# Using Fermats Little Theorem and there are $n$ or less roots of polynomials of degree $n$

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 $$k where $$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.

Hint: Show that if $$m=\exp(G)$$ then, for every $$g\in G,$$ $$g^m=1.$$
Also, how many elements are in $$U(p)?$$
• For this problem, you don’t really need Fermat’s Little Theorem. You know $\exp(G)\mid |G|$ for all finite group $G.$ May 31, 2021 at 0:39