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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 . . .

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    Idea: you know what the cyclic subgroups of $\mathbb{S} ^{1} $ are, now appeal to the fundamental theorem of f. g. abelian groups. – 2017-01-09
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    Hint: Since $\rho(g)^n=1$, you know that $\rho(g)=e^{2\pi i k/n}$ for some integer $k$. The elements of this form form a cyclic group and you're looking at a subgroup of that group. – 2017-01-09

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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.

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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

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    +1; I like that this answers the abstract question asked (and answers from first principles) and not just the question in this context. – 2017-01-09
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    Thanks! Just wondering though, what is the map you use to identify $S^1$ with $\mathbb{R}/\mathbb{Z}$? Is it $e^{i2\pi{r}/n}\mapsto \frac{r}{n} mod \mathbb{Z}$? – 2017-01-09
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    The map $\mathbb{R}/\mathbb{Z}\to S^1$, $\theta+\mathbb{Z}\mapsto e^{2\pi i\theta}$ is an isomorphism. – 2017-01-09
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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.