# Classification of torsion-free nilpotent groups of class 2

## 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.

## Question

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...

• 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. Mar 18 at 21:21
• @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… Mar 18 at 21:27
• 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. Mar 18 at 22:19

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 $$G_n$$.
• 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 $[a, b] \neq 1, [a, c] \neq 1, [b, c] \neq 1$ Mar 19 at 10:55
• 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. Mar 19 at 13:13
• 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. Mar 19 at 13:22
• 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. Mar 19 at 16:14