44

Theorem: The group $(\mathbf Z/(p))^\times$ is cyclic for any prime $p$.

Most proofs make use of the fact that for $r\geq 1$, there are at most $r$ solutions to the equation $x^r=1$ in $\mathbf Z/(p)$, a result which doesn't seem — understandably — to have any group theoretic proofs.

K. Conrad gives ten different proofs of the Theorem — and hints at some others — in his paper here. As all the proofs use the aforementioned fact, none of the proofs are group-theoretic.

I was also able to find a linear algebra based proof in the second chapter of Teoría Elemental de Grupos by Emilio Bujalance García, but still, no group theoretic proof to be found.

Sam
  • 5,208
  • 15
    Given that you are looking at the group of units of a ring, what makes you believe that you can find a purely group theoretic proof? You are dealing with a ring and with properties of primes, after all... – Arturo Magidin Feb 17 '20 at 22:17
  • 39
    If there were one, I'd have included it. :) – KCd Feb 17 '20 at 22:20
  • 30
    More seriously, the unit group of $\mathbf Z/(m)$ is generally not cyclic, so proving it is when $m$ is a prime number (or an odd prime power) will need to use something that distinguishes those choices of $m$ from others, and a very basic one is that $\mathbf Z/(p)$ is a field, which is not a purely group-theoretic issue. – KCd Feb 17 '20 at 22:22
  • 2
    @KCd ... or twice an odd prime power... – Arturo Magidin Feb 17 '20 at 22:23
  • 2
    @ArturoMagidin yeah, or 4 also. – KCd Feb 17 '20 at 22:25
  • 19
    I agree with the previous comments, but maybe we can give a more "group-theoretic flavour" to the question if we ask: "Why is the automorphism group of a simple abelian group cyclic?" – Captain Lama Feb 17 '20 at 22:36
  • 7
    @CaptainLama I agree, the question can be phrased very naturally group theoretically. Another (similar) way: why is the automorphism group of a group of prime order cyclic? – verret Feb 18 '20 at 07:51
  • 1
    @CaptainLama, I think your point is precisely what I had in mind when trying with this: https://math.stackexchange.com/q/4222662/943729 –  Oct 22 '21 at 12:54
  • I just wanted to say that the linked paper by @KCd is great! – Vincent Nov 02 '21 at 15:54
  • I added a version of the linear algebra proof to the end of my document, so now it has 8 proofs, but the version I chose is based on the one in Fedor Petrov's answer to an MO question here. – KCd Nov 03 '21 at 03:19
  • Group Theory and number theory have alot of overlap. I find it hard to separate them out sometimes. – suckling pig Nov 19 '22 at 01:24
  • I've now add Matt Baker's $p$-adic proof, so now it has 9 proofs. – KCd Nov 19 '22 at 15:34
  • The proof using the fact that $x^r-1$ has only at most $r$ solutions seems pretty group-theoretic to me....you are given a group of order $p-1$ and are told it is abelian and that for any $r$, there are at most $r$ elements giving $x^r=e$. How much more group-theoretic can a proof possibly be. – Mike Jan 19 '23 at 21:34
  • To answer the question in @verret 's comment: because every $\psi\in Aut(\mathbb Z_p)$ is given by $\psi([0])=[0]$ and $\psi([i])=[\sigma(i)]$ (for $i=1,\dots,p-1$) where $\sigma\in S_{p-1}$ is of the form: $\sigma=(C)(i_2C)\dots(i_dC)$, being $$(C)=(1,\sigma(1),\sigma(1)^2,\dots,\sigma(1)^{\frac{p-1}{d}-1})$$ for some divisor $d$ of $p-1$, with $$1+\sigma(1)+\sigma(1)^2+\dots+\sigma(1)^{\frac{p-1}{d}-1}\equiv_p 0$$ Then the usual counting argument filtered by $d$ would follow. – Kan't Feb 18 '24 at 15:49
  • And, of course, for ANY field $F$ (including infinite ones), each finite subgroup of $F^\times$ is cyclic, where the proofs referenced above do still pertain. "Fun" fact: The same is true in non-commutative division rings of prime characteristic! Also true in the real quaternions $H$ if the subgroup has odd order. But of course false in general for even order, such as $Q_8 = {\pm 1, \pm i, \pm j, \pm k}$. – Richard_Loves_Music Nov 24 '24 at 21:30
  • 1
    @scanless: You need to get outside and smell the nonabelian groups. – Lee Mosher Feb 02 '25 at 15:24
  • What definition of the multiplicative group of units are we allowed to use to make it a true group theoretic proof? – Daniel Donnelly Feb 02 '25 at 21:31
  • 1
    I am of the opinion that the below three deleted answers should be undeleted based upon how beautifully they are written. They should not be deleted unless someone can provide a canonical answer. – Daniel Donnelly Feb 02 '25 at 21:46
  • @lee ok, but alot of non-abelian groups can be built from abelian, even cyclic, ones – suckling pig Feb 03 '25 at 05:17
  • Can you include the definition of a "group-theoretic proof" in the question? – Gribouillis Mar 28 '25 at 10:36

