Prove that $\text{Aut}(\mathbb{Z}_{p^{n}} )\cong \mathbb{Z}_{\phi(p^{n})}$ where p is a prime and $\phi(.)$ denotes the euler totient function.

Question

Can anyone tell me how to prove this?...I was asked to find the number of homomorphims from $Aut(\mathbb{Z}_{121} )$ to $Aut(\mathbb{Z}_{25} )$ . In the solution they used the result that I asked in the title and then calculated the number of homomorphims from $\mathbb{Z}_{110}$ to $\mathbb{Z}_{20}$ which is easy enough to do as it is just the gcd(20,110)=10 But how so I prove the above theorem?.

Answer 1

An automorphism of $\Bbb Z_n$ is determined by sending a generator, say $1$, to a generator, which is to say to $k$ such that $(k,n)=1$. We pretty easily get the nice result that $\rm{Aut}(\Bbb Z_n)\cong\Bbb Z_n^\times$.

(Note that, in particular, when $n=1,2,4,p^\alpha,2p^\alpha$ for $p\gt2$ prime, we get a cyclic automorphism group. This part is nontrivial. See for instance I.M. Vinogradov's Elements of Number Theory.)