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✓ 19 results · all verified · 11 also independently AI-judged
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Characters and the Orthogonality Relations

1 · Prerequisites

2 · Summary

This page is ordinary character theory: throughout, G is finite, the base field is C, and every representation is finite-dimensional. The opening item fixes that scope, after which the page assembles the objects a character table records — the character χV(g)=tr⁡ρV(g), the space cf(G) of class functions and its standard Hermitian inner product, irreducible characters, the character table itself, and the tensor product, dual, and kernel constructions attached to a character.

The first structural thread derives the basic value properties of a character: χ(1)=dim⁡V, class-function invariance, the description of χ(g) as a sum of roots of unity, the bound ∣χ(g)∣≤χ(1) with its scalar-equality case, and χ(g−1)=χ(g)‾. These immediately identify the kernel of a character with the kernel of the representation, and they feed the three character operations — characters add on direct sums, multiply on tensor products, and conjugate on duals — together with the fixed-point count of a permutation character.

The central thread is orthogonality. The averaging projector onto the fixed subspace, combined with the identification of intertwiners with fixed points in V∗⊗W, converts ⟨χV,χW⟩ into dim⁡Hom⁡G(W,V). Schur's lemma then gives row orthogonality of irreducible characters, and the published count of irreducibles against conjugacy classes upgrades orthonormality to a basis of cf(G). From that basis flow the multiplicity formula, the fact that a representation is determined by its character, the irreducibility test ⟨χ,χ⟩=1, the regular character and its second proof of the sum-of-squares formula, and the column orthogonality relations with their centralizer and square-table consequences.

The closing thread reads group structure off the table. Representations with kernel containing a normal subgroup factor through the quotient with irreducibility preserved; through the faithful regular representation of G/N this shows normal subgroups are exactly intersections of kernels of irreducible characters. The page ends with the criterion that G is abelian exactly when every irreducible complex character has degree 1.

3 · Logical flowchart

4 · Definitions, theorems and proofs

RemarkRemark: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

Standing hypotheses for ordinary character theory: G finite, k=C, and every representation finite-dimensional

Remark

The ordinary-character-theory items on this page work inside the following setting, fixed once here: G is a finite group, the base field is k=C, and every representation is finite-dimensional (A finite-dimensional representation ρ:G→GL⁡(V) over a field, and its degree). This is ordinary character theory in the sense of Webb, Chapter 3 and Etingof et al., Section 3.3: no infinite group, unitary-representation, or modular-character material is load-bearing in the character-theoretic arguments on the page. The quotient-factorisation result A representation with kernel containing a normal subgroup factors through the quotient, and irreducibility is unchanged by inflation is deliberately stated in the greater generality of an arbitrary group, field, and representation because its proof needs none of these standing restrictions.

The choice of k=C is what makes the hypotheses of the published representation-theory spine available. Since char⁡C=0 and G is finite, the characteristic does not divide ∣G∣, so every finite-dimensional representation is completely reducible (If char⁡k∤∣G∣, every finite-dimensional representation of G is completely reducible), and the field is algebraically closed, which feeds the count of irreducibles against conjugacy classes (If k is algebraically closed and char⁡k∤∣G∣, the number of irreducible representations of G equals the number of conjugacy classes). These hypotheses are restated where they are consumed; the present remark fixes the default scope without narrowing an item that explicitly states broader hypotheses.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-29Open item page →

The character χV(g)=tr⁡(ρV(g)) of a finite-dimensional complex representation

Definition

Let ρ:G→GL⁡(V) be a finite-dimensional complex representation (A finite-dimensional representation ρ:G→GL⁡(V) over a field, and its degree). Its character is the function

χV:G⟶C,χV(g):=tr⁡ ⁣(ρV(g)),

where the trace is the basis-independent trace of an endomorphism (The basis-independent trace of an endomorphism of a finite-dimensional vector space); one writes χρ or χ when the representation or the group is fixed.

The definition is well posed in two senses, recorded here because both are used throughout the page. First, the value does not depend on a choice of basis of V: matrices of one endomorphism in two ordered bases are similar, and similar matrices have equal trace (The basis-independent trace of an endomorphism of a finite-dimensional vector space). Second, equivalent representations have equal characters: if T:V→W is an invertible intertwiner, then ρW(g)=TρV(g)T−1, so tr⁡(ρW(g))=tr⁡(ρV(g)T−1T)=tr⁡(ρV(g)) by the identity tr⁡(AB)=tr⁡(BA) (For A∈Mm×n(F) and B∈Mn×m(F), tr⁡(AB)=tr⁡(BA)). Thus the character depends only on the equivalence class of the representation.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

Class functions and the complex vector space cf(G)

Definition

Let G be a finite group. A function f:G→C is a class function when it is constant on every conjugacy class (The conjugacy class Cl⁡G(x) and centralizer CG(x) of an element):

f(gxg−1)=f(x)for all g,x∈G.

The set of all class functions is written

cf(G):={ f:G→C  :  f(gxg−1)=f(x) for all g,x∈G }.

With pointwise addition (f+h)(g):=f(g)+h(g) and pointwise scalar multiplication (λf)(g):=λf(g), the set cf(G) is a complex vector space (Vector space over a field): the axioms (V1)–(V5) are inherited from the field C at every argument g∈G. Because G is finite, the evaluation functions f↦f(x) are complex-linear, and a class function is determined by its values on one representative of each conjugacy class.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-27Open item page →

The standard inner product on cf(G)

Definition

For two class functions φ,ψ∈cf(G) (Class functions and the complex vector space cf(G)), define

⟨φ,ψ⟩:=1∣G∣∑g∈Gφ(g) ψ(g)‾,

the finite sum being that of (A finite sum in a commutative monoid indexed by an arbitrary finite set) for the additive commutative monoid of C. This is the standard Hermitian form on the complex vector space of class functions.

This assignment is an inner product in the exact sense of the published definition (Real and complex inner product spaces, with the inner product linear in the first argument), with the inner product linear in the first argument. Linearity in the first slot is immediate from the pointwise vector-space structure of cf(G) and linearity of the finite sum. The conjugate-symmetry clause holds because conjugation (Real and imaginary parts, complex conjugation, and modulus) is an involution: ⟨ψ,φ⟩=1∣G∣∑gψ(g)φ(g)‾=1∣G∣∑gφ(g)ψ(g)‾‾=⟨φ,ψ⟩‾. For definiteness, ⟨φ,φ⟩=1∣G∣∑g∣φ(g)∣2 is a sum of nonnegative real numbers, hence nonnegative real; it is 0 exactly when every term ∣φ(g)∣2 is 0, and since ∣z∣=0 holds only for z=0 in C, that says exactly φ=0 as a function.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-29Open item page →

An irreducible complex character

Definition

A complex character χ (The character χV(g)=tr⁡(ρV(g)) of a finite-dimensional complex representation) is irreducible when χ=χV for an irreducible representation V of G over C (Subrepresentations, direct sums of representations, and irreducibility).

