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I have seen that several authors say that an infinite group $G$ is an IF-group (or has the IF-property) if every subgroup of infinite index in $G$ is free (for instance, see https://arxiv.org/pdf/1607.02079v1.pdf).

Other than free groups, surface groups, or Fuchsian groups,

What are some examples of IF-groups?

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    $\begingroup$ Olshanskii constructed non-abelian torsion-free groups in which every proper subgroup is cyclic (such a group satisfies your property and is generated by any of its noncommuting pairs). $\endgroup$
    – YCor
    Dec 7, 2016 at 7:37
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    $\begingroup$ What about one-relator groups? Are they all IF or perhaps there is a characterization of the ones which are? $\endgroup$ Dec 7, 2016 at 7:54
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    $\begingroup$ @VictorProtsak Baumslag-Solitar groups are not IF (except those virtually $\mathbf{Z}^2$) $\endgroup$
    – YCor
    Dec 7, 2016 at 8:00
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    $\begingroup$ @YCor and what if the one-relator group has more than two generators? $\endgroup$
    – Pablo
    Dec 7, 2016 at 8:01
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    $\begingroup$ Pablo: you can find a non-IF example... $\endgroup$
    – YCor
    Dec 7, 2016 at 8:02

3 Answers 3

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The following will not answer your question but will (likely) give you some idea about the class of IF groups.

Conjecture (Gromov) A hyperbolic group is either virtually free or contains a surface subgroup.

If one believes this conjecture, the class IF is essentially disjoint from the class of hyperbolic groups.

Conjecture. (Flat closing) If $G$ is a CAT(0) group then either $G$ is hyperbolic or it contains $Z^2$.

If one believes this conjecture, CAT(0) groups will be also useless regarding your question.

Conjecture. (Gersten) Suppose that $G$ is a group which admits a finite $K(G,1)$. Then either $G$ contains a Baumslag-Solitar subgroup or it is hyperbolic.

Remark. The Baumslag-Solitar group $BS(p,q)$ either contains $Z^2$ or $|p|=1$ or $|q|=1$. Furthermore, $BS(p,1)$, $p>1$, contains an infinitely generated abelian subgroup. Hence, apart from the case $|p|=|q|=1$, $BS(p,q)$ cannot be IF. Thus, as Yves observed, instead of saying "do not contain BS subgroups" as Gersten originally formulated, one can equivalently say "do not contain non-virtually cyclic abelian subgroups".

If one weakens the existence of a finite $K(G,1)$ to finite presentability, then, there are several sets of examples of finitely presented groups which are not hyperbolic but contain no Baumslag-Solitar subgroups. The first such examples were constructed by Noel Brady ("Branched coverings of cubical complexes and subgroups of hyperbolic groups"), other examples were constructed by Brady, Clay and Dani ("Morse theory and conjugacy classes of finite subgroups II") and by Lodha ("A hyperbolic group with a finitely presented subgroup that is not of type $FP_3$"). The examples by Brady and Brady--Clay--Dani are not IF. I did not have time to check Lodha's examples but I fully expect them not to be IF either.

Question (C.Y.Tang) (Question FP5 in the database http://www.grouptheory.info/ of group theory problems) Is there a non-free non-cyclic finitely presented group all of whose proper subgroups are free?

In view of the above, you will not be able to find examples of finitely presentable IF groups which are not surface groups or free groups in the existing literature (or, maybe you can but to prove that they are IF you will have to work hard). In particular, Tarsky monsters mentioned by Yves, do not look all that bad.

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  • $\begingroup$ Minor comment: Brady didn't construct a family of examples; he just constructed one example. Other examples have been constructed by Llodha and Kropholler. You can upgrade these examples to families trivially by passing to finite-index subgroups, free products, etc, but there's no sense in which we know infinitely many examples which are interestingly different. $\endgroup$
    – HJRW
    Dec 7, 2016 at 21:36
  • $\begingroup$ Sorry, I misspelled a name: it should be Lodha. $\endgroup$
    – HJRW
    Dec 7, 2016 at 21:46
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    $\begingroup$ I don't really understand the systematic traditional reference to Baumslag-Solitar subgroups in these questions since it reduces to a neater question about abelian subgroups (existence or not of non-virtually-cyclic abelian subgroups). $\endgroup$
    – YCor
    Dec 7, 2016 at 22:09
  • $\begingroup$ @YCor, What do you mean? BS(1,n) doesn't contain a non-cyclic abelian subgroup. $\endgroup$
    – HJRW
    Dec 8, 2016 at 7:00
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    $\begingroup$ @HJRW this is your opinion! (ok, "neater" was an opinion too) $\endgroup$
    – YCor
    Dec 8, 2016 at 7:40
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This is not strictly an answer to your question, but I hope that results proving that certain nice groups are not IF are also of interest.

You point out that free groups and surface groups are IF. Recall that a limit group is a finitely generated group which is fully residually free.

Theorem: A limit group which is IF is a free group or a surface group.

The same also holds for graphs of free groups with cyclic edge groups.

Something similar holds for 3-manifold groups by Kahn--Markovic's proof of the surface subgroup conjecture.

As Misha's answer suggests, these results fit into a general conjectural picture, which suggests that in 'nice' examples, the IF property should characterize free and surface groups.

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  • $\begingroup$ Good to know that! Since I am after something positive, let me ask you if you have in mind some example of a 'nice' group (maybe hyperbolic?) such that there is good control over the subgroups of infinite index. For instance, such that all subgroups of infinite index are free/surface groups (or belong to some other natural family). $\endgroup$
    – Pablo
    Dec 8, 2016 at 6:49
  • $\begingroup$ Well, at the minimum you would expect free products of surface groups as well. But I would certainly conjecture that there are no such hyperbolic groups. This follows for limit groups by the proof of my result above. It can be proved for hyperbolic 3-manifold groups using Agol--Wise. Calegari and I also proved it for random groups. $\endgroup$
    – HJRW
    Dec 8, 2016 at 7:09
  • $\begingroup$ @Pablo: This is not a well-posed question (what is "good control"?). There are some standard restrictions on subgroups one can impose, like require the group to be noetherian, or coherent or local quasiconvex (for hyperbolic groups). If this is what you had in mind then indeed, this was studied quite a bit, but it should be a separate question. $\endgroup$
    – Misha
    Dec 8, 2016 at 9:35
  • $\begingroup$ @Misha, thank you for your answer! Of course, my question about good control is not well defined, but I meant that there will be only very few isomorphism types of subgroups, to the extent that given a subgroup of infinite index we can say that it is necessarily isomorphic to say a free (amalgamated) product of surface/free groups. I thought about that also because in your answer the possibility of non-free surface subgroups arises. $\endgroup$
    – Pablo
    Dec 8, 2016 at 9:41
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Like Misha and Henry, I'll also add to the list of constraints on IF-groups rather than giving any examples. In The Surface Group Conjectures for groups with two generators we prove:

Theorem. Let $G$ be a finitely generated IF-group. If $G$ has vanishing first $L^2$-Betti number and infinite abelianization (e.g. if $G$ is a 2-generator 1-relator group), then $G$ is isomorphic to $\mathbb{Z}$, $\mathbb{Z}^2$ or the Klein bottle group.

Also, an application of recent work of Mutanguha (proving an instance of Gersten's conjecture as mentioned by Misha) shows that a finitely generated IF-group with infinite abelianization is hyperbolic or a surface group, so Gromov's conjecture says it should be free or a surface group (see section 3.2 of our paper).

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