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Nonempty Subset - Abstract Algebra - Exam

Exams, Algebra

Post: February 23th, 2013
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This is the Exam of Abstract Algebra which includes Weighted Equally, Group Theory, Vector Spaces, Linear Algebra, Unique Factoriztaion, Normal Subgroup etc. Key important points are: Nonempty Subset, Group, Subgroup, Multiplication, Element, Isomorphic, Symmetric Group, Alternating Group, Contradiction, Subgroup
This is the Exam of Abstract Algebra which includes Weighted Equally, Group Theory, Vector Spaces, Linear Algebra, Unique Factoriztaion, Normal Subgroup etc. Key important points are: Nonempty Subset, Group, Subgroup, Multiplication, Element, Isomorphic, Symmetric Group, Alternating Group, Contradiction, Subgroup
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MATH 451 FIRST MID-TERM NAME: John Q. Public Question Marks 1 2 3 4 5 1 12 25 28 25 10 2 MATH 451 FIRST MID-TERM Question 1. Let H be a nonempty subset of the group G. Prove that H is a subgroup of G iff a b−1 ∈ H for all a, b ∈ H . First suppose that H is a subgroup of G. Then H is closed under multiplication and taking inverses. Hence if a, b ∈ H , then b−1 ∈ H and so ab−1 ∈ H . Next suppose that ∅ = H ⊆ G is such that ab−1 ∈ H for all a, b ∈ H . Since H = ∅, there exists an element a ∈ H and hence 1 = aa−1 ∈ H . It follows that if a ∈ H , then a−1 = 1 a−1 ∈ H . Finally suppose that a, b ∈ H . Then b−1 ∈ H and so ab = a(b−1 )−1 ∈ H . Thus H is a subgroup of G. MATH 451 FIRST MID-TERM 3 Question 2. (a) Prove that the alternating group A5 does not have a sub- group which is isomorphic to the symmetric group S4 . (b) Prove that the alternating group A5 does not have a subgroup of order 15. (a) Suppose that H A5 is a subg..

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