 And maybe this is a surprise. It shouldn't be a surprise anyway. It's too bold of a statement. Okay, let's see. Some comments on the homework coming back. The homework as a group is looking much better than let's say homework one did. I think you're getting an idea of the sorts of things I'm looking for and that's good. Let's see, I'm going to steal your book here real quick. In section nine, problems ten through... Let's see. Where am I here? Let me make sure I'm in the right problem. Now I'm lost. Bear with me here for a minute. There's a totally insightful comment I need to make. I am lost. It's a question about the way they phrase the question. They want you to take a permutation and they want you to write it as a product of disjoint cycles and then figure out the order. Yeah, thank you. Okay, 13D, thank you. In section nine, 13D. I was looking too quickly at the question here. So section nine, problem 13D. The question was not simply find the order of each of the permutations in exercises ten through twelve. The question is find the order of each of the permutations given in those three exercises by looking at its decomposition into a product of disjoint cycles. And what some of you simply did was you took the given permutation and you wrote it as a product of disjoint cycles, then completely forgot about the fact that you wrote it as a product of disjoint cycles and then just started pounding out the order. And the point of that question was to sort of give you the type of computation that we did in class, which was if you write things as products of disjoint cycles, like something like, I don't know, one, two, five, and then three, four, something like that, if you want the order of this thing, if you're going to start raising it to powers like squared, is, well, the point is what? Well, yeah, you could write this thing out in long hand and then just start pounding it out, but the point is because these are disjoint, they commute. So typically this should just be if we're in a general group. So there it is and here I do it again. But the point is because these are disjoint, these commute, so this is one, two, five, one, two, five, three, four, three, four, and what you learned in a previous piece of this exercise is that that's E. So this is just one, two, five, squared. And then continue to go. And what you're being led to is the result in part E, which says that if you have a transposition written as a product of disjoint cycles, then the order of this thing is going to be the least common multiple to length. But if you didn't do a computation like this on part, on question 13E, if you just started pounding out lengths or orders of permutations without doing a computation like this, then you sort of missed the point of that particular problem. So I nicked you a little bit on that. And, ah, yeah, on the extra problem, I saw some less than optimal notation being used. What you have to distinguish between is the apples and the oranges here. The apples are the underlying elements of the set, maybe the numbers one through five. The oranges, the things that you're interested in are the elements of S5, are permutations. So there are functions, those are the things that are of interest. Those are the things that we're composing, those are the elements of the group. And the description of those functions has to do with what they do on the domain in this particular case, one, two, three, four, five. And so, for example, I saw things like in the extra problem. I mean, I hesitate to write this down because I might see it kicked back and I don't want to see it kicked back. So I write it in a different color. Something like in part two, take the identity element, so E of three equals three, yeah, it does, and you've justified why that's true. And then what some of you said, NH. And that's, you know, fingernails on a chalkboard. It's apples and oranges. Three is something in the underlying set. It's a number from one to five. The things in H are functions, so this completely makes no sense. If you want to write this, therefore E is in H, not three or E of three is in H, E itself is in H, that's the good notation. So in the other parts, too, the same sort of thing happened. I say sigma circle tau of three equals three, therefore sigma circle tau of three is in H. It's not sigma circle tau of three that's in H. It's the composition of the two functions, sigma circle tau that's in H. So please be careful on that. All right, those were the only two commentable details that came up in the assignment from tonight, so that was good. All right, let's see, as I mentioned in the email that I sent out on Friday, I've put the solutions to this stuff on the web. If you want to download those. And I think we're good to go for Wednesday then. All right, I also will, I have a few extra tonight. If you want another blank copy of the practice exam or if you've lost your copy of the info about the exam or something like that, you can come up and grab some after class, that will be no problem. So, at the end of last Wednesday, we got to this really important punchline. Let me remind you of what it said. So, recap, we're going to start with G, any finite group, finite means it has finitely many elements in it. Oh, here was something that actually came up in a question during office hours today. This word, order, is a somewhat unfortunate one because it's used for two related, but technically distinct things. The order of a group is just the number of elements in G. So, if we talk about the order of a group, it would have been nicer or clearer for students if we'd simply call it the number of elements in the group. But we don't. We call it the order of the group. If we talk about the order of an element, let's call it little g in a group, capital G, then what we're talking about is what? Is the smallest positive integer with the property that blah, blah, blah, blah, blah, smallest? I don't want to write it out because you have the definition of the order of an element. So, technically this word order applies to two different things. Obviously, the two are related if the group happens to be a cyclic group and if you happen to have handed me a generator for that cyclic group, then the order of the element, in other words, the order of the generator is the order of the group that it generates. So, the two words make sense when they do