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Induced Representations, Frobenius Reciprocity and Applications
1 · Prerequisites
- Binary Operations, Monoids, Groups and Subgroups
- Chain Conditions, Semisimple Modules and the Wedderburn–Artin Theorem
- Characters and the Orthogonality Relations
- Composition Series, the Jordan–Hölder Theorem and Solvable Groups
- Congruences, the Integers Modulo n and the Chinese Remainder 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
- Cyclic Groups and Direct Products
- 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
- Primes, Euclid's Lemma and the Fundamental Theorem of Arithmetic
- 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
- Sylow's Theorems, p-Groups and Nilpotent Groups
- 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 builds induction in the function model, shows how a choice of left coset representatives turns it into a direct sum, and derives the resulting character formula. From there the adjunction with restriction becomes the structural spine: Frobenius reciprocity, transitivity, the projection formula, and the corollary that an irreducible character of is recovered from an irreducible constituent of its restriction.
The second thread is Mackey theory. Double cosets and conjugate characters make the restriction of an induced character decomposable by double-coset pieces, and that decomposition sharpens into Mackey's irreducibility criterion.
The final thread is arithmetic. Central characters turn class sums into scalars, those scalars are algebraic integers, and that integrality forces the degree of an irreducible complex character to divide both and . The page ends with the character-theoretic applications to prime-power conjugacy classes and Burnside's theorem.
3 · Logical flowchart
4 · Definitions, theorems and proofs
The induced -linear -module as -covariant functions on
Definition
Let be a commutative ring, let be a group, let be a subgroup (Subgroup), and let be an -linear -module (An -linear action of on a left -module, and a -module over ).
The induced -linear -module is the set
where is the set of all functions (The set of all functions ).
Pointwise addition and scalar multiplication make an -module:
The left action of on this module is
This action is well defined on the displayed subset because
and each operator is -linear by the pointwise definitions. Thus is an -linear -module.
Remarks
-
The covariance condition is written on the right, so the values of an induced function are determined by one value on each left coset .
-
When is a field and is finite-dimensional, this construction is the induced representation of from the representation of on .
A left transversal identifies with a direct sum of copies of
Statement
Let be a commutative ring, let be a group, let , let be an -linear -module, and let meet each left coset in exactly one point. Then evaluation on defines an -module isomorphism
In particular .
Facts & Assumptions
Given: A commutative ring , a group , a subgroup , an -linear -module , and a left transversal for .
The induced module consists of the functions satisfying , with pointwise -module structure (The induced -linear -module as -covariant functions on ).
The left cosets of are the subsets of (Left and right cosets and of a subgroup).
For a finite index set, the direct sum is the module of tuples with coordinatewise operations (The direct sum of an indexed family of modules).
Proof
Because meets each left coset in exactly one point, every can be written uniquely as with and .
The map is -linear because [F1] and [F3] define both module structures coordinatewise.
Define by . Step 1.1 makes this well defined, and the displayed formula satisfies the covariance condition of [F1], so .
The map is -linear because the -action on is -linear and the formula of step 2.1 is coordinatewise in the tuple entries.
For , because .
For and as in step 1.1, one has , where the third equality is the covariance condition from [F1]. Hence .
Steps 3.2 and 3.3 show that and are inverse -module isomorphisms. Since has one element on each left coset, its cardinality is .
The dimension of an induced finite-dimensional representation is
Statement
Let be a field, let be a finite group, let , and let be a finite-dimensional representation of over . Then is a finite-dimensional representation of over and
Facts & Assumptions
Given: A field , a finite group , a subgroup , and a finite-dimensional representation of over .
A left transversal identifies with a direct sum of one copy of for each left coset of in (A left transversal identifies with a direct sum of copies of ).
The dimension of a finite-dimensional vector space is the cardinality of any finite basis (Finite-dimensional vector space, and its dimension ; infinite-dimensional means having no finite basis).
A finite-dimensional representation is a finite-dimensional vector space with a linear group action (A finite-dimensional representation over a field, and its degree).
Proof
Choose a left transversal for ; since is finite, . By [F1], as -vector spaces.
