 Today I got yours, but I didn't get it done yet. Sorry, but I've got it not a problem Not to worry. I'll get it back to you on Wednesday Yeah, oh, yeah, this is good. This is good So let me comment first on the homework that just got returned There are Well, there's two There's two things that more than let's say a couple of you had Some issues with the big problem that seemed to challenge people was this one about Show that if you have this condition the condition that if for each element in the group and Each element in the subgroup if you compute g inverse h g for any element in the group H that this winds up back in H. So that's the hypothesis Then let's see then you had to prove Let's see. How was the question officially stated? prove that every left coset's a right coset or prove that a h equals h a I forget prove that every left coset's a right coset right every left coset is a right coset Okay, the only two comments on the homework coming back are comments to be made in the context of this one problem The first comment is this The hypothesis that you're given well two of you wound up proving every left coset's a right coset without using the hypothesis Folks there should be just these huge air raid sirens going off if you're going to prove something without using the given hypothesis then something is I mean I'll guarantee is a hundred percent wrong because in all of the problems that will do this semester I gotta be careful in 99.9% of the problems that will do the semester The hypotheses are required in other words if you remove the hypotheses that are given then the conclusion doesn't necessarily follow So please get in the habit of looking through your work before you submit it Make sure that you've actually used the hypotheses somewhere The point is if you're ever confronted with an expression that looks like this Where G is any element of the group and H is any element subgroup what you get to conclude is that The result lands in the subgroup What some of you tried to do was write down a stronger version of the hypotheses What some of you tried to do was say well I have Something let's call it H1 that's in H So all right at some point in your proof you wound up with some element in H and then you said so there is a G in the group with or and let's call it H2 in H with G inverse H2g equals H1 I'm gonna saw this on three or four of your papers and folks this Hypothesis doesn't at least not on the surface Imply this one all this hypothesis says is if you're confronted with an expression like this That the result lands in H. It doesn't necessarily say that every element in H is of this form It simply says that if you have something of this form It's within the subset of age and what this statement says is if you start with something in H Then it gets hit by something of that form and this is a stronger statement Now it might be the case that you could prove that that implies this But you certainly can't just start here and say this is my hypothesis because this statement is not logically equivalent to this one so is not necessarily the same As this set of hypotheses even though admittedly you'd prefer this hypothesis because this gets you to the Conclusion more directly, but all you get to start with is this So be careful there you can't sort of overstate what the hypotheses are telling you you have to use the hypotheses as given and Most of you saw that the trick here is to sort of maneuver yourself into a position where it doesn't really matter What G is maybe G is the inverse of some element so that if you have something looks like a H a inverse You can write that as a inverse inverse H You know a inverse and so if you let G be a inverse then you're looking at G inverse HG and you wind up with something It was back in H The ones that those of you that use the hint seemed out a little bit easier time with this the hint was to Somehow trade things in in terms of the letter a rather than letter g so that you don't get tunnel vision thinking you have to have The thing called G sitting here. You could have something called a sitting there. You could have something called B sitting there Okay, the second comment is and two of you commented on this and this was really good It's interesting you you actually wind up proving something that on the surface looks a little bit stronger than what you're required to prove all you're required to prove is That every left coset is some right coset in other words all you're asked to prove is that each coset of the form a H is Also of the form Well, that's what a left coset looks like is also of the form a right coset for some In other words, all you have to do is prove that a H is h B So there's nothing explicit in the statement that you're trying to prove that says that you're Required to prove that this left coset is the same as the right coset generated by the same element You started with this really is all you're asked to prove But all of you wind up proving in fact not only if you have this condition is a h equal that it is a h equal to a Right coset in fact you prove that a h is equal to the right coset h a So question is this condition every left coset is a right coset The same as or somehow weaker than the condition that h a is a h as all of you prove and the answer is that turn out They turn out to be the same so here's an observation. I'm going to call it a proposition because we're going to need it in relatively strong form in about 10 minutes proposition is this if Every left coset of h is a right coset of H Then necessarily a H equals h a For all so if you give me the information that if you start with a left coset and It happens to be some right coset I'll tell you which right coset it is It's that right coset where you look at the right coset that's generated by the same element that generates as the left coset and the reason is If the hypothesis is we the hypothesis is That a h is h b for some D in the group That's what we're told that the left coset a h is some right coset The question is why can I conclude that the right coset that I use here is actually the right coset h a And the answer turns out to be this but wait a minute It's always the case that the specific element a is in a h always If you take a specific element in the group and you look at the coset that it generates I don't care what subgroup you start with any subgroup always contains the identity element So you're always at least looking at a e