2 Answers2

3

For the sake of having an answer, since there are now four deleted answers to this question: while I don't know a rigorous argument along these lines, I think we can make a strong informal case that the answer is no. The reason is already contained in KCd's comment from 2020:

More seriously, the unit group of $\mathbb{Z}/(m)$ is generally not cyclic, so proving it is when $m$ is a prime number (or an odd prime power) will need to use something that distinguishes those choices of $m$ from others, and a very basic one is that $\mathbb{Z}/(p)$ is a field, which is not a purely group-theoretic issue.

To my mind the natural level of generality for this result is:

Theorem: Every finite subgroup of the multiplicative group of a field is cyclic.

This is simply not a group-theoretic result! One way or another we crucially make use of the fact that we work in a field, and this hypothesis can't be dropped. I give a version of the proof here (it's a variant of the 8th proof from KCd's note on this subject) which I think highlights the field-theoretic nature of this theorem. Quoting from my answer:

  1. This is really a fact about fields, not a fact about finite abelian groups. Almost no group theory is required in this argument and the hypothesis on a finite abelian group that there are at most $d$ elements of order dividing $d$ only comes up in this case and nowhere else that I know of. The usual statement of this fact as "every finite subgroup of..." is arguably misleading because in fact the first step is to show that the only such finite subgroups are given by the $n^{th}$ roots of unity for some $n$. So this is really a fact about how the roots of unity behave in any field.
  2. We deduce the statement over arbitrary fields from the statement over $\mathbb{C}$, by using the cyclotomic polynomials to "transfer" information about the roots of unity over $\mathbb{C}$ to arbitrary fields. This is a nice and understandable special case of a general strategy with many applications, and our arguments can be understood abstractly in terms of the group scheme $\mu_n$ of $n^{th}$ roots of unity, although this is of course not necessary.
Qiaochu Yuan
  • 468,795
  • 2
    For another survey of proofs of the Theorem, see this MO post: https://mathoverflow.net/q/54735/495174 – Kan't Feb 03 '25 at 06:19
1

$\operatorname{Aut}({\bf Z}/(p))$ acts transitively on $\{1,\dots,p-1\}$ (a fact that, incidentally, has a perfectly group-theoretic proof). This forces every $\varphi\in\operatorname{Aut}({\bf Z}/(p))$ to fulfil the polynomial equation: $$\sum_{k=1}^{m_{\varphi(1)}}\varphi(1)^k\equiv 0\pmod p\tag0$$ where $m_{\varphi(1)}$ is the multiplicative order of $\varphi(1)$. I think that this basic fact necessarily brings every proof of the cyclicity of $({\bf Z}/(p))^\times$ to counting the solutions of polynomial equations in ${\bf Z}/(p)$.

Kan't
  • 4,819