Let $\displaystyle G$ be a finite group and $\displaystyle k(G)$ be the number of conjugacy classes of $\displaystyle G.$ It's easy to see that the probability that two elements of $\displaystyle G,$ chosen randomly, commute is equal to $\displaystyle \frac{k(G)}{|G|}.$

We define $\displaystyle \text{Pr}(G)=\frac{k(G)}{|G|}.$ Obviously $\displaystyle \text{Pr}(G)=1$ if and only if $\displaystyle G$ is abelian. So $\displaystyle \text{Pr}(G)$ measures how "far" the group $\displaystyle G$ is from being abelian. The most basic question is to see whether

or not this "distance" can be made as small as we wish. The answer to this question is surprisingly negative:

Problem: Using the class equation, or any way you like, prove that for any finite non-abelian group $\displaystyle G: \ \text{Pr}(G) \leq \frac{5}{8}.$

Remark: $\displaystyle D_8,$ the dihedral group of order $\displaystyle 8,$ has exactly $\displaystyle 5$ conjugacy classes and thus $\displaystyle \text{Pr}(D_8)=\frac{5}{8}.$

Note: There are many suprising and beautiful results, not all easy to prove, about $\displaystyle \text{Pr}(G).$ I have already posted an easy one here. I will discuss more results later.