ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-30

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Frobenius reciprocity matches multiplicities in the two preceding $S_{3}$ inductions

Example

For the two $S_{3}$ inductions on this page, Frobenius reciprocity gives the same multiplicity on both sides:

$⟨ {Ind}_{A_{3}}^{S_{3}} θ, χ_{2} ⟩_{S_{3}} = ⟨ θ, {Res}_{A_{3}}^{S_{3}} χ_{2} ⟩_{A_{3}} = 1$

and

$⟨ {Ind}_{⟨ (12) ⟩}^{S_{3}} 1, χ_{2} ⟩_{S_{3}} = ⟨ 1, {Res}_{⟨ (12) ⟩}^{S_{3}} χ_{2} ⟩_{⟨ (12) ⟩} = 1.$

Facts & Assumptions

Given: The character $χ_{2} = (2, 0, - 1)$ and the two induced characters from the preceding examples.

[F1]

The induction from $A_{3}$ of a nontrivial linear character is exactly $χ_{2}$ (Inducing a nontrivial character of a three-cycle subgroup of $S_{3}$ gives an irreducible degree-two character).

[F2]

The induction from a subgroup of order two is $1 + χ_{2}$ (Inducing the trivial character of a subgroup of order two in $S_{3}$ gives $1$ plus 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], $⟨ {Ind}_{A_{3}}^{S_{3}} θ, χ_{2} ⟩_{S_{3}} = ⟨ χ_{2}, χ_{2} ⟩_{S_{3}} = 1$ . By [F3], the matching inner product on $A_{3}$ is therefore also $1$ .

F1F3given

1.2

By [F2], $⟨ {Ind}_{⟨ (12) ⟩}^{S_{3}} 1, χ_{2} ⟩_{S_{3}} = ⟨ 1 + χ_{2}, χ_{2} ⟩_{S_{3}} = 1$ . 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∎

Depends on

Used by

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Sources

Pavel Etingof et al., Introduction to Representation Theory, Section 4.11 (standard reference, not scraped)

Frobenius reciprocity matches multiplicities in the two preceding S3S3​ inductions