Is the normalizer of a Sylow $p$-subgroup is a $p$-group?

Question

Given a group $G$ and a Sylow $p$-subgroup $P$ of $G$, is the normalizer $N(P)$ of $P$ a $p$-group?

I think that it may not be a $p$-group, so if there is an element $g \in G$ such that $g \in N(P)$ and $g \notin P$, the order of $g$ is not a power of $p$ since $g \notin P$ and $P$ is the maximal p-subgroup of $G$.

Answer 1

$N(P)$ is a $p$-group if and only if $N(P)=P$, because $P$ is a maximal $p$-subgroup. Any subgroup properly containing $P$ cannot be a $p$-group.

Answer 2

Not true in general. If $G$ is not a $p$-group with a normal Sylow $p$-subgroup, then $N_G(P)=G$. Example $G=S_3$, $P=A_3$.

Answer 3

If $N(P)$ were a $p$-group, since $N(P) \supseteq P$, which is by definition a maximal $p$-group, this would imply that $P=N(P)$.

So the answer is no, it happens if and only if $P$ is self-normalizing (see the other question you asked about that!)