in other words. You're always generating the specific element a within its coset So this is Completely independent of this set of hypothesis. This is always true for any subgroup and any element of the group So in this particular situation if I've got an element in here the hypothesis is that it equals this So here a is actually in h b Because these two sets are equal if it's in this set then necessarily it's in that set But now we're going to make a statement about right cosets as soon as you have an element in the right coset. So h a equals h b Previous work that we did on cosets The work being if you have an element that lives inside a coset Then the two cosets have to be equal because what we know that any two cosets either left cosets or right cosets as long As you're comparing apples to apples any two left or right cosets are either Identical or they're disjoint. Well, these aren't disjoint because a lives inside here So necessarily the two cosets have to be identical So the hypothesis that a h is h b actually leads to the conclusion that h b is h a And so the conclusion is that in fact not only is a h equal to h b a h is actually equal to h a as well I said wait a minute. How can we have two things equal to each other dancers? Remember folks You can very easily find different elements in the group that generate the same left or right coset And that's why I had you in the homework do a whole lot of examples of Compute the right cosets then compute the left cosets and tell me which are equal and tell me which aren't equal and the observation is Sometimes you have different elements that produce the same right coset or different elements that produce the same left coset In fact, that's what's going to make the construction that we're going to start doing today. So well Unfortunately, so mysterious or so uncomfortable at the beginning until you get a feel for it All right, so that's a comment. You weren't necessarily asked to prove That h a is a h all you were asked to prove is that a h is some h b But all of you prove that a h is in fact is in fact a j. All right, that's good Questions there comments All right, so there's sort of the both the wrap up to the homework and an important piece of information that we will wind up using today So let me and this was good what we did last Wednesday For the last 10 minutes was I I sort of just started spewing about What it is that we're going to do for essentially the next week and a half in here. I Gave you the overview without writing anything down what I'm going to do today start writing things down What we'll do on Wednesday is you'll have a quiz on some of the stuff that we'll do today Just so that the ideas start getting pounded in your head You're actually writing them down yourself and then Wednesday we'll do the same thing We'll sort of step back and look at the big picture again and in the context of it all We're just going to keep doing computation after computation after computation so that hopefully you start seeing the patterns and start Understanding what these structures are that we're going to work with So here's the new big idea new Big idea These are usually called factor groups Another phrase for them is groups of cosets and this is section 13. We're skipping section 12 I'm not sure I mentioned that on last Wednesday. We're going to skip section 12. We may come back to it At the end of the semester we have time There's a 14. Sorry, and we're skipping 13. Oh, yeah, okay So so I'll say sections 13 through 15 because homomorphisms is sort of the Lead in to what we're doing with these groups of cosets All right. Here's the goal goal We have a set The given information is this you start with a group G is a group and H is some subgroup of G and Throughout the discussion folks once you write down the group and the subgroup those are meant to stay as they are Throughout the discussion. You're not going to change what subgroup you're going to look at or change What group you're going to look at you're going to build a new group Maybe based on these two pieces of information the original group that you start with and some subgroup that you've chosen Now look you got a lot of choices to begin with You got a lot of choices of groups to start with and you got a lot of choices within that group of what subgroup You want to look at but once you've made those two choices everything's now fixed and all the discussion is given in terms of that context Okay, well what is a group a group is nothing more than a set Together with a binary operation on that set That has three additional properties Associativity the existence of an identity and the existence of inverses what we're about to do is based on this data Produce new groups Now the thing that's the hardest for students to wrap their heads around is that the elements of These new groups that we're about to describe the elements are sets So a thing in this new group is a set In fact more specifically a thing In this group an element of this group is a coset of h in G Well cosets are just sets. They're not groups. They're just sets. Okay, one of them happens to be a group Coset consisting of H itself, but in general they're just sets and What we're about to do is come up with a process by which if you take the group You take the subgroup you then form all the cosets of the subgroup sitting in the group You did that a lot in the homework that just came back you get a set of sets And that's I think where students have a little bit of trouble. It's a set of sets So inside the set you're looking at a bunch of sets and the goal is to Define or provide some sort of binary operation on those sets in other words some sort of operation where if you hand me two of these sets You can combine them together and produce one of the sets and Moreover once we write down that binary operation once we write down this method of taking any two of these sets In other words taking any of these two I'll focus on left cosets for now any of these two left cosets And combining them together to get another left coset once we write that down. That's where the hard work is going to be It'll turn out that that definition of a binary operation on the collection of cosets Actually gives a group structure on the collection So just a little preview of what's about to happen. We're about to write down all the left cosets of a given subgroup That collection of sets where we view each of the cosets individually