 n a n and tau n a n, you have to show sigma tau is an a n. Good. Sigma is an a n doesn't mean that it equals 2m. It means that it can be written as the product of 2m transpositions and tau is the same thing and then blah, blah, blah. Couple of additional comments. If you have sigma as the product of a certain number of transpositions and tau as the product of a certain number of transpositions and you form the product sigma tau, the resulting permutation is going to be the product of the number that you use for sigma plus the number that you use for tau. You're not multiplying the number of permutations together, some of you missed that point. Similarly, I saw this on a couple of papers, it was a little bit unfortunate. When you write sigma as a product of transpositions, or it doesn't matter what you call it, if you write it as something like a1, a2, and then a1, a3, or something like that, and then a1, an, let's say, some of you wrote something that makes completely no sense. Equals, so this is meaningless. In fact, does a1 inverse mean, folks, these are just numbers, one, six, two, three, or something like that. So this notation is absolutely meaningless. Of course, what you're looking to do is put the inverse sign out here, that makes perfect sense. Then you're looking at this transposition and its inverse, and the inverse of the transposition is itself because it's a transposition. But boy, any sort of notation like that just makes no sense here. All right, next in question seven, here was the breakdown of points. Let's see, part A was worth a half. Parts B through E, one each. And then part F was worth a half for a total of five points. See any comments here? No, other than for question F, find a positive integer so that sigma inverse is sigma to the t. Well, because sigma cubed is the identity, sigma inverse is sigma squared, that's no big deal. What some of you said is that sigma to the fifth is sigma inverse. It is, that's OK. So maybe you were thinking geometrically. So there were actually a couple of different answers that were given there, both of which were correct. Two was the one I was sort of anticipating and looking for, but five is a perfectly good answer to question seven F. All right, let's see. The last page, page four, was worth 15 points total. I broke up in question eight. Part A was worth one. B was worth one and a half. C was worth one and a half and D was worth one. There's a total of five on that one. No issues there, you just pounded it out and didn't see any issues coming up. Question nine, both parts were worth one and a half. B, one and a half. What some of you did in part B, which is where you can compute the order, let's see. Using an easy computation, why? Because the two permutations, the two cycles are disjoint. That's why you can use the LCM game, the least common multiple game. If you didn't use the word disjoint somewhere, then you haven't given me appropriate justification as to why things should be true there. So be careful. The only reason you can use the least common multiple in part B, and it doesn't apply in part A, is because the two cycles here are disjoint. And let's see, in the true, false, the only thing that I saw come up more than once is in question E, some of you thought that that was false. Well, I mean, I know why that was true, too, because there was a similar question on the practice exam where the word cyclic didn't appear. So G1 and G2 are groups that have the same order than their isomorphic is not true. But if they're cyclic groups with the same order than they are isomorphic, so that should be true. All right. So in the end, it becomes obvious that when I sit down and write up an exam, I don't really worry about how many points total are on there. This one happened to be worth 47. That's fine. Usually the total number's around 50 or so, but that doesn't necessarily have to be the case. Now, remember the way I assign grades in here. If you're up over 90% when it's all said and done with homework and quizzes and exams and this and the other thing, then you're guaranteed to some sort of a, might be a minus or something like that. That's OK. And if you're up over 80%, you're guaranteed to some sort of b, et cetera. I don't give letter grades on individual exams, but I'll at least give you an idea of how you did as a group, as a group, as a collection of students, just to give you some clue as to where you sit compared to your buddies. And it turns out that as a group, you did extremely, extremely well. So it turns out that the average, the mean on this exam was 39.3, which is right at 83.5%. That's really, really nice. And just for your info, let's see, I've sort of cut the lines here at 90, 80, 70, but again, I'm not implying that one's an a, one's a b, or anything like that. Just 42 and above, there were 1, 2, 3, 4, 5, 6, 7. There were 9 of you. That's really super. 38 to 41 and a half. There were 1, 2, 3, there were 6 of you. 33 and a half to 37. There were 1, 2, 3, 4, 5, another 6 of you. And below 3 of you. And that's the total number of students in here. So again, there's no letter grades implied here. It's just roughly here is how you did as a group. All right, two comments, because this is the first exam that you've gotten back from me. Comment one is there's nothing in concrete here. If you have any questions on what I did, why I marked things off, or why you lost points somewhere, if you can't read my writing or something like that, please feel free to come and see me. I encourage you to come in and ask questions on the specifics. I guess I'd suggest first take your exam, compare it to the solution sheet, and see if you can figure out on your own why it is that maybe you didn't get full credit for a question, or why something sort of went south. And the second comment is there's no statute of limitations here. Because remember, and I'll certainly talk about this in more detail near the end of the semester, the way exams work in here is you get exam one, then you get exam two whenever that happens, end of October, beginning of November, something like that. Then exam three will happen during the final exam period. It's a 75 minute exam that'll cover the last chunk of the course. But then also you have this opportunity to take the sort of optional final if you want to replace your exam one or your exam two score. And for example, when you're perhaps preparing for that optional final if you decide to take it, if you see something on this exam