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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.

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    $\begingroup$ This looks like an assigned exercise from a textbook or a lecture course. If not, could you please say more about where you saw this statement and why you want to know the proof? $\endgroup$
    – Yemon Choi
    Commented Apr 1, 2017 at 22:06
  • $\begingroup$ No it's not any homework $\endgroup$
    – Tree
    Commented Apr 1, 2017 at 22:12
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    $\begingroup$ Am reading character theory by Issacs by myself. It is an exercise from chapter 3 of the book. There is a hint given to the problem in the book but I don't understand it. $\endgroup$
    – Tree
    Commented Apr 1, 2017 at 22:14
  • $\begingroup$ To the OP: why don't you include the hint in the question, and explain where you are stuck? $\endgroup$
    – Yemon Choi
    Commented Apr 1, 2017 at 22:44
  • $\begingroup$ @René you do know that $\chi$ is not necessarily a HM, right? (just checking, sorry; I just wasn't sure what you were getting at in your hint) $\endgroup$
    – Yemon Choi
    Commented Apr 1, 2017 at 22:50

2 Answers 2

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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, where the proof follows Isaac's hint.

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The key to this question is ( I believe) that characters take values in cyclotomic number fields, so that complex conjugation is central in the relevant Galois group. From the fact that $|\chi(g)| = 1,$ it follows that all algebraic conjugates of $\chi(g)$ have absolute value $1$.

It is a standard fact that if all algebraic conjugates of an algebraic integer $\alpha$ have absolute value $1$, then $\alpha$ is a root of unity. I only outline a sketch of the proof of this standard fact. Note that all powers of $\alpha$ have the same property. On the other hand, there are only finitely many possibilities for the minimum polynomials of powers of $\alpha,$ so the powers of $\alpha$ are not all distinct. Hence $\alpha$ is indeed a root of unity.

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  • $\begingroup$ This standard fact is the kind of thing many of us analysts never saw - or perhaps forgot through lack of use :) $\endgroup$
    – Yemon Choi
    Commented Apr 2, 2017 at 1:33
  • $\begingroup$ @YemonChoi : My comments were only made because I wanted to make it clear that this was a well-known fact ( at least to algebraic number theorists, of which I am not one). Richard Stanley attributes it to Kronecker in his answer- I myself was unaware that this was the origin of the result. $\endgroup$
    – Geoff Robinson
    Commented Apr 2, 2017 at 3:58
  • $\begingroup$ @YemonChoi Once we know $\forall \sigma \in Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}),|\sigma(\alpha)| = 1$ we know $\forall m, \forall \sigma \in Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}),|\sigma(\alpha^m)| = 1$. Expressing the coefficients of the minimal polynomial of $\alpha^m$ (at most of degree $n$) in term of its roots, they are bounded by some constant independently of $m$, and what Geoff Robinson said follows : there are finitely many such possible polynomials so $\alpha^l = \alpha^m$ for some $l,m$ $\endgroup$
    – reuns
    Commented Apr 2, 2017 at 4:15
  • $\begingroup$ Thanks @user1952009 -- I could follow Geoff's outline, I was merely remarking that it is not something I remember learning! $\endgroup$
    – Yemon Choi
    Commented Apr 2, 2017 at 14:40

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