ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-11

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 action of $Z / 6$ on the cosets of ${0, 3}$ is transitive with kernel ${0, 3}$ and is not faithful

Example

In the additive group $G = Z / 6$ , let $H = {0, 3}$ . The action of $G$ on $G / H$ by translation is transitive and has kernel $H$ , so it is not faithful.

Facts & Assumptions

Given: The additive group $G = Z / 6$ and the subset $H = {0, 3}$ .

[L1]

The left-coset action is transitive and its kernel is the core of the subgroup (Left multiplication on $G / H$ is transitive, has stabiliser $H$ at $H$ , and has kernel ${Core}_{G} (H)$ ).

[L2]

Addition modulo $6$ makes $Z / 6$ an abelian group (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

[L3]

The residue classes have representatives $0, 1, 2, 3, 4, 5$ (For $n \geq 1$ , every class in $Z / n$ has one representative $r$ with $0 \leq r < n$ , so $∣ Z / n ∣ = n$ ; while $Z / 0$ is in bijection with $Z$ ).

[L4]

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

Verification

technique · direct

1.1

The set $H$ contains $0$ , is closed under addition since $3 + 3 = 0$ , and contains additive inverses; hence $H \leq G$ by [L2], [L3], and [L4]. Its cosets are $H = {0, 3}$ , $1 + H = {1, 4}$ , and $2 + H = {2, 5}$ .

L2L3L4

2.1

Since $G$ is abelian, every conjugate of $H$ is $H$ , so ${Core}_{G} (H) = H$ .

step 1.1L2algebra

3.1

By [L1], the coset action is transitive and has kernel $H$ . Since $3 \neq 0$ lies in the kernel, the action is not faithful.

step 1.1step 2.1L1L3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

25 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

P. Brosnan, Undergraduate Algebra Notes, 3.14: G-Sets, Proposition 3.102 (standard reference, not scraped)

The action of Z/6Z/6 on the cosets of {0,3}{0,3} is transitive with kernel {0,3}{0,3} and is not faithful