 Subtraction might or might not be commutative depends on the underlying set meaning you can't sort of give a general description as to why subtraction Isn't Commutative by simply saying it's not show me that it's not by handing me two integers You know two and three or seven and eighteen. I don't really care and convince me that if you do seven minus 18 You know the same as eighteen minus seven Now if that seems a little bit weak, I mean a good example might be convince me that Matrix multiplication is not commutative Folks sometimes matrix multiplication is commutative depends on the underlying set Heck if you look at those four matrices that we've looked at inside the two by two matrices over the reels the ones with Plus or minus one on the main diagonal and zeros everywhere else There's a set of matrices and multiplication on that set is commutative So if you're going to convince me matrix multiplication isn't commutative in general They actually have to write down some specific examples where things fail Okay, the second comment on that first homework assignment Yeah, what some of you failed to do both in this assignment and on the second assignment was keep your eye on the prize You're handed a set and an operation and you're asked whether or not the set together with the operation forms a group Or maybe forms a subgroup of the given group and in Most of the situations what you're asked to do is show that whatever it is that describes the set in question Maybe the set is the set of even numbers Maybe the set is the set of upper triangular end-by-end matrices having no non-zero entries on the main diagonal Maybe the set is the collection of functions that pass through the point one zero with blah blah whatever the set is The prizes are always Convince me that if you start with two things in the set and you combine them that the results back in the set That's closure convince me that whatever the identity element is is in the set in other words How's the right property convince me that if you start with something in the set? Ie has the right property that it's inverse also has the right property and What some of you were doing was sort of either not fighting the battle at all or sort of fighting the battle halfway For example, this goes for both question The the even integer question and it also goes for the upper triangular no non-zero Values on the main diagram and it also goes for the functions question. You'll say something like pick two things in the set Maybe the set is the even numbers show that when you add them together the results even okay, that's fine But why why is the sum of two even integers even some you just said well the set is closed because the even numbers are closed It doesn't tell me anything why? Because if you take two even numbers one of them looks like two times Acts and one looks like two times why that's what it means to be an even number It can write it as a multiple of two and if you add them together you get two x plus two y which is two times the quantity x plus y Which is of the correct form. It's two times an integer check I say well you're being you know you're being too picky. I'm not being too picky heck I mean if the set was the set of odd integers Why is that any different than the even integers? Well, that's that's not closed Similarly a lot of you just said well the identity element is in the set to z Yeah, it is but why? Not that just you have to check that it is you have to convince me that it is Why is zero an even number because you can write zero as two times an integer happens to be that you can write zero It's two times zero Why can you write the negative of an even integer as an even integer? because negative two times an integer is Two times negative that integer which isn't the correct form. So keep your eye on the form More words about that so for example in the second assignment when you were asked to show that the set of n by n upper triangular matrices having no zeros on the main diagonal is a group What some of you said was well if I take two things in that set and I multiply them I get another n by n matrix I'm done You're not done. Okay, you get another n by matrix. That's good But you have to make sure that the result is in the set you have to make sure that the result is an n by n matrix That's upper triangular because that's what means to be in the set and that has no non-zero entries on the main diagonal You have to check everything You have to check that the identity matrix not just exists of course it exists You have to convince me the identity matrix is a n by n matrix That's upper triangular that has no non-zero entries on the main diagonal. Well, yeah It is because it has all ones on the main diagonal. There's nothing below the main so I'm not sure how to reinforce the The the necessity of when you're showing something as a group always looking back explicitly what the property is that put something in the set or in the subset and Making sure that each of the three things has the right property either the product two things in the set or the identity element of The group or the inverse of something that's already known in the set and I don't know Maybe you might want to just in big bold letters. Here's what this set consists of three properties n by n upper triangular No, non-zero is on the main diagonal and then in each step you got to convince yourself. That's what's gonna happen Some remarks about the function question, I think this was numbers 15 and 16 in section section four or section five I forget number 16 and Number 15. Let's look at number 16. That was a little bit more interesting section five You're looking at this thing called f tilde is what it's the F from the reels to the reels where f of x not equal to zero for all x So that's what f tilde was and then you're looking at a subset of it I forget what the name of the subset was let's just call it s s is the elements in f tilde G in F tilde with the property that G of one one and I do you I get the number right I think so Okay, and the request is to show or disprove that When you look at this subset that you get a subgroup of this where the operation here is multiplication Let's start first by looking at what the heck these things are the elements of the set are elements of this bigger set of this thing that you're told is a group The things in the big set are functions They're functions with the property that the domain of the function is the reels What comes out is in the reels and you're told that any function in the set