If $H$ is a normal subgroup of $G$ with $G/H$ abelian, then the commutator subgroup of $G$ is in $H$. This is part of Exercise 2.7.9 of F. M. Goodman's "Algebra: Abstract and Concrete".

Let $C$ be the commutator subgroup of a group $G$. Show that if $H$ is a normal subgroup of $G$ with $G/H$ abelian, then $C\subseteq H$.

The following seems to be wrong.
My Attempt:
The commutator subgroup $C$ of $G$ is the subgroup generated by all elements of the form $xyx^{-1}y^{-1}$ for $x, y\in G$.
Since $G/H$ is abelian, we have for $x, y\in G$,
$$\begin{align}
xyx^{-1}y^{-1}H&=xyy^{-1}x^{-1}H \\
&=H,
\end{align}$$ so that all elements of the form $xyx^{-1}y^{-1}$ are in $H$. Thus $C\subseteq H$.

But I don't use the fact that $H$ is normal. What have I done wrong and what is the right proof?


The fact that $G/H$ is abelian gives us the second equality of:$$xyH=(xH)(yH)=(yH)(xH)=yxH$$ 
Consequently we find: $$x^{-1}y^{-1}xy=(yx)^{-1}xy\in H$$
This for every $x,y\in G$ so we are allowed to conclude that $H$ contains the commutator subgroup.
