 There's various incendiary properties associated with groups. What we're going to look at tonight is the first of many instances where we've presumably looked at some groups. We've presumably worked pretty hard to show that the given binary operation actually gives a group structure. We've actually checked that we've got a binary operation, that it's well-defined if that's an issue, that presumably it's associative, there's some sort of checking that goes on there, that there is an identity element and that each element has an inverse associated with it in the given set. And what we will do is somehow produce additional groups or new groups from what I'll view as the known groups or the groups that you actually build from scratch. So there's some groups that are built from scratch, like the integers together with addition. Built from scratch means, all right, we know what the set is and we know what the operation plus means, but we actually pounded through all of the details to show that that set together with that binary operation forms a group. It is a binary operation that's sort of property zero. It is associative, there exists an identity element and for each element in there there exists an inverse. The first type of, I don't know, construction in which we take groups that we've presumably already constructed and somehow build additional groups from those known groups is what we'll get to tonight. And that topic is usually referred to as subgroups of groups. And the idea is once you've somehow built or you've somehow verified that a specific set together with a specific binary operation is a group, you can often find many additional examples of groups living inside the given group. So the idea is that sometimes inside groups there are more groups. I'll give you an example, you look at one of these for homework. Example, here's a group. The group is the integers with addition, z with plus. I'll continue at least for another day or two to indicate the group z with plus. But after that, folks, I'm just going to start writing the group is the integers. You might say, well, you haven't told me what the binary operation is. But, yeah, if I've told you that the integers is being viewed as a group, there's only one operation for which it'll be a group, namely addition. It's not a group under subtraction because subtraction is not associative. It's not a group under multiplication. Well, because very few elements have inverses. Like three doesn't have an inverse in the integers because one-third's not in there. So multiplication doesn't work, division's a disaster. So there's really only one operation that will put a group structure on z. So sometimes and very more often throughout the semester just call that group z. Then here's the subset, let's call it s. Just a subset of z, s is the collection of even integers. Two and, in other words, the even integers. So this is a subset of a group. And what you actually showed for homework is that not only is this a subset of a group, this set forms a group in its own right using the same binary operation. Then this set, which is typically denoted by 2z with plus, is a group. Shown for homework. And the notation is this is usually referred to as 2 times z. And so what we've done is we've written down an example of a subset of a known group that happens to be a group in and of itself. And what you did for homework is you actually checked by writing down each of the, what do you want to call them, three properties or four properties. I guess in the homework question you only had to check three properties because you were told that if you look at the even integers together with addition that it gives a binary operation that was given in sort of the preamble to the question. But presumably if somebody just hands you a subset of a group and says, is this a group in its own right? You'd have to first check that it's a binary operation and then you'd have to check the three axioms for a group. The word or the verbiage that we use is we call this set as a subgroup, subgroup of z. And subgroup simply means, means if g is a group, so somehow you've gone through all the torture of checking each of the axioms for g, then, oh and let me be more specific, g with some specific binary operation as a group, then for a subset s we call s a subgroup, subgroup of g in case. Well look folks, you've already got this binary operation defined on the entire set. In other words, for any two elements in g it makes sense to smash them together and get something else in g. We're going to call s a subgroup in case when you look at that subset and you use the same binary operation on the subset as you've used in the entire group in case that is a group in its own right. And we call it a subgroup. I mean the notation makes perfect sense. It's a group that is a subset of another group. So here's a good example. 2z is a subgroup of z, or the operation in addition. It turns out we've seen lots of examples of subgroups. Example, here's the group. g is what we call the general linear group. Pick your favorite value of n. Look at the collection of n by n matrices over the real numbers that have non-zero determinant. So equal the set of n by n matrices with coefficients in the real numbers having non-zero determinant. We spent a while in class showing that this is a group. Here's a subgroup. The subgroup s equal to the collection of let's call it x in g the property that the determinant of x equals plus or minus one. You showed for homework that that is actually a group under the same operation where the operation is understood to be matrix multiplication. Then s is a subgroup of g. A subgroup of g. This was also done, well for homework and I set you up in class as to how to go about doing this. Okay, here's some notation so that I don't have to keep writing out the word subgroup. If s is a subgroup of g, we write, well look, s has to be a subset of g in order for it to be a subgroup of g. Just to avoid continuing to write out s is a subgroup of g, we write s. We take the inclusion sign and we write what looks to be a less than or equal to equal sign, but you're thinking of this as sort of the inclusion becomes sharp and put a point on it. So if I write s less than or equal to g, it doesn't mean less than or equal to, it means it's a sharp subset. It's some sort of special subset, special meaning that it's a group in its own right. Okay. So here's the question if you are working in something that you already know as a group, presumably that's where we're starting, and you're looking at a subset of something that's already known to be a group. Well, you're not just sort of grabbing something out of the ether and asking whether or not you've got a group structure. You're now working inside some sort of concrete structure that you've presumably already checked as a group. So the question is, does it buy you any sort of efficiency knowing that the set that you're looking at lives inside