one that you don't understand or you're not sure why, come in then. I don't care. That's fine. If you deserve points back, I want to give them to you, folks. I'm not trying to play gatekeeper here. Anyway, take them home. And if you have questions, come and see me or send me an email and we can make an appointment. But I was really pleased with this. It's clear that most, if not all of you are working really hard at this stuff. Someone asked, as we were leaving the exam, the exam looked a lot like the practice exam. Do you always do that? I mean, it did look a lot like the practice exam, but those are the important ideas. And if you've got them down, I'm obviously not copying and pasting. I'm throwing the word cyclic in or not cyclic, or I took the permutation in one of them. So you're doing the same flavor of problems, but you're not necessarily doing identical problems. I don't know when exam two comes around or exam three I'll certainly be passing out practice exams. Will they look as similar? Maybe. I mean, they don't have to, but it's not uncommon for that to happen. So there's merit to preparing using the practice exams. Any questions or comments or remarks? OK, very nice. Very, very nice. So we move on. Let me remind you where we were a week ago tonight before the exam. So recap. Is this? Well, we talked about a new way to build groups. The idea is, if you start with a couple of groups, or in fact maybe a whole bunch of groups, that you can somehow hammer them together to get a new group, it's easy to do. You just form what you learn to do back in Math 2.15 and discrete math. You simply form the set Cartesian product of the sets. And you say, oh, but yeah, these are more than sets. They happen to be groups. So can we put a group structure on the resulting Cartesian product? And the answer is yes. In fact, the operation is pretty easy. You just do everything what we say component-wise. If you have a couple of pairs and you want to combine them, you just combine them in each component. And that makes sense because each of the components is assumed to be a group. And so what we did was we looked at the following sort of interesting question. If you know something about each of the pieces that are used to make up this product, whether it's just two pieces or more than that, can you say something about the resulting product in terms of the individual pieces? And the answer is sometimes yes and sometimes no. For example, if all of the pieces that you use to hammer together are each a billion, then the results are going to be a billion, it turns out. But it might be the case that if you hammer together some cyclic groups, the result might be cyclic or might not be cyclic. And that's what we showed at the end of last Monday. So we formed what we called the direct product or the Cartesian product of groups, G1, direct product, G2. Sometimes there's only two of them, but there might be more of them, G sub t, any number of them. Maybe you're hammering six groups together or something like that. And what we saw is you can do this process, folks, in any situation you want, whether the groups are billion or not a billion or finite or not, it doesn't matter. Just hammer them all together. But what we're focusing on for now, focus on for now, is what happens if each of the groups that you use, G sub y, happens to be a billion, ah, happens to be cyclic. And what we observed a week ago, observed this, is that the result might be cyclic or it might not be. Maybe the result is cyclic, the result is cyclic, but maybe not, but maybe not. And the examples were, if we computed z2, direct product with z3, well, there's no issue in trying to figure out how many elements are in the direct product. This is just a statement from Math 215 about sets. The number of elements in this group is six. Question is, is it a cyclic group? The answer turned out to be yes. We actually sat down and found a generator for this group. So it turned out this is cyclic. We just showed that by pounding out a generator. In fact, 1, 1 is a generator. Not the only one, but it is one. But on the other hand, sort of a non-example, example, if I look at the direct product z2 cross z2, this is not cyclic. So sometimes if you take two cyclic groups and you hammer them together, the result is a cyclic group. And sometimes when you take two cyclic groups and you form their direct product, the result is not a cyclic group. And what we're going to do tonight is identify when it's the case that if you take the direct product or you form the direct product of two cyclic groups, that the result is cyclic. And from that, we're not going to prove all of the details of this big result. But from that, what we're going to do is write down, surprisingly, all of the abelian groups that have a specified number of elements in it from this result. So it turns out, proposition, proposition that if I hand you two integers m and n, if we look at, well, let's call them p and q, or how about r and s, z sub r cross z sub s. So if I hand you the cyclic group having that many elements in it, and I hand you the cyclic group having that many elements in it, then it turns out that this is cyclic, and I'm going to put in parentheses before I get to the punch line, and therefore necessarily isomorphic to z sub r times s, if and only if the g, c, d, r and s is 1. Turns out the only time if you take the direct product of two cyclic groups that the result is a cyclic group is when the two numbers that you have handed me are what we call a number theory business, relatively prime. They may not be prime numbers individually, but the point is that there's no number bigger than one device above them. So look, if I hand you two and three, those are relatively prime, and so the result turns out to be cyclic. On the other hand, two and two are not relatively prime, so the result is not cyclic. I'm not going to prove this proposition for you, but the intuition is, yeah, if you're going to try to find the order of this particular element in the group, it turns out if the two numbers that you've handed me are relatively prime, the behavior here is almost similar to the behavior that happened when we were looking at orders of permutations that are written as disjoint cycles. You sort of keep computing, keep computing, and you're trying to figure out when it is the case that you get back to the identity element. Well, for permutations, the identity element looks like I. And one of them sort of worked, but