 Solution

Suppose first that $H$ and $K$ are normal subgroups. For $h\in H$ and $k\in K$, the commutatorlies in $H$, because $k{h}^{-1}{k}^{-1}\in H$, and in $K$, because $hk{h}^{-1}\in K$. Since $H\cap K=\{e\}$, $[h,k]=e$, so $h$ and $k$ commute.

Conversely, suppose every element of $H$ commutes with every element of $K$. Write an arbitrary $g\in G$ as $g=h_0{k}_0$. For $h\in H$,so $H⊴G$. The symmetric calculation proves $K⊴G$.