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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)=trρV(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)=dimV, 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(g1)=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 VW, converts χV,χW into dimHomG(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:GGL(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 charC=0 and G is finite, the characteristic does not divide G, so every finite-dimensional representation is completely reducible (If charkG, 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 charkG, 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:GGL(V) be a finite-dimensional complex representation (A finite-dimensional representation ρ:GGL(V) over a field, and its degree). Its character is the function

χV:GC,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:VW is an invertible intertwiner, then ρW(g)=TρV(g)T1, so tr(ρW(g))=tr(ρV(g)T1T)=tr(ρV(g)) by the identity tr(AB)=tr(BA) (For AMm×n(F) and BMn×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:GC is a class function when it is constant on every conjugacy class (The conjugacy class ClG(x) and centralizer CG(x) of an element):

f(gxg1)=f(x)for all g,xG.

The set of all class functions is written

cf(G):={f:GC  :  f(gxg1)=f(x) for all g,xG}.

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 gG. Because G is finite, the evaluation functions ff(x) are complex-linear, and a class function is determined by its values on one representative of each conjugacy class.

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

The standard inner product on cf(G)

Definition

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

φ,ψ:=1GgGφ(g)ψ(g),

the finite sum being that of (The sum iSai over a finite index set, and its product form). 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=1Ggψ(g)φ(g)=1Ggφ(g)ψ(g)=φ,ψ. For definiteness, φ,φ=1Ggφ(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 charkG, 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 ClG(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 ClG(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:GGL(V) and σ:GGL(W) be finite-dimensional complex representations of a group G (A finite-dimensional representation ρ:GGL(V) over a field, and its degree). Their tensor product representation ρσ is the representation of G on the complex tensor product VCW given on elementary tensors by

g(vw):=(ρ(g)v)(σ(g)w)(gG, vV, wW).

The action is well defined. For fixed gG, the assignment (v,w)ρ(g)vσ(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)λσ(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):VCWVCW with the displayed value on every elementary tensor. Because the two sides of g(h(vw))=(gh)(vw) agree on every elementary tensor and the elementary tensors span VCW, the assignment gρ(g)σ(g) is a group homomorphism into GL(VCW), with ρ(1)σ(1)=id. The underlying space is finite-dimensional: bases of V and W give the basis (viwj)i,j, so dim(VCW)=(dimV)(dimW).

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

The dual or contragredient complex representation

Definition

Let ρ:GGL(V) be a finite-dimensional complex representation of a group G (A finite-dimensional representation ρ:GGL(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

(gf)(v):=f(g1v)(gG, fV, vV),

equivalently gf=fρ(g1).

Each operator ρ(g) is C-linear because f and ρ(g1) are. The inverse appears so that composition is order-preserving: writing gf=fρ(g1),

(gh)f=fρ((gh)1)=fρ(h1)ρ(g1)=g(hf),

and 1f=fidV=f. Hence ρ is a group homomorphism GGL(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:GGL(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

kerχ:={gG  :  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 kerρ of the representation, so the two readings of "kernel" never conflict on this page.

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

For a complex character, χ(1)=dimV, 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,hG:

  1. χ(1)=dimV;
  2. χ(ghg1)=χ(h), so χ is a class function;
  3. χ(g) is a sum of dimV 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(g1)=χ(g).

Facts & Assumptions

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

[F1]

The character is χ(g)=trρ(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(xλi), then its trace is the sum of the eigenvalues iλi (If χT(x)=i<n(xλi) 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 AMm×n(F) and BMn×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 iziizi holds, and equality holds exactly when all the nonzero zi share one argument.

[A6]

A function f:GC 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
1.1

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

F1givenalgebra
1.2

For the same reason ρ(g) and ρ(h) compose as in the group, so χ(ghg1)=tr(ρ(g)ρ(h)ρ(g)1)=tr(ρ(h)ρ(g)1ρ(g))=trρ(h)=χ(h), the middle step being [A3] applied to A=ρ(g) and B=ρ(h)ρ(g)1.

