 And the other way, of course, is the same idea. Star AH is A inverse AH. H, check. Done. So what we've just shown is that if you start with a normal subgroup of a group, that the collection of left cosets of that subgroup inside the group, that collection of sets forms a group in its own right. And I'll say it again for the dozen time. In this particular group, the things that are the elements of the group are inherently sets. They are the cosets of the given subgroup sitting inside the group. So that's the toughest thing to get your head around. You're combining sets to get other sets, but in doing so, the process is giving you a group structure. Now let me give you some of the notation. The more standard notation is this, although I think over the years I've realized that students find it a little bit more interesting or clearer to have separated out the two binary operations or the two proposed binary operations to begin with. The binary operation that you get just by pounding together set products, that's what we called star with a circle around, or the binary operation that you get by simply deeming a h star b h to be a b h. In general, each of those proposed binary operations has problems. This has a problem because if you start with just an arbitrary subgroup, it might be the case that if you just do set product, you might not get another left coset back. This one has the problem in that if you start with an arbitrary subgroup and you simply deem a h double star b h to be a b h, that this might not be well defined. But what we show, and we did most of the details in class on Monday, is that if you start with a normal subgroup, then not only does set product always give you another left coset back, not only does the definition of double star a h star star b h equal to a b h, not only does that become well defined, but in fact the two operations become the same operation. In both cases, the two operations simply become, take the two coset representatives and form their appropriate star. So what do you want to call the binary operation? You want to call it star with a circle? That's fine. You want to call it double star? That's fine. Typically in the literature, they'll just call it star. Or heck, when you work with them after a while, just like we did for groups, they'll just sort of not call it anything anymore. But for now, let's just call it star. So usually, we just denote it by star. We denote the binary operation on the set script s by just a single star. This is technically not a good thing, because I think it's a little bit confusing for students, because the star here is related to the star in the original group. But it's not, because now the star corresponds to asking you to combine left cosets, rather than just asking you to combine elements. So the notation admittedly is not completely precise, but it's completely standard. You just have to keep in mind what set you're working with. And typically, we usually denote the set of left cosets of h in g, this thing that we've been calling script s, by the symbol g diagonal slash h. We sometimes read that g mod h. It's not the greatest of notation either. It's not the greatest of notation when you're first learning about it. It's not unreasonable notation once you're a little bit more comfortable with it. But what we'll need to do is sort of focus on what this thing really is, so that when we're doing some examples and some computations that we sort of keep front and center what it is we're combining, what it is the result should look like, and sort of what it is that actually becomes a group here. Questions, comments? So in retrospect, as I mentioned at the beginning of Monday, in showing that the collection of left cosets of a subgroup inside a group, in showing that that set of sets is a group, really the hardest work was showing that we get a binary operation. And once we get the binary operation, which doesn't happen always, it only happens if we happen to have a normal subgroup. But once we get the binary operation, then showing that the binary operation actually gives a group structure is easy, it took five minutes. All right, and what you've all been waiting for is example after example after example. And that's good, so let's do a bunch here. All right, well, in the end it's just going to be a matter of looking at what we've got, writing down cosets and doing the appropriate combination. So examples, let's see. How about, let's start with example one will be example one, let's start with a group. How about S3, is that the one I want to do first? Yeah, and let's look at a specific subgroup, subgroup generated by row one. I'll tell you what that is as elements. It's row zero, row one, row two. Now, we showed, I think we did, yeah, that not only is H a subgroup of G, that's sort of a non-issue. It is a subgroup just by definition, it's a subgroup generated by row one. We showed actually that H is normal in G. So I'm emphasizing here folks this piece of the statement, not only is it a subgroup, but remember our notation, it's a normal subgroup. That took a little while. I mean, you have to pound out all the left cosets, you have to pound out all the right cosets, and you have to convince me that they're the same. And we'll see when we look at example three or four or something like that, that sometimes knowing that you have a subgroup is enough to conclude that it's normal. But in general, especially if I hand you a group that's not a billion, in order to show that a subgroup is normal, you typically just have to knock out all the left cosets and all the right