**For the group S _{6} with the subgroup consisting of all the even permutations of _{6}. (The even permutations means the permutations can be written as an even number of transposes.)**

So if we left compose the set of even permutations of S_{6} with an odd permutation we get the set of all odd permutation of S_{6}.

(odd)o{(even_{1}),(even_{2}),(even_{3}), … (even_{360})}

Since any odd permutation composed with an even permutation gives us an odd permutations we know we will get a set consisting of all odd

permutations of S_{6} and we know it will have all the odd permutations because the sets of odd and even permutations of a group have the same

number of elements. Also, since our even set includes all the even permutations, we know that is set is forced to contain all the odd permutations.

We know this because the number of cosets are determined by the number of elements in each set and by the total number of elements in the

original group. So S_{6} has 720 elements. The even coset has 360 elements and the odd coset has 360.

720/360=2

So we know we have a maximum of 2 cosets.

So we now have

{(odd_{1}),(odd_{2}),(odd_{3}), … (odd_{360})}

**For the group S _{6} with the trivial subgroup**

The left cosets of the trivial subgroup are all elements(permutations) of S_{6}