Let $n \geq 1$. For any field $E$, define
$$
\mu_n(E):=\{\rho \in E \mid \rho^n=1\} .
$$
Note that the set $\mu_n(E)$ inherits a group structure from $E^*$.
The group $\mu_n(E)$ is a finite cyclic group.
Proof: We’ll use the fact that a finite group $G$ is cyclic iff for any divisor $d\mid\#G$, there is at most 1 subgroup in $G$ of order $d$.
For any $d\mid\#\mu_n$ if there is a subgroup in $\mu_n(E)$ of order $d$, generated by $\zeta$, then $1,\zeta,\dots,\zeta^{d-1}$ are roots of $x^d-1$ in $\mu_n(E)$, but $x^d-1$ has at most $d$ roots in $E$, so these are all the roots, so there is only 1 subgroup in $\mu_n(E)$ of order $d$.
$\Box$