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This is one part of a larger question I'm trying to answer, but I'm stuck on this:

If $u \in \mathrm{GL}_n(\mathbb{F}_p)$, the general linear group of degree $n$ over $\mathbb{F}_p$, and $u$ is unipotent (i.e. there exists a $m\in \mathbb{Z}$ such that $(u-I)^m = 0$), then $u$ has order $p^k$ for some $k \in \mathbb{Z}^+$.

I have already shown that \begin{align*} |\mathrm{GL}_n(\mathbb{F}_p)| &= (p^n-1)(p^n-p)(p^n-p^2)\cdots(p^n-p^{n-1}) \\ &= p^{\frac{n(n-1)}{2}}(p^n-1)(p^{n-1}-1)\cdots(p-1) \end{align*}

and so I was thinking of trying to show $\mathcal{O}(x) \nmid p^j - 1$, so that it must divide $p^{\frac{n(n-1)}{2}}$, but I couldn't show this. Any ideas?

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A unipotent element is an element of the form $I + N$ where $N$ is nilpotent. If $A, B \in M_n(\mathbb{F}_p)$ are matrices that commute then we can show that $(M + N)^p = M^p + N^p$; it follows that

$$(I + N)^{p^k} = I + N^{p^k}$$

which is equal to $I$ as soon as $p^k \ge n$.

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  • $\begingroup$ OH of course. Thanks! $\endgroup$ – Möbius Dickus Nov 27 '17 at 4:05

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