as an element of this big set of cosets That collection is eventually going to become a group In order to make it into a group. We need to step zero to find a binary operation on that collection of cosets That'll be hard. That's going to take us a long time to do But once we've done that hard work of actually defining the binary operation on this collection of sets That binary operation will almost clearly or easily Be shown to be associated of having identity and show that the existence of inverses happens So it's a little bit counterintuitive because typically in the past You'll be handed a set you'll define this thing It's sort of trivial to show that it's a binary operation on the set and the typically the hard work is showing We're usually not associated, but maybe the existence of an identity or the existence of inverses here The hard work is gonna Be in showing that we actually have a binary operation on this collection of subsets and then once we do then things will Be pretty Okay, so here's the goal we seek good word to define a Binary operation operation On the set for now folks. I'm going to call it script s. We're going to give it another I don't know what more standard name once the process is all sudden done of left cosets of H Okay, so I've given the set of left cosets a name. I've called it script s It's the first time we've given that set a name. I Know how big the set is remark How many elements are in this set or another way of asking that question is how many left cosets are there of h ng the index of h and g Now if it happens to be the case that this group is a finite group in that case We happen to know what that number is it's this number But only if You start with a finite group and the point is The construction that we're about to look at holds regardless of whether or not the group happens to be finite The group could be infinite and in fact the number of cosets of h and g could be infinite So I'm not restricting our attention only to Finite groups or only to situations where there's only a finite number of left cosets of h and g It's just if I happen to look at that very important situation I can actually count how many elements are going to be in this group We'll look at some specific examples in a minute So here's what we're going to do. So the goal is rephrased Here's what we want to do We want to define I don't know what to call it. How about I don't know a star with a circle around it or something like that. It's a special binary operation. It's not composition and it's not It's not the star of the group necessarily Something else. So let's just call it something new on This set of left cosets In order to make it into a binary operation In other words, we want to take any two left cosets and somehow define what this means and Spit out some left coset So that's the task somehow to come up with a systematic or reasonable way by which you take any two left cosets of H These Individually are sets. Okay, that's fine and somehow combine them whatever that means to kick out another left coset of H Well, and this is what I was sort of Starting to get at last Wednesday It turns out there there are at least two not unreasonable places to look to try to do that Let me give you one possibility one possibility is all right folks all the things in this set are just well They're elements of the coset But in the end there's simply elements of this group And this is some collection of elements in that group So individually I've got this collection of elements in the group and I've got another collection of elements in the group So the following process at least would make sense just sort of what do they call it foil everything? Multiply everything in here with everything over here and see what you get You know if you got two elements in this left coset and two elements in this left coset Combine that one with that one and that one with that one and that one with that one and that one with that one It's at least a reasonable way of somehow Smashing together the elements in this set And see what you get all right that at least would be a reasonable way of Taking these two subsets of G and smashing them together to get another subset of G It's just what's the possible problem with that you might get more elements Exactly right Jared. You might get elements That don't form a coset and for example we could Look ahead a little bit and say you know Maybe it's the case that I've got two elements in here and two elements in here I can buy now with that and now with that and now with that and that with that Me presumably maybe four elements get kicked out and hey if your cosets each have two elements Every coset then only has two elements and if you kick out four elements You haven't defined a legit binary operation on the set of left cosets you kicked out something. That's not in the set That would be bad In fact, I'm about to show you an example of where that happens Unfortunately So here's possibility one for defining a binary operation on script s possibility one possible Binary operation number one on script s define a H Star with a circle B H to equal I'm going to call it the set product of a H With BH and set product simply means exactly that process I just described in words take everything in this set and simply combine it with every possible thing in this set And see what you get for example Let's do this let GBS three and Let H be the subgroup of G generated by mu one while that turns out to just be a two-element subset I am you let's write out the left cosets left cosets See did you do this one for your homework? I forget if this one actually came up, but whether it did or didn't It's a good exercise here. They are let's see row one H So that's row one I so that's row one mu one so that's row one and Mu three it turns out. I'll let you haul out your table for s3 if you want here So row one H turns it out to be row one times I which is row one and row one times mu one which is mu three So just for what it's worth We also by default have just Realized that this is the same as the left coset mu three because as soon as you find an element in a left coset That is the coset generated by that Here's the second left coset row two H Just do the Appropriate product so that's row two times I so we know row two is in there We always get that one for free and then row two mu one turns out to be what? You somebody have the table in front of them you too. Thank you So