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
Let $p = \left[{G : H}\right]$, $q = \left[{H : K}\right]$.

By hypothesis these numbers are finite.

Therefore, there exist $g_1,\ldots,g_p \in G$ such that $G$ is a disjoint union $\displaystyle G = \bigsqcup_{i=1}^p g_i H$.

Similarly, there exist $h_1,\ldots,h_q \in H$ such that $H$ is a disjoint union $\displaystyle H = \bigsqcup_{j=1}^q h_j K$.

Thus

This expression for $G$ is the disjoint union of $pq$ cosets.

Therefore the number of elements of the coset space is
 * $\left[{G : K}\right] = pq = \left[{G : H}\right] \left[{H : K}\right]$.