 So let's look at Cayley's theorem properly. So I'm going to put it on the board. Cayley's theorem says that if I have a finite group with n elements, it is also morphic to a permutation group in n elements. Let's just remind ourselves what a permutation group is before we go. Remember the symmetric group, all the permutations, any subset of that is a permutation group. So if Cayley's theorem says that I have a group with n elements, remember this one? That was just the, all we had here was the fourth root of one. It has four elements. So I must have a permutation group with four elements. Now lucky for us, we've already looked at this. That if I look down any row, all the elements are unique and they are all there. And if I look down any columns, all of them are unique and they're all there. In other words, I can see each one of these as a permutation of my original set. So I can call this first one permutation 1, the second one permutation 2, the third one permutation 3, the fourth one permutation 4. Permutation 4 for instance takes 1 and turns it into negative i, negative 1 into i, i into 1, and negative i at maps that turns into maps 2 minus 1. And each of these mappings is unique and we've shown that already. So what does it mean remember to be isomorphic to? It means I must find a function that maps my finite group to the permutation group and that mapping must be a bijection. Now, I mean one thing we must have with that bijection at least. Remember bijection if I have 1, 2, 3, 4 in this set and I have a, b, c, d. There's a 1 to 1 and on to. So each one of these is mapped to in the second set and then each one of these is mapped to its own individual. So it's 1 to 1 and on to. So I better have the same number of elements in each and my permutation group here has four elements and my set inside of my finite group has n elements. So at least I have that as far as a bijection is concerned. And these are unique. I mean and we've already shown that each of these are unique. So we don't have any issues there. But what could, what possibly, what could this function be that shows this isomorphism between these two? And I'm going to claim that it is the following. Let's write it there. I'm going to say if I take on my first set which is my set here and I'm going to use just some generic. I'm going to say if I take any one of these elements g i and my binary operation with g j. So it's the binary operation with two of them and I might be equal to j. It doesn't matter. And I take a binary operation with yet another one and I could do this. Remember it is a group so associativity. I'm going to say that that maps to the permutation group i of the permutation group of j of this element here. I'm claiming that that's a bijection and that this mapping shows that there is, that these two sets are isomorphic. That's what I'm trying to show. And this is a composition of permutations. And the way that we're going to define this is to say that I'm going to take all my elements here one, negative one i, negative i. So no matter what these are g1, g2, g3, g4 till gn and I'm done that. And each of those rows I will say that that corresponds to that permutation group. This element corresponds to that permutation group. Let's see. That is my definition. So let's just see if this holds. Let's just take this for an example before we move on to showing that this is so for arbitrary elements hence proof of Cayley's theorem. Now with Cayley's theorem if you want to do that proof remember you first have got to show this uniqueness and completeness in all of this, all of the rows and columns and we've done that. So let's take for instance I'm going to say let's take g2, let's take g2 and g3 and g4. And I want to show that isomorphic to what I'm going, will then be p2 of p3 of g4. And in our instance here g2 I'm just going to call these g1, g2, g3, g4. So that would be g1, g2, g3, g4 and this n equals 4 in this instance. So I'm just using this as an example to illustrate to you what's going on. So g2 is actually minus 1. The binary operation in our instance here is multiplication. So I can actually say multiplication. g3 is i and multiplied by g4 and g4 is minus i. So if I do this minus that that gives me negative i times negative i and that is just equal to negative 1. And there it is negative i times negative i and that just equals 1. So let's see what is the permutation group. This is a, I'm combining these two permutations. So what does, and then we do the inner one first, the right hand one first. So what does p3 do to g4? So yes, so I'm asking what does, what does, what does g4, g4 is, remember that it is negative i. So what does p3 do? Yes, p3, what does it do to g4? It takes negative i, my negative i there, it takes it to 1. So I'm asking what the permutation group, if I take permutation, the permutation 2, what does it do to 1? Permutation 2, what does it do to 1? It maps it to negative 1. And lo and behold, I have negative 1 and I have negative 1. They're actually the same. So at least for this group, for one example, it