In this question you may assume the following result. Let $\chi$ be a character of a finite group $G$ and let $g \in G$. If $\chi(g)$ is a rational number, then $\chi(g)$ is an integer.

(a) If $a$ and $b$ are positive integers, we denote their highest common factor by $(a, b)$. Let $g$ be an element of order $n$ in the finite group $G$. Suppose that $g$ is conjugate to $g^{i}$ for all $i$ with $1 \leqslant i \leqslant n$ and $(i, n)=1$. Prove that $\chi(g)$ is an integer for all characters $\chi$ of $G$.

[You may use the following result without proof. Let $\omega$ be an $n$th root of unity. Then

is an integer.]

Deduce that all the character values of symmetric groups are integers.

(b) Let $G$ be a group of odd order.

Let $\chi$ be an irreducible character of $G$ with $\chi=\bar{\chi}$. Prove that

$$\left\langle\chi, 1_{G}\right\rangle=\frac{1}{|G|}(\chi(1)+2 \alpha),$$

where $\alpha$ is an algebraic integer. Deduce that $\chi=1_{G}$.

$$\begin{aligned} & \sum_{1 \leqslant i \leqslant n,} \omega^{i} \\ & \begin{aligned}&1 \leqslant i \leqslant n \\&(i, n)=1\end{aligned} \end{aligned}$$