I'm looking for a proof of this that does not rely on the Chinese remainder theorem, since the exercise is from a book that has not reached such a point (Aluffi, Algebra).
Show that for $p$ prime, the multiplicative group $G=(\mathbb{Z}/p\mathbb{Z})^*$ is cyclic. You may use that for $p$ prime, the equation $x^d=1$ has at most $d$ solutions in $\mathbb{Z}/p\mathbb{Z}$. (hint: for $g\in G$ of maximal order, $\forall h\in G, |h|\;|\;|g|$, and in particular, $h^{|g|}=1$.)
My only idea for now is constructing an element $a\in G$ such that $|a|=|G|$, using the fact that a group of order $n$ is cyclic $\iff\exists$ an element of order $n$.
Best Answer
If you know the structure theorem for finite Abelian groups, you will be able to prove that either a finite Abelian group $G$ is cyclic, or that its exponent is $<|G|$, that is there is $m<|G|$ with $g^m=e$ for for all $g\in G$.
An alternative attack: show that for $d\mid (p-1)$ the group $G=(\Bbb Z/\Bbb Z)^*$ has exactly $d$ solutions of $x^d\equiv1$. If you compare $G$ to $H=\Bbb Z/(p-1)\Bbb Z$, you find that the number of elements of order dividing $d$ is the same in each, therefore the number of elements of order exactly $d$ is the same in each, and now take $d=p-1$.