Proving N(H) is a Subgroup of G to Normalizers in Group Theory

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SUMMARY

In group theory, for a subgroup H of a group G, the normalizer N(H) is defined as N(H) = {a ∈ G | H^a = H}, where H^a = {x ∈ G | axa^-1 ∈ H}. This discussion establishes that N(H) is indeed a subgroup of G by demonstrating that it satisfies the subgroup criteria: closure, identity, and inverses. Specifically, if elements a and b are in N(H), then H^{ab} can be shown to be equal to H, confirming that N(H) maintains the necessary properties of a subgroup.

PREREQUISITES
  • Understanding of group theory concepts, particularly subgroups and normalizers.
  • Familiarity with the definitions of closure, identity, and inverse elements in groups.
  • Knowledge of the notation and operations involving elements of groups.
  • Basic experience with mathematical proofs and logical reasoning.
NEXT STEPS
  • Study the properties of normalizers in group theory.
  • Learn how to prove subgroup criteria using specific examples.
  • Explore the implications of normalizers on group actions.
  • Investigate the relationship between normalizers and centralizers in groups.
USEFUL FOR

This discussion is beneficial for students and researchers in mathematics, particularly those focusing on abstract algebra, group theory, and anyone looking to deepen their understanding of subgroup properties and normalizers.

wegmanstuna
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For a subgroup H of G and a fixed element a ∈ G,
let H^a = {x∈ G / axa^-1 ∈ H}, it's normalizer N(H) = {a∈G / H^a=H}

Show that for any subgroup H of G, N(H) is a subgroup of G.



I know that for the first one I need to show that closure holds, an identity exists, and inverses exist. But I don't even know where to start with closure!
Help!
 
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If a and b are in N(H), write down what H^(ab) is. Now maybe you can try showing that H^{ab} \subset H and H \subset H^{ab}.
 

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