How to show $\operatorname{GL}_2 (\mathbb Z_2)\cong D_3$

Question

I want to show that $\operatorname{GL}_2 (\mathbb Z_2)\cong D_3$ while $\operatorname{GL}_2(\mathbb Z_2)$ is the group of matrices $2\times 2$ above $\mathbb Z_2$. I tried to show that maybe every $A\in \operatorname{GL}_2(\mathbb Z_2) $ is a rotation or reflection, but I failed to do that.

Answer 1

The group $\operatorname{GL}_2(\mathbb{Z})$ acts on the space $\mathbb{Z}_2\times\mathbb{Z}_2$, and it fixes the origin. The remaining 3 points are permuted, and every permutation is realised (prove it!). This means that $\operatorname{GL}_2(\mathbb{Z})$ has $S_3$ as a homomorphic image, but as both groups have order 6 we have that $\operatorname{GL}_2(\mathbb{Z})\cong S_3$.

Finally, label the points of your triangle with $\{1, 2, 3\}$ and use a similar argument to see that $D_3\cong S_3$. Hence, $\operatorname{GL}_2(\mathbb{Z})\cong D_3$.