Range of Characters

Theorem
Let $G$ be a finite abelian group of order $m$.

Let $\chi: G \to \C^\times$ be a character on $G$.

Then for any $g \in G$, $\chi \left({g}\right)$ is an $m$th root of unity.

If $e$ is the identity of $G$ then $\chi \left({g}\right) = 1$.

Proof
The claim that $\chi \left({e}\right) = 1$ is shown by Group Homomorphism Preserves Identity.

Let $g \in G$ be arbitrary.

Let $k$ be the order of $g$.

By Order of Element Divides Order of Finite Group:
 * $\exists l \in \Z: m = k l$

Therefore:
 * $g^m = \left({g^k}\right)^l = e^l = e$

By the homomorphism property:


 * $1 = \chi \left({e}\right) = \chi \left({g^m}\right) = \chi \left({g}\right)^m$

Hence the result.