# Is it possible to prove that $U(p)$, for prime $p$, is cyclic using only group theory? If not, why not?

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?

• 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? – Lord Shark the Unknown Nov 7 '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 7 '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? – Lord Shark the Unknown Nov 7 '18 at 17:56
• Yea, I mentioned that exact theorem to my prof and he said it was off limits. – farleyknight Nov 7 '18 at 17:56
• Try this root 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 7 '18 at 18:02

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.
• Specifically you rely on the fact that $\psi(d) ≤ \phi(d)$ for $U(p)$. – farleyknight Nov 7 '18 at 18:02