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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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✓ 14 results · all verified · 10 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 4 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

The Group Algebra and Representations of Finite Groups

1 · Prerequisites

2 · Summary

This page builds the finite-group representation language from the group algebra upward. It starts at the arbitrary commutative-ring seam: the free module R[G], the augmentation map, and the dictionary between R-linear G-actions and left R[G]-modules. That bridge is load-bearing for later pages, so the representation and module viewpoints are identified explicitly rather than mixed informally.

Once the dictionary is in place, the page turns to finite-dimensional representations over fields: subrepresentations, intertwiners, the regular and permutation representations, Schur's lemma, splitting fields, and the abelian case. The endpoint is the classification of equivalence classes of degree-one representations by homomorphisms into k×, with the abelianisation factor visible at the statement level.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

The group ring R[G] of finitely supported formal R-linear combinations of group elements

Definition

Let R be a commutative ring (Commutative ring) and let G be a group (Group and abelian group). The group ring R[G] is the free left R-module on the set G (The free module on a set and its standard basis).

For each g∈G, write [g]∈R[G] for the standard basis vector indexed by g. Thus every element of R[G] has a unique expression ∑g∈Frg[g] with F⊆G finite and rg∈R.

The notation is deliberately that of formal finite sums: at this stage R[G] is only the underlying free R-module with its distinguished basis. The multiplication satisfying [g][h]=[gh] and the resulting R-algebra structure are constructed in The group ring R[G] is a unital R-algebra with basis G, and each g∈G is a unit of R[G].

Remarks

  • The support is finite by definition because R[G] is the direct sum, not the full product, of copies of R indexed by G.

  • The basis vectors are written [g] to keep the group element g distinct from its basis image in the free module.

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

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

Statement

Let R be a commutative ring and let G be a group. Write [g]∈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∈G.

There is a unique R-bilinear multiplication on R[G] satisfying [g][h]=[gh](g,h∈G). With this product, R[G] is a unital R-algebra whose underlying R-module has basis {[g]:g∈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 ∑g∈Frg[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→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∈G, the set map ug:G→R[G], ug(h):=[gh], extends uniquely by [L2] to an R-linear map λg:R[G]→R[G] with λg([h])=[gh] for every h∈G.

2.1step 1.1L1L2givenconstruct

For each fixed y∈R[G], the set map vy:G→R[G], vy(g):=λg(y), extends uniquely by [L2] to an R-linear map my:R[G]→R[G]. Define the product by xy:=my(x).

3.1step 1.1step 2.1L1

By construction, [g][h]=m[h]([g])=λg([h])=[gh]. The map x↦xy is R-linear because my is, and if x=∑g∈Frg[g] then xy=∑g∈Frgλg(y), so the R-linearity of each λg makes y↦xy linear as well. Thus the product is R-bilinear.

4.1step 3.1L1givenalgebra

For fixed y,z∈R[G], the maps x↦(xy)z and x↦x(yz) 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↦([g]y)z and y↦[g](yz) 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]=[(gh)k]=[g(hk)]=[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]=[eg]=[g] and [g][e]=[ge]=[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→R[G] defined by η(r)=r[e] is additive and satisfies η(rs)=(r[e])(s[e])=rs[e] and η(1R)=1R[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]=[gh], then for x=∑g∈Frg[g] and y=∑h∈Esh[h] bilinearity forces x∗y=∑g∈F∑h∈Ergsh[gh], 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].

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

The augmentation map ε:R[G]→R and the augmentation ideal IG=ker⁡ε

Definition

Let R be a commutative ring and let G be a group. For an element of R[G] written uniquely as a finite sum x=∑g∈Frg[g] (The group ring R[G] of finitely supported formal R-linear combinations of group elements), define the augmentation map ε:R[G]⟶R,ε(x):=∑g∈Frg.

