Explanation Given information: H 1 = { [ 0 ] , [ 10 ] , [ 20 ] , [ 30 ] , [ 40 ] , [ 50 ] , [ 60 ] } , H 2 = { [ 0 ] , [ 14 ] , [ 28 ] , [ 42 ] , [ 56 ] } and H 3 = { [ 0 ] , [ 35 ] } are subgroups of the abelian group ℤ 70 under addition. Formula used: 1) Let H 1 , H 2 , ...... , H n be subgroups of the abelian group G . The sum H 1 + H 2 + ...... + H n of these subgroups is defined by H 1 + H 2 + ...... + H n = { x ∈ G | x = h 1 + h 2 + ..... + h n with h i ∈ H i } . 2) Let H 1 and H 2 be subgroups of the abelian group G . Then G = H 1 ⊕ H 2 if and only if G = H 1 + H 2 and H 1 ∩ H 2 = { 0 } . Calculations: Let H 1 = 〈 [ 10 ] 〉 = { [ 0 ] , [ 10 ] , [ 20 ] , [ 30 ] , [ 40 ] , [ 50 ] , [ 60 ] } , H 2 = 〈 [ 14 ] 〉 = { [ 0 ] , [ 14 ] , [ 28 ] , [ 42 ] , [ 56 ] } and H 3 = 〈 [ 35 ] 〉 = { [ 0 ] , [ 35 ] }

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ISBN: 9781285965918

Author: Gilbert

Publisher: Cengage

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Chapter 4.7, Problem 6E

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Chapter 4.7, Problem 5E

Chapter 4.7, Problem 7E

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Chapter 4 Solutions

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ℤ 70 = H 1 ⊕ H 2 ⊕ H 3

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