So in this video we're going to prove Cayley's theorem. Now I've made this video before but I was not not really happy with it I wasn't happy with the way that that I explained it So in this video with a lot of noise outside as we're building a new neuroscience unit right outside my window there Let's prove Cayley's theorem first of all some definitions. I want to define Isomorphic there because what we want to prove with Cayley's theorem theorem is that every group is isomorphic To a permutation group Every group is isomorphic to a permutation group. So let's just define What it means to be isomorphic and as I write here with my terrible board handwriting that it defines this I put an inverted comms relationship between two groups and I set my group up a in my group B and The group A consists of a set which I call a subscript set and some binary operation and B Which consists of a set and its own binary operation and I make this Distinction between these two binary operations. They obviously not need not be similar Now I want to show that they're not similar and This relationship that I'm setting up is this bijective function I'm going to call my function f and if map maps the set of group A to the set of group B Such that for all elements in A and I'm just taking two generic ones a subscript one and a subscript two so for all those elements Under this mapping if I take the binary operation between any two of elements in A That is equal to if I take this mapping of the one element And I take separately the mapping of the second element and I do The other groups binary operation that these two things are equal So with that in mind you can clearly see that one thing that you could Possibly see by now is that the order of this set must equal the order of that set because if it's a bijective Mapping it must map one element to one element one to one one to one and that's it There can't be two two year mapping to the same There can't be any there that are not mapped to nothing. This has got to be a bijective function So I went before we do this and then before we actually prove Kayleigh's theorem and look at this Relationship between a group and show that it's isomorphic to a permutation group. Let's look at an example So I've got my two groups here I've called it group G and group H and G is just the fourth roots of one so I've got one negative one I and negative I under the Group operation of multiplication. I'm saying that is isomorphic to another group, which is the integers mod 4 under Modular addition. So is this true? Can I find this? Bijective mapping between those two So let's just look at what the set here is of each set and that is simply going to be 0, 1, 2, and 3 those equivalence classes 0, 1, 2, and 3 and under modular addition It just means I take a representative element of any of these equivalence classes And I add it to a representative of any of the other two and I get back one of these modular Addition, so let's see if I can find a mapping and I'm going to call my mapping here f and You'll forgive me for not writing these square brackets every time I'm just going to write 0, 1, 2, and 3 so you'll know what I mean and I'm going to have the following I'm going to suggest that the f of 1 and That now 1 is an element of G set remember I'm going to say that that is 0 and I'm going to say the f of 2 Is I should not say 2 it's negative 1.  I want you to go through if we look at Kayleigh's table of G Kayleigh's table of G that we have there it's under multiplication. I have my four elements there and 1 minus 1 i minus i and We can fill in the 16 values there and then the same we can do the same for H and Remember that's going to be under modular addition and I have 0, 1, 2, and 3 0, 1, 2, and 3 and I'm going to leave you just with this so let's just do just do one of them Let's suggest we do the f of We're going to do the f of let's do minus 1 times i and I'm saying that this must really equal the f of minus 1 and this is modular addition for the second one Modular addition for the second one so that must be plus f of i And what is minus 1 times i well that was minus 1 times i if you filled in obviously if you filled in the stable This is normal multiplication. We know that's negative i so that's the f of negative i and I'm very I'm hoping very much that this is going to equal the right-hand side So the f of negative 1 in my mapping is 2 2 and the f of i I said is 3 and What is the f of minus i well that's the equivalence class 1 and 2 plus 3 is 5 minus 4 is 1 There we go. Those are equal to each other and I have at least in that one case. I've shown that I do have this fact That I have right here, which is the way that we're going to define this iso more what that we define this iso More for the form most morphism this muscle. So what I would like you to do is Just take some time and do all of these so start with One to all the multiplications negative one with those I with that and just negative i with itself and Just look at this fill in this but and you'll see that this is indeed a bijective mapping and we have shown that the Fourth roots of one under multiplication that group is isomorphic to the group of the integers mod 4 under modular addition So you've seen a nice example of it But we really want to ramp it up now to say that okay, we fully understand this idea of isomorphism Can I show that every group any group that I take any generic group that it is isomorphic to some permutation group? So let's just think of a permutation group