 In the previous video, we learned about cycle notation for permutations, and it's a much more preferred way of representing permutations in this abstract adverse setting. Now, products of permutations can easily be computed using this cycle notation, much easier than we were doing it before, with that two row tableaus. So, for example, suppose we have two cycles, say, sigma tau, much like the two examples you see right here. So, we have two cycles, sigma is 1, 3, 5, 2, and tau is 2, 5, 6. So, we can think of these as two cycles that live inside of the group S6. We first can compute sigma of x, or I guess in this case, tau would come first. So, what if we want to compute something like sigma tau of, say, 2? What does that mean in this situation? Well, sigma tau of 2, what you would do is you would say, well, tau sends 2 to 5. So, this would become sigma of 5, and then 5 sends 2, or sigma sends 5 back to 2. And so, what we see here is that sigma tau of 2 would actually fix 2, and that's what's illustrated right here. Tau sends 2 to 5, and sigma sends 5 back to 2. And so, composition, you just read both cycles the way they are, right next to each other, reading right to left, because again, the function on the right will interact with the element first. So, if we're trying to compute this thing, we might be like, okay, what happens to 1? Well, 1 is sent to 1 by tau. So, notice here in tau, you don't see a 1. The fact you don't see 1 means 1 is fixed, tau doesn't do anything to 1. So, tau sends 1 to 1. Then, sigma will send 1 to 3. And so, then the compositum will give that tau sigma, excuse me, sigma tau will send 1 to 3. And so, then we record that right here. So, sigma tau sends 1 to 3. Well, now, what does sigma tau do to 3? Well, tau will send 3 to 3, because again, we don't see 3 here, so 3 is fixed by it. Then, in this situation, sigma sends 3 to 5. So, then the composite will be that sigma tau sends 3 to 5. So, we then record that right here. What happens to 5? Well, now in this situation, tau sends 5 to 6. And sigma doesn't do anything to 6, so it leaves it fixed. So, then the compositum will send 5 to 6 right here. Nextly, tau sends 6 to 2. We see that right here. And then, sigma will send 2 to 1, which we see right there. So, the composite will send 6 to 1, like so. And then, as 1 was the first number that started this cycle, we actually erase it and put a left parenthesis. Like we already computed, 2 will be sent to itself. And the same thing happens to 4, right? Tau doesn't do anything to 4, nor does sigma. So, 4 is left fixed. And so, the product of this 4-cycle by a 3-cycle is in fact a 4-cycle. And so, this is what we do in general, that we just kind of list the elements one by one and say it out loud if you need to. So, for example, if we switch up the order here, take tau times sigma this time. So, tau comes first. If we want to compute the cycle containing 1, we're like, okay, sigma sends 1 to 3. And then, tau sends 3 back to 3, so the composite would be 3. 3 is sent to 5 by sigma. 5 goes to 6 by tau, so 3 is going to go to 6. 6 is left fixed by sigma. Tau sends 6 to 2, so the composite will be, whoops, 2. Next, sigma will send 2 to 1, and tau will do nothing to 1, so we get back a 1. That starts the cycle over again, so we close it off. Then the next number we haven't used yet would be a 4. What happens to 4? 4, sigma doesn't do anything to 4, neither does tau. So, that would just be a 1-cycle, we're just going to erase that. Then the next one would be 5, which admittedly, that's the last element, so it has to be a 1-cycle. But if you go through the argument here, sigma sends 5 to 2 and tau sends 2 to 5. So, that's a 1-cycle, so we're just going to erase it. And we see that tau sigma is likewise a 4-cycle. But one thing to notice here is that this is not the same 4-cycle. Sigma tau is 1, 3, 5, 6, and tau sigma is 1, 3, 6, 2. This is an example that permutation multiplication is non-commutative. We don't anticipate that switching up the order of permutations will give us the same result. It's actually very rare when permutations will commute with each other. Let's do another example of this, a little bit more complicated. So, we talked about multiplying cycles together. But what if we just take general permutations that are not necessarily cycles? Well, what we saw earlier in the previous video is that every permutation can be written as a product of disjoint cycles. So, if we take a 4-cycle like 1, 6, 1, 6, 2, 4, and then a 2, 3-cycle, 1, 3, and 4, 5, 6, then their product will actually be a product of a 