# Prove $H$ is normal subgroup

I try to prove it, but I can't. Please tell it true or false?

Let a group $G=G_1\times G_2$, i.e. $G$ is direct product of $G_1$ and $G_2$.

If $H$ is normal subgroup of $G$, then $H$ will be direct product of $H_1$ and $H_2$, for $H_1$, $H_2$ are normal subgroup $G_1$ and $G_2$, isn't it?

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No. Read the answer below. –  DonAntonio Sep 5 '12 at 10:34

Counterexample:

$$H:=\{(0,0)\,,\,(1,1)\}\triangleleft \Bbb Z_2\times\Bbb Z_2$$

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No, that is not necessarily true. The mistake to think that it must be true is sometimes called the "Diagonal Fallacy". Take $G_1 = G_2 = \mathbb{R}$ under addition, and think about why it could be called such a thing.
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 $(n_1,n_2) \in Z(G_1) \times Z(G_2) = Z(G_1 \times G_2)$. $\square$