Are all roots of unity (solutions to $a^k \equiv 1$) for a prime modulo $p$, a multiple of $p-1$??

Question

If $a$ is coprime to $p$, and $a \not \equiv 1,-1 \mod p $ then are there any solutions to $a^k \equiv 1 \mod p $ such that $0< k < p-1$?

For any counterexample, it is obvious $GCD(p-1, k) \not = 1$, However, I could not find any counterexamples.

Answer 1

The group $(\mathbf{Z}/p\mathbf{Z})^\star$ is cyclic, hence it admits a generator $g$. All numbers coprime with $p$ can be represented as $g^m$, with $1\le m\le p-1$. Therefore we ask asking whether $$g^{mk}=1 \Longleftrightarrow p-1 \mid mk \Longleftrightarrow \frac{p-1}{\text{gcd}(p-1,m)} \mid k.$$ :)

Answer 2

$(\mathbb Z/p\mathbb Z)^\times\cong \mathbb Z/(p-1)\mathbb Z$. So in general, if $(k, p-1) \ne 1$ there will be multiple examples.

For example, modulo $7$, we have $2^3 \equiv 1 \pmod 7$.