4
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ Why "it seems"? it's a purely algebraic property and could reasonably fail to be bilipschitz/QI-invariant. $\endgroup$ – YCor Oct 22 '17 at 10:21
5
$\begingroup$

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 :)

$\endgroup$
2
  • $\begingroup$ 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 '17 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 '17 at 7:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.