When is a quotient group $G/H$ abelian?

Question

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 :-)

Answer 1

But is there cases when this quotient group is abelian without the group G being abelian?

Ans: Consider $G:=Q_8$, the quaternion group and $H:=\{\pm1,\pm i\}$.Then $G/H$ is Abelian whereas $G$ is not!

Answer 2

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": $[G,G]\le H$.

Answer 3

Use the notion of the commutator subgroup to prove the following characterization:

Let $G$ be any group and let $H$ be a normal subgroup. Denote the commutator subgroup of $G$ by $[G,G]$.

Then the quotient $G/H$ is abelian if and only if $[G,G]\le H$.

This gives an easy way of identifying whether quotients will be abelian or not, given that you know something about $[G,G]$.