has the property that Regardless of what value you plug in you never get zero out, okay? And then what you're asked to do is look at the subset of those things Well, they're again functions with the property that what that when you happen to plug in the number one That the value one comes out and so what I asked you to do as a first Step here is give me some sort of pictorial representation of what the elements of this set look like And let's see did they actually give a name to this set or they just said here's a set, okay? Let's just call it ass. It'll be a little so folks. What does it mean? What does it mean to say that you've got a function where when you plug in one that one comes out? And that the thing goes through on one So can you draw some pictures of some of those sure here's one that's a good one and here's one That's a good one Well, I got to make sure though that f of x is not zero so I got to make sure that whatever I draw here doesn't cross the x-axis For what's worth a couple of you said well therefore it has to be a positive function Well, you're not told that the functions you're interested in are continuous simply that they're zero so technically here's a Valid choice and there's an example goes through There's a lot of other choices Heck here's one in fact let me Put this in a little more boldly here's a function That goes through One one okay now question is this subset actually subgroup Well, what do you got to do you got to make sure that when you take two things in here? And you do whatever the operation is that you get a result that's in there So what do you have to do you have to take two things in there? What are the things in this set? They're functions. So you have to take two functions. What do you want to call them? How about f1 and f2? So pick two functions in the set show that When you do the operation and the operation is simply function multiplication What does it mean to be an s? It means that When you plug in one that you get one out. So how do you do that plug in one? What do you get well folks? What does it mean to multiply two functions good? It means you just plug it in definition of functions f1 times f2 What does it mean to multiply you know x squared times sine x? It means you take the x squared function multiply by sine x function And if you're handed an input you drop it into each one you multiply. Oh, but let's see What does it mean to say f1 is an s? It means that first thing is one and second thing is one and you get one So look I've plugged one into the product and I got one out so check closure I mean I'd ask for another sentence here telling me what you just did, but that's the computation you need to do second Is the identity element in the subset? Well, what is the identity element under function multiplication? Here's the identity element. I just drew it some of you said that the identity element is the number one Identity on it's not the number one. It's the function e Having the property that e of x is one for all x That's the identity Because this is the function that has the property that when you combine it with any other function It doesn't change the other function. So there's the identity function Question is it in the subset? Happens to be in the subset because it happens to go through the right point or rephrase It happens to be in the subset because when you happen to plug in the particular input one then one comes out One comes out for everything But you've got to convince me that this function is in the subset you have to convince me that when you plug One in that one comes out well it does okay, so you have one this one check and The last one is all right If you hand me an element its inverse exists why? That's where this condition gets used folks If I hand you a function and it would have gone through the x-axis here Then there would have been in place where the output was zero and you wouldn't have been able to write down an inverse function for it so the fact that the inverse exists is is The consequence of the hypothesis on the functions that we're looking at in here But then once you've got that it turns out then that f inverse is just one over f of x because The operation here is multiplication now plug one into the inverse one over f of So that's what's going on and it's again just a matter of keeping the iron your eyes on the prize Figuring out what the heck the things are that you're working with and then Secondly deciding what property those things have to have in order that they actually land in the corresponding set or subset last comment on the homework and No, two last comments. They're both stylistic What a lot of you're doing and Quite honestly, I see this a lot for from students Early in the semester the mouth for 14 course is you're writing down lots of facts most of which are true But a lot of which are completely not germane to what you're trying to do And if you say well, I haven't told you anything wrong Yeah, I agree with that, but you know, it's like writing an essay about George Washington and just throwing stuff in about John Kennedy Well, okay, you might be telling me two facts, but it's just not of interest to me So please stylistically go back through your proofs and make sure that all the statements that you're putting in are Really of interest on some of your homework. I wrote, you know, not of interest or not germane or so what or something like that Okay, it's true. I can't put an X through it because you haven't told me anything wrong But you just you sort of started randomly pulling facts out and that's typically not a good thing to do Second stylistic comment is a couple of you wanted to do some proofs by contradiction here Yeah, okay, that's fine For instance in the one show that there's only one item potent in a group or something I Would suggest stylistically any time you do a proof by contradiction read back through your proof and ask whether or not it could have Been done just as easily by a direct proof some of you for instance for that Item potent question said suppose there were two item potents. I Don't know F and G or something like that show that F equals G Or suppose that there's two distinct item potents Let's get to a contradiction and in the end what you wanted to proving basically is that if you have an item potent It is the identity which is the goal Rather than a proof by contradiction. So it's just again that I offered that as a stylistic comment And finally this is more than a stylistic comment on The question that asked if you start