 session. I'm going to put a question mark. It's either going to be, what was the first time, 9 to 1030? 9 to 1030? 30 or 415 to 530. And sort of big question marks to be determined hopefully by email announcement tonight. And I'll try to post it on the website as soon as I find out the exact time. Tomorrow morning. Okay, perfect. Thank you. So no question. Right, she does normally what? Wednesday mornings? Then it's Tuesday afternoon. Then this is no question. Tomorrow? 415 to 530. Thank you. Okay, thank you. It's good that we all know where each other is. Okay, yeah. A couple of quick comments on the quiz coming back. First of all, the homework assignments are worth 10 points total. The quizzes are worth five points total. So quiz total is five points. So the number that you see at the top of the page, sitting by itself is out of five. The only two things that came up on more than, let's say, two papers were these. In your sort of preamble to what a group is, you can call it step zero. A group is, you know, a binary, blah, blah, blah. Saying that you have a closed operation on a group is the right idea, but it's not the same as saying that you have a binary operation. Because binary operation includes the fact that you've got a function, which automatically includes that the operation is well defined. You could just say some sort of binary operation or some sort of map from G cross G to G. You've missed a little bit of the structure of what you need there. So binary operation is what you need to say there. If some of you just said is a map or a rule that gets you from G cross G to G, you need a little bit more than that. And in numbers two and three, you have to make sure that you get your quantifiers correct. The existence of an identity element means there is an element called E with the property that if you hand me any other element and you combine the other element with E, you get the other element. In other words, you start with E and the element E that you've written down works for every other element in the group. That's completely different than saying if you hand me an element in the group that you can find something that behaves like E for each individual element. Conversely, when you talk about an inverse element, you don't want to say, well, if I start with E and any A, I can somehow, there's some sort of inverse for it. The inverse element conversely is specific to the particular element that you start with. So the preambles and the appropriate quantifiers in parts two and three of the definition of a group are different. You start with the element E in step two, and each element in the group works for it. Conversely, in part three, you start with any element A and you can produce for it an element A prime that works. You're not claiming that for each element in the group there's somehow a universal or an inverse element that works for every element simultaneously. The inverse statement is a statement about if you hand me an element, I can find another element in the group that works for it. If you were to hand me a different element of the group, I would have to definitely write down a different element as it's inverse. It seems like a subtlety, but in the end the point is that you need to start parts two and three slightly differently. Questions? Question, David. On questions 15 and 16, presumably you could write down a formal proof of whatever is or isn't a group without any sort of pictures drawn. But in addition to writing down a proof, I want you to tell me what the heck it is you're talking about. The way I want you to tell me what you're talking about is by drawing pictures of what it is you're talking about. Because these are functions, all right, so what do they look like if you have a graph? You can think of the extra statement as, in addition to what you would have already done in the question, do this too. Another question, Lindsay. Extra problem, right? Oh, I see. The question is because we talked about this notion of isomorphism, even though I didn't include that word in the extra problem, I don't think. But yeah, whatever you learned in class that would be appropriate to sort of use in support of an answer to the extra problem is totally appropriate. Just say, we saw in class that this is the same as this or isomorphism, or that we can relabel the elements of this to get to that. However you want to phrase that as well. And the others? Yeah, Lindsay, sure. Explain the identity element, yeah. Sure, it's an element that when you combine it with any other element, it gives you the other element. And the key is, it's up to you to decide whatever the operation is, whether it's addition of functions or multiplication of functions or whatever it is, all right, what function would somehow, when added to or multiplied by or composed with any other function, give the other function? So it's just sort of matter, sort of this little mantra you got to get what function? The significance of n equals two, the significance of n equals two. Let's see, did I have the number 14? Yeah, so Lonnie, the question in number 14 of section 4.4 asks you to analyze all the n by n diagonal matrices that have diagonal entries, one or minus one. Well, you can do that for any n. I only want you to do it for the one particular situation where n equals two. Contrast the, no, there's only three groups that you have to analyze here. And the group that you'll get has four elements. So I'm missing the question. I mean, if you do section 4, problem 14, and you only restrict to n equals two, then you'll get a group, and that group will have four elements in it. Okay, so that now you throw in the list of groups that you want to analyze. And now here's another group. No, sir, what there is only one group that we're describing. The case n equals two, that group will have exactly this many elements in that particular case. Let's chat after. We'll get there. Okay. Okay. Okay. Yeah. So because it's been a week, let me bring you quickly back up to speed as to where we were when we left off a week ago, we were talking about the notion of subgroups. And the idea was, once you've spent a lot of time working to show something as a group, then oftentimes you can find within that group, other groups, groups that somehow live inside the given group. I made a big deal out of that particular way of finding new groups, because later on there will be other ways of finding new groups once you've established some existing groups. Okay, this is just one way look inside. It turns out various groups come up easily. If you look inside known groups, we've looked at a bunch of examples a week ago. It turns out also that if you're trying to establish that some set is really a group, and that set happens to live inside something that's already known to be a group, in other words, that set could potentially be a subgroup, then it's somewhat easier to show that a subset is a subgroup than to simply somehow produce a group out of thin air, as we've been doing up until now. And you only have to invoke what's called the