Does φ(N) Form a Normal Subgroup in G' When φ is a Homomorphism?

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SUMMARY

The discussion focuses on proving that if φ is a homomorphism from group G onto group G' and N is a normal subgroup of G, then φ(N) is a normal subgroup of G'. The key argument involves showing that for any element k in G', which can be expressed as k = φ(g) for some g in G, the conjugation kφ(a)k^{-1} remains in φ(N) for all a in N. The importance of φ being onto is emphasized, as it ensures that every element in G' corresponds to an element in G.

PREREQUISITES
  • Understanding of group theory concepts, specifically normal subgroups.
  • Familiarity with homomorphisms in abstract algebra.
  • Knowledge of the properties of group actions and conjugation.
  • Experience with Herstein's Abstract Algebra, particularly Section 2.7.
NEXT STEPS
  • Study the properties of normal subgroups in group theory.
  • Learn about homomorphisms and their implications on group structure.
  • Explore examples of group actions and their effects on subgroup properties.
  • Review Section 2.7 of Herstein's Abstract Algebra for additional exercises and proofs.
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Students of abstract algebra, mathematicians focusing on group theory, and educators teaching concepts related to homomorphisms and normal subgroups.

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From Herstein's Abstract Algebra. Section 2.7 #7



If φ is a homomorphism of G onto G' and N ◅ G, show that φ(N) ◅ G.



Attempt:
I want to prove that if k ∈ G' then kφ(N)k-1 = φ(N), but k = φ(n) for some n... then idk what.
 
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Take a in N, you need to prove that

[tex]k\varphi(a)k^{-1}\in \varphi(N)[/tex]

Replace k with [itex]\varphi(n)[/itex], what do you get??
 
the fact that φ is onto is important.

this means that EVERY k in G' is the image of some g in G:

k = φ(g). now use the fact that φ is a homomorphism.

what can we say about kφ(n)k-1?
 

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