t so x = y. Thus, x 󳨃→ x−1 x t is a bijection of F. However, x−1 x t = [x, t] ∈ N so that F < N. Another proof. Assume that F and N are nonincident. Set K = F ∩ N; then G/K is a Frobenius group with kernel F/K. In that case, FN/K = (F/K) × (N/K) and N/K is nonidentity subgroup, a contradiction. Theorem 3.6. If F1 and F2 are kernels of a Frobenius group G, then F1 = F2 . Proof. By Lemma 3.5, F1 ≤ F2 and F2 ≤ F1 so that F1 = F2 . Proposition 3.7. Let G be a solvable group. If all Sylow subgroups of G are metacyclic, then {2, 3}󸀠 -Hall subgroup is normal in G. Proof. Assume that G is a counterexample of minimal order. Let R be a minimal normal subgroup of G; then |R| = p k for some prime p. By induction, a {2, 3}󸀠 -Hall subgroup of G/R, say H/R, is normal in G/R. If p > 3, then H is a normal {2, 3}󸀠 -Hall subgroup of G, contrary to the assumption. Thus, p ≤ 3. Since all Sylow subgroups of G are metacyclic, we get R ∈ {Cp , Ep2 }, p ≤ 3. Since Aut(R) is a {2, 3}-group, it follows that R ≤ Z(H) so H = R × F, where F is p󸀠 -Hall subgroup of H so of G. Then F is a normal {2, 3}󸀠 -Hall subgroup of G. (Thus, a solvable Frobenius complement H has a normal {2, 3}󸀠 -Hall subgroup.)

4 Converses to Frobenius’ theorem Let m > 1 be a proper divisor of the order |G| of a group G. Write N [m] =

⋂

ker(χ);

χ∈Irr(G), m∤χ(1)

then N [m] ≤ G󸀠 . Theorem 4.1 ([Ber12]). Let m > 1 be a proper divisor of |G|. Then the following hold: (a) GCD(m2 , |G|) | |G : N [m] | so that |G : N [m] | ≥ GCD(m2 , |G|). In particular, one has m | |G : N [m] |. (b) G is a Frobenius group with kernel of index m if and only if |G : N [m] | = m. Proof. (a) Write A = {χ ∈ Irr(G) | m | χ(1)},

a = ∑ χ(1)2 , χ∈A

X Existence of normal subgroups | 475

and B = Irr(G) − A,

b = ∑ χ(1)2 . χ∈B

Then a ≡ 0 (mod m2 ). As the numbers a and a + b = |G| are divisible by GCD(m2 , |G|), it follows that b = |G| − a ≡ 0 (mod GCD(m2 , |G|)). Since 1G ∈ B, we get b > 0 so that b ≥ GCD(m2 , |G|). As N [m] ≤ ker(χ) for all χ ∈ B, we get |G/N [m] | ≥ ∑χ∈B χ(1)2 = b, completing the proof of (a). (b) If G is a Frobenius group with kernel of index m, then N [m] coincides with this kernel (Theorem 3.2), and so |G : N [m] | = m. Now suppose that |G : N [m] | = m. We have to prove that G is a Frobenius group with kernel N (m) . It follows from (a) that GCD(|G|, m2 ) = m implies GCD( |G| m , m) = 1, so that N = N [m] is a π(m)󸀠 -Hall subgroup of G. Let ϕ ∈ Irr# (N)(= Irr(N) − {1N }) and χ ∈ Irr(ϕ G ). By Clifford’s theorem, ϕ(1) | χ(1). Since, by reciprocity, 1N ≠ ϕ ∈ Irr(χ N ), it follows that N ≰ ker(χ). Therefore, by hypothesis, we have m | χ(1). Since ϕ(1) | |N| (Frobenius–Molien) and GCD(|N|, m) = 1, it follows from GCD(ϕ(1), m) = 1 that ϕ G (1) = |G : N|ϕ(1) = mϕ(1) | χ(1) 󳨐⇒ χ(1) = ϕ G (1). Thus, ϕ G = χ ∈ Irr(G) for any ϕ ∈ Irr# (N). Suppose that CG (x) ≰ N for some x ∈ N # . Then x y = x for some y ∈ CG (x) − N. For −1 −1 z ∈ N, we have (x z )y = (x y )y zy = x y zy and, since y−1 zy ∈ N, the elements x and x zy are conjugate in N. Write K = {x u | u ∈ N}. Then, by the foregoing, K y = K. Therefore, by the Brauer Permutation Lemma 3.1 (c), there exists a character ϕ ∈ Irr# (N) such that ϕ y = ϕ, i.e., IG (ϕ) > N, where IG (ϕ) is the inertia group of ϕ in G. Then ϕ G ∈ ̸ Irr(G) (see Exercise VII.2.2), contrary to what has been proved above. Thus, CG (x) ≤ N for all x ∈ N # . It follows from the Supplement to Frobenius’ theorem (see §X.2) that G is a Frobenius group with kernel N. Corollary 4.2. Let {1} < N < G, |G : N| = m. Then G is a Frobenius group with kernel N if and only if m | χ(1) for all χ ∈ {Irr(G | N). Proof. The implication “⇒” follows from Theorem 3.2. Let us prove the converse implication. By Theorem 4.1, it suffices to prove that N = N [m] . But our N is contained in the kernel of any irreducible character of G whose degree is not divisible by m, so that N ≤ N [m] , and we conclude that |G : N| ≥ |G : N [m] |. Since |G : N| = m | |G : N [m] | (Theorem 4.1), we obtain N = N [m] . Exercise 4.1. Let {1} < N < G. Then G is a Frobenius group with kernel N if and only if ϕ G ∈ Irr(G) for all ϕ ∈ Irr# (N) = Irr(N) − {1N }. Solution. The implication “⇒” follows from Theorem 3.2. Let us prove the converse implication. Let χ ∈ {Irr(G | N) and ψ ∈ Irr# (χ N ). Then ψ G = χ 󳨐⇒ |G : N| | |G : N|ψ(1) = χ(1). Now the result follows from Corollary 4.2 with m = |G : N|.

476 | Characters of Finite Groups 1 Theorem 4.3 ([Ber12]). Let m > 1 be a proper divisor of |G| and set σ m (G) = ∑ χ(1)2 , χ∈B

where B = {χ ∈ Irr(G) | m ∤ χ(1)}

(in the proof of Theorem 4.1 that number was denoted by b). Then σ m (G) ≥ m with equality if and only if G is a Frobenius group with kernel of index m. Proof. Theorem 4.1 implies the first assertion since m | GCD(m2 , |G|) = m. If G is a Frobenius group with kernel of index m, then σ m (G) = m (Theorem 3.2). Now suppose that σ m (G) = m. Then, in the notation of the proof of Theorem 4.1, one has a + b = a + σ m (G) = |G| 󳨐⇒ GCD(m2 , |G|) = m 󳨐⇒ GCD(m,

|G| ) = 1, m

i.e., m is a Hall divisor of |G|. Let π = π( |G| m ) and take P ∈ Sylp (G) for p ∈ π; then m | |G : P|. In that case (in the notation of Theorem 4.1) (1P )G = ∑ a χ χ + ∑ b χ χ, χ∈A

(1)

χ∈B

where a χ , b χ are nonnegative integers. Set u = ∑ a χ χ(1), χ∈A

v = ∑ b χ χ(1). χ∈B

Since u ≡ 0 (mod m) and |G : P| = (1P )G (1) = u + v ≡ 0 (mod m), we conclude that v ≡ 0 (mod m). Since 1G ∈ B, we have v ≥ m. By reciprocity, b χ ≤ χ(1) whence m ≤ v ≤ ∑ χ(1)2 = σ m (G) = m, χ∈B

and therefore for all χ ∈ Irr(G) such that m ∤ χ(1), one has v = m, b χ = χ(1), χ P = χ(1)1P 󳨐⇒ P ≤ ⋂ ker(χ) = D ⊲ G. χ∈B

Since this is true for all p ∈ π, it follows that |G : D| | m. Next, m = ∑ χ(1)2 ≤ |G : D| ≤ m 󳨐⇒ |G : D| = m. χ∈B

Now, by definition, D = N [m] so that |G : N [m] | = m. But then, by Theorem 4.1 (b), G is a Frobenius group with kernel N [m] = D of index m. Exercise 4.2 (see Theorem 9.7). Let |G| ≤ χ(1)2 + 2 ⋅ χ(1) for some χ ∈ Irr1 (G). Then either G is a 2-transitive Frobenius group or |G| = 8. Exercise 4.3. If, in the notation of Theorem 4.3, σ m (G) = 2m, where m > 1 is a divisor of the degree of some nonlinear irreducible character of G, then every minimal normal subgroup of G is abelian and |G : S(G)| ≤ 2m, where S(G) is the solvable radical of G.

X Existence of normal subgroups | 477

Recall that X1 (G) is the first column of the character table of G containing the degrees of all irreducible characters of G (multiplicities are counted). We have |G| = X1 (G) ⋅ X1 (G) (dot product). Corollary 4.4. Let G = (H, N) be a Frobenius group and G1 a group. If X1 (G1 ) = X1 (G), then G1 = (H1 , N1 ), and moreover, X1 (H1 ) = X1 (H), X1 (N1 ) = X1 (N). Proof. Put |H| = m; then σ m (G) = m (Theorem 3.2 or Theorem 4.3). Now, X1 (G1 ) = X1 (G) 󳨐⇒ σ m (G1 ) = m. Then, by Theorem 4.3, G1 = (H1 , N1 ) is a Frobenius group, where |H1 | = m = |H|. Next, X1 (G1 ) = X1 (G) 󳨐⇒ |G1 | = |G| 󳨐⇒ |N1 | = |N|, N [m] (G) = N, N [m] (G1 ) = N. Since X1 (G/N [m] (G)) = X1 (G1 /N [m] (G1 )),

G/N [m] (G) ≅ H,

G1 /N [m] (G1 ) ≅ H1 ,

it follows that X1 (H) = X1 (H1 ). Since X1 (G | N) = X1 (G1 | N1 ) consists of the numbers |H|ϕ(1) (ϕ ∈ Irr# (N)) and |H1 |ϕ1 (1) (ϕ1 (1) ∈ Irr# (N1 )), with appropriate multiplicities, we get X1 (N) = X1 (N1 ). Exercise 4.4 (= Theorem 3.6). If G = (H, N) = (H1 , N1 ), then N = N1 . Give a solution independent of Theorem 3.6.⁵ Theorem 4.5. Let H < G and let m > 1 be a proper divisor of |G : H|. Let χ1 , . . . , χ k be all the irreducible constituents of the induced character (1H )G whose degrees are not divisible by m. Set d = ∑ki=1 χ i (1)2 . Then d ≥ m. If d = m, then D = ⋂ki=1 ker(χ i ) has index m in G. Proof. If H = {1}, Theorem 4.3 implies the result as Irr((1H )G ) = Irr(G). Next, a i = ⟨(1H )G , χ i ⟩, i = 1, . . . , k 󳨐⇒ a = a1 χ1 (1) + ⋅ ⋅ ⋅ + a k χ k (1) ≡ 0 (mod m),

(2)

since, by hypothesis, |G : H| ≡ 0 (mod m),

|G : H| − a = (1H )G (1) − a ≡ 0 (mod m).

Now, 1G ∈ {χ1 , . . . , χ k } implies a ≥ m. By reciprocity, a i ≤ χ i (1), and so d ≥ a ≥ m. Next we study our situation under additional assumption d = χ1 (1)2 + ⋅ ⋅ ⋅ + χ k (1)2 = m.

(3)

5 Of course, this follows immediately from Thompson’s theorem on nilpotency of a group admitting a fixed-point-free automorphism.

478 | Characters of Finite Groups 1 It follows from (3) that a i = χ i (1), i ∈ {1, . . . , k}. Then, by reciprocity, χ iH = χ i (1) ⋅ 1H for all i. This means that H ≤ ⋂ki=1 ker(χ i ) = D ⊲ G. Since we may consider χ1 , . . . , χ k as pairwise different characters of G/D, it follows that |G/D| ≥ ∑ki=1 χ i (1)2 = m. If χ ∈ Irr(G/D), then 1D ∈ Irr(χ D ), so that 1H ∈ Irr(χ H ) since H ≤ D, and therefore χ ∈ (1H )G , by reciprocity. In the notation of Theorem 4.3, σ m (G/D) = m. Supposing that |G/D| > m, we obtain by Theorem 4.3 that G/D = (L/D, F/D), where |G : F| = m. Then F = ⋂ki=1 ker(χ i ), contradicting the definition of D. Thus, |G : D| = m. Theorem 4.6. Assume that {1} < N ⊴ H < G, and suppose that the degrees of all the characters in the set Irr(H/N) are less than the degrees of the characters in the set X = Irr(H | N) = Irr(H) − Irr(H/N). If ϕ G ∈ Irr(G) for all ϕ ∈ X, then N ⊲ G. Proof. By assumption, H is nonabelian since X ≠ 0. Take ϕ1 ∈ X; then ϕ1G = χ ∈ Irr(G), by hypothesis, and χ(1) = |G : H|ϕ1 (1). Let χ H = a1 ϕ1 + ⋅ ⋅ ⋅ + a k ϕ k ,

(4)

where Irr(χ H ) = {ϕ1 , . . . , ϕ k }; then a i > 0 for all i. By reciprocity, a1 = 1. Let i > 1. Then |G : H|ϕ i (1) = ϕ Gi (1) ≥ a i χ(1) = a i |G : H|ϕ1 (1) 󳨐⇒ ϕ i (1) ≥ a i ϕ1 (1) ≥ ϕ1 (1); then ϕ i ∈ X, i = 1, . . . , k. Therefore, ϕ Gi = χ ∈ Irr(G)

(i ∈ {1, . . . , k}),

ϕ1 (1) = ⋅ ⋅ ⋅ = ϕ k (1).

Thus, |G : H|ϕ1 (1) = χ(1) = kϕ(1) 󳨐⇒ k = |G : H|, and we obtain χ H = ϕ1 + ⋅ ⋅ ⋅ + ϕ|G:H| .

(5)

If ψ ∈ X − {ϕ1 , . . . , ϕ|G:H| }, then we have ψ G ∈ Irr(G), and it follows from (5) that ≠ χ. Next, ⟨(ψ G )H , χ H ⟩ = 0. Therefore, X = X 1 ∪ ⋅ ⋅ ⋅ ∪ X s is a partition, where, for i = 1, . . . , s, α, β = 1, . . . , |G : H|, we have

ψG

X i = {ϕ iα | α = 1, . . . , |G : H|},

(ϕ iα )G = χ i ∈ Irr(G),

i = 1, . . . , s,

and the characters χ1 , . . . , χ s are pairwise distinct. We have s

|H| − |H : N| = ∑ ϕ iα (1)2 = |G : H| ∑ ϕ1 (1)2 , α,i

j

(6)

j=1

χ1 (1)2 + ⋅ ⋅ ⋅ + χ s (1)2 = |G : H|2 (ϕ11 (1)2 + ⋅ ⋅ ⋅ + ϕ1s (1)2 ) = |G : H|(|H| − |H : N|) = |G| − |G : N|. {χ1 ,

χ s } and τ

(7)

Let I = Irr(G) − ..., ∈ I. By reciprocity, by assumption, and by what we have proved already, the decomposition of τ H contains only characters in Irr(H/N), and therefore N ≤ ker(τ) for all τ ∈ I. Since it follows from (7) that ∑τ∈I τ(1)2 = |G : N|, we see that N = ⋂τ∈I ker(τ) ⊲ G.

X Existence of normal subgroups | 479

Exercise 4.5. One has N < H in Theorem 4.6. Hint. Assume that N = H. Then, by Corollary 4.2, G = (F, H). It follows from Theorem 4.6 that any nonprincipal character of H is nonlinear. But this is not true since, by Thompson’s theorem, H is solvable, and so H 󸀠 < H. Let CL(G) = {K1 , . . . , K r } be the set of all G-classes, r = k(G). If x i ∈ K i for all i and H = ⟨x1 , . . . , x r ⟩, then ⋃g∈G H x ⊇ ⋃ri=1 K i = G so that H = G since G-conjugates of a proper subgroup do not cover G. We use this fact in the proof of the following theorem. Theorem 4.7 (Saksonov [Sak3]). Let G be a p-solvable group, P ∈ Sylp (G). The degrees of all the characters in Irr((1P )G ) are not divisible by p if and only if P ⊲ G. Proof. First assume that G has a normal p-complement, say H; then G = P ⋅ H. Since (1P )G (of degree |H|) vanishes on H # , it follows that ((1P )G )H = ρ H , where ρ H is the regular character of H. Therefore, for every θ ∈ Irr(H), there exists χ ∈ Irr((1P )G ) such that ⟨χ, θ G ⟩ = ⟨χ H , θ⟩ > 0. Since |G : IG (θ)| | χ(1) (Clifford’s theorem), the number |G : IG (θ)| which divides |G : H| = |P| is thus a power of p, and p ∤ χ(1), by hypothesis, it follows that IG (θ) = G for all θ ∈ Irr(H). Thus, P acts trivially on Irr(H) and, hence, on the set of H-classes (Brauer’s permutation Lemma 3.1 (c)). If K is an H-class, the fact that K x = K for all x ∈ P and GCD(|K|, p) = 1 implies that P centralizes some element y ∈ K. This is true for all H-classes K. But any set of representatives of H-classes generates H (see the paragraph preceding the theorem). Therefore P centralizes H, G = P × H and so P ⊲ G. Turning now to the general case, we prove the theorem by induction on |G|. If Op (G) > {1}, the theorem is true for G, since it is true for G = G/Op (G), which has the faithful character (1P )G . Now let Op (G) = {1}. Then Op󸀠 (G) = T > {1}. Since 1TP ∈ Irr((1P )TP ), it follows that Irr((1TP )G ) ⊆ Irr(((1P )TP )G ) = Irr((1P )G ), and the theorem is true for G/T, so that PT/T ⊴ G/T, by induction. As TP ⊴ G, it follows by reciprocity and Clifford’s theorem that p does not divide the degrees of all the characters in Irr((1P )TP ). Since we have assumed that G has no normal p-complement, we see that TP < G. But the theorem is true for TP. Therefore P is normal so characteristic in TP, and hence P ⊲ G. Now, supposing that P ⊲ G, we prove that any irreducible constituent of (1P )G has p󸀠 -degree. To this end, let χ ∈ Irr((1P )G ). Then, by Clifford’s theorem, χ P = χ(1)1P . It follows that P ≤ ker(χ) so that χ(1), being a divisor of |G/P| (Frobenius–Molien), is a p󸀠 -number. Exercise 4.6. If H < G, then (1H )G = ∑χ∈Irr((1H )G ) χ(1)χ if and only if H ⊲ G. Hint. If (1H )G = ∑χ∈Irr((1H )G ) χ(1)χ and χ ∈ Irr((1H )G ), then χ H = χ(1)1H , by reciprocity, whence H ≤ ker(χ).

480 | Characters of Finite Groups 1 Exercise 4.7. Let m > 1 be a proper divisor of |G|, and let the degrees of all nonlinear irreducible characters of a group G be multiples of m. Then |G : G󸀠 | ≥ GCD(m2 , |G|). If, moreover, |G : G󸀠 | = m, then G = (C, G󸀠 ), a Frobenius group with kernel G󸀠 and cyclic complement C. Hint. Use Theorem 4.1. Exercise 4.8. Let m > 1 be a proper divisor of |G|. Then G = (C, G󸀠 ) with abelian G󸀠 of index m if and only if |G : G󸀠 | = m and cd(G) = {1, m}. Hint. Use Corollary 4.2. Exercise 4.9. Assume that H < G and ⟨χ H , χ H ⟩ = |G : H| for all χ ∈ Irr(G) such that H ≰ ker(χ). Then H ⊲ G. Investigate the situation in greater detail. Hint. Put X = {χ ∈ Irr(G) | H ≰ ker(χ)}. Then χ ∈ X vanishes on G − H. If τ ∈ Irr(G) − X, then H ≤ ker(τ). To prove Theorem 4.8 below, we shall use the properties of S(p a , q b , q c )-groups (see [BZ3, Lemma 11.2.1]). Theorem 4.8. Let G1 be an S(p a , q b , q c )-group, G a group and X1 (G) = X1 (G1 ). Then G = HZ(G), where H is an S(p d , q b , q c )-group for d ≤ a. Proof. Let G1 = P1 ⋅ Q1 ,

P1 ∈ Sylp (G1 ), Q1 = G󸀠1 ∈ Sylq (G1 ).

We have p a q b+c = |G1 | =

∑

χ(1)2 =

χ∈Irr(G1 )

∑

χ(1)2 = |G|,

χ∈Irr(G)

and |P1 | = p a = |G1 : G󸀠1 | = |Lin(G1 )| = |Lin(G)| = |G : G󸀠 |. It follows that G has no normal subgroup of index q. Let G = PQ, where P ∈ Sylp (G), Q ∈ Sylq (G). It follows that Q = G󸀠 . Let Q0 < Q be G-invariant of maximal order. Since c ≤ 2b , where b is the order of q (mod p), it follows that |Q/Q0 | ≥ q b so |Q0 | ≤ q c since |G󸀠 | = |Q| = |G󸀠1 |. Since, in view |Q0 | < q b , the group PQ0 has no minimal nonnilpotent subgroups (Lemma XI.2.1), it follows that PQ0 is nilpotent. In particular, Q0 centralizes all Sylow p-subgroups of G. Since G has no normal subgroup of index q, the Sylow p-subgroups generate G, and we conclude that Q0 ≤ Z(G). Since Q/Q0 is (elementary) abelian, we get cd(G/Q0 ) = {1, p}, by Ito’s Theorem VII.2.3 on degrees, since cd(G/Q0 ) ⊆ cd(G) = cd(G1 ) ⊆ {1, p, q b/2 }, by Lemma XI.2.2 (a)). Let n ∈ cd(G) and let MG (n) denote the set of all characters of degree n in Irr(G). The relations |MG (p)| = |MG1 (p)| ≥ |MG/Q0 (p)|,

cd(G/Q0 ) = {1, p}

X Existence of normal subgroups | 481

imply that |G/Q0 | = |G1 /(Q1 ∩ Z(G1 ))| = p a q b

(Q1 ∈ Sylq (G1 )).

qc .

Therefore, one has |Q0 | = Next, G/Q0 has an abelian subgroup A/Q0 of index p (since cd(G/Q0 ) = {1, p}; see the results of Isaacs–Passman in Chapter XIV on groups X with cd(X) = {1, p}). Since Q0 ≤ Z(G), it follows that A is nilpotent. If A0 = P ∩ A, then CG (A0 ) ≥ ⟨A, P⟩ = G (in fact, P ≅ G/Q = G1 /G󸀠1 ≅ Lin(G1 ) is abelian). Therefore, |G/Z(G)| = pq b and all proper subgroups of G/Z(G) have prime power orders. Let H be a minimal nonnilpotent subgroup of G. Set F = HZ(G). By the properties of minimal nonnilpotent groups, we have |F/Z(G)| = pq b = |G/Z(G)| so that HZ(G) = F = G. We conclude that |G : H| is a power of p. It follows that Q < H. Thus, H is an S(p d , q b , q c )-group for d ≤ a. Exercise 4.10. Let G, G1 be groups such that X1 (G) = X1 (G1 ), let p ∈ π(G), P ∈ Sylp (G) and P1 ∈ Sylp (G1 ). If G1 is p-nilpotent and |P󸀠 | = |P󸀠1 |, then G is p-nilpotent. Hint. Let N/G󸀠 and N1 /G󸀠1 be p󸀠 -Hall subgroups of G/G󸀠 and G1 /G󸀠1 , respectively. Then P ∩ N = P󸀠 . Use Tate’s Theorem VII.10.1. Exercise 4.11 (Kazarin). Let S be a nonabelian simple group of Lie type of order n with the set of character degrees D = cd(S) = {d1 = 1, d2 , . . . , d k }. Prove that if G is the group whose set of character degrees satisfies D󸀠 = (D − {d i }) ∪ {x} for some i ≤ k (then |D󸀠 | = |D|), then |G| ≤ n2 − n and if |S| ∤ d i (d2i − 1), then G󸀠 = G. Solution. One has x | |G| = |S| − d2i + x2 = (n − d2i ) + x2 󳨐⇒ x | n − d2i . Hence

|G| ≤ n − d2i + (n − d2i )2 = (n − d2i )(n − d2i + 1) < (n − 1)n.

Assume that x > 1; then G = G󸀠 because there exists only one character in Irr(G) of degree 1. Suppose that x = 1 and d j ≠ d = d i . Then |G| = |S| − d2 + 1 and d j | |G| − |S|. Hence d j | d2 − 1 for every d j ≠ d. If a prime p divides GCD(d, d j ), then |G| is coprime with p, a contradiction. Hence GCD(d, d j ) = 1 for every d j ≠ d. This implies that |S| = d0 m with d | d0 and d j | m for every d j different from d. Suppose that p is a prime dividing d0 and not dividing d. By the Michler and Willems result, S has an r-block of zero defect for every r dividing its order. This means that the degree of some irreducible character of S is divisible by the highest degree of r dividing |S|. Thus S cannot be a simple group if p ∤ d. Hence, π(d) = π(d0 ) and, moreover, d0 = d. Since LCM{d j | d j ≠ d} = m, in view of the Michler and Willems result cited above, it follows that m divides d2 − 1. Therefore |G| | d(d2 − 1), as asserted. Problem 4.1 (Kazarin). Let S be a simple group and the degrees of irreducible characters of S are 1 < d2 < ⋅ ⋅ ⋅ < d n . Suppose that the degrees of irreducible characters of a group G are 1, d2 , . . . , d n−1 , k. Is it true that k = d n ? (If the answer is “yes”, it is possible to prove, using CFSG, that G ≅ S.)

482 | Characters of Finite Groups 1 It is unknown whether a knowledge of X1 (G) enables one to determine whether G is solvable, p-solvable or simple. It is known (T. Hawkes) that a knowledge of X1 (G) is not sufficient to say whether G is supersolvable. In many cases X1 (G) = X1 (G1 ) implies G ≅ G1 ; this is the case, for example, if G ∈ {Sm , An }, m ≤ 6, n ≤ 6 (probably this is true for all m, n).⁶ Exercise 4.12 (G. Higman). Let G be a nonprimary solvable group all of whose elements have prime power orders. Then |G| = p a q b and one of the following holds: (a) G = (P, Q) is a Frobenius group, P ∈ Sylp (G), Q ∈ Sylq (G). (b) G = (Q1 , P)⋅Q2 = Q1 ⋅(P, Q2 ), P ∈ Sylp (G), Q1 Q2 ∈ Sylq (G), (P, Q2 ) ⊲ G, P is cyclic, p > 2. Solution. Set Q2 = F(G). Then Q2 is a q-subgroup for some q ∈ π(G). Let H be a q󸀠 -Hall subgroup of G. Then HQ2 = (H, Q2 ) is a Frobenius group. If p < r are in π(H), then H contains an element of order pr (see Exercise 3.9), a contradiction. Thus, we obtain H = P ∈ Sylp (G), |G| = p a q b . If Q2 = Q ∈ Sylq (G), then G = (P, Q) satisfies the hypothesis. Let Q2 ∈ ̸ Sylq (G), P1 /Q2 = F(G/Q2 ). Then P1 = (P, Q2 ),

G = NG (P) ⋅ Q2 ,

NG (P) ∩ Q2 = {1},

NPQ2 (P) = P since P is a Frobenius complement of (P, Q2 ). Then, if Q1 ∈ Sylq (NG (P)), then NG (P) = (Q1 , P) and P, Q1 are cyclic. Indeed, if P is not cyclic, it is a generalized quaternion group and then NG (P) has a cyclic subgroup of order 2q, which is a contradiction. Then Q1 , being a subgroup of the abelian group Aut(P), is cyclic. Thus, G = (Q1 , P) ⋅ Q2 = Q1 ⋅ (P, Q2 ), where P ∈ Sylp (G) and Q1 are cyclic, Q2 = Oq (G). (Such a G is said to be a double Frobenius group.) Problem 4.2. A subgroup H < G is said to be principal if Irr((1H )G ) = Irr(G). Since Irr(ρ G ) = Irr(G) (ρ G = (1G )G is the regular character of G), {1} is principal. Is it true that maximal principal subgroups of the group PSL(2, 7) are not conjugate? Find the maximal principal subgroups of known series of simple groups. Is it true that maximal principal subgroups of a solvable group are conjugate? Exercise 4.13. One has H ⊴ G if ⟨(1H )G , (1H )G ⟩ = |G : H|. What can be said about the embedding of H into G if ⟨λ G , λ G ⟩ = |G : H| for some λ ∈ Lin(H)? Exercise 4.14. Let G = HF, H∩F = {1}. Is it true that H ⊴ G if and only if (χ ∈ Irr((1H )G ) implies χ F ∈ Irr(F))? Exercise 4.15. Let H be a π-Hall subgroup of G = H ⋅ F, where H ∩ F = {1}, F ⊲ G, and let ψ ∈ Lin(H). Suppose that χ(1) is a π󸀠 -number for any χ ∈ Irr(ψ G ). What can be said about the embedding of F and H into G?

6 This is the case, see [T-V3].

X Existence of normal subgroups | 483

Let CL(G) = {K1 , . . . , K r } be the set of G-classes of a group G. The notation a

a

h(G) = {h11 , . . . , h s s } means that {|K1 |, . . . , |K r |} = {h1 , . . . , h s } with pairwise different h j , and the set CL(G) contains exactly a i classes of length h i , i = 1, . . . , s. Theorem 4.9. Let G1 = (H1 , F1 ) be a Frobenius group and let G be a group such that h(G) = h(G1 ). Then G = (H, F), h(H) = h(H1 ), h(F) = h(F1 ). Proof. We have |G| = ∑K∈CL(G) |K| = ∑K1 ∈CL(G1 ) |K1 | = |G1 |. By Thompson’s theorem, F1 is nilpotent, so that Z(F1 ) > {1}. Take x1 ∈ Z(F1 )# . Then CG1 (x1 ) = F1 and x1 ∈ K1 , where K1 is a G1 -class of length |G1 : F1 | = |H1 |. Furthermore, Z(H1 ) > {1}, by Theorem 3.3. Take y1 ∈ Z(H1 )# . Then CG1 (y1 ) = H1 , and y1 belongs to a G1 -class L1 of size |G1 : CG1 (y1 )| = |G1 : H1 | = |F1 |. By assumption, there exist G-classes K, L such that |K| = |K1 | = |H1 |, |L| = |L1 | = |F1 |. Let x ∈ K, y ∈ L. Put F = CG (x), H = CG (y). By assumption, |F| = |F1 |, |H| = |H1 |. Since GCD(|F|, |H|) = GCD(|F1 |, |H1 ) = 1,

(8)

we have G = HF = FH. By assumption, the length of any non-one-element G-class is a multiple of |H| or |F|. Hence the elements of F # belong to G-classes whose lengths are multiples of |H| = |H1 | in view of (8). Put F0 = {x ∈ G | x = 1 or |H| divides |G : CG (x)|}. The equality h(G) = h(G1 ) implies that |F0 | = |F1 | = |F|. Therefore, F0 = F and F ⊲ G. But then the inclusion x ∈ F # implies CG (x) ≤ F, and therefore G = (H, F) is a Frobenius group with kernel F. The statements about h(H) and h(F) are now obvious. Exercise 4.16. A group G is called π-separable if the indices of its composition series are π- or π󸀠 -numbers (where π is a set of primes; see [BZ3, Chapter 14, Appendix C]). Using the Schur–Zassenhaus theorem and the theorem on solvability of groups of odd order [FeiT3], one can easily show that G contains a π-Hall subgroup, and, moreover, all maximal π-subgroups of G are conjugate (see [BZ3, Chapter 14, Appendix B, Lemma B.1]). Let now H be a π-Hall subgroup of G. Is it true that, whenever χ(1) is a π 󸀠 -number for any χ ∈ Irr((1H )G ), then H ⊲ G? (See Theorem 4.7.)

