How to prove that $\mathrm{SL}_{2} (\mathbb F_{3})$ is not isomorphic to $S_{4}$? How to prove that $\mathrm{SL}_{2} (\mathbb F_{3})$ is not isomorphic to $S_{4}$?
They are both group of order $24$ and both groups have elements of order $2, 3$ and $4$.
I don't know how to work from here.
 A: $$A:=\begin{pmatrix}2&0\\1&2\end{pmatrix}\implies A^2=\begin{pmatrix}1&0\\1&1\end{pmatrix}\;,\;\;A^3=\begin{pmatrix}2&0\\0&2\end{pmatrix}=-I\implies A^6=I$$
and thus SL$_2(\Bbb F_3)\;$ has an element of order six....but $\;S_4\;$ hasn't any.
A: One can study the subgroups in both cases. The proper subgroups of $SL_2(\mathbb{F}_3)$ are the cyclic groups $C_2,C_3,C_4,C_6$ and the quaternion group $Q_8$. But $S_4$ has no subgroup isomorphic to the quaternions: suppose a subgroup $H \leq S_4$ exists such that $H \cong Q_8$.
Now $Q_8$ contains $6$ elements of order $4$, and $S_4$ contains exactly $6$ elements of order $4$, namely the 4-cycles. Thus $H$ contains all $4$-cycles in $S_4$. But then $H$ also contains all products of two $2$-cycles, so $|H| > 8$, a contradiction. Thus no such $H$ exists, and $S_4$ is not isomorphic to $SL_2(\mathbb{F}_3)$. 
Of course, also $C_6$ does not arise as a subgroup of $S_4$.
A: $SL_2(\mathbb{F}_3)$ has two elements $I, -I$ in its center, but the center of $S_4$ is trivial.
A: The character tables of $S_4$ and of $\mathrm{SL}_2(\mathbb F_3)$ are different. The first group contains only integer entries (as any symmetric group) and the character table of $\mathrm{SL}_2(\mathbb F_3)$ contains a cubic root of unity.
