MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
What's a strict subgroup? – Richard Kent 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. – Misha Oct 3 '12 at 16:48
Search on Google "almost free groups" – Francesco Polizzi 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
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.