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Let $G$ be any group, and $\widehat{G}$ its profinite completion. Is it true that $\widehat{\widehat{G}}=\widehat{G}$, i.e. is it true that $\widehat{G}$ is (canonically isomorphic to) its own profinite completion? It seems that it should follow from the universal property of the profinite completion, but I don't see how.

Thanks in advance for any solutions or suggestions.

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  • $\begingroup$ It could be worth mentioning where the terminology « completion » is from. This is explained in Atiyah & MacDonald's book, p. 102. $\endgroup$ – Watson Aug 2 '17 at 21:55
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I think there are counterexamples concerning the iterated profinite completions of groups, such as the abosulte Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$, which as a profinite group does not equal its own profinite completion (because it has subgroups of finite index that are not open). A deep theorem of Segal and Nikolov from $2007$ (using the classification of finite simple groups) implies that if $\widehat{G}$ is topologically of finite type then $\widehat{G}$ and $\widehat{\widehat{G}}$ are isomorphic.

Edit: Counterexamples to $\widehat{\widehat{G}}\simeq \widehat{G}$ can be found here (section 5, first line) arxiv.org/pdf/0801.2955.

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  • $\begingroup$ I have seen examples of profinite groups that do not equal their own profinite completion before. Bearing this theorem of Segal and Nikolov in mind, my question would be: Is the profinite completion of a group always topologically of finite type? $\endgroup$ – Servaes Sep 17 '13 at 19:59
  • $\begingroup$ No, and counterexamples are cited in "Profinite completion and double-dual: isomorphisms and counter-examples" by Colas Bardavid, section 5, first line. $\endgroup$ – Dietrich Burde Sep 18 '13 at 8:07
  • $\begingroup$ The counterexample is taken from Ribes and Zalesskii's book, see here. $\endgroup$ – Watson Aug 1 '17 at 18:01

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