5 Exercises In this section we present a number of exercises. Exercise 5.1. A group G is nilpotent if and only if for every p ∈ π(G), P ∈ Sylp (G) and χ ∈ Irr((1P )G ), we have p ∤ χ(1). Solution. The implication “⇒” is obvious (see the last paragraph in the proof of Theorem 4.7). We shall prove the converse implication. If G is solvable, the result follows from Theorem 4.7. Now let G be nonsolvable. Thompson’s theorem [Tho2], which describes the minimal simple groups, implies that Theorem V.8.1 character-

484 | Characters of Finite Groups 1 izes solvable groups. Let P1 , . . . , P t be a complete system of representatives of classes of the Sylow subgroups of G. Then, by Theorem V.8.1, there exists χ ∈ Irr# (G) such that 1P i ∈ Irr(χ P i ) for all i ∈ {1, . . . , t}. Suppose that a prime p | χ(1) and P j ∈ Sylp (G). Then χ ∈ Irr((1P j )G ) and p | χ(1), contrary to assumption. Therefore χ ∈ Lin(G). Then χ P i = 1P i for i = 1, . . . , t. This implies χ = 1G , contradicting the choice of χ. Exercise 5.2. Let G1 be an S(p a , q b , q c )-group and G a group such that h(G) = h(G1 ). Study the structure of G1 . Exercise 5.3. Study the groups G containing a nontrivial subgroup N such that, whenever χ ∈ Irr(G | N), then χ vanishes on G − N. Exercise 5.4. Suppose that the degrees of irreducible characters of G are 1, 1, 2, 3, 3. Prove that G ≅ S4 . Exercise 5.5. Suppose that the degrees of irreducible characters of G are 1, 3, 3, 4, k. Prove that k = 5 and G ≅ A5 . Solution. We have |G| = 35 + k2 so that k | 35 hence k ∈ {1, 5, 7, 35}. If k = 1, then |G| = 36, |G/G󸀠 | = 2 and the abelian 3-Sylow subgroup is normal in G. In that case, 3 ∈ ̸ cd(G) (Theorem VII.2.3), a contradiction. Thus, k > 1; then G = G󸀠 is nonsolvable. We have 12 | |G| (Frobenius–Molien) so that k ∈ ̸ {2, 3, 4}, GCD(6, k) = 1. Therefore, we get 12k | |G| = 35 + k2 . If k = 5, then |G| = 60 and G ≅ A5 . Let k = 7; then |G| = 84. By Sylow’s theorem, a subgroup of order 7 is normal in G and G is solvable, a contradiction. Let k = 35; then |G| = 35 ⋅ 36 = 22 ⋅ 32 ⋅ 5 ⋅ 7. Let m = 35; then m | |G|. In the notation of Theorem 4.3, σ m (G) = σ35 (G) = 35 = m. Then, by Theorem 4.3, G is a Frobenius group with kernel of index 35 and order 36, which is not the case: the kernel is abelian (by Theorem 3.4) so we get 3, 4 ∈ ̸ cd(G) (by Theorem VII.2.3). Exercise 5.6. Suppose that the degrees of irreducible characters of a group G are 1, 1, 4, 4, 5, 6, k. Is it true that k = 5? Solution. We have |G| = 95 + k2 . Then we obtain k | 95 so that k ∈ {1, 5, 19, 95}. By Frobenius–Molien, k ≠ 1 so that |G : G󸀠 | = 2. If k = 19, then |G| = 95 + 192 = 24 ⋅ 19 coprime with 5, a contradiction. Let k = 5; then |G| = 120 and G has no normal subgroup of order 5. In that case, G ≅ S5 . Let k = 95; then σ95 (G) = 95 so that G = (A, B), where |B| = 96 and |A| = 95 (by Theorem 4.1). In that case, we have 6 ∈ ̸ cd(G) (by Theorem VII.2.3), a contradiction. Exercise 5.7. Suppose that the degrees of irreducible characters of a group G are 1, 3, 3, 6, 7, k. Prove that k = 8 and G ≅ PSL(2, 7). Solution. We have |G| = 104 + k2 , LCM(6, 7) = 42 | |G| and 2 | k (Frobenius–Molien) so that G = G󸀠 hence k ∈ {2, 4, 8, 26, 52, 104}. Since 42 | |G|, we get k ∈ ̸ {2, 4, 26, 52}.

X Existence of normal subgroups | 485

Assume that k = 104. Then |G| = 104 ⋅ 105. Taking m = 104, we get (in the notation of Theorem 4.3) that σ104 (G) = 104 so, by that theorem, G is a Frobenius group with kernel N of order 105 = 3 ⋅ 5 ⋅ 7 and complement of order 104. By Theorem VII.2.3, 7 ∈ ̸ cd(G), a contradiction. Thus, k = 8 so |G| = 104+82 = 168 and G ≅ PSL(2, 7). Exercise 5.8. Suppose that the degrees of irreducible characters of a group G are 1, 1, 3, 3, 7, 7, 7, k. Prove that k = 1. Solution. We have |G| = 167 + k2 and |G| ≡ 167 (mod k). Hence k | 167. But 167 is a prime number. Hence k ∈ {167, 1}. Set k = 167. Then σ167 (G) = 167. In that case, |G| = 167 ⋅ 168 and G has a normal Sylow 167-subgroup. Using Theorem VII.2.3, we obtain a contradiction. Hence k = 1. Exercise 5.9. Suppose that the degrees of irreducible characters of a group G are 1, 7, 7, 7, 7, 9, 9, 9, k. Prove that k = 8. (In that case, G ≅ SL(2, 8).) Solution. We have |G| = 440 + k2 ≡ 440 (mod k). Since k divides |G|, it follows that k | 440 = 23 ⋅ 5 ⋅ 11. So that k ∈ ̸ {7, 9}. On the other hand, 63 | |G|. If k = 1, then |G| = 441 is odd and |G : G󸀠 | = 2, a contradiction. Thus, k > 1 so that G󸀠 = G. It follows that |G| is even so is k.⁷ By Theorem 4.3, the case k = 440 is eliminated. It remains to consider the following cases: k ∈ {2, 4, 8, 10, 20, 40, 22, 44, 88, 110, 220}. Using now 63 | |G|, we obtain k = 8. In that case, |G| = 504. It is known that then we must have G ≅ PSL(2, 8). Exercise 5.10. Suppose that the degrees of irreducible characters of a group G are 1, 1, 1, 2, 2, 2, 3, k. Prove that k = 24 and G = (SL(2, 3), E25 ). Solution. We have |G| = 24 + k2 . Using Theorem VII.2.3, we get k ∈ ̸ {1, 2, 3}. Next, 6 | k. It is easy to check that k > 12. Let k = 24. Then σ24 (G) = 24, so that G = (H, F), where |H| = 24. By Burnside, F is abelian, and we conclude that F ≅ E25 . It is easily seen that H ≅ SL(2, 3). Exercise 5.11. Let the degrees of irreducible characters of a group G be 1, 10, 10, 10, 11, 16, 16, 44, 45, k. Prove that k = 55 (then |G| = 7920). Hint. In this case, |G| = 4895 + k2 and k divides 4895 = 5 ⋅ 11 ⋅ 89, 55 | k and k is odd. If k = 4895, then, due to Theorem 4.3, G is a Frobenius group with kernel of order 4896 = 25 ⋅ 32 ⋅ 17 and complement of order 4895 = 5 ⋅ 11 ⋅ 89. By Theorem VII.2.3, this is impossible. If k = 5 ⋅ 89 = 445, then 16 ∤ |G|, a contradiction. In all remaining cases, except k = 55, we obtain a contradiction. It remains that k = 55; then |G| = 7920. It is known that then G ≅ M11 . Exercise 5.12. Classify the groups whose irreducible characters have degrees 1, 1, 1, 2, 2, 2, k. 7 We suggest the reader to avoid reference on the Odd Order Theorem.

486 | Characters of Finite Groups 1 Solution. In this case, |G| = 15 + k2 so k | 15, hence k ∈ {1, 3, 5, 15}. If k = 1, then |G| = 16, |G : G󸀠 | = 4 so G is of maximal class; every such G satisfies the condition. Let k = 5. Then |G| = 40, |G : G󸀠 | = 3 ∤ |G|, a contradiction. Let k = 15. Then σ15 = 15 so G is a Frobenius group with kernel of order 16 (Theorem 4.3). In this case, 2 ∈ ̸ cd(G), a contradiction. It remains k = 3; then |G| = 24. In this case, |G : G󸀠 | = 3, G is nonnilpotent, and we conclude that G󸀠 ≅ Q8 . It is clear that then G ≅ SL(2, 3). Exercise 5.13. Let the character degrees of G be 1, 2, 2, 3, 3, 4, 4, 6, k. Find k.

6 Exceptional characters. CA-groups We recall the definition of a TI-subset, introduce exceptional characters and prove their simplest properties. Definition. A nonempty subset S of a group G is called a trivial intersection subset (or simply a TI-subset) if S ∩ S x ⊆ {1} for all x ∈ G − NG (S). It is obvious that if S is a TI-subset of G, so is S ∪ {1}. Moreover, if S is a TI-subset such that S ⊆ H < G, then S is a TI-subset in H. Any G-invariant subset of G we consider as a TI-subset. Lemma 6.1. Let S ⊆ G be a TI-subset and H = NG (S). Then the following statements hold: (a) If s, t ∈ S# are conjugate in G, they are conjugate in H. (b) If t ∈ S# , then CG (t) ≤ H. In particular, S ⊆ H. Proof. If s = t x (s, t ∈ S and x ∈ G), then s ∈ S ∩ S x 󳨐⇒ S x = S (s ≠ 1) 󳨐⇒ x ∈ NG (S) = H, proving (a). If t y = t ∈ S (t ∈ S# and y ∈ G), then t ∈ S ∩ S y and again y ∈ H, so that CG (t) ≤ H and S ⊆ H. In this chapter we have already encountered TI-subsets. A subgroup which is a TI-subset is called a TI-subgroup. Any subgroup of prime order is a TI-subgroup. Lemma 6.2. Let S ⊂ G be a TI-subset and H = NG (S). Suppose that θ ∈ Ch(H) vanishes on the set H − S. Then the following statements hold: (a) θ G (t) = θ(t) for all t ∈ S# and θ G (t) = 0 for all t ∈ G − S G , where S G = ⋃y∈G S y . If, in addition, θ(1) = 0, then (θ G )H = θ. (b) If, in addition, ϕ ∈ Ch(H) vanishes on the set H − S and θ(1) = 0, then ⟨θ G , ϕ G ⟩ = ⟨θ, ϕ⟩.

X Existence of normal subgroups | 487

Proof. (a) Let t ∈ G. Then θ G (t) = |H|−1 ∑ θ(t x ), x∈M t

where M t = {x ∈ G | t x ∈ S}.

If t ∈ S# , then M t = H, by Lemma 6.1 (a), and so, by hypothesis, θ G (t) = |H|−1 ∑ θ(t x ) = |H|−1 ∑ θ(t) = θ(t) x∈H

x∈H

S G , then M

G since θ ∈ CF(H). If t ∈ G − t = 0, by Lemma 6.1 (a) again, i.e., θ (t) = 0, and the last assertion follows. (b) By reciprocity, (a) and the hypothesis, we obtain

⟨θ G , ϕ G ⟩ = ⟨(θ G )H , ϕ⟩ = ⟨θ, ϕ⟩. Theorem 6.3. Let S ⊂ G be a TI-subset and H = NG (S). Assume that ϕ1 , . . . , ϕ n ∈ Irr(H) (n > 1) are pairwise distinct and the generalized character ϕ i − ϕ j vanishes on the set (H − S) ∪ {1} for all i, j. Then the following statements hold: (a) ϕ Gi = εχ i + ∆, where χ i ∈ Irr(G), i ∈ {1, . . . , n}, the coefficient ε ∈ {−1, 1} is independent of i, and ∆ is either the zero function on G or a character of G independent of i. All characters χ j have equal degree. (b) χ i − χ j vanishes on G − ⋃x∈G S x = G − S G for all i, j. Proof. Since 0 = (ϕ i − ϕ j )(1), we get ϕ1 (1) = ⋅ ⋅ ⋅ = ϕ n (1). (a) Let i ≠ j and θ = ϕ i − ϕ j so that θ(1) = 0. Then, by Lemma 6.2 (b), ⟨θ G , θ G ⟩ = ⟨θ, θ⟩ = ⟨ϕ i − ϕ j , ϕ i − ϕ j ⟩ = 2. Therefore, since θ G (1) = |G : H|θ(1) = 0, it follows that θ G is the difference of two distinct irreducible characters of G of equal degree. We shall prove that there exists a set {χ i }1n of irreducible characters of G such that (ϕ i − ϕ j )G = ε ⋅ (χ i − χ j ), i, j ∈ {1, . . . , n}, and ε ∈ {−1, 1} is independent of i, j. By the previous paragraph, the assertion holds for n = 2 with ε = 1. Therefore, one may assume that n > 2. By Lemma 6.2 (b) and reciprocity, ⟨(ϕ1 − ϕ2 )G , (ϕ1 − ϕ3 )G ⟩ = ⟨ϕ1 − ϕ2 , ϕ1 − ϕ3 ⟩ = 1 since ϕ1 , ϕ2 and ϕ3 are pairwise distinct. It follows that (ϕ1 − ϕ2 )G = ε ⋅ (χ1 − χ2 ),

(ϕ1 − ϕ3 )G = ε ⋅ (χ1 − χ3 ),

where χ1 , χ2 , χ3 ∈ Irr(G) are pairwise distinct. Since (ϕ2 − ϕ3 )G = (ϕ1 − ϕ3 )G − (ϕ1 − ϕ2 )G = ε ⋅ (χ2 − χ3 ), this proves the assertion for n = 3.

488 | Characters of Finite Groups 1 Let n ≥ 4. Consider α i = (ϕ1 − ϕ i )G for i ≥ 2. Then, by the foregoing, each α i has exactly one common irreducible constituent with α2 = (ϕ1 −ϕ2 )G and α3 = (ϕ1 −ϕ3 )G . Assume that χ2 is a constituent of α i . Then χ3 is a constituent of α i , i ≥ 4 (indeed, χ1 cannot be constituent of α i : then α1 and α i have the same constituents so ⟨α2 , α i ⟩ = 2, contrary to the above obtained relation) and it is clear that for this index i we have α i = −ε ⋅ χ2 − ε ⋅ χ3 . But 0 = α i (1) = −ε ⋅ (χ2 (1) + χ3 (1)) ≠ 0, which is a contradiction. Thus, there exist characters χ i ∈ Irr(G) such that (ϕ1 − ϕ i )G = ε(χ1 − χ i )

(i = 2, . . . , n).

As before, we have (ϕ i − ϕ j = − for all i, j. But then ϕ Gi − εχ i = ϕ Gj − εχ j so G i that ϕ i − εχ = ∆ is independent of i ∈ {1, . . . , n}. Since ϕ1 (1) = ⋅ ⋅ ⋅ = ϕ n (1), it follows that χ1 (1) = ⋅ ⋅ ⋅ = χ n (1). Assume that ∆ is a nonzero function. If ε = −1, then ∆ is a character. If ε = 1, suppose that ∆ is not a character. In that case, we have χ i ∈ ̸ Irr(ϕ Gi ) for all i, and therefore ∆ = −(χ1 + ⋅ ⋅ ⋅ + χ n ) + τ, where τ is either a character or the zero function and ⟨χ i , τ⟩ = 0 for all i. Then ϕ1G = χ1 − (χ1 + ⋅ ⋅ ⋅ + χ n ) + τ is not a character of G, since n > 1, which is a contradiction. Assertion (b) follows from Lemma 6.2 (a). )G

ε(χ i

χj )

Definition. The irreducible characters χ1 , . . . , χ n of Theorem 6.3 are called the exceptional characters of G, associated with ϕ1 , . . . , ϕ n ∈ Irr(H). Exercise 6.1. We retain the notation of Theorem 6.3. If χ ∈ Irr(G) − {χ1 , . . . , χ n }, then ⟨χ H , ϕ i ⟩ = ⟨χ, ∆⟩ is independent of i. Next, if i ≠ j, then ⟨χ iH , ϕ j ⟩ = ⟨χ i , ∆⟩ is independent of j, and ⟨χ iH , ϕ i ⟩ = ε + ⟨χ i , ∆⟩. Solution. If χ ∈ Irr(G) − {χ1 , . . . , χ n } and i ≠ j, then ⟨χ H , ϕ i ⟩ = ⟨χ, ϕ Gi ⟩ = ⟨χ, εχ i + ∆⟩ = ⟨χ, ∆⟩, ⟨χ iH , ϕ j ⟩ = ⟨χ i , ϕ Gj ⟩ = ⟨χ i , εχ j + ∆⟩ = ⟨χ i , ∆⟩, ⟨χ iH , ϕ i ⟩ = ⟨χ i , ϕ Gi ⟩ = ⟨χ i , εχ i + ∆⟩ = ε + ⟨χ i , ∆⟩. The following exercise is the origin of the notion of exceptional characters. Exercise 6.2. Let A be a nontrivial abelian subgroup of a group G such that CG (x) = A for all x ∈ A# (in that case, A is a Hall subgroup of G). Put a−1 H = NG (A), a = |A|, b = |H : A|, n = . b Then the following statements hold: (a) A is a TI-subset in G. (b) If H = A, then G is a Frobenius group with Frobenius complement A. (c) Let H > A. Then H is a Frobenius group with kernel A. Let ϕ1 , . . . , ϕ n be all the irreducible characters of H induced from characters in Lin# (A). If n ≥ 2, then by Theorem 6.3 there exist exactly n exceptional characters of G, associated with characters ϕ1 , . . . , ϕ n .

X Existence of normal subgroups | 489

Solution. If a ∈ A ∩ A x for x ∈ G − A, then CG (a) ≥ ⟨A, A x ⟩ > A, a contradiction. This proves (a). Now part (b) follows from (a) and Frobenius’ theorem. Part (c) follows easily from the properties of characters of the Frobenius group H (see Theorem 3.2) and from Theorem 6.3. In the following theorem we use the same notation as in Exercise 6.2. Theorem 6.4. Let n ≥ 2 and A < H = NG (A) an abelian subgroup, where A = CG (x) for all x ∈ A# . Then (1A )G − ϕ Gi vanishes on G − ⋃x∈G A x = G − A G and n

(1A )G − ϕ Gi = 1G − εχ i + d ∑ χ j + ∑ c χ χ, χ∈X

j=1

where ϵ ∈ {−1, 1}, c χ , d ∈ ℤ and X = Irr(G) − {1G , χ1 , . . . , χ n }. The coefficients d and c χ satisfy the following relation (below a = |A|): a−1 = b = d2 (n − 1) + (d − ε)2 + ∑ c2χ . n χ∈X Furthermore, given χ ∈ X, one has χ(u) = c χ

(u ∈ A# ),

χ(1) ≡ c χ (mod a),

|c χ | ≤

a−1 2

so that the number c χ is uniquely determined by the number χ(1). In addition, one has χ(1) = x + y(a − 1), c χ = x − y, where the numbers x = ⟨(1A )G , χ⟩ and y are nonnegative. Proof. By Exercise 6.2 (c), H = (B, A) is a Frobenius group. Let Irr1 (H) = {ϕ1 , . . . , ϕ n }. We have ϕ i = λ H for some λ ∈ Lin# (A) (Theorem 3.2). Denoting θ = (1A )H − ϕ i , we see that θ vanishes on H − A since (1A )H and ϕ i = λ H vanish on H − A, and θ(1) = (1A )H (1) − λ H (1) = |H : A| − |H : A| = 0. But then θ G vanishes on G − ⋃x∈G A x (Lemma 6.2 (a)). By Lemma 6.2 (b), ⟨θ G , 1G ⟩ = ⟨θ, 1H ⟩ = ⟨(1A )H − ϕ i , 1H ⟩ = ⟨1A , 1A ⟩ − ⟨λ, 1A ⟩ = 1 − 0 = 1, since λ ∈ Irr# (A). Next, if i ≠ j (for notation χ i , see Theorem 6.3), we obtain ⟨θ G , χ i − χ j ⟩ = ε⟨θ G , (ϕ i − ϕ j )G ⟩ = ε⟨(1A )H − ϕ i , ϕ i − ϕ j ⟩ = −ε since ϕ i , ϕ j ∈ ̸ Irr((1A )H ) and ϕ i ≠ ϕ j . Therefore, n

θ G = 1G − εχ i + d ∑ χ j + ∑ c χ χ, j=1

(1)

χ∈X

where X = Irr(G) − {1G , χ1 , . . . , χ n } and d, c χ ∈ ℤ. Furthermore, by Lemma 6.2 (b), ⟨θ G , θ G ⟩ = ⟨θ, θ⟩ = ⟨(1A )H − ϕ i , (1A )H − ϕ i ⟩ = ⟨ρ H/A , ρ H/A ⟩ + ⟨ϕ i , ϕ i ⟩ = b + 1,

(2)

490 | Characters of Finite Groups 1 where ρ H/A is the regular character of H/A ≅ B, b = |B|, H = (B, A) and ϕ i ∈ ̸ Irr((1A )H ) since A ≰ ker(ϕ i ). On the other hand, it follows from the First Orthogonality Relation and (1) that ⟨θ G , θ G ⟩ = 1 + (d − ε)2 + (n − 1)d2 + ∑ c2χ ,

(3)

χ∈X

and therefore, by (2), b = (d − ε)2 + (n − 1)d2 + ∑ c2χ .

(4)

χ∈X

Let χ ∈ X. Then ⟨ϕ Gi , χ⟩ = ⟨εχ i + ∆, χ⟩ = ⟨∆, χ⟩ = y is independent of i (see Exercise 1). Put x = ⟨(1A )G , χ⟩. Then c χ = ⟨θ G , χ⟩ = ⟨(1A )G − ϕ Gi , χ⟩ = x − y. Let λ ∈ Lin# (A). Then λ H = ϕ i for some i ∈ {1, . . . , n} (Theorem 3.2 (a)). Therefore, by reciprocity, ⟨λ G , χ⟩ = ⟨(λ H )G , χ⟩ = ⟨ϕ Gi , χ⟩ = y. Using reciprocity once again, we obtain ⟨χ A , 1A ⟩ = ⟨χ, (1A )G ⟩ = x, so that χ A = x ⋅ 1A + y

∑

λ i ∈Lin# (A)

λ i = (x − y) ⋅ 1A + y ⋅ ρ A ,

since 1A + ∑λ i ∈Lin# (A) λ i = ρ A is the regular character of A. Therefore, χ(1) = x + y(a − 1) ≡ x − y ≡ c χ (mod a), and for u ∈ A# we have χ(u) = [(x − y)1A + yρ A ](u) = x − y = c χ . By what we have proved, 1 a−1 a−1 ≤ 󳨐⇒ |c χ | ≤ (a − 1). n 2 2 Exercise 6.3. In the notation of Theorems 6.3 and 6.4, let ⟨χ i , (1A )G ⟩ = e, ⟨χ i , ∆⟩ = f . Then χ i (1) = e + (a − 1)f + εb. c2χ ≤ b =

Hint. We have ⟨χ iA , 1A ⟩ = e, by reciprocity. Since ∆ = ϕ Gj − εχ j ,

ϕj = λH js ,

s = 1, . . . , b, j = 1, . . . ,

a−1 = n, b

and, by reciprocity, {f i ⟨χ i , εχ j + ∆⟩ = ⟨χ i , ϕ Gj ⟩ = ⟨χ iH , ϕ j ⟩ = ⟨χ iH , λ H js ⟩ = ⟨χ A , λ js ⟩ = { f +ε { it follows that χ i (1) = e + f(n − 1)b + (f + ε)b = e + fbn + εb a−1 =e+f⋅ b + εb = e + f(a − 1) + εb. b

if j ≠ i, if j = i,

X Existence of normal subgroups | 491

Exercise 6.4. If x ∈ A# , then (in the same notation) χ i (x) = e − f + εϕ i (x). Hint. By Exercise 6.3, one has χ iA = e ⋅ 1A + f ⋅ (ρ A − 1A ) + ε ∑bs=1 λ is . Therefore, if x ∈ A# , then b

(ϕ i )A = ∑ λ is 󳨐⇒ χ i (x) = e − f + εϕ i (x). s=1

Exercise 6.5. In the same notation, ∑ |χ i (x)|2 = [(e − f)2 n − 2ε(e − f)] ⋅ b + b(a − b) ≥ b(a − b).

x∈A#

Hint. Not that ∑x∈A ϕ i (x) = 0 for i ∈ {1, . . . , n}; therefore, ∑ ϕ i (x) = −ϕ i (1) = −b.

x∈A#

Next, ϕ i vanishes on H − A; therefore, ab = |H| = ∑ |ϕ i (x)|2 = ∑ |ϕ i (x)|2 . x∈H

Hence

x∈A

∑ |ϕ i (x)|2 = ab − ϕ i (1)2 = ab − b2 .

x∈A#

But then (see Exercise 6.4), in view of n ≥ 2, we obtain since a − 1 = bn: ∑ |χ i (x)|2 = ∑ |e − f + εϕ i (x)|2 = (e − f)2 (a − 1) − 2(e − f)εb + (ab − b2 )

x∈A#

x∈A#

= [(e − f)2 n − 2ε ⋅ (e − f)]b + b(a − b) ≥ b(a − b). Exercise 6.6 ([Weis1]). If G is a nonabelian solvable group and CG (x) is abelian for all x ∈ G# , then G is a Frobenius group with abelian kernel and cyclic complement. Solution. By hypothesis, Z(G) = {1}. Let F(G) be the Fitting subgroup of G; then one has CG (F(G)) ≤ F(G). Since Z(F(G)) > {1}, it follows that F(G) = CG (Z(F(G))) is abelian, and, by properties of nilpotent groups, F(G) is a Hall subgroup of G. It is easy to show that every maximal abelian subgroup of G is a Hall subgroup of G and coincides with centralizer of all its nonidentity elements. We conclude that G is a Frobenius group with Frobenius kernel F(G), G = (H, F(G)) (see the paragraph following the solution of Exercise 2.3). Since Z(H) > {1} (Theorem 3.3), it follows that H is abelian so cyclic (Lemma 2.4). The following theorem, due to Suzuki [Suz1], was the first important step in the proof of solvability of groups of odd order. Theorem 6.5. If G is a nonabelian group of odd order such that CG (x) is abelian for any x ∈ G# , then G is a Frobenius group with abelian kernel and cyclic complement; in particular, G is solvable.

492 | Characters of Finite Groups 1 Proof. Let G be a counterexample of minimal order. In view of Exercise 6.6, G is not solvable. Let A1 , . . . , A s denote a complete system of representatives of the conjugacy classes of maximal abelian subgroups of G. Set H i = NG (A i ), i ∈ {1, . . . , s}. The following statements are obvious: Claim (i). All A i are TI-subsets. Claim (ii). One has H i > A i , H i = (B i , A i ) is a Frobenius group for all i. Claim (iii). For all i, A i is a Hall subgroup of G, and GCD(|A i |, |A j |) = 1 for i ≠ j. By (i), (ii) and (iii) it is easy to obtain, using induction, that G is nonabelian simple. Write ai − 1 , i = 1, . . . , s. a i = |A i |, b i = |H i : A i |, n i = bi Since |G| = ∏si=1 |A i | is nonabelian, we have s > 1. (Moreover, since G is nonsolvable, s > 2, by Ito’s theorem on solvability of product of two abelian groups.) It is also obvious that: Claim (iv). The union G# = ⋃si=1 (⋃x∈Ti (A#i )x ) is a partition (where Ti is a fixed transversal of H i in G). Using Lemma 6.1 (a), one obtains: Claim (v). By (iv), k(G) = 1 + ∑si=1 n i , where k(G) is the class number of G. are even, and so n i ≥ 2 for all i. For every i, Since |G| is odd, all the numbers n i = a bi −1 i let E i denote the set of exceptional characters associated with A i . Note that |E i | = n i (see Exercise 6.2). Claim (vi). The union {1G } ∪ E1 ∪ ⋅ ⋅ ⋅ ∪ E s = Irr(G) is a partition. Indeed, by (v) and equality |Irr(G)| = k(G), it suffices to prove that E i ∩ E j = 0 for i ≠ j. Let the characters χ, τ be exceptional for A i , χ ≠ τ. Then χ and τ coincide on A j (see Theorem 6.3 (b) and (iii)). Let λ ∈ Lin# (A j ) and λ H j = ϕ (ϕ is irreducible, by Theorem 3.2 (a)). Then, by reciprocity, ⟨χ H j , ϕ⟩ = ⟨χ A j , λ⟩ = ⟨τ A j , λ⟩ = ⟨τ H j , ϕ⟩. Therefore, χ H j = τ H j . Suppose that χ is exceptional for A j too. Then τ is also exceptional for A j (see Exercise 6.1). But then χ A j ≠ τ A j , contrary to what has just been proved. This proves (vi). From now on, we assume that a s ≤ a i for all i and let a = a s , b = b s , n = n s . Let j n E i = {χ1i , . . . , χ i i }; we write χ j instead of χ s . Let Γ mean the same for A s as θ G meant for A in Theorem 6.4. By Theorem 6.4 (see there (1); we use the same notation), we have n

s−1

nk

j=1

k=1

i=1

Γ = 1G − εχ1 + d ∑ χ j + ∑ c k ∑ χ ik .

X Existence of normal subgroups | 493

Claim (vii). If some a i (i < s) divides χ1 (1), then c i ≠ 0 in the expression for Γ (recall that χ1 = χ1s ). Indeed, by (vi) and Theorem 6.4, we have that χ1 (u) (u ∈ A#i ) is a rational integer, and χ1 (1) ≡ χ1 (u) (mod a i ), χ1 (u) ≤ 12 (a i − 1). Then, by assumption, χ1 (u) = 0 for u ∈ A#i (indeed, if a i divides χ1 (1) which is ≡ χ1 (u) (mod a i ) and χ1 (u) ≤ 12 (a i − 1), we must have χ1 (u) = 0). Similarly, χ j (u) = 0 for all j. But if r ∈ {1, . . . , s − 1} and r ≠ i, then χ αr ∈ E r takes the same value y r on A#i . Therefore, if c i = 0 in the expression for Γ, then 0 = Γ(u) = 1 +

cr yr nr .

∑ r∈{1,...,i−1,i+1,...,s−1}

As c r , y r , n r are integers and n r are even, this is impossible. This proves (vii). From now on, we assume that χ1 (1) is divisible by a1 , . . . , a t but not divisible by a t+1 , . . . , a s−1 . Then b ≥ 1 + ∑ c2α n α ≥ 1 + ∑ n i . α 1, contradicting the linearity of χ1 which we have just proved.

494 | Characters of Finite Groups 1 A group G is said to be a CA-group if CG (x) is abelian for all x ∈ G# . It follows from Theorem 6.5 that nonsolvable CA-groups have even orders. As Suzuki has proved, if G is a nonsolvable CA-group, then G ≅ PSL(2, 2n ). See also §V.6. The nonsolvable groups G such that CG (x) is nilpotent for all x ∈ G# have also been classified by Suzuki. Suzuki has proved several stronger results, but we shall not dwell on them.

7 Cossey–Hawkes–Mann’s and Isaacs’ theorems In this section we will prove the theorem of Cossey–Hawkes–Mann, which states that the order of the hypercenter H(G) of a group G (= the last member of the upper central series of G) is uniquely determined by the class sizes vector h(G). In particular, knowing h(G) alone, one can recognize whether G is nilpotent or not. 1°. Let the set ν p (G) be the union of the G-classes whose lengths are powers of p. It is convenient to write h(G) = (h1 ⋅ 1, h2 ⋅ 2, . . . ) (this means that G has h i classes of size i, i = 1, 2, . . .). Then |ν p (G)| = ∑ p α h p α . α≥0

Let m p be the p-part of m ∈ ℕ. Theorem 7.1 (Cossey–Hawkes–Mann [CHM]). One has |H(G)|p = |ν p (G)|p . Let P ∈ Sylp (H(G)) and a ∈ G. Then ⟨a, H(G)⟩ is nilpotent (consider the upper central series of this subgroup!) so [P, a] = {1} provided a is a p󸀠 -element. Since Op (G) is generated by all p󸀠 -elements of G, it follows that [P, Op (G)] = {1}. Lemma 7.2. Let x ∈ G be a p-element. Then x ∈ H(G) if and only if x ∈ CG (Op (G)). Proof. We know that Op (G) is the least normal subgroup K of G such that G/K is a p-group. Let R = CG (Op (G)); then R ⊴ G. If x ∈ H(G) is a p-element, then x centralizes Op (G), by the paragraph preceding the lemma so that a Sylow p-subgroup of H(G) is contained in R. The relations Op (G)R/Op (G) ≅ R/(R ∩ Op (G)),

R ∩ Op (G) = Z(Op (G))

and the fact that G/Op (G) is a p-group imply that R is nilpotent. Let P1 ∈ Sylp (R). Then P1 is G-invariant and so (the p-element) x ∈ P1 . If P1 ≤ P ∈ Sylp (G) and P0 is a subgroup of order p in P1 ∩ Z(P), then P0 ≤ Z(G) (indeed, CG (P0 ) ≥ POp (G) = G). Then, by induction, one has P1 /P0 ≤ H(G/P0 ), so that P1 ≤ H(G), and the result follows since x ∈ P1 . Proof of Theorem 7.1. Let N be a normal p-subgroup of G contained in H(G). Let x ∈ ν p (G) and let q be a prime, q ≠ p. Then Q ≤ CG (x) for some Q ∈ Sylq (G) (see the definition of ν p (G)). But then Q ≤ CG (⟨x, N⟩), by Lemma 7.2. Therefore, if y ∈ xN, then

X Existence of normal subgroups | 495

q ∤ |G : CG (y)|. Since q ∈ π(G) − {p} is arbitrary, we get y ∈ ν p (G), and we conclude that the coset xN ⊆ ν p (G). Hence ν p (G) is a union of cosets of N 󳨐⇒ |N| | |ν p (G)|.

(1)

In particular, |H(G)|p divides |ν p (G)| (take, in the previous argument, N ∈ Sylp (H(G))). We now claim that ν p (G/N) = {gN | g ∈ ν p (G)},

(2)

where N is a G-invariant subgroup contained in H(G), as above. Let xN ∈ ν p (G/N), and let q ≠ p be a prime. Let Q ∈ Sylq (G). As before, one may assume that QN/N ≤ CG/N (xN). But then (xN)g = xN for all g ∈ Q, i.e., Q acts by conjugation on the set xN. Since |xN| is a power of p, it follows that Q centralizes an element z ∈ xN. As xN = zN and Q centralizes N, it follows that Q centralizes the coset xN. Then xN ⊆ ν p (G) (Lemma 7.2) and (2) is proved. Set |ν p (G)|p = p r . Using induction on r, we will show that |H(G)|p = p r . If r = 0, then, by the sentence following (1), |H(G)|p = 1. Now let r > 0. By the formula for |ν p (G)| preceding Theorem 7.1, h1 = |Z(G)| is divisible by p. Let N ∈ Sylp (Z(G)). Then N > {1}. It follows from (1) and (2) and induction that |ν p (G/N)|p =

|ν p (G)|p . |N|

(3)

Since H(G/N) = H(G)/N in view of N ≤ Z(G), it follows that |H(G/N)|p =

|H(G)|p . |N|

(4)

By induction, |H(G/N)|p = |ν p (G/N)|p .

(5)

The result now follows from (3), (4), and (5). It is not known, however, whether a knowledge of class sizes vector h(G) is sufficient to establish the p-nilpotency of G. 2°. Isaacs [Isa17] has proved the following result for X1 (G), the first column of the character table X(G) of the group G: Theorem. If X1 (G) = X1 (G1 ) and G1 is p-nilpotent then G is p-nilpotent as well. A proof of this result is presented below. Recall that if ϕ ∈ Irr(G), then o(ϕ) is the order of det(ϕ) in the group Lin(G), where (det(ϕ))(g) = det(T(g)) (g ∈ G and T is a representation of G affording ϕ). Lemma 7.3. Let G be a group, p a prime, N = Op (G). Write Y(G) = {χ ∈ Irr(G) | p ∤ χ(1)},

Y1 (N) = {ϕ ∈ Y(N) | ϕ is G-invariant},

496 | Characters of Finite Groups 1 and u(G) = ∑ χ(1)2 ,

u1 (N) =

χ∈Y(G)

∑

ϕ(1)2 .

