# Is there a group theoretic proof that $(\mathbf Z/(p))^\times$ is cyclic?

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.

• 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... Feb 17, 2020 at 22:17
• If there were one, I'd have included it. :)
– KCd
Feb 17, 2020 at 22:20
• 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, 2020 at 22:22
• 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?" Feb 17, 2020 at 22:36
• @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? Feb 18, 2020 at 7:51

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.
• I see now that this argument is the particular case for $n=1$ of the more general one given here: math.stackexchange.com/a/3707908/1092170, with which it nicely shares two downvotes :) Apr 5 at 1:53