 With binary operation star is a group, a case, the following three things are true. First, star is associative, and I won't write that out, but you've got the long hand version written out from your notes on Wednesday. There exists an identity element. We'll call it E in G, so that for every A in G, there's a special element that has the property that if you hand me any element in the set and you combine that element with the special one, you get the original element back. And it doesn't matter which order you do it in, E star A equals A, and property three is, once you've identified this thing called E, you can't even talk about step three without having step two, folks, so there is some natural order to things. But once you've identified the special element E then, here's the statement, for every or for each, let's call it A in G, there exists or there is something called, we'll call it B in G, so that the following two equations are true. If you combine the original thing with the new thing, you get the special thing, this thing called the identity, and in the other order as well, B star A is E. You see, I mean, it sounds sort of foolishly cute, but that's this thing and the special thing and the other thing, and I'm trying to get you out of the habit of using phrases like zero or one or inverse or negative or something like that. In certain situations, this special thing called the identity will look like this because it'll happen to be zero, for example, in the integers together with addition. In other situations, this special thing will look like that. In other situations, the special thing will look like one zero zero one. In other situations, it'll look like the flat horizontal line y equals one. So, I'm trying to not use any sort of inflammatory language to describe this thing because I don't want you thinking just about one particular element. It could take on many different forms depending on the setting. Similarly, this process of taking something in the set and trying to pair it up with something else so that when you combine them together, you get the special thing, this thing called the identity. This process might take many different forms. If we're looking in the situation of the integers, let's say together with addition, then the other thing, well, wait a minute, let's see. In addition, the special thing is zero, and so this process of finding the other thing is the negative process. If you start with something, you want to combine it with something else to give you zero. Sure, if you start with something, you just let b be negative a. But in other situations, like if the set is the positive real numbers and the operation is multiplication, then it turns out this special thing. Well, under multiplication, the thing that acts as an identity is the number one so that when we're doing this process in step three of taking an element and trying to pair it up with something so that when we combine them to get one, the process now looks like the one-over process, the reciprocal process. We'll see more examples in a minute. So this is what a group is. And again, the names of the things in special situations might look familiar, but we're going to try to talk about things in enough generality so that the notion of a group can encompass a lot of different things that you've seen. And what we saw on Wednesday was at least the first big example or the main example of a group, so big example. And then we gave a sort of a side example, big example, big meaning this will come up a lot. The collection of integers together with addition is a group. And we proved that on Wednesday. Let me just click through what's required in order to establish that some set together with some operation really forms a group. And David brought up a good point. There's sort of a throwaway here that could jump up and bite you, so you've got to be a little bit careful. The set is called G. This thing, star, is assumed to be a binary operation on the set. So what we're assuming is that somehow you've already established that whatever it is this operation is supposed to stand for, that it's already a binary operation on the set. In fact, let me be even more specific, binary operation on the set. So there's maybe something that you've got to check here. You've got to check that the proposed operation really is a binary operation on the set. Like you've got to check it's well-defined maybe. You at least have to check that if you take two things in the set and you combine them using whatever the proposed operation is that you get something back in the set. So presumably we've done that by the time we're even getting to checking each of these three problems. Okay, so we've done that. Plus is associative, yeah, as we will do most of the time in here. It's an operation that we're familiar with and I'm just going to let you check. Yeah, it is associative. Tell me that you've checked it. Plus is associative, known. Fine. Second, convince me that there's a special thing in here that behaves correctly. Yeah, there is. It's zero. Third, convince me that if you grab an arbitrary thing in there that you can pair it up with some other thing in there so that when you combine the two together you get the special thing from step two. Yeah, the special thing is zero. If you hand me anything in here then you pair it up with its negative. That's the quick proof. Along the way there's some details. Like, hey, if I had changed this to just the positive integers that collection would not form a group. Binary operation, no problem. Associativity is no problem. The existence of an identity element, no problem. Zero's in there. But the point is if I hand you a positive integer, a non-negative integer can you always pair it with something so that when you combine the two things together you get zero and the answer is no. If I hand you three, if you don't have negative three at your disposal that doesn't work. And if the collection that I'm looking at is just the positive integers I don't have negative three. So that one would fail at the third step. Okay, we gave sort of a little example last time too. The little example was, another example was the set of complex numbers. Let's call it s. s was the set one minus one, complex i and complex minus i. And the operation was multiplication. So I'll just write dot for times. And this turned out to be a group as well. It turned out to be a binary operation. We just did it brute force, multiplied each possible pair together. Should we got something back in the set? So binary operation was check. Associativity was okay because it's just multiplication of complex numbers which is already associative. Is there a special element in there? Sure, there it is. Was it the case