 So in the previous video we looked at the orbit of a set. Now we're going to look at the stabilizer. Again I have my set G and my group G. It consists of a set and a group operation and I have my arbitrary set. We're going to act on that. So the stabilizer I'm going to take an element of A and I'm going to define the stabilizer which is a set. It's going to turn out to be a subset of this and I'm going to define this on my element A that I chose from A as all the G's in my G set such that G acting on A just gives me A back. So take the element of A and I look at all my elements of my G set and I see which one if I act on this A is going to give me just the A back. And I claim that this is a group. It's not only a set that under this operation that this is a group and it is a subgroup of G. And to show that it's a subgroup of G, well it is a subset of G because is this going to be some of the elements of G. So we're going to have associativity. Let's look at closure. For us to have closure let's have G1 and we have G2 and we make them an element of the stabilizer of A here in G. And so what we need to show now is that if I take G1 and I compose it with G2 that that is still an element of this stabilizer of A so that I must show this. Now to show this what I really want to what I really need to show then is that G1 composed with G2 under this group operation dot A must still be A. So if that is so then I've shown that there is closure. Now let's assume to the contrary that it's not like that that it's not like that. Well by the first property of group actions I can write three write this quite legally as G.1 and G.2 acting on A and I'm suggesting still that that is then not equal to A. But G2 is an element of GA. So by this property G2 dot A that must just be A. So I'm left here with G1 acting on A and I'm still saying that that's not A. G1 is an element of the stabilizer of A. So by the way that I've defined it this is just A so A does not equal A that's a contradiction. In other words this initial assumption that we did you know assuming to the contrary that it's not as false and this is so. So if G1 and G2 is an element of the stabilizer of A then the composition under the group binary operation is then still an element. So we have closure. Let's see if we have an identity element. Well we know that by the second property of group actions that E dot A equals A and if that just follows the definition of what we have there by the stabilizer therefore E must be the identity element must be an element of the stabilizer of A. Lastly does every element that falls in May does every element have its inverse. So I'm saying if G dot A equals A I need to show that show that G inverse dot A equals A. So the G inverse dot A equals A. So what can we do here? If we have G dot A if we have let's have a look if we have we can just rewrite this as A equals G dot A. A equals G dot A and let's I need to show this G inverse A equals A. So let's take G inverse A and I'm just going to rewrite this instead of A I'm going to use that. So I'm saying G inverse dot G dot A and by the second property by the second by the first property of group actions I can rewrite this as G inverse G acting on A. We've just shown that you know or by the properties by that binary operation that's just E that is just E dot A and E dot A by is just that is just equal to A. So we have shown you very cleverly that G inverse dot A equals A and that obeys the properties that we set the definition that we set for the stabilizer. In other words we have closure, we have the associative property, we have an identity element and we have that for every element in this stabilizer of A we have that the inverse is also there. So indeed we have shown that the stabilizer of A is that is a subgroup that is a subgroup of G. So remember we just think we take a specific A and then we construct the stabilizer that's all the G's that you know that will have that will lead to this property where that element of the G said X on A that we still that we still get A so every A might have a different stabilizer, a difference stabilizer and then this according to the group action if I take all these elements and I apply that binary operation it is still a group and therefore a subgroup of G.