Hyperbolic modules and cyclic subgroups

Maria Loukaki

Dept. of Applied Mathematics, University of Crete, Knosou Av. GR-71409, Heraklion-Crete, Greece

E-mail: loukaki@tem.uoc.gr

Abstract

Let G be a finite group of odd order, F a finite field of odd characteristic p and B a symplectic FG-module. We show that B is FG-hyperbolic, i.e., it contains a self–perpendicular FGsubmodule, iff it is FN-hyperbolic for every cyclic subgroup N of G.

1 Introduction

Let F be a finite field of odd characteristic p, G a finite group and B an FG-module. If B carries a non-singular alternating bilinear form < ·, · > (i.e., a symplectic form) that is invariant by G, then we call V a symplectic FG-module. Following the notation in [3], for any FG-submodule S of B, we write S^{⊥} for the perpendicular subspace of S, i.e., S^{⊥} := {t ∈ B| < S, t >= 0}. We say that S is isotropic if S ≤ S^{⊥}, and B is anisotropic if it contains no non–trivial isotropic FGsubmodules. Furthermore, we say that B is hyperbolic if it contains some self–perpendicular FG-submodule S, i.e., S is an FG-submodule satisfying S = S^{⊥}.

Symplectic modules play an essential role in studying monomial characters. (An irreducible character χ of a finite group G is monomial if it is induced from a linear character of a subgroup of G.) One of the most representative links between symplectic modules and monomial characters can be found in [3]. (For other examples one could look at [2, 7, 8, 9, 10, 11, 12], and [13].) There E. C. Dade proved the following theorem (Theorem 3.2 in [3]):

Theorem 1.1 (Dade). Suppose that F is a finite field of odd characteristic p, that G is a finite p-solvable group, that H is a subgroup of p-power index in G, and that B is a symplectic FGmodule whose restriction B_{H} to a symplectic FH-module is hyperbolic. Then B is hyperbolic.

Using the above theorem, E. C. Dade was able to prove (Theorem 0 in [3]) that, given a p-solvable odd group G, an irreducible monomial character χ of G, and a subnormal subgroup N of G, every irreducible constituent of the restricted character χ_{N} is monomial, provided that χ(1) is a power of p.

In this paper we prove

Theorem A. Suppose that F is a finite field of odd characteristic p, that G is a finite group of odd order, and that B is a symplectic FG–module whose restriction B_{N} to a symplectic FN– module is hyperbolic for every cyclic subgroup N of G. Then B is hyperbolic.

Acknowledgements I am indebted to Professor E. C. Dade for many helpful ideas and suggestions. Also, I would like to thank Professor M. Isaacs for many useful conversations that helped me improve this paper.

2 Symplectic modules

We first give some elementary results about symplectic modules.

Assume that B is a symplectic FG-module, while S is an isotropic FG-submodule of B. Then the factor FG-module S̄ = S^{⊥}/S is again a symplectic FG-module with the symplectic form defined as (see (1.4) in [3]),

< s + S, s_{2} + S >=< s_{1}, s_{2} >, for all s_{1}, s_{2} ∈ S^{⊥}.  (1)

Furthermore, if S is an isotropic FG-submodule of B, then its F-dimension dim_{F} S is at most (1/2) dim_{F} B, (see 19.3 in [1]).

We say that an isotropic FG-submodule S of B is maximal isotropic if S is no properly contained in any larger isotropic FG-submodule of B. Clearly any self–perpendicular FGsubmodule S of B is maximal isotropic. The converse is also correct under the extra assumption that B is G-hyperbolic (see Lemma 3.1 in [3]). Another way to get a self–perpendicular module from a maximal isotropic one is to control its dimension, as the following lemma shows.

Lemma 2.1. Assume that B is a symplectic FG-module, and that S is a maximal isotropic FGsubmodule of B. If dim_{F} S = (1/2) dim_{F} B then S is self–perpendicular and B is G-hyperbolic.

Proof. Let Ŝ denote the dual of S. Then B/S^{⊥} ≅ Ŝ. But dim_{F} Ŝ = dim_{F} S = (1/2) dim_{F} B. Hence dim_{F} S^{⊥} = (1/2) dim_{F} B. Since S ≤ S^{⊥} we conclude that S = S^{⊥}. Thus the lemma holds.

The following is Proposition (2.1) in [3]

Proposition 2.2. Let G be a finite group and B be an anisotropic symplectic FG–module. Then B is an orthogonal direct sum:

B = U_{1} ˙⊥ U_{2} ˙⊥ … ˙⊥ U_{k},  (2)

where k ≥ 0 and each U_{i} is a simple FG–submodule of B that is also symplectic.

Remark 1. If G has odd order then according to Proposition (1.10) and Corollary (2.10) in [3] all the U_{i} that appear in (2) are distinct.

Lemma 2.3. Let U be a simple FG-module that affords a symplectic G-invariant form < ·, · >. Then U is self–dual.

Proof. We write Û for the dual FG-module of U. For every x ∈ U the map α_{x} : U → F defined as:

α_{x}(u) =< u , x > for all u ∈ U

is an element of Hom_{F}(U, F) ≅ Û. Since < · , · > is G-invariant the map α : x → α_{x} is an FG-homomorphism from U to Û. Furthermore the kernel of α is trivial, as U is symplectic. This, along with the fact that U is simple, implies that U ≅ Û. Hence the lemma holds.

Corollary 2.4. Let B be an anisotropic symplectic FG–module. Then each of the simple FG– modules U_{i} that appear in (2) is self–dual.

Proof. Let < · , · >_{U} denote the restriction to U × U of the symplectic bilinear form < · , · > that B × B affords. Then Proposition 2.2 implies that < · , · >_{U} is a symplectic form on U × U. Now Corollary 2.4 follows from Lemma 2.3.

