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 $\rho(h) \in \{\exp((2\pi i k)/n)\mid k \in \{0,1,\dots,n-1\}\}$?
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
"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.
$\begingroup$
$\endgroup$
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 $g^n = 1$.
