Size of the product of two subgroups

Let $(G, \ast)$ be a group and let $H\le G$ and $K\le G$ be subgroups of $G$. Prove that $|HK|$=$\frac{|H|\cdot|K|}{|H\cap K|}$.

Intuitively this is quite obviously true, as otherwise the products of all elements in the intersection of $H$ and $K$ would be counted twice, but no idea how to prove it! Any advice appreciated.

$\newcommand{\Size}[1]{\lvert #1 \rvert}$Consider the set $\mathcal{A}$ of cosets of $K$ in the (subset!) $H K$. The size of $\mathcal{A}$ is $\Size{H K}/\Size{K}$. Consider the set $\mathcal{B}$ of cosets of $H \cap K$ in $H$, of size $\Size{H} / \Size{H \cap K}$.

Now show that for $h \in H$ $$ h K \mapsto h (H \cap K) $$ well-defines a bijection from $\mathcal{A}$ to $\mathcal{B}$.