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.
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。
(a) Let G be a non-commutative simple group, the automorphism group of G and the group of inner automorphisms of G (isomorphic to G). If s is an automorphism of the group , show that (using Exercise 3 above and § 4, no. 9, Proposition 15, note that the intersection cannot consist only of the identity element of ). (b) Show that the only automorphism of leaving invariant each of the elements of is the identity element (use the fact that this automorphism leaves invariant and for all and and use Exercise 3). (c) Let s be an automorphism of , the isomorphism of G onto , the inverse isomorphism and the automorphism of G; show that the automorphism of is the identity automorphism (use (b), noting that, for all , ). 推出的每个自同构都是内自同构。
解答
尚未写出解答。
被以下引用
无。