overlap, but we use the two words even in the situation where they don't overlap. Order meaning either for a set, tell me how many elements in it, or order for an element meaning tell me the smallest positive integer for which you get the identity. So, let me say, let G be any finite group, meaning the order of G is finite. But again, I'll try to use the non-order notation here. H any subgroup of G, then here's what Lagrange's theorem said. Lagrange's theorem, at least part two, second part, that's the one we're going to focus on for now, says this, that the index of H in G, and I'll remind you what this notation stands for, the index of H in G, which is by definition the number of distinct, in other words, different left cosets of H in G. Whatever that number is, multiplied by the, well, the order of H, or the number of elements in H, is the number of elements in G. The first part said if you hand me any two left cosets of H that necessarily they have the same number of elements in them, and that number of elements happens to be the number of elements in H. All right, we didn't technically prove all the details of this, but I showed you by means of an example, and by showing that two left cosets are either disjoint or identical, that we get this result in, it is incredibly powerful, this is powerful. And Fraley, the author here, makes a statement which essentially says any time you have a result that allows you to count something, it's going to be important. And this result allows you to count something. It tells you, for instance, well, it tells you how many cosets there are, number of elements in G divided by the number of elements in H. It tells you, for instance, how many elements are in the subgroup, figure out what the index is, or it tells you how many elements in the group, if you know what the number of elements in the subgroup is and you know how many distinct left cosets are. There are many more reasons why this is powerful. There are many really nice, important consequences of this result of this equation. The first one, for example, if H is a subgroup of G, and throughout this discussion today, folks, we're going to always assume that G is a group that has finitely many elements in it. We can talk about cosets for infinite groups, so we'll do that in a little bit, but at least as far as Lagrange's theorem goes, when you're trying to count something, it turns out there's an arithmetic of infinite sets, but we're not going to get through that this semester. That's sort of a first-year graduate course idea. But the punch line is this. If you take a subgroup of a finite group, then the number of elements in the subgroup is a divisor of the number of elements in G. So if you have a subgroup, then the number of elements in it necessarily divides evenly, and the number of elements in the subset phrase is a divisor of however many elements in the group. Refraised, sort of as a contrapositive, if S is a subset of G and the number of elements in the subset is not a divisor, the number of elements in G, then the subset is not a subgroup. For example, let's look at G is D4. G contains no subgroups having, for example, three elements, because three doesn't divide the number of elements in G, which happens to be eight here. G contains no subgroups having five elements, or six elements, or seven elements. Obviously, it has no subgroups containing nine elements because it has no subsets containing nine elements, because the set already has eight elements in it itself, et cetera. This is incredibly powerful. If you're looking for subgroups, don't bother looking at subsets having three elements, or five elements, or six elements, or seven elements, because those numbers don't divide. Now, you've got to be a little bit careful to not conclude too much from this result. If I hand you a subset of this group, and let's say it happens to have four elements in it, then at least that subset is a candidate to be a subgroup. It doesn't automatically make it a subgroup. I can write down four random elements sitting inside D4. At least as far as Lagrange's theorem goes, that set could be a subgroup. But hey, there are other things that might preclude it from being a subgroup. It might not contain the identity element, so don't go too far. One very interesting question that we will take up in about three weeks or so is this. Okay, give me a group, give me a number that could be the number of elements in a subgroup. If you give me D4, then it's got eight elements. If you give me the number four, at least there could be a four-element subgroup, because four divides eight. Is there a four-element subgroup? It turns out the answer is yes, and we've seen one. The group consisting of row zero, row one, row two, and row three is the subgroup. So inside my eight-element group, there is a subgroup of order four. There are many subgroups of order two that we've seen. There's obviously a subgroup of order one. Every group contains the subgroup just consisting of the identity element. So an interesting question, this is sort of how the, I don't know, I guess the way a mathematician would think is you have now this very powerful theorem, is there some sort of converse to it? If you have a subgroup, then the number of elements in the subgroup divides the number of elements in the group. Here's sort of the converse. If you have a group and you have a number that does divide the number of elements in the group, is there a subgroup having that many elements? And in this example, the answer is yes, and in general it turns out the answer is no. You can have a group, for example, there's a group with 60 elements that doesn't have any subgroups having 30 elements. So, you know, as powerful as this result is, it's not the end all be all, it simply gives a lot of nice information. Here's another example. What's the example? Yeah, let the number of elements in G, so I'm gonna give you some indication of how many elements are in the group I'm interested in. I'm gonna call that number N. I'll give it the numbers 8 or 100. It makes no difference to me. Let G be an element of the group. Let, let's call it M, be the order of G. It's the notation for order. I know what order is. It's the smallest positive integer. All right. It's the smallest positive integer with the property that when you take G and you raise it to that power, you get the