with an abiding group and you look at the elements inside the group That have the property that when you square you get the identity 47 was Some of you got sucked into the trap and the trap is that you prove that result Without using one of the hypotheses of the question Question is if G isn't a billion group show that blah blah blah blah blah And I I can't Strongly enough suggest that once you're done writing your proof down you should always go back through your proof and make sure that Somewhere in your proof you've used every one of the hypotheses in the question that you've been handed Because some of you wrote down proofs and never used the fact that the group was a billion and folks It turns out that the statement is false if you're looking inside a non-abiding group Huh, so if you've proved something and you haven't used all the hypotheses in the given question I mean maybe you've found the hidden secret to this question or something or maybe you've done something incorrect and Unfortunately the latter is probably what happened here All right So again, I'll try to post homework Solutions by tomorrow, but definitely by Friday and we can move on from there. All right all right, so The folks if you have any strong comments one way or the other about having filled in Monday's class by doing a Pre-recorded video. Let me know if you really liked it. I might do it again once or twice in the semester I don't really like to do it on a regular basis, but if it works for you then You know if there's a a day where I have to maybe be out of town again or something I can do that or if you really hated it and said, you know my machine gagged on it I couldn't find time to do it or whatever it was. I just fine Wow, maybe not Okay, so here is what I'll say we did on Monday Here's what I did on Sunday and you viewed between Sunday and now What we're gonna look at is a new type of groups. So new groups These are what are called permutation groups Permutation groups the notation is A little bit better At least I got some today The idea is hand me a set and it doesn't really matter what the underlying set is so that we typically choose the set and denote it at the numbers one through n then s sub n is The set but we showed in the video on Monday is the group of permutations of The underlying set one two up through we showed it's a group We actually noted just by a discrete math result how many elements are in this group There's always n factorial elements. So what we've now done is we've written down a group that has Let's see if n is what then is one Sort of a an interesting group. So we'll typically assume n is bigger than or equal to two Even n equals to give sort of an uninteresting group if you hand me The permutations of a set with two elements while there's only the identity element and the switch So there's only two elements in there and that's sort of not too interesting But once n is bigger than or equal to three these become really interesting For instance s sub three The permutation group with three elements or on a set of three elements has six Members in that group because three factorial six if you look at the group of permutations on the set one through four You get a group of 24 elements if you look at the group of permutations of the set one through five You get 120 elements, etc. Each of these folks are different groups But they sort of are part of the same family So we typically define all of them at once rather than saying here's what s3 is and here's what s4 is and here's what s5 is etc So what this has done for us is For the first time allowed us to write down finite groups groups with finitely many elements that are not Abelian because what we showed in the video is that if n is bigger than or equal to three that you can always find at least two Permutations so that if you do the permutations in one order and then you switch through order that the outcomes are different All right, what we did at the end of that video was to start writing down some ideas Corresponding to permutations and I'll remind you of those briefly and then We'll look at some concepts associated with these permutations We'll look at certain types of subgroups of these permutation groups and and then if we have time at the end talk about some actually relatively deep results regarding some specific subgroups of these Permutation groups, okay, so the idea was if you pick something in here and the Typical notation for permutations are Greek letters Typically the letters sigma or tau or kappa are used so if I pick something in oh Let me make up a sort of easier example start from scratch or make another Maybe sigma is well remember how we can describe what permutations are We simply list out what all the possible input values are and because I've asked you to look at s sub seven Maybe this particular permutation looks like this one one goes to two four one three Five seven six So this means for example sigma of one is two sigma of three is three sigma of six is seven, etc And what we were looking at at the end of Monday at the end of the video was the notion of the orbit In the intuition is if you hand me a permutation You can think about the underlying elements of the set here one through seven is being sort of chunked into subsets And the way you chunk them into subsets is by simply starting with whatever element You'd like in the underlying set and sort of following through what happens to that element under the permutation So that here for example if I start with one one When you run it through the permutation sigma becomes two in turn two becomes four and In turn four becomes one and once you get back to where you started We're going to call this a cycle or an orbit Corresponding to the element one So we get this Subset one two and four notice doesn't really matter where you start if you had started with two you could have played the same game Two goes to four four goes to one one goes back to two. So once you've got this it sort of cycles through What other orbits do we have let's see three? Well three stays at three so that's sort of uninteresting Four oh, I've already analyzed what happens to four it goes to one and then to two five stays at five That's uninteresting six goes to seven which in turn goes back to six. So here's another orbit six now technically The number three itself is its own orbit because it doesn't get moved by sigma So we're not going to really be interested in the effect of sigma on three just because there is no effect of sigma on three It stays at three