Commuting probability of compact groups

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 {G} is a compact group of positive commuting probability then the FC-center {F(G)} is an open subgroup of {G} with finite-index center {Z(F(G))}.

(I actually stated this theorem incorrectly previously, asserting the conclusion {G=F(G)} as well; this is clearly false in general, for instance for {G=O(2)}.)

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 {G} is BFC if and only if {[G,G]} 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 {0} 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 {\mu} be the normalised Haar measure of {G}, and suppose that

\displaystyle \mu(\{(x,y):xy=yx\})\geq\epsilon.

Let {X_n} be the set of elements in {G} with at most {n} conjugates. Then {X_n} is closed, since any element {x} with at least {n+1} distinct conjugates {g_i^{-1}xg_i} has a neighbourhood {U} such that for all {u\in U} the points {g_i^{-1}ug_i} are distinct. Since

\displaystyle \mu(\{(x,y):xy=yx\}) = \int 1/|x^G| \,d\mu(x) \leq \mu(X_n) + 1/n,

we see that {\mu(X_n)\geq\epsilon/2} for all {n\geq 2/\epsilon}. This implies that the group {H_n} generated by {X_n} is generated in at most {6/\epsilon} steps, i.e., {H_n = X_n^{\lfloor 6/\epsilon\rfloor}}, which implies that {H_n} is an open BFC subgroup of {G}. Since {(H_n)} is an increasing sequence of finite-index subgroups it must terminate with some subgroup {F}, and in fact {F} must be the FC-center of {G}. This proves that {F(G)} is an open BFC subgroup of {G}.

In particular in its own right {F} is a compact group with {[F,F]} finite (by the theorem of Bernhard Neumann mentioned at the top of the page). Since the commutator map {[,]:F\times F\rightarrow[F,F]} is a continuous map to a discrete set satisfying {[F,1]=1} there must be a neighbourhood {U} of {1} such that {[F,U]=1}. This implies that {Z(F)} is open, hence of finite-index in {F}. \Box

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 2 If {G} is a compact group of commuting probability {p>0} then there is a finite group {H} also of commuting probability {p}.

We need a simple lemma before proving the corollary.

Lemma 3 For each {n>0} there is a finite group {K_n} of commuting probability {1/n}.

Proof: If {n} is odd then {D_n} has commuting probability {(n+3)/(4n)}. We can use this formula alone and induction on {n} to define appropriate groups {K_n}. Take {K_1=D_1} and {K_2=D_3}. If {n>2} is even take {K_n = K_2\times K_{n/2}}. If {n = 4k+1>2} take {K_n = K_{k+1}\times D_n}. If {n=4k+3>2} take {K_n = K_{k+1}\times D_{3n}}. \Box

An isoclinism between two groups {G} and {H} is a pair of isomorphisms {G/Z(G)\rightarrow H/Z(H)} and {[G,G]\rightarrow [H,H]} which together respect the commutator map {[,]:G/Z(G)\times G/Z(G)\rightarrow[G,G]}. Clearly isoclinism preserves commuting probability. A basic theorem on isoclinism, due to Hall, is that every group {G} is isoclinic to a stem group, a group {H} satisfying {Z(H)\leq [H,H]}. We can now prove the corollary.

Proof: Proof of corollary: By the theorem the FC-center {F} of {G} has finite-index, say {n}, and moreover {F} has finite-index center {Z(F)} and therefore finite commutator subgroup {[F,F]}. Let {E} be a stem group isoclinic to {F}. Then {E} and {F} have the same commuting probability, and {E} is finite since {E/Z(E) \cong F/Z(F)}, {[E,E]\cong [F,F]}, and {Z(E)\leq [E,E]}, so we can take {H=K_{n^2}\times E}. \Box

5 thoughts on “Commuting probability of compact groups

  1. Hi, thanks for your comment! This is a useful lemma. Suppose X is a symmetric subset of a group G containing the identity, and suppose X^{k+1}\neq X^k. Pick x\in X^{k+1}\setminus X^k. Then xX\subset X^{k+2}\setminus X^{k-1}. Thus if G has a Haar measure \mu then \mu(X^{k+2}\setminus X^{k-1})\geq \mu(X) whenever X^{k+1}\neq X^k. Thus if X^{3m}\neq\langle X\rangle then \mu(G) \geq \sum_{k=1}^m \mu(X^{3k+2}\setminus X^{3k-1}) \geq m \mu(X).

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  2. 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 X_n\cdot X_m is a subset of X_{nm}, it follows that F is equal to some X_k for some large-enough k. This is actually explicitly stated in your proof, when you say that F, which is FC-center, is also BFC-subgroup. But that implies that F is also closed (again, this is stated explicitly in the proof, when you say that F is compact). Thus, F is the connected component of the identity in G (which is a normal subgroup of G). Thus, the index t of F in G is just the number of connected components of G. Moreover, the multiplication in G defines a binary operation on the set of connected components of G, making it a finite group B=G/F. Am I right? Second one: why is [F,F] finite? How does it follow from the facts, that F is a BFC-group and that F is compact?

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  3. Being an open subgroup (and therefore in particular a closed subgroup), F contains the connected component of the identity, but F itself need not be connected. The connected component of the identity might even be trivial.That being said, yes F is an open normal subgroup and so G/F is a finite group. Also if G_0\subset F is the connected component of the identity in G then G_0 is a normal subgroup and so G/G_0 is a (not necessarily finite) group.The finiteness of [F,F] is equivalent to the BFC-ness of F. The left-to-right direction of this is easy: if e,f\in F then e^{-1} e^f \in [F,F], so e has at most |[F,F]| conjugates. The other direction is a theorem of Bernhard Neumann. In the compact case one can prove it as follows: Take a commutator c = [e,f]=e^{-1}f^{-1}ef. If one replaces e with any element e of C_F(f)e, and then f with any element f of C_F(e)f, then still [e,f]=c. If F is n-BFC then C_F(f) and C_F(e) both have index at most n, so \mu(\{(e,f)\in F^2 : [e,f]=c\}) \geq 1/n^2, so there are at most n^2 commutators c. Now the task is reduced to showing that if F has finitely many commutators then [F,F] is finite, which is a classical theorem of Schur.

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