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In a note I saw this fact that $PSL(3,q)$ where $q=p^n$ does not have any abelian subgroup of order $q^3$. But I could not prove it or find any reference about it, could you please help me about it? Thanks

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@Beh: You should accept Mark's answer. On the other hand, your question is quite elementary (far from research-level mathematics) and perhaps better suited to a site like math.stackexchange.com. –  Jim Humphreys Dec 19 '12 at 20:40
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up vote 7 down vote accepted

One of the Sylow $p$-subgroups of $PSL(3,q)$ is the subgroup of all unitriangular matrices (i.e. upper triangular matrices with 1 on the diagonal): just compute the order of $PSL(3,q)$ and the order of the unitriangular subgroup which is $q^3$ (or look in Bogopolsky's group theory book). Therefore if $PSL(3,q)$ contained an Abelian subgroup of order $q^3$, the Sylow $p$-subgroup would be Abelian, but it is not (it is a finite version of the Heisenberg group).

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Thank you very much for your very helpful answer. I am really thankful. –  Beh Dec 19 '12 at 11:42
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