now. Here's the idea turns out proposition Now let me give you definition first definition definition a Cycle is a permutation special type of permutation having just one only one Non-trivial orbit Yeah, this is a trivial orbit just three sitting by itself. It's no big deal five sitting by itself So let's see is sigma a cycle no because sigma somehow consists of two non-trivial orbits It has this orbit and this orbit associated with it let me give you an example of a permutation that is a Cycle example tau is And let me indicate where it is in s7 Tau is this permutation one two two four three three four one five five Here's another I haven't indicated that this is a cycle yet, but you'll see in a minute why it is kappa is the permutation of the Numbers from one to seven that you get by doing one to one two to two three to three four to four four to four five to five Six to seven and seven to six all right. Let's see. Let's write out the orbits of these two Permutations well for tau one goes to two two goes to four and four goes back to one So I've found one orbit it involves one two and four Have a look at everything else just stays fixed if I don't look at one two and four and look at everything else 356 and seven remain so the other orbits are trivial so this particular permutation has only one non-trivial orbit So it is what we call a cycle So how is the cycle? Similarly, let's see. What can I say about kappa one two three four and five stay the same So the only orbit here is six to seven back to six So here's the only orbit non-trivial orbit of kappa and kappa is Then a cycle cycle and hopefully it's somewhat clear what I've just done I've in effect taken this thing called sigma which isn't a cycle and I've sort of parsed it into two pieces The piece that corresponds to this first orbit where I do what the orbit Corresponding to sigma did but then I leave everything else alone six to six and seven seven and similarly I then take the other orbit. I do the action or whatever it was that sigma was doing on those two things and leave everything else alone note For the three permutations that I just wrote down Sigma is Tao Circle kappa Why is that? Well, what I'm claiming is regardless of which of the numbers one through seven I run through the function I can either just run it through sigma directly or I could run it through first kappa then tau Well, yeah, if I hand you the number one for example Running the number one through kappa doesn't change it and then it does whatever tau does which of course is the same thing That sigma did to it. How about two? Yeah, if I run two through kappa doesn't change it But then two goes to four Three no three doesn't get changed at all four five Let's try six if I run six through kappa six goes to seven But then seven when I run it through tau remains fixed so six goes to seven. That's exactly what sigma did to six took six to seven So here's gonna be the idea if you hand me a permutation and you break it down in its orbits this way And for each orbit you simply do the permutation that corresponds to the cycle that gets generated Then you can write any permutation as a composition of cycles Now let me give you a little bit more verb use before I write that down. It turns out in this particular case For these two particular permutations. I'm certainly not claiming that I can do this in general It turns out you can switch the order and the reason that you can switch the order is the things that tau moves One two and four and the things that kappa moves six and seven are disjoint So what we typically do is we call these disjoint cycles and I'll write this down in a minute The statement is that in general cycles don't commute in other words if I hand you a cycle and I hand you a cycle And I ask you to compose them doing them in this order might be completely different than if they do in this order But if you have cycles that are disjoint where the things that they're affecting are disjoint in the underlying subset Then they will come so here then is the Results I won't prove it for you, but the punchline is that in effect, I've given you enough information in this example to Allow you to piece things apart appropriately Proposition proposition one Any to disjoint disjoint cycles commute meaning if you hand me any two cycles and The underlying set elements that are affected by those cycles are disjoint As they were here then in fact doesn't matter which order you write them Be careful here folks. This does not say that any two cycles commute It says that any two disjoint cycles commute and I'll show you in a minute a situation where Even if I give you cycles if they're not disjoint, then they might actually Disjoint Richard means this that when you look at the orbits that they affect if you look at the one non-trivial orbit Let's see the things that were included in the one non-trivial orbit here are the elements one two and four The things that were included in the non-trivial element here are six and seven and these two sets of no overlap The intersection of these two sets is empty. That's that's the word disjoint Okay Yes, so the word disjoint means that the intersection of the orbit elements here and the orbit elements here is empty Or there's nothing that appears here that appears down here. That's uh-huh and second proposition is that any element of SN in other words any permutation Can be written as And I'm going to hand you this notation. You might think it's sloppy to begin with but it's totally standard as the product of Disjoint cycles and I said where the heck are you multiplying? Well when I use the word product this in this context means composition and as we've seen already a few times Often when we're talking about General groups. I don't care if they're matrices or functions or permutations or you know integers When we use the general star notation for the binary operation typically that sort of degenerates to dot notation or the standard notation that we Typically associate with multiplication But here the quote-unquote product is inside the group SN. So here it actually corresponds to composition I'm certainly not asking to multiply integers or anything like that But we typically talk about the product of two permutations And we mean there then whatever the appropriate group operation here and happens to be composition All right, I'm going to leave out the proof, but you see exactly how to proceed If I hand you a permutation simply tell me what all of its orbits are For each of the orbits construct a cycle