that if you took anything in there you could pair it with something else? Okay. Let me show you a little trick before we give what will turn out to be at least a half a dozen additional examples of groups today. In certain situations the underlying set is large or infinite. In certain situations though the underlying set isn't too big. Four elements or maybe six elements or ten elements or something like that. When we're in a situation like this, what we can often do is describe what the binary operation is just by making a little table or a little chart. And here's how that happens. So some binary operations on what I'll call small sets, small means half a dozen elements, maybe ten at the most or something like that, can often sort of efficiently be described by using a table. Sometimes be described, well actually they can always be described but sometimes we describe them by using a table and the idea is it's pretty easy. You simply put whatever the name of the operation is, maybe it's plus or maybe it's times or maybe it's just star and you simply list out all of the elements in the underlying set. So one minus one i minus i. Folks I don't care what order you list them in but whatever you order you list them in across the top, list them in the same order down the side. And now here's how you describe the binary operation using a table. You simply take anything on the left side, combine it with anything on the right side and put the result in the row and column where the two numbers happen to live. So one times one is one, one times minus one is minus one, one times i is i. So boy that first row was easy to put in. In fact the first column would be easy to put in too, right? One times one is one, minus one times one, i. So those are both easy. Now it's just a matter of running them together unless you minus one times one is one, minus one times i is minus i, minus one times minus i is i, et cetera. So here's how you can describe a binary operation where the binary operation is given on a small set. Alright, here's many additional examples of groups. More examples of groups. Some of them I'm just going to list out. These would not be sufficient proofs for a homework assignment. Lindsey, questions? Oh yeah, associativity is just painful. So yeah, in theory you could show using the table that the operation is associative, but I mean we could actually probably figure out how many computations that would require because you need to take each triple of numbers, presumably in each order because we don't yet know that this is commutative and so it's like 64 different ways, 64 different computations. There might even be more than that, I don't know. So no tricks. Now later on we'll see that there's an easy way to show whether or not the binary operation is commutative, but it's important and somebody came by my office on Thursday and asked this question. It was a good question. I mean it's important to notice, folks, in the definition of a group, there's absolutely no mention that the binary operation has to be commutative. Some groups will be commutative. That's fine. Some groups won't. That's fine too. So don't get sort of enamored with these first two examples. In fact, a few more of the examples that we'll write down here will all be commutative and then we'll focus on some important non-commutative examples. So here are some more examples of groups. I'm just going to list them out sort of shorthand. The rationals together with addition. The reels together with addition. The complex numbers together with addition. Addition has typically the effect of turning something into a group, but you'll notice you need everything in order to get a group. You can't just look at the positive rationals under addition and get a group because then you're sort of missing the inverses. You're missing the negatives. Here's some more examples of groups. If I look at the positive rationals together with multiplication, now let's just sort of click through and make sure that everything's working here. And again, if some of these were to come up in the homework assignment, I would be looking for a more formal presentation or at least more details as to why these are true. This is a group. Why? Well, we have to, in step zero, make sure that it's actually a binary operation. We've got to check that multiplication on the rationals. Okay, well, is it the case that if you take two positive rationals and multiply them together, you get a positive ration? Yeah, no problem. Is multiplication of the rationals an associative check? What do you even say? Just say associative check. It's a multiplication of rationals. That's all I'm looking for in the associativity other than you know that you need to check something there. Next, is there a special element that behaves in such a way that if you take any other element in the set and you combine it with that special one, that it doesn't change the other element? Yeah, and it's called one here. That's no big deal. Question, if I hand you a positive rational, can you pair it up with some other positive rational so that when you do the operation, when you multiply, you get one? Yeah, because if I hand you a positive rational, now, be careful here. If I hand you a positive rational, it means it's not zero because it's positive, which means that its inverse exists. In other words, one over it makes sense. And when I multiply the two things together, the original one times the new one, I get one. So is that a complete proof? Now, I've left out one important step. Folks, it's not sufficient just to say, yeah, because the number is positive, its inverse exists. You also have to make sure that the inverse is in the set. You have to make sure that this proposed other element lives inside the set that you're interested in. So just because I can flip things over, I also have to make sure that when I flip it over, I get something that's still positive. Of course you do, but you need to note that here. That's a key piece. And that sort of idea will come up in a minute when we do a matrix example. So there's the outline of what you'd need to establish in order to convince me that this is a group. You know, the reals, positive reals with multiplication. Now you're expecting to write the positive complex numbers with, I don't know what the heck positive complex numbers mean. So we'll take that out. But it turns out inside the complex numbers under multiplication, how about these? How about the complex numbers? And I'll use the notation star to denote non-zero. Non-zero. The non-zero complex numbers, oh yeah, the non-zero rational numbers and the non-zero, I'm sorry, little star here, non-zero real numbers under multiplication. Multiplication folks is less, think of it this