Induced blocks have controlled defect

Choose a defect group $E$ of $B$ by F2 and F4. Relative $\mathrm{\Delta }E$-projectivity writes $B$ as a summand of a module induced from $\mathrm{\Delta }E$. Restrict it to $H\times H$ and apply F3. By F1 and finite summand extraction, $b$ is relatively $(H\times H)\cap x\mathrm{\Delta }E{x}^{-1}$-projective for some $x\in G\times G$. Since $\mathrm{\Delta }D$ is a vertex by F4, F3 places it in an $(H\times H)$-conjugate of that intersection, hence in a $(G\times G)$-conjugate of $\mathrm{\Delta }E$.

Write that conjugating element as $(g_1,g_2)$. Projecting $\mathrm{\Delta }D\le (g_1,g_2)\mathrm{\Delta }E(g_1,g_2{)}^{-1}$ onto the first coordinate gives $D\le g_{1}E{g}_1^{-1}$. Conjugating a diagonal vertex simultaneously by $(g_1,g_1)$ shows this conjugate of $E$ is again a defect group of $B$. This proves the required containment. If $D=1$ the conclusion is automatic; if $H=G$, F1 gives $B=b$ and equality is possible. Neither argument infers equality in general. All selections involve finite subgroup sets and finite decompositions, with no additional AC.