 That's true in any set, whenever you have an equivalence relation, you get the set partitioned into chunks, but what we're going to show is that this particular partition of the group actually has some extremely nice properties. So here's the quick reminder of what we did last time at the end, reminder. And the point is here, folks, remember that we've sort of switched gears from looking at specific kinds of groups, permutation groups, to now looking at general groups so don't think that all this has to be done in the context of permutation groups. Everything that we're doing here can be done in the context of any group you want to write down. Ciclic groups, or these V groups, or abelian groups, or not abelian groups, makes no difference. All right, so here's a reminder of the equivalence relation that we wrote down last time. So the setup will always be H is a subgroup of G and the relation is this, and I'm looking at my notes because I'll talk about this maybe the end of today, but possibly on Monday. There's sort of a mirror image relation that we can also talk about and we'll get to that. I mean, we'll definitely get to that eventually because it's important, but I'm not sure of any, but I want to do it in the same order that this author does. The relation is this. It's tilde or a wiggle. A is deemed to be related to B in case, this is the definition, A inverse star B is in the subgroup. That's what the relation means, and what we showed is that this relation actually produces an equivalence relation. Equivalence relation, that's what we proved last time. It's compelling to note that when you prove something is an equivalence relation, you have to show it's reflexive, bless you, symmetric and transitive. When you prove that something is a subgroup of a group, you have to show that it's closed, contains the identity and contains inverses, and it turns out each of the three things that we need for an equivalence relation corresponds exactly to one of the three things that we have because we're assuming that we've got a subgroup of the group. What we did at the end of last time was we looked at a specific example, and I'll remind you of that example quickly. The group was Z6, let's see, G is Z6, and the subgroup was the subgroup generated by three. What we did was we spent a little bit of time asking which elements are related to which other elements, and what I'm trying to do is get in the habit of understanding how an equivalence relation produces a partition of the original set. The partition is simply take any two things that are related and throw them in the same subset, and the beauty of having an equivalence relation to get such a collection of subsets of the original set is that the subsets that you get kicked out are always going to be disjoint, and they're always going to make up the entirety of the original set. So what we got was a partition of G corresponding to H to H. In other words, corresponding to this equivalence relation, the partition turned out to be that was one subset, here was another subset, and here was another subset. Again, all we've done is we've matched up elements inside subsets in such a way that the elements inside a given subset are related to each other, whereas if you take elements from two different subsets, they're not related to each other, where this is the definition of the relation. Let's try another example just to continue the computational process. Let's do this one. The group G is D4, and the subgroup that I want you to look at is, yeah, this one would be a little bit more interesting, is the subgroup that consists of the two elements, row 0 and mu 1. Well, I know what the group is. The thing you need to do first is convince yourself that the subset that I've written down and called H is actually a subgroup, and it might be helpful to haul out these pages that I passed out a couple of days ago so that you've got the group table of D4 right in front of you. Let's see, mu 1 is this element 1, 2, 2, 1, 3, 4, 4, 3. It's the one that swaps 1 with 2 and swaps 3 with 4. So let's see, swaps 1 and 2, and swaps 3 and 4. I guess that's a good thing. So do that and then do that. It's fine. Question, if I look at mu 1, well, is this actually a subgroup? The answer is yes, because look, if I take mu 1 and I square it, mu 1, circle mu 1, well, I run into my table here, mu 1, circle, mu 1 is row 0. It's good. So in fact, this is a subgroup. And now the question is going to be find the equivalence classes. That means the partition classes corresponding to this equivalence relation. Equivalence relation called tilde. So folks, this is the equivalence relation we're going to use regardless of what group you've handed me, regardless of what subgroup you've handed me. The equivalence relation is defined to be deem the element a to be related to the element b in case when you compute a inverse b that you get something in the subgroup. OK, so let's see what elements are related to each other. What elements are related or not related or not? Well, presumably, folks, what you've got to do is take every pair a and b, compute a inverse b, and ask whether or not the result is in the subgroup. So that's tedious. But maybe we can use the table to more efficiently try to compute these things. For example, let's see, I know that any element is related to itself. That's no big deal. How about what's row 0 related to? Now let's just pick something random. How about, I don't know, delta 1? Question is delta 1 related to, pick something else around. And how about mu 2? Just picked an arbitrary pair here. OK. Well, let's see, is it? I know what the test is. Two things are related in case when I compute the inverse of the first thing. So I take the inverse of the thing on the left. And then I combine it with the thing on the right. We have to see what we get. Well, let's see, what is this? What's delta 1 inverse? Well, let's see, delta 1, oh, delta 1 is the thing that takes 1 to 3, and 3 to 1, and leaves 2 and 4 the same. So geometrically, what is that? So 2 and 4 stay fixed, and 1 and 3 get swapped. So what this is is some sort of diagonal rotation, right? Hence the use of the letter D here, at least the Greek letter D for diagonal. So that's a good habit to get into, see if you can interpret what the notation is standing for. So let's see. So delta 1 inverse is just delta 1, mu 2. And now let's compute