 All right then, what else do we need? All right, let's see. Remarks on the homework coming back. A couple of quick things. Let's see, this was section 13. In question 8, that's the one about phi of g is g inverse. Is this or is this not a homomorphism? And if it's not, tell me when it might be. Some of you said, well, in one case you get essentially g1 inverse g2 inverse, in the other case you get g2 inverse g1 inverse, and some of you just said, therefore it's not a homomorphism. Well, if it's not, give me a counter example. If you can tell me something's not true, then convince me it's not true. I mean, sometimes it is true. Let's see, did I have another comment on that? No. Number 47, that's the one with the order of the group is a prime. Convince me that any homomorphism from that group is either one to one or the trivial homomorphism. Most all of you had the right idea. What I was a little bit concerned with though is stylistically you were writing some very imprecise proofs. For example, I saw some of this. Therefore, kernel of phi is one or p. That makes no sense. Staples and oranges. This is a subgroup. That's a number. You can't have a number equaling a subgroup. That makes no sense. I sort of knew what you meant. You meant that, but you didn't say that. Similarly, what some of you would use is the article it. Therefore, it is one or it is one to one or it is trivial or it. I don't know what the heck it is. There's so many it's here. It could be a function. It could be a subgroup. It could be the number of elements in the subgroup. There were a lot of things. If you're going to use this word, that's best to never use this word in mathematics. Just tell me what it is. Is it the function? Is it the subgroup? Is it the number of elements in the subgroup? This was all just sort of style, but the style was un-good in many situations. Please be careful. This is clearly a style comment, but we talk about homomorphisms. Homomorphism is a function. So Phi is a homomorphism. We're eventually going to talk about the notion of two groups being isomorphic. What some of you were doing is putting this one here and this one here. You talked about a homomorphic. Homomorphic is just not a good word. If you use homomorphic, we talk about homomorphic images, but we're not going to talk about that in here. So if you're going to use homomorphism, it's ism and it corresponds to the function itself. When we talk about isomorphisms, we'll do that today. That's going to be a good word. And two groups being isomorphic, that's going to be a good word. But homomorphic is usually just not a good word. At least we won't have chance to use it in this particular course. All right, so please be a little bit more careful with your language, your style here. All right, I want to look ahead a little bit. So let me tell you how things will play out at least as far as the timing goes and then we're going to take a vote in here tonight. Essentially, the material that we'll finish up tonight will be the last material that will appear on exam two. So tonight we're going to do chapter 15 or section 15 and then we're going to go back and pick up some ideas in section three. And that'll be the material through exam two. I had listed on the sheet that I handed out on day one, exam two is happening a week from Wednesday. That's October 24th. Here's the way that the homework will, yeah, here's the way the homework will be scheduled. The assignment that I'm going to give you tonight will essentially just cover the stuff that we do tonight and a little bit of the stuff that we did last Wednesday. So this will be a relatively short assignment that I'm going to give you tonight. You have an assignment due Wednesday that should get back next Monday. The one I'm going to give you tonight will be due on a short turnaround time. We'll do next Monday instead of the usual rhythm of next Wednesday. All right, so everything that will be on exam two you will have seen through tonight and you will have done a homework assignment on and turn it in next Monday. So the question is, do you want to have that exam that Wednesday or do you want to put exam two off until the following Monday so that you might have time to digest that last homework assignment? I mean, as far as I'm concerned, it doesn't really matter. But let's go ahead and vote. So the vote will either be that we'll have exam two on the day it was originally scheduled for, either Wednesday, October 24th, sorry, October 24th or Monday, October 29th. Okay, anybody want to ask any questions before we vote here? Okay, again, the stuff that will be on this exam will essentially be through tonight and you'll have a homework assignment due on it, due next Monday on a short turnaround schedule. We can certainly have the exam Wednesday or the following Monday. So here we go. I prefer it on Wednesday the 24th. Raise your hands relatively high. One, two, three, four, five, six. How about Monday the 29th? One, two, three, four, five, six, seven, eight, nine, ten, eleven. Okay, so officially we'll have it here. I will hand out exam review materials possibly this Wednesday, but definitely by next Monday the 22nd. We can hand for at least a week and we'll be able to put together some SI stuff and comparisons under other things. Okay, let me write that down here. October 29th. Okay, so here's some, I don't know, maybe 20 minutes or so finishing up some of the ideas that we saw in section 15. What we were doing was looking at the structure of certain, excuse me, certain factor groups. So the idea was last time, looked at, given a group G and a normal subgroup of G, H normal in G, form G. Sometimes we say G slash H or G mod H or the factor group of G by H. There's lots of different ways of describing that particular group. What is it? It's the group of cosets, group of cosets of H in G. So the elements of this group are inherently sets, they're the cosets of H in G and we know how to combine them. We have a couple of different ways of viewing what that combination process looks like. You can either view it as just set multiplication. In other words, set combination or you can view it as A H star B H is A B H and we're guaranteed that that's well defined. What I want to do tonight is a few more computations with these particular types of groups. So example, let's look at the following. G to be the group Z 3 cross Z 4. So here's a group and we'll look at the subgroup generated by 1 comma 2. See if I want to do that. No, sorry. 