Restriction to a containing p subgroup retains a vertex

Write $X\mid Y$ to mean that $X$ is a direct summand of $Y$. The source $S$ cannot be relatively $E$-projective for any $E<Q$: otherwise transitivity of induction and $M\mid {\mathrm{Ind}}_Q^{H}S$ make $M$ relatively $E$-projective, contrary to minimality of its vertex $Q$. Hence $S$, as a $kQ$-module, has full vertex $Q$.

Decompose ${\mathrm{Res}}_R^{H}M=⨁U_j$ into nonzero indecomposables. Since $S\mid {\mathrm{Res}}_Q^{H}M=⨁{\mathrm{Res}}_Q^R{U}_j$, [F4] gives one $U=U_j$ with $S\mid {\mathrm{Res}}_Q^{R}U$. This uses finite decomposition after further restricting each $U_j$, not an assertion that its restriction is indecomposable.

Restrict the split inclusion $M\mid {\mathrm{Ind}}_Q^{H}S$ to $R$. The selected $U$ is a summand of its right side. Mackey and indecomposable extraction in [F3] supply an $h\in H$ for which $U$ is relatively $L_h=R\cap hQ{h}^{-1}$-projective. In particular $\mathrm{\mid }{L}_h\mathrm{\mid }\le \mathrm{\mid }Q\mathrm{\mid }$.

Choose an inclusion-minimal subgroup $T$ of $L_h$ relative to which $U$ is projective. The set is finite and nonempty since it includes $L_h$. Any proper subgroup of $T$ would also be a subgroup of $L_h$, so this minimality makes $T$ a vertex of $U$. Thus $\mathrm{\mid }T\mathrm{\mid }\le \mathrm{\mid }Q\mathrm{\mid }$. Put $W={\mathrm{Res}}_T^{R}U$; [F5] gives $U\mid {\mathrm{Ind}}_T^{R}W$.

Restrict this last splitting to $Q$ and use step 1.2. Then $S\mid {\mathrm{Res}}_Q^R{\mathrm{Ind}}_T^{R}W$. A second Mackey decomposition and indecomposable extraction give $r\in R$ such that $S$ is relatively $E=Q\cap rT{r}^{-1}$-projective as a $kQ$-module. Step 1.1 forces $E=Q$, so $Q\le rT{r}^{-1}$.

Now $\mathrm{\mid }Q\mathrm{\mid }\le \mathrm{\mid }T\mathrm{\mid }\le \mathrm{\mid }Q\mathrm{\mid }$ implies $Q=rT{r}^{-1}$. Conjugating a split induction witness inside $R$ preserves its splitting and minimality, and the inner conjugate of a $kR$-module is isomorphic to itself by multiplication by $r$. Thus this conjugate of the vertex $T$ is itself a vertex of the same $U$, consistently with [F2]. This is the claimed literal subgroup $Q$, without changing $U$ by an outside conjugation.