So clearly if $G$ is abelian and $H$ is a normal subgroup of $G$ then $G/H$ is abelian since $$xH.yH =(xy)H=(yx)H=yH.xH$$
But is there cases when this quotient group is abelian without the group G being abelian?
What I came up with is that for $(xy)H=(yx)H$ to be true then $(xy)^{-1}(yx)\in H$ must be satisfied, for all $x,y \in G$. Is this correct ? And any examples to this?
Thanks 🙂
Best Answer
Yes, as mentioned in the comments you can always take the quotient by the so-called commutator subgroup. That is, the group generated by all commutators $[x,y]=xyx^{-1}y^{-1}$ of elements of $G$.
This quotient is always abelian, and is referred to as the abelianization of $G$.
Now, the answer to your title question is that the quotient $G/H$ will be abelian iff $H$ contains the commutator, aka the "first derived subgroup": $[G,G]\le H$.