Fitting's Lemma is:
Let $M$ and $N$ be normal nilpotent subgroups of a group $G$. If $c$ and $d$ are the nilpotent classes of $M$ and $N$, the $L = MN$ is nilpotent of class at most $c+d$.
There is an exercise on Derek J.S. Robinson's A Course in the Theory of Grouops:
If $M,N$ are nontrivial normal nilpotent subgroups of a group, prove that $\zeta(MN) \neq 1$. Hence give an alternative proof of Fitting's Theorem for finite groups.
The proof of the first part is easy. As $M$ is nontrivial and nilpotent, the center of $M$ is nontrivial. Suppose that $a \in \zeta(M)$, $a \neq 1$, if $a$ commutes with every element of $N$, then $a \in \zeta(MN)$ and $\zeta(MN) \neq 1$. If there is some $b \in N$, such that $aba^{-1}b^{-1} \neq 1$, then $aba^{-1}b^{-1} \in \zeta(MN)$, and $\zeta(MN) \neq 1$.
On the book, the proof for the general case of Fittng's theorem shows by induction on $i$ that $\gamma_i L$ is the product of all $[X_i, \cdots, X_i]$ with $X_j = M$ or $N$. The fact that $\zeta(MN) \neq 1$ is not applied obviously in this proof. What can I do with this fact when $M$ and $N$ are finite?
Thank you very much.