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.

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    $\begingroup$ 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, 2020 at 22:17
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    $\begingroup$ If there were one, I'd have included it. :) $\endgroup$
    – KCd
    Feb 17, 2020 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, 2020 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, 2020 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, 2020 at 7:51

1 Answer 1


The fact in the OP ($x^r\equiv 1\pmod p$ has at most $r$ solutions) shows immediately that $({\bf Z}/(p))^\times$ can't have any subgroup isomorphic to $C_q\times C_q$, for any prime $q$ dividing $p-1$ (as otherwise $x^q\equiv 1\pmod p$ would have at least $q^2$ solutions). By the structure theorem for finite abelian groups, $({\bf Z}/(p))^\times$ is cyclic. Another way to show that there cannot be a subgroup isomorphic to $C_q\times C_q$ is recalling that $({\bf Z}/(p))^\times\cong\operatorname{Aut}({\bf Z}/(p))$ acts regularly (by automorphisms) on the set of generators of ${\bf Z}/(p)$. But again this uses the fact the OP would like to avoid. So, take this answer as a long comment, actually.


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