Some simple biset functors

Serge Bouc

Abstract: Let p be a prime number, let H be a finite p-group, and let 𝔽 be a field of characteristic 0, considered as a trivial 𝔽Out(H)-module. The main result of this paper gives the dimension of the evaluation S_{H,𝔽}(G) of the simple biset functor S_{H,𝔽} at an arbitrary finite group G. A closely related result is proved in the last section: for each prime number p, a Green biset functor E_{p} is introduced, as a specific quotient of the Burnside functor, and it is shown that the evaluation E_{p}(G) is a free abelian group of rank equal to the number of conjugacy classes of p-elementary subgroups of G.

AMS subject classification: 18B99, 19A22, 20J15

Keywords: simple biset functors, Green biset functor, p-elementary

1. Introduction

Let R be a commutative ring. The biset category RC over R has finite groups as objects, with morphisms Hom_{RC}(G, H) = R ⊗_{ℤ} B(H, G), where B(H, G) is the Burnside group of (H, G)-bisets. The composition of morphisms is induced by the usual tensor product of bisets. A biset functor over R is an R-linear functor from RC to the category R-Mod of R-modules. Biset functors over R form an abelian category, where morphisms are natural transformations of functors. They have proved a useful tool in various aspects of the representation theory of finite groups (see [12], [7], [8], [9]), and they are still the object of active research ([15], [1], [10], [16], [13], [14], [5], [4], [2], …).

The simple biset functors over R are parametrized ([6], Proposition 2) by equivalence classes of pairs (H, W), where H is a finite group, and W is a simple ROut(H)-module - the simple functor parametrized by (H, W) being denoted S_{H,W}. However for a finite group G, the computation of the evaluation S_{H,W} (G) is generally quite hard: in Theorem 4.3.20 of [9], this evaluation is shown to be equal to the image of a complicated linear map. Assuming that R is a field - which is always possible when dealing with simple functors - the dimension of S_{H,W} (G) is given by Theorem 7.1 of [11], as the rank of a yet complicated bilinear form with values in R.

Let 𝔽 be a field of characteristic 0, let p be a prime number, and H be a finite p-group. The present paper is mainly devoted to the computation of the dimension of the evaluation S_{H,𝔽}(G), where G is an arbitrary finite group, and 𝔽 is the trivial 𝔽Out(H)-module. The result is as follows:

Theorem: Let 𝔽 be a field of characteristic 0, let p be a prime number, and H be a finite p-group. Let moreover G be a finite group.

1. If H = 1, the dimension of S_{H,𝔽}(G) is equal to the number of conjugacy classes of cyclic subgroups of G.

2. If H ≅ C_{p} × C_{p}, the dimension of S_{H,𝔽}(G) is equal to the number of conjugacy classes of non-cyclic p-elementary subgroups of G.

3. If H is any other finite p-group, the dimension of S_{H,𝔽}(G) is equal to the number of conjugacy classes of sections (T, S) of G such that T/S ≅ H and T is p-elementary.

In the last section of this paper, for each prime number p, we introduce a Green biset functor E_{p}, closely related to the two first assertions of the above theorem. Green biset functors have been defined in [9], Section 8.5. They are ring objects in the category of biset functors. For a finite group G, we denote by F_{p}(G) the set of elements of the Burnside group B(G) which vanish when restricted to all p-elementary subgroups of G, and we show that this actually defines a biset subfunctor F_{p} of B. The functor E_{p} is defined as the quotient B/F_{p}, and it then inherits from B a Green biset functor structure (over ℤ). We show moreover that its evaluation E_{p}(G) at a finite group G is a free abelian group of rank equal to the number of conjugacy classes of p-elementary subgroups of G. We also show that the biset functor 𝔽E_{p} = 𝔽 ⊗_{ℤ} E_{p} fits in a non split short exact sequence

0 → S_{(C_{p})^{2},𝔽} → 𝔽E_{p} → S_{1,𝔽} → 0

of biset functors over 𝔽. In the case 𝔽 = ℚ, the restriction of this sequence to p-groups is the short exact sequence of Theorem D of [12], involving the Dade functor ℚD, the Burnside functor ℚB, and the functor of rational representation ℚR_{ℚ}.

2. Preliminary results

The basic definitions and notation on (double) Burnside algebras and their idempotents from Chapters 2.4 and 2.5 of [9] will be freely used throughout this paper.

Recall (Sections 5.3 and 5.4 of [9]) that for a normal subgroup N of a finite group G, the rational number m_{G,N} is defined by

m_{G,N} = \frac{1}{|G|} ∑_{\begin{matrix}X≤G \\ XN=G \end{matrix}} |X|μ(X, G) ,

where μ is the Möbius function of the poset of subgroups of G. The group G is called a B-group if m_{G,N} = 0 for any non-trivial normal subgroup N of G. Any finite group G has a largest quotient B-group β(G), unique up to isomorphism. If N ⊴ G, then m_{G,N} = 0 if and only if β(G) ≅ β(G/N).

2.1. Lemma: [M. Baumann [3] - See also [6], 8), p 713] Let L be a finite group, let p be a prime, and let E be an elementary abelian p-group on which L acts irreducibly, faithfully, and such that H^{1}(L, E) = {0}. Then the group G = E ⋊ L is a B-group.

Proof : First as E is L-simple, it follows that E is a minimal normal subgroup of G. Let N be any normal subgroup of G. Then N ∩E is equal to E or 1. So if N ≱ E, then N∩E = 1, and N centralizes E. But C_{G}(E) = E·C_{L}(E) = E, since L acts faithfully on E. Thus N ≤ E, hence N = 1.

It follows that E is the unique minimal normal subgroup of G. By Proposition 5.6.4 of [9], since E is abelian,

(2.2)  m_{G,E} = 1 − \frac{|K_{G}(E)|}{|E|} ,

where K_{G}(E) is the set of complements of E in G. The group E acts by conjugation on K_{G}(E), and the normalizer in E of K ∈ K_{G}(E) is equal to the group E^{K} of fixed points of K on E. Since E is K-simple, and K-faithful, this is equal to 1. Thus E acts freely on K_{G}(E). Since H^{1}(L, E) = {0}, the set K_{G}(E) is a single conjugacy class, i.e. a single E-orbit. Thus |K_{G}(E)| = |E|, and m_{G,E} = 0. It follows that G is a B-group.

2.3. Recall that a finite group G is called cyclic modulo a prime number p if G/O_{p}(G) is cyclic, and that G is called p-elementary if G ≅ P × C, where P is a p-group and C is a cyclic group.

2.4. Lemma: Let p be a prime number, and G be a finite group.

1. [M. Baumann [3]] The group β(G) is cyclic modulo p if and only if G is cyclic modulo p.

2. The group β(G) is a p-group if and only if G is p-elementary.

Proof : For Assertion 1, use the fact that by a theorem of Conlon, the subspace NC_{p}(G) of ℚB(G) generated by the idempotents e_{H}^{G}, where H is not cyclic modulo p, is equal to the kernel of the morphism ℚB(G) → ℚpp_{k}(G), where ℚpp_{k}(G) is the ring of p-permutation kG-modules. In particular, the correspondence G ↦ NC_{p}(G) is a biset subfunctor of ℚB. It follows that there exists a family B of B-groups such that for any group G, the space NC_{p}(G) is the ℚ-vector subspace of ℚB(G) generated by the idempotents e_{H}^{G}, where β(H) ∈ B. The family B consists of those B-groups H for which e_{H}^{H} ∈ NC_{p}(H), i.e. the B-groups which are not cyclic modulo p. Now for any group G, and any subgroup H of G, the idempotent e_{H}^{G} is in NC_{p}(G) if and only if β(H) ∈ B, on the one hand, but also if and only if H is not cyclic modulo p. Hence β(H) is not cyclic modulo p if and only if H is not cyclic modulo p. This proves Assertion 1.

For Assertion 2, clearly, if G-is p-elementary, one can assume G ≅ P × C, where P is a p-group, and C is a cyclic p′-group. By Proposition 5.6.6 of [9], this implies β(G) ≅ β(P) × β(C) ≅ β(P), since β(C) = 1. Hence β(G) is a p-group.

Conversely, suppose that β(G) is a p-group. In particular, it is cyclic modulo p, hence G is cyclic modulo p, by Assertion 1. The Frattini subgroup Φ(P) of P is a normal subgroup of G, and G/Φ(P) ≅ \overline{P} ⋊ C, where \overline{P} is the elementary abelian group P/Φ(P). Suppose that the 𝔽_{p}C-module \overline{P} admits a simple quotient E with non-trivial C-action (that is, not isomorphic to 𝔽_{p}). Then the action of C on E has a kernel D < C, and the group E ⋊ (C/D) is a quotient of \overline{P} ⋊ C, hence a quotient of G. But E ⋊ (C/D) is a B-group by Lemma 2.1: Indeed E is (C/D)-simple and faithful by construction, and H^{1}(C/D, E) = {0}, since C/D is a p′-group.

Now E ⋊(C/D) is a B-group, which is not a p-group, since D ≠ C, and it is a quotient of G, hence of β(G), which is a p-group. This is a contradiction.

Hence C acts trivially on \overline{P}. But for any p-group P, the kernel of the morphism Aut(P) → Aut(P/Φ(P)) is a p-group. As C is a p′-group, and acts trivially on \overline{P}, it acts trivially on P. Thus G ≅ P × C, as was to be shown.

