# Defect groups and subgroups

I have asked this question on Stack Exchange but had no response; it's been bugging me for a few days. I am struggling to see how to apply Mackey's theorem to prove a certain Lemma in Local representation theory by JL Alperin.

Lemma Let $b$ be a block of the subgroup $H$ of $G$ and let $D$ be a defect group of $b$. If $b^G$ is defined, i.e. if there exists a unique block $b^G$ of $G$ such that $b\mid B_{H\times H}$, then $D$ is contained in a defect group of $b^G$.

Proof. Let $E$ be a defect group of $B=b^G$, so that $B$ has vertex $\delta(E)$ (as a $k(G\times G)$-module). The proof comes down to the following statement which I don't understand: since $b\mid B_{H\times H}$, it follows from Mackey's theorem that the vertex $\delta(D)$ of $b$ is conjugate in $G\times G$ with a subgroup of $\delta(E)$.

Would someone be able to explain this statement to me, i.e. how Mackey's theorem is used here? I understand that we need to prove that $b$ is relatively $\delta(E)$-projective.

Cheers.

This is an exercise in writing out the definitions: since the defect group of $B$ is $E$, we have $B|(B_{\delta(E)})^{G\times G}$. So by assumption, $$b\;|\;B_{H\times H}\;|\;\left((B_{\delta(E)})^{G\times G}\right)_{H\times H}=\bigoplus_{(g_1,g_2)\in \delta(E)\backslash G\times G/H\times H}\left(B_{\delta(E)^{(g_1,g_2)}\cap H\times H}\right)^{H\times H},$$ which means that $b$ divides one of the summands (since $b$ is indecomposable). Say $b|M^{H\times H}$, where $M$ is a $(\delta(E)^{(g_1,g_2)}\cap H\times H)$-module. So the vertex of $b$ is some subgroup of that intersection, and in particular a subgroup of $\delta(E)^{(g_1,g_2)}$, as required.
• Thanks, for some reason I missed the important observation that $B\mid (B_{\delta(E)})^{G\times G}$ -- of course this just comes from the definition of a vertex (since every module $V$ is relatively $Vertex(V)$-projective). – Clinton Boys Sep 10 '11 at 7:57