For a finite subset $S$ of a group write

$$k(S)=\frac{|\{(g,h)\in S^2 \,|\,gh=hg\}|}{|S|^2}$$

for the propability that two elements in $S$ commute. It is known that $k(G)>5/8$ implies that a finite group $G$ is abelian.

Is there any result in this direction for $G$ finitely generated? For example: If we denote by $B_r(G)$ the elements of word-length at most $r$ w.r.t. some finite generating set, is there a condition on the limit

$$\lim_{r\longrightarrow\infty} k(B_r(G))$$

(or maybe on $\liminf$ or $\limsup$) that implies $G$ abelian?

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    $\begingroup$ Many homeomorphism groups such as groups of IETs or subgroups of Thompson's groups, or Grigorchuk's groups have a lot of commuting elements (typically with disjoint support), still I don't see if it's plausible that it takes a positive "probability" inside balls for any of these examples. Actually, in balls of large groups, it is often useful to consider a logarithmic probability, i.e. measure $\frac{\log(\dots)}{\log(|S|^2)}$, to take in consideration subsets that are too small to be detected by additive probability, but are still large in a certain sense. $\endgroup$ – YCor Sep 6 '19 at 21:40
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    $\begingroup$ Yes, log/log is a more natural quantity. $\endgroup$ – Mark Sapir Sep 6 '19 at 22:18
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    $\begingroup$ This question is studied in arxiv.org/abs/1511.07269 $\endgroup$ – Benjamin Steinberg Sep 7 '19 at 3:00
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    $\begingroup$ The results are not definitive but their are some results and conjectures. $\endgroup$ – Benjamin Steinberg Sep 7 '19 at 3:00

Try not to be seduced by balls. When it comes to infinite groups, you get much better theory if you focus on random walks. For example, the most basic thing you might ask of your density function is that every subgroup $H\leq G$ have density $1/[G:H]$. This is true if you measure density with random walks, by the basic theory of Markov chains (I have heard the infinite-index case attributed to Avez, but see also these two previous MO questions.). On the other hand, if you measure density with balls then you can have an infinite-index subgroup with density nearly $1$ (see here for example).

If you measure density with a random walk, then there is a very satisfactory and unsurprising answer to your question. A group has positive commuting probability if and only if it is virtually abelian, and in particular if it has commuting probability more than $5/8$ then it must be abelian. (In fact, if a group has positive commuting probability then its commuting probability is also realized by some finite group.) For all this, see this paper of Tointon. The same results hold for balls, provided the balls have subexponential growth. As mentioned in the comments, it has been conjectured by Antolin, Martino, and Ventura that the same is true of balls of exponential growth.

I am a little sceptical of this conjecture, since as mentioned there can be infinite-index subgroups of positive density. It was admittedly essential in that construction that the subgroup have exponential growth, but I cannot at the moment convince myself that you can't have a positive-density infinite-index abelian subgroup, or even center.


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