# Let $N$ be an Abelian normal subgroup of $G$, if $G/N$ is perfect, then also $G'$ is perfect.

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

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