subgroup there and it says to show that a subset is a subgroup. Then all you need to do is check that the binary operation is closed. In other words, A star B is in the subgroup whenever A and B is the subgroup, you have to show the identity elements in there and you have to show that you grab something in the subgroup that it's inverse not necessarily exists. It does exist because you're living inside a group, but actually also lives in the subset. So there is a thing called the subgroup theorem that's in your notes. I won't repeat what the three properties are. But the point is that there's a relatively easy way to determine when something is or isn't a subgroup. I mean, if you can show, for example, that there are two things in the proposed subset, so that when you combine them and you don't get something in the subset, you're done. You don't have a subgroup. It will be the case, folks, well, it's rare in mathematics, but it's the case at least for this course, it's probably true and all. It's almost always true that if you're gonna be asked to show that something is a subgroup, there's exactly one way to do it. You use the subgroup theorem and you just click through the three things that you have to do. There's not any sort of, well, I'll choose this tool or that tool or the other tool. There's one tool that is the subgroup theorem. All right, what we looked at last Wednesday were examples of specific groups. Excuse me, nah. We're examples of specific groups. And inside those specific groups, we identified some subgroups, like inside gln comma r, we identified what we called s l n comma r. Okay, that's fine. What we're gonna do today, at least for a little while, is ask the question, suppose you hand me any group? I don't even tell you what it is. I don't tell you if it's finite or infinite or a billion or not even anything. It's just here is a group. Are there automatically some subgroups that you can build inside the given group? The answer turns out to be yes. We saw two, I called them silly examples last Wednesday, these things are called trivial subgroups. So the idea here is, given any group G, any group G, we can define special subgroups G. And we looked at this already, example, the trivial ones, the group itself, or just the identity element sitting by itself, there are two subgroups, and you notice I didn't tell you what group it is, just start with any group, finite infinity, not a billion, doesn't matter. Here's another example. This one's a little bit more interesting. I'm not playing it up as the most important example of a subgroup. I am playing it up, though, because I want to show you how to go about proving in this sort of generality, how a subset of a group might be a subgroup, simply by going through the subgroup there. So here's the example. And the example is given in terms that subsets are typically given in terms of I'm going to hand you a group, I'm going to describe a subset, not by telling you what the form of the elements in the subset are, but by telling you what property the elements in the subset have. And then to determine whether or not something's in the subset, you simply determine whether or not the property is established or can be shown in the element that you've got your hands on. So here it is, for any group G define the following, and this is another artifact of the German influence on early group theory, z of G, z of G, this is called the center of G, the German word for centers, centrum, z-e-n-t-r-u-m, so that's where the z comes from. And here's the definition of the center of G. It's the elements, let's call them little z in G, with the property that if you do the binary operation with that element and any other element for every, so the definition of the center in words is this, you're in the center if you have the property that when you take any other element in the group, not just one, but every single one, and you do the binary operation in this order that you get the same thing as doing the binary operation in the other order. And I think you mean the group's a billion? No, I'm not asking that the group be a billion. What I'm asking you to do is in some sense, look at those elements that behave like they're a billion with all the other elements of the group. Just before we start, for example, if I hand you the particular element E, E has this property, because if you take E star G, you get G, regardless of what G you start with by property of E. If you take G star E, you get G. So if I look at the particular element Z equals E, I've got an element in this subset. And the claim is claim. What we'll prove is that the center of G is a subgroup of remember this notation subgroup means where subgroup is denoted by take the subset symbol and just put a little sharp point on it to make it look like a lesser equal sign. I'm going to click through what each of the three steps look like. If you need to do this for a homework problem or something similar to it, I'll ask you to write out more details than I'm about to write out here in class. Well, look, if you're going to convince me or prove for me that the some subset is a subgroup, there's only one tool that you've got at your disposal. It's a subgroup theorem. So okay, use the subgroup theorem. Subgroup theorem says I have to convince you of three things. Step one, I have to convince you that if we take two things in the subset, let's see what have we called the things in the subset we've called them little z. Makes sense given the name of the subset. So let's call two things in the subset. How about z one and z two are assumed to be in the subset, then so is whatever the binary operation gives me. That's what you have to show for for step one, just step one of the subgroup theorem. If you take two things in the subset convince me that when you combine them you get something in the subset. Folks haven't told you anything about the form of z one or the form of z two, but what I have told you something about is the property that an element in the group has to have if it's going to be in the subset. The property is you have to pick any other element in the group, you have to perform this binary operation in this order, then you have to perform the binary operation in the other order and you have to convince me that the two results are equal. So to show you or to prove that this is the case, what we have to do is we have to show, well, we have to show that this is in the subset. In other words, we have to show that if we take this element, that's the element now, and we combine it with any other element in the group that what we get is that other element in the group combined with the element in question. Folks, I don't care if the element you're interested in showing in the subset is called h, if it's called t, if it's called diamond or square, or if it happens to be called z1 star z2, that's the thing that you're trying to play the game with. And playing the game means take it, combine it with g, that's what I've done here, take g and combine it with it in that order and convince me that