 In the previous video, we proved that every cyclic group is up to isomorphism determined by its order. There's only one cyclic group for every possible order you could consider, both finite and infinite. Well, what about other families of groups? Could we classify finite abelian groups up to isomorphism, just like we did with cyclic groups? And the answer is yes, finite abelian groups will be determined up to isomorphism as direct products of cyclic groups. And that's something we'll do later on in this chapter, not in this lecture right now. The question of classifying finite non-abelian groups up to isomorphism is a much bigger task. Up to the present, this is not done. In fact, the state of the art is that only within the last few decades have group theorists been able to classify finite simple groups up to isomorphism, which the simple groups a term will define later on in this lecture series, not in this lecture, though. Simple groups you can think of as the atomic building blocks of finite groups. And so we've only barely determined all of the atoms on the periodic table. Coming up with molecules is gonna be much, much more difficult. Even classifying finite P groups is still a very, very unsettled question in group theory. And so these topics take us way beyond what you would see in a first semester algebra class. Very important topics, but we would be insufficient to be able to prove them right now. In the meanwhile, though, we will prove the very first so-called representation theoretic theorem of group theory, known as Keighley's theorem. I've sprinkled here and there things which I refer to as representation theory, but Keighley's theorem is really the first step towards representation theory, which as I've many times in the past said that representation theory is like equipping group theory with linear algebra. The way we're gonna try to mention it right now is representation theory, a very active research branch of group theory. It aims to represent abstract groups as more familiar groups such as cyclic groups, permutation groups, matrix groups, we could call these concrete groups. And so the advantage of representing a group as a group of matrices allows one to employ linear algebra in the study of groups. By representing a group as a permutation group, we can use things like combinatorics and group actions to help us better understand the abstract group. So similar benefits exist for representing abstract groups as other concrete ones. So Keighley's theorem, which we will prove in this video, allows us to represent every group as a permutation group. This is a very useful result in many situations. In some regard, it's like almost too useful that we actually never have not to use it. It's like that perfect gift you got for Christmas, but it's so perfect. You don't know what to do with it. I know that might not make any sense, but what it basically means for us is that if you can prove something about the symmetric group, it's almost like the Holy Grail, because in some respect, it proves it about every group. And because of that, it almost becomes too grand to prove certain things because otherwise you cover all groups at the same moment. So I digress. Let's prove Keighley's theorem in this video here. We're gonna do it in parts. There's two limits I wanna prove first because the limits actually have some broad ranging consequences. And then we'll prove Keighley's theorem as following those two limos. The first one, essentially, we've already proven it in this series many times, but we're gonna prove it officially right now. Let G be a group with an element little G inside of it. Then we're gonna define a map. We're gonna call it lambda of G, which is gonna be a map from G to G. And it's gonna be given by the rule lambda G of X equals G of X. So this is the left multiplication map by G. Or as I've referred to it in this series earlier, this is the left translation map. All right, translation, I ran out of space there. Analogously, we could talk about the map rho of G of X, which would be defined to be X times G. This is called right translation. And I want you to convince yourself that everything I'm gonna say about left translation, that analogous statements could be set about right translation. We're gonna focus on left translation right here. Again, just for the sake of argument, we need to pick one. So in this lemma, we're gonna prove that left translation by G is a permutation. So what is a permutation, after all? It's a map from a set back into itself, so the domain and codemand are the same. We have to prove this is a bijection. So it's not too difficult of an argument. Let's go through it. So we have to show that phi of G is bijective for any element G. G will be arbitrary so that when we prove it for G, we'll take care of everything at the same time. So let's take two elements inside of the group, X and Y, which might have nothing to do with little G. And let's suppose that their images are the same. We wanna prove that it's one to one. So lambda of G of X is assumed that's equal to lambda G of Y. Well, by definition, that means you're gonna get GX equals GY. But as we have a factor of left on both sides, we can cancel the X and the G, excuse me, we get X equals Y. So left cancellation shows that this map will be injective. Now, if this was a finite set, you'd be like, oh, an injective set here would have to be surjective at the same time. But I wouldn't want this argument also to apply to infinite groups. So we need a supply argument for surjectivity. Well, okay, imagine you have X. This is our target. Who's gonna map onto little X? Well, we're gonna take G inverse of X. Well, notice that lambda G of G inverse of X. This will equal G times