廣西自然科學(xué)基金資助項(xiàng)目(文件)
【正文】 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 that≤[5].To show , suffice it to prove that is an automorphism of for any . 1)(onetoone) For any , if =, then by using cancellation law of group. Thus is onetoone.2)(onto) For any , we take , then, so that is onto.3)(.) 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, .. Hence the proof of ≤is plete. It is easy to see that for every, if and only if where is the center of (short for ).Let be the ma
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