PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck 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.

A permutation group with a regular normal subgroup $G$ embeds in $Hol (G)$

Statement

Let a group $K$ act faithfully on a nonempty set $Ω$ , and suppose $R ⊴ K$ acts regularly on $Ω$ , meaning freely and transitively. After choosing $ω_{0} \in Ω$ and identifying $Ω$ with $R$ by $r \leftrightarrow r ω_{0}$ , the action embeds $K$ in $Hol (R)$ . Under this embedding, $R$ is the subgroup of left translations.

The hypothesis $Ω \neq \emptyset$ is needed rather than automatic: transitivity as defined here is vacuous on the empty set, so without it no base point $ω_{0}$ exists and the displayed identification cannot be made.

Facts & Assumptions

Given: A faithful action of $K$ on a nonempty set $Ω$ , a regular normal subgroup $R$ , and a base point $ω_{0} \in Ω$ .

[L1]

A transitive action carries any chosen point to any other point (Left group actions, transitive actions, and faithful actions), and in a free action only the identity fixes a point (A free group action has no nonidentity element fixing a point). Hence a free transitive action carries any point to any other by a unique group element.

[L2]

An internal semidirect product is recognised by a normal factor, a complement, and trivial intersection ( Recognition theorem: $G = N H$ with $N ⊴ G$ , $N \cap H = 1$ exactly realises an external semidirect product).

[L3]

The holomorph acts faithfully on $R$ by maps $x \mapsto r α (x)$ ( The holomorph acts faithfully on $G$ by affine permutations $x \mapsto g α (x)$ ).

Proof

technique · direct

1.1L1

Let $S = K_{ω_{0}}$ . Regularity gives, for each $k \in K$ , a unique $r \in R$ with $r ω_{0} = k ω_{0}$ . Then $r^{- 1} k \in S$ , so $K = R S$ and $R \cap S = {1}$ .

1.2L1algebra

For $s \in S$ and $r \in R$ , normality gives $s r s^{- 1} \in R$ and $s (r ω_{0}) = (s r s^{- 1}) ω_{0}$ . Thus, under the chosen identification, $s$ acts as the automorphism $r \mapsto s r s^{- 1}$ , while $R$ acts by left translations.

2.1step 1.1L2

Since $R ⊴ K$ , [L2] identifies $K$ with $R ⋊ S$ , where $S$ acts on $R$ by conjugation.

3.1step 1.2step 2.1L3∎

The resulting permutations are precisely of the affine form in [L3], giving a homomorphism $K \to Hol (R)$ . It is injective because the original action is faithful.

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

Peter J. Cameron, The Holomorph of a Group (standard reference, not scraped)

A permutation group with a regular normal subgroup GG embeds in Hol⁡(G)Hol(G)