Let be a basis of with by [F2]. The vectors supported in one summand and equal there to a basis element of form a basis of , so that direct sum has basis vectors. Hence .
The induced module already carries a -linear -action by its definition, and step 2.1 shows that its underlying vector space is finite-dimensional. Therefore it is a finite-dimensional representation of over in the sense of [F3].
Steps 2.1 and 3.1 prove the stated dimension formula and finite-dimensionality claim.
The function model of induction agrees with the tensor-product model
Remark
The present page defines induction by -covariant functions because that model is self-contained in the library's existing module language. When is commutative and is finite, this is the same object as the tensor-product model.
Indeed, the subgroup inclusion makes an -bimodule (-bimodules and commuting left and right scalar actions), so is defined by the universal property of the tensor product (Universal property of the tensor product for balanced maps into abelian groups). The elementary formula
with zero off the left coset , is balanced in the -variable and therefore descends uniquely (A formula on elementary tensors defines a homomorphism exactly when its underlying pairing is balanced). Comparing both sides on a finite left transversal shows that this descended map is a -equivariant isomorphism
Thus, in the finite-index setting used for finite-group character theory, the function model and the tensor-product model are two descriptions of the same induced module. For infinite index, the displayed function model is instead larger: the tensor product corresponds to the finitely supported covariant functions.
The induced character of a complex character
Definition
Let be a finite group, let , and let be a finite-dimensional complex representation of with character (The character of a finite-dimensional complex representation).
The induced character of is the character of the induced representation:
When the representation affording is denoted simply by , one also writes .
Remarks
-
The notation depends only on the character, not on a chosen model of the representation: equivalent -representations induce equivalent -representations by postcomposing every induced function with the intertwiner.
-
The explicit value formula for is proved in Frobenius' formula for the character of an induced representation.
Frobenius' formula for the character of an induced representation
Statement
Let be a finite group, let , and let be the character of a finite-dimensional complex representation of . Then for every ,
Facts & Assumptions
Given: A finite group , a subgroup , a finite-dimensional complex representation of with character , and an element .
The induced character is the character of the induced representation (The induced character of a complex character).
A left transversal identifies with a direct sum of one copy of for each left coset of in (A left transversal identifies with a direct sum of copies of ).
A complex character is constant on conjugacy classes, and on its defining representation (For a complex character, , is a class function, and with equality exactly at scalars).
A finite sum is unchanged by reindexing a finite set bijectively (The sum over a finite index set, and its product form).
Proof
Choose a left transversal for . By [F2], , where each is one copy of indexed by the coset representative .
For , write with and . Under the identification of step 1.1, the action of sends the -summand to the -summand; if , so , then this action on is exactly the action of on . Therefore the contribution of the -summand to the trace is when , and otherwise.
The trace of on the direct sum of step 1.1 is the sum of the traces on the summands fixed by the permutation it induces on . Hence .
Fix with . The elements of the left coset are with , and then . By [F3], the character value is therefore the constant on that whole coset, and every element of contributes to the displayed sum exactly when . So the total contribution of to is .
Summing the identity of step 4.1 over the distinct cosets indexed by , and reindexing by the finite partition , gives . By step 3.1 and [F4], dividing by yields the stated Frobenius formula.
Inducing the trivial representation gives the permutation representation on
Statement
Let be a finite group and let . Inducing the trivial complex representation of to gives the permutation representation of on the left coset set .
Facts & Assumptions
Given: A finite group , a subgroup , and the trivial complex representation of .
The induced module consists of the functions satisfying , with acting by (The induced -linear -module as -covariant functions on ).
The left cosets of are the subsets , and the permutation representation on a finite -set has basis vectors indexed by that set (Left and right cosets and of a subgroup, The trivial representation, the regular representation, and permutation representations from finite -sets).
Proof
In the trivial representation of , every acts as the identity on . So the covariance condition of [F1] becomes for all and . Therefore is constant on each left coset .
Define by . Step 1.1 makes this well defined, and every function on pulls back uniquely to an -covariant function on , so is a linear bijection.