0 comma 2, pardon me. So let's see what the subgroup looks like in here. It's, well, you can always, when you're looking at the cyclic subgroup, you write down the identity element of the group. Then you write down the element itself at 0 comma 2 and then you write down the element star with itself. Well, 0 2 added to 0 2 is what? Well, 0 0. So we already get 0 0 back, so there is the subgroup. So now we can form since H is normal in G, let's see, how do I know that H is normal in G automatically? Why, what do I know about G? G is abelian and we know that any subgroup of an abelian group is necessarily normal. So I don't really even have to worry about writing out all the cosets. I know automatically H is normal in G. So I'll put in parentheses since G is abelian. All this does for me, folks, is it guarantees that I don't have to sit down and check that the left cosets are the right cosets are. I don't have to check that G inverse HG comes for free, comes automatically. All right, we can form, form the factor group G slash H. And what I want to do is practice looking at the orders of various elements inside G slash H. So example, find, here's an element, here is an element in the group G slash H. And what I'm getting used to here, folks, is thinking the things inside the factor group are cosets. So when I write down an element, what I'm about to write down is a coset of H sitting in G. Well, here is H. If you want, I could write it out, but it's easier just to call it H because that's what it is. And what I need to do is star that thing or form the coset with, I don't know, I'm just going to pick an element. How about 1, 2? Now this is a little bit new tonight. What we've been doing is just writing the element of the group next to the subgroup, or in some instances we've been writing little G star H. But in a situation where the binary operation in the given group is usually denoted by a plus, it's typical or it's common to denote the coset by using the plus notation because that's how you go ahead and build the coset. You take whatever the element of the group is and now you're going to combine it with, we'll hear the operation in the given group happens to be additional, you're going to combine it with everything in the subgroup. So here is a coset. Now what I could do is write it out, let's see if I wrote out this coset, you simply combine this thing with everything in the subgroup so I get 1, 2, and let's see what else is in the subgroup. 1, 2 plus 0, 2 gives what? 1, 0. So there it is. Now the following question is reasonable, find the order of this element. Order of this element in the group, G slash H. Now let me sort of preclude a possible question and hopefully steer you correctly on the types of homework questions that'll come up on the stuff that I give you tonight. This word order is a little bit unfortunate, we've wrestled with the issues surrounding this word before. Order, on the surface means two things, if you drill down a little bit deeper the two things turn out to be identical, but order either means how many elements are in the set or it means if you've got an element inside a group, it means if you take that element and you start doing the binary operation to itself, how many times does it take you to get back to the identity element of the group? Well look, I'm asking you for the order of an element. So the way I've asked the question, I've asked you to tell me how do you, well I've asked you to interpret the word order in that second meaning, I've got some element of a group, just take the thing and keep combining it with itself until you get to the identity. Here's what students often misinterpret that word as, as soon as you see this you're thinking, oh how many elements are in some set? And what's unfortunate in this particular context is the element that you're looking at is a set. So what students tend to do is say, oh the order is two because I'm looking at a set with two elements. Not the correct interpretation. You're supposed to view this thing as an element inside some group. Okay, I can do that. And now find its order. I.e. how many times does it take? Times does it take to what? To take the given element to perform the binary operation. The binary operation. Well what's the binary operation? It's the binary operation in this group in G slash H on the element 1, 2 plus H until we see the identity element. We see the identity element of the group of G slash H. Well I know what the identity element of the factor group is. In any factor group it's always you take the identity of the group and you add it to the subgroup. In other words the identity element is always the subgroup itself. Identity is always just H in G slash H. So what we're asked to do here is take this particular element and start combining it with itself and keep going until we see the identity element. So let's go ahead and do that. So I take 1, 2 plus H and I combine it with itself. There's a lot of pluses happening here. This is the plus happening inside G. This is what's allowing you to form this coset. There's that same coset. Of course it's just an element of the factor group. This plus is the binary operation inside the factor group. So I'm asking you to combine that coset to that coset. So how do you do that? Well I know the definition of the operation inside the factor group. You simply combine the two coset representatives. Now here the two coset representatives happen to be the same element. So it makes the computation a little bit easier. Let's see if I add this to that I get 2 inside Z3. If I add this to that I get 4 inside Z4 which is 0. So here's the question. Is this equal to H? Well I'll tell you what it is. It's 2, 0. If I take 2, 0 plus 0, 0 I get 2, 0. If I take 2, 0 then this gives what? 2, 2. So that's not equal to H. So what this says is that the order of that element is not 2 because I combined it with itself and I didn't get the identity. Let's try another one. How about 1, 2, here it is. This is going to get tedious as you can see. 1, 2 plus H. Let's do it three times. I mean I'm writing this out just for emphasis. This is what the question is asking you to do. It has nothing to do with how many elements are in the coset itself. Equals what? Let's see. I already did this chunk of it in the previous line. So let's use that information. That's 2, 0 plus H. So I've combined these already folks. Just take the information that you've already learned in the previous line. Plus 1, 2 plus H. So what do I get? Well you combine the cosets and you get 2, 1. 2 plus 1 is happening inside Z3. So I get in the first coordinate 0. Because that's 2 plus 1 happening inside Z3. And the second coordinate I get 0, 2. Or 0 plus 2 which is 2 plus H. Is that H? Equals H. A couple of ways I know that. One is just pound it out. Take 0, 2 and add to everything in H. You'll get 0, 2 plus you'll get 0, 2 and then here you'll get 0, 0. So you get admittedly in different order, but order doesn't matter. Because we're worried about this thing as a set. You'll simply get, if you pound this thing out, H. Or the approach I'd prefer you to take is, this is in the subgroup. And whenever you have something that's already in the subgroup, it plus the subgroup has to be the subgroup again. So by default, as soon as I see an element as the coset representative that already lives in the subgroup, I'm done. Okay. So now let's answer the question because we've got enough information to. So what is the order of this element? I'm going to start with 1, 2. Is what? 3. Why? Because I took the thing and I combined it with itself twice. In other words, I did it plus it. And I didn't get the identity of the group, but then I combined it with itself three times and I did get the identity. Well, I said, but there's only two elements in there. Yeah, so. But I already knew that. I knew that way back when, three weeks ago. If the question was how many elements are in this coset, the answer is two for every coset that you pick. Because any two cosets have