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Wilson's theorem proof

  1. Dec 12, 2013 #1
    The problem statement, all variables and given/known data.
    This is an exercise I've taken from Rotman's introductory textbook about groups.
    Prove Wilson's theorem: If ##p## is a prime, then ##(p-1)!=-1 \text{ mod } p##. He gives the following hint: The nonzero elements of ##\mathbb Z_p## form a multiplicative group.


    The attempt at a solution.

    I couldn't make use of the hint, until now, all I've deduced is: since ##p-1=-1 \text{ mod } p##, then to prove Wilson's theorem would be equivalent to prove that ##(p-1)!=p-1 \text{ mod } p##. Could anyone guide me on how could I use the hint?
     
  2. jcsd
  3. Dec 12, 2013 #2

    jbunniii

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    If ##p=2## then the theorem is obviously true.

    If ##p## is a prime greater than 2, then ##p## is odd. So the multiplicative group ##\{1, 2, \ldots p-1\}## has an even number of elements. Try pairing each element of the group with its inverse. Can you conclude anything interesting?
     
  4. Dec 12, 2013 #3
    ##p-1=-1 \text { mod } p##. From here, one can deduce that ##(p-1)^{-1}=p-1 \text { mod } p##, so we can consider the set ##\{2,...,p-2\}## and pair each element ##n_{i}## of this set with its inverse ##n_{i}^{-1}##, which is also in the set ##\{2,...,p-2\}##. Then, ##(p-1)!=1.(\prod n_i n_{i}^{-1}).(p-1)=1.1.(p-1)=p-1=-1 \text { mod } p##.

    My notation is awful, but I didn't know how to write properly your suggestion. Thanks for the help!
     
  5. Dec 12, 2013 #4

    jbunniii

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    Yes, you have the right idea. The key is that there are exactly two elements which are their own inverses: ##1## and ##p-1##. (Proof: ##x^2 = 1## iff ##x^2 - 1 = 0## iff ##(x+1)(x-1) = 0## iff ##x=-1## or ##x=1##.) All the other elements cancel each other out in pairs.
     
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