5/8 bound in group theory The odds of two random elements of a group commuting is the number of conjugacy classes of the group
$$  \frac{ \{ (g,h): ghg^{-1}h^{-1} = 1 \} }{ |G|^2}  = \frac{c(G)}{|G|}$$
If this number exceeds 5/8, the group is Abelian (I forget which groups realize this bound).  
Is there a character-theoretic proof of this fact?  What is a generalization of this result... maybe it's a result about semisimple-algebras rather than groups?
 A: There is a beautiful generalization due to Guralnick and Wilson, The Probability of Generating a Finite Soluble Group.  Their results:
1)  if the probability that two randomly chosen elements of $G$ generate a solvable group is greater than $\frac{11}{30}$ then $G$ itself is solvable,
2) If the probability that two randomly chosen elements of $G$ generate a
nilpotent group is greater than $\frac{1}{2}$, then $G$ is nilpotent,
3)  if the probability that two randomly chosen elements of $G$ generate a group of odd order is greater than $\frac{11}{30}$ then $G$ itself has odd order. 
Interestingly, these probabilities are best possible. Note also the elementary McHale article on probability of commutativity again.  
A: One elementary result using character theory, but going in the other direction, which is proved in the paper of R. Guralnick and myself mentioned in my comment above is that if 
$\{\chi_1, \chi_2, \ldots, \chi_c \}$ are the complex irreducible characters of $G$, where $c = c(G)$ is the numberof conjugacy classes of $G,$ then by Cauchy-Schwarz, we have $\sum_{i=1}^{c} \chi_i(1) \leq \sqrt{c}\sqrt{|G|}$, so that $\frac{c(G)}{|G|} \geq \left( \frac{\sum_{i=1}^{c} \chi_i(1)}{|G|} \right)^{2}.$.  
A: If $c(G)> 5|G|/8$, then the average character has a dimension-squared of less than $8/5$, so at least $4/5$ of the characters are dimension $1$ (since the next-smallest dimension-squared is $4$), so the abelianization, which has one element for each 1-dimensional character, is more than half the size of the group, so the commutator subgroup has size smaller than $2$ and so is trivial. 
