Let $p$ be an odd prime and G be a group of order $4p^3$. Prove that G either has a normal subgroup of order $p^2$ or $p^3$.
By Sylow's Theorem, $n_p \equiv 1\pmod{p}$ and $n_p |$ $\mid G \mid$, so $n_p \in \{1,4\}$. If $n_p = 1$ were done so suppose $n_p = 4$. Hence $4\equiv 1\pmod{p}$ which implies $3=0$ mod $p$ and so $p=3$. Thus we have that $|G| = 4 \cdot 3^3$.
I'm not sure where to go from here.