How to prove "a group $G$ of order $72$ can't be a simple group"?

By using Sylow theorem, I can prove that $G$ has either $1$ Sylow $3$-subgroup or $4$ Sylow $3$-subgroup, but I don't know how to continue the proof.

• Most questions of this type (when you first see them) can be solved using one of the following three methods. 1. Show that, for some prime $p$, $G$ must have a unique Sylow-p subgroup. 2. Show that if $G$ fails to have any normal Sylow subgroups, you end up accounting for more elements of Sylow subgroups then there are in your group. 3. Show that, if $G$ has a Sylow p subgroup, then $|G|$ does not divide $n_p(G)!$. This implies that $G$ is non-simple. Try method 3. Jun 22 '14 at 4:11
• math.stackexchange.com/questions/480558/… would be of some help i guess...
– user87543
Jun 22 '14 at 5:28
• @AbrahamFrei-Pearson Sorry? What's $n_p(G)$ here means? I don't quite know. Jun 22 '14 at 6:24
• The number of distinct Sylow-p subgroups in $G$. If $P$ is a Sylow-p subgroup, then $n_p(G)=|G:N_{G}(P)|$. Jun 22 '14 at 18:38

If there is a unique Sylow $3$-subgroup, it is nontrivial and normal, and you are done.
If there are $4$, then $G$ acts on these subgroups by conjugation, inducing a homomorphism $G\to S_4$. Since $|S_4|=24$, the kernel of this map cannot be trivial, otherwise $G$ would inject into $S_4$.
So it remains to show that the kernel is not all of $G$. Mouse over if you want to see the rest.
If it were, then every $g\in G$ normalizes each Sylow $3$-subgroup, implying each is normal, a contradiction. So the kernel is a nontrivial normal subgroup of $G$.