 happening. I went back to the sheet that I handed out on day one in here where I presumably told you when to anticipate exam one and blew it because, well, I had the right day of the week but the wrong date. So what I had put on the info that I handed out on day one was something like Wednesday, but then I had the date of Thursday. So this I think is correct now. It's Wednesday, September 26th, so it's nine days from today. And here's what it'll cover. I've sort of listed out explicitly the topics that you can expect to be on there. I'll give you some suggested study techniques as to how to use the practice exam for what it's worth. The practice exam is identical to the actual exam that I gave a year ago in this course, Fall 2006. So there are a couple of you that have seen this before. Well, that's good. So nobody's got any sort of unfair advantage that they may have seen a recent exam. Now everybody's seen the recent exam. So that's why at the top here I've just, I mean this used to say Fall 2006, exam one solutions. Now it says Fall 2007 practice. Okay, so you have complete faith that this is the sort of exam that you might actually see in class. It should give you some indication as to the length of the exam, the types of questions that I may ask, the sorts of questions that may come up, etc. Okay. All right. Any questions about what to expect? I'll make sure, I'll chat with Jen to see whether or not we can maybe sneak in an extra SI session before the exam, something like that. But for now the SI sessions will just stay as they are. And the SI sessions are sort of loaded up to, you know, to sort of get you to a Wednesday homework submission. So now it'll be loaded up to get you to a Wednesday exam. And as you'll see at the end of class today when I give the homework assignment, there won't be a homework assignment due the exam day, it'll be due the following week. So, and as it turns out for what it's worth, all of the homework assignments that I've given you so far, that's the material that the exam will cover. So the one that I give you today there won't be any exam material for that stuff. So, all right. Just in the interest of time here, I think I'll let the quizzes that you took last Wednesday just circulate. Actually, let me put this in here. The only quick remark on the quiz that's coming back, they look just fine, is a couple of you used the phrase in part one, proof that every cyclic group is a billion. Some of you said something like let G be the generator of the cyclic group, capital G, then blah, blah, blah, blah, blah, blah, blah. And technically, folks, you've used the article the incorrectly here. Because typically, if you have me cyclic group, there isn't just one generator. There are typically many generators. So, rather than the word the, the correct word is let G be a generator of capital G. Because if you have me cyclic group, for example, the cyclic group Z sub, I don't care, take what everyone, Z sub 6, let's say. The element 1 is a generator. Because if you look at all the powers of 1, you eventually get all the elements. But the element 5 is also another generator. So, and as far as your proof goes to show that the cyclic group is a billion, it makes no difference at all which of many possible generators you start with. Of course, the proof looks identical because you just start with some generator, but then you go from there. So, that was the only slight issue that happened to have come up on the quiz. Okay, let's see. My notes say distribute exam materials. It did that. And talk about the quiz. Okay, perfect. So, here is what we were looking at at the end of Wednesday. We were looking inside these permutation groups, these things called S sub n, at various subgroups. And one particular subgroup that we happened to look at inside S4 was what we called D4. And D4 meant the rotations or the rigid motions, motions of the square. And the point is if you somehow just draw a square on the board and label the vertices 1, 2, 3, 4 in some order, then we can talk about taking the square, sort of picking it up, pulling it out of the board, doing whatever we'd like to it other than bending it or tearing it or something. Just sort of rotating or reflecting or flipping or whatever we want to do. Putting it back on the original square and then asking the question, tell me what was originally there. Tell me what number wound up landing on it and describing a permutation that way. And we sort of fidgeted our way through a proof that that actually gave a subgroup that always boils down to the subgroup theorem. It's just a matter of how you show closure or identity or inverses. And here was more of a geometric approach. Show closure if you do a thing and then you do another thing. It's just you've done a bigger thing, but you've done a thing. In general, oh, let's see, we observed this as well at the end of last time. The number of elements in D sub 4, let me get a different pen here, turned out to be 8. We could identify the permutations that come up as rigid rotations of the square. In effect, we could just count them. And what I did for you last Wednesday was I handed out a sheet that gave precisely what the 8 elements of the group D sub 4 looked like. And that's on this sheet. Did anybody not get this sheet last Wednesday? And again, for those of you that are doing this online or via video, I put a scanned copy of this at the website. So you might want to download that and print it off. This isn't anything new. It's in the text. It's just, I think it's good to have this table handy with you just so we can do some group computations. It turns out folks, more generally, more generally. Well, a square is nice, but a square is just a specific example of what we call a regular n-gone. So a regular n-sided figure, here the number of sides of this figure is 4. It's a figure that where all the interior angles are equal and all of the side lengths are equal, that's what a square is. We can talk about a regular 5-sided figure that's called a regular pentagon, or a regular 6-sided figure, a hexagon, or what's it called, a septagon, or an octagon, the stop sign, a regular 8-sided figure, et cetera. And it turns out, regardless of how many sides you want, we can talk about the rigid motions of the regular n-gone. So the regular 4-gone is just a square and the regular 3-gone is just an equilateral triangle, et cetera. We can call that set of permutations d sub n and the two observations or the two propositions are first that dn is always a subgroup of sn. The proof is essentially the same. Folks, it didn't make any difference that I started with a square. Start with a 6-sided figure, a 10-sided figure. It doesn't matter as long as all of the angles and the side lengths are the same. Think about drawing it on the board. Think about then sort of cutting it out of cardboard, pulling it out, flipping it around any way you want, putting it back in. That's then a rigid motion of the regular n-gone. And the proof that that collection of permutations forms a subgroup is the same as the one that we did. Just pull it out, pull it out, do it. And the same sort of proof that we gave to show that the number of elements in d4 is 8 turns out to give essentially the same proof that the number of elements of dn in general is, well remember what the idea was? You pick the thing up, you flip it around any