F1A3given
1.3

By [F2], g has finite order n with nG, so ρ(g)n=ρ(gn)=idV. 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=dimV; each λin=1, so each λi is a root of unity.

F2A1given
2.1

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.

step 1.2A6
2.2

In that basis the characteristic polynomial of ρ(g) is i=1d(xλi), so [A2] gives χ(g)=trρ(g)=i=1dλi, a sum of d=dimV roots of unity, which is claim 3.

F1A2step 1.3step 1.1
2.3

Conversely, if ρ(g)=λidV 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 V0, then the identity ρ(g)n=idV of step 1.3 reads λnidV=idV, and evaluating it on a nonzero vector gives λn=1. Thus λ is a root of unity and λ=1; then χ(g)=λdimV and χ(g)=dimV=χ(1). This closes the biconditional in claim 4.

F1step 1.1step 1.3algebra
2.4

The inverse operator has the inverse eigenvalues, and a root of unity λ satisfies λ1=λ because λ=1=λλ.

step 1.3A4algebra
3.1

Applying [A5] to the eigenvalues of step 1.3 gives χ(g)=iλiiλi=d=dimV=χ(1), which is the inequality in claim 4.

A5step 2.2step 1.3step 1.1
4.1

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)=λidV, a scalar operator.

A5step 1.3algebra
5.1

Hence, using [A2] for ρ(g1)=ρ(g)1 and the additivity of conjugation from [A4], χ(g1)=iλi1=iλi=iλi=χ(g), which is claim 5.

A2A4step 2.4step 2.2
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:GGL(V) be a finite-dimensional complex representation of a finite group G, with character χ=χV. Then

kerχ=kerρ,

where kerρ is the kernel of the group homomorphism ρ and kerχ={gG:χ(g)=χ(1)} is the kernel of the character.

Facts & Assumptions

Given: A finite group G and a finite-dimensional complex representation ρ:GGL(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 kerχ={gG:χ(g)=χ(1)} (The kernel of a complex character).

[F3]

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

Proof

technique · direct
1.1

If gkerρ, then ρ(g)=idV by [F1], so χ(g)=tr(idV)=χ(1) by [F3]. Hence gkerχ by [F2], which proves kerρkerχ.

F1F2F3given
1.2

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

F2F3given
1.3

If dimV=0, then V={0} is the zero space, ρ(g)=idV holds for every g, and χ0, so kerρ=G=kerχ by [F1] and [F2]; the statement is immediate. Hence assume dimV1.

F1F2given
2.1

Since χ(1)=dimV1 by [F3], the equality λχ(1)=χ(1) of step 1.2 forces λ=1; therefore ρ(g)=idV, so gkerρ by [F1]. Together with step 1.1 and step 1.3 this proves kerχ=kerρ.

F1F3step 1.2step 1.3algebra
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 gG:

  1. χVW(g)=χV(g)+χW(g);
  2. χVW(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 gG.

[F2]

The dual action is (gf)(v)=f(g1v) (The dual or contragredient complex representation).

[F3]

The tensor-product action is g(vw)=(gv)(gw) (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 VW 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 (viwj)i,j of the tensor product (The elementary tensors of two bases form the product basis of the tensor product).

Proof

technique · direct
1.1

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 χVW(g)=tr(ρV(g)ρW(g))=trρV(g)+trρW(g)=χV(g)+χW(g) by [F1]. This is claim 1.

F1A1A2given
1.2

By [F3] and [A3], the matrix of the tensor-product action in the basis (viwj) has entry ρV(g)iiρW(g)jj at the position (i,j)(i,j), because g(viwj)=i,jρV(g)iiviρW(g)jjwj.

F3A3given
1.3

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(g1): writing g1vj=iρV(g1)ijvi gives (gvi)(vj)=vi(g1vj)=ρV(g1)ij.