The phrase is well defined for characters rather than for individual representations: if V and W are equivalent, then χV=χW by the invariance clause of (The character χV(g)=tr⁡(ρV(g)) of a finite-dimensional complex representation), and equivalence preserves irreducibility. Throughout the page, "the irreducible characters" means one representative χ1,…,χr from each equivalence class of irreducible complex representations of the finite group G; the number r is finite and equals the number of conjugacy classes of G by the published count (If k is algebraically closed and char⁡k∤∣G∣, the number of irreducible representations of G equals the number of conjugacy classes).

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-29Open item page →

The character table of a finite group

Definition

Let G be a finite group, let χ1,…,χr be the irreducible complex characters of G, one per equivalence class (An irreducible complex character), and let g1,…,gs be representatives of the distinct conjugacy classes of G (The conjugacy class Cl⁡G(x) and centralizer CG(x) of an element). The character table of G is the array whose rows are indexed by χ1,…,χr, whose columns are indexed by g1,…,gs, and whose (i,j) entry is χi(gj).

The array is written with a top row recording the conjugacy-class sizes ∣Cl⁡G(gj)∣ above the column labels, and with the first column recording the degrees χi(1) beside the row labels. The ordering of the rows and of the columns is arbitrary: permuting either does not change the table as data. Because each χi is constant on each conjugacy class, the entry χi(gj) is independent of the choice of representative gj. That the number of rows equals the number of columns is a theorem about the table, proved later on this page as The character table is square and invertible, not part of the present definition.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-29Open item page →

The tensor product of two complex representations

Definition

Let ρ:G→GL⁡(V) and σ:G→GL⁡(W) be finite-dimensional complex representations of a group G (A finite-dimensional representation ρ:G→GL⁡(V) over a field, and its degree). Their tensor product representation ρ⊗σ is the representation of G on the complex tensor product V⊗CW given on elementary tensors by

g⋅(v⊗w):=(ρ(g)v)⊗(σ(g)w)(g∈G, v∈V, w∈W).

The action is well defined. For fixed g∈G, the assignment (v,w)↦ρ(g)v⊗σ(g)w is additive in each variable and C-balanced, because ρ(g) and σ(g) are C-linear: ρ(g)(vλ)⊗σ(g)w=(ρ(g)v)λ⊗σ(g)w. By Universal property of the tensor product for balanced maps into abelian groups there is a unique C-linear map ρ(g)⊗σ(g):V⊗CW→V⊗CW with the displayed value on every elementary tensor. Because the two sides of g⋅(h⋅(v⊗w))=(gh)⋅(v⊗w) agree on every elementary tensor and the elementary tensors span V⊗CW, the assignment g↦ρ(g)⊗σ(g) is a group homomorphism into GL⁡(V⊗CW), with ρ(1)⊗σ(1)=id⁡. The underlying space is finite-dimensional: bases of V and W give the basis (vi⊗wj)i,j, so dim⁡(V⊗CW)=(dim⁡V)(dim⁡W).

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-29Open item page →

The dual or contragredient complex representation

Definition

Let ρ:G→GL⁡(V) be a finite-dimensional complex representation of a group G (A finite-dimensional representation ρ:G→GL⁡(V) over a field, and its degree). Its dual or contragredient representation is the representation ρ∗ of G on the algebraic dual space V∗ (Linear functionals and the algebraic dual V∗=L(V,F)) defined by

(g⋅f)(v):=f(g−1⋅v)(g∈G, f∈V∗, v∈V),

equivalently g⋅f=f∘ρ(g−1).

Each operator ρ∗(g) is C-linear because f and ρ(g−1) are. The inverse appears so that composition is order-preserving: writing g⋅f=f∘ρ(g−1),

(gh)⋅f=f∘ρ((gh)−1)=f∘ρ(h−1)∘ρ(g−1)=g⋅(h⋅f),

and 1⋅f=f∘id⁡V=f. Hence ρ∗ is a group homomorphism G→GL⁡(V∗), a representation of the same degree as ρ.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

The kernel of a complex character

Definition

Let ρ:G→GL⁡(V) be a finite-dimensional complex representation of a finite group G, with character χ=χV (The character χV(g)=tr⁡(ρV(g)) of a finite-dimensional complex representation). The kernel of the character is

ker⁡χ:={ g∈G  :  χ(g)=χ(1) }.

The value χ(1) is the reference point of the definition, not a random number: The kernel of a complex character agrees with the kernel of any representation affording it proves that this set equals the kernel ker⁡ρ of the representation, so the two readings of "kernel" never conflict on this page.

PropositionStatement: Literature-sourcedProof: AI-adaptedverified 2026-09-26 (gpt-6-sol)Open item page →

For a complex character, χ(1)=dim⁡V, χ is a class function, and ∣χ(g)∣≤χ(1) with equality exactly at scalars

Statement

Let G be a finite group, let V be a finite-dimensional complex representation of G, and let χ=χV be its character. Then, for all g,h∈G:

  1. χ(1)=dim⁡V;
  2. χ(ghg−1)=χ(h), so χ is a class function;
  3. χ(g) is a sum of dim⁡V roots of unity, namely the eigenvalues of ρ(g) counted with multiplicity;
  4. ∣χ(g)∣≤χ(1), with equality if and only if ρ(g) is a scalar operator;
  5. χ(g−1)=χ(g)‾.

Facts & Assumptions

Given: A finite group G, a finite-dimensional complex representation ρ:G→GL⁡(V) with character χ.

[F1]

The character is χ(g)=tr⁡ρ(g), the trace of the action operator (The character χV(g)=tr⁡(ρV(g)) of a finite-dimensional complex representation).

[F2]

In a finite group, every element g has finite order dividing ∣G∣ (The order of every element of a finite group divides the order of the group).

[A1]

An element of finite order acts diagonalisably on a finite-dimensional space over an algebraically closed field of characteristic zero (Over an algebraically closed field of characteristic 0, every element of finite order acts diagonalisably in a finite-dimensional representation).

[A2]

If the characteristic polynomial of an endomorphism splits as ∏i(x−λi), then its trace is the sum of the eigenvalues ∑iλi (If χT(x)=∏i<n(x−λi) in F[x], then tr⁡(T)=∑i<nλi: trace is the sum of the eigenvalues counted with algebraic multiplicity).

[A3]

tr⁡(AB)=tr⁡(BA) whenever the products are defined (For A∈Mm×n(F) and B∈Mn×m(F), tr⁡(AB)=tr⁡(BA)).

