= Solution
It suffices to centralize each <minimal normal subgroup> $M$. The <subgroup> $M\cap F(G)$ is normal in $G$, so it is either $1$ or $M$. In the first case
$$
[M,F(G)]\le M\cap F(G)=1.
$$
In the second case $M$ is a <nilpotent group>. Its nontrivial <group center> is a <characteristic subgroup> of $M$, hence normal in $G$, so minimality makes $M$ abelian. The primary components, which are <characteristic subgroups>, and the <subgroup> killed by a prime show that $M$ is <elementary abelian> of exponent $p$. Thus $M\le O_p(G)$.
The <finite p-group> $O_p(G)$ acts by <conjugation> on $M$. Its orbits have prime-power sizes; hence the number of fixed elements is congruent to $|M|=0$ modulo $p$. The identity is fixed, so there is a nonidentity fixed element. Therefore $M\cap Z(O_p(G))\ne1$. This intersection is normal in $G$, and minimality makes it $M$. The other <p-cores> commute with $O_p(G)$, and hence with $M$. Thus every factor of $F(G)$ centralizes $M$. Since the <group socle> is generated by these <minimal normal subgroups>,
$$
\boxed{F(G)\le C_G(\operatorname{Soc}(G)).}
$$
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