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Please help me to solve this problem. Let $P$ is a $p$-Sylow subgroup of the finite group $G$ and $g$ is an element such that $\lvert g \rvert=p^k$ then if $g$ is in $N_G(P)$ then $g\in P$. Where to start?

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Thank you all. I got it right. –  Basil R Feb 28 '13 at 19:18
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up vote 2 down vote accepted

Hint: $P$ is a normal $p$ Sylow subgroup of $N_G(P)$, and because of its order, $g$ must be in a Sylow p-subgroup of this group...

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If $H \leq G$ is a $p$-group, then $H \leq N_G(P)$ if and only if $H \leq P$.

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Let $|G|=p^nm,~~(p,m)=1$ and let $H=\langle g\rangle$. So we see that $|H|=p^k$ and morover $|N_G(P)|=p^n.t$ such that $t\mid m$. Try to use this point that $H\le N_G(P)$ and that $HP$ is a subgroup of the normalizer of $P$. what will you get when you write $|HP|\big||N_G(P)|$...

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Nicely done! +1 –  amWhy Feb 28 '13 at 19:15
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Let $P$ be a $p$-Sylow subgroup of $G$ and recall that $$N_G(P) = \{ n \in G | P^n = P \}$$ so as rschwieb pointed out $P$ is a normal $p$-Sylow subgroup of $N_G(P)$, but that implies it is the only $p$-Sylow subgroup of $N_G(P)$ by the lemma: therefore $g \in P$.

Lemma For a Sylow $p$-subgroup $P$ of $H$: $P$ unique of $H$ iff $P \unlhd H$.
proof: If $P$ is unique then because all Sylow $p$-subgroups are conjugate $P^h = P$ so it is a normal subgroup. If $P \unlhd H$ then $P^h = P$ and every Sylow $p$-subgroup is a conjugate of any other, so there is only one.

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