广西自然科学基金资助项目.doc

Application ofThe Fundamental Homomorphism Theoremof GroupLI Qian-qian LIU Zhi-gang YANG Li-ying（Department of Mathematics and Computer Science, Guangxi Teachers Education University,Nanning Guangxi 530001, P.R.China）Abstract: The fundamental homomorphism theorem is very important consequence in group theory, by using it we can resolve many problems. In this paper we researches mainly about the fundamental homomorphism theorem applied to direct products of groups and group of inner automorphisms of a group G. Keyword: The Fundamental Homomorphism Theorem; Direct Products; Inner AutomorphismsMR(2003) Subject Classification: 16WChinese Library Classification: O153.3Document code: A In the realm of abstract algebra, group is one of the basic and important concept, have extensive application in the math itself and many side of modern science technique. For example Theories physics, Quantum mechanics, Quantum chemistry, Crystallography application are clear certifications. So that, after we study abstract algebra course, go deep into a ground of theories of research to have the necessity very much more. In the contents of group, the fundamental homomorphism theorem is very important theorem, we can use it prove many problems about group theory, in this paper to prove several conclusions as follow with the fundamental homomorphism theorem: These contents are all standard if we not to the special provision and explained.Definition 1. Let be a subgroup of a group with symbol , we say is the normal subgroup of if one of the following conditions hold. To simplify matters, we write . (1) for any ;(2) whenever any ;(3) for every and any .Definition 2. The kernel of a group homomorphism from to a group with identity is the set . The kernel of is denoted by .Definition 3. Let be a collection of groups. The external direct product of , 广西自然科学基金(0447038)资助项目written as , is the set of all m-tuples for which the its component is an element of , and the operation is componentwise. In symbols =,where is defined to be Notice that it is easily to verify that the external direct product of groups is itself a group.4 Definition 4. Let be a group and be a subgroup of . For any , the set is called the left coset of in containing . Analogously is called the right coset of H in containing .Lemma 1.1 ( The fundamental homomorphism theorem) Let be a group homomorphism from to . Then the = is the normal subgroup of , and . To simplify matters, we call the theorem as the FHT.Lemma 2.2 Let be a group homomorphism from to . Then we have the following properties:（1）If is a subgroup of , then is a subgroup of ;（2）If is a normal in, then is a normal in;（3）If is a subgroup of , then is a subgroup of ;（4）If is a normal subgroup of , then is a normal subgroup of Lemma 3.3 Let be a homomorphism from a group to a group , and,. Then .Lemma 4.4 Let H be a subgroup of G and let belong to G, then:（1） if and only if ;（2） if and only if .By using the above lemmas we can obtain the following mainly results.Theorem 1. Let G and H be two groups. Suppose JG and KH, then and .Proof. First we will prove . For any and every . We have:.Since and , we can get , i.e. .Thus . We make use of the FHT to prove that is isomorphic to. Therefore we must look for a group homomorphism from onto and determine the kernel of it. In fact one can define correspondencedefined by . Clearly, , there must be to satisfy. Thus, is onto.Because of JG, we have for, similarly, for .When , there are .For any , we have =.Hence . Therefore is group a homomorphism from ontoand is the identity of. For any , then, according to the property of coset, we can get: if and only if and , i.e. =. Now let we look at our proof: , is a group homomorphism from onto and the kernel of is . According to the FHT, we can get .Theorem 2. Let is a group homomorphism from onto .If and , then where .Proof: According to Lemma 2.2 (2), we know .To establish , we firstly need to construct a mapping and prove is a group homomorphism from onto . We give the mapping defined by where =.For , since is a surjection from to , we must be found such that .Thus is onto.For arbitrary , Therefore is a group homomorphism.We will now show , in fact we know that is identity of , according to Lemma 4, we can get that for, then , say , so that. On the other hand , , that is to say , .Moreover , because of , therefore . That is . According to the FHT, we can obtain .Theorem 1 and Theorem 2 apply Exercise 1 and Exercise 2.Exercise 1. is normal subgroup of , is a normal subgroup of .So that for any and , for a function: we have is a group isomorphism, so that Assume and are sets of all the nonzero real numbers and positive real numbers respectively, it is readily to verify that they are indeed group with ordinary multiplication.Exercise 2. Let be general linear group of 22 matrices over under ordinary matrix multiplication . Then the mapping is a group homomorphism from onto . The group of matrices with determinant 1 over is a normal subgroup of . Moreover .Definition 5. An automorphism of groupis just a group isomorphism from to itself. The set of all automorphisms of groupis denoted by . For any , is called an inner automorphism of and is the set of all inner automorphism of .Theorem 3: Letbe a group and the mapping defined by . Then and.Proof. It is clearly that5.To show , suffice it to prove that is an automorphism of for any . 1)(one-to-one) For any , if =, then by using cancellation law of group. Thus is one-to-one.2)(onto) For any , we take , then, so that is onto.3)(O.P.) For any , we have . Therefore is isomorphism from to .According to the definition of automorphism. We know is an automorphism of . Notice that for any , we have and .In fact for any , it is clearly . Also , Thus .Since , say , we have known . We can obtain, i.e. Hence the proof of is complete. It is easy to see that for every, if and only if where is the center of (short for ).Let be the mapping defined by , we will prove that is a group homomorphism from G onto I(G) and that C is its kernel.For every , we can readily find that , that is to say, is onto. For any , since , so that is a group homomorphism from onto .Notice that for any and every , we have , i.e., , that is . We obtain, hence . Next,