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Problem:

Prove that the subgroup of upper triangular matrices in GL$_3$($\mathbb{F}$$_2$) is isomorphic to the dihedral group of order 8.

I was doing this problem and found a solution but it isn't very elegant and I am hoping there is a better way to solve the problem. I simply listed out the 8 elements of GL$_3$($\mathbb{F}$$_2$) that are upper triangular and then computed their orders then one could go on to compute products to find analogous generators for GL$_3$($\mathbb{F}$$_2$) like D8 has. I stopped after computing orders but I think that general brute force outline would work.

But I was hoping someone could post a more interesting and elegant solution that doesn't use such computationally heavy techniques.

Some quick clarifying definitions:

GL$_3$($\mathbb{F}$$_2$) is the set of $3\times3$ matrices with entries from $\mathbb{F}$$_2$ = $\mathbb{Z}$/$2$$\mathbb{Z}$

$D_8 = \langle r, s | r^4 = s^2 = 1, rs = sr^{-1} \rangle $

Thanks!

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  • $\begingroup$ You could either find matrices $r$ and $s$ satisfying the relations of $D_8$, whihc is not hard, or you could prove that the matrix group is nonabelian of order $8$ and has more than one element of order $2$, and use the fact that any such group is isomorphic to $D_8$. $\endgroup$ – Derek Holt Jun 10 at 7:23
  • $\begingroup$ If you know that there are only non abelian groups of order $8$, then it is very easy to show. Just observe that there at least $2$ matrices of order $2$. $\endgroup$ – Sunny Jun 10 at 7:23
  • $\begingroup$ I did not know that fact on groups of order 8 having more than one element of order 2, thanks! ... I found a matrix with order 4, several with order 2, how would I find an additional matrix of order 2 that satisfies rs = sr^-1 ..... would the only way just be trying out a few candidate matrices in that equation @DerekHolt $\endgroup$ – H_1317 Jun 10 at 7:28
  • $\begingroup$ Any matrix of order $4$, and any mattix of order $2$ that is not in the centre (for example an elementary matrix with $1$ in the $1,2$ position) will work. $\endgroup$ – Derek Holt Jun 10 at 10:41
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Let's use the outline suggested by our commentors.

The order, it is easy to see, is $8$, since the upper triangular invertible matrices have dimension $\dfrac{n^2-n}2=3$.

Now we must find two elements of order $2$, since the quaternions don't have such.

So, $\begin{pmatrix}1&0&1\\0&1&0\\0&0&1\end{pmatrix}$ and $\begin {pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix}$, should work.

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