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Existence of a minimal subgroup follows because $G$ has only finitely many subgroups and every module is relatively $G$-projective. Suppose an indecomposable module $M$ is relatively projective for both $H$ and $K$. Then $M$ is a summand of ${\mathrm{Ind}}_H^{G}U$ and of ${\mathrm{Ind}}_K^{G}V$ for suitable modules $U,V$. Applying the Mackey restriction formula and the Krull–Schmidt theorem shows that $M$ is relatively projective for some subgroupIf $H$ and $K$ are minimal, this forces $H\subseteq {}^{g}K$. Reversing their roles gives the reverse containment after conjugacy; since the groups are finite, $H$ and $K$ are conjugate. Thus vertices form a unique conjugacy class.

Let $Q$ be a vertex and let $P$ be a Sylow p-subgroup of $Q$. Since $[Q:P]$ is invertible in $k$, every $kQ$-module is relatively $P$-projective. Transitivity of relative projectivity makes $M$ relatively $P$-projective, so minimality forces $Q=P$. Hence every vertex is a p-group.

For the trivial module $k$, every $kH$-endomorphism is scalar, and its relative trace to $G$ is multiplication by $[G:H]$. The D. Higman criterion says that $k$ is relatively $H$-projective exactly when $p\nmid [G:H]$. The minimal such subgroups are precisely the Sylow p-subgroups. Therefore the vertex of an indecomposable module gives