Because the expansion in the basis {[g]} is unique, this is a well-defined R-linear map. It is also a ring homomorphism in the sense of Ring homomorphism: additive, multiplicative, and required to send 1 to 1: additivity is immediate, it sends [e] to 1R, and if x=∑grg[g] and y=∑hsh[h], then ε(xy)=∑g,hrgsh=(∑grg)(∑hsh)=ε(x)ε(y) by the multiplication formula supplied by The group ring R[G] is a unital R-algebra with basis G, and each g∈G is a unit of R[G].

The augmentation ideal of R[G] is the kernel IG:=ker⁡ε, which is a two-sided ideal by The kernel of a ring homomorphism is a two-sided ideal.

Remarks

  • The basis element [g] always satisfies ε([g])=1R.

  • When G is finite, the element ∑g∈G[g] is the image under the basis sum of the constant function 1 on G.

CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

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∣.

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

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]=[gh], 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∈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
1.1L1given

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

1.2L1given

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

2.1step 1.1step 1.2∎

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

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

An R-linear action of G on a left R-module, and a G-module over R

Definition

Let R be a commutative ring (Commutative ring), let G be a group (Group and abelian group), and let M be a left R-module (Unital left and right modules over a ring; unqualified module means left module).

A left G-action on M by R-linear maps is a left group action G×M→M, (g,m)↦g⋅m, in the sense of Left group actions, transitive actions, and faithful actions, such that for every g∈G the map m⟼g⋅m is an R-module homomorphism M→M (Module homomorphism and isomorphism, kernel, image and cokernel).

Equivalently, the action satisfies g⋅(m+m′)=g⋅m+g⋅m′,g⋅(rm)=r(g⋅m) for all g∈G, m,m′∈M, and r∈R.

An R-linear G-module, or simply a G-module over R, is a left R-module together with such an action.

Remarks

  • Because g−1 acts as an inverse in the group-action sense, each map m↦g⋅m is automatically an R-module automorphism.

  • When R is a field, this is the module-language form of a representation.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-28Open item page →

A finite-dimensional representation ρ:G→GL⁡(V) over a field, and its degree

Definition

Let k be a field (Field) and let G be a group. A finite-dimensional representation of G over k is a finite-dimensional k-vector space V (Vector space over a field, Finite-dimensional vector space, and its dimension dim⁡FV; infinite-dimensional means having no finite basis) together with a group homomorphism ρ:G⟶GL⁡(V), where GL⁡(V) denotes the group of invertible k-linear maps V→V (Invertible linear maps, linear isomorphisms, and inverse linear maps, Monoid homomorphism and group homomorphism).

The associated action is g⋅v:=ρ(g)(v). This makes V into a G-module over k in the sense of An R-linear action of G on a left R-module, and a G-module over R.

The degree of the representation is the dimension of its underlying vector space: deg⁡(ρ):=dim⁡kV.

Remarks

  • The finite-dimensionality convention is part of the definition on this page, not a later standing assumption.

  • The representation is determined equally well by the homomorphism ρ or by the associated k-linear action of G on V.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-28Open item page →

Subrepresentations, direct sums of representations, and irreducibility

Definition

Let ρ:G→GL⁡(V) be a finite-dimensional representation of G over a field k (A finite-dimensional representation ρ:G→GL⁡(V) over a field, and its degree).

A subrepresentation is a linear subspace W≤V (Linear subspace of a vector space) such that ρ(g)(W)⊆W for every g∈G. Equivalently, g⋅w∈W for all g∈G and w∈W.

If σ:G→GL⁡(W) is another representation, the direct sum representation ρ⊕σ on V⊕W is defined by g⋅(v,w):=(ρ(g)v,σ(g)w).

The representation ρ is irreducible when V≠0 and its only subrepresentations are 0 and V itself.

Remarks

  • The zero representation on the zero vector space is not irreducible, by the explicit nonzero requirement.

