My question is regarding Lemma 11.3 on p82 of Local representation theory by JL Alperin; the Google Books preview unfortunately does not contain this page. I need to prove the following claim:
Claim: If $V$ is a relatively $Q$-projective and a relatively $\mathfrak{Y}$-projective $kL$-module then $V^G$ is relatively $\mathfrak{X}$-projective.
Proof. Let $W$ be an indecomposable summand of $V$. Then $W$ is relatively $Y$-projective for some $Y\in \mathfrak{Y}$. This is as far as I have got -- not sure why we need to bring vertices into it and definitely don't understand why if $W$ has vertex $P$ then $W^G$ is relatively $P$-projective.
Any help would be great!
Notation: $G$ is a finite group, $Q$ is any $p$-subgroup and $L=N_G(Q)$ is its normaliser. There are two collections of subgroups of $G$, $\mathfrak{X}=\{sQs^{-1}\cap Q\mid s\in G\setminus L\}$ and $\mathfrak{Y}=\{sQs^{-1}\cap L\mid s\in G\setminus L\}$. We say a module is relatively $\mathfrak{X}$- (resp. $\mathfrak{Y}$-) projective if it is a direct sum of relatively $X$- (resp. $Y$-) projective modules for $X\in \mathfrak{X}$ (resp. $Y\in \mathfrak{Y}$).
Cheers.