 Solution

A block idempotent is a primitive central idempotent $e\in R$. The module $M$ lies in the corresponding block of a group algebra whenequivalently when every other block idempotent annihilates $M$.

If $V$ lies in $e$, then the centrality of $e$ makes both its submodule $U$ and quotient $W$ lie in $e$. Conversely, suppose $eU=U$ and $eW=W$. Then $(1-e)V$ maps to zero in $W$, so $(1-e)V\subseteq U$. But $(1-e)U=0$, and applying the idempotent $1-e$ once more givesThus $eV=V$, proving