Proof. 1. This is clear, since P is the set of elements of G of order dividing the order of P.

2. The inclusion $Z(P)C_K(P)\leq C_G(P)$ is clear. Conversely, if $xt\in C_G(P)$ , where $x\in P$ and $t\in K$ , then $^ty=y^x$ for any $y\in P$ . It follows that $^{t^n}y=y^{x^n}$ for any $n\in \mathbb {N}$ and any $y\in P$ . Taking $n=|x|$ gives $t^n\in C_K(P)$ , hence $t\in C_K(P)$ since $(n,|t|)=1$ . Then $x\in C_P(P)=Z(P)$ .

3. It is clear that $(b)$ implies $(a)$ .Footnote ² For the converse, assume that $(a)$ holds, and that $C_G(P)=Z(P)$ . Let $x,y\in P$ and $s,t\in K$ . Then

$$ \begin{align*} \varphi(xs\cdot yt)&=\varphi(x\,{^sy}\cdot st)=x\,{^sy}\varphi(st)\\ &=\varphi(xs)\varphi(yt)=x\varphi(s)y\varphi(t)=x\,{^{\varphi(s)}y}\,\varphi(s)\varphi(t). \end{align*} $$

Hence, $^sy={^{\varphi (s)}y}$ for any $y\in P$ . In other words, $z(s):=s^{-1}\varphi (s)\in C_G(P)$ , so z is a map from K to $Z(P)$ .

Now $\varphi (s)=sz(s)$ , so $z(st)=z(s)^t\,z(t)$ for any $s,t\in K$ . In other words, the map z is a crossed morphism from K to $Z(P)$ . Since K and $Z(P)$ have coprime order, it follows that there exists $w\in Z(P)$ such that $z(s)=w^s\cdot w^{-1}$ , for any $s\in K$ . In other words $\varphi (s)=s\cdot w^s\cdot w^{-1}=wsw^{-1}=i_w(s)$ . Since $i_w(x)=x=\varphi (x)$ for any $x\in P$ , it follows that $\varphi =i_w$ .

Proof. 1. This is clear, since $(a,b)\in N$ if and only if $x^a=x^b$ , that is, $x^{ab^{-1}}=x$ , for all $x\in P$ .

2. The group $\overline {N}$ permutes the set $C_G(P)$ transitively, and the stabilizer in $\overline {N}$ of $1\in C_G(P)$ is the group $\{(a,a)\Delta (P)\mid a\in G\}=\Delta (G)/\Delta (P)$ . So $kC_G(P)\cong \mathrm {Ind}_{\Delta (G)/\Delta (P)}^{\overline {N}}k$ , and

$$ \begin{align*} \mathrm{Ind}_{N}^{G\times G}\mathrm{Inf}_{\overline{N}}^NkC_G(P)&\cong \mathrm{Ind}_{N}^{G\times G}\mathrm{Inf}_{\overline{N}}^N\mathrm{Ind}_{\Delta(G)/\Delta(P)}^{\overline{N}}k\\ &\cong \mathrm{Ind}_{N}^{G\times G}\mathrm{Ind}_{\Delta(G)}^N\mathrm{Inf}_{\Delta(G)/\Delta(P)}^{\Delta(G)}k\\ &\cong \mathrm{Ind}_{\Delta(G)}^{G\times G}k\cong kG \end{align*} $$

as $(kG,kG)$ -bimodules.

Proof. We apply Theorem 3.3 to the case $H=G\times G$ and $U=kGb$ , where b is a block idempotent of $kG$ . Then U is a diagonal p-permutation $(kG,kG)$ -bimodule with diagonal vertex $\Delta (D)=\Delta (D,\mathrm {Id},D)$ , where $D\leq G$ is a defect group of b. We can conclude that $[U]$ is a linear combination with integer coefficients of (isomorphism classes of) induced bimodules $\big [\mathrm {Ind}_{H_i}^{G\times G}V_i\big ]$ , where $H_i$ is a subgroup of $G\times G$ such that $D_i=O_p(H_i)\leq \Delta (D)$ and $H_i/O_p(H_i)$ is an elementary $p'$ -group. Let $G_i$ be the first projection of $H_i$ on G. Then the bimodule $\mathrm {Ind}_{H_i}^{G\times G}V_i$ factors as

$$ \begin{align*}\mathrm{Ind}_{H_i}^{G\times G}V_i\cong \mathrm{Ind}_{\Delta(G_i)}^{G\times G_i}\otimes_{kG_i}\mathrm{Ind}_{H_i}^{G_i\times G}V_i,\end{align*} $$

where $H_i$ on the right-hand side is viewed as a subgroup of $G_i\times G$ . Now if $G_i<G$ , then the image of $kGb$ in $\mathcal {E}_{\mathsf {R}}(G)$ is equal to 0. And if $G_i=G$ , then G is a quotient of $H_i$ , so $G/O_p(G)$ is an elementary $p'$ -group. In other words, $G\cong P\rtimes K$ , where P is a p-group, and K is an elementary $p'$ -group.

Now $\mathcal {E}_{\mathsf {R}}(G)$ is non-zero if and only if its identity element is non-zero, that is, if the image of the bimodule $kG$ in $\mathcal {E}_{\mathsf {R}}(G)$ is non-zero. Since $kG$ is the direct sum of the bimodules $kGb$ , when b runs through block idempotents of $kG$ , there is at least one such idempotent b such that the image of $kGb$ in $\mathcal {E}_{\mathsf {R}}(G)$ is non-zero. Hence, $G\cong P\rtimes K$ , where P is a p-group and K is an elementary $p'$ -group.

Proof. 1. If $\mathcal {E}_{\mathsf {R}}(H)=0$ , the identity $(kH,kH)$ -bimodule $kH$ factors through groups of order strictly smaller than $|H|$ , so the same holds for U (by right composition with $kH$ ). Hence, $\mathcal {E}_{\mathsf {R}}(H)\neq 0$ , and Assertion 1 follows from Corollary 3.4.

2. From Assertion 1 follows in particular that the group $G\times H$ is solvable, with a normal Sylow p-subgroup $P\times Q$ . Then there is a diagonal p-subgroup $\Delta _\pi (S)$ of $G\times H$ , where S is a p-subgroup of H (i.e., a subgroup of Q) and $\pi :S\hookrightarrow P$ is an injective group homomorphism, such that

$$ \begin{align*}U\cong\mathrm{Ind}_N^{G\times H}\mathrm{Inf}_{\overline{N}}^NE,\end{align*} $$

where $N=N_{G\times H}\big (\Delta _\pi (S)\big )$ and $\overline {N}=N/\Delta _\pi (S)$ , and E is an indecomposable projective $k\overline {N}$ -module.

