Fitting subgroup is self-centralizing in soluble groups
= Fitting subgroup is self-centralizing in soluble groups
{c}
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 $K'\le D\le Z(K)$, so $K$ is a <nilpotent normal subgroup>, contradicting $K\not\le F$.