Simple modules over Green biset functors

Nadia Romero

                         nadiaro@ciencias.unam.mx

Abstract

We present three examples of Green biset functors for which their simple modules can be parametrized. These are particular cases of a conjecture by Serge Bouc classifying the simple modules over a Green biset functor A, that generalizes the classification of simple biset functors. We also prove this conjecture under certain hypothesis for A.

1 Introduction

A Green functor for a finite group G is defined as a Mackey functor A with an additional multiplicative structure on each A(H), for H subgroup of G, compatible with the structure of Mackey functor. In the context of Mackey functors, Green functors have been extensively studied and many examples and applications of them have been found (see for example Thévenaz [8], Bouc [3] and Panchadcharam and Street [6]). In the context of categories of biset functors, it has been proved by Serge Bouc that if the class of objects in the biset category is closed under direct products, then the category of biset functors defined on it has a symmetric monoidal structure given by tensor product, the identity element is the Burnside functor. A Green biset functor A is then a monoid in this category. That is, A is a biset functor compatible with the monoidal structures of the biset category and that of R-Mod, when R is a commutative ring with unity. This means that A is equipped with bilinear products from A(G) × A(H) to A(G × H), for finite groups G and H, which have a unit element and are associative and functorial in a natural sense. One feature of this practical definition is that it allows us to observe that many known Green functors are Green biset functors too.

The aim of this article is towards the classification of simple Green biset functors, the results are part of my Ph.D. dissertation [7]. As it happens with classical Green functors, the concept of module over a Green biset functor can be defined, and, as we will see in the next section, biset functors are modules over the Burnside functor. If A is a Green biset functor, a left ideal of A is a submodule of the A-module A. Similarly one defines a right ideal, and then a two sided ideal. A is called simple if its only two sided ideals are {0} and A. One natural step is then, the classification of simple modules over Green biset functors.

In order to study the modules over a Green biset functor A, Bouc introduces a category P_{A} such that A-modules correspond to R-linear functors from P_{A} to R-Mod. The class of objects of P_{A} is the same class of groups on which A is defined, and if G and H are groups in P_{A}, then Hom_{P_{A}}(G, H) = A(H × G). Certain composition ∘ is then introduced for these arrows, it is described in the following section. It is a conjecture of Bouc that the simple modules over a Green biset functor A are in correspondence with couples (H, V) for which H is a group such that the quotient algebra Â(H) is different from zero and V is a simple Â(H)-module. Here Â(H) denotes the quotient of A(H × H) over the submodule generated by elements that can be factored through ∘ by groups strictly smaller than H. The classification of simple biset functors turns out to be a particular case of this conjecture. We present here other three examples which satisfy it and we prove it under the assumption that minimal groups for simple A-modules are unique up to isomorphism. The examples are:

1. The functor of rational representations with coefficients in a field of characteristic zero, kR_{ℚ}. Modules over kR_{ℚ} are also known as rhetorical biset functors and simple modules were already classified by Laurence Barker [1]. We present a different proof of this classification, using the fact that kR_{ℚ} is an homomorphic image of the Burnside functor kB.

2. The functor of complex representations with coefficients in the complex field, ℂR_{ℂ}. In this case, ℂR_{ℂ} is the only simple ℂR_{ℂ}-module, and so it is a simple Green biset functor.

3. The Yoneda-Dress construction at a group C of prime order of the Burnside functor, RB_{C}.

2 Notation and settings

By a group we will always mean a finite group. If G is a group, by a G-set, we will always mean a finite G-set.

Let Z be a class of groups. We will say that Z is closed under subquotients if given a group G in Z, then the image of any morphism of groups from a subgroup of G belongs to Z. We will say that Z is closed under direct products if given G and H in Z, their direct product G × H is in Z.

The definitions appearing in this section, as well as more results and theory around them, can be found in Bouc [4].

We will write B(H, G) for the Grothendieck group of the category of finite (H, G)-bisets. If R is a commutative ring with identity, we will write RB(H, G) for R ⊗_{ℤ} B(H, G). Given U an (H, G)-biset and V a (K, H)-biset, the composition of V and U, denoted by V ×_{H} U is the set of H-orbits on the cartesian product V × U, where the right action of H is defined by

∀(v, u) ∈ V × U, ∀h ∈ H, (v, u) · h = (v · h, h^{−1} · u).

This composition has a natural structure of (K, G)-biset.

Notation 2.1. Let Z be a class of groups closed under subquotients. We will write Ω_{R, Z} for the biset category with objects in Z. This means that if G and H are in Z, then Hom_{Ω_{R, Z}} (G, H) = RB(H, G). The composition of morphism is given by the map

×_{H} : RB(K, H) × RB(H, G) ⟶ RB(K, G).

The identity element in RB(G, G) is the (G, G)-biset G.

If Z is the class of all finite groups, we will write Ω_{R} instead of Ω_{R, Z}.

If H and G are groups and L ⩽ H × G, then the corresponding element in RB(H, G) satisfies the Bouc decomposition (2.3.26 in [4]):

Ind_{D}^{H} ×_{D} Inf_{D/C}^{D} ×_{D/C} Iso(f) ×_{B/A} Def_{B/A}^{B} ×_{B} Res_{B}^{G}

with C ⊴ D ⩽ H, A ⊴ B ⩽ G and f : B/A → D/C a group isomorphism.

The biset functors we will consider are R-linear functors from a category Ω_{R, Z} to the category R-Mod. The Burnside functor defined in Ω_{R, Z} is denoted by RB. In the case of Green biset functors, we will always assume that Z is closed under direct products.

Notation 2.2. Let G, G′, H and H′ be groups. If U is an (H, G)-biset and U′ is an (H′, G′)-biset, then the cartesian product U × V has a natural structure of (H × H′, G × G′)-biset. By linearity, this construction yields a map

RB(H, G) × RB(H′, G′) → RB(H × H′, G × G′),

also denoted by (α, β) ↦ α × β.

If α is any element in RB(H, G) and K is a group, α × K will refer to this construction applied to α and the (K, K)-biset K.

The following definition of Green biset functor is 8.5.1 in [4], it is given in terms of the tensor product of bisets functors, defined in 8.4.1 of the same reference.

Definition 2.3. Let Z be a class of groups closed under subquotients and direct products. A biset functor A defined in Ω_{R, Z} is a Green biset functor if there exist maps of biset functors

μ : A ⊗ A → A, y e : RB → A

such that the following diagrams commute

\xymatrix{A ⊗ (A ⊗ A) \ar[dd]_{≅} \ar[r]^{Id⊗μ} & A ⊗ A \ar@{-->}[dr]^{μ} &  & RB ⊗ A \ar[r]^{e⊗Id} \ar[ddr]_{≅} & A ⊗ A \ar[dd]^{μ} & A ⊗ RB \ar[l]_{Id⊗e} \ar[ddl]^{≅} \\  &  & A \\ (A ⊗ A) ⊗ A \ar[r]^{μ⊗Id} & A ⊗ A \ar@{-->}[ur]_{μ} &  &  & A &  & .}

It is shown in Section 8.5 of [4] that this definition is equivalent to the following.

Definition 2.4. A is a Green biset functor if it is a biset functor quipped with bilinear products A(G) × A(H) → A(G × H) denoted by (a, b) ↦ a × b, for groups G, H ∈ Z, and an element ε_{A} ∈ A(1), satisfying the following conditions:

• Associativity. Let G, H and K be groups in Z. If α_{G, H, K} is the canonical group isomorphism from G × (H × K) to (G × H) × K, then for any a ∈ A(G), b ∈ A(H) and c ∈ A(K)

(a × b) × c = A(Iso(α_{G, H, K}))(a × (b × c)).

