Are lower central factors of torsion-free nilpotent groups torsion-free? Let $G$ be a (residually) torsion-free nilpotent group and let $\gamma_i(G)$ denote the $i$-th term in the lower central series, i.e. $G_1 = G$ and $\gamma_{i+1}(G) = [G, \gamma_i(G)]$.
Is the quotient $G/\gamma_i(G)$ torsion-free?
 A: Extended comment: if the lower central series is replaced by the upper central series, the question (without "residually") has a positive answer. Indeed, it reduces to showing that if $G$ is torsion-free nilpotent and $Z$ is its center then $G/Z$ is torsion-free. Equivalently this means that $g^n$ central, $n\ge 1$ implies $g$ central. This then follows from a more general result ($\sharp$): in a torsion-free nilpotent group, the centralizer of $g$ and of $g^n$ are the same for any $g$ and any $n\ge 1$.
Indeed, for ($\sharp$) we can suppose that $G$ is finitely generated; then $G$ embeds into unipotent upper triangular matrices over a field of characteristic zero, so that $g$ belongs to the Zariski closure of the subgroup generated by $g^n$. In particular, they have the same centralizer.
A: No, this is not true. There are torsion-free nilpotent groups, where its  abelianization $G/\gamma_2(G)=G/[G,G]$ has $p$-torsion for every prime $p$. Indeed, let $G$ be the central product of the integer unitriangular matrix group $UT(3,\Bbb{Z})$ with $\Bbb{Q}$, where the center of the former is identified with a copy of $\mathbb{Z}$ in the latter. Then, $G$ is torsion-free, but
$\gamma_2(G)$ is isomorphic to $\mathbb{Z}$, and $G/[G,G]$ is isomorphic to $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Q}/\mathbb{Z}$. This has $p$-torsion for all primes $p$.
