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



If each strict subgroup of G is free, must G be free or cyclic of prime order ?
No. There is a variation of Tarski monster: a nonabelian group whose each proper nontrivial subgroup is infinite cyclic, see [the book of Olshanskii](http://books.google.ru/books?hl=en&lr=&id=jwNkqFQn-GcC&oi=fnd&pg=PR5&dq=olshanskii+geometry+of+defining&ots=XQQkt5vng4&sig=jNhR97XokHHkZU_qaDboCH6EbDc&redir_esc=y#v=onepage&q&f=false).


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](http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=sm&paperid=2990&option_lang=eng).


