# centralizer of a chief factor

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 $G^*$.

• It's part of an article... . – user188323 Oct 29 '14 at 16:37
• Almost every theorem is a part of an article. – mesel Oct 29 '14 at 17:24
• Do you understand why $G^*$ is well-defined? – Derek Holt Oct 29 '14 at 21:06
• @DerekHolt: The OP assumes there exists such smallest normal subgroup. So by well-defined do you mean there is always one? – user795571 Oct 29 '14 at 21:19
• @user795571 I don't see that assumption stated anywhere. Strictly speaking, you should not write "Let $x$ be the object with property $P$" without first proving that there is a unique $x$ with property $P$. But in fact any group does have a smallest (wrt inclusion) normal subgroup with the property in question. – Derek Holt Oct 29 '14 at 22:38

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.