# The 5/8 Theorem – Maximum Probability of Group Elements Commuting

Today we are going to attack a group theory problem that has a little bit of a combinatorial flavour.

Problem. Find an upper bound on the maximum probability that two randomly chosen elements of a finite non-abelian group commute, and show that it is attained.

## Discussion

Suppose that we have a group $G$, and a random element $g \in G$. We have then got two cases.

- If $g$ commutes with everything, then the probability that $g$ commutes with another element is obviously 1.
- If $g$ does not commute with everything, we need to find the probability that it commutes with a random other element.

This gives us two things to compute (or more correctly, to bound): the probability that a random element commutes with everything, and the probability that if it does not commute with everything, then it commutes with a random element. The natural approach is then clearly thinking about centers and centralisers.

## Solution

Let $G$ be a finite group. If it is non-abelian, then its center $Z(G)$ is not the entire group, that is, $G$ has some element $g \not \in Z(G)$. Then $C_G(g)$, centraliser of $g$, also cannot be the entire group, as that would contradict $z \not \in Z(G)$.

Noting that $C_G(g) \leq G$, and $Z(G) \leq C_G(g)$, we can then apply Lagrange’s theorem twice to get

Now let $x$, $y$ be a random elements of $G$. We can bound the probability that they commute with

This bound is achieved by the Quaternion group $Q_8$.