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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

Facts used

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| \le 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} \le \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}.