Math Help - Abelian Factor Group

1. Abelian Factor Group

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

2. Originally Posted by temp31415
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
yes, that was right but you can simplify it..

Let G be a group and N be a normal subgroup of G.
Let $a,b \in G$

G/N is abelian $\Longleftrightarrow aNbN = (ab)N = (ba)N = bNaN$ (let's take the middle..)

$\Longleftrightarrow (ab)(ba)^{-1} \in N$ definition of equal cosets..

$\Longleftrightarrow aba^{-1}b^{-1} \in N$.. QED.

3. Originally Posted by temp31415
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 $aba^{-1}b^{-1}$ is called a 'commutatator'.

Definition:The 'commutatator subgroup' $C$ is generated by all commutatators.

Definition:The factor group $G/N$ is abelian if and only if $N$ contains the commutatator subgroup.