Proposition 2.5. Assume that U is a simple symplectic FG–module. Let N be a normal subgroup of G such that |G : N| is odd. Then any simple FN–submodule of U_{N} is self–dual. Hence any FN-submodule of U_{N} is self–dual.

Proof. As N is a normal subgroup of G, Clifford’s theorem implies that

U_{N} ≅ e(V_{1} ⊕ … ⊕ V_{n})  (3)

where V = V_{1} is a simple FN–submodule of U and V_{1}, …, V_{n} are the distinct G-conjugates of V. So n||G : N| and therefore n is odd.

According to Lemma 2.3 the module U is self–dual. Hence the dual, V̂_{i} of any V_{i} should appear in (3). Therefore we can form pairs among the V_{i}, consisting of a simple FN–module V_{k} and its dual for k ∈ {1, …, n}, where we take as the second part of the pair the module itself if it is self–dual. Since G acts transitively on the V_{i} for i = 1, …, n, either all the V_{i} are self–dual or none of them is. In the latter case we get that any of the above pairs consists of two distinct modules. This implies that 2|n. As n is odd, this case can never occur. Hence any one of the V_{i} is self–dual and the proposition is proved.

Proposition 2.6. Assume that the symplectic FG–module B is hyperbolic. Assume further that B is a semi–simple FG-module. Then every self–dual simple FG–submodule of B appears with even multiplicity in any decomposition of B as a direct sum of simple FG–submodules.

Proof. Because B is hyperbolic it contains a self–perpendicular FG-submodule S. For every FG-submodule V of B we have B/V^{⊥} ≅ V̂. So

B/S ≅ Ŝ  (4)

Now the proposition follows from (4) and the fact that B is semi–simple.

Corollary 2.7. Let B be an anisotropic symplectic semi–simple FG–module. Let N be a normal subgroup of G such that |G : N| is odd. Assume further that B_{N} is a hyperbolic FN-module. Then any simple FN–submodule of B_{N} appears with even multiplicity in any decomposition of B_{N} as a direct sum of simple FN–submodules.

Proof. This is a straightforward application of Propositions 2.2, 2.5 and 2.6.

We close this section with a well known fact that we prove here for completeness.

Lemma 2.8. Assume that U is a self–dual absolutely irreducible FG-module, where G has odd order and F is a finite field whose characteristic does not divide |G|. Then U is trivial.

Proof. Let χ denote the F-absolutely irreducible character that U affords, while ϕ denotes the Brauer character that U affords. Then ϕ is defined for every element of G, since the characteristic of F is coprime to |G|. Because U is self–dual, the character ϕ is real valued. Let ν_{2}(ϕ) = 1/|G| ∑_{g∈G} ϕ(g^{2}) be the Frobenius–Schur indicator (see Chapter 4 in [7]) of ϕ. Then Theorem 4.5 in [6] implies that ν_{2}(ϕ) ≠ 0, since ϕ is real valued. But

ν_{2}(ϕ) = 1/|G| ∑_{g∈G} ϕ(g^{2}) = 1/|G| ∑_{g∈G} ϕ(g),

because G has odd order. Hence ν_{2}(ϕ) is the inner product ν_{2}(ϕ) = [ϕ, 1_{G}], where 1_{G} is the trivial character of G. We conclude that [ϕ, 1_{G}] ≠ 0. Hence ϕ = 1_{G}. Therefore χ = 1_{G}, and the lemma follows.

3 Proof of Theorem A

We can now prove our main result. The proof will follow from a series of lemmas, based on the hypothesis that F, B, G form a minimal counter–example. All the groups considered in this section have odd order. We also fix the odd prime p that is the characteristic of F, and we assume that

Inductive Hypothesis. F, B, G have been chosen among all triplets satisfying the hypothesis but not the conclusion of Theorem A so as to minimize first the order |G| of G and then the F–dimension dim_{F} B of B.

Remark 2. For any proper subgroup H of G the minimality of |G| easily implies that the restriction B_{H} is a hyperbolic FH-module.

Lemma 3.1. B is non–zero and anisotropic.

Proof. If B were zero it would be hyperbolic contradicting the Inductive Hypothesis. So B is non–zero. If B is not anisotropic then it contains a non–zero isotropic FG–module U. Then the isotropic FN-submodule U_{N} of B_{N} is contained in some maximal isotropic FN–submodule V of B_{N}. Since B_{N} is hyperbolic this maximal isotropic is self–perpendicular, i.e., V = V^{⊥}. Hence

U ⊆ V = V^{⊥} ⊆ U^{⊥}.

Therefore the factor module V̄ = V/U is a self–perpendicular FN–submodule of the symplectic FG–module Ū = U^{⊥}/U. Hence F, G, Ū satisfy the hypotheses of the Main Theorem. As dim(Ū) < dim(B), the minimality of dim(B) implies that Ū is a hyperbolic FG–module. So there is a self–perpendicular FG–submodule J̄ in Ū. From the definition of the symplectic form on Ū (see (1)) it follows that the inverse image J of J̄ in U^{⊥} is a self–perpendicular FG– submodule of B containing U. Therefore B is hyperbolic, contradicting the Inductive Hypothesis. So the lemma holds.

Lemma 3.2. p doesn’t divide the order |G| of G.

Proof. Suppose that p divides |G|. Because G is solvable, it contains a Hall p′–subgroup H. If G is a p-group we take H = 1. Since p divides |G|, the subgroup H is strictly smaller than G. Then according to Remark 2 the FH-module B_{H} is hyperbolic. It follows (see Theorem 3.2 of [3] ) that B is a hyperbolic FG-module, contradicting the Inductive Hypothesis. Hence (p, |G|) = 1.