way you want. The point is if you identify where the number 1 goes to, well, it could go to any one of the vertices in the original regular n-gone. There's n such choices. But then the point is the number 2 only has two choices because the number 2 has to go to a vertex that's next to the vertex that the number 1 went to. So once you've identified the n possible choices where the vertex 1 went to, there's only two possible choices where the vertex number 2 goes to. It either goes to the right of it or to the left of it, so you've got two choices there. But folks, once you've set where the number 1 and where the number 2 go to, everything else has to follow because it's a rigid motion. Once you've got 1 and 2 here, then 3 is there, 4 is there, 5 is there all the way over to n. So it turns out that the number of permutations that can be realized as rigid motions of a regular n-gone is just two times n. So in the grand scheme of things, the number of rigid motions of a regular pentagon of a five-sided figure is two times five, which is 10. That sits as a subgroup of s sub five, and s sub five is already pretty big. Remember, there's n factorial elements in this thing. So this thing has 10 elements in it. This has 120. It's five factorial. When n is 6, when we're looking at rigid motions of the hexagon, the number of rigid motions of the hexagon is 12. The number of total permutations of six elements is 720. So somehow in the grand scheme of things, there are very few of these, but these groups in and of themselves are going to be interesting. They're going to give us some nice examples of situations where we've got some structure that we might not otherwise have in other situations. So these are called the, the letter D stands for the dihedral group. A quick remark about notation. Some other algebra texts would call this D sub 2n. I don't like that. I like this D sub n. That's a good notation. That's what this text uses. The number of elements in Dn is 2n. I mean, you don't call this s sub n factorial. So just a notational comment there. All right. Question. Now, what I want to do is a little bit more computation with this notion of taking a permutation, breaking it up into a product where I use the word product in quotes of cycles, disjoint cycles, and then in turn taking the cycles and breaking them down into products of transpositions. So from last time, and what I'm going to do folks, because there was some confusion in class and I apologize for that, is rephrase the results, but I'll rephrase them formally correctly and then we'll feel free to use the sort of jargon from here on out. The first statement is any permutation, any element of Sn, that's the same thing as saying any permutation of the numbers from 1 to n, is either first the identity permutation or a cycle or the product of disjoint cycles. So last time I tried to sort of sneak in the jargon way of phrasing this result. If you hand me a permutation, maybe you handed me the permutation that doesn't do anything. Okay, that's called the identity. Or maybe you've handed me a permutation that is one of these cycles, but if you handed me something that's not that or that, at least in the end I know that I can write the thing as a product. Product here means composition, because that's what the operation is inside S of n, product of disjoint cycles. The way I phrased the theorem last time was just to hand you number three. And folks, this is the way I'm going to, in sort of shorthand or mathematics ease, refer to this result that every permutation is a product of disjoint cycles. Meaning, alright, maybe I have given you the idea, or maybe it's just a cycle itself. And what I tried to do was relate this to the verbies that we use when we talk about integers being products of primes. If I say every integer is a product of primes, technically that's not correct because one's not a product of primes and seven's not really a product of primes. But somehow the primes make everything up. Here, in the context of permutations, disjoint cycles make everything up. And the second comment is that any cycle can be written, written as the product, meaning composition, of transpositions, remember transpositions means a cycle that only switches two elements. It's absolutely essential that you'll note that this word appears here but does not appear here. And that's by design because even though we can write any permutation as a product of disjoint transpositions, we might not be able to write every cycle as a, I'm sorry, disjoint cycles, we might not be able to write every cycle as a product of disjoint transpositions. Alright. I want to do a computation that a couple of people had asked me about in office. Example, I'm going to hand you a cycle. Here's a cycle. Here's a cycle in, I don't know, how about S9? The cycle is called tau. It's the cycle one, four, two, nine, six. So it's a cycle, a single cycle. And the question that we're going to ask is what is the order of tau? Now what some students are having trouble remembering is that these things, S sub n, are just groups and there were some quantities or some properties that we described for all groups and all elements in all groups. And what we're asked to do here is talk about those things that we looked at in general. For example, the notion of the order of an element inside a group, inside this particular context. Well, remember what order means. You take the element and you start raising at the powers. In other words, you start combining the element with whatever the binary operation is with itself. And the question to ask in order to figure out what the order of the element is, is how long does it take you to get back to the identity? Well, there's an identity element in this group. It looks like the identity permutation, one, one, two, two, et cetera, et cetera, et cetera. So what do you have to do? We need to compute. You compute tau squared, in other words, tau, circle tau. Well, let's just do that quickly. That's one, four, two, nine, six. Circle one, four, two, nine, six. And we get, hmm. Well, folks, there's no guarantee that when you do the composition of a cycle with another cycle that you get another cycle. So I'm going to express the result of this composition in this long hand form. Cycles are nice when you've got them, but even if you put two cycles together, the result might not be a cycle. So let's see what happens. Well, one goes to four and in turn four goes to two. So one goes to two. Two goes to nine, which in turn goes to six, so two goes to six. Three is unaffected. Four goes to two, in turn two goes to nine, so four goes to nine. Five is unaffected. Six. Oh, six goes to one, in turn one goes to four, so six goes to four. Seven is unaffected, eight is unaffected. Nine goes to six, six goes to one, so nine goes to one. So there's the result. Is that the identity permutation? Clearly not. So is the order of tau equal to two? No, because I didn't get the identity permutation. So now what do you have to do? You have to keep computing. Tau cubed is, well, it's tau squared, circle tau. That's what tau cubed means. It means do tau with itself three times. Well, folks, you've already done tau squared. There it is. Write it out. I won't do that for you here just in the interest of time. You know, two, six, three, et cetera, et cetera, et cetera, one. There's tau squared, circle tau. You