A group $G$ is equationally Noetherian if every system of group equations with coefficients from $G$ is equivalent to finite sub-system over $G$. It seems that this property must be invariant under geometric equivalences like quasi-isometry or at least bi-Lipschitz equivalence. I have no proof and no counterexample. Is there a pair of f.g. quasi-isometric groups $G_1$ and $G_2$, such that the first is equationally Noetherian and the second is not?

  • $\begingroup$ Why "it seems"? it's a purely algebraic property and could reasonably fail to be bilipschitz/QI-invariant. $\endgroup$
    – YCor
    Oct 22, 2017 at 10:21

1 Answer 1


No. Yes. $(*)$ This is not a QI-invariant, not bilipschitz invariant, and not even (unlabeled) Cayley graph invariant.

Indeed, consider two finite groups $F_1,F_2$ of the same order: then $F_1\wr\mathbf{Z}$ and $F_2\wr\mathbf{Z}$ have isomorphic Cayley graphs (namely with respect to the generating subset $F_i\cup\{1_{\mathbf{Z}}\}$, and in particular are bilipschitz.

On the other hand

  1. if $F_i$ is abelian, then $F_i\wr\mathbf{Z}$ is equationally noetherian: this is due to R. Bryant $[1]$, who proved that more generally every finitely generated abelian-by-nilpotent group is equationally noetherian.
  2. if $F_i$ is not abelian, then $F_i\wr\mathbf{Z}$ is not equationally noetherian: this observation is due to G. Baumslag $[2]$: indeed it is then easy to construct a properly descending sequence of subgroups, each of which is the centralizer of a finite subset.

So we get examples choosing $F_1$ abelian and $F_2$ non-abelian, of the same order, e.g., 6 or 8.

$[1]$ R. Bryant. The verbal topology of a group. Journal of Algebra Volume 48, Issue 2, October 1977, 340-346. Sciencedirect link

$[2]$ G. Baumslag. Two theorems about equationally Noetherian groups Journal of Algebra Volume 194, Issue 2, 15 August 1997, 654-664. Sciencedirect link

$(*)$ Classical joke: you're asking a yes/no question in the title and the opposite question in the core of the question :)

  • $\begingroup$ Thank you dear YCor. Before your reply, I had a wrong image of Cayley graph in my mind: I was supposed that isomorphic (isometric) Cayley graphs imply isomorphism of groups ;-0 $\endgroup$
    – Sh.M1972
    Oct 24, 2017 at 7:23
  • $\begingroup$ Indeed, the Cayley graph of any finite group of order $n$ with respect to the whole group as generating subset, is a complete graph on $n$ vertices, regardless of the group structure. $\endgroup$
    – YCor
    Oct 24, 2017 at 7:27

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