2.5. Lemma: Let p be a prime number, and P be a finite p-group.

1. Let Q be a normal subgroup of P. Then Q ∩ Φ(P) = 1 if and only if Q is elementary abelian and central in P, and admits a complement

in P.

2. Let Q and R be normal subgroups of P, such that |Q| = |R|. Then Q ∩ Φ(P) = 1 = R ∩ Φ(P) if and only if Q and R are elementary abelian and central in P, and admit a common complement in P.

In this case, set H = P/R ≅ P/Q, and denote by γ the rank of the group H/Φ(H). If Q and R have rank m, and if Q ∩ RΦ(P) has rank m − s, the number of common complements of Q and R in P is equal

{to \atop (p^{s} − 1)(p^{s−1} − 1) ⋯ (p − 1)p^{\binom{s}{2}+s(m−s)+m(γ−s)} .}

Proof : For Assertion 1, if Q is elementary abelian and central in P, and admits a complement L, then P = Q × L. Thus Φ(P) = 1 × Φ(L), hence Q ∩ Φ(P) = 1. Conversely, if Q ∩ Φ(P) = 1, then Q maps injectively into P/Φ(P), so Q is elementary abelian. Let L ≥ Φ(P) be a subgroup of P such that L/Φ(P) is a complement of QΦ(P)/Φ(P) in the 𝔽_{p}-vector space P/Φ(P). Then QΦ(P)L = P, thus QL = P, and QΦ(P) ∩ L = Φ(P), i.e. Q ∩ L ≤ Q ∩ Φ(P) = 1. Since L ≥ Φ(P), it follows that L ⊴ P, thus [L, Q] ≤ L ∩ Q = 1, and Q is central in P.

For Assertion 2, let Q and R be normal subgroups of P with |Q| = |R|. If Q and R are elementary abelian central subgroups of P with a common complement in P, then Q∩Φ(P) = R∩Φ(P) = 1 by Assertion 1. Conversely, if Q∩Φ(P) = 1 and R∩Φ(P) = 1, then Q and R are elementary abelian and central in P by Assertion 1. If L is a complement of Q in P, then P = Q × L for Q is central in P, thus L ⊴ P, and P/L ≅ Q is elementary abelian. Thus L ≥ Φ(P), and L/Φ(P) is a complement of QΦ(P)/Φ(P) in P/Φ(P). Conversely if L/Φ(P) is a complement of QΦ(P)/Φ(P) in P/Φ(P), then L is a complement of Q in P, by the argument used in the proof of Assertion 1.

So finding a common complement to Q and R in P amounts to finding a common complement of Q̃ = QΦ(P)/Φ(P) and R̃ = RΦ(P)/Φ(P) in P̃ = P/Φ(P). Moreover |Q̃| = |Q| = |R| = |R̃|. The 𝔽_{p}-vector space P̃ can be split as P̃ = I ⊕E⊕F ⊕V, where I = Q̃∩R̃, where E is a complement of I in Q̃ and F is a complement of I in R̃, and V is a complement of Q̃+R̃ in P̃. Then L = F ⊕ V is a complement of Q̃ in P̃, and all the other complements of Q̃ are of the form {(φ(x), x) | x ∈ L}, where φ : L → Q̃ is a group homomorphism. In other words, any complement L′ of Q̃ is of the form

L′ = {(a(f) + b(v), c(f) + d(v), f, v) | f ∈ F, v ∈ V } ,

where a : F → I, b : V → I, c : F → E and d : V → E are group homomorphisms. The group L′ is a complement of R̃ if and only if its intersection with R̃ is trivial, or equivalently if c is injective, hence an isomorphism, since |E| = |F |.

It follows that the number of common complements of Q̃ and R̃ in P̃ is equal to the number of 4-tuples (a, b, c, d), where c is an isomorphism. Hence

|K_{P} (Q) ∩ K_{P} (R)| = |Aut(E)||Hom(F, I)||Hom(V, I)||Hom(V, E)|
= |Aut(E)||Hom(F, I)||Hom(V, Q̃)| .

Moreover

I = (QΦ(P) ∩ RΦ(P))/Φ(P) = (Q ∩ RΦ(P))Φ(P)/Φ(P) ≅ Q ∩ RΦ(P)

has rank m − s, and F ≅ E ≅ Q̃/I has rank s. Finally

V ≅ P̃/(Q̃R̃) ≅ (P/RΦ(P))/(QRΦ(P)/RΦ(P))

has rank γ − s, since P/RΦ(P) ≅ H/Φ(H), as Φ(P/R) = RΦ(P)/R, and since QRΦ(P)/RΦ(P) ≅ Q/(Q ∩ RΦ(P)). This completes the proof.

2.6. Corollary: Let P be a finite p-group, and M be a normal subgroup

of P. Then ; P/(M ∩ Φ(P)) ≅ E × (P/M) ,

where E = M/(M ∩ Φ(P)) is elementary abelian.

Proof : The normal subgroup \overline{M} = M/(M ∩ Φ(P)) of \overline{P} = P/(M ∩ Φ(P)) intersects the Frattini subgroup Φ(\overline{P}) = Φ(P)/(M ∩ Φ(P)) trivially, hence there exists a subgroup L of \overline{P} such that \overline{P} = \overline{M} ×L. Moreover L ≅ \overline{P} /\overline{M} ≅ P/M.

3. Simple biset functors and bilinear forms

Let 𝔽 be any field. Recall (see [11]) that, given a finite group H, we defined, for any finite group G

𝔽\overline{B}(G, H) = 𝔽B(G, H)/ ∑_{|K|<|H|} 𝔽B(G, K) ∘ 𝔽B(K, H) ,

and that the correspondence G ↦ 𝔽\overline{B}(G, H) is a quotient biset functor of the Yoneda functor G ↦ 𝔽B(G, H) at the group H.

When V is a 𝔽Out(H)-module, we defined an 𝔽-valued bilinear form ⟨ , ⟩_{V,G} on 𝔽\overline{B}(G, H) by

∀α, β ∈ 𝔽\overline{B}(G, H), ⟨α, β⟩_{V,G} = χ_{V} (π_{H}(α̂^{op} ∘ β̂)) ,

where α̂, β̂ are elements of 𝔽B(G, H) lifting α, β ∈ 𝔽\overline{B}(G, H), respectively, where π_{H} : 𝔽B(H, H) → 𝔽\overline{B}(H, H) ≅ 𝔽Out(H) is the projection map, and χ_{V} is the character of V, i.e. the trace function End_{𝔽}(V) → 𝔽. The main property of these constructions is that

𝔽\overline{B}(G, H)/Rad⟨ , ⟩_{V,G} ≅ S_{H,V} (G)^{dim_{𝔽} V} .

Moreover, if L is a finite group, then for any γ ∈ B(L, G), any α ∈ \overline{B}(G, H), and any β ∈ \overline{B}(L, H),

(3.1)  ⟨γ(α), β⟩_{V,L} = ⟨α, γ^{op}(β)⟩_{V,G} .

3.2. Suppose from now on that 𝔽 is a field of characteristic 0. Observe that ẽ_{K}^{G} = (ẽ_{K}^{G})^{op} for any subgroup K of a finite group G. By (3.1), this implies that the decomposition

𝔽\overline{B}(G, H) = ⨁_{K∈[s_{G}]} ẽ_{K}^{G}𝔽\overline{B}(G, H) ,

where [s_{G}] is a set of representatives of conjugacy classes of subgroups of G, is an orthogonal decomposition with respect to the form ⟨ , ⟩_{V,G}. Moreover

ẽ_{G}^{G}S_{H,V} (G)^{dim_{𝔽} V} ≅ ẽ_{G}^{G}𝔽\overline{B}(G, H)/Rad⟨ , ⟩_{V,G} ,

and the isomorphism

𝔽\overline{B}(G, H) ≅ ⨁_{K∈[s_{G}]} (ẽ_{K}^{K}𝔽\overline{B}(K, H))^{N_{G}(K)}

given by Proposition 6.5.5 of [9] induces an isomorphism

(3.3)  S_{H,V} (G)^{dim_{𝔽} V} ≅ ⨁_{K∈[s_{G}]} (ẽ_{K}^{K}𝔽\overline{B}(K, H)/Rad⟨ , ⟩_{V,K})^{N_{G}(K)} .

Now \overline{B}(K, H) is generated by the images of the elements (K × H)/L, where L is a subgroup of K × H. If this image is non-zero, then L is of the form L = {(x, s(x)) | x ∈ X}, where X is a subgroup of K and s : X ↠ H is a surjective group homomorphism.

The (K, G)-biset U = (K × H)/L factors as U = Ind_{X}^{K} ∘ V, for a suitable (X, H)-biset V ([9], Lemma 2.3.26), and by [9], Corollary 2.5.12

ẽ_{K}^{K} ∘ Ind_{X}^{K} = Ind_{X}^{K} ∘ { \atop Res_{X}^{K}e_{K}^{K}} .

Now Res_{X}^{K}e_{K}^{K} = 0 if X is a proper subgroup of K. It follows that ẽ_{K}^{K}𝔽\overline{B}(K, H) is generated by the images \overline{u}_{s} of the elements

u_{s} = ẽ_{K}^{K} ×_{K} (K × H)/Δ_{s}^{⋄}(K) ,

where Δ_{s}^{⋄}(K) = {(x, s(x)) | x ∈ K}, for a surjective group homomorphism s : K ↠ H.