F2given
2.1

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 χVW(g)=χV(g)χW(g), which is claim 2.

F1step 1.2algebra
3.1

A matrix and its transpose have the same diagonal entries, hence the same trace, so χV(g)=trρV(g1)=χV(g1). By [A4], χV(g1)=χV(g), which is claim 3.

F1A4step 1.3algebra
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)={xX:gx=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 gG.

[F2]

The permutation representation has basis (ex)xX and action gex=egx (The trivial representation, the regular representation, and permutation representations from finite G-sets).

Proof

technique · direct
1.1

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

F2given
2.1

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

F1step 1.1algebra
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:GGL(V) be a representation of a group G over a field k (A finite-dimensional representation ρ:GGL(V) over a field, and its degree). The fixed subspace of V is

VG:={vV  :  gv=v for every gG}.

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:GGL(V) be a finite-dimensional representation of a finite group G over C. The averaging operator

P:=1GgGρ(g)

satisfies P2=P and has image exactly VG; consequently trP=dimVG.

Facts & Assumptions

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

[F1]

The fixed subspace is VG={vV:gv=v for every gG} (The fixed subspace VG of a representation).

[A2]

If T is a projection of a finite-dimensional vector space, meaning T2=T, then V=imTkerT, T restricts to the identity on imT, and in a basis adapted to that decomposition the matrix of T is block diagonal with an identity block and a zero block.

Proof

technique · direct
1.1

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

givenalgebra
2.1

Hence for every vV, the vector Pv satisfies h(Pv)=ρ(h)Pv=Pv for every hG, so PvVG by [F1]; thus imPVG.

F1step 1.1given
3.1

If vVG, then ρ(g)v=v for every g by [F1], so Pv=1Ggv=GGv=v. Thus VGimP, and with step 2.1, imP=VG exactly.

F1step 2.1algebra
4.1

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

step 2.1step 3.1algebra
5.1

By [A2] applied to P from step 4.1, V=imPkerP and the matrix of P in an adapted basis has an identity block of size dim(imP) and a zero block. Its trace is therefore dim(imP), and step 3.1 identifies imP with VG.

A2step 4.1step 3.1algebra
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29Open item page →

For finite-dimensional complex V, the intertwiners VW are exactly the fixed points of VW

Statement

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

Φ:VCWHomC(V,W)

be the natural isomorphism of For finite-dimensional V, the canonical map VFWHomF(V,W) is an isomorphism, sending fw to the map vf(v)w. Then Φ is an intertwiner between the diagonal representation on VW and the conjugation representation on HomC(V,W), and it carries the fixed subspace (VW)G bijectively onto HomG(V,W).

Facts & Assumptions

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

[F1]

The dual action is (gf)(v)=f(g1v) (The dual or contragredient complex representation).

[F2]

The diagonal action on the tensor product is g(fw)=(gf)(gw) (The tensor product of two complex representations).

[F3]

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

[F4]

Intertwiners are the linear maps T with T(gv)=gT(v) for all g,v (Intertwiners, the spaces HomG(V,W) and EndG(V), equivalent representations, and faithful representations).

[F5]

The map (f,w)[vf(v)w] induces a natural isomorphism Φ:VCWHomC(V,W) (For finite-dimensional V, the canonical map VFWHomF(V,W) is an isomorphism).

Proof

technique · direct
1.1

For an elementary tensor fw and vV, Φ(g(fw))(v)=Φ((gf)(gw))(v) by [F2], and this equals (gf)(v)(gw)=f(g1v)(gw) by [F1].

F1F2given
2.1

The right-hand side of step 1.1 is g(f(g1v)w)=g(Φ(fw)(g1v)), which is the value at v of the conjugation action on the linear map Φ(fw). Since elementary tensors span the tensor product, Φ is an intertwiner of representations.

F5step 1.1algebra
3.1

An element of VW 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((VW)G)=HomC(V,W)G by [F3].