For , one has , which is exactly the left permutation action of on the coset set from [F2]. Hence is -equivariant.
The bijection of step 2.1 and the equivariance of step 3.1 identify with the permutation representation of on .
Induction is left adjoint to restriction for finite-group modules over a commutative ring
Statement
Let be a commutative ring, let be a finite group, let , let be an -linear -module, and let be an -linear -module. Then there is a natural isomorphism
Facts & Assumptions
Given: A commutative ring , a finite group , a subgroup , an -linear -module , and an -linear -module .
The induced module consists of the functions satisfying , with acting by (The induced -linear -module as -covariant functions on ).
For modules, is an abelian group under pointwise addition, with maps induced by composition (The abelian group and maps induced by pre- and postcomposition).
-equivariant maps are exactly the -module maps, and likewise for (For a commutative ring , -linear -actions are exactly the compatible left -module structures).
Proof
For , define by for and for . If , then for every , so ; if , then . Thus .
Choose a left transversal for with . If is -equivariant, define . The sum is finite because is finite.
If is -equivariant, define . For , one has by the action formula in [F1], so . Hence .
The map is -equivariant. Indeed, for and each , write with and . Then , using the covariance from [F1] and the -equivariance of . Reindexing the finite sum by shows .
For , because for and . Hence .
For and , one has . The function inside equals , because at a point it takes the value by covariance. So .
Steps 3.1 and 3.2 show that and are inverse group isomorphisms, giving the stated adjunction. Via [F3], this is equally the usual -module adjunction.
Frobenius reciprocity for complex characters
Statement
Let be a finite group, let , let be a complex character of , and let be a complex character of . Then
Facts & Assumptions
Given: A finite group , a subgroup , a finite-dimensional complex representation of with character , and a finite-dimensional complex representation of with character .
The inner product of two complex characters equals the dimension of the intertwiner space between the corresponding representations (The class-function inner product equals ).
Induction is left adjoint to restriction: (Induction is left adjoint to restriction for finite-group modules over a commutative ring).
The character is the character of (The induced character of a complex character).
Proof
By [F3] and then [F1], .
By [F2], this dimension equals ; applying [F1] again on gives .
The expressions in steps 1.1 and 1.2 are equal, which is exactly the Frobenius reciprocity identity.
Virtual characters and the character ring of a finite group
Definition
Let be a finite group. A virtual character of is an integral linear combination of irreducible complex characters:
with only finitely many nonzero coefficients. Since irreducible characters are class functions, every virtual character is a class function on (Class functions and the complex vector space , An irreducible complex character).
The set of all virtual characters is the character ring . Addition is pointwise addition of class functions, and multiplication is the bilinear extension of tensor-product multiplication on honest characters:
(Characters add on direct sums, multiply on tensor products, and conjugate on duals).
Remarks
-
The ordinary characters form a subsemiring of , while itself is their Grothendieck group.
-
By The irreducible complex characters form an orthonormal basis of , the irreducible characters are linearly independent as class functions. Since virtual characters are defined as their integral span, each virtual character has a unique decomposition in that basis.
Induction is transitive along subgroup chains
Statement
Let be subgroups of a finite group , and let be an -linear -module over a commutative ring . Then
as -linear -modules.
Facts & Assumptions
Given: A commutative ring , a finite group , subgroups , and an -linear -module .
For a subgroup , the induced module is the space of functions satisfying , with the left action by translation (The induced -linear -module as -covariant functions on ).
The notation means that , , and are subgroup related in the stated order (Subgroup).
Proof
For , define . If , then since as well by [F2], , so .
For , define for and . If , then , so ; and if , then , which is the covariance condition for . Thus lies in that induced module.
For , , so .
For and , one has , so as functions .
Steps 3.1 and 3.2 show that and are inverse -equivariant -module isomorphisms. Therefore induction is transitive along .
Induction and restriction satisfy the projection formula on character rings
Statement
Let be a finite group, let , let , and let . Then
in the character ring .
Facts & Assumptions
Given: A finite group , a subgroup , a complex character of , and a complex character of .
