Alperin p82 Lemma 11.3 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.

As a rule of thumb, if the statement  you want to prove involves at least one restriction and at least one induction, then Mackey decomposition is your best friend.
So you want to show that if $W$ has vertex $Q$ (I guess that's what you mean), i.e. if $W\;|\;M^L$ for some $Q$-module $M$ (and $Q$ is minimal with this property), then $W^G$ is relatively $Q$-projective, i.e. $W^G|((W^G)_Q)^G$.
Now,by Mackey, $(W^G)_Q = \bigoplus_{g\in L\backslash G/Q}((W^g)_{L^g\cap Q})^Q$.
The summand corresponding to the trivial coset representative is $W_Q$, so $((W^G)_Q)^G$ has $(W_Q)^G$ as a direct summand, since induction preserves direct sums. But also, as in your previous question, $M$ being a source of $W$ implies that $M|W_Q$ (this also uses Mackey).
Inducing both sides back to $G$ (and using again the fact that induction preserves direct sums), $M^G\;| \;(W_Q)^G$. Since $W|M^L$, you get $W^G|(M^L)^G = M^G$, and you are done.
