TITLE: Why is a normal subgroup of $G_1\times G_2$ with trivial intersections with $G_1$ and $G_2$ is abelian?
QUESTION [6 upvotes]: Let $G=G_1\times G_2$ be a direct product, and let $H\triangleleft G$ be a normal subgroup such that $H\cap G_1=H\cap G_2=\{1\}.$ Then $H$ is abelian.  

I considered the commutators of two elements in $H$, trying to show that it must be $1,$ but I don't see how this works.
Any hint is very welcomed and appreciated. Thanks in advance.

REPLY [7 votes]: 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 $(n_1,n_2) \in Z(G_1) \times Z(G_2) = Z(G_1 \times G_2)$. $\square$