proof that the multiplicative group of a finite field is cyclic

This method is combinatorial, rather than the more usual algebraic methods. It's based on an exercise from Herstein's book Abstract Algebra.

We note that in any field, the equation

x^m = 1

has at most m roots. Thus, it suffices to prove the following.

Lemma: Let G be a finite group with n elements, in which for every m (which necessarily divides n) there are at most m solutions for the equation x^m = 1. Then G is a cyclic group.

Proof: Denote by a_m the number of elements of G of order m. Then a_m is 0 if m does not divide n. If a_m is non-zero, then there exists some element g∈G of order m. But then the m powers of g are all solutions of x^m = 1, therefore by the conditions on G these are all of the solutions. Now, g generates a cyclic group of order m, and this group has φ(m) generators (where φ(m) is Euler's totient function, the number of numbers in 1,...,m coprime with m). Thus, if a_m is non-zero, then it is precisely φ(m).

We've just shown that a_m=0 if m doesn't divide n, and that if it does divide, then it is either 0 or φ(m). But

∑_m|n φ(m) = n.

(the sum of phi(m) over all factors m of n is precisely n (to see this, just consider orders of elements in the cyclic group!)). On the other hand, the sum of all a_m's is also n: every element has an order. The only way this can be is if a_m=φ(m) for every m which divides n. In particular, a_n = φ(n) > 0, so G contains elements of order n, and is a cyclic group.