cosets. So took some work. But the work is completely straightforward. It's work similar to the work that you did in problems 6, 7, 9, and 10 of section 10. OK, so let's write down the cosets. Here are the, well now folks, in this context I can start getting really sloppy. I don't have to talk about left cosets or right cosets because they're the same beast. So I can just talk about the cosets. Can't do that in general, but I can do it in the context of normal subgroups, which this is one. Here are the cosets of h and g. Just pound them out. One of them is always easy. They call it identity h, so that's just h. Let's see. Oh, you know what? We can actually predict beforehand how many cosets there are. Yeah, because the number of left cosets is what? It's the number of elements in the group divided by the number of elements in the subgroup. That's what Gronys is doing. How many elements in the group? 6, that's 3. How many elements in subgroup 3? So the number of left cosets is 6 divided by 3, which is 2. Well, here's one of them. This also equals for what it's worth for row 1h, which also equals row 2h, because we know if we start with something in the coset, and each of these is in that coset, and it generates the same coset as any other thing. So to get the other one, I'll just write down whatever element you want. What do we call mu1? That's the same as, which is what, mu1. I think that's what they call the other three elements in S3. Actually, you might want to reach into your notes here and grab the S3 table that I handed out. It'll be helpful to have this close by. I don't know what you want to call this. There's lots of different names for it. Heck, there's four different names for that one. Here's three different names for that one. I'm guaranteed that I've got all of them. That one and that one, because they make up all six elements of the group. So let's look at what the binary operation looks like. How to describe that? Well, we can do it as a table. So here is the group g slash h. That's the set of left cosets. So g slash h is these two cosets. Again, I've got a lot of different ways of choosing to write them. Maybe I'll write them this way. Rho 0h, that's a perfectly good name for this first one. If you want to call it Rho 2h, good for you. That's just as correct as calling it Rho 0h. It doesn't matter. What do you want to call this other one? How about mu1h? So there are the two cosets. There are the two elements of this group. And now we're going to teach you how to combine them. Well, let's see. Here's Rho 0h. And here's mu1h. Here's Rho 0h. Here's mu1h. And let's see how to combine these. Well, wait a minute. What we showed when we showed that the collection of left cosets forms a group is that when you look at the coset corresponding to the subgroup itself, the coset corresponding to using the identity element as the coset representative, well, that's this. So that is the identity element of the group. Well, that tells you something already. In fact, that tells you almost all of the information you need to know. That combined with that is, well, it's Rho 0 circle Rho 0, which is just Rho 0h. That was easy. How about Rho 0h mu1h? It's mu1h. Why? Because Rho 0 mu1 is mu1. Remember, that's what the operation is. You simply take the coset representatives, combine them, and look at the left coset that that thing generates. How about mu1 Rho 0? That's easy. It's just mu1h. Now, how about mu1h mu1h? It's just mu1h. It's just Rho 0h. So there is the group G slash h. Now, does anybody want to make a conjecture to what group that you're familiar with might this group be isomorphic? Z2. Isomorphic 2. Remember, this is a symbol we use for isomorphic 2, Z2. Yeah, let me tell you about this group. It's got two elements. It has the property that, well, there's an identity element, and there's a not-the-identity element. And the not-the-identity element has the property that when you combine it with itself, you get the identity element. So which group am I describing? Answer, I don't care. Either one. This is 0, 1, 1, 0. OK, so in this particular case, when we form the factor group, when we form this group of left cosets, or just group of cosets, if that's what you choose to call it, that's fine. Then we get something that we're already familiar with. We get Z2. Not bad. Let's see, is it the case that we necessarily have to get something that we're already familiar with? No. But when we do, it's sort of nice to note that's the top. OK, second example. Example two. Let's see, instead of starting with S3, let's start with D4. And what I want to do this time is let H be the subgroup generated by row two. And in this particular case, that happens to be row zero and row two. Is H a subgroup of G? Yeah, that's a non-issue. It's a subgroup of G because it's the subgroup generated by row two. It happens to be those two elements. Is H a normal subgroup of G? Well, we've got to do some work. So H is a subgroup of G. H is a subgroup of G, no problem. No problem because it's the subgroup generated. Is H a normal subgroup? I'll simply ask, is H normal? Well, pound out left cosets, pound out right cosets. Pound out left and right cosets, just like you did in problem six, seven, nine, and 10. Those might be the four most important problems that you do all semester for homework. Sometimes it works. Sometimes it doesn't. Turns out that they're the same. Out they are the same. So H is, in fact, normal in G. I'm leaving out this computation for you, but you've done a completely similar, completely analogous, nearly identical computation in those previous