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 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.  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.  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.  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.  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.  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.  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.


One of the reasons why I chose Griffith University as my choice of university is the accelerated course option. I also like the fact that it's one subject every six weeks which works well for the way that I learn because I like to focus on one subject at a time.


So, now quickly discuss what is the result we are going to get for this. So, we are going to state this for a particular case. So, let gamma is equals to 0. So, notice that like we are setting this gamma to be 0 in this statement. So, when we have this gamma equals to 0, so we are just going to call this exp4, but when and the case when gamma is greater than 0, when we are this alcohol we are going to call exp4ix, when gamma is greater than 0, but the result we have set it here for the case gamma equals to 0 and you can work out that for gamma greater than 0 also similar bound holds. So, now let us revisit this. What we are saying? Take gamma equals to 0 and you set your eta to be in this fashion and then assume that experts are deterministic. What I mean by deterministic here is, yes, experts are actually giving me a probability vector, but it is going to be the same every time I give a particular context. And we are also assuming that these experts are oblivious. What does that mean? So, that whatever based on whatever the rewards they are going to get right they are not going to change their distribution accordingly. So, in a way you can. So, this means like you can assume that that each expert has already come up with what is the distribution he is going to put on each the arms for each context and he is not going to change with what is the reward he has been observing so far. So, that is his what is the context he observes that is irrelevant to him. He just tell me what is the context and he will just tell you what is the distribution. In that case then the regret we are going to get is under root of 2 tk log m. So, this looks very similar to what we had gotten for exp3 right except the fact that this is now m instead of k. So, the regret bound you have exp3 was 2 tk log k right where k is the number but now that k has become m now where m correspond to the number of experts. So, the proof we are going to skip when I am just there is one lemma which you need to get to prove this I am just going to state this lemma even proof you can work out it says that for all. So, this is a lemma it says that for any m t equals to 1 to t let us take t tilde t m minus. So, what this lemma saying is. So, what is this term giving you here? So, let us take an m expert what is this term giving you here right. So, if you look into this x t tilde right what it give basically give then the expected estimated rewards for that expert m right. So, in that way we can treat it as the total reward that has been obtained by expert m ok and then we are comparing it against what is this quantity over here. So, what is q to n yeah this is the probability with which I am going to choose expert m in round t and what is this sum is going to then give the expected rewards that I would have got from the experts ok. So, it is basically going to say how is the reward gotten from one expert compares with the mean that would have got from all the experts and it is going to bound and eventually this I mean this theorem uses this fact to bound it. So, just to look into the book like how to get this. So, I am using all these things finally we are going to get this is upper bound ok. So, we will just skip it I mean most of the proofs are very similar to what is there in exp 3 again ok. The ideas are all similar except for the fact that now we have to take into account two level of randomization into account one with respect to the experts selection and another with respect to the arm selection ok. So, now let us discuss couple of special cases. Earlier when we started this was by saying that I am going to take my phi to be phi 1, phi 2 some phi or phi could be all mapping from c to k yeah. So, that standard assumption we make that rewards are in the interval 0 1 this x, x t vector right or x t i is in 0 1 ok. Suppose this c is finite I mean the cardinality of c is finite and there are k arms what will be the cardinality of this set phi what is that k to the power. So, it is going to be k to the power cardinality of c right. So, in this case if I took a particular phi it is going to assign one value in k to each context right. So, in terms of if I am going to think that as a probability vector it looks like a unit vector right where for that particular arm it is going to be 1 for others it is going to be 0. We got this point. So, I take a particular map here it is going to assign a unique value to each value each context here, but I can think that assignment as a probability vector as how I can assume that em is equals to. So, let us say I am going to index this all the features let us call them m 1 all that up to k to the power c ok. Then I am going to think that emi in round t is nothing, but indicator phi of