Representation theory and characters I have been studying representation theory for 6 months now. I came across the following question in a graduate course example sheet. 

Let $\chi$ be the character of a representation $\rho$ of dimension $d$ of a finite group $G$. 
1) Is $|\chi(g)|\leq d$?
2) If $|\chi(g)|= d$ then $\rho(g)=\omega I$ for some root of unity $\omega$.
3) If $\chi(g)= d$, is $\rho_{g}$ identity operator?

My attempt: For 1) I was think about proving by contradiction, that is suppose $|\chi(g)|> d$ for some $g$. This means that it would have to be greater than $\chi(1)$ which is a contradiction. 
For 3) My guess would be it is true but have no idea how to prove it. Any help is appreciated!
 A: The statements are false for arbitrary groups.  For example, $\mathbb Z$ can act in lots of ways — choose any invertible matrix for the action of $1\in \mathbb Z$, and invertible matrices can have arbitrary traces.
If $g\in G$ is of finite order, however, then things become straightforward.  Suppose that a $d\times d$ matrix $\rho(g)$ has order $n$.  Then it is diagonalizable, with eigenvalues all $n$th roots of unity.  The trace is then a sum of $d$ $n$th roots of unity (not necessarily primitive), and by the triangle inequality has norm at most $d$.  By the "strict" triangle inequality, the norm is strictly less than $n$ unless all eigenvalues are equal (well, parallel and in the same direction, but they have the same norm).  In this case the matrix is multiplication by an $n$th root of unity $\omega$, and hence has trace $d\omega$, which is $d$ iff $\omega = 1$.
I remark that similar statements hold for representations of compact groups, and more generally for unitarizable representations.  A matrix is unitarizable iff it is diagonalizable with eigenvalues on the unit circle in $\mathbb C$.  Statements 1 and 3 on your list hold without change; statement 2 should be modified by replacing "some root of unity $\omega$" with "some $\omega$ on the unit circle".
