If n is a positive integer and $a^k$ is a primitive root mod n then a is a primitive root mod n.

Question

Show that If n is a positive integer and $a^k$ is a primitive root mod n then a is a primitive root mod n.

suppose that $a^m$ ≡1(mod n) then $(a^k)^m$ ≡1(mod n) so $a^{mk}$ ≡1(mod n)

hence a is a primitive root mod n.

is that correct answer for it?

Answer 1

The idea is mostly right, but it might be clearer if you expressed it more along the lines of: $a^m\equiv1\Longrightarrow a^{mk}\equiv1\Longrightarrow (a^k)^m\equiv1 \Longrightarrow$ the order of $a^k$ cannot exceed the order of $a$, so if $a$ is not primitive, then neither is $a^k$