TITLE: Cyclic group Zp
QUESTION [6 upvotes]: How to show if $p$ is prime, then group $Z_{p}^{*}$ is cyclic.
Tip 

Let $g$ and $h$ of a commutative group $G$ have orders $n$ and $m$ respectively. There exists and element $x \in G$ of order $LCM (n,m)$ 
In any field $\mathbb{K}$ a polynomial $f \in \mathbb{K} [x]$ of degree $n$ has at most $n$ different roots.

I have no idea.

REPLY [3 votes]: We have the much more general
Proposition. Any finite subgroup of the multiplicative group of a field is cyclic.
Proof.
Let $F$ be a field and $G$ a finite subgroup of $F^\times$. Let $n=|G|$.
For $m\in\mathbb N$, let $f(m)$ denote the number of distinct $m$th roots of unity in $F$ and let $g(m)$ denote the number of distinct primitive $m$th roots (i.e., those that are not $d$th roots for any $d<m$). 
Then clearly
$$\tag1 f(m)=\sum_{d\mid m}g(d).$$
By Möbius inversion it follows that
$$\tag2 g(m)=\sum_{d\mid m}\mu(m/d)f(d).$$
Since $m$th roots of unity are roots of $P_m(X):=X^m-1$, we have $f(m)\le m$ for all $m$.
If $d\mid m$ then $$P_m(X)=P_d(X)(1+X^d+X^{2d}+\ldots+X^{m-d})$$ shows that  if $P_m$ has $m$ distinct roots, the same must hoild for $P_d$, i.e.,  if $f(m)=m$ then also $f(d)=d$.
Since $a^n=1$ for all $a\in G$, we conclude $f(n)=n$ and hence $f(d)=d$ for all $d\mid n$. 
Using $(2)$, we obtain
$$ g(n)=\sum_{d\mid n}\mu(n/d)d=\phi(n)>0.$$
Thus $G$ contains at least one element $a$ of order $n$ and hence $G=\langle a\rangle$ is cyclic. $_\square$