Diagonal $p$-permutation functors

TL;DR

This paper introduces diagonal p-permutation functors in a categorical framework, proving the semisimplicity of a key functor and classifying simple objects through pairs of p-groups and generators.

Contribution

It defines and studies diagonal p-permutation functors, establishing their semisimplicity and classifying simple functors parametrized by pairs of p-groups and generators.

Findings

The functor $ ext{F}T^ riangle$ is semisimple in the category of diagonal p-permutation functors.

Simple functors are parametrized by pairs of p-groups and generators of p'-subgroups.

Explicit descriptions of evaluations of simple functors are provided.

Abstract

Let $k$ be an algebraically closed field of positive characteristic $p$, and $\mathbb{F}$ be an algebraically closed field of characteristic 0. We consider the $\mathbb{F}$-linear category $\mathbb{F} pp_k^\Delta$ of finite groups, in which the set of morphisms from $G$ to $H$ is the $\mathbb{F}$-linear extension $\mathbb{F} T^\Delta(H,G)$ of the Grothendieck group $T^\Delta(H,G)$ of $p$-permutation $(kH,kG)$-bimodules with (twisted) diagonal vertices. The $\mathbb{F}$-linear functors from $\mathbb{F} pp_k^\Delta$ to $\mathbb{F}\hbox{-Mod}$ are called {\em diagonal $p$-permutation functors}. They form an abelian category $\mathcal{F}_{pp_k}^\Delta$. We study in particular the functor $\mathbb{F}T^{\Delta}$ sending a finite group $G$ to the Grothendieck group $\mathbb{F}T(G)$ of $p$-permutation $kG$-modules, and show that $\mathbb{F}T^\Delta$ is a semisimple object of $\mathcal{F}_{pp_k}^\Delta$, equal to the direct sum of specific simple functors parametrized by isomorphism classes of pairs $(P,s)$ of a finite $p$-group $P$ and a generator $s$ of a $p'$-subgroup acting faithfully on $P$. This leads to a precise description of the evaluations of these simple functors. In particular, we show that the simple functor indexed by the trivial pair $(1,1)$ is isomorphic to the functor sending a finite group $G$ to $\mathbb{F} K_0(kG)$, where $K_0(kG)$ is the group of projective $kG$-modules.

Equations

$$M[P]:=M^P/\sum _{Q<P}{\mathrm{Tr}}_Q^P(M^Q)$$

$$F_{P,s}^G=\frac{1}{|P||s||C_{{\overline{N}}_G(P)}(s)|}\sum _{\begin{array}{c}\phi \in \widehat{⟨s⟩}\\ L⩽⟨Ps⟩\\ PL=⟨Ps⟩\end{array}}\tilde{\phi }(s^{-1})|L|\mu (L,⟨Ps⟩){\mathrm{Ind}}_L^G{k}_{L,\phi }^{⟨Ps⟩},$$

$$F^{G}_{P,s}=\frac{1}{|C_{N_{G}(P)}(s)|}\sum_{\begin{subarray}{c}\varphi\in\widehat{\langle s\rangle}\\ L\leqslant P\\ L^{s}=L\end{subarray}}\tilde{\varphi}(s^{-1})|C_{L}(s)|\mu\big{(}(L,P)^{s}\big{)}\mathrm{Ind}^{G}_{\langle Ls\rangle}k^{\langle Ps\rangle}_{\langle Ls\rangle,\varphi}.$$

$$\mathbb{F}T(G,H)\circ \mathbb{F}T(H,L)\to \mathbb{F}T(G,L).$$

$$\mathbb{F}{T}^{\Delta }(G,H)\circ \mathbb{F}{T}^{\Delta }(H,L)\to \mathbb{F}{T}^{\Delta }(G,L)$$

$$\displaystyle\mathcal{E}^{\Delta}(G):=\mathbb{F}T^{\Delta}(G,G)/\Big{(}\sum_{|H|<|G|}\mathbb{F}T^{\Delta}(G,H)\circ\mathbb{F}T^{\Delta}(H,G)\Big{)}$$

$$\mathcal{E}(G):=\mathbb{F}T(G,G)/\Big{(}\sum_{|H|<|G|}\mathbb{F}T(G,H)\circ\mathbb{F}T(H,G)\Big{)}$$

$$\mathcal{E}(G)\cong\big{(}\mathbb{F}[X]/\Phi_{n}[X]\big{)}\rtimes\mathrm{Out}(G)$$

$$\Theta :{\mathcal{E}}^{\Delta }(G)\to \mathcal{E}(G).$$

$$a\cdot g\cdot b:=\lambda (b)ag\phi (b)$$

$${\mathrm{Ind}}_{⟨Rt⟩}^{G\times G}{k}_{⟨Rt⟩,\phi }^{⟨Rt⟩}\cong k{G}_{{\eta }^{-1},{\phi }^{-1}}$$

$$F_{Q,t}^{G\times G}=\frac{1}{|C_{N_{G\times G(Q)}(t)}|}\sum _{\begin{array}{c}\phi \in \widehat{⟨t⟩}\\ L⩽Q\\ L^t=L\end{array}}\tilde{\phi }(t^{-1})|C_L(t)|\mu ((L,Q{)}^t){\mathrm{Ind}}_{⟨Lt⟩}^{G\times G}{k}_{L,\phi }^{⟨Qt⟩}.$$

$${\mathrm{Ind}}_{⟨Lt⟩}^{G\times G}{k}_{L,\phi }^{⟨Qt⟩}$$

$$\cong kG{\otimes }_{p_1(⟨Lt⟩)}{\mathrm{Ind}}_{⟨Lt⟩}^{p_1(⟨Lt⟩)\times p_2(⟨Lt⟩)}(k_{L,\phi }^{⟨Qt⟩}){\otimes }_{p_2(⟨Lt⟩)}kG$$

$${\mathrm{Res}}_{G\times p_1(⟨Lt⟩)}^{G\times G}kG\cong {\mathrm{Res}}_{G\times p_1(⟨Lt⟩)}^{G\times G}{\mathrm{Ind}}_{\Delta G}^{G\times G}k\cong {\mathrm{Ind}}_{\Delta (p_1(⟨Lt⟩))}^{G\times p_1(⟨Lt⟩)}{\mathrm{Res}}_{\Delta (p_1(⟨Lt⟩))}^{\Delta (G)}k.$$

$$\bigoplus_{i}\mathrm{Indinf}^{G\times\overline{G}}_{\overline{N}_{G\times\overline{G}}(Q)}\big{(}kC_{G}(P)/C_{\langle s\rangle}(P)\otimes_{k}k_{\alpha_{i}}\big{)}\otimes_{k\overline{G}}\mathrm{Indinf}^{\overline{G}\times G}_{\overline{N}_{\overline{G}\times G}(Q^{\prime})}\big{(}kC_{G}(P)/C_{\langle s\rangle}(P)\otimes_{k}k_{\alpha^{\prime}_{i}}\big{)}$$

$$F_{Q,t}^{G\times G}=\frac{1}{\mid C_{N_{G\times G(Q)}(t)}\mid }\sum _{\begin{array}{c}\phi \in \widehat{⟨t⟩}\\ L⩽Q\\ L^t=L\end{array}}\tilde{\phi }(t^{-1})|C_L(t)|\mu ((L,Q{)}^t){\mathrm{Ind}}_{⟨Lt⟩}^{G\times G}{k}_{L,\phi }^{⟨Qt⟩}$$

$$\sum r_{\phi ,\alpha }k{G}_{\phi ,\alpha }=\sum _{|H|<|G|}{t}_{H,U_H,V_H}{U}_H{\otimes }_{kH}{V}_H$$

$$U_{H}\otimes_{kH}V_{H}\cong\mathrm{Indinf}_{\overline{N}_{G\times G}(\Delta(P))}^{G\times G}\bigoplus_{i}\big{(}kZ(P)\otimes k_{\lambda_{i}}\big{)}^{n_{i}}$$

$${\mathrm{Res}}_H^G{F}_{P,s}^G=\sum _{Q,t}{F}_{Q,t}^H$$

$${\mathrm{Ind}}_H^G{F}_{Q,t}^H=|N_G(Q,t):N_H(Q,t)|F_{Q,t}^G.$$

$${\mathrm{Inf}}_{G/N}^G{F}_{P,s}^{G/N}=\sum _{Q,t}{F}_{Q,t}^G$$

$${\mathrm{Def}}_{G/N}^G{F}_{P,s}^G=m_{P,s,N}\cdot F_{Q,t}^{G/N}$$

$${\mathrm{Def}}_{G/N}^G{F}_{P,s}^G=m_{P,s,N}\cdot F_{PN/N,\overline{s}}^{G/N}.$$

$$H=\{(aN,b):a{b}^{-1}\in N\}⩽(G/N)\times G.$$

$$\displaystyle m_{P,s,N}=\frac{|s|}{|N\cap\langle s\rangle||C_{G}(s)|}\sum_{\begin{subarray}{c}Q\leqslant P\\ Q^{s}=Q\\ \langle Qs\rangle N=G\end{subarray}}|C_{Q}(s)|\mu\big{(}(Q,P)^{s}\big{)}.$$

$$m_{P,s,N}=\frac{1}{|N|}\cdot$$

$$m_{P,s,N}=\frac{|s||C_P(s)|}{|N||C_G(s)|}=\frac{1}{|N|}\cdot$$

$$L_{G,V}(H):=\mathbb{F}{T}^{\Delta }(H,G){\otimes }_{{\mathcal{E}}^{\Delta }(G)}V$$

$$J_{G,V}(H):=\left\{\sum _i{\varphi }_i\otimes v_i\in L_{G,V}:\forall \psi \in \mathbb{F}{T}^{\Delta }(G,H),\sum _i(\psi \circ {\varphi }_i)\cdot v_i=0\right\},$$

$$S_{G,V}:=L_{G,V}/J_{G,V}$$

Full text

We study in particular the functor $\mathbb{F}T^{\Delta}$ sending a finite group $G$ to the Grothendieck group $\mathbb{F}T(G)$ of $p$-permutation $kG$-modules, and show that $\mathbb{F}T^{\Delta}$ is a semisimple object of $\mathcal{F}_{pp_{k}}^{\Delta}$, equal to the direct sum of specific simple functors parametrized by isomorphism classes of pairs $(P,s)$ of a finite $p$-group $P$ and a generator $s$ of a $p^{\prime}$-subgroup acting faithfully on $P$. This leads to a precise description of the evaluations of these simple functors. In particular, we show that the simple functor indexed by the trivial pair $(1,1)$ is isomorphic to the functor sending a finite group $G$ to $\mathbb{F}K_{0}(kG)$, where $K_{0}(kG)$ is the group of projective $kG$-modules.

