Relative projectivity mackey intersections for finite modules

For a double-coset representative $x$, send $a\otimes v$ to $ax\otimes v$. For $t\in L_x$, $atx\otimes v=ax\otimes (x^{-1}tx)v$, so this is balanced. The basis of $kH$ partitions into the disjoint double cosets $AxB$. Representatives $a$ of $A\mathrm{/}{L}_x$ give a right $kB$-basis $ax$ of $k[AxB]$, because $a_{1}xB=a_{2}xB$ holds exactly when $a_2^{-1}{a}_1\in L_x$. Thus the component map is a bijection between the same copies of $V$, and the direct sum is the claimed module isomorphism.

Transitivity is $a\otimes (b\otimes v)\mapsto ab\otimes v$, inverse $a\otimes v\mapsto a\otimes (1\otimes v)$; tensor relations make both maps well-defined. Applying either induction or restriction to split inclusion and projection maps preserves their composite identity, hence summands. Higman supplies a finite inducing witness: if ${\mathrm{i}\mathrm{d}}_M={\mathrm{Tr}}_B^H\alpha$, then $m\mapsto {\sum }_{t\in H\mathrm{/}B}t\otimes \alpha (t^{-1}m)$ splits the counit $h\otimes m\mapsto hm$. This map is $H$-linear by reindexing cosets.

Decompose each term of a finite direct sum into indecomposables by [F4]. If an indecomposable $M$ is a summand of that sum, uniqueness of indecomposable decompositions forces it to be isomorphic to a summand of one term.

Let $Q$ be a vertex of $M$ and suppose $M$ is relatively $B$-projective. Combining the two counit splittings gives $M\mid {\mathrm{Ind}}_Q^H{\mathrm{Res}}_Q^H{\mathrm{Ind}}_B^H{\mathrm{Res}}_B^{H}M$. Steps 1.1–1.2 express the right side as a finite sum induced from $Q\cap xB{x}^{-1}$. Step 1.3 puts $M$ in one term. This intersection is a $p$-subgroup of $Q$; minimality in the definition of vertex forces it to equal $Q$. Hence $Q\le xB{x}^{-1}$, as claimed.