Lemma 3.3. B is an orthogonal direct sum

B = U_{1} ˙⊥ … ˙⊥ U_{k}  (5)

where k ≥ 1 and {U_{i}}_{i=1,... ,k} are distinct, simple FG –submodules of B, that are also symplectic. Furthermore each U_{i} is quasi–primitive (i.e. its restriction to every normal subgroup of G is homogeneous).

Proof. The first statement follows from Lemma 3.1, Proposition 2.2 and Remark 1. For the rest of the proof we fix U = U_{i} for some i = 1, …, k. We also fix a normal subgroup N of G. If the restriction of U to N is not homogeneous then Clifford’s Theorem implies that

U_{N} ≅ e(V^{σ_{1}} ⊕ … ⊕ V^{σ_{r}})

where e is some positive integer, V = V^{σ_{1}} is a simple FN–submodule of U and V^{σ_{1}}, …, V^{σ_{r}} are the distinct conjugates of V in G, with σ_{1}, …, σ_{r} coset representatives of the stabilizer, G_{V}, of V in G.

Let W = U(V) be the V-primary component of U_{N}. Then Clifford’s Theorem implies that W is the unique simple FG_{V}–submodule of U that lies above V and induces U, i.e., that satisfies W^{G} ≅ U and W_{N} ≅ eV. Furthermore the dual Ŵ of W induces in G the dual Û of U as Ŵ^{G} ≅ Ŵ^{G}. Hence Ŵ^{G} = U, because U is self–dual (see Lemma 2.3). On the other hand, the restriction of Ŵ to N is isomorphic to eV as V is self–dual by Proposition 2.5. Hence the unicity of W implies that W is self–dual.

According to Proposition 2.6 the self–dual FG_{V}-module W appears with even multiplicity as a direct summand of B_{G_{V}}, since B_{G_{V}} is hyperbolic (G_{V} < G). This, along with the fact that W appears with multiplicity one in U_{G_{V}}, implies that there is some j ∈ {1, …, k} with j ≠ i such that the V-primary component U_{j}(V) of U_{j} is isomorphic to W. So

W = U(V) ≅ U_{j}(V)

We conclude that

U_{i} = U ≅ W^{G} ≅ U_{j}(V)^{G} ≅ U_{j},

as FG-modules. This contradicts the fact that {U_{i}}_{i=1}^{k} are all distinct, by the first statement of the lemma. Hence the lemma is proved.

From now on and until the end of the paper, we write E for the finite algebraic field extension of F, that is a splitting field of G and all its subgroups.

Lemma 3.4. Assume that U_{i}, for i = 1, …, k, is a direct summand of B appearing in (5). Let N ⊴ G. Then U_{i}|_{N} ≅ e_{i}V_{i}, where V_{i} is an irreducible FN-submodule of U_{i} and e_{i} is an integer. If V_{i} is non-trivial then e_{i} is odd.

Proof. We fix U = U_{i}, for some i = 1, …, k. We also fix a normal subgroup N of G. According to Lemma 3.3, the FG-module U is quasi–primitive. Hence there exists an irreducible FNsubmodule V of U, and an integer e such that U_{N} ≅ eV. Thus, it remains to show that e is odd in the case that V is non-trivial. So we assume that V, and thus U, is non-trivial.

We observe that if U and V were absolutely irreducible modules then it would be immediate that e is odd (even if V was trivial), because for absolutely irreducible modules the integer e divides the order of G (see Corollary 11.29 in [7]). So we assume that F is not a splitting field of G, and we work with the algebraic field extension E of F. We define U^{E} to be the extended EG-module

U^{E} = U ⊗_{F} E.

According to Theorem 9.21 in [6], there exist absolutely irreducible EG-modules U^{i}, for i = 1, …, n, such that

U^{E} ≅ ⨁_{i=1}^{n} U^{i}.

Furthermore the U^{i}, for all i = 1, …, n, constitute a Galois conjugacy class over F. In particular, if E_{U} is the subfield of E that is generated by all the values of the irreducible E-character afforded by U^{i} (the same field for all i = 1, …, n), then n = [E_{U} : F] = dim_{F}(E_{U}). (Note that E_{U} is the unique subfield of E isomorphic to the center of the endomorphism algebra End_{FG}(U).) Clearly n · dim_{E} U^{1} = dim_{F} U. Hence n is even, because dim_{F} U is even (as U is symplectic) and dim_{E} U^{1} is odd (as G is odd and U^{1} is an absolutely irreducible EG-module). In addition, each EG-module U^{i}, for i = 1, …, n, when consider as an FG-module, is isomorphic to a direct sum of [E : E_{U}] copies of U (see Theorem 1.16 in Chapter 1 of [5]). Hence if we denote by U_{F}^{i} the U^{i} regarded as an FG-module, we get

U_{F}^{i} ≅ [E : E_{U}]U^{i},  (6)

for all i = 1, …, n.

We also write V^{E} for the extended EN-module V^{E} = V ⊗_{F} E. Then according to Theorem 9.21 in [6] there exist absolutely irreducible EN-modules V^{j} for j = 1, …, d, such that

V^{E} ≅ ⨁_{j=1}^{d} V^{j}.  (7)

In addition, the absolutely irreducible modules V^{j}, for all j = 1, …, d, form a Galois conjugacy class. Furthermore, d = [E_{V} : F] = dim_{F} E_{V}, where E_{V} is the subfield of E generated by all the values of the irreducible E-character afforded by V^{j} (the same field for all j = 1, …, d). The field E_{V} is the unique subfield of E somorphic to the center of the endomorphism algebra End_{FN}(V). Note that, according to Proposition 2.5, the FN-submodule V of U is self–dual. Hence V^{E} is also a self–dual EN-module. Because V is non-trivial, V^{j} is also non-trivial, for all j = 1, …, d. Therefore the absolutely irreducible EN-module V^{j} can’t be self–dual, because N has odd order and V^{j} is non–trivial (see Lemma 2.8), for all such j. The fact that none of the V^{j} is self–dual, for all j = 1, …, d, while they all appear in (7) in dual pairs, implies that d is even. Even more, if V_{F}^{j} denotes the module V^{j} regarded as an FN-module, then Theorem 1.16 of Chapter 1 in [5] implies that

V_{F}^{j} ≅ [E : E_{V}]V^{j},  (8)

for all j = 1, …, d.