Where all the cycle does is affect whatever Corresponds to the elements in each of the individual orbits So if I know that I have an orbit in sigma that sends one to two to four and then back to one Then simply do one to two two to four four back to one and leave everything else constant If in this second orbit of the original permutation it switches six and seven That's fine Then simply write down the permutation of all of SN that switches the two things in question Leave me everything else blank and in that way Hey, even if you had more orbits corresponding to if you had three orbits or ten orbits Doesn't really matter. You'd be able to write the original permutation as a product of disjoint cycles Alright Now things get a little bit hairier you have to be really careful when you're reading these things again This does not say that any two cycles commute It only says that you're guaranteed that if the two cycles happen to be disjoint Then they commute. I'm not proving that for you But intuitively it's not too bad to see because if they're disjoint and you run it through one and then you run it through the second one If there was an effect by the first one then there's no effect by the second one And whether or not you do the no effect first or the no effect second is irrelevant. Okay This notation for permutations is a little bit cumbersome Yeah, you keep having to write out whatever the underlying set is here one through seven And then you write out what the output for each of the inputs is But it turns out in the specific situation where I've handed you a cycle Rather than just a general permutation in the specific situation where the entire permutation Simply tells or is simply described by telling me what happens to one element and following it through that there's a significantly shorter notation that we can use for those types of permutations and This is usually referred to as cycle notation Notation and all it is folks. It's nothing new. It's just shorthand notation for permutations which happen to be cycles cycles And hey, it's it's really easy to figure out what the notation should be Just tell me what the guts of the permutation is and just conclude or understand that if you haven't mentioned an underlying element of the set that it doesn't get moved So this permutation that I wrote down here tau The permutation that takes one to two two to four and four back to one and leaves everything else the same We simply denote by put a left print one to two two to four four back to one Just write that so the interpretation is this permutation takes the number one to the number two the number two to the number four And since I've gotten to the end of the parentheses here the understanding is that then four goes back to one Karen question. Oh, so the question is can a cycle be a permutation on itself. Yeah, very much can here is a cycle If somebody just dropped that in your lap and said here's a permutation Doesn't necessarily have to refer to anything else It simply isn't the case that the permutation you've written down can be completely described just by looking at What happens to one particular element of the underlying set and following it all the way through good question? How about let's look at kappa? Well, let's see. How would we describe this cycle and so let's see this notation means I've got an underlying set Presumably, I haven't even told you what the underlying set is but at least is one through seven heck It might be one through ten. I don't know but whatever kappa does it simply takes six to seven Well, there's the end takes seven back to six and everything else stays the same It's exactly what this does so it's simply a significantly shorter hand notation for The original permutation which happens to only have one orbit and here we have sigma then is one two four Circle six seven You want to put the circle in grade if not if you just write the two things next to each other The understanding is that you're doing function composition So I've taken the original example sigma and I've written it now as the product quote-unquote or the composition of two cycles If you like to write it in that form that's fine. Of course, that's the same as Whether you need these commas or not some books use it some books don't I'll I like using them So that one two four doesn't look like a hundred twenty four or twelve and four or something. Okay questions there comments question, David. Oh, I see sure Yeah, if you wanted to write this then that completely clarifies what the underlying set is But if if I I'm sorry go ahead. Yeah, well, I see a good question Actually, that's a great question So here the verbiage is all that's coming into play here It's similar to for those of you that saw the number theory course when we say that any integer bigger than two is a Product of primes. Oh, what if you hand me a prime number? Is that a product of primes? It's already prime so when we do this when we make a statement any element can be written as the product of disjoint cycles Technically the implication is or is already a cycle itself? Cycle itself that's a good question And this is just standard verbiage here But it's the same sort of standard verbiage as the fundamental theorem arithmetic says that any integer very or equal to as a product of primes Technically, it's either prime itself or product primes here any element of SN is either already a cycle Or it can be written as a product of disjoint cycles In fact, technically I haven't really included the identity element in this description either So what I should be saying is any element of SN is the identity notice doesn't change anything or is a cycle or It's a product of disjoint cycles, but then that just gets way too cumbersome All right, that's a good question though because the word product typically means you have to have two or more and the implication is not necessarily Okay, so warning Warning if I simply hand you two cycles that aren't necessarily disjoint. They may or may not commute so Two cycles not necessarily disjoint need not commute to disjoint cycles commute and the example is and This will be a good practice with So be a good practice with computing what these things do Let's see if we can run through and figure out What would happen if I take this permutation which happens to be a cycle and this permutation which happens to be a cycle and Sort of run through what the effect of the cycle is or this Composition cycles well look remember function composition is done in the standard order The standard