If $F$ is a free discrete group, then any subgroup $H$ of $F$ is free: this is the well-known theorem of Nielsen-Schreier. Moreover, there is a well-known algorithm, the Nielsen-Schreier method that allows us to determine a set of free generators of $H$ (see e.g. this reference).

Now if $F$ is free pro-$p$-group, then any closed subgroup $H$ of $F$ is pro-$p$-free (cf Ribes and Zaleskii, Profinite groups, Cor. 7.7.5). But is there a method to describe a space on which $H$ is free?

I am interested in the question even in the perhaps much simpler case where $F$ is the pro-p-completion of a free discrete group $F_0$ (with finitely many generators), and $H$ is the closure in $F$ of a subgroup $H_0$ of $F_0$ (of which I happen to know a set of free generators --infinitely many-- explicitly by the Nielsen-Schreier methods). Then by right-exacness of the $p$-completion functor, $H$ is a quotient of the completion of $H_0$, but I can't see by what (if anything).

This question is a follow-up of this one.


1 Answer 1


This doesn't quite answer your question, but... If $F$ is a discrete free group and $H$ is a fg subgroup, Ribes and Zalesskii gave an algorithm, improved by Margolis, Sapir and Weil, to compute the pro-p closure K of H in F. In this case K is fg and its closure in the pro-p completion of F is its pro-p completion. I am not sure when H is not fg what happens.

  • $\begingroup$ So, if I understand you well, if $H \hookrightarrow F$ is an injective morphism of discrete free groups, with $H$ finitely generated, then the map $\hat{H} \rightarrow \hat{F}$ is also injective, where $\hat{X}$ is the pro-$p$-completion of a group $X$. Is that right? $\endgroup$
    – Joël
    Dec 9, 2011 at 3:25
  • $\begingroup$ If the image of H is closed and we are in the pro-C setting with C an extension-closed variety. Closed subgroups inherit their full pro-C topology in this case. $\endgroup$ Dec 9, 2011 at 4:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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