 Let G be a group with subgroups H and K inside of G. And let's suppose that the product, the Frobenius product of H times K is equal to the whole group G. Further, let's suppose that the intersection of H and K is trivial. That is the only thing that's common to both H and K is the identity. And let's further suppose that every element of H commutes with every element of K. So if you take something in H and you times it by something in K, that's the same thing as times it by those same elements in K and H right there. Now, I'm not claiming that H is commutative or that K is Abelian, right? I'm not saying that H commutes with anything in H or K anything with K. I'm just saying they commute with each other. Or oftentimes we say the statement right here that's now underlined. We say that they centralize each other. So G is the product of two subgroups which intersect trivially and centralize each other. In this situation, we say that G is an internal direct product of H and K, which is kind of a curious thing as like internal direct product. What does that mean, right? We've learned about direct products before for which H and K could be 82 groups under the sun and we can build a new group by taking all the possible ordered pairs of H and K as these range over the elements of the two groups. So this is what's commonly referred to as the external direct product. The internal direct product we see here is that you already have a group in hand. We didn't build a group. It's already built. But looking inside of the group, we can find subgroups which in some essence will act like it was a direct product. And we'll be very specific about that in just a second but let's look at some examples of what's going on here. So let's take the group G to be Z8 star. So Z8 star will be the multiplicative group mod eight. We take those integers which are co-prime to eight and we work with them multiplicatively. So two possible subgroups we could consider. You take H, if it's a cyclic subgroup generated by three, well three times three is nine, which is one mod eight. And so therefore the cyclic subgroup generated by three is just cyclic of order two, one and three. The same thing happens when you take the cyclic subgroup generated by five, five times five is 25 which is 24 plus one. So five will be order two, mod eight. So we have this right here. So notice if you take H and K, these are both cyclic subgroups, so they're subgroups. If you take their intersection, the intersection is gonna be one, right? The only thing that's common to both is the identity there. So we have that. And then what does their product look like? If you take H times K, you'll remember from a previous video that the cardinality of the set H times K, this will always equal the order of H times the order of K divided by the order of their intersection for which we then see that H has order two, K has order two, their intersection has order one. So you get two times two divided by one, which is four. So we have a subset HK living inside of G. Well, G has order four and HK itself is a set of size four. So this force is equality, right? Since G is order four right here, that means that HK has to equal G. This is a very common trick that if the size of this Frobenius product is equal to this order of the group, then it has to be the whole group. So HK is equal to G. So boom, we have that, their intersection is trivial. And in this case, G is an abelian group, so everything commutes with each other. So in particular, the subgroups will centralize each other. So this shows that in fact, whoops, we need the check mark there, that shows that G is in fact, the eternal direct product of three and five. Now I do wanna make a comparison here before we go on, right? This group, Z eight star, this is none other than, I mean the climb forward, this is isomorphic to Z four, which is just Z two cross Z two, which notice here, H is a cyclic group of order two, K is a cyclic group of order two. And now we're saying that G is an internal direct product to cyclic groups of order two, that's not a coincidence. We'll see before the end of this video that that's actually to be expected. Let's do another example before that though. This time, let's take the group, D six, so the dihedral group of order 12, the symmetries of a regular hexagon. Let's take as our first subgroup H, the cyclic subgroup generated by R three. This is likewise a subgroup of order two. R three is gonna be the 180 degree rotation, okay? Let's take as K, the subgroup generated by R squared and S. This will contain six elements. It contains one R squared R to the fourth S, R squared S and R four S. This group right here is actually just the dihedral group D three, right? Because if we're working in D six, a single rotation right here would be 30 degrees, for which a, I said that wrong, I'm sorry, it would be 60 degrees, would be a single rotation, right? And so then a double rotation would be 120 degrees. So if we just do the elements squared, you're gonna get 120, 240 and 360. This is just the rotational symmetries of a regular triangle, the equilateral triangle, right? Same thing with the reflections. So this group right here K is just D three inside of D six, or that's also isomorphic to S three, the symmetric group on three letters right there. So in particular, when you look at the intersection, the only thing that's common to both is the identity, right? R three