that by default this also happens to be mu two H and if we do Row three H See what we get here Yeah, there's no real three. Oh, yeah, there's row zero pardon me. So if we do row zero H Well, that's easy. That's I H. So row zero is I here in this notation. So that's row zero and Mu one so that happens to be Mu one H So there are the six left cosets of H Inside G now Let's see whether or not this proposed binary operation on the left cosets of H and G Actually kicks out Another coset. So let's compute How about something like? How about well this yeah, so The request is to take any two of these left cosets. How about this one row zero mu one? There's a left coset That's my AH Of course, I could call it row zero H or mu one H. It doesn't really matter here. There's a left coset Let's combine it with Another left coset. How about Row one mu three and let's see what we get Well, this is by definition the set product in other words what I'm going to ask you to do is simply Pound out all of the possible combinations that you that you can where you take something from this set and combine it with something from this set So the first thing is row zero row one the second thing is row zero mu three the next thing is Mu one row one the next thing is mu one mu three And let's see what we get Well, we get row one because that's row zero row one and we get row zero mu three What's that mu three? That's easy and we get mu one row one somebody have that You one row one Mu two and then we do mu one mu three somebody have that Row two we get a perfectly good subset of the group But we don't get a subset of the correct form We don't get a subset that happens to be a coset of h and g looks. There's only three cosets of h and g here They are If nothing else each of the cosets of h and g have two elements and I've just written down a subset of the group That has four elements. So the issue is problem problem the result is not Another left coset left coset of H in g I'll rephrase that in terms of binary operations on set ie when we did row zero mu one and We start it with Circle started or whatever that we're calling this thing row one mu three that Process did not give us something back in the set. We're interested it didn't give us something back in script s So star defined in this way is not a binary operation on Script s at least in the situation where you happen to start with this particular group in this particular subgroup Now I'm not claiming that Well, it might be the case folks that if you started with a different group in a different subgroup and you started pounding these Products out. Maybe you will get another left coset of subgroup Maybe you will but at least in this particular example The point is if we start with this particular group in this particular subgroup If we try to define a binary operation on the collection of left cosets simply by hammering them together Not an unreasonable thing to do the issue is that what gets kicked out is perhaps not another left coset so in general this process will not Give us a good binary operation on the collection of left cosets so too bad so possibility one Reasonable possibility one for trying to define some sort of binary operation on the collection of left cosets crapped out Because it didn't kick back a left coset All right, so let's reinvent the wheel here and say okay, you know if the person you don't succeed Let's try another approach Here's a good other approach. What was the problem with trying to do set products? The problem was what got kicked out wasn't another left coset Now if somebody just handed you these two left cosets and said all right. Here's what I want you to do Tell me a reasonable left coset That you might want to associate with that left coset and that left coset under some binary operation Anybody got a reasonable guess give me a left coset that might be a Reasonable output when you try to combine these two anyone what should I put here? How about a star B, okay? Good idea. So here's a good definition So let's try possibility number two. So here's possibility number two Because possibility one crapped out possibility two is simply deem forget the set multiplication crap deem Ah, what should we call this one now star star or something like that BH? I'm coming up with some presumed new binary operation equal to ABH Then here's what we've done. We've simply eliminated the problem that came up in possibility one This one said all right take this left coset and combine it with this left coset It's simply deem the output to be this left coset Good plan at least what gets kicked out is another left coset Okay, so now can anybody predict what the issue is going to be with this possible definition Beautiful, that's exactly what's going on So what Jared just suggested is this look if I hand you a left coset There's lots of different ways of writing it down For instance this particular left coset could be viewed both as row one H But equally as good could be viewed as me 3h This particular left coset could be viewed as that or it could be viewed as that So what's about to happen is we have to make sure that regardless of how we name our left cosets That the Combination of the two Producing what you get when you take the two coset representatives and combine them is well defined It's exactly the same issue folks that we looked at in day one in here and the context there was rational numbers Where it was possible to have two different ways of describing the same rational number One half and three sixths here There's for example two different ways of describing this left coset row 1h or mu 3h two different names So if I'm going to try to define a binary operation corresponding to those things I have to be careful that the operations well defined Guess what it's not going to be well defined but Let's do an example consider these consider if I look at uh Let's do this if I do row 1 mu 3 and I ask you to double star it with This one This one row 2 mu 2 So I'm asking you to double star this one with that one what left coset gets kicked out Well, here's the issue folks this thing How do you figure out what gets kicked out? The first step is you have to take the coset and write it in one of these forms All right, how about let's first write it as Row 1h and then I'm asking you to double star that with Let's see row 2h Is that what I want to do here? Yeah, so what should you get? Well, if this is the definition of what double star is