# A Probabilistic Approach to Group Commutativity

Posted on: January 18, 2011

A few weeks ago I went to a talk by Igor Pak. The talk discussed methods of generating random elements in finite groups. During the talk, Professor Pak mentioned an interesting probabilistic result:

Let ${G}$ be a finite group. The probability that two random group elements commute is equal to ${\frac{k}{n}}$, where ${k}$ is the number of conjugacy classes in ${G}$.

The proof is surprisingly straight forward. Indeed, we are trying to calculate the following:

$\displaystyle \begin{array}{rcl} \text{\bf{Pr}}(xy = yx) &=& \frac{\#\{(x, y) \in G \times G : xy = yx\}}{(\# G)^2}. \end{array}$

Let

${C = \{(x, y) \in G \times G : xy = yx\}}$.

For each ${x \in G}$, the number of ${y \in G}$ such that ${(x,y) \in G}$ is equal to the cardinality of the centralizer of ${x}$: ${\#C_G(x)}$. Thus,

${\#C = \sum_{x \in G} \#C_G(x)}$.

If ${x}$ and ${x'}$ are conjugate, then ${C_g(x)}$ and ${C_g(x')}$ are conjugate. Thus,

${\#C= \sum_{x_i} \#C_G(x_i)(G : C_G(x_i)) = \sum_{x_i}|G|}$,

where the sum is over distinct conjugacy classes. Therefore, ${\#C = k\cdot n}$ where ${k}$ is the number of conjugacy classes in ${G}$, as desired.

When ${G}$ is non-abelian, it’s easy to show that ${\frac{k}{n} \leq \frac{5}{8}}$, with equality if, and only if, the index of the center of ${G}$ is ${4}$. See the following paper for some interesting exercises, and the continuous version of this theorem: http://www.jstor.org/pss/2318778

The inequality is correct, but we need to assume that $G$ is non-abelian. Thanks for pointing that out!