Bourbaki

Definition 12

Tag 059O, permanent. Cite it as [Bourbaki, Tag 059O].

Algebra, chapter I, § 4. Groups and groups with operators

The group with operators G=∏i∈IGiG = \prod_{i \in I} G_i defined above is the product group with operators of the GiG_i. The mappings pri:G→Gi\mathrm{pr}_i : G \to G_i are called the projection homomorphisms.

A particular case of the product of groups with operators is the group GEG^E consisting of the mappings of a set EE into a group with operators GG, the laws being defined by:

(fg)(x)=f(x)g(x)(f,g∈GE,x∈E) (fg)(x) = f(x)g(x) \quad (f, g \in G^E, x \in E)
fα(x)=f(x)α(f∈GE,α∈Ω,x∈E). f^\alpha(x) = f(x)^\alpha \quad (f \in G^E, \alpha \in \Omega, x \in E).

Let (ϕi:H→Gi)i∈I(\phi_i : H \to G_i)_{i \in I} be a family of homomorphisms of groups with operators. The mapping h↦(ϕi(h))i∈Ih \mapsto (\phi_i(h))_{i \in I} of HH into ∏i∈IGi\prod_{i \in I} G_i is a homomorphism of groups with operators. It is the only homomorphism Φ:H→∏i∈IGi\Phi : H \to \prod_{i \in I} G_i satisfying pri∘Φ=ϕi\mathrm{pr}_i \circ \Phi = \phi_i for all ii. This justifies the name "product group with operators" (Set Theory, IV, § 2, no. 4).

Let (ϕi:Hi→Gi)i∈I(\phi_i : H_i \to G_i)_{i \in I} be a family of homomorphisms of groups with operators. The mapping ∏i∈Iϕi:(hi)i∈I↦(ϕi(hi))i∈I\prod_{i \in I} \phi_i : (h_i)_{i \in I} \mapsto (\phi_i(h_i))_{i \in I} of ∏i∈IHi\prod_{i \in I} H_i into ∏i∈IGi\prod_{i \in I} G_i is a homomorphism of groups with operators.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

Transcribed from the printed edition, 1998, Springer. Not translated.