Without loss we may assume that V^{1}, …, V^{c} are exactly those among the V^{j}, for j = 1, …, d, that lie under U^{1}, for some c = 1, …, d. Thus Clifford’s theorem implies that

U_{N}^{1} ≅ e′(V^{1} ⊕ ⋯ ⊕ V^{c}),  (9)

where V^{1}, …, V^{c} are the distinct G-conjugates of V^{1}, and e′, c are integers that divide |G| and thus they are odd. (Note that here we are dealing with absolutely irreducible modules so e′ does divide |G|.) If we regard the modules of (9) as modules over the field F then we clearly have U_{F}^{1}|_{N} ≅ e′(V_{F}^{1} ⊕ ⋯ ⊕ V_{F}^{c}). This, along with (6) and (8), implies

[E : E_{U}]U_{N} ≅ e′c[E : E_{V}]V.

Since U_{N} ≅ eV, we have

[E : E_{U}]e = e′c[E : E_{V}].  (10)

If D is the subfield of E generated by E_{V} and E_{U}, then dividing both sides of (10) by [E : D] we obtain

e[D : E_{U}] = e′c[D : E_{V}].  (11)

Assume that e is even. Then (11) implies that [D : E_{V}] is even, as e′ and c are known to be odd. Let Γ be the Galois group Γ = Gal(D/F) of D over F. Because Γ is cyclic, it contains a unique involution ι. Let E_{V}^{∗} and E_{UB}^{∗} be the subgroups of Γ consisting of those elements of Γ that fix pointwise E_{V} and E_{U}, respectively. Then Galois theory implies that [E_{V}^{∗} : 1] = [D : E_{V}] while [E_{U}^{∗} : 1] = [D : E_{U}]. Hence if e is even, then [D : E_{V}] = [E_{V}^{∗} : 1] is also even. So E_{V} has even order. We conclude that the unique involution ι of Γ is an element of E_{V}^{∗}. Therefore, ι fixes the field E_{V} pointwise. So ι fixes, to within isomorphisms, each of the EN modules V^{j}. Because ι acts non–trivially on D and fixes E_{V}, it must act non–trivially on E_{U}. We conclude that ι viewed as an F-automorphism of E_{U} must coincide with the unique involution in the Galois group Gal(E_{U}/F) of E_{U} above F. Furthermore, this unique involution must send U^{i} to its dual Û^{i}, for every i = 1, …, n. (Of course U^{i} is not self–dual, because it is a non–trivial absolutely irreducible module of the odd order group G (see Lemma 2.8).) Hence, applying ι to both sides of (9) we get

Û^{1}_{N} ≅ e′(V̂^{i} ⊕ ⋯ ⊕ V̂^{c}) ≅ e′(V^{1} ⊕ ⋯ ⊕ V^{c}).

Hence the dual V̂^{1} of V^{1} should be among the G-conjugates V^{1}, …, V^{c} of V^{1}. Because V^{1} is not self–dual the G-conjugates of V^{1} should appear in dual pairs. Hence c is even. But c is also odd as a divisor of |G|. This contradiction implies that t is even. So the lemma holds.

Lemma 3.5. The group G is not abelian.

Proof. Assume that G is abelian. Then any cyclic subgroup N =< σ > of G is normal, for every σ ∈ G. Because B_{N} is hyperbolic, Lemma 3.1 and 3.3 along with Corollary 2.7 imply that

B_{N} ≅ 2 · Δ_{(N)},

where Δ_{(N)} is a semi–simple FN-submodule of B. Using the splitting field E of G, we write B^{E} for the extended EG-module B^{E} = B ⊗_{F} E. Then

B_{N}^{E} ≅ 2 · Δ_{(N)}^{E},  (12)

where Δ_{(N)}^{E} is the extended EN-module Δ_{(N)} ⊗_{F} E. Let ϕ be the Brauer character that the EG-module B^{E} affords. Because (p, |G|) = 1, ϕ is defined for every element of G. So ϕ coincides with a complex character of G. In view of (12), for every cyclic subgroup N =< σ > of G, the restriction ϕ_{N} of ϕ to N equals 2 · δ_{(N)}, where δ_{(N)} is a complex character of N. Hence, for every element σ ∈ G, the integer 2 divides ϕ(σ) in the ring ℤ[ω], where ω is a |G|-primitive root of unity. We conclude that 2 also divides ∑_{σ∈G} ϕ(σ) · λ(σ^{−1}), for any irreducible (linear) complex character λ of G. That is, 2 divides |G|· < ϕ, λ >, for any λ ∈ Irr(G). The fact that G has odd order, implies that 2 divides < ϕ, λ > in ℤ[ω], for any λ ∈ Irr(G). Because ϕ = ∑_{λ∈Irr(G)} < ϕ, λ > ·λ, we get

ϕ = 2 · χ,  (13)

where χ is a complex character of G.

On the other hand, Lemma 3.3 implies that B = U_{1} ⊕ ⋯ ⊕ U_{k}, where U_{i} are distinct simple FG-modules, for all i = 1, …, k. Hence the extended EG-module B^{E} will also equal the direct sum of the distinct EG–modules U_{1}^{E}, …, U_{k}^{E}. By Theorem 9.21 in [6], for each i = 1, …, k, there exist absolutely irreducible EG–modules U_{i}^{j}, for j = 1, …, n_{i} such that

U_{i}^{E} ≅ ⨁_{j=1}^{n_{i}} U_{i}^{j}.

