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Math Help - Q is not a free abelian group

  1. #1
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    Q is not a free abelian group

    I have to do the following exercise form the fraleigh:
    Show that Q under addition is not a free abelian group (Hint: Show that no two distinct rational numbers n/m and r/s could be contained in a set satisfying condition 2 of theorem 38.1)

    Condition 2: X generates G, and n_1*x_1 + n_2*x_2 + ... + n_r*x_r = 0 for n_i in Z if and only if n_1 = n_2 =...= n_r = 0
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  2. #2
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    Quote Originally Posted by Diego View Post
    I have to do the following exercise form the fraleigh:
    Show that Q under addition is not a free abelian group (Hint: Show that no two distinct rational numbers n/m and r/s could be contained in a set satisfying condition 2 of theorem 38.1)

    Condition 2: X generates G, and n_1*x_1 + n_2*x_2 + ... + n_r*x_r = 0 for n_i in Z if and only if n_1 = n_2 =...= n_r = 0
    first show that Q cannot be cyclic. then suppose that it's free abelian and let x = n/m and y = r/s be two distinct elements of a basis of Q. see that mrx - sny = 0, which mean that x and y are

    not linearly independent. contadiction!
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