Classification of solvable transitive subgroups of symmetric group of prime degree
Suppose is a prime number, and is the symmetric group on a set of size . Suppose is a solvable subgroup such that the induced action of on is transitive. Then, we have:
- contains, as a normal subgroup, a cyclic group of order , generated by a -cycle.
- is contained in the holomorph of this cyclic group, which is a group of order , isomorphic to the general affine group and more explicitly a semidirect product of the cyclic group generated by the -cycle and a cyclic subgroup of order .
Conversely, any subgroup of satisfying the above two conditions is solvable and the induced action on is transitive.
We assume .
Proof that it contains a cyclic normal subgroup of order
- Let be a minimal normal subgroup of (such a subgroup exists because , being transitive, is a nontrivial solvable group). By fact (1), is an elementary abelian group.
- is also transitive on : Since is transitive on , is transitive on the -orbits. In particular, all the -orbits must have equal size, and this size divides the size of . On the other hand, the size of is prime, so the size of the -orbits is or . Since is nontrivial, the size must be .
- is cyclic of order : Let and be the stabilizer of in . Then, by the fundamental theorem of group actions (fact (2)), has index in . In particular, divides the order of . Since no higher power of divides the order of , is a maximal prime power dividing the order of . Since is elementary abelian, is cyclic of order .
Thus, we have found a normal subgroup of that is cyclic of order .
Proof that it is contained in the holomorph
- Clearly, is normal in if and only if is contained in the normalizer . Thus, it suffices to show that the normalizer is a solvable group of order described as .
- Consider the map induced by the conjugation action. The kernel of this map is , and since has one cycle, . Thus, is isomorphic to a subgroup of , and hence has order at most . So, has size at most .
- On the other hand, identity the elements of with the field of elements in such a way that one of the generating cycles for corresponds to translation by . The general affine group then acts on in the usual way, with as the normal subgroup of translations, and a complement of order acting by dilations (scaling). We thus have a solvable group of size exactly with as a normal subgroup, so this must equal . This completes the proof.