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

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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.
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Pointwise character dual is a finite abelian group

Statement

For a finite abelian group $G$ , the pointwise product, constant-one identity and pointwise inverse on $\hat{G}$ make it a finite abelian group. Its order is $∣ \hat{G} ∣ = ∣ G ∣$ .

Facts & Assumptions

Given: A finite abelian group $G$ and the set $\hat{G}$ of its additive characters.

[F1]

$\hat{G}$ consists of homomorphisms $G \to C^{\times}$ with pointwise product, constant-one function and pointwise inverse (Dual group of a finite abelian group, Additive characters of a finite abelian group).

[F2]

$C$ is a field, so multiplication on $C^{\times}$ is associative, commutative and has identity $1$ ( $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ).

[F3]

A group homomorphism preserves the identity and inverses (A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed).

[F4]

The number of characters is $∣ \hat{G} ∣ = ∣ G ∣ < \infty$ (A finite abelian group and its character dual have equal order).

Proof

1.1F1F2F3

For $χ, ψ \in \hat{G}$ , $x, y \in G$ , and pointwise product as in [F1], $(χ ψ) (x + y) = χ (x) χ (y) ψ (x) ψ (y) = χ (x) ψ (x) χ (y) ψ (y)$ by commutativity in [F2]. Thus $χ ψ$ is again a homomorphism. The constant-one function is a homomorphism because it sends every sum to $1 \cdot 1 = 1$ ; [F3] also confirms each character takes the identity to $1$ .

1.2F1F2algebra

The pointwise inverse is a homomorphism because $(χ (x + y))^{- 1} = (χ (x) χ (y))^{- 1} = χ (x)^{- 1} χ (y)^{- 1}$ in the abelian group $C^{\times}$ . Associativity and commutativity of pointwise product, its identity law, and the inverse laws follow by evaluating at each $x$ and using the corresponding laws in [F2]. Hence $\hat{G}$ is an abelian group.

2.1F4step 1.2∎

The dual is finite by [F4], and its order is $∣ G ∣$ by the same result. Thus it is a finite abelian group of the asserted order.

Depends on

Used by

Dependency tree · two levels

36 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)