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. 




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**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.



Criteria for a character being a root of unity
This problem has little to do with representation theory, other than the fact that
the eigenvalues of a matrix of finite order are roots of unity. If $\alpha$ lies in a cyclotomic extension $K$ of $\mathbb{Q}$ and $\lvert\alpha\rvert=1$, then all the conjugates $\alpha^g$ (where $g$ lies in the Galois group of $K$) of $\alpha$ satisfy $\lvert\alpha^g\rvert=1$ since the Galois group of a cyclotomic extension is abelian, so $g$ commutes with complex conjugation. Now use [Kronecker's theorem](https://books.google.ca/books?id=qIJPxdwSqlcC&pg=PA175), where the proof follows Isaac's hint. 


