 In this video, I want to introduce the notion of a factor group or sometimes called a quotient group. You'll probably hear me say quotient group more often, although in Tom Judson's textbook, he typically calls them factor groups. It's very, very common, both terms. In my opinion, factor group is often introduced first in sort of like introductory abstract algebra courses like the video series you're watching right now. But as mathematicians mature, it tends to gravitate more towards quotient group. But again, maybe that's just my perspective. It's probably a colloquial thing. There's perhaps different regions of the world that use different terms. For me, being in the Intermountain West, quotient group seems to be the most common term when describing the structure we're gonna develop right now. So before we introduce a factor group or a quotient group, let me first introduce the idea of coset multiplication, which is what the factor group's operation is gonna be. Now, as a lima, to help us get to the, the coset multiplication, let's take a subset H of a group G. Let me, I don't wanna prove to you that if you take the product of H times H. So this is the product of sets, the so-called Frobenius product we've introduced earlier in this series. If you take the product of H and H, this gives you back H. So in terms of Frobenius multiplication, a subgroup is actually an idempotent element. Its square is equal to itself. Now, as these are sets, we wanna show two sets are equal to each other. We're gonna show their subsets of each other. That proves the inequality. So we're gonna first prove that H times H is a subgroup of a subset of H. That's what we're gonna first do. So take an arbitrary element of HH that we'll call it X. Well, if you belong to HH, that means there's gonna be elements little H and little H prime, such that X can factor as HH prime. That's what it means to be in the set HH. Everything in HH is a product of something with H with something with H, which those elements do not have to be the same thing. But H is a subgroup, right? So H is closed under multiplication. So if I take a product of two things in H, I'll get you something in H. So this thing belongs to H and that proves the first direction. HH is less than or equal to H. It's contained inside of H. The other direction is fairly simple because if I take something in H, then H can be factored as H times the identity. Because after all, again, it's a subgroup. It'll contain the identity in which case then H times the identity is equal to H. But that's also a product of things inside of H. So HE will be part of HH. And that gives us the other direction. You'll notice here that we assume that H was a subgroup. We use closure under multiplication. We use the inclusion of the identity. We actually don't need inferences to make this property hold. Just sort of as a little side that I just wanna throw out there that this equality held even if we didn't, even if the set's not closed under inverses, which of course it is true for subgroups. All right, so with that definition out of the, excuse me, that limit out of the way, let's now define the idea of a coset multiplication. So now let's suppose we have a normal subgroup. What do we know about normal subgroups again? Normal subgroups, we could define those as those subgroups closed under conjugation. So we could think of this way. So GXG inverse belongs to N for all X inside of N and for all G inside of G. That gives you, you could define that to be a normal subgroup. That's equivalent to be a normal subgroup. This condition of course tells you that GNG inverse is equal to N, right? It's closed under conjugation, which the way we originally defined it, which follows from this equation right here, we say a subgroup is normal if all of the left cosets are equal to their corresponding right cosets. So GN equals NG. That's the one we're gonna use right here. So we have a normal subgroup and take two arbitrary elements of the group G. Now consider the Frobenius products of the cosets AN with BN, all right? So again, what we mean by this is that when you take this AN and this BN, this Frobenius product, we're gonna take all the possible products of X and Y where X comes from AN and Y comes from BN, excuse me. So we're just taking all the possible combinations of these products from AN and BN, okay? So this is just the Frobenius product which we introduced previously. Now, one thing that's true about the Frobenius product is that it is associative. You can redo the parentheses and that doesn't change what the product is. And of course, when you see a single element, really you're thinking of it as a single 10, right? These are single 10s in terms of that Frobenius product. Now this is a statement that I made but we never actually proved for which if you wanna see a proof of it really quick, it's not too difficult of really an argument. So if we have sets like X, Y and Z which are subsets of a group, not necessarily subgroups, consider X times YZ as a set. What this is gonna look like is it's gonna look like all the possible elements of X times YZ, right? As X, well, let's back up a little bit. We'll call this one, we'll call this