G inverse of X, which equals G G inverse of X, which equals X. And so this is gonna turn out to be surjective. And therefore, lambda G is a permutations, both injective and surjective. And like I said, we could also talk about these the same argument would have worked for right translation, which you'll actually notice that row G of X I defined. Despite what I said earlier, I actually could say X G inverse. There's a little bit of advantage of doing an inverse sign there. But don't worry about that too much in this conversation. Again, we're gonna focus on left translation. This G inverse really comes into play if we wanna talk about the regular representation, the right regular representation. What I said earlier about times in by G, that right translation that would also, that's also would turn out to be permutation. It's just later results would be more useful to have this G inverse. And that's because we can't assume the group is abelian. I wanna illustrate this as an example before we go on to the next lemma. Let's take the cyclic group of order four with respect to addition, of course. And so you have the elements zero, one, two, three, and zero, one, two, three, you see them like that. If I were to highlight a row of the Cayley table here, so we have the zero row. If you look at that, what does that mean? So zero will send zero to zero, one to one, two to two, three to three. And so we end up with the permutation, which turns out just to be the identity, okay? If we take the next row, if we take the first row, notice the first row sends zero to one, one to two, two to three, and three to zero. For which if we just record that right here, like if we just record the index row with row one, that actually is the permutation in question right here, for which we can then write that with cyclic notation, the zero goes to one, goes to two, goes to three, go back to zero, okay? If we do this again for the second row, so we take our index row and we take the second row, this gives us the permutation zero, one, two, three, well that's just the index, zero goes to two, one goes to three, two goes to zero, and three goes to one. So we get this permutation to two, two cycle, zero, one are on an orbit, one and three are on an orbit. And then finally, if we take the third row, we see that zero goes to three, one goes to zero, two goes to one, and three goes to two. And so here's the four cycle, zero, three, two, one. So we can represent, we can represent these permutations by really looking at the Cayley table and seeing, oh the row, the rows in the Cayley table correspond to these permutations. Now with our right representation here, row of g-remory is supposed to take the inverse, it's kind of like looking at the columns of this, but again, you have to take inverses. Again, that's a subtlety that we'll worry about a little bit later. So let's look at our second lemma right here. Let g be a group, we're gonna define a new set, we're gonna just call it g-bar for the lack of a better name. Take g-bar to be the set of all left translation maps. Okay, and I claim that g-bar is a subgroup of sg. Now, technically this should be sg really, right? Cause we're talking about permutations on the set g, but these permutations up to isomorphism, it only depends on the size of the set. So I can identify, I can say this is s sub g, the order of g, so it's just the number. So don't worry about that nuance so much right there. So I claim that g-bar is a subgroup. So we have to show it has the identity, it's closed under multiplications, closed under inverses. All right, so let's first show that we have the identity. Well, our candidate, notice over here g, it has an identity element cause it's a group. So let's take left translation by the identity. Notice if you take left translation by the identity, so lambda e of x will just be ex, which is just x, which is just the identity map, right? x didn't matter here. And so the left translation by the identity is the identity permutation. And so therefore we have the identity of the symmetric group. So g-bar has an identity. Great, that's step number one, we have an identity. Next thing, we need to show it's closed under multiplication, okay? Well, let's take two elements of g-bar. They would look like lambda g and lambda r, or lambda h, excuse me. And so what happens when you multiply lambda g with lambda h on x? Now what we have to argue is we have to show that when you multiply these things together, so permutation multiplication, we need to show that this is equal to lambda of something, right? And the candidate actually is gonna be lambda g times h, okay? So when you multiply together these permutations, you actually get the permutation associated to the product g-8. So that's what we wanna show right now. Now how do you show the two functions are equal to each other? You're gonna take an arbitrary element of its domain and show that the two functions agree on that element. If the two functions agree on every element of the domain, that means the functions are actually equal to each other. So consider with the left side over here, lambda g of lambda h of x. Well, by function composition, we're just gonna put lambda h of x inside of lambda g. Well, lambda h of x just means h times x. Lambda g of anything will just be g times that thing. So we get g times h of x. Well, okay, g times h of x, well by associativity, this is gonna be gh times x. But hey, that's what's just what left translation by g times h does, left translation by gh. And so that then proves that lambda g times lambda h is equal to lambda of g times h. Therefore, the product of two of these left translations is in fact the left translations. So g bar is closed under multiplication. So