http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110804_134032.jpg?t=1312484230 [Broken] http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110804_134038.jpg?t=1312484242 [Broken] I found that the order of G/H is 6. According to the Lagrange's Thereom, order of G = order of H * index of H in G = order of H * order of G/H The index of H in G is the number of distinct left cosets. The quotient group G/H is the cosets of H in G. Since H is normal, the left and right cosets are equal. Thus, the distinct left cosets equal to cosets of H in G. Find the distinct elements of G/H? G = S4 has 4! = 24 elements. Do I need to operate each of the 24 elements of G on the 6 elements of H? Or, since it is a symmetric group, is there a faster way to find the distinct elements?