Commuting fraction more than five-eighths implies abelian

This article is about a result whose hypothesis or conclusion has to do with the fraction of group elements or tuples of group elements satisfying a particular condition.The fraction involved in this case is 5/8
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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 appears in a paper by Erdos and Turan.

Statement

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 abelian group, and in particular, a 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

Applications

The fact has applications to the abelianness testing problem, and is the basis for the randomized black-box group algorithm for abelianness testing.

Similar facts

• Commuting fraction equal to five-eighths iff inner automorphism group is Klein four-group
• Commuting fraction more than half implies nilpotent
• Fixed-class tuple fraction is bounded away from one for groups not of that class plays a similar role for the fixed-class nilpotency testing problem
• Randomized black-box group algorithm for normality testing has false positive rate of at most three-fourths plays a similar role for the normality testing problem

Facts used

1. Cyclic over central implies abelian
2. Lagrange's theorem in the special form subgroup of size more than half is whole group

Proof

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 $G$.

Step no.	Assertion	Given data used	Facts used	Previous steps used	Explanation
1	$G/Z$ is not cyclic	$G$ is non-abelian	Fact (1)	--	By fact (1), if $G/Z$ were cyclic, $G$ would be abelian, a contradiction.
2	$|G/Z|$ is at least $4$		--	Step (1)	Any group of order 1, 2, or 3 is cyclic, forcing $G/Z$ to have size at least $4$.
3	For all elements $g$ in $G$ and not in $Z$, the quotient $|C_{G}(g)|/|G|$ is at most $1/2$		Fact (2)	--	If $g$ is not in $G$, then $C_{G}(g)$ is a proper subgroup of $G$. By fact (2), we get that $|C_{G}(g)|/|G|\leq 1/2$.
4	The commuting fraction of $G$ is ${\frac {|Z|}{|G|}}+\sum _{g\in G\setminus Z}{\frac {|C_{G}(g)|}{|G|^{2}}}$		--	--	By definition. Explicitly, the commuting fraction is $\sum _{g\in G}{\frac {|C_{G}(g)|}{|G|^{2}}}$. Those $g$ that are in $Z$ can be separated out and we get the summation in the form presented.
5	The commuting fraction of $G$ is at most ${\frac {|Z|}{|G|}}+{\frac {(|G|-|Z|)}{|G|}}{\frac {1}{2}}$		--	Steps (3) and (4)	By Step (3), ${\frac {|C_{G}(g)|}{|G|^{2}}}\leq {\frac {1}{|G|}}{\frac {1}{2}}$ for each $g\in G\setminus Z$. The sum is thus bounded by the size of $G\setminus Z$ times this individual bound.
6	The commuting fraction of $G$ is at most ${\frac {1}{2}}+{\frac {|Z|}{2|G|}}$		--	Step (5)	Algebraic simplification
7	The commuting fraction of $G$ is at most $5/8$		Fact (2)	Steps (2), (6)	By Fact (2), $|Z|/|G|=1/(|G|/|Z|)=1/|G/Z|$ which by Step (2) is at most 1/4. Plugging this into Step (6), we get that the commuting fraction of $G$ is at most ${\frac {1}{2}}+{\frac {1}{2}}{\frac {1}{4}}={\frac {5}{8}}$.