 In this video, I want to consider some cyclic subgroups of groups that we know. For example, take the rational numbers with respect to addition. We're going to let h be the set of all rational numbers whose denominator is, when written in lowest terms, it divides six. So we would be including things like the following. So we would take like one-half, three-halves, five-halves, notice the denominators two would divide six. We want things like one-third, two-thirds, four-thirds, five-thirds, right? You know, just things like this. We also want integers, you know, like zero, one, two, because their denominators are just one, that divides six as well. We can also take like a sixth, let's see, five-six, you know, just things like that. Basically, we're grabbing those fractions whose denominator in lowest terms is either one, two, three, or six. One can argue that's a subgroup, that when you, that thing will be closed under addition. Clearly has the identity, you can see right here. Inverse isn't too hard to prove, but we want to argue that it's closed under addition. Well, to do that, you know, you could add things together, and given these type of creatures you have, worst case scenario, the LCM, the LCD, I should say, would be six. You add things together, the denominator is six, but it could reduce down. But if it reduces down, then six would be simplified to be one of its divisors, one, two, or three, right? And so one could argue that this construction does give you a group. We can do that directly. But another observation is that this set that I just described, this subgroup, is actually the cyclic subgroup generated by one-sixth. And so I want you to kind of convince yourself of that. If we take the cyclic subgroup generated by one-sixth, what we mean it would be the following. We take as the set zero, one-sixth, two-sixth, three-sixth, four-sixth, five-sixth, six-sixth, just as a, you know, keep on going there, right? And notice that some of these things are going to reduce down, right? If we take, for example, two-sixth, that's just the same thing as one-third. Three-sixth is the same thing as one-half. Four-sixth is the same thing as two-thirds, et cetera. Six-sixth would be the same thing as one. This gives us all the numbers we were seeing earlier. We also have to take things like their inverses. So we have to take like negative one-sixth, negative two-sixth, negative three-sixth. And so we can reproduce this group using a single generator. A single element generates this whole group. H is the cyclic subgroup generated by one-sixth. You're going to get all of these things by just taking multiples of positive or negative one-sixth over and over and over again. And this, of course, strategy works in general that we can form the cyclic subgroup of q generated by the fraction one over n. This cyclic subgroup, the one generated by one over n, this would produce all rational numbers when written in reduced form. Their denominator divides n. And so this is what these cyclic subgroups are going to look like for q with respect to addition. Switching things up a little bit, what if we switched over to q star with respect to multiplication? This is also an abelian group that's infinite. Let's consider the cyclic subgroup generated by two. Now, when we talk about the cyclic subgroup, we're talking about you take all of the iterations of that element. So you take the element with its operation, you repeat it over and over and over and over, including its inverses. So the cyclic subgroup generated by two is going to produce the powers of two. You're going to get one, two, four, eight, 16, 32, 64, 128, 256, 512. You know, you keep on going until you get to 2048 and then you win. But you also take the reciprocals. That is, you take the inverses of these things, the inverse of two, multiplicatively, one half. We also get one fourth, one eighth, one sixteenth, one thirty seconds. You get the idea. And so when it comes to the cyclic subgroup of two, the cyclic subgroup generated by two and q star, you're going to get these multiple, these powers of two. Now, I want to caution you that when you compare the group q star with just the q, the group q with respect to addition, right? The issue in that situation is that these, this is essentially the same set. You're only missing the zero in one of the situations. And so the element two belongs to both of these sets, both of these groups. Now, the cyclic subgroup generated by two with respect to multiplication will give you all these powers of two. On the other hand, if you take the cyclic subgroup generated by two with respect to addition, what you're going to do is you're going to get zero plus two plus two plus two plus two. So you get things like two, four, six, eight. Who do we appreciate? You also get their negatives, negative two, negative four, negative six, negative eight, all the even integers, right? So these cyclic subgroups, even though we use the exact same notation, it's implicit in the notation that the operation in play. The cyclic subgroup generated by two with respect to addition is not the same thing as the cyclic subgroup generated by two with respect to multiplication. So you do have to pay attention to which operation am I using when I talk about this cyclic subgroup? Because it does depend on the group. If there's any ambiguity on which group this element two belongs to, we need to specify, oh, this is the cyclic subgroup of q divided, or generated by two, or this is the cyclic subgroup of q star generated by two, just so we're clear which one we're talking about.