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 four rotations of a square act freely, transitively and faithfully on its vertices

Example

Label the vertices of a square by $Z / 4$ in cyclic order. The rotation group $Z / 4$ acts by $a \cdot x = a + x$ . This action is free, transitive, and faithful.

Facts & Assumptions

Given: The additive group $A = Z / 4$ acting on $X = A$ by translation.

[L1]

A left action is transitive and faithful as defined in Left group actions, transitive actions, and faithful actions.

[L2]

An action is free when only the identity can fix a point (A free group action has no nonidentity element fixing a point).

[L3]

The residue classes modulo $4$ form an additive group (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

[L4]

The four classes have unique representatives $0, 1, 2, 3$ (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 $Z$ ).

Verification

technique · direct

1.1

One has $0 \cdot x = x$ and $(a + b) \cdot x = (a + b) + x = a + (b + x) = a \cdot (b \cdot x)$ , so [L1] and [L3] give an action.

L1L3

2.1

Given vertices $x, y$ , the unique class $a = y - x$ satisfies $a \cdot x = y$ , proving transitivity. If $a \cdot x = x$ , cancellation gives $a = 0$ , so the action is free.

step 1.1L1L2L3L4

3.1

An element fixing every vertex fixes $0$ , so it is $0$ by step 2.1; hence the action is faithful.

step 2.1L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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

T. W. Judson, Abstract Algebra: Theory and Applications, 14.1 (standard reference, not scraped)