 Welcome back to our lecture series math 4220 abstract algebra one for students at Southern Utah University as usual I'll be your professor today. Dr. Andrew Missaldein In lecture 29 here. We're actually gonna talk about one Surprisingly one of my favorite subjects in group theory the idea of the Frobenius product That is the product of sets And so let me kind of explain what I mean by that So we have been talking in this series mostly about groups so far, right? And because it's a group we have some operation which we'll call it multiplication on Elements inside of that group. So if we have elements g and h inside of g We know that g times h will be an element and g of g and this binary operation of elements Will be associative with identity with inverses in the Abelian case. It's commutative We can talk about these talk about these properties of the binary operation on the elements of the set What we want to do today in this lecture is kind of Broaden the broad in the base a little bit that in addition to multiplication of elements, we also want to talk about the multiplication of sets Inside of a group. So take two subsets h and k inside of g now. These are genuine subsets I make no statement that these are cosets or subgroups or there's not necessarily any structure to these sets These are just subsets of g. So these are collection of elements of things who belong These sets belong inside of a group. That's all we're assuming right now And so then we can define the product of the subsets h and k here in the following way We define a new set. So be aware here that if we take a set h and we combined it with this set K, we're gonna produce Some new set hk, which will be a subset of G, right? And so what exactly does that element look like here the set hk will look like the all the possible Products of little h times little k where little h is an arbitrary element of capital H and little k is an arbitrary element of capital K that is we look for every possible combination of products between the elements of h and the elements of k and now this right here will naturally be a subset of g like I was saying earlier since h is in is in g since h is a subset and since k is a Element of g then their product will be in g because we're in a group. So this set makes sense It's a subset of g And so we want to talk about these type of products of sets This is often referred to in the literature as the forbenius product of sets If you want to try to like look it up on Wikipedia or something And I want you to be aware that in the in our previous discussions about cosets This is actually a special case of what we're talking about right now that when you talk about the coset g h That's the left coset. Well g h is really just multiplying the singleton g by the set h Which happens to be a subgroup and so what we're trying to do is broaden the base that in addition to products of elements and in addition to cosets Which are again very important subsets of a group here We can talk about general products and so we can talk about the product of two cosets and things like that Now some things that become immediately clear when we talk about Multiplying sets together is that as the multiplication of sets is defined element wise and g this multiplication will likewise be Associative so if you have say like three sets you have like h k and l it doesn't matter how we do the parentheses It'll automatically be Associative this forbenius product because element wise it is h k little l here Is the same thing as h k little l right because it's associate element wise It'll inherit this associativity on the sets same thing that if if of course the group is like say a billion then Multiplication will be element wise commutative. That'll also imply that subset wise It'll be a cumulative operation as well Now what we of course can see and we'll see some examples of this is that? The sets could potentially commute even though individual elements might not commute. We'll see that in a little bit So let's take as some examples here Let's take the dihedral group d4 and let's take two subs two subsets We're gonna take the cyclic subgroup generated by s the reflection across the x-axis And we're gonna take the cyclic subgroup generated by rs where r of course is a 90 degree rotation there When it comes to this forbenius product the subsets do not have to be subgroups Although those are subsets of particular importance So if we wanted to compute the set h k what that means is we're gonna take the product of The set h which has a subgroup. It's just one and s and we're gonna take the product with the set k Which again as a subset this is actually a subgroup So it's gonna be one and rs So we look at all the possible combinations here So it's kind of like we foil this out a little bit We have to take the product of the identity first. We take the identity in rs outside We take the product of s and one Inside and then we take the product of s and rs last so again, it's kind of like the foil method right here You're gonna get one times one You're gonna get one times rs. You're gonna get s times one and you're gonna take s times rs Now you have to be very careful in the order here because d4 is a non-abelian group The order of multiplication does matter now with the identity It doesn't make much of a difference one times one of course is one one times anything is just that that that thing right there One times rs is rs s times one is s and then you're gonna see that s times rs when you get there remember when it comes to the dihedral group if you need a if you need to commute a Rotation with the reflection s you take the inverse of r. So you're gonna end up with right here You're gonna end up with our inverse s squared now as we're in d4 Our inverse is actually the same thing as r3. So we're gonna put that in normal form right there and s squared is the identity So so h times k is equal to one rs s and r cubed Okay, on the other hand if you take I'll do a different color on this one if you take k h You reverse the order of these things. So we're gonna take one rs And then times that by one s Well again looking at all the possible combinations you get one times one you get one times s You get rs times one and then you're gonna get rs times s For which one times one is the identity one times s is s rs times s is rs So so far, you know, these are the same thing in a different order But the order of a set doesn't actually matter here the final element though Of course, you're gonna get rs squared, which is just an r in that situation So notice that these two sets are not equal to each other Uh hk is not equal to kh Which isn't too surprising right because the multiplication of elements is non-commutative Multiplication of sets might also be non-commutative right it doesn't commute because the elements didn't commute Notice how s rs is not the same thing as rs times s and so that gave us a different product of sets there It should also be mentioned that even if h and k In this case are subgroups the subset hk is not necessarily a subgroup right? I mean if you look at this right here, this isn't a subgroup for example r doesn't have its inverse Which is r cubed on the other hand this one right here is missing its