you get the same thing. If you can convince me that that equals that, then the conclusion is that this thing is in the subset, that's all. Any questions before we do the computation? And once you set up what you have to do, typically the computation is pretty straightforward. So let's do it. Well, let's see. What do we know? We know, and the reason we know it is because we're told that each of z1 and z2 individually are in here, we know that z1 star g is g star z1, we know that because z1 is assumed to be in the subset, and that's what it means to be in the subset. And we also know that z2 star g is g star z2 by the given information, property or definition of z of g, that's what it means to be in there. So now let's do the computation. Well, this is a good connecting word. Here's what I want to show, here's what I know, let's see how we show it. If I do z1 star z2 star g, the goal is to convince you that that expression equals this expression. Let's see what we can do. This is, oh, by associativity, binary operation in a group is associative, this is z1 star z2 star g, which is, any ideas of what to write down next? z2 star g is g star z2. So this is given information, z1 star, we're replacing z2 star g with g star z2, or I'd like to say something about z1 and g. Sure I can do that, but let's use associativity again, star z2, which is, oh, I know something about z1 star g, it's g star z1, which is associativity again, g star z1 star z2. So look what we've shown. So we've shown that if we do the left side here, by writing down half a dozen steps or so, we can melt it into the right side, so check. So instead of spending more time on this example, I'll simply say, all right, we've now completed step one of the subgroup theorem for this particular subset. We've shown that it's closed under the binary operation in order to complete the subgroup theorem, in order to complete the proposition that this subset really is the subgroup. Now I have to show properties 2 and 3, show that e has this property, do it home, show that if a has the property, then so does. And I'll let you think about, really, the hard part is folks just writing down what the heck the task is. Here the task is, I'll give you a hint, show that, well, what does it mean to be in the subset? It means it has the property. The property is that if you take the element in question and you combine it with any other element, that you get the same thing as if you had done that for all. Too far over here, Karen, is that showing up? That's what you have to show, because that's simply what it means to be in the subset. Similarly, here you have to show that if you have an element that has the property a star g is g star a for all or for every element in the subgroup, then you have to show that a inverse has the property. Well, having the property is a non-issue to write down. So here you simply have to show that this is always true. And in the third part, you have to show that if you assume a has the right property that puts it in the subset, that you can somehow conclude that a inverse has the right property. I'm not claiming that they are totally obvious, or even one step. You might have to work at sort of an effort level similar to the one that we used to prove part one. But the task is now relatively straightforward. So here is a subgroup of any group, regardless of what group you start with. Of course, I'm not claiming that this is always some new subgroup. It might be the case, for example, that if you look inside a specific group and you write down what z of g is for that specific group, z of g might actually be g. That's fine. In fact, I'll mumble under my breath and I'll let you think about this and we will talk about it in about a week or so. If z of g is g, it means that every element is in the subset. It means that every element in the group has this property. It means that every element somehow can be switched in order with every other element, which means the group is a billion. So if you have an a billion group, then z of g is g. In fact, if I have not a billion group, then z of g is not g. It might also be the case that z of g is e. So I'm not claiming that we're always producing a new subgroup, but at least we're always, regardless of what group we start with, writing down a subgroup that we can describe. All right. Questions? Let's continue the theme then. The theme is I'm going to write down an arbitrary group or any group, anything you know to be a group, and we're finding subgroups in there. The trivial ones are always in there. The center is always in there. The center is just, again, as I mentioned, maybe not the most important example that comes up, but one that can be used as a good example or tool to show you how to use the subgroup theorem to verify that a subset really is a subgroup. What we'll look at now are subgroups that are significantly more important in the structure of trying to describe what the group itself looks like. And these are usually called the cyclic subgroups, or if you're in England, the cyclic subgroups of a given group. So here's the definition. Start with any group G. So G, any group, that's the theme here. And what I want you to do is pick any element you want. Any element, let's call it little g in capital G. And what I'm going to ask you to do is write down a subset of the group. And what we'll do is show that that subset is actually always a subgroup. And here's how you get that subset. First of all, I'm going to give it a name. The subset is denoted by, you take whatever element is that you started with, I don't care which one you picked, just grab your favorite one, and you put it in diamond brackets like this. And here is its definition. You take all of the elements inside the group. So it's, I guess I'll call it x in G with the property that either x is the identity element or x is G star G star G. As many times as you want, let's call it i times for any i. So it's either G itself. If I ask you to take G star G star G one time by default, I just mean you to, or ask you to write down the element G itself. If I ask you to do G star G twice, it's fine if I ask you to do G star G star G three times, it means just do G star G. Or x is G inverse star G inverse star star G inverse, I'll say J times. And here i and J, I put a big prems around this, here i and J are any positive integers that you want. So here's the subset of the group that I want you to write down. Automatically, just by default, throw in the identity element, throw in everything that we're gonna call a power of G. So throw in G, throw in G star G, G star G star G, but that does many times as you want. If you're asking, you mean infinitely many times? Well, in theory, yes, but after a while, this collection of things might start repeating itself and that's fine. And then finally, take the element that you've started with, find its inverse, and then do the same process with its inverse. Just keep pounding its inverse with itself as many times as you need. And when it's all said and done, just sort of step back