PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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 canonical copy of $N$ is normal, the canonical copy of $H$ is a complement, and conjugation induces the action

Statement

In $N ⋊_{α} H$ , the sets

$\overset{ˉ}{N} = {(n, 1) : n \in N}, \overset{ˉ}{H} = {(1, h) : h \in H}$

are subgroups, $\overset{ˉ}{N}$ is normal, $\overset{ˉ}{N} \cap \overset{ˉ}{H} = {(1, 1)}$ , and every element has a unique factorisation $(n, 1) (1, h)$ . Moreover,

$(1, h) (n, 1) (1, h)^{- 1} = (α_{h} (n), 1) .$

Facts & Assumptions

Given: An external semidirect product $N ⋊_{α} H$ .

[L1]

The semidirect-product law defines a group and gives its inverse formula ( The semidirect-product multiplication makes $N \times H$ a group).

[L2]

A subgroup is a subset closed under the group operations and normality means invariance under conjugation (Subgroup, Normal subgroup: invariance under conjugation).

Proof

technique · direct

1.1

The multiplication and inverse formulas in [L1] show that both displayed sets contain the identity and are closed under products and inverses. Hence both are subgroups by [L2].

L1L2

1.2

Direct multiplication gives $(n, 1) (1, h) = (n, h)$ , while equality $(n, 1) = (1, h)$ forces $n = 1$ and $h = 1$ . This proves existence and uniqueness of the factorisation and the trivial intersection.

2.1

Using the inverse formula gives $(1, h) (n, 1) (1, h)^{- 1} = (α_{h} (n), 1)$ . Conjugation by an element of $\overset{ˉ}{N}$ also preserves $\overset{ˉ}{N}$ because it is a subgroup. By step 1.2 every group element is a product of an element of $\overset{ˉ}{N}$ and one of $\overset{ˉ}{H}$ , so its conjugation preserves $\overset{ˉ}{N}$ ; applying the same argument to its inverse gives equality. Thus $\overset{ˉ}{N}$ is normal by [L2], and the displayed calculation identifies the induced action with $α$ .

step 1.1step 1.2L1L2algebra∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 30 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Keith Conrad, Semidirect Products (standard reference, not scraped)

The canonical copy of NN is normal, the canonical copy of HH is a complement, and conjugation induces the action