Commuting fraction more than half implies nilpotent

Statement
Suppose the fact about::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 fact about::nilpotent group, and in particular, a fact about::finite nilpotent group.

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

Stronger facts

 * Commuting fraction more than half implies direct product of class two 2-group and odd-order abelian group

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) uses::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) uses::Commuting fraction of quotient group is at least as much as that of whole group

Showing that the group has a nontrivial center
We show that there have to be at least two conjugacy classes of size $$1$$.

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.