 Welcome back to our lecture series, Math 42-20, Abstract Algebra 1 for students at Southern Utah University. As usual, I'll be a professor today, Dr. Andrew Missildine. In lecture 16, we're going to continue on the topics of permutation groups that we were talking about in lecture 15, coming from section 5.1, permutation groups from Tom Judson's abstract algebra textbook. Previously, we've been talking about the permutation group S sub X, where X is some arbitrary set here. And so, one thing we're just going to simplify here is we're just going to focus on the set S sub N. We're in this situation, we mean that X, we just want to think of it as the set one, two, three, all the way up to N. I kind of mentioned this in the previous video that there's really not a lot of benefit of thinking about these things for some general set. It's really just the cardinality of the set that matters. And so we're just going to focus on this simple set of integers one through N. So we're going to talk about the symmetric group S N, but be aware that the stuff we're talking about here really generalizes to any permutation group. So in this lecture, we want to focus on the idea of a transposition. A transposition is a permutation of length two, that is, it's just a two-cycle. So you have something like one, two, or three, four, or, you know, one, six, you know, whatever it is, okay? And one thing I should mention is that when people write permutations, we often don't write commas between the elements. Sometimes we do, of course, when it's really necessary because as these things are just letters, we can just put them next to each other without any comma whatsoever. And I say letters, I mean, it's like a character, right? Inside the set X. Of course, when they're numbers, we do realize that when the number's starting to get next to each other, like if I write one, two, three here, it's quite clear. It's like, okay, one goes to two, two goes to three, three goes back to one, but what happens when you start getting two digit numbers so you can be able to confusing. So use commas or spaces when necessary, but it's very common for people to omit either spaces or commas with permutations because it's often clear what's going on here. So a transposition is just a length to permutation. It's a two cycle. Now, what's such a big deal about a two cycle? Well, our first theorem of this lecture, theorem 519 here tells us that any permutation can actually be decomposed as a product of transposition. So this is way of representing permutations. We saw previously that every permutation can be written up to reordering into a unique cycle decomposition. So we're doing something similar here with transpositions. And the idea of transpositions is gonna be the following, that every permutation can be constructed by just reordering two elements at a time. So if, for example, you have this massive movie collection, right, you have all your DVDs lined up in a row on a shelf, well, what happens when a child comes over and messes up that order? It's like, oh no, I had all of the Marvel movies in order of when they're supposed to happen. What's gonna happen now? Well, what you can do is they're all mixed up, right? You just cannot put things, okay, who's supposed to go in the first slot? Oh, Iron Man was the first one. So who's in the first one right now? Spider-Man, a homecoming. No, that's not the first one. So you swap those two movies, right? Then you look at the next one. What is the next one? The Incredible Hulk? I think, who even noticed that one? And so you swap that. You can swap who's in the second spot with who's supposed to be in the second spot, whoop. And then who's in the third spot compared to who's supposed to be in the third spot? You whoop, you swap those. You can always do one transposition at a time going through the list. So that's kind of the idea here, but let's give a more formal argument. So let's first convince ourselves that the identity is, has a product transpositions. I should also mention that for the rest of this chapter, we're also gonna have the assumption that N is greater than or equal to two because there's really no reason to think about the symmetric group, S1. Because S1, you're set X in that situation, it's just a single element. There's only one thing inside of that group and it's the identity map itself. So which is just the trivial group. So we're not gonna talk about the symmetric group on one letter. So for all of these proofs, you can assume, if I never say it, that we'll have at least two letters in play here. So with that, the identity can be written as a product of transpositions because it's a product of A1, A2, which is a transposition times A1, A2, which is a transposition. So hey, that's a product of transpositions, which admittedly you could also describe the identity as the product of an empty product of transposition, so it's a product of no transpositions. But if you don't feel, if you don't like sort of like these vacuous arguments, then sure, the identity permutation is a product of two transpositions. It has a factorization. So the next thing to then consider is that let's suppose that sigma is some non-trivial, non-identity permutation. So since sigma can be expressed as a product of disjoint cycles, if we can write a cycle as a product of transpositions and sigma as a product of cycles, then it would become a product of transposition. So without the loss of generality, we may assume that sigma is a cycle, let's say A1, A2, A3 all the way up to AK, so it's a K cycle, right? Now we can actually convert this into a product of transpositions in the following way. A1, A2, A3 up to AK, this can be factored as A1, AK, all the way down to A1, A4, A1, A3, and then A1, A2. So the idea here is the following because we work right to left in this situation. What we're gonna say is, okay, we're gonna swap A1 and with A2 where A2 is the second element in the list here. Then we're gonna do A1, A3, A1, A4 all the way down to A1, AK. So we kind of move right to left down this list where every two cycle involves an A1. All right, now that's not exactly the way I talked about organizing my Marvel movies from before. Well, that's fine. Basically you look at, the way you wanna think of it, this one is the following. You look at the first two elements. Are they in the right order if not swap them? Then you kinda look at the next one, the first and the third one, are they in the right order swap them if they are? You can kinda talk about that way. So let's give you an example of how to actually do this type of factorization. So consider the permutation one, six and two, five, three. So this is a product of two disjoint cycles. One, six is already transposition, so we don't need to factor it. So let's just worry about this three cycle right here. Using the formula we had right here, two is the first element, so all of the factors, all the cycles, two cycles will have a two in it. But then going right to left, we're going to get, so this one, let me say that again, so all of them will evolve too. So the cycle goes, you read it left to right, but we're gonna put the factors right to left right here. So this one right here will be the first one, so we're gonna get two, five, and then we're gonna get a two, three. So that's how the factorization works there. Just as another example, if you had like one, two, three, we could factor this as, according to this rule, we're gonna go one, two, and then we also get one, three, like so. But I want you to also be aware that this factorization is not unique. We could also factor this as one, two, times two, three, and I'll let you verify that that's actually the case. And how does one get this one? Well, that's because you could actually think of this as two, one, times two, three, and so if you've wrote the cycle instead as two, three, one, then this procedure right here would actually give you the factorization two, one, times two, three, which then, of course, you can reorder these things to get this, you know, so this right here, this is two different factorizations, mind you, of the same thing. And I could produce another one, right? If we wrote this as three, one, two, then I would tell you to do the factorization three, one, and three, two, which we might write as two, three, times one, three, or something. Or we could write this as two, three, times three, one. There's a lot of different ways of writing these things, but essentially there's three different factorizations here. This one, this one, and this one. And I don't claim that's all of them, but the point is there are different ways of factoring a cycle into a product of transpositions. This factorization is not unique. Like the cycle factorization we talked about before, the transposition factorization is not unique. It does exist though. And another issue that you have to also worry about would be something like the following, right? Sigma, we could factor it as three transpositions, okay? And again, here's like a different factorization for the same thing. We could take two, five, three, five, like we did, excuse me, we could do, no, these are two different ones, right? So basically playing the same game I did over here, right? We could factor the two, five, three in a couple of different ways. But then we could also insert other transpositions, like this is a legit factorization, which you can see that since four or five has just joined from everything else, we could commute these around, they'll cancel out. My point here is that the factorization of transpositions is a non-unique process. We can get lots of different ways of factoring a permutation into transpositions. But despite that these factorizations are non-unique, there is some type of uniqueness statement about it. And this is what we're going to talk about in the next video. We can't guarantee we have the exact same transpositions. We cannot guarantee we have even the same number. So what can we guarantee? We can guarantee that these factorizations have the same parity.