Endotrivial irreducible lattices Geoffrey R. Robinson, Institute of Mathematics, University of Aberdeen, September 2010

Abstract

We consider the structure of a finite group which has a faithful irreducible endotrivial RG-module for a suitable complete discrete valuation ring R with R/J(R) of characteristic p.

Introduction: Let G be a finite group containing an elementary Abelian subgroup of order p^{2}. Let R be a complete discrete valuation ring of characteristic 0 with field of fractions 𝕂 containing a primitive |G|-th root of unity, and with residue field F = R/J(R) algebraically closed of characteristic p. An RG-lattice here means an R-free finitely generated RG-module. An RG-lattice is said to be irreducible if it has no proper non-zero R-pure RG-submodule. We are concerned here with irreducible endotrivial RG-lattices, that is, irreducible RGlattices M which also satisfy M ⊗M^{∗} ≅ R⊕P, where P is a projective RG-lattice and G has trivial action on R. We will give a reasonably complete description of the structure of G if there is an irreducible endotrivial RG-lattice of R-rank greater than 1 and the Fitting subgroup F (G) is non-central. If there is such a lattice and F (G) is central, we will prove that there is a single conjugacy class of components of G. If p = 2 and F (G) is central, we can go somewhat further: we will see that G has a single component and a single conjugacy class of involutions, except perhaps when a Sylow 2-subgroup of G has a cyclic subgroup of index 2, in which case O^{2}(G) has a single component and a single conjugacy class of involutions.

1 General Results

Lemma 1.1: Let M be a faithful RG-lattice such that Res_{N}^{G}(M) is indecomposable for some normal subgroup N of G. Then O^{p}(C_{G}(N)) ≤ Z(G). In particular, C_{G}(N) ≤ F (G).

Proof: Let y be a p-regular element of C_{G}(N). For each linear character λ of ⟨y⟩, set

e_{λ} = \frac{1}{|⟨y⟩|} ∑_{x∈⟨y⟩} λ(x^{−1})x.

Then M = ⨁_{λ∈Irr(⟨y⟩)} Me_{λ} as RN-module and y acts as multiplication by λ(y) on Me_{λ} for each λ. Since M is indecomposable as RN-module, there must be some linear character λ such that Me_{λ} = M. Hence y acts as a scalar on M and y is central in G as M is faithful.

The given proof of the following Lemma is due to Gabriel Navarro, and is much shorter than our own original proof.

Lemma 1.2: Let μ be an irreducible character of the normal subgroup N of G. Suppose that the sum of the G-conjugates of μ vanishes on all p-singular elements of N. Then μ itself vanishes on all p-singular elements of N.

Proof: The hypotheses imply that the induced character Ind_{N}^{G}(μ) vanishes on all p-singular elements of G. Hence |G|_{p} divides [G : N]μ(1). Thus |N|_{p} divides μ(1), so that μ lies in a p-block of defect zero of N, and hence vanishes on all p-singular elements of N.

Lemma 1.3: Let N be a normal subgroup of the finite group G. Let M be an irreducible RG-lattice such that Res_{N}^{G}(M) has a projective summand. Then Res_{N}^{G}(M) is isomorphic to a direct sum of projective irreducible RN-lattices. Proof: Let M afford character χ. Let μ = μ_{1} be an irreducible constituent of Res_{N}^{G}(M). Then there are positive integers e and t such that Res_{N}^{G}(χ) = e(μ_{1} + … + μ_{t}), where the μ_{i} are the distinct G-conjugates of μ. Now we may write Res_{N}^{G}(M) = U ⊕ P where U and P are RN-lattices, P is projective and U has no projective summand. If U = 0, we are done by Lemma 2, so suppose otherwise. By the Krull-Schmidt theorem, U^{g} ≅ U for all g ∈ G. In particular, the character, α say, afforded by U is G-stable. Hence there are is an integer f such that α = f(μ_{1} + … + μ_{t}). Then P affords character (e − f)(μ_{1} + … + μ_{t}) and e ≠ f, as P ≠ 0. Since P is projective, (μ_{1} + … + μ_{t}) vanishes on all p-singular elements of N, so μ lies in a block of defect zero of RN by Lemma 2, and we are done.

