If $p$ is prime, then $\mathbb{Z}_{p}^{\times}, \cdot $ is cyclic. How can this be proven using the following statement?: "Let $K$ be a field and let $f(x)$ be a non-constant polynomial of degree $n$ with coefficients in $K$. Then $f(x)$ has at most $n$ roots in $K$."
I don't really see how this could be linked to a group being cyclic. Can someone enlighten me?
If an abelian group has an element $a$ of order $x$ and $b$ of order $y$ then there is an element of order $lcm(x,y)$. To prove this notice that $ac$ has exactly this order, where $c=b^{gcd(x,y)}$.
Let $m$ be the $lcm$ of all the orders of elements in $G$, we have proved that there is an element of order $m$ in $G$.
We now move to the setting where $G$ is the multiplicative group of $\mathbb Z_p$
Clearly $m\leq n$. Now, the polynomial $x^m-1$ has exactly $n$ roots and so $n\leq m$, and so $m=n$.