TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-30

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The degree of an irreducible complex character divides $∣ G ∣$

Statement

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

Facts & Assumptions

Given: A finite group $G$ and an irreducible complex character $χ$ of $G$ .

[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]

The central character satisfies $ω_{χ} (\hat{C}) = ∣ C ∣ χ (g) / χ (1)$ for $g \in C$ (The central character of an irreducible complex character).

[F4]

Central-character values are algebraic integers (The values of a central character are algebraic integers).

[F5]

Character values are algebraic integers, and $χ (g^{- 1}) = \overline{χ (g)}$ (For a complex character, $χ (1) = \dim V$ , $χ$ is a class function, and $∣ χ (g) ∣ \leq χ (1)$ with equality exactly at scalars).

[F6]

Sums and products of algebraic integers are algebraic integers (Integral elements over a nonzero base ring form a subring).

[F7]

A rational algebraic integer is an integer (The rational algebraic integers are exactly the integers).

Proof

technique · direct

1.1

Let $C_{1}, \dots, C_{r}$ be the conjugacy classes of $G$ , and choose $g_{i} \in C_{i}$ . By [F1] and [F2], $1 = ⟨ χ, χ ⟩ = \frac{1}{∣ G ∣} \sum_{i = 1}^{r} ∣ C_{i} ∣ χ (g_{i}) χ (g_{i}^{- 1})$ , where [F5] identifies $\overline{χ (g_{i})}$ with $χ (g_{i}^{- 1})$ and $χ$ is constant on each class.

F1F2F5given

2.1

For each $i$ , [F3] gives $∣ C_{i} ∣ χ (g_{i}) = χ (1) ω_{χ} ({\hat{C}}_{i})$ . Substituting this into step 1.1 yields $∣ G ∣ / χ (1) = \sum_{i = 1}^{r} ω_{χ} ({\hat{C}}_{i}) χ (g_{i}^{- 1})$ .

F3step 1.1algebra

3.1

By [F4] and [F5], each factor $ω_{χ} ({\hat{C}}_{i})$ and $χ (g_{i}^{- 1})$ is an algebraic integer; by [F6], every product and the whole finite sum in step 2.1 are algebraic integers. Therefore $∣ G ∣ / χ (1)$ is a rational algebraic integer, so [F7] makes it an integer.

F4F5F6F7step 2.1algebra

4.1

Since $∣ G ∣ / χ (1) \in Z$ , the degree $χ (1)$ divides $∣ G ∣$ .

step 3.1∎

Depends on

Used by

Dependency tree · two levels

26 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Peter Webb, A Course in Finite Group Representation Theory, Theorem 3.5.4 (standard reference, not scraped)
Anupam Singh, Representation Theory of Finite Groups, Theorem 15.7 (standard reference, not scraped)

The degree of an irreducible complex character divides ∣G∣∣G∣