I've now taken these two cosets I've given them names of the appropriate form And now the way you're supposed to combine cosets given in this form is you're simply supposed to take the two coset representatives That's the definition of what this proposed Binary operation supposed to do which gives what well row 1 row 2h is row 0h Which is H? In other words, which is this left coset the left coset Row 0 mu 1 so if I combine these two left cosets by Calling the first one row 1h and calling or naming the second one row 2h Then the result of that particular combination is just the coset H itself now We're about to see the issue and again I want you to keep in mind the issue that happened when you tried to Like an exam one take two rational numbers and try to Combine them together and then take two other rational numbers that represent the same rational numbers and try to combine them Together and see what happens, but also I can take the same two left cosets row 1 mu 3 Double star with row 2 mu 2 In this particular situation what I'm going to choose to do is write this as mu 3h And I'm going to choose to call this one mu 2h and in this situation. Let's see. What is mu 3 mu 2? Well by definition This is supposed to be mu 3 mu 2h and what is mu 3 mu 2? Is row 2 Which is oh, let's see row 2h is this one right here Is row 2 mu 2? So here's what we just did folks. We took exactly the same two left cosets Row 1 mu 3 and row 2 mu 2 here They are again row 1 mu 3 and row 2 mu 2 if we chose these particular names for those two particular cosets and Ran them through this proposed binary operation Here's the coset that got kicked out on the other hand again We take the same two left cosets, but we choose to name them this time instead by these other names And then run them through the process the issues that what gets kicked out is a left coset That's not the problem. It's just what gets kicked out is the different left coset They're not kicked out the first time so the issue is because this and this are not equal not equal punchline is So this second proposed binary operation is not well defined Too bad. So we're crapping out here. So we've tried on the surface What looked to be the two reasonable possibilities for coming up with a method or an algorithm or definition of? how to take two left cosets of of a subgroup sitting inside a group and combine them to produce Another left coset so we haven't even gotten this process up the ground yet. We haven't even written down a binary operation Well it sort of in rides the hero here The punchline is folks that if you just hand me a group and you hand me a subgroup It's Obviously not always the case that if you try to define either one or the other of these two Proposed binary operations on the collection of left cosets that you get a Another left coset or that you at least get well-defined binary operation on the set But what we're about to see is if you start with a special sort of subgroup Then it is the case That possibility one gives you a binary operation on script s in other words It is the case that possibility one the possibility where you just hammer together all of the elements that you're looking at in the appropriate order Actually kicks out another left coset So that gives you a binary operation script s and the miracle or What's incredibly nice and beautiful about this particular idea is in? Exactly the situations where when you take any two left cosets and hammer them together by this set product And you get another left coset Those will be exactly the same types of subgroups that have the property that if you define this type of Binary operation by simply deeming a h Double-star b h to be a b h that that's going to be well-defined And that's what's going to give us a binary operation on the collection of left cosets So the question is what sort of the magic property that will allow not only the method of Combining left cosets to get another left coset described in possibility one not only for that to give another left coset But for the other one to be well-defined What's the magic property the subgroup that makes that happen answer that every left coset is a right coset? That's why we've been playing up this particular property so much over the last weekend. I So here's the miracle miracle turns out that If you happen to start with with the same sort of data a Subgroup h sitting in G and furthermore H happens to have the property the property that Every left coset of H is a right coset of H then Turns out everything works. So there's the statement of what's about to happen from a From an emotional point of view Now let me write down some of the technical details from a more mathematical point of view What we're gonna show is if we happen to start with Not just your favorite subgroup sitting inside your favorite group But your favorite group and your favorite subgroup sitting inside that group that happens to have this extra property That every left coset of h and g is also a right coset of h and g Then if you go back and try to define a binary operation on the collection of left cosets By simply doing set product You'll always get another left coset. In other words, that'll give you a legit binary operation on the collection of left cosets and But since we're not going to pound through all the details of this I'll just sort of mumble it under breath and if you were to take the same sorts of subgroups and go back to possibility to the situation where you deem the binary operation By combining two left cosets to simply be the product of the coset representatives In fact, that process will give you a well-defined Binary operation on script s as well, but since we're not going to focus on this as much I won't play that up as Being so important the punch line will be this because this is how we'll actually do the operations here more formally Here's the definition. I'm going to give this a name definition if h is a subgroup of g then h is called a normal subgroup and Typically, we'll just say h is normal in g in case of this property holds in case every Left coset of h g is also right coset of H that's what the word normal means that every left coset of h is a right coset a Bit of notation and then a quick comment and then we'll look at some examples Notation if this happens We don't have to write this, but typically we do we write Just for a shorthand h. Well, let's see. We know what the notation for subgroup is It's this less