If $H$ and $K$ are nilpotent normal subgroups then $C(HK)$ is non trivial

Question

I know that this follows from the fact that $HK$ is nilpotent but maybe there is an easier way to proof this? I wanted to show that there is an Element in $HK$ that commutes with $H$ and $K$. I thought about looking at $H\cap K$ wich is a normal subgroup of $H$ and $K$ and because they are nilpotent, we have $x \in C(K) \cap H \neq 1$ as well as $y \in C(H) \cap K \neq 1$. I wanted to try to follow from this that either $x$, $y$ or $xy$ commute with $H$ and $K$ but I don't seem to get the right idea. Am I on the right track?