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PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16
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The canonical semidirect decomposition is an internal direct product if and only if the defining action is trivial

Statement

The canonical factors of NαH form an internal direct product if and only if αh=idN for every hH. In that case NαH is the external direct product N×H.

Facts & Assumptions

Given: An external semidirect product NαH with its canonical factors Nˉ and Hˉ.

[L1]

The canonical factors have trivial intersection, multiply to the whole group, and satisfy (1,h)(n,1)(1,h)1=(αh(n),1) (The canonical copy of N is normal, the canonical copy of H is a complement, and conjugation induces the action).

[L2]

The external direct product has coordinatewise multiplication (The external direct product G×H with componentwise multiplication).

[L3]

A subgroup MG is normal when gMg1=M for every gG (Normal subgroup: invariance under conjugation).

Proof

technique · iff
1.1

[reverse] Suppose the action is trivial. The semidirect law becomes (n,h)(n,h)=(nn,hh), which is the direct-product law from [L2].

L1L2
1.2

[forward] Suppose the canonical decomposition is an internal direct product, so Hˉ as well as Nˉ is normal. For xNˉ and yHˉ, normality gives xyx1y1Hˉ and also xyx1y1=x(yx1y1)Nˉ. Thus this commutator lies in NˉHˉ={1} by [L1], so x and y commute.

L1L3algebra
2.1

The conjugation formula in [L1] now gives (αh(n),1)=(n,1) for every n,h. Hence every αh is the identity.

step 1.2L1

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 15 results over 10 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.

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