# Showing that the image of a one dimensional representation is a subgroup of a cyclic group

Consider the answer given by Michael in what is essentially the title of my question: https://math.stackexchange.com/a/2089754/1027258. I understand everything except why $$\mathrm{Im}(\rho) \subset \{\exp((2\pi i k)/n)\mid k \in \{0,1,\dots,n-1\}\}$$. Namely, we know that 1.) for any $$g \in G$$ there exists $$n \in \mathbb{N}_0$$ s.t. $$g^n = 1$$, the neutral element. Then, if we fix our $$g$$ we know that $$\rho$$ must map $$g$$ to one of the $$n$$th roots of unity, as $$\rho$$ is a homomorphism $$\rho:G\to \mathrm{Aut}(\mathbb{C})\cong \mathbb{C}^\times$$. But then, why must the entire image of $$\rho$$ be contained in the set of the $$n$$ roots of unity? What if we take some other $$h \in G$$ with $$n < n' \in \mathbb{N}_0: h^{n'} = 1$$? Then surely again $$h$$ is $$n'$$th root of unity, but why must $$\rho(h) \in \{\exp((2\pi i k)/n)\mid k \in \{0,1,\dots,n-1\}\}$$?

"Namely, we know that 1.) for any $$g∈G$$ there exists $$n∈\mathbb N_0$$ s.t. $$g^n = 1$$, the neutral element."
You are confused because your hypothesis here is not as strong as it should be. Namely, there exists an $$n\in \mathbb N_0$$ such that $$g^n = 1$$ for every $$g \in G$$. That is, $$n$$ works simultaneously for every $$g \in G$$. This is possible because $$G$$ is finite.
This makes it so that $$\rho$$ maps every element of $$G$$ into the group of complex $$n$$th roots of unity.
Because in the question in the link the group $$G$$ is finite. Let $$\vert G \vert = n$$. Then for all $$g \in G$$ you have $$g^n = 1$$.