Why is a normal subgroup of $G_1\times G_2$ with trivial intersections with $G_1$ and $G_2$ is abelian?

Solution 1:

Theorem If $N \unlhd G_1 \times G_2$, then either $N \subseteq Z(G_1 \times G_2)$ or $N$ intersects one of the factors $G_1$ or $G_2$ non-trivially.

Proof. Assume that $N \cap (G_1 \times$ {$1$}$)$ = {$(1,1)$} = $N \cap $({$1$} $\times$ $G_2)$. We will show that N is contained in the center of $G_1 \times G_2$. Fix an arbitrary $(n_1,n_2) \in N$. Let $g_1 \in G_1$. Since $N$ is normal, $(g_1,1)^{-1}·(n_1,n_2)·(g_1,1) = (g_1^{-1}n_1g_1, n_2) \in N.$ Obviously also $(n_1,n_2)^{-1} = (n_1^{-1},n_2^{-1}) \in N$. It follows that the product $(n_1^{-1},n_2^{-1})·(g_1^{-1}n_1g_1, n_2) = (n_1^{-1}g_1^{-1}n_1g_1, 1) \in N$. However, $N$ intersects $G_1 \times$ {$1$} trivially, whence $n_1^{-1}g_1^{-1}n_1g_1 = 1$, that is, $n_1$ commutes with every $g_1 \in G_1$. Similarly, $n_2$ commutes with every $g_2 \in G_2$. This means that $(n_1,n_2) \in Z(G_1) \times Z(G_2) = Z(G_1 \times G_2)$. $\square$