Commuting fraction in subgroup is at least as much as in whole group

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Statement

In fraction terms

For a group $G$, define $CP(G)$ to be the set:

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

Then, if $H$ is a subgroup of a finite group $G$, we have:

$\frac{|CP(H)|}{|H|^2} \ge \frac{|CP(G)|}{|G|^2}$

We sometimes use the term Commuting fraction (?) for the quotient $|CP(G)|/|G|$ for a given finite group $G$. In those terms, the commuting fraction of a subgroup is at least as much as that of the whole group.

In probability terms

The probability that two elements picked uniformly at random commute cannot increase when we pass from a subgroup to the whole group.

In terms of number of conjugacy classes

For a finite group $G$, let $n(G)$ denote the Number of conjugacy classes (?) in $G$. Then, if $H$ is a subgroup of $G$, we have:

$\frac{n(H)}{|H|} \ge \frac{n(G)}{|G|}$

Equivalently:

$n(H) \ge \frac{n(G)}{[G:H]}$

Related facts

• Abelianness is subgroup-closed: This states that any subgroup of an abelian group is abelian. Note that for finite abelian groups, the statement is a particular case of the one on this page. Namely, if $H \le G$, and the fraction of ordered pairs commuting in $G$ is $1$, then the fraction of ordered pairs commuting in $H$ is $\ge 1$, hence equal to $1$, hence $H$ is also abelian.
• Number of conjugacy classes in a subgroup may be more than in the whole group

Facts used

1. Fraction of tuples satisfying groupy relation in subgroup is at least as much as in whole group (which in turn uses index satisfies transfer inequality)

Proof

The proof follows directly from fact (1) and the observation that the relation of commuting is groupy in both inputs -- for any element, the set of elements commuting with it is a subgroup, called the centralizer of that element.