delta 1, mu 2. Delta 1, mu 2 is row 1. Just delta 2, I'm sorry, delta 1, mu 2, how did I do here? Delta 1, mu 2, oh, it's row 3. Pardon me, went too fast there. It's row 3, but the punchline is whatever got kicked out, is it in the subgroup? Nope, not in the subgroup. So these two things are not related. So delta 1 is not related to mu 2. I just sort of picked those out at random. So presumably, here's what you'd have to do. You'd have to pound through all the possible pairs, and presumably, you'd have to compute A inverse B for each pair A and B. Every time this computation kicked out something in the subgroup, you'd say, yep, those two things are related. Every time those two things, that computation didn't kick out something in the subgroup, then you'd deem those two things as not being related. So this seems tedious. And that's the reason I wanted to do this example. It gets even worse if the groups are bigger. But it turns out, so to avoid the tedium of doing this, to avoid the tedium, but that's all it is. It's just straightforward of determining, determining directly whether or not, whether what pairs or what pairs, which pairs, which elements are related under this equivalence relation. What we do is we instead, we prove a result which says, which gives more direct information about these equivalence classes. And it turns out to be not too hard to do, we need unfortunately one more definition, but this definition will have sort of a nice payoff. Because once we've written down and understood the definition, then the computation that the definition implies will allow us to compute these equivalence classes, these subsets that make up the partition of the group in a significantly more efficient way than simply taking each possible pair, computing A inverse B and deciding which things actually kick out something in the subset. And here's how you do that. So the definition is this. I'm again going to start with a group G and a subgroup H of G. And here is the verbiage. Pick any element. Pick any element, let's call it little a in G. And here's the definition of the left coset of H generated by A. That's the term that we're defining here. That's coset, C-O-S-P-T, is the following set. Here's what I want you to do. I want you to take the set of all elements in the group. So it's the elements in the group with the property that G can be written as A, the element that you started with, times or combined with whatever the operation is, H, for some H in the subgroup. So here's what I want you to do. Take the standard configuration, a group and a subgroup inside it. Now, pick your favorite element in the group. I don't care which one it is. You might have chosen an element that's inside the subgroup. I don't care. Pick anything in the group, in the big group. Now, here's what I want you to do. I want you to tell me all of the elements in the group that you can write or form or produce in this type of way, where you start with the element that you've just picked randomly out of the group, and you combine it with each of the elements in the subgroup. For example, let's look at G is d4, H is rho 0 mu 1. So that's the subgroup situation that we just looked at. Let's look at, let's pick something in your group. How about delta 1? Pick A equals delta 1. Folks, the element that I choose in the group is totally arbitrary. I don't care which one you pick, just pick one. Notice I'm not asking you to necessarily pick an element in the subgroup, although that's certainly loud. Just pick an element in the group. And now, here's what I want you to do. I want you to form all the elements that you can that look like the element that you started with, combined with each of the elements in the subgroup. So I want you to compute this. That's of the correct form. It's the element I started with, together with the first element in the subgroup and this one. So these two elements together, I'll write it in a set, give the left coset of the element delta 1, the left coset of H generated by delta 1. There are the two elements in that coset. Well, let's see what they actually look like. In other words, let's see, I can easily compute this one. Delta 1, rho 0 is just delta 1, because I know rho 0 is the identity element of this group. And let's see delta 1 mu 1, delta 1 mu 1. We look at the table here and we get delta 1 mu 1 is rho 1. So there is the left coset of this subgroup, generated by the particular element delta 1. This seems uninspired. Well, let's just pound out a few more. We're going to start with exactly the same configuration. We get this. And let's pick another element. How about, I don't know, makes no difference. Now pick a new element. How about A equals mu 2 or something like that? It's just sort of randomly going through here, picking elements. But I want you to now compute or find the left coset of H generated by mu 2. And just to make it so that I don't have to keep writing that expression out, this expression, folks, is given a notation or a symbol. And the symbol makes a lot of sense. What I'm asking you to do is take what every element you've picked out of the group and combine it with every one of the elements in the subgroup so that it makes sense to denote this set by, well, A stays fixed throughout the entire discussion, but the elements of capital H, the elements that you're somehow combining on the right, are allowed to roam throughout the subgroup. So when I ask you to find the left coset of H generated by A, what I'm going to do is simply, in shorthand, ask you to find A H. In other words, IE, find mu 2 H. And the computation's just not bad. You just take a table. So what we're supposed to do is we're supposed to write down the set mu 2 times the first element, and I keep saying times. Of course, it means composition here, but since we're using multiplication notation for binary operations in groups, that's the standard jargon, at least. And then mu 2 circle, let's see, what was the other element in the subgroup? It was what, mu 1? Let's see what we get. Well, the first one's always easy. There's not much computation you have to do to get the first one, because this is the identity element of the group. So you just write down the element that you've chosen. And then what did mu 2, mu 1, turn out to be? I'm sorry, say it again? Rho 2? OK, that's good. Rho 2. So here is the left coset of H generated by mu 2. This