Proving Abelian Factor Groups: Group Theory Homework Solution

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SUMMARY

The discussion focuses on proving that a factor group of an abelian group is abelian. The proof begins with the assumption that G is abelian and involves elements a and b in G that commute. It correctly concludes that since H is a subgroup of G, the elements a and b also belong to H, leading to the conclusion that the factor group G/H is abelian. However, the proof requires clarification on the relationship between G and H, particularly that G must not equal H for the factor group to be non-trivial.

PREREQUISITES
  • Understanding of group theory concepts, specifically abelian groups.
  • Familiarity with factor groups and their properties.
  • Knowledge of subgroup definitions and their implications.
  • Basic proof techniques in abstract algebra.
NEXT STEPS
  • Study the properties of abelian groups in detail.
  • Learn about the construction and properties of factor groups.
  • Explore subgroup criteria and their implications in group theory.
  • Review proof strategies in abstract algebra, focusing on direct proofs and counterexamples.
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Students of abstract algebra, particularly those studying group theory, as well as educators and anyone looking to deepen their understanding of the relationship between abelian groups and their factor groups.

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Homework Statement


PRove that a factor group of an abelian group is abelian

Homework Equations


The Attempt at a Solution



Assume G in G/H is abelian. Let there be elements a and b that are in G. such that ab=ba. Since H is a subgroup of G , elements a and b are also in H. Then (aH)(bH)=ab(H) =ba(H)=(bH)(aH) . Therefore , factor group G/H is abelian.

Is my proof correct?
 
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G in G/H is abelian? How about 'G is abelian'. Let there be elements in a and b that are in G such that ab=ba? How about 'Since G is abelian any arbitrary choice of a and b in G will commute'. Since H is a subgroup of G, elements a and b are also in H? There is no injection between G and H unless G=H (H is a subset of G). What you want to say is that every element a in G corresponds to an a+H in G/H.

You have the right idea though.
 
Benzoate said:
Assume G in G/H is abelian.
What do you mean? G is abelian (given), G/H is abelian (trying to prove that). G is not in G/H

Benzoate said:
Let there be elements a and b that are in G. such that ab=ba.
ab = ba is automatically true if a and b are in G, yes.

Benzoate said:
Since H is a subgroup of G , elements a and b are also in H.
If and only if G = H of course, then G/H = {1} which is abelian. So the statement is false, but do you need it?

Benzoate said:
Then (aH)(bH)=ab(H) =ba(H)=(bH)(aH) . Therefore , factor group G/H is abelian.
This looks like the key line in the proof. It's ok, just check what assumptions you need, reviewing the lines above.
 

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