I have a doubt in the proof of the following given in Kurzweil and Stellmacher:
Let $Z$ and $E$ be subgroups of a finite group $G$ such that $Z\leq Z(G)$ and $EZ/Z$ is a component of $G/Z$. Then $E'$ is a component of $G$.
Proof: Since $Z\leq Z(G)$, we have $E'=(EZ)'$ and since $EZ\trianglelefteq\trianglelefteq G$ also $E'\trianglelefteq\trianglelefteq G$. Moreover, 1.5.3 on page 25 shows that $E'$ is perfect. Let $N$ be a normal subgroup of $E'$ and $\overline G=G/Z$. Then either $$\overline N=\overline E=\overline E'\text{ or } \overline N\leq Z(\overline E).$$ The first case gives $N\leq E'\leq NZ$ and thus $N(Z\cap E')=E'$. Hence $N=E'$ since $E'$ is perfect. The second case gives $[E',N]\leq Z$ and $$[E',N,E']=1=[N,E',E'].$$ The Three-Subgroups lemma yields $[E',E',N]=1$ and thus $[E',N]=1$, again since $E'$ is perfect. Hence, $N\leq Z(E')$, and $E'$ is quasisimple.
I don't understand how $N=E'$ is obtained using $E'$ is perfect.