Finite Index of Subgroup of Subgroup Prove the following: If $H$ is a subgroup of finite index in a group $G$, and $K$ is a subgroup of $G$ containing $H$, then $K$ is of finite index in $G$ and $[G:H] = [G:K][K:H]$.
So this is basically a bijective proof? The number of cosets of $H$ in $G$ equals the number of cosets of $K$ in $G$ times the number of cosets of $H$ in $K$ by the multiplication principle? 
Reference: Fraleigh p. 103  Question 10.35 in A First Course in Abstract Algebra

The canonical map $G/H \to G/K$ is surjective. The fiber of $gK$ is $\{gkH : k \in K\}$, which can be identified with $K/H$.