It is convenient to state a well-known fact here for later use.

Lemma 1.4: Let ⟨x⟩ be a finite cyclic p-group, and let R denote the trivial R⟨x⟩- module, ΩR be the kernel of the augmentation map σ : R⟨x⟩ → R. Then ΩR affords a unimodular representation when p is odd, and x acts with determinant −1 on ΩR when p = 2. In either case the determinant of the representation of ⟨x⟩ is the same as the determinant of the regular representation of ⟨x⟩, so that ΩR ⊕ R⟨x⟩ affords a unimodular representation of ⟨x⟩.

Proof: Clearly, the character afforded by ΩR is the sum of all non-trivial linear characters of ⟨x⟩, and the determinant of the representation of ⟨x⟩ on ΩR is the product of all non-trivial linear characters of ⟨x⟩. When p is odd, the non-trivial linear characters of ⟨x⟩ occur in complex conjugate pairs, so their product takes value 1 on x (and all its powers). When p = 2, there are just two real-valued linear characters of ⟨x⟩. One takes the value 1 at x, the other takes the value −1 at x. The other linear characters of ⟨x⟩ pair off in complex conjugate pairs and have trivial product.

Remark: For future reference, we remark that when p = 2 and x has order 2^{e}, ΩR ⊕ R⟨x⟩ is a unimodular representation of ⟨x⟩ of degree 2^{e+1} − 1, and remains unimodular if its direct sum with an even number of copies of the regular representation is taken, while R ⊕ R⟨x⟩ ⊕ R⟨x⟩ is a unimodular representation of ⟨x⟩ of degree 2^{e+1} + 1, and remains unimodular if its direct sum with an even number of copies of the regular representation is taken.

In the former case, the character afforded by the module takes constant value −1 on ⟨x⟩^{#}, while in the latter case, the character afforded by the module takes constant value 1 on ⟨x⟩^{#}.

When p is odd, the action of ⟨x⟩ on ΩR is unimodular, and the character afforded by ΩR takes value −1 on ⟨x⟩^{#}. Addition of direct sums of copies of the regular module does not affect the unimodularity of the action, or the value taken by the character on non-identity elements. Clearly the action of ⟨x⟩ on the trivial module R is unimodular. This allows us to record:

Lemma 1.5: Let χ be a complex character of a finite cyclic p-group ⟨x⟩ such that (detχ) is the trivial character, and such that χ(x^{j}) is a root of unity for each integer j : 1 ≤ j < |⟨x⟩|. Then:

i) If p is odd, there is a sign ε such that χ(1) ≡ ε (mod |⟨x⟩|) and χ(x^{j}) = ε for 1 ≤ j < |⟨x⟩|.

ii) If p = 2, and there is a sign ε such that χ(1) ≡ ε (mod 2|⟨x⟩|), then χ(x^{j}) = ε for 1 ≤ j < |⟨x⟩|.

Proof: Suppose first that p is odd. Since the action of x on the regular representation is unimodular, if ⟨χ, μ⟩ > 0 for each linear character character μ of ⟨x⟩, we may suppose that ⟨χ, μ⟩ = 0 for some linear character μ of ⟨x⟩ (otherwise we can subtract the regular character from χ and still leave a character with trivial determinant). Notice that

0 = ∑_{y∈⟨x⟩} χ(y)μ(y^{−1}) ≥ χ(1) − (|⟨x⟩| − 1),

since χ(y) is a root of unity for each non-identity y ∈ ⟨x⟩. Hence χ(1) < |⟨x⟩|.

