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

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$$\mu(\{(x,y):xy=yx\})\geq\epsilon.$$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$$\mu(\{(x,y):xy=yx\}) = \int 1/|x^G| \,d\mu(x) \leq \mu(X_n) + 1/n,$$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 $[,]:F\times F\to[F,F]$ 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 $F$.$\square$

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.

We need a simple lemma before proving the corollary.Corollary: If $G$ is a compact group of commuting probability $p>0$ then there is a finite group $H$ also of commuting probability $p$.

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 $K_n = K_{k+1}\times D_{3n}$.$\square$Lemma: For each $n>0$ there is a finite group $K_n$ of commuting probability $1/n$.

An

*isoclinism*between two groups $G$ and $H$ is a pair of isomorphisms $G/Z(G)\to H/Z(H)$ and $[G,G]\to [H,H]$ which together respect the commutator map $[,]:G/Z(G)\times G/Z(G)\to[G,G]$. 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 $Z(H)\leq [H,H]$.

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 $H=K_{n^2}\times E$.$\square$

Beautiful proof! One point, which is not clear to me:

ReplyDelete"This implies that the group $H_n$ generated by $X_n$ is generated in at most $6\div\epsilon$ steps" - why?

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 $$\mu(G) \geq \sum_{k=1}^m \mu(X^{3k+2}\setminus X^{3k-1}) \geq m \mu(X).$$

Hello and thank you for your prompt and clear reply! Indeed, a very nice and useful Lemma.

DeleteActually, 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?

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

DeleteBeing 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.

DeleteThat 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.