If m is the number of elements in a finite field, then a^(m-1) = 1 for any element a. Is there a way to prove this without using Fermat's Little Theorem?
what you said is not strictly true.
if a = 0, then 0^(m-1) = 0. it is only true for non-zero a in F.
Fermat's Little Theorem, is actually a special case of a more general theorem, Lagrange's theorem for finite groups, which states:
the order of an element of a group, divides the order (cardinality) of the group.
here, since we have a field F, the non-zero elements F* form a group w.r.t. multiplication. this group has m-1 elements, so the order
of any a is a divisor of m-1, we can call it d, for example, so m-1 = kd.
but then a^(m-1) = a^(kd) = (a^d)^k = 1^k = 1.
the reason this happens, is that any subset S of F* closed under multiplication induces the following equivalence on F*:
a ~ b iff ab^-1 is in S. the equivalence classes are all of the form Sa = {sa : s in S, a fixed in F*}. furthermore, since
fields are cancellative, the correspondence s-->sa is a bijection, meaning all such sets Sa have the same size.
this means in particular, that |S| divides |F*| = m-1. clearly, the set of all non-negative powers of a is such a set S.
Okay, thanks. And if I was going to use Fermat's little theorem, we know b^n= b for n prime in mod n. So is it also true that b^(n^t) = b for n prime? Because I was thinking b^(n^t) = (b^(n*n^(t-1)) = (b^n)^(n^(t-1)) = b^(n^(t-1)) by F.L.T. = b^n^(n^(t-2)) = b^(n^(t-2)) by F.L.T. = b^n^(n^(t-3)) = ... = b. Is that right, or did I mess up somewhere?
a finite field is not necessarily Zp. fermat's litle theorem is a theorem about integers, which may be phrased as a statement about the integers modulo p (Zp).
i can make little sense of what you wrote. in the field GF9, which has the elements {0,1,2,x,2x,x+1,x+2,2x+1,2x+2} (and x^2 = 2) it is not true that b(n^t) = b for n prime.
for example, consider n = 2, and t = 1, and b = x. then x^2 = 2, which is certainly NOT x.