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Characters and the Orthogonality Relations
1 · Prerequisites
- Binary Operations, Monoids, Groups and Subgroups
- Chain Conditions, Semisimple Modules and the Wedderburn–Artin Theorem
- Construction of the Natural Numbers
- Construction of the Real Numbers via Cauchy Sequences
- Construction of the Real Numbers via Dedekind Cuts
- Cosets, Index and Lagrange's Theorem
- Countability and Uncountability
- Determinants of Matrices over a Commutative Ring
- Diagonalisation and the Minimal Polynomial
- Divisibility, Euclidean Domains, Principal Ideal Domains and Unique Factorisation
- Divisibility, Greatest Common Divisors and Bézout's Identity
- Dual Spaces, Bilinear and Quadratic Forms, and Sylvester's Law of Inertia
- Eigenvalues, Eigenvectors and the Characteristic Polynomial
- Finite Counting, Factorials and Binomial Coefficients
- Foundations of the Real Numbers for Analysis
- Free Modules, Exact Sequences, Projective and Injective Modules
- Group Actions, Orbits, Stabilisers and Cayley's Theorem
- Group Homomorphisms and the Isomorphism Theorems
- Ideals, Quotient Rings and the Isomorphism Theorems for Rings
- Inner Product Spaces, Gram-Schmidt, Projections and Adjoints
- Linear Independence, Bases and Dimension
- Linear Transformations, Rank-Nullity and Quotient Spaces
- Maschke's Theorem, Complete Reducibility and the Structure of k[G]
- Matrices, the Matrix of a Linear Map, and Change of Basis
- Modules, Submodules, Quotient Modules and the Isomorphism Theorems
- Normal Subgroups and Quotient Groups
- Order, Zorn's Lemma, and the Axiom of Choice
- Polynomial Rings, the Division Algorithm and Roots
- Relations, Functions, and Quotients
- Rings, Subrings, Integral Domains and Fields
- Roots, Rational Powers, and Classical Inequalities
- Simple Field Extensions and the Construction of the Complex Numbers
- Symmetric Groups, Cycle Decomposition and the Sign Homomorphism
- Tensor Products of Modules
- The Determinant of a Linear Operator, Cofactors and Cramer's Rule
- The Group Algebra and Representations of Finite Groups
- The ZFC Axioms and the Basic Set Constructions
- Triangularisation, Generalised Eigenspaces and Jordan Canonical Form
- Vector Spaces, Linear Subspaces, Span and Direct Sums
2 · Summary
This page is ordinary character theory: throughout, is finite, the base field is , 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 , the space 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: , class-function invariance, the description of as a sum of roots of unity, the bound with its scalar-equality case, and . 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 , converts into . 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 . From that basis flow the multiplicity formula, the fact that a representation is determined by its character, the irreducibility test , 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 this shows normal subgroups are exactly intersections of kernels of irreducible characters. The page ends with the criterion that is abelian exactly when every irreducible complex character has degree .
3 · Logical flowchart
4 · Definitions, theorems and proofs
Standing hypotheses for ordinary character theory: finite, , and every representation finite-dimensional
Remark
The ordinary-character-theory items on this page work inside the following setting, fixed once here: is a finite group, the base field is , and every representation is finite-dimensional (A finite-dimensional representation 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 is what makes the hypotheses of the published representation-theory spine available. Since and is finite, the characteristic does not divide , so every finite-dimensional representation is completely reducible (If , every finite-dimensional representation of is completely reducible), and the field is algebraically closed, which feeds the count of irreducibles against conjugacy classes (If is algebraically closed and , the number of irreducible representations of 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.
The character of a finite-dimensional complex representation
Definition
Let be a finite-dimensional complex representation (A finite-dimensional representation over a field, and its degree). Its character is the function
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 or 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 : 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 is an invertible intertwiner, then , so by the identity (For and , ). Thus the character depends only on the equivalence class of the representation.
