TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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The group ring $R [G]$ is a unital $R$ -algebra with basis $G$ , and each $g \in G$ is a unit of $R [G]$

Statement

Let $R$ be a commutative ring and let $G$ be a group. Write $[g] \in R [G]$ for the basis vector of The group ring $R [G]$ of finitely supported formal $R$ -linear combinations of group elements indexed by $g \in G$ .

There is a unique $R$ -bilinear multiplication on $R [G]$ satisfying $[g] [h] = [g h] (g, h \in G) .$ With this product, $R [G]$ is a unital $R$ -algebra whose underlying $R$ -module has basis ${[g] : g \in G}$ . Its identity is $[e]$ , where $e$ is the identity of $G$ , and every basis element $[g]$ is a unit with inverse $[g^{- 1}]$ .

Facts & Assumptions

Given: A commutative ring $R$ and a group $G$ with identity $e$ .

[L1]

The module $R [G]$ is free on the set $G$ , with basis vectors $[g]$ , and every element has a unique finite expansion $\sum_{g \in F} r_{g} [g]$ (The group ring $R [G]$ of finitely supported formal $R$ -linear combinations of group elements).

[L2]

Every set map from a set $X$ to a left $R$ -module $M$ extends uniquely to an $R$ -module homomorphism from the free module $R^{(X)}$ (Universal property of the free module on a set).

[L3]

An $R$ -algebra is a unital ring equipped with a central unital map $R \to A$ , and the multiplication is $R$ -bilinear (Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

Proof

technique · constructive

1.1L1L2givenconstruct

For each fixed $g \in G$ , the set map $u_{g} : G \to R [G]$ , $u_{g} (h) : = [g h]$ , extends uniquely by [L2] to an $R$ -linear map $λ_{g} : R [G] \to R [G]$ with $λ_{g} ([h]) = [g h]$ for every $h \in G$ .

2.1step 1.1L1L2givenconstruct

For each fixed $y \in R [G]$ , the set map $v_{y} : G \to R [G]$ , $v_{y} (g) : = λ_{g} (y)$ , extends uniquely by [L2] to an $R$ -linear map $m_{y} : R [G] \to R [G]$ . Define the product by $x y : = m_{y} (x)$ .

3.1step 1.1step 2.1L1

By construction, $[g] [h] = m_{[h]} ([g]) = λ_{g} ([h]) = [g h]$ . The map $x \mapsto x y$ is $R$ -linear because $m_{y}$ is, and if $x = \sum_{g \in F} r_{g} [g]$ then $x y = \sum_{g \in F} r_{g} λ_{g} (y)$ , so the $R$ -linearity of each $λ_{g}$ makes $y \mapsto x y$ linear as well. Thus the product is $R$ -bilinear.

4.1step 3.1L1givenalgebra

For fixed $y, z \in R [G]$ , the maps $x \mapsto (x y) z$ and $x \mapsto x (y z)$ are $R$ -linear by step 3.1, so it is enough to compare them on basis elements $[g]$ . For fixed $g, z$ the maps $y \mapsto ([g] y) z$ and $y \mapsto [g] (y z)$ are likewise linear, so it is enough to compare them on basis elements $[h]$ . Repeating once more in the variable $z$ reduces associativity to basis triples, where $([g] [h]) [k] = [(g h) k] = [g (h k)] = [g] ([h] [k])$ by associativity in $G$ . Hence the product on $R [G]$ is associative.

4.2step 3.1L1givenalgebra

The same bilinear reduction shows that $[e]$ is a two-sided identity, because $[e] [g] = [e g] = [g]$ and $[g] [e] = [g e] = [g]$ for every basis element. Likewise $[g] [g^{- 1}] = [e] = [g^{- 1}] [g]$ , so each $[g]$ is a unit with inverse $[g^{- 1}]$ .

5.1step 3.1step 4.2L1givenalgebra

The map $η : R \to R [G]$ defined by $η (r) = r [e]$ is additive and satisfies $η (r s) = (r [e]) (s [e]) = r s [e]$ and $η (1_{R}) = 1_{R} [e] = [e]$ by steps 3.1 and 4.2. For every basis element $[g]$ , one has $η (r) [g] = r [g] = [g] η (r)$ ; bilinearity extends this equality to every element of $R [G]$ .

6.1step 3.1step 4.1step 4.2step 5.1L3discharge-construct∎

If $*$ is any other $R$ -bilinear product with $[g] * [h] = [g h]$ , then for $x = \sum_{g \in F} r_{g} [g]$ and $y = \sum_{h \in E} s_{h} [h]$ bilinearity forces $x * y = \sum_{g \in F} \sum_{h \in E} r_{g} s_{h} [g h]$ , which is exactly the product already constructed in steps 1.1-3.1. Therefore the multiplication is unique, and with steps 4.1-5.1 it makes $R [G]$ a unital $R$ -algebra as in [L3].

Depends on

Dependency tree · 2 levels

Sources

Peter Webb, A Course in Finite Group Representation Theory, Chapter 1 Section 1.1

The group ring R[G]R[G] is a unital RR-algebra with basis GG, and each g∈Gg∈G is a unit of R[G]R[G]