# If each strict subgroup of G is free, must G be free or cyclic of prime order ?

If each strict subgroup of a group G is free, must G be free or cyclic of prime order ?

• What's a strict subgroup? Oct 3 '12 at 16:35
• I think, strict=proper. Then there are even finitely-generated groups $G$ where every proper subgroup is infinite cyclic, but $G$ is not virtually free (Olshansky's central extensions of Tarsky monsters). However, if you add the condition that $G$ contains a free nonabelian subgroup, I do not think there are any know counter-examples. Oct 3 '12 at 16:48
• Search on Google "almost free groups" Oct 3 '12 at 16:49
• It's a well known open question whether there's a non-free word-hyperbolic group with every proper subgroup free.
– HJRW
Oct 3 '12 at 20:03

Concerning Misha's comment. For any countable family of countable involution-free groups $G_1,G_2,\dots$, there is a group $H$ containing all $G_i$ as proper subgroups such that each proper subgroup of $H$ is either infinite cyclic or a conjugate of a subgroup of some $G_i$. This is Obraztsov's embedding theorem.