can write out tau in its cycle form. You could rewrite it in longhand form. I don't care. One, four, two, nine, six equals, well, let's write it out in longhand form. One, two, three, four, et cetera, through nine, and I'll get you started here. What happens to one? One goes to four, and in turn, four goes to nine. So one goes to nine, et cetera. Just keep going. So here's tau cubed. I already know that tau cubed isn't the identity because one goes to nine. So keep going. Tau to the fourth equals, turns out to be not the identity. Tau to the fifth equals, it turns out, to give you the identity. Check it. Check that this works. In other words, actually do the computation. It's just good practice for you with this computation of permutations. All right. Question. Oh, no, let's see. Six goes to one. That's what the first piece does. Sure, because then one goes to two. So one goes to two. Yep. Yeah, so it'll be, you first do this, and it makes no difference that the six happens to appear here, or the nine isn't. Six goes to one. That's the first chunk, and then in turn, one goes to two. So six goes to two. All right. Folks, the order of an element is, well, it's a general group theory idea that says take the element and keep composing with itself until you get to the identity. In some situations, you never get the identity. We call it an element of infinite order, like if you take the number three inside the integers, three plus three is six, plus three is nine, so you never get zero. That's fine. But here you do. And what took, I'm sorry, Lonnie? Can it repeat without going to E? The answer's no. It's, there will be no repeats until you get to E, and then it'll start repeating. Okay. That's a great question. The answer is, yeah. In this list of tau, tau squared, tau cubed, tau of the fourth, when you write those things out longhand, you won't be seeing identical things. And then you'll finally see the identity when you compute tau of the fifth. And once you see the identity, then if you keep computing tau six, it'll be the same as tau one. Tau seven will be the same as tau two. Tau eight will be the same as tau three. You'll want to have getting a cycle. The order is the one where you get the identity, so here the order of tau will be five. So order, this is the notation we used, order of tau, then it's five. So it depends on, yeah. The order of tau is the smallest positive integer, positive mean bigger than zero. That eventually gives you the identity, and that's what we found here at the first time that we see the identity element when we start taking tau and combining it with itself. All right. Now, it turns out that if you hand me a cycle, well, this cycle happened to involve five of the elements of the underlying set. So you can talk about the length of a cycle, meaning how many numbers in the underlying set doesn't involve, or how many elements are in the orbit of that cycle. And it turns out it's not coincidental that the number of things in here is the order of the element. That's not coincidental. And that's one of the questions that came up in the homework. Yeah. But it turns out, then we can ask a similar question. Now, here's a product of two cycles. If I hand you, let's see, sigma equal to one five, composed with two, four, seven. The composition, that's, I'm just making dramatically clear what the operation inside S sub, I don't know, let's say we're working in S sub nine, again, S sub nine is. Do that, then that. So, folks, here's an element of S nine, and we can ask exactly the same question, what's its order? Find the order of sigma. I know exactly what the question's asking me to do. Take sigma, figure out if you do sigma, then sigma circle sigma, and sigma circle sigma, et cetera. In other words, do the binary operation with itself. How many wax does it take you to get back to the identity? Sigma, sigma circle sigma, in other words, sigma squared, sigma circle sigma, circle sigma, et cetera. Well, just do the computation, or you know what it means. Okay, it's a little bit distasteful because you can't use the cycle notation. Not much at least. But, oh, wait a minute. At least if I'm looking at disjoint cycles, disjoint cycles means that the numbers that are affected here don't at all overlap with the numbers that are affected here. What we observed last Wednesday is that when you do products of disjoint cycles that they commute, that they can be put in any order. So the point is, notice here, if I do sigma, circle, sigma, circle, sigma, I'm just going to do this. Let's see, that's one, five, two, four, seven. There's sigma. If I do sigma again, I get one, five, two, four, seven. And then if I do it again, one, five, two, four, seven. So I've simply done sigma, circle, sigma, circle, sigma. The beauty here is this, folks, because the elements that appear in this cycle are disjoint from the elements that appear in this cycle, you can switch those. If there had been any sort of overlap, for instance, if this had been a five instead of a four, or if this had been a two instead of a one, then you'd be stuck. This is as good as it gets. It's not that you can't compute this. It's just it's more painful to compute it if you can't switch things around. But the beauty of being able to switch things around is because the cycles are disjoint, disjoint, we're using that here. That means they commute. This is the same as then one, five, circle, one, five, circle, one, five, circle, two, four, seven, circle, two, four, seven, circle, two, four, seven. Again, the only reason I can switch the order on these is because the elements that appear are disjoint. The question is why bother? Well, the point is, look, this is an element that has a length three, that has three things that it affects, and what I'm asking you to do is do it three times. What do you think the result of this chunk is? It's the identity. Just like when we saw over here, we took an element that had five things in it, and we raised it to the fifth power. We got the identity. If I have something with three in it, and I do it three times, I get the identity. How about this? I do this two times, I get the identity. Yeah, because this is just a transposition. It says if you transpose one in five, and then you transpose one in five again, you're back to where you started. So in the end, if I do this three times, all I'm left with is sort of everything disappears except for one piece. So did I get the identity yet? No. But this at least might allow you some computational flexibility to be able to write down what these things are without having to grind out longhand notation what's going on here. So let's see, sigma, circle sigma three times didn't give the identity because it gave this. If I do sigma, circle, sigma, circle, sigma, circle, sigma four times, that means I'm plopping another copy out here. But then, hey, these cycles are disjoint so I can somehow manipulate them up together. If I had another piece on here, if I had another one five and another two four seven, if I had another one five, that would combine with this one. And all this would be the identity. If I had another two four seven, that's sort of unfortunate because I already had the identity there. But then, all right, but at least the net result is I've just got a two four seven. Now do it five times, do it six times. See, eventually, when it's the case that