though We've looked at s4 or let's just think of s4 or let's make it s3 It's easier to do the symmetric group on three elements That is all the permutations on those three elements So we just if we just looked at cycle notation. We had e That is just identity. We had Tower one two which was this one comma two in cycle notation one goes to two two goes to one and three Stays with three we had tower one three and that is one two three three two one and two stays We had tower two three and that is map two to three and three to two in one state bus It's same we had sigma and Sigma took one to two and two to three and three to one in cycle notation and if we did that twice We had that one goes to three three goes to two and two goes back to one We had that But if we just looked at C3 C3 that was just Some of the permutations We just selectively took e for that and we took sigma and we took sigma squared that gave gave us the three Cycles to give us a cyclic But to give us a cyclic group in three elements. So this is a permutation group We've just taken selectively that is a three is a permutation group It's just the permutations under the upper binary operation of the composition of these of these permutations C3 is just selectively some of those that's a permutation group under the group composition of Discomposing those permutations with each other So that is what we want to say we want to say that any group that I take is isomorphic To a permutation group and we know what a permutation group is and if you think about it That's how we've constructed it this far when we looked at the symmetric groups Well, we constructed it by taking all the permutations of an elements and just giving each of them a name We just gave them a symbol and basically the same for cyclic group for dihedral groups. We actually constructed it from Just a permutation on a set of elements But now we want to go much further with Cayley's theorem and we want to say no matter what group we have if we start with a group And we didn't construct it from there, but it is actually But it is actually isomorphic to a permutation group so I can I can do this So my problem that I'm going to have is I'm going to have a group with elements But I'm not given what this permutation group is So how am I going to construct this permutation group so that I can show that if I do this if I can find this That I can find this function Now to find this function. I better know what this is because remember this if a1 and if a2 and Those are going to be elements of the permutation groups I better construct this permutation group, but the only thing that I have is the group that I'm starting with So clear the board and we'll carry on Okay, so I've cleaned the board one more thing that I want to just tackle here is just this idea of a Bijective we got to find this not only do we have to construct this This permutation group, but we've also really just got to remind ourselves what it means For f to be a function be a bijective mapping. So just want to remind you of that a bijective Let's get the green hood a bijective mapping. So consider for me two sets a and b So as is my usual notation, I'm going to write a sub set and a subscript and b subscript set So if it is injective, what does it mean to be injective is that it means I have this mapping this function f and it maps elements in a to elements in b elements and set a to elements and set b and How does it do that? Well if for all Elements, and I'm going to use just two genetic ones a sub one nice up to elements of a if then if I have the statement that the mapping of a1 Equals the mapping of a2 it implies that a1 equals a2 That is our definition These are our definitions we as human beings decided. This is how we defining an injective mapping so for all a1 and a2 elements of a If we have the statement if of a1 equals f of a2 it implies that a1 equals a2 and For surjective again this mapping of a to b sets a to b if for all b So I'm just taking any b in there they exist in a element in a such that the f of a will give me b So I'm saying that Every b on the second my second set there has this element that maps to it That means it's surjective and if I have something that is both injective and surjective If I have something that's both injective and surjective it is bijective So if I'm if I need to construct this Cayley's proof I've got to find this f and I remember all I'm going to have is this group So I'm gonna know what it's set is and what its group permutation is what is group of binary operation is That's all I have. I have got to find the permutation group in my proof and I've got to find this function that is a bijective so it better have both of those properties and It better Allow for this to happen Then I have shown that that's an isomorphism and if I put all those together I have proven Cayley's theorem. So let me clean the board And we'll carry on good first and let's look at the construction of this and I'm going to use an example again, I'm going to have my Ace a is going to be my group and under this group. I'm just going to have Again, I'm going to use this example the fourth roots that we have of 1 i and negative i and We're going to have that under multiplication So that would be a group that I have and I've got to show that it's isomorphic to a permutation group by constructing this magical Bijective mapping and I'm going to do that for you and we're going to construct this mapping I'm going to call it if subscript a some textbooks use by up