  • A decomposition V=U⊕W into nonzero subrepresentations is exactly a nontrivial decomposition of the representation as a direct sum.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-28Open item page →

Intertwiners, the spaces Hom⁡G(V,W) and End⁡G(V), equivalent representations, and faithful representations

Definition

Let ρ:G→GL⁡(V) and σ:G→GL⁡(W) be representations of G over the same field k.

A linear map f:V→W (Linear map between vector spaces over the same field) is an intertwiner, or G-equivariant map, when f(ρ(g)v)=σ(g)f(v)(g∈G, v∈V). Equivalently, fρ(g)=σ(g)f(g∈G).

Write Hom⁡G(V,W) for the set of all intertwiners. It is a subset of the vector space L(V,W) of all linear maps (The space L(V,W) of linear maps with pointwise addition and scalar multiplication). When V=W, write End⁡G(V):=Hom⁡G(V,V).

The two representations are equivalent if there is an invertible intertwiner V→W.

A representation ρ is faithful when ρ(g)=id⁡V implies g=e.

Remarks

  • Faithfulness says that the action remembers every group element.

  • The equality fρ(g)=σ(g)f is the representation-language form of a module homomorphism condition.

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

For a commutative ring R, R-linear G-actions are exactly the compatible left R[G]-module structures

Statement

Let R be a commutative ring, let G be a group, and let M be a left R-module. Then:

  1. every R-linear action of G on M extends uniquely to a left R[G]-module structure on M satisfying (r[e])⋅m=rm(r∈R, m∈M);
  2. every left R[G]-module structure on M satisfying (r[e])⋅m=rm(r∈R, m∈M) restricts to an R-linear action of G on M;
  3. these two constructions are inverse to each other.

Moreover, if M and N carry the corresponding structures, then an R-linear map f:M→N is G-equivariant if and only if it is an R[G]-module homomorphism.

Facts & Assumptions

Given: A commutative ring R, a group G, and left R-modules M and N.

[L1]

The group ring R[G] has basis vectors [g], multiplication [g][h]=[gh], identity [e], and central scalar copy r↦r[e] (The group ring R[G] is a unital R-algebra with basis G, and each g∈G is a unit of R[G], The group ring R[G] of finitely supported formal R-linear combinations of group elements).

[L2]

An R-linear G-module is a left R-module together with a group action whose each g-operator is R-linear (An R-linear action of G on a left R-module, and a G-module over R).

[L3]

Every set map from G to a left R-module extends uniquely to an R-module homomorphism from R[G] (Universal property of the free module on a set).

[L4]

An R[G]-module homomorphism is a map compatible with addition and with the scalar action of every element of R[G] (Module homomorphism and isomorphism, kernel, image and cokernel, Unital left and right modules over a ring; unqualified module means left module).

Proof

technique · constructive
1.1L1L2L3givenconstruct

Assume first that M is an R-linear G-module. For each m∈M, the map um:G→M, um(g):=g⋅m, extends uniquely by [L3] to an R-linear map u~m:R[G]→M. Define a⋅m:=u~m(a) for a∈R[G].

2.1step 1.1L1L2

For basis elements, [g]⋅m=g⋅m by construction. If a=∑g∈Frg[g], then a⋅m=∑g∈Frg(g⋅m), so the action is additive in a and, because each g-operator is R-linear by [L2], also additive in m and compatible with the scalar action of R on M.

3.1step 2.1L1L2givenalgebra

For g,h∈G, one has ([g][h])⋅m=[gh]⋅m=(gh)⋅m=g⋅(h⋅m)=[g]⋅([h]⋅m). Since both sides are R-bilinear in the two group-ring variables, the equality extends to (ab)⋅m=a⋅(b⋅m) for all a,b∈R[G]. Also [e]⋅m=e⋅m=m, so the identity of R[G] acts as the identity on M, and (r[e])⋅m=r(e⋅m)=rm. Thus the construction of steps 1.1-2.1 gives a compatible left R[G]-module structure on M.

