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This should be easy for the resident representation theory specialists:

Let $F$ be an algebraically closed field of characteristic $p>0$, $G$ a finite group, and $M$ an indecomposable $FG$-module that is a direct summand of $\mathrm{Ind}_{G/H} 1$. Is a vertex of $M$ given by a $p$-Sylow of $H$ or can it be smaller? I am sure that it can be smaller in general (an example would be appreciated). Are there conditions on $H$ that ensure that the vertex is not smaller (e.g. $p$-Sylow normal in $H$; I cannot seem to get Green correspondence to work for me)?

As a reminder, the vertex of $M$ is a minimal subgroup $U$ of $G$ with the property that $M$ is a direct summand of $\mathrm{Ind}_{G/U} 1$. It is always a $p$-group and is well-defined up to conjugation.

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This is false, indeed $k_H\uparrow ^G$ may have projective summands which therefore have vertex $\{1\}$. The vertices of your $M$ are contained in conjugates of $H$, but that's the best that can be said in general.

Example: take the symmetric group $S_4$ in characteristic two. It has two simple modules, namely the trivial module and a 2-d simple. Both have projective cover of dimension 8. To see the non-trivial simple, consider the 4d natural rep on basis elements 1,2,3,4. It has a 3d submodule spanned by 1+2,2+3,3=4 (the "augmentation ideal") which has a 1d submodule spanned by 1+2+3+4. The quotient by this 1d submod doesn't have trivial action, and in fact it is simple.

Let $H$ be the subgroup $\langle (1,2) \rangle$. Then $k_H \uparrow ^G = M \oplus N$ where both $M$ and $N$ are indecomposable, $M$ is projective of dimension 8 (it's the projective cover of the 2d simple module) and $N$ is a copy of the natural module (whose vertex is $H$). Perhaps the easiest way to verify this is with Magma (there's an online Magma calculator if you don't have institutional access).

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Wonderful - thank you, Matthew! A quick follow up question: is it true that at least (exactly?) one summand of such a permutation module will have vertex as big as possible? –  Alex B. Mar 29 '12 at 23:54
    
"Exactly" will be false, but "at least" is true. Take $P$ a Sylow $p$ of $H$ and take $M$ a summand of $k_H\uparrow^G$ such that $M|_P$ contains the trivial summand of $k_H\uparrow^G|_P$. Then if $M$ has vertex $D$, $k_P | (M|_P) | k_D\uparrow ^G |_P = \bigoplus _g k_{D^s \cap P}\uparrow^P$. But the summands are indecomposable by Green's Indecomposability Theorem. If one of them is to be trivial, $D$ must be a conjugate of $P$. –  mt_ Mar 30 '12 at 8:02
    
Good point. Thanks! –  Alex B. Mar 30 '12 at 11:29

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