LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generated

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.

Values of additive characters of finite abelian groups have modulus one

Statement

If $G$ is a finite abelian group, $χ : G \to C^{\times}$ is an additive character, and $x \in G$ , then $∣ χ (x) ∣ = 1$ .

Facts & Assumptions

Given: A finite abelian group $G$ , an additive character $χ$ , and $x \in G$ .

[F1]

In a finite group, each element $x$ has a positive finite order $n = ord (x)$ with $n x = 0$ in additive notation (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

[F2]

An additive character is a homomorphism and sends the identity to $1$ (Additive characters of a finite abelian group).

[F3]

A homomorphism preserves integer powers; in additive notation this gives $χ (n x) = χ (x)^{n}$ (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]

For $n \geq 1$ , every $n$ th root of unity is $\exp (2 π i k / n)$ for some integer $0 \leq k < n$ (The $n$ -th roots of a complex number and the $n$ distinct roots of unity for every $n \geq 1$ ).

[F5]

Every purely imaginary complex exponential has modulus one ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

Proof

1.1F1F2F3

Put $n : = ord (x) \geq 1$ . By [F1] and [F3], $χ (x)^{n} = χ (n x) = χ (0) = 1$ , with $χ (0) = 1$ by [F2].

2.1F2F4F5step 1.1∎

By [F4], $χ (x) = \exp (2 π i k / n)$ for some $0 \leq k < n$ . Therefore [F5] gives $∣ χ (x) ∣ = 1$ . The argument includes $x = 0$ , for which $n = 1$ and $χ (0) = 1$ .

Depends on

Used by

Finite abelian Fourier and convolution examples Example
Fourier sampling for a cyclic subgroup Example
Dual-index character orthogonality Theorem
Fourier samples are uniform in the annihilator Theorem
Fourier transform of a subgroup coset indicator Theorem

Dependency tree · two levels

44 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)