Infinite non-abelian group with every proper subgroup is infinite abelian.

It is well known that there are finite non-abelian groups whose every proper subgroups are abelian. $$S_{3}, Q_{8}$$ are such examples. My query is what would be the case if the group is infinite. That is, are there any infinite non-abelian groups whose every nontrivial proper subgroups are infinite abelian?

I was trying taking infinite direct products of non-abelian groups but not being able to find all subgroups as infinite abelian.

• I'd start with the group generated by letters $x$ and $y$, with the relation $xy=y^nx$ for some $n\ge 2$. Aug 5, 2019 at 9:03
• Such a group is necessarily finitely generated. Indeed any non-commuting pair has to generate the group. In addition it has to be torsion-free. So basically the only examples are the torsion-free "Tarski monsters" (first constructed by Olshanskii). By the way such groups are much easier to construct (using Gromov's methods) than finite exponents analogues.
– YCor
Aug 5, 2019 at 20:58

1 Answer

Such a group exists: the Tarski monster is an example.

https://en.wikipedia.org/wiki/Tarski_monster_group

• Thanks for the comment. But I want every subgroup infinite also. Is it possible to find? Aug 5, 2019 at 9:24
• @SudiptaPurkait There are "infinite cyclic" Tarski monster groups too; finitely generated groups where every proper, non-trivial subgroup is infinite cyclic. However, all constructions of Tarski monster groups are highly non-trivial. Aug 5, 2019 at 10:19
• (Also, I think you have a typo in your question title. Tsemo here has answered the question in the title, which differs from the question in the question body.) Aug 5, 2019 at 10:19
• By the way there's no "the Tarski monster". There's no preferred choice of construction. Actually there are uncountably many (torsion-free, or of large prime exponent).
– YCor
Aug 5, 2019 at 20:59