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?

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Are you asking that if $\,G\cong H\,$ as groups and it $\,G\,$ is vir. nilpotent, then also $\,H\,$ is virt. nilpotent? – DonAntonio Mar 20 '13 at 13:59
I edited the question and hope that it is clear now. – Zeppelin Mar 20 '13 at 21:02

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.