ϕ∈Y1 (N)

Then u(G) = |G : G󸀠 |u1 (N), u1 (N) ≡ |N| (mod p). In particular, p ∤ |N| ⇐⇒ p ∤ u1 (N) = u(G)|G : G󸀠 N|−1 . Proof. If ϕ ∈ Y1 (N), then p ∤ ϕ(1) ⋅ o(ϕ), since p ∤ |N : N 󸀠 | and o(ϕ) | |N : N 󸀠 |. Therefore, by Gallagher’s Extension Theorem VII.6.7 (see also Corollary VII.6.10), there is χ ∈ Irr(G) such that χ N = ϕ. If, in addition, ψ ∈ Irr(G) is an extension of ϕ, then ψ = λχ, where λ is a linear character of G containing N in its kernel. The number of such λ is equal to s = |(G/N) : (G/N)󸀠 | = |G : G󸀠 N|. Therefore, by Theorem XIV.6.2 (c), there exist exactly s = |G : G󸀠 N| such characters χ. If that χ ∈ Y(G), then, by Clifford theory, χ N ∈ Y1 (N) since GCD(χ(1), |G : N|) = 1. Therefore, u(G) = ∑ χ(1)2 = |G : G󸀠 N| χ∈Y(G)

∑

ϕ(1)2 = |G : G󸀠 N|u1 (N).

ϕ∈Y1 (N)

Furthermore, |N| =

∑

ϕ(1)2 ≡

ϕ∈Irr(N)

∑

ϕ(1)2 ≡ u1 (N) (mod p).

ϕ∈Y1 (N)

Indeed, if ψ ∈ Irr(N) − Y1 (N), then either p | ψ(1) or |G : IG (ψ)| = p k for some k > 0 (recall that N = Op (G)). Thus, p ∤ u(G)|G : G󸀠 N|−1 ⇐⇒ p ∤ u1 (N) ⇐⇒ N is a p󸀠 -subgroup. Corollary 7.4 (Isaacs [Isa17]). Let X1 (G) = X1 (G1 ) and let G1 be p-nilpotent. Then G is also p-nilpotent. Proof. Let N1 = Op (G1 ), N = Op (G). By assumption, p ∤ |N1 |. Therefore p ∤ u1 (N1 ) =

u(G) u(G1 ) = = u1 (N), 󸀠 |G : G󸀠 N| |G1 : G1 N1 |

by Lemma 7.3. As u1 (N) ≡ |N| (mod p) implies p ∤ |N|, it follows that N is a normal p-complement of G. By Molien’s theorem on the structure of complex group algebras (see Chapter II), X1 (G) = X1 (G1 ) if and only if ℂG ≅ ℂG1 . Therefore, by Corollary 7.4, if ℂG ≅ ℂG1 , then G and G1 are either both nilpotent or both nonnilpotent. It would be most interesting to investigate group-theoretic properties determined by the structure of the complex group algebra. We have seen that ℂG allows us to recognize whether G is a Frobenius group. T. Hawkes has shown that the group algebra does not allow us to recognize supersolvability. In [BZ3, Chapter 17] it is proved (Nagao) that, if X(G) = X(Sn ), then G ≅ Sn . It is proved recently [T-V3] that ℂG ≅ ℂSn implies G ≅ Sn .

X Existence of normal subgroups | 497

8 Characterization of Frobenius complements Suppose that a π-group V > {1} acts on a π󸀠 -group K > {1} (in that case we speak on a coprime action). Let G = V ⋅ K be the natural semidirect product of V and K, i.e., a product compatible with the action of V on K (then V and K are subgroups of the same group G, and V acts on K by conjugation). Set M(V, K) = {x ∈ K | xv = vx for some v ∈ V # },

D(V, K) = ⟨M(V, K)⟩(≤ K).

Let KV denote the set of all π(V)󸀠 -groups admitting an action of V. A group V is said to be a Frobenius complement if it is isomorphic to a Frobenius complement of some Frobenius group. If G is a Frobenius group with kernel K and complement V, then M(V, K) = {1}, so that D(V, K) = {1}. If G = V ⋅ K is not a Frobenius group, it may happen that D(V, K) < K (let V of order 3 act on K ≅ E8 nontrivially). Moreover, if a π-group V is a Frobenius complement and K0 is an arbitrary π󸀠 -group, then there exists K ∈ KV such that D(V, K) ≅ K0 . In fact, if V ⋅ K1 is a Frobenius group, G = (V ⋅ K1 ) × K0 , K = K1 × K0 , then D(V, K) = K0 . Furthermore, it follows from the theorem below (see Corollary 8.2) that if D(V, K) < K for some K ∈ KV , then V is a Frobenius complement (we do not assert that in this case V ⋅ K is a Frobenius group). Our aim is to prove the following: Theorem 8.1. For V a nonidentity π-group for some nonempty set π of primes, the following assertions are equivalent: (a) D(V, K) = K for every K ∈ KV . (b) V is not a Frobenius complement. Proof. If V is a Frobenius complement, then by what we have said, D(V, K) = {1} for some nonidentity K ∈ KV (indeed, by definition, V is a Frobenius complement of some Frobenius group X; then as K we take the Frobenius kernel of X). Therefore, if (a) holds, then V is not a Frobenius complement, i.e., (a) ⇒ (b). It remains to prove the reverse implication. Suppose that (b) is true. Let a π-group V > {1} be not a Frobenius complement and let K ∈ KV (by the definition of the set KV , GCD(|V|, |K|) = 1 so that K is a π󸀠 -group). We have to prove that D(V, K) = K. Let G = V ⋅ K denote the natural semidirect product of V and K with kernel K. Let x ∈ M(V, K). Then there exists v ∈ V # such that vx = xv. For every y ∈ V, we have v y x y = (vx)y = (xv)y = x y v y . Since v y ∈ V # and x y ∈ K, it follows that x y ∈ M(V, K). Therefore: Claim (i). The set M(V, K) is a V-invariant subset of K. In particular: Claim (ii). The set D(V, K) is a V-invariant subgroup of K.

498 | Characters of Finite Groups 1 First suppose that K is a minimal normal p-subgroup of G for some prime p ∈ π󸀠 . Since V is not a Frobenius complement, it follows that D(V, K) > {1}. As K is abelian, it follows from (ii) that D(V, K) is K- and V-invariant so it is normal in VK = G, and therefore D(V, K) = K by the minimal choice of K. Hence: Claim (iii). Assertion (a) holds if K is an abelian minimal normal subgroup of G. Suppose that K is an elementary abelian p-group for some p ∈ π󸀠 . By Maschke’s theorem, K = K1 × ⋅ ⋅ ⋅ × K s , where K1 , . . . , K s are minimal normal subgroups of G. In view of (iii), one may assume that s > 1. Clearly, K i is a minimal normal subgroup of V ⋅ K i , i = 1, . . . , s, Therefore, by (iii), D(V, K i ) = K i for i = 1, . . . , s. In that case, D(V, K) ≥ D(V, K1 ) . . . D(V, K s ) = K1 . . . K s = K, and so: Claim (iv). Assertion (a) holds if K is an elementary abelian p-group for some p ∈ π󸀠 . Suppose that K is a p-group for p ∈ π󸀠 . Then V acts on the elementary abelian p-group K/Φ(K), so that K/Φ(K) ∈ KV . By (iv), D(V, K/Φ(K)) = K/Φ(K). Let xΦ(K) ∈ M(V, K/Φ(K)). Then V # contains an element v of prime order, say q ∈ π, such that [xΦ(K)]v = xΦ(K) (note that Φ(K) is G-invariant). Consider an action of the cyclic subgroup ⟨v⟩ (of prime order q) on the coset xΦ(K) by conjugation. Since p ≠ q and |xΦ(K)| is a power of p(≠ q), v centralizes some element x1 ∈ xΦ(K), and so xΦ(K) = x1 Φ(K). Let M(V, K/Φ(K)) = {x1 Φ(K), . . . , x n Φ(K)}. By what we have just proved, we may assume that x1 , . . . , x n ∈ M(V, K). By the above, D(V, K/Φ(K)) = ⟨x1 Φ(K), . . . , x n Φ(K)⟩ = K/Φ(K), where x1 , . . . , x n ∈ M(V, K) ⊆ D(V, K). Therefore, since the Φ-subgroup consists of non-generators, we obtain D(V, K) ≥ ⟨x1 , . . . , x n ⟩ = K. Thus: Claim (v). Assertion (a) holds if K is a p-group for some p ∈ π󸀠 . Suppose that |K| is not a power of a prime. Let π(K) = {p1 , . . . , p t }. For every index i ∈ {1, . . . , t}, K contains a V-invariant Sylow p i -subgroup P i (to see this, we can use Frattini’s lemma and Schur–Zassenhaus’ theorem). By (v), D(V, P i ) = P i for all i. Obviously, D(V, P i ) ≤ D(V, K) for all i, and so |P i | | |D(V, K)|. Hence, |K| | |D(V, K)|, and so D(V, K) = K, proving (a) and thereby the theorem. Corollary 8.2. A group V is a Frobenius complement if and only if D(V, K) < K for some K ∈ KV .

X Existence of normal subgroups | 499

Corollary 8.3. Let a group V contain a noncyclic subgroup of order pq, where p, q are primes (not necessarily distinct). If K ∈ KV , then D(V, K) = K. Proof. Indeed, V is not a Frobenius complement (see §X.3). If V ≅ E4 , Corollary 8.3 generalizes the last assertion of Brauer’s Theorem XV.6.3. In the case where V is abelian and noncyclic, the result is also known (see [Gor1, Theorem 5.3.16]). If all subgroups of order pq in V are cyclic, this does not guarantee the fulfillment of the equality D(V, K) = {1} for some K ∈ KV . In fact, if V ≅ SL(2, 257), then all subgroups of V whose orders are products of two primes, are cyclic, however, D(V, K) = K for every K ∈ KV since, as we know, SL(2, 257) is not a Frobenius complement (see §X.3).

9 Groups with few irreducible characters of small degrees Let d > 1 be a proper divisor of |G| and let X d = {χ ∈ Irr(G) | d ∤ χ(1)},

σ d (G) = ∑ χ(1)2 .

(1)

χ∈X d

Then, by Frobenius–Molien, |G| = σ d (G) +

∑

χ(1)2 ≡ σ d (G) (mod d) 󳨐⇒ d | σ d (G).

(2)

χ∈Irr(G)−X d

If G is a Frobenius group with kernel of index d, then σ d (G) = d (Theorem 3.2); the converse assertion is also true (Theorem 4.3). In this section we study the case σ d (G) = 2d in some detail. The set Irr(G) contains a character of degree divisible by d if and only if |G| > σ d (G); then d | |G| implies d | σ d (G). In what follows we assume that |G| > σ d (G). Set K = K d (G) = ⋂ ker(χ).

(3)

χ∈X d

It is clear that |G : K| = σ d (G) if and only if Irr(G/K) = X d . In that case, K is p-nilpotent for any prime divisor p of d (see Remark 9.1 below) and K is solvable (see [BZ3, Proposition 25.9 and Remark 1 following it]). As we have noted, G is a Frobenius group with kernel of index d if and only if σ d (G) = d, i.e., the groups G with σ d (G) = d are classified. In this section we will do the following step and consider the case σ d (G) = 2d. Theorem 9.6 yields the classification in the case when d is odd. The proof of that theorem is partially based on Lemma 9.2. In the general case when d is arbitrary, we prove that the socle of G is abelian (Corollary 9.4). Lemma 9.5 also yields an essential information on the structure of G. Lemma 9.1. Let X ⊆ Irr(G) and assume that, whenever α, β ∈ X, then Irr(αβ) ⊆ X. We have that K = ⋂α∈X ker(α) implies X = {χ ∈ Irr(G) | K ≤ ker(χ)}.

500 | Characters of Finite Groups 1 Proof. Let τ = ∑α∈X α; then τ is a faithful character of G/K. By the Burnside–Brauer Theorem IV.4.3, every irreducible character of G/K is a constituent of some power of τ and thus lies in X, by assumption. Thus, we have Irr(G/K) ⊆ X. The reverse inclusion is obvious. Lemma 9.2. Suppose that an integer d > 1 divides |G| and σ d (G) ≤ 2d. If α, β ∈ X d , then Irr(αβ) ⊆ X d . Proof. One may assume that α, β ∈ ̸ Lin(G) (clearly, Lin(G) ⊆ X d ). If α ≠ β, then 2d ≥ σ d (G) ≥ α(1)2 + β(1)2 > 2α(1)β(1) 󳨐⇒ α(1)β(1) < d. It follows that Irr(αβ) ⊆ X d . If α = β, then αβ = α2 . Assume that α2 has an irreducible constituent ψ of degree divisible by d. Since α(1)2 < 2d (note that 1G ∈ X d ), ψ is the unique irreducible constituent of α2 of degree divisible by d; moreover, ψ(1) = d and ⟨α2 , ψ⟩ = 1. Hence, ⟨α, αψ⟩ = ⟨α2 , ϕ⟩ = 1. Let λ ∈ Lin(G). Then ⟨λ, αψ⟩ = ⟨αλ, ϕ⟩ = 0 since αλ, ϕ ∈ Irr1 (G) have distinct degrees. Since d | α(1)ψ(1), d ∤ α(1)ψ(1) − α(1) = α(1)(d − 1), the character αψ − α has an irreducible constituent γ that is a member of the set X d . As αψ has no linear constituents, we get γ(1) > 1. Since the multiplicity of α in αψ is 1, we obtain γ ≠ α. Now ⟨αγ, ψ⟩ = ⟨γ, αψ⟩ > 0, a contradiction since, by the previous paragraph, Irr(αγ) ⊆ X d in view of the fact that α ≠ γ are members of X d . Thus, by Lemma 9.2, the set X = X d satisfies the hypothesis of Lemma 9.1. As we have noted, |G| > σ d (G) (in that case, X d ⊂ Irr(G) and so there is in Irr(G) a member of degree divisible by d). Set, as above, K = ⋂ ker(χ).

(4)

χ∈X d

Remark 9.1. For a prime p, let G(p󸀠 ) denote the intersection of kernels of those nonlinear irreducible characters whose degrees are not divisible by p; if G has no such characters, set G(p󸀠 ) = G (in that case, the degrees of all nonlinear irreducible characters of G are divisible by p). By Theorem XIV.5.3, the subgroup G(p󸀠 ) is p-nilpotent and solvable (this follows from CFSG) so K = ⋂p∈π(d) G(p󸀠 ) is also p-nilpotent and solvable. (This remark allows us to do some nontrivial assertions on the structure of quotient group G/K.) Corollary 9.3. If σ d (G) ≤ 2d, then the following statements hold: (a) |G/K| = σ d (G). (b) The degrees of all members of the set Irr(G | K) are multiples of d. (c) The subgroup K is p-nilpotent for p ∈ π(d) and solvable. Proof. By Lemmas 9.1 and 9.2, we have Irr(G/K) = X d so |G/K| = σ d (G), proving assertion (a). By assumption, d | χ(1) for χ ∈ Irr(G | K) so (b) holds. Let p ∈ π(d). If p ∤ χ(1) (χ ∈ Irr(G)), then d ∤ χ(1) so K ≤ ker(χ). Thus, K ≤ G(p󸀠 ) for all p ∈ π(d) so that K is p-nilpotent for all such p and solvable (Remark 9.1).

X Existence of normal subgroups | 501

Let σ d (G) = wd, where w ∈ ℕ, and let K be as in (4). Suppose, in addition, that Irr(G/K) = X d so that |G/K| = σ d (G) = wd. Then the same argument as in the proof of Corollary 9.3 (c) shows that K is solvable and p-nilpotent for all p ∈ π(d). Remark 9.2. Assume that |G| > d > 1 and σ d (G) = d | |G|. We will prove that then G is a Frobenius group with kernel of index d (our proof is different from the one of Theorem 4.3). In the above notation, |G : K| = d and d divides the degrees of all characters in Irr(G | K). If ϕ ∈ Irr# (K), then K ≰ ker(χ) for all χ ∈ Irr(ϕ G ) so d | χ(1) and G = (H, K) for some H < G) (Corollary 4.2). Corollary 9.4. Let σ d (G) = 2d < |G|. Assume that L < G centralizes K and intersects K trivially. Then |L| ≤ 2. Proof. By Remark 9.1, K is solvable. Since K > {1}, it has a nonprincipal linear character θ. By assumption, KL = K × L so ψ = θ × 1L ∈ Lin(KL) and K ≰ ker(ψ). Therefore, since the kernels of all irreducible constituents of ψ G do not contain K, it follows that ψ G (1) is a multiple of d. Then d | |G : KL|, and thus |L| ≤ 2 since |G : K| = 2d (Corollary 9.3). Lemma 9.5. Let d > 1 be a proper divisor of |G| and suppose that |G| > σ d (G) = wd. Write K = ⋂χ∈X d ker(χ) and assume that Irr(G/K) = X d (in that case, the degrees of all members of the set Irr(G) − X d are multiples of d). Then K is solvable and has a normal p-complement for any prime divisor p of d. (a) Suppose that K is an elementary abelian p-subgroup of order greater than p and GCD(w, pd) = 1. Then K ∈ Sylp (G). If q ∈ π(d) and Q ∈ Sylq (G), then QK = (Q, K) is a Frobenius group with kernel K. (b) Suppose that w = 2 and K is an elementary abelian 2-subgroup. (1) If 4 | |G/K|, then G is nonabelian of order 8, d = 2. (2) If q ∈ π(d) − {2} and Q ∈ Sylq (G), then QK = (Q, K) is a Frobenius group with kernel K. Proof. By hypothesis, the degrees of all members of the set Irr(G | K) are multiples of d. Therefore, the assertions preceding (a) have been established already (see the paragraph preceding Remark 9.2). (a) Here K is an elementary abelian p-subgroup of order > p. Assume that there is in G/K a subgroup Z/K of order p. Then p | d since p | |G/K| = wd and GCD(w, pd) = 1 implies p ∤ w. We conclude that Z is a noncyclic p-subgroup of order p|K| ≥ p3 so that Z 󸀠 < K. It follows that Z has a linear character ϕ whose kernel does not contain K. Since d (5) d ∤ w ⋅ = |G : Z| = ϕ G (1), p there exists an irreducible constituent χ of ϕ G such that d ∤ χ(1), a contradiction since K ≰ ker(χ), by reciprocity. Thus, Z does not exist so K ∈ Sylp (G) 󳨐⇒ p ∤ |G/K| = wd 󳨐⇒ p ∤ d.

502 | Characters of Finite Groups 1 Let q ∈ π(d) and Q ∈ Sylq (G). To prove that QK is a Frobenius group, it is enough to show that CK is a Frobenius group for any subgroup C of order q in Q. Assume that this is false. Then C centralizes some nonidentity element of K. By Fitting’s lemma (see, for example, Corollary I.1A.4), (CK)󸀠 < K. So there is ϕ ∈ Lin(CK | K), where Lin(CK | K) = {μ ∈ Lin(CK) | K ≰ ker(μ)}. Then d ∤ w ⋅ dq = |G : C| = ϕ G (1), and again, as in the previous paragraph, we obtain a contradiction. Thus QK is a Frobenius group, completing the proof of (a). (b) Since w = 2, we have |G/K| = σ d (G) = wd = 2d (Corollary 9.3). Let p = 2 and suppose that there is T/K ≤ G/K of order 4; then 4 | |G/K| = 2d implies 2 | d. Assume that T 󸀠 < K. In that case, there is a character ψ ∈ Irr(T | K). Then ϕ G (1) = |G : T| = 2d so all irreducible constituents of the character ϕ G have degrees not divisible by d, a contradiction since all those constituents are contained in Irr(G | K), by reciprocity. Thus T 󸀠 = K is of index 4 in T so that T is of maximal class, by Taussky’s theorem [Ber31, Proposition 1.6]. Since K is a normal elementary abelian subgroup of a 2-group T of maximal class and |T : K| = 4, we conclude that |K| = 2. In that case, |G| = σ d (G)|K| = 4d. However, d2 < |G| = 4d so that d = 2 since d is even. Thus, |G| = 8, proving (1). Now let d be odd (then 4 ∤ 2d = |G/K|) and q ∈ π(d). To prove (2), it suffices to show that ZK is a Frobenius group for any Z < G of order q. Assume that this is false. Then, by Fitting’s lemma, K = CK (Z) × V. Therefore, there is μ ∈ Lin(ZK) such that K ≰ ker(μ). Next, μ G (1) = |G : ZK| = 2 ⋅ dq < d implies K ≤ ker(μ G ), a contradiction, and the proof of (2) is complete. Let H ⊲ G and let H be a Frobenius group with kernel K of index d. Then all characters in the set Irr(G | K) have degrees divisible by |H : K| = d. Indeed, take χ ∈ Irr(G | K). Let χ H = e(ψ1 + ⋅ ⋅ ⋅ + ψ t ) be Clifford’s decomposition. Since K ≰ ker(ψ i ) for all i, it follows from Theorem 3.2 (a) that d | ψ i (1); then d | etψ1 (1) = χ(1), as claimed. The case when d is odd and σ d (G) = 2d may be investigated in detail. Namely, the following theorem holds. Theorem 9.6. Let a proper divisor d > 1 of |G| be odd. Then σ d (G) = 2d < |G| if and only if G has a Frobenius subgroup H with kernel K of index d such that |G : H| = 2. Proof. Suppose that there is in G a Frobenius group H with kernel K of odd index d such that |G : H| = 2. It follows from the remark preceding the theorem that all characters from Irr(G | K) have degrees divisible by d. The group G/K has no irreducible character of degree divisible by d since d2 ≥ 3d > 2d = |G/K|. It follows that K = K d = ⋂ ker(χ). χ∈X d

By Corollary 9.3, |G/K| = σ d (G) = 2d. Now suppose that |G/K| = 2d, where K = K d = ⋂χ∈X d ker(χ). Since d is odd, it follows that G/K has a subgroup H/K of index 2 (Burnside). Since, in addition, d > 1, we get Irr(G/K) ⊆ X d . Let us prove that H is a Frobenius group with kernel K (of

X Existence of normal subgroups | 503

index d in H). Let χ ∈ Irr(G | K); then d | χ(1) since K = K d . As d > 1 is odd and |G : H| = 2, all irreducible constituents of χ H have degrees divisible by d (Clifford). If ϕ ∈ Irr(H | K) and τ ∈ Irr(ϕ G ), then d | τ(1) since K ≰ ker(τ), by reciprocity. Then d divides the degrees of all irreducible constituents of τ H so that d | ϕ(1). In that case, H is a Frobenius group with kernel K, by Corollary 4.2. Remark 9.3. Let G be a nonabelian p-group of order p n , d = p s , w ≤ s, σ d (G) = p w+s , n > w+s. Then p n ≡ p w+s (mod p2s ) (this follows from the relation ∑χ∈Irr(G) χ(1)2 = |G|), and we conclude that w = s since p2s | p n . If N ⊲ G is of index p2s , then Irr(G/N) ⊆ X d , and hence N = K = K d is the unique normal subgroup of index p2s in G. Let G = (A, B). Then G is 2-transitive on cosets of A if and only if |A| = |B| − 1. In that case, B# is a conjugacy class of G. Conversely, if B# is a conjugacy class of G, then G is 2-transitive on cosets of A. Note that G = (SL(2, 5), E112 ) is a 2-transitive Frobenius group of degree 112 . In the case when d = χ(1) for some χ ∈ Irr(G) and σ d (G) ≤ 2d, we obtain the complete classification. Note that Theorem 9.7 coincides with Exercise 4.2. Theorem 9.7. Let χ ∈ Irr(G) have degree d > 1 and let |G| = d2 + wd, where w ≤ 2 (i.e., σ d (G) = wd). (a) If w = 1, then G is a 2-transitive Frobenius group. (b) If w = 2, then d = 2 and |G| = 8. Proof. We retain the above notation. (a) If w = 1, then, by Theorem 4.3, G is a Frobenius group with kernel K of index d. Since |G| = d(d + 1), we get |K| = d + 1, so the set K # is a G-class. It follows that G is 2-transitive. (b) Let w = 2. Since |G| = d2 + 2d ≤ 2d2 , it follows that χ is the unique irreducible character of G of degree divisible by d and χ is faithful since |G/ker(χ)| > χ(1)2 = d2 . We have |G| d2 + 2d d χ(1) = d, |K| = = = 1 + ∈ ℕ. σ d (G) 2d 2 Hence, d is even. As we have noted, Irr(G | K) = {χ}. Therefore, it follows, by reciprocity, that all nonprincipal irreducible characters of K are constituents of the restriction χ K and they are G-conjugate (Clifford’s theorem). Let ψ ∈ Irr(χ K ). Therefore, if d > 2, then ψ(1) | GCD(|K|, χ(1)) = GCD(1 + 2d , d) = 1. If d = 2, then |K| = 2 and |G| = d2 + 2d = 8. Next assume that d > 2. Then ψ(1) = 1 so that K is abelian since, in the case under consideration, Irr(K) = Lin(K). In that case K, being a minimal normal subgroup of G, has order p e for some prime p and e ∈ ℕ. We have d , d(d + 2) = |G| = 2d|K| = 2dp e 󳨐⇒ d = 2(p e − 1). 2 We have p > 2 (indeed, d > 2 is even hence GCD(d, 1 + 2d ) = 1 = GCD(d, p e )), and we conclude that K ∈ Sylp (G). Let H < G be of order 2(p e − 1) (Schur–Zassenhaus’ p e = |K| = 1 +

504 | Characters of Finite Groups 1 theorem). Let Q < H be of prime order q. Since k(G) = |Irr(G)| = k(G/K) + 1 = k(H) + 1, it follows that Q ⋅ K is a Frobenius group, and we conclude that G is also a Frobenius group with kernel K, G = (H, K). Then, by Exercise 2.3 (b), 2(p e − 1) = |H| | |K| − 1 = p e − 1, which is a impossible. Thus, |G| = 8. Problem. Suppose that a group G has a nonlinear irreducible character of degree d. Describe the structure of G if σ d (G) = 3d. See also related papers [Sny] and [Isa28] which develop ideas of this section further. Let E24 ≅ A ⊲ G be such that G/A ≅ PSL(2, 5) and k(G) = 9. Note that the degrees of the irreducible characters of G are 1, 3, 3, 4, 5, 15, 15, 15, 15. Let H < A be of index 2 and d = 15; then |G : H| = 120 = 8d. If μ ∈ Irr# (H), then μ G = 2(χ1 + ⋅ ⋅ ⋅ + χ4 ), where Irr(μ G ) = {χ1 , . . . , χ4 } and χ i (1) = 15 = d for i = 1, 2, 3, 4. Then H G = A,

|G : H G | = |G : A| = 60 = 4d = σ15 (G),

Irr(G/H G ) = X15 .

Example. Let G be a semisimple group (i.e., F(G) = {1}). We claim that then Irr(G) contains three distinct faithful characters. Let S = R1 × ⋅ ⋅ ⋅ × R m be the socle of G, where R1 , . . . , R m are nonabelian simple groups. By Isaacs’ Theorem XIV.7.1, Irr(R i ) contains three nonlinear characters μ i1 , μ i2 , μ i3 such that μ i1 (1) < μ i2 (1) < μ i3 (1) for all i = 1, . . . , m. Set λ i = μ i1 × ⋅ ⋅ ⋅ × μ im for all i = 1, 2, 3. Then λ1 , λ2 , λ3 ∈ Irr(S) are faithful and λ1 (1) < λ2 (1) < λ3 (1). Let χ i ∈ Irr(λ Gi ), i = 1, 2, 3. By reciprocity, we have S ∩ ker(χ i ) ≤ ker(λ i ) = {1}, and so ker(χ i ) = {1}, i.e., χ i ∈ Irr(G) is faithful. By Clifford, χ1 , χ2 , χ3 are distinct.

10 Frobenius’ theorem for solvable Frobenius complement We present an elementary proof of the following partial case of Frobenius’ theorem: Theorem. If H < G is solvable with H ∩ H x = {1} for all x ∈ G − H, then there is F ⊲ G such that G = HF and H ∩ F = {1}. O. Grün was the first who gave this proof. The text of this section is taken, with minor changes, from Ito’s lecture notes [Ito7]. We first define a transfer homomorphism of a group G in its subgroup H of index n. Let G = ⋃ni=1 Hr i be the decomposition of G modulo H. For each g ∈ G we define (a) a permutation ρ(g) of the set {1, . . . , n} and (b) the elements g1 , . . . , g n ∈ H by the following equation: r i g = g i r i ρ(g) ; then we have Hr i g = Hg i r i ρ(g) = Hr i ρ(g) . In that case, the mapping V = V G→H → H/H 󸀠 , defined by n

V(g) = ∏ g i H 󸀠 i=1

X Existence of normal subgroups | 505

is said to be the transfer of G into H. Since H/H 󸀠 is abelian, V(g) is independent of the ordering of factors. Proposition 10.1. The following statements hold: (a) The transfer V of G into H is independent of the choice of a transversal. (b) The transfer V is a homomorphism from G into the abelian group H/H 󸀠 . Proof. (a) Let G = ⋃ni=1 Hr i = ⋃ni=1 Hs i with Hr i = Hs i for all i. Then s i = t i r i for some t i ∈ H. Since Hs i ρ(g) = Hs i g = Hr i g = Hr i ρ(g) , the permutation ρ = ρ(g) is independent of the choice of a transversal. Let r i g = g i r i ρ(g) = g i r i ρ ,

s i g = g 󸀠i s i ρ(g) = g 󸀠i s i ρ .

(1)

Then, by (1), we have g󸀠 t iρ r iρ = g󸀠 s iρ = s i g = t i r i g = t i g i r iρ ,

(2)

so that g󸀠i t i ρ = t i g i . It follows that g 󸀠i = t i g i (t i ρ )−1 . Since t i ∈ H, we get n

n

n

n

∏ g󸀠i H 󸀠 = ∏ t i g i (t i ρ )−1 H 󸀠 = (∏ t i H 󸀠 ) ⋅ (∏(t i ρ )−1 H 󸀠 ) ⋅ (∏ g i H 󸀠 ) = ∏ g i H 󸀠 i=1

since

i=1

i=1 n

i=1 n

i=1

i=1

n

∏(t i ρ )−1 H 󸀠 = ∏(t i ρ H 󸀠 )−1 = ∏(t i H 󸀠 )−1 , i=1

i=1

i=1

proving (a). (b) We have (below g i , h i ∈ H for all i and g, h ∈ G) n

n

n

V(g)V(h) = (∏ g i H 󸀠 )(∏ h i ρ H 󸀠 ) = (∏ g i h i ρ(g) H 󸀠 ) = V(gh). i=1

i=1

i=1

Hr i Consider the cycle decomposition of the permutation (Hr ): ig

(Hr1 , Hr1 g, . . . , Hr1 g m1 −1 ) ⋅ ⋅ ⋅ (Hr k , Hr k g, . . . , Hr k g m k −1 ). Here m i is the minimal positive integer such that Hr i g m i = Hr i or, what is the same, i r i g m i r−1 i ∈ H. We have m 1 + ⋅ ⋅ ⋅ + m k = n and m i | n. Besides, the elements r j g for 0 ≤ i < m j and 1 ≤ j ≤ k form a transversal for G in H. For i < m j − 1, we have r j g i g = r j g i+1 ,

r j g m j −1 g = r j g m j = g m j −1,j r j

(g m j −1,j ∈ H),

and we obtain g m j −1,j = r j g m j r−1 j . Now we can compute the transfer from the following formula: k

V(g) = H 󸀠 ∏ r j g m j r−1 j

(g ∈ G).

j=1

Theorem 10.2 (Grün). If a Frobenius complement H is solvable, then the Frobenius kernel F of the Frobenius group G with respect to H exists.

506 | Characters of Finite Groups 1 Proof. Let V be the transfer of G into H. Then k

V(g) = H 󸀠 ∏ r i g m i r−1 i

(g ∈ G),

i=1

where m j ∈ ℕ is minimal such that r j g m j r−1 j ∈ H. If r i ∈ ̸ H = NG (H), then we obtain −1 −1 m i that r i g r i ∈ H ∩ r i Hr i = {1} since H is a Frobenius complement. We can assume that r1 = 1; then, provided g ∈ H, one has m1 = 1 and so r1 gr−1 1 = g ∈ H. Hence, for r i ∈ ̸ H, we have r i g m i r−1 = 1, and now for all g = h ∈ H, one has i k 󸀠 󸀠 V(h) = h ∏ r i hr−1 i H = hH . i=2

Let K = ker(V) = {g ∈ G | V(g) = H 󸀠 }. In that case, K ∩ H = H 󸀠 , by the above formula. Let us prove that G = HK. Indeed, for any g ∈ G, there exists h ∈ H such that V(g) = hH 󸀠 since V : G → H, but V(gh−1 ) = V(g)V(h−1 ) = hH 󸀠 h−1 H 󸀠 = H 󸀠 since V is a homomorphism; hence gh−1 ∈ ker(V) = K and so g = hk ∈ HK for some k ∈ K. Thus, G = HK. If H is abelian, we choose F = K, and we are done. Next suppose that H is nonabelian, i.e., H 󸀠 > {1}. In that case we will show that 󸀠 H is a Frobenius complement of K(= ker(V)). Indeed, as we have noticed, H 󸀠 = K ∩ H. If x ∈ K − H 󸀠 , then x ∈ ̸ H so H 󸀠 ∩ (H 󸀠 )x ≤ H ∩ H x = {1}, and our claim follows. Now we proceed by induction on |G|. In that case, there is a normal complement F to H 󸀠 in K since K < G in view H ∩ K = H 󸀠 < H. Since F is a Hall subgroup of K, it is characteristic in K ⊲ G so normal in G. We have G = HK = H(H 󸀠 F) = (HH 󸀠 )F = HF,

H ∩ F ≤ H ∩ K = H󸀠 .

