8
$\begingroup$

A group $G$ is Noetherian (or slender) if all its subgroups are finitely generated. Does this imply that the minimal number of generators of subgroups of $G$ is bounded above?

For example, if $G$ is a polycyclic group that admits a polycyclic series of length $n$ then every subgroup of $G$ can be generated by $n$ (or fewer) elements. This idea also applies for virtually-polycyclic groups.

It is unknown whether all finitely presented Noetherian groups are virtually-polycyclic. On the other hand, there are finitely generated Noetherian groups that are not virtually-polycyclic, for example the Tarski monster. However, all proper subgroups of the Tarski monster are cyclic and hence there is a bound on the minimal number of generators of its subgroups.

(See this post for a related question.)

Edit: What if we restrict ourselves to finitely presented Noetherian groups?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ A quasi-finite group is a group all of whose proper subgroups are finite. Deryabina-Olshanskii (Russian Math. Surveys 41 (1986), no. 6, 169–170) showed that a finite group is subgroup of a f.g. quasi-finite group iff it's the direct product of a group of odd order and an abelian 2-group. It's possible that the construction (or another one) provides a single quasi-finite group containing a copy of $C^n$ for every $n$, for $C$ a given cyclic group of prime order (this would answer your question negatively), but I don't have access to the paper (iopscience.iop.org/0036-0279/41/6/L10) $\endgroup$
    – YCor
    Nov 3, 2013 at 15:22
  • $\begingroup$ Your question could possibly be true if restricted to finitely presented slender groups, since I think the only known examples are virtually polycyclic. $\endgroup$
    – Ian Agol
    Nov 4, 2013 at 5:05
  • $\begingroup$ Yes, it is either true or open. I will add the question to the original post. $\endgroup$
    – Sebastian
    Nov 4, 2013 at 8:54
  • 2
    $\begingroup$ @Ian, indeed, "Are all finitely presented Noetherian groups virtually polycyclic? is (an open) Question 11.38 (due to S.V.Ivanov, 1990) from the Kourovka Notebook. $\endgroup$
    – Anton Klyachko
    Nov 5, 2013 at 18:59

1 Answer 1

Reset to default
14
$\begingroup$

No.

For any countable family of countable involution-free groups $G_1,G_2,\dots$, there is a 2-generated group $H$ containing all $G_i$ as proper subgroups such that each proper subgroup of $H$ is either cyclic or a conjugate of a subgroup of some $G_i$.

This is Obraztsov's embedding theorem. Clearly, it gives a desired example if we put, e.g, $$ \{G_i\}=\{\hbox{all finite groups of odd orders}\}. $$

Improve this answer
$\endgroup$
1
  • $\begingroup$ This also gives an example of a Slender group that is neither perfect nor solvable! $\endgroup$
    – NWMT
    Nov 4, 2016 at 18:43

Your Answer

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

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