[A4]

Complex conjugation distributes over addition and multiplication, fixes real numbers, and satisfies ∣z‾∣=∣z∣ (Real and imaginary parts, complex conjugation, and modulus).

[A5]

For complex numbers z1,…,zd, the triangle inequality ∣∑izi∣≤∑i∣zi∣ holds, and equality holds exactly when all the nonzero zi share one argument.

[A6]

A function f:G→C is a class function exactly when it is constant on every conjugacy class (Class functions and the complex vector space cf(G)).

Proof

technique · direct
1.1F1givenalgebra

In any ordered basis of V the matrix of ρ(1)=id⁡V is the identity matrix, whose trace is the number of basis vectors. Hence χ(1)=tr⁡(id⁡V)=dim⁡V, which is claim 1.

1.2F1A3given

For the same reason ρ(g) and ρ(h) compose as in the group, so χ(ghg−1)=tr⁡(ρ(g)ρ(h)ρ(g)−1)=tr⁡(ρ(h)ρ(g)−1ρ(g))=tr⁡ρ(h)=χ(h), the middle step being [A3] applied to A=ρ(g) and B=ρ(h)ρ(g)−1.

1.3F2A1given

By [F2], g has finite order n with n∣∣G∣, so ρ(g)n=ρ(gn)=id⁡V. Since C is algebraically closed of characteristic zero, [A1] gives a basis of V in which ρ(g) is diagonal with diagonal entries λ1,…,λd, where d=dim⁡V; each λin=1, so each λi is a root of unity.

2.1step 1.2A6

By step 1.2 the value of χ does not change under conjugation, so χ is constant on each conjugacy class and hence is a class function in the sense of [A6], which is claim 2.

2.2F1A2step 1.3step 1.1

In that basis the characteristic polynomial of ρ(g) is ∏i=1d(x−λi), so [A2] gives χ(g)=tr⁡ρ(g)=∑i=1dλi, a sum of d=dim⁡V roots of unity, which is claim 3.

2.3F1step 1.1step 1.3algebra

Conversely, if ρ(g)=λid⁡V for a scalar λ, first consider the degenerate case V=0. Then χ(g)=0=χ(1), so the equality clause of claim 4 holds. If V≠0, then the identity ρ(g)n=id⁡V of step 1.3 reads λnid⁡V=id⁡V, and evaluating it on a nonzero vector gives λn=1. Thus λ is a root of unity and ∣λ∣=1; then χ(g)=λdim⁡V and ∣χ(g)∣=dim⁡V=χ(1). This closes the biconditional in claim 4.

2.4step 1.3A4algebra

The inverse operator has the inverse eigenvalues, and a root of unity λ satisfies λ−1=λ‾ because ∣λ∣=1=λλ‾.

3.1A5step 2.2step 1.3step 1.1

Applying [A5] to the eigenvalues of step 1.3 gives ∣χ(g)∣=∣∑iλi∣≤∑i∣λi∣=d=dim⁡V=χ(1), which is the inequality in claim 4.

4.1A5step 1.3algebra

If V=0, its only endomorphism is scalar, so the equality case is immediate. If V≠0, equality holds in step 3.1 exactly when the equality clause of [A5] applies: since every ∣λi∣=1, all the eigenvalues share one argument, so all the λi are equal to one root of unity λ. The diagonal form of step 1.3 then shows ρ(g)=λid⁡V, a scalar operator.

5.1A2A4step 2.4step 2.2∎

Hence, using [A2] for ρ(g−1)=ρ(g)−1 and the additivity of conjugation from [A4], χ(g−1)=∑iλi−1=∑iλi‾=∑iλi‾=χ(g)‾, which is claim 5.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

The kernel of a complex character agrees with the kernel of any representation affording it

Statement

Let ρ:G→GL⁡(V) be a finite-dimensional complex representation of a finite group G, with character χ=χV. Then

ker⁡χ=ker⁡ρ,

where ker⁡ρ is the kernel of the group homomorphism ρ and ker⁡χ={g∈G:χ(g)=χ(1)} is the kernel of the character.

Facts & Assumptions

Given: A finite group G and a finite-dimensional complex representation ρ:G→GL⁡(V) with character χ.

[F1]

The kernel of a group homomorphism is the set of elements sent to the identity (The kernel and image of a group homomorphism).

[F2]

The kernel of the character is ker⁡χ={g∈G:χ(g)=χ(1)} (The kernel of a complex character).

[F3]

For the character, χ(1)=dim⁡V, and ∣χ(g)∣=χ(1) holds exactly when ρ(g) is a scalar operator (For a complex character, χ(1)=dim⁡V, χ is a class function, and ∣χ(g)∣≤χ(1) with equality exactly at scalars).

Proof

technique · direct
1.1F1F2F3given

If g∈ker⁡ρ, then ρ(g)=id⁡V by [F1], so χ(g)=tr⁡(id⁡V)=χ(1) by [F3]. Hence g∈ker⁡χ by [F2], which proves ker⁡ρ⊆ker⁡χ.

1.2F2F3given

Conversely, let g∈ker⁡χ, so χ(g)=χ(1) by [F2]. Then in particular ∣χ(g)∣=χ(1), and the equality clause of [F3] gives ρ(g)=λid⁡V for a scalar λ. Evaluating the character at g and at 1 with [F3] gives χ(g)=λχ(1)=χ(1).

1.3F1F2given

If dim⁡V=0, then V={0} is the zero space, ρ(g)=id⁡V holds for every g, and χ≡0, so ker⁡ρ=G=ker⁡χ by [F1] and [F2]; the statement is immediate. Hence assume dim⁡V≥1.

2.1F1F3step 1.2step 1.3algebra∎

Since χ(1)=dim⁡V≥1 by [F3], the equality λχ(1)=χ(1) of step 1.2 forces λ=1; therefore ρ(g)=id⁡V, so g∈ker⁡ρ by [F1]. Together with step 1.1 and step 1.3 this proves ker⁡χ=ker⁡ρ.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

Characters add on direct sums, multiply on tensor products, and conjugate on duals

Statement

Let G be a finite group and let V and W be finite-dimensional complex representations of G. Then, for every g∈G:

  1. χV⊕W(g)=χV(g)+χW(g);
  2. χV⊗W(g)=χV(g) χW(g);
  3. χV∗(g)=χV(g)‾.

Facts & Assumptions

Given: Finite-dimensional complex representations V, W of a finite group G, and an element g∈G.

[F2]

The dual action is (g⋅f)(v)=f(g−1⋅v) (The dual or contragredient complex representation).

[F3]

The tensor-product action is g⋅(v⊗w)=(g⋅v)⊗(g⋅w) (The tensor product of two complex representations).

[A1]

In a basis made of a basis of V followed by a basis of W, the matrix of the direct-sum action on V⊕W is block diagonal with the two action matrices on its diagonal, so its trace is the sum of the two block traces.

