Collecting proofs that finite multiplicative subgroups of fields are cyclic. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T07:39:41Z http://mathoverflow.net/feeds/question/54735 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54735/collecting-proofs-that-finite-multiplicative-subgroups-of-fields-are-cyclic Collecting proofs that finite multiplicative subgroups of fields are cyclic. David Feldman 2011-02-08T08:30:54Z 2011-02-08T18:36:32Z <p>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.</p> <p>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.</p> <p>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).</p> <p>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.<br> 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 $G$. </p> <blockquote> <p>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.</p> </blockquote> http://mathoverflow.net/questions/54735/collecting-proofs-that-finite-multiplicative-subgroups-of-fields-are-cyclic/54739#54739 Answer by Qiaochu Yuan for Collecting proofs that finite multiplicative subgroups of fields are cyclic. Qiaochu Yuan 2011-02-08T09:32:55Z 2011-02-08T09:32:55Z <p>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 $G$.</p> <p>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:</p> <p>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 <em>avoid</em> with his counting argument is that $\Phi_d(x)$ has integer coefficients, so the factorization</p> <p>$$x^n - 1 = \prod_{d | n} \Phi_d(x)$$</p> <p>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.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/54735/collecting-proofs-that-finite-multiplicative-subgroups-of-fields-are-cyclic/54740#54740 Answer by Pete L. Clark for Collecting proofs that finite multiplicative subgroups of fields are cyclic. Pete L. Clark 2011-02-08T09:40:19Z 2011-02-08T09:40:19Z <p>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. </p> <p>For instance, the proof I wrote up for my elementary(ish) number theory course is Theorem 9 <a href="http://www.math.uga.edu/~pete/4400algebra2point5.pdf" rel="nofollow">in these notes</a>. 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 <em>a priori</em> noncommutative finite group to be cyclic -- occupies $11$ lines. (<b>Added</b>: 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.</p> <p>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 <em>reminiscent</em> of an inversion / inclusion-exclusion counting argument, it only made it more complicated to phrase it in that way.</p> http://mathoverflow.net/questions/54735/collecting-proofs-that-finite-multiplicative-subgroups-of-fields-are-cyclic/54741#54741 Answer by Andrea Ferretti for Collecting proofs that finite multiplicative subgroups of fields are cyclic. Andrea Ferretti 2011-02-08T09:45:54Z 2011-02-08T10:10:05Z <p>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.</p> <p>Of course here one uses the classification of finite abelian groups as product of cyclic groups, which you may want to avoid.</p> http://mathoverflow.net/questions/54735/collecting-proofs-that-finite-multiplicative-subgroups-of-fields-are-cyclic/54745#54745 Answer by awllower for Collecting proofs that finite multiplicative subgroups of fields are cyclic. awllower 2011-02-08T10:16:55Z 2011-02-08T10:16:55Z <p>I know less than you know in the topic since you are a teacher now.<br> However, I want to mention two sources you can find the proof which use little prerequisites of algebra.<br> First of all, the classic <strong><em>Basic Nuber Theory</em></strong> by <em>Andre Weil</em> contains a proof in the first section of the first chapter which uses a great method.<br> As for the second, the Chinese mathematician <em>Hua, Lo-keng</em> ( 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.<br> By the way, the first approach is the same as that mentioned by @QiaoChu Yuan in some sense, and the second is mostly elementary.<br> P.S. I hope it is acceptable here using Chinese, thanks.</p> http://mathoverflow.net/questions/54735/collecting-proofs-that-finite-multiplicative-subgroups-of-fields-are-cyclic/54766#54766 Answer by KConrad for Collecting proofs that finite multiplicative subgroups of fields are cyclic. KConrad 2011-02-08T14:25:18Z 2011-02-08T14:25:18Z <p>I once collected six proofs of this theorem, for the field Z/p, and they can be found at <a href="http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/cyclicFp.pdf" rel="nofollow">http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/cyclicFp.pdf</a>. 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. </p> http://mathoverflow.net/questions/54735/collecting-proofs-that-finite-multiplicative-subgroups-of-fields-are-cyclic/54772#54772 Answer by paul Monsky for Collecting proofs that finite multiplicative subgroups of fields are cyclic. paul Monsky 2011-02-08T15:12:24Z 2011-02-08T17:35:47Z <p>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.</p> <p>(I've used this no doubt well-known argument successfully in undergrad courses).</p> <p>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.</p> http://mathoverflow.net/questions/54735/collecting-proofs-that-finite-multiplicative-subgroups-of-fields-are-cyclic/54802#54802 Answer by inkspot for Collecting proofs that finite multiplicative subgroups of fields are cyclic. inkspot 2011-02-08T18:36:32Z 2011-02-08T18:36:32Z <p>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 $z$.</p>