Frobenius' formula gives (Frobenius' formula for the character of an induced representation).
Characters multiply on tensor products, and addition is pointwise (Characters add on direct sums, multiply on tensor products, and conjugate on duals).
A complex character is a class function (For a complex character, , is a class function, and with equality exactly at scalars).
The character ring is the -span of ordinary characters, with addition and multiplication extending -bilinearly (Virtual characters and the character ring of a finite group).
Frobenius reciprocity identifies induction and restriction as adjoint operations on characters (Frobenius reciprocity for complex characters).
Proof
For an honest pair of characters and and any , [F1] gives .
Since is a class function by [F3], for each summand of step 1.1. Factoring that constant out of the finite sum and applying [F1] again yields .
So the identity holds for ordinary characters. By [F4], both induction and multiplication extend -bilinearly to virtual characters, so the same pointwise identity holds for all and .
This pointwise equality is the projection formula in , and [F2] identifies the pointwise product on the right with the character-ring product coming from tensor products. The adjoint viewpoint from [F5] is compatible with it, but step 3.1 already proves the formula.
Every irreducible complex character occurs in the induction of an irreducible constituent of its restriction
Statement
Let be a finite group, let , and let be an irreducible complex character of . Then some irreducible constituent of satisfies
Equivalently, occurs in the induced character .
Facts & Assumptions
Given: A finite group , a subgroup , and an irreducible complex character of .
Over , every finite-dimensional representation of a finite group is completely reducible (If , every finite-dimensional representation of is completely reducible).
The multiplicity of an irreducible constituent is the corresponding character inner product (The multiplicity of an irreducible summand is a character inner product).
Frobenius reciprocity gives (Frobenius reciprocity for complex characters).
Proof
Let be an irreducible representation of affording . Its restriction to is completely reducible by [F1], so for irreducible -representations with characters .
Because is nonzero, some multiplicity is positive. By [F2], this means .
Applying [F3] to gives .
The strict positivity in step 3.1 means that occurs in the induced character , by the multiplicity interpretation of [F2].
The chosen irreducible constituent therefore has the required property.
Double cosets of two subgroups
Definition
Let be a group and let be subgroups (Subgroup). For , the -double coset of is
The set of all such double cosets is written
Remarks
-
A double coset is a union of left cosets of and also a union of right cosets of .
-
The left cosets and right cosets of Left and right cosets and of a subgroup are the special cases and .
Conjugate representations and conjugate characters on conjugate subgroups
Definition
Let , let , and let be a finite-dimensional representation of over a field (A finite-dimensional representation over a field, and its degree).
The conjugate representation is the same vector space, now regarded as a representation of the conjugate subgroup by the rule
If is the complex character of , the conjugate character is the character of :
Remarks
-
Every element of has a unique form with , so the displayed formulas are well defined.
-
Restricting or to a subgroup of will be used in Mackey's formula.
Mackey's double-coset formula for restricting an induced character
Statement
Let be a finite group, let , let be the character of a finite-dimensional complex representation of , and let be a set of representatives for . Then
Facts & Assumptions
Given: A finite group , subgroups , a finite-dimensional complex representation of with character , and representatives for .
The induced character is the character of the induced representation (The induced character of a complex character).
The sets are the -double cosets of (Double cosets of two subgroups).
The conjugate representation of has character in the sense of Conjugate representations and conjugate characters on conjugate subgroups.
Proof
Let . For each , let be the subspace of functions in whose support is contained in the double coset . Because the double cosets partition , every induced function splits uniquely as the sum of its restrictions to those supports, so as -modules.
For , define by . If and , then , where the last action is that of from [F3]. So is well defined in the target induced module.
For , define by on and off . If , then , and the covariance condition in the target induced module gives . Thus is well defined, lies in , and inverts .
Steps 2.1 and 3.1 show that is a -equivariant isomorphism for each . Combining these isomorphisms with step 1.1 gives a -module decomposition of as the direct sum of the stated induced modules.
Taking characters of the -module decomposition in step 4.1 and using [F1] yields the stated Mackey double-coset formula.