seems pretty tedious. It is sort of tedious, but what you'll see after we do another example is the pattern that is going to develop. Now let's pick another one. I don't care. Somebody holler out a choice. Give me an element. Any element? Rho 3. OK, good one. Good one. I don't want to erase this yet. Well, we've got space to do it in here, that's fine. OK, so find the left coset of H generated by Rho 3. Not too bad. You just write down the element that you've chosen, which was Rho 3. And then you combine it with the first element of the subgroup, which is mu 0. And then you take Rho 3 and you combine it with the second element of the subgroup, which is mu 1. And what we get is, let's see what we get. Well, we get Rho 3. And what is Rho 3 mu 1? Delta 2. So let me see what the game is. Let's do a couple more. Here's another one. How about, can I go over here? Mr. Video Man, is this too far over? OK, good. Pick, I don't know, how about another one? A equals Rho, I don't know. Let's do Rho 2 back. OK, so let's see. Find the left coset of H generated by Rho 2. Not too bad. It's Rho 2, Rho 0, and then Rho 2 mu 1, which is Rho 2. Now watch this, folks. I'm not even going to ask for help. I'm not looking at the table. Rho 2 mu 1, I'm 100% guaranteed as muted. Penn never left the hand. That was pretty good. Now you can look at your table and say, yeah, it is. How'd you know that? Here's how I knew that. What we're going to see eventually is if you're cranking out one of these left cosets generated by a subgroup and you identify an element that already appears in one of the other left cosets, then necessarily the left coset generated by this one has to be equal to the left coset generated by the other one. In other words, if there's any overlap at all between two left cosets, and I'll write this down in great detail and with more clarity in a little bit, but the point's going to be, as soon as I saw a Rho 2 in this left coset and realized that I had already seen Rho 2 in another left coset, then immediately I know what the remainder of the coset looks like. It looks like the one that I've already computed. So that was easy. So let's see. Let's try picking another one. Pick A is, I don't know, how about, well, let's go ahead and pick something that's already in the subgroup. That's certainly allowed. When we're talking about the coset generated by some element in the group, but I don't preclude you from picking this element already in the subgroup, that's fine. So find the left coset of H generated by Mu 1. Well, it's easy to do. Let's see. It's Mu 1 Rho 0, Mu 1 Mu 1. I know what Mu 1 Rho 0 is. It's Mu 1. I know what Mu 1 squared is because H happens to be a subgroup, so that's Rho 0. That's kind of interesting. So when I looked at the left coset generated by Mu 1, I happened to get the subgroup back. That is H. That's sort of interesting. So one of the cosets happens to be the subgroup itself. So let's see. How many have we done here? I don't know. We've done like 6 or 7 or something like that. Let's pick another one. Let's see. Have we done Rho 1 yet? We did delta 1. No, let's do Rho 1. A equals Rho 1. But now I want you to think about, given the little trick that I just shared with you, so Rho 1 H is Rho 1 Rho 0 and then Rho 1 Mu 1. Rho 1 Rho 0 is Rho 1. That looks familiar, right? All right, here it is. I've already seen Rho 1 before. So guess what the other element in this coset is? Delta 1. You can check that. Let's go ahead and do the easiest one that there is. How about pick A equal to Rho 0? Perfectly good element in the group. Find this left coset. The left coset of H generated by Rho 0. Rho 0 Rho 0 and then Rho 0 Mu 1. That's totally easy. It's Rho 0 Mu 1. In other words, it's just what? It's just H again. OK. So although that perhaps seemed a little bit tedious and I had sort of played this up as a way of avoiding the tedium, we're going to prove enough properties of this process, of the process of computing or finding the left cosets of a subgroup with respect to each of the elements in the group, that what we'll be able to do is relatively easily write down what the partition of the group is corresponding to this equivalence relation. So here is the theorem. The theorem turns out that the left cosets of the subgroup H. This is always the preamble H, the subgroup of G. The left cosets of H in G form a partition of G. And remember, as a reminder, we looked at this on Monday as well. Partition means you've taken the group or whatever the underlying set is and you've hacked it into pieces in such a way that none of the pieces overlap. In other words, there's no intersection. And that the pieces, when you put them all together, give you the entire set. In other words, any two left cosets of H in G are either disjoint. Disjoint means that they completely miss each other. The intersection is empty or equal. I should have penned one sentence to that, but it'll start me, I spent so much time not, and I did. That's all right. The second sentence I should append to the statement is, well, there's another piece to being a partition. A partition not only means that the sets that you've written down have no overlap that they're disjoint, but it also means that every element in the original set appears in at least one of the subsets. So I'll put that in just for completeness and every element of G appears in some coset, in some left coset. So when it's all said and done, I'm going to have you remind me of what we just computed. Example, I'm simply writing down the data that we just computed. The group was D4. The subgroup was rho 0 mu 1. And what we did was we wrote down the cosets of each of the elements in the group. So if we just look at them again, rho 0 H turned out to be rho 0 mu 1, which turned out to be the same as mu 1 H. What else did we compute? We computed, do I leave any of them here? Probably not. Oh, well. Let's see. We computed rho 1 H. And what did we get? We got rho 1 together with what? Delta 1. Delta 1. And that turned out to be the same as the coset delta 1 H. We computed rho 2 H. And we got rho 2. And what else did we get mu 2? I'm going to guess that. How'd I do? Maybe I lied. Rho 2 was matched up with mu 2. And then rho 3 H turned out to be rho 3, what's left over delta 2, which turned out to be the same as delta 2 H. So here's the point, folks, when we look at all of the left cosets, well, at least on the