we've proven that it's closed under multiplication. So the last thing to prove is that it's closed under inverses. So we need a candidate for the inverse, right? Well, notice here that the identity translation was just to translate the identity. And the product of two translations was equal to the translation of a product. So if I'm going with this analog I see right here, it would seem that the inverse of a translation should be the translation by an inverse. So that's kind of my conjecture here. So notice what happens if I take the inverse of one of these things. So this is the inverse of a function, the inverse of a permutation. That's also a permutation. Now notice that lambda g to the inverse function of x. This is actually gonna map to g inverse of x. How do I know that? We'll notice that, why does that say phi right there? Lambda g of g inverse of x will go, that'll map to g of g inverse x. Whoops. Oh boy, it's falling apart there. Try it again. So lambda g of g inverse of x, well that you'll get g times g inverse of x which gives you x like so. So notice if we wanna pull this thing backwards, right? If we reverse the process because the inverse function goes the other way around. If g inverse of x maps to x, that means x will map to g inverse of x, great. But wait a second, g inverse of x is the image of lambda g inverse of x. And so since these two functions agree on all elements of their domain, excuse me, that means they're actually equal to each other. So in particular what we've shown is that lambda g to the inverse is equal to lambda of g inverse. So our set is closed under inverses and therefore we've proven that g bar is a subgroup of the appropriately sized symmetric group. And so let's see, what do I wanna say here? And so this argument here, it's really this argument right here which is why we define rho g of x as x g inverse. Because the previous lemma about, are these things permutations, those would be permutations, but in order to get g bar to be a subgroup of the symmetric group, the identity would work out the same. Close your under multiplication, you need the inverse map here because if you take rho g times rho h of x, let's actually spell this out for a second. This would look like doing it one step at a time, rho g of rho h of x, you end up with rho g of x h inverse. And then that turns into x h inverse of g inverse. So that looks like x times h inverse g inverse for which that then becomes x times g h to the inverse by the shoe sock principle which would then look like x. Whoops. This would then look like rho of g h of x. So in order to show that this g bar is actually a subgroup, that's why we take the inverse on the right-hand side just so you're aware. The argument is essentially the same left versus right, doesn't make much of a difference. So now we've reached Kayleigh's theorem where we then, we started off with a group g and then using that group g, we produced a subgroup of the symmetric group. Kayleigh's theorem then is gonna argue that these two groups are the same, that g and g bar are the same. And as g bar is a subgroup of SN, this then proves that every group up to isomorphism is a subgroup of SN. This, that is every group can be represented as a group of permutations like the Kayleigh table approach we saw earlier for Z4. All right, so let's prove that this map phi is an isomorphism where the map, we're gonna send phi of g to lambda sub g. So we'll just send g to left translation. You could also send g to rho of g if you prefer right translation, it doesn't matter. We're gonna stick with left translation so we don't have to bother with the extra inverse there. So imagine we have two elements of g, okay? So we have phi of g and phi of h. Let's assume they're equal. We wanna then prove that this is injective, right? So assume, well that is we're gonna improve injectivity by showing that g and h are actually equal to each other. Well, if phi of g equals phi of h, that means lambda g equals lambda h, okay? And then pick an arbitrary element of the group g, like x. Well, if these two functions are equal, that means that their images will be the same for every choice of x. So lambda g of x equals lambda h of x. But what does left translation do? Lambda g of x will give you gx. Lambda h of x will give you hx, like so. And as x here was chosen arbitrarily, we'll just cancel x from the right-hand side and we end up with g equals h. And therefore this shows that phi is an injective map. That's the first step, right? To show surejective, this is fairly clear, right? Because the way that we constructed g bar, everything in g bar looks like phi or, excuse me, lambda of g and therefore g will map onto lambda of g. So surjectivity could chink, it's super easy. Now the homomorphic property, how does this look? Well, we have to show that the product, that phi of a product is equal to a product of phi's, right? So phi of gh, this will look like phi, excuse me, lambda of gh, which we saw in the previous proof that lambda gh is actually the same thing as lambda g times lambda h, which lambda g is just phi of g and lambda h is phi of h. So honestly, my lima's kinda stole all the thunder behind Cayley's theorem here, but this gives us the final proof here that lambda phi is gonna, sorry, lambda phi here is gonna be an isomorphism. So the map, g maps over to lambda g, this is what we refer to in representation theory as the left regular representation. Whoops, left regular representation of the group. And similarly, if you take g mapping to rho of g, this is what we call the right regular representation. And so that's Cayley's theorem, it's pretty, yeah, it's pretty cool. Every group can be realized as a subgroup of the symmetric group.