and say, all right, that's the subset I'm interested in. Proposition, regardless of what element in the group you start with, you always get a subgroup. For any group G, for any G, and any element that you choose to start with, any G in G, this set that I've just described here is always a subgroup of G. Remember, this is a subgroup of notation. And before we prove this, I'll give you the notation. This is called the cyclic. And I said the word cyclic before almost as a joke, but it turns out it's not. I mean, if you go to international meeting and you hear international mathematicians or mathematicians not from the United States talking about these structures because they've learned typically British English, they'll call these cyclic groups. Cyclic or cyclic subgroup generated by G, particular subgroup is called. If we need to be more specific as to what the group G is, sometimes we'll call it the cyclic subgroup of G generated by little G if we need more specificity, but typically we don't. All right, folks, it's a proposition of exactly the type that we've been looking at for the last, well, day and a half. Show that this subset is a subgroup. You have one tool at your disposal, the subgroup theorem, just pound through the three steps. So prove, I'm not gonna go through all three steps, but I'll indicate that there are three steps to go through. Step one, step one's a little bit squirrely, but I won't go through all the details. It says if you take an element of one of these three types and you mash it together with another element of one of these three types that the result is an element of one of these three types. So presumably there's like a lot of choices. If you take an element of this type and you start with an element of this type, do you get an element? Sure, I mean, if you take E and you start with G times G or G star G star G, you just get G star G star G. I mean, starring E with anything certainly isn't gonna change it, so that's a non-issue. If I star one of these with another one of these, if I take G star G star G 12 times and I take G star G star G 18 times and I star them together, then I get G star G star G 30 times or something. I don't care how many times, you just get another element of the same form of the correct form. Same thing with the real sort of interesting one is if you take an element of this type and you start with one of this type, do you get something of the correct type, correct form? The answer again turns out to be yes. Let's say I'm just giving you the intuition here. If I have 10 of these and I have, I don't know, seven of these and you star that with that, well, you're seeing a bunch of things that look like G star G inverse, which is the identity that disappears. So for example, if I do for just an example, so warning, I am not playing this up as a proof. I'm just giving you an intuitive example as to what's going on. If I take G star G, something like this, star G, I don't know, 10 times, and then I ask you to star that with something that looks like the G inverse thing. G inverse star G inverse star star G inverse, I don't know, seven times, what's the result? The result is, if you star those, I'm leaving out all the associativity words, but everything's associative. If I regroup like this, oh, those kill each other, you get the identity. Then I group those together, get the identity, get the identity, get the identity, get the identity. Oh, but let's see, I've got, what, three more here than I have here. So this turns out to be G star G star G. The right form. So that's what's required to do in step one. It's not too bad to do. It's a little bit tedious. Step two is obvious. You have to commit to the identity elements in the subset. That was easy, there it is. Step three, step three is not too bad either. You have to work a little bit, but look, if I hand you G, then I know that it's inverses in there. If I hand you G star G, then guess what? It's inverse is G inverse star G inverse, so it's in there, just sort of by definition. You've rigged things so that everything that's in there also has its inverse in there. So three is also to check, although there's some work to be done here, and so that's presumably what we'd need to go through to show that this is a subgroup. And again, I'm not gonna go through all the details. I'm giving you some idea of what's required or what sort of detail work you'll need to do by showing you this example. The formula that I'll write down though for you is this. It turns out use, and I won't prove this either, but in effect, it's just the generalization of what I showed you here to give you an intuitive idea to more general expressions. If we denote, and this is just notation, if I take G and I put it with a little zero in the upper right-hand corner, well, you're used to thinking of that as something to the zero power and you're probably gonna react, oh, that's just one. Well, for us, the thing that somehow behaves like one is the identity element. So deem this symbol to simply mean the identity element. Deem the symbol G where I is an integer to be, I mean, this makes sense, huh? I times, I times where I is a, where I is an integer and I bigger than zero. So folks, I'm not defining for you what G to the one half power means or what G to the square root of two means, or anything like that. What I'm defining for you is what G, quote unquote, raised to the I power should mean. And remember, folks, this definition of what G to the I power means is completely dependent on the group that you're looking at because it completely depends on what the binary operation is, which is a little bit notationally unfortunate sometimes because if it happens to be that the binary operation in the group that you've got in hand is plus, then G to the seven would mean G plus G plus G plus G seven times, so you gotta be careful. Even though this is suggestive of multiplication, it really is dependent on the specific binary operation of the group. And let's see, if I denote by this, G to the minus J is G inverse star, G inverse star, star G inverse J times J and integer and bigger than zero. So J is bigger than zero, so I've got G raised to a negative power. Then what you can show is that it's always the case that if you do G to the I star G to the K, and I don't care if I is positive or negative or zero or K is positive or negative or zero, that you get the exponent rule that looks like the exponent rule for multiplication. Things are behaving like multiplication. Behaving like simply means as a product of our notation for groups, anything you can typically prove for multiplication inside the non-zero rationals or something like that, you can go ahead and verify. So, I mean look, the point is if I take G to the 10th star, the notation for this thing would be G to the negative seven then the result should be G to the 10 minus seven. And we'll use this for the remainder. I won't go ahead and prove all the details, but the proof roughly looks like. All right, so here then is another way of looking inside a thing that you know