identity. Here's the punchline. This number is a divisor of M. I'm sorry, of N. Let me rephrase that in longhand. Rephrased IE, the order of G, and this is the notation folks for is a divisor of, means properly divides without remainder, the number of elements in G. I try to avoid this notation because there's too many straight lines and too many squiggles, but that's the way you'd probably see this phrased in the textbook. A straight line means is a divisor of. So, for example, before I go ahead and prove this, for example, if you're sitting back inside this group and someone hands you an element and ask what's its order. In other words, what's the smallest positive integer so that when you raise this thing to the power and you see the identity, folks, the only possible answers are divisors of the number of elements in the group. Our divisors of eight. So, if you try to claim for me that you've written down an element inside D4 that has order six, I don't even have to check your arithmetic. I know that it can't be the case because this says that the order of each element has to divide the number of elements in the group and the proof is easy. Proof? Well, let's see. This is powerful. This first consequence says if you have a subgroup, then the number of elements in the subgroup is always a divisor of the number of elements in the group. But, wait a minute, so is the order of G. The order of G, whatever that number is, is the number of elements in this subgroup. We proved that, I don't know, two weeks ago or so. If you want to know how many elements are in the cyclic subgroup generated by G, it's precisely the number of times you have to take in order for G to get back to the identity. Previous result, result, result. My handwriting is going south here, sorry. Previous result. Oh, now use consequence one. The order, oh, is the order of that subgroup and if you have any subgroup, then the number of elements in the subgroup is the divisor of the number of elements in the group. Alright, questions there, comments? So that's pretty powerful. And the final consequence of Lagrange's term that I want to share with you here is, I mean, is one that to me is truly unbelievable. Look, the definition of a group is simply you have a binary operation that's associative, there's an identity element, and each element has an inverse. There's absolutely nothing in definition of groups that's anything about numerical information. It's also the case, folks, that there are very few cyclic groups. I mean, we know what they look like. It's Z sub n for your favorite n. Z2, Z3, Z4, so those are all the cyclic groups. Those are somehow the easiest groups to understand. Here's the third consequence. Consequence of Lagrange's theorem. See the number of these? Here's the first consequence. Oh, this is the second consequence. Sorry. And here's the third consequence. Consequence three. If p is a prime number, if p is prime and the number of elements in g is p, then g is cyclic. Folks, if I hand you a group of seven elements, it's cyclic. I mean, where does that come from? That's truly unbelievable. If I hand you a group of 11 elements, it's necessarily cyclic. Why? Because the number of elements in the group is prime number. Well, prime number is necessarily bigger than or equal to two by definition. So I've handed you a group with at least two elements in it. And I always know what one of the elements in a group is, the identity. So here's all I want you to do. Grab any element in the group that's not the identity element. There's a lot of them. In fact, all but one of them is. So pick any element, let's call it little g in capital G with g not equal to e. You have plenty of choices. Then here's the subgroup. Look at this subgroup. The subgroup generated by g, the cyclic subgroup generated by g. All right? I'm going to tell you a couple things about it. So the first thing is, it's a subgroup of g. That's no big deal. I haven't used anything about the hypothesis yet. In any group, if you take any element in any group and you look at the cyclic subgroup generated by it, you get a subgroup. Okay, that's pretty good. The first thing is the number of elements in this cyclic subgroup. Well, I've asked you to take not the identity. So when you look at cyclic subgroup generated by it, by definition, what did you do? You write down the identity and then you write down g. So I've already written down two elements in there. So there's at least two things in there. I'll put in parentheses because I've asked you to start with. Not the identity, but let's say, oh, the number of elements in that cyclic subgroup divides the number of elements in the group by Lagrange's theorem. It's a subgroup. If I tell you how many elements are in it, I don't know that number yet, but I represent that by number of elements in g and necessarily that number. That was Lagrange, or if you want consequence one of Lagrange's theorem. Lagrange's theorem, consequence one. So wait a minute. What's the hypothesis? But number of elements in g is prime. That's the hypothesis. So I've now written down a number, that number, whatever it is, that divides the prime number. And I haven't written down the number one. I've written down something else. So what's the conclusion? That this thing is the prime number. I have a divisor of a prime that's not one and the only divisor of a prime is the prime number itself. In other words, it's the number of elements in the group. So here's the conclusion now. I have a subgroup is a subgroup of g having the same number of elements as g. So I have a subset of a finite set and the subset has the same number of elements as the set. So the conclusion is that, in fact the subset is the set. So the conclusion that we're able to draw folks is a numerical one. That the number of elements in the subset is equal to the number of elements in the set. And it's from that piece of information that we can conclude just by set results discrete mass stuff, if you have a subset of a set and the subset and the set have the same number of elements then necessarily the subset has to be the entire set. In other words, g is a generator for g. g generates capital G and we're done. So that's incredibly powerful. So if somebody says I'm thinking of a group of 13 elements and in fact you already know what the group is. It's a cyclic group. And hey, there's only one cyclic