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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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  • $\begingroup$ 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
    Commented Jun 20, 2011 at 8:52

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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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  • $\begingroup$ 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 ♦
    Commented 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). $\endgroup$
    – John Goodrick
    Commented 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 ♦
    Commented 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. $\endgroup$
    – John Goodrick
    Commented Oct 16, 2009 at 20:32
  • $\begingroup$ If the action is regular, then your criterion is equivalent to the open problem. $\endgroup$
    – S. Carnahan ♦
    Commented Oct 18, 2009 at 0:25

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