††journal: Journal of Algebra

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[labelstyle=]

1 Introduction

Let $p$ be a prime number. Throughout we denote by $\mathbb{F}$ an algebraically closed field of characteristic zero, and by $k$ an algebraically closed field of characteristic $p$. The $p$-permutation modules play a crucial role in the study of modular representation theory of finite groups. A splendid Rickard equivalence, introduced by Rickard [8], between blocks of finite group algebras is given by a chain complex consisting of $p$-permutation bimodules. Also a $p$-permutation equivalence, introduced by Boltje and Xu [1], and studied extensively later by Boltje and Perepelitsky [7], is an element in the Grothendieck group of $p$-permutation bimodules.

In [5], Ducellier studied $p$-permutation functors: Consider the category $\mathbb{F}pp_{k}$ where the objects are finite groups and the morphisms between groups $G$ and $H$ are given by the Grothendieck group $\mathbb{F}\otimes_{\mathbb{Z}}T(H,G)$ of $p$-permutation $(kH,kG)$-bimodules. A $p$-permutation functor is an $\mathbb{F}$-linear functor from $\mathbb{F}pp_{k}$ to $\mathbb{F}$-$\mathrm{Mod}$. The indecomposable direct summands of the bimodules that appears in a $p$-permutation equivalence between blocks of finite group algebras have twisted diagonal vertices. Therefore, inspired by the work of Ducellier, we consider a category with less morphisms: Let $\mathbb{F}pp_{k}^{\Delta}$ be a category where the objects are finite groups and the morphisms between groups $G$ and $H$ are given by the Grothendieck group $\mathbb{F}\otimes_{\mathbb{Z}}T^{\Delta}(H,G)$ of $p$-permutation $(kH,kG)$-bimodules whose indecomposable direct summands have twisted diagonal vertices. An $\mathbb{F}$-linear functor from $\mathbb{F}pp_{k}^{\Delta}$ to $\mathbb{F}$-$\mathrm{Mod}$ is called a diagonal $p$-permutation functor.

By [2], the simple diagonal $p$-permutation functors are parametrized by the pairs $(G,V)$ of a finite group $G$ and a simple module $V$ of the essential algebra $\mathcal{E}^{\Delta}(G)=\mathrm{End}_{\mathbb{F}pp_{k}^{\Delta}}(G)/I$ at $G$, where $I$ is the ideal generated by the morphisms that factor through groups of smaller order. We show that the essential algebra $\mathcal{E}^{\Delta}(G)$ is isomorphic to the essential algebra studied in [5]. As a result this implies that the essential algebra $\mathcal{E}^{\Delta}(G)$ is non-zero if and only if the group $G$ is of the form $P\rtimes\langle s\rangle$ where $P$ is a $p$-group and $s$ is a generator of a $p^{\prime}$-cyclic group acting faithfully on $P$. Moreover in that case there is an algebra isomorphism \mathcal{E}^{\Delta}(G)\cong\big{(}\mathbb{F}[X]/\Phi_{n}[X]\big{)}\rtimes\mathrm{Out}(G) where $n$ is the order of $s$. See Theorem 3.3.

We also study the functor $\mathbb{F}T^{\Delta}$ that sends a finite group $G$ to the Grothendieck group $\mathbb{F}T(G)$ of $p$-permutation $kG$-modules. We describe the subfunctor lattice (Theorem 5.11) and simple quotients (Proposition 5.15) of $\mathbb{F}T^{\Delta}$. We also give a description for the $\mathbb{F}$-dimension of the evaluations of simple quotients of $\mathbb{F}T^{\Delta}$ at a finite group $G$ (Theorem 5.18). Moreover we prove that the simple functor $S_{1,1}$ that corresponds to the pair $(1,1)$ is isomorphic to the functor that sends a finite group $G$ to the $\mathbb{F}$-linear extension $\mathbb{F}K_{0}(kG)$ of the Grothendieck group of projective $kG$-modules (Theorem 5.20).

2 Preliminaries

Let $G$ and $H$ be finite groups. We denote by $p_{1}:G\times H\to G$ and $p_{2}:G\times~{}H\to H$ the canonical projections. Let $X\leqslant G\times H$ be a subgroup. We define the subgroups $k_{1}(X):=p_{1}(X\cap\mathrm{ker}(p_{2}))$ and $k_{2}(X):=p_{2}(X\cap\mathrm{ker}(p_{1}))$ of $p_{1}(X)$ and $p_{2}(X)$, respectively. Note that $k_{1}(X)\times k_{2}(X)$ is a normal subgroup of $X$. Moreover, $k_{i}(X)$ is a normal subgroup of $p_{i}(X)$ and one has a canonical isomorphism $X/(k_{1}(X)\times k_{2}(X))\to p_{i}(X)/k_{i}(X)$ induced by the projection map $p_{i}$ for $i=1,2$.

Let $\phi:P\to Q$ be an isomorphism between subgroups $P\leqslant G$ and $Q\leqslant H$. Then $\{(\phi(x),x):x\in P\}$ is a subgroup of $H\times G$ and a subgroup of that form is called a twisted diagonal subgroup of $H\times G$. Note that a subgroup $X\leqslant H\times G$ is a twisted diagonal subgroup if and only if $k_{1}(X)=1$ and $k_{2}(X)=1$.

Let $P$ be a subgroup of $G$ and $M$ be a $kG$-module. We denote by $M^{P}$ the $k$-vector space of $P$-fixed points of $M$. If $Q\leqslant P$ is a subgroup, then the map $\mathrm{Tr}_{Q}^{P}:M^{Q}\to M^{P}$ defined by $\mathrm{Tr}(m)=\sum_{x\in[P/Q]}x\cdot m$ is called the relative trace map. The quotient

[TABLE]

is called the Brauer quotient of $M$ at $P$. Note that $M[P]$ is a $k\overline{N}_{G}(P)$-module, where $\overline{N}_{G}(P):=N_{G}(P)/P$. We have $M[P]=0$ if $P$ is not a $p$-group.

A $(kG,kH)$-bimodule $M$ can be viewed as a $k(G\times H)$-module via $(g,h)\cdot m:=gmh^{-1}$, for $(g,h)\in G\times H$ and $m\in M$. Similarly a $k(G\times H)$-module can be viewed as a $(kG,kH)$-bimodule. We will usually switch between these two points of views.

Definition 2.1**.**

Let $G$ be a finite group. A $kG$-module $M$ is called a permutation module, if $M$ has a $G$-stable $k$-basis. A $p$-permutation $kG$-module is a $kG$-module $M$ such that $\mathrm{Res}^{G}_{S}M$ is a permutation $kS$-module for a Sylow $p$-subgroup $S$ of $G$.

For a finite group $G$ we denote by $T(G)$ the Grothendieck group of $p$-permutation $kG$-modules with respect to direct sum decompositions. If $M$ is a $p$-permutation $kG$-module, then the class of $M$ in $T(G)$ will be abusively denoted by $M$. The group $T(G)$ has a ring structure induced by the tensor product of modules over $k$, and $T(G)$ will be called the ring of $p$-permutation modules of $G$, for short. If $H$ is another finite group, we set $T(G,H):=T(G\times H)$. We denote by $T^{\Delta}(G,H)$ the subgroup of $T(G,H)$ spanned by $p$-permutation $k(G\times H)$-modules whose indecomposable direct summands have twisted diagonal vertices.

Let $\mathcal{P}_{G,p}$ denote the set of pairs $(P,E)$ where $P$ is a $p$-subgroup of $G$ and $E$ is a projective indecomposable $k\overline{N}_{G}(P)$-module. The group $G$ acts on the set $\mathcal{P}_{G,p}$ via conjugation and we denote by $[\mathcal{P}_{G,p}]$ a set of representatives of $G$-orbits of $\mathcal{P}_{G,p}$. For $(P,E)\in\mathcal{P}_{G,p}$, let $M_{P,E}$ denote the unique (up to isomorphism) indecomposable $p$-permutation $kG$-module with the property that $M_{P,E}[P]\cong E$. Note that $M_{P,E}$ has the group $P$ as a vertex [4, Theorem 3.2]. We denote by $\mathcal{P}_{G\times H,p}^{\Delta}$ the set of pairs $(P,E)\in\mathcal{P}_{G\times H,p}$ where $P$ is a twisted diagonal $p$-subgroup of $G\times H$.

Remark 2.2**.**

The isomorphism classes of the modules $M_{P,E}$ where $(P,E)\in\mathcal{P}_{G\times H,p}^{\Delta}$ form a $\mathbb{Z}$-basis for $T^{\Delta}(G,H)$.

Definition 2.3**.**

[5, Definition 2.3.1]** Let $(P,s)$ be a pair where $P$ is a $p$-group and $s$ is a generator of a $p^{\prime}$-cyclic group acting on $P$. We denote the semidirect product $P\rtimes\langle s\rangle$ by $\langle Ps\rangle$. Let $(Q,t)$ be another such pair. We say that the pairs $(P,s)$ and $(Q,t)$ are isomorphic if there are group isomorphisms $\phi:P\to Q$ and $\psi:\langle s\rangle\to\langle t\rangle$ such that $\psi(s)=q\cdot t$ for some $q\in Q$ and $\phi(s\cdot u)=\psi(s)\cdot\phi(u)$ for all $u\in P$. In that case we write $(P,s)\simeq(Q,t)$.

Lemma 2.4**.**

[5, Proposition 2.3.3]** Let $(P,s)$ and $(Q,t)$ be two pairs. Then $(P,s)\simeq(Q,t)$ if and only if there is a group isomorphism $f:\langle Ps\rangle\to\langle Qt\rangle$ such that $f(s)$ is conjugate to $t$.

Let $\mathcal{Q}_{G,p}$ denote the set of pairs $(P,s)$ where $P$ is a $p$-subgroup of $G$ and $s\in N_{G}(P)$ is a $p^{\prime}$-element. In that case $\langle Ps\rangle$ denotes the semidirect product $P\rtimes\langle s\rangle$ where the action of $\langle s\rangle$ on $P$ is induced by conjugation. The group $G$ acts on the set $\mathcal{Q}_{G,p}$ and we denote by $[\mathcal{Q}_{G,p}]$ a set of representatives of $G$-orbits. We denote by $\mathcal{Q}_{G\times H,p}^{\Delta}$ the set of pairs $(P,s)\in\mathcal{Q}_{G\times H,p}$ where $P$ is a twisted diagonal $p$-subgroup of $G\times H$.

