Commutator of a group is identity?

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Discussion Overview

The discussion revolves around the properties of the commutator in group theory, specifically addressing the condition under which the commutator of two elements in a group is the identity element. The context includes theoretical aspects of group properties and the implications of the group being abelian.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant asks how to show that the commutator xyx^{-1}y^{-1} equals the identity if the group G/[G,G] is abelian.
  • Another participant seeks clarification on whether it is G or G/[G,G] that is abelian and questions the membership of x and y in G.
  • A subsequent reply asserts that if G/[G,G] is abelian and x, y are from G/[G,G], then the commutator [x,y] equals 1.
  • Another participant emphasizes that G/[G,G] is always abelian and explains that the equation xyx^{-1}y^{-1}=1 is equivalent to stating that x and y commute.

Areas of Agreement / Disagreement

Participants express differing levels of understanding regarding the implications of G/[G,G] being abelian, and there is no consensus on the clarity of the original question or the assumptions involved.

Contextual Notes

There are unresolved assumptions regarding the definitions of the elements x and y, as well as the implications of the group structure on the commutator.

Kanchana
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If the group G/[G,G] is abelian then how do we show that xyx^{-1}y^{-1}=1?

Thanx
 
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I'm not sure I understand your question. Is G or G/[G,G] that is abelian? Are x and y in G?
 
G/[G,G] is abelian and x,y are from G/[G,G]. Then [x,y]=1..?
 
G/[G,G] is ALWAYS abelian, no matter what G is, yet you are phrasing it as if it's an additional assumption. xyx^-1y^-1=1 is the same equation as xy = yx, just by multiplying on the right by y and then by x on both sides of the equation. To put it another way, x and y commute if and only if their commutator, xyx^-1y^-1 is equal to the identity.
 

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