Class functions and the complex vector space
Definition
Let be a finite group. A function is a class function when it is constant on every conjugacy class (The conjugacy class and centralizer of an element):
The set of all class functions is written
With pointwise addition and pointwise scalar multiplication , the set is a complex vector space (Vector space over a field): the axioms (V1)–(V5) are inherited from the field at every argument . Because is finite, the evaluation functions are complex-linear, and a class function is determined by its values on one representative of each conjugacy class.
The standard inner product on
Definition
For two class functions (Class functions and the complex vector space ), define
the finite sum being that of (The sum 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 and linearity of the finite sum. The conjugate-symmetry clause holds because conjugation (Real and imaginary parts, complex conjugation, and modulus) is an involution: . For definiteness, is a sum of nonnegative real numbers, hence nonnegative real; it is exactly when every term is , and since holds only for in , that says exactly as a function.
An irreducible complex character
Definition
A complex character (The character of a finite-dimensional complex representation) is irreducible when for an irreducible representation of over (Subrepresentations, direct sums of representations, and irreducibility).
The phrase is well defined for characters rather than for individual representations: if and are equivalent, then by the invariance clause of (The character of a finite-dimensional complex representation), and equivalence preserves irreducibility. Throughout the page, "the irreducible characters" means one representative from each equivalence class of irreducible complex representations of the finite group ; the number is finite and equals the number of conjugacy classes of by the published count (If is algebraically closed and , the number of irreducible representations of equals the number of conjugacy classes).
The character table of a finite group
Definition
Let be a finite group, let be the irreducible complex characters of , one per equivalence class (An irreducible complex character), and let be representatives of the distinct conjugacy classes of (The conjugacy class and centralizer of an element). The character table of is the array whose rows are indexed by , whose columns are indexed by , and whose entry is .
The array is written with a top row recording the conjugacy-class sizes above the column labels, and with the first column recording the degrees 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 is constant on each conjugacy class, the entry is independent of the choice of representative . 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.
The tensor product of two complex representations
Definition
Let and be finite-dimensional complex representations of a group (A finite-dimensional representation over a field, and its degree). Their tensor product representation is the representation of on the complex tensor product given on elementary tensors by
The action is well defined. For fixed , the assignment is additive in each variable and -balanced, because and are -linear: . By Universal property of the tensor product for balanced maps into abelian groups there is a unique -linear map with the displayed value on every elementary tensor. Because the two sides of agree on every elementary tensor and the elementary tensors span , the assignment is a group homomorphism into , with . The underlying space is finite-dimensional: bases of and give the basis , so .
The dual or contragredient complex representation
Definition
Let be a finite-dimensional complex representation of a group (A finite-dimensional representation over a field, and its degree). Its dual or contragredient representation is the representation of on the algebraic dual space (Linear functionals and the algebraic dual ) defined by
equivalently .
Each operator is -linear because and are. The inverse appears so that composition is order-preserving: writing ,
and . Hence is a group homomorphism , a representation of the same degree as .
The kernel of a complex character
Definition
Let be a finite-dimensional complex representation of a finite group , with character (The character of a finite-dimensional complex representation). The kernel of the character is
The value 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 of the representation, so the two readings of "kernel" never conflict on this page.
For a complex character, , is a class function, and with equality exactly at scalars
Statement
Let be a finite group, let be a finite-dimensional complex representation of , and let be its character. Then, for all :
- ;
- , so is a class function;
- is a sum of roots of unity, namely the eigenvalues of counted with multiplicity;
- , with equality if and only if is a scalar operator;
- .
Facts & Assumptions
Given: A finite group , a finite-dimensional complex representation with character .
The character is , the trace of the action operator (The character of a finite-dimensional complex representation).
In a finite group, every element has finite order dividing (The order of every element of a finite group divides the order of the group).
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 , every element of finite order acts diagonalisably in a finite-dimensional representation).
If the characteristic polynomial of an endomorphism splits as , then its trace is the sum of the eigenvalues (If in , then : trace is the sum of the eigenvalues counted with algebraic multiplicity).
whenever the products are defined (For and , ).
