Cyclic group Zp 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.
 A: (In the given links I did not find a proof using exactly your hints that's why I provided this proof, but you can see other methods too)
Proving that $\Bbb Z_p^*$ is cyclic is equivalent to proving the existence of an element of order $p-1$, take $l=\text{lcm}(\text{ord}(1),\text{ord}(2),\cdots,\text{ord}(p-1))$ so it's clear that:
$$\forall k\in \Bbb Z_p^* \ \ \ \ \ \ \ k^l=1  $$
because $\text{ord}(k)|l$ for every $k$, This means that $x^l-1$ is a polynomial of degree $l$ in the field $\Bbb Z_p$ which has more than $p-1$ roots, as a consequence $l\geq p-1$.
Now because there exists an element of order $LCM(\text{ord(a)},\text{ord}(b))$ for  every two elements (This is the first hint: to prove it take just the product $ab$) there exists an element $g$ of order $l$ hence $l\leq p-1$.
Finally there exits an element $g$ of order $l=p-1$, thus $\Bbb Z_p^*$ is cyclic.
A: 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. $_\square$
