[Math] Fitting Theorem Nilpotent groups

Question

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!!

Answer 1

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.