PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-30

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Induction and restriction satisfy the projection formula on character rings

Statement

Let $G$ be a finite group, let $H \leq G$ , let $χ \in R (H)$ , and let $ψ \in R (G)$ . Then

${Ind}_{H}^{G} ⁣ (χ \cdot {Res}_{H}^{G} ψ) = ({Ind}_{H}^{G} χ) \cdot ψ$

in the character ring $R (G)$ .

Facts & Assumptions

Given: A finite group $G$ , a subgroup $H \leq G$ , a complex character $χ$ of $H$ , and a complex character $ψ$ of $G$ .

[F1]

Frobenius' formula gives ${Ind}_{H}^{G} χ (g) = \frac{1}{∣ H ∣} \sum_{x^{- 1} g x \in H} χ (x^{- 1} g x)$ (Frobenius' formula for the character of an induced representation).

[F2]

Characters multiply on tensor products, and addition is pointwise (Characters add on direct sums, multiply on tensor products, and conjugate on duals).

[F3]

A complex character is a class function (For a complex character, $χ (1) = \dim V$ , $χ$ is a class function, and $∣ χ (g) ∣ \leq χ (1)$ with equality exactly at scalars).

[F4]

The character ring is the $Z$ -span of ordinary characters, with addition and multiplication extending $Z$ -bilinearly (Virtual characters and the character ring $R (G)$ of a finite group).

[F5]

Frobenius reciprocity identifies induction and restriction as adjoint operations on characters (Frobenius reciprocity for complex characters).

Proof

technique · direct

1.1

For an honest pair of characters $χ$ and $ψ$ and any $g \in G$ , [F1] gives ${Ind}_{H}^{G} (χ {Res}_{H}^{G} ψ) (g) = \frac{1}{∣ H ∣} \sum_{x \in G, x^{- 1} g x \in H} χ (x^{- 1} g x) ψ (x^{- 1} g x)$ .

F1F4given

2.1

Since $ψ$ is a class function by [F3], $ψ (x^{- 1} g x) = ψ (g)$ for each summand of step 1.1. Factoring that constant out of the finite sum and applying [F1] again yields ${Ind}_{H}^{G} (χ {Res}_{H}^{G} ψ) (g) = ψ (g) {Ind}_{H}^{G} χ (g) = (({Ind}_{H}^{G} χ) ψ) (g)$ .

F1F3step 1.1algebra

3.1

So the identity holds for ordinary characters. By [F4], both induction and multiplication extend $Z$ -bilinearly to virtual characters, so the same pointwise identity holds for all $χ \in R (H)$ and $ψ \in R (G)$ .

F4step 2.1algebra

4.1

This pointwise equality is the projection formula in $R (G)$ , 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.

F2F5step 3.1∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Peter Webb, A Course in Finite Group Representation Theory, Corollary 4.3.9 (standard reference, not scraped)
Pavel Etingof et al., Introduction to Representation Theory, Section 4.7-4.9 (standard reference, not scraped)