LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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.

Annihilators of subgroups are subgroups

Statement

Let $G$ be a finite abelian group. For every subgroup $H \leq G$ , $H^{⊥}$ is a subgroup of $\hat{G}$ . For every subgroup $K \leq \hat{G}$ , $K^{⊥}$ is a subgroup of $G$ .

Facts & Assumptions

Given: A finite abelian group $G$ , a subgroup $H \leq G$ , and a subgroup $K \leq \hat{G}$ ; the annihilators are as defined in Character annihilator of a subgroup.

[F1]

$\hat{G}$ is a group under pointwise multiplication, with pointwise inverse and identity (Pointwise character dual is a finite abelian group).

[F2]

A subgroup contains the identity and is closed under the operation and inverses (Subgroup).

[F3]

Every character satisfies $χ (x + y) = χ (x) χ (y)$ , $χ (0) = 1$ , and $χ (- x) = χ (x)^{- 1}$ (Additive characters of a finite abelian group).

Proof

1.1givenF1F2F3

The trivial character is in $H^{⊥}$ . If $χ, ψ \in H^{⊥}$ and $h \in H$ , then $(χ ψ) (h) = χ (h) ψ (h) = 1$ , and $χ^{- 1} (h) = χ (h)^{- 1} = 1$ . Thus $H^{⊥}$ contains the identity and is closed under pointwise product and inverse in the group $\hat{G}$ .

1.2givenF2F3

The identity $0$ of $G$ lies in $K^{⊥}$ by [F3]. If $g, h \in K^{⊥}$ , then for every $χ \in K$ , $χ (g + h) = χ (g) χ (h) = 1$ and $χ (- g) = χ (g)^{- 1} = 1$ . Therefore $K^{⊥}$ is closed under addition and inverses in $G$ .

2.1F1F2step 1.1step 1.2∎

By the subgroup criterion in [F2], steps 1.1 and 1.2 show respectively that $H^{⊥} \leq \hat{G}$ and $K^{⊥} \leq G$ .

Depends on

Used by

A Fourier sample cuts a larger common kernel Lemma
Double annihilator returns the subgroup Theorem

Dependency tree · two levels

16 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

Ronald de Wolf, Quantum Computing: Lecture Notes, version 6 (standard reference, not scraped)
Andrew M. Childs, Lecture Notes on Quantum Algorithms (standard reference, not scraped)
John Preskill, Quantum Information, Chapter 6: Quantum Algorithms (standard reference, not scraped)