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 be a finite group. The probability that two random group elements commute is equal to , where is the number of conjugacy classes in . *

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

Let

.

For each , the number of such that is equal to the cardinality of the centralizer of : . Thus,

.

If and are conjugate, then and are conjugate. Thus,

,

where the sum is over distinct conjugacy classes. Therefore, where is the number of conjugacy classes in , as desired.

When is non-abelian, it’s easy to show that , with equality if, and only if, the index of the center of is . 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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Did you mean the inequality the other way around? It’s clearly false if G is abelian.

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