CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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If $G$ is finite then $\dim_{k} k [G] = ∣ G ∣$

Statement

Let $k$ be a field and let $G$ be a finite group. Then the group algebra $k [G]$ is finite-dimensional over $k$ , with $\dim_{k} k [G] = ∣ G ∣ .$

Facts & Assumptions

Given: A field $k$ and a finite group $G$ .

[L1]

The module $k [G]$ is free on the set $G$ , with basis ${[g] : g \in G}$ (The group ring $R [G]$ of finitely supported formal $R$ -linear combinations of group elements).

[L2]

The dimension of a finite-dimensional vector space is the size of any finite basis (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis).

Proof

technique · direct

1.1

By [L1], the set ${[g] : g \in G}$ is a basis of $k [G]$ indexed by the finite set $G$ , so it has exactly $∣ G ∣$ elements.

L1given

2.1

Applying [L2] to that finite basis gives $\dim_{k} k [G] = ∣ G ∣$ .

step 1.1L2∎

Depends on

Used by

Every irreducible representation of a finite group has degree at most |G| Corollary

Dependency tree · two levels

18 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, Chapter 1 Section 1.1 (standard reference, not scraped)

If GG is finite then dim⁡kk[G]=∣G∣dimk​k[G]=∣G∣