Prove that a group of order $540$ is not simple
In this problem I have done the following steps.
Sylow's Theorem gives $n_3=10$ and $n_5=36$. Let $P$ be a Sylow $3$ subgroup and $Q$ a Sylow $5$ subgroup. Then the order of $N(P)$ is $54$ and the order of $N(Q)$ is $15$, so this last one is cyclic.
$N(P)$ and $N(Q)$ have an intersection of order $3.$
Here is the point where I am stuck.
The hints say to prove $45$ dividers the normalizer of this intersection and then that this fact contradicts the order of $N(Q)$ is $15.$
Any help will be very much welcomed.