For any pair $(P,s)\in\mathcal{Q}_{G,p}$ let $\tau^{G}_{P,s}$ denote the additive map $T(G)\to\mathbb{F}$ that sends a $p$-permutation $kG$-module $M$ to the value of the Brauer character of $M[P]$ at $s$. The map $\tau^{G}_{P,s}$ is a ring homomorphism and it extends to an $\mathbb{F}$-algebra homomorphism $\tau^{G}_{P,s}:\mathbb{F}\otimes_{\mathbb{Z}}T(G)\to\mathbb{F}$. The set $\{\tau^{G}_{P,s}:(P,s)\in[\mathcal{Q}_{G,p}]\}$ is the set of all species from $\mathbb{F}T(G):=\mathbb{F}\otimes_{\mathbb{Z}}T(G)$ to $\mathbb{F}$ [3, Proposition 2.18].

The algebra $\mathbb{F}T(G)$ is split semisimple and its primitive idempotents $F^{G}_{P,s}$ are indexed by pairs $(P,s)\in[\mathcal{Q}_{G,p}]$ [3, Corollary 2.19]. If $\phi:\langle s\rangle\to k^{\times}$ is a group homomorphism, we denote by $k_{\phi}$ the $k\langle s\rangle$-module $k$ on which the element $s$ acts as multiplication by $\phi(s)$. Let $\widehat{\langle s\rangle}=\mathrm{Hom}(\langle s\rangle,k^{\times})$ denote the set of group homomorphisms. By [3, Theorem 4.12] we have the idempotent formula

[TABLE]

where $k^{\langle Ps\rangle}_{L,\varphi}=\mathrm{Res}^{\langle Ps\rangle}_{L}\mathrm{Inf}^{\langle Ps\rangle}_{\langle s\rangle}k_{\varphi}$, and $\tilde{\varphi}$ is the Brauer character of $k_{\varphi}$.

By [5, Proposition 2.7.8] we have another formula

[TABLE]

Here \mu\big{(}(-,-)^{s}\big{)} is the Möbius function of the poset of $s$-stable subgroups of $P$.

Lemma 2.5**.**

For finite groups $G$ and $H$, the set $\{F^{G\times H}_{P,s}:(P,s)\in[\mathcal{Q}_{G\times H,p}^{\Delta}]\}$ of primitive idempotents form an $\mathbb{F}$-basis for the split semisimple algebra $\mathbb{F}T^{\Delta}(G,H)$.

Proof.

First we will show that we have $F^{G\times H}_{P,s}\in\mathbb{F}T^{\Delta}(G,H)$ whenever $(P,s)\in[\mathcal{Q}_{G\times H,p}^{\Delta}]$. Let $\varphi\in\widehat{\langle s\rangle}$ and $L\leqslant\langle Ps\rangle$. It suffices to show that $\mathrm{Ind}^{G}_{L}k^{\langle Ps\rangle}_{L,\varphi}\in\mathbb{F}T^{\Delta}(G,H)$. Since $P$ acts trivially on $\mathrm{Inf}^{\langle Ps\rangle}_{\langle s\rangle}k_{\varphi}$, the subgroup $P$ is contained in a vertex of $k_{\varphi}$ considered as a $k\langle Ps\rangle$-module. But since $P$ is the Sylow $p$-subgroup of $\langle Ps\rangle$, it follows that $P$ is the vertex of $k_{\varphi}$. Therefore the module $k^{\langle Ps\rangle}_{L,\varphi}=\mathrm{Res}^{\langle Ps\rangle}_{L}\mathrm{Inf}^{\langle Ps\rangle}_{\langle s\rangle}k_{\varphi}$ has a vertex contained in $L\cap{}^{x}P\leqslant P$ for some $x\in\langle Ps\rangle$. Since a subgroup of twisted diagonal subgroup is again twisted diagonal, this means that $k^{\langle Ps\rangle}_{L,\varphi}$ has twisted diagonal vertices. This shows that $\mathrm{Ind}^{G}_{L}k^{\langle Ps\rangle}_{L,\varphi}\in\mathbb{F}T^{\Delta}(G,H)$ as desired. Now since the $\mathbb{F}$-dimension of $\mathbb{F}T^{\Delta}(G,H)$ is equal to the cardinality of $[\mathcal{P}_{G\times H,p}^{\Delta}]$, which is equal to the cardinality of $[\mathcal{Q}_{G\times H,p}^{\Delta}]$, it follows that the set $\{F^{G\times H}_{P,s}:(P,s)\in[\mathcal{Q}_{G\times H,p}^{\Delta}]\}$ of primitive idempotents form an $\mathbb{F}$-basis for $\mathbb{F}T^{\Delta}(G,H)$. ∎

Let $G,H$ and $L$ be finite groups. If $X$ is a $(kG,kH)$-bimodule and $Y$ is a $(kH,kL)$-bimodule, then $X\circ Y:=X\otimes_{kH}Y$ is a $(kG,kL)$-bimodule. Extending this product by $\mathbb{F}$-bilinearity, we get a map

[TABLE]

Note that this induces a map

[TABLE]

which is used to define the composition of morphisms in the following category.

Definition 2.6**.**

Let $\mathbb{F}pp_{k}^{\Delta}$ be the category with

objects: finite groups

• 2.

$\mathrm{Mor}_{\mathbb{F}pp_{k}^{\Delta}}(G,H)=\mathbb{F}\otimes_{\mathbb{Z}}T^{\Delta}(H,G)=\mathbb{F}T^{\Delta}(H,G)$.

An $\mathbb{F}$-linear functor from $\mathbb{F}pp_{k}^{\Delta}$ to $\mathbb{F}$-$\mathrm{Mod}$ is called a diagonal $p$-permutation functor. Diagonal $p$-permutation functors form an abelian category $\mathcal{F}_{pp_{k}}^{\Delta}$.

3 The Essential Algebra

For a finite group $G$, the quotient algebra

[TABLE]

is called the essential algebra of $G$.

By [5, Proposition 4.1.2 and Theorem 4.1.12] the algebra

[TABLE]

is non-zero if and only if there exists a pair $(P,s)$ in $G$ such that $G=\langle Ps\rangle$ and $C_{\langle s\rangle}(P)=1$. In that case, we also have an algebra isomorphism

[TABLE]

where $n$ is the order of $s$ [5, Theorem 4.1.12].

Note that the inclusion map $\mathbb{F}T^{\Delta}(G,G)\hookrightarrow\mathbb{F}T(G,G)$ induces a map

[TABLE]

We will show that this map is an algebra isomorphism.

Let $\varphi\in\mathrm{Aut}(G)$ be an automorphism and $\lambda:G/O_{p}(G)\to k^{\times}$ be a character, where $O_{p}(G)$ denotes the largest normal $p$-subgroup of $G$. We define a $(kG,kG)$-bimodule structure on $kG$, denoted by $kG_{\varphi,\lambda}$, via

[TABLE]

for $a,b,g\in G$.

Let $\langle Rt\rangle$ be a twisted diagonal subgroup of $G\times G$ with $p_{1}(\langle Rt\rangle)=G$ and $p_{2}(\langle Rt\rangle)=G$. Let also $\eta:p_{1}(\langle Rt\rangle)\to p_{2}(\langle Rt\rangle)$ be the canonical isomorphism. Then by [5, Section 4.1.2] we have an isomorphism

[TABLE]

of $(kG,kG)$-bimodules. Again by [5, Section 4.1.2] the algebra $\mathcal{E}(G)$ is generated by the images of $kG_{\varphi,\lambda}$.

Proposition 3.1**.**

If the essential algebra $\mathcal{E}^{\Delta}(G)$ of a finite group $G$ is non-zero, then there exists a pair $(P,s)$ in $G$ such that $G=\langle Ps\rangle$ and $C_{\langle s\rangle}(P)=1$.

Proof.

Let $(Q,t)$ be a pair contained in $G\times G$ such that $Q$ is a twisted diagonal subgroup and recall the idempotent formula

[TABLE]

By [5, Lemma 2.5.9] we have an isomorphism

[TABLE]

of $(kG,kG)$-bimodules. As $(kG,kG)$-bimodule, we have the isomorphism $kG\cong\mathrm{Ind}^{G\times G}_{\Delta G}k$. Thus as $(kG,kp_{1}(\langle Lt\rangle))$-bimodule we have,

[TABLE]

Therefore as $(kG\times p_{1}(\langle Lt\rangle))$-module, the indecomposable direct summands of $kG$ have vertices contained in $\Delta(p_{1}(\langle Lt\rangle))$. Similary, one can show that the indecomposable direct summands of $kG$ as $k(p_{2}(\langle Lt\rangle)\times G)$-module, have vertices contained in $\Delta(p_{2}(\langle Lt\rangle))$. We also know that the module $k^{\langle Qt\rangle}_{L,\varphi}$, and hence the indecomposable direct summands of $\mathrm{Ind}^{p_{1}(\langle Lt\rangle)\times p_{2}(\langle Lt\rangle)}_{\langle Lt\rangle}(k^{\langle Qt\rangle}_{L,\varphi})$, have twisted diagonal vertices. Now suppose $\mathcal{E}^{\Delta}(G)$ is non-zero. Then there is an idempotent $F^{G\times G}_{Q,t}$ whose image in $\mathcal{E}^{\Delta}(G)$ is non-zero. Therefore the argument above shows that there is a pair $(Q,t)$ in $G\times G$ such that $p_{1}(\langle Qt\rangle)=G$ and $p_{2}(\langle Qt\rangle)=G$. This implies that there is a $p$-subgroup $P$ of $G$ and a $p^{\prime}$-element $s$ of $G$ that normalises $P$ such that $G=\langle Ps\rangle$. Now we will show that in that case we have $C_{\langle s\rangle}(P)=1$.

