Question

I teach elementary number theory and discrete mathematics to students who come with no abstract algebra. I have found proving the key theorem that finite multiplicative subgroups of fields are cyclic a pedagogical speedbump. For example, Serre's proof in A Course in Arithmetic runs a full page, requires introducing Euler's $\phi$-function, and depends on a counting argument that might seem to beginners too clever or magical for a cornerstone result.

I'd like to have a collection of proofs of this fact, to compare their advantages, to match their viewpoints to my various audiences, to contrast for my students, etc.

To get the ball rolling, here's the shortest argument I can think of (and if it's in the literature somewhere I'd love a reference).

Induction on the order of the subgroup. So suppose multiplicative subgroup $G$ of field $F$ has order $n$. If $n=p^k$ with $p$ prime and $G$ isn't cyclic, all $p^k$ elements of $G$ satisfy $x^{p^{k-1}}-1=0$, impossible.
If $n=ab$, $gcd(a,b)=1$, then $(\cdot)^a:G\rightarrow G$ has a kernel $A$ of size at most $a$ and a range $B$ of size at most $b$ (since the $y\in B$ satisfy $y^b=1$), so $|A|=a$, $|B|=b$, and a product $xy$ of cyclic generators $x,y$ for $A,B$ respectively generates $G$.

If you know published proofs distinctly different from either of these, please cite a source. No need to spell out the details, but please mention a key feature to help avoid duplicates. If you have your own favorite approach, please share it.

Answer 1

I don't know how helpful this is for anybody, especially students, but for finite subgroups $G$ of $\mathbb C^*$ you can first observe that every element has modulus $1$, so is on the unit circle and has rational argument, and then choose the element $z$ of least non-zero argument. Then, given $y\in G$, rotate clockwise by dividing by powers of $z$ until the argument lies below that of $z$; this shows that $y$ is a power of $z$.

Answer 2

Let n be the number of elements of F*, p be a prime dividing n, q be the largest power of p dividing n; let r=q/p. Look at the map x-->x^(n/q), F*-->F*. The kernel has order at most n/q, so the image has order at least q, and there are at least q solutions of x^q=1. Since there are at most r solutions of x^r=1, there is an element of exact order q; multiplying these elements together for the various p dividing n gives a generator.

(I've used this no doubt well-known argument successfully in undergrad courses).

Edit: For the final step, let u be the product. Then u^(n/p)=a^(n/p) where a has exact order q. So u^(n/p) is not 1 for all p, and u has exact order n, and is a generator. Looking again at the question, I realize that this is essentially the same as the proposer's short solution, though I've restricted my attention unnecessarily to finite fields. But it combines Lagrange's theorem with the theorem that x^m=1 has at most m solutions in F* in a very simple way.

Answer 3

I once collected six proofs of this theorem, for the field Z/p, and they can be found at http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/cyclicFp.pdf. While Z/p is not the general finite field, since the intent of this MO question is to use proofs in a course to undergraduates without much background, surely Z/p is the only finite field that matters.

Answer 4

I know less than you know in the topic since you are a teacher now.
However, I want to mention two sources you can find the proof which use little prerequisites of algebra.
First of all, the classic Basic Nuber Theory by Andre Weil contains a proof in the first section of the first chapter which uses a great method.
As for the second, the Chinese mathematician Hua, Lo-keng ( in Chinese 華羅庚 ) has published a book entitled Introduction to number theory, which has a proof that uses only elementary terminologies, and I hope it is exactly what you need.
By the way, the first approach is the same as that mentioned by @QiaoChu Yuan in some sense, and the second is mostly elementary.
P.S. I hope it is acceptable here using Chinese, thanks.

Answer 5

Let $n = |G|$ and let $m$ be the l.c.m. of the orders of the cyclic factors of $G$. Then $x^m = 1$ for all $x \in G$; since we are in a field this equation has at most $m$ roots, which shows that $m \geq n$. It follows that $m = n$ and $G$ is cyclic.

Of course here one uses the classification of finite abelian groups as product of cyclic groups, which you may want to avoid.

Answer 6

I actually think it will not be so easy to say when two proofs of this result will be "distinctly different": rather I expect most or all will have common features, including using at least a little bit of group theory.

For instance, the proof I wrote up for my elementary(ish) number theory course is Theorem 9 in these notes. The notes themselves are on finite commutative groups, and Theorem 9 is on page 3, in the section on "cyclic groups". Prior to the statement and proof, a little over a page is spent developing the basic properties of cyclic groups, including a statement involving the Euler $\varphi$-function. The proof of the result itself -- which, note, is a criterion for an a priori noncommutative finite group to be cyclic -- occupies $11$ lines. (Added: sorry, false advertising -- add two more lines to get from Theorem 9 to Corollary 10, which is the statement that any finite subgroup of the multiplicative group of a field is cyclic.) I certainly think it is more or less the proof that any research mathematician is expecting to find.

Let me mention though that I had originally included this argument as an application of the Mobius Inversion Formula. After having looked back at what I'd done, I decided that although the argument was reminiscent of an inversion / inclusion-exclusion counting argument, it only made it more complicated to phrase it in that way.

Answer 7

Let $G$ be a finite subgroup of $F^{\ast}$ of order $n$. Then all the elements of $G$ satisfy $x^n = 1$ in $F$. Since polynomials of degree $n$ over a field have at most $n$ roots, it follows that the roots of $x^n = 1$ in $F$ are precisely the elements of $G$.

The intuitive content of Serre's argument is as follows: if no element of $G$ has order $n$, then they all have to have order less than $n$, so they satisfy various smaller polynomials $x^d = 1$ for $d | n$, and what the counting argument is trying to show is that there isn't enough "room" in these polynomials for all of these roots. I think this is quite intuitive, and it is completely clear for $n$ a prime power, but you want to avoid it, so:

Over $\mathbb{C}$, the roots of $x^n = 1$ are precisely the $n^{th}$ roots of unity. It is natural to organize these by their order, so let $\Phi_d(x) = \prod_{\zeta \text{ has order exactly } d} (x - \zeta)$. The result that Serre is trying to avoid with his counting argument is that $\Phi_d(x)$ has integer coefficients, so the factorization

$$x^n - 1 = \prod_{d | n} \Phi_d(x)$$

makes sense over an arbitrary field. If you can show this, the rest of the proof is trivial: since $x^n = 1$ splits over $F$, it follows that $\Phi_n(g) = 0$ for some $g \in G$, and such an element must have order $n$ and therefore be a generator.

If your students really have no algebra background I think you should consider stating this without proof. It is easy to give examples and hopefully you can give enough to convince them.

The shortest way I can think of to prove that $\Phi_n(x)$ has integer coefficients is by induction and the identity $\gcd(x^n - 1, x^m - 1) = x^{\gcd(n,m)} - 1$, which again 1) is intuitive over $\mathbb{C}$ but 2) makes sense over an arbitrary field. But this is a bit of a detour and precisely why Serre did something trickier. However, I think the larger lesson that "algebraic things that are intuitive over $\mathbb{C}$ are worth generalizing" is worth learning.