then that one was in the order wound up having to be the least common multiple. It turns out here, if you hand me this element, if the two numbers that you've handed me are relatively prime, in other words, if their greatest common divisor is one, then necessarily their least common multiple is the product. So the least common multiple of these two numbers turns out to be six, and you can show that that implies that the order of this element is six. And as soon as you've got an element of order six inside a group of six elements, then necessarily it's a generator. So that's intuitively what's going on here. But again, I don't want to get into the details of the proof. It says, though, that you have a pretty good idea of when it's the case that you can take a group like this and sort of replace it by or trade it in for something like this. So if somebody hands you z2 cross z3 and somebody else hands you z6, those two groups are essentially the same. Of course, they're not written the same, but they're isomorphic, meaning what? That if you relabeled all the elements of z2 cross z3 appropriately, and you relabeled all the elements of z6 appropriately, that the two would look indistinguishable. The same beast. On the other hand, if I hand you z2 cross z2, well, it turned out z2 cross z2 looked indistinguishable from this group that we've been calling v for the last three weeks. It's the group that has the property. It's got four elements. It's got an identity element, of course. And all of the non-identity elements have the property that when you combine them with themselves, you get the identity. In other words, a squared is the identity region. But those you can't somehow relabel or reconstruct so that one looks like the other. z4 and z2 cross z2 are different beasts. But z6 and z3 cross z2 are essentially identical because they're basically. So the question is going to be this. If we keep going, can we somehow, rather than stopping at just the direct product of two cyclic groups, if we somehow hammer together three cyclic groups or four cyclic groups or 18 cyclic groups or something like that, can we devise a way that we can somehow move around or trade one in for the other and say, well, this one, if you were to write out the group table, looks like this one. Or this one, when you write out the group table, doesn't look like this one. And the answer turns out to be yes. And it all boils down to this particular remark. Whenever you hand me the direct product of two groups, if the two numbers are relatively prime, then you can trade it in for one group, the cyclic group. If the two numbers are not relatively prime, then you can't trade it in for that cyclic group. And the point is it turns out that you can play that game regardless of whether you've taken just two groups or whether those two groups somehow appear in the middle of maybe the direct product of three groups or four groups or more. So it turns out this extends to more than just two groups and just the direct product of two groups. Bring a better pen next time. Here's one, maybe this will work. Direct product, a little better, two groups. And let me just explain what that means by presenting an example here. Example, if I look at the group Z2 cross Z3 cross Z4, so it makes sense to form the direct product of these three groups, let's see how many elements are in here. That's easy, six times four, there's 24 elements in here. Question, if I hand you another group with 24 elements like Z6 cross Z4, there's a group with 24 elements. Is that group isomorphic to that group? Well, here's another group with 24 elements. Is this group isomorphic to that one? Or is this group isomorphic to that one? Here's another one, how about Z8 cross Z3? Is this one isomorphic to that one? Here's another one, how about Z2 cross Z2 cross Z3? So what I've done here, folks, is I've written out a bunch of different groups, each of which has 24 elements in it. Heck, I'll give you another one. That has 24 elements in it. And the question we can start asking is, look, here are some naturally occurring groups. They're groups that I can form just by taking some cyclic groups or direct products of cyclic groups or something like that. And the question will be, can we identify which groups on this list, if any, are isomorphic to each other? In other words, can we identify whatever might be viewed as a repeat on this group or on this list? Well, here's the idea. Two and three are relatively prime. So this piece is, and here's the symbol. I don't know if I've used it this semester, but I'm going to start using it more and more. When we talk about two groups being isomorphic instead of having to write the word out all the time, I'll use this symbol, it's an equal sign with a little squiggle on top. This group is isomorphic to Z6. So what this comment means is, if you have a chunk that you can trade in for a cyclic piece, then you can drag along any additional direct products that might be hanging on. So that, because this piece is isomorphic to Z6, this whole thing is isomorphic to Z6 cross Z4. So in fact, these two are isomorphic to each other. Yes. Isomorphic. I need to write down one more here, folks. I forgot this, I had this. I want to write down Z4 cross Z6 as well. All right, let's see. Question is Z6 cross Z4 isomorphic to Z24. No. Reason? The GCD of 6 and 4 is 2. Because the greatest common divisor of these two numbers is not 1, you can't trade that in for this. So this one is not the same as that, which then necessarily means it's not the same as that. How about this one and that one? I don't know. How about this one and that one? Do you like that? Yeah, because the GCD of 8 and 3 is 1. So these two are the same. It's in blue here. So these two are isomorphic by using the same result. But look, folks, we determined this wasn't the same as that. So therefore, this can't be the same as that. Because isomorphism is what we call an equivalence relation. If you have two groups or isomorphic and you have another one that's isomorphic to this one, then these would have to be the same. And because we've established that this isn't the same as that, but this is the same as that. It's just like equality. All right, let's see. How about this one? Z2 cross Z2. Oh, boy. Well, how about this piece? Z2 cross Z3, that looks like Z6. So at least I can trade this in for Z2 cross Z2 cross Z6. But I don't see that up here anywhere. So I don't see how this can be isomorphic to anything else that's on