surface it looks like there should be eight of them because there's eight elements in the group and you can form the left coset based on any one of the eight elements. But the punchline is the only subsets that come out are these. So some of the elements produce the same left coset. This one produces the same as that one. Rho 2 produces the same left coset as mu 2. Rho 1 produces the same left coset as delta 1, et cetera. So when it's all said and done and you stand back and say what are the cosets, the answer is there's only four cosets. There's only four different sets that come out. And you'll notice that if you look at any two of these, there's no overlap between them. And amongst all four of the cosets that come out, you've identified every element of the group. So what this process has done is it's taken the group and it, in effect, has hacked it into smaller pieces. The pieces are these four subsets. Now, and this is looking ahead a little bit, but it's also sort of looking back as to why we were interested in doing various things on days one and two in here. If I hand you this subset of the group, simply say, all right, we're living in the group D4, consider the subset consisting of rho 3 and delta 2. Question, which element generates that subset as a left coset? Answer, there's no good answer. You can either answer correctly, rho 3 does, or you could answer just as correctly, delta 2 does. Refraised, this particular coset has two different, perfectly good names. You can either call it that or you can call it that. Similarly, this coset has two completely different, but legitimate names. You can either call it rho 0H or you can call it mu1H. So this issue of when you're in a situation where you've got the things in the underlying set that somehow might have different names like 1 slash 2 and 3 slash 6 in the rational numbers are two different names for the same beast, we had some issues with well-definedness. And what we're eventually going to get to is situations where the things involved, the things that we're going to be interested in aren't rational numbers anymore, but are these cosets? And the question is going to be, what should we legitimately call a specific coset? Well, hey, there's two totally good and totally appropriate names for it. That might cause some issues. So that's something that we're going to have to be a little bit concerned. We're not looking at what that issue is yet. I just want to bring it up and mention that, look, for a given coset, there might be two perfectly good names for it or two perfectly good ways to describe it. All right. See, what does this theorem say? It says, oh yeah, the form of partition. So I certainly haven't proved the theorem yet. All I've done is shown you in the context of the one example that we worked out that, yeah, in fact, that's exactly what happened. So let's at least sketch a proof of this theorem. So sketch a proof. Sketch means I'm not going to write down all the details, but I'll at least point you in the direction of the computations that you'll have to make. Let me go ahead and at least convince you that this last sort of throwaway statement is legit. The point is if I ask you to compute all of the cosets, the left cosets of a subgroup inside a group, and in the end you write them all down. Yeah, there were some overlap. But it turns out there's four of them. The point is that all, in this particular case, eight elements of the group appear somewhere. Folks, that has to happen. And here's why. Because the thing that you've started with, this thing called age, is a subgroup. So it necessarily contains the identity element of the group. Now if you want me to find a specific element of the group inside one of the cosets, it's easy to do. Tell me which element you want me to find. How about delta 2? Okay, it's easy. I'll tell you which coset it's in. It's in the coset generated by delta 2. Why is that? Because if you take delta 2 and you combine it with each of the elements in the subgroup, one of the elements is the identity element. So you're always computing delta 2 times the identity element. So there's delta 2. Did I find it? Yeah. Did I find delta 2? That's in here. Why? Because one of the computations is delta 2 times identity. You want to find row 1? Sure, it's right there. Why? Because one of the computations you do when you compute the left coset generated by row 1 is row 1, row 0. So there's row 1. So each of the elements in the group has to appear in the coset generated by it. Because when you're generating that coset, you're doing the computation element circle identity element, which is just going to give you the element back. So this second piece is each element appears in a coset. In fact, I'll write it more formally. Each element, let's call it g in the group, appears in the left coset g written next to h. In parentheses, since e is in h and g is of the correct form, g circle e. So the fact that you'll eventually see all the elements in the group by doing those left coset computations, that one's sort of easy. The real issue is, why is it the case that if you hand me two left cosets, that they're either disjoint, in other words, that there's no overlap? Or that they're identical? That one's a lot harder, but here's the idea. So now we show that if, let's see, there's a logically equivalent statement to, if you take two cosets, there's two possibilities. Either they are disjoint, in other words, they miss each other completely, or they're identical. The logically equivalent statement to that is I'll prove that if there's at least one thing in the overlap, then in fact, the two cosets are equal. So logically equivalent, it just makes the proof a little bit shorter. We show that if the two cosets, let's call one of them a h and the other one b h, we'll show that if we're not in the case where they're disjoint, in other words, it's not empty, then in fact, the two are equal. And folks, what I'm about to prove for you is that little trick that I sort of clued in on when we were doing these computations. The point is, as soon as I see a single element that's both in this coset and in this coset, I can conclude actually that the two cosets are identical. And why is that? And this will take us a little bit more time, but let's just see how these things play out. In other words, I assume that there's some element in the group that lives in