or equal to science to this sort of sharp Inclusion sign if the subgroup happens to be normal then we close up this third line here and make a little triangle So if we write this it means a lot it means not only is h a subgroup of g But it has the property that every left coset is a right coset Now by a previous remark The remark that I made right at the beginning of today in the context of the homework question that came back Every left coset is a right coset is the same thing as saying that for every a A h not only is some h b, but in fact a h is h a So this is the same as saying a H is h a for every so group a subgroup of a group is normal in case Not just the statement every left coset is a right coset in fact the statement That every left coset a h is in fact it's sort of mirror right coset h a So I'm going to write down the Key proposition because I want to show that this particular condition leads to where we want to be well question Oh great question. So the question is does this necessarily? Oh? Let me back up make which group of billion No, so the question is Does this condition necessarily give that G is an abelian group and the answer is no Because we did remember a homework problem Which number was that numbers 9 and 10 and the one that came back tonight? section 10 Section 11 was it something like that? section 10 Example Hmm Let's see it. What was G there d4 and what was h? Yeah, the subgroup generated by row 2. So this is row 0 row 2 then You showed in What was it section 10? Section 10 problems number 9 and 10 Uh-huh h is normal in G How'd you prove that you just pounded them out you pound it out all left cosets and you pound it out all the right cosets But I'll put as a remark remark The original group G is not abelian here So will's question Doesn't necessarily require G to be a billion in order for this to happen is no But I hesitate to put this in here because then it might give you the right or give you the wrong impression But hopefully having written down this example first you won't get the wrong impression If it happens to be the case that the original group that you start with this group G is abelian if it is Then it turns out regardless of what subgroup you write down. It'll be normal if G happens to be a billion to be a billion then every Subgroup will H will be normal in G and the reason is well, wait a minute if you give me a Left coset a H if the whole group is abelian if G is abelian then every one of the products that you do here Ah is going to be the same as H a so you just flip it around So if the entire group is a billion then this condition comes for almost, you know almost free. It's almost automatic We'll get back to this observation in a minute I was sort of hoping to keep it under the rug for another couple of minutes, but since it was asked that's fine But I don't want you to get the wrong impression What's going on with the question of a subgroup being normal inside a group is a significantly bigger question Then simply is the group abelian and Here's the good example to keep in mind In fact, let me give you another example just a hammer at home example ah if G is s3 and H is the subgroup generated by Row one So this is row zero row one row two H is normal in G And the reason is I'll put it in friends here. I'm gonna let you do the pounding Just pound them out pound out You know row zero h equals Which happens to equal h row zero compute? Row one h equals Equals h row one compute You know, etc. You got six things to do six things Just pound them all out The last one will be what mu three h or something like that Equals Close h mu three compute I'm leaving out lots of Admittedly tedious computations But it's stuff that at the same level of the stuff that you did in problem six seven nine and ten in section ten for the homework It just came back and they all turn out to be the same So here's another example of a normal subgroup sitting inside a group. That's not abelian This is another good example to write down for the following reason Inside this group. We've just written down a normal subgroup But in this same group if we were to choose a different subgroup it might not be normal But contrast with in s3 Equals s3 so the same group if we choose instead the subgroup consisting of The subgroup generated by mu one so that's row zero mu one then H is not Normal in G And the definition of normal is that every left coset is a right coset That's the definition so in order to show me that some subgroup is not normal All you have to do is produce a single left coset. That's not the same as the corresponding right coset Try to choose wisely here It'll turn out folks if you try to choose The left coset that happens to be the subgroup itself the subgroup itself is always a left coset and it's always a right coset So don't look there if you're trying to show that the subgroup is not normal It turns out you can show that if you compute Mu two no row one H That that's not equal to H row And all you need is one particular situation where the left coset doesn't Equals corresponding right cousin in order to conclude that the subgroup is not normal In fact if you wanted you could have picked the example Provided by problems six and seven in section ten because there you started with D four Right, and you've looked at a specific subgroup. I think it was What was it row zero delta two or something like that? I forget what subgroup you started with in question six there but the point is yeah, so so Row zero mu two. Yeah, so in that particular case if you start with the subgroup row zero and mu two That's a subgroup and it turns out to be not normal. So I'll throw this as a non example contrast with G is D four H is row zero Mu two is Not normal and maybe that's even a little more compelling because in this situation The subgroup I gave you had three elements sitting inside the group This one happened to be normal the subgroup inside that same group, which wasn't normal only had two elements You might think all right well Give me an example where you've got two subgroups and they each have the same number of elements where one's normal and one's not And here's a good example of that. This is a normal subgroup of the group D four It has two elements. This is a subgroup with two elements of D four. It's not normal So let me get to the most important punch line involving normal subgroups Which is what we're about to show is if you start with a normal