Since F ∩ H 󸀠 = {1}, it follows that H ∩ F = {1}, so F is the kernel of the Frobenius group G. Exercise. Let H < G with H ∩ H x = {1} for all x ∈ G − H. Give an elementary proof of the existence of the Frobenius kernel of G with respect to H in the following cases: (a) |G : H| = p a , p a prime. (b) |G : H| = n and |H| = n − 1. Hint. Use the equality |G − ⋃x∈G−H (H # )x | = |G : H|. Now, in case (a), the Frobenius kernel coincides with a Sylow p-subgroup of G. In case (b), all nonidentity elements of the set G − ⋃x∈G−H (H # )x are H-conjugate so have the same order p, a prime. Therefore, (a) ⇒ (b). As a consequence of Exercise 2.3 (g), Frobenius’ theorem follows from the solvability of groups of odd order, in view of Theorem 10.2. (Of course, the Odd Order Theorem depends on Frobenius’ theorem.)

X Existence of normal subgroups | 507

11 On irreducible constituents of induced characters 1°. Given χ ∈ Char(G), we set Irr1 (χ) = Irr(χ) ∩ Irr1 (G),

cd1 (χ) = {τ(1) | τ ∈ Irr1 (χ)}.

Definition. A character χ is said to be a D-character if |cd1 (χ)| = |Irr1 (χ)|, i.e., distinct nonlinear irreducible constituents of χ have pairwise distinct degrees. In this section we study the following class of groups which we call D-groups: Property D. Whenever {1} < N ≤ G󸀠 , N ⊴ G and λ ∈ Irr# (N), then λ G is a D-character. (In that case, we have that all irreducible constituents of λ G are nonlinear and have distinct degrees.) Lemma 11.1. Let G be a nonabelian group and {1} < H ⊲ G, G/H ≅ Cm , m > 1. (a) If λ ∈ Irr# (H) and λ G is a D-character, then λ G ∈ Irr(G). (b) If H = G󸀠 , then G = (Cm , G󸀠 ) is a Frobenius group with kernel H. Proof. (a) By Theorem VII.3.8, all irreducible constituents of λ G have the same degree, which implies |Irr(λ G )| = 1, hence λ G = eχ, χ ∈ Irr(G). Since the ramification of χ over H equals 1 (Exercise VII.3.3), the result follows. (b) This follows from Corollary 4.2. Exercise 11.1. If G󸀠 ≤ N < G and λ ∈ Irr# (N), then |cd(λ G )| = 1, i.e., all irreducible constituents of λ G have the same degree. Hint. Use Theorem VII.3.8. Exercise 11.2. Let {1} < G󸀠 < G with G/G󸀠 ≅ Cm . Then λ G is a D-character for every λ ∈ Irr# (G󸀠 ) if and only if G = (Cm , G󸀠 ) is a Frobenius group with kernel G󸀠 . Solution. The implication “⇒” follows from Lemma 11.1, and “⇐” follows from Theorem 3.2.) Lemma 11.2. Let {1} < H ⊲ G be such that G/H ≅ Q8 and H < G󸀠 (then G/G󸀠 ≅ E4 ). If λ G is a D-character for every λ ∈ Irr# (H), then G = (Q8 , H). Proof. Let

Irr(λ G ) = {χ1 , . . . , χ s },

λ G = e1 χ1 + ⋅ ⋅ ⋅ + e s χ s

(all χ i are nonlinear). By Clifford theory, e1 , . . . , e s are the degrees of irreducible projective representations of the quotient group IG (λ)/H (Lemma VII.3.3 (b)). Since the Schur multiplier of any subgroup of Q8 is trivial, it follows that in fact e1 , . . . , e s are the degrees of the ordinary irreducible representations of IG (λ)/H. Hence e i ≤ 2 for all i. Assume that e i = 2. Then IG (λ)/H ≅ Q8 , i.e., IG (λ) = G and λ is G-invariant. In that case, λ is extendible to G (Corollary VII.4.2), contrary to the reciprocity. Thus, e i = 1 for all i. By Clifford’s theorem, χ1 (1) = ⋅ ⋅ ⋅ = χ s (1) so that s = 1. Thus, λ G = χ1 , and we conclude that G is a Frobenius group with kernel H (Corollary 4.2).

508 | Characters of Finite Groups 1 We suggest the reader to consider the following three cases: G/H ∈ {D2n , Q2n , SD2n } for n > 3. Lemma 11.3 (= Lemma III.9.1). Let p be a prime such that p n = p k + a1 p2c1 + ⋅ ⋅ ⋅ + a s p2c s ,

c1 < ⋅ ⋅ ⋅ < c s ,

where s, n, k, a1 , . . . , a s , c1 , . . . , c s ∈ ℕ. Then the following statements hold: (a) k ≥ 2c1 . (b) If a1 < p2 − p, then s = 1, k = 2c1 , n = 2c1 + 1, a1 = p − 1. Lemma 11.4. Let G be a nonabelian p-group with δ(G) = (p k ⋅ 1, a1 ⋅ p c1 , . . . , a t ⋅ p c t ) (i.e., Irr(G) contains exactly a i characters of degree p c i for all i). If a1 < p2 − p, then G is an extraspecial group ES(m, p) of order p2m+1 , m ∈ ℕ. Proof. This is a corollary of Lemma 11.3. (See also Lemma III.9.1.) Lemma 11.5. Suppose that {1} < N ⊲ G is such that N ≤ G󸀠 and G/N is a p-group. If λ G is a D-character for some λ ∈ Irr# (N), then λ G = eχ, where χ ∈ Irr1 (G). Proof. Let λ G = e1 χ1 +⋅ ⋅ ⋅+e s χ s , where Irr(λ G ) = {χ1 , . . . , χ s }. Assume that s > 1. Since N ≤ G󸀠 and λ ∈ Irr# (N), it follows that all characters χ i are nonlinear, by reciprocity. Let χ iN = e i (λ1 + ⋅ ⋅ ⋅ + λ n ) be the Clifford decomposition, λ1 = λ. Then χ i (1) = e i nλ(1),

|G : N|λ(1) = λ G (1) = nλ(1)(e21 + ⋅ ⋅ ⋅ + e2s ),

so that |G : N| = n(e21 + ⋅ ⋅ ⋅ + e2s ), where n = |G : IG (λ)| is a power of p since N ≤ IG (λ), and we conclude that p α = |G : N|n−1 = e21 + ⋅ ⋅ ⋅ + e2s for some α ∈ N. Since IG (λ)/N is a p-group, the ramifications e1 , . . . , e s are powers of p. If i ≠ j, then ne i λ(1) = χ i (1) ≠ χ j (1) = ne j λ(1) 󳨐⇒ e i ≠ e j since λ G is a D-character. Suppose that e1 < ⋅ ⋅ ⋅ < e s , e i = p β i , i = 1, . . . , s. As s > 1, we get p α−2β1 = 1 + p2(β2 −β1 ) + ⋅ ⋅ ⋅ + p2(β s −β1 ) , which is impossible since β1 < ⋅ ⋅ ⋅ < β s . Hence s = 1 and λ G = e1 χ1 , χ1 ∈ Irr1 (G). Lemma 11.6. Let G = (A, H) be a Frobenius group with kernel H, {1} < N < H and let N ⊲ G be such that H/N is a p-group. If λ ∈ Irr# (N) and λ G is a D-character, then λ G = eψ, where ψ ∈ Irr(G). Proof. Let λ G = e1 χ1 + ⋅ ⋅ ⋅ + e s χ s , where Irr(λ G ) = {χ1 , . . . , χ s }; int that case, the characters χ1 , . . . , χ s of pairwise distinct degrees are nonlinear since N < H ≤ G󸀠 . Assume that s > 1. As above, we have χ i (1) = ne i λ(1) and |G : N| = n(e21 + ⋅ ⋅ ⋅ + e2s ). Since (A, N) is a Frobenius group and G is a Dπ(A) -group in terminology of P. Hall, it follows that IG (λ) ≤ H, and so n = |A| ⋅ n0 , where n0 ∈ ℕ. Therefore, |A||H : N| = |G : H||H : N| = |G : N| = n(e21 + ⋅ ⋅ ⋅ + e2s ) = |A|n0 (e21 + ⋅ ⋅ ⋅ + e2s ),

X Existence of normal subgroups | 509 2 2 and so |H : N| ⋅ n−1 0 = e 1 + ⋅ ⋅ ⋅ + e s . It follows from N ≤ IG (λ) ≤ H that n 0 , as a divisor of |H : N|, is a power of p. Hence, e21 + ⋅ ⋅ ⋅ + e2s = p α for some α ∈ ℕ. As above, the pairwise distinct ramifications e1 , . . . , e s , as divisors of |IG (λ)/N| so divisors of |H/N|, are powers of p. As in the proof of Lemma 11.5, this is G possible only if s = 1.

Lemma 11.7. Let H ⊲ G be a Hall subgroup. Then H ∩ Φ(G) = Φ(H). Proof. Note that Φ(H) ⊲ G so Φ(H) ≤ Φ(G) (Lemma XIV.1.1 (c)), and we may assume without loss of generality that Φ(H) = {1}. Therefore, to complete the proof, it suffices to prove that D = H ∩ Φ(G) = {1}. Assume that D > {1} (note D ⊲ G). The Fitting subgroup F(H) ≥ D > {1} is abelian of square-free exponent since Φ(F(H)) ≤ Φ(H) = {1}. Let A be the least subgroup of H such that AD = H; then D ≰ A since Φ(H) = {1}. Let D1 = A ∩ D. We have D1 ≤ Φ(A) (Lemma XIV.1.1 (b)) and NH (D1 ) ≥ ⟨A, D⟩ = H (D is abelian and D1 ⊲ A), i.e., D1 ⊲ H. Then D1 ≤ Φ(H) = {1} (Lemma XIV.1.1 (c)) and hence {1} = D1 = A ∩ D and H = A ⋅ D (semidirect product). Let {1} ≠ P ∈ Sylp (D), where a prime p | |D|; then P ⊲ G and P is complemented in Sylow p-subgroup of G, by the modular law (recall that H is a Hall subgroup of G). Since P is elementary abelian, it follows from Gaschütz’s theorem on split extension [Gas1] that there is F < G such that G = F ⋅ P and F ∩ P = {1}. Since {1} < P ≤ D ≤ Φ(G), this is a contradiction. Thus D = {1}. Lemma 11.8. Suppose that P is a minimal normal p-subgroup of a group G = C ⋅ P, C ∩ P = {1}, where C ≅ Cb , b > 1, acts on P faithfully.⁸ Then |P| = p m , where m is the order of p (mod b). Proof (Mann). Put E = EndGF(p)C (P). Then E is a finite field (by Schur’s lemma and Wedderburn’s theorem). Since C ⊂ E, all E-subspaces of P are trivial. Therefore, we obtain dimE (P) = 1. Let F be the subfield of E generated by C. As before, we have dimF (P) = 1. Hence |E| = |P| = |F|, F = E. Put |E| = p n . Then b = |C| | |E# | = p n − 1 (Lagrange). Since C generates E as a field, it follows that n is the least natural number such that p n ≡ 1 (mod b). Therefore, n = m, |P| = |E| = p m . It is easy to check that the following groups are nonabelian solvable D-groups: (a) G = ES(m, 2) is extraspecial of order 22m+1 . (b) G = (Q8 , E32 ). (c) G = (Cp n −1 , Ep n ). Theorem 11.9 ([BCH]). Let G be a nonabelian solvable group with |cd1 (G)| = |Irr1 (G)| (i.e., all nonlinear irreducible characters of G have pairwise distinct degrees). Then G ∈ {ES(m, 2), (Q8 , E9 ), (Cp n −1 , Ep n )}. In the following exercises, G is a nonabelian solvable D-group, and R is a minimal normal subgroup of G with |R| = p n , p a prime. 8 In that case, G is a Frobenius group with kernel P since x ∈ C# 󳨐⇒ CG (x) = C.

510 | Characters of Finite Groups 1 Exercise 11.3. The quotient group G/R is either abelian or a D-group. Exercise 11.4. If G is nilpotent, then G ≅ ES(m, 2). Solution. Let G = P × Q, where P ∈ Sylp (G) is nonabelian; then R ≤ P󸀠 be a minimal normal subgroup of G so that |R| = p. Assume that Q > {1}. Let λ ∈ Irr# (R), χ ∈ Irr(λ P ); χ is nonlinear, since R ≤ P󸀠 . Let μ ∈ Lin# (Q). Then χ × 1Q and χ × μ are distinct nonlinear irreducible constituents of λ G of the same degree since their restrictions to R coincide, and this is a contradiction. Thus, G is a p-group. Then, by Lemma 11.5, λ G = eχ, χ ∈ Irr(G). Since λ is G-invariant, we have e = χ(1) (Clifford), and so λ G = χ(1)χ, |G : R| = λ G (1) = χ(1)2 ≤ |G : Z(G)| 󳨐⇒ R = Z(G). Let |G| = p n . Then the set Irr(G) contains exactly p − 1 characters of degree p s , where 2s = n − 1 (since restrictions of these characters to R are distinct and there is no p-th irreducible character of degree p s since p ⋅ p2s + |Lin(G)| > p n = |G|). As G is a D-group, we get p − 1 = 1, hence p = 2. Since n − 1 is even, n − 2 is odd. Hence G/R is abelian (otherwise G/R is a D-group of order 22s , a contradiction). Thus, R = G󸀠 = Z(G) is of order 2, and G = ES(m, p) (indeed, if x, y ∈ G, then 1 = [x, y]2 = [x, y2 ] so that ℧1 (G) ≤ Z(G); then Φ(G) = G󸀠 ℧1 (G) ≤ Z(G) and hence G󸀠 = Z(G) = Φ(G)). Exercise 11.5. If G/R is abelian and G is nonnilpotent, then G = (Cs , R). Solution. We have R = G󸀠 and G = A ⋅ R, where A < G is maximal in G. If λ ∈ Lin# (R), then all irreducible constituents of λ G are nonlinear and have the same degree so λ G = eχ for some χ ∈ Irr1 (G) and e ∈ ℕ. It follows that χ vanishes on G − R hence on A# , and we conclude that |A| | χ(1) so that χ(1) = |A|. In this case, G is a Frobenius group with kernel R (Corollary 4.2). Exercise 11.6. If G/R ≅ ES(m, q), then q = 2 and G = (Q8 , E32 ). Solution. Since G is nonnilpotent (Exercise 11.4), we have q ≠ p, and R < G󸀠 . Indeed, by Lemma 11.2, G = (Q8 , R). Since R is a G-class, we get R ≅ E32 .

12 The Brauer–Suzuki theorem If a Sylow 2-subgroup S of a group G has only one involution, it is either cyclic or generalized quaternion. If S is cyclic, then G is 2-nilpotent. The goal of this section is to prove the following remarkable result of R. Brauer and M. Suzuki [BrS]. Theorem 12.1 (R. Brauer and M. Suzuki [BrS]). Let a generalized quaternion group S be a Sylow 2-subgroup of a group G. If O(G) is a maximal normal subgroup of odd order in G and t ∈ S is an involution, then G = CG (t)O(G). In particular, Z(G/O(G)) = ⟨tO(G)⟩, and if O(G) = {1}, then ⟨t⟩ = Z(G).

X Existence of normal subgroups | 511

All previous non-simplicity criteria, distinct of Frobenius’ theorem, have produced nontrivial normal subgroups with solvable factor-group. The original proof of this theorem is based on ordinary character theory if 16 divides |G|, and based on modular theory if |G|2 = 8. G. Glauberman [Gla3] gave the proof in the second case using ordinary character theory, and this proof is presented in Section 2°. Note that S/Z(S) is either abelian of type (2, 2) or dihedral. The simple groups whose Sylow 2-subgroups are either abelian of type (2, 2) or dihedral are classified and play an important role within the first years of the classification project. In what follows a generalized quaternion group S of order 2n+1 , n ≥ 2, is a Sylow 2-subgroup of our group G. Remark. In the case when G is solvable, the proof of Theorem 12.1 is fairly easy. Indeed, without loss of generality, one may assume that O(G) = {1}. Then O2 (G) > {1}. Let N ≤ O2 (G) be minimal G-invariant. Since O2 (G) is either cyclic or generalized quaternion, we get |N| = 2, and this completes the proof. The proof of the theorem splits into several lemmas. Assuming that G is a counterexample of minimal order, we show that this assumption leads to a contradiction. Lemma 12.2. The group G is simple. Proof. Suppose that O(G) > {1}. Then, by induction, Z(G/O(G)) > {1}. By Sylow’s theorem and Frattini’s lemma, G = CG (t)O(G) (since NG (⟨t⟩) = CG (t)). Thus O(G) = {1}. Suppose that M is a nontrivial normal subgroup of G; then |M| is even, by the previous paragraph. Since S ∩ M ∈ Syl2 (M), it follows that S ∩ M is either cyclic or a generalized quaternion group. Being characteristic in M, the subgroup O(M) ⊲ G; then, by the previous paragraph, O(M) ≤ O(G) = {1}. If S ∩ M is cyclic, then M has a normal 2-complement coinciding with O(M) = {1}. Then ⟨t⟩ ≤ Z(M) is characteristic in the cyclic subgroup M, so that t ∈ Z(G), contrary to the assumption. If S ∩ M is noncyclic, then it is generalized quaternion so, by induction, t ∈ Z(M) (since O(M) = {1}), so that t ∈ Z(G) since ⟨t⟩ is characteristic in M ⊲ G. Since G is a counterexample, M does not exist so G is simple, as was to be shown. Lemma 12.3. If u, v ∈ G are distinct involutions, then o(uv) is odd and so u, v are conjugate in G. Proof. The result follows from the fact that ⟨u, v⟩ is a dihedral group with a cyclic Sylow 2-subgroup, say L (indeed, L is isomorphic to a subgroup of S) so |L| = 2. Then u, v are conjugate in ⟨u, v⟩, by Sylow’s theorem. Let K1 denote the set of all involutions in G; by Lemma 12.3, K1 is a G-class. Let CL(G) = {K0 = {1}, K1 , . . . , K r−1 } be the set of all G-classes, let kν = ∑ z z∈K ν

512 | Characters of Finite Groups 1 be the ν-th class sum, ν = 0, 1, . . . , r − 1, and r−1

(1)

ν k1 k1 = ∑ h11 kν ν=0

ν where h11 ∈ ℕ ∪ {0} (see formula (III.7.1)). Obviously, h01,1 = |K1 |. If the G-class K ν (ν > 1) consists of elements of even order, then Lemma 12.3 ν implies that h11 = 0 (indeed, in our group G each dihedral subgroup is a Frobenius group with kernel of index 2 so all its elements of even order are involutions). If u ∈ K1 , and z ∈ K ν , then, as we know (see the displayed formula in the statement of Lemma III.7.1), ν h11 = |G|−1 |K1 |2

∑ χ∈Irr(G)

χ(u)2 χ(z) . χ(1)

(2)

ν Denoting β(z) = h11 for z ∈ K ν , we see that β is a class function since χ is. We may rewrite formula (2) in the following form:

β = |G|−1 |K1 |2

∑ χ∈Irr(G)

χ(u)2 χ. χ(1)

(3)

As χ(u) and β are real (u is an involution), we may rewrite (3) as follows: β|G| ⋅ |K1 |−2 =

∑ χ∈Irr(G)

χ(u)2 χ. χ(1)

(3’)

Hence, by the First Orthogonality Relation, we obtain |G| ⋅ |K1 |−2 ⟨β, χ⟩ =

χ(u)2 χ(1)

(χ ∈ Irr(G)).

(4)

1°. In this subsection we consider the case n > 2 of Theorem 12.1, i.e., |S| ≥ 16. Write S = ⟨x, y | x2 = 1, y2 = x2 n

and X = ⟨x⟩,

T = ⟨x2 ⟩,

n−1

, x y = x−1 ⟩ ≅ Q2n+1 ,

R = ⟨x4 ⟩,

C = CG (T),

n > 2,

N = NG (T).

Since X ∈ Syl2 (C) is cyclic, we have C = X ⋅ H, a semidirect product with kernel H (Burnside). Since Aut(T) is a 2-group and |S : X| = 2, we obtain |N : C| = 2, so that N = S ⋅ H, a semidirect product with kernel H ⊲ C hence H ⊲ N. Lemma 12.4. The set A = C − RH is a TI-subset in G and NG (A) = N. Proof. The set A consists of all elements of C = X ⋅ H having order 2n−1 or 2n . Let z ∈ A ∩ A w , w ∈ G. Since T is the unique subgroup of order 2n−1 in C and 2n−1 | o(z), it follows that T = ⟨z m ⟩ for some odd m ∈ ℕ. Similarly, T w = ⟨z m ⟩. Then T = T w , and we conclude that w ∈ NG (T) = N. Since N normalizes A (indeed, N normalizes C and RH so it normalizes C − RH = A), A is a TI-subset in G. But if A u = A for some u ∈ G, then T u = T so u ∈ NG (T) = N. Indeed, then u ∈ NG (⟨A⟩) = NG (C) ≤ N = NG (T), and we conclude that NG (A) = N (as we know, NG (A) ⊇ A for a TI-subset A), completing the proof.

X Existence of normal subgroups | 513

Let ϕ ∈ Lin(C/RH) be faithful (note that C/RH is cyclic of order 4); we consider ϕ as a (linear) character of C and let ϕ N = ϕ.̃ Since RH is characteristic in C ⊲ N, we get ̃ = ker(ϕ)N = RH. Since N/RH ≅ S/R ≅ D8 is nonRH ⊲ N, and we conclude that ker(ϕ) ̃ ̃ abelian, we obtain that ϕ ∈ Irr(N/RH) and so ϕ̃ ∈ Irr(N); next, ϕ(1) = |N : C|ϕ(1) = 2. N Since (1C ) = ρ N/C is the regular character of the quotient group N/C of order 2, it follows that (1C )N = 1N + λ, where λ ∈ Irr# (N/C). Put (1C )N = 1̃ C . Lemma 12.5. If θ = 1̃ C − ϕ,̃ then the following statements hold: (a) ⟨θ, θ⟩ = 3. (b) θ(1) = 0 and θ vanishes on N − A. (c) θ G = 1G + χ1 − χ, where χ1 , χ ∈ Irr# (G) are distinct. In particular, χ(1) = 1 + χ1 (1). Proof. (a) Since θ = 1N + λ − ϕ̃ and the irreducible characters 1C , λ and ϕ̃ of N are pairwise distinct (indeed, 1C ≠ λ are linear and ϕ̃ is nonlinear), (a) follows. ̃ ̃ = C ∩ RH = RH, (b, c) We have θ(1) = 1̃ C (1) − ϕ(1) = 2 − 2 = 0. As ker(1̃ C ) ∩ ker(ϕ) ̃ it follows that θ = 1̃ C − ϕ̃ vanishes on RH = C − A since 1̃ C (1) = 2 = ϕ(1). In addition, N θ = (1C − ϕ) vanishes on N − C. Therefore θ vanishes on (C − A) ∪ (N − C) = N − A. By (a), ⟨θ G , θ G ⟩ = ⟨θ, θ⟩ = 3 since A is a TI-subset (see §X.6). Next, by reciprocity, we have ⟨θ G , 1G ⟩ = ⟨θ, 1N ⟩ = 1, and therefore θ G = 1G + χ1 − χ, where χ1 , χ ∈ Irr# (G) are distinct since θ G (1) = |G : N|θ(1) = 0. Next, θ(1) = 0 implies χ(1) = 1 + χ1 (1). This proves (b) and (c). Since θ vanishes on N − A, we see that θ G vanishes on G − ⋃w∈G A w , by Lemma 6.2 (a). But the set A(= N − RH) does not contain involutions and elements of odd order since all such elements of N are contained in RH. Therefore, the following lemma is true: Lemma 12.6. If x ∈ G is either an involution or an element of odd order, then θ G (x) = 0 so that χ(x) = 1 + χ1 (x). Proof. Indeed, (1G + χ1 − χ)(x) = θ G (x) = 0, and the lemma follows. Now we are ready to complete the proof of Theorem 12.1 for n > 2. Proof of Theorem 12.1. Since the class function β (see (3)) vanishes on all elements of even order (see the paragraph following (1)) and θ G vanishes on all elements of odd order, by Lemma 12.6, it follows that β(w)θ G (w) = 0 for all w ∈ G. Summing the last equality over w ∈ G and taking into account that β is real, we obtain ⟨β, θ G ⟩ = 0. Since θ G = 1G + χ1 − χ, it follows from (4) that χ(u)2 χ1 (u)2 =1+ χ(1) χ1 (1)

(u ∈ K1 ).

(5)

Substituting χ1 (u) = χ(u) − 1, χ1 (1) = χ(1) − 1 (u ∈ K1 ) in (5), we obtain (χ(u) − χ(1))2 = 0; hence, χ(u) = χ(1) and so u ∈ ker(χ). Since χ ≠ 1G , it follows that G is not simple, contradicting Lemma 12.2. This proves the theorem for n > 2.

514 | Characters of Finite Groups 1 2°. In this subsection we consider the more difficult case n = 2 of Theorem 12.1; then S ≅ Q8 is the ordinary quaternion group. As we have noted, the presented proof is due to Glauberman [Gla3]. Some lemmas in this subsection have the same proofs as corresponding lemmas in Section 1° (for example, Lemmas 12.4 and 12.4󸀠 ). We introduce some additional notation: C∗ = CG (X),

N ∗ = NG (X),

H ∗ = O(C∗ ),

T = Ω1 (S) = ⟨t⟩;

as above X = ⟨x⟩ is one of cyclic subgroups of order 4 in S. Then C∗ = X × H ∗ and N ∗ = S ⋅ H ∗ , a semidirect product with kernel H ∗ and |N ∗ : C∗ | = 2, by the N/C-theorem. Put A∗ = C∗ − TH and B = ⋃ (A∗ )y . y∈G

All elements of the set have the same order 4. Therefore, by Sylow, B consists of all elements of order 4 in G. A∗

Lemma 12.4󸀠 . The set A∗ is a TI-subset in G, N ∗ = NG (A∗ ). The proof repeats that of Lemma 12.4. Let ϕ ∈ Lin(C∗ ) with ker(ϕ) = TH ∗ ; then o(ϕ) = 2 since |C∗ /TH ∗ | = 2. Denote ∗ 1̃ C∗ = (1C∗ )N ,

∗ ϕ̃ = ϕ N ,

θ = 1̃ C∗ − ϕ.̃

Since N ∗ /C∗ is of order 2, it follows that ϕ̃ = μ1 + μ2 , where μ1 , μ2 ∈ Lin# (N ∗ ) are different. Similarly, we obtain 1̃ C∗ = 1N ∗ + λ, where λ ∈ Lin# (N ∗ /C∗ ). Since the characters 1N ∗ , λ, μ1 , μ2 are pairwise distinct, this proves part (a) of the following lemma. Lemma 12.5󸀠 . The following statements hold: (a) ⟨θ, θ⟩ = 4. (b) θ(w) = 0 for all w ∈ N ∗ − A∗ . In particular, since 1 ∈ ̸ A∗ , we have θ(1) = 0. (c) There exist pairwise distinct χ1 , χ2 , χ3 ∈ Irr# (G) and signs ε i = ±1 (i = 1, 2, 3) such that θ G = 1G + ∑3i=1 ε i χ i . For parts (b) and (c), see the proof of Lemma 12.5. Lemma 12.7. All elements of order 4 are conjugate in G. Proof. By the Frobenius normal p-complement theorem (in our case, p = 2) and Lemma XII.2.1 (b), our (simple) group G possesses a minimal nonnilpotent subgroup H = ⟨w⟩ ⋅ S, where w induces an automorphism of order 3 on S. Since S ≅ Q8 contains exactly six elements of order 4 and |H : CH (z)| = 6 for every z ∈ S of order 4, it follows that all elements of S of order 4 are conjugate in H. But then the lemma follows from Sylow’s theorem. It follows from Lemma 12.7 that B (see the paragraph preceding Lemma 12.4󸀠 ) is the set of elements of G of order 4. Then G − B is the set of all elements of G whose orders are not equal to 4. From now on we let i vary from 1 to 3.

X Existence of normal subgroups | 515

Lemma 12.8. The following statements hold: (a) If w ∈ G − B, then θ G (w) = 1 + ∑3i=1 ε i χ i (w) = 0. (b) If w ∈ A∗ , then o(w) = 4. (c) If w ∈ B, then θ G (w) = 4. (d) For every involution u ∈ G, we have 3

1+∑ i=1

ε i χ i (u)2 = 0. χ i (1)

Proof. (a) This follows from the definition of θ G and parts (b)–(c) of Lemma 12.5󸀠 since B = ⋃y∈G (A∗ )y . ∗ (b) One has θ = (1C∗ − ϕ)N . Since w ∈ A∗ ⊂ C∗ = ker(1̃ C∗ ), it follows that 1̃ C∗ (w) = 1̃ C∗ (1) = 2. Since |C∗ : TH ∗ | = 2, we have ϕ = ρ C∗ /TH ∗ − 1C∗/TH ∗ so that ϕ(w) = −1. Therefore, ̃ ϕ(w) = ϕ(w) + ϕ(w t ) ̃ for some t ∈ N ∗ − C∗ . Since w t ∈ C∗ − TH ∗ , we have ϕ(w t ) = −1. Therefore, ϕ(w) = −2 ̃ and θ(w) = 1̃ C∗ (w) − ϕ(w) = 2 − (−2) = 4, completing the proof of (b). (c) Since θ G is a class function, we may assume that w ∈ A∗ . As A∗ is a TI-subset, it follows from (b) that θ G (w) = θ(w) = 4 (see Lemma 6.2 (a)). (d) Each element of B has order 4. Therefore, by Lemma 12.5󸀠 (a), θ(u) = 0 for u ∈ K1 so that ⟨β, θ G ⟩ = 0. An application of (a) and (4) gives the result. We now introduce further notation. Let x i = χ i (1),

y i = χ i (x2 ),

zi = xi − yi ,

x2

i = 1, 2, 3.

(x2 )

Since is an involution, y i ∈ ℤ (indeed, y i = χ i is a sum of square roots of 1). Therefore z i ∈ ℤ too. Lemma 12.2 implies that z i = x i − y i = χ i (1) − χ i (x2 ) > 0 for all i since G is simple. Lemma 12.9. We have 3

3

3

i=1

i=1

i=1

1 + ∑ εi xi = ∑ εi zi = ∑ εi ⋅

z2i = 0. xi

Proof. The set A∗ (hence also B) has only elements of order 4. Therefore, by Lemmas 12.5󸀠 (c) and 12.8 (a), we have 3

3

3

i=1

i=1

i=1

1 + ∑ ε i x i = 0 = 1 + ∑ ε i χ i (x2 ) = 1 + ∑ ε i y i . Hence, ∑3i=1 ε i z i = 0 since z i = x i − y i . Next, for any i ≤ 3, we have y2i (x i − z i )2 z2i = = − 2z i + x i . xi xi xi

516 | Characters of Finite Groups 1 Therefore, by Lemma 12.8 (d), 3

0 = 1 + ∑ εi ⋅ i=1

3 3 3 3 z2 z2 y2i = 1 + ∑ εi ⋅ i − 2 ∑ εi zi + ∑ εi xi = ∑ εi ⋅ i . xi xi xi i=1 i=1 i=1 i=1

Lemma 12.10. If χ ∈ Ch(G), then ⟨χ, θ G ⟩ = |H ∗ |−1 ∑u∈H ∗ χ(xu). Proof. By Lemma 12.8 (b), θ(a) = 4 for a ∈ A∗ . Furthermore, θ(b) = 0 for b ∈ N ∗ − A∗ (see Lemma 12.5󸀠 (b)). Therefore, by reciprocity, ⟨χ, θ G ⟩ = ⟨χ N ∗ , θ⟩ = |N ∗ |−1 ∑ 4χ(a) = |N ∗ |−1 ∑ (4χ(xu) + 4χ(x−1 u)). a∈A∗

Indeed,

u∈H ∗

A∗ = C∗ − TH ∗ = ⟨x⟩ × H ∗ − ⟨x2 ⟩ × H ∗ = xH ∗ ∪ x−1 H ∗ .

Next, |N ∗ | = 8|H ∗ |. Take w ∈ S − ⟨x⟩. Then (H ∗ )w = H ∗ , since H ∗ is normal in N ∗ ≥ S, and x w = x−1 . Therefore (xH ∗ )w = x w H ∗ = x−1 H ∗ . Hence, ∑ χ(x−1 u) = ∑ χ((x−1 u)w ) = ∑ χ(xu)

u∈H ∗

u∈H ∗

u∈H ∗

and ⟨χ, θ G ⟩ = |N ∗ |−1 ∑ 8χ(xu) = |H ∗ |−1 ∑ χ(xu) u∈H ∗

(recall that 8 =

|N ∗

:

u∈H ∗

H ∗ |).

Lemma 12.11. For i = 1, 2, 3 the following statements hold: (a) The values of the characters χ i are rational integers. (b) χ i (xu) = χ i (xu−1 ) for all u ∈ H ∗ . Proof. Recall that the values of the characters of G lie in the field ℚ(ω), where ω is a primitive |G|-th root of 1, and these values are algebraic integers (see §VIII.1). (a) Let us assume, for example, that χ1 (a) is irrational for some a ∈ G. Since ℚ(ω)/ℚ is a normal extension, there exists α ∈ Gal(ℚ(ω)/ℚ) such that χ1 (a)α ≠ χ1 (a). But then χ1α ≠ χ1 and χ1α ∈ Irr(G). Since θ G takes rational values, it follows that ⟨χ1α , θ G ⟩ = ⟨χ1 , θ G ⟩ = ε1 󳨐⇒ χ1α ∈ Irr(θ G ). Therefore, we get χ1α = χ j for some j ∈ {2, 3}. Then ε1 = ε j . To fix ideas, let j = 2. Since χ1 (x2 ) ∈ ℚ, we have y2 = χ2 (x2 ) = (χ1 (x2 ))α = χ1 (x2 ) = y1 . Similarly, x2 = x1 . But then z2 = x2 − y2 = x1 − y1 = z1 . By Lemma 12.9, 0 = 2ε1 z1 + ε3 z3 = 2ε1 ⋅ Therefore, z3 = −2ε1 ε3 z1 ,

0 = 2ε1 ⋅

z2 z21 + ε3 ⋅ 3 . x1 x3 z21 z2 + 4ε3 ⋅ 1 . x1 x3

X Existence of normal subgroups | 517

Since z1 > 0, it follows that ε3 ε1 +2 = 0, x1 x3

ε1 x3 = −2ε3 x1 .