something that's already known to be a group in trying to determine whether or not that subset, together with the binary operation, is actually a group in its own right. And the answer turns out to be yes. So here's what we usually call the subgroup theorem. It says this. If g is a group, so let g with star be a group. A group. And folks, that's a pretty strong statement. It means somehow you've already done some checking that allows you to conclude that whatever the set is together with whatever the proposed binary operation is actually satisfies the four axioms, that this really is a binary operation on that, that it's associated, that an identity element exists in g, and that for each element a there exists an inverse in g. Then the point is this. Then a subset s, and you'll notice here I haven't used the sharp point, I simply viewed s as a subset, is a subgroup in case these three things are true. First, s is what we call closed. That's a word that came up in the homework assignment or at least in the text. In other words, i.e. for every pair, let's call them s and t in the subset s star t is in capital s. That's what closed or closed under star means. This is maybe the more formal way to put it. Instead of just saying closed, closed under whatever the given binary operation is, but there's only one binary operation floating around here so we don't really need to worry about lack of understanding here. Closure just means if you take any two things in the subset and you do the binary operation, which at least guarantees that you return something back in the subset, what we're requiring is that what you get is something actually back in the subset. Secondly, whatever the identity element for the group is, lives in the subset. And third, if you pick something in the subset, let me write it this way, for every element of the subset, then a inverse is also in s. Here's what the subgroup theorem says. If you want to check whether or not a subset of something that's known to be a group is actually a group in its own right. In other words, if you want to check whether or not a subset is actually a subgroup, well, all you have to do is check, presumably, all of the axioms for a group structure. But what the subgroup theorem says is that you actually don't have to work that hard. For example, you don't have to check, let's see, there's no mention of associativity here. Why? Because if you already know that you're in a group, you know that the binary operation is associative, so you don't have to go about checking that again, because you're inside something where you... Secondly, you don't have to worry about trying to produce an identity element. If this thing is a group, then you know there is a unique identity element already inside there. We call it e, so you don't have to worry about producing it or inventing it or building it. All you have to do is take whatever it is. It's already been identified because you know you're working in a group already, and simply check that it lives inside the subset. And third, you have to check that if you take anything in the subset, folks, you don't have to worry about trying to build an inverse for it. The inverse for it exists, at least it exists inside capital G. All you need to do is check that not only is it inside capital G, that it's actually inside the subset as well. So even though it looks like you still have to check some things, what's required to check that a subset is actually a subgroup is significantly less work than actually trying to show some given set with some given binary operation as a group to begin with. So let me just give you a quick example of this, and then we'll do a bunch of examples of showing some things as a subgroup. For instance, when we showed that this thing was a group, the collection of n-by-n matrices having non-zero determinant, what do we have to do? Well, we had to haul out a 313 result that said if you take something in this set, its inverse exists, and the reason that its multiplicative inverse exists is because there's a theorem that says if you take a matrix with non-zero determinant, then you can somehow go about building its multiplicative inverse. It's pretty hard work. You work really hard in your math 313 course to prove that. If I was to look at some subset of this group, well, here is one, then you'll note in order to show that this subset is actually a group in its own right, in other words, is a subgroup of G, I don't have to worry about hauling out these big theorems anymore. All I have to do is make sure that if I take something in this set, not that its inverse exists. I know its inverse exists because I've already checked that this thing is a group, but simply whether or not its inverse lives in the given subset. So that's how this list of things is much easier than showing something as a group from scratch. So proof, I'll omit it, but I've given you all the ideas. Omitted, the idea is simply that is that all the remaining work, like showing associativity and showing that inverses exist, et cetera, et cetera, et cetera, has been done in showing, or somehow verifying or somehow knowing that G, the big set with stars, is a group. So what you'll do for homework, couple of problems at least, is you'll be handed a group, you'll be handed a subset of the group, and you'll be asked to show that the subset is actually a subgroup, and the way you'll go about doing that is not to start from scratch, but rather to always haul out the subgroup theorem and simply say, alright, there's three things I have to check. Here they are. So let's do an example. Example, let's look at GLNR. So that's the group G. We've, by working relatively hard, we showed that G is a group. You'll notice that I'm already being cavalier with the notation. Technically I'm supposed to tell you what the operation is, too. You're thinking, well, there's two natural operations here. I can define a binary operation instead of n by a matrix. He's by adding them to get another n by a matrix back. Or I can define a binary operation by multiplying them to get n by a matrix back. Yeah, but the issue is, when we're looking at a set like this, it turns out that the collection of matrices having non-zero determinant isn't a binary operation under addition. You can add two matrices having non-zero determinant and get a matrix having zero determinant. In fact, that's a pretty easy thing to do. For example, 1, 0, 0, 1. The identity matrix has determinant 1. Minus 1, 0, 0, minus 1 has determinant 1. There's two matrices that have determinant 1, but if you add them together, you get 0. This set isn't even closed under addition. When I don't stipulate what the operation is, typically folks, there's only one natural operation that has a chance of making a given set of group. I'll typically start blowing that off. That's exactly the same sort of comment as I was making with the integers under addition. Take the integers. Even though there's lots