Tower Law for Subgroups/Proof 1

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
Since the index of $H$ in $G$ is finite, there exists finitely many $g_i \in G$ such that $\displaystyle G = \bigcup_{i=1}^p g_i H$ where $p = \left[{G : H}\right]$.

Similarly, there exists finitely many $h_j \in H$ such that $\displaystyle H = \bigcup_{j=1}^q h_j K$ where $q = \left[{H : K}\right]$.

Thus by substitution, $\displaystyle G = \bigcup_{i=1}^p \bigcup_{j=1}^q g_i(h_j K)$, which is the union of $pq$ disjoint sets. Therefore $\left[{G : K}\right] = pq = \left[{G : H}\right] \left[{H : K}\right]$.