• Identity element. Let G be a group in Z. Let λ_{G} : 1 × G → G and ρ_{G} : G × 1 → G denote the canonical group isomorphisms. Then for any a ∈ A(G)

a = A(Iso(λ_{G}))(ε_{A} × a) = A(Iso(ρ_{G}))(a × ε_{A}).

• Functoriality. If φ : G → G′ and ψ : H → H′ are morphisms in Ω_{R, Z}, then for any a ∈ A(G) and b ∈ A(H)

A(φ × ψ)(a × b) = A(φ)(a) × A(ψ)(b).

Example 2.5. Assigning to a group G the Burnside ring B(G) defines a biset functor in Ω_{ℤ}. In order to define the products B(G) × B(H) → B(G × H), let a be in B(G) and b be in B(H). Then B(Inf_{G}^{G×H})(a) is in B(G × H) and B(Inf_{H}^{G×H})(b) is also in B(G × H), and so a × b is given by the natural multiplication of these two elements in B(G × H). In other words, this product is induced by the bifunctor sending a G-set X and an H-set Y to the (G×H)-set X × Y. Example 2.6. Let 𝔽 be a field of characteristic zero and Z the class of finite groups. As in the previous example, assigning to each group G in Z the Grothendieck group of finitely generated 𝔽G-modules, R_{𝔽}(G), defines a biset functor R_{𝔽}. It also has a structure of Green biset functor given in the following way: Let s ∈ R_{𝔽}(G) and t ∈ R_{𝔽}(K), then R_{𝔽}(Inf_{G}^{G×K})(s) and R_{𝔽}(Inf_{K}^{G×K})(t) are both in R_{𝔽}(G × K) and so the natural product in R_{𝔽}(G × K), given by ⊗_{𝔽}, gives us the product s × t.

These examples show us the way to pass from many known Green functors to Green biset functors. Let Z be closed under subquotients and direct products. Suppose A is a biset functor defined in Ω_{R, Z} such that A(H) is an R-algebra with unity for each H ∈ Z, and which satisfies the following: Let f : K → L be a group homomorphism for K, L ∈ Z, write X for the natural (K, L)-biset L and Y for the (L, K)-biset L. Then A(X) is a morphism of unital R-algebras and A satisfies the Frobenius reciprocity relations, that is, for all b ∈ A(L) and a ∈ A(K)

A(Y)(a) · b = A(a · A(X)(b)) and b · A(Y)(a) = A(Y)(A(X)(b) · a).

In this case, A has also a structure of Green biset functor. To define the products A(G)×A(H) → A(G×H), let a ∈ A(G) and b ∈ A(H). We take A(Inf_{G}^{G×H})(a) and A(Inf_{H}^{G×H})(b), make their product in A(G × H) and define a × b as this product. Then it can be shown that A satisfies the three conditions of Definition 2.4, with ε_{A} being the identity element of the original product in A(1).

Definition 2.7. Let A and C be Green biset functors defined in Ω_{R, Z}. A morphism of Green biset functors from A to C is a morphism of biset functors f : A → C such that f_{G}(a) × f_{H}(b) = f_{G×H}(a × b) for any groups G, H ∈ Z and any a ∈ A(G), b ∈ A(H).

Example 2.8. For a field 𝔽 of characteristic 0, the linearization morphism

lin_{𝔽, G} : B(G) → R_{𝔽}(G)

sending each G-set X to the 𝔽G-module 𝔽X is a morphism of biset functors (Remark 1.2.3 in [4]). By the two previous examples, clearly it is a morphism of Green biset functors.

Definition 2.9. Let Z be a class of groups closed under subquotients and direct products and A be a Green biset functor defined in Ω_{R, Z}. An A-module is a biset functor M from Ω_{R, Z} to R-Mod together with morphism A(G)×M(H) → M(G × H) for groups G, H ∈ Z, denoted by (a, m) ↦ a × m, fulfilling the following conditions.

• Associativity. Let G, H and K be groups in Z. If α_{G, H, K} is the canonical group isomorphism from G × (H × K) to (G × H) × K, then for any a ∈ A(G), b ∈ A(H) and m ∈ M(K)

(a × b) × m = M(Iso(α_{G, H, K}))(a × (b × m)).

• Identity element. Let G be a group in Z. Let λ_{G} : 1 × G → G denote the canonical group isomorphism. Then for any m ∈ M(G)

m = M(Iso(λ_{G}))(ε_{A} × m)

• Functoriality. If φ : G → G′ and ψ : H → H′ are morphisms in Ω_{R, Z}, then for any a ∈ A(G) and m ∈ A(H)

M(φ × ψ)(a × b) = A(φ)(a) × M(ψ)(m).

As it is seen in Section 8.5 of [4], A-modules form a category, denoted by A-Mod.

The category associated to a Green biset functor

The following proposition is point 5 in Proposition 8.6.1 of [4], where it appears without proof.

Notation 2.10. If X is a (G, H)-biset, let \overrightarrow{X} be the (G × H, 1)-biset X with action (g, h)x = gxh^{−1} for g in G, h in H and x in X. This construction induces a map α ↦ \overrightarrow{α} from B(G, H) to B(G × H, 1). For G a group, \overrightarrow{G} refers to this notation applied to the (G, G)-biset G. In this case, \overleftarrow{G} will denote the (1, G × G)-biset \overrightarrow{G}^{op}.

Proposition 2.11. Let A be a Green biset functor over Ω_{R, Z} and let P_{A} be the following category:

• The objects of P_{A} are the groups in Z.

• If G and H are in Z, then Hom_{P_{A}}(H, G) = A(G × H).

• Let H, G and K be groups in Z. The composition of β ∈ A(H × G) and α ∈ A(G × K) in P_{A} is the following:

β ∘ α = A(H × \overleftarrow{G} × K)(β × α).

• If G is in Z, then the identity morphism of G in P_{A} is A(\overrightarrow{G})(ε_{A}).

Then P_{A} is an R-linear category and A-Mod is equivalent to the category of R-linear functors from P_{A} in R-Mod.

We will not present the complete proof here, we will only give the maps of the equivalence, leaving to the reader the details of proving it certainly is.

Proof. We will write Fun(P_{A} → R-Mod) for the category of functors from P_{A} to R-Mod.

First we define a functor

S : A-Mod ⟶ Fun(P_{A} → R-Mod).

Let M be an A-module, define F_{M} : P_{A} → R-Mod in a group G as M(G). For a morphism α ∈ A(G × H), define

F_{M}(α) : M(H) → M(G) m ↦ M(G × \overleftarrow{H})(α × m).

Once proved that F_{M} is a functor, it is not hard to see that if f : M → N is an arrow of A-modules, then f defines a morphism from F_{M} to F_{N}.

Now let F be a functor from P_{A} to R-Mod. Then it is a biset functor through the morphism e : RB → A of Definition 2.3. That is, if α is a (G, H)- biset, then F (e_{G×H}(α)) is a morphism of R-modules from F (H) to F (G). Now, in proving the equivalence between Definitions 2.3 and 2.4, one observes that e_{G×H}(α) = A(\overrightarrow{α})(ε_{A}). With this, it is easy to see that e is a morphism of Green biset functors, and then, that F is functorial in Ω_{R, Z}, using the lemma stated right after this proof. Also, if X_{H} = (H × H)/Δ(H), we have that e_{H×H}(X_{H}) equals A(\overrightarrow{H})(ε_{A}), and so F (X_{H})(m) = m.

To make F an A-module, define the product

A(G) × F (H) → F (G × H) (a, m) ↦ F (A(G × \overrightarrow{H})(a))(m).

After proving this product satisfies the three conditions of Definition 2.9, we have the map in objects of the functor

T : Fun(P_{A} → R-Mod) ⟶ A-Mod,

sending F to itself. Now, if t : F → E is an arrow in the category of the left hand side, because of the way F was defined in bisets, it is clear that t is also an arrow of biset functors. On the other hand, t commutes with morphisms in P_{A}, this implies that t behaves well with the product defined above.

