Block bimodule for the double group

Fix a splitting $p$-modular system for a finite group $G$, with residue field $k$ of characteristic $p$, as in A splitting p-modular system for a finite group is a p-modular system whose fraction and residue fields split the needed group algebras. Let $b\in Z(kG)$ be a primitive central idempotent—specifically, a residue-field block idempotent in the $kG$ branch of p-blocks from primitive central idempotents. The block bimodule is $B=kGb$, viewed as a left $k[G\times G]$-module by $(x,y)v=xv{y}^{-1}$. This is an action because $(x,y)((x^{\mathrm{\prime }},y^{\mathrm{\prime }})v)=x{x}^{\mathrm{\prime }}v(y^{\mathrm{\prime }}{)}^{-1}{y}^{-1}=(x{x}^{\mathrm{\prime }},y{y}^{\mathrm{\prime }})v$ and $(1,1)v=v$. Centrality of $b$ makes $B$ stable and the projection $v\mapsto vb$ splits $kG=B\oplus kG(1-b)$ as bimodules. We write $\mathrm{\Delta }H=\{(h,h):h\in H\}$ for a subgroup $H\le G$.

This bimodule is nonzero and indecomposable. Indeed a bimodule endomorphism $T$ is determined by $c=T(b)$: for $v\in B$, $T(v)=vc=cv$, so $c\in Z(B)=bZ(kG)$, and conversely central multiplication is a bimodule endomorphism. A direct-sum decomposition would give an idempotent endomorphism other than $0,1$, hence a central idempotent $c$ other than $0,b$. Then $b=c+(b-c)$ is a nontrivial orthogonal central decomposition, contradicting primitivity. More precisely the endomorphism algebra is local: for a central $c$, stabilized kernels and images of multiplication by $c$ split $B$ into bimodules, so indecomposability makes $c$ either invertible or nilpotent. The nonunits form an ideal, since in this commutative finite algebra sums of nilpotents are nilpotent by the binomial expansion and multiples of nilpotents are nilpotent. This ideal contains every proper ideal and is the unique maximal ideal.

Webb, A Course in Finite Group Representation Theory, §§5.2, 11.3, 11.6, 12.3–12.5; especially Lemma 12.4.4 and Theorem 12.4.5, pp.240–241. Local argument and conventions as displayed above.