Commuting fraction more than five-eighths implies abelian

History
It is likely that this statement, being quite simple, was known to many people since the beginnings of group theory and/or rediscovered a number of times. The first written proof apppears in a paper by Erdos and Turan.

For a finite group in terms of commuting fractions
Suppose $$G$$ is a finite group such that the commuting fraction of $$G$$ is more than $$5/8$$. Then, $$G$$ is an proves property satisfaction of::abelian group, and in particular, a proves property satisfaction of::finite abelian group.

In particular, if we define:

$$CP(G) := \{ (x,y) \in G^2 \mid xy = yx \}$$

Then:

$$\! \frac{|CP(G)|}{|G|^2} > \frac{5}{8} \implies G$$ abelian

Note that since the commuting fraction of a finite abelian group is $$1$$, this means that the commuting fraction of a finite group cannot take any value strictly between $$5/8$$ and $$1$$.

For a finite group in terms of conjugacy classes
Suppose $$G$$ is a finite group and $$n(G)$$ is the number of conjugacy classes in $$G$$. Then:

$$\frac{n(G)}{|G|} > \frac{5}{8} \implies G$$ abelian.

For a FZ-group
Suppose $$G$$ is a FZ-group whose commuting fraction is more than $$5/8$$. Then, $$G$$ is an abelian group.

Related facts

 * Commuting fraction equal to five-eighths iff inner automorphism group is Klein four-group
 * Fixed-class tuple fraction is bounded away from one for groups not of that class

Facts used

 * 1) uses::cyclic over central implies abelian
 * 2) uses::Lagrange's theorem in the special form uses::subgroup of size more than half is whole group

Proof for a finite non-abelian group
Given: A finite non-abelian group $$G$$.

To prove: The commuting fraction of $$G$$ is at most $$5/8$$.

Proof: Let $$Z$$ be the center of $$G$$.