whatever let us say this ok. So, what I want to write here is let us call my functions here as let them enumerate let them call this f 1 phi 1 phi 2 phi 3 all the way up to phi to the power phi k to the power c ok. I have these many maps. Then can I map it the the value recommended by expert m or the value recommended if I am going to treat it as a map is nothing, but this right. So, let us say each of these experts there are k to the power cardinality of c experts and I am going to treat them these experts are nothing, but these functions ok. So, now experts are nothing, but these functions. So, what are going to they are going to if I am going to give a particular context what they are going to return me it is they are going to return me the arm that should be selected according to that phi in that round right how based on that, but in terms of the probability vector I can write like this right for the nth for the nth expert I can just write it as like this probability vector. So, this is just going to put value 1 on whatever the value this function assigned to that context t and everywhere it is going to put the 0 right. So, in that way I can treat all the all the maps here is as one different different experts and we are already in the setup right where I am going to set my E t function E m i to be exactly like this based on what is the map I am going to use in that round. So, now then even if I have a all my if I am going to define my regret in terms of these policies all possible phi is this regret bound still applies it should apply right it is just like the experts are nothing, but these maps here and they are also even though they are not giving distributions that I can treat that as a distribution. Now, if I am going to apply this point on this what is that I am going to get yeah. So, in that case I am going to replace m by. So, in that case what is this here, but what is this regret bound like we had gotten this regret bound by other method also right what was what was that method yeah by applying exp 3 for each of the context if I maintain my exp 3 for each of this context I would have obtained this then what is big deal about this algorithm then why is this why this I should why this is of any interest okay yes yeah. So, this so when if I have to write this bound it depends on the cardinality of c when I applied all this algorithm when I was in this setup I only cared about the number of experts I did not care about how many possible contexts are there whatever context I have I am just going to give it to an expert and that expert is going to give me a distribution on my arms right. So, this bound only worries about how many experts are there it I do not care about how many context are there as long as I have finally many experts this bound works right and often I may not as I said often we will not be working with all the possible maps we will be working with a restricted set of experts as I said like that as I discussed in the beginning of the class that restricted class could be based on the partition or similarity or you are going to just only fix the number of fees to be finite like what phi 1, phi 2 up to phi m. So, in that case there you have only finitely many m and irrespective of how many context you are going to deal with this bound is valid fine. This number of context is finite this number of context is finite yes, but it could be arbitrarily right which I do not worry about here yes when you are going to deal with all possible maps I am just giving you the worst case when you are going to deal with all of them because see when I worked out with maintaining one context what EXP algorithm for each context right I did not care about how many policies I will be competing against it was just like maintaining one EXP 3 for each one of them and that blind that gave me this bound, but here I am I am deriving it based on how many experts I am going to deal with yes I can take but the number of context could be same here still here though this experts I have only m experts here, but the number of context I am going to deal with could be same as this case also. So, in this case you see that I got log n, but here it could be cardinality into log k it could be much larger if the number of context is very large ok. So, with this we will conclude this discussion on adversarial contextual bandits we will not go into the lower bound proof and for all for this as you see do you expect this bound to be optimal you expect it to be simply square root tk or m should come into picture m should be there or at least order wise if you feel square root t we know that we cannot do better than that fine maybe like k also I cannot get rid of because from the adversarial setting I also know that it is like square root tk right. So, for us now the new term that has popped up is log m is that log m is the best or we can do better I also do not know maybe. Yeah. So, in that there is that log d term is somewhat similar to that log d term where d is the number of experts in this. Yeah, but we do not know weighted majority is the best I mean whatever the bound we got that is the best we could get that is full information whatever it yeah yeah k will not be there because it is a full information case where dealing with the bandit case here right. So, k will come into picture definitely because we are having only one kth of information compared to the full information case here.