order is the reverse order from what you'd expect. We always read our functions from right to left Let's see what this function does well if I plug in one what happens one goes to three It's fine But then three in turn. Oh nothing happens to three in this one. So the net effect of one is it goes to three What happens to two I plug in two into this thing? Stays it to I Then take the output namely to plug it into this thing and it becomes one Next let's plug in three and see what happens plug in three this first function takes three to four in Turn the output value four stays at four when you run it through this function so three goes to four and By default I know what has to come out here because permutation. I need to see every element at least once Let's just make sure it does if I plug in four This permutation says take four to one In turn this second permutation says take one to two so the net effect is to take four to two correct on the other hand and this again will be good practice with computing with Permutations viewed as cycles one three four circle one two Let's see what this thing looks like in long-hand notation now folks You expect me not to be able to write down a cycle here because there's no guarantee that in the end the Product or the composition of two cycles is another cycle. It might be or might not be Here it wasn't if I had the composition of these two cycles. These are each cycles the composition was not a cycle It was this thing called Sigma Let's see what happens here. We'll do this a little bit more quickly Plug in one one goes to two and in turn stays at two So immediately I can conclude that these two cycles don't commute because in this one one goes to three in the second Well one goes to two, but let's just finish up for the sake of practice. What happens to two Two goes to one and in turn one goes to three so the net result is two goes to three What happens to three run through the first one nothing happens run through the second one three goes to four See what happens to four Run four through the first one doesn't change in turn run four through the second one it becomes one Okay, so obviously these are not equal questions there comments Karen Not commuting is exactly what we've indicated here. I took these two cycles I Composed them in this order, and I got that permutation. I've just reversed them here and showed that the results question money We'll be careful because this particular permutation is the composition of only two cycles So the function is made up as a two-step process step one. Tell me what happens when you plug the given input into this and Then step two what happens when you run it through the second piece of the function machine, and then you're done Okay, so cycles are nice. I mean they're obviously notationally nice. It turns out they will be nice from a computational point of view as well the nicest types of cycles are the cycles that correspond to orbits that only have two elements in them Namely the nicest type of cycles are the ones that simply switch two things That's easy like kappa over there is a nice one It's just you swap six and seven and that's the effect of the cycle in some sense If I hand you a cycle that only swaps two things all I need to do is tell you what the two things are I don't even need to indicate the order, you know, do you take six to seven and then seven to six? It doesn't matter On the other hand if I hand you a cycle that involves more than two things like one two and four If I tell you I have a cycle that involves one two and four I really haven't told you enough to tell you exactly what the permutation does All right, if the cycle involves one two and four does one go to four or does one go to two? You know this four go to one or But if I've only handed you a cycle that has two elements that it affects that means I've got a swap and Those are special enough that not only do we give them a special name We can also write down sort of a next step in this list of propositions That says not only can any cycle be written as a product of disjoint cycle Not can not only can any permutation be written as product of disjoint cycles But in fact we can go a little bit further We're going to give up the disjoint part But we're going to be able to conclude that every cycle is in fact a product of these very special types of cycles and therefore that every permutation is a product These very special types of cycles and the definition is this let me do it over here. So I can continue this list Definition transposition that's the word is simply a cycle that transposes two elements a transposition Cycle that have two Elements in other words has only one non-trivial Orbit and that orbit only has two elements for example Kappa is above so the point is rephrased ie transposition Means that you've written down a cycle that only has two things in it. So it looks like a comma B So here for instance kappa is just 6 comma 7 or you might look at the transposition 3 comma 4 or 1 comma 6 Or it doesn't really matter what two elements you have that's what a transposition is Here's a transposition. Here's a not a transposition Happens to be a cycle, but it's a cycle that affects three of the elements in the underlying set transposition means you only affect two So the question is well a why are these special and we'll see that Probably not by the end of today, but certainly next Monday The question is if I hand you Something that's not a transposition is it possible to trade it in for the composition of transpositions or rephrased Is it possible to take permutation and not just viewed as being made up of? Product of disjoint cycles, but actually being made up solely of just swapping things Swapping pairs of things as you go through and the answer turns out to be yes so the proposition is this and it's an interesting one because it'll allow us to Compute not only with cycles, but with transpositions if I hand you a cycle the cycle a1 a2 up through I'll call it a sub t I Don't care how many things are in the cycle two things in other words. It's a transposition three things four things, etc equals The following let me make sure I get this right a1 comma a t Circle dot dot dot dot circle a1 comma a3 Circle a1 comma a2 if you want you can put another So if I hand you a cycle, I don't care how long the cycle is What this proposition says is that you can actually view that cycle as the composition of a bunch of transpositions So any cycle can be written as the composition of transpositions or we'll say the