Normalizer of the normalizer of the sylow $p$-subgroup

Question

If $P$ is a Sylow $p$-subgroup of $G$, how do I prove that normalizer of the normalizer $P$ is same as the normalizer of $P$ ?

Answer 1

We have the following: $P\leq N(P)\leq N(N(P))$. We see that $P$ is also a Sylow $p$-group of $N(P)$ and of $N(N(P))$. If $x\in N(N(P))$, then $xPx^{-1}\leq xN(P)x^{-1}=N(P)$, and since all Sylow $p$-subgroups are conjugate, we have that there exists $y\in N(P)$ such that $xPx^{-1}=yPy^{-1}$. But since $y\in N(P)$, we have that $yPy^{-1}=P$, and so $xPx^{-1}=P$. This shows that $x\in N(P)$, and they must be the same.

Answer 2

Hints ($N(H)$ denotes the normalizer of a subgroup $H\le G$ in $G$):

1) Show that $P$ is the only Sylow $p$-subgroup of$N(P)$. Remember that they are all conjugate in $N(P)$.

2) If $P$ and $P'$ are different Sylow $p$-subgroups, show that $N(P)$ and $N(P')$ are A) conjugate in $G$, B) different.

3) Show that $P$ is the only Sylow $p$-subgroup of $N(N(P))$.

4) Show that $P\unlhd N(N(P))$.

Answer 3

Let $M= N_P(G)$. Clearly, $M\subseteq N_M(G)$.

Now, notice that $P$ is normal in $M$, so it is the unique Sylow $p$-subgroup of $M$. Therefore, if $x\in N_M(G)$, then since $xPx^{-1}$ is a Sylow $p$-subgroup of $xMx^{-1}=M$, then $xPx^{-1} = P$, because $P$ is the only Sylow $p$-subgroup of $M$. That means that $x\in N_P(G) = M$. Therefore, $N_M(G)\subseteq N_P(G)$.

Answer 4

Hint: P is a normal Sylow p-subgroup of $N_G(P)$...