Let ϖ_{H} : Aut(H) → Out(H) denote the projection map. Then:

3.4. Proposition: Let s, t : K ↠ H be two surjective group homomorphisms. Let M = Ker s, and N = Ker t. Then

⟨\overline{u}_{s}, \overline{u}_{t}⟩_{V,K} = m_{K,M∩N} \frac{μ_{⊴ K}(M∩N,M)}{|M:M∩N|} χ_{V} ( ∑_{Y ∈\overline{K}(K,M,N)} ϖ_{H}([s, Y, t]))

= m_{K,M∩N} \frac{μ_{⊴ K}(M∩N,M)}{|M:M∩N|} χ_{V} ( ∑_{{θ∈Aut(H) \atop Δ_{θ}(H)≤(s×t)(K)}} ϖ_{H}(θ)) ,

where μ_{⊴ K} is the Mbius function of the poset of normal subgroups of K, and

\overline{K}(K, M, N) = \{Y ≤ K | Y N = Y M = K, Y ∩ N = Y ∩ M = M ∩ N\}

is the set of subgroups Y of K, containing M ∩ N, such that Y/(M ∩ N) is a common complement of M/(M ∩ N) and N/(M ∩ N) in K/(M ∩ N). Moreover for Y ∈ \overline{K}(K, M, N), the symbol [s, Y, t] denotes the automorphism of H defined by [s, Y, t](t(y)) = s(y), ∀y ∈ Y.

Proof : By definition, and since ẽ_{K}^{K} is an idempotent

⟨\overline{u}_{s}, \overline{u}_{t}⟩_{V,K} = χ_{V} (π_{H}(u_{s}^{op} ∘ u_{t}))
= χ_{V} (π_{H}((H × K)/Δ_{s}(K) ×_{K} ẽ_{K}^{K} ×_{K} (K × H)/Δ_{t}^{⋄}(K))) ,

where Δ_{s}(K) = {(s(x), x) | x ∈ K}. Set

a_{s,t} = (H × K)/Δ_{s}(K) ×_{K} ẽ_{K}^{K} ×_{K} (K × H)/Δ_{t}^{⋄}(K) .

Then

a_{s,t} = \frac{1}{|K|} ∑_{L≤K} |L|μ(L, K)(H × H)/Δ_{s,t}(L) ,

where Δ_{s,t}(L) = {(s(l), t(l)) | l ∈ L}.

This subgroup of H × H is equal to Δ_{θ}(H), for some automorphism θ of H, if and only if

(3.5)  L ∩ M = L ∩ N and LM = K = LN ,

where M = Ker s and N = Ker t. In this case the automorphism θ is defined by θ(t(l)) = s(l), for any l ∈ L. The two conditions 3.5 and the automorphism θ remain unchanged when L is replaced by L(M ∩ N). Moreover the conditions 3.5 are equivalent to saying that the group Y = L(M ∩ N) is in \overline{K}(K, M, N), and in this case θ = [s, Y, t]. Conversely, fix some Y ∈ \overline{K}(K, M, N), and consider all the subgroups L of K such that L(M ∩N) = Y. Recall that

∑_{{L≤Y \atop L(M∩N)=Y}} |L|μ(L, K) = m_{K,M∩N}|Y |μ(Y, K) .

This gives:

π_{H}(a_{s,t}) = m_{K,M∩N} ∑_{Y ∈\overline{K}(K,M,N)} \frac{|Y |}{|K|}μ(Y, K)ϖ([s, Y, t]) .

Now if Y/(M ∩ N) is a complement of M/(M ∩ N) in K/(M ∩ N), and if K/M ≅ H, it follows that |Y | = |M ∩ N||H|. Moreover the poset ]Y, K[ is isomorphic to the poset ]M ∩ N, M[^{Y}. But since M and N are normal subgroups of K, the commutator group [M, N] is contained in M ∩ N. It follows that ]M ∩N, M[^{Y} =]M ∩N, M[^{Y N}=]M ∩N, M[^{K}, and that μ(Y, K) = μ_{⊴ K}(M ∩ N, M). This completes the proof of the first equality of the proposition. The second one follows from the observation that the correspondences

Y ↦ [s, Y, t] and θ ↦ {k ∈ K | θ(t(k)) = s(k)}

are mutual inverse bijections between \overline{K}(K, M, N) and the set of automorphisms θ of H such that Δ_{θ}(H) ≤ (s × t)(K) (see Section 8.3 of [6]).

3.6. Corollary:

1. If ẽ_{K}^{K}S_{H,V} (K) ≠ {0}, the group β(K) is isomorphic to β(L), where L is a subgroup of H × H with the following properties:

(a) p_{1}(L) = p_{2}(L) = H.

(b) k_{1}(L) and k_{2}(L) are direct products of minimal normal subgroups

of H.

(c) There exist an automorphism θ of H such that θ(k_{2}(L)) = k_{1}(L).

2. In particular, if H is a p-group for some prime p, then K is pelementary. If K = P × C, where P is a p-group and C is a cyclic p′-group, then

ẽ_{K}^{K}S_{H,V} (K) ≅ ẽ_{P}^{P} S_{H,V} (P) ,

and this isomorphism is compatible with the action of Aut(K).

Proof : Indeed, if ẽ_{K}^{K}S_{H,V} (K) ≠ {0}, then the bilinear form ⟨ , ⟩_{V,K} is not identically zero on ẽ_{K}^{K}𝔽\overline{B}(K, H). It follows that there exist surjective group homomorphisms s, t : K ↠ H such that ⟨\overline{u}_{s}, \overline{u}_{t}⟩_{V,K} ≠ 0.

Then m_{K,M∩N} ≠ 0, μ_{⊴ K}(M ∩ N, M) ≠ 0, and \overline{K}(K, M, N) ≠ ∅, where M = Ker s and N = Ker t. Hence β(K) ≅ β(K/(M ∩N)). Now K/(M ∩N) is isomorphic to L = (s × t)(K), which is a subgroup of H × H such that p_{1}(L) = p_{2}(L) = H. Moreover k_{1}(L) = s(Ker t) ≅ N/(M ∩ N) and k_{2}(L) = t(Ker s) ≅ M/(M ∩ N). Then μ_{⊴ K}(M ∩ N, M) = μ_{⊴ H}(1, t(Ker s)), and this is non zero if and only if the lattice [1, t(Ker s)]^{H} of normal subgroups of H contained in t(Ker s) is complemented, i.e. if t(Ker s) is a direct product of minimal normal subgroups of H. Finally, let Y ∈ \overline{K}(K, M, N) and θ = [s, Y, t]. If u ∈ k_{2}(L) = t(Ker s), then there exist v ∈ Ker s and y ∈ Y such that u = t(v) = t(y). Then v^{−1}y ∈ Ker t, and θ(u) = s(y) = s(v^{−1}y) ∈ s(Ker t) = k_{1}(L). In other words θ(k_{2}(L)) = k_{1}(L), which completes the proof of Assertion 1.

The first part of Assertion 2 follows from Assertion 2 of Lemma 2.4. Now if H is a p-group, if K = P ×C, where P is a p-group and C is a p′-group, and if s : K ↠ H is a surjective group homomorphism, then C ≤ Ker s. In other words, there is a surjective homomorphism \overline{s} : P ↠ H such that s = \overline{s} ∘ π, where π : K → P is the projection map. Moreover Ker s = Ker \overline{s} × C.

So with the notation of Proposition 3.4, M = \overline{M} × C, where \overline{M} = Ker \overline{s}. Similarly N = \overline{N} ×C, where N = Ker \overline{t}, and \overline{t} : P ↠ H is such that t = \overline{t}∘π. Clearly |M : M ∩ N| = |\overline{M} : \overline{M} ∩ \overline{N}|. Moreover, one checks easily that

m_{K,M∩N} = m_{P ×C,(\overline{M}∩\overline{N})×C} = m_{P,\overline{M}∩\overline{N}}m_{C,C} = m_{P,\overline{M}∩\overline{N}} \frac{ϕ(|C|)}{|C|} ,

since P and C have coprime orders, and C is cyclic.

Also

μ_{⊴ K}(M ∩ N, M) = μ_{⊴ P} (\overline{M} ∩ \overline{N}, \overline{M}) .

Finally, the maps Q ↦ Q × 1 and Y ↦ (Y ∩ P) induce inverse bijections from \overline{K}(P, \overline{M}, \overline{N}) to \overline{K}(K, M, N), and for any Y ∈ \overline{K}(K, M, N),

[s, Y, t] = [ \overline{s}, Y ∩ P, \overline{t}] .

It follows that the matrix of the form ⟨ , ⟩_{V,K} on ẽ_{K}^{K}𝔽\overline{B}(K, H) is equal to the matrix of the form ⟨ , ⟩_{V,P} on ẽ_{P}^{P} 𝔽\overline{B}(P, H), multiplied by the non-zero scalar \frac{ϕ(|C|)}{|C|}. Hence the two forms define isomorphic quadratic spaces. As all the above bijections are obviously compatible with the action of Aut(K) and the canonical group homomorphism Aut(K) → Aut(P), the induced isomorphism

ẽ_{K}^{K}S_{H,V} (K) ≅ ẽ_{P}^{P} S_{H,V} (P)

is compatible with the action of Aut(K).

