How to show $Aut(\mathbb{Z}_{p^{n}})\cong\mathbb{Z}_{p^{n}-p^{n-1}}$ ,p is an odd prime,$n\geq 2.$

Question

How to show $Aut(\mathbb{Z}_{p^{n}})\cong\mathbb{Z}_{p^{n}-p^{n-1}}$ ,p is an odd prime,$n\geq 2.$

I have know that $Aut(\mathbb{Z}_{p^{n}})\cong U(n)$.Here $U(n)=\{k|1\leq k< n,(k,n)=1\}.$But I want to know more about its structure.

Answer 1

Let $G=Z_{p^n}$. How many numbers less than $p^n$ are coprime to $p^n$?

Here coprime numbers are those without $p$ in their factorisation. How many multiples of $p$ are under $p^n$? Alright $p^{n-1}$ multiples. So we end up having $\varphi(p^n) = p^n - p^{n-1}$, since you subtract the $p^{n-1}$ multiples under $p^n$ to obtain the number coprime numbers.

Moreover $G$ is cyclic meaning that there exists a generator where its order equals to $\vert G \vert$ since $\varphi(p^n)=\lambda(p^n)$ as $G$ is isomorphic to a single cyclic group.