somehow you've sort of balanced things out to get the identity altogether, that's the question. Okay. So keep going. It turns out for this one, the hint is if you do it six times, if you do sigma to the sixth, you'll get the identity. So I'll just give you this piece of information turns out that in, remember the notation, the order of sigma is six in this case. And what you're asked to do in one of the homework problems is sort of make a generalization of the process that's going on here to try to compute what the order of the product of disjoint cycles might be. What's interesting, and that's what we're going to look at right now, is if you hand me a situation where you've got a permutation that's written as a product of cycles, but the cycles aren't disjoint, then in effect, all bets are off. If you know the length of the first one, the length of the second one, but somehow there's an overlap between them, then there's absolutely no general formula that'll allow you to conclude the order of the product without just sitting down pounding out and seeing what you get. Because you don't have this luxury of being able to in general switch things if the things aren't disjoint. So, yeah, let's take a look at products of transpositions, transpositions. So something like this, if I hand you, I don't know, one, three, oh yeah, that's right, this way. So here's sigma. Maybe sigma is the following product of cycles. Oops, excuse me. One, four, two, seven, circle one, three, nine. Inside s of nine. Okay, well, first of all, notice that even though this thing is the product of two cycles, the cycles aren't disjoint. So if we're going to try to analyze things about sigma, like what its order is or something like that, it's not going to be as easy as the first example because I can't switch the order here because the cycles aren't disjoint again. The number one occurs in both. But at least that theorem says any cycle can be written as a product of transpositions. Let's go ahead and do that. So if I take this algorithm that I showed you last time, you can take any cycle and write it as a product of transpositions. What you do is you write down the first thing that appears in the cycle and then you simply list the other elements of the cycle in reverse order. So I can trade this in for this and then let's see, I can trade this in for one, nine, one, three. If you're all nervous that you've written things out correctly, let's just check your answer. At least do a little sniff test or plausibility check. Is it the case that this really equals this? Well, what happens, for example, to the number one? One goes to three and in turn stays there. So it better be the case that in rewriting it this way that one goes to three and stays there. Well, one goes to three and stays there. So at least a first approximation shows that I've written down things maybe in the right order. Please note that these transpositions certainly aren't disjoint. The number one appears in all of them. Question, what is the inverse of sigma? Well, I'll answer that two ways. One is, if you were to write this thing out long hand, you know, one, three, five, six, seven, eight, nine, just tell me what each thing went. To figure out the inverse of a trans, you just flip it over. That's no big deal. You're just doing things in opposite order. Typically, though, once you flip it order, you then rewrite the top row in order. But that's easy. It turns out that we proved a result way back when, I think on day two in here, that says, if you hand me an element of a group and the element's written as a product of other things, if you have something that's written as, let's say, AB inside a group and you want the inverse of AB, all you have to do is look at the inverse of each of the individual elements, A inverse and B inverse. But then you write down the product backwards, just like you saw when you did product matrices. So maybe the subtler way of finding the inverse of one of these things is, well, wait a minute, look. The inverse of a transposition is just itself. Of any transposition, this is just itself. In other words, if I hand you something that looks like transposition, because what do you have to combine with AB to get back to the identity? Well, if A goes to B and B goes to A, in other words, if you switched A and B, in order to get back to where you were, you just got to switch A and B and you'll get back to the identity. So here's what this means. If I want to compute sigma inverse, well, here's sigma, I haven't written us the product of a bunch of things in the group, to get the inverse of a product, you take the inverse of each individual thing and then you just switch to the order, you make everything backwards. So is, then, I'm taking the inverse of each individual thing, but each individual thing has the beauty of being its own inverse. And then I'm going to write them all backwards. One, three, one, nine, one, four, one, two, one, seven. That was pretty easy. Let's check our answer. Is it the case that if I combine this with this, did I get the identity? Well, let's make sure we do. If I combine this with this, well look, now I've got one, three sitting next to one, three. But wait a minute, one, three? Circle one, three is the identity. So those go away. Now I've got one, nine sitting next to one, nine. That's the identity. So those go away. It's just like you did when you did the inverse of a product of matrices. When you rig things somehow inverse and then backwards, the advantage of putting things backwards in the product is that when you then wind the results up, you will have then rig things so that the inverse of each element touches the element itself when you do the product and so all those things sort of disappear in order and you indeed get the identity. So it's pretty easy to write down the inverse of a product of transpositions. You just write down the transpositions themselves and reverse the order, the point being that the inverse of each transposition individually is just itself. Now with that computation on inverses having been done, what we're going to do is finish up this discussion of these very special groups, these permutation groups, SN, by looking at a special subgroup that sits inside SN for each N. While we've already done that, we looked at the rigid motions of the regular N-gon. That was a subgroup that sits inside SN. In the grand scheme of things, though, those subgroups weren't very big. Now if you look at S6, that has 720 elements in it. If you look at D6, the rigid rotations of the six-gon, well it's a subgroup but it only has 12 elements. That's pretty small. What we're going to do now is look at some special subgroups that in fact are relatively large inside SN. So here's the definition. No, let me make one more comment before I give the definition just so you can sort of have a little bit better understanding what the issue is. If I hand you a permutation, we can always eventually write the permutation. Either it's the identity or maybe it's already a transposition itself or it's a product of transpositions. So the glib phrase is, any permutation is a product of transpositions. But it turns out there's not any sort of uniqueness to the way I can write a permutation as a product of transpositions. There might be many, many different ways to write the same permutation as a product of transpositions. Or