Converse of Schreier theorem

I know that every subgroup of a free group is free (Schreier theorem).

I'm wondering that a (non-trivial) converse is true, that is, if every proper subgroup of an infinite group $$G$$ is free, then $$G$$ is free.

I think it is false but I cannot find counterexamples.

(I expect that some proper semi-direct product of the free group rank $$n$$ ($$n \geq 2$$) and $$\mathbb{Z}/2\mathbb{Z}$$ is counterexample but I cannot find yet.)

Any comments would be greatly appreciated.

• No, Olshanski constructed counterexamples (in which every proper subgroup is cyclic (trivial or infinite). I guess there are similar constructions with also non-abelian proper subgroups.
– YCor
Oct 28 '20 at 11:35
• For infinitely generated groups, the questions if of very different flavor since then the group is locally free; this was precisely asked this year. In particular, there's no such uncountable group.
– YCor
Oct 28 '20 at 11:38

As @YCor says in comments, there is a finitely generated example due to Ol'shanskii, which is essentially a kind of Tarski monster. However, Ol'shanskii's construction is very complicated. For "nicer" classes of groups, your question remains both important and open. As with most other kinds of Tarski monsters, I believe the answer to the following question remains unknown (though the answer is surely "yes").

Is there a non-free finitely presented group for which every proper subgroup is free?

Indeed, the question remains open even in very nice classes of groups.

Is there a non-free word-hyperbolic group in which every proper subgroup is free?

An example would be huge news, since it would resolve in the negative TWO famous open questions, viz:

Is every word-hyperbolic group residually finite?

and

Does every non-virtually-free hyperbolic group have a surface subgroup?

Since two big questions in one go seems like too much to hope for, I prefer to specialise to the case of subgroups of infinite index.

Is there a non-free, non-surface, infinite, word-hyperbolic group in which every proper, finitely generated subgroup of infinite index is free?

For this last question, there are examples outside of the world of hyperbolic groups. The solvable Baumslag–Solitar groups

$$BS(1,n)=\mathbb{Z}[1/n]\rtimes_n\mathbb{Z}$$

have the property that every nontrivial finitely generated subgroup of infinite index is infinite cyclic.

• From reading Gromov's "Hyperbolic groups" I understood that constructing 2-generated non-abelian groups in which every nontrivial proper subgroup is infinite cyclic (or even finite cyclic, with no uniform bound) is far easier than constructing Tarski monsters of finite exponent.
– YCor
Nov 1 '20 at 8:53
• The 3rd question is answered by surface groups themselves (oriented and non-oriented, negative $\chi$). Any positive answer to the 3rd question implies a counterexample to the second question.
– YCor
Nov 1 '20 at 9:00
• @YCor: you’re quite right about the last question; surface groups also need to be excluded. Thanks for the reminder! I edited the answer. That’s interesting about “infinite cyclic” Tarski monsters.
– HJRW
Nov 1 '20 at 9:15
• Yep (I should have written "any other positive answer"), so now that surface groups are excluded (and counting correctly your questions), the last sentence of my 2nd comment should be read: a positive answer to Q5 gives a counterexample to the well-known hard Q4 (the question on surface subgroups).
– YCor
Nov 1 '20 at 9:24