Now $\overline {N}$ itself also has a normal Sylow p-subgroup X, and there is a $p'$ -subgroup $\overline {T}$ of $\overline {N}$ such that $\overline {N}=X\rtimes \overline {T}$ . Moreover, $\overline {T}$ lifts to a $p'$ -subgroup T of N, that we can assume contained in the $p'$ -Hall subgroup $K\times L$ of $G\times H$ , up to replacing $\Delta _\pi (S)$ by a conjugate subgroup. Finally, $E\cong \mathrm {Ind}_{\overline {T}}^{\overline {N}}\overline {W}$ , where $\overline {W}$ is a simple $k\overline {T}$ -module. Let W be the simple $kT$ -module corresponding to $\overline {W}$ via the isomorphism $\overline {T}\cong T$ . It follows that

$$ \begin{align*} U&\cong \mathrm{Ind}_N^{G\times H}\mathrm{Inf}_{\overline{N}}^N\mathrm{Ind}_{\overline{T}}^{\overline{N}}\overline{W}\\ &\cong \mathrm{Ind}_N^{G\times H}\mathrm{Ind}_{\Delta_\pi(S)\cdot T}^N\mathrm{Inf}_{T}^{\Delta_\pi(S)\cdot T}W\\ &\cong\mathrm{Ind}_{\Delta_\pi(S)\cdot T}^{G\times H}\mathrm{Inf}_{T}^{\Delta_\pi(S)\cdot T}W. \end{align*} $$

Since U is right essential, we have $p_2\big (\Delta _\pi (S)\cdot T\big )=H=Q\cdot L$ . This forces $S=Q$ , and $p_2(T)=L$ , proving Assertion 2.

Proof. Let M be an essential indecomposable diagonal p-permutation $(kG,kG)$ -bimodule. By Lemma 3.5, applied to $H=G$ and $U=M$ , we know that

$$ \begin{align*}M\cong \mathrm{Ind}_{\Delta_\pi(P)\cdot T}^{G\times G}\mathrm{Inf}_{T}^{\Delta_\varphi(P)\cdot T}W,\end{align*} $$

for some $\pi \in \mathrm {Aut}(P)$ , some subgroup T of $N_{K\times K}\big (\Delta _\pi (P)\big )$ with $p_2(T)=K$ , and some simple $kT$ -module W.

Now $T\leq K\times K$ , so T is a $p'$ -group, and $|T|$ divides $|K|^2$ . Moreover, by Artin’s induction theorem, in $R_k(T)\cong R_{\mathbb {C}}(T)$ , we have an equality of the form

$$ \begin{align*}|T|W=\sum_{i=1}^nn_i\mathrm{Ind}_{C_i}^Tk_{\lambda_i},\end{align*} $$

where, for $1\leq i\leq n$ , $C_i$ is a cyclic subgroup of T, $n_i$ is an integer, and $k_{\lambda _i}$ is a one-dimensional $kC_i$ -module. Hence, in $\mathsf {R} T^\Delta (G,G)$ , we have that

$$ \begin{align*} |T|M&=\sum_{i=1}^nn_i\mathrm{Ind}_{\Delta_\pi(P)\cdot T}^{G\times G}\mathrm{Inf}_T^{\Delta_\pi(P)\cdot T}\mathrm{Ind}_{C_i}^Tk_{\lambda_i}\\ &=\sum_{i=1}^nn_i\mathrm{Ind}_{\Delta_\pi(P)\cdot T}^{G\times G}\mathrm{Ind}_{\Delta_\pi(P)\cdot C_i}^{\Delta_\pi(P)\cdot T}\mathrm{Inf}_{C_i}^{\Delta_\pi(P)\cdot C_i}k_{\lambda_i}\\ &=\sum_{i=1}^nn_i\mathrm{Ind}^{G\times G}_{\Delta_\pi(P)\cdot C_i}\mathrm{Inf}_{C_i}^{\Delta_\pi(P)\cdot C_i}k_{\lambda_i}. \end{align*} $$

The image of $|K|^2\,M$ in $\mathcal {E}_{\mathsf {R}}(G)$ is equal to the image of

$$ \begin{align*}\frac{|K|^2}{|T|}\sum_{i=1}^nn_i\mathrm{Ind}^{G\times G}_{\Delta_\pi(P)\cdot C_i}\mathrm{Inf}_{C_i}^{\Delta_\pi(P)\cdot C_i}k_{\lambda_i},\end{align*} $$

which is equal to zero unless there exists $i\in \{1,\ldots ,n\}$ such that

$$ \begin{align*}p_1\big(\Delta_\pi(P)\cdot C_i\big)=p_2\big(\Delta_\pi(P)\cdot C_i\big)=G=P\cdot K.\end{align*} $$

This implies that K is a quotient of $C_i$ . Hence, K is cyclic, which completes the proof.

Proof. Footnote ³ Since G has a normal Sylow p-subgroup, all the blocks of G have defect group P. Moreover, if b is a block idempotent of G, then b is a linear combination of p-regular elements of $C_G(P)=Z(P)\times C_K(P)$ , so $b\in kC_K(P)$ .

Since $C_K(P)\leq Z(G)$ , it means that the block idempotents of $kG$ are exactly the primitive idempotents of the (split semisimple commutative) algebra $kC_K(P)$ . Let e be one of them, and $k_\lambda =kC_K(P)e$ be the corresponding (one-dimensional) simple $kC_K(P)$ -module, where $\lambda :C_K(P)\to k^\times $ is the associated group homomorphism.

Let $\varpi :G\to K$ denote the projection map. Set $N=N_{G\times G}\big (\Delta (P)\big )$ and $\overline {N}=N/\Delta (P)$ . There is a short exact sequence

where

• • $\widetilde {K}=\{(a,b)\in K\times K\mid a^{-1}b\in C_K(P)\}$ .

• • i is the map sending $z\in Z(P)$ to $(z,1)\Delta (P)\in \overline {N}$ .

• • s is the map sending $(a,b)\Delta (P)$ to $\big (\varpi (a),\varpi (b)\big )$ .

So $\overline {N}\cong Z(P)\rtimes \widetilde {K}$ , with the explicit embedding $\widetilde {K}\to \overline {N}$ sending $(a,b)\in \widetilde {K}$ to $(a,b)\Delta (P)\in \overline {N}$ . We consider $\widetilde {K}$ as a subgroup of $\overline {N}$ via this embedding.

The Brauer quotient of the $(kG,kG)$ -bimodule $kG$ at $\Delta (P)$ is isomorphic to $kC_G(P)$ , so $kGe[\Delta (P)]\cong kC_G(P)\mathrm {Br}_P(e)=kC_G(P)e=kZ(P)\otimes _k k_\lambda $ , since $\mathrm {Br}_P(e)=e$ as $e\in kC_K(P)$ . It follows that

$$ \begin{align*}kGe\cong \mathrm{Ind}_N^{G\times G}\mathrm{Inf}_{\overline{N}}^NkC_G(P)e\cong \mathrm{Ind}_N^{G\times G}\mathrm{Inf}_{\overline{N}}^N \big(kZ(P)\otimes_k k_\lambda\big),\end{align*} $$

where $\overline {N}\cong Z(P)\rtimes \widetilde {K}$ acts on $kZ(P)\otimes _k k_\lambda $ by

$$ \begin{align*}(a,b)\Delta(P).(z\otimes 1)=(ab^{-1})_p\,z\otimes (ab^{-1})_{p'}\cdot1=\lambda\big((ab^{-1})_{p'}\big)(ab^{-1})_p\,z\otimes 1\end{align*} $$

for $(a,b)\in N$ and $z\in Z(P)$ , where $(ab^{-1})_p\in Z(P)$ and $(ab^{-1})_{p'}\in C_K(P)$ are the p-part and $p'$ -part of $ab^{-1}\in C_G(P)=Z(P)\times C_K(P)$ , respectively. Then $kZ(P)\otimes _kk_\lambda $ is isomorphic to $\mathrm {Ind}_{\widetilde {K}}^{\overline {N}}\,k_{\widetilde {\lambda }}$ , where $\widetilde {\lambda }:\widetilde {K}\to k^\times $ sends $(a,b)\in \widetilde {K}$ to $\lambda (ab^{-1})\in k^\times $ . So

