ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29

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 character table of $A_{4}$

Example

Let $ω = e^{2 π i / 3}$ and let the classes of $A_{4}$ be represented by $1$ , $(12) (34)$ , $(123)$ , and $(132)$ (sizes $1$ , $3$ , $4$ , $4$ ). The character table is

$\begin{array}{ccccc} 1 & (12) (34) & (123) & (132) \\ 1 & 1 & 1 & 1 & 1 \\ χ & 1 & 1 & ω & ω^{2} \\ χ^{2} & 1 & 1 & ω^{2} & ω \\ ψ & 3 & - 1 & 0 & 0 \end{array}$

Both orthogonality relations hold.

Facts & Assumptions

Given: The group $A_{4}$ , its normal subgroup $V_{4}$ , a primitive cube root of unity $ω$ , and the class representatives $1$ , $(12) (34)$ , $(123)$ , $(132)$ .

[F1]

$A_{4}$ has a normal subgroup $V_{4}$ of order $4$ with quotient of order $3$ , and its conjugacy classes have sizes $1$ , $3$ , $4$ , $4$ ( $A_{4}$ has a normal Klein four subgroup and four conjugacy classes).

[F2]

A group of order $3$ is cyclic (Every cyclic group is isomorphic to $(Z, +)$ or to $(Z / n, +)$ for its finite order $n \geq 1$ ).

[F3]

Degree-one representations are exactly the homomorphisms $G \to C^{\times}$ (Equivalence classes of degree-one representations are exactly homomorphisms $G \to k^{\times}$ ; equivalently they factor through $G / G^{'}$ , and they form an abelian group).

[F4]

The squared degrees of the irreducible characters sum to $∣ G ∣$ (The regular character gives a second proof of the sum-of-squares formula).

[F5]

Row orthogonality: irreducible characters are orthonormal (The first orthogonality relation for irreducible complex characters).

[F6]

Column orthogonality: distinct columns are orthogonal, and a column has squared norm the centralizer size (The second orthogonality relation for irreducible complex characters).

[A1]

A homomorphism $A_{4} \to C^{\times}$ that is trivial on $V_{4}$ factors through the quotient $A_{4} / V_{4}$ ; conversely, a homomorphism from $A_{4} / V_{4}$ pulls back to one on $A_{4}$ that is trivial on $V_{4}$ .

[A2]

The commutator $[(123), (124)]$ equals $(12) (34)$ , and a homomorphism to the abelian group $C^{\times}$ is constant on conjugacy classes.

Verification

technique · direct

1.1

By [F1] and [F2] the quotient $A_{4} / V_{4}$ has order $3$ , hence is cyclic. Its three homomorphisms to $C^{\times}$ send a generator to $1$ , $ω$ , or $ω^{2}$ ; by [F3] and [A1], they pull back to three one-dimensional characters of $A_{4}$ whose values are the first three rows. Conversely, every homomorphism $A_{4} \to C^{\times}$ kills commutators, so [A2] makes it kill $(12) (34)$ and therefore, by conjugacy, every nonidentity element of $V_{4}$ . Hence every degree-one character is trivial on $V_{4}$ and factors through $A_{4} / V_{4}$ by [A1]. Thus these are exactly the three one-dimensional characters of $A_{4}$ .

F1F2F3A1A2given

2.1

By [F4], the remaining irreducible degree $d$ satisfies $1 + 1 + 1 + d^{2} = 12$ , so $d = 3$ .

F4step 1.1algebra

3.1

By [F6], the column of $(12) (34)$ is orthogonal to the column of $1$ : $1 + 1 + 1 + ψ ((12) (34)) = 0$ , so $ψ ((12) (34)) = - 1$ .

F6step 1.1step 2.1algebra

4.1

By [F6], the column of $(123)$ is orthogonal to the column of $(12) (34)$ : $1 \cdot \overline{1} + 1 \cdot \overline{ω} + 1 \cdot \overline{ω^{2}} + (- 1) \cdot \overline{ψ ((123))} = 0$ . Since $1 + \overline{ω} + \overline{ω^{2}} = 1 + ω^{2} + ω = 0$ , this forces $\overline{ψ ((123))} = 0$ , hence $ψ ((123)) = 0$ ; the same argument gives $ψ ((132)) = 0$ .

F6step 1.1step 3.1algebra

5.1

The four rows assembled in steps 1.1 through 4.1 form the displayed table. Row orthogonality holds by [F5]: the first three rows are orthonormal ( $\frac{1}{12} (4 + 3 \cdot 1 + 4 + 4) = 1$ for the identity, with cross terms $1 + 3 \cdot 1 + 4 ω + 4 ω^{2} = 1 + 3 - 4 = 0$ ), and $⟨ ψ, ψ ⟩ = \frac{1}{12} (9 + 3 \cdot 1) = 1$ , while $ψ$ is orthogonal to each of the first three rows.

F5step 1.1step 4.1algebra

6.1

Column orthogonality holds by [F6]: the squared norms are $12$ , $4$ , $3$ , $3$ , matching the centralizer sizes $12$ , $4$ , $3$ , $3$ of the four classes, and distinct columns are orthogonal.

F6step 1.1step 5.1algebra∎

Depends on

Used by

FALSE: every value of an irreducible complex character is real False statement

Dependency tree · two levels

32 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 3.5 (standard reference, not scraped)

The character table of A4A4​