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The unitary group modulo $p$ is also written $(Z/pZ)^*$ and includes the integers $U(p) = \{ 1, 2, …, p - 1 \}$ and is a group under multiplication modulo $p$.

While this is not the exact problem I'm working on, it is most definitely an important piece of a homework problem, so please, hints are prefered.

I have found several proofs of the existence of a primitive root, which show that $U(p)$ is cyclic. However, they all rely on polynomials and field theory. I'm looking for a proof that does not require these subjects. Is one possible? If it's not possible, could someone show why it's not possible?

farleyknight
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    You need to know something about $(\Bbb Z/p\Bbb Z)^*$ in order to prove that it's a cyclic group. What sort of somethings are you willing to entertain? – Angina Seng Nov 07 '18 at 17:50
  • Not quite sure I understand. I guess I'm trying to avoid heavy machinery from other branches of mathematics. You could use Euler's totient function if that helps. But like I said in my OP, field theory and polynomials would be too much. – farleyknight Nov 07 '18 at 17:51
  • Well the usual something is the fact that a polynomial of degree $n$ over a field has at most $n$ zeros. Is that beyond the pale? – Angina Seng Nov 07 '18 at 17:56
  • Yea, I mentioned that exact theorem to my prof and he said it was off limits. – farleyknight Nov 07 '18 at 17:56
  • Try this root https://en.wikipedia.org/wiki/Multiplicative_order, starting from <<As a consequence of Lagrange's theorem, $ord_n(a)$ always divides $\varphi(n)$. If $ord_n(a)$ is actually equal to $\varphi(n)$ ... >> – rtybase Nov 07 '18 at 18:02
  • Yes but that assumes that there exists an element $ord_{p}(a) = p - 1$. There might not be an element of that order to start with. – farleyknight Nov 07 '18 at 18:03
  • FWIW you forgot to mention $p$ is prime. – fleablood Nov 07 '18 at 18:23
  • @farleyknight The polynomial fact may be proved using derivatives without invoking fundamental theorem of algebra. Unless you're taking algebra courses without taking calculus? – Divide1918 Dec 16 '20 at 14:19

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Consider this lemma, which is pure group theory (*):

Let $G$ be a group of order $m$. If, for every divisor $d$ of $m$, there are no more than $d$ elements of $G$ satisfying $x^d=1$, then $G$ is cyclic.

The result follows directly from this lemma because $(\Bbb Z/p\Bbb Z)^*$ is a field and a polynomial of degree $d$ over a field has at most $d$ roots. That is the arithmetic part of the result.

(*) This lemma appears verbatim in chapter X of André Weil's Number theory for beginners, which is a wonderful book.

lhf
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  • Right, but this is not covered in my textbook and to prove this one would need some field theory, correct? I'm trying avoid polynomials and field theory. – farleyknight Nov 07 '18 at 17:55
  • @farleyknight, the lemma has a pure group theory proof. See for instance https://math.stackexchange.com/a/1615033/589. – lhf Nov 07 '18 at 17:56
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    Yes, but in your proof you use part of the hypothesis of the OP on that question. In other words, the part that says: "and so the hypothesis on A implies that ψ(d)=ϕ(d)." I don't have that assumption. – farleyknight Nov 07 '18 at 18:01
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    Specifically you rely on the fact that $\psi(d) ≤ \phi(d)$ for $U(p)$. – farleyknight Nov 07 '18 at 18:02