[A2]

Trace is linear, in particular additive, on the space of square matrices (Trace is a linear functional on Mn(F)).

[A3]

A basis of V and a basis of W give the basis (vi⊗wj)i,j of the tensor product (The elementary tensors of two bases form the product basis of the tensor product).

Proof

technique · direct
1.1F1A1A2given

Writing the direct-sum action in the concatenated basis of [A1], its trace is the sum of the traces of the two diagonal blocks, so χV⊕W(g)=tr⁡(ρV(g)⊕ρW(g))=tr⁡ρV(g)+tr⁡ρW(g)=χV(g)+χW(g) by [F1]. This is claim 1.

1.2F3A3given

By [F3] and [A3], the matrix of the tensor-product action in the basis (vi⊗wj) has entry ρV(g)ii′ ρW(g)jj′ at the position (i,j)↦(i′,j′), because g⋅(vi′⊗wj′)=∑i,jρV(g)ii′vi⊗ρW(g)jj′wj.

1.3F2given

With dual bases (vi) of V and (vi∗) of V∗, the matrix of the dual action of [F2] is the transpose of the matrix of ρV(g−1): writing g−1⋅vj=∑iρV(g−1)ijvi gives (g⋅vi∗)(vj)=vi∗(g−1⋅vj)=ρV(g−1)ij.

2.1F1step 1.2algebra

Its trace is the sum of its diagonal entries: ∑i,jρV(g)iiρW(g)jj=(∑iρV(g)ii)(∑jρW(g)jj). By [F1] the two factors are χV(g) and χW(g), so χV⊗W(g)=χV(g)χW(g), which is claim 2.

3.1F1A4step 1.3algebra∎

A matrix and its transpose have the same diagonal entries, hence the same trace, so χV∗(g)=tr⁡ρV(g−1)=χV(g−1). By [A4], χV(g−1)=χV(g)‾, which is claim 3.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

The character of a permutation representation counts fixed points

Statement

Let X be a finite left G-set and let C(X) be the permutation representation over C. Then the character of C(X) is

χC(X)(g)=∣{ x∈X:g⋅x=x }∣,

the number of fixed points of g acting on X.

Facts & Assumptions

Given: A finite group G, a finite left G-set X, and g∈G.

[F2]

The permutation representation has basis (ex)x∈X and action g⋅ex=eg⋅x (The trivial representation, the regular representation, and permutation representations from finite G-sets).

Proof

technique · direct
1.1F2given

In the basis (ex)x∈X of [F2], the matrix of ρ(g) is the permutation matrix with a 1 in row g⋅x, column x: its x-th diagonal entry is 1 exactly when g⋅x=x, and 0 otherwise.

2.1F1step 1.1algebra∎

The trace of that matrix is the sum of its diagonal entries, counting one for each x∈X with g⋅x=x. By [F1] this is χ(g), which equals the stated number of fixed points.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

The fixed subspace VG of a representation

Definition

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

VG:={ v∈V  :  g⋅v=v for every g∈G }.

The set VG is a linear subspace of V (Linear subspace of a vector space): it contains 0V because every operator ρ(g) is linear, it is closed under addition and scalar multiplication for the same reason, and the defining equations are the required pointwise identities. When G is finite and k=C, The averaging operator projects onto the fixed subspace shows that an explicit projection of V has image exactly VG.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

The averaging operator projects onto the fixed subspace

Statement

Let ρ:G→GL⁡(V) be a finite-dimensional representation of a finite group G over C. The averaging operator

P:=1∣G∣∑g∈Gρ(g)

satisfies P2=P and has image exactly VG; consequently tr⁡P=dim⁡VG.

Facts & Assumptions

Given: A finite group G and a finite-dimensional complex representation ρ:G→GL⁡(V).

[F1]

The fixed subspace is VG={v∈V:g⋅v=v for every g∈G} (The fixed subspace VG of a representation).

[F2]

An idempotent endomorphism has trace equal to the dimension of its image as a field scalar (The trace of an idempotent is its rank as a field scalar).

Proof

technique · direct
1.1givenalgebra

For h∈G, ρ(h)P=1∣G∣∑gρ(hg) =1∣G∣∑gρ(g)=P, because left translation by h permutes G, so the sums run over the same index set.

2.1F1step 1.1given

Hence for every v∈V, the vector Pv satisfies h⋅(Pv)=ρ(h)Pv=Pv for every h∈G, so Pv∈VG by [F1]; thus im⁡P⊆VG.

3.1F1step 2.1algebra

If v∈VG, then ρ(g)v=v for every g by [F1], so Pv=1∣G∣∑gv=∣G∣∣G∣v=v. Thus VG⊆im⁡P, and with step 2.1, im⁡P=VG exactly.

4.1step 2.1step 3.1algebra

For v∈VG, step 3.1 gives P(Pv)=Pv=v=Pv; for general v, Pv∈VG by step 2.1, so P2v=P(Pv)=Pv. Hence P2=P.

5.1F2step 4.1step 3.1∎

Apply [F2] to the idempotent P from step 4.1. Over C this gives tr⁡P=dim⁡(im⁡P), and step 3.1 identifies im⁡P with VG.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

For finite-dimensional complex V, the intertwiners V→W are exactly the fixed points of V∗⊗W

Statement

Let G be a finite group, let V and W be finite-dimensional complex representations of G, and let

Φ:V∗⊗CW⟶Hom⁡C(V,W)

be the natural isomorphism of For finite-dimensional V, the canonical map V∗⊗FW→Hom⁡F(V,W) is an isomorphism, sending f⊗w to the map v↦f(v)w. Then Φ is an intertwiner between the diagonal representation on V∗⊗W and the conjugation representation on Hom⁡C(V,W), and it carries the fixed subspace (V∗⊗W)G bijectively onto Hom⁡G(V,W).

Facts & Assumptions

Given: Finite-dimensional complex representations V and W of a finite group G.

[F1]

The dual action is (g⋅f)(v)=f(g−1⋅v) (The dual or contragredient complex representation).

[F2]

The diagonal action on the tensor product is g⋅(f⊗w)=(g⋅f)⊗(g⋅w) (The tensor product of two complex representations).

[F3]

The fixed subspace of a representation is the set of vectors fixed by every g∈G (The fixed subspace VG of a representation).

[F4]

Intertwiners are the linear maps T with T(g⋅v)=g⋅T(v) for all g,v (Intertwiners, the spaces Hom⁡G(V,W) and End⁡G(V), equivalent representations, and faithful representations).

[F5]

The map (f,w)↦[v↦f(v)w] induces a natural isomorphism Φ:V∗⊗CW→Hom⁡C(V,W) (For finite-dimensional V, the canonical map V∗⊗FW→Hom⁡F(V,W) is an isomorphism).

