 Welcome back to our lecture series, Math 42-20, abstract algebra one for students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Missildine. In this first video in lecture 14, we're actually going to continue on talking about cyclic subgroups and developing some of the theory about cyclic subgroups and subgroups of cyclic subgroups and such. The lecture itself is entitled Repeated Squares, and that's because in section 4.3, which is relatively short, we're going to introduce the algorithm of Repeated Squares, and that'll be coming up in a subsequent video. In the current video, I do want to talk some more about cyclic subgroups. So let's prove a few theorems. They're all relatively short here, but let's prove a few theorems about cyclic subgroups. Suppose we have a group G, which is cyclic, and it's generated by little g. If g has finite order, and we'll say that order is n, then so does g. That is, g would equal the order of g equals n. And furthermore, if little h is in g, and the order of h is n, then h is a generator for g, and therefore g is then generated by n. So there's a few things to be stated there, right there. So when we define the order of an element previously, I kind of did it two definitions at the same time, and we're actually want to establish that these two notions are essentially the same thing, right? So I mentioned that the definition of order, the order of an element is the smallest positive integer for which that element is equal to the identity, g to the n is equal to the identity. If no such integer exists, then we say the order is infinite. But I also said that the order was equal to the size of the cyclic subgroup generated by that element. That is the two words order, one time an order of groups and one order of elements. The two actually coincide with each other. So that's really what we want to establish right now. So imagine g has finite order, then consider the cyclic subgroup generated by g. Well, if it has finite order, what that means is that the nth power of g is equal to the identity. That's what we get. And no smaller power will equal the identity. So if we look at the cyclic subgroup generated by g, we're going to have the identity, g to the first, g squared, g cubed go all the way up. We're going to get g to the n minus one. None of those elements are the identity. And I also want to mention that none of those elements are actually distinct. I mean, excuse me, all of those elements are distinct. So for example, if we take like g to the k times g, could that be equal to g to the l or something like that? What happens that situation? Well, if that were true, then we could times both sides by g l inverse. This would give us that gk times g to the negative l is equal to the identity. That is to say, g to the k minus l is equal to the identity. Now, if we choose, say, k to be the larger of these two, k is larger than l, then that would tell us that k minus l is in fact positive. And this would contradict the minimality of the number n, which is the order. And so this gives us a contradiction. So what we're seeing here is that when you look at this list, the identity, g, g squared, g3, g4, g5, all the way up to g and minus one, there are no repeats in that list whatsoever. So the number of elements in the cyclic subgroup generated by g is itself in many elements. So the smallest power of g being the identity is also the number of elements in its cyclic subgroup. So that claim I made earlier without proof, we've now established that the order of an element is equal to the order of the cyclic subgroup it generates. Now, what about the second statement? Let's say that we have a second, we have a second element, not necessarily g whose order is n. Well, in that situation, let me clean up this screen a little bit. In that situation, if h is an element of g and its order is n, then by the same reason we made a moment ago, the cyclic subgroup generated by h will contain n many elements. So the size of the cyclic subgroup generated by h is itself equal to n. But the cyclic subgroup generated by h is a subset of g. I mean, in fact, a subgroup, we can say better. It's a subgroup of g. And as these are both finite sets, we've already established that the order of g is n. And then the order of the cyclic subgroup generated by h, since h's order is n, this will likewise be n. Since we have two finite sets that are both order n right here, this then forces equality that the subgroup generated by h has to equal g. Thus proving that h is a generator of that group as well. So that approves this first theorem right here. Let's follow it up with another theorem. If g is a group with an element little g, I'm not claiming it's the generator at the moment. What we're going to say is to follow it. If g to the a is equal to the identity, this happens if and only if g, the order of g, there's a typo there. The order of g divides a. So a itself has to be in multiple of the order of g. In particular, if h is g to the m for some power of m, then the order of h divides the order of g. All right, so let's unwrap some of those things we're trying to say right there. So I'm going to erase what I have and put back in the correction we need to have. So the order of g divides a because a is just an integer in this situation. g being an element of some group. All right, so consider a situation. Let's say that the order of g is n. Because after all, if some power of a, some positive integer is equal to the identity, that means by the well-ordered principle, there's some least positive integer which takes g to e. So it's going to have finite