to be a group and finding more subgroups. And again, we're not claiming that whenever you do this that you necessarily get something different than the group itself or whether you get something different than the other trivial subgroup, the identity subgroup. It's just a way of producing subgroups. Let's do some examples, example. And we looked at some, it turns out at the end of last time. See what those looked like. So, let's look at, actually it's funny, it's funny why this is the group I'm going to look at today. So, here is the group from the extra set of problems that gets generated from section four, problem 14, n equals two. Let's call this group G. It's the group 1, 0, 0, 1, minus 1, 0, 0, 1, minus 1, yeah, 0, 0, 1, 0, 0, minus 1, sorry, and minus 1, 0, 0, minus 1. And we looked at this at the end of last week. It is the collection of two by two matrices having as, two by two diagonal matrices having as entries either one or minus one. All right, so let's play this game, this cyclic subgroup generated by game for the elements of capital G. Let me just go through each of them. Let's call this one E and this one A and this one B and this one C. So if I look at the cyclic or cyclic subgroup generated by E, so this means you throw in E by default, just put that in first. Then you put in the element itself, which in this case happens to be the element I've already written down, so I don't need it again, so already there. And then you put in that element, it's already there. And you put in, oh, and then you put in, it's already there because the inverse of the identity is there. So, hey, in the end, all you get is, so the cyclic subgroup generated by the identity element here is just the identity element, and in fact, folks, there's absolutely nothing special about this group. It's always the case that if you look at the cyclic subgroup generated by the identity element, you just get the trivial subgroup consisting of the identity element itself. How about, let's see, if I look at the cyclic subgroup generated by A, again, by default, write down E, part of the definition, then write down the element itself, think that's G to the one, then write down A star A. What is A star A here? The operation here is multiplication. Multiply this thing time itself and you get E, so that's E, so I don't need it because I've already written it down. And then let's see A star A star A, well A star A was already E, so that's A, but I've already got it. And if I keep doing more powers, quote unquote, of A, I either get E, A, E, A, E, A, sort of cycles back and forth, that's where we get this word from. Oh, but let's see, what's A inverse? Is A because why? Well, because if you star A with A, you get E, so that's already A, and that's already in there, so I don't need it. A inverse star A inverse, that's A star A, which is E, but I've already got it. So in the end, all you get is E and A. Well, there is a subgroup, I don't need to prove anything because it is the cyclic subgroup generated by A, and we've shown that the cyclic subgroup thing is actually a subgroup. Okay, now we play the same games here. Similarly, if I take the cyclic subgroup generated by B, the details are exactly the same, the cyclic subgroup generated by C, the details are exactly the same. So it turns out, folks, that if you take any one of the four elements in this group, and you look at the cyclic subgroup generated by it, none of those four elements actually give you the whole group back. One of them always gives you the trivial subgroup back, but neither one of the non-identity elements, or none of the non-identity elements actually gives you a whole group, okay, that's fine. Just something to note. Here's another example. Yeah, this'll be an example that I won't spend a lot of time on because we did most of the details last Wednesday. The example is Z6, so it's 0, 1, 2, 3, 4, 5 with addition mod six, and let's look at, for each of the elements in here, look at the cyclic subgroup generates. Well, hey, we just noted in the last example, if you take any group and you start by taking the identity element and you look at its cyclic subgroup, you just get the trivial subgroup consisting of the identity element. If you do anything to the identity element, it doesn't change anything. How about the cyclic subgroup generated by one? Well, you're supposed to put zero in there by definition. You put one in there, there, then you put one star one in there. So I'm gonna call it that, but let's see, the operation here is addition mod six. So that's one plus one, which is two. And then you put one star one star one in there, and you put one star one star one, and you put, well, and then if you put one star one star one, then if you do it six times, you get six, which is zero in this group, so you get back to where you started. And hey, if you do it again, you get this one and this one. What we'll look at in more detail next Monday is, it turns out, folks, when you do this process, as soon as you get back to the identity element, you're done, exactly right. You've got all the elements. So once it turns out we get to zero again, because we've already got it, not only do we don't need it, we don't need to go any further. We've got the entire subgroup. In fact, here, there's no way to get anything else because we've got everything in the group. So this happens to be the whole group, which is something that didn't happen in this previous example. So we found a specific element that has the property that when you look at the cyclic subgroup generated by that element, you actually get the whole group. So that just makes this group somehow structurally different than this group, even though they both have, no, I won't say that. Okay, cyclic subgroup, I won't say it because it's not true, it's probably a good thing. Cyclic subgroup generated by two, it's the same idea, throw in zero, throw in the element itself, throw in the element star together with itself, then three times two plus two plus two is six, which is zero, and we're done. Cyclic subgroup generated by three, zero and three, hey, three plus three is already zero again. Cyclic subgroup generated by four, throw in zero, throw in the element itself, throw in four star four, four plus four is two, good, then four plus four plus four is 12, which is zero, mod six. So it's interesting, so here we actually get the same subgroup that's starting with different elements, okay, that's possible. How about let's go ahead and finish the example. Well, you throw in zero, you write down the element, then the element star with itself, so five plus five is 10, mod six is four, five plus five plus five is 15, which mod six is five plus five plus five, good, and then zero. So here we actually get the whole group as well. Question, David? Ah, yes, question, so David's question is, is there a name for elements that do that and the answer is yes, in fact, almost taking one step backwards from that question, there is a name, and I'll give it to you right now, for groups that contain elements that do this. So