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RE order of an element

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  • Jose Ramón Brox
    ... From: ali omer I appreciate your concern. Thank you very much.I found by guess that the order of x mod(p^k)=the order of x
    Message 1 of 2 , Jan 27, 2006
      ----- Original Message -----
      From: "ali omer" <ali_1122334455@...>

      I appreciate your concern.
      Thank you very much.I found by guess that the order of x mod(p^k)=the order of x mod(p)
      multiplied by p^{k-1} but I can't prove it and I don't know why it should be multiple of
      it's order mod(p).

      -------------------------------------------

      Counterexample: Order(3,11) = 5, Order(3,11^4) = 605 = 5·11^2 <> 5·11^3 =
      Order(3,11)·11^(4-1)

      What happens here is that Phi(p) = p-1 and Phi(p^k) = p^(k-1)·(p-1), and it's true that
      the order of the element mod n must divide Phi(n), so for small values of p you can expect
      a lot of "simple" behaviours (it's the law of small numbers), like a lot of orders being
      p^(k-1)·e, where e is the order mod p.

      Regards. Jose Brox
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