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CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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If G is finite then dim⁡kk[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⁡kk[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∈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⁡FV; infinite-dimensional means having no finite basis).

Proof

technique · direct
1.1L1given

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

2.1step 1.1L2∎

Applying [L2] to that finite basis gives dim⁡kk[G]=∣G∣.

Depends on

Used by

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