Why are all finite subgroups of the group $S^1$ of complex roots of unity cyclic?

Question

This is the last step in a proof I'm trying to do that the image of any one-dimensional representation of a finite group $G$ is cyclic. Because $g^n=id_G$ for some positive integer $n$ $\forall{g}\in{G}$, we know $\rho(g)^n=1$ as well. So $\textrm{im}(\rho)\subset{S^1}$, and is finite. Apparently this also means it must be cyclic, but I'm not quite sure why . . .

Answer 1

It turns out that any finite subgroup of the multiplicative group of a field must be cyclic. In particular, since $S^1 \subset \mathbb{C}^*$ this implies the result you want.

To see this fact just note that if $G$ is abelian but not cyclic then $\exists n < |G|$ such that $g^n = e$ for all $g \in G$. But then if $G \subset k^*$ for a field $k$, the polynomial $x^n -1$ must vanish on all of $G$, but a degree $n$ polynomial can only have at most $n$ $(< |G|)$ roots, a contradiction.

Answer 2

Let $G$ be a (non-trivial) finite subgroup of $S^1$. Identifying $S^1$ with $\mathbb{R}/\mathbb{Z}$ (and choosing $[0,1)$ as the set of representatives), let $x$ be the minimal non-zero element of $G$.

If $G\neq \langle x\rangle$ then there is some $y\in G$ with $y\not\in\langle x\rangle$. Hence there is some natural number $n$ such that $nx<y<(n+1)x$, or $0<y-nx<x$. This contradicts the minimality of $x$, hence $G=\langle x\rangle$.

Answer 3

Since $\rho(g)^n=1$, you know that $\rho(g)$ is an $n$-th root of unity. In other words, $\rho(g)=e^{2\pi i k/n}$ for some integer $k$. Therefore, $$ \operatorname{im}(\rho)\subseteq\{e^{2\pi i k/n}:k\in\{0,\cdots,n-1\}\}. $$ Since the RHS forms a cyclic group of order $n$, and the LHS is a subgroup, the image is a subgroup of a cyclic group, so it is also cyclic.