group. It's Z13, optisomorphism. Gronch's theorem is incredibly powerful. And again, this is like the third time I mentioned this. And, I don't know, totally surprising. Totally surprising. Just give them the definition of a group. You're impressed hopefully that there is all this numerical data that somehow corresponds to it. All right. Questions or comments? Well, before we move on, I want to spend, I don't know, five to ten minutes talking about, well, an idea that you would expect given what we've been talking about for the last lecture and a half, we've been talking about left cosets of a subgroup inside a group. And the question is what's so special about left cosets? Well, if there's left cosets, there probably should be right cosets and there are. And so let me spend a few minutes talking about right cosets. So, up until now, we have focused, focused on what are called left cosets of a subgroup. A subgroup inside a group. A group, it turns out. Not surprisingly, we can also construct, build what are called the right cosets. H in G. I'll tell you how to do it. Well, let's see. What were the left cosets? The left cosets, just as a reminder. No, I don't want to write it down. Well, yeah, left coset. Oh, well, that's all right. Looks like A H. A H equals all the things that you can write in the form little a times H where H is in the subgroup. Here's what a right coset is. The right coset, coset of H generated by A is simply the set HA. Did I get that backwards? No, I think I'm all right. It's the set of things in the group that you get by putting the element in question, the thing called the little a on the right and multiplying it by all the elements in the subgroup. In effect, everything that we did when we built left cosets can be mimicked for right cosets. There is an equivalence relation on the group. The definition of the equivalence relation is you deem A to be related to B in case AB inverse is in the subgroup rather than deeming A to be related to B by putting A inverse B in the subgroup. You then get these things called right cosets. Any two right cosets are either disjoint or equal. And there's a Lagrange's theorem for right cosets. Lagrange's theorem says the number of right cosets, in other words the right index, times the number of elements in the subgroup is the number of elements in the group. So it turns out there is a lot of similarity between what happens with the left cosets and the right cosets. But the similarity is really only numerical similarity. In general, here's what turns out to be a little bit confusing but true. It turns out that numerically, numerically, left, right cosets share many properties. For example, EG first, the number of elements of elements in any right coset equals the number of elements in any left coset. So somebody sits down and pounds out a right coset. Somebody else sits down and pounds out a left coset. It turns out the number of elements that each person will be looking at is the same. And in fact, both of these numbers simply equal the number of elements in the subgroup. You see, if I write down the subgroup itself, there's no mention of left or right. There simply is. And I'll write it this way. There's sort of a Lagrange's theorem for right cosets. Let's see, what does Lagrange's theorem say? The number of left cosets times the number of elements in the subgroup is the number of elements in the group. Lagrange's theorem phrased in terms of right cosets is the number of right cosets times the number of elements in the subgroup. So folks, the number of left cosets equals the number of right cosets. The number of right cosets of h in G equals the number of left cosets of h in G. So when we talk about this thing that we call the index of a subgroup, technically I define the index to be the number of left cosets of the subgroup in the group. So I probably should have called that the left index or something like that. But the point is, if you want to know how many right cosets there are, it's the same as the number of left cosets. So instead of calling one the right index and the other the left index, we simply use the single word index, the index of the subgroup. But then you're asking, well, okay. So the number of elements in left cosets is the same as the number of elements in the right cosets. And the number of left cosets is, so are they the same? And the answer turns out to be no. But here's the caveat, and we're going to spend at least three chapters talking about this particular observation. In general, the left and cosets of the subgroup might not be the same. I say, wait a minute, they are the same? No, what I've told you folks is that the number of elements in a left coset is the same as the number of elements in a right coset. And the number of left cosets is the same as the number of right cosets. So that's what I mean by numerically, there's a lot of interaction over that. But at the set level, if somebody hands you a left coset and says, is this also a right coset? The answer sometimes is yes, but sometimes is no. The example to look at, example, I'll say, we will see this later. We'll look at this in further detail, detail, much later, but much later, maybe two weeks or so. But I wanted to bring it up at this stage just because one of the homework problems asks you to show that so in certain situations, some situations, we can show that this left coset equals this right coset. Not always, but in some situations and there is one such situation where you ask to show that this equals this in one of the homework problems. I forget which number it was, 28 or 40, I forget. But anyway, it's one where I gave you the hint and the hint was the goal is to show that this left coset equals that right coset and the way to do that is to show that one side is contained in the other. So the point is showing that this left coset equals this right coset can't be done in general because in general, left cosets might not be the same as right cosets. But the punchline is if you have some additional information about the subgroup then you might be able to show that this equals this. So that just gives some sort of context to that particular problem that what you're proving for that particular subgroup sitting inside that particular group holds in that case but definitely does not hold for all possible subgroups of all possible groups. Questions? Okay, so what we're