of different operations, there's only one that turns it into a group. Karen, question. I'm sorry, ask the question again, Karen. Yeah. Yeah, this is by definition the end-by-end matrices. Yeah. Okay. So there's a group. Now let's look at the following subset. As is the collection inside g consisting of the following matrices. The collection of matrices, let's call them m with the property that the determinant of m is bigger than 0. So I want you to look at the collection of matrices having positive determinant. All right, well that's certainly a subset of this, because this is the collection of matrices having non-zero determinant, either positive or negative. I don't care. So this is a subset of g question. Show or prove that s is a subgroup of g, and you'll notice I'm using this notation now, this sharp notation, prove that s is a subgroup of g. Well, here's what the subgroup theorem says. It says in order to verify that this is actually a subgroup of g rephrased that s is actually a group in its own right, here's all we have to do. We have to show that it's closed under the given operation, that the identity element of the group is actually in the subset, and that if we take an element in the subset that it's inverses in the subset as well. So let's do that. So we'll always use the subgroup theorem. So here's what we do. It's three steps. Step one is pick any two things inside s. Let's call them m and n inside capital s. We have to show that when we do the binary operation, and the binary operation is multiplication, so you want to write a little dot in there, that's fine, otherwise I don't really care. You write m next to n, it implies that you're multiplying them, so that the product is in the subset. Well in this particular case, what does it mean for the product to be in the subset? Something's in the subset, if you can convince me that it's determinant is positive. So all I have to do is convince you that it's determinant is positive. But let's see what we're told is the determinant of m is positive. The reason we know that is because we're told that m is in the subset, and we're told that the determinant of n is positive, that's the definition of the subset. So let's compute the determinant of m times, let's give it a, oh and a positive times a positive is positive. So check, we've just shown that m times n is in s, because in order to convince you that m times m is in s, all you need to do is convince me that that thing has positive determinant, we've just done that. So that's the verification of step one there. Any questions on what the heck I'm doing or why I'm doing it here? We just verified the first piece of the subgroup theorem on our way to trying to show that this subset of the group is actually grouping its own right. So let's do parts two and three a little bit more quickly. Folks, I'm not worried about the existence of an identity element. I know that there's an identity element in the given group, because I've already checked that it's there, and it turns out in this group to be the identity matrix of order n. So it's not that I need to verify that it exists. I'm never going to do some sort of computation that says i sub n times a equals a times i sub n equals n. I don't need to do that anymore. I've already verified it. All I need to do though is take that matrix and make sure that it has the right property that actually puts it in the subset. So for number two, we need to check that the identity matrix is in capital s. Check that the identity element of the group, which happens to be that thing. Folks, there's only one thing to do. You have to convince me that it's determined as positive, but the determinant of i n is one, which is bigger than zero, so i n is an s. This is something that students seem to have a little bit of trouble with. A lot of times, folks, in order to describe a subset of a set, you don't specify exactly what the form of the element is. What you do is you specify a property that the element has. Here, you'll see, I haven't told you what the thing has to look like. All I've asked you to do is tell me whether or not it's determined as positive or negative. As soon as you can convince me that it's determined as positive, then the original matrix in the group. These are two separate statements and they live in two different universes. This is a statement about real numbers. This is a statement about matrices. I'm trying to conclude the statement about matrices because I'm trying to show the identity matrix is in this subset. The way I conclude that is by making a statement about real numbers. Third, pick any, any, or every, let's call it a, inside the subset, show that it's inverses in the subset. Folks, the inverse existing is not an issue. I'm not going to multiply a times a inverse. I have to do that. I've already done that. I've already done that because I've already verified that the original set is a group. All I need to do is convince you that if you start with something in the subset that its inverse is in the subset. In other words, pick a matrix whose determinant is positive, that's what it means to be in the subset, show that the determinant of its inverse is positive. Showing something is in a set by verifying a certain property that the given thing has. Here the property is, is it a determinant positive? Well, but let's see. The determinant of a inverse is one over the determinant of a, that's a property from your algebra course, which is, if you take one over a positive number, you get something positive, so check. I'm showing that the determinant of a inverse is positive, so the conclusion is that a inverse is an s, and now if you want you can put a, a sort of finishing touch on what we've just done. We've just proved that this subset is actually a subgroup by using the subgroup theorem, so here's sort of a final line, s is a subgroup of g by the subgroup theorem because we've checked that each of the three properties hold. Questions, comments? Okay. So the big picture, the big philosophy is we start with something that's known to be a group and we look inside the known groups for more groups, and it turns out the subgroup theorem says, alright, you don't have to work as hard, once you've done all the hard work in showing something's a group, you don't have to work as hard to show a subset of a known group, it's actually a group in its own right. Let's do some more examples. Example, this is a key one, here's the group we're going to look at, it's the integers. If you're thinking what's the binary operation, the only binary operation that the integers is naturally a group for is addition. So whenever I write the group of integers that will always mean under addition, and here's what I want you to do. Pick any positive integer, I could do this with negative integers, I could also do it with zero, I guess we'll do that. So how about let n bigger than or equal to zero in z. So pick your favorite integer, I don't care what it is, but