A divisibility identity for Euler’s totient function

In this note I will give a Galois-theoretic proof that for a prime {p} and positive integer {n},

\displaystyle n \mid \frac{\varphi(p^n-1)}{\varphi(p-1)}.

I’d love to see a more elementary proof if you can come up with one.
First we need the following:

Lemma 1 Let {Z_n} be the cyclic group with {n} elements. Let {m} be a positive divisor of {n}, and consider {Z_m} as a subgroup of {Z_n}. Then the number of automorphisms of {Z_n} which fix {Z_m} pointwise is equal to {\varphi(n)/\varphi(m)} (which, in particular, is an integer).

Proof of the Lemma: Note that any automorphism of {Z_n} fixes {Z_m}, though not necessarily pointwise: indeed {Z_n} has a unique subgroup of order {m}, and thus any automorphism of {Z_n} must take this subgroup to itself. Thus we have a group homomorphism {\text{Aut}(Z_n) \rightarrow \text{Aut}(Z_m)} which is easily seen to be surjective; its kernel is precisely the subgroup consisting of those automorphisms of {Z_n} which fix {Z_m} pointwise. The statement follows by comparing orders. {\square}

Now to prove the initial claim, consider the field extension {\mathbf{F}_{p^n}/\mathbf{F}_p}. Basic Galois theory tells that this is a Galois extension of degree {n}. Consider the canonical homomorphism

\psi: \displaystyle \text{Gal}(\mathbf{F}_{p^n}/\mathbf{F}_p) \rightarrow \text{Aut}(\mathbf{F}_{p^n}^\times)

which restricts an {\mathbf{F}_p}-automorphism {\sigma} to the group of units of {\mathbf{F}_{p^n}}. Clearly it is an injective homomorphism since {\sigma} is completely determined by where it sends the units. Moreover for any {\sigma \in \text{Gal}(\mathbf{F}_{p^n}/\mathbf{F}_p)}, \psi(\sigma) lies in the subgroup of {\mathbf{F}_{p^n}^\times} of those automorphisms fixing pointwise the cyclic subgroup {\mathbf{F}_{p}^\times} of order {p-1}, because the Galois group consists of \mathbf{F}_p-homomorphisms. By the lemma the subgroup of these automorphisms has order {\frac{\varphi(p^n-1)}{\varphi(p-1)}}, whereas {\text{Gal}(\mathbf{F}_{p^n}/\mathbf{F}_p)} has order {n}. This does it.

A nice problem in Galois theory

This problem was given to me by my research supervisor. This is the problem and my solution:

Let k be a field of characteristic zero, and k(x) the rational function field in one variable over k. Suppose F_1 and F_2 are subfields of k(x) such that [k(x):F_1] and [k(x):F_1] are finite. Is it possible for [k(x):F_1\cap F_2] to be infinite? Indeed, it is. Note that \mbox{Gal }(k(x)/k) = \mbox{PSL}_2(k), the projective special general linear group over k (we identify its elements with Möbius transformations). Let S, T denote the generators of the modular group (or rather, their image in \mbox{PSL}_2(k)). Note that they are of finite order. Hence, taking F_1=k(x)^{<S>} and F_2=k(x)^{<T>}, we have [k(x):F_i]<\infty for i=1,2. However, F_1 \cap F_2 is fixed by the whole modular group, which is infinite since k has characteristic zero. Hence k(x) cannot be of finite index over F_1 \cap F_2.

I don’t know whether such a construction is possible if k has prime characteristic. I’d be interested to know if you find out!