Fitting group of a soluble group

Question

Let be $G$ a group soluble with center trivial and let $\pi$ the canonical homomorphism from $G$ of $G/F(G)$. Take $$ Z=\pi^{-1}\left(Z\left(F\left(\dfrac{G}{F(G)}\right)\right)\right)\ \ \mbox{and}\ \ J=\pi^{-1}\left(F\left(\dfrac{G}{F(G)}\right)\right). $$ It is true that if $x\notin Z$ then, as $J/F=F(G/F)$ and $G/F$ is soluble, there must exist a prime $p$ such that the $p$-Sylow of $J/F$ is not centralized by $x$ ?

Answer 1

I expect that you are asking whether the image $\bar{x}$ of $x$ in in $G/F$ centralizes the Sylow $p$-subgroups of $J/F$.

If $\bar{x}$ centralizes all Sylow subgroups of $J/F$ then it centralizes $J/F$, but the Fitting subgroup of a solvable group contains its own centralizer, so that would imply that $\bar{x} \in J/F$ and hence $\bar{x} \in Z(J/F)$, so $x \in Z$.