product of transpositions The key observation of folks is be careful. I haven't used this word in the proposition When I've taken this cycle and I've written it as a product of transpositions a product of cycles of length to The element a1 appears in all of these in fact, so they're manifestly not disjoint Proof of the proposition. I won't do it for you. I'll simply run through a few of the details though proof Well, let's check out and make sure that the net result of doing this is the same as the net result of doing this Well, let's check them out. What happens to a1 a1 goes to a2 is That really what happens here? Well, let's see plug in a1 and what do you get? Well, this first thing says take a1 a2 So so far so good and the point is a2 isn't mentioned any more in the rest of them So a1 goes to a2 and quits, so that's good if you plug a1 in a2 comes out here If you plug a1 in a2 comes out here and now what happens if you plug a2 into here a3 comes out It's like what happens if you plug a2 into here? Well in the first step it says a2 goes to a1 Yep, but then in turn a1 goes to a3 So the net result is that a2 has gone to a3 via little, you know D2 or through a1 so it's a2 goes to a1 goes to a3, but then a3 isn't mentioned Past this so a2 goes to a3 Similarly a3 goes to a4 how because a3 doesn't move here a3 gets moved to a1 This one is a1 a4 it's moved so a3 goes a4 Finally what happens to a sub t a t or I don't see it don't see it don't see it Oh here I see it a t goes to a1 and that's exactly what the cycle did So here's the punch line if you hand me any cycle you can always write it as a product of transpositions the way you do it and here you have to be extremely careful in the order because we're not claiming that the transpositions commute in fact they manifestly won't you simply take the cycle You take the first element you write it as the first expression in each of the pairs and Then you simply list out the remaining elements of the cycle in reverse order you list them out from a t back to a2 so proof just follow each of them through follow each element for example then if I hand you Let's say this cycle That we've called tau before example tau is The cycle one two four What this proposition says is that I can actually write tau as a product of a Number of transpositions. Here's how you do it. You write down the first thing Right down the first thing Then you write down the first thing again Let's see how many times well You need to fill in all the remaining things from the original cycle Backwards in the second slots here. So one four one two Check that it works. Just you know at first if you're a little bit unsure What is this thing do it? So it says take one to two and then nothing happens to two So the net result is one goes to two good What happens to two plug in two in this first step two goes to one? In the second step one goes to four so the net result is two goes to four good What happens to four nothing there plug in four one comes out Let's do another example with a cycle. We maybe haven't seen before Let's call it gamma or something like that gamma is two five one seven three Written as a product of transpositions Can be written as I'm going to write the number two How many times however many times I have left over once I've written down to so I'm going to write it down four times so two comma two comma two comma And now I simply lay in the remaining elements in reverse order two three seven a I don't know slightly less computational way to phrase this proposition is Every cycle can be written as a product of transpositions Not a disjoint transpositions, but a product of transpositions So here then is a third proposition to add to this list Look folks any element of SN Can be written as a product of disjoint cycle So if I hand you any permutation as a first step you can always write it as a product of cycles in turn each cycle Can be written as a product of transpositions although not necessarily disjoint. That's fine So those two pieces together give any element of SN is a product of transpositions and Again, I Used this word is a product of or this phrase is a product of Loosely in the same sense that we use it in context of something like the fundamental theorem calculus I should state it as any element of SN is either the identity or is already a transposition or is a product of transpositions Although it's interesting because here Technically I can get by with this phrase without all those caveats It is actually the case that every element of SN is a product of transpositions As long as you don't require me to write them as disjoint transpositions The reason I'll be I'll be able to write the identity element actually as a product of transpositions and I'll be able to Write any transposition actually as a product of transpositions questions comments what we will do next Monday is Look at some very special subgroups of SN that correspond to The way that you write an element of SN as a product of transpositions And we'll look at some of the issues surrounding that next time But what I want to do now is all I've done is put together on one page Some information that I've copied Directly out of the text if you're doing this online. I've put a copy of this sheet on the Course website you just go and copy it off again. It's nothing Out Side of what you've already got in your text. It's just I've put all the information in one place Specifically what I've done here folks is I've written down or put together The group table for the specific group s sub three and this author uses some Specific letters to denote the elements of s sub three. He calls them row zero row one row two this thing is the Greek letter row and Mew one mu two mu three. That's the Greek letter mu And then I've also listed on this sheet a group that I'll talk about for the last five minutes of class here And then we'll pick up again next Monday folks inside s sub n and I'll focus on the specific case n equals four to start with but this will happen and be true for any n bigger than Equal to four technically bigger than equal to three but it's uninteresting when n equals three If I hand you The number four and I ask tell me all the permutations There are of the underlying set one two three four. I know how many there are there's four factorial or 24 Well what I'm about to do is look at a subset of those 24 permutations. I don't want you to look at all The only permutations. I want you to look at are those that arise in the following way and it's sort of an interesting geometry