subgroup of a group and You start doing set products That the set product of any left coset with any other left coset Will kick out another left coset in other words that possibility one of a binary operation on the collection of left cosets Actually works as long as you start with a normal subgroup. So here's the big theorem theorem suppose H is A normal subgroup of G group of G I'll put it in parentheses IE the hypothesis because I'd like you to start using this notation H is a normal subgroup of G. We'll simply write that here, and here's the punch line then the set product of any to Left cosets of H and G Produces another left coset of H and G Let me rephrase that rephrased This binary operation the proposed binary operation number one proposed First binary operation operation on the set S on the set script S of left cosets given by AH I Called it star with a circle around it BH equal to the set product of AH with BH Yields gives a binary operation So if we happen to start with one of these nice subgroups Then the way we'd like to combine these things Actually works Alright, here's the proof. In fact, here's what we're going to show we show that in this case that when we compute AH Multiply everything in there with BH That we actually get That left coset Man say wait a minute. I thought you were worried about well-defined nonsense Well, I was when I simply deemed AH Combined with Whatever we called the thing double star BH to be that but now I'm giving you a method for computing What it means to take that with that and folks the definition of how you combine the two left cosets Is independent of what I call the thing just take this left coset I don't care if we call it ah a prime h a double prime h take this left coset. All you're doing is for me set product So it's independent of the name So the operation is well-defined the question is whether or not what it kicks out is another left coset And the answer is yes, it does kick out another left coset and I'll tell you which one it is that one Okay, here's how to do this do it. Here's the hard part the other part will be easy Hard part meaning Here's the part that actually requires the subgroup to be normal will show That if you do set products That you get something that's always inside That left coset. So here's what we have to do in other words. We show Well, let's see. What is the generic element inside the set product of that left coset with that left coset? Look like it looks like you take something in here that for every element, let's call it ah one That's what a generic element in here looks like and for every for every B h2 in this other guy bh that when we combine them together Ah1 bh2 is actually in a b h Hmm. We're gonna do that you might say well just Flip those two Then you'll get a b and then h1 h2 But since h1 and h2 are each an h and a subgroup is closed then you'll get something in here But folks I can't just flip things because I'm not necessarily a viewing group Of course if I was in the viewing group, I'd be able just to flip them around I'd be done But I'm not in general but the hypothesis that I have is that the subgroup is normal So here's the point but the hypothesis says That every left coset is a right coset in particular this left coset is oh some right coset But not only is it some right coset. I know what right coset it is It's that right coset by hypothesis That's the normal hypothesis. That's hypothesis that h is normal and g the hypothesis that h is normal and g So what in particular? Anything in this set is in that set In particular, I'm gonna identify something in there this element h1 b Well, wait a minute h1 b is by definition in that right coset. That's not an issue It's just something in h times b, but that's bh. That's the hypothesis hypothesis So h1 b Oh is in bh in other words that thing h1 b can be written as b times I don't know h3 for some h3 and h So it means to be in a left coset. It means you can write it as whatever the left coset representative is Combined with something in the subgroup. So here's what we're gonna do so oh a H1 bh2, this is what we're interested in saying something intelligent about Equals well, I'll just slide some associativity in here Societivity of bounds I can group things anywhere I want equals a but wait h1 b is bh3 Just substitute or just substituting what we Observed about h1 b. Oh Which equals a b? I'll combine that with h3 h2 But wait a minute h3 is an h and H2 is an h And did it do and h2 is an h So h3 h2 is an h So the punchline is that this is in a b h And so what we've just shown is that if the subgroup is normal and you do set product with two left cosets Then necessarily the result lives inside the left coset ABH so what we've just shown is this so We can conclude that one direction is true a h do the set product with bh is contained in a bh Notice folks to get there. We needed the hypothesis that the subgroup is normal Now in order to convince you that this set product is actually equal to a bh because that's the goal We just have to show inclusion the other way But inclusion the other way is going to be easy and it's not going to require that the subgroup is normal inclusion The other way always happens for the other direction To show that a bh is contained in The set product of a h with bh. That's easy to do pick something in here pick a bh1 in a bh We have to show that it's in here And I'll do that just in one step here. I have to somehow convince you that I can write a bh1 as Something of the form something in that left coset combined with something in this left coset But a bh1 can be written as a e bh1 No problem there. That's just a so I've just used associativity in the identity element of G because that's just a so that's a bh1. Oh, but wait a minute. This is in Well, that's in a h Why because it's something in it looks like a times something in h How do I know the identities in h because h is a subgroup contains the identity? Oh So this is in a h and I've multiplied it by or combine it with something that's in bh so check and you'll notice to go the other way I Didn't need that the subgroup was normal just to get the Containment going the interesting way or the hard part. That's the easy part that second part So here's the big theorem it says that if you start with a nice subgroup of the group Then if you form set product of two left cosets, you necessarily get another left coset back. Okay Questions comments So we've now written down a couple of things that are equivalent To say that every left coset is a right coset is equivalent to saying that every left coset a h equals its corresponding right coset h a For the homework in problem number 28 what you showed is that if you have this property that G inverse hg is an element of h for every element g in the group then necessarily Every left coset is a right coset So in fact you've written down in problem 28 a condition that implies the subgroup is normal That condition actually happens to be an if and only a statement If you have that condition from problem 28 the subgroup is normal conversely if the subgroup is normal You can show that that condition applies So in fact now there's three equivalent statements the subgroup is normal a H to left coset equals right coset h a for every element in the group Statement three that's equivalent is for every element of the group call it a or call it g I don't care if you take g inverse And you combine it with something in the subgroup h and then you look at g you get something back in the subgroup That's condition three. That's equivalent We've shown now that each of those Conditions implies that if you do the set product of any two left cosets of h you get another left coset That's condition four and what I'm going to leave out is that condition four is actually equivalent to those That's sort of a strange statement So I won't write it down or prove it But the statement is if it's the case that you have a subgroup with the property that whenever you do set product of any Two left cosets as you get another left coset back It turns out you can show that that implies that the subgroup is normal Now there's this other condition floating around which is how do you make possibility to work out? How do you make it or hopefully rig it? So that you're looking at a subgroup sitting inside a group that has the property that if you simply deem Ah Double-star bh to be the left coset a bh so that's well-defined answer you throw that on the stack to that's equivalent to all these conditions So there's a lot of things on the surface look completely different But in fact turn out to be completely equivalent in the end and equivalent to this condition that the subgroup is normal We just go through them once more and then we hand out a sheet about those and we'll call it a night here They are that the subgroup is normal in other words that every left coset is a right coset They are that Every left coset a h equals its corresponding coset h a They are the problem 28 condition The g inverse hg is contained in h for every element g in the group and every element h in the subgroup They are that when you take the set product of any two left cosets you get another of coset back and they are that the Binary operation on the collection of left cosets defined by deeming a h double-star bh to be a bh is well-defined Yeah So let me give you the homework and let me tell you what's in store So we're gonna have a quiz so the first chunk of the homework is that quiz three happens this Wednesday quiz three Wednesday Be the first I don't know five or seven minutes of class And what you'll be asked to do on Wednesday is write down for me the conditions that I just Provided you that are equivalent and what I'm going to do is I'm going to hand you The statement of what I'm going to have you do on Wednesday And I'm going to hand you all the solutions So if nothing else all I'm going to ask you to do on Wednesday is sort of spit back at me Well, I've listed it out as six conditions I've written the subgroup Subgroup being normal I've listed out condition six and we haven't talked about that yet, but I'll have you write it down for me on Wednesday anyway It turns out this is the big punchline folks We know now how to write down a condition on the subgroup Which will ensure that we can write down a well-defined binary operation on the collection of left cosets The condition is you start with a normal subgroup and in the end The punchline is that not only do you get a well-defined binary operation on the collection of left cosets Then in fact that binary operation Turns the collection of left cosets into a group And that's the group that we're going to start studying on Wednesday So statement six is not only do you get a well-defined binary operation, but that makes the collection into a group We haven't proved that yet, but you understand what that statement means. We'll prove that right after you're done taking this quiz on Wednesday But as I mentioned before the hard part Is the part that we just proved the hard part is to show that if you start with a normal subgroup that you at least get a binary Operation you at least have sort of the primordial goo that you could potentially build a group from writing down the three properties Associative existence of an identity and existence of inverses is only going to take us ten minutes at the beginning of Class on Wednesday because we've done all the hard work in actually getting to the point where we've got a binary operation Okay, so when you walk in on Wednesday, I'm going to hand you a sheet with just the first sentence there Let HB subgroup G the following statements are equivalent actually I'll probably give you number one and then I'll let you crank out statements two through five Okay, here is the remainder of the homework and this homework will be due on the standard homework schedule So it'll be due a week from Wednesday this homework assignment includes a Includes a homework sheet together with some problems from the text so here is the collection of problems that I'll have you do And if you're watching on video or online, then I'll post these I should have these up on the website by tomorrow And I'll post the homework there too. So here's home again quiz Wednesday The remainder of the homework assignment is due a week from Wednesday So when the heck is that the 17th? Wednesday October 17th I want you to first do in section 11 problem 1618 and 21 through 26 I want you to turn in 16 and 24 those are questions about That's some more business between direct products of the viewing groups Then I want you to do this sheet that I just handed out sheet on normal subgroups And I want you to turn in the entire sheet So that's a turn in piece of the assignment and Then in section 14 problems 9 through 16 21 24 and 40 and I want you to turn in 12 21 and 40 a it's be relatively large assignment. All right