Index satisfies transfer inequality

Contents

Statement

In terms of index

Suppose $G$ is a group and $H, K$ are subgroups of $G$. Then:

$[K:H \cap K] \le [G:H]$.

In terms of conditional probability

This formulation is valid for finite groups. It says that if $G$ is a group and $H, K$ are subgroups, then:

$\frac{|H \cap K|}{|K|} \ge \frac{|H|}{|G|}$

In other words, what it says is that, for a uniform distribution on a finite group, knowing that a particular element is in the subgroup $K$ either increases or keeps the same the probability that the element is in the subgroup $H$.

Related facts

Applications

The formulation in terms of conditional probability is particularly useful to prove results on the fractions of tuples satisfying a groupy relation. See, for instance:

• Fraction of tuples satisfying groupy relation in subgroup is at least as much as in whole group
• Fraction of ordered pairs commuting in subgroup is at least as much as in whole group

Facts used

1. Product formula: if $H, K \le G$ are subgroups, there is a natural bijection between the left cosets of $H \cap K$ in $K$ and the left cosets of $H$ in $HK$.

Proof

Given: A group $G$ with subgroups $H, K$.

To prove: $[K:H \cap K] \le [G:H]$.

Proof: By fact (1), the number of left cosets of $H$ in $HK$ equals the number of left cosets of $H \cap K$ in $K$. Thus, the number of left cosets of $H$ in $G$ is at least as much as the number of left cosets of $H \cap K$ in $k$, yielding the desired inequality.