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Chapter 3 Solutions
Elements Of Modern Algebra
- 32. Let be a fixed element of the group . According to Exercise 20 of section 3.5, the mapping defined by is an automorphism of . Each of these automorphism is called an inner automorphism of . Prove that the set forms a normal subgroup of the group of all automorphism of . Exercise 20 of Section 3.5 20. For each in the group , define a mapping by . Prove that is an automorphism of .arrow_forward45. Let . Prove or disprove that is a group with respect to the operation of intersection. (Sec. )arrow_forward5. For any subgroup of the group , let denote the product as defined in Definition 4.10. Prove that corollary 4.19:arrow_forward
- Prove that Ca=Ca1, where Ca is the centralizer of a in the group G.arrow_forwardLet a and b be elements of a group G. Prove that G is abelian if and only if (ab)2=a2b2.arrow_forwardExercises 22. Let be a finite cyclic group of order with generators and . Prove that the mapping is an automorphism of .arrow_forward
- Let G be a group. Prove that the relation R on G, defined by xRy if and only if there exist an aG such that y=a1xa, is an equivalence relation. Let xG. Find [ x ], the equivalence class containing x, if G is abelian. (Sec 3.3,23) Sec. 3.3, #23: 23. Let R be the equivalence relation on G defined by xRy if and only if there exists an element a in G such that y=a1xa. If x(G), find [ x ], the equivalence class containing x.arrow_forwardExercises 35. Prove that any two groups of order are isomorphic.arrow_forward9. Suppose that and are subgroups of the abelian group such that . Prove that .arrow_forward
- Let A be a given nonempty set. As noted in Example 2 of section 3.1, S(A) is a group with respect to mapping composition. For a fixed element a In A, let Ha denote the set of all fS(A) such that f(a)=a.Prove that Ha is a subgroup of S(A). From Example 2 of section 3.1: Set A is a one to one mapping from A onto A and S(A) denotes the set of all permutations on A. S(A) is closed with respect to binary operation of mapping composition. The identity mapping I(A) in S(A), fIA=f=IAf for all fS(A), and also that each fS(A) has an inverse in S(A). Thus we conclude that S(A) is a group with respect to composition of mapping.arrow_forwardProve or disprove that H={ hGh1=h } is a subgroup of the group G if G is abelian.arrow_forward
- Elements Of Modern AlgebraAlgebraISBN:9781285463230Author:Gilbert, Linda, JimmiePublisher:Cengage Learning,Linear Algebra: A Modern IntroductionAlgebraISBN:9781285463247Author:David PoolePublisher:Cengage Learning