By Lemma 12.9, 1 + 2ε1 x1 + ε3 x3 = 0. But ε1 x3 = −2ε3 x1 implies ε3 x3 = −2ε1 x1 , and we conclude that 1 + 2ε1 x1 − 2ε1 x1 = 0, which is a contradiction. (b) We have C∗ = X ⋅ H ∗ = X × H ∗ . Let α be an automorphism of the field ℚ(ω) that fixes a primitive 4-th root of 1 and maps a root of unity of odd degree to its inverse. Then χ i (xu) = (χ i (xu))α = χ i (xu−1 ). Lemma 12.12. The integers χ i (x) and ⟨χ2i , θ G ⟩ are odd for i = 1, 2, 3. Proof. Put χ = χ i . By Lemma 12.10, ε i = ⟨χ, θ G ⟩ = |H ∗ |−1 ∑ χ(xu). u∈H ∗

Let I be a subset of H ∗ that contains exactly one element of each pair {u, u−1 } of elements of (H ∗ )# . By Lemma 12.11 (a), χ takes rational integer values, and, by Lemma 12.11 (b), χ(x) + 2 ∑ χ(xu) = χ(x) + ∑ (χ(xu) + χ(xu−1 )) u∈I

u∈I

= ∑ χ(xu) = |H ∗ |⟨χ, θ G ⟩ = |H ∗ |ε i . u∈H ∗

Since |H ∗ | and ε i are odd, χ(x) is odd too. Applying Lemmas 12.10 and 12.11, we obtain |H ∗ |⟨χ2 , θ G ⟩ = ∑ χ2 (xu) = χ2 (x) + 2 ∑ χ2 (xu) ≡ χ2 (x) ≡ 1 (mod 2). u∈H ∗

u∈I

Lemma 12.13. If x ∈ S is of order 4, then χ i (x) = ε i for i = 1, 2, 3. Proof. Suppose that χ i (x) ≠ ε i for some i. Let χ i = χ. By Lemma 12.11 (a), χ takes rational integer values. Due to Lemma 12.12, χ(x) is odd. Therefore, either |χ(x)| > 1 or χ(x) = −ε i . In either case, χ(x)2 > ε i χ(x). Similarly, χ(z)2 ≥ ε i χ(z) for any z ∈ G (here we use Lemma 12.11 (a) again). Therefore, by Lemma 12.10, ⟨χ2 , θ G ⟩ = |H ∗ |−1 ∑ χ(xu)2 > |H ∗ |−1 ε i ∑ χ(xu) = ε i ⟨χ, θ G ⟩ = ε i ε i = 1. u∈H ∗

u∈H ∗

As, by Lemma 12.12, ⟨χ2 , θ G ⟩ is odd, we obtain ⟨χ2 , θ G ⟩ ≥ 3. Due to Lemma 12.2, we get χ(1) > 1. By Lemma 12.8, θ G (w) = 4 for w ∈ B, and θ G (w) = 0 for w ∈ G − B. Since ⟨χ, 1G ⟩ = 0, an argument like that of the previous paragraph gives ∑ χ(w)2 > ∑ (−ε i )χ(w) = ∑ ε i χ(w)

w∈G−B

w∈G−B

w∈B

1 1 1 = ∑ ε i χ(w)θ G (w−1 ) = |G|⟨ε i χ, θ G ⟩ = |G|, 4 w∈G 4 4

518 | Characters of Finite Groups 1 where we have used the equality ∑ ε i χ(w) + ∑ ε i χ(w) = ε i ⟨χ, 1G ⟩ = 0.

w∈B

w∈G−B

Hence, since χ = χ (Lemma 12.11), we obtain |G| = |G|⟨χ, χ⟩ = |G|⟨χ, χ⟩ = |G|⟨χ2 , 1G ⟩ 1 1 = ∑ χ(w)2 + ∑ χ(w)2 > |G| + ∑ χ(w)2 θ G (w−1 ) 4 4 w∈G w∈G−B w∈B 1 1 1 1 |G| + |G|⟨χ2 , θ G ⟩ ≥ |G| + 3 ⋅ |G| = |G|, 4 4 4 4 which is a contradiction. =

Lemma 12.14. Let i = 1, 2, 3. Then the following statements hold: (a) 4 | z i . (b) x i is odd. (c) 8 | z i implies 4 | x i − ε i . Proof. Put χ i = χ. Let ψ ∈ Irr1 (S). Then ψ(1) = 2, ψ(x2 ) = −2, ψ(w) = 0

(w ∈ S − ⟨x2 ⟩).

(The second equality follows since x2 ∈ Z(S)# , the third equality is true since ψ = μ S , where μ ∈ Lin# (S).) Since ⟨ψ, χ S ⟩ ∈ ℕ ∪ {0} and χ(w) = ε i for all w ∈ S − ⟨x2 ⟩ (see Lemmas 12.7 and 12.13), it follows that 8⟨χ S , ψ⟩ = ∑ χ(w)ψ(w−1 ) = 2χ(1) − 2χ(x2 ) = 2x i − 2y i = 2z i . w∈S

Therefore 4 | z i , and (a) is proved. Similarly, since |S| = 8, we have 8⟨χ S , 1S ⟩ = ∑ χ(w) = x i + y i + 6ε i w∈S

≡ 2x i − z i − 2ε i ≡ 2(x i − ε i ) − z i ≡ 0 (mod 8) since ⟨χ S , 1S ⟩ ∈ ℕ ∪ {0}. By (a), 4 | z i , therefore 4 | 2(x i − ε i ), and so x i − ε i is even and x i is odd. Finally, if 8 | z i , then 8 | 2(x i − ε i ) and 4 | x i − ε i . Lemma 12.15. There exists i ∈ {1, 2, 3} such that 8 | z i and 4 | x i + ε i . Proof. Let 2k be the highest power of 2 dividing all z i (i = 1, 2, 3), i.e., 2k is a 2-part of GCD(z1 , z2 , z3 ). Then, by Lemma 12.14 (a), one has k ≥ 2. Let w i = 2zki , i = 1, 2, 3. Then a certain w i , say w3 , is odd. By Lemma 12.9, 3

∑ εi ⋅ i=1

3 z2i w2 = 0 = ∑ εi ⋅ i , xi xi i=1

0 = ε1 x2 x3 w21 + ε2 x1 x3 w22 + ε3 x1 x2 w23 .

(6)

By Lemma 12.14 (b), all the x i are odd. Therefore, a certain w i , say w1 , is even. Since z1 = 2k w1 and k ≥ 2, it follows that 8 | z1 . Furthermore, w2 is odd. Thus, we obtain w2i ≡ 1 (mod 8) for i = 2, 3.

X Existence of normal subgroups | 519

It follows from (6) that 0 ≡ −ε1 x2 x3 w21 ≡ ε2 x1 x3 w22 + ε3 x1 x2 w23 ≡ ε2 x1 x3 + ε3 x1 x2 ≡ ε2 ε3 x1 (ε3 x3 + ε2 x2 ) (mod 4). Therefore, 4 | ε2 x2 + ε3 x3 since x1 is odd. By Lemma 12.9, 1 + ∑3i=1 ε i x i = 0. Hence, 0 ≡ ε2 x2 + ε3 x3 ≡ −(1 + ε1 x1 ) ≡ −ε1 (ε1 + x1 ) (mod 4), so that 4 | x1 + ε1 . As Lemma 12.15 contradicts Lemma 12.14 (c), this completes the proof of Theorem 12.1. Glauberman proved the following stronger version of Theorem 12.1. Theorem 12.16 ([Gla1]). Let G be a group with O(G) = {1}, and let a ∈ P ∈ Syl2 (G). Put a G = {a x | x ∈ G}. Then a ∈ Z(G) if and only if P ∩ a G = {a}. Glauberman also derived the following interesting consequence of Theorem 12.16. Theorem 12.17 ([Gla1]). Let P be a Sylow 2-subgroup of a group G that is the direct product of a generalized quaternion group and a group whose involutions are in its center. Then G is not simple. Yet another corollary of Theorem 12.16 is the following: Theorem 12.18 ([Gla1]). Let G be a group with O(G) = {1}, and let S ∈ Syl2 (G). Let C = {α ∈ Aut(G) | x α = x for all x ∈ S}. Then the following statements hold: (a) A Sylow 2-subgroup of C is abelian. (b) C is 2-nilpotent. In particular, if O(G) = {1} and Aut(S) is solvable, then Aut(G)/Inn(G) is solvable.

XI On sums of degrees of irreducible characters 1 Introduction We begin with some notation, Let G, H be groups of orders g, h, respectively (usually H ≤ G); let G󸀠 , H 󸀠 be their commutator subgroups, of orders g 󸀠 , h󸀠 , respectively; cd(G) = {χ(1) | χ ∈ Irr(G)}. Define the following functions G → ℚ: t(G) = |{x ∈ G | x2 = 1}|,

T(G) =

∑

χ(1),

χ∈Irr(G)

f(G) =

T(G) , g

mc(G) =

k(G) , g

where k(G) is the class number of G. We have |G| =

∑ χ∈Irr(G)

χ(1)2 ≥

∑

χ(1) = T(G)

χ∈Irr(G)

with equality if and only if χ(1) = 1 for all χ ∈ Irr(G) if and only if T(G) = |G| if and only if f(G) = 1 if and only if G is abelian. It is clear that 0 < f(G) ≤ 1. A prime divisor p of the order of a nonabelian group G is essential if a Sylow p-subgroup P of G does not belong to Z(G). If P ∈ Sylp (G) is not essential, then G = A × G1 , where A is abelian, GCD(|A|, |G1 |) = 1 and the least prime divisor of |G1 | is the essential prime divisor of |G| The following theorem holds (for its proof, see [BZ3, §XI.1]). Theorem 1.0 (Berkovich–Nekrasov [NB1]). Suppose that p is the least essential prime divisor of the order g of a nonabelian group G. If f(G) > 1p , then one of the following holds: (a) G contains an abelian subgroup of index p. (b) g󸀠 = p. (c) p = 2, G = G1 ∗ G2 (central product), g1󸀠 = g2󸀠 = 2, where g i = |G i |,

g 󸀠i = |G󸀠i |, f(G i ) =

3 4

(i = 1, 2),

g 󸀠 = 4.

(d) cd(G) = {1, p}, G/Z(G) is of order p3 and exponent p. Conversely, if G is one of the groups (a)–(d), then f(G) > 1p . Exercise 1.1. Check the last statement of the theorem. Solution. Let us check that the group G of part (a) satisfies the conclusion. Set |Z(G)| = z and let A be an abelian subgroup of index p in G; then A ⊲ G. We have g = pg󸀠 z (Lemma 3.1, below) so that |Lin(G)| = gg󸀠 = pz. By Theorem VII.2.3, we get cd(G) = {1, p} so that 1 g g z |Irr1 (G)| = 2 (g − 󸀠 ) = 2 − . g p p p DOI 10.1515/9783110224078-011

522 | Characters of Finite Groups 1 It follows that T(G) = |Lin(G)| + p|Irr1 (G)| = pz + In that case, f(G) =

1 p

+

z(p−1) g

g g − z = + z(p − 1). p p

> 1p .

Let us find k(G) for a group G of Exercise 1.1 (a) (we use the above notation). If u ∈ A − Z(G), then |G : CG (u)| = |G : A| = p, so A is the union of z + 1p ( pg − z) = z + pg2 − pz g G-classes. If u ∈ G − A, then |CG (u)| = pz, so we get |G : CG (u)| = pz , and hence the g−g/p set G − A is the union of g/pz = pz − z conjugate classes. It follows that g z g z k(G) = z + 2 − + pz − z = 2 − + pz. p p p p For example, if G is a p-group of maximal class of order p m with abelian subgroup of index p, then T(G) = p m−1 + p(p − 1),

f(G) =

1 p−1 + , p p m−1

k(G) = p m−2 + p2 − 1.

If G is as in part (b) of the theorem, then |Lin(G)| = pg so that f(G) > pg and mc(G) > pg . The groups G with cd(G) = {1, p} (p is arbitrary) were classified in [Isa11, Theorem 12.11]. Such a G has either an abelian subgroup of index p or G/Z(G) is of order p3 and exponent p. In that case, T(G) =

g − gg󸀠 g g g 1 g + = + 󸀠 (1 − ) > . g󸀠 p p g p p

If G = A × B, then it follows from §IV.8 that T(G) = T(A)T(B),

f(G) = f(A)f(B),

mc(G) = mc(A) mc(B).

Therefore, if A is abelian, then f(G) = f(B) and mc(G) = mc(B). Our notation will differ in one additional respect from that generally accepted. If A, B ⊆ G, we define [A, B] = {[a, b] | a ∈ A, b ∈ B} (usually [A, B] = ⟨[a, b] | a ∈ A, b ∈ B⟩, but in this chapter the above definition is more convenient). In this notation, for x ∈ G, |[x, G]| = |G : CG (x)| ≤ g 󸀠 . Let us recall some known facts of group theory. Lemma 1.1. The following statements hold: (a) (Gaschütz) If N ⊴ G, then N is nilpotent if and only if NΦ(G)/Φ(G) is nilpotent. In particular, F(G/Φ(G)) = F(G)/Φ(G). (b) (Frobenius) If G has no normal p-complement, it contains a minimal nonnilpotent {p, q}-subgroup S with a normal Sylow p-subgroup, q is a prime. (c) (Fitting) If G = H ⋅ A is a semidirect product of Hall subgroups and A is abelian and normal in G, then A = CA (H) × A1 , where A1 = ⟨[h, a] | h ∈ H, a ∈ A⟩. (d) (Wielandt) G is nilpotent if and only if G󸀠 ≤ Φ(G). (e) If N ⊲ G, then Φ(N) ≤ Φ(G).

XI On sums of degrees of irreducible characters | 523

It is known (see Theorem IV.6.1), that t(G) ≤ T(G) with equality if and only if all irreducible representations of G are realizable over ℝ.

2 Minimal nonnilpotent groups In this section we present information on the structure of minimal nonnilpotent groups. The results of Lemma 2.1 are due to O. Y. Schmidt and Y. A. Gol’fand (L. Redei has classified such groups, but we do not use his result). Recall that a group G is said to be minimal nonnilpotent if it is not nilpotent but all its proper subgroups are nilpotent. A p-group Q > {1} is said to be special if Z(Q) = Q󸀠 = Φ(Q) is elementary abelian (in particular, Q is nonabelian). A p-group is said to be extraspecial if it is special with center of order p. In that case, its order is p1+2n for some n ∈ ℕ. Lemma 2.1. Let G be a minimal nonnilpotent group. Then |G| = p a q β , where p and q are distinct prime numbers and a, β ∈ ℕ. Let P ∈ Sylp (G), Q ∈ Sylq (G). The following statements hold: (a) One of the subgroups P, Q (say Q) is normal and coincides with G󸀠 . (b) If Q is abelian, then Q ∩ Z(G) = {1} and exp(Q) = p. (c) P is cyclic and |P : (P ∩ Z(G))| = p. (d) Let Q be nonabelian. Then Q is special and |Q/Q󸀠 | = q b , where b is the order of q (mod p), Q ∩ Z(G) = Z(Q). If L < Z(Q) is of index p, then Q/L is extraspecial so that b is even. (e) If q > 2, then exp(Q) = q. Proof. Let us show by induction on |G| that G is solvable. Assuming that this is false, one may suppose that G is a nonabelian simple group since proper epimorphic images of G are either nilpotent or minimal nonnilpotent. Let H, F be different maximal subgroups of G whose intersection D = H ∩ F has the highest possible order (if G contains only one maximal subgroup, it is cyclic). Suppose that D > {1}. Since H, F are nilpotent, it follows that NH (D) > D, NF (D) > D, and therefore NG (D) is contained neither in H nor in F. Let NG (D) ≤ R, where R is maximal in G (R exists since G is simple); then H ≠ R ≠ F. Since |H ∩ R| > |NH (D)| > |D|, this contradicts the choice of H and F. Thus D = {1} so that the intersection of any two different maximal subgroups of G is trivial. By Sylow’s theorem, there exist in G two maximal subgroups, say H, F again, of different orders. Then, as we have proved, 󵄨󵄨 󵄨󵄨 |G# | ≥ 󵄨󵄨󵄨󵄨 ∑ (H # )x + (F # )x 󵄨󵄨󵄨󵄨 = |G : H| ⋅ |H # | + |G : F| ⋅ |F # | 󵄨 󵄨 x∈G

= 2|G| − |G : H| − |G : F|, i.e., |G : H| + |G : F| ≥ |G| + 1, which is impossible. Thus, G is solvable.

524 | Characters of Finite Groups 1 But then there is (a nilpotent) H ⊲ G of prime index p. By Fitting’s lemma, H = F(G), the Fitting subgroup of G; then H is the only normal subgroup of prime index in G. Let Q be a p󸀠 -Hall subgroup of H. Then Q is a normal p-complement in G. Let P ∈ Sylp (G), |P| = p a . Then G/Q ≅ P, and G/Q contains only one subgroup of index p. Therefore P is cyclic. If P0 = P ∩ H, then CG (P0 ) ≥ ⟨P, Q⟩ = G and P0 ≤ Z(G). Clearly, P0 = P ∩ Z(G). Suppose that |Q| is not a power of a prime. Then Q = Q1 × Q2 , where Q1 ∈ Syl(G). By assumption, PQ1 < G and PQ2 < G are nilpotent; therefore, we have PQ1 = P × Q1 and PQ2 = P × Q2 , and so CG (P) = G (since P is cyclic) and G = P × Q is nilpotent, a contradiction. Thus, Q ∈ Sylq (G), where q ≠ p is a prime. Since G󸀠 ≤ Q and G contains only one normal subgroup of prime index, we obtain Q = G󸀠 . Let Q0 ⊲ G be such that Q0 < Q. Let P, P1 ∈ Sylp (G) be different. As PQ0 , PQ1 < G are nilpotent, it follows that CG (Q0 ) ≥ ⟨P, P1 ⟩ = G (since the subgroup ⟨P, P1 ⟩, containing two different Sylow p-subgroups, is nonnilpotent), and so Q0 ≤ Z(G). Suppose that Q is abelian. Since G󸀠 = Q, it follows from Lemma 1.1 (c) that Q ∩ Z(G) = {1}, and so, by the result of the previous paragraph, Q is a minimal normal subgroup of G. In particular, Q is elementary abelian and P is maximal in G. Let |Q| = q β , and let b be the order of q (mod p), P = ⟨x⟩ (note that x p ∈ Z(G)). If v0 ∈ Q# , p−1 p−1 v = v0 v0x ⋅ ⋅ ⋅ v0x , then v x = v. Since x normalizes Q1 = ⟨v0 , v0x , . . . , v0x ⟩, it follows that Q1 ⊲ G; therefore, by the above, Q1 = Q. The equality v x = v implies that v = 1, since v ∈ Z(G) ∩ Q = {1}. Therefore, β < p. Furthermore, P does not normalize nontrivial subgroups of Q. Let β = kb + t, 0 ≤ t < b. Then the number s of elements of order q in Q is divisible by p. Since s = |Q# | = q β − 1, we get β ≡ 0 (mod b). Thus, t = 0 and β = kb. Suppose that k > 1. The number of subgroups of order q b in Q is equal to d=

(q kb − 1) ⋅ ⋅ ⋅ (q kb − q b−1 ) . (q b − 1) ⋅ ⋅ ⋅ (q b − q b−1 )

Note that we have q kb − q i = q i (q kb−i − 1) ≢ 0 (mod p) for all i = 1, . . . , b − 1, since kb − i ≢ 0 (mod b). Since P does not normalize subgroups of Q of order q b , it follows that p | d, so that q kb − 1 = q b(k−1) + ⋅ ⋅ ⋅ + q b + 1 ≡ k (mod p) 󳨐⇒ k ≥ p. qb − 1 But then β ≥ p, contrary to what has just been said. Thus, k = 1, β = b and |Q| = q b . Now suppose that Q is nonabelian. Then, by the above, we have Φ(Q) ≤ Z(G) so Φ(Q) ≤ Z(Q). Since Q󸀠 ≤ Φ(Q), it follows, by what has been proved already, that the strict inequality Φ(Q) < Z(Q) is impossible (see (b)). Hence, Φ(Q) = Z(Q). In that case, Q󸀠 = Z(Q) so that Q is special. If L is a subgroup of Z(Q) of index q, then, as we have already proved, Q/L is special so extraspecial, and we conclude that b is even.¹ p|

1 Thus, if b is odd, then Q is elementary abelian.

XI On sums of degrees of irreducible characters | 525

Assume that q > 2 and Q is nonabelian. Then Q is regular in Hall’s sense. Let L be as in the previous paragraph. Then Ω1 (Q/L) > Z(Q/L) so that Ω1 (Q/L) = Q/L. It follows, in view of regularity Q/L, that exp(Q/L) = p. Since Q is isomorphic to a subgroup of the direct product of groups Q/M, where M runs over all subgroups of Z(Q) of index p, we conclude that exp(Q) = q. Definition. If G is the group of Lemma 2.1 and |Q ∩ Z(G)| = q c , then G is called an S(p a , q b , q c )-group (where |P| = p a , |Q| = q β = q b+c ). In particular, we have that S(p a , q b , q c )-group is minimal nonabelian if and only if c = 0. Y. A. Gol’fand (see [BZ3, Lemma XI.2]) has shown that c ≤ 2b , but we omit the proof of this result. Write Irr(m) (G) = {χ ∈ Irr(G) | χ(1) = m}. Lemma 2.2. Let G = PQ be an S(p a , q b , q c )-group, P ∈ Sylp (G), Q = G󸀠 ∈ Sylq (G) and χ ∈ Irr1 (G). Then the following statements hold: b b (a) If Q ∩ Z(G) ≤ ker(χ), then χ(1) = p; otherwise, χ(1) = q 2 . Thus, cd(G) ⊆ {1, p, q 2 }. (b) We have, provided c > 0, that mc(G) =

q b + p2 q c − 1 , p2 q b+c

b

f(G) =

b

q b + pq c+ 2 + p − pq 2 − 1 . pq b+c

(c) If H ≤ Z(G), then f(G/H) ≥ f(G). If c > 0 and {1} < H < Z(Q), then b

b

f(G/H) = f(G) ⇐⇒ p = 1 + q 2 , q = 2, f(G) = 2− 2 . (d) If p > q and f(G) ≥ 1q , then p = 3,

q = 2,

G/(P ∩ Z(G)) ∈ {A4 , SL(2, 3)},

f(G) =

1 . 2

Proof. (a) If Q ∩ Z(G) ≤ ker(χ), then G/ker(χ) contains a normal abelian subgroup of index p; then χ(1) = p, by Theorem VII.2.3. Now let Q ∩ Z(G) ≰ ker(χ); in that case, Q ∩ Z(G) > {1}. We have to prove that b χ(1) = q 2 . Without loss of generality, we may assume that ker(χ) = {1}. Then Z(G) is b cyclic so that c = 1, Q is extraspecial of order q1+b , cd(Q) = {1, q 2 }. Since χ is faithb ful and Q is nonabelian, it follows that χ Q has no linear constituents and q 2 | χ(1) b (Clifford’s theorem). As χ(1)2 ≤ |G/Z(G)| = pq b (Lemma III.5.4), we have χ(1) = q 2 , completing the proof of (a). (b) By (a), Irr(G) = Lin(G) ∪ Irr(p) (G) ∪ Irr(q b/2 ) (G) is a partition. As a consequence of Lemma 2.1 (b), |Lin(G)| = p a . By (a), |Irr(p) (G)| =

|G/(Q ∩ Z(G))| − p a p a q b − p a = = p a−2 (q b − 1). p2 p2

Hence, provided c > 0, then |Irr(q b/2 ) (G)| =

|G| − |G/(Q ∩ Z(G))| p a q b+c − p a q b = = p a (q c − 1), qb qb

526 | Characters of Finite Groups 1 and k(G) = p a + p a−2 (q b − 1) + p a (q c − 1) = p a−2 (q b + p2 q c − 1), mc(G) =

k(G) q b + p2 q c − 1 = , |G| p2 q b+c b

b

b

T(G) = p a + p ⋅ p a−2 (q b − 1) + q 2 ⋅ p a (q c − 1) = p a−1 (p + q b + pq c+ 2 − pq 2 − 1), b

b

T(G) q b + pq c+ 2 − pq 2 + p − 1 , f(G) = = |G| pq b+c and the proof of (b) is complete. It follows that mc(G) and f(G) are independent of a. Next we assume that a = 1; then |G| = pq b+c . (c) If H ≤ P ∩ Z(G), then f(G/H) = f(G), by the last sentence in (b). Now suppose that {1} < H ≤ Z(Q). Then G/H is an S(p a , q b , q c1 )-group for some nonnegative integer c1 < c. By (b), b

b

b

f(G/H) − f(G) = p−1 q− 2 (q−c1 − q−c )(1 − q− 2 )(q 2 + 1 − p). b

As we know, b is even. It follows from the definition of b that q 2 + 1 ≡ 0 (mod p), so b that q 2 + 1 ≥ p. Thus, f(G/H) ≥ f(G). b Let f(G/H) = f(G). Then p = 1 + q 2 , q = 2. It is easily checked that in that case, b

b

p2b+c f(G) = |G|f(G) = 2b + p2c+ 2 + (p − 1) − p2 2 b

b

b

b

= 2 2 (2 2 + 1) − p2 2 + p2c+(b/2) = p2c+ 2 . b

It follows that f(G) = 2− 2 . (d) Let p > q and f(G) ≥ 1q . As above, we have a = 1 so |G| = pq b+c . Let c = 0. Then |G| = pq b and, by (b), T(G) = p + q b − 1, and we have T(G) = |G|f(G) ≥ pq b ⋅ q−1 = pq b−1 . It follows that q b + p ≥ pq b−1 + 1. Suppose that p > q + 1. Then b > 1 and q b ≥ p(q b−1 − 1) + 1 > (q + 1)(q b−1 − 1) + 1 = q b + q b−1 − q, and hence q > q b−1 , contradicting b > 1. Thus, by (b), p = q + 1,

q = 2,

p = 3,

b = 2,

f(G) =

1 , 2

G ≅ A4 .

If a > 1, then again f(G) = 12 (by (b)) and G/(P ∩ Z(G)) ≅ A4 . From now on, we assume that c > 0. By (c), we may assume that c = 1; then, |G| = pq b+1 and, by (b), q b + pq1+b/2 + p − pq 2 − 1 . pq b+1 b

f(G) = If b = 2, then q = 2,

p = 3,

G/(P ∩ Z(G)) ≅ SL(2, 3),

f(G) = f(G/(Z(G) ∩ P)) =

1 . 2

XI On sums of degrees of irreducible characters | 527

Now let b > 2. Then b ≥ 4 (since b is even, in view of c > 0), and so p ≥ 5. Since, by hypothesis, |G|(f(G) − 1q ) ≥ 0, it follows that b

b

(p − 1)(q b − 1) ≤ pq 2 (q − 1) < p(q 2 + 1)(q − 1) b

so that (p − 1)(q 2 − 1) < p(q − 1), and we conclude that b

p q2 − 1 > ≥ q + 1 ≥ 3, p−1 q−1 which is impossible. Thus, b = 2, and the proof of (d) is complete. Lemma 2.3. Let G be a minimal nonabelian p-group. Then the following statements hold: 2 (a) mc(G) = p +p−1 . p3 (b) f(G) =

2p−1 . p2

Proof. Let |G| = p n . In that case, we have |G󸀠 | = p, |G : Z(G)| = p2 and |G : CG (x)| = p for all x ∈ G − Z(G). It follows that k(G) = |Z(G)| +

|G − Z(G)| p2 + p − 1 = p n−2 + p n−1 − p n−3 󳨐⇒ mc(G) = . p p3

Since cd(G) = {1, p} (Theorem VII.2.3), we get T(G) = |G : G󸀠 | +

2p − 1 |G| − |G/G󸀠 | ⋅ p = 2p n−1 − p n−2 󳨐⇒ f(G) = . p2 p2

Exercise. Let G be an S(p a , q b , q0 )-group (i.e., G is minimal nonabelian). If b is odd, then G has a trivial Schur multiplier. If b is even, then the Schur multiplier of G has b order p 2 . Hint. Any representation group of G is minimal nonnilpotent.

3 Some properties of functions f(∗) and mc(∗) We omit the very involved proof of Theorem 1.0 in nilpotent case however we present here a lot of results which used in its proof (see [BZ3, Chapter 11]). Lemma 3.1. If G is nonabelian and H ⊲ G is abelian of index p, then g = pg󸀠 |Z(G)|, If x ∈ G − H, then |G : CG (x)| =

g pz

|G : CG (x)| = g 󸀠

(x ∈ G − H).

= g󸀠 .

Proof. By agreement, g = |G|,

h = |H|,

g 󸀠 = |G󸀠 |,

h󸀠 = |H 󸀠 |,

z = |Z(G)|.

528 | Characters of Finite Groups 1 By Theorem VII.2.3, cd(G) = {1, p}. Obviously, Z(G) < H, and for all x ∈ H − Z(G) we have |G : CG (x)| = p. But if x ∈ G − H, then x p ∈ Z(G), and so CG (x) = ⟨x, Z(G)⟩ is of order pz. Therefore, k(G) = z +

g p

−z p

+

(g − pg )pz g

g − pz p2

(1)

g − gg󸀠 g + . g󸀠 p2

(2)

= pz +

(note that h = pg ). On the other hand, k(G) = |Irr(G)| = |Lin(G)| + |Irr1 (G)| = Comparing (1) and (2), we get g = g󸀠 . |G : CG (x)| = pz

g g󸀠

= pz and g = pzg 󸀠 . Hence, if x ∈ G − H, then we have

Lemma 3.1 can be proved without character theory using [Isa11, Lemma 12.12]. Lemma 3.2. Let H ⊲ G be of index p, G󸀠 > {1}. Then the following statements hold: (a) One has mc(H) p + 1 + ⋅ ∑ |CG (x)|. mc(G) = p2 pg2 x∈G−H (b) If H is abelian, then mc(G) =

1 p2

+

p2 −1 . p2 g󸀠

Proof. (a) Define K = {(a, b) | a, b ∈ G, ab = ba} 󳨐⇒ |K| = ∑ |CG (x)|, x∈G

K0 = {(a, b) ∈ K | a, b ∈ H} 󳨐⇒ |K0 | = ∑ |CH (x)|, x∈H

K1 = {(a, b) ∈ K | a ∈ H, b ∈ G − H} so that |K1 | = ∑x∈H |CG (x)| − ∑x∈H |CH (x)| = K − K0 , K1󸀠 = {(a, b) ∈ K | a ∈ G − H, b ∈ H} 󳨐⇒ |K1󸀠 | = ∑ |CH (x)|, x∈G−H

K2 = {(a, b) ∈ K | a, b ∈ G − H} 󳨐⇒ |K2 | = ∑ |CG (x) − CH (x)|, x∈G−H

so that |K1 | = ∑x∈H |CG (x)| − ∑x∈H |CH (x)| = K − K0 , K1󸀠 = {(a, b) ∈ K | a ∈ G − H, b ∈ H} 󳨐⇒ |K1󸀠 | = ∑ |CH (x)|, x∈G−H

K2 = {(a, b) ∈ K | a, b ∈ G − H} 󳨐⇒ |K2 | = ∑ |CG (x) − CH (x)|. x∈G−H

If K ∈ CL(G) is a G-class and y ∈ K, then ∑ |CG (x)| = |CG (y)||K| = |G|. x∈K

XI On sums of degrees of irreducible characters | 529

Therefore, |K| = ∑ |CG (x)| = x∈G

∑

∑ |CG (x)| = gk(G) = g 2 mc(G)

K∈CL(G) x∈K

and |K0 | = h2 mc(H) =

g2 mc(H). p2

Moreover, it is obvious that |K1 | = |K1󸀠 |, and K = K0 ∪ K1 ∪ K1󸀠 ∪ K2 is a partition, so that |K| = |K0 | + |K1 | + |K1󸀠 | + |K2 |. It is clear that |K| = ∑ |CG (x)| = ∑ |CG (x)| + ∑ |CG (x)| x∈G

x∈H

x∈G−H

and ∑ |CG (x)| = ∑ [|CH (x)| + (|CG (x)| − |CH (x)|)]

x∈H

x∈H

= |K0 | + |K1 | = |K0 | + |K1󸀠 | = ∑ |CH (x) + ∑ |CH (x)|. x∈G−H

x∈H

For any x ∈ G − H, we have |CG (x)| = p|CH (x)|. Hence, ∑ |CG (x)| = p ∑ |CH (x)|

x∈G−H

x∈G−H

and, therefore, g2 mc(G) = |K| = ∑ |CH (x)| + ∑ |CH (x)| + p ∑ |CH (x)| x∈H

x∈G−H

x∈G−H

= |K0 | + (p + 1) ∑ |CH (x)| x∈G−H

p+1 g2 = 2 mc(H) + ∑ |CG (x)|. p x∈G−H p Division of the above equality by g2 yields the desired result. (b) By Lemma 3.1, we have (see (2)) g ⋅ mc(G) = k(G) =

g g g g p2 − 1 g − 2 󸀠 + 󸀠 = 2 + , 2 g p p g p p2 g󸀠

completing the proof of (b). Lemma 3.3. Let H < G. Then: (a) (Ernest [Ern1]) One has mc(G) ≤ mc(H). (b) If there are α characters of degree n ∉ cd(H) in Irr(G), then f(H) ≥ f(G) + nα g . In particular, f(H) ≥ f(G). (c) One has f(G) = f(H) if and only if mc(G) = mc(H). Then G󸀠 = H 󸀠 and all normal subgroups of H are normal in G. (d) If H ⊲ G, then f(G) ≤ f(H) − h−1 (1 − f(G/H)).

