Proof of Wilson's Theorem

In the elementary proof here, we solve by pairing an element with its inverse. Why do we know necessarily that this will always happen?

  • 2
    $\begingroup$ Hint. You know every element has an inverse. When can an element be its own inverse (remember your modulus is prime)? $\endgroup$ – Ethan Bolker Feb 16 '16 at 14:43
  • $\begingroup$ when it is congruent to -1 or 1! thanks $\endgroup$ – feng Feb 16 '16 at 14:45
  • $\begingroup$ You didn't read the article correctly. It explains that at length. $\endgroup$ – Yves Daoust Feb 16 '16 at 14:56

Your real question is: how do we know we can pair each element with its own inverse?

The answer to this can be broken into a few pieces.

  1. Every integer relatively prime to $p$ has an inverse mod $p$.
  2. If $a$ has inverse $a^{-1}$ then $a^{-1}$ has inverse $a$.
  3. What about elements which are their own inverse? This amounts to solving $x^2 \equiv 1 \pmod p$, whose only solutions are $x \equiv \pm 1 \pmod p$.

And that's it. If there is a substep which you are not familiar with, then you might consider trying to prove it, or looking it up here or in your text.

  • $\begingroup$ @ThomasAndrews: Could you explain the details about your first comment? $\endgroup$ – nCm Sep 14 '18 at 6:48
  • 1
    $\begingroup$ @Nong In the previous version of this answer, I omitted the "relatively prime" statement. $\endgroup$ – davidlowryduda Sep 14 '18 at 13:27

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