this intersection. I don't know, there is, what should we call it? How about z in a h intersect b h? Assume there's something in there. That's the hypothesis that the intersection is not empty. We want to show that a h is b h. Wow. So what does that mean? Well, what does it mean to say that z is in the intersection? What does it say? Or what does it mean to say that if I look at these two cosets that I've got some element that lives both in the first one and in the second one, it means, well, it means that z is in here. And it means that z is also in the other one. That's nobody does, just a statement about sets. But now let's write down what it means to say that z is in this left coset. In other words, in other words, let's see. z is of the correct form. And what's the correct form? Being in a coset means that you can write the thing as the given element times something in the subgroup. I don't know what the thing in the subgroup is. Let's call it, how about h1 for some h1 in the subgroup. And I suspiciously called that thing h1 and not just h or h prime or something like that for a reason. Because look, I also know that this same element is in this other left coset. What does it mean to be in the left coset generated by b? It means I can write it as b times something in the subgroup. Now folks, there's absolutely no reason to think that the element of the subgroup that gives me z as something combined with a is the same as the element of the subgroup that gives me z when I combine with b. So I've got to call it something else. But at least I know that this thing called h2 is in the subgroup, so that's good. So that's the given information. And what are we interested in doing? We're interested in showing, wow. So what? But then look, that's equal to z. And that's also equal to the same thing called z. So I can conclude at least that a h1 is the same as b h2. That will be convenient, in fact. I haven't even gotten to what I'm trying to prove yet. All I've done is taken the hypotheses that the intersection is empty and distilled out from that hypothesis some information. All right, in fact, here's what else we can conclude. In fact, and it's not clear how I'm getting, well, it's not clear why I want to write down this particular statement, but you'll see exactly that this is the piece of information that we want in order to eventually reach our conclusion. Folks, I have this equation, so I can multiply, quote unquote, I can combine both sides with the same element of the group. As long as I'm careful to do it on the same side, I can't put h2 on this side and h2 on that side or something like that. What I'm going to do is combine both sides with h1 inverse. A h1 h1 inverse is b h2 h1 inverse. Where the heck? Where are these coming from? Don't worry, we'll get there. In other words, i.e., what I've written is that, well, the left-hand side, h1 h1 inverse, no big deal, that's just the identity. So I have a times the identity, which is just the identity, is b combined with h2 combined with h1 inverse. I think that's sort of interesting. All I've done is taken the hypotheses that the intersection of h and b h is empty, I'm sorry, is non-empty, and reached this conclusion. Now let's focus in on what the heck we're trying to prove. We want to show that the left coset a h is the same as the left coset b h. That's the goal. That's where we're headed here. Well, how do you show two sets are equal? You show that this one's contained in this one. You show this one's contained in this one. The standard method to show that two sets are equal. So to do this, here's what we're going to do. We show first that a h is contained in b h, and then we'll show that b h is contained in a h. That's how you show two sets are equal, no big deal. OK. So what does it mean to show that this is a subset of that? It means I have to take an arbitrary element of that and convince you that it's in the other set. So to show that a h is contained in b h, pick any element, what should we call it? How about t in a h? We have to show that t is in b h. Pick an arbitrary element in the first set, show that it's also in the second set. All right, now here's where we get back to the definition of what the left coset generated by a of h is inside g. This set, folks, by definition, a h, yeah, yeah. This set is the collection of things in the group that can be written as the element a combined with something in the subgroup on the right. So the hypothesis that t is in this thing means that t is a h for some h in the subgroup. So what we're given, the definition of this coset, is that t can be written as a times, well, let's see, something in the subgroup. I've already used h1 for something and h2 for something. How about a h3 for some h3 in the subgroup? It's just the definition of what the coset is. And what do we have to do? We have to show that t can be written as, well, does it mean to draw this conclusion? It means that t can be written as b times something in the subgroup, and if I had h4. So at least what we've done is we've reduced the task down to, well, showing that the elements inside the group can be written in the correct form. So here's the given information. Here's where we want to get. And here is a piece of information that we get to use. That's what came from the hypothesis. Well, I love that connector. Well, I'll tell you something about t. t is, what is it? It's a h3. That's given information. But that's equal to, because a is this, we're just going to substitute. Substitute. A is bh2h1 inverse. Just substitute, because that's what I was able to conclude about. This is nice. And here's why. So this is, I'm going to use the associative law. This is b combined with h2h1 inverse h3. What did I know about h1? h1 is in the subgroup. h is a subgroup. If you have an element of a subgroup, its inverse is also in the subgroup. So this is in the subgroup. But if I combine something in a subgroup with something else in the subgroup, closure of a subgroup means the results in the subgroup. So this is in the subgroup. But now I'm combining something in the subgroup with something else in the subgroup. And so this whole mess is in the subgroup. I've used the fact that if I start with something in a subgroup and I compute its inverse, that I necessarily get something back in the subgroup. And then if I combine things in a subgroup, I get back in subgroup