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 $K$.

Proof
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 $$K$$.

Alternatively (assuming $$G$$ is finite):

$$ $$ $$

Since $$K \le H$$, from Lagrange's Theorem we have that $$\frac {\left|{H}\right|} {\left|{K}\right|} = \left[{H : K}\right]$$. Hence the result.