TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03

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$G \times H$ is a group with identity $(e_{G}, e_{H})$ , coordinatewise inverses, and homomorphic coordinate projections

Statement

For groups $G$ and $H$ , the componentwise operation of The external direct product $G \times H$ with componentwise multiplication makes $G \times H$ a group. Its identity is $(e_{G}, e_{H})$ , and

$(g, h)^{- 1} = (g^{- 1}, h^{- 1}) .$

Moreover the coordinate maps $π_{G} (g, h) = g$ and $π_{H} (g, h) = h$ are group homomorphisms.

Facts & Assumptions

Given: Groups $G, H$ with identities $e_{G}, e_{H}$ .

[L1]

$G \times H$ has the componentwise operation $(g, h) (g^{'}, h^{'}) = (g g^{'}, h h^{'})$ (The external direct product $G \times H$ with componentwise multiplication).

[L2]

A group operation is associative, has a two-sided identity, and gives every element a two-sided inverse (Group and abelian group).

[L3]

A map between groups is a group homomorphism exactly when it preserves products (Monoid homomorphism and group homomorphism).

Proof

technique · direct

1.1L1L2givenalgebra

For $(g, h), (g^{'}, h^{'}), (g^{''}, h^{''}) \in G \times H$ , associativity in each factor gives $((g, h) (g^{'}, h^{'})) (g^{''}, h^{''}) = (g g^{'} g^{''}, h h^{'} h^{''}) = (g, h) ((g^{'}, h^{'}) (g^{''}, h^{''}))$ ; thus the componentwise operation is associative.

1.2L1L2givenalgebra

For every $(g, h) \in G \times H$ , $(e_{G}, e_{H}) (g, h) = (g, h) = (g, h) (e_{G}, e_{H})$ ; thus $(e_{G}, e_{H})$ is a two-sided identity.

1.3L1L2givenalgebra

For every $(g, h) \in G \times H$ , $(g, h) (g^{- 1}, h^{- 1}) = (e_{G}, e_{H}) = (g^{- 1}, h^{- 1}) (g, h)$ ; so $(g^{- 1}, h^{- 1})$ is its inverse.

2.1step 1.1step 1.2step 1.3L2

Steps 1.1–1.3 verify the group axioms for $G \times H$ .

3.1step 2.1L1L3givenalgebra

For pairs $(g, h), (g^{'}, h^{'})$ , $π_{G} ((g, h) (g^{'}, h^{'})) = g g^{'} = π_{G} (g, h) π_{G} (g^{'}, h^{'})$ ; the same coordinatewise calculation holds for $π_{H}$ , so both projections are homomorphisms from the group in step 2.1.

4.1step 2.1step 3.1∎

The stated identity, inverse formula, and coordinate homomorphisms follow.

Depends on

Dependency tree · 2 levels

Sources

Sharifi, Abstract Algebra, direct products

G×HG×H is a group with identity (eG,eH)(eG​,eH​), coordinatewise inverses, and homomorphic coordinate projections