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3.3%20Isomorphisms - Math 302 Modern Algebra Homework...

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Math 302 Modern Algebra Spring 2009 Homework Solutions 3.3 ISOMORPHISMS AND HOMOMORPHISMS 1. Write out the addition and multiplication tables for Z 6 and for Z 2 × Z 3 . Use them to show that Z 6 = Z 2 × Z 3 . The addition and multiplication tables for Z 6 are + 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4 · 0 1 2 3 4 5 0 0 0 0 0 0 0 1 0 1 2 3 4 5 2 0 2 4 0 2 4 3 0 3 0 3 0 3 4 0 4 2 0 4 2 5 0 5 4 3 2 1 The addition and multiplication tables for Z 2 × Z 3 are + (0 , 0) (1 , 1) (0 , 2) (1 , 0) (0 , 1) (1 , 2) (0 , 0) (0 , 0) (1 , 1) (0 , 2) (1 , 0) (0 , 1) (1 , 2) (1 , 1) (1 , 1) (0 , 2) (1 , 0) (0 , 1) (1 , 2) (0 , 0) (0 , 2) (0 , 2) (1 , 0) (0 , 1) (1 , 2) (0 , 0) (1 , 1) (1 , 0) (1 , 0) (0 , 2) (1 , 2) (0 , 0) (1 , 1) (0 , 2) (0 , 1) (0 , 1) (1 , 2) (0 , 0) (1 , 1) (0 , 2) (1 , 0) (1 , 2) (1 , 2) (0 , 0) (1 , 1) (0 , 2) (1 , 0) (0 , 1) · (0 , 0) (1 , 1) (0 , 2) (1 , 0) (0 , 1) (1 , 2) (0 , 0) (0 , 0) (0 , 0) (0 , 0) (0 , 0) (0 , 0) (0 , 0) (1 , 1) (0 , 0) (1 , 1) (0 , 2) (1 , 0) (0 , 1) (0 , 2) (0 , 2) (0 , 0) (0 , 2) (0 , 1) (0 , 0) (0 , 2) (0 , 1) (1 , 0) (0 , 0) (1 , 0) (0 , 0) (1 , 0) (0 , 0) (1 , 0) (0 , 1) (0 , 0) (0 , 1) (0 , 2) (0 , 0) (0 , 1) (0 , 2) (1 , 2) (0 , 0) (1 , 2) (0 , 1) (1 , 0) (0 , 2) (1 , 1) In order to determine an isomorphism f : Z 6 Z 2 × Z 3 , first note that f must map the additive and multiplicative identities of Z 6 to the additive and multiplicative identities of Z 2 × Z 3 , so f (0) = (0 , 0) and f (1) = (1 , 1) . Since f must preserve addition, then we must have f (2) = f (1 + 1) = f (1) + f (1) = (1 , 1) + (1 , 1) = (0 , 2) f (3) = f (1 + 2) = f (1) + f (2) = (1 , 1) + (0 , 2) = (1 , 0) f (4) = f (1 + 3) = f (1) + f (3) = (1 , 1) + (1 , 0) = (0 , 1) f (5) = f (1 + 4) = f (1) + f (4) = (1 , 1) + (0 , 1) = (1 , 2) This defines our function. To verify that it is in fact an isomorphism, we can compare the addition and multiplication tables with the correspondence 0 (0 , 0) , 1 (1 , 1) , 2 (0 , 2) , 3 (1 , 0) , 4 (0 , 1) , 5 (1 , 2) . If we replace the elements in the tables for Z 6 with the corresponding elements in Z 2 × Z 3 , we should get the tables for Z 2 × Z 3 shown. This shows that this function is an isomorphism. 33
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3.3 Isomorphisms and Homomorphisms Math 302 Homework Solutions page 34 10. Which of the following functions are homomorphisms? (d) h : R M ( R ), defined by h ( a ) = µ a 0 a 0 . This function is not a homomorphism since it does not preserve multiplication. To show this, we need to show ( a R ) ( b R ) £ h ( ab ) = h ( a ) · h ( b ) § is false. To show this, we need to find a counterexample. Let a = 2 and b = 3. Then h (2 · 3) = h (6) = µ 6 0 6 0 and h (2) · h (3) = µ 2 0 2 0 ∂ µ 3 0 3 0 = µ 6 0 6 0 Since h (2 · 3) 6 = h (2) · h (3), then RH2 does not hold, so h does not preserve multiplication. Therefore h is not a homomorphism. 25. (a) If g : R S and f : S T are homomorphisms, show that f g : R T is a homomorphism. Proof: Suppose g : R S and f : S T are arbitrary functions. Suppose f and g are homomorphisms. We need to show that f g : R T is a homomorphism. To show this, we need to show that f g preserves addition and multiplication. RH1. Preservation of Addition. For all a, b R , ( f g )( a + b ) = ( f g )( a ) + ( f g )( b ).
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