order. Let's call that finite order n. Well, by the division algorithm, if we, by the division algorithm, we're going to have an integer q and an integer r such that a is equal to qn plus r, like so. Where r is going to be strictly less than n, but it's going to be greater than or equal to zero. It could equal zero or it's a positive number less than n. That's the game we're going to play around with here. So this actually kind of saw in a previous argument here. If we take this quotient equation right here and manipulate it, we're going to get that a minus qn is equal to r. And notice here, if you take g to the r, this is the same thing as g to the a minus qn, which this gives us g to the a times g to the n raised to the q. Well, by assumption, g a is equal to, well, let's not get to that assumption right now. g to the n, since it's the order of g, this is going to give you the identity. You get g to the a times the identity to the q, but that's just going to be the identity times g to the a. This is just g to the a, right? So this is the inequality we're trying to establish right here. g to the r, which is the remainder, is the same thing as g to the a. Now, if we assume g to the a is equal to the identity, that implies that g to the r is the identity. But r is a number which is less than or it's less than n. So the minimality of n, which is the order of the group element, that then forces r to be zero. So therefore, if g to the a is equal to the identity, then r would have to be zero in that situation. Now, if r is zero, that means a is divisible by n, which n was the order of g, so that gets what we want. So if g of a equals the identity, then the order of g divides a. But if we go the other way around, if the order of g divides n, we get this a equals qn. But then like we saw before, if g to the a was equal to g to the qn, well, the right-hand side is going to equal the identity, like we saw over here. You're going to get g to the n, which is the identity to the q, which is still the identity. And so therefore, we get this, which then implies this statement above here. All right, so that proves the if and only if statement we started with. Let me kind of clean up the board a little bit. So if a power of g is equal to the identity, that power is divisible by the element's order if and only if. Now the other statement, in particular, if h is equal to g to the m, the order of h divides the order of g. So when you consider powers of h, let me say that again. So if you have a power of an element, the element's order will be a divisor of that group element's order before. And so consider the following. If you have h to the n, well, h to the n here, well, since h is just g to the m, you're going to get g to the mn. Well, if you switch the exponents around by usual exponent rules, you get g to the n to the m. That's e to the m, and that's the identity. So therefore, we see that h to the e, or sorry, h to the n is equal to the identity. So by the previous statement, which we just proved over here, that means that the order of h must divide the n, which was the order of g. We get so. So this now summarizes what we've seen so far here. We've saw in this theorem that if you raise a group element to a power and you get the identity, that means you had a multiple of the divisor. And so particularly as you take powers of elements, the new element you created, its order must be a divisor of the original order. And we're going to use this to talk about cyclic groups, because in a cyclic group, every element is a power of that generator. If g is a cyclic group, then g has exactly one subgroup of order d for every divisor of the order of the group. So let's consider that for a moment. Let's suppose that g has finite order. Because the reason we're assuming that g has finite order, because what does divisibility mean for infinity? That's a topic we won't be talking about right here. So let's say that the order of g is n, and let d be a divisor of n. Consider the element h, which is given as the element g to the n over d power right there. Now I guess I forgot to mention that since the group is cyclic, it needs a generator. We'll say the generator is g in this consideration here. So take h to be the element g to the n divided by d power. Now since d is a divisor of n, n divided by d is in fact a whole number. So it makes sense to take g to that power, define h to be that element. Now, of course, if you take h to the d power, this will then be g to the n divided by d to the d power, which by exponent rules this will be g to the n. And as n is the order of g, we get the identity right here. So notice that h raised to the d is equal to the identity. What we saw in the previous slide is that the order of h must be a divisor then of d. Now on the other hand, what if we chose some number strictly smaller than d, a positive integer strictly smaller than d? Well in that situation, a n divided by d, which is an integer, would be strictly smaller than n. Basically, we just take this inequality right here and we times both sides by n over d. Okay, that's a positive number that inequality is preserved. And so then we get that h to this a, which equals g to the a n over d. That can't equal the identity because a to the n over d, that's strictly smaller than n and no smaller power than n can get g to the identity. That means a is too small to make h be the identity. That is transform h to the identity. And this is showing us that then d is the smallest power of h, say it produces the identity. So the order of h is actually this number d. And therefore the cyclic subgroup generated by