In this note I will give a Galois-theoretic proof that for a prime and positive integer ,
I’d love to see a more elementary proof if you can come up with one.
First we need the following:
Lemma 1 Let be the cyclic group with elements. Let be a positive divisor of , and consider as a subgroup of . Then the number of automorphisms of which fix pointwise is equal to (which, in particular, is an integer).
Proof of the Lemma: Note that any automorphism of fixes , though not necessarily pointwise: indeed has a unique subgroup of order , and thus any automorphism of must take this subgroup to itself. Thus we have a group homomorphism which is easily seen to be surjective; its kernel is precisely the subgroup consisting of those automorphisms of which fix pointwise. The statement follows by comparing orders.
Now to prove the initial claim, consider the field extension . Basic Galois theory tells that this is a Galois extension of degree . Consider the canonical homomorphism
which restricts an -automorphism to the group of units of . Clearly it is an injective homomorphism since is completely determined by where it sends the units. Moreover for any , lies in the subgroup of of those automorphisms fixing pointwise the cyclic subgroup of order , because the Galois group consists of -homomorphisms. By the lemma the subgroup of these automorphisms has order , whereas has order . This does it.