However, the character χ\overline{χ} takes the value 1 on all non-identity powers of x, so that χ(1)^{2} ≡ 1 (mod |⟨x⟩|). As p is odd, there is a sign ε such that χ(1) ≡ ε (mod |⟨x⟩|). Since χ(1) < |⟨x⟩|, we either have χ(1) = 1 or χ(1) = |⟨x⟩| − 1. If χ(1) = 1, then χ must be the trivial character by unimodularity. Suppose then that χ(1) = |⟨x⟩| − 1. Then for the linear character μ above, we must have χ(y) = −μ(y) for each non-identity y ∈ ⟨x⟩. Let θ be the character of ⟨x⟩ afforded by ΩR. Then χ = μθ and (detχ) = μ^{χ(1)}(detθ). But the action of ⟨x⟩ on ΩR is unimodular, and χ(1) is not divisible by p, so μ must be the trivial character, as detχ is trivial.

If p = 2, we may suppose that there is a linear character μ of ⟨x⟩ such that ⟨χ, μ⟩ ∈ {0, 1}, otherwise we may subtract twice a multiple of the regular character from χ and still have a character with trivial determinant (without affecting values on non-identity elements, or the fact that χ(1) ≡ ε (mod 2|⟨x⟩|)). Then [χ(1) − (|⟨x⟩| − 1)] ≤ |⟨x⟩|, so that χ(1) < 2|⟨x⟩|. Hence either χ(1) = 1 (if ε = 1) or χ(1) = 2|⟨x⟩| − 1 if ε = −1. If χ(1) = 1, we are done, since (detχ) is the trivial character. If χ(1) = 2|⟨x⟩| − 1, then for the linear character μ above, we must have ⟨χ, μ⟩ = 1 and χ(y) = −μ(y) for each non-identity element y of ⟨x⟩. Let θ be the character afforded by ΩR ⊕ R⟨x⟩. Then (detθ) is the trivial character, and χ = μθ, so that (detχ) = μ^{χ(1)}(detθ) and μ is the trivial character, since χ(1) is odd and (detχ) and (detθ) are both trivial.

2 General results on endotrivial lattices

Recall that the RG-module M is endotrivial if M ⊗ M^{∗} ≅ R ⊕ P, where R is the trivial RG-module and P is projective. We will say that M is a normalized endotrivial module if, additionally, the action of a Sylow p-subgroup S of G on M is unimodular. First, we point out that several simple groups G do possess faithful irreducible endotrivial lattices.

Lemma 2.1: Suppose that G = G′ and that G has a strongly p-embedded subgroup H such that G acts doubly transitively on the cosets of H. Then for each non-trivial linear character λ of H such that [H : kerλ] does not divide [G : H], the induced character Ind_{H}^{G}(λ) is irreducible, and is afforded by an endotrivial RG-lattice.

Proof: By hypothesis ⟨Ind_{H}^{G}(1), Ind_{H}^{G}(1)⟩ = 2. Set θ_{λ} = Ind_{H}^{G}(λ), and let G be the disjoint union ⋃_{i=1}^{t} Ht_{i}, where t_{1} = 1_{G}. Then for any x ∈ H, we have θ_{λ}(x) = ∑_{j:t_{j}xt_{j}^{−1}∈H} λ(t_{j}xt_{j}^{−1}). Since t_{1} = 1_{H} and x ∈ H, we have |θ_{λ}(x)| ≤

θ_{1}(x), with equality of and only if λ(t_{j}xt_{j}^{−1}) = λ(x) whenever t_{j}xt_{j}^{−1} ∈ H.

Thus we have ⟨θ_{λ}, θ_{λ}⟩ < 2 unless Res_{H}^{G}(θ_{λ}) = λ.Res_{H}^{G}(θ_{1}). Since G = G′, the representations of G affording θ_{1} and θ_{λ} are both unimodular, so equality of these two induced characters forces λ(h)^{[G:H]} = 1 for all h ∈ H and the first claim follows.

Let V_{λ} be a rank 1 RH-lattice affording character λ. Since H is strongly pembedded, Res_{H}^{G}(Ind_{H}^{G}(V_{λ})) ≅ V_{λ} ⊕ W, where W is projective. Thus Ind_{H}^{G}(V_{λ}) is endotrivial.

