# 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.