Proof

technique · direct
1.1F1F2given

For an elementary tensor f⊗w and v∈V, Φ(g⋅(f⊗w))(v)=Φ((g⋅f)⊗(g⋅w))(v) by [F2], and this equals (g⋅f)(v)(g⋅w)=f(g−1⋅v)(g⋅w) by [F1].

2.1F5step 1.1algebra

The right-hand side of step 1.1 is g⋅(f(g−1⋅v)w)=g⋅(Φ(f⊗w)(g−1⋅v)), which is the value at v of the conjugation action on the linear map Φ(f⊗w). Since elementary tensors span the tensor product, Φ is an intertwiner of representations.

3.1F3step 2.1given

An element of V∗⊗W is fixed by G exactly when its image under Φ is fixed by G, because Φ is a bijective intertwiner of step 2.1; so Φ((V∗⊗W)G)=Hom⁡C(V,W)G by [F3].

4.1F4step 3.1algebra∎

A linear map T is fixed by the conjugation action of every g exactly when g⋅T(g−1⋅v)=T(v) for all g,v, i.e. when T(g⋅v)=g⋅T(v); by [F4] this says precisely that T∈Hom⁡G(V,W). Hence Hom⁡C(V,W)G=Hom⁡G(V,W), which combines with step 3.1 into the claim.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

The class-function inner product ⟨χV,χW⟩ equals dim⁡Hom⁡G(W,V)

Statement

Let G be a finite group and let V and W be finite-dimensional complex representations of G with characters χV and χW. Then

⟨χV,χW⟩=dim⁡Hom⁡G(W,V).

Facts & Assumptions

Given: Finite-dimensional complex representations V and W of a finite group G, with characters χV and χW.

[F1]

The inner product of class functions is ⟨φ,ψ⟩=1∣G∣∑g∈Gφ(g)ψ(g)‾ (The standard inner product on cf(G)).

[F2]

For a finite-dimensional complex representation U, the averaging operator PU=1∣G∣∑gρU(g) has image UG and trace dim⁡UG (The averaging operator projects onto the fixed subspace).

[F3]

The fixed points of the diagonal representation on W∗⊗V are exactly the intertwiners W→V, so (W∗⊗V)G≅Hom⁡G(W,V) (For finite-dimensional complex V, the intertwiners V→W are exactly the fixed points of V∗⊗W).

[F4]

Characters add on direct sums, multiply on tensor products, and conjugate on duals (Characters add on direct sums, multiply on tensor products, and conjugate on duals).

[F5]

Trace is linear on finite square matrices (Trace is a linear functional on Mn(F)).

[A1]

The character of a representation is by definition the trace of its action operator, χU(g)=tr⁡ρU(g).

Proof

technique · direct
1.1F2F3given

By [F3], dim⁡Hom⁡G(W,V)=dim⁡(W∗⊗V)G. By [F2] applied to U=W∗⊗V, this equals tr⁡PU for the averaging operator PU=1∣G∣∑gρW∗⊗V(g).

1.2F5A1given

By [F5], tr⁡PU=1∣G∣∑gtr⁡ρW∗⊗V(g)=1∣G∣∑gχW∗⊗V(g), the second equality by [A1].

1.3F4given

By [F4], the tensor-product and dual clauses give χW∗⊗V(g)=χW∗(g)χV(g)=χW(g)‾χV(g).

2.1F1step 1.1step 1.2step 1.3algebra∎

Combining steps 1.1 through 1.3, dim⁡Hom⁡G(W,V)=1∣G∣∑gχW(g)‾χV(g)=1∣G∣∑gχV(g)χW(g)‾, reordering the product of two complex numbers in each summand. This is ⟨χV,χW⟩ by [F1].

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

The first orthogonality relation for irreducible complex characters

Statement

Let G be a finite group and let χ1,…,χr be the irreducible complex characters of G, one from each equivalence class. Then

⟨χi,χj⟩=δij(1≤i,j≤r).

Facts & Assumptions

Given: A finite group G, and irreducible complex representations V1,…,Vr of G with characters χ1,…,χr, one from each equivalence class.

[F1]

Irreducible characters are the characters of irreducible representations (An irreducible complex character).

[F2]

The inner product computes intertwiner dimension: ⟨χi,χj⟩=dim⁡Hom⁡G(Vj,Vi) (The class-function inner product ⟨χV,χW⟩ equals dim⁡Hom⁡G(W,V)).

[F3]

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

[F4]

Over the algebraically closed field C, every intertwiner Vi→Vi is a scalar operator (Over an algebraically closed field, every endomorphism of an irreducible representation is scalar).

Proof

technique · direct
1.1F3given

For i≠j the representations Vi and Vj are inequivalent, because one representative was chosen from each class. If T∈Hom⁡G(Vj,Vi) were nonzero, [F3] would make T an isomorphism, contradicting inequivalence; hence Hom⁡G(Vj,Vi)=0.

1.2F4given

For i=j, [F4] says every element of Hom⁡G(Vi,Vi)=End⁡G(Vi) is a scalar multiple of the identity. The identity operator is nonzero, so the scalars λid⁡Vi form a one-dimensional complex line. Hence dim⁡Hom⁡G(Vi,Vi)=1.

2.1F1F2step 1.1step 1.2algebra∎

By [F2], ⟨χi,χj⟩=dim⁡Hom⁡G(Vj,Vi); steps 1.1 and 1.2 give this dimension to be 0 when i≠j and 1 when i=j. This is exactly ⟨χi,χj⟩=δij.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

The irreducible complex characters form an orthonormal basis of cf(G)

Statement

Let G be a finite group. The irreducible complex characters of G, χ1,…,χr, form an orthonormal basis of the complex vector space cf(G) of class functions on G, with respect to the standard inner product.

Facts & Assumptions

Given: A finite group G and its irreducible complex characters χ1,…,χr.

[F1]

The space cf(G) consists of the functions G→C constant on conjugacy classes, with pointwise operations (Class functions and the complex vector space cf(G)).

[F2]

The standard inner product on cf(G) is the Hermitian form ⟨φ,ψ⟩=1∣G∣∑gφ(g)ψ(g)‾ (The standard inner product on cf(G)).

[F3]

The irreducible characters are orthonormal: ⟨χi,χj⟩=δij (The first orthogonality relation for irreducible complex characters).

[F4]

Over an algebraically closed field of characteristic not dividing ∣G∣, the number of irreducible representations up to equivalence equals the number of conjugacy classes (If k is algebraically closed and char⁡k∤∣G∣, the number of irreducible representations of G equals the number of conjugacy classes).

[A1]

A class function is determined by its values on one representative of each conjugacy class, and the indicator functions of the distinct conjugacy classes form a basis of cf(G); in particular dim⁡Ccf(G) equals the number of conjugacy classes of G.