The following lemma will be used throughout the rest of the article. We leave the easy proof to the reader.

Lemma 2.12. Let A and C be Green functors and f : C → A a morphism of Green functors. If β ∈ C(H × G) and α ∈ C(G × K), then

f_{H×K}(β ∘ α) = f_{H×G}(β) ∘ f_{G×K}(α).

As a first application of the previous proposition, we have the following example.

Example 2.13. Every biset functor is an RB-module. First, observe that the composition in P_{RB} coincides with the known composition for bisets. That is, if β is an (H, G)-biset and α is a (G, K)-biset, there is an isomorphism of (H, K)-bisets between

RB(H × \overleftarrow{G} × K)(β × α) and β ×_{G} α.

The Bouc decomposition for \overleftarrow{G} is

Def_{1}^{Δ(G)} ×_{Δ(G)} Res_{Δ(G)}^{G×G},

so RB(H × \overleftarrow{G} × K)(β × α) is the deflation from H × Δ(G) × K to H × K of β × α having the following action as H × Δ(G) × K-set

(h, g, g, k) · (x, y) = (hxg^{−1}, gyk^{−1}).

This deflation is then the biggest quotient of β × α in which 1 × Δ(G) × 1 acts in the trivial way, and this is precisely β ×_{G} α.

So, the category P_{RB} is equivalent to Ω_{R, Z}, and any biset functor is an RB-module.

A deep study of simple biset functors can be found in Chapter 4 of [4]. There we can find that the isomorphism classes of simple RB-modules are indexed by isomorphism classes of seeds (H, V), where H is a group in Z and V is a simple ROut(H)-module. To each seed (H, V) corresponds the isomorphism class of S_{H, V}, where S_{H, V} (G) is the quotient of L_{H, V} (G) = RB(G, H) ⊗_{RB(H, H)} V over

J_{H, V} (G) = { ∑_{i=1}^{n} φ_{i} ⊗ n_{i} ∈ L_{H, V} (G) | ∑_{i=1}^{n}(ψ ∘ φ_{i}) · n_{i} = 0 ∀ψ ∈ RB(H, G)}.

An important characteristic of H is that it is a minimal group for S_{H, V}, and for a simple biset functor all its minimal groups are isomorphic.

Definition 2.14. Let M be a biset functor defined on Ω_{R, Z}. A group H in Z is called minimal for M if M(H) ≠ 0 and M(K) = 0 for any K in Z with |K| < |H|.

This motivates the following definition.

Definition 2.15. Let A be a Green biset functor on Ω_{R, Z}, and H be a group in Z. We will write I_{A}(H) for the submodule of A(H×H) generated by elements of the form a∘b, where a is in A(H ×K), b is in A(K ×H) and K is a group in Z of order smaller than |H|. We will denote by Â(H) the quotient A(H ×H)/I_{A}(H).

It is clear that Â(H) is an R-algebra. If A = RB, it is known (Proposition 4.3.2 in [4], for instance) that R̂B(H) is isomorphic to ROut(H). Nonetheless, we will see in the following section that this quotient R̂B(H) may vanish.

The classification of simple biset functors is a particular case of the following conjecture.

Conjecture 2.16 (S. Bouc, personal communication). Suppose that A is a Green biset functor. Let S_{A} be the set of equivalence classes of couples (H, V), where Â(H) ≠ 0, V is a simple Â(H)-module and (H, V) ∼ (G, W) if there exists an isomorphism of groups φ : H → G such that V ≅ ^{φ}W.

Then the isomorphism classes of simple A-modules are in one-to-one correspondence with the elements of S_{A}.

The details of the isomorphism V ≅ ^{φ}W are given in Section 4.

In the following two sections we will present other three examples of Green biset functors which also satisfy this conjecture.

If S is a simple A-module, then S ≠ 0, so taking H as a group of minimal order such that S(H) ≠ 0, we have a minimal group for S.

Lemma 2.17. Let S be a simple A-module for a Green biset functor A defined in Ω_{R, Z}. If H ∈ Z is a minimal group for S, then Â(H) ≠ 0 and S(H) is a simple Â(H)-module.

Proof. Suppose we have Â(H) = 0, then in particular A(\overrightarrow{H})(ε_{A}) = ∑_{i=1}^{r} α_{i} ∘ β_{i} where α_{i} ∈ A(H×K_{i}) and β_{i} ∈ A(K_{i}×H) with K_{i} a group of order smaller than |H|. But S(α_{i} ∘ β_{i}) = 0 because H is minimal for S, then S(A(\overrightarrow{H})(ε_{A})) = 0, a contradiction.

Clearly, S(H) is an Â(H)-module, so it suffices to see it is a simple A(H×H)- module. Let W be an A(H × H)-submodule, define

T (G) = {m ∈ S(G) | ∀X ∈ A(H × G), S(X)(m) ∈ W }.

It is easy to see that T is a subfunctor of S. Besides, T (H) = W. Now, if t ∈ T (H), then in particular for X = A(\overrightarrow{H})(ε_{A}) we must have T (X)(t) ∈ W, then T (H) ⊆ W. On the other hand, since W is a A(H × H)-submodule, then S(X)(w) is in W for all w ∈ W and X ∈ A(H × H), and so W ⊆ T (H).

Hence, if T = 0 we have W = 0, and if T = S then W = S(H).

3 kR_{ℚ}-modules or rhetorical biset functors

In this section k will be a field of characteristic 0.

Rhetorical biset functors were introduced by Laurence Barker in [1], there he proves that every rhetorical biset functor over k is semisimple and gives a classification of the simple functors. The definition of rhetorical biset functor we will give next is not exactly the one appearing in [1] but, under the hypothesis we will make, it is equivalent to that one.

Denote by Ω̂_{k} the category having the class of finite groups as objects, and where the morphisms from G to H is the quotient kB(H × G)/Ker(klin_{𝔽 H×G}), with 𝔽 a field of characteristic 0 and

klin_{𝔽} : kB → kR_{𝔽}

being the linearization morphism. Thanks to Lemma 2.12, the composition ∘ can be extended to this quotient. A rhetorical biset functor is defined as a k-linear functor from Ω̂_{k} to k-Mod.

Proposition 2.11 and Artin’s induction theorem gives us that in the case 𝔽 = ℚ, rhetorical biset functors coincide with kR_{ℚ}-modules. An equivalent statement is made in Proposition 3.1 of [1], in view of the following lemma, which describes the composition ∘ in the category P_{kR_{ℚ}}.

Lemma 3.1. Let H, G and K be groups. If [M] is in kR_{ℚ}(H × G) and [N] is in kR_{ℚ}(G × K), then

kR_{ℚ}(H × \overleftarrow{G} × K)([M] × [N]) = [M ⊗_{ℚG} N].

Proof. Recall that the product [M] × [N] is defined as the isomorphism class of the ℚ(H × G × G × K)-module M ⊗_{ℚ} N, where G × K acts in the trivial way in M and H ×G acts in the trivial way in N. From the Bouc decomposition for \overleftarrow{G}, we have that applying A(H × \overleftarrow{G} × K) to this module gives the deflation from H × Δ(G) × K to H × K of the module M ⊗_{ℚ} N restricted to H × Δ(G) × K. But this deflation is the biggest quotient of M ⊗_{ℚ} N in which 1 × Δ(G) × 1 acts in the trivial way, that is M ⊗_{ℚG} N.

Hence, Theorem 1.5 of [1] gives a classification of simple kR_{ℚ}-modules: They are the simple biset functors indexed by seeds (H, V) where H is a cyclic group of order m and V is a primitive k(ℤ/mℤ)^{×}-module.

