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.


1 Answer 1


Hint: Show that if $m=\exp(G)$ then, for every $g\in G,$ $g^m=1.$

Also, how many elements are in $U(p)?$

  • $\begingroup$ there are p-1 elements in U(p) but where do I use FLT or the thing about roots? $\endgroup$
    – Allie
    May 30, 2021 at 23:19
  • $\begingroup$ For this problem, you don’t really need Fermat’s Little Theorem. You know $\exp(G)\mid |G|$ for all finite group $G.$ $\endgroup$ May 31, 2021 at 0:39

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