holds. Now I can prove that it holds for all of them through brute force. But let's just be more, let's just be more generic about this. I'm going to suggest now that I take any gi, binary operation with gj, a binary operation net with gk. And I'm going to say that maps to the permutation group i on the permutation group j. So those two permutations with some element gk. So that's what I want to do. And remember what happens here with these permutation groups. Remember what we would do there is to write 1 minus 1i minus i. And then I would have 1 minus i, a minus 1i and minus i and it's this mapping. So what does, for instance, what does p3 do to g4? Now remember in this order that would be p3 there and that would be p2 there. So what does p2 do to 1? It takes it to negative 1. What does it do to minus 1? It takes it to 1. What does it do to i? It takes it to negative i. And what does it do to negative i? It takes it to i. And then p3, that would be its own. And I'm just asking then what do I do for this one and you follow it through and where it lands up and that's going to land up to negative 1. Remember this is what this composition of these permutations do. Okay. So let's just have a look. If we want this to be a bijection it's got to go, we've got to show this in both ways. So what happens if I do ji and, let's make this then g3 and what did we have here? g2 and g3 with g4. So what is g2 and g3? So there's g2 and g3. So there is that g3. I hope it's not too small, so there it is. And then remember this is a group, so there's closure here. So this g2 binary operation g3 is going to be one of these. It's going to be one of these. Irrespective of which one it is, what do I do if I take this one and I map it to, this is now going to be one of these and I map it to g4. Well it is actually any one of these and it's then going to be mapped to g4. So there's whatever this was, the secret that it was with g4. That's what it is. I mean that's exactly what it is. So p, the same is going to happen here. If this was then p3 what does it do to gk? And that was for us was 4. p3 what does it do to g4? There was g4, what would it map it to? It would map it to 2, so p3, there's g4. It would be g3, g3 and g4. So what would p3 do there which would correspond to g3? What does it map g4 to? Well it maps it to this thing which is g3 times g4. And what does p2 then do to all of that? Well it will be g2 mapped to g3 and g4 and these things are exactly the same. And all you have to just say to yourself now is that if I go from this and instead of taking specific elements 2, 3 and 4 I just take i, j and k and I take i, j and then some element k. I'm asking by definition what does it mean to take the pj of what does this permutation do to that element? Well by definition I've made this into the row that contains the g element of j so that will be gj. So my definition of writing this is gj and the binary operation gk. By definition that is what I'm saying. And by my definition of this because I've equated these numbers to each other, I'm asking by definition this means I'm taking g i and the binary operation with wherever this ended up with but generically it would just be that. So by definition that is what we're saying that is the definition of the permutation group of that element. And by that same argument on this side well this is what I have. I have g i, I have gj and I have gk and that's how I set it up and I land up with exactly the same thing. So on this side g i if I say what is g i times g j, g i times g j that would just be taking the permutation group of p i of this element of this element g i, g j. So I'm just taking the permutation group of that and if I then take this we'll equate to this p i of taking the p j of g k. So it works the other way, it works towards the other side as well. I'm just asking what does it mean, what does this equate to? Well I'm going to take p i whatever that was in our instance here p 2 with that it means turning that into that, turning that what happens to this inside of that permutation and then what happens to this inside of that permutation and I get this. So I can show both ways that it's exactly the same, it's exactly the same thing. So we've got this initial thing, this mapping of this set to this set showing that a finite group maps with n elements maps to a permutation group on n elements. Now remember there's four here, four factorials 24, so they're actually 24, 24 permutations. I'm taking a subset of that symmetric set of 24 permutations, I'm using a subset of that, a subset of the symmetric set is a permutation group and here's my permutation group of four elements and I'm just very clever in the way that I set it up that each of the elements here corresponds to a permutation group with the same number and then it flows very easily that you can show that this holds in both, in absolutely both directions choosing arbitrary permutation groups and an arbitrary element, arbitrary element, arbitrary element, arbitrary element, everything holds proof of Cayley's theorem.