way, if you have an operation that looks like multiplication, it might be multiplication matrices, you typically have to be careful. If you simply say, take the set and do the operation multiplication, you typically are going to have a little bit of trouble. Because, you know, in all of these sets, if you have a multiplication, then the multiplicative identity, the thing called e is one. But if you allow zero to be in your set, then you're sunk, because zero's never going to have a multiplicative inverse. It's never going to give you a solution to number three here. So typically, if you've got a multiplication type operation, if you're going to have a group structure associated with it, you typically have to make sure that you're avoiding the zero element and that's what all these generations are all about. All right, here's another good example of group. Example, I mentioned this on Wednesday, but we never verified that with a group. The collection of what's called the general linear group, gln of r. The definition of this is the collection of n by n matrices with entries in the real numbers having non-zero determinant. So fix the number n, I don't care if it's two or seven or 13 or whatever it is, you're going to look at the square matrices of that size, but you're not going to look at all of them. You're only going to consider those that have non-zero determinant. So the matrices for which the determinant's not zero. Then it turns out that gln and this set, gln r, together with this operation where dot means matrix multiplication is a group. Again, dot here just means usual matrix multiplication. Why? Well, let's just click through the details. Here's a set, here's a proposed operation. Is it a binary operation of the set? If you take two things that are in the set, in other words, that our n by n matrices having non-zero determinant, and you combine them by multiplying them, it's not an issue that you get another n by n matrix. I mean, you do, that's good. Folks, you've got to make sure that when you combine them, you get something back in the set. You have to make sure you get a matrix that has non-zero determinant. And that's already not trivial. I mean, you've got to go back to a result from your linear algebra course. So note this actually is a binary operation and we've noted this before that times is a binary operation on this set on the general linear group GLNR because if I take two things in there, let's call them A and B, if A and B are in the set, if I take two n by n matrices, each of which has non-zero determinant, then A and B is in G and R because, yeah, I mean, well, because it's an n by n matrix, that's not an issue. I guess if you wanted to write that out, that'd be fine because the product of two n by n matrices gives another n by n matrix. Yeah, but the key observation is the determinant of this thing. In order for it to be in the set, it has to have non-zero determinant. Oh, let's see. That is what? The determinant of A times B is the determinant of A times the determinant of B. This is a... In your homework folks are on an exam. If you just write 313, that means you're using the math 313 result from UCCS. If you took your linear algebra course somewhere else, just put L, I, N, A, L, G. You're quoting a linear algebra result. That's fine. Property of determinants. The determinant of A times B is the determinant of A times the determinant of B. Oh, but this is... Wait a minute. This is not zero because we've assumed A is in the set and that's property of the things, and that's not zero, and so this is not zero because you're multiplying two non-zero real numbers together. Check. So here we actually had to do some work to get off the ground. We had to actually do some work to verify that the thing we've got, the set together with the proposed operation, really is the binary operation on the set. All right. Now we can start clicking things off. Next, show that the operation is associative. You get to assume that or believe that. Matrix multiplication on... Yeah, matrix multiplication on square matrices is associative. If you want to say buy a thing that we learned back in our linear algebra course, that's fine. Of course, you probably didn't learn it there. Either your instructor, especially if your instructor was me, would have said, you just do it. You crank it out and you get it, okay? Step three, we have to make sure there's a special thing in the set with the property that. If you take any other thing in the set and you combine the special thing with the other thing, it just gives you the other thing back. In other words, it's going to change the other thing. So we need what we'll call an identity element and that's a suggestive word because what we're going to do is look at the identity matrix. So associativity is okay. Associativity, check, identity. Sure. Let's see. The n by n identity matrix is in the set. I guess I should put a question mark there. There is this thing called the n by n identity matrix and we know that it behaves correctly. Of course, we've got to make sure it's in the set. The set is n by n matrices having non-zero determinant. So why is this in here, not just because it happens to be an n by n matrix, but I better check that its determinant is non-zero. I sub n is n by n matrix, that's no problem, and the determinant of the identity matrix is what? Is one, which is not equal to zero. So I've just checked that it's an n by n matrix that has non-zero determinant and property of the identity matrix, if you take any matrix, equals a, a times n equals a for every matrix a. Now, a quick remark is the identity matrix has the property that if you multiply it by n matrix in GLNOR, you multiply it by n matrix in the set, then you get the original matrix back. Of course, the identity matrix, if you multiply it by n matrix, I don't care if it's a determinant is non-zero or not, multiply it by n matrix to get the other matrix. So this identity matrix is actually an identity matrix for a larger set, but that's fine. All we're worried about is whether or not it's the case that if you take anything in the set of interest and multiply it times this, okay, final, let's see. Now, how about inverses? So here's what we have to show. We have to show that if we pick anything in the set, call it a, so somebody's handed you something in the set that we need to find a matrix called b in the set so that when you multiply the two things together, in other words, you do the operation that you get this, and can we do this? Yeah. If you hand me something in GLNR, if you hand me an n by n matrix having non-zero determinant, well, what does it mean? Because a is chosen to be in this set, we have that the determinant is non-zero. That's part of being