Definition 3.2. Let m be in ℕ − {0}. A simple k(ℤ/mℤ)^{×}-module V is called primitive if given n a divisor of m and π_{m, n} : (ℤ/mℤ)^{×} → (ℤ/nℤ)^{×} the natural projection, if Kerπ_{m, n} acts trivially on V, then n = m.

The rest of this section is devoted to give a different proof of this classification, using the techniques of Green biset functors. First, we observe that this classification is a particular case of Conjecture 2.16. This fact can also be concluded from Proposition 4.2, by proving (as we will do with ℂR_{ℂ} and RB_{C}) that for each simple kR_{ℚ}-module its minimal groups are unique up to isomorphism. On the other hand, using the fact that klin_{ℚ} is an epimorphism of Green biset functors, we can prove directly not only that kR_{ℚ} satisfies the parametrization of Conjecture 2.16, but also that simple kR_{ℚ}-modules are simple biset functors.

In proving the following lemma, we will use the fact, easy to proof, that it f : A → C is an epimorphism of Green biset functors, then the simple Cmodules are the simple A-modules in which Kerf acts in the trivial way.

Lemma 3.3. The simple kR_{ℚ}-modules are the simple biset functors S_{H,V}, where H satisfies k̂R_{ℚ}(H) ≠ 0 and V is a simple k̂R_{ℚ}(H)-module.

Proof. Since the linearization morphism klin_{ℚ} is an epimorphism of Green biset functors, the simple kR_{ℚ}-modules are the simple biset functors S_{H, V} in which Ker(klin_{ℚ}) acts trivially. Hence, by Lemma 2.17, we have that k̂R_{ℚ}(H) ≠ 0 and V is a simple k̂R_{ℚ}(H)-module.

Now let S_{H, V} be a simple biset functor such that k̂R_{ℚ}(H) ≠ 0 and V is a simple k̂R_{ℚ}(H)-module.

Take K and G arbitrary groups, as we said in the previous section, S_{H, V} (G) is the quotient of L_{H, V} (G) = kB(G, H) ⊗_{kB(H, H)} V over its unique maximal subfunctor

J_{H, V} (G) = { ∑_{i=1}^{n} φ_{i} ⊗ n_{i} ∈ L_{H, V} (G) | ∑_{i=1}^{n}(ψ ∘ φ_{i}) · n_{i} = 0 ∀ψ ∈ kB(H, G)}.

Let a be in Ker(klin_{ℚ K}), we shall prove that a acts in the trivial way on S_{H, V} (G). It suffices then, to prove that if m = ∑_{i=1}^{n}φ_{i} ⊗ n_{i} is an element in L_{H, V} (G), the product a × m belongs to J_{H, V} (K × G). By Proposition 2.11, the product a × m is given by S_{H, V} (kB(K × \overrightarrow{G})a)(m), but Ker(klin_{ℚ}) is a subfunctor of kB, so kB(K × \overrightarrow{G})a is in Ker(klin_{ℚ K×G×G}). Then, if ψ is any element in kB(H × K × G), by Lemma 2.12, the composition

x_{i} = ψ ∘ (A(K × \overrightarrow{G})a) ∘ φ_{i}

belongs to Ker(klin_{ℚ H×H}). Finally, since V is a kB(H × H)-module that is also a kR_{ℚ}(H × H)-module, we must have x_{i} · n_{i} = 0 for all i, and hence a × m is in J_{H, V} (K × G).

This proves that Ker(klin_{ℚ}) acts in the trivial way on S_{H, V} and so, this is a kR_{ℚ}-module.

The following result tells us that in order to find the simple biset functors of this lemma, it suffices to consider cyclic groups.

Lemma 3.4.

i) Let G be any non trivial group. Then I_{kR_{ℚ}}(G) is equal to

∑_{K cyclic |K| proper divisor of |G|} kR_{ℚ}(G × K) ∘ kR_{ℚ}(K × G).

ii) If G is not a cyclic group, then k̂R_{ℚ}(G) = 0.

Proof. Let L be any group. By the Artin Induction Theorem, every element x in kR_{ℚ}(G × L) can be written as

∑_{\begin{matrix}C⩽G×L \\ C cyclic \end{matrix}} a_{C}[1_{C} ↑_{C}^{G×L}]

for some a_{C} ∈ k. Now, in Bouc’s decomposition for (G × L)/C

Ind_{D}^{G} ×_{D} Inf_{D/E}^{D} ×_{D/E} Iso_{B/A}^{D/E} ×_{B/A} Def_{B/A}^{B} ×_{B} Res_{B}^{L}

we have that D/E is a cyclic subquotient of G. On the other hand, since the linearization morphism is a morphism of Green biset functors, from Lemma 2.12 we have

[1_{C} ↑_{C}^{G×L}] = [1_{X} ↑_{X}^{G×D_{C}}] ∘ [1_{Y} ↑_{Y}^{D_{C}×L}]

if we write D_{C} for D/E, and suppose that Ind_{D}^{G} ×_{D} Inf_{D/E}^{D} is isomorphic to

(G × D_{C})/X and Iso_{B/A}^{D/E} ×_{B/A} Def_{B/A}^{B} ×_{B} Res_{B}^{L} is isomorphic to (D_{C} × L)/Y.

To prove i) suppose that L is a group of order smaller than |G| and take y in kR_{ℚ}(L × G). If in the previous decomposition we write y_{C} = [1_{Y} ↑_{Y}^{D_{C}×L}] ∘ y, then x ∘ y is equal to

∑_{\begin{matrix}C⩽G×L \\ C cyclic \end{matrix}} a_{C}[1_{X} ↑_{X}^{G×D_{C}}] ∘ y_{C},

which belongs to ∑_{K} kR_{ℚ}(G×K)∘kR_{ℚ}(K ×G) with K cyclic of order a proper divisor of |G|.

To prove ii) suppose that G is not cyclic and take L = G. In this case we must have D_{C} of order smaller than |G| for every C, since G is not cyclic. This proves that k̂R_{ℚ}(G) = 0.

In the rest of the section, H will be a non trivial cyclic group. We will now be interested in finding which simple kOut(H)-modules are also k̂R_{ℚ}(H)-modules.

As we said, k̂B(H) is isomorphic to kOut(H). This isomorphism is given by associating to each automorphism σ of H, the H ×H-set X_{σ} = (H ×H)/Δ_{σ}(H) where

Δ_{σ}(H) = {(a, σ(a)) | a ∈ H}.

The inner automorphisms are all identified with X_{id}. Observe that if H is cyclic, then Out(H) = Aut(H) and Δ_{σ}(H) is cyclic.

Recall that for an abelian group G, the dimension of kR_{ℚ}(G) over k is the number of cyclic subgroups of G. Besides, the elements of the form [1_{C} ↑_{C}^{G}] with C cyclic generate kR_{ℚ}(G), and so they are a basis. We shall denote this basis by β(G).

Lemma 3.5. Let K be a cyclic group with |K| a proper divisor of |H|. Let σ be an automorphism of H. Consider a ℚ(H × H)-module T of the form

1_{C} ↑_{C}^{H×K} ⊗_{ℚK}1_{D} ↑_{D}^{K×H}

with C ⩽ H × K and D ⩽ K × H cyclic subgroups. Then the coefficient of [1_{Δ_{σ}(H)} ↑_{Δ_{σ}(H)}^{H×H}] for [T] in terms of the basis β(H × H) is different from 0 if and only if

i) π_{H}(C) = H = π_{H}(D), with π_{H}(C) and π_{H}(D) being the projections on H of C and D respectively.

ii) C =< (h, x) > ⟺ D =< (x, σ(h)) >.

In this case, the coefficient is |K|/|H|.