closure of the subgroup. So the fact that the subgroup is closer in the inverses and is closed under the multiplication means that this thing is in the subgroup. So what do you want to call it? How about h4? Where h4 is what I've deemed this to be? Check. That's what we had to show. It took us a while to get there. But what we've shown, and this is where I said I'm not going to do the whole computation for you. What we've shown is this chunk of it. So I've shown that any element that's in the left coset generated by a is necessarily in the left coset generated by b. I'm going to leave this piece of the computation to you. So u, not to turn in, show bh contained in ah. The hint is when you get to here, instead of solving for a in terms of b, you instead take this equation, multiply both sides on the right by h2 inverse. That'll give you ah1, h2 inverse is, well, if I'm multiplying by h2 inverse on the right, that's just going to give me b here. And what you'll wind up doing is playing the same sort of game. It's just now substitute for b in terms of a, and you'll wind up essentially writing down the same sort of argument. And that's how we get there. It takes us a little while. But the punchline is, if there's at least one element in the overlap of these two left cosets, then necessarily, in fact, the two left cosets are identical. And that's exactly the trick or the property that I sort of already knew, so that it allowed me to recognize, as soon as I see one of these things, that I've already identified as being one of these left cosets, then I know exactly what the whole left coset looks like. It's the one that I already wrote down. Now, folks, I'm not saying that they happen in the same order. If you're going to run through and actually compute rho 3h, let's say, you'd write down rho 3 circle, rho 0, then you'd write down rho 3 circle, mu 1. And what you get is, well, rho 3 circle, rho 0 is rho 3. And rho 3 circle, mu 1 turns out to be delta 2. That's fine. If you did it in the same order of h, but started with delta 2, if you do delta 2 rho 1, I'm sorry, delta 2 rho 0, you don't get that. You get that one. But we're not claiming that they're the same sort of in order. All we're claiming is that you get the same set out. So if you start off by saying, hey, this one's first going to kick out rho 3, and then you start this one off and say, hey, that one's going to kick out delta 2, you say, well, the first elements that I've written down are different. That's a non-issue. Just keep computing to see whether or not you eventually get something that's in the overlap. OK. Questions, comments? Richard, questions? Those could have had eight elements in them. Great question. So Richard's question is, what if I had started with a different subgroup of g and sort of reinvented the wheel and played the game, but using that different subgroup, and the answer is we would have seen completely different cosets. So no restriction on h. You can play the game starting with any subgroup you want. So the question is, does h have to have the identity yet? Because the hypothesis is always that h is a subgroup of g. But that leads me to a point that I want to make here. What we saw in this particular example is that one of the cosets happens to be the subgroup itself. So the coset generated by rho zero is just h. And that turns out to also be the coset generated by mu one that is just h. So one of the cosets is actually a subgroup. And not only is it some subgroup, it's the subgroup that you started with. But folks, none of the other cosets have anything to do with subgroups. They're simply subsets. And that has to be the case because none of these other ones contain the identity, for example. So that precludes them immediately from being subgroups. So even though we're starting with a group and a subgroup, what we eventually kick out when we kick out these cosets is exactly one coset that is a subgroup. And in fact, it is the original subgroup. But everything else is just a subset. So you think, well, they don't have too much structure. In fact, they'll have a lot of structure. What we're eventually going to show is that it is not coincidental that the sizes of each of these cosets are the same. Every one of them happens to have two elements. And in fact, that's the same as the number of elements in the original subgroup. That's a great question. That's exactly right. So the question is, do they have the same number of elements because that's how we built them? The answer is yes, although we have to keep in mind that when we compute products inside groups, that there's some sort of cancellation that goes on. In other words, that you can't somehow get the same element appearing both as AH1 and also as AH2, that there's some sort of what we called cancellation. Or the idea was, if you look at any row in the group table, that you can't see the same element appearing twice. So we need a property of the groups in order to conclude that. But you're right. It is a property of the way that we're computing or building these left cosets that all of them are going to have the same number of elements. And we wouldn't have eight elements. So, well, so let's do that example. The rich question is, so another example, example. I'm going to start with the same group, but I can choose any subgroup I want. How about if I choose the subgroup that is the whole group? All right, so here's the first left coset. I can start with any one I want. I'll start with row 0. How about row 0H is, hmm, row 0, row, row 0 times. It's just going to mean it is, which is H, I guess, or G. Well, H is G. So there's a coset. You have it, look. Now, what's the coset? Well, what do we prove? They're either equal or disjoint. So if I'm going to write down another coset, the coset that I'm going to write down either misses this completely or is equal to this, it can't miss it completely because row 1 is in there. They're all equal to G. So in this particular case, there's only one coset that gets kicked out, happens to have eight elements in it. Over there, there were four elements, or there were four cosets that got kicked out, and they each had two elements. Heck, I guess we could even talk about the other extreme rather than starting with the subgroup being the whole group. We could start with the trivial subgroup, the one that just consists of the identity. That'd be sort of