Bourbaki

习题 4, § 5

标签 05C0,永久不变。引用为 [Bourbaki, Tag 05C0]。

代数, I, 140

Bourbaki marks this one as harder than the rest.

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(a) Let G be a non-commutative simple group, Γ\Gamma the automorphism group of G and Δ\Delta the group of inner automorphisms of G (isomorphic to G). If s is an automorphism of the group Γ\Gamma, show that s(Δ)=Δs(\Delta) = \Delta (using Exercise 3 above and § 4, no. 9, Proposition 15, note that the intersection Δ∩s(Δ)\Delta \cap s(\Delta) cannot consist only of the identity element of Γ\Gamma). (b) Show that the only automorphism of Γ\Gamma leaving invariant each of the elements of Δ\Delta is the identity element (use the fact that this automorphism leaves invariant α\alpha and σασ−1\sigma \alpha \sigma^{-1} for all α∈Δ\alpha \in \Delta and σ∈Γ\sigma \in \Gamma and use Exercise 3). (c) Let s be an automorphism of Γ\Gamma, ϕ\phi the isomorphism x↦Int⁡(x)x \mapsto \operatorname{Int}(x) of G onto Δ\Delta, ψ\psi the inverse isomorphism and σ\sigma the automorphism ψ∘s∘ϕ\psi \circ s \circ \phi of G; show that the automorphism ξ↦σ−1s(ξ)σ\xi \mapsto \sigma^{-1}s(\xi)\sigma of Γ\Gamma is the identity automorphism (use (b), noting that, for all x∈Gx \in G, s(Int⁡(x))=Int⁡(σ(x))s(\operatorname{Int}(x)) = \operatorname{Int}(\sigma(x))). 推出Γ\Gamma的每个自同构都是内自同构。

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Google 翻译生成的粗略草稿,质量较低。术语依照术语表,但句子可能有误;之后会由仔细的译文取代。