Prove there's no simple group of order $4p^a$ I'm trying to show/prove that there's no simple group of order $4p^a$ with $p \geq 5$ being an odd prime and $a \geq 1$.
Since $p$ is odd prime, $p \nmid 4$. By congruence and divisibility theorem, then $n_p \equiv 1$ (mod p) and $n_p | 4$. But from that $n_p$ can only be 1, which shows there's a Sylow p-subgroup, meaning that there exists a simple group with that order.
Am I doing anything wrong? I would appreciate any help on this.

You showed that $n_p = 1$. This means that there is a unique (hence normal!) $p$-Sylow subgroup. Hence any group of order $4p^a$ has a non-trivial normal subgroup, so it is not simple.