the list. And I say, well, wait a minute, how about this one? No, that's no good. Because this is Z2 cross Z2 cross Z6. This isn't Z2 cross Z2 because the GCD of 2 and 2 and 2. In other words, I can't trade in that piece for this one. So this one sits by itself. Not that there aren't other groups that you could write down that this would be isomorphic to. For instance, this is isomorphic to Z2 cross Z2 cross Z6. But I simply haven't listed that one out here. How about this one? Z4 cross Z6. Yeah, it turns out the following is true, folks. Regardless of whether the GCD is 1 or not 1 or anything, if you take ZR cross ZS or ZS cross ZR, it doesn't matter which order you do things in. So the process of forming direct products is independent of the order you do them in. So let me slide that proposition in here. Proposition ZR cross ZS. And I'm going to use this notation, this squiggly line notation. This is totally standard, folks. Is isomorphic to ZS cross ZR always? Always. So it turns out that the order that you trade things in for doesn't really matter. How about, let's see, Z4 cross Z6. Oh, yeah, that's already up there. So this one is already here. That's isomorphic to this one, which in turn, then, is isomorphic to that one. So what have we got here? We've got what we'll call equivalence classes of isomorphism. This particular group happens to be isomorphic to this one, which happens to be isomorphic to this one. So if somebody asks you to write down a billion groups having 24 elements, folks, it's unreasonable to think that you should write down this one and this one and this one because they represent the same structure. They're isomorphic. So all we're going to do is ask you when we're asking you to write down groups that have the a billion property together with a certain number of elements, we're going to ask you to write down just one of those groups. And we'll phrase that as uptysomorphism. So uptysomorphism here is one group. You might say, well, should I call it that or should I call it that or should I call it that? And the answer is what we typically agree to do is call it whatever you get so that if you can take one of the numbers that appears here and break it up into a product of two relatively prime numbers, you go ahead and do that. We could call this Z6 cross Z4. It's a perfectly good name for this group. But because 6 can be written as 2 times 3 and 2 and 3 are relatively prime, we typically like to trade it in for that. All right, we've got this one. Let's see, we've got Z24 or Z8 cross Z3. Of course I could call this Z3 cross Z8. There's a lot of things. So here's another one. Here's one that can't be isomorphic to anything else in the list, right? Because this is not a billion and everything else is a billion. So here's this process. We can construct a lot of a billion groups using cyclic groups. But here's a good question. If somebody hands you an abelian group and tells you how many elements are in it, do you know which group they're talking about? Definitely not. If somebody says I'm thinking of an abelian group that has 24 elements, question? Megan, question? Please. John? Why it's not? Okay, this chunk of Z2 cross or this chunk of it we can trade in for Z6. So it might be isomorphic to this, but the problem is we're left over with Z2 cross Z2 which we can't trade in for Z4 because these are not relatively prime, two and two. So this chunk can't be traded in for that. Similarly, these three together can't be traded in for that because the twos are not relatively prime to each other. So think the only way you can move around in these diagrams is if you've got two numbers that are relatively prime. So I can trade this in for that but I can't trade that in for that. Now, is that clear things up a little bit? Let's see, so this we can trade in for Z2, Z3, Z4. That's what isomorphic remains. Well, all I'm worried about is that piece there. So the two, Z2 cross Z3 is Z6. I mean two and three are relatively prime, so I can trade. And then the point is if you've got another piece that you're dragging along, you just drag along the same piece. All right, so where from there? Oh yeah, here's where from there. So yeah, think of the question this way. Question, if you're told that you have your friend, this is a better way to think of it, is thinking of an Abelian group, bless you. Thinking of an Abelian group and your friend tells you how many elements are in it and you are told how many elements are in it. In other words, the order of the group. Do you know the group? I have somebody says, all right, I'm thinking of an Abelian group and it's got 24 elements. Do you know which group your friend is thinking of and the answer is no. Because your friend might be thinking of Z2 cross Z3 cross Z4 or your friend might be thinking of Z24 or your friend might be thinking of Z2 cross Z2 cross Z3. I'm sorry, I didn't say enough Z2. Z2 cross Z2 cross Z2 cross Z3, that's got 24 elements. So the answer is in general, no. In general, no. And the example is if the number of elements is 24, if N equals 24, then you don't know because I've given you a lot of examples of Abelian groups. So let me take one step back just to give you some context. The question is how much do you need to know about a group in order to know what the group is? And when we say know what the group is, all right, the group Z2 cross Z3 and the group Z6 are treated as the same group here. They may be written slightly differently but because they're isomorphic because they have the same structure because there's a way of relabeling them to describe the same structure, we don't really wanna call those different. So the first question to ask is, is knowing the number of elements in the group enough to tell you what group it is? And the answer, of course, is definitely no. In general, if somebody says I'm thinking of a group with six elements, do you know what the group is? No. It might be Z6 or it might be S3. There's two different groups with the same number of elements. Of course, you might look out if somebody says I'm thinking of a group with five elements in it, do you know what the group is? The answer to that question is yes, I know what the group is because we proved that if you have a group that has a prime number of elements then necessarily the group is cyclic. So somebody says I'm thinking of a group with five elements, you know exactly what group they're thinking of, Z5, because that's the only possibility. If I ask you to, I don't