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

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For a field $k$ , the group algebra $k [G]$ is commutative if and only if $G$ is abelian

Statement

Let $k$ be a field and let $G$ be a group. Then the group algebra $k [G]$ is commutative if and only if $G$ is abelian.

Facts & Assumptions

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

[L1]

The basis vectors $[g]$ of $k [G]$ satisfy $[g] [h] = [g h]$ , and every element of $k [G]$ is a unique finite $k$ -linear combination of them (The group ring $R [G]$ is a unital $R$ -algebra with basis $G$ , and each $g \in G$ is a unit of $R [G]$ , The group ring $R [G]$ of finitely supported formal $R$ -linear combinations of group elements).

Proof

technique · direct

1.1

If $G$ is abelian, then $[g] [h] = [g h] = [h g] = [h] [g]$ for all $g, h \in G$ . Bilinearity of multiplication then makes every two finite $k$ -linear combinations commute, so $k [G]$ is commutative.

L1given

1.2

Conversely, if $k [G]$ is commutative, then for every $g, h \in G$ one has $[g h] = [g] [h] = [h] [g] = [h g]$ . By the uniqueness of the basis expansion in [L1], this forces $g h = h g$ . Hence $G$ is abelian.

L1given

2.1

Steps 1.1 and 1.2 prove the two directions of the equivalence.

step 1.1step 1.2∎

Depends on

Used by

Every irreducible representation of a finite abelian group over a splitting field is one-dimensional Theorem

Dependency tree · two levels

8 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

Pavel Etingof et al., Introduction to Representation Theory, Section 1.1 (standard reference, not scraped)

For a field kk, the group algebra k[G]k[G] is commutative if and only if GG is abelian