Tower Law for Subgroups

Theorem
Let $\left({G, \circ}\right)$ be a group.

Let $H$ be a subgroup of $G$, and let $K$ be a subgroup of $H$.

Then:
 * $\left[{G : K}\right] = \left[{G : H}\right] \left[{H : K}\right]$

where $\left[{G : H}\right]$ is the index of $H$ in $G$.

Proof 1
Each coset $g H$ of $G$ modulo $H$ contains all the cosets $g \left({h K}\right) = \left({g h}\right) K$ of $G$ modulo $K$, where $h K$ runs through all the cosets of $H$ modulo $K$.

Proof 2
Assume $G$ is finite.

Then:

Since $K \le H$, from Lagrange's Theorem we have that $\dfrac {\left\vert{H}\right\vert} {\left\vert{K}\right\vert} = \left[{H : K}\right]$.

Hence the result.