Proof

technique · direct
1.1F2F3given

By [F3] the family χ1,…,χr is orthonormal, hence linearly independent: any relation ∑iciχi=0 has inner product with χj equal to cj, because ⟨χi,χj⟩=δij and the inner product is linear in its first argument by [F2].

1.2F4given

The group G is finite, C is algebraically closed, and char⁡C=0 does not divide ∣G∣, so [F4] applies and r equals the number of conjugacy classes of G.

2.1A1step 1.1step 1.2

By [A1], dim⁡Ccf(G) equals the number of conjugacy classes, which step 1.2 identified with r. Hence the linearly independent family of r elements from step 1.1 has as many elements as the dimension of the space.

3.1F1step 1.1step 2.1algebra∎

A linearly independent family whose size equals the dimension spans, so the orthonormal family of step 1.1 is a basis of cf(G).

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

The multiplicity of an irreducible summand is a character inner product

Statement

Let G be a finite group, let V be a finite-dimensional complex representation of G. Choose representatives V1,…,Vr of the irreducible representations, write χj:=χVj, and write V≅⨁j=1rmjVj. Then the multiplicity of Vi in V is

mi=⟨χV,χi⟩.

Facts & Assumptions

Given: A finite group G, a finite-dimensional complex representation V of G, representatives Vj of the irreducible representations with characters χj:=χVj, a decomposition V≅⨁jmjVj, and an index i.

[F1]

Every finite-dimensional representation of a finite group over a field of characteristic not dividing ∣G∣ is completely reducible (If char⁡k∤∣G∣, every finite-dimensional representation of G is completely reducible).

[F3]

Irreducible characters are orthonormal: ⟨χj,χi⟩=δji (The first orthogonality relation for irreducible complex characters).

Proof

technique · direct
1.1F1given

Since char⁡C=0 does not divide ∣G∣, [F1] gives a decomposition V≅⨁jmjVj with each Vj irreducible.

2.1F2step 1.1

Applying [F2] iteratively to the decomposition of step 1.1 gives χV=∑jmjχj, a finite sum because V is finite-dimensional.

3.1F3step 2.1algebra∎

Taking the inner product with χi, linearity in the first argument and [F3] give ⟨χV,χi⟩=∑jmj⟨χj,χi⟩=∑jmjδji=mi.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

Finite-dimensional complex representations of a finite group are determined up to isomorphism by their characters

Statement

Let G be a finite group and let V and W be finite-dimensional complex representations of G. Then

V≅W⟺χV=χW.

Facts & Assumptions

Given: A finite group G and finite-dimensional complex representations V and W of G.

[F1]

Every finite-dimensional representation of a finite group over a field of characteristic not dividing ∣G∣ is completely reducible (If char⁡k∤∣G∣, every finite-dimensional representation of G is completely reducible).

[F2]

For irreducible Vi, the multiplicity of Vi in a completely reduced representation U is mi(U)=⟨χU,χi⟩ (The multiplicity of an irreducible summand is a character inner product).

[A1]

Equivalent representations have equal characters.

Proof

technique · direct
1.1F1F2given

By [F1] there are decompositions V≅⨁imiVi and W≅⨁iniVi over the irreducible representations Vi. By [F2] the multiplicities are mi=⟨χV,χi⟩ and ni=⟨χW,χi⟩.

1.2A1given

Conversely, if V≅W via an invertible intertwiner, then [A1] gives χV=χW. This proves the reverse implication.

2.1step 1.1algebra

Assume χV=χW. Then every inner product in step 1.1 agrees, so mi=ni for every i; the two direct sums are built from the same irreducibles with the same multiplicities, hence V≅W. This proves the forward implication.

3.1step 2.1step 1.2∎

Steps 2.1 and 1.2 prove both implications, hence the equivalence.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

A complex character is irreducible if and only if its self-inner-product is 1

Statement

Let G be a finite group and let χ be a complex character of G. Then χ is irreducible if and only if ⟨χ,χ⟩=1.

Facts & Assumptions

Given: A finite group G and a complex character χ=χV of a finite-dimensional representation V, completely reduced as V≅⨁imiVi.

[F1]

The multiplicity of Vi in V is mi=⟨χ,χi⟩ (The multiplicity of an irreducible summand is a character inner product).

[F2]

Characters add on direct sums, so the decomposition V≅⨁imiVi gives χ=∑imiχi (Characters add on direct sums, multiply on tensor products, and conjugate on duals).

[A1]

If χ=∑imiχi and mi=⟨χ,χi⟩, then conjugate-symmetry and linearity in the first argument give ⟨χ,χ⟩=∑imi⟨χi,χ⟩=∑imi⟨χ,χi⟩‾=∑imimi‾=∑imi2, because the multiplicities mi are nonnegative integers.

[A2]

A representation is irreducible exactly when it has one irreducible summand with multiplicity 1 and no others.

Proof

technique · direct
1.1F1F2A1given

Completely reduce V≅⨁imiVi with mi≥0 integers. By [F2] this decomposition gives χ=∑imiχi, and by [F1] each mi=⟨χ,χi⟩; therefore [A1] gives ⟨χ,χ⟩=∑imi2.

2.1A2step 1.1

Assume χ is irreducible. Then by [A2] exactly one multiplicity equals 1 and the rest are 0, so the sum in step 1.1 is 1; hence ⟨χ,χ⟩=1.

2.2A2step 1.1algebra

Conversely, assume ⟨χ,χ⟩=1. The sum ∑imi2 of nonnegative integers in step 1.1 can equal 1 only when exactly one mi equals 1 and all the others are 0. By [A2], V≅Vi is irreducible, so χ is irreducible.

3.1step 2.1step 2.2∎

Steps 2.1 and 2.2 prove the two implications, hence the biconditional.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

The regular character is ∣G∣ at 1 and 0 away from 1

Statement

Let G be a finite group and let χreg be the character of the regular representation C[G] over C. Then

χreg(g)={∣G∣,g=1,0,g≠1.

Facts & Assumptions

Given: A finite group G, an element g∈G, and the regular representation C[G].

[F1]

The regular representation acts on the basis ([h])h∈G by g⋅[h]=[gh] (The trivial representation, the regular representation, and permutation representations from finite G-sets).

[F2]

The character of a permutation representation counts fixed points (The character of a permutation representation counts fixed points).

[A1]

In a group, gx=x holds if and only if g=1, by cancellation.

Proof

technique · direct
1.1F1F2given

By [F1], the regular action is the permutation representation of the left G-set G with basis indexed by the group elements. By [F2], χreg(g) is the number of h∈G with g⋅h=h.

