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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
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The semidirect-product multiplication makes N×H a group

Statement

Let α:H→Aut⁡(N) be an action by automorphisms. The multiplication

(n,h)(n′,h′)=(nαh(n′),hh′)

makes N×H a group with identity (1N,1H) and inverse

(n,h)−1=(αh−1(n−1),h−1).

Facts & Assumptions

Given: Groups N,H and a homomorphism α:H→Aut⁡(N).

[L1]

The external semidirect-product multiplication is (n,h)(n′,h′)=(nαh(n′),hh′) ( The external semidirect product N⋊αH).

[L2]

An action by automorphisms satisfies αhh′=αh∘αh′ and α1=id⁡N, with every αh an automorphism of N (An action of a group H on a group N by automorphisms).

Proof

technique · direct
1.1L1L2algebra

For three pairs, multiplication in either parenthesisation gives (nαh(n′)αhh′(n′′),hh′h′′) because αh is a homomorphism and αhh′=αhαh′. Thus the operation is associative.

1.2L1L2L3

Since α1 is the identity and every αh preserves 1N, the pair (1N,1H) is a two-sided identity.

2.1L1L2L3algebra∎

Put y=αh−1(n−1). Then (n,h)(y,h−1)=(nαh(y),1H)=(1N,1H) and (y,h−1)(n,h)=(yαh−1(n),1H)=(1N,1H). Hence the displayed pair is the two-sided inverse.

Depends on

Used by

Cited to discharge well-definedness by The external semidirect product N⋊_α H.

Dependency tree · two levels

21 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