With p minimal, the normalizer and centralizer of a cyclic Sylow p-subgroup are equal

Let p be the smallest prime dividing |G|, and suppose that some P \in \mathsf{Syl}_p(G) is cyclic. Prove that N_G(P) = C_G(P).

Write |G| = p^km where p does not divide m. Note that P \cong \mathbb{Z}/(p^k). Since P is abelian, we have P \leq C_G(P) \leq N_G(P). Now N_G(P)/C_G(P) \leq \mathsf{Aut}(P) \cong (\mathbb{Z}/(p^k))^\times. Note that \mathsf{Aut}(P) is a group of order \varphi(p^k) = p^{k-1}(p-1), where \varphi denotes the Euler totient function. Moreover, |N_G(P)/C_G(P)| divides m. Since every prime divisor of m is greater than p, N_G(P)/C_G(P) = 1. Thus C_G(P) = N_G(P).

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  • gug  On October 7, 2011 at 4:32 am

    Thank you

    • nbloomf  On October 7, 2011 at 10:59 am

      You’re welcome! Thanks for reading.

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