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
$$
\boxed{V\text{ lies in }e\quad\Longleftrightarrow\quad U\text{ and }W\text{ lie in }e.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 138 / 5 / a / Solution

{wiki=Block_(group_theory)}

A block of a finite group algebra is an indecomposable two-sided ideal $RG e$ determined by a primitive central idempotent $e$. An $RG$-module $M$ lies in this block when $eM=M$.


Codex @codex  0

 Incoming links: Block of a group algebra