3.7. Notation: Let H and P be finite p-groups.

1. Let Q_{H}(P) denote the 𝔽-vector space with basis the set

Σ_{H}(P) = {s | s : P ↠ H}

of surjective group homomorphisms from P to H, endowed with the 𝔽-valued bilinear form ⟨ , ⟩_{V,P} defined as follows: For s, t ∈ Σ_{H}(P), set M = Ker s and N = Ker t. If M ∩ Φ(P) ≠ N ∩ Φ(P), set ⟨s, t⟩_{V,P} = 0. And if M ∩ Φ(P) = N ∩ Φ(P), then the groups M/(M ∩ N) and N/(M ∩ N) are central elementary abelian subgroups of the same rank of P/M ∩ N. In this case, set

⟨s, t⟩_{V,P} = m_{P,M∩N} \frac{μ(M∩N,M)}{|M:M∩N|}χ_{V} ( ∑_{Y ∈\overline{K}(P,M,N)} ϖ_{H}([s, Y, t])) .

2. Let Q_{H}^{♯}(P) be the subspace of Q_{H}(P) with basis the subset

Σ_{H}^{♯}(P) = {s | s : P ↠ H, Ker s ∩ Φ(P) = 1}

of Σ_{H}(P).

{3. Set \atop N_{H}(P) = \{N | N ⊴ P, N ∩ Φ(P) = 1\} .}

4. Denote by E_{H}(P) the set of normal subgroups R of P, contained in Φ(P), and such that P/R ≅ E × H, for some elementary abelian p-

group E.

3.8. Proposition: Let H be a p-group, and K be a p-elementary group.

Set P = O_{p}(K). Then:

1. There is an isomorphism of 𝔽Aut(K)-modules

ẽ_{K}^{K}S_{H,V} (K) ≅ Q_{H}(P)/Rad⟨ , ⟩_{V,P} .

2. Let Γ be a finite group acting on the group K. Then Γ acts on the set E_{H}(P), and there is an isomorphism of 𝔽Γ-modules

ẽ_{K}^{K}S_{H,V} (K) ≅ ⊕_{R∈[Γ\E_{H}(P)]} Ind_{Γ_{R}}^{Γ}(Q_{H}^{♯}(P/R)/Rad⟨ , ⟩_{V,P/R}) ,

where [Γ\E_{H}(P)] is a set of representatives of Γ-orbits on E_{H}(P), and Γ_{R} denotes the stabilizer of R in Γ.

Proof : The map s ∈ Σ_{H}(P) ↦ \overline{u}_{s} ∈ ẽ_{P}^{P} 𝔽\overline{B}(P, H) induces a surjective linear map Q_{H}(P) → ẽ_{P}^{P} 𝔽\overline{B}(P, H). Let s, t ∈ Σ_{H}(P), and set M = Ker s and N = Ker t. Then |M| = |N|. It follows from Lemma 2.5 that \overline{K}(P, M, N) ≠ ∅ if and only if M/(M ∩ N) and N/(M ∩ N) are central elementary abelian subgroups of P/(M ∩ N), which intersect trivially the Frattini subgroup of P/(M ∩ N). But

Φ(P/(M ∩ N)) = Φ(P)(M ∩ N)/(M ∩ N) .

Hence M/(M ∩ N) ∩ Φ(P/(M ∩ N)) = (M ∩ Φ(P))(M ∩ N)/(M ∩ N). This group is trivial if and only if M ∩Φ(P) ≤ M ∩N, i.e. if M ∩Φ(P) ≤ N ∩Φ(P). Hence \overline{K}(P, M, N) ≠ ∅ if and only if M ∩ Φ(P) = N ∩ Φ(P).

This holds in particular if ⟨\overline{u}_{s}, \overline{u}_{t}⟩ ≠ 0. In this case, by Proposition 3.4

⟨\overline{u}_{s}, \overline{u}_{t}⟩_{V,P} = m_{P,M∩N} \frac{μ_{⊴ P} (M∩N,M)}{|M:M∩N|} χ_{V} ( ∑_{Y ∈\overline{K}(P,M,N)} ϖ_{H}([s, Y, t])) .

But μ_{⊴ P} (M ∩ N, M) = μ(M ∩ N, M), since P/(M ∩ N) centralizes both M/(M ∩ N) and N/(M ∩ N). Hence

⟨\overline{u}_{s}, \overline{u}_{t}⟩_{V,P} = ⟨s, t⟩_{V,P}

in this case, and Assertion 1 follows.

Since ⟨s, t⟩_{V,P} = 0 if M ∩ Φ(P) ≠ N ∩ Φ(P), the quadratic space Q = (Q_{H}(P), ⟨ | ⟩_{V,P} ) splits as the orthogonal sum of the subspaces Q_{R} generated by the elements s ∈ Σ_{H}(P) such that Ker s ∩ Φ(P) = R. These subspaces are permuted by the action of Aut(K), and the space Q_{R} is invariant by Aut(K)_{R}.

Let π_{R} : P → P/R be the canonical projection. The map

θ_{R} : \overline{s} ∈ Σ_{H}^{♯}(P/R) ↦ \overline{s} ∘ π_{R}

is a bijection from Σ_{H}^{♯}(P/R) to the set {s ∈ Σ_{H}(P) | Ker s∩Φ(P) = R}, and the map Y ↦ Y/R is a bijection from \overline{K}(P, M, N) to \overline{K}(P/R, M/R, N/R), such that

[\overline{s}, Y/R, \overline{t}] = [θ_{R}(\overline{s}), Y, θ_{R}( \overline{t})] ,

for any \overline{s}, \overline{t} ∈ Σ_{H}^{♯}(P/R).

Moreover, if M ∩ Φ(P) = N ∩ Φ(P) = R, then

m_{P,M∩N} = m_{P,R}m_{P/R,(M∩N)/R} = m_{P/R,(M∩N)/R} ,

as R ≤ Φ(P). Also

M/(M ∩ N) ≅ (M/R)/((M/R) ∩ (N/R)) .

It follows that

∀\overline{s}, \overline{t} ∈ Σ_{H}^{♯}(P/R), ⟨θ_{R}(\overline{s}), θ_{R}( \overline{t})⟩_{V,P} = ⟨\overline{s}, \overline{t} ⟩_{V,P/R} .

Hence there is an isomorphism

Q_{R}/Rad⟨ , ⟩_{V,P} ≅ Q_{H}^{♯}(P/R)/Rad⟨ , ⟩_{V,P/R} ,

of 𝔽Γ_{R}-modules.

To complete the proof of Assertion 2, it remains to observe that if Q is a p-group, the set Σ_{H}^{♯}(Q) is non-empty if and only if the group Q is isomorphic to E × H, for some elementary abelian p-group E: Indeed, if Q = E × H, where E is elementary abelian, then Φ(Q) = 1 × Φ(H), and the projection map s : Q → H = Q/E is an element of Σ_{H}^{♯}(Q). Conversely, if s ∈ Σ_{H}^{♯}(Q), then E = Ker s is an elementary abelian central subgroup of Q, which admits a complement L in Q, by Lemma 2.5. Thus Q = E × L, and L ≅ Q/E ≅ H. Hence Q ≅ E × H.

3.9. Theorem: Let G be a finite group, let H be a finite p-group, and let V be a simple 𝔽Out(H)-module. Then

S_{H,V} (G)^{dim_{𝔽} V} ≅ ⊕_{(K,R)} (Q_{H}^{♯}(K_{p}/R)/Rad⟨ , ⟩_{V,K_{p}/R})^{N_{G}(K,R)} .

where (K, R) runs through a set of G-conjugacy classes of pairs consisting of a p-elementary subgroup K of G, and a p-subgroup R in E_{H}(K_{p}), where

K_{p} = O_{p}(K), and N_{G}(K, R) = N_{G}(K) ∩ N_{G}(R).

Proof : This follows from Equation 3.3, Corollary 3.6 and Proposition 3.8.

4. Proof of the theorem

This section is devoted to the proof of the following theorem, announced in the introduction:

4.1. Theorem: Let 𝔽 be a field of characteristic 0, let p be a prime number, and H be a finite p-group. Let moreover G be a finite group.

1. If H = 1, the dimension of S_{H,𝔽}(G) is equal to the number of conjugacy

classes of cyclic subgroups of G.

2. If H ≅ C_{p} × C_{p}, the dimension of S_{H,𝔽}(G) is equal to the number of conjugacy classes of non-cyclic p-elementary subgroups of G.

3. If H is any other finite p-group, the dimension of S_{H,𝔽}(G) is equal to the number of conjugacy classes of sections (T, S) of G such that T/S ≅ H and T is p-elementary.

Proof : Step 1: Let K be a subgroup of G. Then ẽ_{K}^{K}S_{H,𝔽}(K) = {0}, by Corollary 3.6, unless K ≅ P × C, where P is a p-group and C is a cyclic p′-group, and in this case ẽ_{K}^{K}S_{H,𝔽}(K) ≅ ẽ_{P}^{P} S_{H,𝔽}(P) as 𝔽Aut(K)-modules.