the same number of elements and they have the same number of elements as H. But that's not the question that's being asked. All right, questions? Comments? Dan, a question. Could it be 0, 0, 0, 2? How did I get from here to here? By definition, H is the subgroup generated by that element. So by definition, you write down the identity and then you write down the element itself and then you keep going. But when we kept going, if we did 0, 2 and added it to itself, that was already 0, 0, so we were done. Other questions, comments? Yeah, Lindsay, a question. In G, H is H, is that always true? That's always true. Yeah, this is true for any factor group in E. Because remember, when we proved that G slash H is a group, the identity element looked like that, E, H. E was the identity of the group. Technically, there's whatever the binary operation is in here, but we've sort of quit putting that down. But in the end, when you do E, H, you always just get H back because E is already in the subgroup. Can we just show the element like the 1, 2 with itself? I mean, is it true that we can only just, only have to do just the 1, 2 over there? Oh, that's a good question. And the answer is, that's a great question. Here's why. Okay, so here's a good aha moment. I think, if I understand the question correctly, you know, you can ask, pick just that and ask for its order. So here's what we're going to do. I'm not going to write this down, but I'm going to talk through it. So when it's this question is, well, just find the order of that. Okay, well, what is the order of this inside G? Well, if I add it to itself, I get 2, 0. If I add it to itself again, I get 0, 2. I don't get 0, 0. If I add it to itself again, it turns out that thing inside G has order 6. So the punchline is no. If you can compute or you somehow know or you're interested in the order of the coset representative sitting inside the original group, we could work for another two weeks and come up with some convoluted formula that tells you what the order of the element in the factor group is compared to what the order of the element in the original group is, but no, they're definitely not the same. This one has order 6 in G, but that has order 3 in G mod H. Yeah, so be careful here, folks. You absolutely do need to drag this H along here in general because here, you know, as I just mentioned, this thing sitting by itself in G has order 6, but this thing is an element of the factor group only has order 3. It turns out the order can only go down. It can't ever go up. If you have an element of order something, order p or something like that, p is bad, order r inside the group G, then the order of the element has to not only go down, but has to actually be a divisor of the original one. But it's just not worth the effort to try to figure out what the exact relationship is. Yeah, Susan? Along that same line, could you just take the one element 2 and combine it with itself until you get an element in H? Yes. So the answer to that question is yes. So Susan's question is, suppose you wanted to somehow compute the order of the element in the factor group, could you do it by just considering computations involving the element inside the original group? And the answer is yes. And it boils down to how many waxes it takes, not for this thing to necessarily become 0, 0, but simply to wind up with something in the subgroup. Yeah, exactly right. The first one you get is all you need, because any element inside H, I don't care what it's called, is going to generate the same coset as eH, because 2 lives in the same coset to begin with. So yeah, if you wanted computationally, what you could do is simply say, all right, take this thing, keep beating on it until you see something inside the subgroup, and that's exactly what happened here. It took three waxes to get something inside the subgroup, so it's order three. Good. So this is a great example to write out just to indicate that we're not looking at the number of elements in the coset or anything like that. It's simply the same definition of order as we've always been used to. Now the computation is a little bit more cumbersome because now you're taking a coset and you're combining with itself. Now there's still some, I mean there's some arithmetic that we still know that's still perfectly valid in these cosets because these coset groups are just groups. For example, let's see, I know how many elements are in this group. The coset group G slash H, the original group GS12, the subgroup happens to have two elements. So the number of cosets, so what we call the index of G and H is 12 divided by 2, which is 6. So if I were to write out all the elements of the factor group G slash H, I'd be looking at six elements, a group of six elements. So hey, if I have a group of six elements, what does Lagrange's theorem say? It says that if you're computing the order of any element, that number has to divide 6. So at least the number that we got here satisfies that possibility. Remark, if I had taken an element and I had combined it with itself twice and didn't see the identity or three times and didn't see the identity, in fact I'm done because the order of the element has to divide 6 and if it's not 1, 2, or 3, then the answer is 6. So you might want to play those sort of Lagrange guarantee games when you're doing the homework saying alright, I'm beating on the saying that I'm already passed a certain point. Therefore by the theory I can conclude that the order is something, maybe 6 here, let's just make sure by continuing to pound things out that it actually agrees with what I get in my computations. That might be a good way to convince yourself that you're doing the correct sort of computations here. Good one. Other questions? Comments? Okay, let me show you a slight shorthand notation. If you're comfortable with it, go ahead and use it. If you're not comfortable with it, you can continue to use this sort of coset notation rather than continuing to write out something like that as the coset. Sometimes what we'll do is simply write sometimes written as just G with a line over it. So for instance here when we were talking about the element 1, 2 plus H can be written as you just take the element 1, 2 with a line over it and the interpretation is that the subgroup's been determined, it doesn't change throughout the discussion. You've written it down somewhere, it's understood what the subgroup is and the line simply means you're taking this element and you're writing down the coset that it represents. So that when you start doing the computations, so okay, so this is this coset combined with this coset, well I know how to do that. It's the coset 2 comma 0. Why? Because the computation of combining cosets is you simply take whatever the coset representatives are and you do the combination in there and here because I'm adding in Z4 in the second component this just turns out to be 2, 0 bar. Of course you know I can write out what that is and we wrote that down over there. It turned out to be 2, 0 comma 2, 2. Whether or not that particular coset is of interest to me it probably isn't. Typically all I'm interested in is whether or not I'm looking at the identity which is easy to determine in this case. It's simply a