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May 10, 2015 at 21:09 review Close votes
May 11, 2015 at 4:35
May 10, 2015 at 20:59 vote accept Pablo
May 10, 2015 at 20:55 comment added YCor @DonuArapura: yes it works (that subgroups of finite index in $\hat{\mathbf{Z}}$ can be checked by hand without any fancy theorem. By Nikolov-Segal (and Serre in the pro-$p$-case), any infinite topologically finitely generated profinite group also works. Of course it provides no countable example, but very simple countable examples have been pointed out as well.
May 10, 2015 at 18:52 answer added Yiftach Barnea timeline score: 10
May 10, 2015 at 17:39 comment added Johannes Hahn Isn't $\mathbb{Z}_{(p)}$ still a counterexample?
May 10, 2015 at 17:01 history edited Pablo CC BY-SA 3.0
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May 10, 2015 at 17:01 comment added Pablo @DonuArapura my fault. Edited.
May 10, 2015 at 16:58 comment added Donu Arapura Wouldn't $\hat{\mathbb{Z}}$ be a counterexample? Or am I missing something?
May 10, 2015 at 16:38 history edited Pablo CC BY-SA 3.0
edited title
May 10, 2015 at 16:32 history asked Pablo CC BY-SA 3.0