By Proposition 3.8, there is an isomorphism of 𝔽Aut(P)-modules

ẽ_{P}^{P} S_{H,𝔽}(P) ≅ ⊕_{R} Ind_{Aut(P)_{R}}^{Aut(P)}(Q_{H}^{♯}(P/R)/Rad⟨ , ⟩_{𝔽,P/R}) ,

where R runs through a set of representatives of Aut(P)-orbits of normal subgroups of P contained in Φ(P), such that P/R ≅ E × H, for some elementary abelian p-group E. So the computation of ẽ_{P}^{P} S_{H,𝔽}(P) comes down to the computation of the 𝔽Aut(Q)-module

V_{H}(Q) = Q_{H}^{♯}(Q)/Rad⟨ , ⟩_{𝔽,Q} ,

for a p-group Q = P/R of the form E ×H, where R is some normal subgroup of P contained in Φ(P). Recall that Q_{H}^{♯}(Q) is the 𝔽-vector space with basis

Σ_{H}^{♯}(Q) = {s | s : Q ↠ H, Ker s ∩ Φ(Q) = 1} ,

and that the bilinear form ⟨ , ⟩_{𝔽,Q} is defined for s, t ∈ Σ_{H}^{♯}(Q) by

⟨s, t⟩_{𝔽,Q} = m_{Q,M∩N} \frac{μ(M∩N,M)}{|M:M∩N|}|\overline{K}(Q, M, N)| ,

where M = Ker s and N = Ker t.

This shows that ⟨s, t⟩_{𝔽,Q} depends only on M and N. It follows that Q_{H}^{♯}(Q)/Rad⟨ , ⟩_{𝔽,Q} is also isomorphic to the quotient of the 𝔽-vector space with basis the set N_{H}(Q) = {N ⊴ Q | P/N ≅ H, N ∩Φ(Q) = 1} introduced in Notation 3.7, by the radical of the bilinear form ⟨ , ⟩_{𝔽,Q}^{♮} defined by

⟨M, N⟩_{𝔽,Q}^{♮} = m_{Q,M∩N} \frac{μ(M∩N,M)}{|M:M∩N|} |\overline{K}(Q, M, N)| ,

for M, N ∈ N_{H}(Q) .

Step 2: Now Assertion 1 is well known (see e.g. Proposition 4.4.8 [9]), but it can also be recovered from the argument of Step 1: Indeed, if H = 1, there is a unique normal subgroup N of Q such that Q/N ≅ H, namely Q itself. If moreover N ∩Φ(Q) = 1, then Φ(Q) = 1, and Q is elementary abelian. But as Q = P/R, for some R ≤ Φ(P), it follows that R = Φ(P), and Q = P/Φ(P). Moreover N_{H}(Q) = {Q}, and

⟨Q, Q⟩_{𝔽,Q}^{♮} = m_{Q,Q} ,

which is equal to 0 if Q is non-cyclic, and to 1 − 1/p otherwise. Hence V_{H}(Q) = {0} if Q is non-cyclic, and V_{H}(Q) is one dimensional if Q = P/Φ(P) is cyclic, i.e. if P is cyclic. But P = O_{p}(K) for some p-elementary subgroup K of G. Hence P is cyclic if and only if K itself is cyclic, and this leads to Assertion 1.

We can now assume that H is a non-trivial p-group, of order p^{h}, and make a series of observations:

• Let P be a p-group, and Q be a normal subgroup of P. Example 5.2.3 of [9] shows that m_{P,Q} = m_{P,Φ(P)Q}. By Proposition 5.3.1 of [9], it follows that

(4.2)  m_{P,Q} = m_{P,Φ(P)}m_{P/Φ(P),QΦ(P)/Φ(P)} = m_{E,F} ,

where E is the elementary abelian p-group P/Φ(P), and F its subgroup QΦ(P)/Φ(P). If E has rank n ≥ 2 and F has rank k, then

(4.3)  m_{E,F} = (1 − p^{n−2})(1 − p^{n−3}) ⋯ (1 − p^{n−k−1}) ,

(which is equal to 0 if k ≥ n−1, and non-zero otherwise). This follows from an easy induction argument on k, using Proposition 5.3.1 of [9], and starting with the case k = 1, which is a special case of Equation 2.2.

If E has rank 1 and F = 1, then m_{E,F} = 1. In this case m_{E,E} = 1 − \frac{1}{p}. This is the only case where m_{E,F} is not an integer.

• Let M, N ∈ N_{H}(Q). Then in particular M and N have the same order. Recall that m_{Q,M∩N} is non-zero if and only if β(Q) ≅ β(Q/(M ∩ N)). So either Q and Q/(M ∩N) are both cyclic, or they are both non-cyclic. Equivalently, either Q is cyclic, or Q/(M ∩ N) is non-cyclic. If H is non-cyclic, then Q/(M ∩ N) is non-cyclic, as it maps surjectively on Q/M ≅ H.

So if m_{Q,M∩N} = 0, then H is cyclic, Q is non-cyclic, and Q/(M ∩ N) is cyclic. But then M/(M ∩ N) = N/(M ∩ N), since the cyclic group Q/(M ∩ N) admits a unique subgroup of a given order. Thus M = N. Conversely, if H is cyclic, if Q is non-cyclic, and if M = N, then m_{Q,M∩N} = 0 since Q/(M ∩ N) ≅ H is cyclic.

• If M, N ∈ N_{H}(Q), then the subgroups \overline{M} = M/(M ∩ N) and \overline{N} = N/(M ∩N) are central elementary abelian subgroups of the same order of \overline{Q} = Q/(M ∩ N). If \overline{M} has rank m, then

μ(M ∩ N, M) = (−1)^{m}p^{\binom{m}{2}} .

Now by Lemma 2.5, the product

(4.4)  α_{M,N} = μ(M ∩ N, M)|\overline{K}(Q, M, N)|

is equal to

α_{M,N} = (−1)^{m}(p^{s} − 1)(p^{s−1} − 1) ⋯ (p − 1)p^{\binom{m}{2}+\binom{s}{2}+s(m−s)+m(γ−s)}
= (−1)^{m+s}(1 − p^{s})(1 − p^{s−1}) ⋯ (1 − p)p^{\binom{m}{2}+\binom{s}{2}+s(m−s)+m(γ−s)} ,

where γ is the rank of H/Φ(H), and s is the rank of \overline{M}/(\overline{M} ∩\overline{N}Φ(\overline{Q})) ≅ M/(M ∩ NΦ(Q)).

Step 3: Finally ⟨M, N⟩_{𝔽,Q}^{♮} = 0 if and only if m_{Q,M∩N} = 0, i.e. H is cyclic, M = N, and Q is not cyclic. In all other cases, the groups \overline{M} = M/(M ∩ N) and \overline{N} = N/(M ∩N) are elementary abelian, and central in \overline{Q} = Q/(M ∩N). Moreover \overline{M} ∩ \overline{N} = 1. Let m be the rank of \overline{M}, let s denote the rank of \overline{M}/(\overline{M} ∩ \overline{N}Φ(\overline{Q})), and let γ denote the rank of H/Φ(H). Then

⟨M, N⟩_{𝔽,Q}^{♮} = m_{Q,M∩N} \frac{α_{M,N}}{|M:M∩N|}
= m_{Q,M∩N} \frac{α_{M,N}}{|\overline{M}|}
= (−1)^{m}m_{Q,M∩N}(p^{s}−1)(p^{s−1}−1) ⋯ (p−1)p^{\binom{m}{2}−m+\binom{s}{2}+s(m−s)+m(γ−s)} ,

i.e. finally (4.5) ⟨M, N⟩_{𝔽,Q}^{♮} = (−1)^{m}m_{Q,M∩N}(p^{s}−1)(p^{s−1}−1) ⋯ (p−1)p^{\frac{1}{2} (m−s)(m+s+1)+m(γ−2)} .

Let n denote the rank of Q/Φ(Q). By Equation 4.3

m_{Q,M∩N} = (1 − p^{n−2})(1 − p^{n−3}) ⋯ (1 − p^{n−k−1}) ,

where k is the rank of (M ∩ N)Φ(Q)/Φ(Q) ≅ M ∩ N. Since Q/MΦ(Q) ≅ H/Φ(H) has rank γ, it follows that MΦ(Q)/Φ(Q) ≅ M has rank n − γ. Since M/(M ∩ N) has rank m, it follows that k = n − m − γ. Thus

m_{Q,M∩N} = (1 − p^{n−2})(1 − p^{n−3}) ⋯ (1 − p^{m+γ−1}) .

It follows that

(4.6)  ⟨M, N⟩_{𝔽,Q}^{♮} = A_{M,N}(−1)^{m+s}p^{\frac{1}{2} (m−s)(m+s+1)+m(γ−2)} ,

where

A_{M,N} = (1 − p^{n−2})(1 − p^{n−3}) ⋯ (1 − p^{m+γ−1})(1 − p^{s})(1 − p^{s−1}) ⋯ (1 − p)

is an integer congruent to 1 modulo p.

Step 4: Assume first that H is non-cyclic, i.e. that γ ≥ 2. In this case ⟨M, N⟩_{𝔽,Q}^{♮} is non-zero. If M = N, then m = s = 0, and ⟨M, N⟩_{𝔽,P}^{♮} = A_{M,M} is congruent to 1 modulo p. And if M ≠ N, then m ≥ 1. As γ ≥ 2 and m ≥ s, the exponent

\frac{1}{2}(m − s)(m + s + 1) + m(γ − 2)

of p in the right hand side of 4.6 is non-negative. It is equal to 0 if and only if m = s and γ = 2. In this case \overline{M} ∩ \overline{N}Φ(\overline{Q}) = 1, so \overline{M} maps into \overline{Q}/\overline{N}Φ(\overline{Q}) ≅ H/Φ(H), which has rank γ = 2. It follows that m ≤ 2.

