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

flag	asked Oct 3 at 16:31 js 1,744●1●12

What's a strict subgroup? – Richard Kent Oct 3 at 16:35

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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. – Misha Oct 3 at 16:48

Search on Google "almost free groups" – Francesco Polizzi Oct 3 at 16:49

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It's a well known open question whether there's a non-free word-hyperbolic group with every proper subgroup free. – HW Oct 3 at 20:03

Anton Klyachko · Accepted Answer · 2012-10-04 09:30:52Z UTC

No. There is a variation of Tarski monster: a nonabelian group whose each proper nontrivial subgroup is infinite cyclic, see the book of Olshanskii.

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.

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

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