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Let G be a finite group acting on a finite dimensional vector space V. Let C be a nontrivial subspace of V. Let H be the subgroup of G that fixes C pointwise (the stabilizer of C). I'm fairly sure that H has to be abelian. The argument going along the lines that C contains simultaneous eigenvectors of all the elements of H so these must commute...

I have two questions :

(1)Is the above correct? Is there a reference with a derivation.

(2)Is there a more general statement or setting for this?


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No, this is not correct. Let $G$ be any subgroup, $C=V$ the trivial representation. Then $H=G$ needs not be abelian. –  Joël Jan 3 '12 at 20:15
Nontriviality doesn't help you with anything, either. Suppose you require that no element of the group fixes all the points of the space. Then the first counterexample is in three dimensions. The symmetric group $S_3$ acts on $\mathbb R^3$ by permuting the three coordinates. This means that the subspace generated by $(1,1,1)$ is fixed by this action, so its stabilizer is the whole group $S3$, which is not abelian. –  Will Sawin Jan 3 '12 at 20:55
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closed as too localized by Benjamin Steinberg, Andreas Thom, Matthew Kahle, Kevin Walker, Dan Petersen Jan 3 '12 at 21:36

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1 Answer

At least over the complex numbers, whenever the finite group $G$ has a non-Abelian maximal subgroup $H,$ we can let $U$ be the permutation module induced from the trivial module for $H.$ Then $U$ has a non-trivial irreducible summand $V.$ By Frobenius reciprocity, the fixed-point space $C$ of $H$ on $V$ is non-zero. By maximality of $H$ and non-triviality of $V$, $H$ is precisely the stabilizer of $C$ indicated in the question. It can be arranged that $V$ is faithful, by taking $G$ to be a non-Abelian simple group, for example (and all maximal subgroups of non-Abelian simple groups are themselves non-Abelian).

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