# Prove that $FS_4$-module is simple

I am solving the following problem:

Consider a field $F$ with $\operatorname{char} (K)=0$, let $\sigma = (1,2)$ and $\pi = (1,2,3,4)$.

An $FS_4$-representation $\rho$ is given by \begin{align} \rho(\sigma) &= \left(\begin{array}{ccc} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \\ \end{array}\right), \\ \rho(\pi) &= \left(\begin{array}{ccc} 1 & 1 & 1 \\ -1 & 0 & 0 \\ 0 & -1 & 0 \\ \end{array}\right). \end{align} Let $V$ denote the $FS_4$-module corresponding to $\rho$. Prove that the $FS_4$-module $V$ is simple.

Well, I feel that the actual proof leads to showing that some particular matrix has no eigenvectors with elements in a field, but I still do not know how to start. And could you please explain what is a "module corresponging to $\rho$"? Is it just the codomain of it?

• I'm not sure what you know already, but you could find the character of the representation and prove it is irreducible. Also the module is just the vector space $V$ on which $S_4$ acts (through $\rho$) – mv3 Nov 22 '13 at 1:26
• Unfortunately, I do not know characters yet. – kraken kraken Nov 22 '13 at 1:35

Hint: If $V$ is reducible, the hypotheses imply that it has a subrepresentation of dimension $1$, i.e., you can find a common eigenvector for $\rho(\sigma)$ and $\rho(\pi)$.