Furthermore, the U_{i}^{j}, for j = 1, …, n_{i}, constitute a Galois conjugacy class over F, and thus they are all distinct. In addition, for all i = 1, …, k and all j = 1, …, n_{i}, the above absolutely irreducible EG-module U_{i}^{j} are all distinct, because, for all i = 1, …, k, the corresponding simple FG-modules U_{i} are distinct. We conclude that

B^{E} ≅ ⨁_{i=1}^{k} ⨁_{j=1}^{n_{i}} U_{i}^{j},

where U_{i}^{j} are all dinstinct absolutely irreducible EG-modules. So the Brauer character ϕ that B^{E} affords equals

ϕ = ∑_{i=1}^{k} ∑_{j=1}^{n_{1}} χ_{i}^{j},

where, for all i = 1, …, k and all j = 1, …, n_{i}, the character χ_{i}^{j} is the Brauer character that U_{i}^{j} affords. So all these characters are distinct. This contradicts (13). Hence the group G is not abelian, and the lemma is proved.

Lemma 3.6. G acts faithfully on B

Proof. Suppose not. Let K denote the kernel of the action of G on B and Ḡ = G/K. Thus |Ḡ| ⪇ |G| (as K ≠ 1 ).

If Ḡ is not itself cyclic, then any cyclic subgroup N̄ of Ḡ is the image N̄ = N/G of some proper subgroup N of G. Since B is FN -hyperbolic, it is clearly FN̄-hyperbolic. Hence the triplet F, B, Ḡ satisfies the hypothesis of the Main Theorem. The minimality of |G| implies that B is a hyperbolic FḠ–module, and therefore a hyperbolic FG–module, because any FḠ–submodule of B is also an FG–submodule of B. This contradicts the Inductive Hypothesis.

If Ḡ is cyclic, then Ḡ =< σ̄ >, for some σ ∈ G. Let M =< σ >. Then M is a proper subgroup of G, because G is not cyclic. In addition, the image of M in Ḡ is Ḡ. So G = MK with M ⪇ G. Then Remark 2 implies that B is FM–hyperbolic and thus FG–hyperbolic. This last contradiction implies the lemma.

Lemma 3.7. Suppose M is a minimal normal subgroup of G. Then M is cyclic and central.

Proof. According to Lemma 3.3 and 3.4 for each i = 1, …, k there is a simple FM–submodule V_{i} of U_{i} and an odd integer e_{i}, such that U_{i}|_{M} ≅ e_{i}V_{i}. As G acts faithfully on B, there is some i ∈ {1, …, k} such that V_{i} ≠ 1 is non-trivial. Let K_{M}(V_{i}) be the kernel of the action of M on V_{i}. The fact that V_{i} is G-invariant implies that K_{M}(V_{i}) is a normal subgroup of G contained in M. Hence K_{M}(V_{i}) = 1. Therefore M admits a faithful simple representation. In addition, M is a q-elementary abelian group, for some prime q that divides |G|, because G is solvable. We conclude that M ≅ ℤ_{q} is a cyclic group of order q.

It remains to show that M is central. If F is a splitting field of M (that is, it contains a primitive q-root of 1), then the fact that there exists a faithful, simple and thus one–dimensional, G-invariant FM-module V_{i} implies that M is central in G. If F is not a splitting field of M, we work with the extension field E of F. The extended module B^{E} = B ⊗_{F} E equals the direct sum of the extended EG–modules U_{1}^{E}, …, U_{k}^{E}, because B is the direct sum of U_{1}, …, U_{k}. As we have already seen, for each i = 1, …, k, there exist absolutely irreducible EG–modules U_{i}^{j}, for j = 1, …, n_{i}, that constitute a Galois conjugacy class over F and satisfy

U_{i}^{E} ≅ ⨁_{j=1}^{n_{i}} U_{i}^{j}.  (14)

Since U_{i}|_{M} ≅ e_{i}V_{i} we have U_{i}^{E}|_{M} ≅ e_{i}V_{i}^{E}. In addition,

V_{i}^{E} ≅ ⨁_{r=1}^{s_{i}} V_{i}^{r},

where V_{i}^{r}, for r = 1, …, s_{i}, are absolutely irreducible EM-modules, and thus one–dimensional, that form a Galois conjugacy class over F. Therefore,

U_{i}^{E}|_{M} ≅ ⨁_{j=1}^{n_{i}} U_{i}^{j}|_{M} ≅ e_{i} ⨁_{r=1}^{s_{i}} V_{i}^{r},  (15)

for all i = 1, …, k.

As we have already seen, there exists i ∈ {1, …, k} such that V_{i} is a faithful FM-module. Without loss, we may assume that i = 1. Then it is clear that the V_{1}^{r} are faithful EM-modules, for all r ∈ {1, …, s_{1}}. If V_{1}^{r} is G-invariant, for some r ∈ {1, …, s_{1}} (and thus for all such r) we are done.

Thus we may assume that the stabilizer G_{V} of V = V_{1}^{1} in G is strictly smaller than G. Then G_{V} = C_{G}(M), because V is EM-faithful. Let C := G_{V} = C_{G}(M). Note that C is a normal subgroup of G, since M ⊴ G. Furthermore, C is also the stabilizer of V_{1}^{r}, for all r = 1, …, s_{1}. According to Lemma 3.3, for all i = 1, …, k, we have U_{i}|_{C} = m_{i} · Y_{i}, where Y_{i} is a simple FC-module, and m_{i} is an odd integer. For the extended EC-modules Y_{i}^{E} we have

U_{i}^{E}|_{C} ≅ m_{i}Y_{i}^{E} ≅ m_{i} ⨁_{l=1}^{t_{i}} Y_{i}^{l},

where the Y_{i}^{l}, for l = 1, 2, …, t_{i}, are absolutely irreducible EC-modules that constitute a Galois conjugacy class over F. Hence

U_{i}^{E}|_{C} ≅ ⨁_{j=1}^{n_{i}} U_{i}^{j}|_{C} ≅ m_{i} ⨁_{l=1}^{t_{i}} Y_{i}^{l},  (16)

for all i = 1, …, k. We remark here that, because U_{i} is quasi–primitive, all the group conjugates of Y_{i}^{1} are among its Galois conjugates, for every i = 1, …, k.

