Block defect is an intersection of two sylow subgroups

Put $R=P\times P$. The source existence in [F6] and the vertex in [F1] allow [F3] with ambient group $G\times G$. Thus ${\mathrm{Res}}_{R}B$ has an indecomposable summand $U$ with vertex $\mathrm{\Delta }D$. Multiplication by the central idempotent $b$ splits $B$ from $kG$ as a bimodule; restriction preserves this splitting, as in [F2]. Hence $U\mid {\mathrm{Res}}_{R}kG$.

The action on the basis $G$ is $(a,c)g=ag{c}^{-1}$, so its orbits are $PgP$ and the stabilizer of $g$ is $L_g=\{(a,g^{-1}ag):a\in P\cap gP{g}^{-1}\}$. Indeed $ag{c}^{-1}=g$ is equivalent to $c=g^{-1}ag\in P$. Mapping a coset in $R\mathrm{/}{L}_g$ to its translate of $g$ gives an equivariant bijection. Thus ${\mathrm{Res}}_{R}kG={⨁}_{g\in P\mathrm{\backslash }G\mathrm{/}P}k[R\mathrm{/}{L}_g]$.

By [F4] each summand is indecomposable with vertex $L_g$. By [F5], $U$ is isomorphic to one such summand. Vertex conjugacy in the group $R$, using [F6], gives $r,s\in P$ with $\mathrm{\Delta }D=(r,s)L_g(r,s{)}^{-1}$. Projecting to the first coordinate gives $D=r(P\cap gP{g}^{-1})r^{-1}$.

For $d\in D$, membership of $(d,d)$ in this conjugate stabilizer says $r^{-1}dr=g(s^{-1}ds)g^{-1}$. Put $h=rg{s}^{-1}$. Rearranging gives $d=hd{h}^{-1}$ for every $d$, so $h\in C_G(D)$. Since $r,s\in P$, step 3.1 yields $D=P\cap rgP{g}^{-1}{r}^{-1}=P\cap hP{h}^{-1}$. This calculation explicitly handles the twisted diagonal stabilizer.