Thread: 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

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

G/N is abelian (let's take the middle..)

definition of equal cosets..

.. QED.

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 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.