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.

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 $G$.