4.1step 3.1L1L2L4givenalgebra

Conversely, assume M is a left R[G]-module compatible with the given R-module structure, meaning that (r[e])⋅m=rm for all r∈R and m∈M. Define g⋅m:=[g]⋅m. Then e⋅m=[e]⋅m=m, and (gh)⋅m=[gh]⋅m=([g][h])⋅m=[g]⋅([h]⋅m)=g⋅(h⋅m), so this is a group action. For r∈R, one has g⋅(rm)=[g]⋅((r[e])⋅m)=(([g](r[e]))⋅m)=(((r[e])[g])⋅m)=(r[e])⋅([g]⋅m)=r(g⋅m). Thus the restricted action is R-linear.

5.1step 2.1step 4.1L1L4givenalgebra

Let f:M→N be R-linear. If f is an R[G]-module homomorphism, then f(g⋅m)=f([g]⋅m)=[g]⋅f(m)=g⋅f(m), so f is G-equivariant. Conversely, if f is G-equivariant and a=∑g∈Frg[g], then f(a⋅m)=∑g∈Frgf(g⋅m)=∑g∈Frg(g⋅f(m))=a⋅f(m), so f is an R[G]-module homomorphism.

6.1step 2.1step 4.1step 5.1L1discharge-construct∎

In the first construction, the recovered action of a basis element [g] is the original action of g by step 2.1, so the recovered G-action is unchanged. In the second construction, the recovered R[G]-action agrees with the original one on each basis element [g], and it has the same scalar action because the compatibility condition fixes (r[e])⋅m=rm. Therefore the two constructions are inverse.

CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-28Open item page →

Under the dictionary, subrepresentations are exactly submodules and irreducible representations are exactly simple modules

Statement

Let k be a field, let G be a group, and let V be a finite-dimensional representation of G over k. Under the correspondence of For a commutative ring R, R-linear G-actions are exactly the compatible left R[G]-module structures:

  1. the subrepresentations of V are exactly the k[G]-submodules of V;
  2. V is irreducible if and only if it is simple as a k[G]-module.

Facts & Assumptions

Given: A finite-dimensional representation V of G over k.

[L1]

A subrepresentation is a linear subspace stable under every group element, and irreducible means nonzero with no proper nonzero subrepresentation (Subrepresentations, direct sums of representations, and irreducibility).

[L2]

A G-action on a k-vector space is the same thing as a left k[G]-module structure, and G-equivariant maps are exactly k[G]-module homomorphisms (For a commutative ring R, R-linear G-actions are exactly the compatible left R[G]-module structures).

[L3]

A simple module is a nonzero module whose only submodules are 0 and the whole module (Simple module: a nonzero module with no proper nonzero submodule).

Proof

technique · direct
1.1L1L2given

A linear subspace W≤V is stable under every g∈G if and only if it is stable under every basis element [g] of k[G] under the action of [L2]. Because the action of an arbitrary a=∑grg[g] is the corresponding k-linear combination of the actions of the basis elements, this is equivalent to stability under every element of k[G]. So the subrepresentations of V are exactly the k[G]-submodules.

2.1step 1.1L1L3∎

By [L1], irreducibility means that V is nonzero and has no proper nonzero subrepresentation. By step 1.1 those are exactly the proper nonzero k[G]-submodules, and [L3] is the same condition in module language. Therefore V is irreducible if and only if it is simple as a k[G]-module.

PropositionStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Hom⁡G(V,W) is a k-vector space and End⁡G(V) is a k-algebra

Statement

Let V and W be representations of a group G over a field k.

  1. The intertwiner space Hom⁡G(V,W) is a k-vector space.
  2. The endomorphism space End⁡G(V) is a k-algebra.

Facts & Assumptions

Given: Representations ρ:G→GL⁡(V) and σ:G→GL⁡(W) over a field k.

[L1]

An intertwiner f:V→W satisfies fρ(g)=σ(g)f for every g∈G, and End⁡G(V)=Hom⁡G(V,V) (Intertwiners, the spaces Hom⁡G(V,W) and End⁡G(V), equivalent representations, and faithful representations).