And this is going to sort of introduce us to this connection between group theory and geometry Which is one of the historical motivators for looking at groups Take a square and What I want you to do is label the sides of the square In a standard order doesn't really matter which order have about one two three four Label them either clockwise or counterclockwise the issue Well, it won't be an issue. We'll put it that way And now what I'm going to ask you to do is view this thing as a piece of cardboard where you you know labeled the four corners Take it out of the board Take the piece of cardboard flip it around any way you want you'll flip it or rotate it or don't do anything to it I don't really care then put it back on the board and the question I want you to ask is follow through what the original numbering was Versus the number that it lands on so for example if I picked the thing up and I Somehow diagonally flip it along this axis and I put it back down Then what would have happened is let's see the number four would land on the number two the number two would land on the number Four the numbers one and three would stay the same Here's another possibility. I pull this thing out of the board I sort of flip it around a vertical axis and I put it back Let's see what that would mean that would mean that one would eventually land on two that two would Well when you flip the thing around wind up on one three would land on four four would land on three In this way what we're going to get is permutation as long as you're not bending or folding or you know Crumbling up the square if you pick it up Flip it around or do anything you want to it without changing its shape and then put it back down You've affected or produced a permutation of the numbers one through four hmm and It turns out that the collection of permutations that you get that you can somehow view is being generated by the permutations of a square Doesn't give you all the permutations you're not going to get 24 different ones Intuitively why because if I pick the square up Let's see regardless of how I set it back down The corner one is always going to be next to the corner two For instance the corner one's never going to be next to the corner three Regardless of what I do to the square because they're already at opposite corners. They got stayed opposite corners So if I've told you where one goes I can't put three next to it because three isn't next to it So the indication is that the permutations that arise this way Don't reflect or correspond to all the possible permutations only some of them you only get a subset of the permutations and we'll call that subset the subset of Rigid rotations or rigid motion of the square I'm not going to write that down because we're not going to use that notation here But that's typically what this thing would be called the most textbooks the Standard letter notation for the collection of permutations that arise from motions rigid motions of the square is d sub four rigid motions Or rigid motions of a square Here's the proposition proposition The permutations that arise from the rigid motions of the square Give a subgroup of S4 subgroup statement folks How do you prove a subgroup statement? You have one way to do it subgroup theorem exactly right What do you have to do? You have to convince me that if you take two things that correspond to rigid motions of a square and You compose them compose them means do one then the next That you get a rigid motion of the square sure do one Then do the next one and what's the net result? There's the rigid motion of the square. So that was easy There's the proof of closure Is the identity permutation origin motion of the square sure? There it is okay Don't do anything There's a rigid motion of the square if you have a rigid motion of the square Is it's inverse origin motion of the square? Well, yeah, just whatever the steps were just do them backwards You know get back to where you started so subgroup theorem check Now obviously you can't write that out in your notes, but that's the idea If you have something that's a rigid motion of the square Then the collection of those actually form a subgroup and we'll go ahead and Observe what the size of this subgroup is and we'll call it a day question All right, how many ways are there doing that? How many different ways are there picking the square up? Moving it around if you want anyway, which way you aren't putting it back down Look if I take the square And I you know move it around as much as I want and I put it back down The first observation is how many different places can the corner corresponding to the number one land? Well, you could put the square down so that one is on there or on there or on there or on there So as a first step to describe all these things there's four places where this corner could go Then here's the point folks once you've told me which corner you're gonna put this one down on Then there's only two possibilities left to choose You're either gonna put this one to its right or you would have flipped the thing over and you're gonna put this one to its right So once you've told me where you've set down the corner corresponding to the number one There's only two choices. Actually. There's only one choice left and there's only two choices to be made Are you then putting things down think clockwise or you putting things down counterclockwise? Have you put down the number two to its right or have you put down the number two to its left? Okay, so how many choices or how many ways to complete the first part? Four different ways How many ways of determining the second piece two different ways? You've either put the number two to its right or to its left so it turns out the number of elements in this subgroup is And what I have done for you is I've listed out the eight permutations That correspond to rigid motions of the square And in this table I've given you well in this second column I've given you actually the group table of these things what we'll wind up doing is Interpreting why these letters are used for instance. He uses the row notation. That's going to stand for rotation the row the row notation that's going to stand for rotation and The muses and the deltas are going to stand for mirror images or diagonal Flipped depending on what the corresponding permutation looks like All right. This is a good place to quit If you have Any questions on your homework that you got back today? Go ahead and check the web tomorrow to see whether or not the solutions are posed