Is being virtually nilpotent a profinite propert? If a virtually nilpotent group $G$ has  isomorphic profinite completion with another group $H$, i.e. $\hat G \cong \hat H$, does it follow that $H$ is  virtually nilpotent?
 A: The following is true: If $G$ is a finitely generated virtually nilpotent group and $H$ is a residually finite group such that $G$ and $H$ have isomorphic profinite completion then $H$ is virtually nilpotent.
In fact, since $G$ is residually finite (virtually nilpotent+finitely generated implies residually finite) then $G$ can be viewed as a dense subgroup of $\widehat{G}$. If $N$ is a normal subgroup of $G$ of finite index which is nilpotent then it can be proved that the closure $\overline{N}$ of $N$ in $\widehat{G}$ is its profinite completion which is also a normal nilpotent subgroup of $\widehat{G}$ of finite index. I do not remember the details of the proof of the fact that $\overline{N}$ is nilpotent. Instead we can choose $N$ to be torsion free (easy excersise) and then the upper central series of $\overline{N}$ is just the closure of the upper central series of $N$. 
Then we have proved that $\widehat{G}$ is virtually nilpotent. Now if $H$ is residually finite then $H$ can be viewed as a dense subgroup of $\widehat{H}\cong\widehat{G}$. Then $H$ is virtually nilpotent because is isomorphic to a subgroup of a virtually nilpotent group.