matter of deciding whether or not the element that you cranked out there is in the subgroup. So that's just some notation you'll typically see that. I think the author of this text uses it relatively often. But at least when I was learning this stuff I was just more comfortable pulling this old notation along because it sort of kept front and center what it is I was working with, working with cosets. Alright you've got to be a little more careful and my computation is a little more cumbersome. Alright good question so far. Other questions, comments? Alright let's see. Yeah so the examples, all of the examples that I've written down so far of a group and a normal subgroup and then the formation of the corresponding group of cosets or factor groups whichever notation or rubies you prefer has always started with the group G being a finite group but all the computations and definitions that we've made for computations inside these coset groups are perfectly valid even if the group is infinite and in fact there's a very important example of a situation where you write down an infinite group and a subgroup of it which also happens to be infinite where when you look at the factor group you actually get a finite group and you actually get a group that's relatively familiar or something that we'll be able to recognize right away so here's an example of the situation where G is infinite G is infinite is perfectly valid still is okay as far as forming factor groups go forming factor groups the issue is the same you start with a group G you write down a subgroup of it you have to make sure the subgroup is normal but the definition of normal doesn't depend on whether or not the group is finite or infinite or anything like that and then form G slash H here's a good example example your favorite infinite group the group of integers of course under addition that's the only legit binary operation that turns this particular set into a group obviously there's infinite things in there let's look at a subgroup how about H is the subgroup cook one up 4 Z so I'm multiples of 4 let's see the group is a billion because it's just the integers under addition so any subgroup and we prove that if you hand me any fixed integer here I've handed you 4 that the collection of all multiples of it is a subgroup so that's not an issue we learned that a month ago or so because that's a billion this is by default normal so H is normal in G a little bit nicer H is normal in G so we can form the factor group G slash H now all sort of numerical bets are off here if you want to somehow interpret the number of elements in here as infinity divided by infinity or something like that but it turns out there's only going to be 4 elements in there so I'll tell you what they are let's see here's a coset the coset consisting of the subgroup itself well there's the identity element of the group so if I look at the coset consisting of itself I just get 4 Z I just get the subgroup back here's another coset yeah it's add 1 to everything in the subgroup so it's 1 5 9 I don't know minus 3 minus 7 blah blah blah blah it's more convenient to write it that way that's perfectly good notation it's just a coset here's another coset here's another coset that's all the cosets how do I know that? because I've captured every possible integer by writing down these 4 subsets every integer is either a multiple of 4 or one more than a multiple of 4 or two more than a multiple okay so let's see so I have this group group's got 4 elements here's a reasonable notation for the elements in this group instead of calling it 0 plus h and 1 plus h let's use this bar notation so these are each individually infinite sets they are these cosets but wait a minute I know that regardless of whether they're finite or infinite makes no difference happens to be finite and there's only 4 of them but this process forms a group and I can write out the group table let's see 0 plus 0 0 bar, 1 bar, 2 bar now the first row and the first column of a group table is always easy because you're just combining 0 alright let's do some more how about 1 bar plus 1 bar that's easy, that's 2 bar because I'm just combining 1 and 1 and let's see 1 bar plus 2 bar is 3 bar and how about 1 bar plus 3 bar 0 bar the reason being if I simply do the computation 1 plus 3 I get 4 and the question is which coset does 4 represent answer it's that one because 4 is in here let's do another one let's see 2 bar plus 1 bar is 3 bar 2 bar plus 2 bar is 4 bar which is 0 bar a little more interesting when 2 bar plus 3 bar you want to say 2 plus 3 is 5 so that's 5 bar but where does 5 bar live? right there it's the same as 1 bar 0 bar, 1 bar, 2 bar so there's the group table for this group that has 4 elements does that look familiar? yeah, what group does it look like? that one group with 4 elements is it cyclic? yeah, let's see if I look at 0 bar if I look at 1 bar 1 bar plus 1 bar 2 bar plus 1 bar so there is the subgroup generated by is all of g slash h so what we've shown is if we take the group g and the subgroup 4z g is z and the subgroup h is 4z and we form the factor group that what we get is a group that locks and talks like z4 we get a cyclic group having 4 elements in it here's a generator turns out 3 is also a generator 3 bar but in order to verify that this group is cyclic suffices to just write down a generator and that's easy to do 1 bar works and so let's see if we can together come to this sort of general conclusion based on this particular example I won't run through the details in general but in general if I start with this group if g is the group of integers under addition and h is the subgroup nz so multiples of this fixed integer n maybe n is 4, 10 or 2 I don't know just tell me what you want it to be then this group the group of cosets of nz sitting inside z it turns out there are n elements in that group 0 bar, 1 bar, 2 bar up through n minus 1 bar and this thing is in fact cyclic with n elements it has n elements in exactly the same way that z slash 4z had 4 elements you simply write out the numbers 0 through n minus 1 and that will generate all the possible cosets because then things cycle back it's cyclic because it will always be the case that if you look at 1 bar well 1 bar plus 1 bar is 2 bar which in turn gives 3 bar etc and you wind up generating all the elements in here in fact i.e. this thing z slash nz is isomorphic too well hey as soon as I've told you that you've got a cyclic group and I've told you how many elements are in it z what was this, this is the number 0 through n minus 1 with addition mod n so in fact mod n arithmetic is really nothing more than manipulating cosets it's just we prefer it's easier to manipulate the numbers 0 through n minus 1 and simply interpret what the operation is as mod n but the way that most authors will phrase this is if you're interested in the group with n elements that is cyclic in other words a cyclic group of n elements typically it's presented this way first but most authors would view this as what they'd call the elegant way to construct a cyclic group with n elements you take the cyclic group having infinitely many elements and you look at the factor group by another subgroup having well infinitely many elements but you