Prove $aba^{-1}b^{-1}\in{N}$ for all $a,b$

Question

I am trying to prove the above with the conditions that N is normal to G and $G/N$ is abelian.

So then for $Na,Nb\in{G/N}, NaNb=NbNa$. But $NaNb=Nab=Nba$, so then we know $ab=ba$, so G is abelian.

So if $ab\in{G}, N=abN(ab)^{-1}=abN(ba)^{-1}=baNa^{-1}b^{-1}....$ and i'm getting lost...

Answer 1

Since $G/N$ is abelian, for all $a,b\in G$ we have $NaNb=NbNa$, so $Nab=Nba$. It does not follow that $ab=ba$, but rather that $Naba^{-1}b^{-1}=N$, i.e. that $aba^{-1}b^{-1}\in N$.