is not inside this set. So their intersections trivial. In terms of their product, right, H times K, well H has order two, K has order six and their intersections one. So two times six divided by one is 12. That is the order of the dihedral group, right? D six, its order is equal to two times six, which is 12. So again, by counting argument, K, HK is equal to the whole group. Well, do they centralize each other, right? Well, if you look at the element of H, there's only two elements there, right? You have the identity. The identity will commute with everything because whether you have one times the identity or excuse me, one times the permutation pi or pi times one, right? It's always just equal to pi there. So the identity centralizes everything, right? It commutes with everything. What about R three, for example? Well, this one we might have to consider using normal forms. So if you take R three times the biopermutation pi, well, let's say pi is a rotation. Well, R three times pi will then be R three times RK for which as an exponent, that's R to the three plus K but the exponent's commute, we get R K R three. So yeah, rotations commute with rotation. So R three would commute with any rotation. But in terms of reflections, right? The typical reflections in D six look like R K S. So if you take R three times pi, you get R three, R K S. Well, like we saw before, R K will commute with R three. And then in terms of S here, commuting S with R three, there's a toll that has to be paid. You have to take the inverse of R to pass it by S. But as we're working mod six, negative three and positive three are actually the same thing. So you get R three again. And so it does in fact commute. So recall that the center of a group, right? This is the set of all elements Z, all elements Z inside of G, such that Z G is equal to G Z for all G inside G. So the center of the group is the set of all elements which commute with everything, okay? And so this turns out to be a subgroup, right? It turns out to be a normal subgroup, something we'll define a little bit later. So what we see, what you can see here is that the center of D six is actually equal to this set H that we've introduced. And so in particular, H and K centralize each other since H is the center, it centralizes everything. And so we got all the conditions. Let's see intersection of H and K is trivial. Their product is the whole group and they centralize each other. So again, D six is the internal direct product of its center with D three. Let's look at one more example of this, of this internal direct product. This time let's take S three, we're gonna take H to be the cyclic subgroup generated by one, two, three. So this is actually the alternating group A three which contains the identity one, two, three and one, three, two. Let's take the subgroup K which is just a cyclic subgroup generated by one, two. So it'll contain one and the transposition one, two like so. When you take their product, a simple counting argument gives you this again, right? So the size of HK, this is gonna equal three times two divided by, well their intersection is gonna be one, right? The only thing common to both. So this gives you six which is three factorial which is the size of S three right here. But notice that these subgroups do not actually centralize each other. So we have their product is the whole group, their intersection is trivial but these things don't actually centralize with each other. Notice what happens, right? When you take one, two, three times one, two, that's gonna equal one, three, right? So one goes to two, two goes to three and then three goes back to one, two will be fixed so that product's true. On the other hand, if you take one, two times one, two, three, right? One goes to two, two goes to one. So one is fixed, two goes to three and then three goes back to two. So these two products are not, they don't commute with each other. So H and K do not centralize one another. So S three is not an internal direct product of H and K. It's close, it's really close. In this situation, this is what one actually would call an internal semi-direct product but that's a topic for another lecture. I don't wanna get into that anymore here. So let's actually explain why we call this structure the internal direct product. So suppose G is an internal direct product of subgroups H and K. So remember our axioms here that G is equal to H times K. We have that H intersect K is trivial and we have that they centralize each other. Centralize each other. I should come up with some shorthand for something like that. So they centralize each other. Then in that situation, then we can prove that G is actually isomorphic to H cross K which this right here, H cross K, this is referred to as the external direct product as we mentioned beginning this video because H and K are genuine subgroups of G, right? They're inside of G hence the internal right here as opposed to H and K might have anything to do with each other in general when you take the direct product you form G. So from externally we build G or internally can we factor G? These two notions really are the same thing because up to isomorphism is the same thing and for group theories we don't distinguish between things that are isomorphic. So since G is an internal direct product we have our assumption that G equals H times K. Ka-chingk, that's the first one right there. And as such we're