Complex conjugation distributes over addition and multiplication, fixes real numbers, and satisfies (Real and imaginary parts, complex conjugation, and modulus).
For complex numbers , the triangle inequality holds, and equality holds exactly when all the nonzero share one argument.
A function is a class function exactly when it is constant on every conjugacy class (Class functions and the complex vector space ).
Proof
In any ordered basis of the matrix of is the identity matrix, whose trace is the number of basis vectors. Hence , which is claim 1.
For the same reason and compose as in the group, so , the middle step being [A3] applied to and .
By [F2], has finite order with , so . Since is algebraically closed of characteristic zero, [A1] gives a basis of in which is diagonal with diagonal entries , where ; each , so each is a root of unity.
By step 1.2 the value of does not change under conjugation, so is constant on each conjugacy class and hence is a class function in the sense of [A6], which is claim 2.
In that basis the characteristic polynomial of is , so [A2] gives , a sum of roots of unity, which is claim 3.
Conversely, if for a scalar , first consider the degenerate case . Then , so the equality clause of claim 4 holds. If , then the identity of step 1.3 reads , and evaluating it on a nonzero vector gives . Thus is a root of unity and ; then and . This closes the biconditional in claim 4.
The inverse operator has the inverse eigenvalues, and a root of unity satisfies because .
Applying [A5] to the eigenvalues of step 1.3 gives , which is the inequality in claim 4.
Equality holds in step 3.1 exactly when the equality clause of [A5] applies: since every , all the eigenvalues share one argument, so all the are equal to one root of unity . The diagonal form of step 1.3 then shows , a scalar operator.
Hence, using [A2] for and the additivity of conjugation from [A4], , which is claim 5.
The kernel of a complex character agrees with the kernel of any representation affording it
Statement
Let be a finite-dimensional complex representation of a finite group , with character . Then
where is the kernel of the group homomorphism and is the kernel of the character.
Facts & Assumptions
Given: A finite group and a finite-dimensional complex representation with character .
The kernel of a group homomorphism is the set of elements sent to the identity (The kernel and image of a group homomorphism).
The kernel of the character is (The kernel of a complex character).
For the character, , and holds exactly when is a scalar operator (For a complex character, , is a class function, and with equality exactly at scalars).
Proof
If , then by [F1], so by [F3]. Hence by [F2], which proves .
Conversely, let , so by [F2]. Then in particular , and the equality clause of [F3] gives for a scalar . Evaluating the character at and at with [F3] gives .
If , then is the zero space, holds for every , and , so by [F1] and [F2]; the statement is immediate. Hence assume .
Since by [F3], the equality of step 1.2 forces ; therefore , so by [F1]. Together with step 1.1 and step 1.3 this proves .
Characters add on direct sums, multiply on tensor products, and conjugate on duals
Statement
Let be a finite group and let and be finite-dimensional complex representations of . Then, for every :
- ;
- ;
- .
Facts & Assumptions
Given: Finite-dimensional complex representations , of a finite group , and an element .
The character is (The character of a finite-dimensional complex representation).
The dual action is (The dual or contragredient complex representation).
The tensor-product action is (The tensor product of two complex representations).
In a basis made of a basis of followed by a basis of , the matrix of the direct-sum action on is block diagonal with the two action matrices on its diagonal, so its trace is the sum of the two block traces.
Trace is linear, in particular additive, on the space of square matrices (Trace is a linear functional on ).
A basis of and a basis of give the basis of the tensor product (The elementary tensors of two bases form the product basis of the tensor product).
For a complex character, (For a complex character, , is a class function, and with equality exactly at scalars).
Proof
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 by [F1]. This is claim 1.
By [F3] and [A3], the matrix of the tensor-product action in the basis has entry at the position , because .
With dual bases of and of , the matrix of the dual action of [F2] is the transpose of the matrix of : writing gives .
Its trace is the sum of its diagonal entries: . By [F1] the two factors are and , so , which is claim 2.