Let $\overline{G}:=G/C_{\langle s\rangle}(P)$, $Q:=\{(u,\overline{u}:u\in P\}\leqslant G\times\overline{G}$ and $Q^{\prime}:=\{(\overline{u},u):u\in P\}\leqslant\overline{G}\times G$. Then by [5, Proof of Proposition 4.1.2] we have an isomorphism between $kG$ and

[TABLE]

as $(kG,kG)$-bimodules, where $\mathrm{Indinf}^{G\times\overline{G}}_{\overline{N}_{G\times\overline{G}}(Q)}=\mathrm{Ind}_{N_{G\times\overline{G}}(Q)}^{G\times\overline{G}}\circ\mathrm{Inf}_{\overline{N}_{G\times\overline{G}}(Q)}^{N_{G\times\overline{G}}(Q)}$. Here $\alpha_{i}$ and $\alpha^{\prime}_{i}$ run over the irreducible characters of $\langle s\rangle$. Again by [5, Proof of Proposition 4.1.2] for each $i$, the modules $kC_{G}(P)/C_{\langle s\rangle}(P)\otimes_{k}k_{\alpha_{i}}$ and $kC_{G}(P)/C_{\langle s\rangle}(P)\otimes_{k}k_{\alpha^{\prime}_{i}}$ are projective indecomposable $k\overline{N}_{G\times\overline{G}}(Q)$-modules and $k\overline{N}_{\overline{G}\times G}(Q^{\prime})$-modules respectively. Now since $kC_{G}(P)/C_{\langle s\rangle}(P)\otimes_{k}k_{\alpha_{i}}$ is projective indecomposable, it has the trivial group as vertex. So \mathrm{Inf}^{N_{G\times\overline{G}}(Q)}_{\overline{N}_{G\times\overline{G}}(Q)}\big{(}kC_{G}(P)/C_{\langle s\rangle}(P)\otimes_{k}k_{\alpha_{i}}\big{)} has the group $Q$ as a vertex. Note that the group $Q$ is twisted diagonal. Therefore indecomposable direct summands of \mathrm{Indinf}^{G\times\overline{G}}_{\overline{N}_{G\times\overline{G}}(Q)}\big{(}kC_{G}(P)/C_{\langle s\rangle}(P)\otimes_{k}k_{\alpha_{i}}\big{)} have twisted diagonal vertices, i.e. \mathrm{Indinf}^{G\times\overline{G}}_{\overline{N}_{G\times\overline{G}}(Q)}\big{(}kC_{G}(P)/C_{\langle s\rangle}(P)\otimes_{k}k_{\alpha_{i}}\big{)}\in\mathbb{F}T^{\Delta}(G,\overline{G}). Similarly, we have \mathrm{Indinf}^{\overline{G}\times G}_{\overline{N}_{\overline{G}\times G}(Q^{\prime})}\big{(}kC_{G}(P)/C_{\langle s\rangle}(P)\otimes_{k}k_{\alpha^{\prime}_{i}}\big{)}\in\mathbb{F}T^{\Delta}(\overline{G},G). Now since $\mathcal{E}^{\Delta}(G)\neq 0$, the image of identity element $kG\in\mathbb{F}T^{\Delta}(G,G)$ in $\mathcal{E}^{\Delta}(G)$ is non-zero. Hence we have $\overline{G}=G$, i.e. $C_{\langle s\rangle}(P)=1$. ∎

Suppose we have $G=\langle Ps\rangle$ and $C_{\langle s\rangle}(P)=1$. The essential algebra $\mathcal{E}^{\Delta}(G)$ is generated by the images of the primitive idempotents

[TABLE]

where $Q$ is a twisted diagonal subgroup of $G\times G$. By [5, Lemma 2.5.9], if the image of $\mathrm{Ind}^{G\times G}_{\langle Lt\rangle}k^{\langle Qt\rangle}_{L,\varphi}$ is non-zero, then we must have that $p_{1}(\langle Lt\rangle)=G=p_{2}(\langle Lt\rangle)$. Write $t=(u,v)$. Then $p_{1}(\langle Lt\rangle)=\langle p_{1}(L)u\rangle$ and $p_{2}(\langle Lt\rangle)=\langle p_{2}(L)v\rangle$. Therefore we have $|u|=|v|=|s|$. Being a subgroup of twisted diagonal subgroup $Q$, the group $L$ itself is also twisted diagonal. Since $k_{1}(L)=k_{2}(L)=1$ and $|u|=|v|=|s|$, we have $k_{1}(\langle Lt\rangle)=k_{2}(\langle Lt\rangle)=1$. This shows that the subgroup $\langle Lt\rangle$ is twisted diagonal and $p_{1}(\langle Lt\rangle)=G=p_{2}(\langle Lt\rangle)$. Since the images of $\mathrm{Ind}^{G\times G}_{\langle Lt\rangle}k^{\langle Qt\rangle}_{L,\varphi}$ in $\mathcal{E}(G)$ with $\langle Lt\rangle$ satisfying these properties, generate the non-zero algebra $\mathcal{E}(G)$, this shows that the algebra $\mathcal{E}^{\Delta}(G)$ is also non-zero and the map $\Theta:\mathcal{E}^{\Delta}(G)\to\mathcal{E}(G)$ is surjective. Thus we have proved the following:

Proposition 3.2**.**

The essential algebra $\mathcal{E}^{\Delta}(G)$ is non-zero if and only if there is a pair $(P,s)$ in $G$ such that $G=\langle Ps\rangle$ and $C_{\langle s\rangle}(P)=1$. Moreover the map $\Theta:\mathcal{E}^{\Delta}(G)\to\mathcal{E}(G)$ is surjective.

Suppose we have $G=\langle Ps\rangle$ for some pair and $C_{\langle s\rangle}(P)=1$. We will show that the map $\Theta:\mathcal{E}^{\Delta}(G)\to\mathcal{E}(G)$ is also injective.

Suppose an element $\sum\overline{r_{\varphi,\alpha}kG_{\varphi,\alpha}}\in\mathcal{E}^{\Delta}(G)$ is mapped to zero by $\Theta$. Then the element $\sum\overline{r_{\varphi,\alpha}kG_{\varphi,\alpha}}$ of $\mathcal{E}(G)$ is zero. Write

[TABLE]

for some $(kG,kH)$-bimodule $U_{H}$ and $(kH,kG)$-bimodule $V_{H}$ and some constants $t_{H,U_{H},V_{H}}\in\mathbb{F}$. Suppose the coefficient $t_{H,U_{H},V_{H}}$ is non-zero for some group $H$. Then as in [5] we can assume that $H=\langle Rt\rangle$ for some pair $(R,t)$ and that the modules $U_{H}$ and $V_{H}$ are indecomposable. By [5, Section 4.1] one has

[TABLE]

where $\lambda_{i}$ is a character of $\langle s\rangle$ and $n_{i}\in\mathbb{N}$. Again by [5, Section 4.1] each summand $kZ(P)\otimes k_{\lambda_{i}}$ is a projective indecomposable $k\overline{N}_{G\times G}(\Delta(P))$-module. This shows that if the the coefficient $t_{H,U_{H},V_{H}}$ is non-zero, then the indecomposable direct summands of the bimodule $U_{H}\otimes_{kH}V_{H}$ have twisted diagonal vertices. Therefore the element $\sum\overline{r_{\varphi,\alpha}kG_{\varphi,\alpha}}$ is zero in $\mathcal{E}^{\Delta}(G)$. This proves that the map $\Theta:\mathcal{E}^{\Delta}(G)\to\mathcal{E}(G)$ is injective. We summarise our results as a theorem below.

Theorem 3.3**.**

*The essential algebra $\mathcal{E}^{\Delta}(G)$ is non-zero if and only if there is a pair $(P,s)$ in $G$ such that $G=\langle Ps\rangle$ and $C_{\langle s\rangle}(P)=1$. In that case, the algebra $\mathcal{E}^{\Delta}(G)$ is isomorphic to the algebra \big{(}\mathbb{F}[X]/\Phi_{n}[X]\big{)}\rtimes\mathrm{Out}(G) where $n$ is the order of $s$.

4 $D^{\Delta}$-pairs

Let $H\leqslant G$ be a subgroup. The $(kG,kH)$-bimodule $kG$ is denoted by $\mathrm{Ind}_{H}^{G}$ and $(kH,kG)$-bimodule $kG$ is denoted by $\mathrm{Res}_{H}^{G}$. Similarly, if $N\unlhd G$ is a normal subgroup, the $(kG/N,kG)$-bimodule $kG/N$ is denoted by $\mathrm{Def}^{G}_{G/N}$ and $(kG,kG/N)$-bimodule $kG/N$ is denoted by $\mathrm{Inf}^{G}_{G/N}$. This notation is consistent with our previous use of induction, restriction, inflation and deflation symbols, in the sense that for example, if $M$ is a $kH$-module, then the induced module $\mathrm{Ind}_{H}^{G}M$ is isomorphic to $\mathrm{Ind}_{H}^{G}\otimes_{kH}M$.

We have the following lemma due to [3] and [5].

Lemma 4.1**.**

1. (i)

Let $(P,s)\in\mathcal{Q}_{G,p}$ be a pair and $H\leqslant G$ be a subgroup. Then we have

[TABLE]

where $(Q,t)$ runs over a set of representatives of $H$-conjugacy classes of $G$-conjugates of $(P,s)$ contained in $H$. 2. (ii)

Let $(Q,t)\in\mathcal{Q}_{H,p}$ be a pair and $H\leqslant G$ be a subgroup. Then we have

[TABLE] 3. (iii)

Let $N\unlhd G$ and $(P,s)\in\mathcal{Q}_{G/N,p}$. Then

[TABLE]

where $(Q,t)$ runs over a set of representatives of $G$-conjugacy classes of pairs in $\mathcal{Q}_{G,p}$ such that $QN/N={}^{\overline{g}}P$ and $\overline{t}={}^{g}s$ for some $\overline{g}\in G/N$. 4. (iv)

Let $N\unlhd G$ and $(P,s)\in\mathcal{Q}_{G,p}$. Then

[TABLE]

for some pair $(Q,t)\in\mathcal{Q}_{G/N,p}$ and a constant $m_{P,s,N}\in\mathbb{F}$.

If $G=\langle Ps\rangle$ then

[TABLE]

Proof.

See [3, Proposition 3.1. and Proposition 3.2.] for (i) and (ii), [5, Proposition 3.1.3] for (iii) and [5, Lemma 3.1.4 and Proposition 3.1.5] for (iv). ∎

Lemma 4.2**.**

Let $N\unlhd G$ be a normal subgroup of $G$.

1. (i)

We have $\mathrm{Def}^{G}_{G/N}\in\mathbb{F}T^{\Delta}(G/N,G)$ if and only if $N$ is a $p^{\prime}$-group. 2. (ii)

We have $\mathrm{Inf}^{G}_{G/N}\in\mathbb{F}T^{\Delta}(G,G/N)$ if and only if $N$ is a $p^{\prime}$-group.

Proof.

(i) Let $Q\leqslant(G/N)\times G$ be a maximal vertex of an indecomposable direct summand of the $(kG/N,kG)$-bimodule $kG/N$. Equivalently $Q$ is a maximal $p$-subgroup having a fixed point on the set $G/N$. Suppose $(aN,b)\in Q$ stabilises a basis element $gN$ of $kG/N$. Then we have $(aN)gNb^{-1}=gN$ which implies that $a^{g}\cdot b^{-1}\in N$. Since the vertices of an indecomposable module are conjugate, we may assume that $g=1$. Thus, up to conjugacy, $Q$ is a Sylow $p$-subgroup of

[TABLE]

Note that $k_{1}(Q)=k_{1}(H)=1$ and $k_{2}(Q)$ is a Sylow $p$-subgroup of $N$. Hence $Q$ is twisted diagonal if and only if $N$ is a $p^{\prime}$-group. The result follows.

