= Solution
A block idempotent is a primitive central <idempotent> $e\in R$. The module $M$ lies in the corresponding <block of a group algebra> when
$$
eM=M,
$$
equivalently 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 gives
$$
(1-e)V=(1-e)^2V=0.
$$
Thus $eV=V$, proving
$$
\boxed{V\text{ lies in }e\quad\Longleftrightarrow\quad U\text{ and }W\text{ lie in }e.}
$$
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