Mfcttrf

This is an extension of a question I asked here on Math.SE

Assume that I have a finitely generated residually finite centerless group $G$. Is it true that the profinite completion $hatG$ also has trivial center?

In the linked question, user YCor was able to show that this fails in general if you do not assume either finite generation or residually finite. However, the result happens to be true if $G$ is a surface group. I’d like to know if this is a phenomenon specific to surface groups, or if this is a more general fact.

The answer is No in general. Let $ngeq 3$ be odd (it is not necessary that $n$ be odd) and suppose $G=mathrmSL_n(mathbb Z)$. There exists a subgroup $Gamma subset mathrmSL_n(mathbb Z)$ of finite index which is torsion-free and centreless (the centre can only be $pm 1$ and because $n$ is odd the centre can only be trivial). However, $mathrmSL_n(mathbb Z)$ has the congruence subgroup property which means that the profinite completion of $Gamma$ contains a group of the form $$prod _pin S U_p times prod _ ell notin S mathrmSL_n(mathbb Z_ell),$$ where $S$ is a finite set of primes, $U_p$ is an open subgroup of finite index in $mathrmSL_n(mathbb Z_p)$, and $ell$ runs through primes in the complement of $S$. Since for infinitely many $ell$ (for example, all $ell$ with $ellequiv 1 ; (mathrmmod;n)$), the group $mathrmSL_n(mathbb Z_ell)$ has $n$-th roots of unity in the centre, it follows that the profinite completion of $Gamma $ is not centreless.