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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.

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First, does $\zeta(G)$ denote the center of $G$? If so, I don't understand your proof of the first part. You seem to be claiming that if $c=aba^{-1}b^{-1} \ne 1$, then $c$ is in the center of $MN$. Why is that? –  Ted Oct 8 '11 at 19:32
    
@Ted: Yes, that is the center of $G$. As $M$ and $N$ are normal subgroups of $G$, for any $a' \in M$, $aba^{-1}b^{-1}a'=aba^{-1}b^{-1}a'=aba^{-1}(b^{-1}a'b)b^{-1}=ab(b^{-1}a'b)a^{-1}‌​b^{-1}=aa'ba^{-1}b^{-1}=a'aba^{-1}b^{-1}$, similarly, for any $b' \in N$, $aba^{-1}b^{-1}b'=ab(a^{-1}b'a)a^{-1}b^{-1}=a(a^{-1}b'a)ba^{-1}b^{-1}=b'aba^{-1}‌​b^{-1}$. If I have made any mistake, please point out for me. Thank you very much. –  ShinyaSakai Oct 9 '11 at 14:35
    
The first part (starting with $a' \in M$) is fine but I don't understand the calculation for the second part (starting with $b' \in N$). In the first step, you seem to have assumed that $b$ commutes with $b'$, which you don't know. –  Ted Oct 9 '11 at 16:26
    
@Ted: Thank you very much for pointing out this. I was wrong because I had left behind the case in which $\zeta(M)$ and $\zeta (N)$ didn't intersect, and were both contained in $C_G(\zeta(M))$ and $C_G(\zeta(N))$. I have thought for quite a while and begin to doubt whether this elementary method really works... Please tell me if you know the proof. Thanks again. –  ShinyaSakai Oct 10 '11 at 11:09

1 Answer 1

up vote 1 down vote accepted

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 $c+d+1$.

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I realize that the final nilpotency class is off by one. After a quick survey, I'm not sure how to change the argument to fix this. I would welcome any input on this matter. –  Miha Habič Oct 8 '11 at 14:54
    
Thank you very much for this near-perfection proof~ I will also think about descending the possible nilpotent class by $1$. Welcome any input on this matter~ –  ShinyaSakai Oct 9 '11 at 15:09

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