4-cycle, a 2-cycle, and a 3-cycle. And we just go through all three pieces together. So, if we ask what happens to 1, moving right to left, the first cycle doesn't do anything to 1, the next one goes to 3. So, this would send 1 to 3, and then the last cycle doesn't do anything to 3. So, the composite would be 1 goes to 3. The next step, well, where does 3 go? Well, the first one doesn't have a 3, the second one will send 3 to 1, and then the last one will send 1 to 6. So, the composite will be that 3 goes to 6. How about the next one? Well, the first cycle again going right to left, 6 goes to 4, this doesn't have a 4, 4 goes to 1. And so, this then closes up the cycle to get a 3-cycle. Next, we'll do the cycle of the 2. This doesn't have a 2, this doesn't have a 2, 2 goes to 4. So, the composite will be a 4. Then we do 4, 4 goes to 5, this doesn't have a 5, this doesn't have a 5. So, we then record a 5. Which then, that should finish it up for us, because I think we've done all six elements. But if you have any doubts, just go through it again. 5 goes to 6, this doesn't have a 6. 6 goes to 2, so that closes up the cycle. So, the product of Sigma Tau right here is a 3-3 cycle, 1-3-6 and 2-4-5 that we see here on the screen. Let's do another example here. Let's do Tau Sigma, so we're just going to switch things around here. So, this time Sigma comes on the right and then Tau comes on the left. We'll compute the product the same way. So, we'll do 1, 1 goes to 6, 6 goes to 4, this doesn't have a 4. So, then we record a 4. 4 goes to 1, this doesn't have a 1. 1 goes to 3, so we end up with a 3. And the next one we're going to get is 3, this doesn't have a 3, this doesn't have a 3. 3 goes to 1, and that closes the cycle. We'll start the next cycle with a 2. 2 goes to 4, 4 goes to 5, this doesn't have a 5. So, we record 5. The next one we're going to have is 5, this doesn't have a 5. 5 goes to 6, this doesn't have a 6. So, we record 6. That's all 6 elements, so that should finish it up. But again, for due diligence here, if we want to finish it up to make sure we made no errors, 6 goes to 2, this doesn't have a 2, this doesn't have a 2. So, then we finish it up. So, we see that Tau Sigma is the 3, 3 cycle, 1, 4, 3 and 2, 5, 6. Notice again, this is not the same element as 1, 3, 6 and 2, 4, 5. I mean, you'll notice that they're both 3, 3 cycles. Which, they have the same cycle structure, but Sigma Tau is not equal to Tau Sigma. Again, this is non-commutative. But this idea of cycle structure, something we'll come back to, that although the elements don't commute with each other, their cycle structure is preserved even when you commute things. Now, I do want to, of course, tell us that just because most permutations don't commute, doesn't mean that there's never commutations. Sometimes they do commute. Take, for example, the permutations Pi, which will be 1, 2, 3. And Rho is the 3 cycle, 4, 5, 6. So, notice here that if we were to do this, if you take Pi times Rho, you get something like this. But what happens when you go through the elements, right? If you go through the elements, you'll end up with, okay, let's see, 1, this doesn't have a 1, 1 goes to 2. So, the product is 1 and 2. Next, 2, let's see, this doesn't have 2, 2 goes to 3. So, we get a 3. Then, the next one, this doesn't have a 3, 3 goes to 1. So, we close off that cycle, then start the next cycle, 4. 4 goes to 5, this doesn't have a 5. Then, 5 goes to 6, this doesn't have a 6. And then, that's the last one, so we close it off as well. Wait a second, when I took the product of these two guys, I just ended up with the exact same thing I had before, interesting. All right, well, what about the other side? I don't need to write that on the screen twice. What about this one right here? So, if we do 1 goes to 2, this doesn't have a 2. Then, we'll do 2 goes to 3, this doesn't have a 3. Then, we get 3 goes to 1, this doesn't have a 1. So, that closes up the cycle, so another 1, 4. This doesn't have a 4, 4 goes to 5. This doesn't have a 5, 5 goes to 6. And, that'll close it up. That's just this guy again, right? So, these 2, 3 cycles actually commute with each other. And, if you're kind of following off the argument there, the main reason that we are able to commute with each other is that these 2 cycles are in fact disjoint. As I was going through the cycle constructions, I kept on being like, oh, this cycle doesn't involve that element, this doesn't use that one, this doesn't use that one. That happened every single time. And so, what we actually see in greater generality is that if