# 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?

• That formula can't be quite right, since the term on the left is $<1$ and the term on the right is $>1$. Presumably you're off by a factor of $|G|$? The quaternions and $D4$ realize that bound. Mar 20, 2012 at 4:40
• A nice version would be to ask this for finite loops or quasigroups. Gerhard "Ask Me About System Design" Paseman, 2012.03.19 Mar 20, 2012 at 5:00
• Yeah, I dug up the article by Gustafson. This question appears as an exercise. Both groups you mention have order 8. Mar 20, 2012 at 5:03
• Any group with center of index 4 realizes this bound. (This is an iff). Mar 20, 2012 at 5:21
• Robert Guralnick and I have a Journal of Algebra Article called "On the Commuting Probability in Finite Groups" (~2006) where we discuss at some length links between the commuting probability and character theory among other things. Much of the paper is reasonably elementary, including a proof that that the commuting probabilty tends to $0$ as $[G:F(G) \to \infty,$ where $F(G)$ is the largest nilpotent nomal subgroup of the finite group $G.$ I am not sure whether this paper would help for other algebraic systems though. Mar 20, 2012 at 7:27

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.

• Neat! Small correction: the average dimension-squared is less than $8/5$. Mar 20, 2012 at 5:28
• The fact that more than 4/5 of the characters have degree 1 does NOT imply that the group is abelian. Look, for example, at an extraspecial group of order 2^{2n+1}. This nonabelian group has 2^n degree 1 characters, but only one of larger degree. The argument in this answer is thus, at best, incomplete. Feb 14, 2014 at 21:33
• @MartyIsaacs Recall that $(4/5) \times (5/8)=(1/2)$, and so the number of $1$-dimensional characters is more than half the number of elements of the group, as desired. The implication you suggest is not stated in the argument. Feb 14, 2014 at 23:36
• @WillSawin Right. Sorry, I misunderstood your argument. Feb 16, 2014 at 17:42
• @pre-kidney One can take products of those two with abelian groups. My argument shows that any non-abelian group meeting the bound must have the kernel of the abelianization of order $2$, and the traditional argument shows that the center has index $4$, which constrains the structure of the groups quite a bit. Feb 20, 2016 at 16:21

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.

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 $\frac{c(G)}{|G|} \geq \left( \frac{\sum_{i=1}^{c} \chi_i(1)}{|G|} \right)^{2}.$.