530 | Characters of Finite Groups 1 Proof. (a) If λ ∈ Irr(H), then |Irr(λ G )| ≤ hg , by reciprocity (see also Exercise V.8.3 (a)). Therefore, the equality ⋃λ∈Irr(H) Irr(λ G ) = Irr(G) yields g g ⋅ mc(G) = k(G) ≤ ∑ |Irr(λ G )| ≤ k(H) = g ⋅ mc(H), h λ∈Irr(H) which implies mc(G) ≤ mc(H). (b) One has g g ⋅ T(H) = ⋅ ∑ ϕ(1) = ∑ ϕ G (1) h h ϕ∈Irr(H) ϕ∈Irr(H) =

χ(1)⟨χ,

∑ χ∈Irr(G)

ϕG ⟩ =

∑

χ(1)⟨χ H ,

∑ χ∈Irr(G)

ϕ∈Irr(H)

∑

ϕ⟩.

ϕ∈Irr(H)

Let {n1 , . . . , n s } = cd(G) − cd(H), and suppose that there are exactly α i characters of degree n i , i ∈ {1, . . . , s}, in Irr(G). Denote by X the set of all χ ∈ Irr(G) such that χ(1) ∈ {n1 , . . . , n s }. Then ∑χ∈X χ(1) = α1 n1 + ⋅ ⋅ ⋅ + α s n s . If χ ∈ X, then χ H is reducible, by the choice of χ, so that ⟨χ H , ∑ϕ∈Irr(H) ϕ⟩ ≥ 2. This implies g ⋅ T(H) = ∑ χ(1)⟨χ H , ∑ ϕ⟩ h χ∈Irr(G) ϕ∈Irr(H) ≥

∑

χ(1) + ∑ χ(1)(⟨χ H , χ∈X

χ∈Irr(G)

∑

ϕ⟩ − 1)

ϕ∈Irr(H)

≥ T(G) + ∑ χ(1) = T(G) + (n1 α1 + ⋅ ⋅ ⋅ + n s α s ), χ∈X

and (b) is proved since, taking n = min{n1 , . . . , n s }, we get n1 α1 + ⋅ ⋅ ⋅ + n s α a ≥ nα. (c) Note that mc(G) = mc(H) if and only if k(G) = hg ⋅ k(H). Therefore, if ϕ ∈ Irr(H), then |Irr(ϕ G )| = hg , so that T(G) = hg ⋅ T(H) if and only if f(G) = f(H). Conversely, let f(G) = f(H), i.e., T(G) = hg ⋅ T(H) = ∑ϕ∈Irr(H) ϕ G (1) and, therefore, if μ ∈ Irr(H) − {ϕ}, then ⟨μ G , ϕ G ⟩ = 0. It follows that |Irr(ϕ G )| = hg for all ϕ ∈ Irr(H), and so k(G) = hg ⋅ k(H) if and only if mc(G) = mc(H). g It remains to prove the last statement. Let λ ∈ Lin(H). Then λ G = μ1 + ⋅ ⋅ ⋅ + μ h , where the linear characters μ i are pairwise distinct (indeed, |Irr(λ G )| = hg = |G : H|). If ϕ ∈ Lin(H) − {λ}, then ⟨ϕ G , λ G ⟩ = 0. Therefore, g g g h g = |Lin(G)| = ⋅ |Lin(H)| = ⋅ 󸀠 = 󸀠 . g󸀠 h h h h We conclude that g 󸀠 = h󸀠 , i.e., G󸀠 = Hg󸀠 . Now let M ⊲ H and Irr(H/M) = {ϕ1 , . . . , ϕ k }. Then, by the above, ϕ Gi = χ1i + ⋅ ⋅ ⋅ + χ ih , i ∈ {1, . . . , k}, where χ i ∈ Irr(G) are pairwise j distinct. Since (χ i )H = ϕ i , it follows that j

j

H ∩ (⋂ ker(χ i )) = ⋂(H ∩ ker(χ i )) = ⋂ ker(ϕ i ) = M, i,j

i,j

i

and therefore M ⊲ G. Thus, all normal subgroups of H are G-invariant. Since we shall not need part (d) in the sequel, its proof is left to the reader.

XI On sums of degrees of irreducible characters | 531

Remark. It follows from Lemmas 3.3 (a)–(b) and 2.3 (a)–(b) that, if a p-group G is 2 and f(G) ≤ 2p−1 . nonabelian, then mc(G) ≤ p +p−1 p3 p2 Lemma 3.4. One has f(G)2 ≤ mc(G), with equality if and only if G is abelian. Proof (A. Mann). Let Irr(G) = {χ1 , . . . , χ r }, where r = k(G). Consider the two r-dimensional vectors a = (χ1 (1), . . . , χ r (1)) and b = (1, . . . , 1). By the Schwarz inequality, we have (a ⋅ b)2 ≤ ‖a‖2 ⋅ ‖b‖2 with equality if and only if a = γb for some γ ∈ ℝ. Next, a ⋅ b = T(G), ‖a‖2 = |G|, ‖b‖2 = k(G) 󳨐⇒ T(G)2 ≤ g ⋅ k(G), which is equivalent to f(G)2 ≤ mc(G). If f(G)2 = mc(G), then a = γb (γ ∈ ℝ), and hence χ1 (1) = ⋅ ⋅ ⋅ = χ r (1) and G is abelian since 1 ∈ cd(G). Denote cd1 (G) = {χ(1) | χ ∈ Irr1 (G)}. Lemma 3.5. Suppose that all subgroups of G󸀠 > {1} are G-invariant. Then all χ ∈ Irr(G) with G󸀠 ∩ ker(χ) = {1} are induced from the same abelian subgroup, so all of them have the same degree. Proof. By hypothesis, such χ ∈ Irr1 (G). Since G is supersolvable, it is an M-group (Blichfeldt), and therefore χ = λ G , where λ ∈ Lin(H) for some H ≤ G of index χ(1). Since G󸀠 ∩ ker(χ) = {1}, ker(χ) = ⋂ (ker(λ))x ≥ ⋂ (H 󸀠 )x = H 󸀠 , x∈G

x∈G

we get H 󸀠 = {1}, and hence H is abelian. Let χ(1) = min{n ∈ cd1 (G)}. If ψ ∈ Irr(G) − {χ} with ker(ψ) ∩ G󸀠 = {1} (in that case, ψ is nonlinear), then, by the choice, we have ψ(1) ≥ χ(1) = |G : H|. However, ψ(1) ≤ |G : H|, and we conclude that ψ(1) = χ(1). If ν ∈ Irr(ψ H ), then ψ = ν G . Lemma 3.6. Let {1} < H ⊲ G. (a) If H ∩ G󸀠 = {1}, then f(G/H) = f(G), mc(G/H) = mc(G). (b) If l = min{χ(1) | χ ∈ Irr(G | H)}, l > 1, f(G) ≥ l−1 , then f(G/H) ≥ f(G). Proof. (a) We have [H, G] ≤ H ∩ G󸀠 = {1} so that H ≤ Z(G). Write G = G/H,

g = |G|,

h = |H|,

g = |G|,

󸀠

G󸀠 H/H = G ≅ G󸀠 ,

󸀠

g 󸀠 = |G |.

Let us prove that there exist pairwise distinct λ1 , . . . , λ h ∈ Lin(G) such that Irr(G) = λ1 Irr(G) ∪ ⋅ ⋅ ⋅ ∪ λ h Irr(G)

(∗)

is a partition. Indeed, Lin(G) = λ1 Lin(G) ∪ ⋅ ⋅ ⋅ ∪ λ h Lin(G)

(λ1 , . . . , λ h ∈ Lin(G)).

Let χ ∈ λ i Irr(G) ∩ λ j Irr(G). Then χ = λ i μ = λ j τ for some μ, τ ∈ Irr(G). Therefore, we get τ = λ j λ i μ. Note that τ(1) = μ(1). For any x ∈ H, one has τ(1) = τ(x) = (λ j λ i )(x)μ(x) = (λ j λ i )(x)μ(1).

532 | Characters of Finite Groups 1 Hence (λ j λ i )(x) = 1 so that λ j λ i ∈ Lin(G), and we conclude that i = j. Therefore, the right-hand side of (∗) is a partition. By Gallagher’s Corollary VII.8.5, equality (∗) is true. Hence, g ⋅ f(G) = T(G) = h ⋅ T(G) = h ⋅ gf(G) = g ⋅ f(G) 󳨐⇒ f(G) = f(G). Similarly, g ⋅ mc(G) = k(G) = h ⋅ k(G) = h ⋅ g mc(G) = g ⋅ mc(G) 󳨐⇒ mc(G) = mc(G). (b) We may assume that H is a minimal normal subgroup of G. Since l > 1, we have H ≤ G󸀠 so that gg󸀠 = g󸀠 . Therefore, g

g=

∑

χ(1)2 +

∑

χ(1)2

χ∈Irr(G|H)

χ∈Irr(G)

g g ≥ + l ⋅ ∑ χ(1) = + l ⋅ (T(G) − T(G)) h h χ∈Irr(G|H) =

g f(G) g + l ⋅ (gf(G) − gf(G)) = + lg(f(G) − ), h h h

so that

1 f(G) ≥ l(f(G) − ). h h Suppose that f(G) > f(G). Then 1 − 1h > l ⋅ f(G)(1 − 1h ), and so f(G) < l−1 , a contradiction. Hence, f(G) ≤ f(G), completing the proof of (b). 1−

Later we shall see that it is impossible to omit the condition f(G) ≥ l−1 in Lemma 3.6 (b) (see §XI.5). Lemma 3.7. Let G = A ∗ B be a central product. If A󸀠 ∩ B󸀠 = {1}, then mc(G) = mc(A) mc(B),

f(G) = f(A)f(B).

Proof. Let G = A ∗ B = (A × B)/D, where A ≅ A, B ≅ B, and D is isomorphic to a subgroup of Z(A) and Z(B). Since A󸀠 ∩ B󸀠 = {1}, it follows that 󸀠

󸀠

󸀠

G󸀠 = A󸀠 × B󸀠 ≅ A × B = (A × B)󸀠 = G ,

where G = A × B.

󸀠

Hence, D ∩ G = {1}, and the result follows from Lemma 3.6 (a). Indeed, G ≅ G/D and so f(G/D) = f(G) = f(A)f(B) = f(A)f(B) = f(A × B) = f(G), as desired. It follows from Lemma 3.7 that if G = G1 ∗ G2 is a group of Theorem 1.0 (c), then 9 f(G) = f(G1 )f(G2 ) = ( 43 )2 = 16 > 21 , i.e., G satisfies the condition. Let u < v be real and let ]u, v[ = {x ∈ R | u < x < v}. If u = v, then ]u, v[ = 0.

XI On sums of degrees of irreducible characters | 533

Lemma 3.8. Let u, v (1 < u ≤ v) be real numbers such that ]u, v[ ∩ cd(G) = 0. Then (u − 1)(v − 1) + (u + v)f(G) g󸀠

uv ⋅ mc(G) + 1 ≥

with equality if and only if cd(G) = {1, u} ∪ {v}. Proof. Let 1 < n ∈ cd(G). Then (n − u)(n − v) ≥ 0 or n2 + uv ≥ n(u + v), with equality if and only if n ∈ {u} ∪ {v}. Let d n be the number of characters of degree n in Irr(G). Then d1 +

d n = |Irr(G)| = k(G) = g ⋅ mc(G)

∑

(3)

n∈cd(G)−{1}

and d1 +

n2 d n =

∑ n∈cd(G)−{1}

∑ χ(1)2 = g.

(4)

χ∈Irr(G)

Multiplying (3) by uv and adding (4), one obtains (uv + 1)d1 +

∑

(n2 + uv)d n = (uv ⋅ mc(G) + 1)g.

(5)

n∈cd(G)−{1}

Applying the inequality n2 + uv ≥ n(u + v) to (5), we get g(uv ⋅ mc(G) + 1) ≥ (uv + 1)d1 + (u + v)

(6)

nd n

∑ n∈cd(G)−{1}

= (uv − u − v + 1)d1 + (u + v) ∑

nd n

n∈cd(G)

= (u − 1)(v − 1) ⋅

g + (u + v)g ⋅ f(G) g󸀠

since ∑n∈cd(G) nd n = T(G), and the first statement of the lemma is proved. Equality holds in (6) if and only if n2 + uv = (u + v)n for all n ∈ cd(G) − {1}, hence cd(G) ⊆ {u} ∪ {v, 1}. Lemma 3.9. Let G be a p-group and let H < G be nonabelian of index p, f(G) > 1p . If 󸀠 |CG (x)| ≤ pg2 for all x ∈ G − H, then hg 󸀠 > p, and |CG (y)| = pg2 for some y ∈ G − H. Proof. By Lemma 3.2 (a), taking into account that |G − H| = mc(G) = ≤

p−1 p g,

we obtain

1 p+1 mc(H) + ∑ |CG (x)| p2 pg2 x∈G−H 1 p+1p−1 g 1 p2 − 1 mc(H) + g = mc(H) + . p2 pg2 p p2 p2 p4

Since h=

∑ ϕ∈Irr(H)

=

ϕ(1)2 ≥

h h + p2 (k(H) − 󸀠 ) h󸀠 h

h h h + p2 (h ⋅ mc(H) − 󸀠 ) = p2 h ⋅ mc(H) − (p2 − 1) 󸀠 , h󸀠 h h

534 | Characters of Finite Groups 1 it follows that mc(H) ≤

p2 −1 p2 h󸀠

mc(G) ≤

+

1 . p2

Therefore,

p2

1 −1 1 p2 − 1 p2 − 1 1 + + = 4 󸀠 + 2. ( ) 2 2 󸀠 2 4 p p h p p p h p

On the other hand, by Lemma 3.8 with u = p, v = p2 (note that cd(G) ∩ ]p, p2 [ = 0 and by f(G) > 1p , one has mc(G) >

(p − 1)(p2 − 1) 1 (p − 1)(p2 − 1) p + p2 1 + − 1] = + 2. [ g󸀠 p p3 p3 g󸀠 p

Comparing the last two inequalities, we obtain (p − 1)(p2 − 1) p2 − 1 < 4 󸀠 p3 g󸀠 p h

or

g󸀠 > p(p − 1) ≥ p, h󸀠

and the first statement is proved. Now suppose that |CG (y)| < pg2 for all y ∈ G − H. Then |CG (y)| ≤ pg3 for all y ∈ G − H (Lagrange since G is a p-group). Hence, by Lemma 3.2 (a) (note that |G − H| = g(p−1) p ), mc(G) ≤

mc(H) p + 1 g(p − 1) g mc(H) p2 − 1 + ⋅ ⋅ 3 = + . 2 2 p p pg p p2 p5

Since H is nonabelian, we have h ≥ |Z(H)| + p(k(H) − |Z(H)|) = p ⋅ k(H) − (p − 1)|Z(H)|, and thus, since |Z(H)| ≤

h , p2

we obtain

ph ⋅ mc(H) = p ⋅ k(H) ≤ h + (p − 1)|Z(H)| ≤ h + (p − 1) Hence mc(H) ≤

1 p

+

p−1 p3

mc(G) ≤

=

p2 +p−1 . p3

h . p2

Therefore,

p2 + p − 1 p2 − 1 2p2 + p − 2 1 + = ≤ 2 < f(G)2 , 5 5 5 p p p p

contrary to Lemma 3.4. Lemma 3.10. Suppose that q is a minimal prime divisor of |G| = g and f(G) ≥ 1q . Then the following statements hold: (a) If f(G) > 1q , then G is q-nilpotent. (b) If G is not q-nilpotent, then one has q = 2, f(G) = 12 and G = HZ(G), where H is an S(3a , 22 , 2ϵ )-group, ϵ ∈ {0, 1} (in that case, G/Z(G) ≅ H/Z(H) ≅ A4 ). Proof. Suppose that G is not q-nilpotent. Then, by Lemma 1.1 (b), it contains an S(r a , q b , q c )-subgroup F. However, f(F) ≥ f(G) ≥ 1q (Lemma 3.3 (b)). Therefore q = 2, r = 3, f(F) = 12 (Lemma 2.2 (d)). Thus, 12 = f(F) ≥ f(G) ≥ 12 , hence f(G) = 12 . Thus, if f(G) > 1q , then G is q-nilpotent, and (a) is proved. Next suppose that G is not q-nilpotent. By statement (a) and the assumption, one has q = 2 and f(G) = 12 . Let F be chosen as in the previous paragraph. Then F = PQ is

XI On sums of degrees of irreducible characters | 535

an S(3a , 22 , 2ϵ )-group with ϵ ∈ {0, 1}, P ∈ Syl3 (F),

Q ∈ Syl2 (F),

F 󸀠 = Q ∈ {E4 , Q8 }.

Then Q = F 󸀠 = G󸀠 (Lemma 3.3 (c)) hence Q ⊲ G. By Lemma 1.1 (d), Q ≰ Φ(G) as F is nonnilpotent, and hence G = MQ, where M is a maximal subgroup of G. (i) Suppose that ϵ = 0, i.e., Q = E4 . Since Q is a minimal normal subgroup of G, we have M ∩ Q = {1}. Hence M is abelian (indeed, Q = G󸀠 ), CM (Q) = Z(G), and hence G/Z(G) ≅ A4 , G = FZ(G). (ii) Suppose that ϵ = 1, i.e., Q = Q8 , F/℧1 (P) ≅ SL(2, 3). In that case, Q/Z(Q) is a minimal normal subgroup of G/Z(Q). Then M 󸀠 ≤ M ∩ G󸀠 = M ∩ Q = Z(Q) ≤ Z(G), so that M is nilpotent of index |Q/Z(Q)| = 4 in G since G = MQ and M ∩ Q = Z(Q). Next, G/M G ≅ F/Z(F) ≅ A4 . Let K be a 3󸀠 -Hall subgroup of M; then K ⊲ G and G = KF. Put C = CG (K); then C ⊲ G. If Q < C, then K ≤ Z(G) and G/Z(G) ≅ A4 . Let R ∈ Syl3 (M). In any case, R < C. Then F < C so Q < C. Thus, in any case G/Z(G) ≅ A4 . Lemma 3.11. Let G be a nonabelian p-group with f(G) > 1p . Then the following statements hold: (a) One has p−1 g󸀠 ≤ pf(G) − 1 with equality if and only if cd(G) = {1, p}. 2 , then one of the following holds: (b) If f(G) ≥ p +p−1 p3 (1) g󸀠 = p, G contains an abelian subgroup of index p and f(G) = 2p−1 . p2 (2) g󸀠 = p, G contains an abelian subgroup of index p2 but does not contain abelian 2 subgroups of index p, f(G) = p +p−1 . p3 (3) g󸀠 = p2 , cd(G) = {1, p} and either G contains an abelian subgroup of index p or 2 G/Z(G) ≅ Ep3 and G󸀠 is noncyclic, f(G) = p +p−1 . p3 Proof. (a) Since g=

∑

χ(1)2 ≥

χ∈Irr(G)

g g g g + p ⋅ ∑ χ(1) = 󸀠 + p(T(G) − 󸀠 ) = pg ⋅ f(G) − (p − 1) 󸀠 , g󸀠 g g g χ∈Irr (G) 1

p−1 it follows that g󸀠 ≤ pf(G)−1 since pf(G) − 1 > 0. The second assertion follows from the above computation. As for the proof of (b), we shall not need it in the sequel and therefore leave it to the reader.

Lemma 3.12. Let G = P ⋅ K be a semidirect product with kernel K, P ∈ Sylp (G), Then mc(G) ≤

1 p2

|P| = p2 ,

CG (K) = K.

(and so, by Lemma 3.4, f(G) < 1p ).

536 | Characters of Finite Groups 1 Proof. By Theorem VII.2.3, cd(G) ⊆ {1, p, p2 }. We may assume that G has no nontrivial abelian direct factor. Let d i = |Irr(p i ) (G)|, i = 0, 1, 2. It follows from Lemma 1.1 (c) that d0 = gg󸀠 = p2 and Z(G) = {1}. Then NG (P) = P. Suppose that mc(G) > p12 . Then χ(1)2 = p2 + p2 d1 + p4 d2 = g,

∑

p2 + d1 + d2 = k(G) = g ⋅ mc(G) >

χ∈Irr(G)

g , p2

and so p2 (p2 − 1) − p2 (p2 − 1)d2 > 0. Therefore, d2 = 0 and cd(G) = {1, p}. Since G is nonnilpotent, it has an abelian subgroup of index p (see [Isa11, Theorem 12.14]), contrary to the hypothesis. Thus, mc(G) ≤ p12 . Lemma 3.13. Let G be a nonabelian p-group, p > 2 and f(G) > 21 . Then p = 3 and f(G) = 59 . Proof. If H ≤ G is minimal nonabelian, then 2p − 1 1 = f(H) ≥ f(G) > , 2 p2 hence p2 < 4p − 2 < 4p, and we get p = 3. Furthermore, g=

g g g g g g 3g 2g + ∑ χ(1)2 ≥ 󸀠 + 3(T(G) − 󸀠 ) > 󸀠 + 3( − 󸀠 ) = − 󸀠. g󸀠 χ∈Irr (G) g g g 2 g 2 g 1

2 g󸀠

1 2,

Hence > and so g 󸀠 < 4, and we conclude that g 󸀠 = 3 since G is nonabelian. By Lemma 3.5, all the characters in Irr1 (G) are of the same degree 3t , which is the index of a maximal abelian subgroup of G. Then T(G) =

g g 2g 1 2 g 1 5 + (g − 󸀠 )3−t = + t+1 󳨐⇒ < f(G) = + t+1 , t = 1, f(G) = , g󸀠 g 3 3 2 3 3 9

as was to be shown. It follows that the group G of Lemma 3.13 has an abelian subgroup of index 3 so |G : Z(G)| = 3|G󸀠 | = 32 (Lemma 3.1). Lemma 3.14. Let p be the smallest essential prime divisor of g = |G|, G is a nonnilpotent group and f(G) > 1p . Then G has a normal abelian p󸀠 -Hall subgroup. Proof. In view of Lemma 3.7, one may assume that p is the minimal prime divisor of g. By Lemma 3.10 (a), G contains a normal p󸀠 -Hall subgroup H. Suppose that H is nonabelian. From now on, let us assume that G is a counterexample of minimal order. Let x ∈ P ∈ Sylp (G) be of minimal order such that x ∈ ̸ CG (H). Since f(⟨x, H⟩) ≥ f(G) (Lemma 3.3 (b)), it follows by induction that G = ⟨x, H⟩ (otherwise, H is abelian), and so P = ⟨x⟩ is cyclic, G󸀠 ≤ H, x p ∈ Z(G). Lemma 3.6 (a) implies that f(G/⟨x p ⟩) = f(G). Therefore, by induction, x p = 1, so that |G : H| = p. Let {0 ϵ = ϵ(p) = { 1 {

if p = 2, if p > 2.

XI On sums of degrees of irreducible characters | 537

As there are no characters of degrees less than p +1+ ϵ in Irr1 (H)) (Frobenius–Molien), we conclude, keeping in mind that f(H) ≥ f(G) > 1p , that h≥

h h h h h + (p + 1 + ϵ)(T(H) − 󸀠 ) > 󸀠 + (p + 1 + ϵ)( − 󸀠 ). 󸀠 h h h p h

It follows that p+ϵ 1+ϵ , > h󸀠 p

h󸀠
2, then ϵ = 1, h󸀠 < (p+1)p 2 , and so h󸀠 = r is a prime > p. Let α be the number of characters of degree p in Irr(G). Then, by Lemma 3.3 (b) (since p ∈ ̸ cd(H)), we have f(H) ≥ f(G) +

pα pα 󳨐⇒ ≤ f(H) − f(G). g g

(8)

(i) Suppose that p = 2. In that case, by what has just proved, h󸀠 = 3. Since 3 is the least prime divisor of |H|, it follows that H 󸀠 ≤ Z(H) so H is nilpotent. We have g g + 22 α + 3(T(G) − 󸀠 − 2α) g󸀠 g g g 3g 2g g > 󸀠 + 4α + 3( − 󸀠 − 2α) = − 󸀠 − 2α, g 2 g 2 g

g≥

and hence 1 2α 2 > − . 󸀠 g 2 g

(9)

Since f(H) ≥ f(G) > 12 , we get f(H) = 95 (Lemma 3.13). Therefore, it follows from (8) that 2α 5 5 1 1 ≤ − f(G) < − = . g 9 9 2 18 By (9),

2 1 2α 1 1 4 > − > − = , 󸀠 g 2 g 2 18 9

and so g󸀠 ≤ 4. Since 3 = h󸀠 divides g󸀠 , it follows that then g 󸀠 = 3 and G󸀠 = H 󸀠 . By assumption, G is nonnilpotent; then G󸀠 ≰ Φ(G) (Lemma 1.1 (d)). On the other hand, since H is nilpotent, we get G󸀠 = H 󸀠 ≤ Φ(H) ≤ Φ(G) (Lemma 1.1 (e)), a contradiction. Thus, H is abelian if p = 2. (ii) Suppose that p > 2. In that case, |H 󸀠 | = r, where r < 21 p(p + 1) is a prime, r > p. Since there are no characters of degree p + 1 in Irr(G) (Frobenius–Molien), it follows that g g + p2 α + (p + 2)(T(G) − 󸀠 − pα) 󸀠 g g g g g 2 > 󸀠 + p α + (p + 2)( − 󸀠 − pα), g p g

g≥

538 | Characters of Finite Groups 1 so that p + 1 2 2pα > − . g󸀠 p g

(10)

Also, it follows from (8) and (1) that 2 2 4 p+1 2 > − 2(f(H) − f(G)) > − 2f(H) + = − 2f(H). g󸀠 p p p p

(11)

Suppose that H is nonnilpotent. By Lemma 3.5, all the characters in Irr1 (H) have the same degree m. Therefore (recall that h󸀠 = r, a prime number), we have T(H) =

h h h h h − hr + m= + − r r m rm m2

so that, since m being a divisor of h, is > p, we obtain f(H) =

1 1 1 1 1 2 + − < + < , r m mr r m p

3 and hence 4p − 2f(H) > 0. We claim that f(H) < 2p . Indeed, set |H : CH (H 󸀠 )| = s. Then s > 1 divides r − 1 and |H|. However, as a divisor of h, s > p, and the even number r − 1 is composite (indeed, r > p > 2 so r ≥ 5). It follows that r − 1 ≥ 2p so r ≥ 2p + 1, and we conclude that, since m > p,

f(H)
− = , r m 2p p 2p p p p p

and from (11) we obtain g󸀠
p exceed p(p + 1) > g 󸀠 , and we have g󸀠 = r = h󸀠 (since r = h󸀠 | g 󸀠 ), G󸀠 = H 󸀠 . By Lemma 1.1 (d), G󸀠 ≰ Φ(G) so G = M ⋅ G󸀠 (semidirect product), where M < G is maximal (obviously, M is abelian). Since CM (G󸀠 ) = Z(G), it follows by induction and Lemma 3.6 (a) that Z(G) = {1} since Z(G) ∩ G󸀠 = {1}. Therefore, G = (Cpm , Cr ) is a Frobenius group. Here pm | r − 1 (see Exercise X.2.3),

m > p,

p ∤ m, r − 1 ≥ 2pm.

Therefore, since cd(G) = {1, pm} (Theorem VII.2.3), we obtain T(G) = pm +

r−1 pm = pm + r − 1, pm

where pm = |Lin(G)|,

r−1 = |Irr1 (G)|, pm

so that f(G) =

pm + r − 1 1 1 1 1 3 3 1 1 < + < + = = ⋅ < , pmr r pm 2pm pm 2pm 2m p p

contrary to the hypothesis. Thus, H is nilpotent. It follows from h󸀠 = r that R ∈ Sylr (H) is nonabelian and G-invariant. Since f(PR) ≥ f(G) > 1p (Lemma 3.3 (b)), we obtain, by induction, that either PR = G or PR = P × R.

XI On sums of degrees of irreducible characters | 539

Let PR = P × R. Then G = R × F, F is nonnilpotent with an abelian subgroup of index p, by induction. By the remark preceding Lemma 3.4, one has f(R) =

2 2r − 1 2 < ≤ , 2 r p+2 r

and f(G) = f(R)f(F)
− 2f(R) ≥ − = 󳨐⇒ g 󸀠 < < r3 , 󸀠 g p p p + 2 p(p + 2) 8 so that g󸀠 ≤ r2 since G󸀠 ≤ R. Then PG󸀠 = P × G󸀠 (we use the inequality 2p > r and equality PR󸀠 = P × R󸀠 ). Since P is characteristic in P × G󸀠 ⊲ G, it follows that P ⊲ G. In that case, G = P × R is nilpotent, a final contradiction. Thus, H is abelian, as was to be shown. The identity (a) in Lemma 3.15 was proved by D. Rusin [Rus]. Lemma 3.15. Let G be a p-group, G󸀠 ≤ Z(G), exp(G󸀠 ) = p. Then mc(G) =

p 1 g󸀠 − 1 + mc(G/K)} { − ∑ 󸀠 g p p(p − 1) K {1} is a p󸀠 -Hall subgroup of G. Since |SL(2, p)| = p(p2 − 1), it follows that: Claim (ii). One has |G : P| = |F| | p2 − 1. By construction, Z(G) = Z(P). Put G = G/Z(P). Claim (iii). The group G is a Frobenius group with kernel P. Indeed, assume that σ ∈ F # and CP (σ) > {1}. By Maschke’s theorem, σ can be represented by a matrix a 0 ( ) (a ∈ GF(p) − {0}) 0 1 relative to an appropriate basis of the linear GF(p)-space P. Since det(σ) = 1, it follows that a = 1, σ = 1 and so G is a Frobenius group, as claimed. In particular, if F 󸀠 = F, then F ≅ SL(2, 5) (see Chapter X). Claim (iv). One has k(G) = k(G) + (p − 1)k(F). Indeed, every character ϕ ∈ Irr1 (P) admits extension to G (Corollary VII.4.8). If χ is one of such extensions, then μχ ∈ Irr1 (G) for all μ ∈ Irr(G/P). Then number of the so-obtained characters is equal to (p − 1)|Irr(G/P)| = (p − 1)k(F), and the sum of their degrees is T(G/P)p(p − 1) = T(F)p(p − 1). This proves (iv). We also have proved that T(G) = T(G) + p(p − 1)T(F) = T(F) + p2 − 1 + (p2 − p)T(F) = p2 − 1 + (p2 − p + 1)T(F). Therefore, f(G) =

and we get

T(G)

=

|G| T(G) (p2 − p + 1)T(F) + p2 − 1 f(G) = = , |G| p3 ⋅ |F| f(G) − f(G) =

Hence:

T(F) + p2 − 1 , p2 ⋅ |F|

(p − 1)2 ⋅ [T(F) − (p + 1)]. p3 |F|

Claim (v). The natural semidirect product G = F p ⋅ P = F ⋅ P satisfies f(G) > f(G/Z(G)) if and only if T(F) > p + 1.

XI On sums of degrees of irreducible characters | 543

Below we present examples of pairs H ⊲ G such that f(G/H) < f(G). Example 5.1. Let F p = Q8 , G = Q8 ⋅ P < D. By (v), we get p = 3,

p + 1 < T(F p ) = 6,

|G| = 23 ⋅ 33 .

Example 5.2. Let p > 3, F p = SL(2, 3). Then p + 1 < T(F p ) = 12,

p ∈ {5, 7},

|G5 | = 23 ⋅ 3 ⋅ 53 ,

|G7 | = 23 ⋅ 3 ⋅ 73 .

Example 5.3. Let p ≡ ±1 (mod 8), and let F p be a subgroup of order 48 in SL(2, p); obviously, F p is the representation group of the symmetric group S4 with generalized quaternion Sylow 2-subgroup. Now, p + 1 < T(F p ) = 18 󳨐⇒ p = 7, |G7 | = 24 ⋅ 3 ⋅ 73 . Example 5.4. Let p ≡ ±1 (mod 5), F p = SL(2, 5). As p + 1 < T(F p ) = T(SL(2, 5)) = 30, it follows that p = 11 or p = 19, so that G11 and G19 are the only two counterexamples with such F p . Example 5.5. Let p > 3, and let F p = C4 ⋅ C3 be a nonabelian semidirect product. Then p + 1 < T(F p ) = 8, i.e., p = 5, and G5 of order 22 ⋅ 3 ⋅ 53 is the only counterexample with such F p . Example 5.6. Let F p be a subgroup of order 2(p + 1) with a cyclic subgroup of index 2, p > 2, F p < SL(2, p). Then T(F p ) = p + 3 > p + 1, so this G p is a counterexample of order 2(p + 1)p3 . It is interesting to find a pair H ⊲ G such that f(G/H) < f(G) and either (a) H ≰ Z(G), or (b) G is a p-group. Exercise 5.1. Find all F < SL(2, p) with p ∤ |F| such that f(F ⋅ P) > f(F ⋅ P) (here we are using the previous notation). Exercise 5.2. The following statements hold: (i) If f(G) > 23 , then G is nilpotent. (We have f(S3 ) = 23 .) 2 (ii) If f(G) > 34 , then G is abelian. (We have f(D8 ) = 34 .) In particular, if t(G) |G| > 3 , then t(G) 3 G is nilpotent, and the inequality |G| > 4 implies that G is abelian. Hint. (i) Assume that f(G) > 32 and G is nonnilpotent. Let F < G be minimal nonnilpotent; then f(F) ≥ f(G) > 32 (Lemma 3.3 (b)). In view of Lemma 2.2 (c), one may assume that Z(F) = {1}; then F = P ⋅ Q, where Q = F 󸀠 ∈ Sylq (F), In that case, f(F) =

P ∈ Sylp (F),

|F| = pq b .

