 In the previous video, we've learned that there's two types of permutations in SN. There's even permutations, those that can be factored as an even product, that is a product of an even amount of transpositions, two cycles, and odd permutations. Those permutations can be factored as a odd number of transposition factors. And you can't get both, right? We use an argument using the determinant. So we're gonna take the set of A sub N, and we're gonna call this the set of even permutations in SN. The reason why we call A sub N is that this is known as the alternating group on N letter. Now, why do we call it the alternating group? Why not the alternating set? Well, it's because this subset is in fact a subgroup of N. And there's a couple of ways of seeing that. One argument I really like based upon the previous proof, we basically saw that with the language of permutation matrices, the alternating group consists of permutation matrices with a determinant equal to one. Therefore, the alternating group would be the intersection of the permutation group inside of the general linear group with the special linear group. And the intersection two subgroups is a subgroup, right? So there's a lot of stuff we haven't proven about matrix groups yet. So we'll try a more elementary approach again as a subgroup of N. So actually we have two transpositions, S and sigma and tau, excuse me. Well, if these are even permutation sigma, can they be factors, a product of two S mini transpositions? And tau can be factored as a product of two T mini transpositions. So then when we consider the product sigma and tau, well, sigma will give us two S mini transpositions, tau will give us two T mini transpositions. And when you put all of those together, list there, you're gonna get two times S plus T mini transpositions. So the product will then be an even permutation. So we took like an even amount right here and then we concatenate it with an even amount right here. So we had two S, two T. So collectively you're gonna get two times S plus T. There you go. So the alternating group is closed under multiplication. We've also talked about previously that the identity is an even permutation, right? We can multiply it at the product of two transpositions. Personally, I like to think of it as a product. If you don't like that argument, we can also say two, that works great. So what about inverses, right? Well, if you have sigma, if the sigma has a factorization as sigma one, sigma two up to sigma two S, then the inverse of this permutation by the shoe sock principle will be the product of all of these same transpositions in reverse order and you take all of their inverses. This is the shoe sock principle. But transpositions are their own inverses, they're order two. And so you see that sigma inverse will likewise be a product of two S mini transpositions. And therefore, because again transpositions are equal to their own inverses. So this then shows that the alternating group is closed under multiplication, identity and inverses. So about SN. How big is the alternating group? So recall that we've proven previously that SN, we just proved this as a set argument, right? Just a combinatorial argument, even before we had it there, we proved that the size of SN is gonna be in fact, or you'll get in fact, or many permutations. We will see that with the exception of N equals one, when N is greater than equal to two, because when you just take S one, this is just the trivial group itself, which is actually equal to the alternate group. So there's no distinction there. So throughout this argument, N will be great equal to two here. We're gonna show that half of the symmetric group is even, thus the alternating group has its order in fact, twirled divided by two. This is gonna be a nice combinatorial argument, because what we're gonna do is we're gonna do the following. So we're gonna introduce another set B sub N, which is gonna be a set of all odd permutations. Now notice B sub N is not gonna be a sub group of SN because it doesn't contain the identity which is an even permutation, all right? The odd permutations B sub N, what we're gonna do is we're gonna find a function between A sub N, B sub N, and what this is gonna be is the map that sigma will be mapped to sigma times two, cycle one, two. And so what this is called in group theory, this map right here is called the translation map, translation by one, two. Or sometimes they call this the right multiplication map. So we just multiply on the right by this element one, comma two right here. That's all that we're doing here. And so there's gonna be some consequences of that. First of all, right multiplication is, and we're here, so notice that you're going from, as you're going from A N to B N, sigma, the element of the domain is gonna be an even permutation. Notice that this does work out though because if sigma is an even permutation, it has a factorization with an even number of permutations adding transpositions, excuse me, adding one more two-cycle then increase the number of transpositions in the factorization by one, you go from even to odd. So this is going to be an odd permutation. So this is a well-defined map. This map is going to be one to one. All right, how do we see that? Well, suppose there were two permutations, tau and sigma, they're even permutations which have the same image, all right? So these are actually things that are gonna be an A N, we could say that even better. I mean, although this statement's true in general, these are gonna be permutations in A N. Well, if you times both sides on the right by one, two, we could cancel that. This is a group, right? We're in S N. So in a group, you can cancel out the one, two and so this would give you that sigma equals tau. And so this is a one to one map. And so this in general shows us that right multiplication by any element is a one to one map. So if you have any group, right, you wanna function from G to G right here and let's have a specific element X inside of G which then sends G to GX. This map's gonna be one to one, right? Cause if you have GX is equal to HX, then you can cancel out the X's and get that G equals H. Right multiplication, likewise left multiplication is a one to one map, all right? I also wanna argue that it's onto. Why is it onto in this situation? Well, if we take row inside of BN, that means it's an odd permutation. If I add another two cycle to the factorization of row, that would make it an even permutation cause it's two K plus one plus one, so two K plus two. Then, so row one, two is an even permutation. What's the image of row