I’m trying to show that if $G$ is a solvable finite primitive group and $M$ is a core-free maximal subgroup of $G$, then $O_p(M)=1$ where $p$ divides $|Soc(G)|$ ($Soc(G)$ is the socle of $G$).
I know that a minimal normal subgroup $N$ of a solvable group will be elementary abelian, and then, if $G$ is a primitive group, it can be shown that $C_G(N)=N$. I also know that for a solvable group $G$, if $N$ is a minimal normal group of $G$ such that $C_G(N)=N$, then there is some prime $p$ such that $F(G)=N=O_p(G)$, where $F(G)$ is the Fitting subgroup of $G$, $N$ is complemented in $G$ by a maximal core-free subgroup $M$ and there exists a unique conjugacy class of such maximal subgroups in $G$.
I guess I could now use the definition of the Fitting subgroup as the direct product of $p$-cores for all the primes $p$ that divide the order of $G$ and try to conclude from there... I’ve also read here that for this kind of group, the minimal normal subgroup $N$ will be unique and the socle will then coincide with $N$ (and then $Soc(G)=p^n$). I’m not sure how to show this, though.
Am I on the right track? Is there an easier way to do this?