rephrase, I might hand you and your friend exactly the same permutation. I'll put you in separate rooms and say, all right, write this permutation as a product of transpositions and you and your friend might come back with what on the surface look like completely different answers. But both could be correct. For example, here's a really cheap way of showing that there might be many different ways of writing a given permutation in different ways as a product of transpositions. I'm going to take the same thing and just tack that on the end. Well, wait a minute. Heck, that's just the identity, right? So there's certainly no harm in tacking it on the end. I haven't changed anything. But what I've just done is I've taken sigma and I've now written it in a different way as a product of transpositions. I originally had it written as the product of five transpositions. Now I've got it written as the product of seven transpositions. Boy, there's certainly nothing special about one, too. I could put a three, six. Let's see. Let me take two things that don't appear here. I could have put the transposition five, seven here and then slid another version of five. No, seven's about them. Five, eight here and a five, eight here because five and eight don't appear in any of the other ones. So the five, eight transposition would actually commute with all of these and that it could eventually have it touch the other five, eight that I've slid in there and that would give the identity, wouldn't have changed anything. So folks, there's a whole lot of sort of elbow room, there's a whole lot of flexibility that you have in taking a permutation and trying to write it as a product of transpositions. But this is not something I'm going to prove for you. But it turns out that if you and your friend take the same permutation, I don't care what it looks like, and you each complete the task of writing that permutation as a product of transpositions that you may not come out with the same answers, you may not even come out with the same number of transpositions that you use. And I could write this as a product of five transpositions or seven transpositions or nine transpositions or 101 transpositions. But what you will come back with, and this is what turns out to be interesting in these permutation groups, is you'll come back with the same parity in the number of transpositions that you've used. So that for instance, if you use an odd number of transpositions to write the original permutation as a product of, this one for instance I originally wrote as a product of five transpositions, and your friend completes the same task, the number of transpositions that your friend will use has to also then be odd. Or rephrased, if you can write a permutation as the product of an odd number of transpositions, then you can't also write it as the product of an even number of transpositions, and vice versa. So the definition is this, we call a permutation, I mean I didn't even call it a permutation, an element of the group s and even in case, I don't know what it's called, sigma. In case sigma can be written as the product, an even number of transpositions. Leave to your imagination what an odd transposition, I'm sorry an odd permutation is. And the theorem which I'm not going to prove for you is that each element of Sn is either even or odd but not both. That's, it's certainly not beyond the scope of this course to do, but it would take me about 45 minutes to explain all the details. The proof is in the text if you're interested in reading it, it's basically just a sort of a housekeeping approach, you've got to keep track of how many things move and how many things you've already accounted for, etc. But the punchline is that we can talk about the notion of an even permutation or an odd permutation, and that label simply depends on when you take the given permutation and you write it as the product of transpositions, how many transpositions does it take? And it'll always be either an even number or an odd number, regardless of how you write it. Now you have to be a little bit careful, so here's the warning. Let's do an example. So example tau is the cycle 1, 3, 2. Is tau even or odd? Well, I can't answer that question, at least not yet, because the task in order to determine whether or not a permutation is even or odd is you have to take the permutation and write it as a product of transpositions and then look at the product that you've written down and ask whether or not you've used an even number or an odd number of transpositions. So let's do it. Well, let's see. If I write tau as a product of transpositions, well, hey, tau is at least a cycle, so I know how to do that. I write down the number 1, whatever appears first, and then I write down the other things in reverse order. This is the algorithm that we looked at last time. Well, now I'm ready to answer the question. How many transpositions did I use? Two. And so the answer is tau is even. So the warning statement is, you see, if I hand you a cycle that has an odd number of things in it, it turns out to be an even permutation. So just be really careful. The best way, I think, to be careful is to just keep in mind what the definition of the words even and odd mean in these permutation groups. It means something about writing things as a product of transpositions. And when it's handed to you in this form, you haven't even gone close to answering my right question yet, which is, all right, write the things of product of transpositions and then give me the answer. I don't want to write that down. I mean, there's some general statement that says, if you hand me a cycle that has an odd number of things in it, it's an even permutation. If you hand me a cycle with an even number of things in it, it's an odd permutation, but I don't want to go there. Just the point is, if I hand you some element on a cover to cycle or not, you can eventually always write it as a product of transpositions, do that, and then just look up and see whether or not you've got an even or odd number. Okay. The subset of Sn consisting of the even permutations is denoted by the expression A sub n. This is called the alternating, I'll give you that word in a minute. So throw all the even permutations inside a hat and the proposition is that the subset A sub n is a subgroup of Sn. If you throw all the even ones together in a subset, it turns out that subset of Sn is actually a subgroup of Sn. Proof, I'm not going to do it for you, but we know exactly how to proceed. If the statement is something is a subgroup of something, there's exactly one thing to use. It's a subgroup theorem. Let's talk about the subgroup theorem. If I hand you two things in here and I compose them, in other words, do the product, is the result in there. Well, if I take two things in here, it means the first one is written as the product of an even number of transpositions. The second one is the product of an even number of transpositions so that when I product the two things together, I've got an even number of transpositions here, maybe six of them here. I've got an even number of transpositions here. In the end, folks, when you step back and look at it, how many transpositions do I have? This many plus this many. Be careful. If this one was the product of six transpositions and this is the product of 10 transpositions, then when it's all