simply look at the multiples of n one final sort of philosophical comment about this look at another quick example of an infinite group intuitively here's what you're doing folks when you form the factor group so g slash h the identity element of the factor group is h itself that's e or e bar or whatever you want to call it any element that's in the subgroup then generates the identity element of the factor group so whether you want to look at zero bar or in that situation that we looked at in the first example zero comma two bar as long as you have something in the subgroup it is the identity element of the factor group and at least if the operation is written as an addition the identity element is typically you know interpreted as zero so the intuition is if you take a subgroup then forming the factor group in effect makes everything in the subgroup tan amount to zero so if you're going to form something like z slash four z then what you've essentially done is ask me to take anything that's in the subgroup and view it as zero you know make four zero make eight zero and that's precisely what's happening when you do this sort of operation four becomes zero eight becomes zero so if you ever look at four or eight or twelve or negative four et cetera you're treating it just as if it was zero to begin with because that's the identity element of the addition alright hey that's all mod n arithmetic does you view n as zero alright let me just sneak this final comment in about factor groups for infinite groups we can talk about factor groups for infinite groups even if the infinite group isn't a billion let me give you one important example of where that happens another example is the group is this thing that we call G L N R it's the collection of n by n matrices with entries in the real numbers whose determinant is not zero so n by n matrices with non zero determinant here is a subgroup we call it SL N R but let me just write it out this way it's the elements of G L N R so the matrices in G with the property that the determinant of the matrix is one so if I look inside the general linear group if I look inside the collection of matrices having non zero determinant which under matrix multiplication we've shown to be a group if I look at this subset we know how to show that H is a subgroup H is a subgroup of G C essentially exam one that was a bigger than sign but it turns out so roughly exam one or it looks like a homework we know how to show this it's just subgroup theorem it's pretty straightforward actually in fact H is actually a normal subgroup of G that's a little bit trickier but typically if you're asked to show something it's a normal subgroup not necessarily the only thing to do but typically you run for that one equivalent condition on the list that asks you to compute G inverse HG for everything in the subgroup and see whether or not you get something back in the subgroup how do we show this reason let's see if I take something in the group let's call it little G and I take something in the subgroup and I take something in the group and I compute G inverse HG where this is in G and this is in the subgroup H so let's see what does this mean this means that G has non-zero determinant that's what puts it in the group H has determinant one and what we're asking is whether or not when we perform this whether we get something back in H in other words where the result gives us determinant one so let's compute its determinant determinant of G inverse HG math 313 it's the determinant of G inverse times the determinant of H times the determinant of G because determinant is multiplicative but wait a minute this is one over determinant of G that's also math 313 result times determinant of H if H is in the subgroup then its determinant is one that's convenient so I get one over determinant G times determinant of G which is one check so this thing G inverse HG is in fact in the subgroup so what does it mean it means then we can form G slash H and this is a really interesting group but we're not going to spend any additional time picking it apart alright oh for what it's worth this group that's infinite there's certainly infinitely many matrices having non-server determinant coefficient there's infinitely many matrices having determinant one so each is infinite and when you form the factor unlike what happened when we did Z slash 4 Z this is still infinite so it turns out you can't list out all the of course my example well question let's see ask it again okay yeah so yeah so the question becomes how many different cosets are there and it turns out there will be a coset I'm going to be careful here we know how to determine when two elements of a group determine different cosets you have to determine whether or not A B inverse is in the subgroup and so it turns out it's relatively easy to write down infinitely many different cosets because all you have to do is for instance take two and a bunch of ones and zeros everywhere else there's a matrix it'll generate a coset and then if you change the matrix to a three and ones and zeros and then four and five so it turns out for every real number if I put it in a one one slot and a bunch of ones and zeros everywhere else that those will generate different cosets for this particular subgroup not that that's all of them but at least there's a relatively easy way to show that there are infinitely many or there are incredibly many there are incredibly many there are uncannily many there's at least as many as are so before we start I'll spend the last 15 minutes talking about isomorphisms tonight but this is a good time to talk about a really big problem that was around in algebra that I can now state for you relatively easily because you've now got the background and the verbiage to get there and whose solution is a really I think sort of monumental achievement in mathematics and is an achievement that was completed within the last 20 years or so so here's the word definition definition is this a group G is called this is an interesting word to use simple in case the only normal subgroups subgroups of G R the trivial ones plus E and G let's see we know if we start with any group that not only do we always have the two trivial subgroups those are always subgroups in fact it was totally trivial to notice that not only are these subgroups they're actually normal subgroups so if I hand you any group and it's not just the group consisting one element there's always inside that group two normal subgroups for some groups that's all the normal subgroups for other groups there are additional normal subgroups here's a big question question find examples of simple groups well one type of example is totally easy answer one Z sub p where p is prime sort of a cheating example here's why let's see can I prove that the only normal subgroups of this group are E and G yeah because we actually proved that the only subgroups period of Zp are the identity subgroup that was Lagrange's theorem so in fact this word never even came into play if the only subgroups are E and G then those are the only normal subgroups so the question is are there any more interesting examples of groups that have the property that if you hunt around inside them that the only normal