Home » Galois Theory » A divisibility identity for Euler’s totient function

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.

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6 thoughts on “A divisibility identity for Euler’s totient function

  1. Hi; I only just saw this. For any positive integer n, couldn’t you just argue as follows? In the group of automorphisms of \mathbb{Z}/(n^m - 1) that fix multiples of (1 + n + \ldots + n^{m-1}) \pmod (n^m-1) pointwise (which is of order \frac{\phi(n^m - 1)}{\phi(n-1)} by your lemma 1), the automorphism given by multiplication by n generates a subgroup of order m, so m divides \frac{\phi(n^m - 1)}{\phi(n-1)} by Lagrange’s theorem.

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