We first show that the map is one-to-one. Suppose that for elements We must show that but and So and by left cancellation To show that is onto is easy. By definition every element of is of the form for some and
The group is partitioned into distinct left cosets. Each left coset has elements; therefore,
Let be in such that Then by Corollary 6.11, the order of must divide the order of the group. Since it must be Hence, generates
Since there are only two cosets of in Inasmuch as one of the cosets is itself, right and left cosets must coincide; therefore, or for every Since there are eight -cycles in at least one -cycle must be in Without loss of generality, assume that is in Then must also be in Since for all and all and
we can conclude that must have at least seven elements
Therefore, has no subgroup of order
Suppose that
Define to be the permutation
Then
Conversely, suppose that is a -cycle and If and then Hence,
Since is one-to-one and onto, is a cycle of the same length as