if $g^k=e$ then $\chi(g)=\sum_j^n \zeta_k$ 
Let $G$ be a group. Let $g \in G$ and $g^k=e.$ Let $\chi$ be an
   $n$-dimensional character of the group $G.$ Let $\zeta_k$ be  $k$-th root of unity.
  Prove that $\chi(g)$  is equal  to sum of a $k$-th roots of unity.

My trying. Consider the cyclic subgroup $C_k \in G,$ Such  that $C_k=\{g^i| i<k\}.$
 Then any representation  $\rho$ of $C_k$ is equal to  prime sum of 1-dimensional representations $\rho_j$ of the group $C_k.$ Them number  is $n.$  and $\rho_j(g)=\zeta_k^r$ as $r|k.$
I'm not sure of the correctness of my proof

If $g^k = e$ then since the function $\rho: G \to GL_n(\mathbb C)$ that constitutes your representation is a homomorphism you have that $\rho(g^k) = \rho(g)^k = $ I, $ $thus in particular you know that all of the eigenvalues of $\rho(g)$ must be kth roots of unity (as for any $v \in \mathbb C^n$ you have that $\rho(g)^kv = v$, including its eigenvectors). But the character $\chi_\rho(g) = \text{Trace}(\rho(g))$ and the trace of a matrix is the sum of its eigenvalues as you know from linear algebra. Thus it must be a sum of kth roots of unity.
