Subgroups of infinite solvable groups

I'm looking for results of the form "every infinite solvable group contains <...> as a subgroup". Specifically, I believe:

If $$G$$ is infinite solvable, finitely generated and not virtually cyclic, then $$G$$ contains as subgroup either the metabelian group $$\mathbb Z[1/mn]\rtimes\mathbb Z$$, with $$\mathbb Z$$ acting by multiplication by $$m/n$$ [possibly $$m=n=1$$], or the wreath product $$\mathbb Z/p\wr\mathbb Z$$ for some prime $$p$$.

(I know that every infinite solvable group contains one of these two subgroups as a section [= quotient of two subgroups], by Kropholler's 1984 result; but really need them as subgroups. In fact, if the answer to the question is "no", I'm very much interested in a corrected statement with as few extra cases as possible.)

• Are you missing some conditions in 1? Finitely generated? Non-abelian? Nov 2, 2020 at 9:52
• Ouch, yes! I'll edit the question
– grok
Nov 2, 2020 at 9:58
• Both questions are non-trivial and I don't expect an answer to any of them to solve the other one: I'd suggest to post the second one separately.
– YCor
Nov 2, 2020 at 10:13
• Clarified $m=n=1$ allowed. Hopefully "2." is useful in understanding "1."; one could weaken "2." in requiring $A$ to have infinite index in its normalizer, and then I think it's really an intermediate step towards "1.".
– grok
Nov 2, 2020 at 11:12
• Anyway I'm close to an answer to 1 and it won't answer 2, so I'd be reluctant to post a long answer complete answer to 1 if it doesn't answer the whole question.
– YCor
Nov 2, 2020 at 11:14

It's not true: the lamplighter group $$L=\mathbb Z/p\wr\mathbb Z$$ has a universal central extension (by $$H_2(L,\mathbb Z)=(\mathbb Z/p)^\infty$$) which does not contain any lamplighter group as a subgroup.