2.1A1step 1.1algebra∎

By [A1], gh=h has a solution in G only when g=1, in which case every h is fixed. Hence the count of step 1.1 is ∣G∣ for g=1 and 0 for g≠1.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

The regular character gives a second proof of the sum-of-squares formula

Statement

Let G be a finite group and let χ1,…,χr be its irreducible complex characters, of degrees ni=χi(1). Then

∑i=1rni2=∣G∣.

Facts & Assumptions

Given: A finite group G with irreducible characters χ1,…,χr, the regular representation C[G], and the regular character χreg.

[F1]

The regular character is ∣G∣ at 1 and 0 elsewhere (The regular character is ∣G∣ at 1 and 0 away from 1).

[F2]

The multiplicity of an irreducible summand is an inner product: mi=⟨χreg,χi⟩ (The multiplicity of an irreducible summand is a character inner product).

[F3]

Characters add on direct sums: χV⊕W=χV+χW (Characters add on direct sums, multiply on tensor products, and conjugate on duals).

[A1]

The degree of a character is its value at 1, χi(1)=ni.

Proof

technique · direct
1.1F1F2A1given

By [F2], the regular representation decomposes as C[G]≅⨁imiVi with mi=⟨χreg,χi⟩. By [F1] and the definition of the inner product this is 1∣G∣⋅∣G∣⋅χi(1)=χi(1)=ni, because the only nonzero term is at g=1.

2.1F1F3step 1.1algebra∎

Characters add on direct sums by [F3], so evaluating both sides of the decomposition of step 1.1 at 1 gives χreg(1)=∑imini. By [F1] the left side is ∣G∣, and step 1.1 gives mi=ni, so ∣G∣=∑ini2.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

The second orthogonality relation for irreducible complex characters

Statement

Let G be a finite group with irreducible complex characters χ1,…,χr. For g,h∈G,

∑i=1rχi(g) χi(h)‾={∣CG(g)∣,h∈Cl⁡G(g),0,h∉Cl⁡G(g).

Facts & Assumptions

Given: A finite group G, its irreducible characters χ1,…,χr, and elements g,h∈G.

[F1]

The irreducible characters form an orthonormal basis of the class functions cf(G) (The irreducible complex characters form an orthonormal basis of cf(G)).

[F2]

The inner product of class functions is ⟨φ,ψ⟩=1∣G∣∑xφ(x)ψ(x)‾ (The standard inner product on cf(G)).

[A1]

The indicator function δg of the conjugacy class of g, defined by δg(x)=1 for x∈Cl⁡G(g) and δg(x)=0 otherwise, is a class function, and its class has ∣Cl⁡G(g)∣=∣G∣/∣CG(g)∣ elements.

[A2]

If f=∑iciχi in the orthonormal basis of [F1], then ci=⟨f,χi⟩.

Proof

technique · direct
1.1F1A1A2given

The function δg of [A1] is a class function, so by [F1] it expands as δg=∑iciχi with ci=⟨δg,χi⟩ by [A2].

1.2F2A1givenalgebra

By [F2], ⟨δg,χi⟩=1∣G∣∑xδg(x)χi(x)‾. By [A1] the only nonzero terms are at x∈Cl⁡G(g), where δg(x)=1 and χi(x)‾=χi(g)‾ because characters are class functions. Hence ci=∣G∣/∣CG(g)∣∣G∣χi(g)‾=χi(g)‾/∣CG(g)∣.

2.1step 1.1step 1.2algebra

Evaluating the expansion of step 1.1 at h and substituting the coefficients of step 1.2 gives δg(h)=1∣CG(g)∣∑iχi(g)‾χi(h).

3.1A1step 2.1algebra∎

By [A1], δg(h)=1 exactly when h∈Cl⁡G(g), and 0 otherwise. Multiplying the identity of step 2.1 by ∣CG(g)∣ and taking complex conjugates gives ∑iχi(g)χi(h)‾=∣CG(g)∣δg(h), which is the stated formula.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

For g∈G, the sum of ∣χi(g)∣2 over the irreducible complex characters is ∣CG(g)∣

Statement

Let G be a finite group with irreducible complex characters χ1,…,χr. For every g∈G,

∑i=1r∣χi(g)∣2=∣CG(g)∣.

Facts & Assumptions

Given: A finite group G, its irreducible characters χ1,…,χr, and an element g∈G.

[F1]

For g,h∈G, ∑iχi(g)χi(h)‾=∣CG(g)∣ when h is conjugate to g, and 0 otherwise (The second orthogonality relation for irreducible complex characters).

Proof

technique · direct
1.1F1given

The element g is conjugate to itself, so applying [F1] with h=g gives ∑iχi(g)χi(g)‾=∣CG(g)∣.

2.1step 1.1algebra∎

Each summand equals ∣χi(g)∣2, so the sum of step 1.1 is exactly ∑i∣χi(g)∣2, which proves the claim.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

The character table is square and invertible

Statement

Let G be a finite group. The character table of G has as many rows as columns, and the table matrix is invertible.

Facts & Assumptions

Given: A finite group G with irreducible characters χ1,…,χr and conjugacy-class representatives g1,…,gs.

[F1]

The irreducible characters form an orthonormal basis of the class functions, so there are as many of them as there are conjugacy classes (The irreducible complex characters form an orthonormal basis of cf(G)).

[F2]

The columns of the character table satisfy the second orthogonality relation: ∑iχi(ga)χi(gb)‾=∣CG(ga)∣δab (The second orthogonality relation for irreducible complex characters).

Proof

technique · direct
1.1F1given

By [F1], r equals the number of conjugacy classes of G, which is the number s of columns; hence the table is square.

1.2F2givenalgebra

Rescale the a-th column of the table by the positive number 1/∣CG(ga)∣, forming the matrix Uia=χi(ga)/∣CG(ga)∣. By [F2], distinct columns of U are orthogonal and each column has squared norm 1, so U has orthonormal columns.

2.1step 1.1step 1.2algebra∎

A square matrix with orthonormal columns has linearly independent columns, hence is invertible. Since U is obtained from the table by multiplying columns by nonzero scalars, the table matrix is invertible too.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

A representation with kernel containing a normal subgroup factors through the quotient, and irreducibility is unchanged by inflation

Statement

Let G be a group, let N⊴G be a normal subgroup (Normal subgroup: invariance under conjugation), and let ρ:G→GL⁡(V) be a representation over a field k with N⊆ker⁡ρ. Then:

  1. the formula ρ‾(gN):=ρ(g) defines a well-posed representation ρ‾:G/N→GL⁡(V) with ρ=ρ‾∘π, where π:G→G/N is the canonical quotient map;
  2. V is irreducible as a representation of G if and only if it is irreducible as a representation of G/N.

Facts & Assumptions

Given: A group G, a normal subgroup N⊴G, a field k, and a representation ρ:G→GL⁡(V) with N⊆ker⁡ρ.