Proof. Let τ be the character associated with T. By Theorem 15.4 in Curtis and Reiner [5], the coefficients of τ in terms of the basis β(H × H) are given in the following way: For G ⩽ H × H cyclic, we have

q_{G} = \frac{1}{[H × H : G]} ∑_{G⩽G^{∗}} μ([G^{∗} : G])τ(z^{∗})

where {G^{∗}} runs over all cyclic subgroups containing G, μ is the Möbius function and z^{∗} is a generator of G^{∗}.

Since the exponent of (H × H) is |H|, then Δ_{σ}(H) is a maximal cyclic subgroup, that is, it is not properly contained in any other cyclic subgroup. Hence, if q_{σ} denotes q_{Δ_{σ}(H)}, then q_{σ} = (1/|H|)τ(z) for z a generator of Δ_{σ}(H).

Suppose that H =< a >. Let τ_{C} and τ_{D} be the characters of 1_{C} ↑_{C}^{H×K} and 1_{D} ↑_{D}^{K×H}, respectively. By Lemma 7.1.3 in Bouc [4], we have

τ(a, σ(a)) = \frac{1}{|K|} ∑_{x∈K} τ_{C}(a, x)τ_{D}(x, σ(a)).

So, q_{σ} is different from 0 if and only if there exists x ∈ K such that (a, x) belongs to C and (x, σ(a)) belongs to D, and in this case

τ_{C}(a, x) = [H × K : C] and τ_{D}(x, σ(a)) = [K × H : D].

Now, since |K| is a proper divisor of |H|, then (a, x) and (x, σ(a)) are elements of order |H|, and we must have < (a, x) >= C and < (x, σ(a)) >= D. For the same raison, x is the only element of K such that (a, x) is in C (respectively (x, σ(a)) is in D). From this we have q_{σ} = |K|/|H|.

If σ is an automorphism of H, we will write 1_{σ} for the trivial ℚΔ_{σ}(H)- module. Suppose that |H| = m. We take the smallest non negative representatives of (ℤ/mℤ)^{×} and denote by σ_{t} the automorphism of H corresponding to the class of t in (ℤ/mℤ)^{×}. By composition with the isomorphism between k̂B(H) and k(ℤ/mℤ)^{×}, we can consider the following extension of the linearization morphism

ℓ_{H} : k(ℤ/mℤ)^{×} ⟶ k̂R_{ℚ}(H),

which sends [t] to the class of [1_{σ_{t}} ↑_{Δ_{σ_{t}} (H)}^{H×H}].

Proposition 3.6. Let x_{n} be

∑_{[t]∈Kerπ_{m, n}}[t].

The kernel of ℓ_{H} is the ideal generated by {x_{n} | n proper divisor of m}.

Proof. First suppose n is a proper divisor of m, let us see that x_{n} is in the kernel of ℓ_{H}. Let K be a cyclic group of order n and suppose that H =< a > and K =< x >. Let C =< (a, x) > and D =< (x, a) >. If [t] is in the kernel of π_{m, n}, then

< (x, a) >=< (x, a)^{t} >=< (x, σ_{t}(a)) > .

So, by the previous lemma, the isomorphism class of

∑_{[t]∈Kerπ_{m, n}} 1_{σ_{t}} ↑_{Δ_{σ_{t}} (H)}^{H×H}

appears with coefficient |K|/|H| in the decomposition of

[1_{C} ↑_{C}^{H×K} ⊗_{ℚK}1_{D} ↑_{D}^{K×H}]

in terms of β(H × H). Clearly, this element is 0 in k̂R_{ℚ}(H). Besides, if any other basic element of the form [1_{σ_{r}} ↑_{Δ_{σr} (H)}^{H×H}] appears in this decomposition, then there exists an integer d such that (x, σ_{r}(a))^{d} = (x, a), but it is easy to see that this implies [r] ∈ Kerπ_{m, n}. This means that the rest of the elements appearing in this decomposition come from (H ×H)-sets which are 0 in k̂B(H). Hence, ℓ_{H}(x_{n}) is 0 in k̂R_{ℚ}(H).

Suppose now that

∑_{i=1}^{d} a_{i}[s_{i}]

with [s_{i}] ∈ (ℤ/mℤ)^{×} and a_{i} ∈ k is in the kernel of ℓ_{H}. Using Lemma 3.4 and the Artin Induction Theorem, this means that

∑_{i=1}^{d} a_{i}[1_{σ_{s_{i}}} ↑_{Δ_{σs_{i}} (H)}^{H×H}]  (1)

can be written as

∑_{j=1}^{e} b_{j}[1_{C_{j}} ↑_{C_{j}}^{H×K_{j}}] ∘ [1_{D_{j}} ↑_{D_{j}}^{K_{j}×H}],  (2)

with C_{j}, D_{j} and K_{j} cyclic groups, and |K_{j}| a proper divisor of |H|. By Lemma 3.5, if [1_{σ_{s_{i}}} ↑_{Δ_{σs_{i}} (H)}^{H×H}] is in the decomposition of [1_{C_{j}} ↑_{C_{j}}^{H×K_{j}}] ∘ [1_{D_{j}} ↑_{D_{j}}^{K_{j}×H}], we must have C_{j} =< (a, x) > and D_{j} =< (x, σ_{s_{i}}(a)) > for some x ∈ K_{j}. Now, if n_{j} is the order of x, then, such as we did before, the isomorphism class of

∑_{[t]∈Kerπ_{m, n_{j}}} 1_{σ_{t}σ_{s_{i}}} ↑_{Δ_{σ_{t}σs_{i}} (H)}^{H×H}  (3)

appears with coefficient |K_{j}|/|H| in this decomposition. This means that if any element in this sum has coefficient different from 0 in (2), then the whole sum has this same coefficient. Since all the elements in (1) and (3) are basic elements, then all the elements of (1) must be of the way described in (3), which proves the claim.

Note that if Kerπ_{m, n} = 1 for some n proper divisor of m, then [1_{Δ(H)} ↑_{Δ(H)}^{H×H}]

is 0 in k̂R_{ℚ}(H), and hence k̂R_{ℚ}(H) = 0.

Corollary 3.7. Let H be a group such that k̂R_{ℚ}(H) ≠ 0, and V a simple kOut(H)-module. Then V is a k̂R_{ℚ}(H)-module if and only if V is primitive.

Proof. We know that H must be cyclic. Suppose that it has order m > 1 and let n be a proper divisor of m.

Suppose that V is a simple k(ℤ/mℤ)^{×}-module that is also a k̂R_{ℚ}(H)-module. If Ker π_{m, n} acts trivially on V, then there exists v ≠ 0 in V such that x · v = v for all x ∈ Ker π_{m, n}, but this implies x_{n} · v ≠ 0, a contradiction.

Now suppose that V is a simple primitive k(ℤ/mℤ)^{×}-module. If there exists v ∈ V such that x_{n} · v = w with w ≠ 0, then for all x ∈ Kerπ_{m, n} we have

x · w = x · (x_{n} · v) = xx_{n} · v = x_{n} · v = w

since xx_{n} = x_{n}. But this is a contradiction.

4 Other examples of the conjecture

In this section we will prove that ℂR_{ℂ} and RB_{C} for C of prime order, also satisfy Conjecture 2.16. In order to do this, we will first prove that if A is a Green biset functor such that for any simple A-module its minimal groups are isomorphic, then A satisfies Conjecture 2.16. Then we will prove that this is the case of ℂR_{ℂ} and RB_{C} for C of prime order.

Notation 4.1. Observe that if G and H are isomorphic groups in the category grp, then they are also isomorphic in P_{A} for any Green functor A. To prove it, suppose that φ : H → G is a group isomorphism. Consider X = Iso(φ) and Y = Iso(φ^{−1}), then define α_{1} = A(\overrightarrow{X})(ε_{A}) and α_{2} = A(\overrightarrow{Y})(ε_{A}). It is not hard to prove that α_{1} ∘ α_{2} = A(\overrightarrow{G})(ε_{A}) and that α_{2} ∘ α_{1} = A(\overrightarrow{H})(ε_{A}).

