Can I say that $\mu_{d}=(\mathbb{F}^*_q)^{(q-1)/d}$?

Question

Asume that $\mu_d=\{x\in\mathbb{F}^*_q:x^d=1\}$, I want to proof that if $d|(q-1)$ then $\mu_d=(\mathbb{F}^*_q)^{(q-1)/d}$. I all ready do that proof if we have $\zeta\in\mu_d\Leftrightarrow\zeta^d=1\Leftrightarrow\left(\zeta^{\frac{d}{q-1}}\right)^{q-1}=1\Leftrightarrow\zeta^{\frac{d}{q-1}}\in\mathbb{F}^*_q\Leftrightarrow\zeta\in(\mathbb{F}^*_q)^{\frac{q-1}{d}}$. But the problem I have in the existance of $\zeta^{\frac{d}{q-1}}$.

Answer 1

Your claim is correct, $\mu_d=(\Bbb{F}_q^*)^{(q-1)/d}$.

To show that every element of $\mu_d$ is of the form $x^{(q-1)/d}$ for some $x\in\Bbb{F}_q^*$ it is probably easiest to use the fact that $\Bbb{F}_q^*$ is cyclic. If $g$ is a generator (aka a primitive element), then it is not difficult to deduce that if $\zeta\in\mu_d$, then $\zeta=g^j$ where $j$ must be divisible by $(q-1)/d$. For otherwise $\zeta^d\neq1$ contradicting $\zeta\in\mu_d$.

Anyway, the claim would be false in general if it weren't for the fact that $\Bbb{F}_q^*$ is cyclic. Therefore it is not surprising that this fact plays a role in any proof.