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Let $G$ be an abstract finitely generated residually finite group, and suppose that it's $p$-completion $\widehat{G_p}$ is a pro-$p$ free group. Does this implies that $G$ is a free group?

The converse is indeed true. For this proposition though, I'm unsure how to proceed. I'm using this result to prove that $H^2(G, \mathbb{F}_p) \neq 0 \implies H^2(\widehat{G_p},\mathbb{F}_p) \neq 0$. There seems to be no need for a basis of $\widehat{G_p}$ to be contained in $G$, or for one such basis to exist. A density argument may do something here, but I'm a bit out of ideas.

I believe this to be true given that pro-$p$ completion commutes with group presentations in a sense - the abstract presentation of $G$ is a topological presentation of $\widehat{G_p}$. The converse of such statement would be a proof of the result I'm looking for.

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  • $\begingroup$ web.ma.utexas.edu/users/areid/StAndrews3.pdf This paper seems to pose this as an open problem for the profinite case. $\endgroup$ – Henrique Augusto Souza Mar 8 at 17:24
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    $\begingroup$ It's clearly false, just take $G=\mathbf{Z}/q\mathbf{Z}$ for $q\ge 2$ coprime to $p$. You should at least assume that $G$ is residually-$p$. $\endgroup$ – YCor Mar 8 at 21:25
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    $\begingroup$ If it's true (assuming residually-$p$), it will certainly not be an easy argument. I'm rather inclined to believe that it's false (although it might be hard to disprove), i.e. construct a counterexample. $\endgroup$ – YCor Mar 8 at 21:27
  • $\begingroup$ @YCor Yes, that is indeed a counterexample... However, in this case we have $\widehat{G_p} = \{0\}$! While the trivial group is indeed free, it is a trivial type of free group. What if we suppose that $\widehat{G_p}$ is a free pro-$p$ group of rank $\geq 1$? Also, Corollary 4.13 of that paper establishes conditions of when is this valid for the profinite completion, but it doesn't seem to work in general either. $\endgroup$ – Henrique Augusto Souza Mar 9 at 23:28
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    $\begingroup$ It's trivial to convert my example to another one yielding a nontrivial free group. $\endgroup$ – YCor Mar 9 at 23:30
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This is false. All parafree groups have this property. Look at https://en.wikipedia.org/wiki/Parafree_group

One of the easiest example of a parafree group is

$\langle x,y,z: x^2y^2=z^3\rangle$.

This group is also residually-$p$ for all primes $p$.

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