know, if I tell you I'm thinking of a group with six elements, forget it. You don't know if it's S3. If I'm asking you to think of a group with 24 elements, then it's big, forget it. I mean, there's lots of different groups up here. So the answer in general is no. Just knowing the number of elements in the group doesn't tell you what the group is. But here's a more reasonable question. If I tell you that I'm thinking of an abelian group and I tell you how many elements are in it. So now I've got two sort of pieces of information about the group I'm thinking of. A, that it's a billion and B, how many elements are in it? Is that information alone enough to tell you what group you're looking at? And the answer turns out to still be no. Here's an example. But here's the incredibly nice thing about the adjective abelian. If you just focus your attention on abelian groups, which means admittedly folks that you're leaving out a lot of things. But if you just focus on the groups that happen to be abelian, there's sort of a miraculous thing that happens and we're not gonna prove it in here. The miraculous thing that happens is the following. I know how to construct. I know how to explicitly, concretely write down a lot of abelian groups. Take cyclic groups and form their direct products. So think, I've got this sort of building block process which allows me to produce a lot of abelian groups. The miracle is, in fact, that gives you all the possible finite abelian groups. The process of taking cyclic groups, those are special. And forming direct products of those, that's a very concrete process. In fact, in the end, not only gives you a lot of possible abelian groups, it gives you, up to isomorphism, all the possible finite abelian groups. And that statement, and again, it's not really beyond the scope of this course, but it would take us a full class period to run through the proof and I don't wanna spend the time doing that. Other than just have you understand the, I don't know if the miraculousness of the result is, turns out, turns out, the following is true. Any finite abelian group is isomorphic to to the direct product of cyclic groups. Now, this is usually referred to as the fundamental theorem of abelian groups, theorem of, I'm gonna call it finite abelian groups. And we're, this is the third time this semester that my notation and statement is gonna be on the surface sloppy, but the interpretation will mean exactly what you would hope. Folks, I'm not saying that if you hand me an abelian group, it's necessarily the direct product of more, of two or more groups. For instance, if I hand you the abelian group Z7, it's not the direct product of cyclic groups. It's already a cyclic group. It's just like when we said, if you have a permutation, it can be written as a product of cycles. It might be a cycle already, or if you have an integer bigger than two, that it's a product of prime numbers, except for if it's a prime number already here. If you have an abelian group, if we say it's the isomorphic to the direct product of cyclic groups, it might be a cyclic group. So I could put this in, but I'm just trying to get you sort of speaking mathematics ease here. It should technically say any finite abelian group is either already a cyclic group or is isomorphic to the direct product of two or more cyclic groups, but that just gets too cumbersome, right? So it's a miracle. It says that these very basic abelian groups, the cyclic ones, together with this very concrete construction, direct product, allows you to describe all possible abelian groups. Now there's some significant extensions to this. We can talk more generally about what are called finitely generated abelian groups, rather than just finite abelian groups. When we talk about finitely generated abelian groups, that allows us to include the integers and direct products of the integers and these are bands and those sorts of things. So there is a more general statement that we could make that, again, is not beyond the scope of this course, but would take us too much time to get there. And then what happens beyond what are called finitely generated abelian groups? Well, then there's other abelian groups, infinitive abelian groups, like ones that you gave on the exam, maybe the rationals with addition or the complex numbers, non-zero complex numbers under multiplication or something like that. Those might not be isomorphic cyclic groups. In fact, the structure of abelian groups in general is really intricate and really interesting. In fact, that's what Professor Rangaswamy does for a living. He studies the structure of abelian groups and you can talk about certain properties of some that are shared by some and not by others. All right. So look, if somebody says, I'm thinking of an abelian group and it has a certain number of elements in it, 24 elements or 120 elements or 50 elements, do you necessarily know what group they're thinking of? No. But what this allows you to do is say, all right, you've got an abelian group, therefore it's the direct product of cyclic groups by this fundamental theorem. Can I write down what the cyclic groups are? Can I write down what the possibilities are? And the answer is yes, because we can just systematically go through this process and say, hey, how are or how is it possible to take a direct product of cyclic groups and get 24 elements in it? Well, it looks like there's a lot of different ways, but we can do it in a systematic way and I'll show you how to do that. So example, let's see example. Find all abelian groups having, let's do, 36 elements. And I'm going to put in parentheses, uptie some morphism. Don't list two isomorphic copies of the same group. So in fact, what we're asking is the same question that we looked at some examples of, but instead of asking it for groups having 24 elements, we're going to look at a question for groups having abelian groups having 72 elements. Well, I can think of one right off the bat, Z sub 72. The cyclic group of 72 elements is an abelian group having 72 elements. So when it's all said and done, at least that group's going to come up, but hey, right off the top of my head, I can think of a lot more, Z8 cross Z9 or Z2 cross Z2, oh, I'm sorry, 36. Okay, let me do that again. I can think of one that has 36 elements. Z sub 36. I have 72 in my notes, but this will be a little cleaner example. So Z sub 36 works. Z sub 4 cross Z sub 9 works. Z sub 2 cross Z sub 2 cross Z sub 9 works. Z sub 6 