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9 results · all verified · 3 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 6 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Induced Representations, Frobenius Reciprocity and Applications — Examples

1 · Prerequisites

2 · Summary

The companion examples keep the computations inside one legal witness family. The subgroup A3S3 supplies the nontrivial induction whose character is already irreducible of degree two, while a subgroup of order two supplies the contrasting induction that splits as 1+χ2. Those two witnesses are enough to see Frobenius reciprocity numerically and to refute the common mistakes that induction and restriction preserve irreducibility or compose to the identity.

The last example changes theme from multiplicities to divisibility. The cyclic group C4 shows that the theorem χ(1)G is only a necessary condition for an irreducible degree: divisors of the group order need not all occur.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-30Open item page →

Inducing a nontrivial character of a three-cycle subgroup of S3 gives an irreducible degree-two character

Example

Let H=A3=(123)S3, let ω=e2πi/3, and let θ be the nontrivial linear character of H with θ((123))=ω and θ((132))=ω2. Then IndA3S3θ has values

(2,0,1)

on the conjugacy classes {e}, the transpositions, and the 3-cycles respectively. Its self-inner-product is 1, so it is irreducible of degree 2.

Facts & Assumptions

Given: The subgroup A3=(123)S3 and the nontrivial character θ defined in the Example.

[F1]

The induced character is computed by Frobenius' formula (Frobenius' formula for the character of an induced representation).

[F2]

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

[F3]

The notation IndA3S3θ is the induced character from The induced character IndHGχ of a complex character.

Verification

technique · direct
1.1

Since [S3:A3]=2, Frobenius' formula [F1] at the identity gives IndA3S3θ(e)=2. If g is a transposition, no conjugate of g lies in A3, so Frobenius' formula gives IndA3S3θ(g)=0.

F1F3given
1.2

If g is a 3-cycle, then A3 is normal in S3, so every xS3 satisfies x1gxA3. Exactly three of those conjugates equal (123) and three equal (132), so [F1] gives IndA3S3θ(g)=(3ω+3ω2)/3=ω+ω2=1.

F1givenalgebra
2.1

Therefore the induced character has values (2,0,1) on the three class types of S3. Its self-inner-product is (1/6)(22+302+2(1)2)=1, so [F2] makes it irreducible; the value at e shows that its degree is 2.

F2step 1.1step 1.2algebra
ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-30Open item page →

Restricting that degree-two S3 character to the three-cycle subgroup gives the two nontrivial linear characters

Example

Let χ2 be the degree-two irreducible character of S3 from the previous example. Then

ResA3S3χ2=θ+θ,

the sum of the two nontrivial linear characters of A3.

Facts & Assumptions

Given: The degree-two character χ2=(2,0,1) of S3 from the previous example, and the two nontrivial linear characters θ and θ of A3.

[F1]

The previous example identifies χ2 with the induced character IndA3S3θ (Inducing a nontrivial character of a three-cycle subgroup of S3 gives an irreducible degree-two character).

[F2]

Frobenius reciprocity identifies multiplicities before and after induction (Frobenius reciprocity for complex characters).

Verification

technique · direct
1.1

Restricting the values (2,0,1) from [F1] to A3={e,(123),(132)} gives ResA3S3χ2=(2,1,1).

F1given
2.1

Frobenius reciprocity [F2] gives θ,ResA3S3χ2A3=IndA3S3θ,χ2S3=χ2,χ2S3=1, and the same computation with θ gives multiplicity 1 for θ.

F1F2step 1.1algebra
3.1

The restriction has degree 2 at the identity, while the two nontrivial linear characters already account for degree 1+1=2. Hence no trivial summand occurs, and ResA3S3χ2=θ+θ.

step 1.1step 2.1algebra
ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-30Open item page →

Inducing the trivial character of a subgroup of order two in S3 gives 1 plus an irreducible degree-two character

Example

Let H=(12)S3. Then IndHS31H is the permutation character on the three left cosets of H, so it has values (3,1,0) on the class types e, transpositions, and 3-cycles. Subtracting the trivial character gives the irreducible degree-two character (2,0,1).

Facts & Assumptions

Given: The subgroup H=(12)S3 and its trivial character 1H.

[F1]

Inducing the trivial character gives the permutation representation on S3/H (Inducing the trivial representation gives the permutation representation on G/H).

[F2]

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

[F3]

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

Verification

technique · direct
1.1

By [F1] and [F2], the induced character counts fixed cosets of the left action on the three cosets of H. The identity fixes all three cosets, a transposition fixes exactly one coset, and a 3-cycle fixes none, so IndHS31H=(3,1,0).

F1F2given
2.1

Subtracting the trivial character (1,1,1) gives the class function (2,0,1). Its self-inner-product is (1/6)(22+302+2(1)2)=1, so [F3] makes it irreducible; its value at the identity is 2, so it has degree 2.

