Let be a group of prime order . It will be shown that for any with it holds that where is the set generated by and is the identity element of .

Since the order of group is , the order of element is also finite, i.e. for some positive integer .

Note that the set generated by is a subgroup of , since is closed with respect to multiplication and with respect to inverses. Indeed, for any , the division algorithm of integers ensures that there exist integers with such that . Moreover, the inverse of any is .

The order of the cyclic subgroup of is equal to the order of generator , that is .

By Lagrange’s theorem, is a factor of . Since is a prime, or . It follows from that , so , which means that .

and .

## Leave a Reply