Commutator of a group is identity?

Kanchana · May 25, 2014

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

Thanx

HallsofIvy · May 25, 2014

I'm not sure I understand your question. Is G or G/[G,G] that is abelian? Are x and y in G?

Kanchana · May 25, 2014

G/[G,G] is abelian and x,y are from G/[G,G]. Then [x,y]=1..?

homeomorphic · May 26, 2014

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.

Commutator of a group is identity?

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