F3step 2.1given
4.1

A linear map T is fixed by the conjugation action of every g exactly when gT(g1v)=T(v) for all g,v, i.e. when T(gv)=gT(v); by [F4] this says precisely that THomG(V,W). Hence HomC(V,W)G=HomG(V,W), which combines with step 3.1 into the claim.

F4step 3.1algebra
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The class-function inner product χV,χW equals dimHomG(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=dimHomG(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=1GgGφ(g)ψ(g) (The standard inner product on cf(G)).

[F2]

For a finite-dimensional complex representation U, the averaging operator PU=1GgρU(g) has image UG and trace dimUG (The averaging operator projects onto the fixed subspace).

[F3]

The fixed points of the diagonal representation on WV are exactly the intertwiners WV, so (WV)GHomG(W,V) (For finite-dimensional complex V, the intertwiners VW are exactly the fixed points of VW).

[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).

[A1]

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

Proof

technique · direct
1.1

By [F3], dimHomG(W,V)=dim(WV)G. By [F2] applied to U=WV, this equals trPU for the averaging operator PU=1GgρWV(g).

F2F3given
1.2

Trace is linear, so trPU=1GgtrρWV(g)=1GgχWV(g), the second equality by [A1].

A1algebragiven
1.3

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

F4given
2.1

Combining steps 1.1 through 1.3, dimHomG(W,V)=1GgχW(g)χV(g)=1GgχV(g)χW(g), reordering the product of two complex numbers in each summand. This is χV,χW by [F1].

F1step 1.1step 1.2step 1.3algebra
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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(1i,jr).

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=dimHomG(Vj,Vi) (The class-function inner product χV,χW equals dimHomG(W,V)).

[F3]

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

[F4]

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

Proof

technique · direct
1.1

For ij the representations Vi and Vj are inequivalent, because one representative was chosen from each class. If THomG(Vj,Vi) were nonzero, [F3] would make T an isomorphism, contradicting inequivalence; hence HomG(Vj,Vi)=0.

F3given
1.2

For i=j, [F4] says every element of HomG(Vi,Vi)=EndG(Vi) is a scalar multiple of the identity. The identity operator is nonzero, so the scalars λidVi form a one-dimensional complex line. Hence dimHomG(Vi,Vi)=1.

F4given
2.1

By [F2], χi,χj=dimHomG(Vj,Vi); steps 1.1 and 1.2 give this dimension to be 0 when ij and 1 when i=j. This is exactly χi,χj=δij.

F1F2step 1.1step 1.2algebra
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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 GC 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=1Ggφ(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 charkG, 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 dimCcf(G) equals the number of conjugacy classes of G.

Proof

technique · direct
1.1

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].

F2F3given
1.2

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

F4given
2.1

By [A1], dimCcf(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.

A1step 1.1step 1.2
3.1

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

F1step 1.1step 2.1algebra
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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 Vj=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 VjmjVj, 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 charkG, 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
1.1

Since charC=0 does not divide G, [F1] gives a decomposition VjmjVj with each Vj irreducible.

F1given
2.1

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

F2step 1.1
3.1

Taking the inner product with χi, linearity in the first argument and [F3] give χV,χi=jmjχj,χi=jmjδji=mi.

F3step 2.1algebra
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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

VWχV=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 charkG, 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
1.1

By [F1] there are decompositions VimiVi and WiniVi over the irreducible representations Vi. By [F2] the multiplicities are mi=χV,χi and ni=χW,χi.

F1F2given
1.2

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

A1given
2.1

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 VW. This proves the forward implication.

step 1.1algebra
3.1

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

step 2.1step 1.2
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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 VimiVi.

[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 VimiVi 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=imiχi,χ=imiχ,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
1.1

Completely reduce VimiVi with mi0 integers. By [F2] this decomposition gives χ=imiχi, and by [F1] each mi=χ,χi; therefore [A1] gives χ,χ=imi2.

F1F2A1given
2.1

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.

