Let $p>2$ be a prime, $(a,p)=1,$ and suppose that $a^k$ is a primitive root mod $p$ for some $k>0$. Prove that $a$ itself is a primitive root.

Question

Let $p>2$ be a prime, $(a,p)=1,$ and suppose that $a^k$ is a primitive root mod $p$ for some $k>0$. Prove that $a$ itself is a primitive root.

I looked at the definition of primitive root but I wasn't able to figure out a proof for this. Would really appreciate it if someone could construct one as I am very new to number theory. Many thanks.

Answer 1

Hint: Suppose $a$ were not a primitive root, so that there exists $n < p - 1$ such that $a^n \equiv 1 \pmod{p}$. Then, try to show that $(a^k)^n \equiv 1 \pmod{p}$, which would be a contradiction because $a^k$ is a primitive root modulo $p$.

Answer 2

If $(a^{k l})_{l\ge 0}$ gives all numbers relatively prime to $p$, then $(a^{m})_{m \ge 0}$ also will.

Answer 3

Just note that $p-1 = ord(a^k) \le ord(a) \le p-1$ implies $ord(a)=p-1$.