cross Z sub 6 works. I mean, so I've written down a lot of examples. The miracle, again, is by simply writing down all those possible direct products of cyclic groups, the fundamental theorem says I'm guaranteed to have written down all the possible abelian groups that they all have to arise as a direct product of cyclic groups. So that's an incredibly nice simplification that we can make. Now the only extra step in the process is we have to come up with a systematic way of listing all these out so that we've made sure we've captured all of them and that we did any more than once. And the systematic approach typically is this, folks, take the number that you're given, write it in its prime factorization, so 36 is 4 times 9, there's its prime factorization, and then simply ask how many ways are there of taking 2 squared times 3 squared and writing it in chunks that are somehow, well, I won't say relatively prime, that are either identical or distinct as far as the factorization goes. For example, I can write 36 as, well, I can write it as 2 times, yeah, okay, so 2 times 2 times 9, I can also write it as 2 times 2 times 3 times 3 and those two I'm going to view as different because, well, 3 times 3 and 9, well, 3 times 3 is obviously 9, but if I had Z3 cross Z3, I can't trade that in for Z9 because these aren't relatively prime. I can write this as 4 times 9, I can write it as 4 times 3 times 3, so here are a bunch of factorizations of 36. In the end, folks, there's lots of other ways to do it. I mean, I can write it as 6 times 6 or I can write it as 18 times 2, but in the end, there are only 4 factorizations of the number where you've written things either as primes or prime powers. So I've got 9 or 3 times 3, I've got 2, I've got 4 or 2 times 2 and so in the end, here are all 4 of them, so here they are. Z2 cross Z2 cross Z9 corresponds to this first factorization. Z2 cross Z2 cross Z3 cross Z3 Z4 cross Z9 and Z4 cross Z3 cross Z3 and here are all of them. So what happened to Z6 cross Z6? Well, the claim is that I've already got it captured somewhere on there. Well, yeah, let's see. This is the same as what? Z2 cross Z3 I can trade this piece in and Z6 is Z2 cross Z3 I can trade that piece in but let's see the order that I write these things down is irrelevant. So Z6 cross Z6 is actually already on the list and it's isomorphic to that one. Z2 cross... I'm just making some up and asking, alright, here is one. Here's a direct product of a billion groups. It's got 36 elements. You've claimed that this is all of them. You better make sure this one's already on the list. It is, let's see. Z18 is isomorphic to what? Z2 cross Z9. It's not isomorphic to... Let's see how it's going to do this. It's not isomorphic to Z3 cross Z6 because 3 and 6 aren't relatively prime but it is isomorphic to Z2 cross Z9 and so let's see. Is that one already on the list? Yep, right there. Well, how about Z sub 36? There's going to be a group of 36 elements in it. Is it on the list? Where is it? 4 and 9. So in the end, all you need to do is take whatever integer you're given, write it in all its possible prime power factorizations where the prime powers are viewed as different if you write 9s 3 times 3 and our 4 is 2 times 2, etc. Write it out in all possible configurations you can and then simply write down direct products of cyclic groups based on those factorizations. What's unfortunate is there's no nice formula that says if you give me the number of groups that there are, I'm sorry, if you tell me how many elements you want to be in the group like 36 here that you can somehow predict how many of being groups there are having that order. I mean, there is a formula for it but it's nasty enough that it's not worth trying to memorize or use or something like that. So, look, there's four groups, four being groups that uptystomorphism having 36 elements. There's only one of being group uptystomorphism having 37 elements. There turns out to be only one of being group uptystomorphism having 38 elements because 38 is 2 times 19 and then there's only one factorization of that. It's 2 times 19 and 2 and 19 are relatively prime. So, let's see what would be next. 39, oh yeah, 39 is 3 times, there's only one of being group uptystomorphism having 39 elements. There's a lot more having 40 elements because there's a lot of ways of writing down 40 in a prime factorization. Yeah, David, a question? Okay, here you go. So, unfortunately, there's no, oh, the number 36 means that I'm looking for four groups or something but what you see after doing enough examples of these and those are included in the homework is that, all right, I see the pattern. I sort of have to break down each of the prime powers and all its appropriate factorizations and go from there. All right, that is good, good, good. So, let me give you, yeah, let me give you a little bit of an overview and we'll start on the next topic tonight because we've still got 15 minutes left. What we're gonna do is put, this is chapter 11, put chapter 11 to bed. This was the fundamental theorem of finite and being groups and how to describe all the being groups having a certain number of elements in it. We're gonna skip chapter 12 for now. We might pick up some of those ideas later on in the semester but what we're gonna do is start essentially a new idea. This is chapter 13, it's called homomorphisms. This is actually a new section. Intuitively, homomorphisms are the analog to groups of what linear transformations are to vector spaces. Homomorphisms are ways of getting from one group to another. All right, so you're saying, well, you mean we're just looking at functions? Yeah, we are looking at functions but not just any sorts of functions. The intuition here is the same as the intuition was for linear transformations in vector spaces. You know, linear transformation between two vector spaces and nothing more than a function between the two vector spaces but it's a function that somehow understands that what's going on is not just two sets being connected but two vector spaces being connected. In other words, it's a function that says whatever the operations of the vector space are over here are somehow preserved by the function corresponding to the operations of the vector space over here. So linear transformation is what? T of v plus w is T of v plus T of w. In other words, the transformation respects the one operation called plus and it also respects the scalar operation. Well, a homomorphism is nothing