Bourbaki

Definition 3

Tag 058L, permanent. Cite it as [Bourbaki, Tag 058L].

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

Let GG and G′G' be groups with operators in Ω\Omega. A homomorphism of groups with operators of GG into G′G' is a homomorphism of the group GG into the group G′G' such that

f(xα)=(f(x))α f(x^\alpha)=(f(x))^\alpha

for all α∈Ω\alpha\in\Omega and all x∈Gx\in G.

An endomorphism of the group with operators GG is an endomorphism of the group GG which is permutable with all the homotheties of GG.

As two homotheties of a group with operators GG are not necessarily permutable, a homothety of GG is not in general an endomorphism of the group with operators GG.

The identity mapping of a group with operators is a homomorphism of groups with operators; the composition of two homomorphisms of groups with operators is also one. For a mapping to be an isomorphism of groups with operators, it is necessary and sufficient that it be a bijective homomorphism of groups with operators and the inverse mapping is then an isomorphism of groups with operators.

More generally, let G (resp. G') be a group with operators in Ω\Omega (resp. Ω′\Omega'). Let ϕ\phi be a mapping of Ω\Omega into Ω′\Omega'. A ϕ\phi-homomorphism of G into G' is a homomorphism of the group G into the group G' such that

f(xα)=(f(x))(ϕ(α)) f(x^\alpha) = (f(x))^{(\phi(\alpha))}

for all α∈Ω\alpha \in \Omega and all x∈Gx \in G.

In the rest of this paragraph we shall be given a set Ω\Omega. Unless otherwise mentioned the groups with operators considered will admit Ω\Omega as set of operators.

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

Cited by

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Cites

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