If m = 2, then \overline{M}\overline{N}Φ(\overline{Q}) = \overline{Q}, thus \overline{M}\overline{N} = \overline{Q}, and H ≅ \overline{Q}/\overline{N} ≅ \overline{M} (since \overline{M} ∩ \overline{N} = 1), so H is elementary abelian of rank 2.

If m = 1, then as \overline{M} ≅ C_{p} maps into \overline{Q}/\overline{N}Φ(\overline{Q}) ≅ C_{p} × C_{p}, the group \overline{Q}/(\overline{M}\overline{N}Φ(\overline{Q})) is cyclic, so \overline{Q}/\overline{M}\overline{N} is cyclic. But \overline{M}\overline{N} is a central subgroup of \overline{Q}. It follows that \overline{Q} is abelian, so H ≅ \overline{Q}/\overline{M} is abelian. Hence \overline{Q}/\overline{M} is non-cyclic, and it has a subgroup \overline{M}\overline{N}/\overline{M} of order p such that the corresponding quotient \overline{Q}/\overline{M}\overline{N} is cyclic. It follows that \overline{Q}/\overline{M} ≅ H ≅ C_{p} × C_{p^{h−1}}, for some h ≥ 2.

Step 5: Assume that H is neither cyclic nor isomorphic to C_{p} × C_{p^{h−1}}, for some h ≥ 2. Then the matrix of the bilinear form ⟨ , ⟩_{𝔽,Q}^{♮} is congruent to the identity matrix modulo p. In particular, it is non-singular, and the 𝔽Aut(Q)- module V_{H}(Q) = Q_{H}^{♯}(Q) is isomorphic to the permutation module on the set N_{H}(Q).

It follows that the Aut(P)-module ẽ_{P}^{P} S_{H,𝔽}(P) is isomorphic to the permutation module on the set of normal subgroups M of P such that P/M ≅ H. Going back to Step 1 and to the p-elementary subgroup K = P × C of G, it follows that the space ẽ_{K}^{K}S_{H,𝔽}(K)^{N_{G}(K)} has a basis in one to one correspondence with the N_{G}(K)-orbits of normal subgroups M of K such that K/M ≅ H. Now the isomorphism (3.3) shows that S_{H,𝔽}(G) has a basis in one to one correspondence with the G-conjugacy classes of sections (K, M) of G such that K is p-elementary and K/M ≅ H. This proves the theorem, in the case where H is neither cyclic nor isomorphic to C_{p} × C_{p^{h−1}}, for some h ≥ 2.

Step 6: Suppose now that H is cyclic, of order p^{h} > 1. Assume first that Q is cyclic. Then since Q ≅ E × H for some elementary abelian p-group E, it follows that E = 1, i.e. Q ≅ H. In this case N_{H}(Q) = {1}, and ⟨1, 1⟩_{H,𝔽}^{♮} = 1. Hence V_{H}(Q) is isomorphic to the trivial 𝔽Aut(Q)-module in this case.

If Q is non-cyclic, let M, N ∈ N_{H}(Q). Recall that

⟨M, N⟩_{𝔽,Q}^{♮} = m_{Q,M∩N} \frac{α_{M,N}}{|M:M∩N|} ,

where α_{M,N} is defined in (4.4).

The groups \overline{M} = M/(M ∩ N) and \overline{N} = N/(M ∩ N) are non-trivial elementary abelian central subgroups of \overline{Q} = Q/(M ∩N), and have a common complement in \overline{Q}. As \overline{M} is isomorphic to the subgroup MN/N of the cyclic group Q/N ≅ H, it follows that \overline{M} ≅ C_{p}. Moreover \overline{M} has a complement in \overline{Q}, so \overline{Q} ≅ C_{p}×C_{p^{h}}. Hence if Q/Φ(Q) has rank n, then Φ(Q)(M ∩N)/Φ(Q) has rank n − 2 since

Q/(Φ(Q)(M ∩ N)) ≅ \overline{Q}/Φ(\overline{Q}) ≅ C_{p} × C_{p} .

By Equations 4.3 and 4.2, it follows that

m_{Q,M∩N} = (1 − p^{n−2})(1 − p^{n−3}) ⋯ (1 − p) .

Moreover since m = 1 and γ = 1, Equation 4.5 gives

⟨M, N⟩_{𝔽,Q}^{♮} = −m_{Q,M∩N}(p^{s}−1)(p^{s−1}−1) ⋯ (p−1)p^{\frac{1}{2} (1−s)(2+s)−1} .

Since 0 ≤ s ≤ m = 1, there are two cases:

• If s = 1, then \overline{M} maps into \overline{Q}/\overline{N}Φ(\overline{Q}) ≅ H/Φ(H) ≅ C_{p}, hence MN = Q as above, and Q/N ≅ C_{p^{h}} ≅ \overline{M} ≅ C_{p}, so h = 1. In this case

⟨M, N⟩_{𝔽,Q}^{♮} = −(1 − p^{n−2})(1 − p^{n−3}) ⋯ (1 − p)(p − 1)/p
= (1 − p^{n−2})(1 − p^{n−3}) ⋯ (1 − p^{2})(1 − p)^{2}/p .

• If s = 0. Then \overline{M} ≤ \overline{N}Φ(\overline{Q}), so \overline{M}Φ(\overline{Q}) = \overline{N}Φ(\overline{Q}). If h = 1, then \overline{Q}/\overline{M} ≅ C_{p}, so \overline{M} ≥ Φ(\overline{Q}), and it follows that \overline{M} = \overline{N}, a contradiction. Thus h > 1 in this case. Moreover

⟨M, N⟩_{𝔽,Q}^{♮} = −(1 − p^{n−2})(1 − p^{n−3}) ⋯ (1 − p) .

So in any case, there is a non-zero rational number ρ, depending only on Q (and H), such that ⟨M, N⟩_{𝔽,Q}^{♮} = ρ when ⟨M, N⟩_{𝔽,Q}^{♮} ≠ 0. Moreover ⟨M, N⟩_{𝔽,Q}^{♮} ≠ 0 if and only if M ≠ N.

So the matrix of the form ⟨ , ⟩_{𝔽,P}^{♮} is equal to ρJ, where J is a matrix of size |N_{H}(Q)|, with zero diagonal, and non-diagonal coefficients equal to 1. Hence this matrix is non-singular if and only if |N_{H}(Q)| > 1.

But Q = E × L, where L ≅ H and E is a non-trivial elementary abelian p-group. The elements of N_{H}(Q) are exactly the groups

E_{φ} = {(e, φ(e)) | e ∈ E} ,

where φ is a group homomorphism from E to L. There are |E| such homomorphisms, hence |N_{H}(Q)| = |E| > 1.

It follows that the matrix of the form ⟨ , ⟩_{𝔽,Q}^{♮} is non-singular, hence the form ⟨ , ⟩_{𝔽,Q}^{♮} is non-degenerate.

So either when Q is cyclic, or when it is not, the form ⟨ , ⟩_{𝔽,Q}^{♮} is nondegenerate. By the same argument as at the end of Step 4, this proves that S_{H,𝔽}(G) has a basis in one to one correspondence with the G-conjugacy classes of sections (K, M) of G for which K is p-elementary and K/M ≅ H. This proves the theorem in the case where H is cyclic.

Step 7: Suppose now that H ≅ C_{p} × C_{p^{h−1}}, for some h ≥ 2. Note that if h = 2, then H is elementary abelian, so Q = P/R ≅ E × H is elementary abelian. Since R ≤ Φ(P), this forces R = Φ(P).

Now if M, N ∈ N_{H}(Q), since γ = 2 in this case,

⟨M, N⟩_{𝔽,Q}^{♮} = (−1)^{m}m_{Q,M∩N}(p^{s}−1)(p^{s−1}−1) ⋯ (p−1)p^{\frac{1}{2} (m−s)(m+s+1)} ,

and moreover

m_{Q,M∩N} = (1 − p^{n−2})(1 − p^{n−3}) ⋯ (1 − p^{m+1}) ,

where n is the rank of Q/Φ(Q), where m is the rank of the elementary abelian subgroups \overline{M} = M/(M ∩ N) and \overline{N} = N/(M ∩ N) of \overline{Q} = Q/(M ∩ N), and s is the rank of M/(M ∩ NΦ(Q)) ≅ \overline{M}/(\overline{M} ∩ \overline{N}Φ(\overline{Q})). Since the exponent \frac{1}{2}(m−s)(m+s+1) of p is non-negative, it follows that ⟨M, N⟩_{𝔽,Q}^{♮} is an integer. Moreover, if m > s, this integer is a multiple of p. On the other hand m = s if and only if \overline{M} ∩ \overline{N}Φ(\overline{Q}) = 1, or equivalently if M ∩ NΦ(Q) = M ∩ N. Since M ∩ Φ(Q) = N ∩ Φ(Q) = 1, this is equivalent to MN ∩ Φ(Q) = 1. In this case

(4.7)  ⟨M, N⟩_{𝔽,Q}^{♮} = (1 − p^{n−2})(1 − p^{n−3}) ⋯ (1 − p)

is congruent to 1 modulo p. It follows that the matrix of the form ⟨ , ⟩_{𝔽,Q}^{♮} is congruent modulo p to the incidence matrix of the relation ∼ on N_{H}(Q) defined by M ∼ N if and only if MN ∩ Φ(Q) = 1. There are now two cases:

• Case 1: Assume first that h ≥ 3, i.e. that H is not elementary abelian of rank 2.