In the case i = 1, equations (15) and (16) imply

U_{1}^{E}|_{M} ≅ m_{1} ⨁_{l=1}^{t_{1}} Y_{1}^{l}|_{M} ≅ e_{1} ⨁_{r=1}^{s_{1}} V_{1}^{r}.  (17)

Without loss we may assume that U_{1}^{1} lies above Y_{1}^{1}, and that Y_{1}^{1} lies above V_{1}^{1} = V. Because C is the stabilizer of V in G, Clifford’s theory implies that Y_{1}^{1} is the unique simple EC-module that lies above V_{1}^{1} and induces irreducibly to U_{1}^{1} in G. Note that Y_{1}^{1} appears with odd multiplicity m_{1} as a summand of U_{1}^{E}|_{C}, because the EC-modules Y_{1}^{l} are distinct for distinct values of l, as they form a Galois conjugacy class over F. Furthermore, if Y_{1}^{1} lies under some U_{i}^{j}, for i ≠ 1, then it induces U_{i}^{j}. So U_{i}^{j} ≅ U_{1}^{1}. Hence the sum of the Galois conjugates of U_{i}^{j} is isomorphic to the sum of the Galois conjugates of U_{1}^{1}. Therefore

U_{i}^{E} ≅ ⨁_{j=1}^{n_{i}} U_{i}^{j} ≅ ⨁_{j=1}^{n_{1}} U_{1}^{j} ≅ U_{1}^{E}.

The above contradicts the fact that U_{1} and U_{i} are non-isomorphic simple FG-modules (see Lemma 3.3). We conclude that Y_{1}^{1} appears with odd multiplicity m_{1} in the decomposition of

B^{E}|_{C} ≅ ⨁_{i=1}^{k} U_{i}^{E}|_{C} ≅ ⨁_{i=1}^{k} m_{i} ⨁_{l=1}^{t_{i}} Y_{i}^{l}.

On the other hand, in view of Corollary 2.7 every simple FC-submodule of B appears with even multiplicity in any decomposition of B_{C}, as C is a normal subgroup of G and B_{C} is hyperbolic as an FC-module, by Remark 2. Hence every absolutely irreducible EC-submodule of B^{E} should also appear with even multiplicity in any decomposition of B^{E}|_{C}. This contradicts the conclusion of the preceding paragraph. So we must have G_{V} = C = G. Hence the lemma is proved.

Clearly the proof of Lemma 3.7 implies

Corollary 3.8. Suppose that M is a minimal normal subgroup of G and E a splitting field of G and all its subgroups. Assume further that U_{i}|_{M} ≅ e_{i}V_{i}, for some i = 1, …, k, while the extended EM-module V_{i}^{E} satisfies V_{i}^{E} ≅ ⨁_{r=1}^{s_{i}} V_{i}^{r}, where V_{i}^{r}, for r = 1, …, s_{i}, are absolutely irreducible EM-module that form a conjugacy class over F. Then V_{i}^{r} is a G-invariant EM-module, for all r = 1, …, s_{i}.

We can now show

Lemma 3.9. Suppose that M is a minimal normal subgroup of G. Then the restriction B_{M} is homogeneous. Furthermore B_{M} ≅ eV, where V is a simple faithful G-invariant FM-submodule of B_{M} and e is a positive integer.

Proof. As in the previous lemma we write U_{i}|_{M} = e_{i}V_{i}, where i = 1, …, k, the integer e_{i} is odd, and V_{i} is a simple G-invariant FM-submodule of U_{i}. If FM is not homogeneous, then there are at least two non-isomorphic simple FM-submodules of B_{M}, say V and W. We may suppose that V is non-trivial. Assume that V_{i} ≅ V as FM–modules, for all i = 1, …, l for some l such that 1 ≤ l ≤ k, and that V_{i} ≇ V for i = l + 1, …, k. Let U be the orthogonal direct sum

U = U_{1} ˙⊥ … ˙⊥ U_{l}.

of the corresponding FG–submodules of B. We also write

R = U_{l+1} ˙⊥ … ˙⊥ U_{k},

for the orthogonal direct sum of the remaining simple FG-submodules of B. Clearly

B = U ˙⊥R,

while U_{M} and R_{M} have no simple FM-submodules in common.

We will show

Claim 1. U is FN-hyperbolic for every cyclic subgroup N of G.

We first prove Claim 1 in the case that the product NM is a proper subgroup of G. In this case Remark 2 implies that B_{NM} is hyperbolic. Hence there exists a self–perpendicular FNM-submodule S > 0 of B. Then S is a maximal isotropic FNM-submodule of B_{NM}. Furthermore, B_{NM} = U_{NM} ˙⊥R_{NM}, where U_{NM} and R_{NM} have no simple FNM-submodule in common (otherwise U_{M} and R_{M} would have some common simple FM-submodule). Hence

S = (S ∩ U_{NM}) ˙⊥(S ∩ R_{NM}).

