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.
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.
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 $N(N(P))=N(P)$.
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 $N(N(P)) = \{g \in G: gN(P)g^{-1} = N(P) \}$.