Prove that if p is a prime then there exist $\ ϕ(ϕ(p^2 )) = (p − 1)ϕ(p − 1)$ primitive roots modulo $p^2.$
I know how to prove the theorem Let p be prime and let d ∈ N be a divisor of p − 1. Then there are exactly ϕ(d) elements a mod p such that ordp(a) = d. In particular, there are ϕ(p − 1) primitive roots modulo p.
Not sure how to go about proving the first statement, any ideas what theorems I would use?
Let $g$ be a primite root modulo $p^n$, which exists. Then $\{1,\ldots,\varphi(p^n)\}$ is equal to $\{g^1,\ldots,g^{\varphi(p^n)}\}$ modulo $p^n$. If $g^x$ is a primitive root, for some $x \in \{1,\ldots,\varphi(p^n)\}$, then it is equivalent to ask that the map $\{1,\ldots,\varphi(p^n)\}\to \{1,\ldots,\varphi(p^n)\}$ defined by $y\mapsto xy$ is a bijection modulo $\varphi(p^n)$. Such $x$ can be chosen in $\varphi(\varphi(p^n))$ ways.
