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Multiplication of primitive roots

  1. May 5, 2007 #1

    I noticed that multiplication of all primitive roots modulo p ,p>3,
    congruent to 1 modulo p...

    I have tried some examples (13,17,19...) but i couldn't prove the general case
    (let g1....gk be primitive roots modulo p,p>3 ==> g1*g2*...*gk=1(p))

    I need help to prove or disprove it...

    thanks in advance
  2. jcsd
  3. May 5, 2007 #2


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    Try multiplying them in a clever order.
  4. May 6, 2007 #3
    thanks for the replay
    I'll try to follow your lead ...
  5. May 6, 2007 #4
    I would suggest looking at detail, leave no stone unturned!
  6. May 7, 2007 #5
    I will rephrase my question...
    How can i explain that if g is a primitive root then g^-1 is primitive root?
  7. May 7, 2007 #6


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    I would have expected it to be straightforward, although I haven't worked it out explicitly.

    Where, when proving the inverse of a primitive root is primitive, do you get stuck?
    Last edited: May 7, 2007
  8. May 7, 2007 #7

    matt grime

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    It's a group. The order x and x^-1 are trivially shown to be the same.
  9. May 7, 2007 #8
    If 1/g^k ==1, Mod M, then of course, g^k==1 Mod M.
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