going to do here is another change in topic. What we've done up until now once we got past the general notion of binary operation and we got into the the specific discussion of groups was, alright, we we've had sort of two methods that we've looked at of building groups. One is sort of the from scratch method. Look at this set. Look at this binary operation. Go through all of the computations to convince me that that set together with the binary operation is a group. It convinced me it's a legit binary operation. You know, at least some lip service to making sure it's associative. Convince me that an identity exists. Convince me that for each element an inverse exists, et cetera. Groups from scratch. We did that a lot right at the beginning of our discussion of groups and then we did it again when we built these permutation groups and we built the SN groups, alright. We sort of built them from scratch. The other sort of point of view that we've taken is once you've got something that you've worked hard to show me as a group and look inside those groups to find other groups and presumably those other groups are a little bit easier to verify because to show that something sitting inside a known group is also a group. You just got to show it by the subgroup there. You don't have to worry about existence of this or anything like that. Just show it's close. Show the identities in there. What we're about to do now is play another sort of game and the game, I guess, that are known to be groups and build new groups from them. We've done that in some sense already. Just by you take something that's known to be a group look inside it and see whether or not there are other groups living inside. Here we're going to take things that are known groups and find new groups not by looking inside but somehow piecing them together. In effect, this process is really nothing more than a set process back in the discrete map of course you talked about taking two sets and forming their Cartesian product or their direct product. It's simply the ordered pairs where you take the first coordinates coming from the first set and the second coordinates coming from the second set. It's something you're totally familiar with. When you do R cross R Cartesian product with R you get R2 which is just ordered pairs of real numbers and that's something you're familiar with. We can talk about doing that for all possible sets just ordered pairs and what we're about to see it's pretty easy to see it will be pretty easy to see that at the end if you take two sets that happen to be groups and you simply hammer them together in a Cartesian product then the result is going to give you a group. This is going to be another way of building new groups from known groups and we'll see many more ways in at least the first 10 weeks of this semester. So, new subject this is called direct products of groups and the idea is if we start with two or more groups two or more and we'll see how to do more in a minute groups we can form the direct product of the sets direct product of these things as sets it turns out the resulting set result will be a group direct product so remember what this looks like specifically let's just do it with two of them if I hand you two groups let's call them G1 and G2 and you know what let me be even more specific well I'm going to ask you to start with two groups so one of them I'll call G1 G1 will have its own binary operation well just to distinguish it from the binary operation that might somehow be defined on the second group I'll call it star sub 1 and star sub 2 this might be S5 together with composition this might be Z6 together with addition mod 6 but I don't really care the two operations in the underlying groups that are going on might be completely different of course it might be the same group or in the same operation I haven't precluded that possibility but just start with two groups and here's what you do you form the set G1 cross G2 so what does this consist of it consists of things that look like ordered pairs you can see that little G1 is in capital G1 and little G2 is in capital G2 so that's all we're doing here well it's just some math 215 it's just some basic set theory form the Cartesian product of the two underlying sets now what I have to do is teach you how to make this set into a group with some binary operation it's pretty easy to do now define a binary operation on this set G1 cross G2 it's pretty easy to do let's see so what might it look like take something in G1 cross G2 what does it look like it looks like an ordered pair and folks the things in this set by definition include the parentheses and include the comments just pairs of things where the first thing is from the first set second thing is from the second set now I need to teach you how to combine that with something else in this set well what does something in that set look like it looks like something in the first group but I've already called this thing G1 how about let's call it G1 prime G2 slash or G2 hat or G2 something and I have to teach you how to combine these well there's sort of an obvious way to do it in the first coordinate simply produce G1 star G1 prime it makes sense to take G1 and combine it with G1 prime because both of those things live in this particular set and this operation called star one is a binary operation and then G2 G2 prime in other words I mean the phrase would be just do things coordinate wise then the punchline is then proposition proposition this gives a group structure G1 star G2 with this I'm sorry cross G2 with this binary operation gives a group won't prove it in gory detail but in some sense folks what we've done is we've produced a new group now this isn't a subgroup of a known group so technically to convince you that this is a group have to go through the you know the gory details of have to convince you first it's a binary operation that's a binary operation that's no big deal if you take any two things in this set and the things in this set are inherently ordered pairs here are two things in that set they always get something well defined back in the set yeah because well we're using some properties of the binary operation already we know that if we take two things in the first group and compute their star whatever that is we get something back in the first group and we know that this back in the second group so at least we have a chance of writing down a binary operation that's good is it associative well that's sort of painful but it is what does that mean you got to take three things in the underlying set and convince me that if you combine them in this order and then do that one that you'd get the same thing if you had switched the brackets well look folks in the first case where are you going to get you're going to get G1 prime star G1 double prime that's what we get in the first slot in the second slot yeah but wait a minute each of these two things individual is a group so you can slide them over individually in each coordinate and then things are good third thing we have to show that an identity exists we're back to basics here folks can you identify the element in here that you think might be the identity element just you know take the identity element of the first group and the identity element of the second group put them in the same ordered pair and that should give you the identity element of this new group and it does I mean there's something to show but it's not too hard to do final question is let's see if I hand you an element in here is there something that behaves as it's inverse in other words there's something in here that you can write down so that when you combine the two things in definition of a binary operation that you get the identity sure if I hand you something in here it means I've handed you something in the first group and something in the second group you want the inverse of that thing sure just write down the inverse of that one the inverse of that one it works so this thing g1 g2 inverse turns out to be g1 inverse comma g2 inverse so it turns out turns out that the identity element is that and the inverse so again I think I could spend 15 minutes running through all the gory details but I think our time is better spent just saying yeah exactly where you'd expect to look for the identity element in the direct product and exactly where you expect to look for inverses that's exactly where we should be so let's just do a quick example example here's a group is d4 the second one is z6 and I've just built a group now I haven't carried along the standard notation that we used when we were writing down technically the definition of the product of the two groups the direct product or cartesian product of the two groups but we know what the operation is in d4 it's composition and we know what the operation is in z6 so this is a group I don't know here's an element here's the identity element e is this thing you put the identity of d4 in we call it that you put the identity of z6 in so there's the identity element of the group you just put the identity element of that one notice here folks it makes no sense to talk about switching these things around for example the pair 0 comma row 0 is a complete non-issue makes no sense here because it's always the case that the thing you write down the first coordinate has to be in d4 and the second thing has to be z6 let's do another example maybe something like if I give you a row 1 comma 4 and I ask you for the inverse of that well the inverse of an element in the product is you simply take the inverse of each one individually for inverse and of course you've got to interpret this in the context of the group that it lives in turns out the inverse of row 1 is row 3 because row 1 circle row 3 is the identity the quote unquote inverse of 4 in here it's the thing that you have to add to 4 to get 0 which is 2 because 4 plus 2 is 0 so there's the inverse of that thing let's do one more quick computation if I ask you to compute row 1 comma 3 we are with I don't know mu1 comma 5 let's see what we get well the definition folks of combining two things is you get row 1 mu1 comma 3 plus 5 is 8 but we're doing arithmetic mod 6 so we get 2 and row 1 mu1 turns out to be what so we have the table out say it again d1 delta 1 okay thank you delta 1 so working in these things doing computations in these direct product groups is actually pretty easy now the definition I gave you was the definition that you use if you're only asked to take two groups and form their product there's absolutely nothing special about just taking two groups we can do the same thing the same product process process with more than two groups that's no big deal two groups I won't write down all the details but it's the same idea if somebody hands you seven groups it makes sense to form their Cartesian product g1 cross g2 cross g7 that's a complete non-issue of sets but then it turns out that that has a group structure you just look at the philosophy is you just do everything coordinate-wise and things work out just fine okay questions come in so far okay so here's what we're going to focus on let's see if that's all I need to say about products in general the the philosophy is this process of taking the direct product of two or more groups to produce new groups is somehow viewed as a straightforward sort of building block process that allows you to take things that you already know about and somehow extend them to other things like we knew about the group D4 before tonight we knew about the group Z6 before tonight but until tonight we didn't have information about this new sort of group this D4 cross group even though we know each of the pieces individually and the question will be alright here's this new group can you say something about the new group in terms of what you might already know about each of the pieces that make it up so the goal is goal say something about the properties of properties of G1 cross G2 and I'll do the more general situation where you're taking more than two groups and forming their direct product maybe up through G sub t in terms of or as related to the properties of each of the groups that make it up of individual individual groups G1 G2 up through Gt individually let me give you an example for example what's the order of this group how many elements in it that turns out to be easy because it has nothing to do with group theory it's simply a fact about sets and you form the Cartesian product the number of elements in the Cartesian product is simply the product of you multiply the number of elements in each of the unknowing sets so for example the number of elements in that group D4 cross Z6 is the number of elements in D4 is 8 and the number of elements in Z6 is 6 so 6 times 8 is 48 I've built a group with 48 elements in it example first in the direct product you just multiply these numbers together G1 times G2 and the times here is as whole numbers times Gt so there's an example I can say something about the number of elements in this group by saying something about the number or the property of each of the individual groups that make it up I can tell you how many elements there are so for example the number of elements in this group that we looked at briefly D4 cross Z6 is the number of elements in D4 times the number of elements in Z6 which is 8 times 6 which is 48 so there's sort of a brief glimpse as to the sort of flavor of questions we can ask here's another one that's more interesting suppose I hand you two groups and each of them is properties how we study, how about a billion if I hand you two groups or three groups or some number of groups and each one individually is a billion is the product still a billion that's a fairly interesting question turns out the answer to that one is yes if I hand you a bunch of groups and they're each cyclic is it the case that the direct product is cyclic turns out to be true sometimes but not always and we'll focus for the last 15 minutes tonight on that question but there's other questions we can ask you know if I hand you some non-Abelian groups and I form the direct product is the corresponding group or the subsequent group is still non-Abelian so example two if each G1 G2 through G sub T is a billion then it turns out then so is the direct product G1 cross G2 G sub T that works out nice it's pretty easy to show just because when you do the binary operation in the product it means that you're simply combining each corresponding term with the next one so you have all these products here but if each one of the individual groups is a billion it means you can switch this product and you switch this product and switch this product in the end you simply switch the whole thing so that's no big deal but it turns out this turns out to be the interesting question question but it turns out in general only sometimes is it the case that if G1 and G2 are cyclic cross G2 is also cyclic every cyclic group is a billion yet you prove that for quiz two so if I hand you two cyclic groups and I've certainly handed you two Abelian groups that's no big deal and let's see this property two says because each of these is Abelian then at least this thing is Abelian but the question is whether or not it's cyclic and the answer is sometimes yes and sometimes no let's go ahead and look at two particular examples and then we'll try to make some general conjectures based on the two examples as to when it's the case that if you start with two or more cyclic groups and you form their direct product the result is again cyclic so example let's look at this direct product of and again I'm only taking two groups here just because the idea is only a half hour old and I want to show you a lot of small examples before we look at bigger issues so here's Z2 cross Z3 so what are the elements look like well let's list them out Z2 only has two elements Z3 only has three elements so the product has only six elements so you know this is small enough that we can actually list out things explicitly so one of them looks like zero zero and then what have a zero one and zero two and one zero and one one and one two there they are there's the six elements I'll list them out in some sort of systematic order doesn't matter what system you use what I did of course was I listed all the elements and I listed the first element in Z2 and paired it with each of the elements in Z3 and then heck if you wanted to have listed the elements in Z3 first it doesn't matter so there they are now look at the following look look if I take this particular element and I look at the subgroup that it generates so folks all this stuff is the same it's just the group that we're considering now happens to be that one element that we happen to start with is that let's see what we get well we get the identity element because we always throw the identity element in for free and there is the identity element there's E why because it's paired the two identity elements of the two underlying groups and then you put the element in okay now let's see what's one one I guess we'll use star notation here one one what are you supposed to do you're supposed to take the thing in the first coordinate and combine them of course the operation in the first coordinate is addition mod 2 and then what you're supposed to do is take the two things in the second coordinate and combine them of course the operation in the second coordinate is mod 3 arithmetic so what do we get well here we get 1 plus 1 in the first coordinate which in Z2 is 0 because we're in Z2 1 plus 1 in Z3 is 2 so that's what we've just gotten so what we've just shown is that if we take this thing and we combine it with itself we get that game now let's do it three times one one one one one one in other words do the binary operation on this element with itself three times well we know how to compute these sorts of things because we've got some practice with it just you know don't reinvent the wheel there's one one star one one and I've already computed that from what we just did right now I know how to do this that's then zero plus one in the first coordinate and two plus one in the second coordinate and what does the result of that computation give in the context of Z2 cross Z3 let's see the first coordinate is one and the second coordinate is zero exactly right because the computation in the second coordinate is happening inside Z3 did we get the identity yet nope keep going let's do it four times you know one one well let's use that notation that's one one star one one star one one star one one and heck I know how to do that we'll do it by taking one one three times that's what we already got and doing it one more time and we get what we got one plus one in the first coordinate and we get zero plus one in the second coordinate one plus one in the first coordinate is zero okay so we still haven't seen the identity element yet how about let's see one one is zero one star one one what's this going to give oh let's see it's zero plus one one plus one and so we get one comma two still haven't seen the identity element yet hmm I know the answer