Faithful characters of finite groups Related to an answer to a previous question. The answer assume the following result:
Let $G$ be a finite group and $\rho : G \rightarrow \text{GL}(\mathbb{C}, n)$ be a faithful representation of $G$ (ie. $\text{Ker}(\rho) = 1_G$). Let $\chi$ be the character associated to $\rho$. Then, for all $g \in G$ such that $g \not= 1_G$ we have $|\chi(g)| < n$.
Is this true? If yes, why? I couldn't find any proof and I can't understand the small justification given in the previous answer.
 A: The other result in this direction is that if $\rho$ is irreducible, then $|\chi(g)| = n$ if and only if g is in the center of G/Ker.  The proof is to start wiht what Qiaochu said, namely that by the triangle inequality and the fact that the eigenvalues are roots of unity you get that $|\chi(g)|=n$ if and only if g is a scalar matrix.  Thus they commute with everything in End(V), and hence lie in the center of G/Ker.  Conversely, use irreducibility to show that commuting with everything in G/Ker implies that you commute with everything in End(V).
When I took representation theory with Lenstra this argument was very memorable.  He had started out in complete generality (arbitrary fields etc.) and as the course went on we needed more and more assumptions (algebraically closed, characteristic prime to the size of the group, etc.).  When he got to this argument he said "Now this is the only time that we need to assume that the field is the complex numbers.  This argument doesn't work over an arbitrary algebraically closed field of characteristic zero.  (Although it's still true for such fields by model theoretic arguments.)"
A good related theorem to try to prove when you're thinking about the question you asked is that a representation is faithful if and only if every representation appears inside one of its tensor powers.
A: It is a well known fact that $Ker(\rho)$ is the set of elements $g$ such that $\chi(g)=\chi(1) $ -this can be found in Isaacs character theory or any other book in character theory- Notice that, as pointed out in the above example, $\chi(g)=\chi(1) $ is not the same as $|\chi(g)|=\chi(1)$. So the point of the previous answer is that for faithful $\rho$ one has that $\chi(g)=n$ if and only if $g=1$.  
Edit: See Darij's comment below.
The proof I know is algebraic(I think).  Let $\alpha$ be the arithmetic mean of the root of units in question. Then, for all $\beta$ which is conjugated to $\alpha$ over $\mathbb{Q}$ we have that $|\beta| \leq 1$. In particular the product of all such $\beta$'s has absolute value less than equal to $1$. On the other hand the product must be an integer, by the hypothesis on $\alpha$, hence it is either $1$ or $0$. If it is non-zero then each term in the product must be equal to $1$, hence $\alpha =1$.  The last can only happen if all the root's of unity are the same(we have equality in the triangle inequality.) 