one, two? The image would be row one, two, one, two but one, two cancels I use to end up with row. So this shows you that in fact you're an onto. It's onto function. I wanna show that this is actually something we can kinda do in general. If you have right multiplication, if you wanna land on H, basically you're just gonna start off with G X inverse and then you times it on the right by X, right? Not G, sorry, this here should be a G. So if you take G X inverse times X, that's equal to G. So this right multiplication is an onto map. So I should mention right multiplication and left multiplication is a bijection from a group onto itself. It's gonna be important bijection but what we've done is actually something a little bit more specialized. So multiplication by one, two is a bijection on SN to SN but we've also argued that if you start with an even permutation, then the output will be odd. So we've now shown that this map F is a bijection from A N to B N, this is a bijection. And this is the critical combinatorial fact. Whenever you have a bijection between finite sets, this means that the cardinalities are the same. I mean, it's also true for infinite sets. I mean, that's actually how we essentially define cardinality. Two sets have the same cardinality if there exists a bijection between them. This is extremely, and so whenever you try to make a counting argument, this is something you should consider. Can we find a bijection between two sets because then the two sets will have the same cardinality. But the problem is I want to count the number of elements in A N. If I knew the number of elements of B N, we'd be done but we don't know either of those but we do know that the same. So now we're gonna cover to SN. We're gonna bootstrap off of our knowledge of SN right here. So what we see is the following. We've learned previously that every element of SN is either even or odd but not both. So SN is the union of A N and B N but it is a disjoint union because there's nothing that belongs to both. And so since SN is a disjoint union of A N, the size as a set of SN will be the size of A N plus the cardinality of B N. But as these two sets are the same thing, this becomes two times the cardinality of A N. Divide both sides by two. You then get that half of the symmetric group is the ultraman group. Since the symmetric group has order in factorial, the order of the ultraman group then becomes in fact. So in the remaining minutes of this video, let's look over a specific example. If we look at A four. So this would be the ultraman group of four letters. This should be a group of 12 elements. It should be four factorial divided by two. Four factorial is 24, of course. And I've actually listed the 12 even permutations or 24 permutations in S four. I listed the 12 even permutations. There's the identity. There's gonna be three cycles. Three cycles are in fact are even because the three cycle one, two, three can be factors one, two times two, three. And this is true for all three cycles. So three cycles are going to be even and there's also two, two cycles. That is those functions, those permutations which are written as a product of two disjoint transposition. If you want to count these things, there's only one identity, that's pretty clear. In terms of the two, two cycles, what here is gonna happen is there's gonna be four chews, let's think about this. Basically, if you wanna count the two, two cycles, you have to decide who's gonna go with one because you have a two, two cycle, you have a one, you have a blank and then you have these other guys right here. So you have to decide who goes with the blank. So there's some choice right here of X. But whoever goes with one, then the other two elements, Y and Z, will have to be paired together and the order doesn't matter because it's just a two cycle. So there's gonna give you three options for the two, two cycles. So we see all three of them right here. There's one, two and three, four. There's one, three and two and four and then there's one, four and two, three. So there's three, two cycles right there. In terms of, let's see, in terms of, what do I wanna say next? If you were to count transpositions, right? In S4, the number of transpositions, there are gonna be, as transposition, you just have two elements, A and B, which if we count those, let's see, you gotta pick who goes in the first one, you gotta pick in the second one, but the order doesn't matter. You're counting, excuse me, you're counting the cycles. So you have four total and you have to choose two for a two cycle. That gives you four choose two, which is going to be six. So there's six transpositions that are in S4. So if we kind of keep track of these things, so we have this one identity, we have these three, two, two cycles, kind of keep all of the cycle structures, one identity, if you count two cycles, I'd better do odd ones in like red, two cycles, there's gonna be six of those. If we do two, two cycles, there's three of them like we just counted. The next thing to count would be three cycles, which are even. According to the list right here, it looks like it should be eight. How do we get this eight figure? Basically, you have the following for a two, two cycle. You have to decide who's not going to be in there. So you're gonna get four options for the element who's gonna be fixed. And then of the remaining elements, pick the smallest one to go in the first spot, we get a one, and then you have to decide who comes next. And so there's two options for who comes next, first and last. So you get four times two, so there's gonna be eight, three cycles. Again, like we see here on the screen. Next you can't, I mean three one cycle is just a three cycle. And since there's only four, the only thing left will be four cycles, which four cycles are going to be odd permutations. And so I mean, we could actually try and count what's left, but when it comes to the four cycles, basically you could always put one first, and then who's the next one, who's the next one? You're gonna get three factorial options right there, which is likewise six. So we should get six options for our four cycles. And so if we put these together, you'll notice that six and six gives you 12, which is how many we expected for BN, because it should be the same side as AN. And then if you get one and three, that's four with another eight, that's 12. So that counts all of them. That gives us all 24 right there. Just kind of see how these things broke up by their types. The A4, though, of course, will have the identity, the two two cycles, and then the three cycles there.