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So exactly what am I doing here?

  1. Dec 2, 2004 #1
    to show that U(14) = <3> = < 5 >.

    i.e, U(14) is cyclic.

    ok, so U(14) = 1, 3, 5, 9, 11, 13

    and <3> = 3^1 = 3, 3^2 = 9, 3 ^3 = 13, 3^4 = 11, 3^5 = 5, and 3^6 = 1.

    So what am I doing to show that it is Cyclic? Am I showing that with <3>, I can generate the same elements over and over? is that what it means that it is cyclic?
     
  2. jcsd
  3. Dec 2, 2004 #2

    matt grime

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    Surely your notes define cyclic?

    A finite group is cyclic if there is an element whose order is the same as the order of the group.

    Or a group is cyclic if there is a single element x such that {x^r | 1<=r<=ord(x)}=G. Ie all elements are a power of x for some x.
     
  4. Dec 2, 2004 #3

    mathwonk

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    i.e. if the map taking n to r^n is surjective, then your group is cyclic with generator r.
     
  5. Dec 5, 2004 #4
    I'm sorry, but I can't see the problem. You've shown that U(14)=<3>. You can show similarly that <3>=<5> in U(14).
     
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