Let $G$ be a finite solvable group. Call $J\subseteq G$ a nilpotent injector if it is a nilpotent subgroup that contains $\mathbf{F}(G)$, and that is maximal with this property (not properly contained in a larger nilpotent subgroup).
I would like to show the following:
Theorem: If $M$ is subnormal in $G$, then $J\cap M$ is a nilpotent injector of $M$.
It suffices to assume that $M$ is maximal normal in $G$, then apply induction to get to any subnormal subgroup. I have only managed to prove the theorem in the case that $M$ doesn't contain $\mathbf{F}(G)$. Here's how it goes:
PARTIAL PROOF:
Let $F=\mathbf{F}(G)$, and suppose $F\not\subseteq M$. Then by maximality of $M$, $G=MF$. Since both $F$ and $M$ are normal in $G$, $[M,F]\subseteq F\cap M\subseteq\mathbf{F}(M)$.
Let $J$ be a nilpotent injector of $G$. Let $K=J\cap M$. Dedekind's Lemma tells us that $J=FK$.
We want to show that $K$ is a nilpotent injector of $M$.
$\mathbf{F}(M)=\mathbf{F}(G)\cap M\subseteq J\cap M=K$, so $K$ is contained in some nilpotent injector $D$ of $M$. We aim to show that $D=K$.
Since $[M,F]\subseteq\mathbf{F}(M)\subseteq D$, we have that $[D,F]\subseteq D$, so $F$ normalizes $D$. Since $D$ and $F$ are both normal nilpotent subgroups of $\mathbf{N}_G(D)$, $DF$ is nilpotent.
Since $K\subseteq D$, we must have that $J=KF\subseteq DF$. By maximality of $J$, this forces $J=DF$, so $D\subseteq J\cap M=K$.
Hence, we have that $K=D$ as desired.
END OF PARTIAL PROOF
I feel that this is a promising approach, but I haven't been able to complete this proof. Does anyone know how to settle the remaining case of $F\subseteq M$?
Edit: Notice that I haven't used the assumption that $G$ is solvable!