A matrix and its transpose have the same diagonal entries, hence the same trace, so . By [A4], , which is claim 3.
The character of a permutation representation counts fixed points
Statement
Let be a finite left -set and let be the permutation representation over . Then the character of is
the number of fixed points of acting on .
Facts & Assumptions
Given: A finite group , a finite left -set , and .
The character is (The character of a finite-dimensional complex representation).
The permutation representation has basis and action (The trivial representation, the regular representation, and permutation representations from finite -sets).
Proof
In the basis of [F2], the matrix of is the permutation matrix with a in row , column : its -th diagonal entry is exactly when , and otherwise.
The trace of that matrix is the sum of its diagonal entries, counting one for each with . By [F1] this is , which equals the stated number of fixed points.
The fixed subspace of a representation
Definition
Let be a representation of a group over a field (A finite-dimensional representation over a field, and its degree). The fixed subspace of is
The set is a linear subspace of (Linear subspace of a vector space): it contains because every operator is linear, it is closed under addition and scalar multiplication for the same reason, and the defining equations are the required pointwise identities. When is finite and , The averaging operator projects onto the fixed subspace shows that an explicit projection of has image exactly .
The averaging operator projects onto the fixed subspace
Statement
Let be a finite-dimensional representation of a finite group over . The averaging operator
satisfies and has image exactly ; consequently .
Facts & Assumptions
Given: A finite group and a finite-dimensional complex representation .
The fixed subspace is (The fixed subspace of a representation).
If is a projection of a finite-dimensional vector space, meaning , then , restricts to the identity on , and in a basis adapted to that decomposition the matrix of is block diagonal with an identity block and a zero block.
Proof
For , , because left translation by permutes , so the sums run over the same index set.
Hence for every , the vector satisfies for every , so by [F1]; thus .
If , then for every by [F1], so . Thus , and with step 2.1, exactly.
For , step 3.1 gives ; for general , by step 2.1, so . Hence .
By [A2] applied to from step 4.1, and the matrix of in an adapted basis has an identity block of size and a zero block. Its trace is therefore , and step 3.1 identifies with .
For finite-dimensional complex , the intertwiners are exactly the fixed points of
Statement
Let be a finite group, let and be finite-dimensional complex representations of , and let
be the natural isomorphism of For finite-dimensional , the canonical map is an isomorphism, sending to the map . Then is an intertwiner between the diagonal representation on and the conjugation representation on , and it carries the fixed subspace bijectively onto .
Facts & Assumptions
Given: Finite-dimensional complex representations and of a finite group .
The dual action is (The dual or contragredient complex representation).
The diagonal action on the tensor product is (The tensor product of two complex representations).
The fixed subspace of a representation is the set of vectors fixed by every (The fixed subspace of a representation).
Intertwiners are the linear maps with for all (Intertwiners, the spaces and , equivalent representations, and faithful representations).
The map induces a natural isomorphism (For finite-dimensional , the canonical map is an isomorphism).
Proof
For an elementary tensor and , by [F2], and this equals by [F1].
The right-hand side of step 1.1 is , which is the value at of the conjugation action on the linear map . Since elementary tensors span the tensor product, is an intertwiner of representations.
An element of is fixed by exactly when its image under is fixed by , because is a bijective intertwiner of step 2.1; so by [F3].
A linear map is fixed by the conjugation action of every exactly when for all , i.e. when ; by [F4] this says precisely that . Hence , which combines with step 3.1 into the claim.
The class-function inner product equals
Statement
Let be a finite group and let and be finite-dimensional complex representations of with characters and . Then
Facts & Assumptions
Given: Finite-dimensional complex representations and of a finite group , with characters and .
The inner product of class functions is (The standard inner product on ).
For a finite-dimensional complex representation , the averaging operator has image and trace (The averaging operator projects onto the fixed subspace).
The fixed points of the diagonal representation on are exactly the intertwiners , so (For finite-dimensional complex , the intertwiners are exactly the fixed points of ).
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).
The character of a representation is by definition the trace of its action operator, .
