Exercise 4.11 Allufi

Question

This exercise is on page 69 of Algebra chapter 0.

Problem:

In due time we will prove the easy fact that if $p$ is a prime integer, then the equation $x^d = 1$ can have at most $d$ solutions in $\mathbb{Z}/p\mathbb{Z}$. Assume this fact, and prove that the multiplicative group $G =(\mathbb{Z}/p\mathbb{Z})^{\ast}$ is cyclic.

Attempt:

Using the this fact along with another exercise one can show that $h^{|g|} = 1$ for $g \in G$ element of maximal order. This shows that for all $|h| \leq |g|$ for all $h \in G$. We can also show that $|g| \leq |G| = p - 1$. I am not sure how can we show that $|G| \leq |g|$ though ? I don't want to use Lagrange theorem as we don't know this result yet in the book.

Answer 1

If $\mid g\mid\lt \mid G\mid=p-1$, then $x^{\mid g\mid}=1$ has $p-1\gt\mid g\mid$ solutions. This contradicts the fact you were given about $x^d=1$.