(ii) Similar. ∎

Let $(P,s)$ be a pair and suppose $G=\langle Ps\rangle$. Then by [5, Corollary 3.1.9] for any normal subgroup $N$ of $G$, we have the following formula for the constant $m_{P,s,N}$:

[TABLE]

Lemma 4.3**.**

Let $(P,s)$ be a pair and suppose $G=\langle Ps\rangle$. Then for any normal $p^{\prime}$-subgroup $N$ of $G$ we have

[TABLE]

Proof.

First observe that since $N$ is a $p^{\prime}$-group, we have $N\leqslant C_{\langle s\rangle}(P)$. For any subgroup $Q$ of $P$ the condition $\langle Qs\rangle N=\langle Ps\rangle$ implies that $|Q|=|P|$ and hence $Q=P$. Therefore the formula above becomes

[TABLE]

∎

Definition 4.4**.**

A pair $(P,s)$ is called $D^{\Delta}$-pair if $\mathrm{Def}^{\langle Ps\rangle}_{\langle Ps\rangle/N}F^{\langle Ps\rangle}_{P,s}=0$ for any nontrivial normal $p^{\prime}$-subgroup $N$ of $\langle Ps\rangle$.

Lemma 4.5**.**

Let $(P,s)$ be a pair. Then $(P,s)$ is a $D^{\Delta}$-pair if and only if the group $\langle Ps\rangle$ does not have any nontrivial normal $p^{\prime}$-subgroup, that is, if and only if $C_{\langle s\rangle}(P)=\nolinebreak 1$.

Proof.

By Lemma 4.3, for any normal $p^{\prime}$-subgroup $N\unlhd\langle Ps\rangle$ we have $m_{P,s,N}=1/|N|$. Therefore $(P,s)$ is a $D^{\Delta}$-pair if and only if the group $\langle Ps\rangle$ does not have any nontrivial normal $p^{\prime}$-subgroup. The result follows. ∎

5 The functor $\mathbb{F}T^{\Delta}$

By [2], the simple diagonal $p$-permutation functors are parametrized by the pairs $(G,V)$ where $G$ is a finite group and $V$ is a simple $\mathcal{E}^{\Delta}(G)$-module. Note that this implies $\mathcal{E}^{\Delta}(G)\neq 0$.

For a simple $\mathcal{E}^{\Delta}(G)$-module $V$, we define two functors in $\mathbb{F}pp_{k}^{\Delta}$ by:

[TABLE]

and

[TABLE]

for any finite group $H$. The action of morphisms in $\mathbb{F}pp_{k}^{\Delta}$ on these evaluations is given by left composition. The functor $J_{G,V}$ is the unique maximal subfunctor of $L_{G,V}$, so the quotient

[TABLE]

is a simple functor [2].

Let $\mathbb{F}T^{\Delta}:\mathbb{F}pp_{k}^{\Delta}\to\mathbb{F}$-$\mathrm{Mod}$ be the functor given by

$\mathbb{F}T^{\Delta}(G):=\mathbb{F}\otimes_{\mathbb{Z}}T(G)=\mathbb{F}T(G)$,

• 2.

$\mathbb{F}T^{\Delta}(X):\mathbb{F}T(G)\to\mathbb{F}T(H),M\mapsto X\otimes_{kH}M$ for any $X\in\mathbb{F}T^{\Delta}(H,G)$.

For any $kG$-module $X$, we denote by $\widetilde{X}$ the $(kG,kG)$-bimodule $k(G\times X)$ where the action of $kG$-$kG$ is given by

[TABLE]

for all $a,b,g\in G$ and $x\in X$. We have an isomorphism of $(kG,kG)$-bimodules

[TABLE]

where $\delta:G\to G\times G^{op}$, $g\mapsto(g,g^{-1})$. See [5, Definition 2.5.17]. Note that the image $\delta(G)$ of $G$ in $G\times G^{op}$ is a twisted diagonal subgroup. If $X$ is an indecomposable $p$-permutation $kG$-module with a vertex $Q$, then any vertex of an indecomposable direct summand of $\widetilde{X}$ is contained in $\delta(Q)$, up to conjugation. Therefore for any $X\in\mathbb{F}T(G)$ we have $\widetilde{X}\in\mathbb{F}T^{\Delta}(G,G)$.

Lemma 5.1**.**

Let $F$ be a subfunctor of $\mathbb{F}T^{\Delta}$. Then for any finite group $G$, the $\mathbb{F}$-vector space $F(G)$ is an ideal of the algebra $\mathbb{F}T^{\Delta}(G)$ of $p$-permutation modules.

Proof.

Let $Y\in F(G)$ and assume $X$ is a $p$-permutation $kG$-module. By [5, Proposition 2.5.18] we have an isomorphism $X\otimes_{k}Y\cong\widetilde{X}\otimes_{kG}Y$ of $kG$-modules. Since $\widetilde{X}\in\mathbb{F}T^{\Delta}(G,G)$ and $F$ is a functor, we have $\widetilde{X}\otimes_{kG}Y\in F(G)$. This shows that $F(G)$ is an ideal of $\mathbb{F}T^{\Delta}(G)$. ∎

Definition 5.2**.**

For any pair $(P,s)$ let $\mathbf{e}_{P,s}$ denote the subfunctor of $\mathbb{F}T^{\Delta}$ generated by the idempotent F^{\langle Ps\rangle}_{P,s}\in\mathbb{F}T^{\Delta}\big{(}\langle Ps\rangle\big{)}.

Proposition 5.3**.**

Let $F$ be a subfunctor of $\mathbb{F}T^{\Delta}$. Then we have

[TABLE]

Proof.

Since $F$ is a subfunctor, we have

[TABLE]

Now let $G$ be a finite group, and $u=\sum_{(P,s)}\lambda_{P,s}F_{P,s}^{G}$, where $(P,s)$ runs in a set of representatives of $G$-conjugacy classes of $\mathcal{Q}_{G,p}$, and $\lambda_{P,s}\in\mathbb{F}$. Then $F_{P,s}^{G}\cdot u=\lambda_{P,s}F_{P,s}^{G}\in F(G)$, since $F(G)$ is an ideal of $\mathbb{F}T^{\Delta}(G)$. Hence $F_{P,s}^{G}\in F(G)$ if $\lambda_{P,s}\neq\nolinebreak 0$. In this case we have \mathrm{Res}^{G}_{\langle Ps\rangle}F^{G}_{P,s}\in F\big{(}\langle Ps\rangle\big{)}, which implies by Lemma 4.1 that F^{\langle Ps\rangle}_{P,s}\in F\big{(}\langle Ps\rangle\big{)}. This shows that $\mathbf{e}_{P,s}\leqslant F$. By Lemma 4.1 again, $F_{P,s}^{G}$ is a non zero scalar multiple of $\mathrm{Ind}_{\langle Ps\rangle}^{G}F^{\langle Ps\rangle}_{P,s}$, so $F_{P,s}^{G}\in\mathbf{e}_{P,s}(G)$, which gives finally

[TABLE]

Therefore we have

[TABLE]

as desired. ∎

Proposition 5.4**.**

Let $(P_{i},s_{i})_{i\in I}$ be a set of pairs for an indexing set $I$. Then for any pair $(Q,t)$ we have $\mathbf{e}_{Q,t}\leqslant\sum_{i\in I}\mathbf{e}_{P_{i},s_{i}}$ if and only if $\mathbf{e}_{Q,t}\leqslant\mathbf{e}_{P_{i},s_{i}}$ for some $i\in I$.

Proof.

If $\mathbf{e}_{Q,t}\leqslant\mathbf{e}_{P_{i},s_{i}}$ for some $i\in I$, then we obviously have $\mathbf{e}_{Q,t}\leqslant\sum_{i\in I}\mathbf{e}_{P_{i},s_{i}}$. Conversely assume we have $\mathbf{e}_{Q,t}\leqslant\sum_{i\in I}\mathbf{e}_{P_{i},s_{i}}$. Then \mathbf{e}_{Q,t}\big{(}\langle Qt\rangle\big{)}\leqslant\sum_{i\in I}\mathbf{e}_{P_{i},s_{i}}\big{(}\langle Qt\rangle\big{)} and so F^{\langle Qt\rangle}_{Q,t}\in\sum_{i\in I}\mathbf{e}_{P_{i},s_{i}}\big{(}\langle Qt\rangle\big{)}. Since $F^{\langle Qt\rangle}_{Q,t}$ is a primitive idempotent and since \mathbf{e}_{P_{i},s_{i}}\big{(}\langle Qt\rangle\big{)} is an ideal of \mathbb{F}T^{\Delta}\big{(}\langle Qt\rangle\big{)} it follows that we have F^{\langle Qt\rangle}_{Q,t}\in\mathbf{e}_{P_{i},s_{i}}\big{(}\langle Qt\rangle\big{)} for some $i\in I$ and hence $\mathbf{e}_{Q,t}\leqslant\mathbf{e}_{P_{i},s_{i}}$. ∎

Let $G$ be a finite group and $(P,s)\in\mathcal{Q}_{G,p}$ be a pair such that $G=\langle Ps\rangle$. Let also $(Q,t)\in\mathcal{Q}^{\Delta}_{H\times G,p}$ for a finite group $H$. Suppose that $\eta:p_{1}(Q)\to p_{2}(Q)$ is the canonical isomorphism. Up to conjugation in $H\times G$, we can assume $t=(u,s^{j})$. By [5, Section 3.2] if $p_{2}(\langle Qt\rangle)\neq G$, then the product $F^{H\times G}_{Q,t}\otimes_{kG}F^{G}_{P,s}$ is zero. So assume that we have $p_{2}(\langle Qt\rangle)=G$. This implies that we have $p_{2}(Q)=P$ and $|s^{j}|=|s|$. Then since $k_{1}(Q)=k_{2}(Q)=1$, this implies that we have $p_{1}(Q)\cong P$. Since the group $Q$ is $t$-stable, the isomorphism $\eta:p_{1}(Q)\to P$ commutes with conjugations by $u$ and $s^{j}$. Now [5, Equation (3.3), Section 3.2] implies that as $kH$-module the product $F^{H\times G}_{Q,t}\otimes_{kG}F^{G}_{P,s}$ is equal to

[TABLE]

where \sigma(J):=\sum_{\begin{subarray}{c}L\leqslant P\\ L^{s}=L\\ \eta(J)=L\end{subarray}}|C_{L}(s)|\mu\big{(}(L,P)^{s}\big{)} and $\phi(u):=\varphi(u,s^{j})\psi(s)^{j}$.

