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 $G'/N \cap G' \cong G/N$?

  • $\begingroup$ It is sufficient to show that $G'/N\cap G'\cong G/N$, since $G/N$ is perfect by hypothesis. $\endgroup$ – Shaun Dec 29 '18 at 10:48
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    $\begingroup$ This looks like a job for an isomorphism theorem. $\endgroup$ – Shaun Dec 29 '18 at 10:50

It's the second isomorphism theorem applied here: $$G'/(N\cap G') \cong G'N/N$$ and we already know $G'N/N=G/N$ which is perfect by hypothesis.

Consequently, by the same reasons as in the first part of the proof, we get $G'=G''N$, and thus $G=G'N=G''NN=G''N$.

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