Remark: The groups PSL(2, p^{n}), for example, satisfy the hypotheses, with H a Sylow p-normalizer. A linear character λ of H has order dividing \frac{p^{n}−1}{2}, so the induced character θ_{λ} is irreducible unless perhaps \frac{p^{n}−1}{2} is even and λ has multiplicative order 2. This exception does occur when p^{n} ≡ 1 (mod 4), but there is at least one choice of λ such that θ_{λ} is irreducible, except for p^{n} = 3 and p^{n} = 5.

Remark: A familiar example of a finite simple group G which has no strongly p-embedded subgroup, yet has a faithful endotrivial RG-module is G = GL(3, 2) (for p = 2). There is an irreducible RG-lattice M of rank 3, and M ⊗ M^{∗} ≅ R ⊕ St, where St is a projective irreducible RG-lattice of rank 8.

Lemma 2.2: Let M be an irreducible endotrivial RG-lattice affording character χ, and let N be a normal subgroup of G of order divisible by p. Then Res_{N}^{G}(M) and Res_{N}^{G}(M̄) are both indecomposable. Furthermore, Res_{N}^{G}(χ) is multiplicity free.

Proof: We know that Res_{N}^{G}(M) is not projective, so that Res_{N}^{G}(M) is projective free by Lemma 1.3. Hence Res_{N}^{G}(M) is endotrivial and indecomposable. If Res_{N}^{G}(M̄) is not indecomposable, then it has a projective summand. But this summand then lifts to a projective summand of Res_{N}^{G}(M), a contradiction.

Notice that |χ(x)|^{2} = 1 for each non-identity p-element x of N. But by Clifford’s theorem, Res_{N}^{G}(χ) = e(μ_{1} + … + μ_{t}), where e is a positive integer, μ_{1} is an irreducible constituent of Res_{N}^{G}(χ) and μ_{1}, …, μ_{t} are its distinct Gconjugates. Hence e = 1.

The following Lemma uses a variant of a well-established technique in Clifford theory of characters:

Lemma 2.3: Whenever M is an endotrivial RG-module, there is a unique RGmodule V of R-rank 1 such that V ⊗ M is normalized endotrivial and O^{p}(G) acts trivially on V. Furthermore, if S does not have a cyclic subgroup of index 2, then the character χ afforded by V ⊗ M takes constant integral value on S^{#}, so that rank_{R}(S) ≡ ±1 (mod |S|).

Proof: Notice that the rank of |M| is prime to p. Let λ be the linear character of G afforded by the determinant of its action on M. There are integers h and k such that hrank_{R}(M) + k|S| = 1. Then the RG-module V of R-rank 1 defined by v.g = λ(g_{p})^{−h}v for all g ∈ G, v ∈ V is the required module for the first claim.

For the second part, we might as well suppose that G = M and that the action of S on M is unimodular and indecomposable. Suppose first that p = 2. Then there is a sign ε such that the rank of M is congruent to ε (mod \frac{|S|}{2}). We may replace M by ΩM and assume that ε = −1. Let Q be a cyclic subgroup of S and let U be the regular RQ-module. Then the RQ-module U ⊕ ΩR (where the action on R is trivial) is a normalized endotrivial RQ-module. For each odd integer h, it follows that hU ⊕ ΩR affords the unique character of degree (h + 1)|Q| − 1) which may be afforded by a normalized endotrivial RQ-module. Now rank_{R}(M) ≡ −1 (mod 2|Q|) since 2|Q| < |S|. Thus Res_{Q}^{S}(M) affords the same character as hU ⊕ ΩR for some odd integer h.

If p is odd, then for any cyclic subgroup Q of S both the regular module and the first Heller translate of the trivial module are endotrivial and afford unimodular representations. Hence we can replace M by ΩM if necessary, and suppose that rank_{R}(M) ≡ −1 (mod |S|). Then Res_{Q}^{S}(M) affords that same character as ΩR ⊕ hU for some integer h, where U is the regular RQ-module.