Proof
By [F3], . By [F2] applied to , this equals for the averaging operator .
Trace is linear, so , the second equality by [A1].
By [F4], the tensor-product and dual clauses give .
Combining steps 1.1 through 1.3, , reordering the product of two complex numbers in each summand. This is by [F1].
The first orthogonality relation for irreducible complex characters
Statement
Let be a finite group and let be the irreducible complex characters of , one from each equivalence class. Then
Facts & Assumptions
Given: A finite group , and irreducible complex representations of with characters , one from each equivalence class.
Irreducible characters are the characters of irreducible representations (An irreducible complex character).
The inner product computes intertwiner dimension: (The class-function inner product equals ).
Every nonzero intertwiner between irreducible representations is an isomorphism, and in particular is a division ring (Schur's lemma for irreducible representations: a nonzero intertwiner is an isomorphism, and is a division ring).
Over the algebraically closed field , every intertwiner is a scalar operator (Over an algebraically closed field, every endomorphism of an irreducible representation is scalar).
Proof
For the representations and are inequivalent, because one representative was chosen from each class. If were nonzero, [F3] would make an isomorphism, contradicting inequivalence; hence .
For , [F4] says every element of is a scalar multiple of the identity. The identity operator is nonzero, so the scalars form a one-dimensional complex line. Hence .
By [F2], ; steps 1.1 and 1.2 give this dimension to be when and when . This is exactly .
The irreducible complex characters form an orthonormal basis of
Statement
Let be a finite group. The irreducible complex characters of , , form an orthonormal basis of the complex vector space of class functions on , with respect to the standard inner product.
Facts & Assumptions
Given: A finite group and its irreducible complex characters .
The space consists of the functions constant on conjugacy classes, with pointwise operations (Class functions and the complex vector space ).
The standard inner product on is the Hermitian form (The standard inner product on ).
The irreducible characters are orthonormal: (The first orthogonality relation for irreducible complex characters).
Over an algebraically closed field of characteristic not dividing , the number of irreducible representations up to equivalence equals the number of conjugacy classes (If is algebraically closed and , the number of irreducible representations of equals the number of conjugacy classes).
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 ; in particular equals the number of conjugacy classes of .
Proof
By [F3] the family is orthonormal, hence linearly independent: any relation has inner product with equal to , because and the inner product is linear in its first argument by [F2].
The group is finite, is algebraically closed, and does not divide , so [F4] applies and equals the number of conjugacy classes of .
By [A1], equals the number of conjugacy classes, which step 1.2 identified with . Hence the linearly independent family of elements from step 1.1 has as many elements as the dimension of the space.
A linearly independent family whose size equals the dimension spans, so the orthonormal family of step 1.1 is a basis of .
The multiplicity of an irreducible summand is a character inner product
Statement
Let be a finite group, let be a finite-dimensional complex representation of . Choose representatives of the irreducible representations, write , and write . Then the multiplicity of in is
Facts & Assumptions
Given: A finite group , a finite-dimensional complex representation of , representatives of the irreducible representations with characters , a decomposition , and an index .
Every finite-dimensional representation of a finite group over a field of characteristic not dividing is completely reducible (If , every finite-dimensional representation of is completely reducible).
Characters add on direct sums (Characters add on direct sums, multiply on tensor products, and conjugate on duals).
Irreducible characters are orthonormal: (The first orthogonality relation for irreducible complex characters).
Proof
Since does not divide , [F1] gives a decomposition with each irreducible.
Applying [F2] iteratively to the decomposition of step 1.1 gives , a finite sum because is finite-dimensional.
Taking the inner product with , linearity in the first argument and [F3] give .
Finite-dimensional complex representations of a finite group are determined up to isomorphism by their characters
Statement
Let be a finite group and let and be finite-dimensional complex representations of . Then
Facts & Assumptions
Given: A finite group and finite-dimensional complex representations and of .
Every finite-dimensional representation of a finite group over a field of characteristic not dividing is completely reducible (If , every finite-dimensional representation of is completely reducible).