uninteresting too, but guess what? We would get eight different cosets, each of which would contain one element. So let's see, if I start with a subgroup having two elements, I get four cosets, two times four is eight, or I start with a subgroup containing eight elements, I get one coset, one times eight is eight. If I were to start with a subgroup that contains four elements, and there is one, I won't run through the computation, but for instance, the subgroup generated by row one. So it's a subgroup, row zero, row one, row two, row three, that turns out to form a subgroup. It's a subgroup corresponding to the permutations that can be viewed as just rotations. There's a subgroup, it's got four elements. It turns out if you do all these computations, what you'll show is that there's only two cosets, two left cosets. They each have four elements, two times four is eight. What we're seeing, or starting to see, is this sort of numerical stuff. Start with a group, take a subgroup, do this computation of left cosets, ask how many elements were in the subgroup, ask how many different left cosets got kicked out, and somehow in all the examples that we can write down, the product of those two numbers, the size of the subgroup times the number of different left cosets seems to always give the size of the group, at least in this particular case. What I want to do now is relate this left coset stuff back to the original equivalence relation that we wrote down. So here's the connection between the equivalence relation and left cosets. And the connection is a very tight one. Proposition, same setup, h is a subgroup of g. Then a, take elements a, b, elements of g, picking two elements of g, then a is related to b, meaning a inverse b is in the subgroup, if and only if a, h is the same as b, h. In other words, the equivalence relation that we wrote down on Monday, the relation that a inverse b is in h, that relation is exactly the same as saying that the left coset of h generated by a is the same as the left coset of h generated by b. So if the question is, write down all of the equivalence classes corresponding to the relation that we looked at on Monday, in other words, ask which pairs of things are related to each other, well, what we did at the end of Monday was just ground out an example. And what I did at the beginning of today was I started grinding out those computations in the context of g is d4 and h is the subgroup rho 0 and mu 1. And that was getting totally tedious, because the idea was take a pair of things, compute a inverse b, decide whether or not it's in the subgroup. But what this proposition does for us, I'm going to omit the proof. It's not too bad, but just in the interest of time. The point is, if you want to figure out which things are related to each other, just figure out what the cosets are that each of them generate. And hey, the cosets that they generate are relatively easy to work with. Why? Because if you start generating cosets, you know something about the two possibilities associated with this coset and this coset. They're either equal or they miss each other completely. They're either equal or disjoint. So if you start computing this coset and you recognize an element that's already in this other coset, then you're done. Then they're equal, and that means that the original things were related. In particular, IE, the equivalence classes that came up in the partition, classes from the equivalence relation, the equivalence relation are precisely the cosets. Precisely the left cosets. So if you were to go back, and boy, I don't suggest that you do this, but if you were to go back and try to tediously compute A inverse B for each pair of elements A and B and D4 with the goal being to try to figure out which things are related to each other in this equivalence relation, here's what you'll wind up getting. Rho 0 is related to mu1, and that's it. Rho 1 is related to delta 1, Rho 2 is related to mu2, and Rho 3 is related to delta 2. That's it. Rho 2 is not related to delta 2. Delta 1 is not related to Rho 3. So the point is the equivalence classes that you'll kick out are precisely the cosets. So now let's go back to the example that we actually did at the end of Monday and convince you that at least in that particular case, things work out the way they should. So recall, and this is what I erased that I didn't want to erase, in the example G is Z6, the one that we did on Monday and that I rewrote momentarily today. Here were the equivalence classes computed in here on this very board 48 hours ago. Turned out to be what? 0, 3, 1, 4, 2, 5. Notice these are exactly the cosets. These are precisely the cosets. Let's make sure that that's the case. The cosets should be pretty easy to compute. Well, here's the subgroup. Let's compute what each of the cosets are. Now you have to be a little bit careful. When we talk about the definition of the coset, when we're talking about the left coset is the elements in the group that can be written as AH for some element H in the group. I'm sorry, H in the subgroup. Yeah, remember what's implied here folks is that there's a binary operation going on. We've chosen to sort of sweep it under the rug, but there's a binary operation and that depends on the group that you're living in. So it might not be multiplication. It might be composition, like it is in D4. Or it might be multiplication if this was happening inside the non-zero rational numbers. Or it might be that the binary operation is addition mod six. That's what the binary operation is in this particular group. So when we talk about computing AH, this means A star H where the star is whatever the binary operation is. So for example, if we take the element zero in the group, that's certainly in the group, and I want you to compute the coset zero H. Well, what we often do if the binary operation happens to be addition, rather than just writing it as zero H, because that would look like a multiplication, we sort of emphasize the fact that the binary operation is in addition by describing or calling the coset A plus H. Technically what we could have done is called the cosets living over in D4, rho zero circle H because the operation inside G is composition of functions. So I could have put a little circle in there. But if you have an operation that looks like a composition or it looks like a multiplication, we typically just