I had been given the following question and answer ( in the image)
However i do not understand, why for example: (1,4,2,3)K does not belong the the quotient group?
Is there any faster way of finding all the cosets for H on a Symmetric group, rather than computing every single one? It takes a very long time for me to do so.
Actually, $(1,4,2,3)K$ is one of the elements of the quotient group. Where you're getting is confused is the fact that it is the very same element as $(1,3,2,4)K$ – we're just picking different representatives.