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.

Facts used

 * 1) uses::Minimal normal implies elementary abelian in finite solvable
 * 2) uses::Fundamental theorem of group actions

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.