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