If each strict subgroup of a group G is free, must G be free or cyclic of prime order ?
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$\begingroup$ What's a strict subgroup? $\endgroup$– Autumn KentCommented Oct 3, 2012 at 16:35
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7$\begingroup$ 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. $\endgroup$– MishaCommented Oct 3, 2012 at 16:48
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$\begingroup$ Search on Google "almost free groups" $\endgroup$– Francesco PolizziCommented Oct 3, 2012 at 16:49
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2$\begingroup$ It's a well known open question whether there's a non-free word-hyperbolic group with every proper subgroup free. $\endgroup$– HJRWCommented Oct 3, 2012 at 20:03
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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.
