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Does anybody have an example of a normal subgroup N of a group $G\times H$ such that N is not equal to some product $N_1\times N_2$ while $N_1$ is normal in G and $N_2$ is normal in H.

I've already proofed the analog statement for rings-ideals, but I've understood that this is not necessary so using groups-normal subgroups and I'm having a hard time to find an example.

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  • $\begingroup$ Hint: Consider a diagonal subgroup in a direct product of two copies of the same abelian group. $\endgroup$ – Tobias Kildetoft Jan 15 '14 at 10:46
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    $\begingroup$ As others have pointed out, there is an example in which $G$ and $H$ have order $2$, so why are you finding it so hard to find an example? $\endgroup$ – Derek Holt Jan 15 '14 at 10:54
  • $\begingroup$ @AndreasCaranti: I believe the ideals (and even modules) of $R\times S$ are all of the form $I \times J$, since $M = (e_1 M) \oplus (e_2 M)$. From that point of view, the group theory result is surprising, since ideals and normal subgroups are vaguely similar. From the “groups of order 4” perspective, it is pretty obvious though. :-) $\endgroup$ – Jack Schmidt Jan 15 '14 at 20:41
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Even in the simplest of cases: take $\;N_1=N_2=\Bbb Z_2=\{0,1\}\pmod 2\;$, and now take the Klein group $\;\Bbb Z_2\times \Bbb Z_2$ and in it the diagonal subgroup $\;N:=\langle\,(1,1)\;\rangle\;$ ...

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Hint: What are the subgroups of the Klein 4-group?

There are three proper, non-trivial subgroups. List them.

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