What's going down with y'all man It's going down What up? What up? What up? Can you hear me good? Let me know if y'all can hear me good. Let me know if the audio is good all that good stuff Ah, man, it's been a minute since I've been on here, bro It has been a while since I've been on here I had to get back on some call of duty since Vanguard is dropping this week I want to say it's dropping Thursday Thursday or Friday and I've been hearing some Some pretty good pretty good information They've definitely from what I've heard is really listen to the community on Changing some things from the beta because the beta even though I you know, I was enjoying it for what it was It was very glitchy very buggy, so You know, there was a lot of things they needed to work on but I did hear they've made some major updates major changes So it kind of got me somewhat interested in this game. I heard dead silence.  What's the name of that call of duty cold war But I think modern warfare would have benefited from just it being a perk So I'm interested to see there's other changes they've made and I'm looking forward to it, bro I think I think we are in for a good treat Call of duty or battlefield. I'm getting both of them.  I don't know how I'm gonna balance the time But I'm gonna be playing all three. So I'm looking forward to it, bro I am looking forward to it Yeah, man. Oh Let's see what playlist I got think I headquarters I think I'm gonna try to just just running gun.  Yeah, so But a comment in the chat. What's your favorite call of duty of all time? I just want to know what's your favorite call of duty of all time Put it in the chat. Let me know. What is your favorite call of duty of all time for me? Bruh, I gotta give it to modern warfare 2 will always be my one of my favorites, bro It was just so much fun back then. It was toxic as hell, but it was fun, bro. I See some black ops 2. That's a good one black ops 2 was solid. So I said modern warfare Which modern warfare are we talking about 2019 when we talk about ghosts because of the story I Played ghosts. I played I've had many hours in ghosts. It's just the maps was just Maps wasn't it I see modern warfare 2 Black ops 2 it was a toxic time. Yep modern warfare 2. I see a lot of modern warfare 2 and black ops 2 Hey, I love the content keep going here to support you always. Thank you so much, man Thank you for all y'all in here black ops one Ghost was the worst car.  It's I think that caller duty would have been fine If it was the maps were better. I do not want to play on Piccadilly, bro I do not want my first game to be on Piccadilly. I just don't Modern warfare 3 block ops 2 not a bad combo New subscriber was happening brother. Hey, I appreciate you subscribing Robert Brown, man Hey, everyone that's in the chat right now. If you could do me a solid you can go ahead and hit that like button for me That really helps me out. You know I'm saying so if you hit that like button for me I really appreciate that everyone that's in the chat right now.  Yeah Man, I think this is this is an L for the city of Houston once again 2018 is really when car games were getting bad Modern warfare 2019 really for me is one of the better caller duties It's just there's certain things they could have did with this caller duty that made it one of the best ones But I think them switching to the new engine that they have now a no updated engine The graphics on this game is some of the best we've had ever in caller duty The guns the way they sound the way they look some of the best we've ever had in caller duty That's all the reason why I'm still playing this game because I feel like Cold War looks like shit That's why I'm not playing Cold War.  I don't want to be one. Nobody, bro What attachments I have on here. I think this is cool. I'm gonna try to do something with the bison man Wait, it's to be. Oh, no, I'm not playing them bro. Nope I've played against these guys too many times. Nope. I was not about to play them, bro Nope, I've already yeah, I've played those guys too many times Nope, I think I may have to go there like so I guess I'll go to kill confirm for now You on the series X.  So Ross I'm going to survivor series. Hey, man, where are you gonna be sitting at?  How you doing Ross doing pretty good, bro doing good since you guys are in here I prefer Cold War over modern warfare because Cold War stuck to the classic arcade feel that I do a lot I know a lot of people say Cold War is more arcadey and I can appreciate that It's just I feel like for me if I'm gonna pay that much for a game. I want it to at least Yeah, I'm a little rusty y'all Told y'all I'm a little rusty. I've played this game in a while. I think I played last week I think it was like last week one time. I don't even think I played it Back into the rhythm of things. So let me get back into the rhythm of things That's why I'm playing this so I can get ready for Vanguard kind of get back in the rhythm of get my get my Reflexes back up if I didn't kill him at a time. I was dead The only reason I don't like playing these modes because it takes so long to spawn spawn in that's another reason I don't like playing these game modes Damn it. Ah See this is why oh man, we can win this game.    No I don't know how y'all play hardcore. I'm not even they're just lining up for him. I am awful.  