Is there any example of group $G$ satisfying the following properties?

  1. $G$ is non-abelian, infinitely generated (i.e. it is not finitely generated) and not a free group.
  2. $H< G$ implies that $H$ is a free group.

Clearly such an example should be torsion-free and countable.

(Added, from comments) For completeness, here's why it has to be countable. First, it is enough to construct a subgroup of index 2. Indeed, a theorem of Swan asserts that a torsion-free group with a free finite-index subgroup is free. Next, if a group $G$ has no subgroup of index 2 (i.e. every element is product of squares), it is easy to construct a nontrivial countable subgroup $H$ with the same property). In the current setting, $G$ is uncountable so $H$ is a proper subgroup, and hence has to be free, contradiction.

  • 5
    $\begingroup$ Olshanskii constructed a non-abelian group all of whose proper subgroups are infinite cyclic and so this group meets your property in a somewhat degenerate way since they are free on one-generator. $\endgroup$ Jan 27 '20 at 22:56
  • 3
    $\begingroup$ Also why must such groups be countable? I am not seeing it immediately $\endgroup$
    – user35370
    Jan 27 '20 at 23:28
  • 2
    $\begingroup$ @PaulPlummer: Locally free but non-free groups are easy to construct. Take a proper ascending HNN extension of the free group $F_k$, and the normal subgroup $N$ generated by $F_k$. For example $\langle a,b,t \mid a^t=ab, b^t=ba>$. The normal closure $N$ of $a,b$ is locally free but not free. The OP does not want this example. He wants all proper subgroups to be free, not just finitely generated subgroups. $\endgroup$
    – user6976
    Jan 27 '20 at 23:31
  • 2
    $\begingroup$ @PaulPlummer: I do not think it is clear (or even true) that these groups must be countable, $\endgroup$
    – user6976
    Jan 27 '20 at 23:37
  • 3
    $\begingroup$ Higman constructed non-free groups of cardinality $\aleph_1$ in which every countable subgroup is free. But every proper subgroup being free sounds hard. $\endgroup$
    – IJL
    Jan 28 '20 at 15:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.