homework questions. So now we have a normal subgroup. So that means that we can form the group of cosets, this thing, G slash H, which I'm going to start using a name for, and then I'll write down officially that we call it. This is typically called the factor group of G by H. So let's see what it looks like. So what is, at least what are the elements of G slash H? Well, what I need to do is write out the cosets. Let's do that. We can do it pretty quickly. If I look at rho 0H, well, that's always a good coset to look at, the coset generated by the identity. It's always the subgroup, which in this case is rho 0, rho 2. Let's look at another one. How about rho 1H? Well, that turns out to be, let's see, rho 1, rho 0, which is just rho 1, and rho 1, rho 2, which is rho 3, which turns out to be rho 3H, but that's all right. If I look at rho 2H, folks, I know I'm not going to get anything different. Because if I look at rho 2H, I'll simply get the same coset as rho 0H, because rho 2 is already in that coset. If I look at rho 3H, I'm not going to get anything different. So how about let's look at, I don't know, mu 1H. That was a good one. So what do we get? We get mu 1, because that's mu 1, rho 0, and then we get mu 1, rho 2, and mu 1, rho 2 is mu 2. And then let's look at another one. How about delta 1H? We get delta 1, and if we look at delta 1, rho 2, we get delta 2. That's convenient. So there is the collection of left cosets of H. It also happens to be the collection of right cosets of H. This was, I think, problem 9 and 10. So let's look at the group table for the group G slash H. And we write out the elements here. So the first one is rho 1H, I'm sorry, rho 0H. Second one's rho 1H. I mean, in general, it doesn't matter technically what order you write them out in, but certainly it's customary to write the identity element first. Then mu 1H and delta 1H. And now we'll write this down the side too, rho 0H, rho 1H, mu 1H, delta 1H. Well, I know always how to fill in the first row and column of any group table. You just use the fact that the first thing that you've written down is the identity element. So that's rho 0H, rho 1H, mu 1H, and delta 1H. Too bad. And down this first column, rho 1H, mu 1H, delta 1H. All right, now let's see what happens here. If I combine rho 1H with rho 1H, I get rho 1, rho 1, which is rho 2H. Folks, I don't see rho 2H here anywhere. At least not in that form, but I do see rho 2H here because rho 2H is the same as rho 0H. So what I get here is rho 0H. Again, the observation is that rho 1H, rho 1H, if you do the binary operation, by definition gives you rho 1 squared H, which is rho 2H. But as a coset, I need to identify it as one of the four things that I've written my group as. And the punch line is that rho 2H is the same as rho 0H because rho 2 is in that same coset. So it equals rho 0H, and that's what we then enter in the table there. Similarly here, let's see if I do rho 1, mu 1. What does that give me? Rho 1, mu 1 gives, yeah, rho 1, mu 1 gives delta 1, and that one's fortunately already in the table, so that's delta 1H. And rho 1, delta 1, let's see, rho 1, delta 1 gives mu 2. So rho 1H, delta 1H gives rho 1, delta 1H, that's the definition of the binary operation on cosets, which is mu 2H. But look, folks, mu 2H is the same as mu 1H, the reason being that mu 2 is in the coset generated by mu 1, so the two cosets are the same. So I will list this as mu 1H, even though when you do the operation inside the group, the result gives the element mu 2, but I'm not worried about elements here, I'm worried about cosets. All right, and we'll fill in the rest of the table a little bit more quickly. Mu 1, rho 1 is what? Mu 1, rho 1 is delta 2, but delta 2H is the same as delta 1H. So I'm going to write it as delta 1H, because that's how I've described it here. Let's see if we can fill these in appropriately. Well, I know it has to be here. Why does that have to be here? Because in any group table, if you look in any row or column, you see each element exactly once. So that's no big deal. Let's do another one, mu 1, mu 1. Well, mu 1 star mu 1 is rho 0. So this one's rho 0H, that's not an issue, because mu 1, mu 1 is rho 0. Now let's complete this row. The only element that I don't see yet is rho 1. So that makes life a little bit easy. Now let's look in this column. The only element I don't see yet is rho 1 here. And the only element I don't see here is rho 0. Well, the original group was not a billion. The subgroup was a billion, but that's OK. The subgroup happened to be normal, that's fine. And here then is the group table. So we get a group with four elements. Question is the group of cosets is script S or G slash H a billion. Turns out to be yes. And how can we determine that? Because if you're handed the group table, the group being a billion simply asks whether or not it's somehow symmetric around the main diagonal here, the main axis. So yeah, I get a rho 1 there and mu 1s and delta 1s and delta 1s. So it turns out this group of cosets happens to be a billion. There's no guarantee that that should work out, because the original group G was not a billion. So I have a billion group with four elements. And now the question becomes a little bit more interesting. Can you identify this group as something you already know? Well, at least on the surface, we know that there's two different groups having order four for elements that are a billion. Either V, which we've now identified in a different form of Z2 cross Z2, or Z4. Those two groups are inherently different. Is there any way to determine which one of