# $p$-completion is pro-$p$ free

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.

• web.ma.utexas.edu/users/areid/StAndrews3.pdf This paper seems to pose this as an open problem for the profinite case. Mar 8, 2019 at 17:24
• 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-$p$.
– YCor
Mar 8, 2019 at 21:25
• 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.
– YCor
Mar 8, 2019 at 21:27
• @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. Mar 9, 2019 at 23:28
• It's trivial to convert my example to another one yielding a nontrivial free group.
– YCor
Mar 9, 2019 at 23:30

$$\langle x,y,z: x^2y^2=z^3\rangle$$.
This group is also residually-$$p$$ for all primes $$p$$.