Q: Is $U_{7^{n}}$ cyclic? 
Let $(U,\cdot)$ be a group such that $U=\{z\in \mathbb{C}|\exists n \in \mathbb{N}, z^{7^{n}}=1\}$. Prove that any proper subgroup of $U$ is cyclic. Is $U$ cyclic?

I know that our group $U$ is $U_{7^{n}}$ and that any $z\in U$ can be written with deMoivre formula, but I am not sure how to handle it from here on.
I am new to this type of problems and I do not have many examples, could you provide a full proof, or at least in the form of an answer, such that it would serve as a model for similar problems I encounter? Thank you very much!!!

A: It's easy to show that $U$ can't be cyclic.  First, note that $U$ is infinite.  But $z \in U \Rightarrow z \in U_{7^n}$ for some $n \in \Bbb N$, so $z$ has finite order $7^n$ and the group it generates is necessarily finite.  Thus, $z$ can't generate all of $U$.
Now assume that $H \subseteq U$ is a proper subgroup and assume $z \in U \setminus H$.  Choose $n$ such that $z \in U_{7^n}$.  If $m \geq n$ and $w \in H$ with $\operatorname{order}(w) = 7^m$, then $\exists k ~ (w^k=z)$, which contradicts the assumption that $z \notin H$.  Thus, $H \subseteq U_{7^m}$ for some $m \lt n$, which means $H$ is a subgroup of a cyclic subgroup, and therefore must itself be cyclic.

A: 
$U$ is not cyclic.

$U$ is infinite and contains non-trivial elements of finite order; the infinite cyclic group does not.