this group doesn't have any elements that sort of work. This group doesn't contain any elements with the property that if you look at the cyclic subgroup generated by it, you get the whole group, that group does. And here is the verbiage that distinguishes between those two possibilities. A group G is called cyclic, or cyclic in case, there's at least one, there is at least one, I don't care how many, element little g in the group, so that, or with the property of that, when you look at the cyclic subgroup generated by G, you get the definition of what the word cyclic or cyclic group means. So notice now we've used this word cyclic in two technically different, but obviously very related senses. We use the word cyclic in the sense of the cyclic subgroup generated by an element, and now we've used the adjective cyclic on a group. The group is cyclic if, and we can use the original language, if there exists an element in the group for which the cyclic subgroup generated by that element is the whole group. And in this case, when this happens, any such G is called a generator. So let's just go back and look at some verbiage, and if I look at this group, so this group then is not cyclic. And we showed it was not cyclic because we specifically went through every element in the group, there was only four of them. We pounded out what each cyclic subgroup was generated by each of the elements, and we never got the whole group back, so it's not cyclic. Just as a piece of verbiage, when we talk about this group, if we instead of using the notation with matrices, use the E, A, B, C notation with the table that I gave you last time, we typically call that group V, and I'll give you the historical reason. Maybe today. This clock is right, yeah? Relatively right, okay. So what we've just shown is this group called V is a group that has four elements that's not cyclic. Quick remark, and somehow students seem to get these backwards or get confused about this. Folks, this group V is an abelian group. I'd say it's multiplication matrices. Yeah, but remember when we have diagonal matrices, multiplication turns out to be commutative, so even though this group is commutative, it turns out to be not cyclic. Z-cic turned out to be a cyclic group. So Z-6 is cyclic, and in fact it has two different generators. It has both one and five as generator. So the element one is a generator for Z-6. The element five is also a generator for Z-6, okay? So what we're about to do is look at specific examples of groups that are cyclic or are not cyclic. The first thing that we're gonna note though is the following property. It turns out proposition, proposition. If G is a cyclic group, then necessarily G is abelian. So if somehow you've hunted around inside your group and you happen to have come up with an element, at least one, but one is sufficient, that has the property that when you look at the cyclic subgroup generated by it, you get the whole group. In other words, if the group is a cyclic group, then necessarily it's the case that if you take any two elements in the group that they commute, the intuition is not too bad behind that. To say that I have a cyclic group means that every element in the group can be written as well, either the identity or G to some positive power or G to some negative power. But in our notation, it just means that every element in the group can be written as G to some power. If the power happens to be negative, then it just means that you've looked at the product of a bunch of inverses of G. If the power happens to be zero, it means you're just looking at the identity element. So the proof is actually pretty easy. If you prove, let's see, how do you show that something is a billion? You have to convince me that if you take any pair of elements in there and you do their operation in one order, that you get the same thing as doing it in the other order. Well, let's see. G is cyclic means, the definition means, that there is some element, some... I don't care what it looks like. Let's call it little G, having the property that if you look at the cyclic subgroup generated by G, you get everything in the group. That's what it means to be cyclic, that there's some element so that when you run through all of the powers, positive, negative, and zero of this particular element G, that you get everything in the group. All right, now, that's the hypothesis. The conclusion I'm trying to draw is that G is the billion, in other words, if I take any two elements in here and I combine them in either order, that I always get the same result. So let's see, here's what I want to do. I can now pick any two elements. Let's call them x and y in the group G. Here's what we have to show. Is that x star y? Is y star x? You're thinking, well, I don't have a chance. Well, you do have a chance. Here's y. Look, anything in the group, oh, looks like G to some power. That's what it means to be cyclic group generated by this particular element. But G equals the cyclic subgroup generated by G means that if you take something in the group, that you can write it as G to some power. Let's call it i for some integer i. In fact, I'll write it this way, folks. For some element of capital Z, remember, capital Z denotes integers positive, negative, or zero, and that's precisely what x has to look like. If it looks like G star G a bunch of times, then i is positive. If x happens to be the identity element, i is zero. If x happens to be G inverse star G inverse a bunch of times, i happens to be negative. That's no big deal. And similarly, y is G to the J for some integer J. So now let's compute. Let's compute. You're still maybe thinking you don't have a chance because you have no idea what G looks like or what capital G looks like or what anything looks like. Yeah, but wait a minute. What's x star y? It's then G to the i, G to the J. But wait a minute. G to the i star G to any power is G to the starting with G. Again, this should look familiar. It's what you teach your eighth graders and how to multiply exponents. G to a power times G to another power is you add the exponents together. And similarly, if I do y star x, I get G to the J star G to the i. And that's G to the J plus i. And that's equal to G to the i plus J y since addition in Z is commutative in the basics. Folks, these expressions in the exponents are just adding whole numbers, not anything with the operation in the group or anything like that. The way that you combine G to the i star G to the J is you just add the two exponents together. And that's all we did in each case. In the first case, it's that plus that. In the second case, it's that plus that. But obviously, those two are the same. So check. So y star x equals x star y. x star y, y star x. And we're done. And the conclusion is that the group is then a billion. So the phrase is this. Every cyclic group is a billion. Every cyclic group is a billion. This is a nice situation where your math 2, 15 year logic of course comes in. But the converse is not true. Exactly right. It's not the case that every abelian group is cyclic. Here's an example of an abelian group that's not cyclic. So not conversely, it's a warning. And I'm not sure why students have a little bit of trouble with this warning. The converse is not true. It's not true. There are many, many, abelian groups which are not cyclic. Here's an example. Example v. I mean, there's also obviously many groups that aren't cyclic. If I hand you a non-abelian group, it has to be not cyclic by what would just prove. But if I hand you an abelian group, they might not be cyclic like v. Let me give you another somewhat more interesting example. If I hand you the group of rational numbers under addition, rational numbers under addition, it is not cyclic. So it's the group of fractions under addition. So I'm including 0 here. That's fine. The operation now is plus. Certainly an abelian group addition of rational numbers is perfectly commutative. The claimant is it's not cyclic. And let me just walk you through the idea. The claim is that it's impossible to pull out a single element. I don't care which element you pick. Pick out a single element, and you look at the cyclic subgroup generated by it. It's impossible to get all of the rational numbers. You'll get a bunch of them, but you'll never get all of them. And the reason is this. If you hand me any rational number, well, it looks like a over b, where b is not 0. You can choose b to be positive if you want. And the claim is, if you look at the identity element of this group, which is 0, if you look at the element that you've chosen, that's a over b, and a over b plus a over b, that's 2a over b, and a over b plus a over b plus a over b, that's 3a over b, and da, da, da, da, da, da. And then all of its inverses, which is all the negatives, then in effect, folks, what you've got is all of the elements of q that look like some multiple of the original element that you started with. And the claim is that that can never give you all the rational numbers. And there's lots of different proofs to the one I like the best is this. Look at your denominator b. Tell me what primes make b up. In other words, write b as a product of primes. There's infinitely many primes. Pick your favorite prime that doesn't appear in the factorization of b. There's obviously at least one effect. There's obviously at least an infinite number. Just pick one. Then look at the rational number 1 over p. And it turns out that you'll never be able to write a multiple of a over b as 1 over b, because the prime p just won't work out for you. So there's one of many possible proofs that is not cyclic. I won't ask you to write that out formally for me, but just the idea is that for any a over b and q, if p is a prime, and I'm going to start denoting the prime numbers by capital p with an extra line in it, a set of prime numbers, those that don't factor non-trivial a, then you can show that if p is a prime number which doesn't appear in the factorization of b, then it's not too hard to show, and I'll leave out the details here, but the point is that this particular fraction is never in the subgroup generated by that element. If you think about it just size-wise, I guess. Size-wise is not a good way. You wind up using the fundamental theorem of arithmetic or unique factorization of integers that if you try to write 1 over p in the correct form, you'd get a contradiction to the uniqueness of factorization in the prime. All right, questions? Comments? So there are lots of cyclic groups that aren't immediate. Here are the, I'll say the most important ones, and then I'll actually use the word that we used last Wednesday, and then we'll call it a break. Here are key examples, examples of cyclic groups. We've already given one. z sub 6 was an example of a cyclic group. It turns out z sub n, and I'll just remind you what the operation on z sub n is to make it into a group. Addition mod n for any n is a cyclic group. So the specific example we wrote down is z sub 6, turned out to be a cyclic group. In fact, z sub 6 had generators 1 and 5. It will be of interest later on to try to write down how many generators each of the z n groups have, but at least to verify that each one is a cyclic group, all I need to do is pick out at least one element that acts as a generator, and that's easy to do since for any n, z sub n is always the cyclic subgroup generated by 1. Because if you start with 1, then you do 1 plus 1, which is 2, and you do 1 plus 1 plus 1, which is 3, and you do it, and you just get all the way to zero. Secondly, in this one's maybe a little less obvious, but certainly turns out to be a cyclic group. If you look at z, and here I'm not writing down an operation because it's always understood when I'm talking about z as a group that the operation is addition, z is cyclic. Reason? Well, what's the cyclic subgroup generated by, how about 1? And this is why I play cyclic subgroups up this way. Cyclic subgroup generated by an element is by definition the following. You automatically throw in the identity element of the group, you throw in the element, you throw in the element, you start with itself, then 1 plus 1 plus 1, then 1 plus 1, blah, blah, blah, blah, blah, but then you start over. And you put in, by definition, the inverse of the element that you started with, there's its inverse, and the inverse start with itself, and the inverse start with itself, start with itself, et cetera, et cetera, et cetera. So this one's a little bit maybe uncommon because in all these other examples, when you started with this particular element, all you had to do was start walking through its powers and eventually got all the elements of the group because you end up getting back to 0 by looking at powers. Here, you start with this element, and if you keep looking at, it started with itself. You never get the identity element back, but that's OK. The definition of a cyclic subgroup is, hey, if you don't get the identity element, that's fine. I don't care. Now go back and throw in all of the quote unquote powers of the inverse of the given element. So there's a generator for z. And all I had to do is write down one element that works. In fact, there's exactly one other element that works as a generator. Here's another generator. One is a generator. Minus one is also a generator, are the generators. I only need one in order to establish that z is a cyclic group, but this one happens to have two. The number of generators in z sub n is going to depend on n, and we'll talk about that. Questions there, comments. Now here's the statement. Theorem, proposition, result. I'm going to use this phrase. I'm not going to require you to fully understand exactly what it's meaning yet, but when we get to all the details of what isomorphism means and we've opened this canworms already by some discussion that we did last Wednesday. Up to isomorphism, the groups z n and n, in other words, the cyclic groups having n element and z are the only cyclic groups. So it's sort of interesting. If somebody tells you they have a cyclic group, however they've determined that, I