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If $G_1$ and $G_2$ are simple groups, what can we say about normal subgroups of $G_1 \times G_2$? I remember when I was taking Algebra I this was brought up in the class but at the time the professor left it for us to think over it.

Well, I remember I conjectured that $\{e\}, G_1 \times \{e\}, \{e\}\times G_2$ and $G_1 \times G_2$ are the only normal subgroups of $G_1 \times G_2$. I'm trying to see whether I am right or not. I haven't studied Sylow theorems yet and at the time the professor brought up this question I remember that we were studying direct products of groups. Is it possible to answer this question with elementary theorems in abstract algebra? Notice that $G_1$ and $G_2$ are not restricted to finite groups in my question.

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Hints: If $N$ is a normal subgroup of $G_1\times G_2$, as the projections $p_i:G_1\times G_2\to G_i$ are surjective, they map $N$ onto a normal subgroup.

For the case both $p_i(N)=G_i$, let $M:=\{g_1\mid (g_1,1)\in N \}$. Then $M$ is again a normal subgroup (of $G_1$). If $M=G_1$ we are ready soon.
Finally, if $M=\{1\}$ (by symmetry it also means $\{g_2\mid (1,g_2)\in N\}=\{1\}$), conclude that $N$ is a graph of an isomorphism (e.g. $(g_1,g_2)\in N$ and $(g_1,g_2')\in N$ implies $g_2=g_2'$), in other words $N$ is just the diagonal of $G_1\times G_2$ applying the isomorphism $G_1\cong G_2$.

On the other hand, unless $G$ is commutative, the diagonal of $G\times G$ for a simple group is not a normal subgroup.

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  • $\begingroup$ You beat me to it by few seconds. I was applying the correspondence theorem to the projection maps in my mind just when you wrote this. I have trouble following the reasoning in your answer by the way, would you please rephrase it if possible? $\endgroup$ – user66733 Nov 15 '13 at 3:26
  • $\begingroup$ There are a few cases you have to consider only: if either $p_i(G_i)$ is trivial, it is easy to get the conclusion what exactly $N$ is. ... $\endgroup$ – Berci Nov 15 '13 at 3:36
  • $\begingroup$ Yes. I already did it. I said let's assume $A_1 \times A_2$ is a normal subgroup of $G_1 \times G_2$ then, by using the projection maps, we conclude $A_1$ must be either $\{e\}$ or $G_1$ and we conclude the same thing for $A_2$. So, the only possibilities are those subgroups that I have mentioned in the problem and it's easy to see that all of them are normal so those are the only ones. Right? I was a bit confused by that diagonal thing and the graph of an isomorphism thing because I couldn't see how they were relevant to the solution but now I'm fine. Thanks. $\endgroup$ – user66733 Nov 15 '13 at 3:47
  • $\begingroup$ @some1.new4u No, no, they are relevant. You started out of a normal subgroup of the form $A_1\times A_2$, but noone said that $G_1\times G_2$ cannot have a normal subgroup of other form than a full 'rectangle'. In case, $\phi:G_1\to G_2$ is an arbitrary isomorphism, and $G_i$ are commutative, the set $\{(a,\phi(a))\mid a\in G_1\}$ is a normal subgroup of $G_1\times G_2$. $\endgroup$ – Berci Nov 15 '13 at 18:02

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