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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.

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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.

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    Sorry, I'm not sure how this proves what I asked. We want to show that $W^G$ is relatively $\mathfrak{X}$-projective, not relatively $Q$-projective. Let $W$ be an indecomposable summand of $V$. Then $W$ is relatively $Y$-projective for some $Y\in\mathfrak{Y}$, say $Y=sQs^{-1}\cap L$ for some $s\in G\setminus L$. Hence if we let $P$ be the vertex of $W$, $sQs^{-1}\cap L$ contains a conjugate of $P$ -- i.e. $sQs^{-1}\cap L\supseteq_G P$ or $P\subseteq_G \mathfrak{Y}$. Why can we change this $\subseteq_G$ to a $\subseteq_L$ (I think this is the fact that $V$ is relatively $Q$-projective?2011-08-31
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    Ah right it is a $\subseteq_L$ since $W$ is a $kL$-module. But why does this imply $W^G$ is relatively $P$-projective?2011-08-31
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    In your answer above, how do we know $W=\mathrm{Ind}_{L/Q}M$ for some $M$? (And what do you mean by this notation)2011-08-31
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    @user That happens if you don't state clearly what your question is (e.g. there is not a single question mark in your post). I read your post as asking "why is $W^G$ $P$-projective if $W$ has vertex $P$?" But since $P$ has never appeared in the setup, I interpreted $P$ as wanting to be $Q$. Of course the letter is not important, but it _is_ important that $W$ is an $L$-module for some $L$ _containing_ _the_ _vertex._ Apart from that, there is a typo in my answer, let me correct it ($\rm{Ind}_{L/Q}$ is induction from $Q$ to $L$; but it's not what I wanted to say).2011-08-31
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    In your answer above, if $W\mid M^L$ why does $W_Q=(M^G)_Q$? Do you mean $W_Q\mid (M^L)_Q$?2011-08-31
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    Sorry, I must have gotten confused with too many open windows in my browser. I though I had corrected it. Now I have.2011-08-31