I am struggling with a proof of a two part question:
Let $\chi$ be a character of a finite group $G$.
a) If $g$ has order 2, then $\chi(g) \in \mathbb{Z}$ and $\chi(g) \equiv \chi(1)$ (mod 2)
a) If $g$ has order 3 and $\chi(g) \in \mathbb{R}$, then $\chi(g) \in \mathbb{Z}$ and $\chi(g) \equiv \chi(1)$ (mod 3)
For a), I said $g^2 = 1$ so $g = g^{-1}$. Then, if $\rho$ is a matrix representation of $G$, we have $\rho(g) = \rho(g^{-1})$. Then the entries of both matrices will be the same, and I already know the matrices will be diagonal, with entries $\lambda_1, ... , \lambda_n$ for $\rho(g)$. Then $\lambda_i = \lambda_i^{-1}$, so all the entries are either 1 or -1. Noting that $\chi(1) = n$, we get that $\chi(1) - \chi(g) = n - (m - (n - m)) = 2(n - m) \equiv 0$ (mod 2) for some $1 \leq m \leq n$.
For b), I've tried to proceed in a similar way, using the fact that $g^3 = 1$ and $\chi(g) = \overline{\chi(g)}$ since $\chi(g) \in \mathbb{R}$, therefore $\chi(g^{-1}) = \overline{\chi(g)} = \chi(g)$. However I'm not sure where to go from here. I guess you could say $g^{-1} = g^2$, although I'm not sure if it would help.
Any help at all would be much appreciated!
Let $\omega$ be a complex cube root of unity, and suppose $\chi$ is the character of representation $\rho$.
If $|g|=3$ and $\chi(g) \in {\mathbb R}$, then the number of eigenvalues of $\rho(g)$ equal to $\omega$ must be the same as the number equal to $\bar{\omega}=\omega^{-1}$.
But $\omega + \bar{\omega} = -1 \equiv 2 \bmod 3$, and b) follows.