Let $G$ be a group and $H$ be a normal subgroup of $G$ such that $G/H$ is abelian. Then $G' \subseteq H$

Question

Let $G$ be a group and $H$ be a normal subgroup of $G$ such that $G/H$ is abelian. Prove trat $G' \subseteq H$.

I start by taking an arbitrary element $g' \in G'$. Then I can write $g'$ as an arbitrary product of elements of $G'$, that is: $g' = g_1^{-1}h_1^{-1}g_1h_1 \ldots g_n^{-1}h_n^{-1}g_nh_n$. Now, how do I show that $g' \in H$?