(3.7.1) $$ \begin{align} kGe&\cong\mathrm{Ind}_{N}^{G\times G}\mathrm{Inf}_{\overline{N}}^{N}\mathrm{Ind}_{\widetilde{K}}^{\overline{N}}k_{\widetilde{\lambda}}\nonumber\\ &\cong \mathrm{Ind}_{N}^{G\times G}\mathrm{Ind}_{\Delta(P)\widetilde{K}}^{N}\mathrm{Inf}_{\widetilde{K}}^{\Delta(P)\widetilde{K}}k_{\widetilde{\lambda}}\nonumber\\ &\cong \mathrm{Ind}^{G\times G}_{\Delta(P)\widetilde{K}}\mathrm{Inf}_{\widetilde{K}}^{\Delta(P)\widetilde{K}}k_{\widetilde{\lambda}}. \end{align} $$

Now we set $\overline {G}=G/C_K(P)$ . We denote by $g\mapsto \overline {g}$ the projection map, and by $\delta :G\to G\times \overline {G}$ the map $g\mapsto (g,\overline {g})$ . We will show that the $(kG,kG)$ -bimodule $kGe$ factors through the group $\overline {G}$ , that is, there is a diagonal p-permutation $(kG,k\overline {G})$ -bimodule U and a diagonal p-permutation $(k\overline {G},kG)$ -bimodule V such that $kGe\cong U\otimes _{k\overline {G}}V$ .

The group P embeds in $G\times \overline {G}$ via $\delta $ . Its image $\delta (P)$ is a diagonal subgroup of $G\times \overline {G}$ , and its normalizer is

$$ \begin{align*}N_\delta:=N_{G\times\overline{G}}\big(\delta(P)\big)=\big\{(a,\overline{b})\in G\times \overline{G}\mid \overline{x^a}=\overline{x^b},\;\forall x\in P\big\}.\end{align*} $$

In other words, $(a,\overline {b})\in N_\delta $ if and only if $\overline {x^{ab^{-1}}}=\overline {x}$ for all $x\in P$ , or equivalently, if the commutator $[x,ab^{-1}]$ is in $C_K(P)$ . But since $P\trianglelefteq G$ , we have that $[P,ab^{-1}]\subseteq P$ . Hence, $(a,\overline {b})$ normalizes $\delta (P)$ if and only if $[P,ab^{-1}]\subseteq P\cap C_K(P)=1$ , that is, $ab^{-1}\in C_G(P)$ . Thus,

$$ \begin{align*}N_\delta=\{(a,\overline{b})\in G\times \overline{G}\mid ab^{-1}\in Z(P)\times C_K(P)\}.\end{align*} $$

Recall that $\varpi :G\to K$ denotes the projection map. We have a surjective group homomorphism $\sigma :N_\delta \to K$ sending $(a,\overline {b})$ to $\varpi (a)$ . It induces a surjective group homomorphism

$$ \begin{align*}\overline{\sigma}:\overline{N}_\delta:=N_\delta/\delta(P)\to K\end{align*} $$

sending $(a,\overline {b})\delta (P)$ to $\varpi (a)$ . The kernel of this morphism consists of the elements $(a,\overline {b})\delta (P)$ such that $a\in P$ and $ab^{-1}\in Z(P)\times C_K(P)$ . Since

$$ \begin{align*}(a,\overline{b})\delta(P)=(1,\overline{ba^{-1}})(a,\overline{a})\delta(P)=(1,\overline{ba^{-1}})\delta(P),\end{align*} $$

and since $\overline {ba^{-1}}\in C_G(P)/C_K(P)=Z(P)$ , we get a short exact sequence

where $\iota (z)=(1,\overline {z})\delta (P)$ for $z\in Z(P)$ . This sequence is split, via the morphism $a\in K\mapsto (a,\overline {a})\delta (P)$ , for $a\in K$ , so $\overline {N}_\delta \cong Z(P)\rtimes K$ .

Now since K is cyclic, we can extend $\lambda :C_K(P)\to k^\times $ to a group homomorphism $\beta :K\to k^\times $ . This gives a one-dimensional $kK$ -module $k_\beta $ , that we can induce to $\overline {N}_\delta =Z(P)\rtimes K$ . We get a projective $k\overline {N}_\delta $ -module, and we set

$$ \begin{align*}U:=\mathrm{Ind}_{N_\delta}^{G\times \overline{G}}\mathrm{Inf}_{\overline{N}_\delta}^{N_\delta}\mathrm{Ind}_{K}^{\overline{N}_\delta}k_\beta.\end{align*} $$

This is a diagonal p-permutation $(kG,k\overline {G})$ -bimodule, and

$$ \begin{align*}U\cong \mathrm{Ind}_{N_\delta}^{G\times \overline{G}}\mathrm{Ind}_{\delta(P)K}^{N_\delta}\mathrm{Inf}_{K}^{\delta(P)K}k_\beta\cong \mathrm{Ind}_{\delta(P)K}^{G\times\overline{G}}\mathrm{Inf}_{K}^{\delta(P)K}k_\beta,\end{align*} $$

where K is viewed as a subgroup of $N_\delta $ via the map $a\in K\mapsto (a,\overline {a})\in N_\delta $ . We observe that $\delta (P)K$ is equal to $\delta (G)$ , so

$$ \begin{align*}U\cong \mathrm{Ind}_{\delta(G)}^{G\times\overline{G}}\mathrm{Inf}_{K}^{\delta(G)}k_\beta.\end{align*} $$

We define similarly a $(k\overline {G},kG)$ -bimodule V by

$$ \begin{align*}V:=\mathrm{Ind}_{\delta^o(G)}^{\overline{G}\times G}\mathrm{Inf}_{K^o}^{\delta^o(G)}k_{\beta^{-1}},\end{align*} $$

where $\delta ^o:G\to \overline {G}\times G$ sends x to $(\overline {x},x)$ , and $K^o=\{(\overline {a},a)\mid a\in K\}$ .

