 Let's have a chat today about one of the very interesting groups types of groups at least and that is the symmetric group Symmetric group And all of us to look at the symmetric group first a few things the first thing we need to look at is just permutations That's a very interesting thing permutations So imagine I have a set My set is called s and it contains three elements one two three alpha beta gamma Apples oranges pears doesn't matter it contains three elements. What are the permutations? The permutations are a Bijection it's a function first of all It's called a function Which means actually it's a mapping and I'm going to map the set onto itself Well, let's say I'm going to map this set to itself and that mapping is a bijection remember those lectures Put them up in the link in the little card at the top click on that eye and you'll you can watch this these lectures on bijections So it's injective and surjective in other words if I have One two three I have my three elements so my set and Again to my set and that means the mappings that I do Each one of these has to be mapped to and they have to be mapped to them only once I can't have one map to one and two map to one and three map to one and They all have to be used and they all have to be unique. So that's a bijection. So how many of these? Bijections can I make and those will be my permutations? We see here that I have three elements and if I use the counting numbers This and in this instance n equals three turns out I will have three factorial In other words if they were in I would add in factorial of these mappings So I would have in factorial of these mappings now Let's do a few of them and the first notation that we're going to have I can I hope you see why you will have six Let maybe I'll get this to array notation She's we called a ray notation and that is where I just write one two three And I see what what what does one map to well one can map to one two can map to two three can map to three and That would be one permutation and usually we call this permutation I That's usually what it's called this permutation I so that's not a matrix a system mapping in a ray notation one maps to one Two maps to two three maps to three the second one is usually called a Transcription and therefore they the book shows towel and that's one two So that's where we have one two three and one maps one maps to two two maps to one and three maps to three and Because you know, there's just these two that swap around when there's usually when there's just two that swap around It's this transposition We have another one called towel one three usually that's one two three and that's where one maps to three and three maps to one and we leave two alone and then there's Two three and that's where we have one two and three and that's usually where we leave one alone and two turns to three and three turns to two and then we have one called sigma one and two two three and three and That's where One maps to two two maps to three and three maps back to one And then if I repeat this one twice I have the last one already have one two three four five So I need one more six and that is just where we have Two three one is that right? Let me just check If we have sigma squared if we do this one twice and I have one two three That's the last one that is left. So let's do this one. Let's just do this one twice So let's imagine that I have one two three I have one two three and I have one two three So what I'm doing here is I'm saying sigma Sigma of X and I am combining these two functions I'm combining these two functions. So let's just do sigma sigma says one goes to two two goes to three and Three goes all the way back to one and I'm do that again one goes to two Two goes to three and three goes all the way back to one. So what do we have left? Let's see where one two and three goes One two and three so one goes to two and two goes to three so one goes to three Two goes to three and three goes to one so three goes to One that can't be one two three one two three one two three one goes to two Two goes to three so one goes to three two goes to three So sorry two goes to three and three goes to one in other words two goes to one There we go. 3 goes to 1 and 1 goes to 3. In other words, 3 goes to 1 and 1 goes to 2. In other words, 3 goes to 2. 3 goes to 2. So there we have it. 1, 2, 3. So 1 goes to 3. And 1 goes to 3. 2 goes to 1. 2 goes to 1. 1 goes to 3. 3 goes to 2. And 2 goes to 1. There we go. That took me a while, didn't it? Anyway, so there we have sigma and sigma squared. So if you look at it, those are the six permutations that we have. So that's a ray notation. And we've got sigma squared through the composition of functions. And that's going to become important. And I just want to show you another notation to write this. And that is just the cycle notation. And the cycles that I'm going to for, so that was a ray notation. This is a cycle notation. And what I want to do here with cycle notation is just to write these cycles. 1 goes to 1 and then it just cycles to 1. So that stays on its own. 2 goes to 2 and it just cycles to 2 and 3 goes to 3 and it just cycles to 3. So it's not very impressive there. This one is those. 1 goes to 2. 2 goes back to 1. 1 goes to 2. 2 goes back to 1. So that cycle just carries on there. 1 goes to 2. 2 goes to 1. 1 goes to 2. And just carry on. And then 3 just goes to itself. So there's nothing there. So tau 1, 2 is usually written like that. Tau 1, 3, remember we have 1 goes to 3. But 3 goes back to 1. 1 goes to 3. 3 goes back to 1. It just cycles around there. And then we have 2 all the way out there. Here we have 1 just cycles to 1 all the way around. But 2 goes to 3. And 3 goes to 2. So that way around. This one is 1 goes to 2. 2 goes to 3. 3 cycles back to 1. 1 goes to 2. So it's just that I don't have this redundancy here. And then lastly I have 1 goes to 3. 3 goes to 2. 2 goes to 1. So I'm back at 1. 1 goes to 3. 3 goes to 2. 