 so let's have a look at an example and here we're going to have a right coset because A is on the right of H so I'm going to define this as H binary operation with A so A on the right side the right coset of H remember H is a subgroup of some G I'm going to define H as this permutation this set of these different permutations on the symmetric group s4 so I have 1 if I just write it in this notation it means 1 goes to 1 2 to 2 3 to 3 4 to 4 the identity and this goes 1 goes to 2 2 goes to 1 3 goes to 3 and 4 goes to 4 and this one is 1 goes to 1 2 goes to 2 and then 3 goes to 4 and 4 goes to 3 1 goes to 2 2 goes to 1 3 goes to 4 4 goes to 3 and I have this composition of these permutations as my as my group operation and I have this A is an element, it goes one goes to four, four goes to three, three goes to two, two goes to one, that's an element of the symmetric group on four elements. So I'm suggesting now that HA, that is going to equal, this must go on the right, so it's going to be one composed with one, four, three, two, and I have one, two composed with one, four, three, two, and I have three, four composed with one, four, three, two, and lastly I have one and two and three and four composed with one, four, three, two, and that is it. So how do we do this composition? Remember, if I have this composition of these permutations, what happens on the right happens first and then on the left. And if you've got some experience, you can do this all in your head. So let's just try that first one, just if we can. So one goes, just imagine that is one, two, three, four, one, two, three, four, so one goes to, so we start with one, one goes to four on the first side and then four, four goes to four. In other words, one goes to four. If I think of four, four goes to three and on that side three goes to three, so four eventually went to three. Three goes to two and two goes to two, so three actually went to two and two goes to one and one goes to one, so two goes to one, so that's done and that's exactly as I would have had with my identity permutation there. Let's do this one. Let's just do this one the long way, one, two, three, four, one, two, three, four. Again, I'm going to start with one. Remember what happens on the right is first. So one goes to four, four goes to three, three goes to two and two goes to one and on this side one goes to two, two goes to one, three goes to three, three goes to three. If I start with one, one goes to four. One goes to four, four goes to three, three goes to one. So I'm back there and eventually two is just going to go to two, so we just leave that out. It's just for simplicity's sake we leave that. So let's look at this one. I start again with one, let's try and do this one in our heads. So if I think of what happens on the right-hand side first, one goes to four and four goes to three. In other words, one goes to three in the end and then what happens to three? Three goes to two and two goes to two, so three goes to two and then two, two goes to one, two goes to one and one goes to one, so two goes to one. So two goes to one, so we are done with that one and that just leaves four which goes to four and on this last one, let's see what happens to one and this last one, one goes to four and four goes to three, so one goes to three and now three goes to two, three goes to two and two goes to one, so three goes to one, so we are done with that one. That goes nowhere. Let's see what happens to two, two goes to one and one goes to two, so two goes to two, so two just goes to two, we leave that and four will go to four because four goes to three and three goes to four, in other words, four goes to four and we just left with that. So that's our right coset of H with this A being this one of these permutations on the symmetric group of four elements, that would be AH. See if you can work out what the right coset of H is going to be with this same element A.