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.
Best Answer
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.