now it's fixed throughout the entire discussion, maybe it's four, that's fine. Then here's what I want you to do, look at this set, the set nz, that's the notation, it's all the multiples of that fixed integer. So typically that's written like this, z in capital Z, it's the collection of multiples of n. So for example we've already looked at a specific case of this type of subset, we looked at the case right at the beginning of today, where n equals two, asks you to look at all the multiples of two, better known as the even integers, that's a subset of all the integers, but we can play the same game if you want to look at all multiples of four, then that would be the set, you know, zero, four, eight, twelve, sixteen, blah blah blah blah and also include the negatives, negative four, negative eight, negative twelve, etc. Then here's the punch line, then s is the subgroup of z. What I've done here folks is I've made infinitely many different statements all at once. The set consisting of all the multiples of two, there's a subset, turns out to form a group. Now, completely new statement, the set consisting of all the multiples of three forms a group, it's a subgroup of z, completely different statement, the second, I don't want to have to prove all of those individually, but I can prove them all in one fell swoop, we'll do that here. Reason, subgroup theorem, it's fair to say folks that anytime you're asked to verify that a subset of a known group is actually subgroup, you'll always run to the subgroup theorems, alright, there's three things I need to do, always. Step one, show that the subset is closed under the given operation. Step two, show that whatever the identity of the group is, is in the subset. Step three, show that if you take something in the subset, that it's inverse is also in the subset. Alright, so let's do that, let's see property one, use the subgroup theorem, subgroup theorem, property one, here's what we need to do, pick any two elements, let's call them a and b inside the subset, we have to show that when you combine the two elements using the binary operation, and again I haven't explicitly written this down, but it's understood that the binary operation in addition here is in the set, that's what closure means. What does it mean to say that you've taken two things in the set? The set consists of multiples of n. So a can be written as n times, let's call it z one, and b can be written as n times z two where these two things z one and z two are integers, that's what it means by definition to be in this set, set of all multiples. Oh, so what do we need to do? We need to show that a plus b also has that property. Namely that a plus b is n times some integer, but what is a plus b? I mean the computation is often just trivial, and it'll be the case here too, and z two which is, oh, arithmetic n times z one plus z two, but the point is this folks, since z one and z two are integers, the sum of two integers is an integer, and so check. So equals n times some integer which means that we've shown that the sum a plus b is of the correct form so is in capital s check. So that's what it's required. We're required to show that the sum satisfies whatever the property is to put in the set. The property to put in the set is that you can write it as n times some integer, we've just verified that and we're done. Questions I wanted to just did there. I mean obviously the arithmetic is straightforward, it's just you need to verify whatever it is that takes an element and shows that it's in the subset and this equation does exactly that. Step two, I'm not worried about building an identity element. I'm not going to write zero plus blah blah blah, I already know that. I know what the identity element in the integers looks like, it looks like that. All I need to do is make sure that that element is actually inside the subset. In other words I have to make sure that I can write it in the correct form. Can I? Sure. What do you want to call this? Trivial arithmetic? That's all it is. Zero is certainly n times zero. It's not news to anybody. But the reason I'm interested in writing this equation is because now I've verified that the identity element of the group the identity element in the group z can be written in the correct form n times some integer puts it in the subset. Equals n times some integer. It happens to be the integer zero, but that's neither of you know there. So it is in s, check. Looks like I don't have to convince you that zero works. We already know zero works. The question is whether or not zero lives in the subset. We've just verified that it does. Property three, pick something in the subset a and s show let's see what is the inverse? The inverse of an element in the group of integers under addition, the inverse looks like it's negative. But let's see. a is n times little z1 for some integer z1. That's what it means to be in the subset. So now I have to convince you that minus a is in there. Well, that's minus n times z1. Folks, if you leave that you haven't convinced me that you know what's going on. You need to convince me that this thing is in this set. The things in this set look like n times some integer. In this form I've written it as negative n times some integer. No good. But it's easy to algebraically fix. This is n times negative z1. Of course it is. Now the point is I've written the thing in question as n times some integer and the reason that minus z1 is an integer is because z1 is an integer. Since z1 is an integer so this then looks like n times some integer and we're done. Check. There's the third step. So, what we've just verified by using the subgroup theorem is that if you hand me any integer you want and I've asked you to take something bigger than or equal to zero although that turns out to not be necessary. This works perfectly well with negative integers but we typically don't form those subgroups or those subgroups can be realized as the same collection as the corresponding positive integer then you get a subgroup. So there's lots of examples that get then kicked out of groups. Here's one. If you look at all the multiples of one that's the whole set. Any integer is a multiple of one. If you look at all the multiples of zero, in other words an n equals zero well there's not too many of those just zero. So that's sort of weird. What it says is if you just look at the set consisting of zero sitting inside the integers that you get a subgroup. Alright. And what? Yeah. If you just look at the collection consisting of zero inside the integers you get a group. Seems a little bit odd but it works out. What do you have to do? You have to show it's a binary operation. So take two things in there. Not very many choices. Zero and zero. So do zero plus zero you get something back in the set. Yeah. Cause you get zero. That's easy. Does the set have an identity element? Zero. That's pretty easy. If I hand you anything in there does it have an inverse? Well