For irreducible , the multiplicity of in a completely reduced representation is (The multiplicity of an irreducible summand is a character inner product).
Equivalent representations have equal characters.
Proof
By [F1] there are decompositions and over the irreducible representations . By [F2] the multiplicities are and .
Conversely, if via an invertible intertwiner, then [A1] gives . This proves the reverse implication.
Assume . Then every inner product in step 1.1 agrees, so for every ; the two direct sums are built from the same irreducibles with the same multiplicities, hence . This proves the forward implication.
Steps 2.1 and 1.2 prove both implications, hence the equivalence.
A complex character is irreducible if and only if its self-inner-product is
Statement
Let be a finite group and let be a complex character of . Then is irreducible if and only if .
Facts & Assumptions
Given: A finite group and a complex character of a finite-dimensional representation , completely reduced as .
The multiplicity of in is (The multiplicity of an irreducible summand is a character inner product).
Characters add on direct sums, so the decomposition gives (Characters add on direct sums, multiply on tensor products, and conjugate on duals).
If and , then conjugate-symmetry and linearity in the first argument give , because the multiplicities are nonnegative integers.
A representation is irreducible exactly when it has one irreducible summand with multiplicity and no others.
Proof
Completely reduce with integers. By [F2] this decomposition gives , and by [F1] each ; therefore [A1] gives .
Assume is irreducible. Then by [A2] exactly one multiplicity equals and the rest are , so the sum in step 1.1 is ; hence .
Conversely, assume . The sum of nonnegative integers in step 1.1 can equal only when exactly one equals and all the others are . By [A2], is irreducible, so is irreducible.
Steps 2.1 and 2.2 prove the two implications, hence the biconditional.
The regular character is at and away from
Statement
Let be a finite group and let be the character of the regular representation over . Then
Facts & Assumptions
Given: A finite group , an element , and the regular representation .
The regular representation acts on the basis by (The trivial representation, the regular representation, and permutation representations from finite -sets).
The character of a permutation representation counts fixed points (The character of a permutation representation counts fixed points).
In a group, holds if and only if , by cancellation.
Proof
By [F1], the regular action is the permutation representation of the left -set with basis indexed by the group elements. By [F2], is the number of with .
By [A1], has a solution in only when , in which case every is fixed. Hence the count of step 1.1 is for and for .
The regular character gives a second proof of the sum-of-squares formula
Statement
Let be a finite group and let be its irreducible complex characters, of degrees . Then
Facts & Assumptions
Given: A finite group with irreducible characters , the regular representation , and the regular character .
The regular character is at and elsewhere (The regular character is at and away from ).
The multiplicity of an irreducible summand is an inner product: (The multiplicity of an irreducible summand is a character inner product).
Characters add on direct sums: (Characters add on direct sums, multiply on tensor products, and conjugate on duals).
The degree of a character is its value at , .
Proof
By [F2], the regular representation decomposes as with . By [F1] and the definition of the inner product this is , because the only nonzero term is at .
Characters add on direct sums by [F3], so evaluating both sides of the decomposition of step 1.1 at gives . By [F1] the left side is , and step 1.1 gives , so .
The second orthogonality relation for irreducible complex characters
Statement
Let be a finite group with irreducible complex characters . For ,
Facts & Assumptions
Given: A finite group , its irreducible characters , and elements .
The irreducible characters form an orthonormal basis of the class functions (The irreducible complex characters form an orthonormal basis of ).
The inner product of class functions is (The standard inner product on ).
The indicator function of the conjugacy class of , defined by for and otherwise, is a class function, and its class has elements.
If in the orthonormal basis of [F1], then .
Proof
The function of [A1] is a class function, so by [F1] it expands as with by [A2].
By [F2], . By [A1] the only nonzero terms are at , where and because characters are class functions. Hence .
Evaluating the expansion of step 1.1 at and substituting the coefficients of step 1.2 gives .
