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A group $G$ is defined to have the Bergman property if for any subset $X=X^{-1}$ generating $G$ there exists $n$ such that $X^n=G$.

By a result of Bergman, the permutation group of any set has the Bergman property. Later this result was extended to some other automorphism groups (of sufficiently homogeneous structures).

Very often the Bergman property appears in combination with uncountable cofinality.

A group $G$ is defined to be of uncountable cofinality if it can not be written as the countable union of a strictly increasing sequence of proper subgroups.

An example of a group of uncountable cofinality without the Bergman property is any infinite finitely-generated group (for example, $\mathbb Z$).

Problem. Is there a group of countable cofinality possessing the Bergman property?

I came to this question studying macro-uniform functions on finitary balleans of groups, but quickly have found that it was asked by Bergman himself in 2004 or earlier.

In 2006 Anatole Khelif announced a counterexample to this problem and in his paper in C.R. Acad. Paris promised: Les démonstrations des résultats ci-dessus seront publiées ultérieurement.

But I cannot google any papers of Khelif containing the promised proof. What does it mean? Is there a counterexample or not? Does anybody know the real situation?

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  • $\begingroup$ Do you mean a countable union of strictly increasing etc. in the definition of uncountable cofinality? $\endgroup$
    – Apollo
    Oct 24, 2018 at 14:01
  • $\begingroup$ @Apollo Yes I had in mind countable. But "sequence" means countable (usually). $\endgroup$ Oct 24, 2018 at 14:34
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    $\begingroup$ It seems that this paper of Khelif sciencedirect.com/science/article/pii/S1631073X06000276 contains the required example. So, I think it is better to remove this question from MO. $\endgroup$ Oct 24, 2018 at 21:36
  • $\begingroup$ Note that this property is very inconvenient to deal with and another stronger property (combination with uncountable cofinality), much more practical (passes to quotients, finite direct products, finite index subgroups, extensions...) is usually referred as Bergman property in the literature. I'm not aware of any use of this property, and the only specific result about it is this highly technical proof of Khelif; I'm not sure it has been seriously read at all: the CRAS announcement you mention contains no proof, and the promised paper has never appeared. $\endgroup$
    – YCor
    Oct 24, 2018 at 22:42
  • $\begingroup$ @YCor Thank you for your quick answer. Indeed, the stronger Bergman property is an invariant of the coarse structure and I was curious if its components (the Bergman property and the countable cofinality) also are coarse invariants. The Khelif example (if exists) says that the Bergman property is not a coarse invariant. What about the countable cofinality? $\endgroup$ Oct 24, 2018 at 22:48

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