[L2]

The space of all linear maps L(V,W) is a k-vector space (L(V,W) is a vector space over the common scalar field).

[L4]

A k-algebra is a unital ring whose multiplication is k-bilinear and whose scalar copy of k is central (Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

Proof

technique · direct
1.1L1L2givenalgebra

The zero map satisfies 0ρ(g)=0=σ(g)0 for every g∈G. If f,h∈Hom⁡G(V,W) and λ∈k, then (f+h)ρ(g)=fρ(g)+hρ(g)=σ(g)f+σ(g)h=σ(g)(f+h) and (λf)ρ(g)=λ(fρ(g))=λ(σ(g)f)=σ(g)(λf). So Hom⁡G(V,W) contains 0 and is closed under the pointwise operations of L(V,W).

1.2L1L3givenalgebra

When V=W, the identity map satisfies id⁡Vρ(g)=ρ(g)=ρ(g)id⁡V, and if f,h∈End⁡G(V) then (fh)ρ(g)=f(hρ(g))=f(ρ(g)h)=ρ(g)fh. Thus End⁡G(V) is closed under composition and contains the identity.

2.1step 1.1L2

Step 1.1 exhibits Hom⁡G(V,W) as a linear subspace of the vector space L(V,W) of [L2]. Therefore Hom⁡G(V,W) is itself a k-vector space.

3.1step 2.1step 1.2L3L4∎

By step 2.1, End⁡G(V) is a k-vector space. By step 1.2 and [L3], it is also a unital subring of End⁡k(V). For λ∈k and f,h∈End⁡G(V), the usual identities (λf)h=λ(fh)=f(λh) show that multiplication is k-bilinear and the scalar copy of k is central. Therefore [L4] makes End⁡G(V) a k-algebra.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

The trivial representation, the regular representation, and permutation representations from finite G-sets

Definition

Let k be a field and let G be a finite group.

The trivial representation of G over k is the one-dimensional representation on the vector space k in which every g∈G acts as the identity map.

The regular representation of G over k is the representation on the vector space k[G] (The group ring R[G] of finitely supported formal R-linear combinations of group elements) given by left multiplication by the basis units: g⋅x:=[g]x(g∈G, x∈k[G]). Equivalently, on the basis vectors of k[G], g⋅[h]=[gh]. This uses the ring structure and unit property proved in The group ring R[G] is a unital R-algebra with basis G, and each g∈G is a unit of R[G].

More generally, if X is a finite left G-set (Left group actions, transitive actions, and faithful actions), the free k-module k(X) on X (The free module on a set and its standard basis) becomes a representation by g⋅ex:=eg⋅x(g∈G, x∈X). This is the permutation representation attached to the action of G on X.

Remarks

  • The regular representation is the permutation representation of G acting on itself by left translation.

  • Because G and X are finite, these constructions are finite-dimensional.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

The sign representation of Sn and the restriction Res⁡HG(V) of a representation to a subgroup

Definition

Let k be a field.

For the symmetric group Sn (The symmetric group Sym⁡(X): the bijections of a set X under composition), the sign representation over k is the one-dimensional representation on k in which σ⋅a:=sgn⁡(σ)a. This is a representation because sgn⁡:Sn→{±1} is a group homomorphism (The sign is a homomorphism Sn→{+1,−1}, surjective exactly when n≥2).

Now let ρ:G→GL⁡(V) be a representation of a group G over k, and let H≤G be a subgroup (Subgroup). The restriction of ρ to H is the representation Res⁡HG(V):=ρ∣H:H⟶GL⁡(V),h⟼ρ(h). It has the same underlying vector space V and only forgets the action of the elements outside H.

Remarks

  • In characteristic 2, the sign representation coincides with the trivial representation because −1=1 in the field.

  • Restriction changes the acting group, not the underlying vector space.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28Open item page →

The regular representation is faithful

Statement

Let G be a finite group and let k be a field. Then the regular representation of G over k is faithful.

