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Math Help - Show that this group is cyclic.

  1. #1
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    Show that this group is cyclic.

    Ok, I have this:
    U_n = \sqrt[n]{1} where n \in [1, \infty].
    Show that U_n is cyclic.
    It is clearly cyclic as all it's elements are equal to 1, but how do I prove it.
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  2. #2
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    Re: Show that this group is cyclic.

    Quote Originally Posted by Legjendat View Post
    Ok, I have this:
    U_n = \sqrt[n]{1} where n \in [1, \infty].
    Show that U_n is cyclic.
    It is clearly cyclic as all it's elements are equal to 1, but how do I prove it.
    Are we to assume that n\in\mathbb{Z}^+, i.e. a positive integer?
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  3. #3
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    Re: Show that this group is cyclic.

    Quote Originally Posted by Plato View Post
    Are we to assume that n\in\mathbb{Z}^+, i.e. a positive integer?
    Yes. I mean n is an integer, we don't assume it as one, I didn't mention that.
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  4. #4
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    Re: Show that this group is cyclic.

    Quote Originally Posted by Legjendat View Post
    Ok, I have this:
    U_n = \sqrt[n]{1} where n \in [1, \infty].
    Show that U_n is cyclic.
    OK, then U_n\text{ is simply the multiplicative group of the }n, ~~n^{th}\text{-complex roots of unity.}.

    So U_n is generated by \rho=\exp\left(\frac{2\pi i}{n}\right)
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  5. #5
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    Re: Show that this group is cyclic.

    Quote Originally Posted by Legjendat View Post
    Ok, I have this:
    U_n = \sqrt[n]{1} where n \in [1, \infty].
    Show that U_n is cyclic.
    It is clearly cyclic as all it's elements are equal to 1, but how do I prove it.
    IF all its elements were "equal to 1" this would be trivial. But it's not true. What is true is that the absolute value of all elements of U_n is 1.
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  6. #6
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    Re: Show that this group is cyclic.

    Actually, given the notation used by the OP, U_n = 1 for all n. Even if the OP wrote U_n = \{x \in \mathbb{C} \mid x = \sqrt[n]{1}\}, then U_n = \{1\} since \sqrt[n]{1} = 1 is distinct in \mathbb{C}. Now, if U_n = \{x\in \mathbb{C} \mid x^n = 1\}, then we still would only have U_n being the n-th roots of unity in the case that n is rational. But the OP is allowing n \in [1,\infty], which is an interval of the extended reals. We must assume that \sqrt[\infty]{x} is defined and \sqrt[\infty]{1} = 1.
    Last edited by SlipEternal; October 19th 2013 at 03:24 PM.
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  7. #7
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    Re: Show that this group is cyclic.

    Quote Originally Posted by SlipEternal View Post
    Now, if U_n = \{x\in \mathbb{C} \mid x^n = 1\}, then we still would only have U_n being the n-th roots of unity in the case that n is rational. But the OP is allowing n \in [1,\infty], which is an interval of the extended reals. We must assume that \sqrt[\infty]{x} is defined and \sqrt[\infty]{1} = 1.
    It is clear that you did not read the whole thread.
    That issue is addressed in replies #2 & #3.
    He clearly states that n\in\mathbb{Z}^+.
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  8. #8
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    Re: Show that this group is cyclic.

    Quote Originally Posted by Plato View Post
    It is clear that you did not read the whole thread.
    That issue is addressed in replies #2 & #3.
    He clearly states that n\in\mathbb{Z}^+.
    Sorry, I skimmed the beginning. Still, U_n is not a set as it is written and \sqrt[n]{1} = 1 remain valid issues.
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  9. #9
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    Re: Show that this group is cyclic.

    Quote Originally Posted by SlipEternal View Post
    Still, U_n is not a set as it is written and \sqrt[n]{1} = 1 remain valid issues.
    Well, we are dealing with beginning students.

    And this is such a well known result for the n complex nth roots of unity.

    I am confident that is what U_n is meant to be.
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