Commutator of a group is identity?

  • Thread starter Kanchana
  • Start date
  • #1
3
0
If the group G/[G,G] is abelian then how do we show that xyx^{-1}y^{-1}=1?

Thanx
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
956
I'm not sure I understand your question. Is G or G/[G,G] that is abelian? Are x and y in G?
 
  • #3
3
0
G/[G,G] is abelian and x,y are from G/[G,G]. Then [x,y]=1..?
 
  • #4
1,765
125
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.
 

Related Threads on Commutator of a group is identity?

  • Last Post
Replies
4
Views
1K
Replies
2
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
2K
Replies
3
Views
1K
Replies
6
Views
887
  • Last Post
Replies
5
Views
668
Replies
2
Views
2K
Replies
1
Views
2K
Replies
5
Views
932
Top