in the set. It's part of the property that puts it in the set. That's good. Oh, so by a linear algebra result, so by math 313 result, a inverse exists. There is an inverse matrix. In other words, there is a matrix that has the property that when you multiply it times a, you get the identity matrix. Not enough, though. Okay, there is an inverse, but we got to make sure that the inverse is in the set. The question is a inverse in the underlying set. What I'm doing here, folks, is checking, just as we checked when we did the multiplication of positive real numbers, I have to make sure that the thing that we found not only exists sort of out in the ether, but it actually lives in the set that we're interested in. So I'm worried about this particular seemingly small piece, but actually a huge piece of the definition. Well, yeah, what do we need to do? A inverse is an n by n matrix. It is n by n. That's not an issue. I mean, it is an issue, but not the interesting one. So check, but we have to make sure that the determinant of A inverse, if it's going to be in the set, we have to make sure that the determinant is not zero. Oh, 313 result. What's the determinant of the inverse of a matrix? We want a determinant of A. That's convenient. Wait a minute. We picked A to be in GLNR. In other words, determinant of A is not zero. So this is not zero. It's not only not zero. It actually exists because that's not zero. So A inverse is in there. It's in GLNR because its determinant exists and is not zero. So what? So we have found B equals A inverse works as the inverse of A. Nice. And so we're done. So the conclusion is that the general linear group consisting of matrices having non-zero determinant of size n by n towards a group. So this actually gives infinitely many different groups. The collection of 2 by 2 matrices over R having non-zero determinant, that is A group. Here's another group. The collection of 3 by 3 matrices over R having non-zero determinant, et cetera. The flavor here, folks, is similar to the flavor of what we just looked at when we were looking at various examples of groups under multiplication. You know, if I hand you the rationals or the reels, let's say, and say, all right, look at the operation multiplication. That's a perfect, good binary operation. I'm going to form a group structure on the reels because of this pesky zero thing. You know, if I hand you zero and the operation's multiplication, you just, there's, number three is just not going to happen. You can't take zero and multiply it by something else. You can't pair zero up with anything and hope that when you multiply zero times it, you get one. So in order to get some sort of group structure sort of focused around multiplication, you've got to throw zero things out. When you're looking at matrices, you don't just throw the zero matrix out. That's not enough. What you need to throw out are the things that behave like zero and the things that behave like zero are the things having zero determinant. So the things that behave like not zero elements, the reels are the things having non-zero determinant. So it's sort of the same role, just jacked up to matrices here. All right, here's a comment that may help you with one of the homework problems that's still on Wednesday. You know, we can cut this down even further. Instead of looking at the collection of all n by n matrices having non-zero determinant, we could look at the collection of n by n matrices having determinant even more specified than being non-zero, like maybe determinant equal to one or determinant equal to one or minus one, something like that. To show that such a collection really forms a group, you basically have to run through the same things. The key step in that is going to be this. If you take something in whatever set you're looking at, maybe the collection of matrices having determinant equal to one or minus one, all right, well the fact that the determinant is one or minus one means the determinant is not zero, which means the inverse exists. So you're close to being able to establish number three, but you need to convince me more than just the inverse exists. You have to convince me that the inverse is in the set you're interested in, and if the set you're interested in is those matrices having determinant one or minus one, you have to convince me that the inverse also has determinant one or minus one. But it's easy to do, and it's not the issue that it's not zero, you have to convince me that if you start with a matrix that has determinant one or minus one, and you do this, that you get another matrix having determinant one or minus one, and it's not that it's hard to do, it's just you have to check that. So for example, if I were to change, or the book were to have changed the homework problem to look at the collection of matrices having determinant equal to two or minus two, something like that, that wouldn't form a group for a lot of reasons. Okay. Ooh, I didn't want to recent. Questions, comments? Okay. Presumably any time you're handed a set, together with a proposed binary operation, you and there exist so that, okay. Then you have to march through these three steps in order to convince me that you get a group. Let's look at, oh. So, look folks, here's an example of what we'll call a non-commutative group. So GLNR is certainly a group. We just proved it. But it's certainly the case that it's possible to find a pair of things in there so that when you do A star B, it's not the same as B star A. So remark. So GLNR gives example, an example of a group which is not commutative. Of course, the other examples of groups that we've looked at up until now, the integers with addition or any of the things with addition or even any of the things with multiplication, non-zero reals or the positive rationals or the subset of the complex numbers consisting of those four elements that are those that have all been commutative. It turns out in the context of groups, context of groups, the word we use for commutative is replaced by, is replaced by the word abelian, a-b-e-l-i-a-n. So we talk about an abelian group or a non-abelian group. So this would be a non-abelian group. And the genesis of that word is Niels Henrik Abel, a-b-e-l. It was a Norwegian mathematician early to mid-1800s. And Abel did a significant amount of work on these structures. He didn't officially call them groups, but he looked at, well, he looked at binary operations that have these three properties. Okay, that's good. So that's groups, even though he never used that word. And you know you've arrived to mathematics not only when they use your name as an adjective, but when they use your name as an adjective and they lowercase the first letter, that's when you're