Given $H \lhd G$ and $G/H$ is abelian then the derived group $G'$ or $[G,G]$ satisfies $G'\subseteq H$
I've approached this problem the following way:
As $H$ is abelian then given $a,b \in G$,
$$\begin{align} aH \cdot bH=bH \cdot aH &\Rightarrow (a \cdot b)H= (b \cdot a)H\\ & \Rightarrow H=(b^{-1}\cdot a^{-1}\cdot b \cdot a)H \end{align}$$
This means that for any given $a,b \in G$ and $h \in H$ we have that $(b^{-1}\cdot a^{-1}\cdot b \cdot a)\cdot h \in H$
Making $h:=1_G$ we have that $(b^{-1}\cdot a^{-1}\cdot b \cdot a) \in H$ $\forall a,b \in G$
Therefore, as $H$ is a subgroup $\langle \{b^{-1}\cdot a^{-1}\cdot b \cdot a: a,b\in G\}\rangle \subseteq H$ thus $G'\subseteq H$
My problem with this is that I don't use the condition $H \lhd G$ and this is usually a bad sign in a proof. I would appreciate if anyone could verify this proof or help me see where its wrong
Thanks in advance.