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Prove that U(2^N) is cyclic

  1. Dec 1, 2012 #1
    Prove that U(p^k) is cyclic

    p^k is an odd prime power.

    I've been working on this problem for a while and can't figure it out. The professor's hint is "to think about the solutions to x2 =1." (pk - 1)2 mod pk = 1 but I'm unsure how that is helpful.

    I know that that 2 generates every set by trial and error, and I'm reasonable sure that any prime less than p generates the set. The order of the set is pk - pk - 1.

    If |q| = n, then q|U(pk)| = 1 mod pk and n | p - 1 or n | pk - 1 or n | (p - 1)pk - 1. I'm unsure of how to solve the modular arithmetic from here though.

    Edit: I wrote down wrong question for title, apologies. U(2n) is not cyclic for n>2 because it will contain two elements of order 2.
     
    Last edited: Dec 1, 2012
  2. jcsd
  3. Dec 2, 2012 #2
    I'm still stuck on this. Anyone able to lend a hand?
     
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