Tensoring preserves relative projectivity for finite-group modules

By F1 and F2 take a finite-dimensional $kR$-module $W$ and split $X$ from ${\mathrm{Ind}}_R^{H}W$; one may take $W={\mathrm{Res}}_R^{H}X$ using F2's counit splitting. Tensor the inclusion and retraction with $1_Y$. Their composite remains $1_{X\otimes Y}$, and both are $H$-linear for the diagonal actions. Thus it suffices to identify the tensor of the inducing module.

Define $\mathrm{\Phi }((h\otimes w)\otimes y)=h\otimes (w\otimes h^{-1}y),\mathrm{\Psi }(h\otimes (w\otimes y))=(h\otimes w)\otimes hy\mathrm{.}$ These map between $(kH{\otimes }_{kR}W){\otimes }_{k}Y$ and $kH{\otimes }_{kR}(W{\otimes }_k{\mathrm{Res}}_R^{H}Y)$. The balancing relation $hr\otimes w=h\otimes rw$ is respected by $\mathrm{\Phi }$: moving $r$ from the first tensor factor on its right acts diagonally on $w\otimes r^{-1}{h}^{-1}y$, giving $rw\otimes h^{-1}y$. The same identity verifies $\mathrm{\Psi }$ is balanced. Applying $a\in H$ on the left replaces $h,y$ by $ah,ay$, leaving $h^{-1}y$ unchanged, so $\mathrm{\Phi }$ is $H$-linear. The displayed formulas compose to the identity in both orders. Therefore the tensor in step 1.1 is a summand of an $R$-induced module, proving relative $R$-projectivity.

For the consequence, decompose $X={⨁}_j{X}_j$ using F3 and choose a vertex $R_j$ for each nonzero indecomposable by F4. Step 2.1 makes $X_j\otimes Y$ relatively $R_j$-projective. Any indecomposable summand $Z$ of their finite sum is a summand of one term by F2. F2 then puts a vertex $T$ of $Z$ inside an $H$-conjugate of $R_j$. If $T$ contained a conjugate of $Q$, so would that conjugate of $R_j$, contradicting the hypothesis. This proves the consequence. If either tensor factor is zero the sum has no indecomposable summands. For $R=H$ the assertion is automatic, and for $Q=1$ its hypothesis forces $X=0$. All decompositions, bases and subgroup choices here are finite.