subscript a doesn't matter and g equals a g That is going to be our magical mapping. Okay, and this is for all a comma g Elements in this whatever and in this instance. It is this so let me just show you how we construct then this is very easy Let's start off with the a being one They being one so what I'm going to have is one times one I'm going to have one times negative one and I'm going to have one times I I'm going to have one times negative i and that's going to give me this permutation one negative one i a negative i There's a permutation on the set of my elements. That's all I have So let's make a equal to negative one. What are we going to have? Negative one times one is we're going to have negative one we're gonna have one Negative one is we can have negative i and we can have I There's another permutation Let's make a equal to i i times one is i i times negative one is negative i i times i is negative one and one there's another permutation and I'm going to let i equals negative i and Now I'm going to have a negative i but I'm not making a mistake negative i negative i It's this i and i times i is negative one times one is just one and negative one and lo and behold I have four permutations so it's very easy if I use this mapping If I use this mapping to give me a permutation set now, I can call these permutations p1 P2 say p3 and p4 and what I have now is this P my permutation group and I haven't shown that it's a group, but it is So let's just do that. It's P1, P2, P3, P4 and under the Binary operation there of just the combination of these permutations, so I have constructed this I Came construct this permutation. I'm not going to show now that it's a permutation group I just want to show you if you see in your textbooks what this is What this mapping is this bijective mapping that we're going to have that this is how we would construct it if you see it you just take each one of these and use its its Binary operation there with all of the others all of the others all of the others and you are going to construct different permutations on This set and we've looked at those proofs say of the uniqueness and the completeness of all the rows and all the columns in Kayleigh's table, so we've looked at that Not to worry now. We can construct from that We can construct at least some elements that are permutations of just that of course We have we'll have 24 permutations on four so not using all of that and definitely we can't if it's isomorphic We only need four so we know that there somehow we've got to show that one of these is going to map to one of those You know one of each one of these is going to map to each one of those and under this I'm going to show you now, you know that we can construct this Proving Kayleigh's theorem again clean board. Let's have a look so we've done all of that So let's have a look I have this magical function. I've shown you how what what it does So if you give me any group I can construct some permutations on the set of that group And it will be equal in number to the elements by this function So I'm saying let's have this now. Let's just prove that it's injective and prove that it's surjective So let's have a look at it remember That I have my group G it contains a set and a binary operation I'm saying let this mapping and I've shown you how to create this one equals that one So how is it created? Well, it's right there. It says a composed with G1 and On this side we have a composed of G2 and remember this is for all a G1 and G2 elements of GC Very easy now. This is a group if a is in there. It's in this must be in there I can left multiply. I'm just using multiply, but that's this binary operation G1. I do the same on this side P and G2 I can use associativity. It is a group and The binary operation with an element and its inverse is the identity element and Identity element composed with any of the elements is just that so I have G1 equals G2 So if this is true, then that implies that that is true. I have shown that it is injective now a bit more difficult Let's just think about how to prove that it's surjective Well, let's take an element of the set any element of the set for all a and G and that And I'm just going to write a G in a slightly different way. I'm going to say a Composed with a in composed of a inverse composed with G and if I do that associativity Element and its inverse identity element identity element and the element is just the element so I can really rewrite that Now, let's just use the associative property so I can rewrite that as a inverse in G No problem there and let's just call this put let's just call this bit G prime and By closure property, it's still an element of G set. So I'm writing that G equals a composed with G prime But isn't that just the definition that I've got here So what have I got here? It says that for all G elements of G set they exist to this element G prime element of G set So that I have that the f a of G prime is going to give me G Because remember, how do I write this? It's a composed with G prime and I've just shown it is there because that is G and in other words Everyone that you give me a G I can map to By this function in other words, it is subjective and what I will have I have this magical mapping that is by objective You've given me some arbitrary group. I've shown you how to construct a Same number of permutations on that by that and I've proven Kayleigh's theorem for every group Or every group at least is isomorphic to a permutation group as simple as that