(This is an extremely useful fact with a straightforward proof. It follows from this, for instance, that any finite extension of a finite field is simple).
Let be a finite field, so for some prime and . Denote by the multiplicative group consisting of the non-zero elements of . We show that is cyclic, i.e., generated by a single element.
Let be the lowest common multiple (LCM) of the orders of all the elements . Then , and by Lagrange’s theorem, for all , i.e. Hence, , since can have at most roots. Thus i.e. .
It remains to show that there exists an element whose order is the LCM of the orders of all group elements. Indeed, this is generally true for abelian groups, as shown by the following.
Proposition: Let be an abelian group and let be elements with finite orders (respectively). Then there exists of order .
Proof: First, suppose and are co-prime and set . Then if , then and so so which implies by co-primality. Similarly, . So since and are co-prime, so has order .
Now consider the general case and write where the are distinct primes. For each , define by and let Then and , while and are co-prime and . We have Thus by the coprime order case first considered, their product has order .