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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 $G$.

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$N(P)$ 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.

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

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  • $\begingroup$ I'm not sure I follow your answer. There's no mention of the subgroup being normal. $\endgroup$ – Matt Samuel Nov 23 '16 at 22:59
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    $\begingroup$ I am just giving an example where it is NOT the case. The normalizer does not have to be a $p$-group. As simple as that, $N_{S_3}(A_3)=S_3$. $\endgroup$ – Nicky Hekster Nov 24 '16 at 6:36
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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!)

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