Showing $N_1\times N_2\unlhd G_1\times G_2$ iff there exist $N_1\unlhd G_1$ and $N_2\unlhd G_2$.

Question

Let $G=G_1\times G_2$ with $G_1$ and $G_2$ as groups.

Show that $N=N_1\times N_2$ is normal subgroups of $G$ if and only if there exist $N_1$ and $N_2$ are normal subgroups of $G_1$ and $G_2$ respectively

My proof for 2nd direction, $(g_1,g_2)\in G_1\times G_2$ and $(n_1,n_2)\in N_1\times N_2$ then
$(g_1,g_2)(n_1,n_2)(g_1^{-1},g_2^{-1})=(g_1n_1g^{-1}_1,g_2n_2g^{-1}_2)\in N_1\times N_2$

Answer 1

The question is to show that for any groups $G_1$, $G_2$, $N$ is a normal subgroup of $G_1\times G_2$ iff $N=N_1\times N_2$ for some normal subgroups $N_1$, $N_2$ of $G_1$, $G_2$ respectively.

The claim is false. In your question, you've outlined a proof of one direction of the implication. Here is a counter-example that shows that the other direction is false:

Let $G_1=G_2=\mathbb{Z}$, and let $N=\left\{(x,x)\,|\,x\in\mathbb{Z}\right\}$. Clearly $N$ is normal in $G_1\times G_2$, but $N$ is not expressible as a direct product of subgroups of $\mathbb{Z}$.