F3step 1.1algebra
3.1

Therefore IndHS31H=1+χ2 with χ2 irreducible of degree 2.

step 2.1
ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-30Open item page →

Frobenius reciprocity matches multiplicities in the two preceding S3 inductions

Example

For the two S3 inductions on this page, Frobenius reciprocity gives the same multiplicity on both sides:

IndA3S3θ,χ2S3=θ,ResA3S3χ2A3=1

and

Ind(12)S31,χ2S3=1,Res(12)S3χ2(12)=1.

Facts & Assumptions

Given: The character χ2=(2,0,1) and the two induced characters from the preceding examples.

[F1]

The induction from A3 of a nontrivial linear character is exactly χ2 (Inducing a nontrivial character of a three-cycle subgroup of S3 gives an irreducible degree-two character).

[F3]

Frobenius reciprocity matches the two inner products (Frobenius reciprocity for complex characters).

Verification

technique · direct
1.1

By [F1], IndA3S3θ,χ2S3=χ2,χ2S3=1. By [F3], the matching inner product on A3 is therefore also 1.

F1F3given
1.2

By [F2], Ind(12)S31,χ2S3=1+χ2,χ2S3=1. Again [F3] makes the inner product after restriction equal to the same value.

F2F3givenalgebra
2.1

So both preceding inductions realize Frobenius reciprocity numerically with multiplicity 1 on each side.

step 1.1step 1.2
ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-30Open item page →

C4 shows that divisibility of irreducible degrees by G is not an equivalence

Example

The divisor 2 of C4=4 is not the degree of any irreducible complex character of C4. So the theorem χ(1)G is a necessary condition for irreducible degrees, not a characterization.

Facts & Assumptions

Given: The cyclic group C4.

[F2]

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

[F3]

Irreducible complex character degrees divide the group order (The degree of an irreducible complex character divides G).

Verification

technique · direct
1.1

By [F1], C4 is a finite cyclic group of order 4, hence finite abelian.

F1given
2.1

Over C, every irreducible representation of the finite abelian group C4 has degree 1 by [F2]. Therefore no irreducible complex character of C4 has degree 2, even though 24.

F2step 1.1algebra
3.1

This shows that [F3] is not an equivalence: a divisor of G need not occur as an irreducible degree.

F3step 2.1
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

An induced irreducible complex character is always irreducible

Statement

False claim: if φ is an irreducible complex character of a subgroup HG, then IndHGφ is irreducible.

Facts & Assumptions

Given: The subgroup H=(12)S3 and its trivial character 1H.

[F1]

The induced character IndHS31H equals 1+χ2, where χ2 is irreducible of degree 2 (Inducing the trivial character of a subgroup of order two in S3 gives 1 plus an irreducible degree-two character).

Refutation

technique · direct
1.1

The trivial character 1H of the order-two subgroup H is irreducible because every one-dimensional character is irreducible.

givenalgebra
2.1

But [F1] shows that its induction to S3 is 1+χ2, a nontrivial sum of two characters. So the induced character is reducible.

F1step 1.1
3.1

This single witness refutes the claim that induced irreducible characters are always irreducible.

step 2.1
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Induction followed by restriction is the identity on complex representations

Statement

False claim: for every subgroup HG, the composite ResHGIndHG is the identity on complex representations of H.

Facts & Assumptions

Given: The subgroup A3S3 and the nontrivial linear character θ of A3.

[F1]

Inducing θ to S3 gives the irreducible degree-two character χ2 (Inducing a nontrivial character of a three-cycle subgroup of S3 gives an irreducible degree-two character).

[F2]

Refutation

technique · direct
1.1

By [F1], IndA3S3θ=χ2.

F1given
2.1

Applying restriction and then [F2] gives ResA3S3IndA3S3θ=ResA3S3χ2=θ+θθ.

F2step 1.1
3.1

So induction followed by restriction is not the identity in general.

step 2.1
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-30Open item page →

Restriction of an irreducible complex representation is always irreducible

Statement

False claim: if a complex representation of G is irreducible, then its restriction to every subgroup is irreducible.

Facts & Assumptions

Given: The irreducible degree-two character χ2 of S3 and the subgroup A3S3.

Refutation

technique · direct
1.1

The character χ2 is irreducible on S3 by the example that constructs it.

givenalgebra
2.1

But [F1] writes its restriction to A3 as the sum of two distinct nontrivial characters, so the restricted representation is reducible.

F1step 1.1
3.1

Therefore restriction does not preserve irreducibility in general.

step 2.1
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30Open item page →

Every divisor of G is an irreducible character degree

Statement

False claim: if d divides G, then d is the degree of some irreducible complex character of G.

Facts & Assumptions

Given: The cyclic group C4.

[F1]

In C4, the divisor 2 of C4=4 is not an irreducible character degree (C4 shows that divisibility of irreducible degrees by G is not an equivalence).

Refutation

technique · direct
1.1

The integer 2 divides C4=4.

givenalgebra
2.1

But [F1] shows that no irreducible complex character of C4 has degree 2.

F1step 1.1
3.1

Hence not every divisor of the group order occurs as an irreducible character degree.

step 2.1

Sources