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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?

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    $\begingroup$ 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
    Commented 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
    Commented 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
    Commented Nov 4, 2013 at 8:54
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    $\begingroup$ @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. $\endgroup$ Commented Nov 5, 2013 at 18:59

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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, $$ \{G_i\}=\{\hbox{all finite groups of odd orders}\}. $$

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

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