# Vertex of a trivial source module

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).
"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 $P$. – m_t_ Mar 30 '12 at 8:02