Normalizer of the normalizer of the sylow $p$-subgroup 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 $P$ ?
 A: 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 $P\unlhd N(N(P))$.
A: Hint: P is a normal Sylow p-subgroup of $N_G(P)$...
A: Another proof
We have that $P\in\text{Syl}_p(\mathbf{N}_G(P))$ and $\mathbf{N}_G(P)\trianglelefteq \mathbf{N}_G(\mathbf{N}_G(P))$. By Frattini's Argument:
$$\mathbf{N}_{\mathbf{N}_G(\mathbf{N}_G(P\,))}(P)\cdot\mathbf{N}_G(P)=\mathbf{N}_G(P)=\mathbf{N}_G(\mathbf{N}_G(P)),$$ because $\mathbf{N}_{\mathbf{N}_G(\mathbf{N}_G(P\,))}(P)\subseteq \mathbf{N}_G(P)$.
A: Let $N=N_G(P)$. Let $x\in N_G(N)$, so that $xNx^{-1}=N$. Then $xPx^{-1}$ is a Sylow $p$-subgroup of $N\leq G$. Since $P$ is normal in $N$, $P$ is the only Sylow $p$-subgroup of $N$. Therefore $xPx^{-1}=P$. This implies $x\in N$. We have proved $N_G(N_G(P))\subseteq N_G(P)$.
Let $y\in N_G(P)$ Then certainly $yN_G(P)y^{-1}=N_G(P)$, so that $y\in N_G(N_G(P))$. Thus $N_G(P)\subseteq N_G(N_G(P))$.
A: 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.
A: There are many answers already but I'll give anoter one (I didn't see it somewhere above).
It is based on the following proposition:

Let $G$ be a finite group and $P$ a Sylow $p-$subgroup of $G$. If $N_G(P)\leq H\leq G$ then $H=N_G(H)$.

Proof: Let $g\in N_G(H)$. Then $P\leq H\Rightarrow gPg^{-1}\subseteq gHg^{-1}=H$. So $P,gPg^{-1}$ are Sylow $p-$subgroups of $H$. Hence they are conjugate, i.e. there exists $h\in H$ s.t. $P=hgPg^{-1}h^{-1}\Rightarrow hg\in N_G(P)\leq H\Rightarrow g\in H\Rightarrow N_G(H)\subseteq H\checkmark$.
Now, apply the proposition for $H=N_G(P)$.
A: Let $M= N_G(P)$. Clearly, $M\subseteq N_G(M)$.
Now, notice that $P$ is normal in $M$, so it is the unique Sylow $p$-subgroup of $M$. Therefore, if $x\in N_G(M)$, 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 $x\in N_G(P) = M$. Therefore, $N_G(M)\subseteq N_G(P)$. 
