Group Actions, Transitive, Doubly Transitive

Transitive, Doubly Transitive

A group action is transitive if there is only one orbit.  In other words, G maps every point in S to every other point.

The action is doubly transitive if some permutation takes any pair of elements to any other pair.

Assume the action of G on S is transitive, and |S| = p (prime).  If G has a kernel that maps to the identity permutation, mod out by the kernel and call the factor group H.  Now H acts effectively on S.

since the orbit has size p, |H| is divisible by p.  By Macay's theorem, H contains a subgroup Zp.  The only possible permutation with period p is a p cycle.  Thus the action of G includes a p cycle.

If S is divisible by p, G contains an element of order p, whose action on S is one or more disjoint p cycles.

Equal Orbits Principle

Let a possibly infinite abelian group G act transitively on a finite set S.  Pick an element c in G and let it act on S.  Let c move 0 to 1 to 2 to 3 to 4 to 5 and back around to 0, for example.  This is an orbit of length 6.  If there is more to S, let d move 0 to 6.  Since d induces a permutation, the images of 0 through 5, under d, are distinct.  If one of these images lies in the first orbit, say d maps i to j, then c6-id maps i to 6, while dc6-i keeps i within the first cycle.  This is a contradiction.  So let d map the first cycle onto elements 6 through 11.

If there is more to S, let e map 0 to 12.  Again, the images of 0 through 5 are distinct under e, and none of these images lie in the first cycle.  Suppose e maps i to 6+j.  Then c6-ie moves i to 12, while ec6-i moves i into the second cycle.  This is a contradiction, hence e moves the first cycle onto a third, which is disjoint from the previous two.

continue by induction until S is partitioned into cycles of length 6.  In other words, c moves S around in parallel cycles of the same length.

If c fixes anything in S, it fixes all of S.

This principle holds if S is countably infinite.  Build a renumbering map based on c: 0 through 5, 6 through 11, 12 through 17, and so on.  If you like transfinite induction, we can probably do the same for any well ordered set.

Note that this fails if G is not abelian.  Let G = S3, and let G act on 3 elements by permuting them in the obvious way.  One of the six actions swaps 1 and 2, and leaves 3 alone.