Let $G$ be a group. We will denote the order of $g\in G$ by $o(g)$. Assume that $\{o(a):a\in G\}$ is finite (which is true if $G$ is finite). We define the exponent of $G$ by ${\rm exp}(G)={\rm lcm}\{o(g):g\in G\}$.
Recall that a polynomial of degree $n$ over a field has at most $n$ roots. Use this in conjunction with Fermat's little theorem to prove that for a prime $p$ we have ${\rm exp}(U(p))=p−1$.
So I know that $U(p)$ will contain every number $k<p$ where $k$ is a natural number. I also know that there is a theorem that states "In a finite cyclic group, the order of an element divides the order of the group" which will help with finding the lcm of the group is going to be $p-1$ but I don't know exactly how to use the fact that a polynomial of degree $n$ has $n$ or less roots. I also don't know where to use Fermat's Little Theorem or how to get to $U(n)$ being cyclic. So basically I don't know how to start but I know how to finish out the proof.
Please help.
