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](https://mathoverflow.net/questions/26059/example-of-noetherian-group-every-subgroup-is-finitely-generated-that-is-not-f) post for a related question.)




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



Minimal number of generators of subgroups of Noetherian groups
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](http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=sm&paperid=2990&option_lang=eng). Clearly, it gives a desired example if we put, e.g, 
$$
\{G\_i\}=\{\hbox{all finite groups of odd orders}\}.
$$ 


