廣西自然科學(xué)基金資助項(xiàng)目-wenkub
2023-07-15 00:15:07 本頁面
【正文】 can get , . .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 , . =. 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 .
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