Index satisfies transfer inequality

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 $$H$$.

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) uses::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.