Is G/H Always an Abelian Group if H is Normal in G?

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SUMMARY

In the discussion, it is established that if H is a normal subgroup of G, then the factor group G/H is an abelian group. The proof involves demonstrating that for elements x and y not in H, the relation xyH = yxH holds, leading to the conclusion that (xyH)(yxH)^{-1} = id. Additionally, it is clarified that the factor group G/H is not a subgroup of G, particularly when H is the trivial subgroup {e}.

PREREQUISITES
  • Understanding of group theory concepts, specifically normal subgroups.
  • Familiarity with factor groups and their properties.
  • Knowledge of abelian groups and their characteristics.
  • Basic proof techniques in abstract algebra.
NEXT STEPS
  • Study the properties of normal subgroups in group theory.
  • Learn about the structure and properties of factor groups.
  • Explore the implications of abelian groups in various mathematical contexts.
  • Investigate examples of groups where H is the trivial subgroup and analyze their factor groups.
USEFUL FOR

This discussion is beneficial for students and professionals in mathematics, particularly those studying abstract algebra, group theory, and anyone interested in the properties of normal subgroups and factor groups.

Kanchana
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Let H be a normal subgroup of G. Then factor group G/H is an abelian subgroup.
For x, y not in H
xHyH=yHxH
and xyH=yxH
(xyH)(yxH)^{-1}=id
xyx^{-1}y^{-1}=id

Are these steps correct?


thnx
 
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Kanchana said:
Let H be a normal subgroup of G. Then factor group G/H is an abelian subgroup

Are you trying to prove this?

What happens if ##H=\{e\}##?

Also, the factor group ##G/H## is not a subgroup of ##G##.
 

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