more than a function between two groups that respects or preserves or keeps intact the underlying binary operation. And what we're going to do is look at examples of homomorphisms between groups, understand what they might represent maybe geometrically and then look at various ideas or various quantities associated with these special sorts of functions and it turns out those special quantities like certain subgroups that get kicked out of these things will play an important role so that eventually by the end of this new section not only will we have some sort of idea of what goes on with functions between groups but we'll actually come up with a bit of producing yet another group. Not just subgroups or factor groups. No, I'm sorry, not just subgroups or direct product groups or this that. And the other thing we're going to eventually write down what are called factor groups and that'll take us a little while to develop. So here's the idea. I'll call one of them G1 and G prime and you know what folks, I'm going to revert back to our technically correct notation. Our groups. So the binary operation on G and the binary operation on G prime might be completely different beasts than a function. And we usually use Greek letters here. Phi from G to G prime is called a homomorphism. Homo meaning same and morphism meaning structure. In case, well let's see, given the little introduction I just gave you the hope would be that you'd be able to actually write down what it means. It's a function that somehow preserves the binary operation. In other words, if I take two elements of G let's call them x and y. So take two elements in the domain. Domain is just a set. But hey, it's a set with more. It's a set with a binary operation. What I'm going to ask you to do is perform the binary operation on these two elements. So you get another element of G and then run it through the function so that what gets spit out is something over in G prime. I don't know what it is. Well folks, here's another way of producing something in G prime. Take each of these two elements individually. Don't worry about somehow combining them in the context of the binary operation yet. Take them individually. Run each of them individually through the function. Well that's going to give you two elements over here. And then combine them over in G prime. So that's two ways of somehow taking these elements and hooking them up with the given function and the requirement is in order for G to be called a homomorphism that somehow the function has to behave correctly regardless of which pair of elements you start with. For all x comma y in G. That's what it means for a function to be a homomorphism. Structure preserving. That whatever the binary operation is if you decide to combine the two things first inside the domain group G and then move the result over into the group G prime you get something in G prime. Another way of attacking the problem is take the two things that you're handed in the input group run each of them individually over to the new group G prime and then combine them over there. If you get the same answer both ways then we call the function a homomorphism. Let's write down some examples. Example. Let's write down the easiest one first. Yeah. Here's a homomorphism. Define phi from this group. So by default I've told you what the operation is because it's the integers. There's only one operation. To some other group in this case the two groups are going to be the same. There's certainly nothing that precludes G and G prime from being the same or star and star prime from being the same as follows by setting phi of any little z equal to 4 times z. And let's show that phi is a homomorphism. I'll abbreviate that HOM homomorphism. Well it's certainly a function from the integers to the integers that's no big deal. If you take an integer and you multiply it by 4 you get another integer. The question is, is it the case that multiplying by 4 preserves whatever the group operation is in the underlying group? And again, the operation here is plus and the operation here is plus. Same group. So it showed it is. Well what do we have to do? To do this, pick any two x and y in the input group. We have to show or prove that phi of x plus y plus because that's the operation of the group equals what you get if you were to do the operation to the two of them individually. Now here folks, this plus is being interpreted as happening inside the domain. This plus is being interpreted as happening inside the range. So we have to show that. Well let's see. Well, phi of x plus y, by definition, this function does what? It multiplies by 4. It's 4 times x plus y. But phi of x plus phi of y is 4x plus 4y. Of course this is just 4x plus 4y. So the left-hand side equals the right-hand side. Check. So phi is the homomorphism. Supplying by 4 gives a homomorphism. Let's do a non-example just to keep you honest. Non-example, define, let's call it something else. How about Greek letter psi from z to z? Psi of little z equals 4 plus z. So it looks like I'm doing the same sort of thing as before. It certainly is a function from the integers to the integers. You take an integer and you add 4. You certainly get another integer. But let's see. If I do psi of x plus y, x plus y is what? Is 4 plus x plus y? On the other hand, if I do psi of x plus psi of y, that's 4 plus x, and that's 4 plus y, which gives 8 plus x plus y. Are these two equal? Technically, what should I do at this stage? It doesn't look like these are going to be the same. Better write down a specific counter example. So example, if x equals, it doesn't matter what you pick. How about x equals y equals zero? Just make them both zero. Then psi of zero plus zero. Psi of zero plus zero is psi of zero, which is 4. But psi of zero plus psi of zero is 4 plus 4, which is 8. And so psi is not a homework. I've written down a specific counter example by giving you two integers x and y. They happen to both be zero. That's fine. And I've shown that when I run those two integers through psi of x plus y, I get something different than if I had to run them through psi of x plus psi of y. So I'm just trying to clean up and drive home again this same sort of idea. If you're going to convince me something's not true, write down a specific counter example and convince me that's not. Here's the specific counter example. You want to write down a different counter example? Sure, there are many. In fact, most anything will work. Questions? Comments? All