oct18 - θ a 6 = a 3 where a ∈ Z is arbitrary Suppose a 6...

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Math 3124 Tuesday, October 18 October 18, Ungraded Homework Problem 19.3 on page 101 Determine whether the groups Z 6 and S 3 are isomorphic. Z 6 and S 3 are not isomorphic because Z 6 is abelian and S 3 is nonabelian. Another reason is that Z 6 has an element of order 6, whereas the elements of S 3 only have orders of 1,2,3. Problem 19.5 on page 101 Determine whether the groups Z 4 × Z 2 and Z 2 × Z 2 × Z 2 are isomorphic. All nonidentity elements of Z 2 × Z 2 × Z 2 have order 2, whereas Z 4 × Z 2 has an element of order 4, namely ([1],[0]). Therefore the groups are not isomorphic. Problem 21.5 on page 109 Define θ : Z 6 Z 3 by θ ([ a ] 6 ) = [ a ] 3 for each [ a ] 6 Z 6 . (a) Prove that θ is well-defined. (b) Prove that θ is a homomorphism. (c) What is ker θ ? (a) Really the problem should say that
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Unformatted text preview: θ [ a ] 6 = [ a ] 3 , where a ∈ Z is arbitrary. Suppose [ a ] 6 = [ b ] 6 . Then 6 | ( a-b ) , hence 3 | ( a-b ) and we deduce that [ a ] 3 = [ b ] 3 . Thus if we write [ a ] 6 in different ways, we always get the same result when we apply the given formula for θ . Therefore θ is well-defined. (b) Let a , b ∈ Z . Then θ ([ a ] ⊕ [ b ]) = θ [ a + b ] = [ a + b ] = [ a ] ⊕ [ b ] = θ [ a ] ⊕ θ [ b ] . (c) [ a ] ∈ ker θ if and only if [ a ] 3 = [ ] ; that is if and only if 3 | a . Therefore ker θ consists of the elements of the form [ a ] 6 where 3 | a , in other words { [ ] , [ 3 ] } ....
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