- #1
Dead Metheny
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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.
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.
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