By [A1], exactly when , and otherwise. Multiplying the identity of step 2.1 by and taking complex conjugates gives , which is the stated formula.
For , the sum of over the irreducible complex characters is
Statement
Let be a finite group with irreducible complex characters . For every ,
Facts & Assumptions
Given: A finite group , its irreducible characters , and an element .
For , when is conjugate to , and otherwise (The second orthogonality relation for irreducible complex characters).
Proof
The element is conjugate to itself, so applying [F1] with gives .
Each summand equals , so the sum of step 1.1 is exactly , which proves the claim.
The character table is square and invertible
Statement
Let be a finite group. The character table of has as many rows as columns, and the table matrix is invertible.
Facts & Assumptions
Given: A finite group with irreducible characters and conjugacy-class representatives .
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 ).
The columns of the character table satisfy the second orthogonality relation: (The second orthogonality relation for irreducible complex characters).
Proof
By [F1], equals the number of conjugacy classes of , which is the number of columns; hence the table is square.
Rescale the -th column of the table by the positive number , forming the matrix . By [F2], distinct columns of are orthogonal and each column has squared norm , so has orthonormal columns.
A square matrix with orthonormal columns has linearly independent columns, hence is invertible. Since is obtained from the table by multiplying columns by nonzero scalars, the table matrix is invertible too.
A representation with kernel containing a normal subgroup factors through the quotient, and irreducibility is unchanged by inflation
Statement
Let be a group, let be a normal subgroup (Normal subgroup: invariance under conjugation), and let be a representation over a field with . Then:
- the formula defines a well-posed representation with , where is the canonical quotient map;
- is irreducible as a representation of if and only if it is irreducible as a representation of .
Facts & Assumptions
Given: A group , a normal subgroup , a field , and a representation with .
The kernel of a group homomorphism is the preimage of the identity (The kernel and image of a group homomorphism).
The canonical map , , is a surjective group homomorphism (The canonical projection , , is a surjective group homomorphism).
The cosets form the group under , with identity and inverse (For , the cosets form a group with identity and inverse ).
Proof
If , then by the coset laws of [F3], so by [F1] and the hypothesis ; hence . Therefore is independent of the chosen coset representative.
By [F3], ; applying gives , so is a group homomorphism into , with by [F2]. This proves claim 1.
A subspace is -stable exactly when it is -stable, because of [F2] is surjective and : the two stability conditions quantify over the same operators.
A representation is irreducible exactly when its only stable subspaces are and . By step 3.1 the stable subspaces for and for coincide, so the two irreducibility statements are equivalent, which is claim 2.
The normal subgroups of a finite group are exactly the intersections of kernels of irreducible complex characters
Statement
Let be a finite group. A subgroup of is normal if and only if it is an intersection of kernels of irreducible complex characters of .
Facts & Assumptions
Given: A finite group and a subgroup of .
Every finite-dimensional representation of a finite group over a field of characteristic not dividing is completely reducible (If , every finite-dimensional representation of is completely reducible).
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).
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).
The regular representation over a field is faithful (The regular representation is faithful).
The kernel of a direct sum of representations is the intersection of the kernels of its summands.
A kernel of a group homomorphism is a normal subgroup, and an intersection of normal subgroups is normal.
Proof
Assume . Let be the regular representation of the quotient. Since is faithful by [F4], its kernel in is the trivial subgroup .
Conversely, if for irreducible characters of , then by [F3] each is the kernel of a group homomorphism, hence normal by [A2], and the intersection of normal subgroups is again normal by [A2]. Thus .
Since does not divide , [F1] decomposes as a direct sum of irreducible representations of ; by [A1] the trivial kernel of step 1.1 is the intersection of their kernels, so .
For each , let be the inflation of to . By [F2], each is irreducible as a representation of , and , which contains . By [F3] each equals the kernel of the corresponding irreducible character of .
By [A1] and steps 2.1 and 3.1, . Hence a normal subgroup is an intersection of kernels of irreducible characters.
Steps 4.1 and 1.2 prove the two implications, hence the equivalence.
