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

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Actions changed by automorphisms of the kernel and complement give isomorphic semidirect products

Statement

Let $α, β : H \to Aut (N)$ be actions. If $u \in Aut (N)$ and $v \in Aut (H)$ satisfy

$β_{v (h)} = u α_{h} u^{- 1} (h \in H),$

then

$N ⋊_{α} H ≅ N ⋊_{β} H$

by $(n, h) \mapsto (u (n), v (h))$ .

Facts & Assumptions

Given: Actions $α, β$ and automorphisms $u, v$ satisfying the displayed compatibility.

[L1]

The multiplication in $N ⋊_{α} H$ is $(n, h) (n^{'}, h^{'}) = (n α_{h} (n^{'}), h h^{'})$ ( The external semidirect product $N ⋊_{α} H$ ).

[L2]

Both semidirect products are groups ( The semidirect-product multiplication makes $N \times H$ a group).

[L3]

A bijective homomorphism is an isomorphism (Group isomorphisms, automorphisms and the set $Aut (G)$ ).

Proof

technique · direct

1.1

Let $F (n, h) = (u (n), v (h))$ . Then $F ((n, h) (n^{'}, h^{'})) = (u (n) u (α_{h} (n^{'})), v (h) v (h^{'}))$ . The compatibility gives $u α_{h} = β_{v (h)} u$ , so this is $F (n, h) F (n^{'}, h^{'})$ under the $β$ multiplication. Thus $F$ is a homomorphism.

L1algebra

2.1

The map $(m, k) \mapsto (u^{- 1} (m), v^{- 1} (k))$ is the set-theoretic inverse of $F$ . Therefore $F$ is bijective and is an isomorphism by [L2] and [L3].

step 1.1L2L3∎

Depends on

Used by

For primes p<q, nontrivial actions of Cₚ on C_q exist exactly when p∣(q-1) and are unique up to automorphisms Lemma

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 28 results over 14 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)