Let $G = H_1 \times H_2$. Let $N$ be a normal subgroup of $G$ such that $N \cap H_1 = N\cap H_2 = \{e\}$. Prove that $N$ lies in the center of $G$. I can see that $G$ is a direct product of $H_1$ and $H_2$. How can $N \cap H_1 = N \cap H_2 = \{e\}$ lead to the conclusion that $N$ lies in the center? Do I need to use the fact that the center of a group is a normal subgroup (which would be $N$ here)?  I feel like I can't use that assumption to prove this...
Any insight will be greatly appreciated!

Let $(h_1,h_2)\in N$ and $h\in H_1$, $(h,1)(h_1,h_2)(h,1)^{-1}=(hh_1h^{-1},h_2)\in N$, since $N$ is normal
$(hh_1h^{-1},h_2)(h_1,h_2)^{-1}=(hh_1h^{-1}{h_1}^{-1},1)\in N$ since $N$ is a group.
Since $N\cap H_1=1$, $hh_1h^{-1}{h_1}^{-1}=1$. This implies that $hh_1=h_1h$. Thus $N$ commutes with $H_1$ similarly, we show that $N$ commutes with $H_2$
