Let $H$, $K$ be subgroup of $G$. Does $|HK|$ always divide $|G|$?

Question

I know that if either $H$ or $K$ is normal, it is true. But i want to know that in condition $H$ and $K$ are any group. If not, could i have a counterexample?

Answer 1

Let $G=S_3$ be the symmetric group of order six and let $H$ and $K$ be distinct subgroups of $G$ of order 2. Then $$|HK|=\frac{|H|\cdot|K|}{|H\cap K|}=\frac{2\cdot2}{1}=4$$ which does not divide $|G|=6$.

More generally, if $H$ and $K$ are distinct Sylow $p$-subgroups of a finite group $G$ then $|HK|$ will not divide $|G|$.