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

Solution 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.

Solution 2:

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, $N_G(M)\subseteq N_G(P)$.