If W is an A(G × G)-module, we denote by ^{φ}W the A(H × H)-module W with action of a ∈ A(H × H) on w ∈ W given by a · w := (α_{1} ∘ a ∘ α_{2})w.

The following proposition will provide us two other examples of Green biset functors that satisfy the conjecture.

Proposition 4.2. Suppose that A is a Green biset functor for which the minimal groups of each simple A-module form a single isomorphism class. Let S_{A} be the set of equivalence classes of couples (H, V), where Â(H) ≠ 0, V is a simple Â(H)-module and (H, V) ∼ (G, W) if there exists an isomorphism of groups φ : H → G such that V ≅ ^{φ}W.

Then the isomorphism classes of simple A-modules are in one-to-one correspondence with the elements of S_{A}.

Proof. If S is a simple A-module and H is a minimal group for S, then it is minimal for any other A-module isomorphic to S. Also, as we have seen before, in this case Â(H) ≠ 0 and S(H) is a simple Â(H)-module. To S we associate the pair (H, S(H)). The hypothesis over the isomorphism class of H, justifies that this choice is well defined, because if G is a group isomorphic to H then S(H) is isomorphic to S(G) as stated in the assertion. On the other hand, if (H, V) is a couple satisfying the hypothesis, then we will assign to it the module S_{H, V}^{A} defined as the quotient L_{H, V}^{A} /J_{H, V}^{A}, where

L_{H, V}^{A} (K) = A(K × H) ⊗_{A(H×H)} V, L_{H, V} (a)(x ⊗ v) = (a ∘ x) ⊗ v

for a ∈ A(G × K), and

J_{H, V}^{A} (K) = { ∑_{i=1}^{n} x_{i} ⊗ n_{i} | ∑_{i=1}^{n}(y ∘ x_{i}) · n_{i} = 0 ∀y ∈ A(H × K)}.

Since V is a simple A(H × H)-module, then by Lemma 1 in Bouc [2], L_{H, V}^{A} has a unique simple quotient, this is S_{H, V}^{A}. This quotient satisfies S_{H, V}^{A} (H) = V.

Let us prove that if (H, V) ∼ (G, W), then S_{H, V}^{A} ≅ S_{G, W}^{A}. Let φ : H → G be the group isomorphism which makes V and ^{φ}W isomorphic as A(H × H)- modules, then L_{H, V}^{A} ≅ L_{H,^{φ}W}^{A}. We prove now that L_{H,^{φ}W}^{A} ≅ L_{G, W}^{A}.

Let K ∈ Z, define

μ_{K} : A(K × G) × W ⟶ A(K × H) ⊗_{A(H×H)} ^{φ}W,

by μ_{K}(a, v) = a∘α_{1} ⊗_{A(H×H)} v, where α_{1} is defined as in the previous notation. Observe that μ_{K} can be extended to A(K × G) ⊗_{A(G×G)} W, since if b is in A(G × G), then

μ_{K}(a, bv) = a ∘ α_{1} ⊗_{A(H×H)} bv
= a ∘ α_{1} ⊗_{A(H×H)} (α_{1} ∘ α_{2} ∘ b ∘ α_{1} ∘ α_{2})v

where α_{2} is defined as in the previous notation. Then

μ_{K}(a, bv) = a ∘ α_{1} ⊗_{A(H×H)} (α_{2} ∘ b ∘ α_{1}) · v
= a ∘ (α_{1} ∘ α_{2} ∘ b ∘ α_{1}) ⊗_{A(H×H)} v
= a ∘ b ∘ α_{1} ⊗_{A(H×H)} v
= μ_{K}(a ∘ b, v).

Clearly μ is natural in K. The inverse of μ is defined in a similar way. So L_{H, V}^{A} is isomorphic to L_{G, W}^{A} and hence, their unique simple quotients S_{H, V}^{A} and S_{G, W}^{A} are isomorphic.

Finally, we prove these assignments define a bijection.

The functor from A(H × H)-mod to P_{A} that sends V to L_{H, V}^{A} is left adjoint of the evaluation P_{A} → A(H × H)-mod sending each functor M to M(H). With this we can prove that if S is a simple A-module and S(H) ≠ 0, then S is isomorphic to S_{H, V}^{A} with V = S(H). Since V is a simple A(H × H)-module, then

Hom_{P_{A}}(L_{H, V}^{A} , S) ≅ Hom_{A(H×H)}(V, S(H))

implies that S is a simple quotient of L_{H, V}^{A} and hence it is isomorphic to S_{H, V}^{A}.

Now take a couple (H, V) with Â(H) ≠ 0 and V a simple Â(H)-module. We see that H is minimal for S_{H, V}^{A}: Let K be a group of smaller order than H. Since V is a Â(H)-module, from the definition of S_{H, V}^{A}, we have that S_{H, V}^{A} (K) = 0.

ℂR_{ℂ}-modules

Proposition 4.3. ℂR_{ℂ} is the only simple ℂR_{ℂ}-module. In particular, it is a simple Green biset functor.

Proof. By Theorem 10.33 in Curtis and Reiner [5], for any G and K groups

ℂR_{ℂ}(G × K) ≅ ℂR_{ℂ}(G) ⊗_{ℂ} ℂR_{ℂ}(K)

which, according to Lemma 3.1, is equivalent to

ℂR_{ℂ}(G × K) ≅ ℂR_{ℂ}(G × 1) ∘ ℂR_{ℂ}(1 × K).

This implies that if M is a ℂR_{ℂ}-module different from 0, then M(1) ≠ 0. So, by Proposition 4.2, the only simple module corresponds to (1, ℂ). Now, if V ⩽ ℂR_{ℂ} is a left ideal different from 0, then V (1) ≠ 0, and hence ε_{ℂR_{ℂ}} = ℂ ∈ V (1). From this we conclude V ≅ ℂR_{ℂ}.

In Chapter 7 of [4], Bouc proves that ℂR_{ℂ} is semisimple as a biset functor, its components are the simple ℂR_{ℚ}-modules. We know that ℂR_{ℚ} ≅ S_{1, ℂ} is not a ℂR_{ℂ}-module. This proposition tells us that, in fact, none of these components is a ℂR_{ℂ}-module.

RB_{C}-modules

The following results concern the Yoneda-Dress construction at C of RB, for a group C. It is denoted by RB_{C}. If F is a biset functor, then F_{C} sends each group G to F (G × C). In an element φ ∈ RB(G × H) it is defined as F (φ × C). More results on this construction can be found in Section 8.2 of [4].

Lemma 4.4. If A is a Green biset functor and C is a group, then A_{C} is a Green biset functor.

Proof. That A_{C} is a biset functor is Lemma 8.2.2 in [4].

Let K and G be groups. Define D = {(c, g, c) | c ∈ C, g ∈ G} and let f be the natural isomorphism from D to G × C. The product for A_{C} is given by the following composition

\xymatrix{A_{C}(K) × A_{C}(G) \ar[r]^{×_{A}} & A(K × C × G × C) \ar[r]^{A(K×X_{C}^{G})} & A_{C}(K × G)}

where X_{C}^{G} = Iso(f) ×_{D} Res_{D}^{C×G×C} and ×_{A} is the product of A. The identity element of A_{C} is ε_{A_{C}} = A(Inf_{1}^{1×C})(ε_{A}) ∈ A_{C}(1).

Let us see that this composition defines a product in A_{C}.

