ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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.

$C [Z / 3 Z] ≅ C \times C \times C$

Example

Let $G = Z / 3 Z = {0, 1, 2}$ . Then

C [G] ≅ C \times C \times C .

Facts & Assumptions

Given: The cyclic group $G = Z / 3 Z$ .

[L1]

Over an algebraically closed field of characteristic prime to $∣ G ∣$ , the number of irreducible representations equals the number of conjugacy classes (If $k$ is algebraically closed and $char k ∤ ∣ G ∣$ , the number of irreducible representations of $G$ equals the number of conjugacy classes).

[L2]

Under the same hypotheses, the irreducible degrees satisfy the sum-of-squares formula (If $k$ is algebraically closed and $char k ∤ ∣ G ∣$ , then $\sum_{i} (\dim_{k} V_{i})^{2} = ∣ G ∣$ ).

[L3]

Under the same hypotheses, the group algebra is a product of full matrix algebras over the base field (If $k$ is algebraically closed and $char k ∤ ∣ G ∣$ , then $k [G] ≅ \prod_{i = 1}^{r} M_{n_{i}} (k)$ ).

Verification

technique · direct

1.1

The group $G$ is abelian and has three elements, so each element forms its own conjugacy class. Hence [L1] gives exactly three irreducible complex representations, with degrees $d_{1}, d_{2}, d_{3}$ , and [L2] gives $d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = 3.$

L1L2givenalgebra

2.1

Each $d_{i}$ is a positive integer, so the only way three positive squares can sum to $3$ is $d_{1} = d_{2} = d_{3} = 1$ . Applying [L3], all three Wedderburn factors are $1 \times 1$ matrix algebras, so $C [G] ≅ M_{1} (C) \times M_{1} (C) \times M_{1} (C) = C^{3} .$

step 1.1L3givenalgebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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

Pavel Etingof et al., Introduction to Representation Theory, Section 3.3 (standard reference, not scraped)

C[Z/3Z]≅C×C×CC[Z/3Z]≅C×C×C