Difference between revisions of "Commuting fraction more than half implies nilpotent"

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 1/2
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Statement

Suppose the commuting fraction of a finite group is strictly greater than $1/2$. In other words, the probability that two elements, picked independently uniformly at random, commute, is more than half. Then, the finite group is a nilpotent group, and in particular, a finite nilpotent group.

In other words, if the number of conjugacy classes is more than half the order of the group, the group is nilpotent.

Related facts

Tightness

The group symmetric group:S3 is an example of a non-nilpotent group (in fact, a centerless group) whose commuting fraction is exactly $1/2$.

Converse

The converse is not true -- a nilpotent group, even a group of nilpotency class two, can have an arbitrarily low commuting fraction.

Facts used

1. Equivalence of definitions of commuting fraction: We use the definition that characterizes it as the quotient of the number of conjugacy classes to the order of the group.
2. Commuting fraction of quotient group is at least as much as that of whole group

Proof

Showing that the group has a nontrivial center

We show that there have to be at least two conjugacy classes of size $1$. PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

Using induction

We use the fact that the quotient of the group by its center has a commuting fraction at least as large (fact (2)) to argue, by induction on the order, that the quotient group is nilpotent, hence so is the original group.