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

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What's a strict subgroup? –  Richard Kent Oct 3 '12 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 '12 at 16:48
    
Search on Google "almost free groups" –  Francesco Polizzi Oct 3 '12 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. –  HJRW Oct 3 '12 at 20:03
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1 Answer

up vote 25 down vote accepted

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.

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Very nice, Anton, I forgot about Obraztsov's theorem! –  Misha Oct 3 '12 at 17:24
    
Thanks. I thought this had already been answered, but I had no keyword. –  js21 Oct 3 '12 at 21:10
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