Prove that the order of a group $G$ must equal $36$ if $|G| = 4p^2$ for some odd prime and it has a non-normal Sylow $p$-subgroup

Question

The question is the following:

Let $p$ be an odd prime, and let $G$ be a group of order $4p^2$. Assume that a Sylow $p$-subgroup of $G$ is not normal in $G$. Prove that $|G| = 36$.

I understand that a Sylow $p$-subgroup is normal iff $|\mathrm{Syl}_p(G)| = 1$, and I can prove that if $G$ has order $|G| = 4p^2$ then it isn't simple. However, I cannot find anything wrong about having all Sylow $p$-subgroups normal for $p\geq 5$. Could someone provide a hint?

Answer 1

By Sylow's third theorem, the number of Sylow $p$-subgroups $n_p$ satisfies $n_p\equiv1\pmod p$. Also $n_p\mid|G|=4p^2$. So $n_p$ is one of $1$, $2$, $4$, $p$, $2p$, $4p$, $p^2$, $2p^2$ and $4p^2$. Which of these could be $\equiv1\pmod p$?