Classification of solvable transitive subgroups of symmetric group of prime degree

Statement

Suppose $p$ is a prime number, and $G = \operatorname{Sym}(p)$ is the symmetric group on a set $S$ of size $p$. Suppose $H$ is a solvable subgroup such that the induced action of $H$ on $S$ is transitive. Then, we have:

1. $H$ contains, as a normal subgroup, a cyclic group of order $p$, generated by a $p$-cycle.
2. $H$ is contained in the holomorph of this cyclic group, which is a group of order $p(p-1)$, isomorphic to the general affine group $GA(1,p)$ and more explicitly a semidirect product of the cyclic group generated by the $p$-cycle and a cyclic subgroup of order $p-1$.

Conversely, any subgroup $H$ of $G$ satisfying the above two conditions is solvable and the induced action on $S$ is transitive.

Proof

We assume $S = \{ 1,2,\dots, p \}$.

Proof that it contains a cyclic normal subgroup of order $p$

1. Let $N$ be a minimal normal subgroup of $H$ (such a subgroup exists because $H$, being transitive, is a nontrivial solvable group). By fact (1), $N$ is an elementary abelian group.
2. $N$ is also transitive on $S$: Since $H$ is transitive on $S$, $H$ is transitive on the $N$-orbits. In particular, all the $N$-orbits must have equal size, and this size divides the size of $S$. On the other hand, the size of $S$ is prime, so the size of the $N$-orbits is $1$ or $p$. Since $N$ is nontrivial, the size must be $p$.
3. $N$ is cyclic of order $p$: Let $a \in S$ and $M$ be the stabilizer of $a$ in $N$. Then, by the fundamental theorem of group actions (fact (2)), $M$ has index $p$ in $N$. In particular, $p$ divides the order of $N$. Since no higher power of $p$ divides the order of $G$, $p$ is a maximal prime power dividing the order of $N$. Since $N$ is elementary abelian, $N$ is cyclic of order $p$.

Thus, we have found a normal subgroup $N$ of $H$ that is cyclic of order $p$.

Proof that it is contained in the holomorph

1. Clearly, $N$ is normal in $H$ if and only if $H$ is contained in the normalizer $N_G(N)$. Thus, it suffices to show that the normalizer $N_G(N)$ is a solvable group of order $p(p-1)$ described as $GA(1,p)$.
2. Consider the map $N_G(N) \to \operatorname{Aut}(N)$ induced by the conjugation action. The kernel of this map is $C_G(N)$, and since $N$ has one cycle, $C_G(N) = N$. Thus, $N_G(N)/N$ is isomorphic to a subgroup of $\operatorname{Aut}(N)$, and hence has order at most $|\operatorname{Aut}(N)| = p-1$. So, $N_G(N)$ has size at most $p(p-1)$.
3. On the other hand, identity the elements of $S$ with the field of $p$ elements in such a way that one of the generating cycles for $N$ corresponds to translation by $1$. The general affine group $GA(1,p)$ then acts on $S$ in the usual way, with $N$ as the normal subgroup of translations, and a complement of order $p-1$ acting by dilations (scaling). We thus have a solvable group of size exactly $p(p-1)$ with $N$ as a normal subgroup, so this must equal $N_G(N)$. This completes the proof.