Suppose we have $H=\langle P^{\prime}s^{\prime}\rangle$ for a pair $(P^{\prime},s^{\prime})$. Then by [5, Lemma 2.7.6] if $\tau^{H}_{P^{\prime},s^{\prime}}(F^{H\times G}_{Q,t}\otimes_{kG}F^{G}_{P,s})\neq 0$, then we must have $p_{1}(Q)=P^{\prime}$ and $|u|=|s^{\prime}|$. This implies in particular that we must have $P^{\prime}\cong P$. Moreover again by [5, Lemma 2.7.6] we have \tau^{H}_{P^{\prime},s^{\prime}}\big{(}\mathrm{Ind}^{H}_{\langle Ju\rangle}(k^{\langle p_{1}(Q)u\rangle}_{\langle Ju\rangle,\phi})\big{)}=0 if $J\neq P^{\prime}$. Therefore if we have $P^{\prime}\cong P$ then $\tau^{H}_{P^{\prime},s^{\prime}}(F^{H\times G}_{Q,t}\otimes_{kG}F^{G}_{P,s})$ is equal to

[TABLE]

This shows that if we have \mathbb{F}T^{\Delta}\big{(}\langle P^{\prime}s^{\prime}\rangle,\langle Ps\rangle\big{)}\otimes_{k\langle Ps\rangle}F^{\langle Ps\rangle}_{P,s}\neq 0, then there is an isomorphism $\eta:P^{\prime}\to P$ and a $p^{\prime}$-element $(u,s^{j})\in\langle P^{\prime}s^{\prime}\rangle\times\langle Ps\rangle$ such that $\eta\circ c_{u}=c_{s^{j}}\circ\eta$ and $|u|=|s^{\prime}|$, $|s^{j}|=|s|$. In that case, assume further that $C_{\langle s\rangle}(P)=1$. Then we have $|c_{s}|=|s|$ and $|c_{s^{j}}|=|s^{j}|$. Since we have $\eta\circ c_{u}=c_{s^{j}}\circ\eta$ it follows that $|c_{u}|=|c_{s^{j}}|$. Therefore we have $|s|\mid|s^{\prime}|$. But then [5, Proposition 2.3.6] implies that there is a surjective group homomorphism $\overline{\eta}:\langle P^{\prime}s^{\prime}\rangle\to\langle Ps\rangle$ that induces an isomorphism of pairs \big{(}P^{\prime}\ker(\overline{\eta})/\ker(\overline{\eta}),s^{\prime}\ker(\overline{\eta})\big{)}\simeq(P,s). Note that since $|P^{\prime}|=|P|$ the order of $\ker(\overline{\eta})$ is coprime to $p$. We have the following:

Lemma 5.5**.**

Let $(P,s)$ be a pair with $C_{\langle s\rangle}(P)=1$ and set $G:=\langle Ps\rangle$. Let $H$ be a finite group. The following statements are equivalent:

1. (i)

$\mathbb{F}T^{\Delta}(H,G)\otimes_{kG}F^{G}_{P,s}\neq 0$. 2. (ii)

There exists a pair $(P^{\prime},s^{\prime})$ contained in $H$ such that the pair $(P,s)$ is isomorphic to a $p^{\prime}$-quotient of the pair $(P^{\prime},s^{\prime})$, that is, there exists a normal $p^{\prime}$-subgroup $K$ of $\langle P^{\prime}s^{\prime}\rangle$ such that $(P,s)\simeq(P^{\prime}K/K,s^{\prime}K)$.

Proof.

(i) $\Rightarrow$ (ii) Suppose we have $\mathbb{F}T^{\Delta}(H,G)\otimes_{kG}F^{G}_{P,s}\neq 0$. Then there exists a pair $(P^{\prime},s^{\prime})$ in $H$ such that

[TABLE]

Via the restriction map this implies that we have

[TABLE]

Therefore by the argument above we have an isomorphism $(P^{\prime}K/K,s^{\prime}K)\simeq(P,s)$ of pairs where $K$ is a normal $p^{\prime}$-subgroup of $\langle P^{\prime}s^{\prime}\rangle$.

(ii) $\Rightarrow$ (i) Suppose $\Phi:(P^{\prime}K/K,s^{\prime}K)\to(P,s)$ is an isomorphism of pairs where $K$ is a normal $p^{\prime}$-subgroup of $\langle P^{\prime}s^{\prime}\rangle$. Then we have

[TABLE]

This shows (i). ∎

Proposition 5.6**.**

Let $(P,s)$ be a pair. The following are equivalent:

1. (i)

$(P,s)$* is a $D^{\Delta}$-pair.* 2. (ii)

For any finite group $H$ with $|H|<|\langle Ps\rangle|$, we have $\mathbf{e}_{P,s}(H)=\{0\}$. 3. (iii)

If $H$ is a finite group with $\mathbf{e}_{P,s}(H)\neq\{0\}$, then the pair $(P,s)$ is isomorphic to a $p^{\prime}$-quotient of a pair $(P^{\prime},s^{\prime})$ contained in $H$. 4. (iv)

For any nontrivial normal $p^{\prime}$-subgroup $N$ of $\langle Ps\rangle$, we have $\mathrm{Def}^{\langle Ps\rangle}_{\langle Ps\rangle/N}F^{\langle Ps\rangle}_{P,s}=0$. 5. (v)

The group $\langle Ps\rangle$ does not have any nontrivial normal $p^{\prime}$-subgroup. 6. (vi)

We have $C_{\langle s\rangle}(P)=1$.

Proof.

(vi)$\Leftrightarrow$(v)$\Leftrightarrow$(i) : This follows from Lemma 4.5.

(iv)$\Leftrightarrow$ (i): This follows from the definition of $D^{\Delta}$-pairs.

(i)$\Rightarrow$ (iii): Since $(P,s)$ is a $D^{\Delta}$-pair, we have $C_{\langle s\rangle}(P)=1$. So (iii) follows from Lemma 5.5.

(iii)$\Rightarrow$ (ii): Assume that (iii) holds and $\mathbf{e}_{P,s}(H)\neq 0$ where $H$ is a finite group with $\lvert H\rvert<\lvert\langle Ps\rangle\rvert$. Then by the assumption, we have $\lvert H\rvert\geq\lvert\langle P^{\prime}s^{\prime}\rangle\rvert\geq\lvert\langle Ps\rangle\rvert$. Contradiction.

(ii)$\Rightarrow$ (iv): Clear. ∎

Proposition 5.7**.**

Let $(P,s)$ and $(Q,t)$ be two pairs.

1. (i)

If the pair $(Q,t)$ is isomorphic to a $p^{\prime}$-quotient of the pair $(P,s)$, then we have $\mathbf{e}_{P,s}=\mathbf{e}_{Q,t}$. 2. (ii)

If $(Q,t)$ is a $D^{\Delta}$-pair, and if $\mathbf{e}_{P,s}\leqslant\mathbf{e}_{Q,t}$, then $(Q,t)$ is isomorphic to a $p^{\prime}$-quotient of $(P,s)$.

Proof.

(i) Assume we have an isomorphism $\phi:(PK/K,sK)\to(Q,t)$ of pairs for some normal $p^{\prime}$-subgroup $K$ of $\langle Ps\rangle$. Then we have

[TABLE]

Therefore we have F^{\langle Ps\rangle}_{P,s}\in\mathbf{e}_{Q,t}\big{(}\langle Ps\rangle\big{)} which implies that $\mathbf{e}_{P,s}\leqslant\mathbf{e}_{Q,t}$.

Now we also have

[TABLE]

which implies that F^{\langle Qt\rangle}_{Q,t}\in\mathbf{e}_{P,s}\big{(}\langle Qt\rangle\big{)}. Therefore we have $\mathbf{e}_{Q,t}\leqslant\mathbf{e}_{P,s}$ and so $\mathbf{e}_{Q,t}=\mathbf{e}_{P,s}$ as desired.

(ii) Since $\mathbf{e}_{P,s}\leqslant\mathbf{e}_{Q,t}$, we have F^{\langle Ps\rangle}_{P,s}\in\mathbf{e}_{Q,t}\big{(}\langle Ps\rangle\big{)}. Since $(Q,t)$ is a $D^{\Delta}$-pair, by the proof of Lemma 5.5, there exists a normal $p^{\prime}$-subgroup $K$ of $\langle Ps\rangle$ such that $(Q,t)\simeq(PK/K,sK)$. ∎

Proposition 5.8**.**

Let $F$ be a nonzero subfunctor of $\mathbb{F}T^{\Delta}$. If $H$ is a minimal group of $F$, then $H=\langle Qt\rangle$ for some $D^{\Delta}$-pair $(Q,t)$. Moreover we have

[TABLE]

and $\mathbf{e}_{Q,t}\leqslant F$.

In particular, if $F=\mathbf{e}_{Q,t}$ for some $D^{\Delta}$-pair $(Q,t)$, then we have

[TABLE]

Proof.

Let $F$ be a nonzero subfunctor of $\mathbb{F}T^{\Delta}$ and assume $H$ is a minimal group of $F$. Since $F(H)\neq 0$, there exists a pair $(Q,t)\in\mathcal{Q}_{H,p}$ such that $F^{H}_{Q,t}\in F(H)$. This implies, via the restriction map, that we have F^{\langle Qt\rangle}_{Q,t}\in F\big{(}\langle Qt\rangle\big{)}. Since $H$ is a minimal group, this implies that we have $H=\langle Qt\rangle$. Now if $N$ is a normal $p^{\prime}$-subgroup of $\langle Qt\rangle$, then $\mathrm{Def}^{\langle Qt\rangle}_{\langle Qt\rangle/N}F^{\langle Qt\rangle}_{Q,t}=\frac{1}{|N|}F^{\langle Qt\rangle/N}_{QN/N,tN}\neq 0$. Again since $H$ is a minimal group this means that $N$ is trivial and hence the pair $(Q,t)$ is a $D^{\Delta}$-pair. It follows moreover that

[TABLE]

For the last part, consider the functor $\mathbf{e}_{Q,t}$ for some $D^{\Delta}$-pair $(Q,t)$. If F^{\langle Qt\rangle}_{Q^{\prime},t^{\prime}}\in\mathbf{e}_{Q,t}\big{(}\langle Qt\rangle\big{)} for some $D^{\Delta}$-pair $(Q^{\prime},t^{\prime})$, then by the second part of Proposition 5.7, the pair $(Q,t)$ is isomorphic to a $p^{\prime}$-quotient of the pair $(Q^{\prime},t^{\prime})$. But the pair $(Q^{\prime},t^{\prime})$ is contained in $\langle Qt\rangle$. Thus we have $(Q^{\prime},t^{\prime})\simeq(Q,t)$.