Now we compute the tensor product $U\otimes _{k\overline {G}}V$ using Theorem 1.1 of [Reference Bouc3]. Since $p_2\big (\delta (G)\big )=\overline {G}$ , there is a single double coset $p_2\big (\delta (G)\big )\backslash \overline {G}/p_1\big (\delta ^o(G)\big )$ . Moreover, $k_2\big (\delta (G)\big )=1$ , so we have

(3.7.2) $$ \begin{align} U\otimes_{k\overline{G}}V\cong\mathrm{Ind}_{\delta(G)*\delta^o(G)}^{G\times G}\big(\mathrm{Inf}_{K}^{\delta(G)}k_\beta\otimes_k\mathrm{Inf}_{K^o}^{\delta^o(G)}k_{\beta^{-1}}\big), \end{align} $$

where

$$ \begin{align*} \delta(G)*\delta^o(G)&=\big\{(a,b)\in G\times G\mid \exists \overline{c}\in\overline{G},\;(a,\overline{c})\in \delta(G)\,\text{and}\,(\overline{c},b)\in \delta^o(G)\big\}\\ &=\big\{(a,b)\in G\times G\mid \overline{a}=\overline{b}\big\}\\ &=\big\{(a,b)\in G\times G\mid ab^{-1}\in C_K(P)\big\}. \end{align*} $$

Now if $a=xu$ and $b=yv$ , with $x,y\in P$ and $u,v\in K$ , we have

$$ \begin{align*}ab^{-1}=xuv^{-1}y^{-1}=x\cdot{^{uv^{-1}}(y^{-1})}\cdot uv^{-1},\end{align*} $$

so $ab^{-1}\in C_K(P)$ if and only if $uv^{-1}\in C_K(P)$ and $x=y$ , that is, $(u,v)\in \widetilde {K}$ and $(x,y)\in \Delta (P)$ . It follows that $\delta (G)*\delta ^o(G)=\Delta (P)\widetilde {K}$ .

The action of $(a,b)\in \delta (G)*\delta ^o(G)$ on the tensor product

$$ \begin{align*}T:=\mathrm{Inf}_{K}^{\delta(G)}k_\beta\otimes_k\mathrm{Inf}_{K^o}^{\delta^o(G)}k_{\beta^{-1}}\end{align*} $$

is obtained as follows: Let $\overline {c}\in \overline {G}$ such that $(a,\overline {c})\in \delta (G)$ and $(\overline {c},b)\in \delta ^o(G)$ , that is, $\overline {c}=\overline {a}=\overline {b}$ . Then for $v\in \mathrm {Inf}_{K}^{\delta (G)}k_\beta $ and $w\in \mathrm {Inf}_{K^o}^{\delta ^o(G)}k_{\beta ^{-1}}$ , we have $(a,b)\cdot (v\otimes w)=(a,\overline {c})\cdot v\otimes (\overline {c},b)\cdot w$ . Here, T is one-dimensional, with basis $1\otimes 1$ , and

$$ \begin{align*} (a,b)\cdot(1\otimes 1)&=\beta\big(\varpi(a)\big)\beta\big(\varpi(b)\big)^{-1}(1\otimes 1)\\ &=\beta\big(\varpi(ab^{-1})\big)(1\otimes 1)\\ &=\lambda\big(\varpi(ab^{-1})\big)(1\otimes 1), \end{align*} $$

since the restriction of $\beta $ to $C_K(P)$ is equal to $\lambda $ .

Now for $(a,b)\in \delta (G)*\delta ^o(G)=\Delta (P)\widetilde {K}$ , we have $\lambda \big (\varpi (ab^{-1})\big )=\widetilde {\lambda }\big (\rho (a,b)\big )$ , where $\rho :\Delta (P)\widetilde {K}\to \widetilde {K}$ is the projection map. Then $T\cong \mathrm {Inf}_{\widetilde {K}}^{\Delta (P)\widetilde {K}}k_{\widetilde {\lambda }}$ , and by (3.7.1) and (3.7.2), we get that

$$ \begin{align*}U\otimes_{k\overline{G}}V\cong \mathrm{Ind}_{\delta(G)*\delta^o(G)}^{G\times G}T\cong \mathrm{Ind}_{\Delta(P)\widetilde{K}}^{G\times G}\mathrm{Inf}_{\widetilde{K}}^{\Delta(P)\widetilde{K}}k_{\widetilde{\lambda}}\cong kGe,\end{align*} $$

as was to be shown.

So if $C_K(P)\neq 1$ , all the bimodules $kGe$ are mapped to 0 in $\mathcal {E}_{\mathsf {R}}(G)$ , and the image of their direct sum $kG$ is also 0. Now as its identity element is equal to 0, the algebra $\mathcal {E}_{\mathsf {R}}(G)$ itself is equal to 0.

Proof. Indeed if $\mathcal {E}_{\mathsf {R}}(G)\neq 0$ , we know by Theorem 3.6 that K is cyclic, and by Theorem 3.7, that K acts faithfully on P.

Notation 3.10. For a $D^\Delta $ -pair $(L,u)$ , we denote by $\mathrm {Aut}(L,u)$ the group of automorphisms of the semidirect product $L\langle u\rangle =L\rtimes \langle u\rangle $ which send u to a conjugate of u, and by $\mathrm {Out}(L,u)$ the quotient $\mathrm {Aut}(L,u)/\mathrm {Inn}(L\langle u\rangle )$ of this group by the group of inner automorphisms of $L\langle u\rangle $ .

Notation 4.1. For a finite group G, denote by $G^\natural $ the group $\mathrm {Hom}(G,k^\times )$ . For $\lambda \in G^\natural $ , let $k_\lambda $ denote the corresponding one-dimensional $kG$ -module. For $\gamma \in \mathrm {Aut}(G)$ , denoteFootnote ⁵ by $kG_{\gamma ,\lambda }$ the $(kG,kG)$ -bimodule equal to $kG$ as a vector space, with action given by

$$ \begin{align*}\forall (x,y,z)\in G^3,\;x\cdot z\cdot y=\lambda(y)^{-1}xz\gamma(y).\end{align*} $$

Proof. 1. Let S be a Sylow p-subgroup of G. Then $\lambda (x)=1$ for any $x\in S$ , so the restriction of $kG_{\gamma ,\lambda }$ to the Sylow p-subgroup $S\times S$ of $G\times G$ is the permutation bimodule $kG$ , with action $x\cdot z\cdot y=xz\gamma (y)$ , for $x,y\in S$ and $z\in G$ . Moreover, this action is free on both sides, so $kG_{\gamma ,\lambda }$ is a diagonal p-permutation bimodule. Finally, the map $g\in G\mapsto (g,1)\Delta _\gamma (G)$ is a bijection from G to $(G\times G)/\Delta _\gamma (G)$ , and using this bijection, it is easy to check that $kG_{\gamma ,\lambda }\cong \mathrm {Ind}_{\Delta _\gamma (G)}^{G\times G}k_{\lambda }$ .

2. The map $(g\otimes g')\mapsto \lambda (g')^{-1}g\gamma (g')$ , for $g,g'\in G$ , from $kG_{\gamma ,\lambda }\otimes _{kG}kG_{\delta ,\mu }$ to $kG_{\gamma \circ \delta ,(\lambda \circ \delta )\times \mu }$ induces a well-defined isomorphism of $(kG,kG)$ -bimodules.

