Symmetric and Alternating Groups Generated by a Full Cycle and Another Element

Working in the symmetric group on the positive integers, let pk denote the ?-cycle (1,2,... , &), where k > 1. It is fairly well known (and is proved in [2]) that if n > k, then the subgroup generated by pk and pn is either the full symmetric group Sn or the alternating group An. In this paper, we determine all the permutations a e Sk such that (cr, pn) is either An or Sn for all sufficiently large integers n. Of course, there is no mystery about when we get An and when we get Sn. If a set of permutations is known to generate one of these two groups, then the result is An if all of the generating permutations are even, and otherwise it is Sn. In particular, if 1 = S2oThus in TopSpin®, every permutation of the objects can be effected by legal moves of the puzzle. (See [3] for more about puzzles of this type.) We present an easily-checked condition on a permutation a e Sk that is necessary for (a, pn) to be either An or Sn for all sufficently large n. (The necessity is proved in Lemma 1, below.) Our main theorem is that this condition is also sufficient: if the condition is satisfied and n > 2k 1, then (cr, pn) is either An or 5n. To state our results, we define a numerical invariant for nonidentity permutations on a finite set of positive integers. Given a € Sk, write g (a) to denote the greatest common divisor of the integers cr(i) i for 1 < / < k. Since the GCD of a list of integers is unaffected by adjoining some extra zeros to the list, g (a) depends only on the numbers actually moved by