Question: Let G be a group and N be a normal subgroup of G. Show that the factor group G/N is abelian iff aba^-1b^-1 is in N for all a,b in G.
If G/N is abelian then Na*Nb = Nb * Na and Naa^-1bb^-1 is in N and so Naba^-1b^-1 is in N
If aba^-1b^-1 is in N then Nb*Na = Nba = aba^-1b^-1ba which canceles to ab which I would like to say is Nab = Na*Nb.
I feel like I'm missing something though.
Thanks
Try to prove a more useful related result.
Definition:An element expressable as is called a 'commutatator'.
Definition:The 'commutatator subgroup' is generated by all commutatators.
Definition:The factor group is abelian if and only if contains the commutatator subgroup.