Product formula for the order of HK

Statement

Let $G$ be a group and let $H$ and $K$ be finite subgroups of $G$. Then the product $HK$ has exactly $\dfrac{|H| \, |K|}{|H \cap K|}$ elements.

Remarks

The formula holds also when $HK$ is not a subgroup. The proof counts the pairs $(h, k)$: two pairs $(h, k)$ and $(h', k')$ give the same product exactly when $h' = hx$ and $k' = x^{-1}k$ for some $x \in H \cap K$. For example, two distinct subgroups of order $2$ in $S_3$ meet in $\{e\}$, and their product has $2 \cdot 2 / 1 = 4$ elements.

Explanations

L1 human

Each product $hk$ can be written in several ways, because you can move an element of $H \cap K$ from one factor to the other. Each element of $HK$ has exactly $|H \cap K|$ such ways. So you divide the number of pairs, $|H| \, |K|$, by $|H \cap K|$.