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With reference to John B Fraleigh's Abstract Algebra 7th ed page 164, Theorem15.20: Let G be a group.The set of all commutators generates a subgroup C (the commutator subgroup) of G.This subgroup C is a normal subgroup of G.Furthermore, if N is a normal subgroup of G, then G/N is abelian iff C is a subgroup of N.

In the proof why G/C is proved to be abelian.I cannot understand why this is proved here.Is this the part of the theorem?

Khan
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    Well, if you want to show that $G/N$ is abelain iff $C$ is a subgroup of $N$ you'll need to show in particular that it is true for $C=N$ that $G/N= G/C$ is abelian. I am not sure if this answers your question though. – quid Nov 30 '16 at 18:30
  • As it is proved later there G/N is abelian.Why do we need to prove G/C is abelian? Does this help in proving G/N abelian? – Khan Nov 30 '16 at 18:46
  • It can help yes as you have under the assumtions a surjective homorphism $G/C \to G/N$ you can derive directly commutativity of the later from the former. I don't know the book so I do not know what specfically the author does. – quid Nov 30 '16 at 23:19

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You are correct, that is a redundant part in the proof. Probably he is trying to develop further intuition about the factor groups. There is also an emphasis to the fact that factoring out the commutator subgroup of an arbitrary group kills the pairs that do not commute and turns the object abelian: $G/C$ is the abelianization of $G$.

Alp Uzman
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