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?


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 $p$.

  • $\begingroup$ Variant: the subgroup of the integral Heisenberg group for which the 12-entry is in $n\mathbf{Z}$: then the abelianization is $\simeq\mathbf{Z}^2\times(\mathbf{Z}/n\mathbf{Z})$. $\endgroup$ – YCor Jun 6 '18 at 9:31
  • $\begingroup$ Yes, thank you. I think this also comes up as an example that there are infinitely many affine crystallographic groups in dimension $3$, i.e., with the groups $\Gamma_n=\langle a,b,c\mid [b,a]=c^n,[c,a]=[c,b]=e\rangle$ and $\Gamma_n/[\Gamma_n,\Gamma_n]=\Bbb{Z}^2\oplus \Bbb{Z}/n\Bbb{Z}$. $\endgroup$ – Dietrich Burde Jun 6 '18 at 9:40
  • $\begingroup$ Actually these groups (the one in your comment and the one in mine) are isomorphic. $\endgroup$ – YCor Jun 6 '18 at 9:45
  • $\begingroup$ Thanks Dietrich and Yves. I was going to ask whether the situation changes with $G$ being finitely generated but I believe that is covered by Yves' example. $\endgroup$ – Michal Ferov Jun 6 '18 at 9:50

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.

  • $\begingroup$ I'm currently dealing with the same problem. First: Why does it reduce to showing $G/Z$ is torsion-free? Second: do you have a reference or a short proof to why the centralizers of $g$ and $g^n$ are the same in a torsion-free nilpotent group? Thank you! $\endgroup$ – Arwid Sep 20 '18 at 13:27
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    $\begingroup$ @Arwid First: by induction. Second: I already said that the proof I'm aware of, consists in embedding $G$ into the group of unipotent upper triangular matrices over some field of characteristic zero (Malcev; see the book "discrete subgroup of Lie groups" by Raghunathan for instance). Then, using a Jordan form, one can check by hand that the subalgebra generated by $g$ and $g^n$ coincide for all $n\ge 1$. In particular, $g$ and $g^n$ have the same centralizer. $\endgroup$ – YCor Sep 20 '18 at 13:31
  • $\begingroup$ To first: By induction over what? to Second: I've posted a question regarding that topic earlier here: math.stackexchange.com/questions/2923936/… . Do you see that the proof of Robinson uses an easier method to show the equality of the centralizer instead of embedding $G$ into unipotent $UT_n$ or does he use exactly this fact? I don't yet fully understand every step of the proof. $\endgroup$ – Arwid Sep 20 '18 at 14:03
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    $\begingroup$ Induction on the nilpotency length, since for $G\neq 1$, the nilpotency length of $G/Z(G)$ is strictly smaller than that of $G$. $\endgroup$ – YCor Sep 20 '18 at 14:18
  • $\begingroup$ I'm confused. I show that $G/Z(G)$ is torsion-free and then by induction on the the nilpotency length it follows $G$ is torsion-free, correct? $\endgroup$ – Arwid Sep 20 '18 at 14:29

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