Let $G = GL(2,3)$, the group of all invertible $2 \times 2$ matrices over the field of $3$ elements. Show that $G/Z(G) \cong S_4$.
I know that $G/Z(G)\cong Inn(G)$, where $Inn(G)$ is the inner automorphism group of $G$. But it's difficult for me to construct an isomorphism from $Inn(G)$ to $S_4$ -- how to enumerate the inner automorphisms of the matrix group, for example, is a tough idea for me to see.
I'd rather construct a surjective homomorphism $\phi: G \longrightarrow S_4$ such that $ker(\phi) = Z(G)$, so that the result follows from the First Isomorphism Theorem. But I'm not sure how such a map can be constructed. I do know that, for the desired result to hold, $Z(G)$ has order $2$ -- so I think that $ker(\phi)$ should only contain two matrices, the identity matrix along with one other matrix in $G$ (which, I believe, would have to be the identity matrix multiplied by $2$).
How can I construct the above homomorphism ?
Thanks!
Group $\mathrm{GL}_2(\mathbb{F}_3)$ acts on one-dimensional subspaces of space $\mathbb{F}_3^2$. There are precisely four one-dimensional subspaces of $\mathbb{F}_3^2$. This gives a homomorphism $\phi:\mathrm{GL}_2(\mathbb{F}_3) \rightarrow S_4$. You can verify that $\phi$ is surjective and $\mathrm{ker}(\phi) = Z\left(\mathrm{GL}_2(\mathbb{F}_3)\right)$.