Mackey's irreducibility criterion for finite groups
Statement
Let be a finite group, let , let be an irreducible complex character of , and let be representatives for with . Then is irreducible if and only if for every ,
Facts & Assumptions
Given: A finite group , a subgroup , an irreducible complex character of , and representatives for with .
A complex character is irreducible if and only if its self-inner-product is (A complex character is irreducible if and only if its self-inner-product is ).
Frobenius reciprocity gives (Frobenius reciprocity for complex characters).
Mackey's formula expands as a sum over the double cosets in (Mackey's double-coset formula for restricting an induced character).
Proof
Because is irreducible, [F1] gives . Applying [F2] with gives .
Apply [F3] to the restriction on the right side of step 1.1. The term for the identity double coset is exactly , so it contributes . Every other term is , whose inner product with is, by another use of [F2], precisely .
Therefore . Each summand is a multiplicity and hence a nonnegative integer.
The self-inner-product in step 3.1 equals if and only if every nonidentity summand vanishes. By [F1], that is equivalent to being irreducible.
The central character of an irreducible complex character
Definition
Let be a finite group, let be an irreducible complex character of , and let be a conjugacy class of with class sum (The class sum of a conjugacy class ).
The central character of is the function on class sums defined by
This is well defined because is a class function (For a complex character, , is a class function, and with equality exactly at scalars), so the value does not depend on the chosen , and for an irreducible character.
Class sums act on an irreducible representation by central-character scalars
Statement
Let be a finite group, let be a conjugacy class of , let be its class sum, and let be an irreducible complex representation of with character . Then acts on as the scalar :
Facts & Assumptions
Given: A finite group , a conjugacy class of , its class sum , and an irreducible complex representation of with character .
The class sums form a basis of the center of , so each class sum lies in the center (For a finite group, the class sums form a basis of ).
The central character is for (The central character of an irreducible complex character).
On an irreducible complex representation, every -endomorphism is scalar (Over an algebraically closed field, every endomorphism of an irreducible representation is scalar).
Proof
Because is central by [F1], the operator commutes with for every , so it is a -endomorphism of the irreducible representation . By [F3], there is a scalar with .
Taking traces gives for any , because is constant on . Hence by [F2].
Substituting the scalar from step 2.1 into step 1.1 yields .
The values of a central character are algebraic integers
Statement
Let be a finite group, let be an irreducible complex character of , and let be a conjugacy class of . Then is an algebraic integer.
Facts & Assumptions
Given: A finite group , an irreducible complex character of , and a conjugacy class of .
The class sum acts on the irreducible representation affording as the scalar (Class sums act on an irreducible representation by central-character scalars).
An element is integral over exactly when it lies in a faithful module that is finitely generated over (Integrality and finite-module characterizations for one element).
Integral elements over a nonzero base ring form a subring (Integral elements over a nonzero base ring form a subring).
Proof
Let be the conjugacy classes of , and let be the -subring of generated by the class sums . If , then each coefficient counts pairs with , so . Hence the additive group of is contained in the free abelian group on the finitely many class sums, so is finitely generated as a -module.
The ring is a faithful -module over itself, and step 1.1 makes it finitely generated over . Therefore [F2] shows that each generator is integral over . Then [F3] implies that every element of is integral over .
Let be an irreducible representation affording . By [F1], the action of on sends each to a scalar , and is a ring homomorphism because the action of is by endomorphisms. If is monic with , then applying gives the monic equation . So is integral over , that is, an algebraic integer.
The degree of an irreducible complex character divides
Statement
Let be a finite group and let be an irreducible complex character of . Then divides .
Facts & Assumptions
Given: A finite group and an irreducible complex character of .
An irreducible complex character has self-inner-product (A complex character is irreducible if and only if its self-inner-product is ).
The class-function inner product is (The standard inner product on ).
The central character satisfies for (The central character of an irreducible complex character).
Central-character values are algebraic integers (The values of a central character are algebraic integers).
Character values are algebraic integers, and (For a complex character, , is a class function, and with equality exactly at scalars).
