Let $p$ be a prime dividing $|G|$, and let $S_p$ be a $p$-Sylow subgroup of $G$. Show that $N(N(S_p))=N(S_p)$.

Question

Attempt: Trivially $N(S_p) \subseteq N(N(S_p))$ because a subgroup is a subset of its normalizer. Then, let $g\in N(N(S_p))$. Then we have $gN(S_p)g^{-1}=N(S_p)$ by the definition of normalizer. But $S_p \unlhd N(S_p)$, and since $S_p$ is also a $p$-Sylow subgroup of $N(S_p), gS_pg^{-1}=S_p$, and so $g\in N(S_p)$, then we are done. My question is whether it is okay I go from $S_p \unlhd N(S_p)$ and $S_p$ a $p$-Sylow subgroup of $N(S_p)$ to it is conjugate to itself. This reminds me of if there is only one $p$-Sylow subgroup, it is equivalent of saying this $p$-Sylow subgroup is normal. Is the inverse statement of this true?

Answer 1

Any subgroup containing $N(S(P))$ is self-normalizing. For instance, $N(N(S(P)))$.

You can find more detailed explanations in https://en.wikipedia.org/wiki/Frattini%27s_argument.