leave those out and call them times or products. But when we have an addition, we usually slide the addition symbol back in. All right, so what is this? You're supposed to combine zero with everything in the subgroup. Well, there's the subgroup. And the last thing we'll do today, I mean, you're sort of seeing what the obvious statement is. If the element of the group that you happen to start with happens to already be in the subgroup, then when you do the coset, you just get the subgroup back. That's no big deal. All right, how about one plus H? I'm just gonna list out usually the elements of G and see what we get. So one plus zero comma one plus three. So what do we get? One, four. That's a good one. How about two plus H? You take two and you add it to each of the elements in the subgroup. And you get, lo and behold, two, five. Should we keep computing three plus H? Probably no reason to do that. This is three plus H. Why is that? Well, look, if you take an element, the element's always in the coset that it generates because the first thing that you're gonna compute is three plus the identity element. Well, there's three plus zero is three. So as soon as I locate three inside here, that is three plus H. Guess what, this is, guess what this is. So the cosets, if we were to compute them this way, are precisely what we get when we look at the equivalence classes relationship. So let's finish up today with the following observation. I'll call it a proposition as well, but it's pretty easy to prove. The proposition is this. H is a subgroup of G. Let A be an element of the group. Then I'll tell you which elements are in the subgroup. A is in H, if and only if A, H is H. So if you start with an element that happens to be in the subgroup, then the coset that the subgroup generates will be the entire group. And conversely, if the coset is, so I won't put this for you. All right, now, here is the observation. And we'll run through some more of the details of the observation, but I wanna at least slide this in at the end of today so that if you're gonna work on the homework assignment that I gave you on Monday, you'll understand what some of the words mean. What we've seen in these two examples is the following. Actually, we did a few more examples, those sort of trivial examples, is if you figure out how many elements are in the subgroup, then the number of elements in any coset is the same as the number of elements in the subgroup. And because there's no overlap between these subsets, in other words, because every element in the group appears in exactly one of these things, you can view the number of elements in the group as the product of the following two numbers. Tell me how many elements are in each of the cosets. Well, hey, I know how many are in each of them, it's just the number of elements in the subgroup. And then multiply that by the number of different cosets that you've got. So, here is the theorem. And again, we'll look at the proof of the theorem a little bit more on Monday. This is usually referred to as Lagrange's theorem. I'll talk a little bit about the history on Monday as well. Lagrange's theorem comes in two pieces. The first piece is any two left cosets of H in G have the same number of elements. The same size, namely this many elements, the number of elements in the subgroup. Remember if I enclose a set in vertical lines like this, it means the number of elements in the set. So any two left cosets have exactly that many elements. And secondly, we get the following formula. The number of distinct or different left cosets of H in G figure out how many different things comes out times the number of elements in the subgroup equals the number of elements in the group. So it's always the case that if you somehow determine how many distinct left cosets there are, I mean, don't count the same coset twice. You might say, look, this coset is both of the same cosets twice, you might say, look, this coset is both rho 3H and delta 2H. Yeah, you're right, I can give it two different names, but it is one left coset, and there it is. Tell me how many different left cosets came out. Tell me what the size of the subgroup is. If you multiply those two numbers together, you get all the elements in the group. It turns out this thing is so cumbersome to write out every time that we give it a notation. We call it the index of H in G. Index of H in G. And the notation for it is you put round brackets or parenz G with a colon on it H. And so Lagrange's theorem rephrased is, number two says if you take the index of H in G and you multiply that number by the number of elements in H that you get the number of elements, which is, I mean, unbelievable. Here's why. Out of nothing related to numerical information in the definition of a group, what this says is if you ever write down a subgroup of a group, then the number of elements in the subgroup has to be a divisor of the number of elements in the group because it says the number of elements in the subgroup times this whole number, this positive integer tells me how many different luck cosets there are, gives the number of elements in the group. So for example, if you have a group with eight elements, any subgroup has to have a number of elements in it that divides eight. So if I write down a subgroup of a group with eight elements, like D4, the subgroup has to either contain one element or two elements or four elements or eight elements. You can't have, for example, a five element subgroup of D4. You can't have a three element subgroup of D4. And where that comes from is somehow looking at this partition of the group that gets generated regardless of which subgroup you're looking at, and the point is you've taken the subgroup and you've used it to somehow slice the group up, not only into a bunch of pieces, but into a bunch of equal sized pieces. And as soon as you can make that observation that each of these cosets has the same number of elements, and that's what we'll end up proving on Monday. Then the punchline is then you've captured all of the elements in the group, you've counted them because each element of the group appears in exactly one of these cosets and you've counted them simply by sort of regrouping them as take each of the subsets and then somehow multiply them all together. Questions, comments? I'll give some of the details of the proof on Monday. Proof next time.