TITLE: Minimal number of generators of subgroups of Noetherian groups
QUESTION [8 upvotes]: 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?

REPLY [14 votes]: 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, 
$$
\{G_i\}=\{\hbox{all finite groups of odd orders}\}.
$$