A finite group is abelian if and only if all its irreducible complex characters have degree
Statement
Let be a finite group. Then is abelian if and only if every irreducible complex character of has degree .
Facts & Assumptions
Given: A finite group with irreducible characters of degrees .
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).
A field is a splitting field for when every irreducible representation has (A splitting field for a finite group: every irreducible representation has scalar endomorphism ring).
Every irreducible representation of a finite abelian group over a splitting field has degree (Every irreducible representation of a finite abelian group over a splitting field is one-dimensional).
The degrees of the irreducible characters satisfy (The regular character gives a second proof of the sum-of-squares formula).
The number of irreducible representations of up to equivalence equals the number of conjugacy classes, when the field is algebraically closed of characteristic not dividing (If is algebraically closed and , the number of irreducible representations of equals the number of conjugacy classes).
The degree of an irreducible character is the dimension of any representation affording it.
A finite group is abelian exactly when every conjugacy class has one element.
Proof
Assume is abelian. For every irreducible representation of , [F1] gives ; by [F2], is a splitting field for .
Conversely, assume every irreducible character has degree , so for all . By [F4], .
Since is algebraically closed and does not divide , [F5] applies: equals the number of conjugacy classes of .
By [F3] applied over this splitting field, every irreducible representation of the abelian group has degree ; by [A1] every irreducible character has degree . This proves the forward implication.
Steps 1.2 and 1.3 show equals the number of conjugacy classes, so the conjugacy classes each have exactly one element; by [A2], is abelian. This proves the reverse implication.
Steps 2.1 and 2.2 prove the two implications, hence the equivalence.
5 · Examples, counterexamples and false statements
None yet.
Sources
- Peter Webb, A Course in Finite Group Representation Theory, Chapter 3
- Pavel Etingof et al., Introduction to Representation Theory, Section 3.3
- Peter Webb, A Course in Finite Group Representation Theory, Proposition 3.1.1
- Peter Webb, A Course in Finite Group Representation Theory, Section 3.2
- Pavel Etingof et al., Introduction to Representation Theory, Section 3.5
- Pavel Etingof et al., Introduction to Representation Theory, Theorem 3.7
- Peter Webb, A Course in Finite Group Representation Theory, Section 3.1
- Pavel Etingof et al., Introduction to Representation Theory, Section 3.4
- Shani Meynet and Robert Moscrop, McKay quivers and decomposition, Appendix A.3
- Peter Webb, A Course in Finite Group Representation Theory, Proposition 3.1.3
- Peter Webb, A Course in Finite Group Representation Theory, Example 4.3.4
- Peter Webb, A Course in Finite Group Representation Theory, Lemma 3.2.2
- Peter Webb, A Course in Finite Group Representation Theory, Lemma 3.2.1 and Proposition 3.1.3
- Pavel Etingof et al., Introduction to Representation Theory, Theorem 3.8
- Peter Webb, A Course in Finite Group Representation Theory, Theorem 3.2.3
- Peter Webb, A Course in Finite Group Representation Theory, Section 3.3
- Peter Webb, A Course in Finite Group Representation Theory, Corollary 3.3.1
- Peter Webb, A Course in Finite Group Representation Theory, Corollary 3.3.3
- Peter Webb, A Course in Finite Group Representation Theory, Corollary 3.3.4
- Peter Webb, A Course in Finite Group Representation Theory, Lemma 3.3.6
- Peter Webb, A Course in Finite Group Representation Theory, Corollary 3.3.7
- Peter Webb, A Course in Finite Group Representation Theory, Corollary 3.4.4
- Pavel Etingof et al., Introduction to Representation Theory, Theorem 3.9
- Peter Webb, A Course in Finite Group Representation Theory, Section 3.4
- Peter Webb, A Course in Finite Group Representation Theory, Theorem 3.4.3
- Peter Webb, A Course in Finite Group Representation Theory, Section 4.2
- Peter Webb, A Course in Finite Group Representation Theory, Theorem 4.1.5