廣西自然科學(xué)基金資助項目(存儲版)
【正文】 toand 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 .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 ont
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