计算机问题求解一论题43 -群同态基本定理 2019年3月20日
计算机问题求解 – 论题4-3 - 群同态基本定理 2019年3月20日
问题1：我们为什么定义“同构”函数？ iso-morphology Two groups (G,)and (H,o)are isomorphic if there exists a one-to-one and onto map GH such that the group operation is preserved;that is, (a·b)=(a)o(b) 同构其实可在 for all a and b in G.If G is isomorphic to H is called an isomorphism. 任何代数结构 (系统)上讨论
问题1：我们为什么定义“同构”函数 ？ iso-morphology 同构其实可在 任何代数结构 (系统)上讨论
问题2：从这个定理中，你能解释我们为 什么研究“同构”吗？ Let GH be an isomorphism of two groups.Then the following statements are true. 1.-1:HG is an isomorphism. 2.1G=H 3.If G is abelian,then H is abelian. 4.If G is cyclic,then H is cyclic. 5.If G has a subgroup of order n,then H has a subgroup of order n
问题2：从这个定理中，你能解释我们为 什么研究“同构”吗？
如何判断两个系统的同构？ Theorem 9.2 All cyclic groups of infinite order are isomorphic to Z. PRoOF.Let G be a cyclic group with infinite order and suppose that a is a generator of G.Define a map:ZG byo:n a".Then 观察 o(m+n)amtn a"a"o(m)o(n). 构造 To show that o is injective,suppose that m and n are two elements in Z, where m n.We can assume that m >n.We must show that am a". 证明 Let us suppose the contrary;that is,am=a".In this case am-=e,where m-n >0,which contradicts the fact that a has infinite order.Our map is onto since any element in G can be written as an for some integer n and o(n)a". ▣
观察 构造 证明 如何判断两个系统的同构？
如何判断两个系统不同构？ Example 9.5.Even though S3 and Z6 possess the same number of elements,we would suspect that they are not isomorphic,because Z6 is abelian and S3 is nonabelian.To demonstrate that this is indeed the case,suppose that Z6-S3 is an isomorphism.Let a,b e S3 be two elements such that ab ba.Since o is an isomorphism,there exist elements m and n in Z6 such that o(m)=a and o(n)=b. However, ab =o(m)o(n)=o(m+n)=o(n+m)=o(n)o(m)=ba, which contradicts the fact that a and b do not commute
如何判断两个系统不同构？