Normalizer of a Sylow subgroup

Question

In this question Sylow subgroups of soluble groups Jack Schmidt mentions that the normalizer of $P$ in $S_p$ is solvable. Suppose $P$ is generated by a cycle of length $p$. Could you provide any hints, why is it true, because I do not quite see the solvability of the normalizer and why actually $G$ (which is transitive) lies in it.

Answer 1

By counting the number of elements of order $p$ in $S_p$ and by using the fact that elements of order $p$ are conjugate in $S_p$, you can prove that $|C_{S_p}(P)| = p$ and $|N_{S_p}(P)| = p(p-1)$.

Then prove that $N_{S_p}(P) / C_{S_p}(P)$ is cyclic of order $p-1$, so $N_{S_p}(P)$ is solvable.

The second part of your question is answered by Jack in the other question, but here it is again with more detail:

Let $G \leq S_p$ be solvable and transitive. Then $G$ is primitive, since any stabilizer has prime index $p$ (so every stabilizer is maximal). Hence any nontrivial normal subgroup $N$ of $G$ is transitive, since the orbits under $N$ form a block system.

Take $N \neq 1$ to be a minimal normal subgroup of $G$. Now $N$ is transitive, so $p$ divides the order of $N$. On the other hand $G$ is solvable, so $N$ is elementary abelian of prime power order. Thus $N = P$ and $P$ is a normal subgroup.