central to the entire universe. So Abel has arrived. Okay. Questions, comments? So here's lots of examples. Let's do another example. An example is this. For each N, for each positive integer N, or natural number N, so I'll use the notation that we wrote down on day one in here. Oh, good. Here's the set. The set Z with an N underneath it. Denotes the set 0, 1, 2 up through N minus 1. So it's the first N integer starting at 0. And the operation plus of N equals addition mod N. So mod N arithmetic. Then it turns out if you look at the set Z, N, together with the operation mod N arithmetic is a group. For those of you that maybe haven't seen the number theory, of course, mod N arithmetic is like it's clock arithmetic where you don't necessarily need to use 12. So if I add 7 hours to 6 hours, I get, well, 1 o'clock. Just go around the horn. Go around the horn on base 12 or base 2. It doesn't really matter. Let's see. I won't prove this in gory detail, but let's at least march through the steps. I have to make sure if we take any two things in here and we combine them, that the result's back in here. In other words, we've got to make sure that we've actually got a binary operation. That's not too bad. If you look at addition mod N, mod N arithmetic is simply, you take whatever number you got. You take whatever the, what's called modulus is, this thing called N here, and you divide out as much of that as you can and you simply look at the remainder. So if we're doing something like mod 12 arithmetic and you take something like 6 plus 9, then you get 15 and the way you determine what the result really is, is you take 15, you take as many as 12s out as you can, well, there's one 12 you can take out and what's left over and what's left over is 3. So do you get something? Yeah. When you do mod N arithmetic, you get something that's less than or equal to N minus 1 and bigger than or equal to 0. I mean, if you take something like 5 plus 7, mod 12, you get 0 because the result is 12. For those of you that saw the number theory course in effect, it's the division algorithm coming into play here. But I won't push that too hard here. So let's see, it's a binary operation. I'll let you believe it's associative. It turns out there's a little twist there. You've got to make sure that if you add things and do it mod N, that doesn't matter which order your group things in, but I'll let you play with that. It turns out by the end of week 7 in here, we'll give a nice sort of elegant proof that this operation is associative. Is there a special element in there that when you do this operation, doesn't change anything? Sure, 0's in there. If you add anything to 0, it stays. The final one is a little bit subtle. You have to make sure that if you hand me anything in here, you can find something else in there with the property that when you add the two things together, that you get 0. So let's think about it in the particular case like N is 12 and clock arithmetic. If I hand you something, you need to somehow make something up so that when you add the two things together, you get 0, and the answer is just simply subtract from whatever the modules is. Maybe you subtract from 12. So if I'm doing something like mod 12, I need to figure out what I need to add to, let's say, 3 to get 0, and the answer is you just take 12 minus 3. You take 9. 3 plus 9 is 12, which is 0. If I hand you 4, what do you take? You take 8. 4 plus 8 is 0. So the inverse operation isn't too bad here either. Let me just comment on that remark. If I hand you something in here, if A is in Z sub N, then N minus A is also in the set, and A plus that thing equals 0 in set when you're doing arithmetic mod N. All right. So this thing forms a group. Heck, let's quickly do a group table. How about this one? I don't want to do that one. Yeah. A table example. How about Z sub, I don't know, Z sub 3 maybe. So by definition, this is the set 0, 1, and 2. And what I'm interested in doing is having you present for me a table that describes the group operation on these three elements. 0, 1, 2, 0, 1, 2. That's not too bad. Let's see, 0 plus it. Oh. You know, if you list out the identity element first, by definition, the identity element does what? It combines with anything else to give you the other thing. So the first row and column are actually pretty easy. 1 plus 1 is 2. How about 1 plus 2? 0 good, because we're doing mod 3 arithmetic. 2 plus 1 is 0, and 2 plus 2 is 1. Exactly right, because 2 is 4, but mod 3 is the same as 1. So there is a table that describes the operation of the group Z3, Z sub 3. All right. There are comments. So there's another example of a group. These groups are a billion, regardless of what value of n you pick here. Okay. So the next little section sometimes causes difficulty for students, and I'm going to try to sort of present it in such a way that we bring the difficulty up to the surface immediately and then try to deal with it. I'm going to do some examples of what it is I'm talking about. You see, we're talking about general, we're talking about situations that could describe many different types of sets. Could be matrices, could be integers mod n, could be rational numbers, could be complex numbers. Later on we'll see it could be functions, it could be permutations, it could be blah, blah, blah. And we've realized that in these various sets there might be a special thing called e. Could take on different forms depending on what set we're in. What we often do is ask once you're in a group and once you've identified this thing, I don't care what form it takes on, is there some property that describes what that thing is? Let me give you an example. The thing called the identity element has the following property. It has the property that if you combine it with any other thing in the set, put this back here and e star a. Okay, so here's a property that it has. It has the property that if you take it and you combine it with anything else in the set, you get the other thing. Here's a reasonable question. Suppose you've proved for me that the collection together with the binary operation is a group and you've identified an element that works in the role of this thing called e. And suppose your friend has gone through and taken the same set and the same binary operation and your friend has verified that that set with binary operation is a group. Question is it possible that your friend has identified a different element that works? There's nothing in the definition that says that there only has to be one element that works. Maybe there's two different elements, two of which have the property, not that they're called the letter e, but that have the property that when you combine them with anything else in the set, you get