ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-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.

Common-kernel subgroup recovery in a product of cyclic groups

Example

Take the invariant-factor group $G = Z / 2 Z \times Z / 6 Z$ and the trivial subgroup $H = {(0, 0)}$ . The characters are $χ_{a, b} (x, y) = \exp (2 π i (a x / 2 + b y / 6))$ . Two possible Fourier samples, $χ_{1, 0}$ followed by $χ_{0, 1}$ , reduce the initial candidate group $G$ to $H$ .

Facts & Assumptions

Given: The character labels of a product of cyclic groups and the common-kernel output convention.

[F1]

Cyclic characters are indexed by residues and product characters factor by coordinates. (A cyclic group of order n has exactly n additive characters, Characters of a finite product factor coordinatewise)

[F2]

A sample is uniform on $H^{⊥}$ , and independent repetitions use fresh samples. (Fourier samples are uniform in the annihilator, Finite abelian hidden-subgroup common-kernel reconstruction)

[F3]

The output list represents the intersection of the sampled character kernels. (Finite abelian hidden-subgroup common-kernel reconstruction)

Verification

Given: The group and subgroup in the Example, with additive coordinates reduced modulo $2$ and $6$ .

1.1givenF1algebra

Since $H$ is trivial, every character is trivial on $H$ . The labels are $a \in {0, 1}$ and $b \in {0, 1, 2, 3, 4, 5}$ , so $∣ H^{⊥} ∣ = 12$ and both $(1, 0)$ and $(0, 1)$ label allowed outcomes.

1.2F2algebra

By [F2], the probability that two independent samples occur in the specified order is $(1 / 12) (1 / 12) = 1 / 144$ . The first sampled character has kernel ${0} \times Z / 6 Z$ , since $χ_{1, 0} (x, y) = (- 1)^{x}$ .

1.3F3algebra

The second character has kernel $Z / 2 Z \times {0}$ , since $χ_{0, 1} (x, y) = \exp (2 π i y / 6)$ equals $1$ exactly for $y \equiv 0 (m o d 6)$ . Intersecting the two kernels gives ${(0, 0)} = H$ .

2.1F1F3step 1.3∎

Each constraint kernel contains $H$ , and the ordered sample list therefore presents the exact hidden subgroup without converting it to generators.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

35 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

Andrew M. Childs, Lecture Notes on Quantum Algorithms (17 April 2025) (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, arXiv:1907.09415v6 (standard reference, not scraped)