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The matrices $$\begin{pmatrix} 1 & 1 & 1\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} $$ with entries in $\Bbb Z_2$ generate the symmetric group $S_4$. Is this representation irreducible, and if not how does it decompose? I tried to use charactertables $\mod{p}$ but this fails when the order of the group is a multiple of $p$.

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    Do you think of your matrices as acting on the left on column vectors, or on the right on row vectors? In the first case there is an obvious 2-dimensional invariant subspace, and in the second an obvious 1-dimensional invariant subspace. (${\rm GL}(3,2)$ has two conjugacy classes of maximal subgroups isomorphic to $S_4$, these being the stabilizers of $1$- and $2$-dimensional subspaces.) – 2017-01-14
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    I obtaiin these matrices as a list of rows $(e_1,e_2,e_3)$ for the actioni of a group element $g$ giving a result $a^{e_1}b^{e_2}c^{e_3}$ where $a = (2,7)(4,5),$, $b = (3,6)(4,5)$ and $c=(1,8)(2,7)(3,6)(4,5)$. as generators of $\Bbb Z_2^3$ and where g runs in $[ (1,2)(7,8), (2,3)(6,7), (3,4)(5,6) ]$ and the action is the conjugate action. So I suppose the action is on the right on row vectors (please correct me if I'm wrong). Can you point out how to find these subspaces? – 2017-01-14
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    What I did was to determine the orbits of the group on the $8$ points of the vector space giving two orbits of size $1$ and an orbit of size $6$ which tells me nothing at all. – 2017-01-14
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    If they are acting on the right, just look at the third row of each of the three matrices. – 2017-01-14
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    The action is indeed on the right, and the invariant subspace is the one dimensional subspace generated by $c$, but how do I find an invariant complement? – 2017-01-14
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    You don't because there isn't one. The $1$-dimensional submodule is the only nonzero proper submodule. So the representation is reducible but indecomposable. – 2017-01-14
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51764/discussion-between-marc-bogaerts-and-derek-holt). – 2017-01-14

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