the other thing in the set. Now, all the examples we've looked at, well, it was easy to identify immediately what the thing was that worked in the collection of integers with addition. The thing that worked happened to be this. And we didn't worry about, well, maybe somebody else could find another thing that worked. Where worked means has this property. So in all the examples we've looked at, it's the case that when we identified a thing that worked, it, in fact, was the only thing that worked. Again, where worked means has this property. So here's a reasonable question. Is it the case that if we show that a certain set, together with the binary operation is a group, that there's only one choice for that or might there be other things that behave that way? Surface we don't know. What we're about to show though is that there is only one element that behaves this way. And you notice I've completely avoided the language, E is unique. E isn't unique, E is E. F is unique because F is F. And G is unique because G, I mean, I'm not worried about letters and I'm not worried about notation. What I'm worried about here is something called the uniqueness of an element that has a property. The property that we talk about here is this property. There are a lot of other properties we can talk about to decide whether or not they're unique. For example, if I've identified this, well, your friend presumably has identified a pair, something else, so that when you do the combination, you get E. Question, is it possible that there's more than one choice for B? There's no stipulation that says there has to only be one. All that's required is that there exists something that works. It's not required that the thing be unique, at least not in the definition. Give me another example of a property. The thing called E, much as this one's a little bit subtle, the thing called E is in the set. So that means because the property is supposed to hold for everything in the set, it means in particular it's supposed to hold for E itself. In other words, it's supposed to hold if I just, by coincidence, happen to choose A equal to E. It's perfectly allowable. In other words, this equation says that if you do E star E, you get E. So the thing E, whatever it is, 0, if it's the integer's under addition, or 1, if it's the non-zero reals under multiplication, or 1, 0, 0, whatever it is, has the property that when you do the operation with itself, you get itself back. E star E is E. So there's a property. There are other things in the group that have the property that when you do the operation with itself, that you get itself. In other words, that have the property that F star F equals F, or G star G equals G. Maybe there is, maybe there isn't. It turns out what we'll be able to show is that at least the three properties that I just mentioned turn out to be what we call uniqueness with respect to that property, or have the property of uniqueness with respect to the given condition. For example, proposition turns out given if G with star is a group. So presumably what you've checked is that you've got a binary operation on a set that makes those three properties true. So you're at the stage where you've done some work already, then here are some things that are true. First, the identity element, element E is the only property, is the only element in G which makes property number two true, or property number two, this property number two, property number two in the definition of the group true. Once you've found a thing that works to make property number two true, it is the only thing that makes property number two true. So the identity property turns out to be unique. I'll give you another problem. Second property, maybe I shouldn't have listed them out this way, but that's all right, given for each element, little a in G, the element B in G described in property three in three of a group is unique. Somebody hands you an element in the group and they ask you to verify for them what the element B is that works. Well, if your friend writes down an element B that works and you write down an element B that works, then you and your friend have written down the same element. Property one is that we say it very cavalierly and I think what I was getting at when I said students sometimes misinterpret what's going on here is it's not that the thing called E is unique, but E is unique with the property that if you combine it with everything else, you get the other thing. If you hand me an element, this thing called B is unique in the sense that if you combine A with something else and you get E in both directions, you and your friend have written down the same thing. Let's prove number two. Proof of number two. Proof of number one is similar. I'm just giving you the flavor here. I just want you to see what the structure of a proof of the uniqueness of an element with respect to a certain property looks like. So let's see. We want to show that let's see if your friend has written down an element, let's see if B is an element of the group and B star A equals E. In other words, if you have identified something called B that acts as the inverse to the element A and your friend has identified something called B prime, that works, and B prime. So think, you've written down an inverse for the element A and yours happens to be called B. Your friend has written down an inverse for the element A and your friend happens to call it B prime. The question is, do you and your friend have the same solution and the answer turns out to be yes, then B equals B prime. That's the given information. So how are we going to do this? Well, let's see. A star B equals E and what? Let's see. B star A equals E. That's given information. Oh, here's what I'm going to do. I'm going to take this equation and I'm going to multiply both sides by B. So B star A star B is B star E. Let's see. Do I want to do this? Yeah. Oh, let me do it this way. This is quicker. What I'm going to do is multiply both sides by B prime. It's B prime star E. But wait a minute. Anything star E is what? That thing. Okay. So I've started with this equation and I've starred both sides of that equation with the thing called B prime. But wait a minute. Now I'm going to take this same expression, B prime star A star B. So I'm looking at the exact same expression that I just analyzed to equal B prime equals, because group operations are associative, this is B prime star A star B. I haven't changed the order. I've not used commutativity. I've simply used associativity. I've chosen to group them together this way. Oh, but wait a minute. What's B prime star A? It's E, because that's the hypothesis on B prime. But anything star with E is that thing. So what we've just shown is that B