I am pretty new to group theory, and need a little head-start on a proof.
Let $G = GL_2(\mathbb{Z}/2\mathbb{Z})$ be the group of two-by-two invertible matrices over the field $\mathbb{Z}/2\mathbb{Z}$ and $V = \{e,a,b,c\}$ be the Klein $4$-group Let $α = \begin{pmatrix} u & v\\ w & x \end{pmatrix}$ be in $G$. Show that $\alpha$ defines a permutation of the set $\{a,b, c\}$ by the rules: $\alpha (a) = a^u b^w, α(b) = a^v b^x$. Show that $G$ can be identified with $S_3$, the symmetric group on 3 letters.
I suppose it wants me to show $GL_2(\mathbb{Z}/2\mathbb{Z})$ is isomorphic to $S_3$, but I'm not too sure how to do it.
Thanks again.
So by counting the number of possible elements, $GL_2(\mathbb{Z}/2)$ is a group of order $6$, as is $S_3$. Thus to show that the map $GL_2(\mathbb{Z}/2)\to S_3$ is an isomorphism it suffices to show that it is injective. But if we assume that we have some $\alpha \in GL_2(\mathbb{Z}/2)$ that is sent to the identity permutation, then we have by definition that $\alpha(x)=x$, for $x\in \{a,b,c\}$. Since we also have that $\alpha(e)=e$ by definition, we see that $\alpha$ is the identity on $V$. Thus we have injectivity and the map must be an isomorphism.