I mentioned before the following theorem of Hofmann and Russo, extending earlier work by Levai and Pyber on the profinite case.

Theorem 1 (Hofmann and Russo)If is a compact group of positive commuting probability then the FC-center is an open subgroup of with finite-index center .

(I actually stated this theorem incorrectly previously, asserting the conclusion as well; this is clearly false in general, for instance for .)

Here the *FC-center* of a group is the subgroup of elements with finitely many conjugates. In general a group is called FC if each of its elements has finitely many conjugates, and BFC if its elements have boundedly many conjugates. A theorem of Bernhard Neumann states that a group is BFC if and only if is finite.

I noticed today that one can prove this theorem rather easily by adapting the proof of Peter Neumann’s theorem that a finite group with commuting probability bounded away from is small-by-abelian-by-small. Some parts of the argument below are present in scattered places in the above two papers, but I repeat them for completeness.

*Proof:* Let be the normalised Haar measure of , and suppose that

Let be the set of elements in with at most conjugates. Then is closed, since any element with at least distinct conjugates has a neighbourhood such that for all the points are distinct. Since

we see that for all . This implies that the group generated by is generated in at most steps, i.e., , which implies that is an open BFC subgroup of . Since is an increasing sequence of finite-index subgroups it must terminate with some subgroup , and in fact must be the FC-center of . This proves that is an open BFC subgroup of .

In particular in its own right is a compact group with finite (by the theorem of Bernhard Neumann mentioned at the top of the page). Since the commutator map is a continuous map to a discrete set satisfying there must be a neighbourhood of such that . This implies that is open, hence of finite-index in .

For me, the Hofmann-Russo theorem is a negative result: it states that commuting probability does not extend in an interesting way to the category of compact groups. To be more specific we have the following corollary.

Corollary 2If is a compact group of commuting probability then there is a finite group also of commuting probability .

We need a simple lemma before proving the corollary.

Lemma 3For each there is a finite group of commuting probability .

*Proof:* If is odd then has commuting probability . We can use this formula alone and induction on to define appropriate groups . Take and . If is even take . If take . If take .

An *isoclinism* between two groups and is a pair of isomorphisms and which together respect the commutator map . Clearly isoclinism preserves commuting probability. A basic theorem on isoclinism, due to Hall, is that every group is isoclinic to a *stem group*, a group satisfying . We can now prove the corollary.

*Proof:* Proof of corollary: By the theorem the FC-center of has finite-index, say , and moreover has finite-index center and therefore finite commutator subgroup . Let be a stem group isoclinic to . Then and have the same commuting probability, and is finite since , , and , so we can take .

Beautiful proof! One point, which is not clear to me:”This implies that the group generated by is generated in at most steps” – why?

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Hi, thanks for your comment! This is a useful lemma. Suppose is a symmetric subset of a group containing the identity, and suppose . Pick . Then . Thus if has a Haar measure then whenever . Thus if then

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Hello and thank you for your prompt and clear reply! Indeed, a very nice and useful Lemma.Actually, after carefully working-out all the steps of the proof, I realized that there are two other points there, which I wanted to understand. Could you, please, help me with them?First one: Since is a subset of , it follows that is equal to some for some large-enough . This is actually explicitly stated in your proof, when you say that , which is FC-center, is also BFC-subgroup. But that implies that is also closed (again, this is stated explicitly in the proof, when you say that is compact). Thus, is the connected component of the identity in (which is a normal subgroup of ). Thus, the index of in is just the number of connected components of . Moreover, the multiplication in defines a binary operation on the set of connected components of , making it a finite group . Am I right? Second one: why is finite? How does it follow from the facts, that is a BFC-group and that is compact?

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Again, thank you so much for your time and help!

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Being an open subgroup (and therefore in particular a closed subgroup), contains the connected component of the identity, but itself need not be connected. The connected component of the identity might even be trivial.That being said, yes is an open normal subgroup and so is a finite group. Also if is the connected component of the identity in then is a normal subgroup and so is a (not necessarily finite) group.The finiteness of is equivalent to the BFC-ness of . The left-to-right direction of this is easy: if then , so has at most conjugates. The other direction is a theorem of Bernhard Neumann. In the compact case one can prove it as follows: Take a commutator . If one replaces with any element of , and then with any element of , then still . If is -BFC then and both have index at most , so , so there are at most commutators . Now the task is reduced to showing that if has finitely many commutators then is finite, which is a classical theorem of Schur.

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