Associativity:

It is enough to prove that, for groups K, G and L, the exterior square in the following diagram is commutative, considering the corresponding group isomorphisms in vertices in the middle and bottom right:

\xymatrix{A_{C}(K) × A_{C}(G) × A_{C}(L) \ar[d]^{a}_{(id, ×_{A})} \ar[r]^{(×_{A}, id)} & A_{C}(K × C × G) × A_{C}(L) \ar[d]^{b}_{×_{A}} \ar[r]^{(A(α),id)} & A_{C}(K × G) × A_{C}(L) \ar[d]_{×_{A}} \\ A_{C}(K) × A_{C}(G × C × L) \ar[d]^{c}_{(id, A(G×X_{C}^{L}))} \ar[r]^{×_{A}} & A_{C}(K × C × G × C × L) \ar[d]^{d}_{A(K×C×G×X_{C}^{L})} \ar[r]^{A(β)} & A_{C}(K × G × C × L) \ar[d]_{A(K×G×X_{C}^{L})} \\ A_{C}(K) × A_{C}(G × L) \ar[r]_{×_{A}} & A_{C}(K × C × G × L) \ar[r]_{A(K×X_{C}^{G×L})} & A_{C}(K × G × L)}

with α = K × X_{C}^{G} and β = α × L × C. The square a is commutative since the product of A is associative. Squares b and c commute because the product of A is natural with respect to bisets. Finally, it is not hard to see that the composition of bisets in square d are isomorphic, and so square d is commutative.

Identity element:

If a belongs to A_{C}(G), then ε_{A_{C}} ×a = A(1×X_{C}^{G})(A(Inf_{1}^{1×C})(ε_{A})×_{A}a). Now, A(Inf_{1}^{1×C})(ε_{A}) ×_{A} a = A(Inf_{1}^{1×C} × G × C)(ε_{A} ×_{A} a), but it is straightforward to prove that (1 × X_{C}^{G}) ×_{1×C×G×C} (Inf_{1}^{1×C} × G × C) is isomorphic to the identity biset 1 × G × C.

Functoriality with respect to bisets:

Let Z be a (G, L)-biset and Y be a (K, D)-biset. We should prove the commutativity of the exterior square in

\xymatrix{A_{C}(L) × A_{C}(D) \ar[d]_{(A_{C}(Z), A_{C}(Y))} \ar[r] & A_{C}(L × C × D) \ar[d]_{A(Z×C×Y ×C)} \ar[r] & A_{C}(L × D) \ar[d]_{A_{C}(Z×Y)} \\ A_{C}(G) × A_{C}(K) \ar[r] & A_{C}(G × C × K) \ar[r] & A_{C}(G × K)} ,

where in the rows we have the product × of A_{C}. The square on the left commutes since the product of A is natural with respect to bisets. To verify the commutativity of the square on the right, we must only verify that the bisets involved are isomorphic.

Following the ideas of Example 2.13, it is not hard to see that with this product, the composition in P_{RB_{C}}, which we will denote by ×^{d} is the following. Let X be a (K × G × C)-set, and Y be a (G × L × C)-set. Then we can consider X as a (K, G)-biset endowed with an action of C, which we will suppose on the right and which commutes with those of K and G. In the same way, Y is a (G, L)-biset with an action of C that commutes with those of G and L. Hence X ×_{G}^{d} Y is the (K, L)- biset X ×_{G} Y with a diagonal action of C,

[x, y]c = [xc, yc].

The identity in RB_{C}(G × G) is RB_{C}(\overrightarrow{G})(\{•\}) which, by Bouc’s decomposition for \overrightarrow{G}, equals (G × G × C)/(Δ(G) × C).

Let E be a subgroup of G × L × C and D be a subgroup of L × K × C, we will use the Notation 2.3.19 of [4], E ∗ D to denote

{(g, k, c) ∈ G × K × C | ∃l ∈ L s. t. (g, l, c) ∈ E and (l, k, c) ∈ D}.

With this notation, we have a sort of a Mackey formula for 3-sets.

Lemma 4.5. Let G, L, K and C be groups. If E is a subgroup of G×L×C and D is a subgroup of L × K × C, then we have an isomorphism of G × K × C-sets

((G × L × C)/E) ×_{L}^{d} ((L × K × C)/D) ≅ ⨆_{(l, c)}(G × K × C)/(E ∗^{(l, 1, c)}D)

where (l, c) is running through a set of representatives of the double cosets p_{2, 3}(E)\L × C/p_{1, 3}(D), with p_{2, 3}(E) and p_{1, 3}(D) being the projections over L × C of E and D respectively.

Proof. If [(g, l, c)E, (l′, k, c′)D] is an element on the left, then it is easy to see that its orbit under the action of G × K × C is the same as the orbit of [(1, 1, 1)E, (l^{−1}l′, 1, c^{−1}c′)D]. With this observation, the proof that there exists a bijection between the orbits of (G × L × C)/E ×_{L}^{d} (L × K × C)/D under the action of G × K × C and p_{2, 3}(E)\L × C / p_{1, 3}(D), is analogous to the proof of Lemma 2.3.24 in Bouc [4]. Similarly, the stabilizer of [(1, 1, 1)E, (l, 1, c)D] equals E ∗^{(l, 1, c)}D.

We will proceed now to prove:

Proposition 4.6. If C is a group of prime order, then the simple RB_{C}-modules are in one-to-one correspondence with the equivalence classes of couples (H, V) where H is a group and V is a simple R̂B_{C}(H)-module.

The proof will be given by Corollary 4.9 and Lemma 4.10. The first will tell us that RB_{C} with C of primer order satisfies the hypothesis of Proposition 4.2, that is, every simple RB_{C}-module has isomorphic minimal groups. It is not hard to see that this hypothesis on a functor A is fulfilled if A has the following property: Suppose G and H are groups of order n. Let I_{n} be the submodule of A(G × H) generated by elements of the form a ∘ b with a ∈ A(G × K), b ∈ A(K × H) and K a group of order smaller than n. If G and H are not isomorphic, then A(G × H) = I_{n}. All the functors we have considered have this property and we will prove next that RB_{C} with C of prime order has it. Nonetheless, we will also see that this property cannot be extended to an arbitrary group C. For instance, if C is a cyclic group of order 4, we will find two non isomorphic groups G and H of order 8 and an element X ∈ B_{C}(G×H) such that X does not factor through a group of order smaller than 8.

Until otherwise is stated, we may suppose C is any group.

Notation 4.7. Let D be a subgroup of H × K × C. We will write p_{1}(D), p_{2}(D) and p_{3}(D) for the projections of D in H, K and C respectively; p_{1, 2}(D) will denote the projection over H × K, and in the same way we define the other possible combinations of indices. We write k_{1}(D) for {h ∈ p_{1}(D) | (h, 1, 1) ∈ D} which is a normal subgroup of p_{1}(D). Similarly, we define k_{2}(D), k_{3}(D) and k_{i, j}(D) for all possible combinations of i and j.

Lemma 4.8. If D is a subgroup of H × K × C, then as (H × K × C)-sets, (H × K × C)/D is isomorphic to

i) X ×_{D_{1}}^{d} Y where D_{1} = p_{1}(D)/k_{1}(D), for some X ∈ RB_{C}(H × D_{1}) and Y ∈ RB_{C}(D_{1} × K).

ii) W ×_{D_{2}}^{d} Z where D_{2} = p_{2}(D)/k_{2}(D), for some W ∈ RB_{C}(H × D_{2}) and Z ∈ RB_{C}(D_{2} × K).

Proof. The proof of point 2 in Proposition 2.3.25 of [4], gives us the following group isomorphisms

\frac{p_{1}(D)}{k_{1}(D)} ≅ \frac{p_{2, 3}(D)}{k_{2, 3}(D)} and \frac{p_{2}(D)}{k_{2}(D)} ≅ \frac{p_{1, 3}(D)}{k_{1, 3}(D)},

the first one sends ak_{1}(D) to (b, c)k_{2, 3}(D) if (a, b, c) is in D, the second one sends bk_{2}(D) to (a, c)k_{1, 3}(D).