Corollary 2.4: Let M be a faithful irreducible endotrivial RG-lattice of rank greater than 1 and suppose that N = O_{p}(G) > 1. Then C_{G}(N) = N × Z(G). Furthermore, N is elementary Abelian, M has rank |N| − 1 and either N is a Sylow p-subgroup of G or else N is a Klein 4-group and G ≅ Z(G) × S_{4}.

Proof: Since M is endotrivial of rank greater than 1, M ⊗ M^{∗} is a faithful RN-module, so that N ∩ Z(G) = 1. Hence we may, and do, suppose that M is normalized. Now N is Abelian by Clifford’s theorem, since the rank of M is prime to p. By Lemma 2.2, Res_{N}^{G}(M) is indecomposable, so the second claim follows by Lemma 1.1. Notice also that G has no normal p-complement, since G is not Abelian. Let χ be the character afforded by M.

If N has a cyclic subgroup of index 2, then N must be a Klein 4-group, for otherwise N would contain a characteristic subgroup of order 2, which would be central in G. In that case, all non-identity elements of N must be conjugate in G, since G/C_{G}(N) is isomorphic to a subgroup of S_{3}, and G has no normal 2-complement. Thus χ takes a constant value ±1 on N^{#}. If N has no cyclic subgroup of index 2, then Lemma 2.3 applies, and χ takes a constant value ±1 on N^{#}.

In all cases, then, χ takes a constant value on N^{#}, and this value is 1 or −1. Since, by Clifford’s theorem and the fact that χ is faithful, the trivial character must occur with zero multiplicity in Res_{N}^{G}(χ) this forces χ(1) = |N| − 1 and χ(n) = −1 for all n ∈ N^{#}. Hence all non-trivial linear characters of N are in the same G-orbit, so N is elementary Abelian in all cases. If N is not a Klein 4-group, then a Sylow 2-subgroup S of G does not have a cyclic subgroup of index 2. In that case, χ takes the constant value −1 on S^{#}, so that |S| divides χ(1) + 1. Hence S = N in that case. If N is a Klein 4-group, but is not a Sylow 2-subgroup of G, then we have seen that G/Z(G) must be isomorphic to S_{4}. But as G has no normal 3-complement, it easily follows that G ≅ Z(G) × S_{4}.

From now, unless the contrary is stated, we will suppose that M is a faithful irreducible normalized endotrivial RG lattice affording character χ and that O_{p}(G) = 1.

Lemma 2.5: If M is induced from an RH-lattice W, where H is a proper subgroup of G, then W is irreducible and endotrivial, and H is strongly p-embedded in G.

Proof: Clearly N must be irreducible as Ind_{H}^{G}(N) is irreducible. That N is endotrivial and H is strongly p-embedded follows by a slight modification of the proof of Lemma 1, part iv) of [2].

Notation: We set A = O_{p′}(G).

Lemma 2.6: A is Abelian, and is central in G if M is primitive.

Proof: If M is primitive, then, as in [3], there is a p′-central extension Ĝ of G and an irreducible RÂ-lattice U such that U extends to an RĜ-lattice, and as such M ≅ U ⊗ Hom_{RN̂}(U, M). Then U is endotrivial, and its reduction (mod J(R)) is endotrivial and simple, so since Ĝ contains an elementary Abelian subgroup of order p^{2}, U has rank 1 by [2]. Thus A acts as scalars on M.

If M is imprimitive, it is induced from a primitive RH-module V, where H is a strongly p-embedded proper subgroup of G. Then A ≤ O_{p′}(H) since G contains an elementary Abelian subgroup of order p^{2}. Thus A acts as scalars on V, so by Clifford’s theorem, all irrreducible constituents of Res_{A}^{G}(χ) are linear, where χ is the character afforded by M. Thus A′ acts trivially on M, so A is Abelian as M is faithful.

Lemma 2.7: Suppose that C = C_{G}(A) ≠ A. Then A ≤ Z(G), and G has a single conjugacy class of components, each of order divisible by p.

