Proving Wilson's Theorem: Finite Field Product of Non-Zero Elements is -1

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In a finite field F, the product of all non-zero elements is shown to be -1, aligning with Wilson's Theorem. The multiplicative group of non-zero elements, G, is cyclic and generated by an element a. The discussion highlights that if G has order p-1, it can be related to Z*_p, reinforcing the conclusion through isomorphism. The reasoning includes the fact that each element a_i has an inverse a_i^(-1) in the product, and only 1 and -1 are their own inverses. Ultimately, the argument concludes that no other element can have order 2, confirming the product is indeed -1.
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Homework Statement



Let F be a finite field. Show that the product of all non-zero elements of F is -1.



Homework Equations



An example of this is Wilson's Theorem.



The Attempt at a Solution



Let G be the multiplicative group of non-zero elements of F. Then G is cyclic. Let a be the generator of G. Here I get stuck. I thought that by just taking the product of each element represented by some power of the generator would be enough, but hey, it isn't. Not sure what to do now. Thnx for any help.
 
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Can we just say that if F has characteristic p, then |G| = p - 1. Since G is cyclic then it isomorphic to Z*_p (which is also cyclic), and then use Wilson's Theorem and the isomorphism to conclude the product is -1.
 
actually it's p^n for some positive integer n, but the same thing...
 
There's a more direct way. Take the product of all the nonzero members a_1*a_2*...*a_n. For each a_i, a_i^(-1) is also in the list. Only +1 and -1 are their own inverses.
 
but why isn't it possible for an element besides 1, -1, to have order 2 so that its inverse is itself?
 
If a^2=1 then a satisfies a^2-1=0. Factor to (a-1)(a+1)=0. Since F is a field, one of those factors must be zero.
 
awesome :) just what i needed. thnx
 

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