# Fitting Theorem Nilpotent groups

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!!

I agree that this is confusing. I'll number your parenthetical remarks.

1. You're performing strong induction on

$P(n) :$ If $G$ is a group of order $n$ and $H, K \lhd G$ are nilpotent, then $HK \lhd G$ and $HK$ is also nilpotent.

So $P(1)$ is trivially verified, which is likely why the author didn't mention it.

2. I think you know how to show that $N \lhd G$. That $N \subset H$ follows because $N = [H, Z(K)]$ is generated by elements of the form $h(zh^{-1}z^{-1})$, where $h \in H$ and $z \in Z(K)$, and because $H$ is normal in $G$.

3. As in (2) we have $N \subset Z(K)$ by normality, and $L \subset N$.

4. This means that there is some surjective group homomorphism $P \to HZ(G)/Z(G)$ with $P$ nilpotent. The point is that $HZ(G)/Z(G)$ is (isomorphic to, if you like) a quotient of $H$, namely $H/(H \cap Z(G))$. The map comes from restricting the quotient homomorphism $G \to G/Z(G)$ to a map $H \to G/Z(G)$, and the image of this is $HZ(G)/Z(G)$; similarly for $K$.

Does everything else make sense? I always have to remind myself that it's very important to quotient out by a nilpotent subgroup contained in the centre, and not just some nilpotent normal subgroup.

• So let's check if I understand: 1. By induction I suppose that every group with order $<n$ is nilpotent, so if $HK<G$ then $HK$ is nilpotent .The base of induction is $n=1$ and in this case $G=1$ and so is nilpotent. Then I have to prove that $G=HK$ is nilpotent. 2. and 3. are claer thanks! But for 4. I still have doubts: in this case which is $P$?I don't understand the utility of the last steps. – stacy Aug 2 '11 at 9:28
• @stacy Let's write $\overline{H}$ for the image of $H$ in $G/Z(G)$, and so on. Then $\overline{G} = \overline{H}\overline{K}$, and $\overline{H}$ is nilpotent (because it's a homomorphic image of $H$); similarly for $\overline{K}$. – Dylan Moreland Aug 2 '11 at 14:18

I think this statement remains true for infinite groups, though I don't know if this was proved in the book you refer to. But to prove this version, you have to be more careful with the induction. You can use induction on the nipotency class of $H$ plus the nilpotency class of $K$ (considering the trivial group to have nilpotency class 0 and an Abelian group to have nilpotency class $1$ (and, in general, class($X$) = 1 + class($X/Z(X))$ by definition, where we consider a non-trivial group with trivial center to have infinite class). The base cases where this integer is $0$ or $1$ are easy. Also, it is clear that $HK \lhd G$ when $H \lhd G$ and $K \lhd G$. Furthermore, it is quite easy to see that $[H,K] = \langle h^{-1}k^{-1}hk : h \in H, k \in K\rangle \lhd G$, since the generating commutators for $[H,K]$ are permuted under conjugation by $G$ (for $g^{-1}(h^{-1}k^{-1}hk)g = (g^{-1}hg)^{-1}(g^{-1}kg)^{-1}(g^{-1}hg)(g^{-1}kg)$ and $H,K \lhd G$). Also, note that $[H,K] \subseteq H \cap K$, for we have $h^{-1}k^{-1}hk = h^{-1}(k^{-1}hk) \in H$ as $H \lhd G$ and $h^{-1}k^{-1}hk = (h^{-1}k^{-1}h)k \in K$ as $K \lhd G$. We may suppose that $H$ and $K$ are both nontrivial, or there is nothing to do. Now $Z(H) {\rm char} H \lhd G$ and $Z(K) {\rm char }K \lhd G.$ By induction, $HK/Z(H)$ and $HK/Z(K)$ are both nilpotent (the sum of the nilpotence classes has dropped by at least one in each case). It follows that $HK/(Z(H) \cap Z(K))$ is nilpotent (the lower central series for $HK$ terminates in a group contained in $Z(H)$ because $HK/Z(H)$ is nilpotent, and terminates in a subgroup contained in $Z(K)$ since $HK/Z(K)$ is nilpotent). But $Z(H) \cap Z(K) \leq Z(HK)$, so that $HK$ is nilpotent.