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Is the group of positive rational numbers under * cyclic?

  1. Feb 15, 2017 #1
    1. The problem statement, all variables and given/known data
    Is the group of positive rational numbers under multiplication a cyclic group.

    2. Relevant equations


    3. The attempt at a solution
    So a group is cyclic if and only if there exists a element in G that generates all of the elements in G.
    So the set of positive rational numbers would be cyclic if we could find a fraction a/b such that (a/b)^n, where n is an integer, generates all elements of positive Q.

    I feel like the group is not cyclic, which means that I would have to prove that for all elements in positive Q, they don't generate positive Q. I am not sure exactly how to approach this... For (a/b)^n, do I have to find another fraction in terms of a and b such that (a/b)^n can't equal that expression for any n?
     
  2. jcsd
  3. Feb 15, 2017 #2
    What does ##<\frac{1}{2}\; , \;\frac{1}{3}>## mean? I thought that one only looks at what subgroup is generated by a single element, in this case a single fraction
     
  4. Feb 15, 2017 #3

    fresh_42

    Staff: Mentor

    This should work. Solve ##(\frac{a}{b})^n=2## and show that you can't get ##(\frac{a}{b})^m=3## .
     
  5. Feb 15, 2017 #4

    fresh_42

    Staff: Mentor

    You beat me. I confused addition and multiplication. In general ##<a>## meant to be the group generated by ##a##, so ##<a>=\{a^n\,\vert \,n\in \mathbb{Z}\}## but I was wrong with the example.
     
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