# Infinitely generated non-free group with all proper subgroups free

Is there any example of group $$G$$ satisfying the following properties?

1. $$G$$ is non-abelian, infinitely generated (i.e. it is not finitely generated) and not a free group.
2. $$H< G$$ implies that $$H$$ is a free group.

Clearly such an example should be torsion-free and countable.

(Added, from comments) For completeness, here's why it has to be countable. First, it is enough to construct a subgroup of index 2. Indeed, a theorem of Swan asserts that a torsion-free group with a free finite-index subgroup is free. Next, if a group $$G$$ has no subgroup of index 2 (i.e. every element is product of squares), it is easy to construct a nontrivial countable subgroup $$H$$ with the same property). In the current setting, $$G$$ is uncountable so $$H$$ is a proper subgroup, and hence has to be free, contradiction.

• Olshanskii constructed a non-abelian group all of whose proper subgroups are infinite cyclic and so this group meets your property in a somewhat degenerate way since they are free on one-generator. Jan 27 '20 at 22:56
• Also why must such groups be countable? I am not seeing it immediately
– user35370
Jan 27 '20 at 23:28
• @PaulPlummer: Locally free but non-free groups are easy to construct. Take a proper ascending HNN extension of the free group $F_k$, and the normal subgroup $N$ generated by $F_k$. For example $\langle a,b,t \mid a^t=ab, b^t=ba>$. The normal closure $N$ of $a,b$ is locally free but not free. The OP does not want this example. He wants all proper subgroups to be free, not just finitely generated subgroups.
– user6976
Jan 27 '20 at 23:31
• @PaulPlummer: I do not think it is clear (or even true) that these groups must be countable,
– user6976
Jan 27 '20 at 23:37
• Higman constructed non-free groups of cardinality $\aleph_1$ in which every countable subgroup is free. But every proper subgroup being free sounds hard.
– IJL
Jan 28 '20 at 15:08