1 1 1 2 1 1 2 − > > . + 󳨐⇒ + p q b pq b 3 p qb 3

544 | Characters of Finite Groups 1 Since p < q b , we get 1p + 1p > 32 so that p < 3, and we get p = 2. Then one has b = 1 and 12 + 1q > 32 so q < 6. If q = 3, then f(F) = 21 + 13 − 16 = 23 , a contradiction. If q = 5, 1 then f(F) = 21 + 15 − 10 = 35 < 23 , a final contradiction. (ii) Assume that f(G) > 34 and G is nonabelian. Let F ≤ G be minimal nonabelian. By (i), G is nilpotent so that F is a p-group for some prime p. In that case, one has f(G) ≤ f(F) = 2p−1 ≤ 43 , a contradiction. p2 Exercise 5.3. Let p ∈ π(G) be minimal. Find the least numbers ν = ν(p) and α = α(p) such that the following statements hold: (a) If f(G) > ν(p), then G is nilpotent. (b) If f(G) > α(p), then G is abelian. 3 Exercise 5.4. Let i(G) = t(G) − 1 be the number of involutions in G. If i(G) |G| > 4 , then G 3 2 is abelian (and the number 4 cannot be replaced by a smaller number). If i(G) |G| ≥ 3 , 2 then G is nilpotent (and the number 3 cannot be replaced by a smaller number). Note that for G = S3 × E2n , one has

for G = D8 × E2n , one has

i(G) |G|

i(G) 2 1 = − ; |G| 3 3 ⋅ 2n+1 =

3 4

− 2−(n+3) .

Hint. We have T(G) > t(G). Use Exercise 5.2. Exercise 5.5. If mc(G) > 58 , then G is abelian (and 85 cannot be replaced by a smaller number). If mc(G) > 12 , then G is nilpotent (and 21 cannot be replaced by a smaller number). Hint. Assume that mc(G) > 58 and G is nonabelian. Let F < G be minimal nonabelian. Then mc(F) ≥ mc(G) > 85 (Lemma 3.3 (a)). Let F be nonnilpotent; then |F| = p a q b , so that f(F) = Since

1 qb

< 1p , we get

1 p2

+

|F 󸀠 | = q b ,

k(F) = p a + p a−1

qb − 1 , p

p a + p a−2 (q b − 1) 5 1 1 5 > or 2 + b > . a b 8 8 p p q q 1 p

> 58 . Thus (here we use Lemma 2.1),

5p2 − 8p − 8 < 0 󳨐⇒ p = 2, b = 1, |G| = 2a q. In that case,

7 5 1 1 1 1 < + ≤ + = , 8 4 q 4 3 12 a contradiction. It remains to consider the case where F is a p-group. Then mc(F) =

p2 + p − 1 5 p+1 5 > 󳨐⇒ > 󳨐⇒ 5p2 − 8p − 8 < 0. 8 8 p3 p2

It follows that p = 2 and mc(F) =

22 +2−1 23

= 58 , contrary to the hypothesis.

XI On sums of degrees of irreducible characters | 545

Exercise 5.6. Let p be the least prime divisor of |G| = g. Find the least numbers ν0 = ν0 (p) and α0 = α0 (p) such that the following statements hold: (a) If mc(G) > ν0 , then G is nilpotent. (b) If mc(G) > α0 , then G is abelian. (c) Similarly, find the least number α1 = α1 (p) such that, if mc(G) > α1 , then G is p-nilpotent.

6 Further properties of functions f(∗) and mc(∗) We omit consideration of very involved nilpotent case of Theorem 1.0 (see [BZ3, §11.6]. It is easy to reduce this case to p-groups. Proposition 6.1. Let G be a p-group of order g = p n . , then g󸀠 = p, |G : Z(G)| = p2 . (a) If f(G) = 2p−1 p2 2 p +p−1 (b) If mc(G) = p3 , then g󸀠 = p, |G : Z(G)| = p2 . Proof. (a) Let H ≤ G be minimal nonabelian. Then f(H) = 2p−1 = f(G) (Lemma 2.3 (b)). p2 By Lemma 3.3 (c), G󸀠 = H 󸀠 , so that |G󸀠 | = g 󸀠 = h󸀠 = p. Set |Irr1 (G)| = s. By Lemma 3.5, cd(G) = {1, p d } for some natural d. We have p n = p n−1 + sp2d , which implies

p n−2 (2p − 1) = |G|f(G) = T(G) = p n−1 + sp d ,

p n−1 (p − 1) = sp2d ,

p n−2 (p − 1) = sp d .

These equalities imply d = 1, so that cd(G) = {1, p}. By Lemma 3.5 again, G contains an abelian subgroup of index p. By Lemma 3.1, |G : Z(G)| = p|G󸀠 | = p2 . (It is easy to check that if |G : Z(G)| = p2 and |G󸀠 | = p, then f(G) = 2p−1 .) p2 (b) Let H ≤ G be minimal nonabelian. Then mc(G) =

p2 + p − 1 = mc(H) p2

(Lemma 2.3 (a)) so that g 󸀠 = h󸀠 = p (Lemma 3.3 (b)). In that case, cd(G) = {1, p t } for some natural t. Then k(G) = |Irr(G)| = p n−1 +

2 p n − p n−1 n p +p−1 = |G| mc(G) = p ⋅ , p2t p3

and we conclude that t = 1. Then, by Lemma 3.5, G contains an abelian subgroup A of index p. It follows that |G : Z(G)| = p|G󸀠 | = p2 . Exercise 6.1 (Amitsur for p = 2 ([Ami]), Isaacs–Passman for p > 2 ([IP2, IP3])). Let us suppose that cd(G) = {1, p}. Then either G contains a normal abelian subgroup of index p or G/Z(G) is of order p3 and exponent p. Conversely, if G is one of the above groups, then cd(G) = {1, p}.

546 | Characters of Finite Groups 1 Proposition 6.2. Let G be a nonabelian group of order p m and let δ(G) = {a0 ⋅ 1, a1 ⋅ d1 , . . . , a s ⋅ d s }, the degrees vector of G (i.e., cd(G) = {d0 = 1, d1 , . . . , d s } and Irr(G) contains exactly a i characters of degree d i for all i). Then the following statements hold: (a) (Mann) One has p − 1 | a i for i = 1, . . . , s. (b) If T(G) = p m−1 , then p = 2. Proof. (a) Let χ ∈ Irr1 (G), χ(1) = d > 1, and let x ⋅ ker(χ) ∈ Z(G/ker(χ)) be of order p. If R is a representation of G with character χ, then R(x) = ϵ ⋅ Iχ(1) is a scalar χ(1) × χ(1) matrix. Then the number of algebraic conjugates of χ is divisible by the number of algebraic conjugates of ϵ, and the latter is p − 1 since ϵ is a p-th root of 1. Summing over the set of all characters of degree d, we obtain the desired result. (b) Let δ(G) = {p k ⋅ 1, a1 ⋅ p c1 , . . . , a s ⋅ p c s } be the degree vector of G. By (a), one has a i = (p − 1)a󸀠i , where a󸀠i ∈ ℕ, i = 1, . . . , s. Let 0 < c1 < ⋅ ⋅ ⋅ < c s . It follows from |G| = p m = p k + (p − 1)[a󸀠1 p2c1 + ⋅ ⋅ ⋅ + a󸀠s p2c s ], T(G) = p m−1 = p k + (p − 1)[a󸀠1 p c1 + ⋅ ⋅ ⋅ + a󸀠s p c s ] that |G| − T(G) = p m−1 (p − 1) = (p − 1)[a󸀠1 p c1 (p c1 − 1) + ⋅ ⋅ ⋅ + a󸀠s p c s (p c s − 1)], and so p m−1 = a󸀠1 p c1 (p c1 − 1) + ⋅ ⋅ ⋅ + a󸀠s p c s (p c s − 1) ≡ 0 (mod 2), hence p = 2. Thus, if G is a p-group such that f(G) = 1p , then p = 2. K. G. Nekrasov [Nek1] has classified all the groups G such that f(G) = 12 . Exercise 6.2. Let p be the least prime divisor of |G|. Classify all the groups G with 1 . mc(G) ≥ 1p . Classify all the groups G with f(G) ≥ p+1 The second part of Exercise 6.2 is rather difficult (more precisely: this is an open problem). But the first question of this exercise does not seem to be difficult. A character χ ∈ Irr(G) is said to be normally monomial if there exist H ⊴ G and λ ∈ Lin(H) such that λ G = χ. By Corollary VII.8.5, if N ⊲ G, then k(G) ≤ k(N)k(G/N), or, dividing this by |G| = |N||G/N|, one obtains mc(G) ≤ mc(N) mc(G/N) ≤ mc(G/N).

(1)

Exercise 6.3. The following statements hold: (a) Let G be a solvable group, p the minimal prime divisor of |G|. Suppose that every irreducible character of G of prime degree is normally monomial. If mc(G) > p12 , then G󸀠 is abelian. (b) (A. Mann and C. Scoppola, a partial case) If G is a p-group and mc(G) ≥ p12 , then G󸀠 is abelian.

XI On sums of degrees of irreducible characters | 547

Solution. (a) Let G be a counterexample of minimal order. Since the hypothesis is inherited for epimorphic images, it follows from (1) that G is a monolith. Let N be a (unique) minimal normal subgroup of G; obviously, N < G󸀠 (by assumption, G󸀠 is nonabelian). Set |G| = g, |G󸀠 | = g 󸀠 , |N| = n; then n is a prime power since G is solvable. Let χ ∈ Irr(G) be faithful (note that χ exists since, in the case under consideration, ⋂ψ∈Irr1 (G) ker(ψ) = ker(ρ G ) = {1}; here ρ G is the regular character of G). Assume that χ(1) is a prime. By condition, there exist H ⊲ G and λ ∈ Lin(H) such that χ = λ G . By Clifford’s theorem, Irr(χ H ) ⊆ Lin(H) so that H is abelian since H 󸀠 ≤ ker(χ H ) and χ H is faithful. It follows from G󸀠 ≤ H (|G/H| is a prime) that G󸀠 is abelian, contrary to the assumption. Thus, if χ ∈ Irr(G) is faithful, then χ(1) is composite so that χ(1) ≥ p2 since p is the minimal prime divisor of |G|. It follows from g < k(G) = |Irr(G/N)| + |Irr(G | N)| p2 g g g 1 g g 1 p2 − 1 g p2 − 1 ⋅ 󸀠 + 4 ⋅g+ 4 ≤ 󸀠 + 2 ( − 󸀠 ) + 4 (g − ) = 2 g n g p n g p p p n p 2

p n that g󸀠 < n−1 . Suppose that n > 2. Then g󸀠 < 32 p2 so g 󸀠 is a product of two primes. Since 󸀠 N < G , n is a prime so G󸀠 ≤ CG (N) hence G󸀠 is abelian, contrary to the assumption. If n = 2, then g󸀠 < 2p2 and again g󸀠 is a product of two primes. Then G󸀠 is again abelian, a final contradiction. (b) The same argument yields the result.

Note that mc(SL(2, 3)) =

7 24

>

1 22

and SL(2, 3)󸀠 ≅ Q8 is nonabelian.

Remark. A. Mann and C. Scoppola [MS2] have proved that if a p-group G, p > 2, satisfies mc(G) ≥ p12 , then either G contains an abelian subgroup of index p or |G󸀠 | ≤ p3 (then G󸀠 is abelian) but this is not true for p = 2.

7 Groups with many involutions 1°. In this subsection we classify all the groups G such that t0 (G) =

t(G) 1 > , g 2

where t(G) = |{x ∈ G | x2 = 1}|. This classification was obtained by G. A. Miller (but the authors are not familiar with his work); the same result was obtained by C. T. C. Wall [Wall] and Liebeck–McHale. Another proof was given by K. G. Nekrasov [Nek1]; that proof is presented in [BZ3, Theorem 11.24]. Throughout this subsection g = |G|, T = {x ∈ G | x2 = 1}; then |T| = t(G). Let E be a group whose exponent divides 2. By the Frobenius–Schur formula (Chapter IV), we have t(G) =

∑ χ∈Irr(G)

ν2 (χ)χ(1),

548 | Characters of Finite Groups 1 where ν2 (χ) is the Frobenius–Schur indicator of χ ∈ Irr(G) (see §IV.6). Therefore, t(G) ≤ T(G) =

∑

χ(1);

χ∈Irr(G)

moreover, the equality holds if and only if all the irreducible representations of the group G are realizable over ℝ (see formula (IV.6.2)). Thus, t0 (G) =

t(G) 1 1 > 󳨐⇒ f(G) > g 2 2

since t0 (G) ≤ f(G). Therefore, it remains to choose those groups G among those mentioned in Theorem 1.0 for which t0 (G) > 21 . Theorem 7.1. If t0 (G) > 21 , then one of the following holds: (a) G contains an abelian subgroup A of index 2, G − A ⊂ T (such groups are called generalized dihedral). (b) Φ(G) ≅ C2 with G = G1 ∗ ⋅ ⋅ ⋅ ∗ G n × E, where G i ≅ D8 , G i ∩ G j = Φ(G) for i ≠ j (thus, G1 ∗ ⋅ ⋅ ⋅ ∗ D n is extraspecial). (c) G = D(8) × D(8) × E. (d) G = ⟨x1 , y1 , . . . , x r , y r , t⟩ × E, where x1 , . . . , x r , y1 , . . . , y r , t are involutions and [x i , x j ] = [x i , y j ] = [t, x i ] = 1,

i ≠ j,

[t, y i ] = x i ,

i, j ∈ {1, . . . , r}.

For a proof of Theorem 7.1, see [BZ3, Theorem 11.24]. Below we consider functions i(G) = t(G) − 1, the number of involutions in G, 1 1 and i0 (G) = i(G) |G| . We study the nonsolvable groups G with mc(G) ≥ 16 , f(G) ≥ 4 and 1 i0 (G) ≥ 4 and prove the following theorem. Theorem 7.2. Let G be a nonsolvable group. (a) We have 1 󳨐⇒ G = G󸀠 Z(G), G󸀠 ∈ {PSL(2, 5), SL(2, 5)}. 16

mc(G) ≥ (b) We have f(G) ≥

1 󳨐⇒ G = G󸀠 Z(G), G󸀠 ∈ {PSL(2, 5), SL(2, 5)}. 4

t0 (G) ≥

1 ⇐⇒ G = PSL(2, 5) × E, where exp(E) ≤ 2. 4

(c) We have

Recall that 1 , 12 4 f(PSL(2, 5)) = , 15 26 t0 (S5 ) = . 120

mc(PSL(2, 5)) =

3 , 40 1 f(SL(2, 5)) = , 4

mc(SL(2, 5)) =

1 , 28 1 f(PSL(2, 7)) = , 6

mc(PSL(2, 7)) =

XI On sums of degrees of irreducible characters | 549

The following lemma contains some known results. Lemma 7.3. The following statements hold: (a) (see [Bli]) If G is a nonabelian simple group and χ ∈ Irr1 (G) is such that χ(1) < 4, then χ(1) = 3 and G ∈ {PSL(2, 5), PSL(2, 7)}. (b) (Isaacs; see Theorem XIV.7.1) If G is nonsolvable, then |cd(G)| ≥ 4. (c) (Exercises 5.2 and 5.5) If G is nonabelian, then mc(G) ≤ 85 and f(G) ≤ 43 . Lemma 7.4. Let G = G󸀠 > {1} and d ∈ {4, 5, 6}. If mc(G) ≥ acter χ ∈ Irr1 (G) such that χ(1) < d.

1 , d2

then there exists a char-

Proof. Let G be a counterexample. Then, by virtue of Lemma 7.3 (a)–(b), we obtain |G| =

∑

χ(1)2 ≥ 1 + d2 (k(G) − 3) + (d + 1)2 + (d + 2)2

χ∈Irr(G)

≥ 1 + d2 (

|G| − 3) + 2d2 + 6d + 5 = |G| − d2 + 6d + 6 > |G| d2

since d ∈ {4, 5, 6}, a contradiction. Exercise. Let G = P ⋅ Q be a group of order 36, where P ∈ Syl2 (G) is abelian of type (2, 2) and Q ∈ Syl3 (G) is abelian of type (3, 3). If G has no abelian subgroup of prime index, then G ≅ S3 × S3 , where S3 is the symmetric group of degree 3. Solution. By hypothesis, CG (P) = P and CG (Q) = Q. It follows that Z(G) = {1}. Since 9 ∤ |Aut(P), we get P ⋬ G. Therefore, by Burnside’s theorem on normal p-complement, we get NG (Q) > Q, and so Q ⊲ G; in that case, Q = G󸀠 (if |G󸀠 | = 3, then G󸀠 , by Gaschütz’ theorem, has a complement in G which is abelian of index 3). As G is not a Frobenius group, there are x ∈ P# and y ∈ Q# such that xy = yx. Then one gets ⟨y, P⟩ ≤ CG (x) < G (here “ i(G) = 14 |G| so that |G : H| = 3. We have f(H) > i0 (H) =

i(H) 3 i(G) 3 = = , |H| |G| 4

hence H is abelian (Exercise 5.2); then CG (P) = P (recall that P is the centralizer of all its involutions) so H = P, and we conclude that G ≅ A4 is as in part (a). Let |H| be odd; then |G : H| is even. Since PH is not a Frobenius group, there is x ∈ P# such that CG (x) > P, a contradiction. (ii) Suppose that G = P ⋅ K is 2-nilpotent. Assume that P is cyclic. Then all involutions are G-conjugate, and for involution x ∈ P one has CG (x) = P. It follows that G is a Frobenius group with kernel K of index |P| = 4 so G is as in (c). Now assume that P = ⟨α⟩ × ⟨β⟩ ≅ E4 . We have P = {1, α, β, αβ} and any two elements from P# are not conjugate in G. It follows that |K| = |G : P| = i(G) = |G : CG (α)| + |G : CG (β)| + |G : CG (αβ)| (the number |G : CG (α)| is the size of the G-class, containing α). Note that CG (x) = P ⋅ CK (x) for all x ∈ P# . Therefore, |CK (α)|−1 + |CK (β)|−1 | + CK (αβ)|−1 = 1.

(1)

Since |CK (x)| > 1 is odd for all x ∈ P# , it follows from (1) that |CK (α)| = |CK (β)| = |CK (αβ)| = 3.

(2)

By the Brauer formula (see Theorem XVI.6.3), we have |K||CK (P)|2 = |CK (α)||CK (β)||CK (αβ)| = 33 .

(3)

If CK (P) > {1}, then (3) implies |K| = 3. In that case, G = P × K is an abelian noncyclic group of order 12. Now assume that CK (P) = {1}. In that case, |K| = 33 , by (3). Then (2) implies that K is noncyclic (otherwise, by the Frobenius normal p-complement theorem (p = 3), P is a direct factor of G, a contradiction). It follows from exp(P) = 2 that G is supersolvable (see Exercise III.8.5). Let R < K be G-invariant of order 3. Then K/R is noncyclic. Let U/R and V/R be G-invariant of order 3 (here we use Maschke’s theorem and supersolvability of G/R). Considering the subgroups PU and PV, we see that U and V are noncyclic (since the centralizer of every involution in G has order 12) so exp(K) = 3. Now G/R ≅ S3 × S3 , by the exercise preceding the lemma. We see that G is as in (e).

XI On sums of degrees of irreducible characters | 551

Proof of Theorem 7.2. Let S(G) be the solvable radical of G. (a) Let us prove that G = G󸀠 Z(G). Claim (i). If G is simple, then G ≅ PSL(2, 5). Indeed, take d = 4 in Lemma 7.4. By that lemma and Lemma 7.3(a), there is a character χ ∈ Irr(G) with χ(1) = 3. In that case, by Lemma 7.3 (a), G ∈ {PSL(2, 5), PSL(2, 7)}. 1 1 6 = 28 < 16 , it remains to consider the case G ≅ PSL(2, 5). Since As mc(PSL(2, 7)) = 168 5 1 1 mc(PSL(2, 5)) = 60 = 12 > 16 , this group satisfies the hypothesis. Claim (ii). If S(G) = {1}, then G ≅ PSL(2, 5). Indeed, take in G a minimal normal subgroup D. Then D = D1 × ⋅ ⋅ ⋅ × D s , where D i are 1 isomorphic nonabelian simple groups. Since, by Lemma 3.3 (a), mc(D i ) ≥ mc(G) ≥ 16 , 1 we get D i ≅ PSL(2, 5), by (i), and so mc(D i ) = 12 . Next, by Lemma 3.3 (a), mc(D) = mc(D1 )s = (

1 s 1 , ) ≥ mc(G) ≥ 12 16

hence s = 1 so that D ≅ PSL(2, 5). Since the quotient group G/CG (D) is isomorphic 1 7 , < 16 to a subgroup of Aut(PSL(2, 5)) ≅ S5 containing PSL(2, 5) and mc(S5 ) = 120 we get G/CG (D) ≅ PSL(2, 5). However, one has D ∩ CG (D) = {1} so G = D × CG (D). Assume that CG (D) > {1}. Since S(G) = {1}, it follows that CG (D) is nonsolvable and so 1 mc(CG (D)) ≤ 12 , by Theorem 7.1. We have mc(G) = mc(D) mc(CG (D) ≤

1 1 1 ⋅ < , 12 12 16

a contradiction. Thus, CG (D) = {1} so G = D ≅ PSL(2, 5). Claim (iii). We have G/S(G) ≅ PSL(2, 5). Indeed, this follows from mc(G/S(G)) ≥ mc(G) ≥

1 16

(Corollary VII.8.5) and (ii).

Claim (iv). If G󸀠 = G, then G ∈ {PSL(2, 5), SL(2, 5)}. Indeed, in view of (iii), one may assume that S(G) > {1}. Suppose that our assertion holds for all proper epimorphic images of G. Let R < G be an abelian minimal normal subgroup of G, and put |R| = p n . Then, by Corollary VII.8.5 and induction, one obtains G/R ∈ {PSL(2, 5), SL(2, 5)}. Case 1. Suppose that G/R ≅ PSL(2, 5) and |G| = 60p n ; then R = S(G). If Z(G) > {1}, then R = Z(G) is isomorphic to a subgroup of the Schur multiplier of G/R so |R| = 2 and G ≅ SL(2, 5) (Schur). In the sequel we suppose that Z(G) = {1}. In that case, CG (R) = R so n > 1. If x ∈ R# , then |G : CG (x)| ≥ 5, where 5 is the minimal index of a proper subgroup in PSL(2, 5). Let kG (M) denote the number of G-classes containing elements from M ≤ G (so that kG (G) = k(G)). Then kG (R) ≤ 1 +

|R# | p n + 4 = . 5 5

552 | Characters of Finite Groups 1 If x ∈ G − R, it follows from CG (R) = R that |G : CG (x)| ≥ 12p (indeed, x does not centralize R so that |G/R : CG (xR)| ≥ 12). Hence, 60p n p n + 4 71p n − 16 − = kG (G − R) = k(G) − kG (R) = |G| mc(G) − kG (R) ≥ 16 5 20 and 12p(71p n − 16) p(213p n − 48) |G − R| = 59p n ≥ 12pkG (G − R) ≥ = . 20 5 The above inequality implies 5 ⋅ 59p n−1 = 295p n−1 ≥ 213p n − 48 ≥ 426p n−1 − 48 󳨐⇒ 131p n−1 ≤ 48, a contradiction. Case 2. Suppose that G/R ≅ SL(2, 5). Then S = S(G) has order 2|R|. Assume that R1 ≠ R is a minimal normal subgroup of G. As S(G) = R × R1 is of order 2|R|, we get |R1 | = 2. By induction, G/R1 ≅ SL(2, 5), and then G󸀠 < G since the Schur multiplier of SL(2, 5) is trivial (Theorem VII.4.6 (b)), a contradiction. Thus, R is a unique minimal normal subgroup of G so that CG (R) = R, and we conclude that n > 1. Then Z(G) = {1}. Suppose that p > 2. In that case, S(G) is a Frobenius group with kernel R (indeed, take in S an involution x and consider CG (x) using Frattini’s lemma). As in Case 1, we have kG (S) = kG (S − R) + kG (R) ≤ 1 +

pn + 4 pn + 9 = . 5 5

If x ∈ G − S, then |G : CG (x)| ≥ 12p, and so kG (G − S) = k(G) − kG (S) = |G| mc(G) − kG (S) ≥ and |G − S| = 118p n ≥ 12pk(G − S) ≥ It follows that

120p n p n + 9 73p n − 18 − = 16 5 10 6p(73p n − 18) . 5

295p n−1 ≥ 219p n − 54 ≥ 666p n−1 − 54 󳨐⇒ 54 ≥ 371p p−1 , a contradiction. Now assume that p = 2. By the above, R is a unique minimal normal subgroup of G and Z(G) = {1}. We have kG (S) ≤ 1 + and kG (G − S) ≥ so that

2n+1 − 1 2n+1 + 4 = 5 5

120 ⋅ 2n 2n+1 + 4 71 ⋅ 2n−1 − 4 − = , 16 5 5

|G − S| = 59 ⋅ 2n+1 ≥ 12 ⋅ 2kG (G − S) ≥ and we get 24 ≥ 131 ⋅ 2n−1 , a contradiction.

24 ⋅ (71 ⋅ 2n−1 − 4) , 5

Claim (v). If D is the last term of the derived series of G, then D ∈ {PSL(2, 5), SL(2, 5)}. Indeed, as D󸀠 = D and mc(D) ≥ mc(G) ≥

1 16 ,

the result follows from (iv).

XI On sums of degrees of irreducible characters | 553

Claim (vi). The subgroup D from (iv) coincides with G󸀠 , i.e., G/D is abelian. Indeed, we have D ∈ {PSL(2, 5), SL(2, 5)}, by (v) and (iv). Assume that G/D is nonabelian. Then, by Gallagher result (Corollary VII.8.5), we get mc(G) ≤ mc(D) mc(G/D). We have mc(G/D) ≥ 85 . If D ≅ PSL(2, 5), then mc(G) ≤

5 1 1 5 ⋅ = < ≤ mc(G), 12 8 96 16

a contradiction. Now let D ≅ SL(2, 5). Then 9 5 3 1 mc(G) ≤ ⋅ = < ≤ mc(G), 120 8 64 16 a contradiction. Claim (vii). One has G = SG󸀠 , where S = S(G). This follows from (iii) and (iv). Claim (viii). One has |S󸀠 | ≤ 2. In particular, S is nilpotent and all its Sylow subgroups of odd orders are abelian. In fact, S󸀠 ≤ G󸀠 ∩ S ≤ Z(G). Now the result follows from (vii). Claim (ix). One has G = S ∗ G󸀠 . Indeed, let x ∈ G󸀠 be of order 5. Since G󸀠 ∩ S ≤ Z(G), then G/(G󸀠 ∩ S) = G󸀠 /(G󸀠 ∩ S) × S/(G󸀠 ∩ S) 󳨐⇒ ⟨x, S⟩ is nilpotent. Hence, ⟨x, S⟩ = P × A, where P ∈ Syl2 (S) and A is abelian. As G󸀠 = ⟨x ∈ G󸀠 | x5 = 1}, it follows that [G󸀠 , S] = {1} so that G = SG󸀠 = S ∗ G󸀠 . It follows from (viii) and (ix) that 2󸀠 -Hall subgroup of S is an abelian direct factor of G so, without loss of generality, one may assume that S is a 2-subgroup. Claim (x). The subgroup S is abelian. Assume that S is nonabelian; then mc(S) ≤ 58 (Exercise 5.5). If G󸀠 ≅ PSL(2, 5), then G = G󸀠 × S and 1 1 5 5 1 mc(S) ≤ ⋅ = < ≤ mc(G), mc(G) = mc(G󸀠 ) mc(S) = 12 12 8 96 16 a contradiction. Now let G󸀠 ≅ SL(2, 5). Then G ≅ (A × B)/Z, where A ≅ G󸀠 and B ≅ S. By Corollary VII.8.5, mc(G) ≤ mc(A) mc(B) =

9 5 3 1 ⋅ = < 16 ≤ mc(G), 120 8 64 0

a contradiction. Thus, S is abelian, and now it follows from (ix) that S = Z(G). (b) Since G is nonabelian, we have, by Lemma 3.4, G = G󸀠 Z(G), where G󸀠 ∈ {PSL(2, 5), SL(2, 5)}.

1 16

≤ f(G)2 < mc(G). By (a), we get

554 | Characters of Finite Groups 1 1 (Lemma 3.4). Therefore, we (c) We have f(G) ≥ t0 (G) ≥ 41 so that mc(G) > f(G)2 ≥ 16 󸀠 󸀠 get G = G Z(G), where G ∈ {PSL(2, 5), SL(2, 5)}. Since mc(G󸀠 ) ≥ mc(G) (Lemma 3.2), we get G󸀠 ≅ PSL(2, 5). Now clear that G = G󸀠 × E, where exp(E) ≤ 2.

2°. In this subsection we find all groups G of order 2n and exponent > 2 with maximal possible number i(G) of involutions. We use the following facts. If G is of order 24 and exponent > 2, then i(G) ≤ 11 with equality if and only if G ≅ D8 × C2 . If |G| = 25 and exp(G) > 2, then i(G) ≤ 23 with equality if and only if G ≅ D8 × E4 . Note that i(G) = c1 (G), the number of subgroups of order 2 in G. Theorem 7.6. If G is a group of order 2m , m > 2 and G ≇ E2m , then i(G) ≤ 3 ⋅ 2m−2 − 1 with equality if and only if G ≅ D8 × E2m−3 . Proof. As we have noticed, the theorem is true for m = 4, 5. By the Frobenius–Schur formula, 1 + i(G) ≤ T(G) =

χ(1) ≤ |G : G󸀠 | +

∑ χ∈Irr(G)

=

|G| − |G : G󸀠 | ⋅2 22

1 1 |G| + |G : G󸀠 | ≤ 2m−1 + 2m−2 = 3 ⋅ 2m−2 . 2 2

Indeed, the contribution of one irreducible character χ of degree 2k > 2 in the sum ∑χ∈Irr(G) χ(1) equals 2k ; on the other hand, the contribution of 22k−2 irreducible characters of degree 2 in that sum equals 22k−1 ≥ 2k (the sum of squares of degrees of those characters is 22 ⋅ 22k−2 = (2k )2 ). Now suppose that i(G) = 3 ⋅ 2m−2 − 1. Then it is easy to show that in this case cd(G) = {1, 2}, |G󸀠 | = 2 and every irreducible character of G is afforded by a real representation. It follows that exp(G/G󸀠 ) = 2. We have |Irr(G)| = 2m−1 +

2m − 2m−1 = 2m−1 + 2m−3 = 5 ⋅ 2m−3 . 22

Let |Z(G)| = 2s . Then k(G) = 2s +

2m − 2s = 2m−1 + 2s−1 . 2

We get 5 ⋅ 2m−3 = |Irr(G)| = k(G) = 2m−1 + 2s−1 , or 2m−3 = 2s−1 and so s = m − 2. Thus, |G : Z(G)| = 4. If A is a minimal nonabelian subgroup of G, then |A| = 8 (since A󸀠 = G󸀠 and exp(G/G󸀠 ) = 2) and G = AZ(G), hence G/A ≅ E2m−3 . Assume that exp(Z(G)) = 4. Then i(AC) = 7, where C is a cyclic subgroup of order 4 in Z(G). In that case, we have G = (AC) × E, where E2m−4 ≅ E < Z(G). Then, however, i(G) = 8 ⋅ 2m−4 − 1 = 2 ⋅ 2m−2 − 1 < 3 ⋅ 2m−2 − 1, a contradiction. Thus, we obtain that exp(Z(G)) = 2 and G = A × E, where E ≅ E2m−3 . If A ≅ Q8 , then i(G) = 2m−2 − 1 < 3 ⋅ 2m−2 − 1, a contradiction. Thus, A ≅ D8 ; in that case, i(G) = 3 ⋅ 2m−2 − 1.

XI On sums of degrees of irreducible characters | 555

8 Irreducible constituents of induced characters Given τ ∈ Char(G), put s(τ) = |Irr(τ)|, w(τ) = ∑χ∈Irr(G) ⟨τ, χ⟩; then w(τ) is the sum of coefficients in decomposition of τ in irreducible constituents, and so w(τ) ≥ s(τ). The main result of this section is the following Theorem 8.1 ([Ber19]). Let H ⊲ G, let p be the least prime divisor of |G : H| and let ϕ ∈ Char(H). Suppose that s(ϕ G ) |G : H| > . s(ϕ) p2 Then G/H is solvable. Lemma 8.2. If p is the least prime divisor of |G| and mc(G) ≥

p+1 , p2

then G is abelian.

