TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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If $k$ is algebraically closed and $char k ∤ ∣ G ∣$ , the number of irreducible representations of $G$ equals the number of conjugacy classes

Statement

Let $G$ be a finite group and let $k$ be an algebraically closed field with $char k ∤ ∣ G ∣$ . Then the number of irreducible representations of $G$ over $k$ , up to equivalence, is exactly the number of conjugacy classes of $G$ .

Facts & Assumptions

Given: A finite group $G$ and an algebraically closed field $k$ with $char k ∤ ∣ G ∣$ .

[L1]

Under these hypotheses, $k [G] ≅ \prod_{i = 1}^{m} M_{n_{i}} (k)$ for some positive integers $m, n_{i}$ (If $k$ is algebraically closed and $char k ∤ ∣ G ∣$ , then $k [G] ≅ \prod_{i = 1}^{r} M_{n_{i}} (k)$ ).

[L2]

For such a product ring, there is exactly one simple left module isomorphism class per factor (Simple modules over a product of matrix rings over division rings).

[L3]

Under the dictionary, irreducible representations are exactly simple left $k [G]$ -modules (Under the dictionary, subrepresentations are exactly submodules and irreducible representations are exactly simple modules).

[L4]

The center of $M_{n} (k)$ consists exactly of the scalar matrices (The center of $M_{n} (k)$ consists of the scalar matrices).

[L5]

The dimension of $Z (k [G])$ is the number of conjugacy classes of $G$ (The dimension of $Z (k [G])$ is the number of conjugacy classes of $G$ ).

Proof

technique · direct

1.1

By [L5], the dimension of $Z (k [G])$ is the number of conjugacy classes of $G$ .

L5given

1.2

Write the product decomposition of [L1]. An element of a direct product is central exactly when each coordinate is central in its own factor, so [L4] gives $Z (k [G]) ≅ Z ⁣ (\prod_{i = 1}^{m} M_{n_{i}} (k)) ≅ \prod_{i = 1}^{m} Z (M_{n_{i}} (k)) ≅ k^{m} .$ Hence $\dim_{k} Z (k [G]) = m$ . By [L2], the product ring has exactly $m$ simple left module isomorphism classes, and [L3] translates these exactly into the irreducible representations of $G$ . So the number of irreducible representations is also $m$ .

L1L2L3L4givenalgebra

2.1

Steps 1.1 and 1.2 are the two computations of the same dimension, so the number of irreducible representations equals the number of conjugacy classes.

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

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Sources

Peter Webb, A Course in Finite Group Representation Theory, Theorem 3.4.3 (standard reference, not scraped)
Pavel Etingof et al., Introduction to Representation Theory, Corollary 3.6 (standard reference, not scraped)

If kk is algebraically closed and char⁡k∤∣G∣chark∤∣G∣, the number of irreducible representations of GG equals the number of conjugacy classes