Conversely, if the pairs $(Q^{\prime},t^{\prime})$ and $(Q,t)$ are isomorphic via a map $\phi$, then we have $F^{\langle Qt\rangle}_{Q^{\prime},t^{\prime}}=\mathrm{Iso}(\phi)F^{\langle Qt\rangle}_{Q,t}$. Therefore we have

[TABLE]

∎

Let $(P,s)$ be a pair and $N$ a normal $p^{\prime}$-subgroup of $\langle Ps\rangle$. Then the pair $(PN/N,sN)$ is a $p^{\prime}$-quotient of the pair $(P,s)$ and so by Proposition 5.7 we have $\mathbf{e}_{P,s}=\mathbf{e}_{PN/N,sN}$.

Proposition 5.9**.**

Let $(P,s)$ be a pair. Then the group $\langle Ps\rangle/C_{\langle s\rangle}(P)$ is the unique, up to isomorphism, minimal group of the functor $\mathbf{e}_{P,s}$. Moreover there is a unique isomorphism class of $D^{\Delta}$-pairs $(P^{\prime},s^{\prime})$ such that $\langle P^{\prime}s^{\prime}\rangle\cong\langle Ps\rangle/C_{\langle s\rangle}(P)$ and we have $\mathbf{e}_{P^{\prime},s^{\prime}}=\mathbf{e}_{P,s}$. Furthermore we have (P^{\prime},s^{\prime})\simeq\big{(}PC_{\langle s\rangle}(P)/C_{\langle s\rangle}(P),sC_{\langle s\rangle}(P)\big{)}.

Proof.

Let $(P^{\prime},s^{\prime})$ be a $D^{\Delta}$-pair such that $\langle P^{\prime}s^{\prime}\rangle$ is a minimal group of the functor $\mathbf{e}_{P,s}$. By Proposition 5.8, we have $\mathbf{e}_{P^{\prime},s^{\prime}}\leqslant\mathbf{e}_{P,s}$. Let $N:=C_{\langle s\rangle}(P)$. Then the pair $(PN/N,sN)$ is a $D^{\Delta}$-pair, and we have $\mathbf{e}_{P,s}=\mathbf{e}_{PN/N,sN}$. Since $(PN/N,sN)$ is a $D^{\Delta}$-pair, by Proposition 5.7 there exists a normal $p^{\prime}$-subgroup $K$ of $\langle P^{\prime}s^{\prime}\rangle$ such that $(P^{\prime}K/K,s^{\prime}K)\simeq(PN/N,sN)$. This means that the idempotent $F^{\langle P^{\prime}s^{\prime}\rangle/K}_{P^{\prime}K/K,s^{\prime}K}$ is in the evaluation at $\langle P^{\prime}s^{\prime}\rangle/K$ of the functor $\mathbf{e}_{PN/N,sN}=\mathbf{e}_{P,s}$. Since the group $\langle P^{\prime}s^{\prime}\rangle$ is a minimal group of $\mathbf{e}_{P,s}$ it follows that we must have $K=1$. Thus we have $(P^{\prime},s^{\prime})\simeq(PN/N,sN)$. Therefore we have $\mathbf{e}_{P^{\prime},s^{\prime}}=\textbf{e}_{PN/N,sN}=\mathbf{e}_{P,s}$.

Now we will show the uniqueness of the isomorphism class of the minimal groups of $\mathbf{e}_{P,s}$. Let $H$ be a minimal group of $\mathbf{e}_{P,s}$. It suffices to show that $H$ is isomorphic to $\langle P^{\prime}s^{\prime}\rangle$. By Proposition 5.8 the group $H$ is of the form $H=\langle Qt\rangle$ for some $D^{\Delta}$-pair $(Q,t)$. By the first part of the proof we have $\mathbf{e}_{Q,t}=\mathbf{e}_{P,s}=\mathbf{e}_{P^{\prime},s^{\prime}}$. Since both $(Q,t)$ and $(P,s)$ are $D^{\Delta}$-pairs, the equality $\mathbf{e}_{Q,t}=\mathbf{e}_{P^{\prime},s^{\prime}}$ implies that $(Q,t)$ is isomorphic to a $p^{\prime}$-quotient of $(P,s)$, and vice versa. Therefore we have $(Q,t)\simeq(P^{\prime},s^{\prime})$ which implies that $H=\langle Qt\rangle\cong\langle P^{\prime}s^{\prime}\rangle$ as desired. ∎

For any pair $(P,s)$ we denote by $(\tilde{P},\tilde{s})$ a representative of the isomorphism class of the pair $(PC_{\langle s\rangle}(P)/C_{\langle s\rangle}(P),sC_{\langle s\rangle}(P))$.

Theorem 5.10**.**

Let $(P,s)$ be a pair.

1. (i)

If $(Q,t)$ is isomorphic to a $p^{\prime}$-quotient of $(P,s)$ and if $(Q,t)$ is a $D^{\Delta}$-pair, then $(Q,t)$ is isomorphic to the pair $(\tilde{P},\tilde{s})$. In particular, for any normal $p^{\prime}$-subgroup $N\unlhd\langle Ps\rangle$, we have $(PN/N,sN)\simeq(\tilde{P},\tilde{s})$ if and only if $(PN/N,sN)$ is a $D^{\Delta}$-pair. 2. (ii)

Let $N\unlhd\langle Ps\rangle$ be a normal $p^{\prime}$-subgroup. Then the pair $(\tilde{P},\tilde{s})$ is isomorphic to a $p^{\prime}$-quotient of $(PN/N,sN)$ and we have $(\tilde{P},\tilde{s})\simeq(\widetilde{PN/N},\widetilde{sN})$.

Proof.

(i) Since the pair $(Q,t)$ is isomorphic to a $p^{\prime}$-quotient of the pair $(P,s)$, by Proposition 5.7, we have $\mathbf{e}_{\tilde{P},\tilde{s}}=\mathbf{e}_{P,s}\leqslant\mathbf{e}_{Q,t}$. Since $(Q,t)$ is a $D^{\Delta}$-pair, again by Proposition 5.7, the pair $(Q,t)$ is isomorphic to a $p^{\prime}$-quotient of $(\tilde{P},\tilde{s})$. But since the pair $(\tilde{P},\tilde{s})$ is a $D^{\Delta}$-pair, it follows that the pair $(Q,t)$ is isomorphic to the pair $(\tilde{P},\tilde{s})$.

(ii) Since the constant $m_{P,s,N}$ is non-zero, we have F^{\langle Ps\rangle/N}_{PN/N,sN}\in\mathbf{e}_{P,s}\big{(}\langle Ps\rangle/N\big{)}=\textbf{e}_{\tilde{P},\tilde{s}}\big{(}\langle Ps\rangle/N\big{)}. Therefore we have $\mathbf{e}_{PN/N,sN}\leqslant\textbf{e}_{\tilde{P},\tilde{s}}$ and since $(\tilde{P},\tilde{s})$ is a $D^{\Delta}$-pair, by Proposition 5.7, $(\tilde{P},\tilde{s})$ is isomorphic to a $p^{\prime}$-quotient of $(PN/N,sN)$. Again since the pair $(\tilde{P},\tilde{s})$ is a $D^{\Delta}$-pair, by part (i), it is isomorphic to the pair $(\widetilde{PN/N},\widetilde{sN})$. ∎

Let $[D^{\Delta}\textit{-pair}]$ denote a set of isomorphism classes of $D^{\Delta}$-pairs. Then the subfunctor lattice of the functor $\mathbb{F}T^{\Delta}$ is isomorphic to the lattice of subsets of the set $[D^{\Delta}\textit{-pair}]$ ordered by inclusion.

Theorem 5.11**.**

Let $\mathcal{S}$ be the lattice of subfunctors of $\mathbb{F}T^{\Delta}$ ordered by inclusion of subfunctors. Let $\mathcal{T}$ be the lattice of subsets of $[D^{\Delta}\textit{-pair}]$ ordered by inclusion of subsets. Then the map

[TABLE]

that sends a subfunctor $F$ to the set $\{(P,s)\in[D^{\Delta}\textit{-pair}]:\mathbf{e}_{P,s}\leqslant F\}$, is an isomorphism of lattices with inverse

[TABLE]

that sends a subset $A$ to the functor $\sum_{(P,s)\in A}\mathbf{e}_{P,s}$.

Proof.

We need to show that the maps $\Theta$ and $\Psi$ are inverse of each other. Let $F\in\mathcal{S}$ be a subfunctor. By Proposition 5.3 we have

[TABLE]

where $\Gamma$ is a set of representatives of the isomorphism classes of pairs. But for any pair $(P,s)$ we have $\mathbf{e}_{P,s}=\mathbf{e}_{\tilde{P},\tilde{s}}$ and $(\tilde{P},\tilde{s})$ is a $D^{\Delta}$-pair. Therefore we have

[TABLE]

This shows that $\Psi(\Theta(F))=F$.

Now let $A\in\mathcal{T}$ be a subset and let $(Q,t)\in\Theta(\Psi(A))$ be a $D^{\Delta}$-pair. Then we have $\mathbf{e}_{Q,t}\leqslant\sum_{(P,s)\in A}\mathbf{e}_{P,s}$ and so by Proposition 5.4 this implies that we have $\mathbf{e}_{Q,t}\leqslant\mathbf{e}_{P,s}$ for some $(P,s)\in A$. Since both $(P,s)$ and $(Q,t)$ are $D^{\Delta}$-pairs, it follows that $(P,s)\simeq(Q,t)$ and hence $(Q,t)\in A$. This shows that $\Theta(\Psi(A))\subseteq A$. The inclusion $A\subseteq\Theta(\Psi(A))$ is trivial. Therefore we have $\Theta(\Psi(A))=A$. ∎

The following corollary follows immediately from Theorem 5.11.

Corollary 5.12**.**

We have $\mathbb{F}T^{\Delta}=\bigoplus_{(P,s)\in[D^{\Delta}\textit{-pair}]}\mathbf{e}_{P,s}$.

The first statement of Proposition 5.8 can also be made stronger.

Corollary 5.13**.**

Let $F$ be a nonzero subfunctor of $\mathbb{F}T^{\Delta}$. If $H$ is a minimal group of $F$, then $H=\langle Qt\rangle$ for some $D^{\Delta}$-pair $(Q,t)$ and we have

[TABLE]

Proof.

Since $H$ is a minimal group of $F$, by Proposition 5.8 it follows that $H=\langle Qt\rangle$ for some $D^{\Delta}$-pair with the property that $\mathbf{e}_{Q,t}\leqslant F$. By Theorem 5.11 we have

[TABLE]

Therefore by Proposition 5.8 again we have

[TABLE]

as desired. ∎

Corollary 5.14**.**

Let $(P,s)$ be a $D^{\Delta}$-pair. Then the subfunctor $\mathbf{e}_{P,s}$ of $\mathbb{F}T^{\Delta}$ is isomorphic to the simple functor $S_{\langle Ps\rangle,W_{P,s}}$ where $W_{P,s}=\oplus_{\begin{subarray}{c}(Q,t)\simeq(P,s)\\ \langle Qt\rangle=\langle Ps\rangle\end{subarray}}\mathbb{F}F^{\langle Ps\rangle}_{P,s}$.