h, its order is the same as the order of the element, so this is equal to d. So we've now produced a subgroup, a subgroup of g whose order is d. That's half of the statement. So when you're trying to prove a statement about uniqueness, there's exactly one. We have to prove existence, there's at least one, and then the other half of uniqueness is proving there's at most one. So that's, we've done the first half already, we've proven there's at least one subgroup of order d. Why can't there be another one? Well, let's take k to be an element of the group g, whose order is also d. Now remember, for a cyclic group, every subgroup of a cyclic group is cyclic itself. So if we're looking for subgroups in g, we only have to consider the cyclic subgroups. So we're going to consider the cyclic subgroup, for example, generated by k. Notice that the subgroup generated by k, because the order of k is d, the order of this cyclic subgroup likewise has to be d. So this is a potential counter example to the theorem, right? What about like the cyclic subgroup generated by h versus the cyclic subgroup generated by k? These are both subgroups of order d. Why aren't there two of them? Well, we're going to argue these are the same thing. Now there exists some integer m such that k equals g to the m because g itself is cyclic. And therefore we have the property that g to the dm power is going to be g to the md, which is kd. And then by assumption, that's the identity because the order of k is d. Now by the previous theorem, we've learned is that n must divide dm. So since g raised to the dm power is equal to the identity, the order of g divides dm. You can see this theorem in the previous video, of course. Actually, I'm sorry. It's the current video we're in right now. What am I talking about? So you're going to get n divides dm right there. Okay. So since n divides dm, there's some integer we'll call it l so that nl equals dm. We also know that d is a divisor of n. So if you divide both sides by d, you're going to get that n divided by d times l is equal to m. So therefore n divided by d, which is an integer, divides m. All right. It's n divided by d, divides m. Therefore, m is a multiple of nd. Therefore, k, which is equal to gm, is actually contained in the cyclic subgroup generated by h. Because remember, h is the element g to the n over d. And so what we've now seen here is if I take h to the l power, this is equal to g to the n divided by d to the l power, multiplying those things together. You get g to the ln over d, but that's equal to m. And so you get g to the m, which is equal to k, right? So as h to the l is contained in the subgroup generated by h, this shows that k is in the subgroup generated by h. And as we've established earlier, any subgroup that contains the generator will contain the entire cyclic group. So the cyclic group containing the cyclic group generated by k is then a subgroup of the cyclic subgroup generated by h. That's what we've now established. But wait a second. These are two finite sets with the same order, the same size. And so if you have a finite set, if you have a subset of a finite set, and both have the same size, the two sets are actually equal to each other. This improves the fact we wanted that a cyclic group has a subgroup for every, has exactly one subgroup for every divisor of the group, okay? And then a corollary to this, two corollaries actually. These ones will kind of go without proof here. Let g be a cyclic group with order n, and let h be an element where h equals g to the m. And so again, we're using the assumption that g is cyclic generated by g right here. I should have mentioned that, but oh well. Then in fact, the order of h is going to be the n, the order of g, divided by the gcd of n and m. And so I want you to try to convince yourself why this is true. This follows from a very basic, really just a division algorithm, type of gcd type argument right here, that the order of h is going to equal the order of g divided by the greatest common divisor of n and m right there. And that's kind of the same thing that we saw going on over here when we did this proof right here of that uniqueness of the subgroup. Another corollary, let h be the cyclic subgroup generated by little h, and that's a subgroup of zn. Then the set of generators of h are exactly those elements of the form h to the m is given as the gcd of m and the order of h. That is, we were looking for those numbers which are co-prime. In particular, the set of generators for zn is zn star. These are results that follow from the previous theorem right here. And these might actually be things you're going to prove in the homework, but I want to provide one example of this. Let's take the cyclic subgroup generated, the cyclic group of order 12, so z12 right here. What does, what does its hausei diagram look like? Well, you're going to have the trivial subgroup which is generated by zero. Notice it contains only zero right there. What about the element one? Well, one is always going to be a generator. You're going to get one right there, and let's actually be specific. You know, we have all these elements, zero, one, two, all the way up to 11, like so. One will generate it. Two, two is not co-prime with 12. In fact, 12 divided by two is six. So the subgroup generated by, the subgroup generated by two is going to give us a subgroup of order six. So two is going to contain six elements. You're going to have zero, two, four, six, eight, and 10, and then two more gives you 12 again. So those are the six elements we predicted. The subgroup generated by three, three goes into 