Given $p$ prime and $H\le S_p$ s.t. $|H| = p$, show if $H\subseteq N\subseteq S_p$ for some nilpotent subgroup $N$, then $H=N$

Question

Problem Statement:

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 $H=N$.

By working with conjugacy classes and orbit-stabilizer theorem, we know that $C_{S_p}(H) = H$. Hence if we can show that $N$ is abelian, then we are done. But I only know that $H$ is a cyclic subgroup, it is normal in $N$ by properties of nilpotent group. Can I extract something from $N/H$ or do I miss some techniques?

Answer 1

We know that $H$ is a subgroup of order $p$ in $G=S_p$, and the question already states that $C_G(H)=H$ (it is clear from the conjugacy classes of $S_p$ that $|C_G(H)|=p$).

Let $N$ be any subgroup of $G$ containing $H$. If $N$ is nilpotent then $Z=Z(N)$ is non-trivial. But $H\leq N$, so $Z(N)$ centralizes $H$. Thus $1\neq Z(N)\leq C_G(H)=H$.

Thus $Z(N)=H$ as $H$ has no subgroups other than $1$ and $H$. But then $N\leq C_G(H)=H$, and so $H=N$.