Show that the groups $SL_2(\mathbb{F}_3)$ and $S_4$ are two nonisomorphic groups of order 24.

Question

I proved that $$SL_2(\mathbb{F}_3)=\left\langle \begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix}, \begin{pmatrix} 1 & 0\\ 1 & 1\end{pmatrix}\right\rangle$$ and know that $$S_4=\langle (1\ 2),(1\ 2\ 3\ 4)\rangle.$$ Is it enough to say that the order of each generator is different so any isomorphism $$\varphi:SL_2(\mathbb{F}_3)\to S_4$$ sending generators to generators would fail?

Answer 1

Generators are not unique, so it's not so simple. I'll give a hint for a solution: $SL_2(\mathbb{F_3})$ has an element of order $6$.

Answer 2

Since $1\neq-1$ in $\Bbb F_3$ the group ${\rm SL}_2({\Bbb F_3})$ has a non-trivial centre.

On the other hand the centre of ${\cal S}_4$ is trivial.

Answer 3

It is possible to turn your idea into a correct argument.

To do this, you need to show that $S_4$ cannot be generated by elements of order $3$. And this follows by noting that all such elements are contained in the proper subgroup $A_4$.