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Let $G$ be a countable (that is edit) residually-$p$ group and let $\hat{G}_p$ be its pro-$p$ completion.

  1. If $\hat{G}_p$ is finitely generated does it mean that $G$ is finitely generated?

  2. If $\hat{G}_p$ is finitely presented does it mean that $G$ is finitely presented? (I think the Grigorchuk group is not finitely presented, what about its pro-$p$ completion?)

Edit: Following @Ycor comment. Let me add that in Question 2, I would like $G$ to be finitely generatd. Also, let me be explicit:

  1. Is the pro-$p$ completion of the Grigorchuk group finitely presented?
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    $\begingroup$ 1,2 have a trivial negative answer. Consider $G=\mathbf{Z}[1/q]$ for some prime $q\neq p$. Then the embedding of $\mathbf{Z}$ in $G$ induces an isomorphism of pro-$p$ completions. $\endgroup$ – YCor Jan 20 '18 at 13:27
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As I said in a comment, the answer to 1 is trivially no.

The answer to 2 (as edited) is also no. Platonov and Tavgen produced a finitely generated, infinitely presented subgroup $H$ in the square $F\times F$ of a free groups such that the inclusion induces an isomorphism of profinite completions (and hence of pro-$p$-completions for all $p$).

Here, $F$ can be constructed as follows. Consider any epimorphism $p:F\to P$ from $F$ to an infinite finitely presented group $P$ with no nontrivial finite quotient; then $H$ is defined as the fibre product $\{(g,h)\in F\times F:p(g)=p(h)\}$. That $P$ is finitely presented is used to ensure that $H$ is finitely generated. That $P$ has no nontrivial proper quotient is used by Platonov and Tavgen to prove that the inclusion induces an isomorphism of profinite completions. That the fibre product $H$ is infinitely presented is a classical fact on fiber products.

Reference: V. Platonov O.Tavgen. On the Grothendieck problem of profinite completions of groups. (Russian) Dokl. Akad. Nauk SSSR 288 (1986), no. 5, 1054-1058. English translation: Soviet Math. Dokl. 33 (1986), no. 3, 822-825. (MR link)

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The answer to question 3 is "no" (for $p=2$) : https://arxiv.org/abs/1011.3880 and "yes" for every other $p$.

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  • $\begingroup$ What do you mean "yes" for every other $p$? Did you mean I should have said pro-$2$ completion? $\endgroup$ – Yiftach Barnea Jan 20 '18 at 19:53
  • $\begingroup$ I think so: Mark mentioned the trivial case of odd $p$. $\endgroup$ – YCor Jan 20 '18 at 21:44
  • $\begingroup$ I can only accept one answer, so I choose Yves' one as he answer two of my questions. However, Mark answered my third one. $\endgroup$ – Yiftach Barnea Jan 20 '18 at 23:06

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