3. It is clear that for any $(kG,kG)$ -bimodule M, any $\delta \in \mathrm {Aut}(G)$ , and any $\mu \in G^\natural $ , the k-vector space $M\otimes _{kG} kG_{\delta ,\mu }$ is isomorphic to M. So if $kG_{\gamma ,\lambda }$ splits as a direct sum of non-zero $(kG,kG)$ -bimodules M and $M'$ , the tensor product $kG_{\gamma ,\lambda }\otimes _{kG} kG_{\delta ,\mu }$ splits as the direct sum of $(kG,kG)$ -bimodules $M\otimes kG_{\delta ,\mu }$ and $M'\otimes kG_{\delta ,\mu }$ , none of which is equal to zero. Then $kG_{\gamma \circ \delta ,(\lambda \circ \delta )\times \mu }$ splits as a direct sum of non-zero bimodules. Taking $\delta =\gamma ^{-1}$ and $\mu =(\lambda \circ \gamma ^{-1})^{-1}$ , we get that the $(kG,kG)$ -bimodule $kG_{\mathrm {Id},1}\cong kG$ splits non-trivially. So $kG$ is not indecomposable, that is, $kG$ has more than one block.

Hypothesis 4.3. The group G is of the form $P\rtimes K$ , where P is a p-group and K is a cyclic $p'$ -group of order invertible in $\mathsf {R}$ , acting faithfully on P.

Proof. 1. The p-subgroup P of G is normal, and $C_G(P)=Z(P)\times C_K(P)=Z(P)\leq P$ since $C_K(P)=1$ as K acts faithfully on P. So $kG$ has only one block [Reference Benson1, Proposition 6.2.2]. Then all the bimodules $kG_{\gamma ,\lambda }$ are indecomposable by Lemma 4.2.

Let $\gamma \in \mathrm {Aut}(G)$ . Set $N:=N_{G\times G}\big (\Delta _\gamma (P)\big )$ and $\overline {N}=N/\Delta _\gamma (P)$ . Then

$$ \begin{align*}N=\{(a,b)\in G\times G\mid a^{-1}\gamma(b)\in C_G(P)=Z(P)\},\end{align*} $$

so the second projection $p_2$ induces a short exact sequence

which is split (by the map $x\in K\mapsto \big (\gamma (x),x\big )\Delta _\gamma (P)$ ). Now the indecomposable projective $k\overline {N}$ -modules are the modules $\mathrm {Ind}_{K}^{\overline {N}}k_\lambda $ , for $\lambda \in K^\natural $ . Moreover,

$$ \begin{align*} \mathrm{Ind}_{N}^{G\times G}\mathrm{Inf}_{\overline{N}}^N\mathrm{Ind}_{K}^{\overline{N}}k_\lambda&\cong \mathrm{Ind}_{N}^{G\times G}\mathrm{Ind}_{\Delta_\gamma(P)\cdot\Delta_\gamma(K)}^Nk_\lambda\\ &\cong\mathrm{Ind}_{\Delta_\gamma(G)}^{G\times G}k_\lambda\cong kG_{\gamma,\lambda} \end{align*} $$

by Lemma 4.2. This gives another proof that $kG_{\gamma ,\lambda }$ is indecomposable and also shows that it has vertex $\Delta _\gamma (P)$ and Brauer quotient

(4.4.1) $$ \begin{align} kG_{\gamma,\lambda}[\Delta_\gamma(P)]\cong \mathrm{Ind}_K^{\overline{N}}k_\lambda\cong kZ(P)\otimes_kk_{\lambda}. \end{align} $$

2. First, if $\lambda =\mu $ and $\delta =i_x\circ \gamma $ , where $i_x$ is conjugation by $x\in G$ , then one checks easily that the map $g\in G\mapsto gx\in G$ induces a bimodule isomorphism $kG_{\gamma ,\lambda }\cong kG_{\delta ,\mu }$ .

For the converse, let $\gamma '=\gamma ^{-1}$ and $\lambda '=(\lambda \circ \gamma ')^{-1}$ . Then by Lemma 4.2, we have an isomorphism of $(kG,kG)$ -bimodules

$$ \begin{align*}kG_{\gamma,\lambda}\otimes_{kG}kG_{\gamma',\lambda'}\cong kG_{\gamma\circ\gamma',(\lambda\circ \gamma')\times \lambda'}=kG_{\mathrm{Id},1}=kG.\end{align*} $$

So if $kG_{\delta ,\mu }\cong kG_{\gamma ,\lambda }$ , we have an isomorphism of $(kG,kG)$ -bimodules

$$ \begin{align*}kG_{\delta,\mu}\otimes_{kG}kG_{\gamma',\lambda'}\cong kG,\end{align*} $$

that is,

$$ \begin{align*}kG_{\delta\circ\gamma',(\mu\circ\gamma')\times \lambda'}\cong kG_{\mathrm{Id},1}.\end{align*} $$

So if we know that an isomorphism of bimodules $kG_{\theta ,\rho }\cong kG_{\mathrm {Id},1}$ , where $\theta \in \mathrm {Aut}(G)$ and $\rho \in G^\natural $ implies that $\theta $ is inner and $\rho =1$ , we are done, since we can conclude that $\delta \circ \gamma '=\delta \circ \gamma ^{-1}$ is inner, and that $(\mu \circ \gamma ')\times \lambda '=1$ , that is, $\mu =\lambda $ . In other words, we can assume $\gamma =\mathrm {Id}$ and $\lambda =1$ , and that $kG_{\delta ,\mu }\cong kG$ .

Now if $kG_{\delta ,\mu }$ is isomorphic to $kG$ , then its vertex $\Delta _\delta (P)$ is contained in—hence equal to - $\Delta (P)$ , up to conjugation in $G\times G$ . It means that the restriction of $\delta $ to P is equal to the conjugation by some element of G. Up to composing $\delta $ with some inner automorphism of G, we can assume that this restriction is equal to the identity, and then $\delta $ is equal to the conjugation by some element of $Z(P)$ , by Lemma 3.1. This shows that $\delta $ is inner.

Then we have a bimodule isomorphism $kG_{\delta ,\lambda }\cong kG_{\mathrm {Id},\lambda }$ by the first remark in the proof of Assertion 2. In other words, we can assume $\delta =\mathrm {Id}$ and $kG_{\mathrm {Id},\lambda }\cong kG_{\mathrm {Id},1}$ . Then the Brauer quotients at $\Delta (P)$ of these bimodules are isomorphic. Hence, $kZ(P)\otimes k_\lambda \cong kZ(P)\otimes k\cong kZ(P)$ . Now the fixed points of $Z(P)$ on $kZ(P)\otimes k_\lambda $ form a $kK$ -module isomorphic to $k_\lambda $ , so $k_\lambda \cong k$ as K-module, hence $\lambda =1$ , as was to be shown.

3. Suppose conversely that M is an indecomposable diagonal p-permutation $(kG,kG)$ -bimodule with vertex $\Delta _\gamma (P)$ , where $\gamma \in \mathrm {Aut}(G)$ . Then $M[\Delta _\gamma (P)]$ is an indecomposable projective $k\overline {N}$ -module of the form $\mathrm {Ind}_K^{\overline {N}}k_\lambda $ for some $\lambda \in K^\natural $ , and then

$$ \begin{align*}M\cong\mathrm{Ind}_N^{G\times G}\mathrm{Inf}_{\overline{N}}^N\mathrm{Ind}_K^{\overline{N}}k_\lambda\cong kG_{\gamma,\lambda}.\end{align*} $$

4. As in the proof of Theorem 3.6, we apply Lemma 3.5, in the case $H=G$ and $U=M$ . We know that

$$ \begin{align*}M\cong \mathrm{Ind}_{\Delta_\pi(P)\cdot T}^{G\times G}\mathrm{Inf}_{T}^{\Delta_\varphi(P)\cdot T}W,\end{align*} $$

for some $\pi \in \mathrm {Aut}(P)$ , some subgroup T of $N_{K\times K}\big (\Delta _\pi (P)\big )$ with $p_2(T)=K$ , and some simple $kT$ -module W.

