# Can a profinite completion be free pro-p?

Is there a prime number $p$ and a finitely generated residually finite group whose profinite completion is a free pro-$p$ group on a nonempty finite set?

Thanks to YCor we see that we cannot take the group to be finitely generated.

• – Yiftach Barnea Jan 8 '15 at 15:32
• The answer to the main question is trivially no, because if a f.g. group has abelianization $\mathbf{Z}^k\oplus F$ for $F$ finite, then the abelianization of its profinite completion is $\hat{\mathbf{Z}}^k\oplus F$. While the abelianization of a free pro-$p$-group of rank $\ell$ is $\mathbf{Z}_p^\ell$. – YCor Jan 8 '15 at 17:31
• @YCor this is very simple, and I have missed this. – Pablo Jan 8 '15 at 17:40
• If $G$ is the localization of $\mathbf{Z}$ at the prime $p$ (i.e. the subring of $\mathbf{Q}$ generated by all $\ell^{-1}$ for $\ell$ ranging over primes $\neq p$). Then its profinite completion is free pro-$p$ of rank 1. – YCor Jan 8 '15 at 18:15
• Something similar can be done with higher rank: consider the smallest subgroup of the pro-$p$-completion $G$ of the free group $F_X$ containing $F_X$ and stable by extracting $\ell$-roots for all primes $\ell\neq p$. Then it's easy to see that its profinite completion is canonically $G$, hence free pro-$p$ on $X$. – YCor Jan 8 '15 at 18:21