4.8. Lemma: Let H = C_{p} × C_{p^{h−1}}, for h ≥ 3, and Q = E × H, where E is an elementary abelian p-group of rank e. Let S denote the incidence matrix of the relation ∼ on N_{H}(Q) defined by

M ∼ N ⇔ MN ∩ Φ(Q) = 1 .

Then:

1. if e = 0, the matrix S is the matrix (1).

2. if e ≥ 1, the eigenvalues of S are p^{e+1} − p + 1, p^{e} − p + 1, and 1 − p, with respective multiplicities 1, p^{e+1} − p, and p^{2e} − p^{e+1} + p − 1.

In both cases M is invertible modulo p.

Proof : If e = 0, then E = 1 and Q ≅ H, so N_{H}(Q) consists of the trivial subgroup E of Q. Since E ∼ E, Assertion 1 follows.

If e ≥ 1, then N_{H}(Q) consists of the subgroups

E_{φ} = {(x, φ(x)) | x ∈ E} ,

where φ : E → H is a group homomorphism. Since E is elementary abelian, the image of φ is contained in the subgroup C_{p} × C_{p} of H = C_{p} × C_{p^{h−1}}. So there are group homomorphisms a, b : E → C_{p} such that φ = (a, b), i.e. φ(x) = (a(x), b(x)) for any x ∈ E.

Let φ = (a, b) and φ′ = (a′, b′) be two group homomorphisms from E to H. Then, with an additive notation

E_{φ}E_{φ′} = {(x − x′, a(x) − a′(x′), b(x) − b′(x′)) | x, x′ ∈ E} ≤ E × C_{p} × C_{p} .

The element (x − x′, a(x) − a′(x′), b(x) − b′(x′)) is in Φ(Q) = 1 × 1 × C_{p^{h−2}} if and only if x = x′ and a(x) = a′(x′). Thus

E_{φ} ∼ E_{φ′} ⇔ Ker(a − a′) ≤ Ker(b − b′) .

Identifying E with the vector space (𝔽_{p})^{e}, and C_{p} with 𝔽_{p}, the homomorphisms a, b, a′, b′ become elements of the dual vector space E^{∗}, and the condition Ker(a − a′) ≤ Ker(b − b′) means that there is a scalar λ ∈ 𝔽_{p} such that b − b′ = λ(a − a′). Hence the incidence matrix S is the matrix indexed by pairs ((a, b), (a′, b′)) of pairs of elements of E^{∗}, defined by

S((a, b), (a′, b′)) = \{ \begin{matrix}1 & if ∃λ ∈ 𝔽_{p}, b − b′ = λ(a − a′) \\ 0 & otherwise \end{matrix} .

Let T be the rectangular matrix indexed by the set of pairs ((a, b), (c, λ)), where a, b, c ∈ E^{∗}, and λ ∈ 𝔽_{p}, defined by

T ((a, b), (c, λ)) = \{ \begin{matrix}1 & if c = b − λa \\ 0 & otherwise \end{matrix} .

Then for a, b, a′, b′ ∈ E^{∗}, consider the sum

s = ∑_{{c∈E^{∗} \atop λ∈𝔽_{p}}} T ((a, b), (c, λ))T ((a′, b′), (c, λ)) .

The non-zero terms in this summation correspond to pairs (c, λ) such that c = b − λa = b′ − λa′. Hence s is equal to the number of λ ∈ 𝔽_{p} such that b − λa = b′ − λa′. This is equal to 1 if a′ ≠ a and if b′ − b is a scalar multiple of a′ − a, to p if a = a′ and b = b′, and to 0 if a = a′ and b ≠ b′. In other words

T ·^{t}T = S + (p − 1)Id .

Since T ·^{t}T is symmetric, it is diagonalizable over ℂ, with real eigenvalues. Let μ be an eigenvalue of T ·^{t}T, and u be a corresponding eigenvector. Then T ·^{t}T u = μu, thus ^{t}T · T ·^{t}T u = μ^{t}T u. So either ^{t}T u = 0, and then μ = 0. And if μ ≠ 0, then ^{t}T u is an eigenvector of ^{t}T · T for the eigenvalue μ. Moreover, the map u ↦ ^{t}T u is an injection of the μ-eigenspace of T ·^{t}T into the μ-eigenspace of ^{t}T · T. The same argument applied to ^{t}T · T instead of T ·^{t}T shows that these two matrices have the same non-zero eigenvalues, and the same multiplicities.

Now for c, c′ ∈ E^{∗} and λ, λ′ ∈ 𝔽_{p}

tT · T ((c, λ), (c′, λ′)) = ∑_{a,b∈E^{∗}} T ((a, b), (c, λ))T ((a, b), (c′, λ′)) .

The right hand side is the number of pairs (a, b) of elements of E^{∗} such that c = b − λa and c′ = b − λ′a, i.e. the number of elements a ∈ E^{∗} such that c + λa = c′ + λ′a, or c − c′ = (λ − λ′)a. This is equal to 1 if λ ≠ λ′, to |E| if λ = λ′ and c = c′, and to 0 if λ = λ′ and c ≠ c′. Hence the matrix ^{t}T · T is a block matrix of the following form

tT · T = ( \begin{matrix}|E|Id & Ω & ⋯ & Ω \\ Ω & |E|Id & ⋯ & Ω \\ ⋮ & ⋮ & ... & ⋮ \\ Ω & Ω & … & |E|Id \end{matrix} ) ,

where all the p^{2}-blocks are square matrices of size |E|, and Ω is a matrix with all entries equal to 1. Let μ be an eigenvalue of this matrix, and

v = ( X_{1} ; X_{2} ; ⋮ ; X_{p} )

be a corresponding eigenvector, where X_{1}, …, X_{p} are column vectors of size |E|. Equivalently, for each i ∈ {1, …, p}

|E|X_{i} + ∑_{j≠i} ΩX_{j} = μX_{i} .

But ΩX = s(X)ω for any column vector X of size |E|, where s(X) denotes the sum of the entries of X, and ω is a column vector of size |E| with all entries equal to 1. Setting σ = ∑_{j=1}^{p} s(X_{j}), this gives, since |E| = p^{e}

p^{e}X_{i} + (σ − s(X_{i}))ω = μX_{i} .

Hence if μ ≠ p^{e}, the vector X_{i} is a multiple of ω, i.e. X_{i} = α_{i}ω for some scalar α_{i}. Then s(X_{i}) = α_{i}p^{e}, thus σ = τp^{e}, where τ = ∑_{j=1}^{p} α_{j}. Finally

p^{e}α_{i} + (τ − α_{i})p^{e} = τp^{e} = μα_{i} .

Thus if μ ≠ 0, all the α_{i}’s are equal to α, say, and then τ = pα, thus μ = p^{e+1}. Conversely, if X_{i} = ω for all i, then v is an eigenvector of ^{t}T · T with eigenvalue p^{e+1}. So p^{e+1} is an eigenvalue of ^{t}T · T, with multiplicity 1.

If μ = 0, then the vector v corresponding to X_{i} = α_{i}ω for i ∈ {1, … p} is in the kernel of ^{t}T · T if and only if ∑_{j=1}^{p} α_{i} = 0. Hence 0 is an eigenvalue of tT · T with multiplicity p − 1.

Finally, if μ = p^{e}, then s(X_{i}) = σ for i ∈ {1, … p}, hence σ = pσ = 0. The vector v is in the p^{e}-eigenspace of ^{t}T · T if and only if s(X_{i}) = 0 for all i. Thus p^{e} is an eigenvalue of ^{t}T · T, with multiplicity p(p^{e} − 1).

It follows that T · ^{t}T has eigenvalues p^{e+1}, p^{e}, and 0, with respective multiplicities 1, p^{e+1} − p, and p^{2e} − p^{e+1} + p − 1. This completes the proof, since S = T ·^{t}T − (p − 1)Id.

Lemma 4.8 shows that the form ⟨ , ⟩_{𝔽,Q}^{♮} is non-degenerate whenever H is a quotient of Q. By the argument of the end of Step 4, or the end of Step 6, this shows that S_{H,𝔽}(G) has a basis in bijection with the G-conjugacy classes of sections (T, S) of G such that T is p-elementary and T/S ≅ H.

