Show that there are no nonabelian simple groups with order smaller than $60$. (The methods we have discussed up to this point are sufficient to deal with every order except for $24$, $36$ and $48$. Here is a hint if $|G| = 36$: Suppose that $G$ has distinct Sylow $3$-subgroups $H$ and $K$. What is $|H \cap K|$? What can you say about $|N(H \cap K)|$? Find a nontrivial proper normal subgroup of $G$.)
This is my last group theory exercise from the book "Abstract Algebra" by gregory T. Lee. I have already managed to cover all the cases except those mentioned in the comment of the problem. In this and in this answer they use Burnside's theorems and in others that I have seen, group actions. But this material is not in the book and I would like to do it by following the hint that is in it, as this is an elementary way of doing it. I just need an extra hint to get started, any help would be appreciated.
This doesn't use the hint, but it's an answer.
Let $\vert G \vert=36=2^2 \cdot 3^2$. For $G$ to have any chance to be simple, we must have $n_3(G)=4$. Let $H_1, H_2, H_3, H_4$ be the $3$-Sylow subgroups of $G$. Then we can define a homomorphism $\varphi:G \to S_4$ via $\varphi(g)(n)=k$, where $g^{-1}H_n g=H_k$. In other words, we let $G$ permute the Sylow subgroups via conjugation.
Consider $\ker(\varphi)$. We know that $\varphi$ has non-trivial image because the Sylow subgroups are not normal, and therefore can't all be fixed by every element of $G$. Thus, $\ker(\varphi) \neq G$. But $36 \gt 24 = \vert S_4 \vert$, so $\varphi$ cannot be injective and it must have non-trivial kernel, which is therefore a non-trivial normal subgroup of $G$.