one W for the moment, right? All of the X times W's for which X is in X and W is in YZ, okay? But you can then equate that with all the elements X times YZ after all where X is in X, Y is in Y and Z is in Z. And the connection here of course is that what is the set YZ? It's just all the possible products of things from Y and Z, like so. So that's what X times YZ looks like. On the other hand, if you look at the set X, Y times Z, this is equal to, well, it's gonna look like W times Z where W belongs to X, Y and Z belongs to Z. But like we observed above, that W could be replaced with things of the form Y, X, Y times Z where X is in X, Y is in Y and Z is in Z, right? Cause after all, what is the set X, Y? It's all the possible products of things from X and things from Y. And since you're in a group, these things are associative. So element-wise, there's this correspondence between X, Y, Z and X, Y, Z. So that'll force equality of these sets. So the fermionist product is an associative operation. I just wanna make that very explicit here. So as I redo parentheses in situations like this, right, you redo the parentheses, there's no concern whatsoever. The Frobenius product is associative because the group product is associative. If the group was abelian, this would also prove that the Frobenius products are also abelian. The products of these sets is abelian. If that was the case, we don't need that right here. All right, so when you look at A N times B N, so this is a product of two cosets. I can redo parentheses, so I get a singleton A times the coset NB times the set N. Right, so now this is where normality comes into play. Because the subgroup is normal, we get that NB is the same thing as BN. I can commute those things around and there's no problem there whatsoever, okay? So that's, again, I'm gonna leave this on the screen. This is where normality is actually used. Then we redo parentheses again. This becomes ab times NN. And the previous limit we just showed that NN is the same thing as N right here. And so then we see that when we highlight what happened here, the product of two cosets is itself a coset, okay? So this actually defines a multiplication. So if we take the set G mod N, so the G slash N right there, it's often pronounced G mod N. That's the set of all left cosets, which for normal subgroups, there's no difference between left cosets and right cosets. So you could call it right cosets if you want to. And this turns out we actually have now a binary operation. I can take a left coset and I can times that by a left coset and this produces a left coset. So on the set of cosets, I now have a binary operation. What type of operation do we have here? Well, it's an operation that forms a group. And that's this theorem 10.1.12 right here. If you have a normal subgroup, then coset multiplication as defined above forms a group on the set of cosets. So the set of cosets itself is a group. We made a new group using a group in one of its normal subgroups. And this subgroup, this group is called the factor group or the quotient group. And like I said, it's pronounced G mod N when you read this G mod N right here. Okay, so why is it a group? It has to be associative, has to have an identity, has to have inverses, okay? Associativity, I've already argued to you, right? The Frobenius product is associative as we just saw and coset multiplication is just a special case of Frobenius multiplication. So in the Frobenius product, we take any product of two subsets of a group. For the coset multiplication, we're just restricting our attention to only Frobenius products of cosets. That's an associative operation as we've already argued. This operation has an identity, okay? How do we know it has an identity? Well, the idea is the following. If you take EN and you times it by any coset, AN, right? Well, the rule from above gives you that you take EA times N, but as easy as the identity, you'll just get back AN. And since you have an associative operation, right? You have an associative operation if someone walks like an identity and quacks like an identity, it's gonna be the identity, right? It's gonna be the identity, it's gonna be unique, great. Same thing with inverses, right? I actually claim that the inverse of a coset GN is just gonna be the coset represented by its inverse. How do we know that? Well, take GN and times it by GN inverse, what happens? Well, by the coset multiplication, you're gonna get GN inverse N, which those become the identity and the identity is the identity coset is actually the identity of the quotient group right here. And so again, if it walks like an inverse and quacks like an inverse, then it's an inverse. So it has inverses as identities and it's associative, boom, it is a group. And that's what we call the quotient group, of course. So I want you to note that we've defined coset multiplication above using the subset multiplication, the Frobenius multiplication in a group and not based upon the representative of the coset, right? So like I mentioned earlier that since the Frobenius product is associative, we can redo parentheses. And so that's irrelevant of the representation. So when you talk about like AN