Transitive p-group permutation modules have point-stabilizer vertices

Every nonzero finite-dimensional $kR$-module $W$ has a nonzero fixed vector: choose the largest $j\ge 0$ with $I_R^{j}W\mathrm{\ne }0$, which exists by [F4]. Every vector in this nonzero space is annihilated by $I_R$ and hence by each $r-1$. On the other hand $V^R$ is one-dimensional, since invariance means constant coefficients on the transitive basis $R\mathrm{/}L$. If $V$ split into two nonzero submodules, their nonzero fixed subspaces would be in direct sum inside $V^R$, a contradiction. Thus $V$ is indecomposable.

The map $kR{\otimes }_{kL}k\to V$, $r\otimes 1\mapsto rL$, is an isomorphism on coset bases and respects the $R$-action. Therefore $V$ is relatively $L$-projective by [F1]. To prove minimality suppose $T<L$ and ${\mathrm{id}}_V={\mathrm{Tr}}_T^R(\alpha )$ with $\alpha \in {\mathrm{End}}_{kT}(V)$, as required by [F2] for relative $T$-projectivity.

Let $e$ denote the basis vector $L$ and $\lambda$ its coefficient functional. For a left coset $rT$ define $c(rT)=\lambda (r\alpha (r^{-1}e))$. Replacing $r$ by $rt$ does not change this value, since $\alpha (t^{-1}v)=t^{-1}\alpha (v)$. For $l\in L$, $l^{-1}e=e$ and $\lambda (lv)=\lambda (v)$, so $c(lrT)=c(rT)$. Each $L$-orbit on $R\mathrm{/}T$ has size $[L:L\cap rT{r}^{-1}]$. Since $\mathrm{\mid }T\mathrm{\mid }<\mathrm{\mid }L\mathrm{\mid }$, that index is a positive power of $p$, and is zero in $k$.

The coefficient of $e$ in ${\mathrm{Tr}}_T^R(\alpha )(e)$ is ${\sum }_{rT}c(rT)$. Grouping by the orbits in step 2.1 makes each contribution zero. The coefficient of $e$ in ${\mathrm{id}}_V(e)$ is $1$, a contradiction. No proper $T<L$ is therefore a relative-projectivity subgroup. Together with steps 1.1–1.2 and [F3], this says $L$ is a vertex.