Facts & Assumptions

Given: A finite group G and its regular representation on k[G].

[L1]

In the regular representation, g⋅[h]=[gh] for all g,h∈G (The trivial representation, the regular representation, and permutation representations from finite G-sets).

[L2]

A representation is faithful when the only group element acting as the identity linear map is the identity element of the group (Intertwiners, the spaces Hom⁡G(V,W) and End⁡G(V), equivalent representations, and faithful representations).

Proof

technique · direct
1.1L1L2given

Suppose g∈G acts as the identity in the regular representation. Applying that operator to the basis vector [e] gives [g]=g⋅[e]=[e] by [L1].

2.1step 1.1L2∎

The basis vectors of k[G] are indexed by the elements of G, so [g]=[e] forces g=e. By [L2], this is exactly faithfulness.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Every irreducible representation of a finite group is a quotient of the regular representation

Statement

Let G be a finite group and let V be an irreducible representation of G over a field k. Then there is a surjective morphism from the regular representation of G over k onto V.

Facts & Assumptions

Given: A finite group G, a field k, and an irreducible representation V of G over k.

[L1]

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

[L2]

The regular representation is the action of G on k[G] by left multiplication on the basis vectors [g] (The trivial representation, the regular representation, and permutation representations from finite G-sets).

Proof

technique · direct
1.1L1given

By [L1], the representation V is a simple k[G]-module.

2.1step 1.1L1givenchoose

Choose 0≠v∈V. The submodule k[G]v is nonzero, so simplicity from step 1.1 forces k[G]v=V.

3.1step 2.1L3givenalgebra

Define ϕ:k[G]→V by ϕ(a):=a⋅v. Then ϕ is a k[G]-module homomorphism, and its image is exactly k[G]v=V by step 2.1. So ϕ is surjective.

4.1step 3.1L2L3∎

By [L2] and [L3], the map ϕ is a surjective morphism from the regular representation onto V. Hence V is a quotient of the regular representation.

CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

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

Statement

Let G be a finite group and let V be an irreducible representation of G over a field k. Then deg⁡(V)≤∣G∣.

Facts & Assumptions

Given: A finite group G, a field k, and an irreducible representation V of G over k.

[L1]

The group algebra k[G] has dimension ∣G∣ over k (If G is finite then dim⁡kk[G]=∣G∣).

[L2]

The representation V is a quotient of the regular representation (Every irreducible representation of a finite group is a quotient of the regular representation).

Proof

technique · direct
1.1L2given

By [L2], there is a surjective linear map q:k[G]→V.

2.1step 1.1L1

The images under q of the ∣G∣ basis vectors of k[G] span V, so V has a spanning set with ∣G∣ elements by [L1].

3.1step 2.1L3L4∎

Let B be a basis of V. Then B is linearly independent, and step 2.1 together with [L3] shows that ∣B∣≤∣G∣. By [L4], ∣B∣=deg⁡(V). Therefore deg⁡(V)≤∣G∣.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

A splitting field for a finite group: every irreducible representation has scalar endomorphism ring

Definition

Let G be a finite group and let k be a field. The field k is a splitting field for G when every irreducible representation V of G over k satisfies End⁡G(V)=k, meaning that every G-endomorphism of V is a scalar operator λid⁡V with λ∈k (A finite-dimensional representation ρ:G→GL⁡(V) over a field, and its degree, Intertwiners, the spaces Hom⁡G(V,W) and End⁡G(V), equivalent representations, and faithful representations).

Remarks

  • This is the representation-theoretic condition used on the next pages. It is weaker than algebraic closedness, and Brauer's theorem records one sufficient roots-of-unity criterion for it without making that criterion the definition.

  • The definition quantifies only over irreducible representations of the fixed finite group G.

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28Open item page →

Schur's lemma for irreducible representations: a nonzero intertwiner is an isomorphism, and End⁡G(V) is a division ring

Statement

Let V and W be irreducible representations of a group G over a field k. Then every nonzero intertwiner f:V→W is an isomorphism. Consequently End⁡G(V) is a division ring.

