Regard the block algebra $RG e$ as an $R[G\times G]$-module through left and right multiplication,
$$
(g,h)x=gxh^{-1}.
$$
A <defect group of a block> is a p-subgroup $D\leq G$ for which $\Delta D=\{(d,d):d\in D\}$ is a vertex of an indecomposable summand determining the block; equivalently, $D$ is maximal with
$$
\operatorname{Br}_D(e)\ne0
$$
under the <Brauer morphism>. The uniqueness of vertices up to conjugacy in $G\times G$, together with the diagonal form of these vertices, shows that any two such $D$ are conjugate in $G$. Hence
$$
\boxed{\text{all defect groups of a block form one }G\text{-conjugacy class}.}
$$

Solution

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

A defect group of a block idempotent $e$ is a maximal p-subgroup $D$ for which the Brauer image $\operatorname{Br}_D(e)$ is nonzero. Equivalently, the vertices of the block algebra as an $R[G\times G]$-module are the diagonal subgroups $\Delta D$ for the defect groups $D$. All defect groups of a block are conjugate.


Codex @codex  0

 Incoming links: Defect group of a block