2 goes to 1. So that is the cycle notation that we do have. So we have this array notation. And we have cycle notation. So I'm going to leave those there. So that's all our permutations. So let's suggest that there's a group in here somewhere. There's a group in here somewhere. And it's called the symmetric group. So I'm going to suggest that I have this group. And it's going to have a set. And my set is going to be this set. Let's call this set S on three elements. And I have my group operation. Where S3 equals this I, tau12, tau13, tau23, sigma and sigma squared. And my binary operation is this composition of any of these two. So we usually write that. In other words, I might say what is I and tau13? Composition of those two mappings. And remember the way that we do that is we do this one first and then that one. So it will sort of be I of the tau13 of X. And X is any one of these, one, two, three. So I would say get the tau13 first of each of these. And then get the tau of that. So let's just do one of those. So we're going to have the I of tau13 of one. And that's going to equal something. We're going to have the I of tau13 of two. And we're going to have the I of tau13 of three. So we do that side first and then this side. So let's have a look at what that is. So tau13, so this is going to be I of something now. I of something now. And I of something now. Let's do that. And that something is the tau of one three. Well, the tau of one three, one goes to three. And that's where that cycle notation works very well. So that's the I of three. tau13 of, what is the tau13 of two? If we look at one three of two, two just stays on its own. So that's of two. And the tau13 of three, three goes to one. So that's just one. And the I of three, well, that's just three. And that's just two. And that's just one. So we have that this composition is of this is, I'm just left with tau13. I'm just left with, if I have this composition of functions that is just going to be, it seems like this I does serve in this instance as, this I serves as an identity function. Okay. And it is one three because remember one went to three, two stayed with two, and three went to one. Which is exactly tau13. So remember the composition of these methods. So that's what I have. So my binary operation is the composition of functions. I have tau12, I have tau13, I have tau23, I have sigma, and I have sigma squared. I have I, I have tau12, tau13, tau23, sigma, and sigma squared. And it is for you to fill this in. Now these ones are going to be the easy I and I is that. That's going to be tau12, tau13, tau23, sigma, sigma squared. And this is going to be tau12, tau13, tau23, sigma, and sigma squared. And we have to fill in all the rest of these 36 in total. So maybe we should just do, do a few of them. And then at your leisure, you can do the rest. But I'll do at the end is I'll just fill it in and I'll skip in the video to that so that you can see that you get the same, same solution as what I, as what I get. So let's do, let's do tau12 and tau12. So that's tau. Let's get another piece of chalk here, tau12 with tau12. So I'm going to do that one first and then that one, but they are the same so it doesn't really matter. So 1, 2, 3, 1, 2, 3, 1, 2, 3. So let's do tau12. First one goes to 2, 2 goes to 1, and 3 goes to 3. And then I do the 1. 1 goes to 2, 2 goes to 1, and 3 goes to 3. So in the end what am I left with? 1, 2, and 3, and 1, 2, and 3. So 1 goes to 2, goes to 1. So 1 goes to 1. I have 2 goes to 1, and 1 goes to 2. So 2 goes to 2. And I have 3 goes to 3, goes to 3. So I'm actually left with the identity here. The tau12 composed of tau12 is the identity. It's our first i. And I want to call it the identity because we have to look on our table in the end if we want it to be a group. Remember we have to have our famous four properties. Closure, associativity, an identity element, and an inverse for everyone. And an inverse for everyone really means that if I look down any row, each one of these have to be unique. Any column, if I have to look down any row, each element has to be unique as well. And then we'll get those inverses. So let's do one more. So from these two, we've got i. In other words, tau12 is its own inverse. Let's just do another one. Let's do tau13 composed with tau13. Let's compose that with tau12. So we do this one first. 1, 2, 3, 1, 2, 3, 1, 2, 3. You can also do it just by doing it with the cycle notation. So tau12 takes 1 to 2, 1 to 2, 2 to 1, and 3 to 3. tau13 takes 1 to 3, it takes 3 to 1, and at least 2 to 2. So let's see what we have left. 1, 2, 3, 1, 2, 3. So 1 is going to 2, 2 goes to 2, so 1 goes to 2. 2 goes to 1, and 1 goes to 3. In other words, 2 goes to 3. 2 goes to 3. And we have 3 goes to 3, 3 goes to 1, so 3 goes to 1. So I have this 1 goes to 2, 2 goes to 3, 3 goes back to 1, and that's just sigma. And that is just sigma in there. So I'm going to fill in the rest, pause the video, and see if you get exactly the same. And what we have here is this group of permutations on this set. The set with 3 elements gives us this set of permutations, all of the permutations on the set with 3 elements to combine with the binary operation of this composition of functions. It's going to give me a group, and you can check if all 4 properties hold for this, and indeed it does. So the symmetric group on n elements is going to be, well, it is a group. Let me fill this in, and in the end I'll show you. So there you go, I've completed this table, and we can clearly see that we do have our identity element, e, which is just going to be our i permutation. And it seems that everyone does have an inverse because the composition of any one of these is eventually going to give me the identity element. So if I have any 2 of the permutations, permutation 1 and permutation 2, I'm going to get e, every one of them, well, every one of them will have its inverse. So if I look at tau 23 times tau 23, that's its own inverse. But when I get to sigma, sigma's inverse is sigma squared, and sigma squared's inverse is sigma. And that gives me, so they are the inverses of each other. So check the associativity, there's inverses, there's the unique identity element, and this closure, all of these do belong to my original set of permutations. So we have a symmetric group.