Let G be a finitely generated group. Suppose we have two families F1 and F2 of finite index subgroups of G, and each family has trivial intersection and is filtered from below (i.e. for any two elements in the family their intersection contains some third element).

These families generate two profinite topologies on G. (Taking the subgroups in the families as basis of open neighborhoods around identity).

Suppose the completions wrt to these families produce isomorphic profinite groups.

Can we say that these families generate the same topology on G?

(Equivalently, given any N∈F1 is there N2∈F2 such that N1≤N2 and vice versa.)

What if one family is a subfamily of the other?


1 Answer 1


No. $\widehat{\mathbb{Z}} \times \mathbb{Z}_p$ is isomorphic to $\mathbb{Z}_p \times \widehat{\mathbb{Z}}$, and the two topologies on $\mathbb{Z} \times \mathbb{Z}$ are different, though homeomorphic.

Am I interpreting your question too narrowly?

Also, in the case of a subfamily, you have a map $G'' \to G'$ from one completion to the other. If the subfamily is cofinal in the other, then the topologies are the same, and otherwise there will be a nontrivial kernel. Even then, I don't see why $G''$ and $G'$ couldn't still be isomorphic.

EDIT: By Proposition 2.5.2 of Ribes-Zalesskii's book "Profinite Groups," a finitely generated profinite group $G$ is hopfian in the sense that any continuous epimorphism $G \to G$ is an isomorphism.

So, in the situation of a subfamily, it turns out the answer is yes.

  • $\begingroup$ Richard, my motivation for this question was exactly the situation you describe in your last paragraph. I have two families and I know one is not cofinal in the other hence we have a nontrivial kernel. I was curios maybe they still could be isomorphic. $\endgroup$ Aug 12, 2011 at 20:00
  • 4
    $\begingroup$ Actually, that situation does not occur. See Proposition 2.5.2 of Ribes-Zalesskii, "Profinite Groups," which says that finitely generated profinite groups are hopfian. $\endgroup$ Aug 12, 2011 at 20:05
  • $\begingroup$ I didn't know this until now, so I learned something! I'll edit to address this. $\endgroup$ Aug 12, 2011 at 20:05
  • $\begingroup$ Richard, this is very nice, thanks a lot. $\endgroup$ Aug 12, 2011 at 20:14
  • $\begingroup$ Sure thing, Mustafa. $\endgroup$ Aug 12, 2011 at 20:25

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