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 $P$ ?

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 $\mathbf{N}_{\mathbf{N}_G(\mathbf{N}_G(P\,))}(P)\subseteq \mathbf{N}_G(P)$.