Moreover, $k_1(T)\leq k_1\Big (N_{G\times G}\big (\Delta _\pi (P)\big )\Big )=C_G(P)=Z(P)$ . So $k_1(T)=1$ since T is a $p'$ -group. Then $p_2:T\to K$ is an isomorphism, with inverse $\theta $ , and $T=\Delta _\theta (K)$ . Then T is cyclic, and $W\cong k_\lambda $ for some $\lambda \in T^\natural $ .

Now T normalizes $\Delta _\pi (P)$ if and only if $^{\theta (x)}\pi (y)=\pi (^xy)$ for any $x\in K$ and any $y\in P$ . Then the map $\gamma :y\cdot x\mapsto \pi (y)\cdot \theta (x)$ , where $y\in P$ and $x\in K$ , is an automorphism of G, such that $\gamma (K)=K$ and $\Delta _\pi (P)\cdot L=\Delta _\gamma (G)$ . Then

$$ \begin{align*}M\cong\mathrm{Ind}_{\Delta_\gamma(G)}^{G\times G}k_\mu,\end{align*} $$

where $\mu =\lambda \circ \varpi \in G^\natural \cong \Delta _\gamma (G)^\natural $ . Now $M\cong kG_{\gamma ,\mu }$ by Lemma 4.2.

Proof. We know from Lemma 3.5 that $H=Q\rtimes L$ , where Q is a p-group with an embedding $\pi :Q\hookrightarrow P$ , and L is an elementary $p'$ -group. Moreover, there is a subgroup T of $N_{K\times L}\big (\Delta _\pi (Q)\big )$ with $p_2(T)=L$ , and a simple $kT$ -module W such that

$$ \begin{align*}U\cong\mathrm{Ind}_{\Delta_\pi(Q)\cdot T}^{G\times H}\mathrm{Inf}_{T}^{\Delta_\pi(Q)\cdot T}W.\end{align*} $$

Then a vertex of U is contained in $\Delta _\pi (Q)\cdot T$ up to conjugation, so it has order at most $|Q|\leq |P|$ . And if it has order $|P|$ , the embedding $\pi :Q\hookrightarrow P$ is an isomorphism. Moreover, $k_1(T)\leq k_1\Big (N_{G\times H}\big (\Delta _\pi (P)\big )\Big )=C_G(P)$ , so $k_1(T)\leq C_K(P)=1$ since K acts faithfully on P. Then the projection map $p_2:T\to L$ is an isomorphism, and then $T=\Delta _\tau (L)$ , for some injective group homomorphism $\tau :L\hookrightarrow K$ . In particular, L and T are cyclic.

Moreover, since $T=\Delta _\tau (L)\leq N_{G\times H}\big (\Delta _\pi (P)\big )$ , we have $^{\tau (l)}\pi (x)=\pi ({^lx})$ for any $l\in L$ and $x\in Q$ . Then the map $\sigma :H=Q\cdot L\to G$ sending $x\cdot l$ , for $x\in Q$ and $l\in L$ , to $\pi (x)\cdot \tau (l)$ , is an injective group homomorphism, such that $P\leq \gamma (H)\leq G$ . Moreover, $\Delta _\pi (Q)\cdot T=\Delta _\sigma (H)$ . Finally, T is cyclic, so there is $\lambda \in T^\natural =L^\natural \cong H^\natural $ such that $W=k_{\lambda }$ , and $U\cong \mathrm {Ind}_{\Delta _\sigma (H)}^{G\times H}k_\lambda $ , completing the proof.

Proof. 1. Let X be a vertex of U. Then X is a diagonal subgroup of $P\times H$ , so $|X|\leq |P|$ , and U is a direct summand of $\mathrm {Ind}_{X}^{G\times H}k$ . Similarly, if Y is a vertex of V, then Y is a diagonal subgroup of $H\times P$ , hence $|Y|\leq |P|$ , and V is a direct summand of $\mathrm {Ind}_{Y}^{H\times G}k$ . It follows that $U\otimes _{kH}V$ is a direct summand of

$$ \begin{align*}\bigoplus_{h\in p_2(X)\backslash H/p_1(Y)}\mathrm{Ind}_{X*{^{(h,1)}Y}}^{G\times G}k.\end{align*} $$

So the vertices of the indecomposable summands of $U\otimes _{kH}V$ are contained (up to conjugation) in some group $X*{^{(h,1)}Y}$ , which has order at most $\min (|X|,|Y|)$ . If one of them has order $|P|$ , then $|P|\leq |X|\leq |P|$ , hence $|X|=|P|$ , and $|P|\leq |Y|\leq |P|$ , so $|Y|=|P|$ .

By Lemma 4.5, it follows that there is an embedding $\sigma :H\hookrightarrow G$ with $P\leq \sigma (H)\leq G$ , and $\lambda \in H^\natural $ , such that $U\cong \mathrm {Ind}_{\Delta _\sigma (H)}^{G\times H}k_\lambda $ . Similarly, swapping G and H, there is an embedding $\tau :H\to G$ with $P\leq \tau (H)\leq G$ , and $\mu \in H^\natural $ , such that $V\cong \mathrm {Ind}_{\Delta _\tau ^o(H)}^{H\times G}k_\mu $ , where $\Delta _\tau ^o(H)=\Big \{\big (h,\tau (h)\big )\mid h\in H\Big \}$ . Then

$$ \begin{align*}U\otimes_{kH}V\cong \mathrm{Ind}_{\Delta_\sigma(H)*\Delta_\tau^o(H)}^{G\times G}(k_\lambda\otimes_kk_\mu).\end{align*} $$

Now $\sigma (H)/P$ and $\tau (H)/P$ are subgroups of the same order of the cyclic group K, so $\sigma (H)=\tau (H)$ . Set $I:=\tau (H)$ . There is a unique automorphism $\psi $ of I such that $\psi \big (\tau (h)\big )=\sigma (h)$ for all $h\in H$ . Then

$$ \begin{align*}\Delta_\sigma(H)*\Delta_\tau^o(H)=\Big\{\big(\sigma(h),\tau(h)\big)\mid h\in H\Big\}=\big\{(\psi(x),x)\mid x\in I\big\}=\Delta_\psi(I).\end{align*} $$

Moreover, $k_\lambda \otimes k_\mu $ is one-dimensional, so there is a unique $\zeta \in I^\natural $ such that $k_\lambda \otimes k_\mu \cong k_\zeta $ as $kI$ -modules, defined by $\zeta (x)=(\lambda \mu )\big (\tau ^{-1}(x)\big )$ for $x\in I$ . This completes the proof of Assertion 1.