said and done, you've listed 16 transpositions all together, you're adding. The point is, if there were an even number there and an even number there, you're adding two even numbers so you used an even number of transpositions. Closed, checked. Second is the identity element in here. Can you write the identity as the product of an even number of transpositions? Sure. Identity is, let's say after we chose one, two, I don't care, choose any two numbers you want. Three six, circle three six. Take any transposition you want and simply write it next to itself. That's the identity. And how many have you used? Two. And two is even. That's all. And finally, if I hand you a permutation that can be written as the product of an even number of transpositions, I have to convince you that its inverse can be written as the product of an even number of transpositions. But wait a minute, that's why we did this example. If I hand you any product of transpositions, an even number or an odd number, I don't really care. There's a product of transpositions. What does the inverse look like? It looks like the same transpositions. Just reverse order. When I handed you the original sigma and wrote it this way, how did you get its inverse? You simply listed out exactly the same transpositions just in reverse order. Folks, when we're asking is the element inside this subset or asking you to count the number of transpositions that you've used. And if you've used a certain number here, here I've used five, you can use exactly the same number when you write down its inverse. Whether they're an opposite order or not, I don't really care. I'm just worried about how many transpositions you're using. So if you hand me a permutation that you can write as a product of an even number of transpositions and you ask about its inverse, you're writing its inverse as a product of the same number of transpositions. And if the original number is even, then it's even. So the fact that a sub n is closed under inverses. In other words, if you start with something in an and you compute its inverse, you get something back in an is a byproduct of the fact that the inverse of the product of a bunch of transpositions is just the same transpositions in reverse order. So there's at least sort of heuristic proof as to why a sub n should be a subgroup of s sub n. Now I'll give you the name. This is called the alternating subgroup. So a sub n is called the alternating subgroup of s n or just the alternating group. Or the alternating group on n elements. Remark. All right. So what's so special about the even ones? Answer, the even ones are special because if you look at the, you know, at the sort of other natural subset, the collection of odd permutations, that collection does not form a subgroup. Nothing else, the identity element's not in that set. The fact that I can write the identity, that's the product of an even number of transpositions, means by this theorem that I can't write it as the product of an odd number of transpositions. So immediately we know that the collection of odd permutations doesn't form a subgroup, but it turns out, proposition, and I'm not going to prove this one for you either. Again, it's not beyond the scope, but it would take me 20 minutes or so, and I think our time is better spent looking at some other topics. If you want to know how big this thing is, inside s n, it's actually relatively large. It's a half of the entire group. In other words, half of the permutations inside s n are even in half or odd. In other words, the number of elements in a n is one half of n factorial. So for example, the number of elements in a sub five is a half of five factorial, so it's a half of 120, which is 60, and it turns out that particular group a sub five is interesting for many, many, many reasons, some of which we'll get to this semester, and some of which, the more compelling interesting reasons as to why the particular group a sub five, that group with 60 elements, is of interest, will come up in the context of the math four slash five 15 course, which I hope some of you will stick around for spring semester, so that was a quick advertisement of what we might do in the spring. All right. Questions there? Comments? All right. That then concludes the material on the s n groups, on the permutation groups, and various subgroups of those. What we're going to do for the last 20 minutes of tonight or so is start a new topic. This new topic won't be on the exam that you'll see a week from Wednesday, but the ideas that we will look at will certainly be based on all the stuff that we've done up until now. So it's not that we're doing something completely different, it's just we're doing a slightly different topic, which won't come up. We're going to start chapter 10 tonight. Here is the overall idea. It is, I don't want to say it's a miracle, or it's unbelievable, but it is sort of a miracle. What we're about to do is look at the following question. If you hand me a group, and we got lots of examples of groups in mind now. We've got the z n's, and we've got the s n's, and we've got the d n's and the a n's. Well, we've looked at subgroups of those groups. We've got lots of examples of those. The question is somehow, how does the subgroup relate to the group itself? I ask the question sort of nebulously to begin with, but what we're going to see is that even though the notion of a group sort of bases itself just on those three axioms that you need associativity and the existence of an identity and the existence of inverses or something like that, that apparently minimal amount of structure leads to an incredible, an incredible amount of structure vis-a-vis the numerical characteristics of the relationship between the group and its subgroups. So for example, what we'll eventually show is that if you hand me a group that has, let's say, 10 elements, that there can't be a subgroup that has six elements. If you hand me a group with 20 elements, there can't be a subgroup with eight elements. And there's going to be enough numerical information that we're going to get about groups that will preclude those sorts of configurations from happening. And to me that's totally surprising, simply because the definition of a group on its surface has nothing to do with the number of elements in the group or the way that the group has to behave or something like that, but what we will get out is this really powerful sort of numerical structure that each group happens to contain. So that's where we're headed. And the idea is, yeah, we now look at, so I'll put a line here, sort of a new topic, we now consider some, the relationship between groups and their subgroups. All right, let me do a quick review of an idea that you saw back in the discrete math course, the notion of an equivalence relation, and we touched on that briefly on day one and here, and then show you that whenever you have a group and a subgroup of the group that we're naturally led to an equivalence relation, and it's that equivalence relation that we're going to exploit on Wednesday. So here's the idea, the setup is, setup, we're given a group G, and we're given some subgroup H of G. And what we're about to show is that this configuration, as sort of minimal structure as it might seem, gives a partition, a way of slicing the group into subsets. I remember partition means you take a set and you