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

Dear Bruno: Doesn’t it work if p is a power of a prime?

Bruno Joyal says:

April 21, 2012 at 2:26 pm | Reply

Dear Pierre-Yves: indeed, the same proof works! In fact it appears to work for any positive integer, but some other proof is needed for the general case.

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.