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.

  • $\begingroup$ 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. $\endgroup$
    – Colin Reid
    Jun 20, 2011 at 8:52

1 Answer 1


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.

  • $\begingroup$ 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). $\endgroup$
    – Ben Webster
    Oct 13, 2009 at 15:33
  • $\begingroup$ 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). $\endgroup$ Oct 14, 2009 at 0:01
  • $\begingroup$ 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. $\endgroup$
    – S. Carnahan
    Oct 14, 2009 at 2:54
  • $\begingroup$ 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. $\endgroup$ Oct 16, 2009 at 20:32
  • $\begingroup$ If the action is regular, then your criterion is equivalent to the open problem. $\endgroup$
    – S. Carnahan
    Oct 18, 2009 at 0:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.