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Maybe this is just a very fundamental problem, but I am not too sure the answer. It is well-known that $SL(2,5)$ is contained in $SL(2,q)$ iff $q$ is odd and $5\mid q(q^2-1)$. My question is whether we can always say $SL(2,5)$ is irreducible on the vector space $GF(q)^2$ ?

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I doubt that this a research level question (see FAQ - the link is at the top). The proper place to ask it is Anyway, if $SL(2, 5)$ is reducible, what dimension does a proper invariant subspace have? How does the stabilizer in $SL(2, q)$ of the invariant subgroup look like? – j.p. Jan 13 '13 at 10:48
It has to be irreducible. For, otherwise, after a change of basis, $SL(2,5)$ will be embedded in upper triangular matrices, and will be solvable. And yes, it is not a research level question and I vote to close. – Venkataramana Jan 13 '13 at 10:50
I agree that SL$(2,5)$ can be embedded into a unipotent group and hence soluble, which leads to a contradiction. But I suspect it would be isomorphic to $\mathcal{Z}_p:\mathcal{Z}_{q-1}$ but not just a $p$-group. – Easy Jan 13 '13 at 13:38
No, it woudl be a $p$-group, because all its elements $x$ have $(I-x)^{2} = 0$ on the underlying space, where $q=p^{a},$ so certainly we have $(I-x^{p}) = 0$. The unipotent action comes because ${\rm SL}(2,5)$ is a perfect group, so acts trivially whenever it acts on $1$-dimensional space. – Geoff Robinson Jan 13 '13 at 18:39
$SL(2,5)$ is not just irreducible, it is absolutely irreducible, i.e. irreducible in its natural action on the the 2-d vector space over $k$, the algebraic closure of $GF(q)$. Again, the way to see this is using @Aakumadula's comment - the stabilizer in $GL_2(k)$ of a $1$-dimensional subspace is solvable and $SL(2,5)$ is not. – Nick Gill Jan 15 '13 at 10:04

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