[proofplan] Set $\alpha = \chi(g)/\chi(1)$. Our goal is to apply the [Roots of Unity Average Lemma](/theorems/2466) to $\alpha$, which requires showing that $\alpha$ is an algebraic integer expressible as an average of $\chi(1)$ roots of unity. The averaging is automatic: $\chi(g)$ is the trace of $\rho(g)$, which equals the sum of its $\chi(1)$ eigenvalues (each a root of unity since $g$ has finite order). The integrality is the genuine content: although $\chi(g)$ and $\omega_\chi(\mathcal{C}) = |\mathcal{C}|\chi(g)/\chi(1)$ are individually algebraic integers, the ratio $\chi(g)/\chi(1)$ has $\chi(1)$ in the denominator and is not visibly integral from these facts alone. We extract integrality using Bezout's identity on the coprime pair $(\chi(1), |\mathcal{C}|)$: writing $1 = a\chi(1) + b|\mathcal{C}|$ and dividing by $\chi(1)$ produces $\alpha$ as an integer combination of $\chi(g)$ and $\omega_\chi(\mathcal{C})$, both algebraic integers. [/proofplan] [step:Express $\chi(g)$ as a sum of $\chi(1)$ roots of unity] Let $\rho: G \to \operatorname{GL}(V)$ be a representation affording $\chi$, with $\dim V = \chi(1)$. Since $G$ is finite, the element $g$ has finite order, say $|g| = d$. Then $\rho(g)^d = \rho(g^d) = \rho(1) = I_V$, so the matrix $\rho(g)$ satisfies $X^d - 1 = 0$. The polynomial $X^d - 1$ has $d$ distinct roots in $\mathbb{C}$ (the $d$-th roots of unity), so $\rho(g)$ is diagonalisable over $\mathbb{C}$ and its eigenvalues $\lambda_1, \ldots, \lambda_{\chi(1)}$ are $d$-th roots of unity. In particular each $\lambda_i$ satisfies $\lambda_i^N = 1$ where $N = |G|$ (a multiple of $d$). Taking the trace, \begin{align*} \chi(g) = \operatorname{tr}\rho(g) = \sum_{i=1}^{\chi(1)} \lambda_i. \end{align*} Therefore \begin{align*} \alpha := \frac{\chi(g)}{\chi(1)} = \frac{1}{\chi(1)} \sum_{i=1}^{\chi(1)} \lambda_i \end{align*} is the average of $m = \chi(1)$ roots of unity (each a root of $X^N - 1 = 0$). This puts $\alpha$ in the form required by the [Roots of Unity Average Lemma](/theorems/2466). [/step] [step:Show $\omega_\chi(\mathcal{C}) = |\mathcal{C}|\chi(g)/\chi(1)$ is an algebraic integer] By [Central Character Algebraic Integrality](/theorems/2463) applied to the irreducible character $\chi$ on the finite group $G$, the central character value \begin{align*} \omega_\chi(\mathcal{C}) = \frac{|\mathcal{C}|\, \chi(g)}{\chi(1)} \end{align*} is an algebraic integer. The hypotheses required there — $G$ finite and $\chi$ irreducible — are exactly the hypotheses of the present theorem. [/step] [step:Apply Bezout's identity to the coprime pair $(\chi(1), |\mathcal{C}|)$ to express $\alpha$ as an algebraic integer] By hypothesis, $\gcd(\chi(1), |\mathcal{C}|) = 1$. By Bezout's identity in $\mathbb{Z}$, there exist integers $a, b \in \mathbb{Z}$ with \begin{align*} a\, \chi(1) + b\, |\mathcal{C}| = 1. \end{align*} Dividing by $\chi(1) \neq 0$, \begin{align*} \frac{1}{\chi(1)} = a + b\, \frac{|\mathcal{C}|}{\chi(1)}. \end{align*} Multiplying by $\chi(g)$, \begin{align*} \alpha = \frac{\chi(g)}{\chi(1)} = a\, \chi(g) + b\, \frac{|\mathcal{C}|\, \chi(g)}{\chi(1)} = a\, \chi(g) + b\, \omega_\chi(\mathcal{C}). \end{align*} We verify each term on the right is an algebraic integer: 1. $\chi(g) = \sum_{i=1}^{\chi(1)} \lambda_i$ is a $\mathbb{Z}$-linear combination of roots of unity. Each root of unity is an algebraic integer (root of $X^N - 1 \in \mathbb{Z}[X]$), and algebraic integers are closed under $+$ and integer scalar multiplication, so $\chi(g)$ is an algebraic integer. Multiplying by $a \in \mathbb{Z}$ keeps it an algebraic integer. 2. $\omega_\chi(\mathcal{C})$ is an algebraic integer by Step 2. Multiplying by $b \in \mathbb{Z}$ keeps it an algebraic integer. Algebraic integers form a subring of $\mathbb{C}$, so the sum $a\chi(g) + b\omega_\chi(\mathcal{C}) = \alpha$ is also an algebraic integer. [guided] The Bezout step is the engine. We want to show $\alpha = \chi(g)/\chi(1)$ is an algebraic integer, but $\chi(1)$ in the denominator obstructs a direct argument. The trick is to exhibit $1/\chi(1)$ as an integer combination of $1$ and $|\mathcal{C}|/\chi(1)$: \begin{align*} \frac{1}{\chi(1)} = a \cdot 1 + b \cdot \frac{|\mathcal{C}|}{\chi(1)}, \end{align*} which is precisely Bezout's identity rearranged. This *requires* coprimality: if $\gcd(\chi(1), |\mathcal{C}|) = d > 1$, then no integers $a, b$ satisfy $a\chi(1) + b|\mathcal{C}| = 1$, only $a\chi(1) + b|\mathcal{C}| = d$, and dividing by $\chi(1)$ would give $1/\chi(1) = a/d + b|\mathcal{C}|/(d\chi(1))$ — irrational denominators reappear. So coprimality is exactly the hypothesis needed. Multiplying through by $\chi(g)$ converts the rational equation into an equation in algebraic integers: each piece $a\chi(g)$ and $b\omega_\chi(\mathcal{C})$ is integer-times-algebraic-integer. The point is that we have *constructed* $\alpha$ as a $\mathbb{Z}$-combination of two known algebraic integers ($\chi(g)$ and $\omega_\chi(\mathcal{C})$), so $\alpha$ inherits algebraic integrality from the closure of the ring of algebraic integers under addition and scalar multiplication. [/guided] [/step] [step:Apply the Roots of Unity Average Lemma to conclude] We apply [Roots of Unity Average Lemma](/theorems/2466) to $\alpha$. The required hypotheses are: 1. *$\alpha$ is an algebraic integer*: established in Step 3. 2. *$\alpha = (1/m) \sum_{j=1}^m \lambda_j$ where each $\lambda_j$ is an $n$-th root of unity for some fixed $n$*: from Step 1, with $m = \chi(1)$, $n = N = |G|$, and $\lambda_j$ the eigenvalues of $\rho(g)$. The lemma's conclusion gives \begin{align*} \alpha = 0 \quad \text{or} \quad |\alpha| = 1. \end{align*} In the first case, $\chi(g)/\chi(1) = 0$, so $\chi(g) = 0$ (since $\chi(1) \ne 0$). In the second case, $|\chi(g)/\chi(1)| = 1$, so $|\chi(g)| = \chi(1)$. [/step]