TITLE: $\operatorname{Aut}(A_4)\simeq S_4$
QUESTION [7 upvotes]: I have to prove that $\operatorname{Aut}(A_4)\simeq S_4$.
If $G$ is a finite group, $p$ divides $|G|$, $n=n_p(G):=|\operatorname{Syl}_p(G)|$ then we know that $\operatorname{Aut} G\le S_n$. I used this to prove that $\operatorname{Aut}(A_4)\le S_4$, since $n_3(A_4)=4$.
But now, how can I show the inverse inclusion, i.e. that $S_4\le \operatorname{Aut}(A_4)$?
I know that $S_4\simeq \operatorname{Aut}(S_4)$ but I don't know if it could help.
Any hint would be appreciated! Thanks!

REPLY [2 votes]: Let $G=A_4$,
By the fact you mentioned, $\operatorname{Aut}(G)\leq S_4 $ and since $\operatorname{Inn}(G)\cong G/Z(G)\cong G\leq\operatorname{Aut}(G)$;
we have $$A_4\leq\operatorname{Aut}(G)\leq S_4$$.
Since $A_4$ is only subgroup of $S_4$ with order $12$ it is characteristic in $S_4$ which means there is a homomorphism from $
\operatorname{Aut}(S_4)\cong S_4$ to $
\operatorname{Aut}(G)$ and since $
\operatorname{Aut}(G)$ has at least $12$ elements and $S_4$ has no normal subgroup of order $2$, we have isomorphism from $S_4$ to $
\operatorname{Aut}(G)$.