those two it might be? Aha. OK, so Jared's coming is, look, the main diagonal, we see the identity element each time. In other words, each element in this group has a property that when you combine it with itself, you always get the identity or rephrase to each element. That's not the identity, it has order two, which means the group can't be cyclic, because the group itself has four elements. So we see it's not cyclic, and the only not-cyclic group having four elements. So it's isomorphic to V. Or if you want to call it isomorphic to Z2 cross Z2, sure, you're saying exactly the same thing, because V and Z2 cross Z2 are recognized as being the same these. Oh, it's not cyclic, because if you take each element, and you combine it with itself, you already get the identity element. So if you take this thing combined with itself, it's the identity. This thing combined with itself, it's the identity. And this thing combined with itself, it's the identity. So if you start with any element in the group and you start looking at its powers, you get the identity before four wax. And so you don't have any elements of order four, so it can't be cyclic. All right, so there's a second example. That example's a little bit more intriguing for two reasons. One is the group that you get out in the end, you can't automatically say, well, it has to be this because it's got a certain number of elements in it. In this first example, we got a group with two elements, so it had to be Z2. So that's the first reason why it's a little bit intriguing. The second reason it's a little intriguing is that when you're doing the appropriate binary operation here, sometimes you don't get something that's already on the list of things that you've written out. You don't get the original name of the element that you've written down, and you have to look a little bit harder, all right? Hey, when I did the operation, I got rho 2h. I don't see rho 2h here, but you gotta look a little bit harder. Yeah, actually rho 2h is the same coset as rho 0h, so we listed it as rho 0h. All right, a little bit more verbiage or notation. We sometimes call this g slash h is sometimes called called the factor group of g by h. Some people call it the coset group. Some people call it the factor group, and I'll use both terminologies in here. Both are relatively standard. OK, so here's what we've done. We've looked at a couple of examples. Question, David? Yeah, OK, so, right. So, David's question is, let's talk about the relationship between abelianness and normalness. I'll rephrase your question that way. It's a good question. Yeah, let's talk about that right now. That's OK. Question, how does the property abelian relate to property normal? And the answer is, technically, this is apples and oranges. Technically, and here's what I mean by that. We talk about a group being abelian. Any group, I don't care which one. The property being normal is not a property solely inherent to a group. The property being normal means you need two players involved, a group and a subgroup. And you need to determine, essentially, how the subgroup sits inside the group. So if you say, is it abelian, you have to describe what it is. And then if you ask, is it normal, you're really asking a different question. Now, let me write down some things that are relatable, apples to apples. But in a more specific case, we can say the following. If the group happens to be abelian, then two things are true. First of all, any subgroup of g is necessarily normal. If I hand you an abelian group, and I hand you a subgroup, now I'm going to mumble something under my breath that's true but not related here, just so that we can look at it and nod our head and move on. The subgroup also happens to be an abelian group, because it sits inside an abelian group. That's a true piece of information but totally irrelevant here. What I am interested in, though, is whether or not when you look at a subgroup of this group, whether or not, it's a normal subgroup of the group. And we observed on Monday that the answer is always yes. If the group that things are living in happens to be abelian, then the subgroup automatically is normal. Reason, well, I'll re-prove something we did on Monday because it allows us to sort of keep sort of going back to one nice reason to have many different equivalent conditions for the same result. Well, if you want to share every left coset, the right coset, you could. But hey, all I really need to do is show that any one of the numerous properties that you listed on your quiz is true. And typically, folks, this is usually the case. If you're asked to show that a subgroup is normal, I mean, one way is just to bang out all the left cosets and bang out all the right cosets and show that they're the same. But what you might be able to do more easily is that, what it turned out to be, property three or property four, whichever it is, show me that G inverse HG is always back in H for every G in the group. So pick any G in the group and any H in the subgroup. Now compute G inverse HG. If the group is abelian, G abelian means that this is the same as HG inverse G, which is HE, which is H, which is certainly in capital H because it's little h. So H is normal in G by property three or something like that. Property three of being a normal subgroup. So I've just verified that any subgroup of an abelian group is a normal subgroup. And moreover, if you happen to start with an abelian group, property two is that necessarily when you form the factor group that the factor group is also abelian. Now I won't be able to say it enough, at