Recall that a group $G$ satisfies max (or is said to be Noetherian) if all its proper subgroups are finitely generated. Similarly $G$ satisfies max-n if all its normal subgroups are normal closures of finite subsets. Note that property max is "closed with respect to extension", i.e if $N \unlhd G$ and $G \backslash N$ have this property than so does $G$ (in particular this implies that polycyclic groups satisfy max). However max-n is not inherited by subgroups. Nonetheless we have the following Theorem:

If a group $G$ satisfies min-n (resp. max-n) and $H$ is a subgroup of $G$ with finite index, then $H$ satisfies min-n (resp. max-n).

My question is, if these properties max, min, max-n and min-n are preserved under quasi-isometries (of f.g. groups).

Edit 1: as YCor pointed out in a comment the answer is negative for max-n and min-n. The question for max and min seems still open (in general), however we may have a lack of examples.

In this sense we may collect examples of groups satisfying max resp. min.

  • 2
    $\begingroup$ Not max-n: $F_2\times F_2$ being quasi-isometric to irreducible lattices in $SL_2(\mathbf{Q}_p)^2$, which are just-infinite hence max-n. They are also QI to Burger-Mozes groups which are simple and hence min-n, so min-n is also not QI-invariant. For max, it's probably open (too few known examples). For min, I guess there are too few known examples as well. $\endgroup$
    – YCor
    Sep 9, 2015 at 9:10
  • 1
    $\begingroup$ This is many years later, but the property "Max" for Noetherian surely ought to be "Emmy" :-) $\endgroup$ Feb 16, 2021 at 8:38

1 Answer 1


For finitely presented groups, you're close to two well known open questions/conjectures.

Question 1: If a finitely presented group is Noetherian, is it virtually polycyclic?

(This question is FP11 here, where it's attributed to S. Ivanov.)

Question 2: The class of virtually polycyclic groups is closed under quasi-isometry.

(Alex Eskin conjectures this here.)

Putting these together, I think it's fair to say that, for finitely presented groups, it's unknown whether the property of being Noetherian is a qi-invariant.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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