Tower Law for Subgroups/Proof 2

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
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.