Fitting subgroup is self-centralizing in soluble groups

In a finite soluble group, $C_G(F(G))\le F(G)$. Put $C=C_G(F)$ and $D=C\cap F=Z(F)$. If $C>D$, choose an elementary abelian minimal normal subgroup of $G/D$ inside $C/D$, and let $K$ be its preimage. Then $K^{\prime }\le D\le Z(K)$, so $K$ is a nilpotent normal subgroup, contradicting $K\nleqq F$.