Integral Mackey decomposition and Higman's criterion for group lattices

Decompose $H$ into its finite $A$-$B$ double cosets. The summand of $\mathcal{O}H{\otimes }_{\mathcal{O}B}V$ supported on $AxB$ is identified with $\mathcal{O}A{\otimes }_{\mathcal{O}{L}_x}{}^{x}V,a\otimes v⟼ax\otimes v,$ where $q\in L_x$ acts on ${}^{x}V$ as $x^{-1}qx$ acts on $V$. These maps and their inverses are well-defined on the tensor relations, and their finite direct sum is the displayed Mackey isomorphism. Tensor associativity gives ${\mathrm{Ind}}_A^H{\mathrm{Ind}}_B^A\cong {\mathrm{Ind}}_B^H$ for $B\le A\le H$. Since $\mathcal{O}H$ is finite free as a right subgroup algebra, these operations preserve finite-free lattices; functoriality preserves split inclusions and retractions.

First connect F1's summand definition to a split induction counit. Put $Y={\mathrm{Ind}}_Q^{H}V=\mathcal{O}H{\otimes }_{\mathcal{O}Q}V$. The counit ${\epsilon }_Y:{\mathrm{Ind}}_Q^H{\mathrm{Res}}_Q^{H}Y\to Y$ has the explicit $H$-linear section $s_Y(h\otimes v)=h\otimes (1\otimes v)\mathrm{.}$ It respects the relation $hq\otimes v=h\otimes qv$ because $q\otimes v=1\otimes qv$ in $Y$, and ${\epsilon }_Y{s}_Y=1_Y$. If $M$ is a summand of $Y$ with inclusion $i:M\to Y$ and retraction $r:Y\to M$, then $s_M=({\mathrm{Ind}}_Q^H{\mathrm{Res}}_Q^{H}r)s_{Y}i$ splits the counit for $M$, by naturality of the counit and $ri=1_M$. Conversely, a split counit displays $M$ as a summand of its induced module. Now let $T$ be left-coset representatives for $H\mathrm{/}Q$. The induction counit $\epsilon :\mathcal{O}H{\otimes }_{\mathcal{O}Q}M\to M$ is $\epsilon (h\otimes m)=hm$. Given an $H$-linear section $s$, write $s(m)$ in the direct sum indexed by $T$ and let $\alpha (m)$ be its coefficient in the identity-coset component. Equivariance makes $\alpha$ $Q$-linear, and $\epsilon s=1$ says $1_M={\sum }_{t\in T}t\alpha t^{-1}={\mathrm{Tr}}_Q^H(\alpha )\mathrm{.}$ Conversely this trace identity makes $s(m)={\sum }_{t\in T}t\otimes \alpha (t^{-1}m)$ an $H$-linear section of $\epsilon$. This proves the integral Higman criterion.

The image of every relative trace is a two-sided ideal of $E={\mathrm{End}}_{\mathcal{O}H}(M)$: an $H$-endomorphism can be moved inside either side of the finite trace sum. Suppose now that $M$ is indecomposable, relatively $P$-projective and relatively $R$-projective. By step 1.2 choose trace expressions for $1_M$ from $P$ and from $R$ and multiply them. The diagonal $H$-orbits on the finite set $H\mathrm{/}P\times H\mathrm{/}R$ regroup the product as a finite sum of relative traces from their stabilizers $sP{s}^{-1}\cap tR{t}^{-1}\mathrm{.}$ The element inside each orbit trace is fixed by that stabilizer, so every summand belongs to the corresponding trace ideal.

By F2 the ring $E$ is local. If every summand from step 2.1 were a nonunit, their sum could not be $1_M$; hence one is a unit. Its two-sided trace ideal then contains $1_M$, and step 1.2 makes $M$ relatively projective for the associated intersection. Conjugating that subgroup by $s^{-1}$ shows that $M$ is relatively $P\cap s^{-1}tR{t}^{-1}s$, a subgroup of $P$. If $P$ is a vertex, its minimality forces this intersection to be $P$. Thus $P\le s^{-1}tR{t}^{-1}s$, as required. Empty double-coset sets cannot occur because $H$ is nonempty; trivial subgroups and $P=1$ are included. All coset sets and sums are finite, so no choice principle is used.