12 four times. And so the subgroup generated by three is going to have four elements. So three is generated. The subgroup generated three will contain three, zero, three, six, nine, and then 12. So we have that one. Next four, the cyclic subgroup generated by four, four goes into 12 exactly three times. And so we're going to get a subgroup generated by four. It's going to contain three elements. You get zero, four, and eight. Four plus eight is 12. And in terms of, in terms of ordering here, I do want to mention that the subgroup generated by four is contained inside the subgroup generated by two. You can see it right here. So that's a haze diagram. You're going to get something like this. You get something like this as well. All right, keep on going with this. We are, let's consider the subgroup generated by five. Well, if you take the subgroup generated by five, you would get five, then 10, then three, then eight, then let's see, eight plus five, is that where we're at right now? That's going to give me 13, which is going to give me one, which, hey, if you have one, you actually get everything. The subgroup generated by five is actually the entire thing. And look at that. Five was our first number besides one, which was co-prime to 12. All right, just like we predicted it would. Then if you try six, the subgroup generated by six, six goes into 12 two times. And so six, you're actually, this subgroup can be order two. You're going to get zero and six, like so. It's contained in the subgroup generated by three. Notice you have a zero and a six right there. It's also contained in the subgroup generated by two, zero and two. So in terms of a haze diagram, you're seeing something like the following. All right, and that actually is going to give us all of the subgroups. But let's mention why, what the other ones are going to be. If you take the subgroup generated by seven, seven is co-prime to 12. So I claim that seven is going to produce the whole thing. And if you go through that, you're going to get like seven plus seven is 14, which reduces to two. Two plus seven is nine. Seven plus nine is 16, which reduces to four. Four plus seven is 11. Seven plus 11 is 18, which reduces down to six. Six plus seven is going to equal where we at right now. It's going to give us 13, which is one up. You've got another generator. It's going to give you the rest of it, right? If you ever grab a generator, you're going to get the whole thing. So seven's there too. What about eight? What the cyclic subgroup generated by eight? Well, eight plus eight is 16, which reduces to four. Four plus four, four plus eight, excuse me, is 12. The cyclic subgroup generated by eight is actually the same thing as the cyclic subgroup generated by four. Let's consider that one for a moment, right? What is the GCD between eight and 12? It's equal to four. And what is 12 divided by four, that is going to be three. The order of eight is three in this group. And therefore, eight will generate a cyclic subgroup of order three. And as that cyclic subgroup is unique, it's going to be this one. Okay. What about nine? Nine does not divide into 12, but they do have a common divisor, right? The GCD of nine and 12 is equal to three, which means that the order of nine is going to equal 12 divided by three, which is four. The subgroup generated by nine is going to equal the subgroup of order four, which actually tells us instead then, oh, I kind of erased part of my thing, this is the subgroup generated by eight. This over here is going to be the subgroup generated by nine. And notice three and nine are inverses. They'll generate the same group. Four and eight are inverses. They'll generate the same group. All right. What about 10? 10 and 12 have a GCD of two, which means that the order of 10 is going to equal 12 divided by two, which is six. The subgroup generated by 10 is going to be the subgroup of order six, which you see right here, two and 10. These are inverses of each other. They generate the same group. And so this is the subgroup generated by 10. And then lastly, 11, which is co-prime to 12, its order will be 12. And so we can see 11 is also going to be a generator there. Notice that one and 11 are inverses, and four and seven are inverses. If we look at those integers, which are co-prime to 12, that is exactly one, five, seven, and 11. These four elements of Z12 star are exactly the generators of the cyclic subgroup of order 12. And so this was kind of a little bit longer video than I usually like to make, but I wanted to give you not just all these proofs, but an example of what they actually mean. When it comes to a finite cyclic group, it'll have a subgroup, exactly one subgroup for every divisor of the order. And then we can, and then also every subgroup, it's cyclic. We've seen that. And then finally, the order of an element will exactly be that number's GCD with the order divided. So like what we were saying earlier, like 10, its order was the GCD between 10 and 12 divided, or divide that from 12, which is where we got the six from. And so we can calculate without having to go through all of these calculations here, we can go and calculate and determine exactly which element will generate which cyclic groups. It really comes down to calculating GCDs. And so this is a calculation we can do even for very large moduli, which becomes a very important problem to be able to solve when you start getting into computer algebra when working with these very large, but still finite cyclic groups, is actually important for things like computer security and other types of computer algebra problems.