[F1]

The kernel of a group homomorphism is the preimage of the identity (The kernel and image of a group homomorphism).

[F2]

The canonical map π:G→G/N, π(g)=gN, is a surjective group homomorphism (The canonical projection π:G→G/N, π(g)=gN, is a surjective group homomorphism).

[F3]

The cosets form the group G/N under (gN)(hN)=ghN, with identity N and inverse g−1N (For N⊴G, the cosets form a group with identity N and inverse (gN)−1=g−1N).

Proof

technique · direct
1.1F1F3given

If gN=hN, then g−1h∈N by the coset laws of [F3], so ρ(g−1h)=e by [F1] and the hypothesis N⊆ker⁡ρ; hence ρ(h)=ρ(g). Therefore ρ‾(gN):=ρ(g) is independent of the chosen coset representative.

2.1F2F3step 1.1given

By [F3], (gN)(hN)=ghN; applying ρ gives ρ‾(gN)ρ‾(hN)=ρ(g)ρ(h)=ρ(gh)=ρ‾(ghN)=ρ‾((gN)(hN)), so ρ‾ is a group homomorphism into GL⁡(V), with ρ=ρ‾∘π by [F2]. This proves claim 1.

3.1F2step 2.1given

A subspace U⊆V is ρ-stable exactly when it is ρ‾-stable, because π of [F2] is surjective and ρ(g)=ρ‾(gN): the two stability conditions quantify over the same operators.

4.1step 3.1algebra∎

A representation is irreducible exactly when its only stable subspaces are 0 and V. By step 3.1 the stable subspaces for ρ and for ρ‾ coincide, so the two irreducibility statements are equivalent, which is claim 2.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

The normal subgroups of a finite group are exactly the intersections of kernels of irreducible complex characters

Statement

Let G be a finite group. A subgroup N of G is normal if and only if it is an intersection of kernels of irreducible complex characters of G.

Facts & Assumptions

Given: A finite group G and a subgroup N of G.

[F1]

Every finite-dimensional representation of a finite group over a field of characteristic not dividing ∣G∣ is completely reducible (If char⁡k∤∣G∣, every finite-dimensional representation of G is completely reducible).

[F2]

A representation with kernel containing a normal subgroup factors through the quotient, irreducibility being preserved in both directions (A representation with kernel containing a normal subgroup factors through the quotient, and irreducibility is unchanged by inflation).

[F3]

The kernel of a character agrees with the kernel of any representation affording it (The kernel of a complex character agrees with the kernel of any representation affording it).

[F4]

The regular representation over a field is faithful (The regular representation is faithful).

[A1]

The kernel of a direct sum of representations is the intersection of the kernels of its summands.

[A2]

A kernel of a group homomorphism is a normal subgroup, and an intersection of normal subgroups is normal.

Proof

technique · direct
1.1F4given

Assume N⊴G. Let ρ‾:G/N→GL⁡(C[G/N]) be the regular representation of the quotient. Since ρ‾ is faithful by [F4], its kernel in G/N is the trivial subgroup {N}.

1.2F3A2given

Conversely, if N=⋂jker⁡χj for irreducible characters χj of G, then by [F3] each ker⁡χj is the kernel of a group homomorphism, hence normal by [A2], and the intersection of normal subgroups is again normal by [A2]. Thus N⊴G.

2.1F1A1step 1.1given

Since char⁡C=0 does not divide ∣G/N∣, [F1] decomposes C[G/N] as a direct sum of irreducible representations U1,…,Um of G/N; by [A1] the trivial kernel of step 1.1 is the intersection of their kernels, so ⋂jker⁡(ρ‾∣Uj)={N}.

3.1F2F3step 2.1given

For each j, let Vj be the inflation of Uj to G. By [F2], each Vj is irreducible as a representation of G, and ker⁡ρVj=π−1(ker⁡(ρ‾∣Uj)), which contains N. By [F3] each ker⁡ρVj equals the kernel of the corresponding irreducible character χj of G.

4.1A1step 2.1step 3.1algebra

By [A1] and steps 2.1 and 3.1, ⋂jker⁡χj=⋂jπ−1(ker⁡(ρ‾∣Uj))=π−1({N})=N. Hence a normal subgroup is an intersection of kernels of irreducible characters.

5.1step 4.1step 1.2∎

Steps 4.1 and 1.2 prove the two implications, hence the equivalence.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29Open item page →

A finite group is abelian if and only if all its irreducible complex characters have degree 1

Statement

Let G be a finite group. Then G is abelian if and only if every irreducible complex character of G has degree 1.

Facts & Assumptions

Given: A finite group G with irreducible characters χ1,…,χr of degrees ni=χi(1).

[F1]

Over an algebraically closed field, every intertwiner from an irreducible representation to itself is a scalar (Over an algebraically closed field, every endomorphism of an irreducible representation is scalar).

[F2]

A field k is a splitting field for G when every irreducible representation V has End⁡G(V)=k (A splitting field for a finite group: every irreducible representation has scalar endomorphism ring).

[F3]

Every irreducible representation of a finite abelian group over a splitting field has degree 1 (Every irreducible representation of a finite abelian group over a splitting field is one-dimensional).

[F4]

The degrees of the irreducible characters satisfy ∑ini2=∣G∣ (The regular character gives a second proof of the sum-of-squares formula).

[F5]

The number of irreducible representations of G up to equivalence equals the number of conjugacy classes, when the field is algebraically closed of characteristic not dividing ∣G∣ (If k is algebraically closed and char⁡k∤∣G∣, the number of irreducible representations of G equals the number of conjugacy classes).

[A1]

The degree of an irreducible character is the dimension of any representation affording it.

[A2]

A finite group is abelian exactly when every conjugacy class has one element.

Proof

technique · direct
1.1F1F2given

Assume G is abelian. For every irreducible representation V of G, [F1] gives End⁡G(V)=C; by [F2], C is a splitting field for G.

1.2F4given

Conversely, assume every irreducible character has degree 1, so ni=1 for all i. By [F4], ∣G∣=∑ini2=r.

1.3F5given

Since C is algebraically closed and char⁡C=0 does not divide ∣G∣, [F5] applies: r equals the number of conjugacy classes of G.

2.1F3A1step 1.1given

By [F3] applied over this splitting field, every irreducible representation of the abelian group G has degree 1; by [A1] every irreducible character has degree 1. This proves the forward implication.

2.2A2step 1.2step 1.3

Steps 1.2 and 1.3 show ∣G∣ equals the number of conjugacy classes, so the ∣G∣ conjugacy classes each have exactly one element; by [A2], G is abelian. This proves the reverse implication.

3.1step 2.1step 2.2∎

Steps 2.1 and 2.2 prove the two implications, hence the equivalence.

5 · Examples, counterexamples and false statements

None yet.

Sources