centralizer of a chief factor

Question

Let $G$ be a finite solvable group and $p$ be a prime. Let $G^*$ be the smallest normal subgroup of $G$ for which the corresponding factor is abelian of exponent dividing $p-1$. Show that every chief factor of order $p$ is central in $G^*$.

Answer 1

If $H/K$ is a chief factor of order $p$, then $G/C_G(H/K)$ is isomorphic to a subgroup of ${\rm Aut}(C_p)$, which is abelian of order $p-1$, so this quotient of $G$ is abelian of exponent dividing $p-1$. Hence $G^* \le C_G(H/K)$, QED.