If $G$ group of order $52$ includes a normal group of order $4$ then $G$ is abelian.
I did like this $ |G|=52=2^2\cdot 13 $ let $H$ be normal group of order $4$. $n_{13}=1$ thus $G$ has a $K$ normal group order $13$. $|H|=4,|K|=13$, $H∩K=\{e\}$, $|H \cap K|=1$ , $H$ and $K$ finite groups $52=|HK|=|H|||K|/|H∩K|$ thus $HK=G$
I dont know either what to do next or if it's correct with this way