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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 — and hints at some others — in his paper here. The first six make use of the previously mentioned fact, while the seventh proof makes extensive use of cyclotomic polynomials and is thus still not 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.

citadel
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Sam
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    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
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    If there were one, I'd have included it. :) – KCd Feb 17 '20 at 22:20
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    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
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    @KCd ... or twice an odd prime power... – Arturo Magidin Feb 17 '20 at 22:23
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    @ArturoMagidin yeah, or 4 also. – KCd Feb 17 '20 at 22:25
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    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
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    @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
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    @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. – calc ll 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. – citadel Feb 18 '24 at 15:49

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