Some background

Let $G$ be a torsion-free nilpotent group of class $2$ and rank $2$ (i.e., generated by two elements). Then, $G$ has to be isomorphic to the Heisenberg group.

This is relatively easy to see: if $G$ has any relations besides those required for it to be nilpotent of class $2$, then such a relation can be expressed as $\lambda^k x^a y^b = 1$, where $G$ is generated by $x, y$ and $\lambda = [x, y]$. Thus, $y^{-b} = \lambda^k x^a \implies 1 = [\lambda^k x^a, y] = [x^a, y] = \lambda^a$. If $G$ is torsion-free, this forces $a = 0$, implying $y^{-b} = \lambda^k \implies 1 = [x, y^{-b}] = \lambda^{-b}$. Thus, $b$ is also equal to $0$ and $\lambda^k = 1 \implies k = 0$, meaning no such relation exists.


Going one step further, the paper "A Note on Finitely Generated Torsion-Free Nilpotent Groups of Class $2$", by Fritz J. Grunewald and Rudolf Scharlau, the authors give multiple examples of torsion-free nilpotent groups of class $2$ and rank $4$ which are non-isomorphic, leading me to ask:

Does there exist any kind of classification of torsion-free nilpotent groups of class $2$? For any given rank, are there finitely many distinct isomorphism classes?

The paper I cited above constructs arbitrarily large collections of pairwise non-isomorphic torsion-free nilpotent groups of class $2$, but I'm not sure about their ranks...

I gratefully accept any references or databases, as I was unable to find much and, maybe I'm wrong, but I think GAP only has a finite group database...

Thanks in advance!

  • $\begingroup$ They are all quotients of the corresponding relatively free nilpotent group of class $2$ of the same rank, whose structure is known (it is the 2-nilpotent product of infinite cyclic groups). One can probably work out if there are only finitely many nonisomorphic such quotients for a given fixed rank. $\endgroup$ Mar 18 at 21:21
  • $\begingroup$ @ArturoMagidin Indeed, and any one of them that is not free has to have a relation involving purelly the commutators of the generators. Still, even for rank $3$ I’m at a loss as to how many there are… $\endgroup$
    – Gauss
    Mar 18 at 21:27
  • $\begingroup$ Well, I'm not certain off the top of my head (though it sounds plausible and I could probably prove it if I sit down and work through a few calculations) that it has to be a quotient modulo a subgroup of the commutator subgroup; that commutator subgroup is free abelian, and we "know", more or less, the subgroups that have torsionfree quotient. E.g., from the theorem here it will amount to finding a basis and moding out by it. But not every choice of basis for the commutator subgroup corresponds to basic commutators of a "basis" for the group. $\endgroup$ Mar 18 at 22:19

1 Answer 1


The groups $$G_n = \langle a,b,c \mid [a,c]=[b,c]=1,[a,b]=c^n \rangle $$ are nilpotent of class $2$ and have rank $3$ for $n>1$.

Their abelianizations $G_n/[G_n,G_n] \cong {\mathbb Z}^2 \oplus {\mathbb Z}/n$ are pairwise non-isomorphic and hence so are the groups $G_n$.

  • $\begingroup$ Interesting - even with $3$ generators there are going to be infinitely many… I’m going to try to see if the same remains true if I impose $[a, b] \neq 1, [a, c] \neq 1, [b, c] \neq 1$ $\endgroup$
    – Gauss
    Mar 19 at 10:55
  • $\begingroup$ That won't help, because you can just take for example $a,b,abc$ as generators of the same groups. It might be possible to classify the $3$-generator examples. In addition to the $G_n$ you would get $H \times {\mathbb Z}$, where $H$ is the Heisenberg group. $\endgroup$
    – Derek Holt
    Mar 19 at 13:13
  • $\begingroup$ I'm not sure I follow the first part of your comment... If $a$ and $b$ don't commute and neither commutes with $c$, then they don't commute with $abc$, unless $[a, b] = [a, c]^{-1}$ and $[a, b] = [b, c]$. As for the second part, now that you say it, I wouldn't be surprised if $G_n, H \times \mathbb{Z}$ and the free nilpotent group of class $2$ and rank $3$ were the only possible examples. $\endgroup$
    – Gauss
    Mar 19 at 13:22
  • 1
    $\begingroup$ The point of the first part of my comment was to say yes, the same remains true if you impose $[a,b] \ne 1$, $[a,c] \ne 1$, $[b,c] \ne 1$. The free nilpotent group of class $2$ and rank $3$ also has quotients in which the derived group has rank $2$, so a complete classification might be more complicated. $\endgroup$
    – Derek Holt
    Mar 19 at 16:14

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