# When does a transitive action of a profinite group have an infinite orbit?

That is: suppose G is a profinite group acting 1-transitively (but maybe not regularly) on a set X. Is there a reasonable criterion for when there is a g in G and a point a in X such that the g-orbit of a is infinite?

I wonder if it's enough to have a family (g_i, a_i) of pairs in G times X such that the g_i-orbit of a_i has size at least i.

Also, does anybody study these things much? A google search for "profinite group action" yields only a few hits; "profinite permutation group(s)" yields none.

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You might find this article interesting: arxiv.org/abs/1008.3062 I don't know if many people are looking at profinite group actions in complete generality, but there is plenty of work on actions on locally finite rooted trees, for instance. – Colin Reid Jun 20 '11 at 8:52

## 1 Answer

The answer depends on whether the action map GxX -> X is continuous (where I'm assuming X has the discrete topology). If so, then I think transitivity implies X is finite. If not, then you might as well view G as some abstract infinite group. If X is not discrete, e.g., given by a profinite system of sets, then I think you can have more interesting actions.

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I think you definitely want to consider sets with inverse limit topologies. I mean, where's the fun in finite actions of profinite groups (I mean, aside from all of Galois theory). – Ben Webster Oct 13 '09 at 15:33
Yes, the case I'm interested in is when X is a projective limit of finite discrete spaces and G acts continuously on X. Another way to think about it is you have a 1-transitive infinite permutation group (G,X) which is a projective limit of finite permutation groups (G_i, X_i). – John Goodrick Oct 14 '09 at 0:01
Your proposed criterion smells a lot like the open problem you mentioned in that other thread: whether profinite groups with elements of arbitrarily large order can be torsion. – S. Carnahan Oct 14 '09 at 2:54
Indeed, it does smell a lot like it. If anybody has a proof that the two statements are actually equivalent, I'd love to see it. – John Goodrick Oct 16 '09 at 20:32
If the action is regular, then your criterion is equivalent to the open problem. – S. Carnahan Oct 18 '09 at 0:25