ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck 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.

$A_{4}$ has no subgroup of order $6$

Example

The group $A_{4}$ has no subgroup of order $6$ .

Facts & Assumptions

Given: The explicit list of the elements of $A_{4}$ and a hypothetical subgroup $H \leq A_{4}$ with $∣ H ∣ = 6$ .

[L1]

$A_{4}$ consists of the identity, eight three-cycles, and three products of disjoint transpositions ( $A_{4}$ consists of the identity, eight $3$ -cycles, and three products of disjoint transpositions).

[L2]

Every subgroup of index $2$ is normal (Every subgroup of index two is normal).

[L3]

If $H$ is a subgroup of a finite group $G$ , then $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Verification

technique · contradiction

1.1

Suppose, for contradiction, that $∣ H ∣ = 6$ . Since $∣ A_{4} ∣ = 12$ by [L1], [L3] gives $[A_{4} : H] = 2$ , and [L2] makes $H$ normal.

assume-contraL1L2L3

2.1

The complement of $H$ in $A_{4}$ has six elements, so it cannot contain all eight three-cycles from [L1]; hence $H$ contains a three-cycle $g$ .

step 1.1L1

3.1

Write $g = (a b c)$ and let $d$ be the fourth symbol. The three-cycle $u = (a b d)$ belongs to $A_{4}$ by [L1], so normality and direct evaluation give $u g u^{- 1} = (b d c) = : h \in H$ . Likewise $h \in A_{4}$ and $h g h^{- 1} = (a d b) = : k \in H$ . Subgroup closure also puts $g^{- 1}, h^{- 1}, k^{- 1}$ in $H$ .

step 2.1L1

4.1

The seven elements $e, g, g^{- 1}, h, h^{- 1}, k, k^{- 1}$ are distinct: their displayed supports or orientations differ. This contradicts $∣ H ∣ = 6$ , so no subgroup of order $6$ exists.

step 3.1L1discharge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

19 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, §6.2, Proposition 6.9 (standard reference, not scraped)

A4A4​ has no subgroup of order 66