right, let's try one more example. This one's going to be interesting. These homomorphisms, well, in the situation that I wrote down, the one that worked was a homomorphism from a group to itself. Here is a significantly more interesting one. For example, I'm going to write down a function. Let's call it tau from Sn to Z2. So what I want you to do is start with any n bigger than or equal to 3. I don't care which one you pick. Pick n equals 7 or n equals 4, n equals on or on. I don't really care. And what I'm about to do is write down a function from Sn to Z2. And here is what the function does. Tau of, let's see, what am I looking to put in here? A permutation. So the notation looks sort of strange. So here tau denotes a function from Sn to Z2. So it's waiting for you to input something in Sn. That's what the things in Sn are typically denoted as. I don't want to use a tau here because I don't want to use it twice. Equal to, there's two possibilities. I want you to spit out 0 if Sn is even, and I want you to spit out 1 if Sn is odd. Piecewise defined function. If you input an even permutation, then I want 0 to come out, and if you put in an odd permutation, I want 1 to come out. So at least it's a function from here to here. Why is the function well-defined? Well, it's this big theorem that I apologize for not proving for you. This is a big theorem. It says, if you hand me a permutation, there might be many different ways of writing it as a product of transpositions. But either, I don't care how you do it, either each way that you do it is always going to involve an even number of transpositions, or it's always going to involve an odd number of transpositions, but it's never going to be both. So there can't be any argument about which of these two numbers is supposed to come out. So the well-definedness of this particular function is actually a relatively deep result. But we're using that as known. It's really going on under the scenes. Okay, that's fine. Now the claim is going to be that this particular function is a homomorphism. So we'll prove that tau is a homomorphism. And the reason is, well, you have to show me that regardless of what two permutations you hand me, sigma one and sigma two, that tau of sigma one. Okay, well, let's see. So pick any two permutations, any two elements of Sn of Sn. What do you want to call them, X and Y? Yeah, if you want, that's fine. But typically we use Greek letters to denote elements of Sn. How about tau and sigma? Well, I've already used tau for something else. How about sigma and kappa? That's fine. Let's call one of them sigma one and the other one sigma two. There's two elements of Sn. Now, in this particular case, it turns out that there are few enough possible outputs that we can go ahead and pound this thing out just by brute force. So there's really only four possibilities. There are what? Let's see. Sigma one even and sigma two even or sigma one is even and sigma two is odd or sigma one is odd and sigma two is even or finally that they're both odd and sigma two is odd. So in this particular situation, if you hand me two permutations because all you're interested in is whether or not they are even and odd that's all you need to define what the function does. Let's just look at what happens in each of the situations. So if I compute in this case, let's say in case one here this is good. So let's call this case one, case two, case three and case four. So really all we have to do is check that in each of the four situations that the equation that we need holds true. And let's see if it does. Well, in the first situation, what is tau of sigma one, sigma two? Let's see. If sigma one is even and sigma two is even then what is sigma one, sigma two? Is it even or odd? It's even because the product of two evens is even. That's what you're proving in the exam. So because this is even, this gives zero. On the other hand, what's tau of sigma one? Well, let's see what are we supposed to do? We're supposed to do the operation as whatever it is inside the target group and the target or the codomain or the range group is z two. So when I do the operation to the two things that come out what I'm asking that in this particular case is add in z two. Well, let's see. Sigma one is even, so tau of it is zero. Sigma two is even, so tau of it is zero. And the question is, does zero equal zero plus zero? Yes, it does. So it worked out fine. Not bad. Case two and case three and case four, I'll just knock out these. These will be totally quick. We'll get out of here in about a minute or so. Case two, tau of sigma one, sigma two. Well, wait a minute. If sigma one is even but sigma two is odd, folks, if I do the product of an even permutation with an odd permutation what do I get? It's odd because I've got an even number added to an odd number and an even plus an odd is an odd. So this is odd. So tau of it is one. Well, let's see. On the other hand, if I do tau of sigma one plus tau of sigma two, let's see. If sigma one is even, that's zero. If sigma two is odd, that's one. I get zero plus one, which is one. That worked out good too. Case three, I won't go through it similar to case two. Case four is the interesting one. What happens if I do tau of sigma one, sigma two, where both sigma one and sigma two are odd? Let's see, folks, if you take two odd permutations and you multiply them, what do you get? An even. If I've got, this is the product of seven transpositions and this is the product of five transpositions and I nailed them together and I've got the product of twelve transpositions. An odd plus an odd is an even. So in this particular case, this is even because the product of two odd permutations is even. Well, let's see what happens over here. Well, let's see. If that's even, then by definition, tau of it's zero. But on the other hand, tau of sigma one plus tau of sigma two, sigma one is odd. So tau of it is one. Sigma two is odd. So tau of it is one. What happens if you add one plus one? Yeah, you zero, because the group that we're working in is z two. Perfect. So did that one work? Yeah, it did. And that was the interesting one. So in the end, tau's a homomorphism. So it's highly possible to have homomorphisms from non-abiding groups to abiding groups. It's possible to have homomorphisms from abiding groups to non-abiding groups. And we'll see a whole bunch of examples on Wednesday. All right, sorry to run over a little bit here. If you have specific questions on your...