Facts & Assumptions

Given: Irreducible representations V and W of G over a field k.

[L3]

A nonzero homomorphism between simple modules is an isomorphism, and the endomorphism ring of a simple module is a division ring (Schur's lemma for simple modules).

Proof

technique · direct
1.1L1L2given

By [L1], the irreducible representations V and W are simple k[G]-modules, and by [L2] a nonzero intertwiner f:V→W is a nonzero k[G]-module homomorphism between them.

2.1step 1.1L3∎

Applying [L3] to that module homomorphism shows that f is an isomorphism. Applying [L3] with V=W shows that End⁡G(V) is a division ring.

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Over a splitting field, every G-endomorphism of an irreducible representation is scalar

Statement

Let G be a finite group, let k be a splitting field for G, and let V be an irreducible representation of G over k. Then every endomorphism in End⁡G(V) has the form λid⁡V with λ∈k.

Facts & Assumptions

Given: A finite group G, a splitting field k for G, and an irreducible representation V of G over k.

[L2]

By definition, a splitting field for G is a field over which every irreducible representation has endomorphism ring exactly k (A splitting field for a finite group: every irreducible representation has scalar endomorphism ring).

Proof

technique · direct
1.1L2given

By [L2], the endomorphism ring End⁡G(V) is exactly the scalar copy of k inside End⁡k(V).

2.1step 1.1L1∎

Therefore every G-endomorphism of V is λid⁡V for a unique λ∈k. Step 1.1 is compatible with the division-ring conclusion of [L1], which is why the scalar copy is a field.

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28Open item page →

Over an algebraically closed field, every endomorphism of an irreducible representation is scalar

Statement

Let k be an algebraically closed field, let G be a group, and let V be an irreducible representation of G over k. Then every T∈End⁡G(V) is of the form T=λid⁡V for some λ∈k.

Facts & Assumptions

Given: An algebraically closed field k, a group G, an irreducible representation V of G over k, and an endomorphism T∈End⁡G(V).

[L1]

Every finite-dimensional endomorphism over an algebraically closed field is triangularisable (Every finite-dimensional endomorphism over an algebraically closed field is triangularisable).

[L2]

For an irreducible representation, every nonzero G-endomorphism is an isomorphism and End⁡G(V) is a division ring (Schur's lemma for irreducible representations: a nonzero intertwiner is an isomorphism, and End⁡G(V) is a division ring).

Proof

technique · direct
1.1L1given

The representation V is nonzero because it is irreducible, so [L1] applies to T and yields a basis in which T is upper triangular. Let λ be one diagonal entry. Then T−λid⁡V is upper triangular with a zero diagonal entry, hence is not invertible.

2.1step 1.1L2givenalgebra

For every g∈G, the identity map commutes with the action of g, so T−λid⁡V is still a G-endomorphism. By [L2], a nonzero G-endomorphism of an irreducible representation must be invertible. Therefore the noninvertible endomorphism T−λid⁡V is zero.

3.1step 2.1∎

Hence T=λid⁡V, as required.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-28Open item page →

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

Statement

Let G be a finite abelian group and let k be a splitting field for G. Then every irreducible representation of G over k has degree 1.

Facts & Assumptions

Given: A finite abelian group G, a splitting field k for G, and an irreducible representation ρ:G→GL⁡(V) over k.

[L1]

Over a splitting field, every endomorphism of an irreducible representation is scalar (Over a splitting field, every G-endomorphism of an irreducible representation is scalar).

[L2]

A representation is irreducible exactly when every nonzero invariant subspace is the whole space (Subrepresentations, direct sums of representations, and irreducibility).

Proof

technique · direct
1.1L1L2givenalgebra

Because G is abelian, for every g,h∈G one has ρ(g)ρ(h)=ρ(gh)=ρ(hg)=ρ(h)ρ(g). So each operator ρ(g) commutes with every ρ(h) and therefore lies in End⁡G(V).

