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Math Help - cyclic group

  1. #1
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    Unhappy cyclic group

    Need some help on the following 2 related problems:
    Show that any finitely generated subgroup of the additive group of rationals (Q,+,0) is cyclic.
    Prove that this group is not isomorphic to the direct product of two copies of it.
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  2. #2
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    Quote Originally Posted by MathBird View Post
    Need some help on the following 2 related problems:
    Show that any finitely generated subgroup of the additive group of rationals (Q,+,0) is cyclic.
    Remember some facts about the structure of \mathbb{Z}. For x_1,...,x_l\in \mathbb{Z} define (x_1,...,x_l) to be the ideal generated these elements, i.e. \{\Sigma_i a_ix_i | a_i \in \mathbb{Z} \}. Since ideals are principal it means (x_1,...,x_l) = (d), where d\geq 0. We call d the greatest common divisor of x_1,...,x_l.

    Now given, a finitely generated subgroup H of \mathbb{Q}^+. This means, by definition, H = \left< \frac{p_1}{q_1},...,\frac{p_2}{q_2} \right>. The subgroup generated by these elements is precisely \sum_i a_i \frac{p_i}{q_i} where a_i\in\mathbb{Z}. Define, q=\Pi_i q_i and Q_i = \frac{q}{q_i}. Thus, \sum_i a_i\frac{p_i}{q_i} = \frac{1}{q}\sum_i a_ip_iQ_i. Define d to be the greatest common divisor of p_1Q_1,...,p_nQ_n. Then, as seen in the first paragraph, H=\left< \frac{d}{q}\right>. Thus, we see that H must be cyclic.

    Prove that this group is not isomorphic to the direct product of two copies of it.
    Since it is an infinite cyclic group (assuming d>0) it is isomorphic to \mathbb{Z}.
    It is trivial to show that \mathbb{Z} is not isomorphic to \mathbb{Z}\times \mathbb{Z}.
    Last edited by ThePerfectHacker; September 6th 2009 at 07:54 PM.
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  3. #3
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    This is great!
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