TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-30

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The degree of an irreducible complex character divides $[G : Z (G)]$

Statement

Let $G$ be a finite group and let $χ$ be an irreducible complex character of $G$ . Then $χ (1)$ divides $[G : Z (G)]$ .

Facts & Assumptions

Given: A finite group $G$ , an irreducible complex character $χ$ of $G$ , and an irreducible complex representation $V$ of $G$ affording $χ$ .

[F1]

An irreducible complex character has self-inner-product $1$ (A complex character is irreducible if and only if its self-inner-product is $1$ ).

[F2]

The class-function inner product is $⟨ φ, ψ ⟩ = \frac{1}{∣ G ∣} \sum_{g \in G} φ (g) \overline{ψ (g)}$ (The standard inner product on $c f (G)$ ).

[F3]

A central class sum acts on an irreducible representation by a scalar (Class sums act on an irreducible representation by central-character scalars).

[F4]

The degree of an irreducible complex character divides the order of the group (The degree of an irreducible complex character divides $∣ G ∣$ ).

[F5]

The center is $Z (G) = {z \in G : z g = g z for every g \in G}$ (The center $Z (G)$ of a group).

Proof

technique · direct

1.1

Let $n = χ (1)$ . For each integer $N \geq 1$ , let $G^{N}$ act on $V^{\otimes N}$ by $(g_{1}, \dots, g_{N}) \cdot (v_{1} \otimes \dots \otimes v_{N}) = (g_{1} v_{1}) \otimes \dots \otimes (g_{N} v_{N})$ . Its character is $Ξ_{N} (g_{1}, \dots, g_{N}) = \prod_{i = 1}^{N} χ (g_{i})$ , so by [F2] and [F1] its self-inner-product is $⟨ Ξ_{N}, Ξ_{N} ⟩_{G^{N}} = ⟨ χ, χ ⟩_{G}^{N} = 1$ . Hence $Ξ_{N}$ is irreducible by [F1].

F1F2givenalgebra

2.1

If $z \in Z (G)$ , then the conjugacy class of $z$ is the singleton ${z}$ by [F5], so [F3] gives $ρ_{V} (z) = λ (z) {id}_{V}$ for some scalar $λ (z) \in C^{\times}$ . Therefore $(z_{1}, \dots, z_{N}) \in Z (G)^{N}$ acts on $V^{\otimes N}$ as the scalar $λ (z_{1}) \dots λ (z_{N})$ .

F3F5step 1.1algebra

3.1

Let $T_{N} : = {(z_{1}, \dots, z_{N}) \in Z (G)^{N} : z_{1} \dots z_{N} = e}$ . Since $λ$ is multiplicative, step 2.1 shows that every element of $T_{N}$ acts trivially on $V^{\otimes N}$ . Thus the irreducible representation of step 1.1 descends to an irreducible representation of the quotient $G^{N} / T_{N}$ , still of degree $n^{N}$ .

step 1.1step 2.1algebra

4.1

The map $Z (G)^{N - 1} \to T_{N}$ , $(z_{1}, \dots, z_{N - 1}) \mapsto (z_{1}, \dots, z_{N - 1}, (z_{1} \dots z_{N - 1})^{- 1})$ , is bijective, so $∣ T_{N} ∣ = ∣ Z (G) ∣^{N - 1}$ . Hence $∣ G^{N} / T_{N} ∣ = ∣ G ∣^{N} / ∣ Z (G) ∣^{N - 1} = ∣ G ∣ ([G : Z (G)])^{N - 1}$ . Applying [F4] to the irreducible quotient representation from step 3.1 yields $n^{N} ∣ ∣ G ∣ ([G : Z (G)])^{N - 1}$ for every $N \geq 1$ .

F4step 3.1algebra

5.1

Fix a prime $p$ and write $n = p^{a} u$ and $[G : Z (G)] = p^{b} v$ with $p ∤ u v$ . If $a > b$ , then choosing $N$ so large that $N (a - b) > v_{p} (∣ G ∣) - b$ makes $p^{N a}$ too large to divide $∣ G ∣ p^{(N - 1) b}$ , contradicting step 4.1. Therefore $a \leq b$ for every prime $p$ , so $n$ divides $[G : Z (G)]$ .

step 4.1choosealgebra

6.1

Since $n = χ (1)$ , the degree of the irreducible character divides the index of the center.

step 5.1∎

Depends on

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Sources

K. Conrad, Degrees of irreducible representations (standard reference, not scraped)
John Tate's tensor-power argument, as summarized in expository sources (standard reference, not scraped)

The degree of an irreducible complex character divides [G:Z(G)][G:Z(G)]