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- Let G be an abelian group of order 2n, where n is odd. Use Lagranges Theorem to prove that G contains exactly one element of order 2.25. Prove or disprove that every group of order is abelian.27. a. Show that a cyclic group of order has a cyclic group of order as a homomorphic image. b. Show that a cyclic group of order has a cyclic group of order as a homomorphic image.
- Find two groups of order 6 that are not isomorphic.15. Assume that can be written as the direct sum , where is a cyclic group of order . Prove that has elements of order but no elements of order greater than Find the number of distinct elements of that have order .Prove part c of Theorem 3.4. Theorem 3.4: Properties of Group Elements Let G be a group with respect to a binary operation that is written as multiplication. The identity element e in G is unique. For each xG, the inverse x1 in G is unique. For each xG,(x1)1=x. Reverse order law: For any x and y in G, (xy)1=y1x1. Cancellation laws: If a,x, and y are in G, then either of the equations ax=ay or xa=ya implies that x=y.
- True or false Label each of the following statements as either true or false, where is subgroup of a group. 6. If a subgroup of a group is abelian, then must be abelian.True or false Label each of the following statements as either true or false, where is subgroup of a group. 5. Any subgroup of an abelian group is abelian.1.Prove part of Theorem . Theorem 3.4: Properties of Group Elements Let be a group with respect to a binary operation that is written as multiplication. The identity element in is unique. For each, the inverse in is unique. For each . Reverse order law: For any and in ,. Cancellation laws: If and are in , then either of the equations or implies that .
- Let H be a subgroup of a group G. Prove that gHg1 is a subgroup of G for any gG.We say that gHg1 is a conjugate of H and that H and gHg1 are conjugate subgroups. Prove that H is abelian, then gHg1 is abelian. Prove that if H is cyclic, then gHg1 is cyclic. Prove that H and gHg1 are isomorphic.Lagranges Theorem states that the order of a subgroup of a finite group must divide the order of the group. Prove or disprove its converse: if k divides the order of a finite group G, then there must exist a subgroup of G having order k.Prove or disprove that H={ hGh1=h } is a subgroup of the group G if G is abelian.