Let $G$ be a finite group and let $\chi$ be its character. Suppose $g \in G$ as order $3$ and $\chi(g) \in \mathbb{R}$. Show that, in fact, $\chi(g) \in \mathbb{Z}$ and $\chi(g) \equiv \chi(1) \pmod3$.
Now, since $\chi(1)$ is the dimension of the matrix representation of $g$, then $\chi(g)$ is the sum if $\chi(1)$ cubed roots of unity: $\; 1, \dfrac{-1-i\sqrt3}{2}, \dfrac{-1 + i\sqrt3}{2}$, but I'm not sure where to proceed from here?
I know that if $\dfrac{-1-i\sqrt3}{2}$ appears in the diagonal then so must $\dfrac{-1+i\sqrt3}{2}$, but I still can't see how to prove the statement?
$\chi(g) = tr(\rho(g))$ with $\rho$ a representation of the cyclic group $\{1,g,g^2 \} \to GL_n(\mathbb{C})$, so $\rho$ is equivalent to some representation $\rho'(g) =\text{diag}( \zeta_3^{e_1}, \ldots,\zeta_3^{e_m})$.
Let $K = \mathbb{Q}(\zeta_3)$, $O_K$ its ring of integers, $\mathfrak{p}$ a prime ideal of $O_K$ above $(3)$. There is a "Frobenius" $\sigma \in Gal(K/\mathbb{Q})$ such that $\forall a\in O_K, \sigma(a)- a^3 \in \mathfrak{p}$, in particular $\chi(1) = \chi(g^3) =\sum_{m=1}^n \zeta_3^{3e_m} \equiv \sum_{m=1}^n \sigma(\zeta_3^{e_m})\equiv \sigma(\sum_{m=1}^n \zeta_3^{e_m})\equiv\sigma(\chi(g)) \bmod \mathfrak{p}$.
It is said that $\chi(g) \in \mathbb{R}$ therefore $\chi(g) \in O_K \cap \mathbb{R}=\mathbb{Z}$, so that $\sigma(\chi(g)) = \chi(g))$ and $\chi(1) \equiv \chi(g) \bmod \mathfrak{p} \cap \mathbb{Z} = (3)$.
Also note for the same reason $\chi(g^3) \equiv \chi(g)^3 \bmod 3$.