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Consider $G=GL_n(\mathbb{Z}/2\mathbb{Z})$. What is the smallest $n$ for which $G$ has an element of order 8? Give an example of an element of order 8.

I've thought about just considering what happens to powers of Jordan blocks since each matrix in $G$ is similar to a matrix in Jordan normal form. I've managed to convince myself that the smallest $n$ is 4, but I have no idea how to go about finding a specific element of order 8. Is there a better way than guess and check? There are far too many elements in $GL_4(\mathbb{Z}/2\mathbb{Z})$ just to guess and check.

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By your own argument, you don't need to try all matrices, just the Jordan normal forms (of which there are very few over $\mathbb F_2$). – Henning Makholm Apr 9 '12 at 15:47
@Henning Makholm's hint will get you the right answer, but please don't be fooled into thinking that every matrix over $\mathbb{F}_2$ has a Jordan normal form. This is true for matrices over an algebraically closed field, which $\mathbb{F}_2$ isn't. For example, $\left( \begin{smallmatrix} 0 & 1 \\ 1 & 1 \end{smallmatrix} \right)$ can't be put into Jordan normal form over $\mathbb{F}_2$. – David Speyer Apr 9 '12 at 16:07
The hint I would give (after a bit more thought). You want the minimal polynomial to be $x^8-1$. This will be particularly useful if you note that $x^8-1 \equiv (x-1)^8 \mod 2$. – David Speyer Apr 9 '12 at 16:18
Try writing down the entire Sylow $2$-subgroup. (I promise this is easier than it sounds! Compute the power of $2$ dividing the order of the group first.) – Qiaochu Yuan Apr 9 '12 at 16:27
You want the minimum polynomial to divide x^8-1, but not x^4-1. (x-1)^5 is the first candidate, and its companion matrix has order 8. – Jack Schmidt Apr 9 '12 at 18:29

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