Classification of solvable transitive subgroups of symmetric group of prime degree

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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. Minimal normal implies elementary abelian in finite solvable
  2. Fundamental theorem of group actions


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.