Bourbaki

Example

Tag 05BM, permanent. Cite it as [Bourbaki, Tag 05BM].

Algebra, chapter I, § 5. Groups operating on a set

(1) Let G operate on itself by left (resp. right) translation. Thus a left (resp. right) homogeneous principal G-set structure is defined on G, which is sometimes denoted by GsG_s (resp. GdG_d).

(2) Let E be a homogeneous set under a commutative group G. If G operates faithfully on E, the latter is a homogeneous principal G-set.

(3) Let E and F be two isomorphic sets with structures of the same species and let Isom(E, F) be the set of isomorphisms of E onto F (with the given structures). The group Aut(E) of automorphisms of E (with the given structure) operates on Isom(E, F) on the right by the law (σ,f)↦f∘σ(\sigma, f) \mapsto f \circ \sigma and Isom(E, F) is a right homogeneous principal Aut(E)-set. Similarly, the group Aut(F) operates on Isom(E, F) on the left by the law (σ,f)↦σ∘f(\sigma, f) \mapsto \sigma \circ f and Isom(E, F) is a left homogeneous principal Aut(F)-set.

(4) *A homogeneous principal set under the additive group of a vector space is called an affine space (cf. II, § 9, no. 1).*

The group of automorphisms of the homogeneous principal G-set GsG_s (Example 1) is the group of right translations of G which is identified with G0G^0 (no. 5, Proposition 5). Let E be a homogeneous principal G-set and aa an element of E. The orbital mapping ωa\omega_a defined by aa is an isomorphism of the G-set GsG_s onto E. By transporting the structure an isomorphism ψa\psi_a is derived of G0G^0 onto Aut(E). It should be noted that ψa\psi_a in general depends on a; more precisely, for α∈G\alpha \in G,

ψαa=ψa∘Int⁡G0(α)=ψa∘Int⁡(α−1). \psi_{\alpha a} = \psi_a \circ \operatorname{Int}_{G^0}(\alpha) = \psi_a \circ \operatorname{Int}(\alpha^{-1}).

For, writing δa\delta_a for the translation x↦xαx \mapsto x \alpha on G,

ωαa=ϕωa∘δα \omega_{\alpha a} = \phi \omega_a \circ \delta_\alpha
and
ψa(x)=ωa∘δx∘ωa−1,x∈G, \psi_a(x) = \omega_a \circ \delta_x \circ \omega_a^{-1}, \quad x \in G,
whence
ψαa(x)=ωa∘δα∘δx∘δα−1∘ωa−1=ωa∘δα−1xα∘ωa−1=ψa(α−1xα). \psi_{\alpha a}(x) = \omega_a \circ \delta_\alpha \circ \delta_x \circ \delta_\alpha^{-1} \circ \omega_a^{-1} = \omega_a \circ \delta_{\alpha^{-1} x \alpha} \circ \omega_a^{-1} = \psi_a(\alpha^{-1} x \alpha).

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

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