Given $p$ odd prime and $G=S_p$ and $P\leq G$ $p$-Sylow and $H=N_G(P)$ find $|H|$ and prove if $p=5$ then $H \cong C_4 \ltimes C_5 $

Question

Given $p$ odd prime and $G=S_p$ and $P\leq G$ $p$-Sylow and $H=N_G(P)$ find $|H|$ and prove if $p=5$ then $H \cong C_4 \ltimes C_5 $.

For the first part we know that if we set $n_p$ the number of Sylow groups in $G$ then $n_p=1(p)$ and $n_p\mid (p-1)!$ so $n_p\in\{1,(p-2)!\}$ if $n_p=1$ then $P$ is normal and $H$ is all $G$ and otherwise we can use the stabilizer orbit thm. and get $|H|=p(p-1)$ not sure how to should I procced and say that $n_p\neq1$ and not sure if the calculations are right. any help please

Answer 1

Hints.

1. $P$ is a cyclic group. Let $P=\operatorname{gr}(\sigma)$.

2. For every $k$, $0<k<p$ there exists $\alpha\in S_p$ such that $\alpha\sigma\alpha^{-1}=\sigma^k$.

3. If $\alpha,\beta\in S_p$ and $\alpha\sigma\alpha^{-1}=\beta\sigma\beta^{-1}$, then $\alpha P=\beta P$.

4. It follows that $|H/P|=p-1$. In fact $H/P\cong\mathbb{Z}_p^*$.