Normal subgroup of a product of simple groups

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SUMMARY

The discussion focuses on proving that every proper normal subgroup K of the group G = G1 x G2, where G1 and G2 are simple groups, is isomorphic to either G1 or G2. The initial attempt identifies that the intersections of K with G1 x {1} and {1} x G2 are normal and can be either the trivial group or isomorphic to G1 and G2, respectively. The solution is clarified by considering three cases regarding the nature of these intersections, leading to a definitive conclusion about the structure of K.

PREREQUISITES
  • Understanding of group theory concepts, particularly normal subgroups.
  • Familiarity with simple groups and their properties.
  • Knowledge of the direct product of groups.
  • Experience with isomorphism in the context of algebraic structures.
NEXT STEPS
  • Study the properties of normal subgroups in group theory.
  • Learn about simple groups and their significance in algebra.
  • Explore the concept of direct products of groups in detail.
  • Investigate isomorphism theorems in abstract algebra.
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Students of abstract algebra, particularly those studying group theory, and educators seeking to deepen their understanding of normal subgroups in the context of simple groups.

john_nj
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Homework Statement



This is an exercise from Jacobson Algebra I, which has me stumped.
Let G = G1 x G2 be a group, where G1 and G2 are simple groups.
Prove that every proper normal subgroup K of G is isomorphic to G1 or G2.

Homework Equations





The Attempt at a Solution


Certainly the intersection of K with G1 x {1} is normal, and so is isomorphic to the trivial group {1} or to G1. Similarly, the intersection of K with {1} x G2 is isomorphic to {1} or G2. But anyway this falls well short of a solution.

Thanks,

John

 
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Well, you've basically got it. Try each of the three cases... if both intersections are trivial, then what is K? If both intersections are the full group, then what is K? And if only one intersection is trivial, what is K?
 
Thank you. It's now clear.
 

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