subgroups are the identity subgroup and the group itself if you look inside S5 the subgroup that we called A5 the collection of even permutations on the set 1, 2, 3, 4, 5 turns out to be a group it's a group of 60 elements it's half of 5 factorial and it turns out and this results not be on the scope of this course but would take us an extra lecture to show why it's true this turns out to be simple it turns out there are actually many many subgroups of A5 just take any element and look at the subgroup it generates you don't get all A5 so there's a whole lot of subgroups but you can actually show that none of those subgroups are normal even if you look at some of the subgroups that don't just look like subgroups generated by various elements none of them are normal it's an interesting result there's another one around there it turns out in fact A4 is not simple turns out the group consisting of the even permutations on the set 1 through 4 it's a non-abiding group with 12 elements but it's not simple you can find a normal subgroup in there but somehow the difference between A4 and A5 is dramatic there are normal subgroups of A4 there are no normal subgroups of A5 and when I say that folks would be no normal subgroups other than the ones that you have to have or A6 or in fact if you look at the collection of even permutations on any set 1 through N as long as N is bigger than or equal to 5 you get a simple group a quick comment about why these should be simple in effect folks the idea is this if you hand me a group that's not simple in other words if you hand me a group that has a normal subgroup then the intuitive idea is you can then form or consider two groups you can consider the subgroup itself it's a group but if the subgroup is normal in the group then you can also consider the factor group G slash H so what you've done is you've taken the group G and I don't know what a good phrase is maybe trade in you sort of traded in the group G for the two smaller groups H and G slash H and they're each smaller because if this is a subgroup then alright it sits inside the big one and G slash H if G starts as finite G slash H is also smaller and the intuition is somehow it will be easier to study H and G mod H because those are smaller groups and somehow if you can sort of weave the information that you get about those two groups together that you could somehow conclude something about the original group G so the idea is if I have a group that's not simple that I can somehow study it by building it up from these pieces piece H and G mod H which presumably are simpler easier simple intuitive of the means you somehow can't break it down any further if there's no normal subgroups then the group that you're handed you're sort of stuck with breaking down process too so you can almost think of the simple groups as the groups that you're going to have to use as the building blocks to describe all the other groups if there's some chance of using this breaking down process so it would be of interest to sort of know what all these look like so here's a big question, question are there more the answer turns out to be yes I'm not going to describe them for you though this is what we'll call an infinite class of simple groups so there's infinitely many on this list even though they sort of come up in the same way it's just the even permutation sitting inside the group one there are there are some other infinite classes just like the alternating groups are classes of simple groups there's some what are called sporadic simple groups simple groups in other words groups that just by good luck or bad luck depending on your outlook on life I guess I mean there is one there's one with 168 elements or here's one with almost 10 to the 34th number of elements there's some small ones and some big ones but there are some simple groups that don't sort of follow some sort of pattern so here's the sort of gigantic question big question can we somehow describe all the simple groups and let me put one disclaimer in there all finite simple groups is it possible to come up with a list saying if you have a finite simple group it has to be on this list well the list admittedly would be infinite because I've already written down infinitely many simple groups there's infinitely many there and it turns out there's a couple more lists that are that contain an infinite number of groups but they're all sort of the same structure and then there's some that don't so this was a huge question that the mathematical community worked on for about 50 years because intuitively these somehow describe the building blocks for all groups and for a while people thought well this is just too big a project and how could you ever hope to get your not on your head around it but how would you hope to be able to say if I start with a simple group that it has to be one on this list and by the mid sixties it turned out enough work had been done on this question that people thought you know what we might actually have a chance of getting to an answer to this thing we don't really know what the answer is yet but if we sort of keep working and sort of going down this road and looking for some more examples and doing a little bit of stuff in geometry and doing a little stuff with sort of finite structures maybe we'll be able to get a handle of what more of them look like and eventually hopefully get some handle on where they all have to be and eventually the answer turned out to be yes, done it's not clear when victory was really declared but there was one gentleman named Dan Gornstein who was at Rutgers who sort of acted as the clearing house for all this stuff because there were literally hundreds of mathematicians working on this question each working on a little piece where they sort of get a little piece and then hand that little piece off to someone else and they sort of modify the little piece and see where they got and you know there'd be a big push to say oh yeah in the end here I've got another simple group or this thing that I was looking at that has all the makings of a simple group turned out to not be simple or turned out to be a simple group that we already know something like that and eventually by about 1985 it sort of all congealed if you look at the literature all of the work that was done under this list of all finite simple groups takes about 10,000 pages in the literature there's pieces everywhere and you might say well you know why bother who cares and that's a fair question but in the end I think it's a testament to people sort of focusing on a question that was of interest and B it turns out there's been some nice consequences of knowing where all the simple groups are you're able to sort of get a handle on how it's possible to build other types of structures not just groups some crystallographic type constructions have come out of these things and you know mathematicians sort of looked at this proof as well this is a testament to just working hard sort of the Mount Everest question that people eventually were able to put together there are a couple of mathematicians now that are working on taking this sort of you know massive uber project and trying to distill out of it sort of more streamlined or more intuitively clear