Sums and products of algebraic integers are algebraic integers (Integral elements over a nonzero base ring form a subring).
A rational algebraic integer is an integer (The rational algebraic integers are exactly the integers).
Proof
Let be the conjugacy classes of , and choose . By [F1] and [F2], , where [F5] identifies with and is constant on each class.
For each , [F3] gives . Substituting this into step 1.1 yields .
By [F4] and [F5], each factor and is an algebraic integer; by [F6], every product and the whole finite sum in step 2.1 are algebraic integers. Therefore is a rational algebraic integer, so [F7] makes it an integer.
Since , the degree divides .
The degree of an irreducible complex character divides
Statement
Let be a finite group and let be an irreducible complex character of . Then divides .
Facts & Assumptions
Given: A finite group , an irreducible complex character of , and an irreducible complex representation of affording .
An irreducible complex character has self-inner-product (A complex character is irreducible if and only if its self-inner-product is ).
The class-function inner product is (The standard inner product on ).
A central class sum acts on an irreducible representation by a scalar (Class sums act on an irreducible representation by central-character scalars).
The degree of an irreducible complex character divides the order of the group (The degree of an irreducible complex character divides ).
The center is (The center of a group).
Proof
Let . For each integer , let act on by . Its character is , so by [F2] and [F1] its self-inner-product is . Hence is irreducible by [F1].
If , then the conjugacy class of is the singleton by [F5], so [F3] gives for some scalar . Therefore acts on as the scalar .
Let . Since is multiplicative, step 2.1 shows that every element of acts trivially on . Thus the irreducible representation of step 1.1 descends to an irreducible representation of the quotient , still of degree .
The map , , is bijective, so . Hence . Applying [F4] to the irreducible quotient representation from step 3.1 yields for every .
Fix a prime and write and with . If , then choosing so large that makes too large to divide , contradicting step 4.1. Therefore for every prime , so divides .
Since , the degree of the irreducible character divides the index of the center.
A finite group with an irreducible complex character of degree greater than is nonabelian
Statement
If a finite group has an irreducible complex character with , then is nonabelian.
Facts & Assumptions
Given: A finite group with an irreducible complex character such that .
A finite group is abelian if and only if all its irreducible complex characters have degree (A finite group is abelian if and only if all its irreducible complex characters have degree ).
Proof
If were abelian, then [F1] would force every irreducible complex character of to have degree .
That contradicts the hypothesis . Therefore is nonabelian.
An algebraic-integer average of roots of unity is either or a common root of unity
Statement
Let be roots of unity, and put . If is an algebraic integer, then either or .
Facts & Assumptions
Given: Roots of unity and their average , with an algebraic integer.
A rational algebraic integer is an integer (The rational algebraic integers are exactly the integers).
The modulus is the usual complex absolute value (Real and imaginary parts, complex conjugation, and modulus).
An algebraic integer is a complex number integral over (Integral elements over a commutative ring and algebraic integers).
Every algebraic conjugate of a root of unity is again a root of unity.
The average of complex numbers of modulus has modulus at most , with equality only when all of them are equal.
Proof
If , then and the conclusion holds.
Assume now that the are not all equal. By [A2], . Every algebraic conjugate of has the form with each a root of unity by [A1], so by [A2].
Suppose also that . Let be the monic minimal polynomial of over ; then is the product of the algebraic conjugates of . Because is an algebraic integer by [F3], that product is a rational algebraic integer, hence an integer by [F1]. But step 2.1 gives its modulus strictly between and , impossible. So .
Under the assumption that the roots are not all equal, step 3.1 forces . Together with step 1.1, this proves that is either or a common root of unity.
A conjugacy class of size coprime to forces either or scalar action
Statement
Let be a finite group, let be an irreducible complex representation of with character , and let . If the size of the conjugacy class of is coprime to , then either or acts as a scalar on .
Facts & Assumptions
Given: A finite group , an irreducible complex representation of with character , and an element .
Central-character values are algebraic integers (The values of a central character are algebraic integers).