least historically. I haven't been able to get it across, but I'll say it a lot. There are some nice properties of factor groups when the original group that you start with happens to be abelian. But the abelian property is just not that important here. It's not that big an issue. That's why the first two examples that I wrote out of a group and a normal subgroup, in those two examples, the original group was not abelian. But it still made sense to form the factor group because the subgroup that we wrote down happened to be normal. And it happened to be that in the two examples we did, it happened to be that the factor group was abelian. So hey, you can start with something not abelian, form the factor group, and get something abelian. Now why is the factor group also abelian? That's easy. Let's see. Take two things in the factor group. What do they look like? They look like cosets. Left, right, I don't care what you call them. They're the same. If you combine them, what's the binary operation on left cosets? That's the definition of the binary operation. So that's the definition. That's what it means to combine two left cosets when the subgroup is normal. You just combine the appropriate coset representatives. But wait a minute. If the group happens to be abelian, then inside the group, ab is the same as ba, because g is abelian. So abh is the same as bh, but that's bh star ah by definition. So what I've just shown is that if you take any two left cosets and you combine them in that order, you get the same result as if you combine them in the other order, so the group is abelian. Well, question. Oh, that's a great question. So let me, let's see, yeah, we're actually ready to take that question. No, let me hold off just for about two minutes on that one. Any other questions on the observations about abelian? OK, now. Will's question was this. If you remember way back in week two or so, we looked at inside any group a special sort of subgroup called the center of the subgroup. Intuitively, the center of the subgroup was those elements that you could switch the order with any other element in the group. So like the identity elements always in the center, because the identity element times any element of the group is the same as it doesn't matter which order. Sometimes there are elements that commute with all the other elements, even though the group itself isn't abelian. Inside d4, it turns out the element rho2 has that property. If you look at the table for d4 and you look at rho2, start with anything else. If you look at rho2, rho1, you get rho3. Or if you look at rho1, rho2, you get rho3. If you look at mu1, rho2, you get mu2. If you look at rho2, mu1, you get, I didn't say that. Rho2, mu1, you get mu2, et cetera. So it turns out inside d4, the center, these special elements that commute with everything else happens to contain rho2. So the question is, is there a relationship between a subgroup living inside the center and the subgroup being normal? And the answer in general is, if your subgroup happens to live inside the center, then it'll be normal. But in general, your subgroup might be completely far removed from the center. And in fact, a good example of that will write down in a minute. So this is sort of a side remark. Side remark, if you happen to remember this remark, it turns out that if you have a subgroup of g and h happens to be contained in the center of g, then necessarily h is normal in g. But that, too, is not really germane to what we're interested in doing. Let me give you a good example. I'm going to give you an example of a group that's really, really, really not a billion. And a big subgroup, a subgroup that's totally not related to the center of it, that turns out to be normal. Proposition, let g be this particular group. Sn. Yeah, Davia. Please. I'm going to pass on that. Just in the interest of time, if you want to go back to your notes in like week two or something, we talked about this idea really, really on. But I don't want to spend too much time judging now. All right. But here's the point. This notion of being commutative versus the notion of being normal really, I mean, there's some relationship, but the relationship is pretty tentative. And here's why. This is a good example. Let h be this subgroup, the subgroup of even permutations. Folks, this is ridiculously not a billion for n bigger than or equal to 3. That's ridiculously not an ambiguous subgroup for n bigger than or equal to 3. But it turns out then we know that h is a subgroup of g, already known. In fact, you proved it on exam one. Exam one for the proof. In fact, we can say more, h is a normal subgroup of g. So the collection of even permutations is in fact a normal subgroup. Now, I'll prove it in the way that I would hope that you approach most of the proofs that ask you to show a subgroup is normal. It's this computational one. Convince me that if you take something in the group, take something in the subgroup, and you compute g inverse hg that you get something back in h. And then I'll show you once we do the proof that way, I'll show you in this particular situation a somewhat more direct proof, a somewhat more subtle proof for elegant proof. Proof, I don't have to spend any time convincing you that h is a subgroup. We already know that. So all we have to do is show that h is normal. We'll use the computational oriented condition. We have to show that if we pick something called, let's call it sigma in g, and something called, I don't know, tau in h, we have to show that if we compute sigma inverse tau sigma that we get something back in h. But we're going to use some properties of permutations that we developed back in section, what was that? Section 8 or something like that? All right, let's see. If you write sigma as a product of transpositions, I don't know how many are in there. Could be any number. I mean, all I'm assuming about sigma is that some element of Sn, it might be even, it might be odd, who knows? But if nothing else, suppose that sigma can be written can be written as the product of some number of, I don't know, how many have a t transpositions. And folks, I have no idea whether t is even or odd, some number bigger than or equal to 1. Then suppose tau can be written as the product of, I'm going to call it w transpositions, but I do know something about w here. The number that we pick is even, because that's what it means to be in this subgroup. That's the definition of An. So now the question becomes, how many transpositions does it take if we try to write out sigma inverse tau sigma? Well, that's pretty easy. Then it turns out, by what we did back when we studied permutations, sigma inverse is also the product of the same number of transpositions as sigma is of t, where t is the same number here, transpositions. I won't write out the proof for you, but remember, if you take sigma, it's the product of t transpositions. If you take its inverse, then you reverse the order on everything. You take everything inverse, but the transpositions inverse is itself, so you've written down the same number. So sigma inverse tau sigma requires, well, let's see, that one required t, but then I put that next to something that required w, but then I put that next to something that requires t, and so the total number of transpositions required is that. Remember, when you do products of transpositions, the number of transpositions that you're using is getting added together. In other words, it's 2t plus w, but 2 times t is even, because it's 2 times some integer, and w is even, that's hypothesis. And so I get the sum of two even numbers, which is even. And so the conclusion is sigma inverse tau sigma is in h, and we're done. So in fact, inside Sn, whether I'm looking at S3 or S4, S5, or whatever it is, it turns out the subgroup consisting of the even permutations is not only a subgroup, but it's a normal subgroup. And in this particular case, here's the follow-up to this observation, look, if it's a normal subgroup, it means that I can form the factor group. See, that's a big headache. What the heck are you going to do there? Well, it turns out that this factor group isn't 2, hard to understand. Well, let's see, how big is it? Well, I know how many elements are in the group. I know how many elements are in the subgroup. We showed that the number of even permutations is half the number is 2. So in fact, the factor group that we can form looks like this. What are the cosets? Well, one coset is always identity times An. I'll call it identity. Identity times An. Identity times An. There's the coset itself. Now we need another one. How about the 1, 2, An? 1, 2, An. I know that works. Well, look, I've got all the even permutations in here. Now all I need to do is find all the odd permutations, just take all the even ones and multiply by whatever transposition you want. Everything in there is going to require an odd number of transposition. And what's the observation? This is going to give identity. This is going to give 1, 2. You should get a group of two elements, folks. You know what group it is. It's Z2. So isomorphic to Z2. So this is yet another example of a situation where you can start with a non-abelian group and get an abelian factor group. So all the examples that I've done so far of factor groups, all of them have turned out to be abelian. Is it the case that anytime you write down a group and a normal subgroup and you form the factor group, you get an abelian group? The answer is definitely no. I'm going to write down two trivial examples of normal subgroups. And in one of those, we'll find many examples of factor groups that are not abelian. Before I do that, let me make a quick comment, this sort of elegant proof that the collection of even permutations is a normal subgroup. Remember way back when, when we looked at kernels of homomorphisms, this sort of was the first, that was our sort of lead in to the notion of a normal subgroup because what we showed is that if you have a kernel of a homomorphism, that it has this one property on this list, this computational property, where we're able to relatively easily show that the kernel of any group homomorphism is a normal subgroup because it has this property that G inverse HG is always back in H, where H is the kernel. So that we proved we could go tonight. But we also proved the first night that we looked at group homomorphisms, that we could write down a group homomorphism from Sn to Z2. And the homomorphism was, remember we sort of defined it element-wise, if you throw in an even permutation, I want zero to come out, and if you throw in an odd permutation, I want one to come out. We showed that that was a group homomorphism. And we identified what its kernel was. It was An. So we already know, in fact, that An is a normal subgroup because we've already identified An inside Sn as the kernel of some homomorphism. It happens to be a homomorphism to Z2. Questions, comments? All right, more examples slash information about normal subgroup. So it turns out, observation, this isn't too deep. Remember, folks, inside any group, we always have these two things that the author calls trivial subgroups. If you have a group G, then the identity subgroup, then the identity element sitting by itself is always a subgroup. And the group itself sitting inside the group is always a subgroup. Sort of silly. Well, it turns out those two trivial examples of subgroups always turn out to be by default normal subgroups. So if G is a group, then the group itself sitting inside itself as a subgroup is always a normal subgroup. That's pretty easy to see. There's lots of ways to see it. If you take anything in the group and anything in the subgroup, you do G inverse HG to get something back in the... Sure you do. You get something back in the subgroup because the subgroup's the group. Or here's another way to think of it. What are all the left cosets? There's only one of them. It's identity G because you get the whole group already. And that's the same as G times identity. Left cosets are right cosets. There's only one of them. So that's not a big deal. The other is maybe a little bit more intriguing. Not only is the identity a subgroup, it's actually a normal subgroup of the group. I'll tell you how to prove that one. Use this computational approach. Is it the case that I take anything in the group and I take anything in the subgroup? There's no choice there. And I multiply or I do the product G inverse something in the subgroup G, do I get something back in the subgroup? Yeah, because it happens to be that the only thing I can choose in the subgroup is that and that thing is easy to work with. Which then gives the following. It turns out, turns out, and I'm not going to run through all the details. All right. If you've got a normal subgroup, it means you can form this. How many elements in that? One. The number of left cosets inside here is Lagrange's theorem. It's the number of elements in the group divided by the number of elements in the subgroup. So this is a pretty uninteresting group. In fact, this group then is isomorphic to, I don't know, how about that group? Zero under addition or one under multiplication or E or whatever you want to call it. So it's certainly not very interesting to look at. On the other side, if you look at the factor group of G by that normal subgroup, let's see how many elements are in this. I shouldn't have put an equal sign here. The number of elements is one. Yeah, I did apples and oranges, but that was bad by me. So as a group, G slash G just looks like the identity. Pardon me. How about the number of elements in this thing? Here's the number of elements in the group divided by the number of elements in the subgroup. That's easy. There's just one element in the subgroup. So the punchline is it turns out the number of elements in the factor group is the same as the number of elements in the group. That doesn't automatically allow me to conclude what I'm about to conclude, but at least it's a good plausibility argument. It turns out, in fact, in fact, if you look at G slash E, you get something that just looks like the original group back. So if you mod out, if you do the factor process and the thing that you're factoring out is the trivial subgroup E, then you haven't really created anything new at all. You're just looking at the original group. So what that means, folks, is that if you start with a non-abelian group and you use the identity subgroup as the subgroup that you're going to use to factor out, then you still get a non-abelian group back because you get the original group back. All right, lots of different examples here. So questions, comments. Okay. Yeah, last 10 minutes here, let me do the following. I'm going to write down a couple more examples that I'm going to have you work out the details on. What turns out to be interesting, and this is maybe surprising the first time you hear it, is this. I'm going to write down a group and I'm going to write down a subgroup. The group's going to be a billion, the group's automatically normal, so it makes sense that we'll be able to form the factor group. And I'll be able to ask the question, all right, do you know or can you recognize what this factor group looks like up to isomorphism? Well, if the original group is a billion, then the factor group is a billion. So you have an abelian group, you'll be able to count the number of elements in it. This fundamental theorem that we talked about said, all right, well, hey, I now have an abelian group with a certain number of elements, so there's only a certain number of choices and if the abelian group happens to have four elements, there's only two choices, it's either z4 or z2 cross z2 and here's what I'm going to write out for you. A group that has eight elements and is a billion, a subgroup that has two elements, which is by default normal because the original group's a billion, so when you form the factor group, you get eight divided by two or four elements and you write out the group table and the group table will look like z4. Then I'm going to write down the same group, the exact same group. I'm going to write down a different subgroup that has two elements in it. Don't think about that, it's a different subgroup, but it's still got two elements in it. So the new subgroup that I'm looking at, in some sense, looks like the old subgroup, it's just a group of two elements. But when you form the factor group of the original group by this new subgroup, you again get a group with four elements, but it'll be the other group with four elements. So lots of things can happen when you form factors. So that's what I want you to do. Consider, so here's a big example. The group G is z2 cross z4. If I'm not mistaken. No, let me write it this way. G is z4 cross z2. So there's G. And in the first situation, what I want you to do is let H be this subgroup. The subgroup generated by 0, 1. I'll tell you what that is. It's just 0, 0 and 0, 1. That's a complete non-issue. I mean, you take this thing, you combine it with itself, you get 0 in the first coordinate, you get 1 plus 1, which is 0 in the second coordinate. Form G slash H. I'll tell you already something about it. The original group is a billion. It's a billion? No problem. It has, well, let's see. The number of elements in G is 8, 4 times 2. That's not an issue. The number of elements in the subgroup is 2. So the number of elements in the factor group is 8 divided by 2, which is 4. So here's the conclusion. So by this fundamental theorem of finite a billion groups, I have an a billion group with a certain number of elements. So there's only two possibilities. Either G slash H is isomorphic to Z2 cross Z2 or isomorphic to Z4. Can we predict based on this information which one it is? Answer, no. You have to sit down and actually pound out the group table to figure out which one it is. And here's why example revisited part 2 of the same sort of example. Same group, same G, in other words, Z4 cross Z2. But I'm going to pick a different subgroup. Now, let the subgroup be this one. The subgroup generated by 2 comma 1. I'll tell you what that looks like. It looks like 0, 0 and 2, 1. Why is that? Because if you take this thing and you combine it with itself, 4 comma 2, which inside Z4 cross Z2 is 0 comma 0. Now I want you to form G slash K. G slash K, sorry. Now I'm going to make the same statements. It's a billion because G is, no big deal. The number of elements in the factor group is 4 because the original group had 8 elements and the subgroup had 2 elements. So either G slash K is isomorphic to Z2 cross Z2 or is isomorphic to Z4. Can we determine which one it is on the surface? The answer is no. And the punchline in this example is the following. Even though the data on the surface looks almost identical in each case, in one of these situations, the factor group is isomorphic to Z2 cross Z2 and in the other situation the factor group is isomorphic to Z4. So what I'm going to have you do as an exercise that's going to come up in this homework assignment that I gave you on Monday, determine which. Determine what you get. Which. Right out of group table. Let's see, do I have time to do this? Yeah, last five minutes, this will be perfect way to sort of wrap things up here. Okay, lots of things can happen. When you're looking at normal subgroups you form these factor groups and what we're about to do is sort of make things come full circle now with this final comment. The original motivation, the original idea that eventually led us to this notion of a normal subgroup was the kernel of a homomorphism. If you have a kernel of a homomorphism we showed that it has the nice property that G inverse HG is contained in H where H is the kernel, therefore the kernel is normal, therefore we can form G slash the kernel. So now the question is, is there some symmetry to the process? If we have the kernel of a homomorphism it's necessarily a normal subgroup. Question, if I hand you a normal subgroup is it necessarily the kernel of a homomorphism? So do things have some sort of symmetry or some sort of mirror image and the answer turns out to be yes and I'll conclude this description and discussion of normal subgroups with that observation. Turns out, look, turns out the following is true. Originally we showed way back when that if phi from G to G prime where G and G primes are groups is a homomorphism then kernel of phi is normal in G. Turns out the converse is true too, also true. Namely that is if H is a normal subgroup of G a normal subgroup of G then there exists a group G prime and a homomorphism phi from G to G prime with the property that the kernel of phi is the normal subgroup that you started with equal to H. So every kernel of a homomorphism is normal and let me rephrase that so it really sounds symmetric. Every kernel of a homomorphism is a normal subgroup and this last statement is and conversely every normal subgroup is the kernel of a homomorphism and I'll tell you where to find it. Here's how to find it, here. This seems like a really esoteric statement. I'm simply asking you to write down whatever normal subgroup you want of the group G and then I'm asking you to build out of thin air some other group G prime and some other group G prime possibly and some homomorphism from G to G prime so that the kernel is precise in the subgroup that you started with that's a really tall order but it turns out to be easy to do. Here's what you do. You let G prime be G slash H. G factor H. Is this a group? It is. Why? Because we've assumed H is normal. So I can build this group. I'm going to give it a name. I'm going to call it G prime. Then I want you to define the following function. I'm going to call it pi from G to G prime and here's what the definition of pi is. Pi of G is... I need to spit out something in here. That's easy to do. G H. And what we'll show next Monday is that by playing all these games by defining this factor group and defining the function that takes any element in the group to its left coset that it generates that not only is pi a homomorphism that'll be pretty easy but it turns out we'll be able to show that the kernel of pi is precisely H. So that'll complete the circle. That'll sort of give us the symmetry between kernels of homomorphisms and normal circles. All right. That's a good place to quit.