I am trying to prove a statement from Isaacs' book "Character theory of Finite Groups" (exercise $8.5$).
Let $\chi \in \operatorname{Irr}(G)$ and suppose $|G| = mn$, with $(m, n) = 1$. Assume that $\chi(x) = 0$ for all $1 \neq x \in G$ such that $x^{n} = 1$. Suppose $y \in G$ and $y^{m} \neq 1$. Show that $\chi(y) = 0$.
I tried to solve this problem like this:
Since $y^m\neq 1$, we can say that there is at least one prime divisor of $n$ that is a divisor of $|y|$.
$[\chi, \chi] = \frac{1}{|G|}\sum\limits_{x \in G}\chi(x)\overline{\chi(x)} = \frac{1}{|G|}\sum |\chi(x)|^{2} = 1$
In the second equality, the sum is taken over those $x\in G : |x|\not\mid \; n$.
Let 's try to use the following theorem:
Let $\chi \in \operatorname{Irr}(G).$ Let $\exists \; p \in \mathbb{P}$ such that $p$ doesn't divide $\frac{|G|}{\chi(1)}$. Then $\chi(g) = 0$ for all $g \in G$ such that $p \; | \; |g|$.
If $n \; |\; |y|$, then our statement will be fulfilled.
If $\exists \; p \in \mathbb{P}$ such that $p$ doesn't divide $\frac{|G|}{\chi(1)}$ and $p \; | \; |y|$, then our statement will be fulfilled.
If $\chi(1) = |G|$, then our statement will be fulfilled
Unfortunately, I'm stuck in this place, and I don't know what to do next to solve this problem.
I will be grateful for any help!