slice it into subsets and the subsets have the property that there's no overlap between them and taken together, they give you all of the elements in the original set. So here's what we do. We define a relation. I'm going to call the relation sort of wiggle or tilde or whatever you want to call this symbol, this is a standard notation for a relation on a set, on G as follows by defining the following. A, squiggle or tilde B, so what does it mean for one element of the group to be related to another? Here's what it means in case, here's the computation I want you to make. When you compute A inverse B that the result is inside the subgroup. So I've deemed the element A to be related to the element B in case when you do this computation, when you compute A inverse times B that the result lives in capital H. So it seems like this sort of is dropped out of the sky. But it turns out, proposition, this is actually an equivalence relation. When I do this tilde relation I get an equivalence relation on the group G. Equivalence relation, remember what an equivalence relation is, it's a relation that's reflexive, symmetric, and transitive. Reason, this merits proof because of the following. Remember to show that something is a subgroup, we need to show three things. We need to show closure, we need to show the identities in there and we need to show inverses in there. In order to show that something is an equivalence relation we need to show three things. We need to show that the relation is reflexive, symmetric, and transitive. And it turns out that the fact that there are three things to show in both of those situations is not coincidental. What we'll show is that each of the three things that we have to show in order to show this is an equivalence relation will correspond to each of the three properties that we know are true because H is a subgroup of G. So first, what do we have to do? We have to show that this thing is reflexive. In other words we have to show that for each element of the group that the element is related to itself for every element of the group, reflexivity means. In other words we have to show, well what is the definition of the relation? We say that one thing is related to another in case when you write down the inverse of the first one, the inverse of the left one, and then you multiply it by the thing that appears on the right, well the thing that appears on the right in this relation happens to be A itself, is NH. That's what it means for two things to be related. I.e. we have to show, yeah but what's A inverse A? D, is the identity element in H? Sure is, why? Because it's a subgroup, H is a subgroup and property two of being a subgroup is that the identity element's in there. So the reflexivity of this relation boils down to the fact that property two of a subgroup holds, namely that the identity element's in there. Second, show that the relation is symmetric. In other words we have to show that if two things are related in this order then also they're related in the opposite order. That's what reflect, I'm sorry, symmetry of a relation means that if A is related to B, you have to convince me that B is also related to A. What does that mean in this context? I.e. show that if, well what does it mean for A to be related to B? The definition is that if we compute A inverse B and we get something in H then also the reverse relation holds. Well what does it mean to say that B is related to A? The definition of the relation is you take the first thing, the thing that appears on the left, you take its inverse, you take the thing that appears on the left, take its inverse, you take the thing that appears on the right and multiply it and you get something in H. That's what I have to show. But wait a minute, but let's see, if A inverse B is in H, that's the hypothesis, since H is a subgroup, property three of being a subgroup means that if you have something in there that its inverse is in there. Then the inverse of this thing is in there. Groups are closing their inverse. But wait a minute, but what is A inverse B inverse? Well I know how to get the inverse of the product to these two things, we've already looked at that today. Remember what do you do? You take the two things individually, you take the inverses individually, but then you reverse the order. You take the inverse of each one individually, so I'm going to have A inverse inverse and B inverse, and then I write them down in the opposite order. What's A inverse inverse? A, the inverse of the inverse is, here's what we've just shown. So A inverse B inverse is in H, but A inverse B inverse is B inverse A. In other words, by definition B is related to A. So what we've just shown is that the fact that subgroups are closed, there it is, the fact that subgroups are closed under inverses is precisely the piece of information that we need to show that this relation is symmetric. And guess what? The third piece of information we need about a relation in order to show it's an equivalence relation is transitivity, and the third property of being a subgroup, the one that we haven't used yet, is closure under the operation. Not surprisingly, those are exactly the two that get matched up here. Third, show that the relation is transitive. In other words, IE show that if A is related to B, and in turn B is related to C, then the conclusion is that A is related to C. That's what we have to show. And hey, we'll just write down the definition of what each of these pieces means, and then see where we can get. A is related to B means, by definition, A inverse B is in H. That's what this means. B related to C means the thing on the left, inverse times the thing on the right is in H. That's the definition of this. So here are the two pieces of information that I'm giving. The conclusion that I want to draw, what we want to show is that when you do A inverse C that you get something in H. That's the goal. But wait a minute. I've got this is in H, and I've got this is in H. It's hypothesis. Subgroups are closed. So by closure of a subgroup, that means the product is in there. A inverse B, B inverse C is in H. Closure of the operation of a subgroup. Alright, so technically I've got that with that in H, but now use the associativity. B, B inverse C is in H, associativity, but hey, guess what? That's identity. I get A inverse identity, C is in H, A inverse C is in H, and that's exactly what I needed to show. So the three properties of being a subgroup, that the identity is in there, that inverses are in there, and that the operation is closed, are precisely the three pieces of information we needed to show that this relation is in fact an equivalence relation. Now, anytime you have an equivalence relation, that automatically leads to what we call a partition of the set. So we get a partition of G into subsets. What we call the equivalence classes, equivalence classes of the equivalence relation. In other words, we split the group up into chunks. The things that appear in each chunk are the things that are related to each other in the given relation. So let's look at this partition. Yeah, this will be good. Let's look at a specific example. Let's look at a relatively interesting group. How about the group G is, I'm going to do S3 to start with. No, let's do Z6. Start with a community of example. And here's the subgroup that I want you to consider, the