Ok, I have this:
$\displaystyle U_n = \sqrt[n]{1}$ where n $\displaystyle \in [1, \infty]$.
Show that $\displaystyle U_n$ is cyclic.
It is clearly cyclic as all it's elements are equal to 1, but how do I prove it.
Actually, given the notation used by the OP, $\displaystyle U_n = 1$ for all $\displaystyle n$. Even if the OP wrote $\displaystyle U_n = \{x \in \mathbb{C} \mid x = \sqrt[n]{1}\}$, then $\displaystyle U_n = \{1\}$ since $\displaystyle \sqrt[n]{1} = 1$ is distinct in $\displaystyle \mathbb{C}$. Now, if $\displaystyle U_n = \{x\in \mathbb{C} \mid x^n = 1\}$, then we still would only have $\displaystyle U_n$ being the $\displaystyle n$-th roots of unity in the case that $\displaystyle n$ is rational. But the OP is allowing $\displaystyle n \in [1,\infty]$, which is an interval of the extended reals. We must assume that $\displaystyle \sqrt[\infty]{x}$ is defined and $\displaystyle \sqrt[\infty]{1} = 1$.