Help with primitive root modulo $p^r$

Question

Let $p\ge3$ be a prime number, $r$ be a natural number and $x$ be a primitive root modulo $p^r$. Show that $x$ is a primitive root modulo $p$.

Answer 1

Suppose that $b$ is not divisible by $p$. Then $b$ is relative prime to $p^r$.

Since $x$ is a primitive root of $p^r$, it follows that $b\equiv x^k\pmod{p^r}$ for some positive integer $k$. Thus $b\equiv x^k\pmod{p}$.

Since every $b$ not divisible by $p$ is congruent to some power of $x$ modulo $p$, it follows that $x$ is a primitive root of $p$.

Remark: Essentially the same argument shows that if $x$ is a primitive root of $p^r$, then $x$ is a primitive root of $p^i$ for all positive $i$ less than $r$.