and BN, right? You could choose different representatives. That would change this product AB, but I wouldn't change the coset ABN. Like the representatives could change, but the coset would still be the same. And so we've defined coset multiplication using the Frobenius product, which is well-defined, is irrelevant of the representation here, but it does require that be a normal subgroup in order for this equality to be validated. Now oftentimes, in like a first semester abstract algebra course, when they define coset multiplication, they don't introduce the Frobenius product and all of this intermediate stuff is missing and they're just like, oh, we're just gonna say that this coset times this coset is equal to this coset. And then they have to go through this long tedious argument that this multiplication is well-defined, that it doesn't depend on the representative. We of course, I mean, it's gonna take some effort to develop, but we took the approach that we defined at using the Frobenius multiplication for which no tedious, well-defined argument is necessary here because we've already established that, but we also have the benefit of we can use the Frobenius product or other things in addition to coset multiplication. So we don't have to prove that at all. Also another thing to be aware of is when you were talking about the quotient group, g mod n, the elements of the quotient group are cosets, right? Not elements of g. So g mod n is a set of sets, those sets being of course called cosets. So the elements of the factor group are not elements of the group. They're actually subsets of the group itself. So that distinction is important. Let's look at a couple of examples here. So we have to take groups with normal subgroups and then we can talk about a quotient group. So let's take the group S3 and let's take its normal subgroup, the alternating group A3. We've proven previously that the alternating group is normal inside of S3. The easiest way to see that it has index two, therefore it's normal. The cosets associated to the alternating group are first itself, one of the cosets is always the subgroup itself. Now, because you're index two, the other coset is actually gonna be everything else. So you have the even permutations and the odd permutations. And for S3, that is you're gonna take the identity with the three cycles and then the two cycles as well. Okay, those are gonna be our cosets. And so to form the factor group, these are then gonna be the two elements of the factor group, okay? So the factor group contains two elements for which I'm gonna show you the Cayley table with the coset multiplication over here. The alternating group is gonna serve as the identity element. If you take the alternating group times itself, you get back the alternating group. We saw that in our first lemma. If you take the alternating group times the other coset, according to coset multiplication, you take the identity times one two, which is one two. And this is gonna be the coset right here. And I want you to kind of verify that. If I take A3, if I take A3 and I times it by one two A3, right? This is a product of sets. So you're gonna take one, one two three, one three two. And then you're gonna times that by, I'm gonna scooch this over. And you're gonna times that by one two, one three, and two three. And so when you look at all the possible products right here, you're gonna get the identity distributed through, that's easy. So you're gonna get one two three, one three two. And then you're gonna hit two three, like so, okay? What I want you to commit yourself now is when you take the product of one two three and one two, you're gonna get that one goes to two, two goes to three. So you end up getting this friend right here. When you do one two three with one three, you're gonna see that one goes to three, three goes to one, so one is fixed, and you end up getting a two three right here. And when you take one two three times two three, what you're gonna see happening there is one goes to two, and then two is gonna go to three, which then goes back to one. So you end up with this element right here. So you got the same three elements, it just showed up a second time. And now if you take one three two times one two, you get one goes to two, two goes to one, so one is fixed, and you're gonna see that one is fixed. So you're gonna get a two three right there. Then you take one two three times one three, one goes to three, three goes to two. So you're gonna get one goes to two, which is this one right here. And then finally you get one two three times two three. You're gonna get that two goes to three, but then three goes back to two. So two is fixed, and you get this element right here. All right, so when you take the product of a three with one two three a three, you get the three two cycles, and they actually show up three times. But you know, we don't really care about the multiplicity of such things. Well, I mean I do. I certainly as an algebraic commentatorist, I really care about the multiplicity of those things. But you know, we don't need to worry about that in this lecture right here. But the point is when we take the product of a three and