Now let X = (H × D_{1} × C)/U with

U = {(h, α(h)) | h ∈ p_{1}(D), α : p_{1}(D) ↠ D_{1}} × C.

Let σ be the isomorphism from D_{2, 3} := p_{2, 3}(D)/k_{2, 3}(D) to D_{1}, we define Y = (D_{1} × K × C)/V with

V = {(σ(β(k, c)), k, c) | (k, c) ∈ p_{2, 3}(D), β : p_{2, 3}(D) ↠ D_{2, 3}}.

By Lemma 4.5 we have

X ×_{D_{1}}^{d} Y ≅ ⨆_{\begin{matrix}(t, c) in \\ [p_{2, 3}(U)\D_{1}×C/p_{1, 3}(V)] \end{matrix}}(H × K × C)/(U ∗^{(t, 1, c)}V).

Since p_{2, 3}(U) = D_{1} × C, this union reduces to the element (H × K × C)/U ∗ V. Now, U ∗ V is by definition

{(h, k, c) | ∃t ∈ D_{1} s. t. (h, t, c) ∈ U, (t, k, c) ∈ V },

so (h, k, c) is in U ∗ V if and only if α(h) = σ(β(k, c)), which happens if and only if (h, k, c) is in D.

Second decomposition is obtained in a similar way.

Corollary 4.9. Let C be a group of prime order and H and K be groups of order n. If there exists a transitive (H × K × C)-set X that does not factor through a group of order smaller than n, then G and H are isomorphic.

Proof. Let X = (H × K × C)/D. If it does not factor through a group of order smaller than n, then we must have p_{1}(D) = H, p_{2}(D) = K, k_{1}(D) = 1 and k_{2}(D) = 1. From this we obtain that there exists a surjective morphism α : p_{2, 3}(D) → H. If C is a group of prime order, we have only two choices for p_{2, 3}(D). If p_{2, 3}(D) = K × C_{1} for C_{1} ⩽ C, then α(k, c) = α_{1}(k)α_{2}(c), where α_{1} is a morphism from K to H. If k is in Kerα_{1}, then taking (k, 1) in p_{2, 3}(D) we have α(k, 1) = 1, hence k ∈ k_{2}(D) = 1. So we have that α_{1} is injective and H is isomorphic to K. If p_{2, 3}(D) = {(k, t(k)) | t : K → C}, then again through α we can define a surjective homomorphism from K to H and so they are isomorphic.

This allows us to see that if S is a simple RB_{C}-module for C of prime order and H and K are two minimal groups for S, then they are isomorphic. First, they have the same order n. Also, if A = RB_{C} and V = S(H), then S ≅ S_{H, V}^{A} and S(K) ≠ 0. Hence there exists a transitive element X in RB_{C}(H × K) such that S(X) ≠ 0 and in particular X does not factor through a group of order smaller than n.

For the proof the next lemma we will use the following observations. Here C is again any group.

Let G, H and K be any groups. Take A ⩽ G × K × C and B ⩽ K × H × C. For A ∗ B we can define the following morphism

ρ_{A, B} : A ∗ B → (p_{2}(A) ∩ p_{1}(B))/(k_{2}(A) ∩ k_{1}(B)) (a, b, c) ↦ t(k_{2}(A) ∩ k_{1}(B))

where t is an element in K such that (a, t, c) ∈ A and (t, b, c) ∈ B (that exists for each (a, b, c) in A∗B), it is clearly unique modulo k_{2}(A)∩k_{1}(B). The kernel of this morphism is

N = {(a, b, c) ∈ G × H × C | (a, 1, c) ∈ A, (1, b, c) ∈ B}.

Suppose now that A∗B = D satisfies p_{1}(D) = G, p_{2}(D) = H, k_{1}(D) = 1 and k_{2}(D) = 1. Then there exists α : p_{2, 3}(D) → G such that D = {(α(h, c), h, c) | (h, c) ∈ p_{2, 3}(D)}. In this case, the kernel N of ρ_{A, B} is of the form

{(α(f_{1}(w), w), f_{1}(w), w) | w ∈ C′ ⩽ C, f_{1} : C′ → H}.

To see this, take (r, s, q) in N. First we must have r = α(s, q). Also, if there is another s′ such that (α(s′, q), s′, q) is in N, then (1, s′s^{−1}, 1) is in B, but taking (1, 1, 1) in A we have (1, s′s^{−1}, 1) in D, hence s = s′. So we can define a morphism f_{1} from C′ = p_{3}(D) to H, and N has the desired form.

Lemma 4.10. For any groups G and C we have R̂B_{C}(G) ≠ 0.

Proof. Let θ be an automorphism of G and ζ : C → Z(G) be a morphism of groups. We define D_{θ, ζ} = {(θ(g)ζ(c), g, c) | (g, c) ∈ G × C}, and prove that X_{θ, ζ} = (G × G × C)/D_{θ, ζ} is different from 0 in the quotient R̂B_{C}(G).

Suppose there exists K of order smaller than |G| such that (G×G×C)/D is in RB_{C}(G×K)×_{K}^{d} RB_{C}(K ×G). Then by Lemma 4.5, there exist A ⩽ G×K ×C and B ⩽ K × H × C, such that D_{θ, ζ} = A ∗ B. According to the previous observations, if N is the kernel of the morphism ρ_{A, B}, then (A∗B)/N has order |G|[C : C_{1}] for a subgroup C_{1} of C. This is a contradiction, since |(A ∗ B)/N| must also divide |K|.

Finally, we present the following example, showing that the property of Corollary 4.9 cannot be extended to an arbitrary group C.

Example 4.11. Let C =< c > be a group of order 4, G the quaternion group

< x, y | x^{4} = 1, yxy^{−1} = x^{−1}, x^{2} = y^{2} >

and H the dihedral group of order 8

< a, b | a^{4} = b^{2} = 1, bab^{−1} = a^{−1} > .

We will find D a subgroup of G × H × C such that the corresponding element in B(G × H × C) does not factor through a group of order smaller than 8.

Consider T_{1} =< (a, c^{2}) > and T_{2} =< (b, c) >, subgroups of H × C of order 4. It is easy to observe that T_{1} is a normal subgroup, so T = T_{1}T_{2} is a subgroup of H × C. Moreover, T_{1} ∩ T_{2} = {1} and if α = (a, c^{2}) and β = (b, c), then βαβ^{−1} = α^{−1}, so T = T_{1} ⋊ T_{2}. Finally, there is clearly a surjective morphism τ : T → G whose kernel is the subgroup of order 2 generated by α^{2}β^{2}. So, we define D = {(τ(t), t) | t ∈ T }.

Observe that p_{1}(D) = G, p_{2}(D) = H, k_{1}(D) = {1} and k_{2}(D) = {1}.

Suppose there exists K of order smaller than 8 and A ⩽ G × K × C and B ⩽ K × H × C, such that D = A ∗ B. Consider the morphism ρ_{A, B}. From the observations before Lemma 4.10 we have that the kernel N can only have order 1, 2 or 4. Since [D : N] must also be smaller than 8, we conclude that N has order 4 and it is generated by an element of the form (r, s, c) ∈ D. There are only four elements of this form in D, and none of them generates a normal subgroup, so we have a contradiction.

Acknowledgements

I would like to thank Serge Bouc and Peter Webb for enlightening conversations, and Dağhan Yaylıoğlu for his help with GAP.

I also would like to thank the following institutions, for their financial support: ECOS - ANUIES - CONACYT for project “Biconjuntos y funtores asociados” and Fundación Sofía Kovalévskaia - Sociedad Matemática Mexicana.

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