# 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 $\mathcal{B}$.

Show that for each $g \in HK$ the number of couples $(h,k) \in H \times K$ such that $hk = g$ is $|H \cap K|$. It might help to start with the case $g=1$.

Consider $f$: $H\times K\rightarrow HK$ where $(h,k)\mapsto hk$, which elements go to $e$? $hk=e\iff h=k^{-1}\implies h\in H\cap K$. Once $h$ is chosen $k$ is $h^{-1}$. So there are $H\cap K$.

Notice if $(h,k)$ maps to $g$ then for every $l\in H\cap K$ $(hl,l^{-1}k)$ goes to $l$, so there are at least $H\cap K$ elements that go to $g$. Now suppose $(h_1,k_1)$ maps to $g$. Then $h_1k_1=hk\implies h_1=hkk_1^{-1}$ and $k_1=h_1^{-1}hk$ and $h^{-1}h_1=kk_1^{-1}$

So let $l=kk_1^{-1}$, then $l^{-1}=k_1k^{-1}=h^{-1}{h_1}$. So that

$(h_1,k_1)=(hl,l^{-1}k)$. so $(h_1,k_1)$ is of the intial form, hence there are $H\cap K$ elements that map to $g$ for each $g$. This tells us $|HK|=\frac{|H|\times |K|}{|H|\cap |K|}$.