 In this video, I want to introduce the notion of a direct product of two groups. This is a way of building a new group from two already known groups. And it turns out that in Judson's textbook, the notion of the direct product is not officially defined until much, much later, but there was sort of like a naive principle used earlier in the text. And so I just actually want to officially define it right now so we can talk about it officially. So imagine we have two groups, G and H. And I've included their operations here to emphasize the fact that the operations between G and H don't have to have anything to do with each other. Circle is an operation for G and asterisk is a operation for H. They might have nothing to do with each other. But given these two groups, we can define a new group on the Cartesian product of H, of G and H right here. So remember, G cross H as a set is the Cartesian product. We're gonna be taking the set of all ordered pairs, G comma H, where G belongs to G and H belongs to H. So this Cartesian product is a set. Now to make it a group, we have to define a binary operation on that set. And what we're gonna do is we're gonna define what we call component-wise multiplication. So when you have two ordered pairs, you have GH and G prime, H prime, we're gonna define component-wise multiplication be the following. Well, since G and G prime both belong to G, we can operate on them by using the circle. So that's what we're gonna do for the first component, G circle G prime. Likewise, H and H prime both belong to the set H and therefore we have the operation asterisk that we can use. So we're gonna operate on those elements in that way, H asterisk H prime. And so this is what we call component-wise multiplication. I claim that this binary operation attached to the Cartesian product gives you a group structure and this is referred to as the direct product of G and H. So let's now prove that this binary operation satisfies the three actions of a group. It needs to be associative, it needs to have an identity and it needs to be, and these have inverses. So to show associativity, we have to take three arbitrary elements of the group. And as the group is the set of ordered pairs here, we're gonna just take three arbitrary ordered pairs. We can take a GH, a G prime, H prime and a G double prime, H double prime. And so then we have to operate them all together. So to do associativity, we're gonna do the second two first, the last two first and then the first one. So according to the component-wise multiplication, I should operate the G prime and G double prime using circle. I'm gonna operate together H prime and H double prime using asterisk, like so. This now gives us an ordered pair, which if we do component-wise multiplication again, we've put the first coordinate together. So we're gonna get G times G prime circle, G double prime, which does require the appropriate use of parentheses here. Next, we're also gonna take, for the second component, we're gonna take H to times H prime, H double prime. So you see this right here again. Make sure you have the parentheses. Now we have to be really careful about parentheses because again, we're trying to show associativity. If we're too careless with our parentheses, we might fall into the danger of circular reasoning. So we don't want to do that. Don't do that, that would be bad. Okay, so with that in mind though, if you look at just the first component, this is an element that belongs to G and because it belongs to G, G has an associative operation, we can redo the parentheses like we do here. And the same thing for the second component, this second component lives entirely inside of the group H and in H, the operation is associative, so we can redo the parentheses like we did here. Now it just comes down to unraveling this thing, right? So notice we're operating on the right now by a G double prime, H double prime. So in terms of component wise multiplication, we can take this G double prime and H double prime out into its own ordered pair. And if you're not convinced at all, take this thing and do component wise multiplication, you'll get back this thing right here. And then the same thing right here, since we're operating, operating, we can rip this thing apart by component wise operations. You get this thing right here. If you're not convinced, push it forward and you'll see that the two things are equal to each other. And so now what we've seen here is that if you operate on the last two first, that's the same thing as operating on the first two first. And that's the associativity axiom proven right there. So component wise multiplication is associative because the original two operations are associative. What about identities? Well, we need a candidate for the identity. And that candidate is actually gonna be the identity of G with the identity of H. So if you take G's identity as the first component and you take H's identity, which could be totally different things, I'm just gonna call both of them E in this context, abusing some notation there. But if you take the identities of G and H and put them together in order pair, that's gonna be the identity of G cross H. And so what do we get here? Well, E E times G H by component wise multiplication, you're gonna get E operated on G and E operated on H. That just gives you the element G H, right? So that's what an identity should do. But what if we check on the right as well? Well, by component wise multiplication you get G E and H E, but that's again, gives you back G H. So we have an identity element on G cross H, just as a quick textbook moment right here. If you wanna put that cross symbol, this is just backslash times in latex right there. So now inverses, we need a candidate for the inverses. Well, if you have an element G H, we're actually just gonna take as its first component, G inverse, and it's the second component, H inverse. I claim this is the inverse element. Well, by component wise multiplication, you put together the first components and the second components. Well, looking at just the first component, G inverse and G with respect to the circle operation, they're inverses, so you're gonna get back the identity. For the second component, H inverse times H, when you operate those together, you're gonna get back the identity. So G inverse H inverse is the left identity, as the left inverse, excuse me, but the same thing happens on the right. If you multiply this element on the right, you're gonna get a G circle G inverse, which is the identity, and you're gonna get an H star H inverse, which is the identity as well. So we see that G H inverse is just G inverse H inverse, and therefore G cross H has inverses and that establishes that G cross H is a group. The direct product is a very important way of building groups using smaller groups, and these do give us something that's potentially new from groups we haven't seen before. So for example, perhaps one of the most famous of all direct product groups is the so-called Klein four group, named after Felix Klein. And so the Klein four group is formed by the following way. We're gonna take the group Z two cross Z two, which that is we take this group of order two and we take all the possible pairings, right? So Z two, it's itself the group one and zero. And so we take all the possible pairings, we see four of them. So you get zero, zero, one, zero, zero, one and one, one. This is what's called the Klein four group. It's a group of order four. Now some things to notice about this group. Now, if we look at its subgroups, so you see the Haase diagram illustrated right here, the Klein four group is often denoted as V sub four. I'm not exactly sure why it's a V. I imagine that's probably a German thing. Don't speak German myself, maybe I ought to, but it's V four there. That'll be a subgroup. You'll have the trivial subgroup itself, which in this case, the identity would be zero, zero. Then you're gonna have three subgroups of order two. There's gonna be the subgroup that contains the identity and the element one, zero. There's the subgroup that contains the identity and the element one, one. And there's a subgroup that contains the identity and zero, one. They're really distinct subgroups of order two. They're distinct from each other. They're subgroups in their own right and there is no intermediate subgroups. So you get a Haase diagram that looks like the following. This is supposed to be seen in contrast to the cyclic, well, to the group Z four, right? Sometimes we prefer to as a cyclic group of order four. It is a group of order four and it's a billion. The Klein four group is also a billion. I should mention that. But when you look at the subgroup structure at the Haase diagram, you see something very different. The identity can make a subgroup, the trivial subgroup. There's the whole group itself, Z four. But as you search for subgroups of order two, there's only one subgroup of order two that's contained zero and two. The problem is if you try to take like one, right? If you're looking for a subgroup that contains one, if you have one, you're gonna have to contain one plus one, which is two. If you have one and two, you're gonna have to have one plus two, which is three, because it should be closed. But then you have everything at that point, right? The same thing happens if you do three. If you take a subgroup that contains three, you have to contain three plus three, which is two. You have to have three plus two, which is one, and then you get the whole group again, right? So when you compare these things side by side, right? The Haasei group for Z four, it'll contain the subgroup containing zero and two, and it'll contain the trivial subgroup. So there's only three subgroups. But with the Klein four group, you get these three distinct subgroups of order two. And so this tells us that these two groups are fundamentally different from each other. In group theory, we'd say that these groups are non-isomorphic, which is something we'll talk about in the future. But basically these two groups are not different representations of the same structure. They're two very different groups, even though they both are a billion of order four.