Commutator of a group is identity?

[Kanchana]

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

Thanx

 

[HallsofIvy]

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

 

[Kanchana]

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

 

[homeomorphic]

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.

 

[blue_raver22]

for the interesting question! The commutator of a group, denoted [x,y], is defined as the element xyx^{-1}y^{-1}. If the commutator is equal to the identity element, denoted as 1, then it means that x and y commute, or in other words, they can be rearranged without changing the result. This is a property of abelian groups, where the order of multiplication does not matter.

To show that xyx^{-1}y^{-1}=1 in an abelian group G/[G,G], we can use the fact that this group is abelian. This means that [x,y] = xyx^{-1}y^{-1} = yxy^{-1}x^{-1} = [y,x]. Since the commutator is equal to its own inverse, we can rewrite this as [x,y][y,x] = 1. Using the associative property, we can also write this as [x,[y,x]] = 1. Since [y,x] is an element of the commutator subgroup [G,G], this means that [x,[y,x]] is also an element of [G,G]. However, since G/[G,G] is abelian, the commutator subgroup is equal to the identity element, so [x,[y,x]] = 1. Therefore, xyx^{-1}y^{-1} = 1, and we have shown that if the group G/[G,G] is abelian, then xyx^{-1}y^{-1} = 1.

In conclusion, the fact that the group G/[G,G] is abelian allows us to show that the commutator xyx^{-1}y^{-1} is equal to the identity element. This is because in an abelian group, the commutator subgroup is equal to the identity element, making it easier to prove this property.