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Here is a question I stumbled upon in my research.

Question: Given a finitely generated nilpotent group $G$, do its subgroups fall into finitely many abstract commensurability classes?

Recall that two groups $H_1,H_2$ are abstractly commensurable if there exists finite-index subgroups $K_1\le H_1$ and $K_2\le H_2$ such that $K_1$ and $K_2$ are isomorphic.

One can suppose $G$ torsionfree (replace $G$ by a finite-index torsionfree subgroup). Then I believe we can rephrase the question as "Do the Lie subgroups with lattices of the Malcev completion of $G$ fall into finitely many isomorphism classes?". An idea is to prove that there are only finitely many isomorphism classes of simply connected nilpotent Lie groups in each dimension, but this seems to be wrong. (There are a few lists in small dimension, for instance the article A Cornucopia of Carnot groups in Low Dimensions by Le Donne and Tripaldi. This list contains infinite families, for instance (147E). Do these contain lattices?)

For the application I have in mind, a positive answer for nilpotent groups of class 2 is sufficient.

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  • $\begingroup$ Maybe I do not understand your question, but $\mathbb{Z}^2$ already contains infinitely many subgroups up to commensurability, say the $\langle (1,n) \rangle$ for $n \geq 0$. $\endgroup$
    – AGenevois
    Commented Dec 30, 2023 at 6:01
  • $\begingroup$ Right, I didn’t realize “commensurability” and “subgroups” usually combine in another way. I should put more emphasis on abstractly commensurable. All your examples are commensurable to $\mathbb Z$. $\endgroup$
    – Corentin B
    Commented Dec 30, 2023 at 8:00

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No. Let $U_n(R)$ be group upper triangular $n\times n$ matrices with identity diagonal over the ring $R$.

The groups $U_3(\mathbf{Z}[\sqrt{d}])$ are pairwise non-abstractly-commensurable, when $d\ge 2$ ranges over squarefree numbers. And they all embed into the upper triangular unipotent $U_6(\mathbf{Z})$.

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  • $\begingroup$ Here $G$ has Hirsch length $15$, or $12$ if we observe that the diagonal $2\times 2$ blocks are constant (which also makes $G$ 2-step nilpotent). But I expect one can choose, without difficulty, $G$ of Hirsch length $7$ using another construction. For Hirsch length $\le 6$, infinite index subgroups have Hirsch length $\le 5$ and there are only finitely many abstract commensurability classes at all. $\endgroup$
    – YCor
    Commented Dec 30, 2023 at 9:19
  • $\begingroup$ Is it obvious that the $U_3(\mathbb{Z}[\sqrt{d}])$ are pairwise non-abstractly-commensurable? $\endgroup$
    – AGenevois
    Commented Dec 30, 2023 at 10:40
  • $\begingroup$ @AGeneovois Sort of, they have (infinitely many) elements with different (linearly independent over rationals) transition lengths on their real Malcev completions: central subgroup generated by the square root. Commensurable nilpotent groups cannot have this. $\endgroup$
    – Denis T
    Commented Dec 30, 2023 at 11:57
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    $\begingroup$ Thanks, Yves. (Meanwhile, I managed to tweak my argument to make it work using only $G=N_{2,k}$ free $2$-step nilpotent of (large) rank $k$, and there every subgroup is virtually $N_{2,m}\times \mathbb Z^n$.) $\endgroup$
    – Corentin B
    Commented Dec 30, 2023 at 16:15
  • $\begingroup$ @AGenevois that's equivalent to the fact that their rational Malcev completions are pairwise non-isomorphic. That's not too hard to prove (and in anyway the fact is contained in the known classification of 6-dimensional Lie algebras over an arbitrary field). $\endgroup$
    – YCor
    Commented Dec 30, 2023 at 16:22

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