 Solution

Write $F=F(G)$, $C=C_G(F)$ and $D=C\cap F=Z(F)$. Both $C$ and $D$ are normal in $G$. Suppose $C>D$. Choose a minimal normal subgroup $K/D$ of $G/D$ contained in $C/D$. In a finite soluble group, such a minimal normal subgroup is elementary abelian: the last nontrivial term of its derived series is an abelian characteristic subgroup, so minimality makes the whole subgroup abelian; its primary components and subgroup of elements killed by $p$ then force one prime and exponent $p$.

It follows that $K^{\prime }\le D$. Moreover $K\le C$ centralizes $D\le F$, so $K^{\prime }\le Z(K)$. Therefore $K$ is a nilpotent group of class at most two. Since $K$ is normal in $G$, the definition of the Fitting subgroup gives $K\le F$. But then $K\le C\cap F=D$, a contradiction. This proves the Fitting subgroup is self-centralizing in soluble groups: