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Abstract Algebra - Q/Z

  1. Sep 11, 2008 #1
    The question:

    If G is the additive group Q/Z, what are the elements of the subgroup G(2)? Of G(P) for any positive prime P?

    Where G(n)={a e G| |a| = n^(k) for some k is greater than or equal to 0}...That is the set of all a in G, s.t. the order of a is some power of n. (But since it is the additive group, I suppose it would just a be a multiple of n)

    How do I even begin with this? Aren't the elements of Q/Z sets? The collections of right cosets? and don't they have infinite order?....
     
  2. jcsd
  3. Sep 12, 2008 #2

    morphism

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    I don't see why you would suppose that. The additive notation has no bearing on this matter.

    Sure.

    No. Take an element r+Z in Q/Z. What does it mean for n(r+Z) to be the zero element of Q/Z? If you can answer this correctly, then you can easily deduce that all the elements of Q/Z have finite order.

    My suggestion is to actually think about what Q/Z is, as a group. Make sure you understand how the group operation works, and what order means in this setting.
     
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