[proofplan] By [Schur's Lemma](/theorems/2414), every element of $Z(\mathbb{C}G)$ acts on the irreducible representation $V$ affording $\chi$ as a scalar; the resulting algebra homomorphism $\omega_\chi: Z(\mathbb{C}G) \to \mathbb{C}$ sends $C_i \mapsto \frac{|\mathcal{C}_i|\,\chi(g_i)}{\chi(1)}$. By [Class Sums Form a Basis for the Center](/theorems/2462), the products $C_i C_j$ expand as $C_i C_j = \sum_\ell a_{ij\ell} C_\ell$ with non-negative integer structure constants $a_{ij\ell}$. Applying $\omega_\chi$ converts these multiplication relations into integer-coefficient relations among the values $\omega_\chi(C_1), \ldots, \omega_\chi(C_k)$. Hence the $\mathbb{Z}$-subalgebra of $\mathbb{C}$ generated by these values is finitely generated as a $\mathbb{Z}$-module, which is precisely the third equivalent characterisation of an algebraic integer. [/proofplan] [step:Identify $\omega_\chi$ as the scalar by which $C_i$ acts on the irreducible $V$] Let $\rho: G \to \operatorname{GL}(V)$ be an irreducible representation affording $\chi$, with $\dim V = \chi(1)$. Extend $\rho$ linearly to an algebra homomorphism (also denoted $\rho$) from the group algebra to operators on $V$: \begin{align*} \rho: \mathbb{C}G &\to \operatorname{End}(V), \\ \sum_{g \in G} \alpha_g g &\mapsto \sum_{g \in G} \alpha_g\, \rho(g). \end{align*} Restricting to the centre $Z(\mathbb{C}G)$, we observe that for any $z \in Z(\mathbb{C}G)$ and any $g \in G$, \begin{align*} \rho(z) \rho(g) = \rho(zg) = \rho(gz) = \rho(g) \rho(z), \end{align*} so $\rho(z)$ commutes with every $\rho(g)$. Hence $\rho(z)$ is a $G$-equivariant linear map $V \to V$. By [Schur's Lemma](/theorems/2414) (applied to the irreducible representation $V$), every $G$-equivariant endomorphism of $V$ is a scalar multiple of the identity. Therefore there is a unique scalar $\omega(z) \in \mathbb{C}$ with \begin{align*} \rho(z) = \omega(z) \cdot I_V. \end{align*} The map $\omega: Z(\mathbb{C}G) \to \mathbb{C}$ defined by $z \mapsto \omega(z)$ is the **central character** $\omega_\chi$ associated with $\chi$. From the definition, \begin{align*} \rho(z_1 z_2) = \omega(z_1 z_2) I_V \quad \text{and} \quad \rho(z_1)\rho(z_2) = \omega(z_1) \omega(z_2) I_V, \end{align*} and the equality $\rho(z_1 z_2) = \rho(z_1)\rho(z_2)$ gives $\omega(z_1 z_2) = \omega(z_1)\omega(z_2)$. Linearity of $\omega$ is immediate from linearity of $\rho$ and the fact that $z \mapsto \omega(z) I_V$ is the composition with the (injective) inclusion $\mathbb{C} \hookrightarrow \operatorname{End}(V)$, $\lambda \mapsto \lambda I_V$. So $\omega_\chi$ is an algebra homomorphism $Z(\mathbb{C}G) \to \mathbb{C}$. To compute $\omega_\chi(C_i)$, take traces on both sides of $\rho(C_i) = \omega_\chi(C_i) I_V$: \begin{align*} \operatorname{tr}(\rho(C_i)) = \omega_\chi(C_i)\, \dim V = \omega_\chi(C_i)\, \chi(1). \end{align*} On the other hand, expanding $C_i = \sum_{x \in \mathcal{C}_i} x$ and using that $\chi$ is a class function (so $\chi(x) = \chi(g_i)$ for every $x \in \mathcal{C}_i$), \begin{align*} \operatorname{tr}(\rho(C_i)) = \sum_{x \in \mathcal{C}_i} \chi(x) = |\mathcal{C}_i|\, \chi(g_i). \end{align*} Solving, \begin{align*} \omega_\chi(C_i) = \frac{|\mathcal{C}_i|\, \chi(g_i)}{\chi(1)}. \end{align*} [guided] Schur's Lemma is the central engine. The centre $Z(\mathbb{C}G)$ acts on the irreducible $V$ by $G$-equivariant operators (because central elements commute with every group element), and Schur's Lemma forces every $G$-equivariant endomorphism of an irreducible to be a scalar. So central elements act as scalars, and the assignment $z \mapsto \text{(scalar)}$ is the central character. The trace identification works because the trace of $\lambda I_V$ is $\lambda \dim V$. So we just need to compute $\operatorname{tr}(\rho(C_i))$ directly from the definition of $C_i$ as a sum of group elements: trace is linear and $\operatorname{tr}(\rho(x)) = \chi(x)$. Class invariance of $\chi$ collapses the sum to $|\mathcal{C}_i|\chi(g_i)$. The hypothesis that $\chi$ is irreducible is used precisely once: to invoke Schur's Lemma. For a reducible $\chi$, $Z(\mathbb{C}G)$ does not act by scalars on the underlying representation, and there is no central character to speak of. [/guided] [/step] [step:Translate the integer multiplication relations of $C_i C_j$ into relations on $\omega_\chi(C_i)$] By [Class Sums Form a Basis for the Center](/theorems/2462), for each $i, j \in \{1, \ldots, k\}$, \begin{align*} C_i C_j = \sum_{\ell=1}^k a_{ij\ell} C_\ell \end{align*} where the structure constants $a_{ij\ell}$ are non-negative integers. Since $\omega_\chi: Z(\mathbb{C}G) \to \mathbb{C}$ is an algebra homomorphism (Step 1), \begin{align*} \omega_\chi(C_i)\, \omega_\chi(C_j) = \omega_\chi(C_i C_j) = \omega_\chi\left(\sum_{\ell=1}^k a_{ij\ell} C_\ell\right) = \sum_{\ell=1}^k a_{ij\ell}\, \omega_\chi(C_\ell). \end{align*} Set $\alpha_i := \omega_\chi(C_i) \in \mathbb{C}$ for $i = 1, \ldots, k$. Then \begin{align*} \alpha_i\, \alpha_j = \sum_{\ell=1}^k a_{ij\ell}\, \alpha_\ell, \qquad i, j \in \{1, \ldots, k\}, \quad a_{ij\ell} \in \mathbb{Z}_{\geq 0}. \end{align*} [/step] [step:Show the $\mathbb{Z}$-module $M = \mathbb{Z}\alpha_1 + \cdots + \mathbb{Z}\alpha_k$ is finitely generated and closed under multiplication by each $\alpha_i$] Define \begin{align*} M := \sum_{\ell=1}^k \mathbb{Z}\alpha_\ell = \{n_1 \alpha_1 + \cdots + n_k \alpha_k : n_\ell \in \mathbb{Z}\} \subseteq \mathbb{C}. \end{align*} By construction, $M$ is a finitely generated $\mathbb{Z}$-submodule of $\mathbb{C}$ (with generators $\alpha_1, \ldots, \alpha_k$). We verify $M$ is closed under multiplication by each $\alpha_i$. Fix $i$ and let $m = \sum_{\ell} n_\ell \alpha_\ell \in M$ with $n_\ell \in \mathbb{Z}$. Then \begin{align*} \alpha_i \cdot m = \sum_{\ell=1}^k n_\ell\, (\alpha_i \alpha_\ell) = \sum_{\ell=1}^k n_\ell\, \sum_{p=1}^k a_{i\ell p}\, \alpha_p = \sum_{p=1}^k \left( \sum_{\ell=1}^k n_\ell\, a_{i\ell p} \right) \alpha_p, \end{align*} using the relation from Step 2 with $(i, j) = (i, \ell)$. The coefficient of $\alpha_p$ is $\sum_\ell n_\ell\, a_{i\ell p} \in \mathbb{Z}$ (a