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 seven 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.

  • 7
    $\begingroup$ 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... $\endgroup$ Feb 17 '20 at 22:17
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    $\begingroup$ If there were one, I'd have included it. :) $\endgroup$
    – KCd
    Feb 17 '20 at 22:20
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    $\begingroup$ 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. $\endgroup$
    – KCd
    Feb 17 '20 at 22:22
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    $\begingroup$ 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?" $\endgroup$ Feb 17 '20 at 22:36
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    $\begingroup$ @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? $\endgroup$
    – verret
    Feb 18 '20 at 7:51

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