Do you think nxt is gonna be like, huh way back around? I'm gonna try to get into some hardcore. That's not hardcore or headquarters. That's what I'm trying to get into Of course didn't even see him call the duty vanguard. That's the one that's dropping this week Oh awful But where's my shots go? Oh my god. I'm trash. I don't even know where my shots was going It was just going everywhere where except they weren't supposed to go.   Well, my shots are so bad Look at my bra.  I was trash. I gotta try this again. I just did not get nothing going Pop mobile is too easy sometimes. Yeah, I've only played it like once Are you going to purchase the scream skin or no? No What's the purpose? How long he's already over he tried this? I don't want to use this Because that's the easy way out, but I may pull this bad boy out. I may pull it out. I'm gonna track her And then I'm probably back out I Could go headquarters or hard point I'm gonna do a hard point black ops for in Even in warfare to the most underrated cards in my opinion black ops 4 was actually fun It actually was fun. I actually had a good time in black ops 4 man. It was actually pretty fun Ross, have you ever been to a wrestling show? I don't know. I actually never been to a live wrestling show at some point I will have to go sometime soon preferably if it's a smackdown show Nothing Rob related right now. Although I have been keeping up with some of the stuff that's going on with raw. I know a Seth Rollins has a opportunity now There was like a little a ladder match and it's like a contract where you get a title opportunity It was basically like a money in a bank ladder match on Monday night raw So I did see that like here about that and stuff. Is it really getting better? You said raw is actually getting better, huh? The rock now returning. Okay, it makes sense because I know he has some movie obligations so that makes sense and if he's not returning then I Don't know man. It thought the scream skin was free. Was it is it free? I didn't even know I Didn't play it. I haven't played this game. I didn't play it none last week actually What's up, bro clutch? You be giving me through the long days for sure. Hey, man. It's Tyson TV. I'm glad I can do that for you bear But I can do that for you man. Hey, what's up? What's up, Ross? Y'all be on apex now I don't play apex. I still say smack down is the best. It's for me. I've just been enjoying it a lot more Why can I not get in this match? See it was doing this the other day when I was playing this game. I Couldn't get into a match not no, I Think it was like a hard point Hard point or headquarters. I couldn't get in the match. It would say I was loading in and it wouldn't let me in the match Let's see what else y'all saying It's gonna be roaming versus rock at mania. That's what it's looking like I know a lot of people are probably not gonna be bigger fans of that cuz how many times we have to see that but This is the problem with WWE not booking someone or having someone to potentially be on that side of of Beating Roman at some point they haven't built up any new stars Rock made smack down. So it's s-plus. It's an s-plus. I Actually played with you before it probably the best experience I've had in this game. Oh for real Well, I'm glad I was able to make that experience for you that much greater They're all in a party and they all sweat it's not fun. Trust me. I've played with them too many times I don't even feel like dealing with that. I think they only What I was in hard point. They usually in headquarters. I think they just have it on both Like if I'm by myself I am not about to go against a party of sweaty individuals that I know are gonna be sweating Putting in the best work shooting dead bodies. That's how much I've played this game I know certain people's names and if I see their names, they're never so low I'm the only one that usually plays solo What do y'all think about Roman versus Brock? 2015 WrestleMania That was one of their better matches for sure That that match was very good Surprisingly, it was very good. I enjoyed that match Wait, which way? 2015 WrestleMania, where WrestleMania was that? Huh Good the one I'm thinking of is WrestleMania 31 That was the one where that it was actually pretty good. Yeah, 31 was the one that was actually really good I didn't I believe they had one I think it was at the I don't know if it was the Dallas stadium or something like that I think I was WrestleMania and I was Where it was trash like dummies had a match people were booing people expected Roman to win and he didn't win Brock just busted him open and then that was it.   I know the one he had against Triple H when they were trying to Re-coronate him as the guy that still didn't work because No one cared and they still booed him. I think they should have went with Dean Ambrose I think what they should have did if you guys remember it was a pay-per-view right before WrestleMania I think it was called roadblock or some shit and It was a match between Dean Ambrose and Triple H Fantastic match. I think they should have gave the type the championship to Triple H. That would have been great I mean I triple H or Dean Ambrose at the time because he was hot and Then what they could have did is set up a match between Roman and Dean Ambrose Granny everyone would have chose Roman Everyone would have chose Roman, but I think Dean Ambrose should have definitely been a champion sooner What I stuck him I stuck him in my nape.  That was a risky play. I'm cooking up. I'm cooking up with grease Preferably Crisco grease and I died Damn it, I died.        Oh My god Niggas just camp so much. No one wants to move out there spawn.   We got one dude. They have no kids Jesus, bro The camping was annoying in this game does Got one dude with no kills. Thank you, buddy. We needed help Damn it dog We had something I had to pull out the shotgun because they were camping in one in multiple Little corners, so I'm all right, bro. Anytime I'm in the building.    What's good Steve one down, bro What is going down smack down versus Rob 2006 2011 was lit you talking about the game What up family? I wish they would bring back Undisputed title and replace it with the universal title. That would be cool That would be cool The universal championship does need a new look Editing for a client and playing Xbox. Hey, sounds like a good day to me my guy.  I want to play headquarters or hard point What's going down was popping in these streets Was popping Yeah, yeah Hey Also, we reached another milestone nine million views on my on this channel. So Hey, man, thank you all so much for everyone that Watched any of my videos. Thank you guys. You guys are amazing 9 Million of them things I appreciate that 9 million of them things You gotta say it like that them things 9 million views bro That's a lot of views and y'all could have watched anything else So, thank you They're there. Oh, I came in cooking Came in cooking Cooking baby someone tell me did my bullets disappear Did I not pay my rent for my bullets because they definitely just disappeared Yeah, okay, I gotta push back. I gotta push back. I gotta push back Yeah, I'm pushing around back. Yeah, so they know where I'm at. Okay, just push back Push back this way. I'm just trying to get my streets off Okay, I do not want to challenge that I want to kind of push around. Well, at least I got my top of gun for the next point, so All right, they're all dead All right, I gotta see what the next point gonna be at Yeah, I did sound like I was falling off a cliff. I was falling off a cliff Load out. Oh, yeah, I got on cold blooded. All right, I'm back All right, that's perfect. Let's go. They're not gonna get this point. They shouldn't get this one Baby, it's a party now. It's a party. It's a party.  That was fun I don't know if y'all know there's a glitch where if you you can still see people through the wall It's like a aim. It's like a wall height glitch so I can see him there I melted bro. I just one shot dead. I got six shots Oh We kind of have a good spawn I didn't I didn't know he went back I know one should probably be coming this way right here right here. I let my teammate tell Well, yeah, my aim is awful y'all. I am I am a guy get my aim right What is that roaches?  Congrats the braves Congrats to the braves They beat us soundly 7-0 Jesus, bro Congrats to the braves Hey, they definitely Don't know what happened there It just kind of kicked me out. Yeah, that was weird. That was weird Hmm part of me wants to I don't know Should I stay with the the fire shotgun as a secondary? Let me know. All right, Prometheus. Appreciate you stopping my Detections are not going to the Superbowl So we can hang that up.  And that's my team Yep, shout out to the braves one in the world series, man better team one Who's going to the Superbowl? Um, I don't know. It's it's it's really kind of a toss up this season.  None of my teammates actually went for it Bro, what? Oh my god, bro.   They spawned right there. I can't get nothing going y'all. I am Watered and the teammates backing out. Good job. The teammates are so bad, bro Look at my teammates, man. They're not even go We're getting trounced right now. We're getting destroyed Bro, I'm the only one that's over there and then I spawned on the opposite side of the map No one else is pushing that side No one Geez, bro As soon as I kill somebody. Oh, I'm dead.   This shit's just too much movement.  I was stunned And this is game is over.  I'm kidding. Yeah, we're getting destroyed right now. Yeah, my team is Everybody pulled out shotguns, but they don't even matter Look at this. I'm the only one that has a capture on my team. I want y'all to see that Where was he? Where was he? Even though at this point they spawned And I can't get nothing going Appreciate the 499 Yeah, they got it all sold up. I'm backing out of it because there's no point Jesus, there's no point. There's nothing I can do.  I'm the only one that was getting kills, bro And they had the fire shotty too. Like there was nothing I can do there. That's it's annoying, bro That's the only bad thing about playing solo is you ain't got really no help unless you get a good team I had to quit, bro. I wouldn't this is what there's no point in we didn't get a single point.  That was a waste of my time Gotta try it again Oh, yeah, you're not a mod because I thought you was actually dub But you're not a mod. So that's why I took you off That's the only reason why I made you a mod because I thought you was dub, but you're not dub You're impersonating dub. Come on. Find me a match Find me a match I could use that. Um, what's that other shot? I don't like the ads on that one.  I'm me. You're not bro You want to know how I know you're not You're definitely not Because one I think he's chilling with his family. I think he was I actually was on the phone with him before I started streaming So, yeah You're not him Ask ask a dub question Dub don't even talk like that. That's how I know you're not him I've had conversations with him. So yeah, I I would know I mean, all I have to do is just go to your profile and it would show that you're not him So you're definitely not him bro. What's moriah's middle name? Hell What's my middle name?  Well, I'm just dead. I'm here I What's an issue middle name? So nobody from my team Decided to to stop them from capping Nobody you said now i'm getting personal, but you just put moriah's middle name up on on youtube. So It's not a dub thing That's not something dub would do Just saying it's not a dub thing Trust me. Okay. So the points back here. Why is only three of us? Where is everybody, bro? I'm out to stay up here. I don't know why he didn't see him And then we spawn out because the biggest I can't get nothing going. I need help He's over there. They don't know that No one's pushing out. I mean pushing into the objective It's just me there's someone behind me There's someone over here There's someone over there. There was literally people everywhere. God damn I leave one game to join another game where the same thing is happening That's a good place Look at bro. Look at my team. Look at me and then look at my team.  What is he doing there? There's someone behind me. Why you want me to check my phone?  If he would have shotgun, I would have been dead, bro Okay So you're pushing back this way It should be pushing back this way exit out of the chopper you get like this constant wall hack is a glitch See so I can see where they are through the walls and watch our back. Watch our back. Watch our back But it's only for certain people I was in his own man.  He guy whatever he's going off right now.  So I'm trying to run and play defense This is actually pretty nice I'm not saying anything, bro. It's brilliant. I don't want to drop 70 moves, bro Fuck.  That was actually a nice call. Push my push my picture 81 and 22 man shout out to Scarface man. He held it down Bro, you're not the real dove in Asia, bro. You're not that's how I All right, hold on. I'ma back out just to prove to you that you're not hold on Hold on It's real simple Let me go ahead and do this All right, hold on Doing this live What's good, bro? Are you in the chat right now? No, you're not bro type one right now Type one right now type one right now if you in the chat right now Thought it was the other guy damn But you never No No, bro, because you said look at your phone, but you never text me. So that's how I know. I was like, oh I did not know that bro. I didn't know that. Hold on. Let me go to instagram Come on, but that's not that nigga.  Hold on No, I saw it. I don't see it on instagram. Yeah, I don't see it, bro No Yeah, I don't see it Oh, well, yeah, I did not know I thought you was still out and about with the family That's the dude that be pretending to be dove in Asia Oh now I see it Damn I didn't believe you bro. I thought I Yeah, because I thought it was the other guy, bro It was 81, but yeah, you know All right, now would I drop on in this bad boy?  Yeah, 81 in 81 and 22 Yep 81 and 22 No, yeah, I I couldn't get nothing going I can get nothing going, bro. I did not know. Well, I was wrong Once you said money. I was like, hold on. Wait, wait What's this nigga start talking about money and betting that may be him Yeah, that's why I had to call bro But I'm about to get back on here, man Appreciate you stopping by Oh, yeah, I still got you on speaker. Yeah. Yeah That was that you was right. I was wrong. So appreciate you stopping by Yeah Well, that was him That was him y'all.  I legit thought that was somebody else. I thought it was the imposter, man I thought you was imposter my bad, bro Oh, man, that was that was on me. Who am I going against? Oh, somebody from ancient That just made me mad, bro Thank y'all for joining in Thank you for stopping by should have believed me Now that I know that's the real me Move on my way team, man They spawn back here. I'm with you. I'm with you That was the only bad thing about hardcore They got the good spawn if they get a good spawn and you die, you're gonna keep spawning out That's the only thing I don't like about hardcore It's all spawn No way, you know, I just saw you. Yeah, they're not gonna push it There's one in mid, so I'm probably not gonna play with subs tonight Oh, plenty objective as a team. Oh, that's the dude from the last game Thanks, coffee. Yeah, this happened to be the revenge that I'd spawn these this way Bro, I should have had that man He's gonna be And I still died still died Oh, it's the zanky guy. He's pretty good. He was already pre-firing before I came around the corner That ain't a little weird to y'all He was pre-firing before I got to that corner, which is kind of stuff Why would he be pre-firing right there? Unless he's hard to play There's no way he could have hurt me. Yeah, we're getting we're getting ran through I killed somebody. He doesn't see him up there.  What's they do they got 23 kills? Shoot right ahead There's someone behind me. Is he gonna push back by defense? Why does he get so mad that I'm killing him? He's camping.  I'm trying bro.    I appreciate the old mega one. We're trying to make an epic comeback, bro Hey, that's far-fetched dude solid man. He's solid teammate I'm glad I ran into him, bro Yeah, so they're already here probably that's a good spot They got it set up. That's a good one just push out I don't know if we're gonna win this one y'all. I don't know if we're gonna win this one y'all I'm okay with my team though We should be winning.  So I pick up somebody else, young Oh You dropped 1644 My Scarface is doing this thing man. Huge fan on channel. Hey, thank you so much, bro Huh trying I gave it a my best shot y'all But if y'all haven't already you're in the stream go ahead hit that like button for me if you haven't thank you so much You know I'm saying this is this has been cool. I can't wait to do some more I will be live streaming This week call of duty vanguard cannot wait to play it. It's gonna be great It is going to be great I'm gonna have a good time Are we playing the same people? I don't think so No, you're not playing the same people playing somebody different. Hey, I'm glad I can make you laugh every day, man Be honest with you I'm glad y'all find me funny One of them Adam Pierce does next For Brock Lesnar He's talking about Teases down Classic classic classic He's doing that weird stuff again Where I'm already I guess some matches already started, but it's not loading me in He's doing that weird stuff now. I don't play 2k Try it again You still here?   Like he was good or something like he's just crazy good So I'm staying in this one Well, you don't play You're gonna play between two when it comes out. Uh, I think we're gonna do that on the main channel We're gonna play some matches on the main channel for you guys because I know you guys would love that so That's what we'll probably end up doing We'll probably try to do another Booker T video for Brandon Also, uh, also you should be at the triple-h second match rest many 28 all-time favorite match Um, I wish we could but it's gonna be hard to do that because you know You can't post that type of stuff on youtube though. Damn their block it. Why no wwe would probably I mean, he already punished him. He fined them last week and he suspended So it would make sense to punish him more like Alan Pierce. He already punished him from a story point perspective. So yeah Brock will be off tv until probably the royal rumble where he'll probably win the royal rumble where he'll probably face Brock less I mean Roman Reigns and I hope and pray Roman doesn't lose Because it wouldn't make sense Uh, I don't know if I'm gonna watch the Chuckie series I'm not really interested in it I think the one thing I'll do want to check out is that new paranormal activity Mexican. I heard it's actually pretty Pretty solid pretty good Like I heard the it's actually pretty good. So I may check that out. Well, I should I am gonna check that out. Uh, Jessica Supposed to be coming back in time So I'm gonna check that out with her She's supposed to be coming back in town. I want to say next week Yeah next Thursday she's coming back in town And I still will be recording y'all. So we will be making videos even though she'll be here So and y'all may see a video With her, you know or her me in it if y'all want to so if y'all want to Be formally introduced to Jessica y'all more than welcome. Let me know in the chat We discussed it uh Few weeks back potentially her just showing up on the page just saying hi I would love to see more of broad getting beat up You and Dub both But we can't get nothing and I want to stay in this lobby because this dude was talking shit the lamp erode Whatever the name is. I want to stay in this lobby girl. Just some more man a couple more people getting here So we can you know saying get this going Actually, I want to see something. I don't know if y'all can see me. Well if I turn off the lights Guys usually play with the lights off I don't know. Let me see See if y'all can see. Can y'all see me good. Ah, no, it's a little bit too dark Nevermind It's a little bit too dark Here's a little bit too dark I heard raw was getting better. I heard raw was decent last night. You thought it was born. Why you thought it was born Why would I react to the best senior punk promos again? I'm kind of kind of redundant to react to it again Oh, he's on the Is he in here?    Oh, he's coming this way.  Yeah, they're already over there. That Nate didn't hit him. Why do you think see him sucks? I mean they should be He used to be in the open again Why he just he gave me so Do y'all know we actually got some good team name He's getting mad. I don't have enough bullets Yeah, they're gonna have to cap it.  I couldn't do nothing What I literally just spawned in I wasn't trying to do that y'all No, I ain't pulling out the shoddy. I'm really saying too much. I'm focusing on Did he hear me? Right at the last second.  Don't blame me. Sorry Going back to this side of the map It's me. She's gonna be coming from this way. He's still right there. All I had to do I'm surprised he didn't get mad. I killed him that time. They spawned him back there So I can hit them with the biggest flink He's right over there. He ain't got them Then the spawn flip that's a gg y'all He didn't go that way.  I don't know what you're talking about Yo shotgun somebody had a riot show We got the dub man. We got the dub. Sorry. I was focusing on pulling your shoddy. No, I didn't want to do that What game modes is this? This is headquarters This is headquarters, so Bye y'all man. I think I'm done for the night I'm definitely We'll be streaming some more this week. I will be streaming call of duty vanguard for you guys As soon as it drops, I'll be playing that. It'll probably be during the day because I know During the night, I'm probably gonna be busy and stuff like that. So Appreciate all the love and support man. Thank y'all so much. We're almost at 60k And you guys are amazing man, but uh, I'm about to Call it a night. I will catch you guys later on the next one Probably next time I stream will be a call of duty vanguard later on this week Catch you guys name