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I have this representation of $S_3$ obtained in the usual way $$\varrho\left(\sigma\right)e_i=e_{\sigma_i}$$. Being more explicit the representation is this one: $$\varrho\left(e\right)=\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right),\,\,\varrho\left(\sigma_{1}\right)=\left(\begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0 \end{array}\right),\,\,\varrho\left(\sigma_{1}\right)=\left(\begin{array}{ccc} 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{array}\right),$$ $$\varrho\left(\sigma_{1}\right)=\left(\begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1 \end{array}\right),\,\,\varrho\left(\sigma_{1}\right)=\left(\begin{array}{ccc} 0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0 \end{array}\right),\,\,\varrho\left(\sigma_{1}\right)=\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{array}\right).$$ As far as I can see this should be reducible... but I really don't understand how since the matrices above don't share any eigenvector. Can anybody help me with this?

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    $\begingroup$ Hint: How does the group act on the sum of the standard basis vectors $e_1 + e_2 + e_3$? $\endgroup$ May 25, 2016 at 11:20

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The idea here is that we want to find an invariant subspace, that is, a subspace $W$ of $V$ such that if $w \in W$ we have that $gw \in W$ for all $g \in S_3$. As Tobias mentioned, we want to pick a subspace where no matter how we permute the basis vectors, we get the same subspace. The subspace $W = <(1,1,1)>$ is invariant, and it's complement is a two dimensional representation which is irreducible.

I mention this all in such detail as I would like to point out that the permutation representation of any subgroup of $S_n$ is not irreducible (N.B. Cayley's theorem tells us that this means the permutation representation of any finite group). For all of these groups we have the same invariant subspace, and some subgroups might have other invariant subspaces.

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