prime star, the quantity A star B is B prime. And we've also shown that that same quantity B prime star, A star B is B, so B equals B prime. And we're done. Check. To prove uniqueness with respect to the property, you write out what it means for the thing to have the property. You hear the thing having the property means that when you combine it with A, you get E in each order. You write it out and then you convince me that the two presumed things are equal. All right, questions there? Comments? Let me set up how you'd go about proving number one and also set up how you'd go about proving one of the problems that will appear on the homework assignment that I'll give you tonight. So to prove one, prove number one, what you need to do is assume that you've got something in the group that for every A in the group that A star E is A and E star A is A. In other words, assume that the thing called E is an identity element and assume that there's something called F that behaves the same way. So assume that you've got something else that does exactly what presumably this letter E does. In other words, has the identity property. In order to show that the identity is unique, you have to show that E equals F and the hint would be proceeding exactly the same way that we proceeded over here. It's not that big a deal. One more property. Property, property. It's a funny word, but I'll give it to you because it's used relatively often in algebraic settings. If I take zero inside the integers under addition, zero plus zero is zero. Zero has the property that when you combine it with itself, you get itself back. Inside the real numbers under multiplication, the non-zero real numbers under multiplication. If you take one and you combine it with itself, you get one. One times one is one for two by two matrices. One, zero, zero. If you combine one, zero, zero, one with one, zero, zero, one, you get one, zero, zero, one. That property of taking an element and combining it with itself under whatever the appropriate binary operation is and getting itself back is a property of, is what's called an item potent property. So the word item potent, I-D-E-M, P-O-T-E-N-T, item potent means combine an element, an element with itself and get itself back. Now I'm going to give you a couple of additional examples of item potent things. If I'm looking at a collection of sets and the operation is intersection, if you take a set and you intersect it with itself, you get itself. So there's an item potent in the operation. Or maybe the operation is union. If you take a set and you union it with itself, you get itself back. Here's another example of an item potent element. If you look at two-by-two matrices and the operation's multiplication, if you look at one, zero, zero, zero, and you multiply one, zero, zero, zero times one, zero, zero, zero, you get one, zero, zero, zero. It's easy. Just pound it out. It's easy to do that. So I've just given you a lot of examples of item potent elements corresponding to various operations. But it turns out whenever you're in a group, so if you've established that you've got a binary operation on a set that forms a group, there's always one item potent element in there because E star E is E. But the point is in a group, that's the only item potent element in the operation. So it turns out that if G with star is a group, group, then the identity element, then E is the only item potent in G. You think, wait a minute, you just gave me a bunch of examples. In fact, you gave me an example of a two-by-two matrix that's not the identity of the item potent. One, zero, zero, zero. So what's up with that? And the answer is folks, one, zero, zero, zero, that matrix doesn't live inside any group. It certainly doesn't live inside the GLA groups because it's determined in a zero. So I'm not saying that there's very few item potent out there. The punchline is there's very few item potent out there that live inside groups. So here's what you need to do. Group, show that if you have an item potent, let's call it little f, such that it has the property, in other words, if you do the operation with itself, then f equals e, that's what you have to show. And I'll let you think about how to do that again. That property or at least something similar to that will arise on the homework that I give you this evening. Questions, comments? All right, let me make a quick comment about this property number three, this inverse property. I'll talk a little bit about subgroups, and then we'll get out of here. So look, I mean, I'm not, I haven't hidden anything from you. But notationally, this turns out to be a little bit uncomfortable. At this stage in the game, when we talk about a group, we ask that regardless of which element you start with, that you can somehow come up with a second element so that when you combine them, you get whatever this identity thing is. And the thing that we've been looking at, whether it was the context of a matrix or a non-zero real number or maybe even a whole number where we're looking at addition, this thing is always called the inverse of that thing, you know, it was the inverse matrix or it might have been an inverse, meaning reciprocal in a multiplication, or it might have been additive inverse, meaning negative. So we typically use the inverse notation, the negative one notation in this context. But you have to be really careful. The negative one notation doesn't always just mean flip it over, negative one notation is used in the context of whatever the binary operation is. So the quote, unquote, negative one notation, if the operation is addition, the negative one denotes minus sign because that's how you get the inverse when the operation is addition. Similarly, when we talk about a negative one, maybe in the context of the z sub n groups, the addition mod n groups, the negative one notation might be n minus the thing. So, you know, if they were to rewrite some of these books, I think they might do things a little bit differently. But we usually, not always, we usually denote the thing called b in property number three of groups of the definition by a inverse. But warning, the notation might be bad. This could lead to some, I don't know, unfortunate notation. Especially when the operation happens to be addition. So, for example, in z under addition, if I want you to figure out what the inverse of two is, I'm now using this notation in the context of general groups. And the point is, if in general groups we call the inverse element a with a negative one, then in this context, the thing that gets you back to the identity under this operation is negative two. So, it's, I mean, it's unfortunate. But in most other situations, it works out just fine. For example, in, how