you have 2 disjoint cycles in your permutation group, then disjoint cycles always commute with each other. And, this is actually part of the unique factorization of permutations and cycle notation. The ordering of the cycles doesn't matter because they actually commute with each other. And, we're going to prove this fact right here. So, imagine we have 2 cycles. Sigma, let's say it's the cycle containing elements a1, a2, up to ak, it has a length k. Then, say that tau is the cycle b1, b2, up to bl. I do not claim that k and l are the same lengths. These can be different lengths. So, what we want to do is we need to show that sigma tau equals tau sigma. Now, these are permutations and as such, they are functions. How do you show that 2 functions are equal to each other? Well, you pick an arbitrary element of its domain and then argue that the images are the same. So, we have to prove that sigma tau equals x, sorry, sigma tau of x equals tau sigma of x. If you choose x to be an arbitrary element of the domain and argue that the images are equal, this will actually prove that the assignment by the 2 functions is the same. And if the assignments are the same, then the functions are the same because after all, functions is the relationship. So, in this direction, we're going to choose an arbitrary element x and argue that sigma tau and tau sigma do the same thing to x. Now, there's a couple possibilities, right? The first possibility is that x could be an element ai that is involved in the permutation sigma. So, we can still see it on the screen right here. Now, by assumption, sigma and tau are disjoint cycles. So, if x is equal to a1, then x can't equal any of the bj's. Therefore, tau will do nothing to sigma. It'll be a fixed element. And so, then we see that if you take sigma tau of x, well, because x is a1, tau won't do anything to it. So, this will just become sigma of ai. And then, sigma will send that to ai plus one where that number i plus one will reduce mod k. So, like, if i was k, then i plus one, we just mean one in that situation. So, it'll cycle around. Okay, so that's what sigma tau of x is. But on the other hand, if we take tau sigma of x, well, x is ai, which sigma, what does it do to ai? It sends it to ai plus one, but tau doesn't do anything to an a. I mean, it only does stuffs to b, so we end up with ai plus one. So, it does the same thing. If x was equal to ai, well, what if x was equal to bj? Well, if x is equal to bj, again, if the cycles are disjoint, sigma won't do anything to a b, but then tau will just progress bj to the next one, bj plus one, reducing, of course, the index mod l. And so, it's the exact same type of argument. You're gonna get, as we saw just a moment ago. Well, what if x is neither a nor b? Because those are only possibilities because a and b are disjoint. In that case, both sigma will leave x fixed because it's not an a, and tau will also leave x fixed because it's not a b. And so, you end up seeing that sigma tau of x just becomes sigma of x, which is an x, which is the same thing as tau sigma of x, which is a tau of x, which is an x. And so, in all three possibilities, we see that sigma tau of x equals tau sigma of x. Therefore, because the function agrees for every element of the domain, the two functions are equal to each other. This proves that disjoint cycles will commute with each other. And that's not to say those are the only times that permutations commute with each other. Like, for example, if you take one, two, three, and you times that by one, three, two in this situation, these are not disjoint cycles, but I would claim that they commute with each other because what happens is one goes to three, three goes to one, so one is fixed. Two goes to one, one goes to two, so two is fixed. And then in this case, we also have a three goes to two and two goes to three. So this is actually the identity. That's because one, two, three and one, three, two are permutation inverses of each other. But if we switch this thing around, you're gonna see the same thing happening. One goes to two, two goes to one, so one's fixed. Two goes to three, three goes to two, so two is fixed. And then three goes to one, one goes to three, so three is fixed. These are, this is also the identity. So in this situation, these things are not disjoint cycles, they can commute. So there is some commutation in the symmetric group. We can guarantee it for disjoint cycles, but we don't anticipate it in general. But the cycle notation is very, very useful nonetheless in helping us compute the product of permutations.