• Case 2: Suppose finally that H = C_{p}×C_{p}. As observed earlier, in this case, if R is a normal subgroup of P contained in Φ(P) such that P/R ≅ E × H for some elementary abelian p-group E, then in fact R = Φ(P). The group Q = P/R is elementary abelian, and decomposes as Q = E×L, where L ≅ H is elementary abelian of rank 2. The set N_{H}(Q) is the set of complements M of L in Q, and MN ∩ Φ(Q) = 1 for any M, N ∈ N_{H}(Q). Equation 4.7 shows that

⟨M, N⟩_{𝔽,Q}^{♮} = (1 − p^{n−2})(1 − p^{n−3}) ⋯ (1 − p) ,

where n is the rank of P/Φ(P). This is non-zero, and does not depend on M, N ∈ N_{H}(Q). Hence the form ⟨ , ⟩_{𝔽,Q}^{♮} has rank 1 in this case. Thus ẽ_{P}^{P} S_{H,𝔽}(P) is one dimensional if P is non-cyclic, and it is zero otherwise. Saying that P is non-cyclic is equivalent to saying that the p-elementary group K = P × C of Step 1 is non-cyclic. Hence S_{H,𝔽}(G) has a basis in bijection with the conjugacy classes of non-cyclic p-elementary subgroups of G. This completes the proof of Theorem 4.1.

4.9. Remark: As C_{p} × C_{p} is a B-group, Case 2 above also follows from Proposition 11 of [6]: for a B-group H and a finite group G, the dimension of S_{H,𝔽}(G) is equal to the number of conjugacy classes of subgroups K of G such that β(K) ≅ H. Now by Lemma 2.4, if β(K) ≅ C_{p} × C_{p}, then K is p-elementary, and non cyclic (for otherwise β(K) = 1). Conversely, if K is p-elementary and non cyclic, then β(K) is a non trivial p-group, and also a B-group, hence β(K) ≅ C_{p} × C_{p}.

5. A Green biset functor for p-elementary groups

The following theorem is closely related to Theorem 4.1. In particular, it yields an alternative proof of its Assertions 1 and 2. We refer to Section 8.5 of [9] for the basic definitions on Green biset functors.

5.1. Theorem: Let p be a prime number.

1. For a finite group G, let El_{p}(G) denote the set of p-elementary subgroups of G. Set

F_{p}(G) = {u ∈ B(G) | ∀H ∈ El_{p}(G), Res_{H}^{G}u = 0} .

Then the assignment G ↦ F_{p}(G) is a biset subfuntor of the Burnside functor B, and the quotient functor

E_{p} = B/F_{p}

is a Green biset functor (over ℤ).

2. For a finite group G, the evaluation E_{p}(G) is a free abelian group of rank equal to the number of conjugacy classes of p-elementary subgroups

of G.

3. Let 𝔽 be a field of characteristic 0. Then the biset functor 𝔽E_{p} = 𝔽 ⊗_{ℤ} E_{p} has a unique non zero proper subfunctor I, isomorphic to S_{(C_{p})^{2},𝔽}, and the quotient 𝔽E_{p}/I is isomorphic to S_{1,𝔽} ≅ 𝔽R_{ℚ}. In other words there is a non split short exact sequence

(5.2)  0 → S_{(C_{p})^{2},𝔽} → 𝔽E_{p} → S_{1,𝔽} → 0

of biset functors over 𝔽.

Proof : Let 𝔽B = 𝔽 ⊗_{ℤ} B be the Burnside functor over 𝔽. If we forget the 𝔽-structure on 𝔽B, we get an inclusion B → 𝔽B of biset functors over ℤ. In particular, for each finite group G, we get an inclusion

f_{p} : F_{p}(G) → 𝔽B(G) .

Now saying that u ∈ B(G) lies in F_{p}(G) amounts to saying that the restriction of f_{p}(u) to any p-elementary subgroup of G is equal to 0. Since any subgroup of a p-elementary group is again p-elementary, this amounts to saying that |f_{p}(u)^{H}| = 0 for any H ∈ El_{p}(G). In other words f_{p}(u) is a linear combination of idempotents e_{K}^{G} of 𝔽B(G), where K is a subgroup of G which is not p-elementary. By Lemma 2.4, we get that u ∈ 𝔽_{p}(G) if and only if f_{p}(u) is a linear combination of idempotents e_{K}^{G}, for subgroups K such that β(K) is not a p-group, that is β(K) is non trivial and not isomorphic to (C_{p})^{2}.

Let G_{p} be the class of B-groups which are non trivial, and not isomorphic to (C_{p})^{2}. Then G_{p} is a closed class of B-groups ([9], Definition 5.4.13), that is, if a B-group L admits a quotient in G_{p}, then actually L ∈ G_{p} (this is because the only quotient B-groups of (C_{p})^{2} are the trivial group and (C_{p})^{2}, up to isomorphism). By Theorem 5.4.14 of [9], this closed class G_{p} is associated to a subfunctor N_{p} of the Burnside functor 𝔽B, defined for a finite group G by

N_{p}(G) = ∑_{\begin{matrix}K≤G \\ β(K)∈G_{p} \end{matrix}} 𝔽e_{K}^{G} .

This shows that f_{p}(F_{p}(G)) = f_{p}(B(G)) ∩ N_{p}(G), and since N_{p} is a biset subfunctor of 𝔽B, it follows that F_{p} is a biset subfunctor of B. As biset subfunctors of B are also ideals of the Green biset functor B (see Lemma 2.5.8, Assertion 4 in [9]), we get that F_{p} is an ideal of B. It follows that the quotient E_{p} = B/F_{p} is a Green biset functor. This completes the proof of Assertion 1.

Moreover, the ghost map

Φ_{G} : B(G) → ∏_{\begin{matrix}K≤G \\ mod. G \end{matrix}} ℤ

sending u ∈ B(G) to the sequence |u^{K}|, is injective by Burnside’s theorem. The above discussion shows that Φ induces an injective map

E_{p}(G) = B(G)/F_{p}(G) → ∏_{\begin{matrix}K∈El_{p}(G) \\ mod. G \end{matrix}} ℤ ,

which becomes an isomorphism after tensoring with 𝔽. Assertion 2 follows.

Finally, it follows from Theorem 5.4.14 of [9] that the lattice [0, 𝔽E_{p}] of biset subfunctors of 𝔽E_{p} is isomorphic to the set of closed classes of B-groups which contain G_{p}. There are exactly three such classes: the class G_{p}, the class of non-trivial B-groups, and the class of all B-groups. So [0, 𝔽E_{p}] is a totally ordered set of cardinality 3. Hence 𝔽E_{p} admits a unique non zero proper subfunctor I. The quotient 𝔽E_{p}/I is the unique simple quotient of 𝔽B, hence it is isomorphic to S_{1,𝔽} ≅ 𝔽R_{ℚ}. Now I is a simple biset functor, which is a subquotient of 𝔽B. By Proposition 5.5.1 of [9], it follows that I ≅ S_{H,𝔽} for some B-group H. Since the group K = (C_{p})^{2} has a unique non cyclic subgroup, it follows that I(K) is one dimensional, and a trivial 𝔽Out(K)- module. Moreover K is a group of minimal order such that I(K) ≠ {0}. Hence H ≅ K, and I ≅ S_{(C_{p})^{2},𝔽}. This completes the proof of Assertion 3, and the proof of Theorem 5.1.

5.3. Remark: One can show that the exact sequence (5.2) is essentially unique as a non split exact sequence in the category F_{𝔽} of biset functors over 𝔽: more precisely, one can show that Ext_{F_{𝔽}}^{1}(S_{1,𝔽}, S_{(C_{p})^{2},𝔽}) ≅ 𝔽.

5.4. Remark: If G is a p-group (or even if G is p-elementary), then F_{p}(G) = {0}, so E_{p}(G) ≅ B(G). So if we restrict the exact sequence (5.2) to finite p-groups, we get an exact sequence

0 → S_{(C_{p})^{2},𝔽} → 𝔽B → 𝔽R_{ℚ} → 0

of p-biset functors over 𝔽. This (restricted) exact sequence was introduced in [12], where is was shown that for a finite p-group P, the evaluation S_{(C_{p})^{2},𝔽}(P) is isomorphic to 𝔽D(P), where D(P) is the Dade group of endopermutation modules. It was also shown that the dimension of S_{(C_{p})^{2},𝔽}(P) is equal to the number of conjugacy classes of non cyclic subgroups of P, which is also the number of conjugacy classes of non-cyclic p-elementary subgroups of P. So this agrees with Assertion 2 of Theorem 4.1.

5.5. Remark: For a finite group G, let M_{p}(G) be the ℤ-submodule of B(G) generated by the classes of the transitive G-sets G/H, where H is a pelementary subgroup of G. One can check easily that M_{p}(G) ∩ F_{p}(G) = {0}, so comparing ranks, one might hope that B(G) = M_{p}(G) ⊕ F_{p}(G). This is false in general: for p = 2, when G is the symmetric group S_{3}, there are three p-elementary subgroups in G, up to conjugation, namely the proper subgroups of G (that is the trivial group, the alternating subgroup A = A_{3}, and the subgroup C of order 2). Hence if B(G) = M_{p}(G) ⊕ F_{p}(G), then in particular G/G ∈ M_{p}(G) ⊕ F_{p}(G) so there exist integers a, b, c such that the element u = G/G − (b G/1 + a G/A + c G/C) is in F_{p}(G). Taking fixed points by A then gives |u^{A}| = 0 = 1 − 2a, a contradiction. One can show more precisely that M_{p}(G) ⊕ F_{p}(G) has index 2 in B(G) in this case.

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Serge Bouc LAMFA-CNRS UMR 7352 Université de Picardie-Jules Verne 33, rue St Leu, 80039 Amiens Cedex 01 France

serge.bouc@u-picardie.fr
http://www.lamfa.u-picardie.fr/bouc/