Proof: If C > A, then |C| is divisible by p and C ⊲ G. Thus C acts indecomposably on M and O^{p}(C_{G}(C)) ≤ Z(G). Thus A ≤ Z(G). In that case, we now have F (G) ≤ Z(G), so that E(G) ≠ 1. All components of G must have order divisible by p as A is Abelian.

Let X be a the normal closure of a component of G. Then O^{p}(C_{G}(X)) ≤ Z(G), so that G has no component not contained in X for if L is a component of G not contained in X, then L = O^{p}(L) and [L, X] = 1. Also p divides [L : Z(L)], since O_{p′}(G) is Abelian.

Lemma 2.8: Suppose that A = C_{G}(A) ≠ G. Then A = F^{∗}(G) and M is induced from a primitive irreducible endotrivial RH-lattice N of a strongly p-embedded subgroup H of G such that A acts as scalars on N.

Proof: Under the stated hypothesis, we clearly have O_{p}(G) = 1 = E(G). Since A is not central in G, Res_{A}^{G}(M) is not homogeneous ( otherwise A would act as scalars). Thus M must be induced from an irreducible endotrivial primitive RH-lattice M_{0} for some strongly p-embedded subgroup H of G. We have A ≤ H as G contains an elementary Abelian subgroup of order p^{2}. Now O_{p′}(H) acts as scalars on M_{0}, by Lemma 2.6, so A certainly does.

Lemma 2.9: If M is a faithful irreducible endotrivial RG-lattice, then M restricts irreducibly to O^{p}(G).

Proof: By induction, it suffices to prove that if N ⊲ G and [G : N] = p, then M restricts irreducibly to N. Let χ be the character afforded by M. Since the rank of M is not divisible by p, the character χ is not induced from N, so that Res_{N}^{G}(χ) is irreducible by Clifford’s theorem.

Lemma 2.10: If M is faithful, endotrivial and irreducible and A ≤ Z(G), then G has a unique conjugacy class of components.

Proof: Let X/A be a minimal normal subgroup of G/A. Then Y = X′ is the central product of a unique conjugacy class of components. Also p divides |Y |. Hence O^{p}(C_{G}(Y)) ≤ Z(G) by Lemma 1.1 and 1.2. Thus there are no components of G outside Y since A = O_{p′}(G). Thus F^{∗}(G) = Y Z(G).

3 Special arguments for the prime 2

In this section we will suppose that R/J(R) has characteristic 2.

We first state some general results which may have interest in a wider context than that of endotrivial modules.

Lemma 3.1: Let G be a finite group of even order. Suppose that G has a strongly real element of order 2^{e}, but no strongly real element of order 2^{e+1}. Suppose also that G has a complex irreducible character χ with χ(1) ≡ ±1 (mod 2^{e+1}). Then one or more of the following occurs:

i) χ is linear. ii) G has one conjugacy class of involutions. iii) There is a strongly element x of even order in G such that χ(x) is not a root of unity.

Proof: We recall that a strongly real element of G is an element which is inverted by an involution of G. It is well-known and easy to check that such an element is either itself an involution, or else is the product of two involutions. If the Sylow 2-subgroups of a finite group X are not cyclic or generalized quaternion, then every strongly real element of X is expressible as a product of two involutions.

Suppose that for some sign ε and some irreducible character χ of G, we have χ(1) ≡ ε (mod 2^{e+1}) and suppose also χ(x) is a root of unity for each strongly real element, x of even order in G. Notice, in particular, that χ has odd degree. We may multiply χ by a linear character μ if necessary and suppose that (detχ)[x] = 1 for each 2-element x ∈ G, and we do so. The linear character μ is defined (if needed) by μ(g) = (detχ)[g_{2}]^{−a} where g_{2} is the 2-part of g and a, b are integers with aχ(1) + b|⟨g_{2}⟩| = 1. We first claim that χ(x) = ε for each strongly real 2-element x of G.

