# 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. Beautiful proof! One point, which is not clear to me:”This implies that the group $H_n$ generated by $X_n$ is generated in at most $6\div\epsilon$ steps” – why?

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2. 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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3. 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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4. Again, thank you so much for your time and help!

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5. 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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