 Welcome back to our lecture series Math 42-20, abstract algebra one for students at Southern Utah University. As usual, I'll be a professor today, Dr. Andrew Misteldine. Lecture 28 is going to be entitled the Chinese remainder theorem in the statement proof and some applications of the Chinese remainder theorem we'll actually see in the next video of lecture 28. In the current video, I want to talk a little bit more about cyclic groups and their connection to direct products. So actually section 9.2, which is what we'll, we'll attach ourselves to that in Tom Judson's textbook here, because our series follows Judson's textbook, but this lecture 28 is actually all over the place. This section is actually in 9.2. Our discussion on the Chinese remainder theorem, we want to attach to section 9.2, but actually in Judson's textbook, the Chinese remainder theorem is actually talked very briefly about in chapter 16. We're not going to do a full coverage chapter 16. And then we'll also talk about the fundamental theorem of finite a billion groups, which is talked about in chapter 13, which again, we're going to just introduce all of these ideas here in section 9.2 about direct products. All of these are quite relevant to direct products. Now, formally in our series, we defined direct products a long time ago, but this is the official location in Judson's textbook we've defined, so we've already introduced that. What I want to do in this video is in fact, prove a very useful statement about the order of direct products. In particular, suppose we have the direct product of two groups, G times H. And if we pick an arbitrary element of G times H, let's call it G comma H, little G comma little H. If we know the order of little G, let's say it's R inside of capital G, and let's say little H has order S inside of capital H, right? And of course, R and S are integers here. If we know the orders of the two terms in the ordered pair, then the order of the pair in the direct product will be the least common multiple of R and S. So in other words, the order of G and H, excuse me, in the direct product, this will be the LCM of the order of G times the order of H individually, whoops, like so. So that's what we want to compute here. So for the sake of argument, I guess just for the sake of simplicity, let's call the LCM, least common multiple of R and S, let's call it little M. And so let's consider what happens if we raise G to the M power. Well, M is a common multiple of R and S, which is to say that R divides M. We know by Lagrange's theorem that any multiple of the order of G will give you the identity. So G to the M is equal to the identity. And by similar reasoning, we know that S divides M, therefore same argument as before, H to the M will also equal the identity of H. Now the identities in play here might not be the same identity. We have the identity inside of G and we have the identity inside of H, but Lagrange's theorem would apply to all of these groups here. So what we've now shown here is if we take G, H and raise it to the M power, this will equal, well, just by usual composite multiplication in the direct product, G, H to the M will equal G to the M comma H to the M, which those are both identities and that is the identity element of G cross H. So we see that the order, the order of the element is going to be dividing M right here, okay? Now let's say that the order of G comma H, some manager of course, let's say it's equal to K for a moment, then we know that K is gonna divide M, great, which these are both positive integers, so two things are gonna happen here, one of two things I should say, either K is equal to the M, which is what we're trying to show, or K is actually less than M. So let's consider that situation for a moment. What if K is strictly less than M? Well, since K is less than M and M was the least common multiple of R and S, that means that either R doesn't divide K or S doesn't divide K. And so without the loss of generality, let's assume that R doesn't divide K. Well, that would mean that G to the K is not the identity because it's not a multiple of R. And so if we take G H to the K power, well, that's G K, H K, this won't equal the identity, it won't, because the first parameter G to the K is not equal to the identity. So that would sort of contradict the fact that K was the order of the element right here. Therefore, M is the smallest possible positive integer, which takes G H to the identity. And therefore it will have to be the, it has to be the order of the element, proving the theorem we then are trying to prove right here. As a very quick corollary, if we have an arbitrary direct product, right? That is, we have N many terms. So we have like G1 times G2 times G3 all the way up to, you know, GN right here. So if we have N factors in this direct product, how do you compute the order? Well, it's a simple induction argument that I'll leave it for the viewer to convince themselves that convince themselves of, if you take the order of this N tuple, the order will be the least common multiple of the orders of each of the individual terms inside of that N tuple. So let's see an example of this real quick. Consider the group Z12 crosses Z60. So we take the cyclic group of order 12 and cross it with the cyclic group of order 60. And let's take the specific elements, eight comma 56 inside of that group. Well, how do we compute the order of eight inside of Z12? As we've seen previously, we want to compute the least common or the greatest common divisor, excuse me, between the element eight and the order of the cyclic group 12, that's gonna be four. And therefore the order of eight inside of Z12 will be the order 12 divided by that GCD4, which says that eight is an element of order three. And that's not, you know, if you actually look at it, it's not too hard, right? So you have eight here, the cyclic subgroup generated by eight, you get zero, you get eight, you get, if you add eight to eight, you get 16, which is congruent to four. And then eight plus four is 12, which is zero again. So that's the cyclic group generated by eight, it's order three, great. We could have done that by hand, but we also have a formula for it. Likewise, let's consider the order of 56 inside of Z60 by a similar computation, the GCD of 56 and 60 is likewise four. Therefore the order of 56 will be 60 divided by four, which is 15. And so then highlighting those things right there, the order of eight inside of Z12 is three, the order of 56 inside of Z60 is 15. And so the least common multiple of three and 15 will be 15. Notice that in this situation, three actually divides 15, since it's three times five right there. So the LCM would be 15. And so the order of the element eight and 56, that's an element of order 15 inside of Z12 cross Z60. Let's look at another example. If we take Z2 cross Z3, and let's take the element, so Z2 is of course the cyclic group order two, Z3 is the cyclic group of order three. So their product will form an abelian group of order six. Let's consider the element one one inside of Z2 cross Z3. Now to compute the order of one, we probably don't need to do it since we know one generates these cyclic groups with respect to modular addition every time. But you know, if you went through the argument, the GCD of one and two of course is one, two divided by one is equal to one. And therefore the order of one, the order of one, let me make that look like a one, the order of one is gonna be one, excuse me, it's gonna be two. What's going on there? That should be two. And we're viewing this as an element inside of Z2. Likewise, if we compute the GCD of one and three, that of course is also one, right? And so by similar reasoning, you see that we see that in Z3, the element one will likewise have order three. And that should also be a one. Sorry about the typos here. So then, when you look at that together, the LCM, excuse me, least common multiple of one and one, well, that's just the order of one and one, which will be the least common multiple of two and three, that's gonna equal the element six. So one comma one is an element of order six, all right? But like I said earlier, Z2 times Z3, that's a group of order six, two times three. If you're looking for the order of a direct product, G and H, this is always just equal to the order of G times the order of H. So Z2 cross Z3 is a group of order six, but it contains an element of order six. That would suggest that the group is actually cyclic. And as we've seen previously, all cyclic groups are isomorphic up to their orders. That means Z2 cross Z3, it's a cyclic group of order six, it has to be isomorphic to Z6. And in fact, the way that isomorphism is if you go from Z6, so we're gonna go from Z6 to Z2 cross Z3, the map is just send N to N comma N, right? And reduce mod two and three respectively. So if you were to go through all the possible elements, let's actually look at the details of this really quickly. You get zero, one, two, three, four, five right here. So zero will map to zero, zero, great. One would map to one, one, no reduction there. Two would map to two, two, which that reduces to zero and two. Three would map to three, three, which reduces to be one and zero. Then we get four would map to four, four, which reduces down to be zero comma one, excuse me. And then the last one, if you take five comma five, that would reduce down to be one comma two. And so we see here the six possible pairs between Z2 and Z3. So it turns out the product of these two cyclic groups, direct product turned out to be cyclic. That's not always the case though. One can in fact argue that Z12 cross Z60 is not a cyclic group, and that's what will lead us to the Chinese remainder theorem in the very next video.