I'm reading Kurzweil & Stellmacher's "The Theory of Finite Groups", its 1.5.3 says:
Let $N$ be an Abelian normal subgroup of $G$. If $G/N$ is perfect, then also $G'$ is perfect.
Proof. From 1.5.1, applied to the natural epimorphism, we obtain $$G/N = (G/N)' = G'N/N$$
and thus $G = G'N$. Since also $G'/N \cap G'$ ($\cong G/N$) is perfect, the same argument gives $G' = G''(N \cap G')$. It follows that $G = G''N$ and $G/G'' \cong N/N \cap G''$. Now 1.5.2 implies $G' = G''$ since $N$ is Abelian. $\square$
I'm a bit lost here: why $G'/N \cap G'$ is perfect and $G'/N \cap G' \cong G/N$?