Let G be a finite group , g$\in$G and $\chi$ be a character of G. If |$\chi(g)$|=1 then show that $\chi(g)$ is a root of unity. $\\$

Hint: Let |G|=n and consider E=$\mathbb{Q}(\zeta_{n})$ where $\zeta_{n}$ is a primitive nth root of unity. Now let $\alpha$ be an algebraic integer in E with $|\alpha|$=1 and consider the minimal polynomial $f_{\alpha}$ and show that there are only finitely many such polynomials. But the question how do I show that and even if I show that how does it solve the problem.

Let G be a finite group , g$\in$G and $\chi$ be a character of G. If |$\chi(g)$|=1 then show that $\chi(g)$ is a root of unity. $\\$

Hint: Let |G|=n and consider E=$\mathbb{Q}(\zeta_{n})$ where $\zeta_{n}$ is a primitive nth root of unity. Now let $\alpha$ be an algebraic integer in E with $|\alpha|$=1 and consider the minimal polynomial $f_{\alpha}$ and show that there are only finitely many such polynomials. But the question how do I show that and even if I show that how does it solve the problem.

note added by Y. Choi, in case of confusion: the question comes from Chapter 3 of Isaacs's book, and as is usual in the representation theory of finite groups, "character" means the trace of a finite-dimensional representation, not a homomorphism from $G$ into the multiplicative group of some field.

Let G be a finite group , g$\in$G and $\chi$ be a character of G. If |$\chi(g)$|=1 then show that $\chi(g)$ is a root of unity.

Let G be a finite group , g$\in$G and $\chi$ be a character of G. If |$\chi(g)$|=1 then show that $\chi(g)$ is a root of unity. $\\$

Hint: Let |G|=n and consider E=$\mathbb{Q}(\zeta_{n})$ where $\zeta_{n}$ is a primitive nth root of unity. Now let $\alpha$ be an algebraic integer in E with $|\alpha|$=1 and consider the minimal polynomial $f_{\alpha}$ and show that there are only finitely many such polynomials. But the question how do I show that and even if I show that how does it solve the problem.