Let $A$ be a normal subgroup of $G\times H$. If the identity of $G$ is $1_G$ and the identity of $H$ is $1_H$, $(x,y)\in A$ has the property that $x\ne1_G$ and $y\ne1_H$ unless $(x,y)=(1_G,1_H)=1_A$. The problem is to show that $A$ is abelian.
I believe that the way to approach this problem is to look at the commutator subgroup of $A$ and somehow show that it consists of just the identity element, but I'm not sure how.
Suppose $(a,b)$ and $(c,d)$ are in $A$. Then $(ac,bd)\in A$. Conjugating by $(c,1)$ shows that $(ca,bd)$ is in $A$ because $A$ is normal. Then $(ac,bd)^{-1}(ca,bd)=((ac)^{-1}ca,1)$. Since the second coordinate is 1, so is the first one. Thus $ac=ca$. By a similar argument, $bd=db$. It follows that $(a,b)(c,d)=(c,d)(a,b)$, so $A$ is abelian.