to this one already because I remember Lagrange is there the order of any element in any group is a divisor of the number of elements in the group the number of elements in this group is six so the order of this element has to be divisor of six well the order isn't two because I didn't get the identity when I did the operation with itself twice and the order is not three so I didn't get the identity element so in fact at this stage I could have run immediately to the order is six but let's just go ahead and verify that our arithmetic is on target one one which is one plus one inside z two two plus one inside z three which is zero comma zero which is the identity element so that checks out that's nice so the order of that particular element is six or rephrased if I list out all the things that appeared here let's go ahead and do that I got zero two I got one zero I got zero one and I got one two which is all of the group z two cross z three the whole thing so there is at least one element inside this group with the property that when you look at it and all of its quote unquote powers you actually get the whole group back so the group is cyclic so z two cross z three is cyclic proof to show that something is cyclic is all you have to do is somehow come up with what generates it and I've done that now it turns out there is one other element that there's one other element that is also a generator of this group but at least for the purposes of what we're trying to get across here that's either here or there there is at least one and therefore it works on the other hand example here's the group g is z two cross z two it's easier to write out what the four elements of this group look like zero zero zero one one zero and one one that's what they look like a set that's no big deal and what I'm about to convince you hopefully is that this group turns out even though just like the example we did previously is the direct product of two cyclic groups that one turned out to be cyclic this one will turn out to be not cyclic show something's not cyclic just take everything in the group look at the subgroup that they generate and convince me that you never get the entire group that's pretty easy to do look if I look at the subgroup generated by folks this is true in any group the subgroup generated by the identity element is always just the identity element just keep beating on it the subgroup generated by one zero I mean these will be pretty easy to see is well zeros in there are always by default the elements in there by definition now if I combine this thing with itself one zero plus one zero you get two comma zero that's happening inside z two and so that's it you get back to zero zero this has ordered two similarly zero one you get just these two things zero zero zero one just do the computation why because if you do this with itself you get zero two which is zero zero and then finally if you take one one you get zero zero and you get itself now the difference here and I think this is intuitively what makes it clearest in this situation if I take one comma one and I combine it with itself I get two comma two which in this particular group is zero comma zero on the other hand when I took what looks to be the same element over here and I combined it with itself I didn't get zero zero because somehow it sort of skewed off a little bit it's not just excuse me it's not just two comma two it's two comma three so it sort of miss and then alright so you keep going and then oh you keep going to the point where yeah now things have worked out so that you get a zero here but by getting a zero here you didn't get a zero here so you sort of had to keep going so that when you finally saw zeros in both slots it took you six wax to get there where it's here in order to get zero in both slots it didn't take you four wax you wound up there after two steps so not cyclic not cyclic well but it is a billion so that's no big deal this turns out to be an abelian group it's got four elements it's got the property that each element at least the element that's not the identity you take each element that's not the identity and you combine it with itself you get the identity and we've seen a group like that what do we call it V this is the group V you take four elements abelian but if you take any of the non-identity elements and you combine it with itself you get back to the identity that's precisely what we call V it had a different form as well we also realized it as the collection of four matrices each two by two with plus or minus one in the main diagonals so this group now is sort of arising a little bit more naturally so Z2 cross Z2 is isomorphic to V so just as we've realized in a relatively concrete way as the direct product or the Cartesian product of the group Z2 with itself so let me go ahead and give you a homework assignment it won't be due for another couple weeks but it's sort of the right rhythm to give you an assignment on Monday and let you start on it if you choose what we'll do after the exam next Monday is we'll continue looking at this question of when when you have a a direct product of cyclic groups when is the resulting direct product also cyclic and can we say something about the structure of these things so let's see let me make sure I'm not doing the wrong thing yeah here is home and this is due not until October 10th because the assignment that I gave you last Monday isn't due until October 3rd because of the exam in between this is in section 13 problems 1 through 16 28 to 32 and 47 to 51 the ones I'm going to have you turn in are 6 8 29 47 problem if this is Greek letter phi from G to G prime and this is a word that we didn't use tonight but we'll use next Monday after the exam homomorphism show and here's a notation that we haven't used yet but we'll use next Monday KER phi is the subgroup of G so that's a question that we have indeed asked before and then I'll say and I'll repeat these with the appropriate let me just do these on number 8 what I want you to do add the following instruction what property of the group G would ensure that phi is a homomorphism so that's an added piece on number 8 and on number 29 I want you to change the instructions slightly 29 is phrased as is it or is it not a homomorphism it is so I want you to prove that G is a homomorphism and again we slightly touched on some of the ideas