 So we get a chance to get back to some group theory. I want to talk to you about group isomorphisms. So that's a group-automorphism. So what a group-automorphism is, is really just this mapping. We're going to take a mapping and from a group onto itself, from a group onto itself. So that in some essence it makes it very easy and in some instances it makes it a bit more difficult to understand or to at least do the proof. So we have a group G and that's going to be a set and some group operation. So just a group onto itself. One of the easiest mappings that you can think of is just the identity mapping and that is just the way we map every element onto itself and we'll call that, we'll call that f e. So that's just identity mapping. So everything onto itself. Now if we have this definition of it what we're really saying is that this group-automorphism and let's call it, well if you can think of a group you can think perhaps that there'd be more than one, that there really will be more than one automorphism and you can actually put all of them together in a set and we call that set the automorphisms of the group G. So not as any a ut. That's the way it's written in most textbooks. So we have the set of the automorphisms and that is going to be a set of all these mappings that are elements of the symmetric group on G. So why this is this element of the symmetric group of G? Because you think about what a group is really. We've shown that we can construct a group as just the set and all the possible permutations on a set and if we give all of those permutations a symbol those symbols make up the elements that go inside of that group and if we look at the symmetric group that refers to all the possible permutations of a set. So it's got to be the element of this such that we obey something. So if we take for all G1 and G2 elements of the set there. So if I take my set there and I take any two of the elements I have this property of isomorphisms, this property of isomorphism applying here. So the fact that we have if we have this mapping of this composition of these two so you take any two you compose them your binary operation between them and that is going to equal must equal this mapping on one of them composed with the same mapping on the second one and that is the set of all the automorphisms of this group and what we are saying here is that it's all these mappings that we can do of the set unto itself so then look at that that's one of them but there can be more of them and we're saying that it's an element it all these mappings are element of one of the symmetric of the symmetric group which is this huge thing it's all the possible permutations. Now what we actually want to show is that this set here is actually a group the fact that the set of all automorphisms of G is this group is a subgroup at least of the symmetric group on G so it makes up a group and for it to make up a group it really has to obey all the properties of a group and what we need to show of course is closure and we need to show that identity element exists so can we show closure so we have the factor that if one and if two are elements of the automorphisms of G so that is what they look like and I they need to show that if they are so then we have this fact that the composition of them must also be an automorphism so if I compose the two of them what I really have is the f1 of the f2 of G1 composed of G2 and that is really for all G1 and G2 two elements of G set so that's really what I have that what I have there and what I want to show is if I compose them now that that that that composition of these two that this must also obey this definition of what this isomorphism criterion that it must obey so if I have f1 composed with f2 what I really have is this f1 of G1 composed with G2 and I'm composing that with f2 of G2 a G1 composed with G2 and what I have there by this property of isomorphisms is I have the f1 of G1 composed with the f1 of G2 composed with the f2 of G1 and composed with the f2 of G2 that's really what I'm saying if I'm composing these two and all I'm using is this definition of the isomorphisms but if I look at this composition that I really have there this because it is an element of that would be the f1 of the f2 of G1 composed with the f2 of G2 and really what I have here is nothing other than what I have what I have there I have the f1 of of this would be the f1 of G1 composed with the f2 of G2 composed with the f1 of G which one have I got not have I got etc you can see that these two things are going to be exactly the same so by this definition if I have f1 and I have two and they're part of this automorphism of G so they obey this isomorphism property here that composing them is going to leave me with exactly the same thing if if one of if two of this if one of if two of that write them out and I land exactly there and I land exactly there the because it is a subset of course I don't have to worry about associativity the identity is is definitely it obeys this property so it I definitely is there so this identity mapping identity group isomorphism is there and all I'm concerned about now is to show is to show this idea of the inverse being there and the way that we're going to show go about this is just to show that it obeys this property of the isomorphisms so what we can really show is the fact that if I have the group and the onto the group I can really find if I have G1 on the second part remember so if I if I can find if I if I have G1 there what we have here remember is a bijective mapping and it's been nice that it's bijective because I can find G1 prime and I can make that G1 and I can find G2 prime such that that is G2 so I can really I can really have that by this I really need to show that the F inverse of G1 composed of G2 equals the F inverse of G1 composed of the F inverse of G2 if I can show that that means this inverse you know it does obey this and if it obeys this it really is one of the automorphisms so if we have the F of G1 composed of G2 I can write that as or at least the inverse of that I can write that as the F inverse of G1 I can just have this F of G1 star and for G2 I can write the F of G2 star because because this is a bijective mapping I can find in this I can find G1 and G2 here and it will have a corresponding F there's a corresponding element in here that should I take the F of that I'm going to end up with that and if we look at this and if we look at this it's the inverse mapping of the mapping of that so I'm just left with G1 star composed with G2 star because the mappings it's the mapping of the inverse mapping of this the inverse of this and the inverse of the inverse of the mapping composed of the mapping is just going to leave me with that the same for those two it's just going to leave me it is just going to leave me with that and if we look at this right hand side here if we have this combined here the F inverse we have F inverse of G1 and we have composed with F inverse of G2 and once again it's the same issue that I have here this is the F inverse of G1 is the F of G star 1 and this is the F inverse of the F of G2 star and the F inverse and the F that is just G1 star composed of G2 star and these two things are exactly the same so if I really if I have F1 and F2 in there and I look at the inverse the inverse the inverse does obey this so it must be one of the automorphisms of G so to base this property of the automorphisms of G these two things are exactly the same so really this group of automorphisms the set of automorphisms is a group of itself and it's a subgroup of the symmetric group it's some of the elements of this symmetric group on G all the possible permutations of G so that in short is a group automorphism I think in the next video I'll just show you a short example of one of an easy at least to put this into perspective an easy example of of of these group automorphisms