My group is A_{5} which is an alternating cyclic subgroup of S_{5}. An alternating group is the group of even permutations of a finite set.

The alternating group on the set {1,…,n} is called the alternating group of degree n. For n > 1, the group A_{n} is the commutator subgroup of the symmetric group S_{n} with index 2 and has therefore $5!/2$ elements.

(I will be honest I have no idea how or why it is what it is. I just found this info and regurgitated it here.)

The identity of this group I believe is 1^{5}. I have no idea how to find the inverse of this group.

Pretty much this group is the number of different ways you can arrange the numbers {1,2,3,4,5} in different positions and orders without duplicating any number. The best way I have found to think of it is if you had a combination lock with the digits 1-5 on each dial and there were five dials how many different combinations would there be? This is the order of the group.

This is about all I could come up with thus far.