A2step 1.1
2.2

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], VVi is irreducible, so χ is irreducible.

A2step 1.1algebra
3.1

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

step 2.1step 2.2
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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,g1.

Facts & Assumptions

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

[F1]

The regular representation acts on the basis ([h])hG 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
1.1

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 hG with gh=h.

F1F2given
2.1

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 g1.

A1step 1.1algebra
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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: χVW=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
1.1

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 1GGχi(1)=χi(1)=ni, because the only nonzero term is at g=1.

F1F2A1given
2.1

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.

F1F3step 1.1algebra
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The second orthogonality relation for irreducible complex characters

Statement

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

i=1rχi(g)χi(h)={CG(g),hClG(g),0,hClG(g).

Facts & Assumptions

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

[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=1Gxφ(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 xClG(g) and δg(x)=0 otherwise, is a class function, and its class has ClG(g)=G/CG(g) elements.

[A2]

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

Proof

technique · direct
1.1

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

F1A1A2given
1.2

By [F2], δg,χi=1Gxδg(x)χi(x). By [A1] the only nonzero terms are at xClG(g), where δg(x)=1 and χi(x)=χi(g) because characters are class functions. Hence ci=G/CG(g)Gχi(g)=χi(g)/CG(g).

F2A1givenalgebra
2.1

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

step 1.1step 1.2algebra
3.1

By [A1], δg(h)=1 exactly when hClG(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.

A1step 2.1algebra
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For gG, 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 gG,

i=1rχi(g)2=CG(g).

Facts & Assumptions

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

[F1]

For g,hG, 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
1.1

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

F1given
2.1

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

step 1.1algebra
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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
1.1

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

F1given
1.2

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.

F2givenalgebra
2.1

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.

step 1.1step 1.2algebra
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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 NG be a normal subgroup (Normal subgroup: invariance under conjugation), and let ρ:GGL(V) be a representation over a field k with Nkerρ. Then:

  1. the formula ρ(gN):=ρ(g) defines a well-posed representation ρ:G/NGL(V) with ρ=ρπ, where π:GG/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 NG, a field k, and a representation ρ:GGL(V) with Nkerρ.

[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:GG/N, π(g)=gN, is a surjective group homomorphism (The canonical projection π:GG/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 g1N (For NG, the cosets form a group with identity N and inverse (gN)1=g1N).

Proof

technique · direct
1.1

If gN=hN, then g1hN by the coset laws of [F3], so ρ(g1h)=e by [F1] and the hypothesis Nkerρ; hence ρ(h)=ρ(g). Therefore ρ(gN):=ρ(g) is independent of the chosen coset representative.

F1F3given
2.1

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=ρπ by [F2]. This proves claim 1.

F2F3step 1.1given
3.1

A subspace UV 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.

F2step 2.1given
4.1

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.

step 3.1algebra
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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 charkG, 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
1.1

Assume NG. Let ρ:G/NGL(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}.

F4given
1.2

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

F3A2given
2.1

Since charC=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(ρUj)={N}.

F1A1step 1.1given
3.1

For each j, let Vj be the inflation of Uj to G. By [F2], each Vj is irreducible as a representation of G, and kerρVj=π1(ker(ρUj)), which contains N. By [F3] each kerρVj equals the kernel of the corresponding irreducible character χj of G.

F2F3step 2.1given
4.1

By [A1] and steps 2.1 and 3.1, jkerχj=jπ1(ker(ρUj))=π1({N})=N. Hence a normal subgroup is an intersection of kernels of irreducible characters.

A1step 2.1step 3.1algebra
5.1

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

step 4.1step 1.2
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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 EndG(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 charkG, 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
1.1

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

F1F2given
1.2

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

F4given
1.3

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

F5given
2.1

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.

F3A1step 1.1given
2.2

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.

A2step 1.2step 1.3
3.1

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

step 2.1step 2.2

5 · Examples, counterexamples and false statements

None yet.

Sources