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TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26
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A transitive imprimitive action embeds modulo its kernel in an imprimitive wreath product

Statement

Let G act transitively on Ω, and let BΩ be a nontrivial block. Put Σ:={gB:gG},GB:={gG:gB=B}. Let K be the permutation group induced by G on Σ, and let H be the permutation group induced by GB on B.

Choose for each CΣ an element tCG with tCB=C,tB=e. Then there is a homomorphism Φ:GHΣK whose kernel is exactly the kernel of the given action on Ω.

In particular, if the action of G on Ω is faithful, then Φ is an embedding.

Facts & Assumptions

Given: A transitive action of G on Ω, a block BΩ, the block system Σ:={gB:gG}, and a choice of tCG with tCB=C and tB=e.

[L1]

A block B satisfies: for every gG, either gB=B or (gB)B= (Blocks and block systems for a group action).

[L2]

The imprimitive wreath product HΣK is the semidirect product HΣK acting on B×Σ by (f,k)(b,C)=(f(kC)b, kC). (The imprimitive wreath product of permutation groups).

Proof

technique · constructive
1.1

For each gG, let kg be the permutation of Σ induced by g, so kg(C)=gC. For each CΣ, the element tC1gtg1C stabilizes B setwise because tC1gtg1CB=tC1g(g1C)=tC1C=B. Let fg(C)H be the induced permutation of B defined by this element.

L1construct
2.1

Define Φ(g):=(fg,kg). For CΣ, the function component of Φ(g)Φ(h) at C is fg(C)fh(g1C), while tC1ght(gh)1C=(tC1gtg1C)(tg1C1hth1g1C), so it induces the same permutation of B as fgh(C). Also kgh=kgkh. Hence Φ(gh)=Φ(g)Φ(h).

step 1.1L2algebra
3.1

Identify Ω with B×Σ by Ψ(b,C):=tCb. Then for every gG one has Ψ(Φ(g)(b,C))=tgC(fg(gC)b)=g(tCb)=gΨ(b,C). So Φ(g)=1 exactly when g fixes every point of Ω.

step 1.1step 2.1L2algebra
4.1

Step 3.1 shows that kerΦ is the kernel of the given action. Therefore a faithful action makes kerΦ=1, so in that case Φ is an embedding into HΣK.

step 3.1discharge-construct

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Sources