3
$\begingroup$

Question: For which "interesting" classes of finitely generated groups is it known whether every infinite group in the class has an element of infinite order?

Some examples:

  1. For finitely generated abelian groups the question trivially has a positive answer.

  2. For recursively presented groups the answer is negative. An example of a periodic such group is the Grigorchuk group.

  3. For finitely presented groups, if I recall correctly, the question is still open.

The motivation for this question is that I am trying to find out whether the class of finitely generated subgroups of the group discussed here has this property. Extensive systematic searches by computer have been inconclusive so far, i.e. the results found so far neither point to a reason why there are always elements of infinite order nor do they reveal a way to construct a periodic group. Knowing suitable other classes of groups for which the answer is known might help me in getting further.

Improve this question
$\endgroup$
7
  • 2
    $\begingroup$ Linear groups are an example by Schur. $\endgroup$
    – Benjamin Steinberg
    Commented Jul 16, 2013 at 19:21
  • 2
    $\begingroup$ Hyperbolic groups. Among periodic groups, there are both amenable (Grigorchuk) groups and groups with property T. I don't know if a finitely-generated periodic group can have the Haagerup property though. $\endgroup$
    – Ian Agol
    Commented Jul 16, 2013 at 22:30
  • 2
    $\begingroup$ Automatic groups also. $\endgroup$
    – Benjamin Steinberg
    Commented Jul 17, 2013 at 0:35
  • 2
    $\begingroup$ Like hyperbolic groups, automatic groups (or any group with a regular language of unique normal forms) has no infinite finitely generated periodic group. $\endgroup$
    – Benjamin Steinberg
    Commented Jul 17, 2013 at 11:02
  • 2
    $\begingroup$ @Ian: About your comment:" I don't know if a finitely-generated periodic group can have the Haagerup property though": I'm sure this is an open question. $\endgroup$
    – Alain Valette
    Commented Jul 17, 2013 at 16:26

2 Answers 2

Reset to default
5
$\begingroup$

Since the comments are already answering the question I am copying here the answers so far as Community Wiki. Please add new answers here.

  • Linear groups
  • Groups with a regular language of unique normal forms. This includes hyperbolic groups, automatic groups and groups with a finite complete rewriting system.

(This is very easily proved. By the Pumping Lemma for regular languages, such a language would contain a subset of the form $\{ uv^nw∣n≥0 \}$ for words $u,v,w$, and so the group element represented by $v$ must have infinite order.)

  • Relatively hyperbolic groups with "non-trivial" peripheral structure. See Agol's comment below.

  • Elementary amenable groups - the smallest class of groups that contains finite and abelian groups and closed under taking extensions, direct unions, subgroups and factor groups. In particular all solvable groups belong here.

Improve this answer
$\endgroup$
7
  • $\begingroup$ I would add here relatively hyperbolic groups. On the other hand, for CAT(0) groups, this seems to be an open problem (due to Swenson). $\endgroup$
    – Misha
    Commented Jul 17, 2013 at 12:17
  • $\begingroup$ Misha, please add. That's why it is CW! $\endgroup$
    – Benjamin Steinberg
    Commented Jul 17, 2013 at 12:54
  • 1
    $\begingroup$ The claim about regular language of unique normal forms is very easy to prove. Such a language would contain a subset of the form $\{ uv^nw \mid n \ge 0 \}$ for words $u,v,w$, and so the group element represented by $v$ must have infinite order. $\endgroup$
    – Derek Holt
    Commented Jul 17, 2013 at 12:58
  • 1
    $\begingroup$ @DerekHolt, yes it is an immediate consequence of the pumping lemma. This was Bob Gilman's nice observation. You can add this to the answer. It is CW. $\endgroup$
    – Benjamin Steinberg
    Commented Jul 17, 2013 at 13:06
  • $\begingroup$ I find the reference to the Pumping Lemma particularly interesting -- though I am not sure whether it is of any use for obtaining progress with the class of groups I am looking at. $\endgroup$
    – Stefan Kohl ♦
    Commented Jul 17, 2013 at 16:11
2
$\begingroup$

Obviously, all free Burnside groups $B(m,n)$ are recursively presentable, so examples with the uniform torsion are possible here. Of course, the proof is much more complicated than for Grigorchuk's group.

For finitely presented groups the question is opened, even for the case of unbounded exponents. The closest result is by I.Ivanov-Pogodaev and A.Kanel-Belov, that gives an example of finitely presented infinite nil-semigroup $\Pi_0$, i.e. a semigroup where every element is equal in some power to an element $0$, for which identities $0x=0$, $x0=0$ hold for every $x \in \Pi_0$.

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ Another candidate for a "closest result" is the construction of finitely presented torsion-by-cyclic groups by Olshanski and Osin. $\endgroup$
    – YCor
    Commented Jul 17, 2013 at 22:04
  • 1
    $\begingroup$ Yves, is there a short version of such a group when the exponents of the group are not necessarily bounded (the paper of Olshanskii and Osin is ~100 pages)? As I remember, in I-P-K-B result the bound is non-uniform, and the proof is long (not sure if one can find it in the web). $\endgroup$
    – Al Tal
    Commented Jul 18, 2013 at 9:14
  • $\begingroup$ @ Al Tal: yes, there is a f.p. ascending HNN extension of the first Grigorchuk group, due to Grigorchuk as well. [R.I. Grigorchuk, An example of a finitely presented amenable group that does not belong to the class EG, Sb. Math. 189 (1–2) (1998) 75–95; Russian original: Mat. Sb. 189 (1) (1998) 79–100.] $\endgroup$
    – YCor
    Commented Oct 12, 2013 at 18:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .