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.
Best Answer
Induct on the number of elements of $G$. Suppose we have proved Fitting's lemma for groups with less than $|G|$ elements. Since $\zeta(MN)$ is nontrivial, the group $MN/\zeta(MN)$ has less than $|G|$ elements and the result holds for it.
We can write $$MN/\zeta(MN)=(M\zeta(MN)/\zeta(MN))(N\zeta(MN)/\zeta(MN))$$ where each of the factors on the right is nilpotent of class at most $c$ or $d$, since it is the homomorphic image of a nilpotent group. The factors are also normal subgroups. Apply the induction hypothesis to conclude that $MN/\zeta(MN)$ is nilpotent of class at most $c+d$. Therefore, $MN$ is nilpotent of class at most $c+d+1$.