Exercise 2.4.11 ($\mathrm{SL}_2(\mathbb{F}_3) \not \simeq S_4.$)

Proof. We know that $|S_4|=4!=24$, and by the first part of the solution of Exercise 10, $|{\mathrm{SL}}_2({𝔽}_3)|=24$.

I use the Sylow Theory of Section 4.5 to show that these two groups are not isomorphic.

By Exercise 10, ${\mathrm{SL}}_2({𝔽}_3)$ contains a group $Q$ isomorphic to $Q_8$ as a $2$-Sylow subgroup.

Moreover, by Exercise 7, $S_4$ has a subgroup $H=⟨(12),(13)(24)⟩$ isomorphic to $D_8$, which is a $2$-Sylow.

Suppose for the sake of contradiction that ${\mathrm{SL}}_2({𝔽}_3)\simeq S_4$. Then $S_4$ contains a subgroup isomorphic to $Q_8$ and a subgroup isomorphic to $D_8$. By the second Sylow Theorem, all the $2$ subgroups are conjugate. This implies $D_8\simeq Q_8$, but this is false, since $D_8$ has $5$ elements of order $2$, and $Q_8$ only one.

In conclusion,

□