finite sum of products of integers). Hence $\alpha_i m \in M$, and $M$ is closed under multiplication by $\alpha_i$. In particular, taking $m = \alpha_1$ if $1 \in \{1, \ldots, k\}$ (it is, by convention), or more carefully: since $C_1 = 1$ is the class sum of the identity class $\mathcal{C}_1 = \{1\}$, we have $\alpha_1 = \omega_\chi(1) = 1$ (as $\omega_\chi$ is an algebra homomorphism, sending the multiplicative identity of $Z(\mathbb{C}G)$ to $1 \in \mathbb{C}$). So $1 = \alpha_1 \in M$, which means $\mathbb{Z} \subseteq M$. [guided] This step is the heart of the integrality argument. The "right" notion of algebraic integer for this proof is the third equivalent characterisation: $\alpha \in \mathbb{C}$ is an algebraic integer iff there exists a finitely generated $\mathbb{Z}$-submodule $M \subseteq \mathbb{C}$ with $1 \in M$ (or more generally $\alpha M \subseteq M$ with $M$ a faithful $\mathbb{Z}[\alpha]$-module). We have constructed exactly such an $M$. The integer relations $\alpha_i \alpha_j = \sum_\ell a_{ij\ell}\, \alpha_\ell$ are the crucial ingredient: they say that multiplication by any $\alpha_i$ takes the generators $\alpha_1, \ldots, \alpha_k$ of $M$ to integer combinations of the same generators. This is exactly what is needed to conclude that $M$ is invariant under multiplication by $\alpha_i$, which in turn pins down $\alpha_i$ as a root of a monic integer polynomial — namely, the characteristic polynomial of the integer matrix representing multiplication by $\alpha_i$ on $M$ in the generating set $\alpha_1, \ldots, \alpha_k$. [/guided] [/step] [step:Conclude $\omega_\chi(C_i)$ is an algebraic integer for each $i$] We use the following standard characterisation: a complex number $\beta \in \mathbb{C}$ is an algebraic integer if and only if there exists a finitely generated $\mathbb{Z}$-submodule $N \subseteq \mathbb{C}$ with $1 \in N$ and $\beta N \subseteq N$. Apply this with $\beta = \alpha_i = \omega_\chi(C_i)$ and $N = M$ (from Step 3). The hypotheses are verified: 1. *$M$ is a finitely generated $\mathbb{Z}$-submodule of $\mathbb{C}$*: by definition $M$ has generators $\alpha_1, \ldots, \alpha_k$. 2. *$1 \in M$*: shown at the end of Step 3 since $\alpha_1 = 1$. 3. *$\alpha_i M \subseteq M$*: shown in Step 3. Therefore $\alpha_i = \omega_\chi(C_i)$ is an algebraic integer. Equivalently, expressing the multiplication-by-$\alpha_i$ map on $M$ in the (possibly redundant) generating set $\alpha_1, \ldots, \alpha_k$, we obtain an integer matrix $A_i \in \mathbb{Z}^{k \times k}$ with $(A_i)_{p\ell} = a_{i\ell p}$ such that $\alpha_i \cdot \alpha = A_i\, \alpha$, where $\alpha = (\alpha_1, \ldots, \alpha_k)^\top$. Hence $\alpha$ is a non-zero vector in the kernel of $\alpha_i I_k - A_i$ over $\mathbb{C}$ (it is non-zero because $\alpha_1 = 1 \neq 0$), so $\det(\alpha_i I_k - A_i) = 0$. The polynomial $\det(X I_k - A_i) \in \mathbb{Z}[X]$ is monic of degree $k$ with integer coefficients (since $A_i$ has integer entries), and $\alpha_i$ is one of its roots. So $\alpha_i$ is an algebraic integer. [/step]