Normal subgroups of a product of simple groups

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Every normal subgroup of the direct product of two simple groups G = G1 × G2 is isomorphic to G, G1, G2, or the trivial subgroup. The discussion highlights attempts to prove this by examining the structure of normal subgroups and using quotient groups. It notes that normal subgroups cannot simply be formed by combining normal subgroups from G1 and G2 due to counterexamples like the Klein four-group. The isomorphism theorem is referenced to analyze the implications of normality in relation to the simple groups. Understanding the significance of the "simple" condition is crucial for concluding that no other normal subgroups exist.
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Homework Statement


Let G = G1 × G2 be the direct product of two simple groups. Prove that every normal subgroup of G is isomorphic to G, G1, G2, or the trivial subgroup.

The Attempt at a Solution


I tried proving that the normal subgroups would have to be of the form Normal subgroup X Normal subgroup. However, that's false because, e.g., <(1,1)> is a normal subgroup of the Klein four-group.
 
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Consider the quotients ##(G_1\times G_2)/ G_1## and ##(G_1\times G_2)/ G_2##.
 
I tried looking at the 1st isomorphism theorem with φ: (projection onto G1), but I didn't get anywhere. Clearly the quotient groups that micromass gives are homomorphic to the simple groups ... But how would that fact lead to the conclusion that no other subgroups can exist?

I feel like I could make more progress if I understood the difference that the "simple" requirement makes ...
 
So you have an isomorphism ##\varphi: (G_1\times G_2)/(G_1\times \{e\}) \rightarrow G_2##. Take a normal subgroup ##N## of ##G_1\times G_2##. Look at ##\varphi(N)##. Is this normal?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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