Because S is isotropic, both S ∩U_{NM} and S ∩R_{NM} are also isotropic. Hence their F-dimensions are at most 1/2 the dimension of U_{NM} and R_{NM}, respectively. But S is self–perpendicular and thus its F-dimension is exactly (1/2) dim(B_{NM}). We conclude that the F-dimensions of S∩U_{NM} and S ∩R_{NM} are exactly 1/2 the dimension of U_{NM} and R_{NM}, respectively. Therefore S ∩U_{NM} is a maximal isotropic FNM-submodule of U_{NM} of dimension 1/2 the dimension of U_{NM}. So S ∩ U_{NM} is self–perpendicular, by Lemma 2.1. Thus U_{NM} is hyperbolic as an FNM-module. Hence it is also hyperbolic as an FN-module. So Claim 1 holds when NM < G.

Assume now that N is a cyclic subgroup of G such that NM = G. Because M is minimal, Lemma 3.7 implies that M ≅ ℤ_{q} is central. Hence G = MN is an abelian group. This contradicts Lemma 3.5. Therefore NM < G, for every nilpotent subgroup N of G. Thus Claim 1 holds.

If U < B, then the Inductive Hypothesis, along with Claim 1, implies that U is FGhyperbolic. Hence U contains a self–perpendicular FG-submodule T . Let T^{⊥} be the submodule of B that is perpendicular to T . Then R lies inside T^{⊥}. We conclude that T is an isotropic FG-submodule of B. Hence B is not anisotropic. This last contradiction implies that U = B, and completes the proof of Lemma 3.9.

Lemma 3.10. Every abelian normal subgroup of G is cyclic.

Proof. Let A be an abelian normal subgroup of G. By Lemma 3.3 there is a simple FA– submodule R_{1} of U_{1} and an integer e_{1} such that

U_{1}|_{A} ≅ e_{1}R_{1}.

It follows from Lemma 3.9 that R_{1} is non-trivial, since its restriction to any minimal normal subgroup of G in non–trivial. Let K_{1} denote the corresponding centralizer of R_{1} in A . Then K_{1} equals the centralizer C_{A}(U_{1}) of U_{1} in A, and therefore is a normal subgroup of G. If K_{1} is not trivial then it contains a minimal normal subgroup M of G. In view of Lemma 3.7 the restriction U_{1}|_{M}, cannot be trivial, contradicting the definition of K_{1}. Hence K_{1} is trivial. Thus A is cyclic and the lemma is proved.

Let F = F (G) be the Fitting subgroup of G. We write F = T_{1} × T_{2} × ⋯ × T_{r}, where T_{i} is a Sylow subgroup of F, for every i = 1, …, r. Every characteristic abelian subgroup of F is cyclic, according to Lemma 3.10. Hence (see Theorem 4.9 in [4]) either T_{i} is cyclic or T_{i} is the central product T_{i} = E_{i} ⊙ Z(T_{i}) of the extra special q_{i}–group E_{i} = Ω(T_{i}) of exponent q_{i}, for some prime q_{i}||G|, and the cyclic group Z(T_{i}). We complete the proof of the Main Theorem exploring the two possible types of T_{i}.

Assume first that T_{i} is a cyclic group, for all i = 1, …, r. In this case F = T_{1} × ⋯ × T_{r} is also a cyclic group. Let C/F be a chief factor of G. So C̄ = C/F is an elementary abelian q-group, for some prime q, because G is solvable. Then C̄ acts coprimely on T_{i} for all i such that q does not divide |T_{i}|. But T_{i} is cyclic, and the minimal subgroup of T_{i} is central in G. Hence C_{T_{i}}(C̄) ≠ 1. We conclude that T_{i} = [T_{i}, C̄] × C_{T_{i}}(C̄) = C_{T_{i}}(C̄). So the q-Sylow subgroup C_{q} of C centralizes the q′-Hall subgroup R of F that is also a q′-Hall subgroup of C. We conclude that M = C_{q} × R. But R is nilpotent as a subgroup of F. So C is a nilpotent normal subgroup of G bigger than the Fitting subgroup F of G. Therefore G = F is a cyclic group, contradicting the Inductive Hypothesis. Hence there exists a Sylow subgroup T_{i} of F = F (G) that is not cyclic.

Let T be a non-cyclic q-Sylow subgroup of F. Then T = E ⊙ Z(T), where E = Ω(T) is an extra special q-group of exponent q and Z(T) is the center of T. Of course E is a normal subgroup of G, since it is a characteristic subgroup of F. Furthermore, Z(E) is a central subgroup of G because it is a minimal (it has order q) normal subgroup of G. According to Lemma 3.10, there exists a faithful G-invariant FZ(E)-module V so that the restriction B_{Z(E)} of B to Z(E) is a multiple of V.

Using the extension field E of F, we write V^{E} for the extended EZ(E)-module V ⊗_{F} E. Then

V^{E} ≅ ⨁_{j=1}^{s} V^{j}

where V^{j} is an absolutely irreducible EZ(E)-module, for all j with j = 1, …, s. Furthermore, V^{j} constitute a Galois conjugacy class over F, and thus they are all distinct. As we have already seen (see Corollary 3.8 and Lemma 3.9), the module V^{j} is a non-trivial G-invariant EZ(E)-module. Because E is extra special, there exists a unique isomorphism class of absolutely irreducible EEmodule W^{j} lying above V^{j}, for every j = 1, …, s. Note that for all such j the EE-module W^{j} is G-invariant because V^{j} is G-invariant. According to Theorem 9.1 in [7] (used for modules) there exists a canonical conjugacy class of subgroups H ≤ G such that HE = G and H ∩ E = Z(E). Furthermore, for this conjugacy class there exists a one–to–one correspondence between the isomorphism classes of absolutely irreducible EG-modules lying above W^{j} and those classes of absolutely irreducible EH-modules lying above V^{j}. In addition, the fact that G has odd order implies that if U and Ψ are representatives of the above two isomorphism classes, then they correspond iff U_{H} ≅ Ψ ⊕ 2 · Δ, where Δ is a completely reducible EH-submodule of U_{H}.

