Given a finite group $G$, and a non-identity representative $g$ in a conjugacy class of prime order $p$, I'm trying to show that some nontrivial irreducible character of $G$ must have $\chi(g) \neq 0$ and $\chi(1) \neq 0 \text{ mod } p$.
The suggestion was to show that the nonexistence of such a character would imply that $1/p$ is an algebraic integer.
So far, I've used the fact that $\chi(g) \cdot \frac{p}{\chi(1)} \in \overline{\mathbb{Z}}$ for any character. With some manipulation (and closure of algebraic integers over sums and products), I can get this into a form that allows me to sum over all irreducible characters $$p \sum \chi_i(g)^2 \in \overline{\mathbb{Z}}.$$ My hope was to then use the orthogonality relations on the columns of a character table, but doing so leaves a pesky factor of the order of the group, i.e. $|G|/p \in \overline{\mathbb{Z}}$ instead of $1/p$.
Am I going about this the wrong way? Is there something else I am overlooking?
The column orthogonality relations for character tables give that $\sum_i \chi_i(g) \overline{\chi_i(1)} = 0$. But if every nontrivial character has either $\chi_i(g) = 0$ or $\chi_i(1) \equiv 0 \text{ mod } p$, then we have that $$\sum_{\text{nontrivial characters}} \chi_i(g) \overline{\chi_i(1)} \equiv 0 \text{ mod } p,$$ and then $\sum_i \chi_i(g) \overline{\chi_i(1)} \equiv 1 \text{ mod } p$ due to the trivial character - a clear contradiction.