Let $G$ be a (topologically) simple Hausdorff topological group. Let $H$ be a dense subgroup of $G$. Now throw away the topology. What restrictions are known on the structure of $H$ as an abstract group? I imagine not much can be said if $G$ has a very coarse topology, but I am particularly interested in the case where $G$ is totally disconnected and locally compact, that is, the intersection of all open compact subgroups of $G$ is trivial.

A related question: two (t.d.l.c.) topological groups $G$ and $K$ have a dense subgroup $H$ in common. Suppose $G$ is (topologically) simple. What does this say about $K$?

I don't have a precise question I want to answer here, this is more of an appeal for references on the subject.

  • $\begingroup$ You're probably well aware if this, but if G is compact (and totally disconnected) then it is profinite and so never topologically simple. How does your G arise? $\endgroup$
    – HJRW
    Commented Aug 12, 2010 at 14:44
  • $\begingroup$ My G is only locally compact; it has open profinite subgroups, but none of them are normal. Finding simple t.d.l.c. groups with a specified open profinite subgroup (or showing that none exist) is an active area of research. An example involving the profinite Grigorchuk group can be found in the following paper of Barnea, Ershov and Weigel: arxiv.org/abs/0810.2060 $\endgroup$
    – Colin Reid
    Commented Aug 12, 2010 at 16:49

1 Answer 1


In many cases a simple topological group contains a nonabelian free group which is dense. This is true for Lie groups and for many examples of locally compact totally disconnect groups. In many of these cases every limit group can be found as a dense subgroup. Now, I do not rememebr the details, but you can find more informtaion in the webpage of the Workshop on "The Algebraic Structure of Profinite Groups" see https://www.ma.rhul.ac.uk/profinite_groups/.

  • $\begingroup$ Ah yes, this sounds familiar. It looks like to get anywhere with this idea, I would have to show the dense subgroup I am considering embeds in some special way. $\endgroup$
    – Colin Reid
    Commented Aug 12, 2010 at 17:45

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