A Probabilistic Approach to Group Commutativity

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

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