Conjugacy of nilpotent injectors in soluble groups

Question

Hello,

I was wondering if anyone is aware of an elementary proof of the claim in the title, assuming the existence of nilpotent injectors in soluble groups. By elementary I mean a proof that does not involve any recourse to Fitting classes etc.

Thanks in advance.

Answer 1

I take a nilpotent injector of a finite solvable group $G$ to be a maximal nilpotent subgroup $M$ of $G$ such that $M \cap N$ contains a maximal nilpotent subgroup of $N$ of whenever $N$ is subnormal in $G$. Assuming existence of $M$ , I think uniqueness up to conjugacy follows inductively. Notice that $M \cap H$ is then a nilpotent injector of $H$ whenever $H$ is normal in $G.$

We may suppose that $Z(G) = 1$. Now let $p$ be a prime divisor of $|F(G)|$. Since $F(G) \leq M$, we have $O_{p'}(M) \leq C_{G}(O_{p}(G)).$ Thus $O_{p'}(M) = O_{p'}(L)$, where $L = M \cap C_{G}(O_{p}(G))$ is a nilpotent injector of $C_{G}(O_{p}(G))$. For notice that $O_{p'}(L) \lhd M$ so that $O_{p'}(L) \leq O_{p'}(M)$, while $O_{p'}(M) \leq M \cap C_{G}(O_{p}(G)) =L$ and $O_{p'}(M) \leq O_{p'}(L)$.

Now $L$ is unique up to conjugacy within $C_{G}(O_{p}(G))$, so certainly unique up to conjugacy in $G$. Hence $O_{p'}(M)= O_{p'}(L)$ is unique up to conjugacy within $G$. By maximality as a nilpotent subgroup, $ M = P \times O_{p'}(M)$, where $P$ is a Sylow $p$-subgroup of $C_{G}(O_{p'}(M)).$ Hence we see that $M$ is unique up to conjugacy. in $G$.

Answer 2

Another good reference here is "Injectors and Normal Subgroups of Finite Groups" by Avinoam Mann. Israel Journal of Mathematics, Vol 9.