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Let $G$ be a finite group and $P$ is a p-Sylow subgroup of $G$.
Show that $N(N(P))=N(P)$ where $N(P)$ is the normalizer.

Not sure where to begin... Any hints will be appreciated.

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This is a very standard result, and its proof is by a standard technique. (I say this just to give a psychological context in which to approach the question.) The proof is an application of the Sylow theorems, but you will have to apply them not just in $G$, but in a well-chosen subgroup of $G$ as well. –  Matt E Feb 2 '11 at 15:42

2 Answers 2

up vote 5 down vote accepted

Since $P$ is normal in $N(P)$ it is the unique normal $p$- Sylow subgroup of $N(P)$. However if $x \in N(N(P))$ then $xN(P)x^{-1}=N(P)$ so $xPx^{-1} \subseteq xN(P)x^{-1}=N(P)$. This forces $xPx^{-1}=P$, so $x \in N(P)$. Hence $N(N(P)) \subseteq N(P)$. But $N(P) \subseteq N(N(P))$. Hence $N(N(P))=N(P)$.

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why a downvote without any reason :x) –  anonymous Feb 16 '11 at 11:01

Show that $P$ is normal in $N(N(P))$. We know that $N(P) = \{g \in G: gPg^{-1} = P \}$. So $N(P)$ is the largest subgroup of $G$ for which $P$ is normal. Now $N(N(P)) = \{g \in G: gN(P)g^{-1} = N(P) \}$.

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