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Math Help - cyclic

  1. #1
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    cyclic

    how do you prove that the set of positive real numbers under multiplication is not cyclic

    in general, how do you prove that something is not cyclic?

    thanks
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    how do you prove that the set of positive real numbers under multiplication is not cyclic

    in general, how do you prove that something is not cyclic?

    thanks
    Every cyclic group is countable. Let G=\langle g\rangle be cyclic and define \theta:\mathbb{N}\to G:n\mapsto g^n. Clearly \theta is surjective and so G is countable.

    So, why's that help us?
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  3. #3
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    So if a subgroup is not countable, then it is not cyclic.
    so in this case, if the element in the group has infinite order, then it is not cyclic..but im not sure how does that help in writing a proof for this question

    in the function that you have defined, wont there be no identity under multiplication? im not sure how to define functions and always wonder how are they defined.
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  4. #4
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    Quote Originally Posted by alexandrabel90 View Post
    So if a subgroup is not countable, then it is not cyclic.
    so in this case, if the element in the group has infinite order, then it is not cyclic..but im not sure how does that help in writing a proof for this question
    No, infinite is not enough- it must be uncountable. The set of positive real numbers is uncountable, therefore any group having the set of positive real number as underlying set, in particular the "set of positive real numbers under multiplication" is not cyclic.

    in the function that you have defined, wont there be no identity under multiplication? im not sure how to define functions and always wonder how are they defined.
    I'm not sure what you mean by that. You will have to include 0 in \mathbb{N}, as is often done. Then g^0= 1 is the identity. Note that this is simply a "one-to-one" function from the non-negative integers to the group. It is not claimed to be an "isomorphism".
    Last edited by HallsofIvy; May 29th 2010 at 11:19 AM.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    No, infinite is not enough- it must be uncountable. The set of positive real numbers is uncountable, therefore any group having the set of positive real number as underlying set, in particular the "set of positive real numbers under multiplication" is not cyclic.


    I'm not sure what you mean by that. You will have to include 0 in \mathbb{N}, as is oftern done. Then g^0= 1 is the identity. Note that this is simply a "one-to-one" function from the non-negative integers to the group. It is not claimed to be an "isomorphism".
    so to form a proof for it, i have to show that it is uncountable? how do i do that?
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  6. #6
    Senior Member roninpro's Avatar
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    Usually the uncountability of the real numbers is a standard fact that you can use. However, if you need to prove it, see Wikipedia:

    Cantor's diagonal argument - Wikipedia, the free encyclopedia
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