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 $G^*$.
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$\begingroup$ It's part of an article... . $\endgroup$– user188323Oct 29, 2014 at 16:37
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1$\begingroup$ Almost every theorem is a part of an article. $\endgroup$– meselOct 29, 2014 at 17:24
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$\begingroup$ Do you understand why $G^*$ is well-defined? $\endgroup$– Derek HoltOct 29, 2014 at 21:06
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$\begingroup$ @DerekHolt: The OP assumes there exists such smallest normal subgroup. So by well-defined do you mean there is always one? $\endgroup$– Minimus HeximusOct 29, 2014 at 21:19
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$\begingroup$ @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. $\endgroup$– Derek HoltOct 29, 2014 at 22:38
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1 Answer
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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.
answered Oct 29, 2014 at 21:05