An algebraic-integer average of roots of unity is either or constant (An algebraic-integer average of roots of unity is either or a common root of unity).
The character value is the sum of the eigenvalues of , which are roots of unity, and equality in the modulus bound means scalar action (For a complex character, , is a class function, and with equality exactly at scalars).
For the conjugacy class of , the central character satisfies (The central character of an irreducible complex character).
Proof
Let and . By [F4], for the class sum of , so [F1] makes an algebraic integer. Since , choose integers with . Then is an algebraic integer because [F3] shows that is an algebraic integer.
Let be the eigenvalues of . By [F3], they are roots of unity and . Applying [F2] to this average and step 1.1 gives either or .
If , then the diagonalizable operator from [F3] is that common root of unity times the identity, so acts as a scalar on .
Steps 2.1 and 3.1 prove the stated dichotomy.
A conjugacy class of prime-power size forces a proper nontrivial normal subgroup
Statement
Let be a finite group, and let be a conjugacy class in whose size is a positive power of a prime . Then has a proper nontrivial normal subgroup.
Facts & Assumptions
Given: A finite group and a conjugacy class with for some prime and integer .
The second column orthogonality relation gives whenever (The second orthogonality relation for irreducible complex characters).
If the class size of is coprime to and , then acts as a scalar on the irreducible representation affording (A conjugacy class of size coprime to forces either or scalar action).
Character values are algebraic integers (For a complex character, , is a class function, and with equality exactly at scalars).
A rational algebraic integer is an integer (The rational algebraic integers are exactly the integers).
The subgroup generated by a set is the smallest subgroup containing it, and normality means invariance under conjugation (The subgroup generated by a subset, the cyclic subgroup , and cyclic groups, Normal subgroup: invariance under conjugation).
Proof
Choose . Since , one has . Split the irreducible characters into the trivial character, the set of nontrivial irreducibles whose degree is divisible by , and the set of nontrivial irreducibles whose degree is not divisible by . Applying [F1] at gives .
Write for . Because each is an algebraic integer by [F3], the sum is an algebraic integer. If the sum over were zero, step 1.1 would give , so would be a rational algebraic integer, contradicting [F4]. Therefore some satisfies .
For that chosen , the class size is coprime to by the definition of , so [F2] shows that acts as a scalar on the irreducible representation affording . Every element of the conjugacy class is conjugate to , so each of them acts by the same scalar on .
Let . Because conjugation permutes , the set is stable under conjugation, so [F5] makes a normal subgroup of . By step 3.1 every generator acts trivially on , so acts trivially on the nontrivial irreducible representation ; hence .
Since , choose distinct ; then lies in , so is nontrivial. Together with step 4.1, this makes a proper nontrivial normal subgroup of .
Burnside's theorem
Statement
Let and be distinct primes, and let . Every finite group of order is solvable.
Facts & Assumptions
Given: Distinct primes , integers , and a finite group of order .
A conjugacy class of prime-power size forces a proper nontrivial normal subgroup (A conjugacy class of prime-power size forces a proper nontrivial normal subgroup).
Every nontrivial normal subgroup of a finite -group meets the center nontrivially (Every nontrivial normal subgroup of a finite -group meets the center nontrivially).
If and are solvable, then is solvable (Extensions and finite direct products of solvable groups are solvable).
Every subgroup and quotient of a solvable group is solvable (Subgroups and quotients of solvable groups are solvable).
Sylow's first theorem gives a Sylow -subgroup (Sylow I: every finite group has a Sylow -subgroup).
The conjugacy class of has size ( is a bijection, so whenever these cardinalities are finite).
Solvability means that some derived subgroup is trivial (The derived series, solvable groups, and derived length).
Proof
Suppose the statement false, and choose a counterexample of minimal order. If one of or is zero, then is a finite -group for or . If is trivial or abelian, then its derived subgroup is trivial, so is solvable by [F7]. Otherwise [F2], with in place of its generic prime, applied to gives a nontrivial central subgroup . Then both and are smaller finite -groups, so minimality makes them solvable, and [F3] makes solvable, contradiction. So any minimal counterexample has .