about in z 12, or let me give you another bad one and then the rest of them will be good. If I want three, what I'm asking is, find the thing inside z 12 so that when you combine it with this thing, where the operation is addition mod 12, so that when you do this thing, you combine it with something else to get zero, it'll be nine here. Why? Because three plus nine is 12, which is zero mod 12. But let's see in, how about the non-zero real numbers under multiplication, if I ask for three inverse, then you're sort of happy with that. Because the operation is multiplication, which means that the identity element is one. And the question is, what do you combine with three to get one? The answer is one-third. In GL2R, let's see, what's the inverse of, oh boy, I'm going to have to compute this thing. I forget how to do inverses. One, three, two, four. Inverse is, help me here, let's see. So what do you do? You do one over the determinant, right? So one over four minus six times, let's see, you switch these, and then when you put minus signs on these, something like that, yeah. What does that look? One over minus two times four, which is what? Minus two, three-halves, one. I think, let's check that it works here. This is also a good idea. That's minus two plus three, which gives one, that's good, that's three-halves, minus three-halves, that's zero, that's minus four plus four, that's zero, that's good, that's three, minus two is one. Perfect. So we can talk about the inverse of an element. If you hand me something called A, then there's some other thing in the group called B for which this equation holds and this equation holds. What we've proved is that there is only one thing that works, so it's legit to call it the inverse. And what you can prove is this. It turns out, proposition, if I hand you two elements, if G is a group, is a group. And let's look at two things. How about little g and little h are elements of g? Then the following makes sense. You can take the two elements of g and h, combine them, and you can ask, what's the inverse of that? Well, this is in the group because the group is a binary operation. So because it's in the group, it has an inverse. And the question is, what does this thing look like? And the answer is, it looks like this. h inverse star g. So a property that you learned was true of matrices, that the inverse of A times B is B inverse times A. It turns out to be true in all groups. And the proof is... Well, let's see, what are we claiming? We're claiming that this is the special thing so that when you combine it with this, that you get the identity. That's what it means to be the inverse of something. It's the thing that you combine with the given element to produce the identity. So if I want to convince you that that thing is this, all I need to do is convince you that if I combine this with g star h, that I get the identity. Well, combine the thing that's given with the proposed thing and see if e results. And it will. So I'll let you do that at home. And in the other order. I'll say both directions. Both directions. Because in order to be the inverse of an element, it has to be the case that when you combine it with the other thing in either order that you presumably get the identity. Questions, comments? Yeah, to show something as the inverse of something else, it's not too bad to do. Just combine the two proposed things and make sure the identity comes out. And as soon as you've done that, then you can conclude that what you've got here. Okay. Yeah, rather than opening the subgroup can of worms, I think this will be a good place to quit. So because it's Monday, here's a homework assignment. But folks, next Monday is Labor Day. And because I want to give you typically the same amount of time between when I give an assignment and when it's due, instead of making it do what would be usual on Wednesday, I'm going to push it to Friday. So if you want extra time to work on it, that'll be fine. So this will officially be due, this is an unusual one, only because of Labor Day holiday, due Friday, September 7th, by 4.30 p.m. at my office. Hey, if you want to turn it in in class, a week from Wednesday or something like that, that's fine. But if you want to spend Thursday and maybe Friday working on it, that'll be fine too. Officially, there's no class next Tuesday either, so what I'm going to see is whether or not Jen can do an extra SI session, maybe on Thursday or something like that, but I'll get back to you on that. Oh, first of all, no, this is pre-homework. On Wednesday, there'll be a quiz, and the quiz will be give the definition of a group precisely the definition of a group. So you can either pull it off the... straight out of the text if you want. I'd prefer you not pulling it directly out of here. I'd prefer you pull it off of your notes from last Wednesday if you're going to grab it from your notes because last Wednesday I gave the complete description. This is just sort of the abridged version here. What I'll do on Wednesday is I'll simply hand you a sheet of paper as you walk in the class. You'll have five minutes just to write it down. There's no mystery here. I want you to be able to... Okay, so I'm sorry. So here now is the homework assignment. In section four, problems 14, 20, 23, and 31, I want you to turn in 31 and I'm going to hand in an extra sheet of problems, and here that is the extra sheet and I want you to turn all these in for grading and then as an additional part of this assignment in chapter or section five, section five, problems eight through 19 and 26 through 40 and here are the ones I want you to turn in there. 11, 12, 15 and 16, 34, 40 and 47. You should be able to start all of the section four problems and the first two section five problems. A couple of quick notes on the section five problems. One is in one of the questions, the verb that's used is determine whether or not the following is a group or something like that. Determine folks means prove if yes and disprove if no. Yes, say why not if no. And the second comment on these section five problems is in problems 15 and 16, it turns out the sets that you'll be working with there are actually functions. What I want you to do is give either pictures or some sort of graphical interpretation of what these things are in terms of the set. So you'll be looking at a collection of functions that have a certain property. It's easy to draw graphs that describe what that property is. Go ahead and do that. I've got a little bit, a somewhat more lengthy description of what I want you to do on these problems, at least 15 to 16, if you go to the website, but that should give you enough information to get started on these assignments. Okay, so let's see, Jen will have...