one two a three, we get back one two a three. And then if you take one two a three times one two a three, you get back a three. And so this is a product, right? If I take this set right here and I square it, you're gonna get back these three elements and they're gonna show up with multiplicity three times. That can get a little bit tedious as you saw on that. The cool thing about coset multiplication is that you don't have to draw out all of the combinations. You only have to look at the representatives. So you get one times one is gonna give you one. You're gonna get one times one two because you want two. You just have to look at the representatives, right? One two times one two gives you one. And so that speeds up this calculation so quickly. Now, when you look at this Cayley table right here, this Cayley table looks a lot like the, well the cyclic group mod two. And that's because we see that a three mod a, sorry, excuse me, s three mod a three is actually isomorphic to z two. It's a cyclic group of order two. And in fact, this is something we see in more generality. If you take sn and you mod out an, you always end up with z two. One important thing to know about factor groups here is that if you have a group g mod n and you look at its order, well, the order is the number of elements inside the group, but this is this group of all the cosets. So the order is gonna be the number of cosets. The order of this group is gonna be the index of this normal subgroup. Well, since a n has index two, this group will always be order two. And there's only one group of order two, which is z two right there. Let's look at another example. Let's take this time the example. We're gonna take the integers and we're gonna look at the normal subgroup three z. So we're taking multiples of three. Well, this group has index three, right? There's three options, right? You're either gonna have multiples of three. So those numbers which are congruent to zero mod three, one mod three and two mod three, okay? And so when you start combining these things together, what do you get? Well, if you take a number, which is a multiple of three and you add it to multiple of three, you get a multiple of three. If you add it, if you take zero, that is when the remainder is zero and you add it to the remainder one, you're gonna get a remainder one. And when you take the remainder zero and combine it with a number which has remainder two when you divide by three, you're gonna get a remainder two, okay? And when you go through all of these things, right? So this here is our identity. We get zero plus three X right here. So this should just look like the row right above it, okay? If you take one and zero, you get one. If you take one and one, you get two. If you get one and two, you're gonna get three, a three plus a multiple of three. But hey, you can actually suck that into the three Z there and actually this is the same thing, right? And when you look again, go through all these combinations. This Cayley table right here is the exact same Cayley table as working addition mod three, up to relabeling of the symbols. And we in fact see that, that Z mod three Z is isomorphic to the cyclic group of order three. And this happens in greater generality, right? If you take Z mod NZ, this is equal to, this is isomorphic to the, this quotient group is isomorphic to the cyclic group mod N. Let's see, we'll do one more example and then we'll finish up the video there. So let's look at the dihedral group, right? The dihedral group Dn has a very natural normal subgroup that is the rotational subgroup. We mentioned that in the previous video that I mean, you could see that because it's index two. You can also see that every rotation, the only conjugates are itself and it's inverse. So the rotation subgroup is a union of consciousness classes. So it's gonna be normal, it's a normal subgroup. So we can mod out the normal, this normal subgroup. If we take the dihedral group and we mod out the rotations, well then we're gonna get back Z2, okay? Cause like again, we saw this above with the, with the symmetric group. If you mod out, this is subgroup, the rotation subgroup has index two. So when you mod it out, you have to get a group of order two. There's only one possibility, which is the cyclic group of order two. Now you'll notice that of course that this rotational subgroup R is itself a cyclic group of order N, it's isomorphic to N. And so sometimes in the dihedral group, you actually denote it as ZN or more commonly ZN without the blackboard font there. And so you'll see this, that DN mod out its cyclic subgroup turns out to be Z2 as well. And so this gives you some examples of computations involving these quotient groups. We'll of course do some more of these in the future. In the next chapter of Judson's textbook, chapter 11, we're gonna introduce the notion of a homomorphism, which will eventually lead to our, what's sometimes called the fundamental theorem of homomorphisms or what we'll call the first isomorphism theorem that tells you that there is a deep connection between the quotient groups we introduced in this video and homomorphism which we'll introduce in the next chapter.