inverse, which is r So that is that's a realistic possibility the product of two subgroups is not necessarily a subgroup But all we're saying right now is the product of sets is going to be a set inside of the group Let's do another example. Let's look at the quaternion group right here on eight elements Let's take the set ijk and let's take the We'll call that h and we'll also take the set of their inverses negative i negative j negative k Notice these are not subgroups. Um, they're just they're just you know, they're just Uh, they're subsets. Um, if we take their product what we want is we want the product of i j and k With negative i negative j and negative k Like so looking at the possible combinations. You're going to get i times negative i, which is one You're going to get i times negative j, which is a negative k You're going to get i times Negative k, which is a j Next you're going to get j times negative i, which is a k you're going to get Let's see j times negative j, which is one Uh, you're going to get j times negative k, which is a negative i And then lastly you're going to get k times negative i, which is a negative j You're going to get k times negative j, which is an i and you're going to get negative k k times negative k which is equal to one, right? And so then rewriting this thing, I mean, you'll notice that the identity showed up three times, which as this is a subset, the multiplicity doesn't really matter, right? So we just list, we're just gonna list it once. Then when you look at everyone else, you'll see you end up with an i and a negative i, you end up with a j and a negative j, and then you end up with a k and a negative k. Negative one doesn't show up anywhere. So we end up with plus or minus i, plus or minus j, plus or minus k. This right here is just the quaternion group, take away negative one. That's the only thing that's missing in this consideration right here. I want you to convince yourself that if we go the other way around, if we take k, h, right? So now you're gonna get negative i, negative j, negative k, and you times that by i, j and k. You end up with all of these same elements, right? Let's convince ourselves of that. You're gonna get negative i times i, which is one. Negative i times j is negative k. Negative i times k is positive j. You're gonna get negative j times i, which is a k. You're gonna get negative j times j, which is a one. You're gonna get a negative j times a k, which is a negative i. And then lastly, you get negative k times i, which is a negative j. You're gonna get negative k times j, which is a positive i, and then a negative k times k is a one, right? And so you'll notice when you compare these things, right? Well, the identity showed up three times again, which admittedly, for the set, the multiplicity doesn't matter, so we get the identity. And then let's say we have an i and a negative i. They showed up each one time. You get a j and a negative j. They showed up each one time as well. And then you get k and negative k, right there. And so everyone showed up except for negative one again. So in particular, we see that kh is equal to hk. So in this example, it's even possible that a product of subsets could be, that even in an onabillion group, right? It's still possible that product of two subsets can commute. So h and k actually equals kh. The sets commute even though the elements didn't. That's an important thing to realize when we talk about this Frobenius product. Let's do one more example here. Let's look inside of the symmetric group S3, and let's take the subgroups, h equals one, one, two, three, one, three, two. That's the cyclic subgroup generated by, of course, one, two, three. It's also, so happens to be the alternating group A3, right? We don't need that observation though. And then k here is gonna be the set one and then one, two. This likewise is a subgroup. It's the cyclic subgroup generated by one, two. So we have a cyclic subgroup Z3 and a cyclic subgroup Z2 in play right now, okay? So let's consider their product. Let's take h times k right here. So if you look at all the possible products, one, one, two, three, one, three, two. We times that by one and one, two, okay? So in this situation, if you distribute, of course, the one, that's pretty easy. You get one times one, which is the identity, one times one, two, which will be one, two. Next, you're gonna go one, two, three times the identity, which is one, two, three, all right? This one takes a little bit more effort, right? So think about this one right here. We have one, two, three times one, two, right? So one goes to two, two goes to three. So we see that one goes to three. Now we're gonna see that three goes to one. So we're gonna bring that thing off. And then as this is a permutation, two would have to go to two. So that's gonna be that element right there. Then the next one, we're gonna take one, two, three times the identity, what I say, one, three, two times the identity. And then the last one, we have to take one, three, two times one, two. And you see what happens here is one goes to two, two goes to one, so one's fixed, okay? Two goes to one, what did I say there? So one goes to two, two goes to one, so one is fixed, right? Two goes to one, one goes to three. And so what we see here is the last element is gonna be two, three. You'll notice that, hey, we actually got every single element of the group. In this situation, it's even possible that the product of subgroups can in fact equal every element of the group, right? The whole group when you multiply these two sets together. And what you're gonna see is if I go the other direction that these two sets in fact commute, even though it doesn't happen element-wise. And again, we're gonna get the whole group. Let's convince ourselves of this. So you get one times the identity, one times one, that's a one. You're gonna get one times, that's a one, two, three there, what did I write down? Try that again, one, two, three, and one, three, two. When you take one times this, you'll get, of course, the permutation one, two, three. And then likewise, when you take one times the last element, you end up with one, three, two, that's pretty easy. If you take one, two times the identity, you'll get one, two. And then so we have to do one, two times one, two, three, right, what happens here? Notice one goes to two, two goes to one, so one is fixed. You're gonna see that two goes to three and then three will go to two as well. So we end up with two, three. And then the last possible product, one, two times one, three, two, you'll see one goes to three, and then three goes to two, but two goes to one, so three goes back to one, and so two has to be fixed as we only have three elements. And so yeah, we got the whole group again. These two sets, these two subgroups in fact, commute with each other. And so this gives you a variety of the type of products one gets when you start multiplying various sets together. And I hope this gives you a good feel that even in an on abelian setting, the sets can commute, even though it doesn't commute element-wise, subgroups can actually multiply together to give you the whole group. And we'll see some other interesting things coming up in the next video.