# A group of linear isomorphisms of $\mathbb C^n$ must have an invariant subspace

Let $G$ be a finite group acting linearly on $\mathbb C^n$, and suppose that $|G| < n^2$. I am trying to show that there is a nonzero invariant subspace $W\subset\mathbb C^n$, i.e. $g(w) \in W$ whenever $w\in W$.

I know that if there is an $x\in \mathbb C^n$ with $\sum_{g\in G} g(x) \not=0$, then the linear span of $x$ is the desired subspace. However, I am not sure how to produce such an $x$ or if this is even the right way to go about the problem. Any suggestions?

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That would give a vector which is fixed, and in general there is none. – Mariano Suárez-Alvarez Jan 6 '13 at 18:41
Are you familiar with Burnside's irreducibility criterion? – Jyrki Lahtonen Jan 6 '13 at 18:43

## 2 Answers

It is an elementary property of finite groups that the sum of the squares of the dimensions of the irreducible representations is equal to the order of the group. (Consider decomposing the regular representation). Since here $|G| < n^2$, this representation certainly cannot be irreducible, so by definition there is an invariant subspace. (In fact, we see that the result holds for $|G| = n^2$ as well, as long as $n>1$, since there is always the trivial representation).

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Pretty and effective! – Geoff Robinson Jan 6 '13 at 19:51

Every simple $G$-module is of dimension at most $|G|$, so your $\mathbb C^n$ cannot be simple.

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