2. By Lemma 4.4, an essential diagonal p-permutation bimodule M is isomorphic to $kG_{\gamma ,\lambda }$ for some $\gamma \in \mathrm {Aut}(G)$ and $\lambda \in G^\natural $ . Then M has vertex $\Delta _\gamma (P)$ , of order $|P|$ , so the conclusion of Assertion 1 holds. Hence, there is a subgroup I of G, containing P, an automorphism $\psi $ of I, and $\zeta \in I^\natural $ , such that

$$ \begin{align*}U\otimes_{kH} V\cong\mathrm{Ind}_{\Delta_\psi(I)}^{G\times G}k_\zeta.\end{align*} $$

In particular, $\Delta _\gamma (P)$ is contained in $\Delta _\psi (P)$ , up to conjugation, and we can assume that $\Delta _\gamma (P)=\Delta _\psi (P)$ , that is, that $\psi $ is the restriction of $\gamma $ to P. We have $\Delta _\psi (P)\leq \Delta _\psi (I)\leq N:=N_{G\times G}\big (\Delta _\psi (P)\big )=N_{G\times G}\big (\Delta _\gamma (P)\big )$ . So N fits in a short exact sequence of groups

Similarly, the group $\overline {N}:=N/\Delta _\psi (P)$ fits in the sequence

This sequence splits via $x\in K\mapsto \big (\gamma (x),x\big )\Delta _\psi (P)$ , and we view K as a subgroup of $\overline {N}$ via this map.

The group $\overline {\Delta }_\psi (I)=\Delta _\psi (I)/\Delta _\psi (P)$ is a subgroup of $\overline {N}$ and intersects $Z(P)$ trivially. So $\overline {\Delta }_\psi (I)$ is isomorphic to a subgroup J of K. Let $\theta :J\to k^\times $ be the image of $\overline {\zeta }$ under this isomorphism.

Since $\Delta _\psi (P)$ acts trivially on $k_\zeta $ , we have $k_\zeta =\mathrm {Inf}_{\overline {\Delta }_\psi (I)}^{\Delta _\psi (I)}k_{\overline {\zeta }}$ , where $\overline {\zeta }$ is the homomorphism $\overline {\Delta }_\psi (I)\to k^\times $ corresponding to $\zeta $ . Hence, we have

$$ \begin{align*} U\otimes_{kH}V&\cong \mathrm{Ind}_{\Delta_\psi(I)}^{G\times G}k_\zeta\\ &=\mathrm{Ind}_{N}^{G\times G}\mathrm{Ind}_{\Delta_\psi(I)}^N\mathrm{Inf}_{\overline{\Delta}_\psi(I)}^{\Delta_\psi(I)}k_{\overline{\zeta}}\\ &\cong \mathrm{Ind}_{N}^{G\times G}\mathrm{Inf}_{\overline{N}}^N\,\mathrm{Ind}_{\overline{\Delta}_\psi(I)}^{\overline{N}}k_{\overline{\zeta}}\\ &\cong\mathrm{Ind}_{N}^{G\times G}\mathrm{Inf}_{\overline{N}}^N\,\mathrm{Ind}_{\overline{\Delta}_\psi(I)}^{\overline{N}}\mathrm{Iso}_{J}^{\overline{\Delta}_\psi(I)}k_\theta\\ &\cong \mathrm{Ind}_{N}^{G\times G}\mathrm{Inf}_{\overline{N}}^N\mathrm{Ind}_K^{\overline{N}}\mathrm{Ind}_J^Kk_\theta. \end{align*} $$

Now K is cyclic, so $\mathrm {Ind}_J^Kk_\theta =\mathop {\sum }_{\substack {\alpha \in K^\natural \\\mathrm {Res}_J^K\alpha =\theta }} k_\alpha $ , and if $\alpha \in K^\natural $ , then $\mathrm {Ind}_K^{\overline {N}}k_\alpha $ is an indecomposable projective $k\overline {N}$ -module. Then $\mathrm {Ind}_{N}^{G\times G}\mathrm {Inf}_{\overline {N}}^N\mathrm {Ind}_K^{\overline {N}}k_\alpha \cong kG_{\gamma ,\alpha }$ . Now

$$ \begin{align*}U\otimes_{kH}V\cong\bigoplus_{\beta\in \mathcal{I}_\theta}kG_{\gamma,\beta},\end{align*} $$

as was to be shown.

Notation 4.7. Assume that Hypothesis 4.3 holds.

1. 1. We abuse notation identifying $\lambda \in K^\natural $ with $k_\lambda \in R_k(K)$ .

2. 2. We set $\overline {R}_k(K)=R_k(K)/\sum _{L<K}\mathrm {Ind}_L^KR_k(L)$ , and we let $\alpha \mapsto \overline {\alpha }$ denote the projection map.

3. 3. Let $\gamma \in \mathrm {Aut}(G)$ . Then $\overline {N}_{G\times G}\big (\Delta _\gamma (P)\big )\cong Z(P)\rtimes K$ , so taking coinvariants by $Z(P)$ yields an isomorphism

   $$ \begin{align*}v\in \mathrm{Proj}\Big(k\overline{N}_{G\times G}\big(\Delta_\gamma(P)\big)\Big)\mapsto v_{Z(P)}\in R_k(K).\end{align*} $$
   For $u\in T^\Delta (G,G)$ , let $r_\gamma (u)$ denote $\overline {u[\Delta _\gamma (P)]_{Z(P)}}\in \overline {R}_k(K)$ .

Notation 4.8. We denote by $\mathrm {Out}(G)\ltimes \overline {R}_k(K)$ the semidirect product of $\mathrm {Out}(G)$ with $\overline {R}_k(K)$ , that is,

$$ \begin{align*}\mathrm{Out}(G)\ltimes \overline{R}_k(K)=\bigoplus_{\gamma\in\mathrm{Out}(G)}\gamma\ltimes \overline{R}_k(K),\end{align*} $$

where $\gamma \ltimes \overline {R}_k(K)$ denotes a copy of $\overline {R}_k(K)$ indexed by $\gamma $ .

Then $\mathrm {Out}(G)\ltimes \overline {R}_k(K)$ is a ring for the product defined by

$$ \begin{align*}\forall \gamma,\delta\in\mathrm{Out}(G),\,\forall \lambda,\mu\in R_k(K),\;(\gamma\ltimes \overline{\lambda})\cdot(\delta\ltimes\overline{\mu}):=(\gamma\circ\delta)\ltimes\overline{\big((\lambda\circ\delta)\times\mu\big)}.\end{align*} $$

We set

$$ \begin{align*}\mathrm{Out}(G)\ltimes\mathsf{R} \overline{R}_k(K):=\mathsf{R}\otimes_{\mathbb{Z}} \big(\mathrm{Out}(G)\ltimes \overline{R}_k(K)\big)\cong \bigoplus_{\gamma\in\mathrm{Out}(G)}\gamma\ltimes \mathsf{R} \overline{R}_k(K).\end{align*} $$

Proof. Proving that the map $\mathbf {T}$ is well-defined amounts to proving that if $u\in K^\natural $ is induced from a proper subgroup J of K, and if $\gamma \in \mathrm {Out}(G)$ , then $\mathbf {T}(\gamma \ltimes \overline {u})=0$ . Let $J<K$ and $\theta \in J^\natural $ . Then $u=\mathrm {Ind}_J^K\theta =\sum _{\alpha \in \mathcal {I}_\theta } k_\alpha $ , so

$$ \begin{align*}\mathbf{T}(\gamma\ltimes u)=\mathcal{E}_{\mathsf{R}}\Big(\bigoplus_{\alpha\in \mathcal{I}_\theta}kG_{\gamma,\alpha}\Big).\end{align*} $$