subgroup consisting of zero and three. So I've just handed you a specific group, and I've handed you a specific subgroup. They're sort of randomly chosen, but I don't want them to be too big so that we can actually do some of the computations here. And what I want you to do is find the partition of G generated by or produced by the equivalence relation. In other words, find the collection of things that are related to each other under this particular relation. All right, well, the first thing we have to do is make sure that we interpret what this means in the context of this particular group. It means that what you're supposed to do is deem two things to be related. In case you take the inverse of the first element, well, inverse is in the context of whatever the group is. Here inverse is going to be, well, I don't know, inverse mod six, so we're going to have to do the computation on each of these things. And then you're supposed to combine that with whatever the second element is. Well, even though I haven't written it here, remember, you have to put whatever the binary operation is. Here the binary operation is addition mod six. So let's go ahead and ask for those things that are related. Well, let's see, how about zero? What's zero related to? Is zero related to, well, of course related to itself because this particular relation has been shown to be an equivalence relation, so everything's related to itself. Let's just start looking at other possibilities. Is zero related to one? Well, what's the computation we have to do to determine whether or not zero is related to one? We have to look at the inverse of zero. Well, let's see, the inverse of zero in this group is zero itself because zero is the identity, so I'm just going to write it as zero. Then I'm supposed to combine that with, well remember the operation in this group is addition mod six with one and ask is that in the subgroup. That's the definition of what it means for two things to be related. This one's related to that one in case when you do the computation a inverse b, you get something in the subgroup. So question is one in the subgroup? No, it's not, so no. So I'll say that was easy, it was tedious, but you just sit down and start pounding them out. How about is zero related to two? Do the same sort of computation. You have to take the inverse of the thing on the left, well the inverse of zero is zero. You have to then combine that with the thing on the right. Is that in the subgroup? Is two in the subgroup? No, no, so no. Everything right, you may have to run through all possible pairs and sort of. Let's look at one that sort of gives a winner. Is zero related to three? We have to ask is it the case that when you take the inverse of zero, which is zero, and you combine it with three, do you get something in the subgroup? Yeah, we do. Is three in H? It is. So yes. So in this partition of z six, in this partition of the group that this thing generates, wherever I locate zero, I'm also going to locate three. Now importantly, because we've got a partition here, that's coming from an equivalence relation, folks, the fact that I've shown that zero is not related to one, and that zero is related to three, means that one is not related to three. Because when you've got an equivalence relation, if you've got two things that are related, then it has to be related to the third. So automatically, once I've done some of these computations, some other ones will come for free. Let me at least show you what happens here. It turns out, how about one? Is one related to, I don't know, three? I'll say no, it can't be since one is not related to zero. And because one is not related to zero, and one is related to three, one can't be related. Because one is not related to zero, and three is related to zero, one can't be related to three. But, for example, is one related to four? What do you have to do? You have to take the inverse of one. Well, the inverse of one in this group, in Z6, the inverse of one is five, because one plus five is the identity, one plus five is zero. I have to add that to four. There's A inverse combined with B. I get nine, but the group is Z6, so I get three, which is in the subgroup. And so the answer to this is yes. So it turns out one's related to four. So when I write down this partition, one and four will appear in the same subset. It turns out, and I won't cut to the chase here because we're running low on time here, also, it turns out you can show two is related to five for the same reason. The inverse of two is four. If I add that to five, four plus five is nine, which is three, which is in the subgroup. So two is related to five. And what you can show by continuing to compute is that one is not related to two, so that these two things, while related to each other, aren't related to these things. You can show that two is not related to zero. So here's the partition that we get. Partition of G, which is Z6, turns out to be these three subsets. Zero's related to three, one's related to four, and two is related to five. So what we've done is we've split the six elements of Z6 into three non-overlapping subsets, and each of the elements of Z6 appears in exactly one of these subsets. So I'm going to call this the first subset, the second subset, and the third subset. Questions? Comments? Now the computation we just did, we will wind up doing for any group and any subgroup, what we'll see on Wednesday is that even though the definition of how you're supposed to deem things to be related is, I mean it's easy, it's one equation. Just take the thing on the left, compute its inverse, combine it with the thing on the right, and ask whether or not the results in the subgroup, that that very basic definition leads to some incredibly nice structure. As it turns out, folks, it is not coincidental that each subset has the same number of elements in it. It's also not coincidental that one of the subsets is exactly the original subgroup, and so what we're going to start doing is seeing why that's the case and exploiting that in general. Okay, here is home, that's Monday, so it's time for a homework assignment, but this won't be due until a week after normal, due Wednesday, October 3rd, because of the exam that is happening on the day that it would otherwise be due, and here's what it looks like. In section 10, problems one through seven, nine and 10, 12 through 14, 27 to 33, most of these are computational, 37, 39 and 40, here are the ones I want you to turn in. Six and seven, those are actually a pair, nine and 10, those are actually a pair, 28 and 40, and a couple of quick hints on number 28, the hint is show that this, and this symbol will make sense as of Wednesday, AH equals HA for all A in G, do this by showing that HA is contained in AH, then by showing that AH is contained in HA, and again the notation will make sense as of Wednesday. On number 40, the hint is use, and this will make more sense on Wednesday as well, use Lagrange, that's the person's name, so that's the stuff from section 10, and then in section 11, problems one through seven, 14 and 46 through 49, and the ones I want you to turn in there are one, two and 14, parts A, B, C and D. Six, seven, nine and 10 if you're interested, you can actually start out, but I think your time might be better spent now in starting to prepare or review for the exam. All right.