gonna then define a function from H cross K to HK by the following rule. Phi of HK is equal to HK. Now I want you to be aware that this is the same function we used in the proof of theorem 9216 in this lecture series. It's actually was just the previous video of this lecture series right here. And so some things we proved about that function is that first of all it's well-defined surjective function that's important to know. Also be by the assumption G equals HK. Or G equals HK. Notice this is actually a function from H cross K to G. It's a surjective function. But we also mentioned in the previous video that this function right here that every pre-image, every pre-image of an element, phi inverse of HK, right? This thing has as its set size the intersection of H and K. Basically two things will map to the same element in a one-to-one correspondence to elements of HK. And so we made the statement that this map will be injective if and only if H intersect K is trivial, but could chink, that's the assumption we have right here. So based upon the results of that previous theorem we have that this map phi is in fact a bijection. Well that's a pretty good direction to go for an isomorphism. So we have that the map phi from H cross K to G, so far it's a bijection. To show it's an isomorphism we need to have the homomorphic property. So let's consider that. Which again, I want you to be aware that this is gonna be a fairly simple argument here. Let's take two arbitrary elements of H cross K. So take two elements of H, we'll call them little h and h prime and take two elements of K, we'll call them little K and K prime. And so therefore, if we were to, if we were to take phi of HK times H prime K prime. So this is the product of two arbitrary elements in HK. This is gonna equal this element right here because again, you just component wise multiplication, multiply together H and H prime and multiply together K and K prime. So we get here, but then by the map phi, this will become H prime times K K prime. And by re-associativity, you know, this is a group after all, we can do the parentheses like this. And so this gets us to our last condition, kachinka, right, they centralize each other. So H and K commute with one another. So H prime K is equal to K H prime, right? For which then if you re-associate, you're gonna get HK H prime K prime for which the first one will be phi of HK, the second will be phi of H prime K prime. And so that then proves the homomorphic property. So we do in fact have an isomorphism between G and H cross K. So if you have an internal direct product, that means your group is actually isomorphic to this external direct product that I've mentioned a couple of times. So an example of this, if you take for example, the group Z six, you can argue that Z six is isomorphic to Z two cross Z three, where basically your generator is gonna be the element one comma one because the GCD of two and three is, excuse me, the LCM of two and three is six. So this will give us a cyclic group of order six. So it's gotta be isomorphic to Z six because all cyclic groups are isomorphic if they have the same order. Now if this is right here in an external direct product, we think of Z six. So if we were to elaborate here, what is Z two crosses Z three here? We think of it as the six elements, you get zero, zero, one, zero, zero, one. We're gonna get one, one. Who have we forgotten here? We're gonna get one, two. Feels like I'm going through this randomly. Zero, two, one, two, three, four, five, six. So there's six elements right there. On the other hand, Z six, we really think of it as like zero, one, two, three, four, and five. And so these groups are isomorphic. We've established that though. And so since Z six is isomorphic to the external direct product of Z two crosses Z three, then there should be some groups of Z six, one of which looks like Z two up to isomorphism. One of them looks like Z three up to isomorphism. And it should be then Z six should be then a product of all these things. So our candidates are gonna be the following. Take the cyclic subgroup generated by three inside of Z six and take the cyclic subgroup generated by two inside of Z six. Well, H here will contain two elements, zero and three, known as three plus three is six. So this is gonna be isomorphic to Z two. The subgroup generated by two on the other hand, this will contain zero, two, and four, right? Two plus two is four, two plus four is six, which is zero. And this is isomorphic to Z three. So we have our candidates here, H and K. Their intersection, of course, is the identity, right? So it's just, it's just zero is what's common to both of those things. And it's an abelian group, so they will commute with each other. And then their product, right? If you take H, in this situation, since Z six is an additive group, you probably would instead of writing H times K, you write as H plus K. But notice what goes on right here. You're gonna get the order of H is two, the order of K is three, divided by the intersection, which one? This is gonna use six, which is the order of Z six right here. So we do in fact see that Z six is the product of H and K, and it's an internal direct product. And so this is what the previous theorem was showing us here, that internal and external direct products are in fact one of the same thing. We can find the factor groups inside of the group up to isomorphism here.