LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The cyclic induction subgroup is an ideal of the character ring

Statement

Let $G$ be a finite group. Then the cyclic induction subgroup $I_{c y c} (G) \subseteq R (G)$ is an ideal of the character ring $R (G)$ .

Facts & Assumptions

Given: A finite group $G$ , an element $x \in I_{c y c} (G)$ , and an element $ψ \in R (G)$ .

[F1]

By definition, $I_{c y c} (G)$ consists of finite sums $\sum_{i} {Ind}_{C_{i}}^{G} (θ_{i})$ with each $C_{i} \leq G$ cyclic and each $θ_{i} \in R (C_{i})$ (The cyclic induction subgroup of the character ring).

[F2]

Induction and restriction satisfy the projection formula: ${Ind}_{H}^{G} (χ {Res}_{H}^{G} ψ) = ({Ind}_{H}^{G} χ) ψ$ (Induction and restriction satisfy the projection formula on character rings).

Proof

technique · direct

1.1F1given

By [F1], write $x = \sum_{i = 1}^{r} {Ind}_{C_{i}}^{G} (θ_{i})$ with each $C_{i} \leq G$ cyclic and each $θ_{i} \in R (C_{i})$ .

2.1F1F2step 1.1algebra

Multiplying by $ψ$ and applying [F2] termwise gives $x ψ = \sum_{i = 1}^{r} ({Ind}_{C_{i}}^{G} (θ_{i})) ψ = \sum_{i = 1}^{r} {Ind}_{C_{i}}^{G} ⁣ (θ_{i} {Res}_{C_{i}}^{G} ψ)$ . Each factor $θ_{i} {Res}_{C_{i}}^{G} ψ$ lies in $R (C_{i})$ , so every summand on the right again belongs to $I_{c y c} (G)$ by [F1].

3.1step 2.1∎

Therefore $x ψ \in I_{c y c} (G)$ . Since $x \in I_{c y c} (G)$ and $ψ \in R (G)$ were arbitrary, $I_{c y c} (G)$ is an ideal of $R (G)$ .

Depends on

Used by

Artin induction for rational characters Theorem

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Tammo tom Dieck, Representation Theory, Section 4.5 (standard reference, not scraped)
Janos Kramar, Artin's and Brauer's Theorems on Induced Characters, Lemma 1 (standard reference, not scraped)