Green vertex retention and inducing lift

Choose a source $S$ at $Q$. Thus $S\mid {\mathrm{Res}}_Q^{G}M$ and $M\mid {\mathrm{Ind}}_Q^{G}S$. If $S$ were relatively $D$-projective for some proper subgroup $D<Q$, transitivity would make $M$ relatively $D$-projective, contradicting vertex minimality. Thus $S$ has vertex $Q$ as a $kQ$-module.

Decompose ${\mathrm{Ind}}_Q^{L}S={⨁}_j{U}_j$ into indecomposables. Transitivity gives $M\mid {⨁}_j{\mathrm{Ind}}_L^G{U}_j$, so $M\mid {\mathrm{Ind}}_L^G{U}_j$ for some $j$. Put $U=U_j$. It is relatively $Q$-projective. Vertex containment and conjugacy let us choose a vertex $T$ of $U$ with $T\le Q$. Then $M$ is relatively $T$-projective by transitivity. Vertex containment for $M$ gives $\mathrm{\mid }Q\mathrm{\mid }\le \mathrm{\mid }T\mathrm{\mid }$, whereas $T\le Q$ gives the reverse inequality. Hence $T=Q$, proving (b). The decompositions here use the inherited finite-length chain under AC.

Independently decompose ${\mathrm{Res}}_L^{G}M={⨁}_j{V}_j$. From $S\mid {\mathrm{Res}}_Q^{G}M={⨁}_j{\mathrm{Res}}_Q^L{V}_j$, extract a summand $V=V_j$ with $S\mid {\mathrm{Res}}_Q^{L}V$. Because $M\mid {\mathrm{Ind}}_Q^{G}S$, Mackey expresses ${\mathrm{Res}}_L^{G}M$ as a summand of a finite sum of modules induced from $L\cap {}^{g}Q$. Extracting $V$ from one term and applying vertex containment shows that any vertex $T$ of $V$ has $\mathrm{\mid }T\mathrm{\mid }\le \mathrm{\mid }Q\mathrm{\mid }$.

Relative $T$-projectivity supplies the counit splitting $V\mid {\mathrm{Ind}}_T^L{\mathrm{Res}}_T^{L}V$ from F4. Restrict to $Q$ and use $S\mid {\mathrm{Res}}_Q^{L}V$. Mackey and summand extraction make $S$ relatively $Q\cap {}^{l}T$-projective for some $l\in L$. Its vertex is the whole group $Q$, by 1.1, so $Q=Q\cap {}^{l}T$. The inequality in 2.2 forces $Q={}^{l}T$. Conjugacy of vertices now makes $Q$ itself a vertex of $V$, proving (a).

All decompositions and coset sums used above are finite, and the two nonzero witnesses are extracted separately. If $L=G$, the module $M$ itself witnesses both conclusions. If $L=Q$, the source witnesses both, and its full vertex was checked in 1.1. The same reasoning covers $Q=1$; zero $M$ is excluded from the statement. Thus both assertions hold with precisely the stated inherited AC assumption. [step 1.1, step 2.1, step 3.1, given] QED