If you have a non-abelian group $G$ with some normal subgroup $K$, is it possible to have a non-abelian quotient group $G/K$?
Besides actually sitting down and trying to generate quotient groups through exhaustion, I have been thinking about using the fundamental theory of homomorphisms to pick a small non-abelian group like $D_6$ and find the quotient group it is isomorphic to. Does this seem like a good tactic?
I'm not looking for answers, just confirmation that this is a useful way to be thinking about it.
Best Answer
In general, there is the following fact:
If $N$ is a normal subgroup of $G$, then $G/N$ is abelian if and only if $[G,G]$ (the commutator subgroup) is a subgroup of $N$.
So the quotient will not be commutative precisely when your normal subgroup does not contain the commutators.