2.1step 1.1L1L2given

By [L1], each ρ(g) is a scalar operator. Hence for every nonzero v∈V, the line kv is stable under every ρ(g) and is therefore a nonzero subrepresentation.

3.1step 2.1L2∎

Irreducibility from [L2] forces that nonzero line to be all of V. Therefore V is one-dimensional.

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

Equivalence classes of degree-one representations are exactly homomorphisms G→k×; equivalently they factor through G/G′, and they form an abelian group

Statement

Let k be a field and let G be a group.

  1. choosing a basis of a degree-one representation of G over k produces a homomorphism χ:G→k×, and every such homomorphism produces a normalized degree-one representation on the one-dimensional space k;
  2. two degree-one representations are equivalent if and only if they produce the same homomorphism, so equivalence classes of degree-one representations are in bijection with the homomorphisms G→k×;
  3. equivalently, those equivalence classes are in bijection with the homomorphisms that factor through the abelianisation quotient G/G′;
  4. under pointwise multiplication and inversion of the corresponding homomorphisms, these equivalence classes form an abelian group.

Facts & Assumptions

Given: A field k and a group G.

[L1]

A degree-one representation is a finite-dimensional representation whose underlying vector space has dimension 1 (A finite-dimensional representation ρ:G→GL⁡(V) over a field, and its degree).

[L3]

Every homomorphism from G to an abelian group factors uniquely through the quotient G/G′ (The derived subgroup is characteristic and the abelianization is universal, Monoid homomorphism and group homomorphism).

[L4]

Two representations are equivalent when there is an invertible intertwiner between them (Intertwiners, the spaces Hom⁡G(V,W) and End⁡G(V), equivalent representations, and faithful representations).

Proof

technique · constructive
1.1L1L2givenconstruct

Let ρ:G→GL⁡(V) be a degree-one representation. By [L1], choose a basis vector v of the one-dimensional space V. For each g∈G there is a unique scalar χ(g)∈k with ρ(g)v=χ(g)v, and because ρ(g) is invertible, that scalar lies in k×.

2.1step 1.1L2givenalgebra

The representation law gives χ(gh)v=ρ(gh)v=ρ(g)ρ(h)v=χ(g)χ(h)v, so χ(gh)=χ(g)χ(h). Thus χ:G→k× is a group homomorphism. Under the identification V≅k sending v to 1, the representation becomes the normalized action g⋅a=χ(g)a on the one-dimensional space k.

3.1step 2.1L2givenconstruct

Conversely, if χ:G→k× is a group homomorphism, define ρχ(g):k→k by ρχ(g)(a)=χ(g)a. Each ρχ(g) is invertible with inverse multiplication by χ(g)−1, and the homomorphism law for χ gives ρχ(gh)=ρχ(g)ρχ(h). So ρχ is a degree-one representation.

4.1step 2.1step 3.1L4givenalgebra

If two degree-one representations are equivalent, an invertible intertwiner sends a chosen basis vector of one space to a nonzero scalar multiple of a basis vector of the other, so the character extracted in step 2.1 is unchanged; conversely, if two degree-one representations produce the same homomorphism χ, then step 2.1 identifies each of them with the same normalized representation ρχ of step 3.1, so they are equivalent by [L4]. Therefore equivalence classes of degree-one representations are in bijection with homomorphisms G→k×.

5.1step 4.1L2L3discharge-construct∎

Because k× is abelian by [L2], [L3] identifies the same equivalence classes with the homomorphisms factoring through G/G′. If χ,ψ:G→k× are homomorphisms, then (χψ)(g):=χ(g)ψ(g) and χ−1(g):=χ(g)−1 are again homomorphisms. Thus the equivalence classes of degree-one representations form an abelian group under pointwise multiplication and inversion of their corresponding homomorphisms.

5 · Examples, counterexamples and false statements

None yet.

Sources