Index satisfies transfer inequality

Statement

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

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

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.