there's only one thing to hand you zero. Is there something you can add to zero to get to zero? So I mean it seems silly on the surface but it works out perfectly. Well in fact we'll make this more general coming in a minute. So there's an example. All the even integers form a subgroup. All the multiples of four form a subgroup. All the multiples of whatever integer you want form a subgroup. The reason I say pick n bigger than equals zero is if I ask you to write down all the multiples of negative two. So you know negative two times one, negative two times two, negative two times zero, negative two times negative seven. Because you get to multiply by any integer you want if you look at all the multiples of negative two you get the same set as if you looked at all the multiples of positive two. Because it's ranging over all possible integers. So there's really no issue in letting n bigger than equals zero although we don't typically have to. Here's the proposition. I'm not going to prove this for you but this one turns out to be relatively interesting. We've just written down inside the group of integers under addition infinitely many different subgroups. The multiples of two form a subgroup, the multiples of three form a subgroup, the multiples of four form a subgroup, etc. It turns out there's some relations between those like if you look at any multiple of four four, eight, twelve, blah blah blah hey anything in there is a multiple of two necessarily. So it turns out that that subgroup lives inside this subgroup which in turn lives inside the entire group and then I might like look at the multiples of eight. Those live inside the multiple of four which is so the subgroups sort of line up but in the end we've got infinitely many different ones and it turns out that's all of them. The only subgroups of the group g equals and I'll emphasize that z with plus here are of the form nz for some positive. So somebody says I'm thinking of a subgroup of the integers well there's infinitely many different possibilities as to what the subgroup might be but each one of them looks like the same form, it looks like multiples of some fixed integer. I'm going to leave the proof out it's certainly not beyond the scope of this course it would take us 20 minutes or something like that and I think our time can be better spent doing something a little bit different but just for those of you that have seen the number theory course the proof of this result uses the division algorithm for integers which some of you are familiar with. So proof omitted but uses the division algorithm uses the division algorithm. Alright that's not too bad. Questions comments? Let's look at some more examples. Example example oh yeah it turns out folks that if I hand you a group but I don't care what the group looks like maybe it's the integers, maybe it's this thing maybe it's the complex numbers under maybe it's the non-zero complex numbers under multiplication, maybe GLNR I don't care if it's a billion, not a billion makes no difference to me. Inside any group there are always two subgroups at least. So let me not call this an example this is just sort of an observation I'll call it a lemma it doesn't deserve the name proposition because it's straightforward. If G is any group then G always contains at least two subgroups contains these two subgroups one is well folks if you take any set you can view it as a subset of itself this is going to look stupid too but technically we got nope this. Any subset is a set is a subset of itself so the question is if I look at this subset does it form a group? Oh yeah because it's a group that's no big deal and secondly what's sort of interesting is if you just look at the identity element sitting by itself that one element set also forms a subgroup and the proof of this second one the first one is just check it is the proof of the second one is essentially what I just walked you through the question is it the case that this set satisfies these three things is it closed in other words if you take two things in there well there's only one thing in there so you've got to take it twice E and E if you do the binary operation E star E you get E we noted that on Monday so it's closed question is the identity element in there? Yeah it is it's there question if you take something in there is it's inverse in there well the inverse of the identity element is the identity element so it's its own so these are sometimes called the trivial subgroups this notation is not standard this author calls either one of these the trivial subgroup other authors would simply call this one the trivial subgroup other authors might call that the full subgroup or the group itself or something like that so don't get too caught up in the in the verbiage here but there might be one or two questions that ask you to say something about the trivial subgroup or the trivial subgroups or something like that okay let's look at some more examples of groups and subgroups here's a group here's a group it's actually a subgroup of GL2 R but I'm simply going to present this group to you as four elements and you can check that all of the group axioms are satisfied it's the group that I'm going to call and I sort of want to call it V but I won't let's just call it G for now I'll give a special name to something that looks very much like it probably by the end of today it's the following collection of four matrices one zero zero one minus one zero zero one one zero zero minus one and minus one zero zero minus one so there's a set of four matrices now admittedly this sits inside GL2 R because if you look the determinant of each of these matrices is non-zero that's got determinant one determinant minus one so in fact this collection sits inside that special subgroup that we looked at last but that's not the point to be made here it turns out that G is a group in its own right G is a group you want to check that we can use the subgroup theorem because this thing oh sorry I have to be a little bit more careful G is a group where the operation is multiplication when I've indicated that it's a subset of GL2 R I've implied that the operation that I want to use is a group sure we can check that just pound out one of those tables there's only four elements or use the subgroup theorem subgroup theorem says to check that it's a group you just got to check that it's closed in other words if you take any two things in this set and you multiply them together do you get something back in the set sure I mean you just got to check but how about this times this I mean that times anything is going to be back in the set that's not an issue how about this times itself yeah that's that how about this times this so yeah that's that you just go through all the possibilities in fact let's write out one of those tables let's call this thing E because it's the identity element I'll call it a capital letter because well matrices are usually referred to as capital letters E, A, B and C and let's write out one of these tables we have this opportunity because there's only four elements in this set and see what it looks like E, A, B, C and let's just do the operation well let's see E is the identity so as we've seen before if you multiply E times whatever the other element is it doesn't change the other element and it doesn't matter what order you do it in so when you're doing these tables let's take forward to write down the first row and call them all right now let's see if I do A times A in other words if I multiply A times itself well look what's happening here is just only the main diagonal so if I do minus 1001 times minus 1001 I get minus 1 times minus 1 in the upper left and I get 1 times oh so that's E and how about A times B A times B so that's a minus sign there and a minus sign there so A times B is C and let's see A times C what do we get here A times so that times that gives a positive 1 in the upper left and a negative 1 in the lower right and that's B let's keep going real quick B times A well now you got to be careful you might say well I already computed A times B does that help me with B times A sometimes it does if the original group is a billion but of course the collection of this thing sits inside is itself not a billion so you got to be a little bit careful so I'll do B times A oh but it turns out that these matrices actually do commute with each other in other words B times A is C I checked that just do the multiplication and how about B times B if I multiply this thing times itself we get E and I'll let you sort of fill in the rest of the table here let's see C times A C times A so I do that times that when I get B if you're not seeing these multiplications just sort of write them next to each other and count them out it's not too bad C times B turns out to be A and C times C turns out to be E so there's the group table for this group and I want to use this opportunity to play up an interesting property of group tables if you've written down the table of a group of a group not just some binary operation but something that you've shown as a group what will always be the case is that every element of the group appears in each row and every element of the group appears in each column in some order what that also means is that by some pigeonhole principle is that you can't have any repeats in any rows or columns so you see E, A, B, C, A, E, C, B oh and the same is true with rows that property is typically called the cancellation property so note turns out in any group table that in any group G G we can cancel on either side here's what that means in other words if you're in a group and you hand me three elements let me call them X, Y and Z if I hand you three elements of the group and I've somehow computed X star Y and that equals to X star Z so think what that would mean folks if I'm in the X row this would mean that presumably I've computed X star Y and I've gotten an answer and then I computed with that same element X star Z and gotten an answer if the two answers are the same then in fact the entries had to be the same in other words they can't happen in two different columns that can only happen if you pick the same column twice then Y equals Z and the proof is pretty easy proof this is why it's nice to know that there are inverses for each element in a group if I hand you this piece of information and you want to try to conclude that Y equals Z it's not too bad to do just take this equation and combine both sides on the right with X inverse which exists because I'm living in a group X star Y equals X inverse star X star Z why is that the case? because this equals this so I can combine with any element in the group I want I happen to choose X inverse but what does this say? X inverse star X star Y is then X inverse star Z Y associativity law so associativity and now you see how we end up here X inverse star X E but E star Y is Y and E star Z is Z so check so what we get in a group and this is one reason why it's nice to be working in a group is you can't take a single element and multiply it by two different things and get the same output this is another reason why we like to avoid zero when we're working with groups at least under multiplication this is certainly not true if you look in the set of let's say all rational numbers all rational numbers why because zero times one equals zero times seven does one equals seven? No but hey if I told you that X is not zero then the point is you can multiply both sides by the inverse of it and conclude that the original expression is equal to what the group says at least this cancellation in a group says is that if you're going to write out the group table you'll necessarily see each element at most once and in fact then what that tells you is you'll see each element exactly once as long as the set is finite by some sort of pigeon alright so there's some groups let's look at this group sort of interesting it is sort of interesting you know we've seen actually we've seen two other groups each of which have four elements already so go ahead and call this one g sub one I'm just going to use this notation for tonight and then we'll rehash notation when we get back after Labor Day so here is a group it has four elements in it here's another group that has four elements in it let's see it's the group oh yeah one minus one i minus i there was a group with four elements in it we looked at that one a week ago Wednesday here's another group with four elements in it remember when we used this notation z sub n it means addition mod n so this is addition mod four and now here's what I'm going to do suppose I label the elements the elements of one of these groups using whatever letters I want using some collection of letters I don't know of letters how about let's call them e f g and h folks you know I don't have to call it one minus one i and minus i I could call that e f g and h or I could call that e f g and h or I could call this e f g h the specific letters or the specific names I use for the elements in a group is completely irrelevant I care what you call the thing but what I do care is how they go about combining with each other and suppose I tell you the group table now I present the group table and here it is e f g h e f g h I'm going to tell you what this thing looks like let's see this is e this is f this is g this is h this is f g and h let's see this is f plus f is g f plus g is h and f plus h is e or star here I should call this star g with f is let's see I want to put an h here and I want to put an f here I want to put an e here sorry and I want to put an f and then I'm going to put let's see h and I'm going to put an e here and I'm going to put oh wait a minute let's see remember if I'm going to hand you a group table what did we just prove we just proved that if you look at any row or column that each entry has to appear exactly once so if this is a group this is going to have to be f and this is a group this is going to have to be g so now I'm going to give you about two minutes and what I'd like you to do is answer the question here is the group it's a group with four elements which of the three groups that I've presented here is it so take about two or three minutes and do some computations if necessary and see whether or not you can determine how I've labeled the elements from one of these groups to get that particular table so you're thinking well let's see if I label them the way that I labeled them group in group one here if e f g h is really e a b c does that work out it doesn't look like it I'll give you a hint here e is whatever the identity element is and I can recognize that because I've taken whatever e is and I've combined it with each of the things and gotten the thing back in either order so if you're going to try to identify what e corresponds to well e either corresponds to zero if it turns out to be that group or corresponds to one if it happens to be that group or corresponds to capital E this group in about one minute we're going to vote to see whether or not you can recognize which group I've written down and if nothing else if you can you might want to use some sort of process of elimination too maybe you can figure out which group it can't be or which groups it can't be okay here we're going to take a vote we've got three things to vote for group one, group two, and group three this is g sub three so question how many people think that I have somehow relabeled the elements of group one and gotten that group table raise your hand okay so g one gets zero votes g one g two and g three you have to vote for one folks so g one gets zero votes how about g two how many people think that I've taken the elements of g two and relabeled them somehow to get that table hands up high here one, two, three, four, five, six okay how many people think that I've taken the elements of g three and relabeled them one, two, three, four, five, six, seven, eight, nine okay that's good so so they're about eighteen and the answer is everybody's correct it turns out this table this table represents both g two and g three look if I were to take the elements one minus one i and minus i if I were to call this e and I were to call this here it is if I were to call this f and I were to call this g and I were to call this h what I just mumbled under my breath we'll talk about after the break here it turns out if you were to label the elements this way then what you would get is oh I have to be careful yeah that's good if I were to label the elements this way then what you would get is exactly this table hmm of course on the other hand if I took this group and I were to label the elements this way what you'd get is exactly that table so here's a word that we're going to use for the rest of the semester the word is isomorphic iso means same and morph somehow means structure these two groups on the surface look completely different g2 and g3 look completely different this contains complex numbers the operations multiplication this contains whole numbers the operation is addition mod 4 but in the end folks if I label the elements correctly and I write down the corresponding group table the table is indistinguishable between the two the two really are the same group just express slightly differently here we happen to prefer the multiplication notation and we happen to prefer using complex number notation here we happen to prefer the addition mod 4 notation and we happen to be using whole numbers but when it's all said and done these two groups have the same structure meaning that there's a way of labeling the elements in the two groups so that when you write down the corresponding group table using those elements you're actually writing down the same thing now here's a question is it possible to somehow relabel the elements of this group to get that same table and the answer is no and here's why you see in these groups of course I really could say in this group because it really just represents one group and here it is it's the case that it's possible to find an element that works with the property that when you combine f with itself you don't get the identity one plus one is two i times i is negative one on the other hand in this group this group has the property that regardless of which element you pick if you combine it with itself you always get the identity element a times a is e b times b is e c times c is e so folks if you were to start relabeling this thing using whatever letters you want what's always going to appear down the diagonal is e because if you combine anything with itself you get e but the table that we wrote down that represents this group or these groups however you want to view that doesn't have that property so there's no way you can relabel that first group g1 and make it look like this because regardless of how you relabel things you're always going to see e e e e down the main diagonal so it turns out that that's one of many reasons so g1 is not isomorphic to those other two groups morphic to the groups g2 or g3 doesn't have the same structure as g2 and g3 it doesn't because here's the intuition every element in this group has the property that when you combine it with itself you get the identity there are some elements in this group or these groups however you want to view that that don't have that property so no matter how you furiously relabel the elements you'll never be able to line up with the same group table for these things so that's sort of interesting if I present to you some groups it turns out we may at least under the surface have presented you with the same group twice in effect just using different colors or you know I sort of written it using circles and squares and diamonds or I've written it using complex numbers and complex multiplication questions, comments I've given you a rough idea of what this notion of isomorphic is so that's good everybody in here nailed this question because there were two separate correct answers okay let me make one more comment and then we'll call it a night yeah okay if I hand you a group so given a group g given a group g what we can ask is for some sort of indication as to what all the subgroups look like question or request or task task find all the subgroups comment in general in other words if somebody just drops a group in your lap and asks you to complete this task this is impossible there's no algorithm that will allow you to take any group start with any group and somehow identify what all the subgroups look like so the fact that we could do that with z well it turns out we worked pretty hard I left out the proof but it turns out we can do it with z there happen to be infinitely many different subgroups but they're all of the same form they're all multiples of some fixed integer whatever energy you want but in some cases cases this is doable those cases are example if the group