Trace criterion for defect groups

For a $p$-subgroup $D$ of a finite group $G$, set $I_D={\mathrm{Tr}}_D^G((kG{)}^D)$, a transfer ideal of conjugation-fixed elements in $Z(kG)$. A block of a group algebra idempotent $b$ lies in $I_D$ if and only if a defect group of a block of $b$ is conjugate to a subgroup of $D$. In particular, defect groups can equivalently be defined as minimal subgroups $D$ with $b\in I_D$.

Here is a proof using the Brauer morphism. A subgroup of smallest order with $b\in I_D$ exists, since a Sylow subgroup $P$ gives $b={\mathrm{Tr}}_P^G(b/[G:P])$. Write $b={\mathrm{Tr}}_D^G(a)$ and replace $a$ by $ba$. If ${\mathrm{Br}}_D(b)=0$, then $ba$ belongs to the kernel of the Brauer morphism, which is the sum of proper-subgroup relative traces. Hence $b\in {\sum }_{Q<D}{I}_Q$. The algebra $bZ(kG)$ is a commutative Artinian ring and a local ring, so multiplying this equality by $b$ forces some ideal $b{I}_Q$ to contain its identity $b$. This contradicts minimality. On the other hand, ${\mathrm{Br}}_E(I_D)=0$ unless $E$ is conjugate into $D$: the $E$-action on $G/D$ has no fixed coset otherwise, and nonfixed orbit sizes vanish in characteristic $p$. Thus these minimal subgroups are precisely the maximal subgroups with nonzero Brauer image. They are all conjugate. Trace transitivity proves the stated criterion for larger $D$.