understanding of why it is that all of the simple groups look like the following things I think I don't quite mind this I think there's 5 infinite families and 26 sporadic groups or something like that why is that I mean it's just unbelievable that answer could even be achieved and then it's a question of well how do you go about determining that this really is everything of the quest along the way as well it turned out and this I think is what drove a lot of mathematicians to work quite hard on this particular project that there were these totally unexpected connections made between the structure of groups and many different structures that were coming up in other areas of mathematics like in geometry or in crystallographic stuff or in coding theory etc that when the group there started looking at these structures they started saying yeah you know this numerically looks like something over here or there's too much numerical coincidence between the structure that we're looking at here and the structure that you folks are looking at in what seems to be a completely unrelated field there must be some connection and they start working on connections and not only were they able to establish a connection but by sort of understanding what the groups brought to the table it gave information about the geometry and vice versa what the geometers had already known about this structure they engaged some information about the group so there was this really nice sort of back and forth that was happening between many different disciplines and the finite group theorists there was a really nice article written in Scientific American by Dan Gorenstein who died in the early 90s he wrote a SIAM article that you folks should read I've posted it on the course website I've posted a link to it on the screen pages or something like that it's certainly aimed at a general audience but now that you know what a group is and you know what subgroups are you know what normal subgroups are you know what factored groups are etc I think that article will really make a lot of sense to you so I would strongly recommend if you're looking for some bedtime reading download it or take a look at it it's pretty fun it's just compelling the way that this thing played out ok let me just for the last ten minutes talk a little bit about this word that we've been using almost since week one in here what it means for two groups to be isomorphic and what I've been doing is playing up this word from an intuitive point of view isomorphic means that the two groups that you're looking at have the same structure so isomorphic and the way we've been sort of playing that up is if you write out the group table for one of them and you write out the group table for the other if you simply somehow relabel things by some common language that the two tables really represent the same sort of structure that's what isomorphism means and that's the intuition that I've that I've been hoping you to develop well now it turns out we can actually write down the technical definition of what it means for two groups to be isomorphic the definition of isomorphic means that there is a nice homomorphism from one group to the other and if you think about it homomorphism is a function from one group to another group that somehow preserves structure and if in the end that homomorphism turns out to pair the elements of the two groups up in a one to one and onto or in a bijective way then we're going to call the homomorphism an isomorphism and we're going to say that the two groups are isomorphic so the definition is that the two groups well let me start with the definition of isomorphism first a homomorphism phi from one group to another is called an isomorphism isomorphism so that's the word isomorphism notice that the two endings are the same here in case this is just a function requirement phi is both one to one and onto we're in slightly higherfalutin language phi is a bijection one to one correspondence so a homomorphism a homomorphism is an isomorphism in case as a function it happens to be both one to one and onto oh we can actually measure when it's one to one in terms of the kernel so this happens if and only if the kernel of phi is just the identity and phi is onto phi is onto there's no real good subgroup measure of whether or not a homomorphism is onto in terms of any sort of easy test but at least we have that for one to one so that's what an isomorphism is and then the second word is isomorphic we say the groups G and G prime are isomorphic and we've already used that phrase this semester here's what it means technically in case there's an isomorphism from one to the other there exists an isomorphism phi from G to G prime so two groups are isomorphic in case you can get from one to the other by an isomorphism in terms by a function that preserves the structure in these things that's what homomorphism means but that at the same time simply pairs things up in a one to one and onto way that's a bijective correspondence between the elements and if you think intuitively that's exactly what it should mean to say that the group tables of these two can simply be relabeled to some sort of common table where they look the same the relabeling process is simply what your isomorphism does if you've got this thing relabeled to something else then simply ask well if you've got a common relabeling what is this relabeled to and what is that relabeled to and then just match those up then match those up match those up that'll be a one to one and onto in other words a bijective correspondence but it's something that preserves the structure in other words that labeling will in fact be a homomorphism so here's a pretty interesting question if I hand you two groups can you determine whether or not they are isomorphic whether they have the same structure up until now the test was simply to try to write out the group tables and see if you could somehow come up with some sort of correspondence that made the two tables look the same or conversely you look at the two tables and say you know what there's something about these two tables that inherently makes them not isomorphic and for example when we looked at the group let's say Z4 in this group that we've been calling V well inside Z4 it was the case that there were elements that when you did the binary operation to themselves you didn't get the identity inside Z4 if you do one plus one you get two which isn't zero but inside the group V if you ever do the binary operation on an element to itself you always got the identity so those two couldn't be isomorphic you couldn't somehow relabel these to get these because over here whenever you did A star A you got E and over here there were things such that A star A wasn't E so question in general if somebody simply plots two groups on the table in front of you and says are they isomorphic or not it turns out to be a relatively hard question general so what you do is this will be perfect I mean this is sort of philosophy then about mathematics so how do you answer this question well if you don't know the answer to begin with if they yes no type question you sort of push one way and see if you can get an answer and if you can't then maybe start pushing from the other way