Notice that χ(x) = ±1 for strongly real element x of even order, since χ(x) is a real root of unity. By hypothesis, χ(1) ≡ ε (mod 4,) and for each involution t ∈ G, we have χ(1) ≡ χ(t) (mod 4), since −1 occurs with even multiplicity as an eigenvalue of t. Hence χ(t) = ε. Suppose then that x has order 4 or more. We have detχ(x^{j}) = 1 for each integer j. Notice that x^{2} is also strongly real. Suppose by induction on the order of x that χ(x^{2j}) = ε for each positive integer j less than \frac{|⟨x⟩|}{2}. If χ(x) = ε, we are done, so suppose that χ(x) = −ε. There is a linear character, λ, of ⟨x⟩ with λ(x) = −1, and λ(x^{2}) = 1. Then λRes_{⟨x⟩}^{G}(χ) is the character afforded by ΩR ⊕ hR⟨x⟩ for some odd integer h if ε = −1 and is the character afforded by R ⊕ hR⟨x⟩ for some even integer h if ε = 1, where R denotes the trivial module. In each case, the action of x on the given module is unimodular. Hence

det(λRes_{⟨x⟩}^{G}(χ)) = 1,

a contradiction as

det(λRes_{⟨x⟩}^{G}(χ)) = λ^{χ(1)}det(Res_{⟨x⟩}^{G}(χ))

and χ(1) is odd.

We next claim that if x is a non-identity 2-element of G and y is an element of odd order which commutes with x such that xy is strongly real, we have χ(xy) = ε. For x is strongly real, so we have established that χ(x) = ε. We proceed by induction on the order of y, there being nothing to prove if y has order 1. Let q be a prime divisor of the order of y, and write y = zw = wz. where z is a q-element, w has order prime to q. Then xy = (xw)z and xw is strongly real. By induction, we have χ(xw) = ε. Also, χ(xwz) is a real root of unity, since xzw is a strongly real element of even order. Now χ(xzw) is also real, since xzw is conjugate to its inverse in G. Thus χ(xwz) = ±1. But if χ(xwz) = −ε = −χ(xw) we have ε ≡ −ε (mod q) since z is a q-element commuting with the q-regular element xw. This is a contradiction since q is odd. Hence χ(xy) = ε, as claimed.

Hence we have now established that χ(x) = ε whenever x ∈ G is a strongly real element of even order. Now suppose that t and u are involutions of G which are not conjugate. Then tu^{g} is strongly real of even order for each g ∈ G. Thus χ(tu^{g}) = ε for each g ∈ G (and χ(t) = χ(u) = ε).

Since we have

∑_{g∈C_{G}(u)\G} χ(tu^{g}) = \frac{[G : C_{G}(u)]χ(t)χ(u)}{χ(1)},

where C_{G}(u)\G is a full set of representatives for the right cosets of C_{G}(u) in G, we obtain χ(1)ε = 1, so that χ(1) = ε. Thus ε = 1 and χ is linear.

Theorem 3.2: Let G be a finite group and let M be a faithful irreducible endotrivial RG-lattice. Let S be a Sylow 2-subgroup of G. If S has no cyclic subgroup of index 2, then G has a unique conjugacy class of involutions. In any case O^{2}(G) has a unique conjugacy class of involutions.

Proof: We have seen that M restricts irreducibly to O^{2}(G). If S has a cyclic subgroup of index 2, then so does S_{0} = S ∩ O^{2}(G). By the Thompson transfer Lemma, every involution of O^{2}(G) is conjugate to the unique involution of the cyclic subgroup of index 2 of S_{0}. Hence we may suppose that S has no cyclic subgroup of index 2. If M has R-rank 1, then G is cyclic, and the result is clear. Hence we may suppose that M has R-rank greater than 1.

Let χ be the character afforded by M. Since M is endotrivial of rank greater than 1, Z(G) has odd order. Hence we may suppose that M is normalized endotrivial, and we do so, since λχ is still faithful whenever λ is a linear character such that G/kerλ is a 2-group. Then there is a sign ε such that χ(s) = ε for all s ∈ S^{#}, since χ(1) ≡ ±1 (mod \frac{|S|}{2}) and S has no cyclic subgroup of index 2.