don't know. It doesn't matter. And they tell you how many elements are in the group, then in effect you know what the group is. They may not have used the lettering 0, 1, 2, 3 up through n minus 1. They may have labeled it using other expressions. But in the end, their group will be exactly one of those groups, and you can determine exactly which group it is just by figuring out how many elements are in there. So for example, so proof, I'm going to omit this, omitted. It would take us a half hour, so it's not beyond the scope of this course, but it's just essentially a lot of details in keeping track of elements, et cetera. The idea, though, is this. Example, here's a group. It's one that we've looked at before. The group 1 minus 1 i minus i, I'm going to smile because you've done the additional sheet. OK, so look. Let's call this, I don't know, k or something like that. That's the name of the group. I'm just making it up for this particular example. k is cyclic. Why? Well, all I have to do is convince you that there's at least one element in there. I'm going to choose wisely. It's not the case that all elements work, but this one happens, too. Why? Let's see. Put the identity element of this group in. There it is. You put the element you're interested in. You put it star with itself in, which is i squared is negative 1. You put it star with itself, star with itself in, i cubed is minus i. And you put it star with it. Well, folks, hey, you're done. In fact, by default, you have to get something that you already got because there's only four elements in the group, and you've just written down, which is all k. So here is a cyclic group. It's got four elements. So it has to be, so this group k is isomorphic to, whatever the cyclic group is on my list of all cyclic groups that happens to have four elements, in other words, z4, meaning that there is a way of labeling the elements of z4 and labeling the elements of this group called k. And I think I may have called it something slightly different in the homework sheet. That's all right. So that when you write down the group table of each of them, they are indistinguishable that you end up getting the same group table. Questions there? Good deal. OK. Let's see. You got that? Got that? Yeah. Let me conclude with one last property of cyclic groups. And that's this. What will be of interest are questions of the following flavor. Suppose somebody hands you a group. Suppose somebody hands you a subgroup of that group. Question is, if you know the group has certain properties, is it necessarily the case that the subgroup has the same property? So the question phrased in sort of mathematical ease is, if you have a property of a group, is that property necessarily inherited by every subgroup? Well, let me give you sort of an obvious one. If the original group is a billion, then every subgroup has to be a billion. Because you're starting with the information that you take any two elements in the entire group and you do them in this order, then you get the same as doing that. So the question is, does that same property necessarily held in the subgroup? Well, yeah, you're taking the same elements. So the property of being a billion translates to subgroups. Property of being not a billion doesn't necessarily translate to subgroups at MITRE. It might not, depending on which subgroup you look at. A good example is this group of four matrices. The group of matrices, GLNR, is not a billion. But the group consisting of these four matrices happens to be a billion. So the property of being not a billion doesn't necessarily. So that's the sort of idea. Here's a question. If the group is cyclic, and I hand you a subgroup, is the subgroup necessarily cyclic? The answer turns out to be yes. So this is actually a relatively deep result. Proposition, again, I'm not going to give you the details of the proof, because I think our time is better spent looking at more topics or more ideas. But the proposition is this. The property of being cyclic is inherited by subgroups. So I've given you the informal, but the way that the mathematicians would speak language. In other words, i.e., if G is cyclic and H is a subgroup of G, then necessarily H is proof omitted. So every subgroup of cyclic group is necessarily cyclic. And folks, that's definitely not an obvious statement. Not an obvious statement. But again, I'm going to leave out the proof just for what it's worth. The flavor of the proof is essentially this. Take your cyclic group. So there's at least one element in there called G with the property of that. Every element in the group looks like this particular element raised to some power. In other words, the powers of G, positive, negative, or zero, gives everything in the group. Now what has happened is, presumably, you've been handed a specific subset of the group, namely a subgroup. And what you need to do is find inside that subgroup a special element with the property that everything in the subgroup is if to some power. And the algorithm for finding it, something that works, finding a generator for the subgroup is, all you do is you take the original element that generated the entire group, and you start asking the following question. Is the original element itself in the subgroup? Well, if it is, then the subgroup is the original group. And you're done. Cyclic, the original group is cyclic. So that was easy. All right, now, if that's not in the subgroup, then you look at G star G. Either is or it isn't. If it is, what you can show is that G star G is a generator for the subgroup. And that takes 20 minutes for the work, which you can do then. And if that's not in, then you look at G star G star G. Eventually, you have to get something in there. That's the idea. So what that means is, as a consequence, and this is something that we somewhat mentioned in the number theory course, as a consequence, the subgroups of this cyclic group are of the form N, Z. In other words, if somebody asks you to pick a subgroup of Z, then the punchline is, if you've been asked to pick a subgroup of Z, you can automatically view the subgroup as a cyclic subgroup. Because the original group, Z is cyclic. Every subgroup is cyclic. So, well, what does it mean to be a cyclic subgroup? It means that you've got a special element in there with the property that when you look at all of its quote unquote powers. Questions, comments? Good, this is a good place to call it quits then. If you happen to have the homework that's due Friday, if you happen to have it with you today, you just want to turn it in today, that's fine. And we'll take it. Otherwise, it is due Friday at 4.30, 5 o'clock. I don't care. You can just turn it in. Right afternoon sometime. SI session with Jen tomorrow, 4, 15 to 5.30. And sorry for any misinformation. And folks, at this particular time, please don't make a special trip to campus to turn in your homework. If you want to fax it in, email me. I'll give you the fax number or it's 262-3605. I'm not writing that down on the board, Karen, for public consumption. 262-368.