I find it hard to understand some steps of the Fitting Theorem's proof, by which I mean "The product of two normal and nilpotent subgroups $H$ and $K$ of a group $G$ is normal and nilpotent." I'll write what I read in my book, with some of my comment in round brackets and if someone could help me I would be grateful: the product $HK$ is a group, since $H$ and $K$ are normal; $H$ and $K$ are normal in $HK$ and nilpotent. If $HK<G$, then $HK$ is nilpotent by induction. Then suppose that $G=HK$. (I don't understand what is the base and what is the hypothesis of induction, maybe on the cardinality of G??). If $K=1$ there is nothing to prove. Then let $K\neq1$ . The nilpotence of $K$ shows that $Z(K)\neq1$. Now $Z(K)$ is characteristic in $K$ and so $Z(K)$ is normal in $G$. Then $N=[H,Z(K)]\trianglelefteq G$. If $N=1$, $Z(K)$ centralizes $H$ and $K$, and so $Z(K)$ centralizes $G$, $Z(K)\subseteq Z(G)$ and so $Z(G)\neq 1$. If $N\neq1$, $N$ is a normal subgroup of $H$ (why?) and so by nilpotence of $H$ we have that : $L=N\cap Z(H)\neq1$. Then $L$ centralizes $H$ and $K$, and so $L$ centralizes $G$. (I understand the reason why $L$ centralizes $H$, but why $K$ too?). And so $Z(G)\neq1$. It follows that $HZ(G)/Z(G)$ and $KZ(G)/Z(G)$ are homomorphic images of nilpotent groups. (What does it mean?) and so they are nilpotent. Since $Z(G)\neq1$ , by induction $G/Z(G)$ is nilpotent and so $G$ is nilpotent.
Thanks!!