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. 



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


