Order of Subgroup Product

Theorem
Let $$g$$ be a group.

Let $$H$$ and $$K$$ be subgroups of $$G$$

Then $$\left|{H K}\right| = \frac {\left|{H}\right| \left|{K}\right|} {\left|{H \cap K}\right|}$$.

Proof
From Intersection of Subgroups, we have that $$H \cap K \le H$$.

Let the number of left cosets of $$H \cap K$$ in $$H$$ be $$r$$.

Then the left coset space of $$H \cap K$$ in $$H$$ is $$\left\{{x_1 \left({H \cap K}\right), x_2 \left({H \cap K}\right), \ldots, x_r \left({H \cap K}\right)}\right\}$$.

So each element of $$H$$ is in $$x_i \left({H \cap K}\right)$$ for some $$1 \le i \le r$$.

Also, if $$i \ne j$$, we have $$x_j x_i^{-1} \notin H \cap K$$.

Let $$h k \in H K$$.

We can write $$h = x_i g$$ for some $$1 \le i \le r$$ and some $$g \in K$$.

Thus $$h k = x_i \left({g k}\right)$$.

Since $$g, k \in K$$, this shows $$h k \in x_i K$$.


 * Now suppose the cosets $$x_i K$$ are not all disjoint.

Then $$x_i K = x_j K$$ for some $$i, j$$ by Coset Spaces form a Partition.

So $$x_j^{-1} x_i \in K$$ by Congruence Class Modulo Subgroup is Coset.

Since $$x_i, x_j \in H$$, we have $$x_j^{-1} x_i \in H \cap K$$, which is contrary to the definition.

Therefore, the cosets $$x_i K$$ are disjoint for $$1 \le i \le r$$.

This leads us to $$\left|{H}\right| / \left|{H \cap K}\right| = \left|{H K}\right| / \left|{K}\right|$$

whence the result.