is z then we know what all the subgroups look like see previous proposition they look like multiples of some fixed integer there's infinitely many of them but they all somehow conform to the basic structure of multiples of some fixed integer another example of the situation where we're able to do this is if g is small small means not too many elements in it like 10 or fewer or something like that then what you can do is just roll up your sleeves and get your hands dirty it's alright what do the subgroups look like here's an example let's look at inside z4 and let's just start mucking around and see if we can identify some subgroups of z4 well we already know one in fact we already know two for free here's one how about z4 itself because any group is a subgroup of itself here's another one the other trivial subgroup we always get for free just the identity element setting by itself question are there any others well let's see suppose I have a subgroup suppose one is in that subgroup so if one is in the subgroup then because every subgroup is closed then one plus one has to be in the subgroup but a subgroup is closed so now that I have one and two in the subgroup then one plus two has to be in the subgroup so if one was in there I can then conclude that two is also in there and that three is also in there but the identity has to be in there because it's a subgroup so the point is this folks if we have a subgroup that contains one it's already on the list okay how about a subgroup containing two well let's see if I just look at two and I add two to itself I get two plus two which is zero and if I take zero and I add it to itself I get zero and if I take zero and I add it to two I get something back in the set so here's a possibility if I just look at the collection zero and two is that a subgroup? well let's check that it is is it closed under the addition? yeah take any two things in here be careful there's a little bit to check well one possibility is zero and zero is zero plus zero zero another possibility is zero and two zero plus two is two which is in there another possibility is take two and two but two plus two is zero so that's in there so the set's closed is the identity element in there? yeah there it is if you hand me anything in here it's inverses in there well if you hand me identity element it's inverses itself it's no big deal if you hand me two what's it's inverse? two because you take two what do you add it to to get the identity you add it to itself so this is a subgroup so here's another one we've identified how about if I have a subgroup that contains three and watch what happens the same thing as what happened when we looked at subgroups containing one if I have a subgroup containing three because it's a subgroup that's what we're assuming that means three plus three has to be in there but three plus three is two that means two is in there but then three plus two has to be in there but then the identity has to be in there if it's a subgroup so if I have a subgroup that contains one it's that if I have a subgroup that contains three it's that if I have a subgroup that contains two it might be that or that groups. And what we often do is we list them out in some sort of picture. This is called the subgroup diagram. And the subgroup diagram of a group, folks, is simply a way of presenting in sort of easy-on-the-eye fashion all of the subgroups of a given group in situations that you have the opportunity to do that, which isn't often, but when we do it we take it. And what you do is you list things out in such a way that the things that live below a given group are contained in that subgroup. So there are three subgroups here. The trivial subgroup consisting of zero sits inside this one, which in turn sits inside the entire group. What you'll do for one of the homework problems is you'll make the same sort of analysis, but instead of in Z4 you'll wind up doing that analysis inside Z sub 6. And in Z sub 6 the subgroup structure winds up being sort of amenable to this analysis where you just try to figure out what all the subgroups look like. Of course you'll get the two trivial subgroups, but inside Z6 you actually get two non-trivial subgroups. Turns out to be the multiples of two and the multiples of three. And when you draw the diagram it turns out the multiples of two turn out to be zero, two and four. The multiples of three turn out to be zero and three. Each of those is going to form a subgroup, but these neither one is contained in the other. So when you draw the subgroup diagram it'll look like some sort of diamond. Zero will be on the bottom, the entire group will be on the top and things will end up looking like that. So when you see that particular question coming up in the homework you'll know how to attack. Okay, questions, comments? Good. This is a good place to quit then. So officially folks there's no class next Monday because it's Labor Day. Please be safe and have a good holiday. If you have your homework I'll let you put it on the table here. Oh also please, which normally is supposed to happen on Tuesday, isn't going to happen on Tuesday because campus is closed or there's no school on Tuesday, but will happen next Thursday instead at the usual time, let's see what's the morning time, 9.15 to 10.30 or something like that. In EAS, 9 o'clock, 9 o'clock, 10.30 in EAS, 1.77. And what's the date next Thursday, September 6th? So that'll at least keep with the same rhythm of trying to get a couple of SI sessions in before the homework is due and then remember the homework will be due the next. Listen. Any questions? Under our guess or systematic method we arrive at this result by just replacement. In other words we didn't touch this problem directly. We didn't touch this problem in this direction we just guessed and then plug them in right and then afterwards let's extract the common factor from this equality. If the alpha t is the common factor then the remaining terms are alpha squared, 5 alpha, and then 6. And as you might remember in calculus 1 or 2 you learn that this function can never become zero. This is never become zero. That means the only possibility that the equality becomes zero is this becoming zero right? So this should be zero because this cannot be zero. So we now arrive at quadratic equation. Initially we had differential equation but suddenly we are handling a quadratic equation and by handling this quadratic equation we can find alpha. Once alpha is found we return back to our guess. We return back to our guess to determine the solution. This is it right? So after guess. So what are the solutions to this problem? Alpha should be, alpha should be negative 2 or negative 3 right? Alpha should be negative 2 or negative 3. If you don't believe plug them in then because of this we get the solution like this. So we get the solution like this and in chapter 2 we learn about systematic approach of doing these things and today let's stop here.