Proof.

By Theorem 5.11 the lattice of subfunctors of $\mathbf{e}_{P,s}$ is isomorphic to the lattice of subsets of the set $\Theta(\mathbf{e}_{P,s})=\{(Q,t)\in[D^{\Delta}\textit{-pair}]:\mathbf{e}_{Q,t}\leqslant\mathbf{e}_{P,s}\}=\{(P,s)\}$. Therefore the subfunctor $\mathbf{e}_{P,s}$ is simple. By Proposition 5.9 the group $\langle Ps\rangle$ is a minimal group of the functor $\mathbf{e}_{P,s}$. By Proposition 5.8 we have \mathbf{e}_{P,s}\big{(}\langle Ps\rangle\big{)}=W_{P,s}. Moreover, by [5, Theorem 4.2.5], the module $W_{P,s}$ is a simple module for the essential algebra \mathcal{E}^{\Delta}\big{(}\langle Ps\rangle\big{)}. Thus we have $\mathbf{e}_{P,s}\simeq S_{\langle Ps\rangle,W_{P,s}}$ as desired. ∎

Proposition 5.15**.**

If $F\leqslant F^{\prime}$ are subfunctors of $\mathbb{F}T^{\Delta}$ such that $F^{\prime}/F$ is simple, then there exists a unique $D^{\Delta}$-pair $(P,s)\in[D^{\Delta}\textit{-pair}]$ such that $\mathbf{e}_{P,s}\leqslant F^{\prime}$ and $\mathbf{e}_{P,s}\nleqslant F$. In particular, we have $\mathbf{e}_{P,s}+F=F^{\prime}$, $\mathbf{e}_{P,s}\cap F=\{0\}$, and $F^{\prime}/F\simeq S_{\langle Ps\rangle,W_{P,s}}$

Proof.

The existence of a pair $(P,s)$ with the property that $\mathbf{e}_{P,s}\leqslant F^{\prime}$ and $\mathbf{e}_{P,s}\nleqslant F$ is clear. Suppose $(P^{\prime},s^{\prime})$ is another pair with these properties. Since $F^{\prime}/F$ is simple, we have

[TABLE]

Thus $(P^{\prime},s^{\prime})\simeq(P,s)$ as $(P^{\prime},s^{\prime})\notin\Theta(F)$. Now since $\mathbf{e}_{P,s}\nleqslant F$ and $F^{\prime}/F$ is simple, we have $\mathbf{e}_{P,s}+F=F^{\prime}$. Thus the quotient $\mathbf{e}_{P,s}/(\mathbf{e}_{P,s}\cap F)\simeq F^{\prime}/F$ is simple and so $\mathbf{e}_{P,s}\cap F=\{0\}$. Therefore we have $F^{\prime}/F\simeq S_{\langle Ps\rangle,W_{P,s}}$. ∎

Proposition 5.16**.**

Let $F\leqslant F^{\prime}$ be subfunctors of $\mathbb{F}T^{\Delta}$ such that $F^{\prime}/F$ is simple. Let $H$ (respectively $H^{\prime}$) be a finite group and $W$ (respectively $W^{\prime}$) be a simple $\mathcal{E}^{\Delta}(H)$-mod (respectively $\mathcal{E}^{\Delta}(H^{\prime})$-mod) such that $S_{H,W}\simeq S_{H^{\prime},W^{\prime}}\simeq F^{\prime}/F$. Then $H\cong H^{\prime}$. Moreover $W\cong W^{\prime}$, after identification of $H$ and $H^{\prime}$ via the previous isomorphism.

Proof.

By Proposition 5.15 there exists a unique $D^{\Delta}$-pair $(P,s)$ such that $F^{\prime}/F\simeq S_{\langle Ps\rangle,W_{P,s}}$. Therefore it suffices to prove that $H\cong\langle Ps\rangle$. Since $(F^{\prime}/F)(H)\neq 0$ there exists a pair $(Q,t)$ contained in $H$ such that $F^{H}_{Q,t}\in F^{\prime}(H)\setminus F(H)$. Since $H$ is a minimal group of $F^{\prime}/F$, it follows that $H=\langle Qt\rangle$ and $(Q,t)$ is a $D^{\Delta}$-pair. Moreover we have $\mathbf{e}_{Q,t}\leqslant F^{\prime}$ and $\mathbf{e}_{Q,t}\nleqslant F$. But the pair $(P,s)$ is the unique $D^{\Delta}$-pair with these properties. Therefore we have $(Q,t)\simeq(P,s)$. Thus $H\cong\langle Ps\rangle$ as desired. The last assertion follows from the fact that $S_{H,W}(H)\cong W$. ∎

Proposition 5.17**.**

Let $(P,s)$ be a pair. Then for any finite group $H$, the $\mathbb{F}$-vector space $\mathbf{e}_{P,s}(H)$ is the subspace of $\mathbb{F}T(H)$ generated by the set of primitive idempotents $F^{H}_{Q,t}$ where $(Q,t)$ runs over a set of conjugacy classes of pairs in $H$ with the property that $(P,s)$ is isomorphic to a $p^{\prime}$-quotient of $(Q,t)$.

Proof.

Since the pair $(\tilde{P},\tilde{s})$ is isomorphic to a $p^{\prime}$-quotient of the pair $(P,s)$ and since $\mathbf{e}_{P,s}=\textbf{e}_{\tilde{P},\tilde{s}}$, we may assume that the pair $(P,s)$ is a $D^{\Delta}$-pair. Since $\mathbf{e}_{P,s}(H)$ is an ideal of $\mathbb{F}T(H)$, it has a $\mathbb{F}$-basis consisting of a set of primitive idempotents $F^{H}_{Q,t}$. If $F^{H}_{Q,t}\in\mathbf{e}_{P,s}(H)$, then F^{\langle Qt\rangle}_{Q,t}\in\mathbf{e}_{P,s}\big{(}\langle Qt\rangle\big{)} and so $\mathbf{e}_{Q,t}\leqslant\mathbf{e}_{P,s}$. Since $(P,s)$ is a $D^{\Delta}$-pair, by Proposition 5.7, it is isomorphic to a $p^{\prime}$-quotient of the pair $(Q,t)$. Conversely, if $(P,s)$ is isomorphic to a $p^{\prime}$-quotient of the pair $(Q,t)$, then again by Proposition 5.7, we have $\mathbf{e}_{Q,t}\leqslant\mathbf{e}_{P,s}$. So we have F^{\langle Qt\rangle}_{Q,t}\in\mathbf{e}_{P,s}\big{(}\langle Qt\rangle\big{)} and hence $F^{H}_{Q,t}\in\mathbf{e}_{P,s}(H)$. The result follows. ∎

Theorem 5.18**.**

Let $(P,s)$ be a $D^{\Delta}$-pair. Then for any finite group $H$, the $\mathbb{F}$-dimension of $S_{\langle Ps\rangle,W_{P,s}}(H)$ is equal to the number of conjugacy classes of pairs $(Q,t)$ in $H$ such that $(\tilde{Q},\tilde{t})\simeq(P,s)$.

Proof.

By Proposition 5.17 $\mathbf{e}_{P,s}(H)$ is generated by the idempotents $F^{H}_{Q,t}$ where $(Q,t)$ is a pair in $H$ with the property that the pair $(\tilde{P},\tilde{s})\simeq(P,s)$ is isomorphic to a $p^{\prime}$-quotient of the pair $(Q,t)$. Since $(P,s)$ is a $D^{\Delta}$-pair, Theorem 5.10 implies that $(\tilde{Q},\tilde{t})\simeq(P,s)$. The result follows. ∎

Corollary 5.19**.**

Let $H$ be a finite group. The $\mathbb{F}$-dimension of $S_{1,\mathbb{F}}(H)$ is equal to the number of isomorphism classes of simple $kH$-modules.

Proof.

By Theorem 5.18, $\mathrm{dim}_{\mathbb{F}}S_{1,\mathbb{F}}(H)$ is equal to the number of conjugacy classes of pairs $(Q,t)$ in $H$ such that $(\tilde{Q},\tilde{t})\simeq(1,1)$. Suppose $(Q,t)$ is a pair with $(\tilde{Q},\tilde{t})\simeq(1,1)$. Then we have $\tilde{Q}=1$ and $\tilde{t}=1$. So there exists a normal $p^{\prime}$-subgroup $N$ of $\langle Qt\rangle$ such that $(QN/N,tN)\simeq(1,1)$. Since $|Q|$ and $|N|$ are coprime, this implies that $Q=1$. We also have $t\in N$. But then $N\unlhd\langle t\rangle$ implies that $N=\langle t\rangle$. Therefore the number of conjugacy classes of pairs $(Q,t)$ in $H$ such that $(\tilde{Q},\tilde{t})\simeq(1,1)$ is equal to the number of conjugacy classes of $p^{\prime}$-elements in $H$. The result follows. ∎

Theorem 5.20**.**

The functor $S_{1,\mathbb{F}}$ is isomorphic to the functor that sends a finite group $H$ to the subspace $\mathbb{F}K_{0}(kH)$ of $\mathbb{F}T^{\Delta}(H)$ generated by the projective indecomposable $kH$-modules.

Proof.

Let $H$ be a finite group. We have

[TABLE]

where $J_{1,\mathbb{F}}(H)=\{\phi\in\mathbb{F}T^{\Delta}(H,1):\forall\psi\in\mathbb{F}T^{\Delta}(1,H),(\psi\circ\phi)\cdot 1=0\}$. Now $\mathbb{F}T^{\Delta}(H,1)$ is isomorphic to the subspace $\mathbb{F}K_{0}(kH)$ of $\mathbb{F}T(H)$ generated by the isomorphism classes of projective indecomposable $kH$-modules. Similarly any $W\in\mathbb{F}T^{\Delta}(1,H)$ can be identified with $W^{*}\in\mathbb{F}K_{0}(kH)$. As in [6] we have the following:

For any $p$-permutation $kH$-modules $V$ and $W$ we have

[TABLE]

Therefore $J_{1,\mathbb{F}}(H)$ is the right kernel of the bilinear form

[TABLE]

defined as $<W,V>:=\mathrm{dim}_{k}(\mathrm{Hom}_{kH}(W,V))$. But the matrix that represents this bilinear form is the Cartan matrix of $kH$. Since the Cartan matrix of a group algebra is non-degenerate, it follows that $J_{1,\mathbb{F}}(H)=0$. Therefore we have

[TABLE]

Note that both of these isomorphisms are functorial in $H$. The result follows. ∎