Brauer–Green block compatibility

By F6 choose a defect group $D_b$ of $b$ containing the literal $Q$, after conjugating inside $H$. Then $C_G(D_b)\le C_G(Q)\le H$, so F2 defines $b^G$. Suppose for contradiction $b^G\mathrm{\ne }B$. F1 implies $b$ is not a summand of ${\mathrm{Res}}_{H\times H}B$.

Set $M={⨁}_{HtH\mathrm{\ne }H}k[HtH]$. The double-coset decomposition gives $kG=kH\oplus M$ as $H\times H$-modules under F7's double action, and left multiplication by $e$ gives $ekG=b\oplus eM$. This multiplication is an $H\times H$-linear idempotent because $e$ is central in $kH$. Similarly $eB$ is a direct summand of both $B$ restricted and $ekG$. The first conclusion in step 1.1 excludes $b$ from $eB$. By F5 every indecomposable summand of $eB$ must therefore come from $eM$, so $eB$ is a summand of $eM$, hence of $M$.

Restrict $eB$ to $\mathrm{\Delta }H$. For each of its indecomposable summands with vertex $T$, F3's Mackey decomposition puts every vertex of a restricted indecomposable inside a $\mathrm{\Delta }H$-conjugate of $\mathrm{\Delta }H\cap xT{x}^{-1}$ for some $x\in H\times H$. If such a vertex contained a conjugate of $\mathrm{\Delta }Q$, then $T$ would contain an $(H\times H)$-conjugate of $\mathrm{\Delta }Q$, contrary to step 2.1. Identifying $H$ with $\mathrm{\Delta }H$ gives the conjugation action $h:a\mapsto ha{h}^{-1}$ on $eB$. Thus this $kH$-module has no summand whose vertex contains an $H$-conjugate of $Q$. F4 gives the same exclusion for $eB{\otimes }_{k}eV$ with diagonal action.

Define $i:eV\to eB\otimes eV$ by $i(v)=ef\otimes v$, and $r:eB\otimes eV\to eV$ by $r(a\otimes v)=av$. The first is $H$-linear because $h(ef)h^{-1}=ef$; the second is $H$-linear because $(ha{h}^{-1})(hv)=h(av)$. Its image lies in $eV$ since $ea=a$. Since $fV=V$ and $ev=v$ on $eV$, their composite is $ri(v)=efv=v$. Hence $eV$ is a summand of this tensor. Also $eU=U$ because $U$ belongs to $b$, so the given splitting of $U$ from the restriction of $V$, after applying $e$, splits $U$ from $eV$. Consequently $U$ is a summand of $eB\otimes eV$.

This contradicts step 3.1, since $U$ has vertex $Q$. Therefore $b^G=B$. If $Q=1$, the centralizer condition forces $H=G$ and the conclusion is the identity block assignment. If $H=G$ directly, the same conclusion holds. The nonzero module $U$ ensures the splittings in step 4.1 cannot be vacuous. All tensor maps and decompositions are finite and require no AC.