We next claim that χ(x) is a root of unity for each element x of even order in G. Since M is endotrivial, we have |χ(x)|^{2} = 1 for each element x of even order in G. Then for each integer j with gcd(j, |⟨x⟩|) = 1, we have |χ(x^{j})|^{2} = 1. Hence χ(x) is a root of unity for each element x of even order in G. By Lemma 3.1, G has one conjugacy class of involutions.

Corollary 3.3: Let G be a finite group of even order with F (G) ≤ Z(G). Suppose that M is a faithful endotrivial irreducible RG-lattice of rank greater than 1. Then F^{∗}(G) = Z(G)E(G) and E(G) is quasisimple. Furthermore, G has one conjugacy class of involutions unless perhaps G ≠ O^{2}(G) and a Sylow 2-subgroup of G has a cyclic subgroup of index 2. In any case, O^{2}(G) has a unique conjugacy class of involutions and E(G) ≤ O^{2}(G).

Proof: Since F (G) ≤ Z(G), we must have E(G) ≠ 1. Now M has rank greater than 1, so that Z(G) has odd order, as M is endotrivial, and hence restricts faithfully to a Sylow 2-subgroup. Thus O_{2}(G) = 1. We have also seen that G has a single conjugacy class of components, and (unless a Sylow 2-subgroup of G has a cyclic subgroup of index 2) a single conjugacy class of involutions. If S ∈ Syl_{2}(G) has a cyclic subgroup of index 2, then S can’t be expressed as a direct product of two or more non-cyclic groups, so since each component of G has centre of odd order, G has a unique component. Suppose then that S has no cyclic subgroup of index 2, so that G has one conjugacy class of involutions.

Let L_{1}, L_{2}, …, L_{r} be all the distinct components of G and suppose that r > 1. Let π_{i} denote the homomorphism from E(G) onto L_{i}/Z(L_{i}) with kernel ∏_{j≠i} L_{j}. Then an involution t of E(G) with tπ_{i} ≠ 1 for exactly one i is not G-conjugate to an involution u with uπ_{i} ≠ 1 for all r values of i (and there are such involutions t and u). Thus E(G) is quasisimple.

Since M restricts irreducibly to N = O^{2}(G) by Lemma 2.9, and the restriction is still endotrivial, there are two possibilities: a Sylow 2-subgroup of N has no cyclic subgroup of index 2- in which case N has one conjugacy class of involutions- or a Sylow 2-subgroup S of N does have a cyclic subgroup T of index 2, in which case every involution of N is conjugate to the unique involution of T by the Thompson transfer Lemma as N = O^{2}(N).

Now F (N) ≤ Z(N), so that E(N) ≠ 1. Since G has a unique component, E(G) ≤ N = O^{2}(G).

Remark: When p = 2, our results have established that if G has even order and has a faithful irreducible endotrivial RG=lattice M of rank greater than 1, then O^{2}(G) has a single conjugacy class of involutions without exception (if G contains no Klein 4-subgroup, then G certainly has one conjugacy class of involutions). If M is imprimitive, M has a strongly embedded subgroup, so one conjugacy class of involutions. If O_{2}(G) ≠ 1, then either O_{2}(G) is an elementary Abelian Sylow 2-subgroup of G and G acts transitively by conjugation on its non-identity elements, or else G ≅ Z(G) × S_{4} and Z(G) has odd order. Otherwise, we are in the case that O_{2}(G) = 1 and O_{2′}(G) ≤ Z(G).

References

[1] J. F. Carlson, N. Mazza and J. Thévenaz, Endotrivial modules for p-solvable groups, to appear in Transactions of the American Mathematical Society.

[2] G. Navarro and G.R. Robinson, On endo-trivial modules for p-solvable groups, preprint, July 2010.

[3] G.R. Robinson, On simple endotrivial modules, preprint July 2010.