A Group of Order $540$ is not simple

Question

Why is a group of order $540$ not simple? The hints I have been given are not helpful.

Here's what I have been told.

Let $G$ be such a group. Then there are $36$ Sylow $5$-subgroups; let $H$ be one of them. Also, there are $10$ Sylow $3$-subgroups; let $K$ be one of them.

Then $[G: N(H)] = 36$, whence $N(H)$ has order $15$; also, $[G: N(K)] = 10$, whence $N(K)$ has order $54$. We can show that $N(H)$ and $N(K)$ intersect in a subgroup of order $3$.

Then $N(N(H)\cap N(K))$ has order divisible by $45$.

All is fine up to here.

The next hint I get says that a group of order $45$ is abelian (which it indeed is).

But how does this solve the problem?

Answer 1

Hint in light of Jack's second comment:

Lemma: Let $G$ be a simple group and let $H< G$ such that $[G:H]=n$. Then $G\hookrightarrow A_n$.

Why we can use here. Because if $|G|<\infty$ and $H< G$ such that $[G:H]=n$ provided $|G|\nmid n!/2$ so $G$ is not simple. Now think of using this lemma while you're led to a subgroup of order $108$.