and see if you can get an answer and then hopefully you sort of squeeze what's left over till you finally get an answer at least from yes no so here's two groups are they isomorphic well you could maybe try to answer no just by looking at some general principles like if I hand you a group with five elements and a group with eight elements can they be isomorphic no because to be isomorphic it means there has to be an isomorphism one to the other in other words if nothing else there has to be a one to one and on to function from one to the other and hey if there's a one to one and on to function from one set to another set it means the two sets have to have the same number of elements in it to begin with so if I hand you two groups and they have a different number of elements immediately they can't be isomorphic out alright so it's a more interesting question here's two groups with eight elements can they be are they isomorphic well you sort of gross level hope fails here they have the same number of elements so they could be so your first sort of filter fails you alright now you got to look a little bit deeper maybe they are maybe they aren't okay well what we develop is a list of things similar to the number of elements where if one group has it and another group doesn't then they can't be isomorphic if this has five elements and this has eight elements they can't be isomorphic just for function reasons let's see if this group is cyclic and this group is not cyclic then they can't be isomorphic why because if this group has a generator then regardless of how you relabel the elements here the other group that you relabel it to is always going to have a generator because this thing it's a generator over here I don't care what you call it it's always going to be the case that when you keep combining it with itself you'll get all the other so if you have a cyclic group and not cyclic group they're not isomorphic if you have an abelian group and a not abelian group then they're not isomorphic if you have a group that has let's say exactly two elements with the property that when you combine them with itself you get the identity think of it as if you have a group with the property that x star x equals e only has two solutions maybe e and one other thing that's true in z sub four for example and you hand me another group and there's maybe three solutions to that quote-unquote equation x star x equals e then the groups are not isomorphic think because all the things that we've been talking about cyclic versus not cyclic, abelian versus not abelian the number of solutions to things that look like combine an element with itself to get the identity all those things are what we call structural properties of the group there are things that don't depend on how you label the elements there are things that are inherently true about the elements in the group independent of how you label them and those are typically referred to as isomorphism invariance isomorphism invariance and I'll just list a few out for you here for example being abelian is an isomorphism invariant meaning if you hand me two groups one which is abelian and one which isn't abelian then the two groups are not isomorphic being cyclic is an isomorphism invariant the number of elements is an isomorphism invariant the number of solutions to an equation of the form x star x you know x star x star star x equals e where I've handed you a fixed number of elements maybe m elements here or something like that is an isomorphism invariant isomorphism invariant let me give you an example of this for example here's two groups the group of non-zero real numbers under multiplication and the group of non-zero complex numbers under multiplication there's two groups are they isomorphic well they're both infinite and they're both with that multiplication you might say that one sits inside that one so they can't be isomorphic oh yeah being a subgroup of doesn't preclude two groups of being isomorphic what you'll show for homework for example is if you look at the group of integers and you look at it subgroup consisting of multiple to four as groups those are actually isomorphic groups that have the same structure even though one sits inside the other how do I know that? because it's pretty easy to write down an isomorphism from one to the other it's the function that simply multiplies everything by four so being a subgroup doesn't preclude this one from being isomorphic or not isomorphic to that one but it turns out they're not let's see how many solutions are there to the equation inside this group how many wait a minute let me do one better here if I do x star x star x equals e let's see inside the nonzero reals have the property that when you raise them to the fourth power you get one two elements in there one and minus one both do this how many elements in here have that property? four one and minus one do but so do I and minus I so because the number of elements that are solutions to that completely binary operation inside the group equation so those two numbers are different these are not isomorphic so not isomorphic you know what happens sometimes which makes things a little bit harder than you'd hope is sometimes you go through this list of isomorphism invariance and they're all still matching up but it's maybe not clear how to write down a function from one to the other that behaves as an isomorphism that's where the interesting question is coming okay here is home again this will be due on a shorter turnaround time due Monday a week from today excuse me let's see what's the 22nd in section 15 problems one through five I want you to do two and three to turn in but change the instructions at least just slightly to let's see number one compute the order of each of the elements of the factor group element in the factor group just like we did in class for the first 15 minutes or so tonight and then secondly answer the yes no question is the factor group because that group cyclic that's what I want you to do on those two then a question to turn in prove if G is cyclic and H is a subgroup of G then the factor group G slash H is cyclic to turn that one in and then finally let's see in section three question about isomorphisms problems 29 through 33 I want you to turn in three parts A and B and on these two questions actually write down actually produce the isomorphism produce the isomorphism the way the book asks the question it's simply a yes no question are these two groups isomorphic well I want you to either say yes and here's why because here's the isomorphism or no because it's an isomorphism invariant and I want you on 33B to change the following change C dot to C star dot in other words I want you to put a star there and change H dot to H star remember star means non-zero elements of this author does things a little bit backwards he talks about isomorphisms in the context of binary operations not necessarily in the context of groups for what we're interested in I'm more concerned that you understand what isomorphism means in the context of groups so back here he looking at two different sorts of structures neither which is a group but if you throw out the zero element then you do get groups and that's what I'm more interested in