If is proper and nontrivial, then both and have smaller order of the form , so minimality makes them solvable; then [F3] makes solvable, contradiction. Hence a minimal counterexample must be simple.
By [F5], choose a Sylow -subgroup . Since is a nontrivial finite -group, [F2] applied to gives a nonidentity element . Because is simple and , it is not abelian, so .
Every element of commutes with , so . Therefore the full -part of lies in , and [F6] gives for some . Applying [F1] to this prime-power class yields a proper nontrivial normal subgroup of , contradicting simplicity from step 2.1.
The contradiction shows that no counterexample exists. Therefore every finite group of order is solvable.
5 · Examples, counterexamples and false statements
None yet.
Sources
- Pavel Etingof et al., Introduction to Representation Theory, Definition 4.28
- Peter Webb, A Course in Finite Group Representation Theory, Section 4.3
- Peter Webb, A Course in Finite Group Representation Theory, Proposition 4.5.1
- Anupam Singh, Representation Theory of Finite Groups, Chapter 17
- Pavel Etingof et al., Introduction to Representation Theory, Remark 4.30
- Peter Webb, A Course in Finite Group Representation Theory, Proposition 4.3.1
- Peter Webb, A Course in Finite Group Representation Theory, Proposition 4.3.2
- Anupam Singh, Representation Theory of Finite Groups, Section 18.1
- Peter Webb, A Course in Finite Group Representation Theory, Proposition 4.3.5
- Pavel Etingof et al., Introduction to Representation Theory, Theorem 4.32
- Peter Webb, A Course in Finite Group Representation Theory, Example 4.3.4
- Pavel Etingof et al., Introduction to Representation Theory, Section 4.8
- Peter Webb, A Course in Finite Group Representation Theory, Lemma 4.3.7
- Pavel Etingof et al., Introduction to Representation Theory, Theorem 4.33
- Peter Webb, A Course in Finite Group Representation Theory, Corollary 4.3.8
- Anupam Singh, Representation Theory of Finite Groups, Chapter 19
- Pavel Etingof et al., Introduction to Representation Theory, Definition 4.26
- Pavel Etingof et al., Introduction to Representation Theory, Problem 4.31
- Peter Webb, A Course in Finite Group Representation Theory, Corollary 4.3.9
- Pavel Etingof et al., Introduction to Representation Theory, Section 4.7-4.9
- Peter Webb, A Course in Finite Group Representation Theory, Section 5.1
- Anupam Singh, Representation Theory of Finite Groups, Section 20.1
- Peter Webb, A Course in Finite Group Representation Theory, Section 5.2
- Anupam Singh, Representation Theory of Finite Groups, Theorem 20.6
- Anupam Singh, Representation Theory of Finite Groups, Theorem 20.9
- Anupam Singh, Representation Theory of Finite Groups, Chapter 15
- Peter Webb, A Course in Finite Group Representation Theory, Section 3.5
- Peter Webb, A Course in Finite Group Representation Theory, Proposition 3.5.3
- Peter Webb, A Course in Finite Group Representation Theory, Proposition 3.5.2
- Anupam Singh, Representation Theory of Finite Groups, Proposition 15.5
- Peter Webb, A Course in Finite Group Representation Theory, Theorem 3.5.4
- Anupam Singh, Representation Theory of Finite Groups, Theorem 15.7
- K. Conrad, Degrees of irreducible representations
- John Tate's tensor-power argument, as summarized in expository sources
- Peter Webb, A Course in Finite Group Representation Theory, Theorem 4.1.5
- Pavel Etingof et al., Introduction to Representation Theory, Lemma 4.22
- Pavel Etingof et al., Introduction to Representation Theory, Theorem 4.21
- Anupam Singh, Representation Theory of Finite Groups, Lemma 16.2
- Pavel Etingof et al., Introduction to Representation Theory, Theorem 4.23
- Peter Webb, A Course in Finite Group Representation Theory, Theorem 3.7.1
- Pavel Etingof et al., Introduction to Representation Theory, proof after Theorem 4.23