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?



Is being equationally Noetherian a quasi-isometry invariant?
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](http://www.sciencedirect.com/science/article/pii/002186937790312X)


$[2]$ G. Baumslag. Two theorems about equationally Noetherian groups Journal of Algebra
Volume 194, Issue 2, 15 August 1997, 654-664. [Sciencedirect link](http://www.sciencedirect.com/science/article/pii/S0021869397970253)


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


