Let F be a finite field with n elements. Prove that $x^{n-1}=1$ for all nonzero x in F.
I'm not understanding where this proof is going. So we have a ring $Z_{7}$, and we know there are 7 elements in $Z_7$ which are $\{0,1,2,3,4,5,6\}$ (a complete system of residues modulo 7), obviously every nonzero element to the power of 6, (from n-1 where n=7) is congruent to 1 modulo 7. However, I'm not understanding what they want me to show. I could show this by induction but I get the feeling that there is an easier way to show this?
Best Answer
As noted by Ragib if $F$ is field then $F - \{0\}$ is a multiplicative group of order $n-1$. Therefore the order of any element $x$ in here must be a divisor of $n-1$, viz. if $m$ is the order of $x$, then $m | n-1$. Hence $qm = (n-1)$ for some $q \in \Bbb{N}$. It follows that
$$x^{n-1} = x^{mq} = (x^m)^q = 1^q = 1.$$
Q.E.D.