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I'm stuck with the following problem:

Let $G$ be a finite group and $p$ be the smallest prime dividing $|G|$. Suppose the Sylow p-subgroup $H$ of $G$ is normal and cyclic. Show that $H$ lies in the center of $G$. Hint: Consider the order of automorphism group of $\mathbb{Z}/p^n \mathbb{Z}.$

My thought until now is assume $|H|=p^n.$ Let $G$ act on $H$ by conjugation and consider the class equation. Since $H$ is assumed to be normal and cyclic, each action on $H$ can be seen as an automorphism on $\mathbb{Z}/p^n \mathbb{Z}$. The order of the automorphism group of $\mathbb{Z}/p^n \mathbb{Z}$ should be $\phi(p^n)$, where $\phi$ is the Euler phi-function. But I don't know how to use this to show that all elements of $H$ are fixed points, and hence lies in the center of $G$. Are there any mistakes above? Could I ask for some help here? Thanks a lot.

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  • $\begingroup$ The automorphism group of $\mathbb{Z}/p^n\mathbb{Z}$ has order $\phi(p^n)=p^{n-1}(p-1)$ $\endgroup$ Jan 2, 2016 at 0:03
  • $\begingroup$ It's been corrected. Thanks! $\endgroup$
    – qwe0912
    Jan 2, 2016 at 0:27

1 Answer 1

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Every element of $G$ induces an automorphism on $H$ by conjugation since $H$ is normal, and since $H$ is abelian $H$ itself acts trivially. This means there is an induced homomorphism $G/H\to \mathrm{Aut}(H)$. All prime divisors of $|G/H|$ are larger than $p$, so by divisibility considerations this homomorphism must be trivial. Thus every element induces the trivial automorphism, meaning $H$ is in the center.

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  • $\begingroup$ Just wondering why by divisibility considerations this homomorphism must be trivial? $\endgroup$ Jun 12, 2017 at 23:42
  • $\begingroup$ @Sid None of the orders of the elements of $G/H$ except the identity divide the order of elements in the codomain. $\endgroup$ Jun 12, 2017 at 23:57

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