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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.

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    $\begingroup$ My previous comment had overlooked a vital detail (the normality of $H$). Your idea is the right one. Just remember that anything here can be written uniquely as $g_1g_2$ with $g_1\in G_1$ and $g_2\in G_2$. Also remember that things from $G_1$ commute with things from $G_2$. $\endgroup$ – Tobias Kildetoft Feb 17 '14 at 14:48
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    $\begingroup$ Let $(a,b),(c,d) \in H$. Then $[(a,1),(c,d)] = ([a,c],1) \in H$, so $[a,c]=1$ and similarly $[b,d]=1$. $\endgroup$ – Derek Holt Feb 17 '14 at 14:50
  • $\begingroup$ @DerekHolt Why is the commutator of $(a,1)$ and $(c,d)$ in $H?$ $\endgroup$ – awllower Feb 17 '14 at 14:52
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    $\begingroup$ @awllower: Because $(c,d) \in H$ and $H$ is normal. Rewrite it as $\bigl( (a,1) (c,d) (a,1)^{-1} \bigr) (c,d)^{-1}$ $\endgroup$ – Najib Idrissi Feb 17 '14 at 14:54
  • $\begingroup$ OK. Thanks. How about writing it as an answer? :) $\endgroup$ – awllower Feb 17 '14 at 14:56
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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 $(n_1,n_2) \in Z(G_1) \times Z(G_2) = Z(G_1 \times G_2)$. $\square$

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  • $\begingroup$ it is really a nice result... $\endgroup$ – mesel Feb 18 '14 at 8:11
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    $\begingroup$ Thanks. It also shows that if $G_1$ and $G_2$ are both non-abelian simple, the only normal subgroups of $G_1 \times G_2$ are $\{(1,1)\}$, $G_1 \times \{1\}$, $\{1\} \times G_2$ and $G_1 \times G_2$. $\endgroup$ – Nicky Hekster Feb 18 '14 at 8:23

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