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

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!

• 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 $D_8$. – Derek Holt Jun 10 at 7:23
• 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 $2$. – Sunny Jun 10 at 7:23
• 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 – H_1317 Jun 10 at 7:28
• 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. – Derek Holt Jun 10 at 10:41

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.