Defect groups are maximal Brauer support

Choose a defect group $D$ and $a\in B^D$ with $b={\mathrm{Tr}}_D^G(a)$. If ${\mathrm{Br}}_D(b)=0$, [F2] writes $b={\sum }_{Q<D}{\mathrm{Tr}}_Q^D(u_Q)$. Multiplying by $b$ if necessary puts $u_Q$ in $B^Q$. Since $b$ is central and $a$ is $D$-fixed, trace transitivity gives $b=b^2={\mathrm{Tr}}_D^G(ab)={\sum }_{Q<D}{\mathrm{Tr}}_Q^G(a{u}_Q)$. Each term lies in its trace ideal of $Z(B)$. By [F3], $b$ belongs to one proper-$Q$ trace ideal, contradicting minimality of $D$. Thus its image is nonzero.

For any $P$ with ${\mathrm{Br}}_P(b)\mathrm{\ne }0$, the same trace expression for $b$ and [F2] imply $P\le gD{g}^{-1}$ for some $g$. A subgroup containing $D$ with nonzero image therefore has order at most $\mathrm{\mid }D\mathrm{\mid }$, proving maximality of $D$. Conversely a maximal $P$ with nonzero image lies in $gD{g}^{-1}$, whose image is nonzero since it too is a defect group by [F4] and step 1.1. Maximality gives equality, and [F4] makes $P$ a defect group.