But setting $N:=N_{G\times G}\big (\Delta _\gamma (P)\big )$ and $\overline {N}:=N/\Delta _\gamma (P)$ , we have

$$ \begin{align*} \bigoplus_{\alpha\in \mathcal{I}_\theta}kG_{\gamma,\alpha}&\cong\bigoplus_{\alpha\in \mathcal{I}_\theta}\mathrm{Ind}_{N}^{G\times G}\mathrm{Inf}_{\overline{N}}^N\mathrm{Ind}_K^{\overline{N}}k_\alpha\\ &\cong\mathrm{Ind}_{N}^{G\times G}\mathrm{Inf}_{\overline{N}}^N\mathrm{Ind}_K^{\overline{N}}\Big(\bigoplus_{\alpha\in \mathcal{I}_\theta}k_\alpha\Big)\\ &\cong \mathrm{Ind}_{N}^{G\times G}\mathrm{Inf}_{\overline{N}}^N\mathrm{Ind}_K^{\overline{N}}\mathrm{Ind}_J^Kk_\theta\\ &\cong \mathrm{Ind}_{N}^{G\times G}\mathrm{Ind}_{\Delta_\gamma(P)\cdot\Delta_\gamma(K)}^N\mathrm{Inf}_K^ {\Delta_\gamma(P)\cdot\Delta_\gamma(K)}\mathrm{Ind}_J^Kk_\theta\\ &\cong \mathrm{Ind}_{\Delta_\gamma(G)}^{G\times G}\mathrm{Inf}_K^{\Delta_\gamma(G)}\mathrm{Ind}_J^Kk_\theta\\ &\cong \mathrm{Ind}_{\Delta_\gamma(G)}^{G\times G}\mathrm{Ind}_{\Delta_\gamma(P)\cdot\Delta_\gamma(J)}^{\Delta_\gamma(G)}\mathrm{Inf}_J^{\Delta_\gamma(P)\cdot\Delta_\gamma(J)}k_\theta\\ &\cong \mathrm{Ind}_{\Delta_\gamma(P\cdot J)}^{G\times G}\mathrm{Inf}_J^{\Delta_\gamma(P\cdot J)}k_\theta. \end{align*} $$

But $p_2\big (\Delta _\gamma (P\cdot J)\big )=P\cdot J<G$ since $J<K$ . Hence, $\mathcal {E}_{\mathsf {R}}\Big (\bigoplus _{\alpha \in \mathcal {I}_\theta }kG_{\gamma ,\alpha }\Big )=0$ , as was to be shown, so the map $\mathbf {T}$ is well defined.

Now comparing the products in Lemma 4.2 and Notation 4.8, we get that $\mathbf {T}$ is a homomorphism of $\mathsf {R}$ -algebras. Moreover, the identity element of $\mathrm {Out}(G)\ltimes \mathsf {R} \overline {R}_k(K)$ , which is $\mathrm {Id}\ltimes \overline {1}$ , is mapped by $\mathbf {T}$ to $\mathcal {E}_{\mathsf {R}}(kG_{\mathrm {Id},1})=\mathcal {E}_{\mathsf {R}}(kG)$ , which is the identity element of $\mathcal {E}_{\mathsf {R}}(G)$ .

2. Proving that the map $\mathbf {S}$ is well-defined amounts to proving that if H is a finite group with $|H|<|G|$ , if U (resp. V) is a right (resp. left) essential diagonal p-permutation $(kG,kH)$ -bimodule (resp. $(kH,kG)$ -bimodule), then $\mathbf {S}(U\otimes _{kH}V)=0$ . So let $\gamma \in \mathrm {Aut}(G)$ such that $r_\gamma (U\otimes _{kH}V)\neq 0$ . Then, in particular, $(U\otimes _{kH}V)[\Delta _\gamma (P)]\neq 0$ , so $U\otimes _{kH}V$ admits an indecomposable summand with vertex $\Delta _\gamma (P)$ . By Lemma 4.4, this summand is isomorphic to $kG_{\gamma ,\lambda }$ , for some $\lambda \in G^\natural $ . By Theorem 4.6, there is a subgroup J of K with $P\cdot J\cong H$ , and $\theta \in J^\natural $ such that

$$ \begin{align*}U\otimes_{kH}V\cong\bigoplus_{\alpha\in \mathcal{I}_\theta}kG_{\gamma,\alpha}.\end{align*} $$

Now by (4.4.1)

$$ \begin{align*}(U\otimes_{kH}V)[\Delta_\gamma(P)]\cong \bigoplus_{\alpha\in\mathcal{I}_\theta}\mathrm{Ind}_K^{\overline{N}}k_\alpha\cong\mathrm{Ind}_K^{\overline{N}}\big(\bigoplus_{\alpha\in I_\theta}k_\alpha\big)\cong kZ(P)\otimes_k\mathrm{Ind}_J^Kk_\theta.\end{align*} $$

It follows that in $\overline {R}_k(K)$ , we have $r_\gamma (U\otimes _{kH}V)=\overline {\mathrm {Ind}_J^Kk_\theta }=0$ since $J<K$ as $P\cdot J\cong H$ and $|H|<|G|=|P||K|$ . This contradiction shows that $\mathbf {S}$ is well defined.

We postpone the proof that $\mathbf {S}$ is an algebra homomorphism at the end of the proof of Assertion 3.

3. Let $\gamma \in \mathrm {Aut}(G)$ and $\alpha \in K^\natural $ . Then $kG_{\gamma ,\alpha }$ has vertex $\Delta _\gamma (P)$ , and $kG_{\gamma ,\alpha }[\Delta _{\gamma }(P)]\cong kZ(P)\otimes k_\alpha $ by (4.4.1). We also get from Lemma 4.4 that $kG_{\gamma ,\alpha }[\Delta _{\delta }(P)]=0$ if $\delta \in \mathrm {Aut}(G)$ and $\delta \neq \gamma $ in $\mathrm {Out}(G)$ : Indeed, if $\Delta _\gamma (P)$ and $\Delta _\delta (P)$ are conjugate in $G\times G$ , then the restriction of $\delta $ to P is equal to the restriction of $\delta $ to P, up to an automorphism given by conjugation by some element of G, which we may assume to be trivial. Then $\delta ^{-1}\gamma $ is inner, by Lemma 3.1.

It follows that $\mathbf {S}\big (\mathcal {E}_{\mathsf {R}}(kG_{\gamma ,\alpha )})\big )=\gamma \ltimes \overline {\alpha }$ . Since $\mathbf {T}(\gamma \ltimes \overline {\alpha })=\mathcal {E}_{\mathsf {R}}(kG_{\gamma ,\alpha })$ , the maps $\mathbf {S}$ and $\mathbf {T}$ are inverse to each other. In particular, they are bijections, so $\mathbf {S}$ is a map of $\mathsf {R}$ -algebras, as $\mathbf {T}$ is. This completes the proof.