sol09 - MAT2200, SPRING 2009 SOLUTIONS TO THE MANDATORY...

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Unformatted text preview: MAT2200, SPRING 2009 SOLUTIONS TO THE MANDATORY ASSIGNMENT Problem 1 (a) Consider the integers Z as a subgroup of the additive group Q of rational num- bers. Show that every element of Q / Z has finite order, and that Q / Z has exactly one subgroup H n isomorphic to Z n for each natural number n . Using additive notation for the group operation, the elements of Q / Z have the form m n + Z , where m and n are integers, n 6 = 0. Then n ( m n + Z ) = m + Z = Z , which represents the identity element in Q / Z . The element 1 n + Z has order n , since n 1 n ∈ Z and k 1 n / ∈ Z for 1 6 k < n . Hence the cyclic subgroup H n = < 1 n + Z > is isomorphic to Z n . Any cyclic subgroup of Q / Z has the form K = < a b + Z > , where we can choose a and b to be positive integers that are relatively prime. If K ' Z n , we must have n a b ∈ Z , hence b | n , and a b + Z = an/b n + Z ∈ H n . Thus K ⊆ H n , and since both groups have n elements, K = H n . (b) Show that the formula φ n ( x + Z ) = nx + Z defines a surjective homomorphism φ n : Q / Z → Q / Z . What is the kernel of φ n ? φ n is a homomorphism since φ n ( a + Z ) + φ n ( b + Z )) = ( na + Z ) + ( na + Z ) = ( na + nb ) + Z = φ n (( a + b ) + Z ) . Since every element a + Z ∈ Q / Z can be written n a n + Z = φ n ( a n ), φ n is surjective. a b + Z ∈ ker φ n ⇐⇒ na b ∈ Z ⇐⇒ a b = k n , for some k ∈ Z . Hence ker φ n = { k n | k ∈ Z } = H n . (c) Show that there is no nontrivial homomorphism from Q / Z to a finite group....
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sol09 - MAT2200, SPRING 2009 SOLUTIONS TO THE MANDATORY...

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