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Math Help - Q is not finitely generated

  1. #1
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    Q is not finitely generated

    Hello all Mathematicians and lover of math:

    I need your help in verifying this statement :

    Show that every finitely generated subgroup of ( Q , + ) is cyclic, show also thet ( Q , + ) is not isomorphic to the set of positive rational numbers with usual multiplication

    What I know is that the generator for any finitely-generated subgroup of Q is the smallest non-zero member of that subgroup
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by fuzzy topology View Post
    Hello all Mathematicians and lover of math:

    I need your help in verifying this statement :

    Show that every finitely generated subgroup of ( Q , + ) is cyclic, show also thet ( Q , + ) is not isomorphic to the set of positive rational numbers with usual multiplication

    What I know is that the generator for any finitely-generated subgroup of Q is the smallest non-zero member of that subgroup
    Suppose that G\subseteq Q is finitely generated. Then G=\left\{n_1\cdot\frac{p_1}{q_1}+\cdots+n_k\cdot\f  rac{p_k}{q_k}:n_k\in\mathbb{Z}\right\}\subseteq \left\{\frac{n}{q_1\cdots q_k}:n\in\mathbb{Z}\right\}=\left\langle\frac{1}{q  _1\cdots q_k}\right\rangle. Thus, G is a subgroup of a cyclic group, and thus cyclic.

    For the second part, use this.
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