Let U = U_{1} be one of the simple FG-submodules of B appearing in (5). Then U^{E} ≅ ⊕_{j=1}^{n_{1}}U^{j}, where U^{j} are absolutely irreducible EG-modules that form a Galois conjugacy class. As earlier, we write E_{U} for the extension field of F generated by all the values of the absolutely irreducible character that U^{1} affords. Let Γ = Gal(E_{U}/F) be the Galois group of that extension. Then (see Theorem 9.21 in [6]),

U^{E} ≅ ⨁_{j=1}^{n_{1}} U^{j} ≅ ⨁_{τ∈Γ}(U^{1})^{τ} ,  (18)

Clearly U^{1} lies above W^{j}, for some j = 1, …, s, since U = U_{1} lies above V. Let Ψ be a representative of the isomorphism class of absolutely irreducible EH-modules that corresponds to U^{1} and lies above V^{j}. Then

U_{H}^{1} ≅ Ψ ⊕ 2 · Δ,  (19)

for some completely reducible EH-module Δ. If F(Ψ) is the extension field of F generated by all the values of the absolutely irreducible character that Ψ affords, and Gal(F(Ψ)/F) the Galois group of that extension, then

Gal(F(Ψ)/F) = Γ = Gal(E_{U}/F).  (20)

Indeed, for any element σ in the Galois group Gal(E/F) of E above F we get

(U^{1})_{H}^{σ} ≅ Ψ^{σ} ⊕ 2 · Δ^{σ}.

Hence (U^{1})^{σ} corresponds to Ψ^{σ}, as Ψ^{σ} is the only absolutely irreducible EH-module that appears with odd multiplicity in (U^{1})_{H}^{σ}. Therefore, (U^{1})^{σ} ≇ U^{1} iff Ψ^{σ} ≇ Ψ. This is enough to guarantee that (20) holds. We conclude that

Π := ⨁_{τ∈Γ} Ψ^{τ}

is the extension to E of an irreducible FH-module. In addition, (18) and (19) imply that Π appears with odd multiplicity as a summand of of U_{H} = U_{1}|_{H}.

Next we observe that if Π appears as a summand of U_{i}|_{H}, for some i = 2, …, k, then it appears with even multiplicity. The reason is that U_{1} ≇ U_{i} for all such i. In addition, the fact that U_{1} ≇ U_{i} also implies that U_{i}^{j} ≇ U^{1}, for all i = 2, …, k and all j = 1, …, n_{i}, where (see (14)) U_{i}^{E} ≅ ⊕_{j=1}^{n_{i}}U_{i}^{j}. So the EH-module Ψ can’t correspond to U_{i}^{j}, for any such i, j. Therefore if Ψ appears as a summand of the restriction U_{i}^{j}|_{H} of U_{i}^{j} to H, then it appears only with even multiplicity. Hence the same holds for Π, i.e., Π appears only with even multiplicity as a summand of U_{i}|_{H}, whenever i = 2, …, k. We conclude that Π appears with odd multiplicity as a summand of B_{H} = U_{1}|_{H} ⊕ ⋯ ⊕ U_{k}|_{H}.

We complete the proof of the Main Theorem with one more contradiction, that follows the fact that Π is a self–dual FH-module. That we get a contradiction if Π is self–dual is easy to see, because according to Proposition 2.6, Π should appear with even multiplicity as a summand of the hyperbolic FH-module B_{H}. Thus it suffices to show that Π is self–dual.

The fact that U = U_{1} is self–dual implies that U^{E} is also self–dual. Hence the dual Û^{1} of U^{1} is a Galois conjugate (U^{1})^{τ} to U^{1}, for some τ ∈ Γ. Furthermore, (19) implies that

Û^{1}_{H} ≅ Ψ̂ ⊕ 2 · Δ̂.

Thus the dual Û^{1} corresponds to the dual Ψ̂ of Ψ. Therefore the dual Ψ̂ of Ψ is a Galois conjugate of Ψ. Hence Π = ⊕_{τ∈Γ}Ψ^{τ} is a self–dual FH-module.

This completes the proof of Theorem A.

References

[1] M. Aschbacher, “Finite Group Theory,” Cambridge Studies in Advanced Mathematics, 10.

[2] T. R. Berger, Representation Theory and Solvable Groups: Length Type Problems, The Santa Cruz Conference on Finite Groups, Proc. Sympos. Pure Math. 37 (1980), 431–441.

[3] E. C. Dade, Monomial Characters and Normal Subgroups, Math.Z. 178, 401–420 (1981).

[4] D. Gorenstein, “Finite Groups,” New York: Harper and Row 1968.

[5] B. Huppert, N. Blackburn, “Finite Groups II,” Berlin–Heidelberg–New York : Springer 1982.

[6] I. M. Isaacs, “Character Theory of Finite Groups,” Academic Press, New York, 1976.

[7] I. M. Isaacs, Characters of Solvable and Symplectic Groups, Amer. J. Math. 95 (1973), 594–635.

[8] I. M. Isaacs, Abelian Normal Subgroups of M-Groups, Math. Z. 182 (1983), 205–221.

[9] I. M. Isaacs, Primitive Characters, Normal Subgroups and M-Groups, Math. Z. 177 (1981), 267–284.

[10] M. L. Lewis, Characters of Maximal Subgroups of M-Groups, J. Algebra 183 (1996) 864– 897.

[11] A. E. Parks, Nilpotent by Supersolvable M-Groups, Canad. J. Math. 37 (1985), 934–962.

[12] R. W. van der Waall, On Clifford’s Theorem and Ramification Indices for Symplectic Modules over a Finite Field, Proc. Edinburgh Math. Soc. 30 (1987), 153–167.

[13] R. W. van der Waall and N.S. Hekster, Irreducible Constituents of Induced Monomial Characters, J. Algebra 105 (1987), 255-267.