Given $p$ prime and $H \le S_p$ such that $|H| = p$, show that if $H \leq N \leq S_p$ for some nilpotent subgroup $N$, then $H=N$.
The question was already asked. But I didn't understand the hint. So far, I know that $H$ is normal in $N$ and $H<Z(N)$. From this, how can I conclude $H =N$? I can't see. Could you give me a hint?
You first need to note that $C_{S_{p}}(H)=H$ ($H$ has order $p$, so you can assume that (a conjugate of) $(1 2 3\cdots p)$ is a generator; by the way, in general, if $\sigma$ is an $n$-cycle in $S_n$, then $\langle \sigma \rangle$ is the centralizer of $\sigma$ in $S_n$). Now $N$ is nilpotent, so $Z(N) \gt 1$, and we can pick an $x \in Z(N)$, $o(x)=p$. Since $x$ is in the center of $N$ and $H \subseteq N$, we have $x \in C_{S_{p}}(H)$, whence $x \in H$, and this forces $\langle x \rangle = H \subseteq Z(N)$. But then $N$ centralizes $H$, so $N \subseteq C_{S_{p}}(H)=H$ and you are done.