 So we left things off by wanting now to show that we have this equivalence class on G containing A. So there's some element A in the set that makes up the group and we define it by this equivalence relation such that if I have A and B elements of my set G that I have A inverse composed with A with B that that A inverse composed with B that will be an element of H and I want to show that this set is the same as this set with A this composition with H such that H is an element of my subgroup that these two are equivalent and I said if I want to show that two sets are equivalent I've got to show that one is a sub this is a subset of that and that that is a subset of that. So in our proof first let's show that this is indeed a subset of A composed with H such that H is an element of my subgroup. Let's show that that is so. Now first of all let's just let choose an element in this equivalence class and we'll just say let X be an element of this equivalence class this equivalence set and if that is so it does mean that X inverse composed with A must be an element of must be an element of this and let's say then it's it's an element of this and I'm just going to say that by the symmetric property because we've shown that it is an equivalence class I also have to have a inverse X being an element of that and I've defined it such that remember that this relation at the moment I'm dealing I remember I'm dealing with that left relation isn't such that if I created two of these it must be an element of H it must be an element of H let's choose an arbitrary element in H so that is going to let's say equal H an element of H quite legitimate to do because that is how I defined my equivalence relation and because H is a group A inverse is in that group that means its inverse which is this A must also be in that group so I would have A inverse A X equals A should have put that one there so put if that's an element this must be an element left multiply it or addition whatever this binary operation is on that side this gives me this by the by the associative law this gives me the identity element so I'm showing X is equal to A H which is an element of H and that will only happen if X is then an element of this if X is in there so I've just shown now that if X is in this equivalence class it is a subset of that X is a subset then of this side in other words X must be in this when it's in there it must be a subset of there so I've shown it in this direction so that's quite easy to do let's just look at this side where I have my set such that H is an element of H I must now show that this is a subset of the equivalence class on that once again I'm going to choose an X let's X be an element of this set A H such as H is an element of H and if it is an element of that if it is an element of that I can just say choose it to be one of those choose it to be one of those if I multiply on the left or binary operation on the left with X that is going to equal H which is an element of H if I've shown it in this way it means it's an equivalence relation that is how I defined my equivalence relation in other words what I have here what I've shown you is this is the same as this equivalence relation that we had and if that is so it means it must be a subset it must be inside of it must be inside of A as well so I've shown this way around as well that this must be a subset of that and and if I've shown both of these I have shown both of those being a subset it means these two are equal to each other and remember this is the left I've just shown the left coset here and I must also again you can look at that yourself if you turn all of these around show that the right coset exists as well so this would be the left coset on H and this would be the right coset so not a difficult proof to show that one is indeed a subset of another and this turns out also to be very important as we will move eventually to Lagrange's theorem