Proof. Let G be a counterexample of minimal order. Then, by Lemma 3.3 (a), every proper subgroup A of G with p | |A| is abelian. By Frobenius’ normal p-complement theorem, G is a minimal nonabelian group. If G is a p-group, then, by Lemma 2.3 (a), one has p2 + p − 1 p2 + p p + 1 mc(G) = < = , p3 p3 p2 contrary to the hypothesis. Now, let G be nonnilpotent. Then we have G = RQ, where b 2 −1 Q = G󸀠 ∈ Sylq (G), R ∈ Sylr (G), |Q| = q b . By Lemma 2.2 (b), mc(G) = q r+r . If r < q, 2 qb then r = p and p + 1 p2 − 1 − pq b mc(G) − = < 0, p2 p2 q b a contradiction. If r > q, then q = p and b > 1 hence r2 p b (mc(G) −

p+1 ) = p b−1 (p − r2 ) − r2 (p b−2 − 1) − 1 < 0, p2

and we again obtain a contradiction. Proof of Theorem 8.1. (a) First assume that H = {1}. Since Irr(H) = {1H }, one may assume that ϕ = 1H ; then ϕ G = ρ G is the regular character of G. Let Irr(ϕ G ) = Irr(G) = {χ1 , . . . , χ k }, We have s(ϕ) = 1,

s(ϕ G ) = k,

s(ϕ G ) = k, s(ϕ)

Thus, k>

k = k(G).

mc(G) =

k 1 > 2. g p

g . p2

We claim that if G satisfies the above relation, then it is solvable. Let G be a counterexample of minimal order. For L ≤ G we have mc(L) ≥ mc(G) > p12 (Lemma 3.3 (a)) so all proper subgroups of G are solvable. Then G󸀠 = G. Hence, if χ ∈ Irr(G) with χ ≠ 1G ,

556 | Characters of Finite Groups 1 then χ(1) ≥ p (Frobenius–Molien). Let 1 < χ2 (1) ≤ ⋅ ⋅ ⋅ ≤ χ k (1). Since |cd(G)| ≥ 4 (see Theorem 7.1), we have χ k (1) ≥ p + 2. Therefore, k

k−1

|G| = ∑ χ i (1)2 = 1 + ∑ χ i (1)2 + χ k (1)2 i=1

i=2 2

≥ 1 + (k − 2)p + (p + 2)2 = 1 + (k − 1)p2 + 4p + 4 = 4p + 5 + (k − 1)p2 . Since p is the least prime divisor of |G| and G󸀠 = G, it follows from the Burnside normal ∈ ℕ. By hypothesis, k − 1 ≥ |G| . Hence, p-complement theorem that |G| p2 p2 |G| ≥ 4p + 5 +

|G| 2 ⋅ p = 4p + 5 + |G|, p2

a contradiction. Thus, if H = {1}, then G is solvable. (b) Now let the G-invariant subgroup H > {1}. We proceed by induction on ϕ(1). for all To this end, let Irr(ϕ) = {ϕ1 , . . . , ϕ s }. Suppose that s > 1 and s(ϕ Gi ) ≤ |G:H| p2 i ∈ {1, . . . , s}. Then k

s(ϕ G ) ≤ ∑ s(ϕ Gi ) ≤ i=1

s(ϕ)|G : H| s(ϕ G ) |G : H| 󳨐⇒ , ≤ 2 s(ϕ) p p2

a contradiction. If, say, s(ϕ Gi ) > |G:H| for some i, then, by induction, G/H is solvable. p2 Therefore, we may assume that ϕ ∈ Irr(H). Let Irr(ϕ G ) = {χ1 , . . . , χ s },

ϕ G = e1 χ1 + ⋅ ⋅ ⋅ + e s χ s .

If χ iH = e i (ϕ1 + ⋅ ⋅ ⋅ + ϕ t ) is the Clifford decomposition, ϕ1 = ϕ, t = |G : IG (ϕ)|, then χ i (1) = e i tϕ(1) for all i, which implies |G : H|ϕ(1) = ϕ G (1) = tϕ(1)(e21 + ⋅ ⋅ ⋅ + e2s ) = |G : IG (ϕ)|ϕ(1)(e21 + ⋅ ⋅ ⋅ + e2s ), and so |IG (ϕ) : H| = e21 + ⋅ ⋅ ⋅ + e2s ≥ s >

|G : H| 󳨐⇒ t = |G : IG (ϕ)| < p2 . p2

Since t | |G : H| and p is the least prime divisor of |G : H|, it follows that t is either a prime number or t = 1. Case 1. Suppose that t is a prime. Then IG (ϕ) is maximal in G, and hence IG (ϕ)/H is maximal in G/H. We have ϕIG (ϕ) = e1 ψ1 + ⋅ ⋅ ⋅ + e s ψ s ,

Irr(ϕIG (ϕ) ) = {ψ1 , . . . , ψ s }.

Suppose that e i > 1 for all i. Then e i ≥ p for all i, since e1 , . . . , e s are divisors of |IG (ϕ)/H| as the degrees of irreducible projective representations of IG (ϕ)/H. Hence, |IG (ϕ) : H| = e21 + ⋅ ⋅ ⋅ + e2s ≥ p2 s >

p2 |G : H| = |G : H|, p2

XI On sums of degrees of irreducible characters | 557

a contradiction. Assuming that e1 ≤ ⋅ ⋅ ⋅ ≤ e s , we have e1 = 1. Then, by reciprocity, (ψ1 )H = ϕ since ψ is IG (ϕ)-invariant, and, by Gallagher’s theorem, Irr(IG (ϕ)/H) = {ψ1 , . . . , ψ s }, ψ i (1) = e i , i = 1, . . . , s. Since |G| = t|IG (ϕ)|, it follows that mc(IG (ϕ)/H) = s|IG (ϕ)/H|−1 = s|G/H|−1 t >

1 t > , p2 p2

and so IG (ϕ)/H is solvable, by (a). If t = p, then IG (ϕ)/H ⊲ G/H and G/IG (ϕ) is cyclic. Therefore G/H is solvable. Assume that t > p. Then p+1 t . mc(IG (ϕ)/H) > 2 ≥ p p2 By Lemma 8.2, IG (ϕ)/H is abelian so that G/H is solvable, as a group containing an abelian maximal subgroup (the Herstein theorem). Thus the theorem is proved for prime t. Case 2. Suppose that t = 1; then ϕ is G-invariant. Since e1 = 1, it follows that χ1H = ϕ. Let Irr(G/H) = {χ1 , . . . , χ s }. By Gallagher’s Theorem VII.4.5, |Irr(ϕ G )| = |Irr(G/H)|. By assumption, s > |G : H|/p2 . Then mc(G/H) >

s 1 , > |G : H| p2

and G/H is solvable, by (a). Similarly (but in a simpler way), one can prove the following result: Exercise. Let H ⊲ G, let p be the least prime divisor of |G : H| and let ϕ ∈ Char(H). Suppose that w(ϕ G ) |G : H| > , w(ϕ) p Then G/H is solvable.

9 Mann’s theorem The following nice theorem (the answer on a question of the first author) was proved by A. Mann. Theorem 9.1. Let G be a nonsolvable group. Then T(G) > 2|G : G󸀠 |. Proof. Subtracting |G : G󸀠 | from both sides of the claimed inequality, we see that the theorem is equivalent to the claim that T1 (G) = ∑χ∈Irr1 (G) χ(1) > |G : G󸀠 |. Let N be the last member of the derived series of G, and let M be maximal among the normal subgroups of G that are contained in N. Since N = N 󸀠 ≤ G󸀠 , it suffices to prove the theorem for G/M, and thus we may assume that M = {1} and N is a minimal

558 | Characters of Finite Groups 1 normal subgroup of G. Then N is a direct product of a certain number, say k, of isomorphic nonabelian simple groups. Let S be one of this simple factors. Write s = |S| and C = CG (N). First suppose that C = {1}. Then G/N is isomorphic to a subgroup of Out(N). Let L be the intersection of the normalizers of the simple factors of N. Then L ⊲ G, and G/L is the group of permutations that G induces on the set of these simple factors, and thus G/L is isomorphic to a subgroup of Sk . By a result of L. G. Kovacs and C. E. Praeger [KP], k |G/L : G󸀠 L/L| ≤ 3 3 . Since N ≤ L, we have |G : G󸀠 | = |G/N : G󸀠 N/N| ≤ |G/L : G󸀠 L/L| ⋅ |L/N|. Since L/N is isomorphic to the direct product of k copies of a certain subgroup 1 of Out(S), we have, by [CCNPW], |L/N| < (log s)k , and so |G : G󸀠 | < (3 3 log s)k . On the other hand, by reciprocity, T1 (G) ≥ T1 (N) > √ s k − 1 (we use here the inequality T1 (S) > √s − 1 which is true for any nonidentity perfect group S; see Lemma 3.4), and thus our theorem follows since for s ≥ 60 we have √ s k − 1 > (3 31 log s)k ≥ |G : G󸀠 |. In fact, we have proved the following inequality: T1 (N) > |G : G󸀠 |. Now let C > {1}. We know already that the theorem holds for G/C, so T1 (G) ≥ T1 (N × C) ≥ T1 (N) ⋅ |C : C󸀠 | > |G/C : G󸀠 C/C| ⋅ |C : C󸀠 | = |G : G󸀠 C||C : C󸀠 | ≥

|G||C||G󸀠 ∩ C| |G||C󸀠 | |G||C| = ≥ 󸀠 󸀠 = |G : G󸀠 | 󸀠 󸀠 󸀠 󸀠 |G C||C | |G ||C||C | |G ||C |

since N ≅ (N × C)/C ⊴ G/C. For further results in this direction, see [MT-V].

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Author index A Alperin J. L., VI.10 Amberg B., Preface Amitzur S. A., XI.6 Arad Z., Preface, IV.10 Artin E., V.4 B Baer R., III.11 Bender, Preface Berger T., V.10 Berman S. D., V.10, VII.15 Berkovich Y., Preface, I.1, V.11, V.14, VI.11, VI.13, VII.12, VII.17, VII.18, VIII, IX.7, X, XI, I.A Blichfeldt H. F., IV.4, VII.13 Blau H., Preface, IV.10 Brauer R., Preface, I.B, II.7, IV.4, IV.7, IV.10, V.1, V.4, V.6, V.7, V.11, VI.10, VIII, X.6, X.11, X.12 Brenner, Preface Burnside W., I.1, II, III.5, III.6, III.7, III.11, IV.4, IV.6, V.4, V.7, V.9, VII.3, VIII.5, X.2, XI C Cameron P. J., I.B Casolo C., III.11 Cayley A., I.1 Cauchy A.-L., I.1, V.4, V.7 Chankov E. I., Preface, III.9 Chillag D., Preface, IV.4, IV.10 Cunikhin S. A., III.6 Clifford A. H., Preface, VII Cossey J., X.7 D Dade E. C., II.9, V.10, VII.13, VII.15, VIII.6 Dornhoff L., Preface, V.13, VII.9 DOI DOI 10.1515/9783110224078-013

E Ernest L., XI.3 Euler L., IX.3 F Feit W., Preface Ferguson P., III.11, VII.18 Fisman E., IV.10 Frattini G., I.9, II.7 Frobenius G., Preface, I.1, I.9, 2, III.2, III.3, IV.6, IV.8, IV.11, V.1, V.4, V.7, V.11, VIII.1, VIII.4, VIII.5, X.2, X.3, X.4 Frucht R., Preface, VI.6 Fowler K. A., Preface, I.B G Gagola S., I.9 Gallagher P. X., Preface, III.12, V.8, VII.4, VII.6, VII.8 Galois E, III.4, V.11 Garrison S., Preface, II.7, IV.4, IV.10 Gaschütz W., Preface, VI.11, IX.5, X.3, X.11 Glauberman G., VII.17, X.11 Goldschmidt, Preface Golfand Y. A., XI.2 Gorenstein D., Preface, III.8 Gow R., III.8 Grün O., X.3, X.10 H Hall P., III.6, V.8, V.11 Hawkes T. O., X.7 Herstein I., X.2, XI.8 Herzog M., Preface, IV.10 Higman G., III.7, X.3, X.4 Hölder O., I.10 Howlett R. B., III.5 Huppert B., Preface, VII.13

588 | Characters of Finite Groups 1 I Isaacs I. M., Preface, VII.5, VII.18, X.7 Itô N., Preface, VII.2, XI.5 K Karpilovsky G., Preface Kazarin L. S., Preface, III.9 Kronecker L., III.1 Kuo T.-N, X.6 L Lam T. Y., V.4 Leitz M., IV.9 Liebeck H., XI.7 M MacHale D, XI.7 Mackey G., V.3, V.12 MacWilliams A., IV.6 Mann A., Preface, III.9, III.11, IV.12, V.14, X.7, XI.6, XI.9 Maschke H., Preface, I.8, VIII.2 Matsuyana, Preface Miller G. A., III.5 Möbius A., IV.11, V.11, IX.3 Molien T., II.5, III.3, IV.8 Moore E. H., I.1 N Nagao H., II.8 Nakayama T., V.15 Navarro G. O., Preface Nekrasov K. G., XI Neubüser J., Preface, VI.11 Neumann B. H., I.B Neumann P. M., I.B, XI.7 O Osima M., V.15 Ostrovskaya S., Preface

P Passman D., VII.14, VII.17, XI.6 Price D., V.3 R Redei L., XI.2 Reynolds W. F., Preface, VI.10 Roitman M., Preface, I.1 Roquette P., VII.10 Rusin D., XI.3 S Sagirov I. A., Preface Saksonov A. I., II.9, III.7, X.3, X.4 Schumyatsky P., VII.17 Schur I., Preface, I.7, I.12, II.2, IV.6, IV.8, VI Schmidt O. Y., X.12, XI.2 Scoppola C., XI.6 Shoda K., Preface, V.3, X.5 Seiz G., III.5 Snyder N., X.9 Solomon L., II.9, V.11 Stickelberger L., I.9 Strunkov S. P., I.B Suzuki M., Preface, V.6, X.2, X.6, X.11, X.12 T Taketa K., V.10, V.13 Tate J., Preface, IV.8, V.1, VII.10 Taussky O., VI.6 Thompson J. G., III.5, V.8, X.2, X.3 Ti Yen, VI.11 V Vishnevetsky, A. V., II.6, X.3 W Wall C. T. C., XI.7 Wall G. E., V.6

Author index

Waall van der R., V.10, VII.15 Walter J. H., III.8 Wedderburn J. H. M., I.11, II Weisner L., X.5 Weiss M., X.6 Wiegold, J. VI.11 Wielandt H., III.6, V.11, VII.10, VIII.5, X.2 Wigner E. P., Preface Witt E., X.3 Y Yamazaki K., VI.6 Yanishevskii V., Preface Z Zalesski A. E., Preface Zassenhaus H., V.6, X.3 Zisser I., Preface, IV.8 Zhmud E. M., Preface, I.1, III.5, III.6, V.1, VI, VII.6, VII.7, VII.A, IX, X.3

| 589

Subject index A abelian group of symmetric type, VI.6 action by conjugation, I.1 action by multiplication, I.1 action, faithful, I.1 action of a group, I.1 action, transitive, I.1 algebra (associative), I.1 algebra EndA (M) of endomorphisms of homogeneous A-module M, I.10 algebra of G-layer character, IX.4 algebra, semisimple, I.11 algebra, twisted, VI.2 algebraic integer, III.1 algebraic number, III.1 algebraically closed field, I.7 algebraically conjugate characters, III.4 Alperin–Kuo Tze-Nan theorem, VI.10 alternating group, I.1 annihilator, I.3, I.9 annihilator of a subgroup of an abelian group, VI.6 antikernel, IX.1 antisocle, IX antisymmetric bicharacter, VI.6 antisymmetric m-form, IV.2 Arad–Mann–Fisman theorem, IV.10 Artin exponent, V.4 Artin index, V.4 Artin’s induction theorem, V.4 autodual lattice, I.10 averaging method, I.8 B basis of an abelian group, I.9, VI.6, basis of a vector space, I.4, IV.1 Berkovich–Nekrasov theorem, XI bicharacter, VI.6 biprimary group, III.3 DOI DOI 10.1515/9783110224078-014

Blau–Cillag theorem, IV.10 Brauer–Burnside theorem, IV.4 Brauer elementary (p-elementary) group, VIII.1 Brauer’s characterization of characters, VIII.1 Brauer’s induction theorems, VIII Brauer’s paper on quotient groups, VIII.5 Brauer’s permutation lemma, X.3 Burnside’s p a -lemma, III.6 Burnside’s p-nilpotency theorem, V.9 Brauer’s realization theorem, VIII.2 Brauer–Suzuki theorem on groups with generalized quaternion Sylow 2-subgroup, Glauberman’s proof of, X.12 Brauer–Suzuki–Wall theorem, V.6 Burnside’s theorem on Frobenius complements, X.3 Burnside’s theorem on subgroups of small index in doubly transitive groups, V.7 Burnside’s theorems on commutators, III.7 Burnside’s two-prime theorem, III.6 C ℂ-monolith, IX CA-group, Suzuki’s theorem, X.6 Carter subgroup, VII.10 cd(G), the set of irreducible character degrees of G, XI center of a group admitting a faithful irreducible matrix representation, I.7 center of a group algebra, II.5 center of an algebra (of full matrix algebra), I.10, I.12

592 | Characters of Finite Groups 1 center of an irreducible group, I.7 center of the twisted group algebra, VI.3 centerless CM-groups, IX.6 central character, II, III.4 central extension, VI.3 central (= class) function, II.1, VII.1 central idempotents, I.11 central product, XI character, II character degree, II character, irreducible, number of, II.2, II.5 character of a permutation representation, V.4 character of a representation, II.3 character of a representation of a group (algebra), II.3, I.12 character of a tensor product, exterior and symmetric powers of representations, IV.4, IV.5 character of defect 0, VIII.4 character restriction property, VII.18 character table, II.2, II.6, II.7 character values are algebraic integers, III.2 characterization of characters, VIII.1 characterization of extraspecial p-groups, III.9 characterization of Frobenius complements, X.8 characterization of kernel and quasikernel of character in the language of its values, II.5 characterization of p-groups on the language of degrees and kernels irreducible characters, IX.7 characterization of p-groups which are CM-groups, on the language of degrees and kernels irreducible characters, IX.7 characterizations of Frobenius groups, X.4

characters of defect 0, vanishing of, IV.9, VIII.4 characters of Frobenius groups, X.3 characters of p-groups, III.9, VII.18 class equation, III.10 class number, II.5, XI class number of a group containing a normal subgroup of index p, VII.3 class number of extension, Gallagher theorem, VII.8 Clifford’s decomposition, VII.1 Clifford’s theorem, VII.1 Clifford’s theorem on ramification, VII.2 CM-groups, structure of, IX.6 CM2 - and CM3 -groups, solvability of, IX.7 cocycle, VI.2 codimension, IX commutator, I.1 commutator subgroups of representation groups, VI.7 compatible factor set, VI.5 completely reducible modules, I.6 completely reducible representations, I.6 complex conjugate characters, II.4 complex general linear group, I.5 conjugacy class, II.5 conjugate numbers, III.1, III.4 connection between linear representation of the twisted group algebra ℂπ G and projective π-representations the group G, VII.2 constituents irreducible of a character, II.4 contragredient characters, III.4, IV.6 convolution, I.4, V.11 Cossey–Hawkes–Mann theorem, X.7 counterexamples to the inequality f(G) ≤ f(G/H), XI.5 covering group, VI.3 CR- and NR-subgroups, VII.18

Subject index | 593

criterion for the existence of a faithful character (Gaschütz, Weisner, Shoda), IX.5 criterion of the simplicity of an algebra (group algebra), I.11 cyclic group, I.1 cyclotomic field, III.2, VIII.2 D Dade’s example, VII.15 Dade’s theorems, VII.12, VIII.6 decomposable module, I.3 decomposition of semisimple algebra in direct sum of two-sided ideals, I.11 decomposition of the regular representation of a semisimple algebra over a (algebraically closed) field, I.12 defect zero character, IV.9, VIII.4 degree of a character, II.3, XI degree of a linear (projective) representation, I.4, I.5, VI.1 degree of a permutation group, I.1 degree of induced character, V.1 degrees of irreducible projective representations, divisibility property, VI.3 derived length of an M-group, V.10 descent (lowering) of a linear representation, VII determinant of a character, VII.4 dihedral group, III.3, VI.4 dimension of a vector space, I.3 dimension of an algebra, I.3 direct product of groups, IV.8 direct sum of modules, I.4 Dornhoff’s theorems, V.13, VII.9 dual space, I.9, IV E element of the first (second) type, VIII.5 endomorphism ring, I.4

equivalent factor sets, VI.1 equivalent representations (linear, projective), IV.1, VI.1 Euler function, X.3 Euler totient function, III.4 exceptional character, X.6 existence of a representation group, VI.3 exponent of a group, Appendix I.A exponent of Schur multiplier, VII.4 extension of characters, VII.4, VII.6 extension of groups, VII.8, VII.11 exterior power of spaces, operators, modules, representations, IV.2, IV.4 extraspecial p-group, II.5, III.3, III.9 F 𝔽-algebra, I.3 factor set, VI.1 faithful central extension, VI.4 faithful character, II.1, IX.1, X.12 faithful homomorphism, I.1 faithful representation, IX FG-module, I.3 f(G), XI.1 finiteness of the Schur multiplies, VI.3 first orthogonality relation, II.3 Fitting subgroup, II.7 fixed-point-free automorphism, properties of, X.2 Frattini subgroup, II.7, Appendix I.A Frobenius complement, properties, X.2, X.3 Frobenius groups, properties of, X.3 Frobenius kernel, properties, VIII.5, X Frobenius–Molien theorem, III.3 Frobenius reciprocity, V.1 Frobenius–Schur for the number of involutions, IV.6 Frobenius–Schur indicator, IV.6 Frobenius theorem on Frobenius groups, VIII.5, X.2

594 | Characters of Finite Groups 1 Frobenius theorem on the number of solution of x n = 1, IV.11 (L. Solomon’s proof and elementary proof), V.11 Frobenius theorem on the solvable groups with exactly n solutions of x n = 1, V.11 Frobenius-type group, III.9 Frucht’s theorem, VI.6 full matrix algebra, I.3 G G-antikernel, IX.4 G-conjugate, V.1 G-group, IX.1 G-kernel, IX.4 G-module, ℂG-module, I.2 G-orbit, I.1 G-transitive, I.3 Gallagher’s theorems, III.12, V.8, VII.4, VII.6, VII.8, VII.11 Galois group, I.1, II.9, V.8, VII.4, VII.6, VII.8, VII.11 Garrison’s theorems, II.7, IV.4 Gaschütz’ criterion of existence of a faithful irreducible character, IX.5 Gaschütz–Neibüser–Ti Yen theorem, VI.11 general matrix algebra, II.5 generalization of M-groups, V.13 generalized characters, characterization of, IV.5, V.4, VIII generalized quaternion group, multiplier of, VI, X.11 Glauberman’s theorem, VII.17, X.12 group, I.1 group (abelian) of symmetric type, VI.6 group algebra, I.4 group of Frobenius type, III.9 group of linear characters, I.9 group-theoretical functions of Euler (of the first and the second type), IX.2

group-theoretical function of Möbius, IX.2 groups in which two greatest sizes of the conjugacy classes are consecutive integers, Mann’s theorem, III.11 groups with induced character having a large number of irreducible constituents, XI.8 groups with many involutions, Nekrasov’s proof, XI.7 groups with metacyclic Sylow 2-subgroups, II.8 groups with partitions of special type, VIII.3 H Hall subgroup, III.6 Hall’s theorem on solvable groups, III.6 height of a character, X.12 height of a group, X.12 homocyclic p-group, Appendix I.B homogeneous completely reducible modules, I.10 homogeneous components of completely reducible modules, I.10 homomorphisms groups, algebras, nodules, I Huppert’s monomiality criterion, VII.13 hyperbolic abelian p-groups, VI.6 I ideal (left, right, two-sided), I idempotent basis of the algebra Z(ℂG), III.4 idempotent of algebra, I.11, III.4, VI.9 image of a homomorphism, I.1 indecomposable module representation, I.3 indicator of Frobenius-Schur, IV.6 induced character, V.2 induced function, V.1 induced representation, V.5

Subject index | 595

inertia subgroup of an irreducible character of a normal subgroup, its index, VII.1 inflation of a character, I.9 inner product, II.1 intersection of all maxima left (right) ideal of an algebra, I.10 intersection of kernels and quasikernels, IV.9, V.14 intersection of kernels of projective representations, VI.8 invariant character, VII.1 invariant of the group, IV.6 Irr(G), the set of irreducible characters of G, II.2 Irr(G | N) = Irr(G) − Irr(G/N), II, V, X Irr1 (G), the set of nonlinear irreducible characters of G, II, X, XI irreducible character, II irreducible linear (projective) representations, I.5, VI.1 irreducible polynomial, III.1 irreducible (= simple) module, I.3, I.6 Isaacs’ restriction theorem, VII.5 Isaacs’ theorems, VII.5, VII.9, VII.11, X.7 Isaacs–Zisser theorem, IV.8, Appendix VII.A isometric mapping, VII.1 Ito’s theorem on degrees, VII.2

kernel of induced character, V.2 kernels of characters and representations, II, III.5 kernels of irreducible projective π-representations of the abelian group, coincidence of, VI.6 Kronecker’s lemma, III.1

K k-antikernel of a group, IX K-character k(G), class number of G, II.5, XI.1 k-kernel of a group, IX K-representation, VIII.2 Kazarin, III.6, III.9, VII.1 kernel of a homomorphism, I.1 kernel of a linear representation, I.5 kernel of a projective representation, VI.1

M m-linear function, mapping, IV.1, IV.2 m-th outer power of a linear operator, IV.2 m-vector, IV.2 Mackey’s irreducibility criterion, VI.3 Mackey’s theorems, V.3, V.12 MacWilliams theorem, IV.7 Mann’s theorems, III.9, III.11, IV.12, V.14, XI.4, XI.6, XI.9 Maschke’s theorem, I.8

L lattice of submodules of homogeneous completely reducible module, I.10 left action, I.1 left ideal of a ring (an algebra), I.11 left module, I.2 length of completely reducible module, I.10 lexicographic ordering, IV.2 lifting of a projective representation, VI.2 linear character ω χ (z) of Z(ℂG), II.9, III.3 linear characters, the group of, I.9 linear independence of irreducible characters of a semisimple algebra over algebraically closed field, I.12 linear representation, I.9 linear representations of abelian groups, I.9 linearly equivalent projective representations, VI.1 Lin(G), the set of linear characters of G, I.9, II

596 | Characters of Finite Groups 1 Maschke’s theorem for abelian p-groups, Appendix I.A matrix elements of a representation, II.2 matrix orthogonal, unitary, I.8 matrix representation, I.5 measure of commutativity mc(∗), XI metacyclic group, VI.13 M-group, V.10, VII.13 minimal ideal, I.10 minimal idempotent, I.10 minimal nonabelian group, VI.4, XI.8 minimal nonnilpotent groups, degrees of their irreducible characters, other properties, XI.2 Möbius function, V.11, IX.3 module, I.2, I.3 monic polynomial, III.1 monomial representations and characters, V.9 monomorphism, I.1 multiplicative table for class sums, III.7 multiplicity of irreducible constituent of a character, II.4 multiplier (Schur multiplier), VII.3, VII.4, VII.5 VII.9, VII.11, VII.12 multiplier of direct product of groups, VI.9 N n-transitive, I.1 Nagao’s theorem, II.8 Nakayama’s theorem, V.15 nilpotent class-two p󸀠 -subgroups of GL(n, p), VII.17 nilpotent elements of an algebra, I.11 nilpotent ideal, I.10 non-degenerate bicharacter, VI.6 nonexistence of finite groups for which the two maximal sizes of two noncentral conjugacy classes are consecutive odd numbers, Mann’s theorem, III.11

nonlinear character, II nonprincipal character, II nonreal character, IV.4 normal p-complement, VII.10 normal subgroup, I.1 normalized factor set, VI.3 number of classes of linearly equivalent projective representations with the same factor set, VI.6 number of equivalence classes of irreducible representations of a semisimple algebra, I.12 number of involutions, IV.6, XI.7 number of irreducible constituents of induced character, XI.8 number of irreducible characters of a group, II.5 O operator representation, I opposite algebra, I.12 orbit of a point, size of, I.1 order of the Schur multiplier, VI.11, VI.12 orthogonal complement, IV.6 orthogonal idempotents, I.11 orthonormal basis, IV Osima’s theorem, V.15 P p-conjugacy, VIII.1 p-defect of a character, VIII.4 p-groups with large derived subgroups, VI.11 p-groups with large multipliers, VI.11 p-groups with small number of nonlinear characters of minimal degree, III.9 p-groups with small number of nonlinear irreducible characters, III.9 p-nilpotent group, VIII.1

Subject index | 597

p- and p󸀠 -part of an element, VIII.1 p-rational character, VII.7 permutation, I.1 permutation character, V.4, V.7 permutation representation, I.1, V.4, V.7 π- and π󸀠 -part of an element, VIII.5 π-central function, VI.3 π-character, VI.5 π-class sums, VI.3 π-element, VI.3 π-kernel, criteria, VI.5, VII.3 {π, π󸀠 }-partition of a group, VIII.3 π-representation, VI.1 Price’s theorem, VII.16 primitive module, V.5 primitive root of 1, III.2 principal character, I.9, II.3 product of characters, IV.4 projection, I.6 projective representation, VI.1 projectively equivalent projective representations, VI.1 Prol, VII.4 Q ℚ-conjugate element, III.8 ℚ-group, III.8 quasikernel Z(χ) of a character χ, III.5 R radical Rad(A) as the annihilator of all irreducible A-modules, I.11 radical Rad(A) of an algebra A, I.11 ramification, VII.1 rational character, III.4 rational element, III.4 rational group (= ℚ-group), III.8 real character, IV.6 real element, class, II.3, IV.7, VIII realization theorem for linear (projective) representations, VIII.2, VI.10

reducibility, I.2–I.6, II.3 reducible module, I.3 reducible representation, I.3 regular A-module A A, I.11 regular action, I.1 regular character, II.4 regular character, decomposition of, II.4 regular representation, I.3 representation group, VI.3 representation group of a nonabelian metacyclic p-group, VI.13 representation linear, I.5 representation matrix, I.5 representations (linear, projective) of abelian groups, I.9, VI.6 representations of semisimple algebras, I.12 restriction of characters, II.6 Reynolds’ theorem on realization of projective representations, VI.10 right ideal of a ring (an algebra), I.11 right module, I.2 right (left) multiplication of an algebra, I.10 Roquette, VII.11 Rusin, XI.3 S Saksonov’s example, II.9 Saksonov’s theorems, X.4 same type modules, I.10 scalar (= inner) product, I.8 scalar (= inner) product of characters, II.6 Schur’s lemma, I.7 Schur multiplier, VI Schur multiplier of a direct product, VI.9 Schur multiplier of a group and its quotient group, VI.5 Schur multiplier of an abelian p-group, VI.6 Schur’s relations, II.2

598 | Characters of Finite Groups 1 Schur’s theorem on the structure of multiplier of the abelian group, VI.6 Schur’s theorem: Schur multiplier of a group is a direct product of multipliers its Sylow subgroups, VI.4 second cohomology group, VI.3 second orthogonality relations, II.6 section of a group, VIII.6 semisimple algebra (over algebraically closed field), I.11, I.12 semisimplicity criterion, I.11 Shoda’s irreducibility criterion, V.3 sign of a permutation, I.1, IV.2 similar matrices, II.3 simple algebra, I.11 simplicity of matrix algebra K n over skew field K, I.11 sizes of classes, III.10, X.4 skew field corresponding to a module, I.10 socle of a group, IX.5 Solomon’s theorem, II.9 solvability criteria, III.6, V.10, V.11 special p-group, III.9, XI.2 squares of characters, Appendix VII.A stabilizer of a point, index of, I.1 standard permutation regular representation, I.1 strong order, I.11 structure of the group algebra ℂG, II.5 subalgebra, I.3 subgroups with the character and normal restricted properties, VII.18 submodule, I.2 subspace of imprimitivity, V.5 sum of representations, I.5 sum of squares of degrees of irreducible characters, II.3 sums of character degrees, XI supersolvable groups, III.8 Suzuki’s theorems, VIII.5

Sylow subgroups of Frobenius complements, X.2 symmetric group, I.1 symmetric power of spaces, operators, matrices, IV.3, IV.5 symplectic abelian group, VI.6 T T(G), the sum of degrees of irreducible characters of G, XI.1 Taketa’s theorem on solvability of M-groups, V.10 Tate’s normal p-complement theorem, VII.10 Taussky theorem on 2-groups, III.5 tensor product of vector spaces, representations of a group, IV.1, IV.4 tensor power, IV.1 tensor product of abelian groups, VI.9 Thompson’s theorems, X.3 TI-subset, TI-subgroup, properties, VIII.5, X.1, X.2, X.6 transfer, X.10 transitivity of induction, V.1 trivial factor set, VI.1 twisted group algebra, VI.2 2-transitivity, V.7 type of a homogeneous completely reducible module, I.10 W W-triple (= Wielandt triple), kernel of, VIII.5, X.2 Wedderburn theory, I.11 Wedderburn’s theorem on the structure of a simple algebra, I.11 Weisner irreducibility criterion, IX.4 Wielandt’s generalization of Frobenius’ theorem, X.2 Wielandt’s theorems (first, second, third), X.2

Subject index | 599

Witt’s theorem on nilpotence of solvable group admitting fixed-point-free-automorphism, X.3 Y Yamazaki’s theorem on existence of non-degenerate bicharacter of an abelian group, VI.6 Z zero-form, IV.1 Zhmud’s criterion for existence of a faithful irreducible character with given number of irreducible constituents, IX.5

Zhmud’s theorem on equality of numbers of kernels and antikernels, IX.4, IX.5 Zhmud’s theorem on existence of faithful projective representation with at most two irreducible components, VI.6 Zhmud’s theorem on existence of a cyclic subgroup Z in the abelian group G such that G/Z is of symmetric type, VI.6 Zhmud’s theorems, VI, VIII.3, IX Zisser’s lemma, V.6

De Gruyter Expositions in Mathematics Volume 42 Helmut Strade Simple Lie Algebras over Fields of Positive Characteristic. Volume II: Classifying the Absolute Toral Rank Two Case, 2nd Edition, 2017 ISBN 978-3-11-051676-0, e-ISBN 978-3-11-051760-6, Set-ISBN 978-3-11-051761-3 Volume 38 Helmut Strade Simple Lie Algebras over Fields of Positive Characteristic. Volume I: Structure Theory, 2nd Edition, 2017 ISBN 978-3-11-051516-9, e-ISBN 978-3-11-051544-2, Set-ISBN 978-3-11-051545-9 Volume 62 Yakov Berkovich, Zvonimir Janko Groups of Prime Power Order, 2016 ISBN 978-3-11-029534-4, e-ISBN 978-3-11-029535-1, Set-ISBN 978-3-11-029536-8 Volume 61 Yakov Berkovich, Zvonimir Janko Groups of Prime Power Order, 2016 ISBN 978-3-11-028145-3, e-ISBN 978-3-11-028147-7, Set-ISBN 978-3-11-028148-4 Volume 60 Benjamin Fine, Anthony Gaglione, Alexei Myasnikov, Gerhard Rosenberger, Dennis Spellman The Elementary Theory of Groups, 2014 ISBN 978-3-11-034199-7, e-ISBN 978-3-11-034203-1, Set-ISBN 978-3-11-034204-8 Volume 59 Friedrich Haslinger The d-bar Neumann Problem and Schrödinger Operators, 2014 ISBN 978-3-11-031530-1, e-ISBN 978-3-11-031535-6, Set-ISBN 978-3-11-031536-3