 Welcome back to our lecture series Math 4220 Abstract Algebra 1 for students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Missildine. This is the first video on lecture 12, which begins our discussion of cyclic groups, which is the main topic for chapter 4 in Tom Judson's Abstract Algebra textbook. This lecture will follow the majority of the content from section 4.1 entitled cyclic subgroups. And so we want to actually define now this titular topic, what is a cyclic subgroup? Well, it turns out that oftentimes a subgroup will depend entirely on just a single element of the group. That is knowing that the particular element, knowing that particular element will allow us to compute any other element in the group in the subgroup, I should say. So suppose we do have a group in play, call it G. And let's take an element inside of the group G. We'll also call it G, but lowercase G. We then will denote the set angle G angle. And just to be clear in terms of the latex, this would be backslash L, A, N, G, L, E, langle G, and then backslash R, A, N, G, L, E, wrangle. Don't use the less than or greater than symbols to work on this one. So we're going to take, the angle G angle, and this is defined to be all of the integer powers of G. So this would include things like G to the first, this will have G squared, G cubed, G to the fourth, right? I mean, it has all integer powers. So it would have G to the zero, which is just the identity, it would have G to the negative one, which is its inverse, G to the negative two, which is its inverse squared, etc. Now, admittedly, it could be that there is some finite number that makes this thing equal to the identity, like maybe G to the fourth is equal to E. So some of these could be repetition, that's okay. So this set that you take the powers of G, we claim that this set is in fact an abelian subgroup of G. And this is referred to as the cyclic subgroup generated by G. And we're going to in fact see that this cyclic subgroup is the smallest subgroup containing the element G inside of the group G. So for this discussion, we're going to use capital H to denote the set, this set generated by G. And we want to argue that this is a subgroup. So let's first, let's first show that it is in fact a subgroup. So let's take two different elements that live inside of H, let's call them H and K. Now, because they live inside of this set, H and K can both be expressed as integral powers of G. So let's say that H is G to the N, and let's say that K is equal to G to the M. Now, to show that this set H is in fact a subgroup, we have to show it's closed under the multiplication here, it's closed under binary operation. So H and K were arbitrarily chosen from the set H. And because they belong to H, we're able to use these identities here, right? What happens when we multiply H by K? Well, since H is G to the N and K is G to the M, we multiply those powers of G together. But by the exponential rules that work for every group, G to the N times G to the M is the same thing as G to the N plus M. This of course being a consequence of the usual associative law of a group. Notice then that H times K can be written as a power of G. So it belongs to the set H itself. This shows us that H is closed under multiplication. Now, when one's proving that a subset is a subgroup, you don't have to prove associativity because an associativity will be inherited from the larger group. But as that group isn't necessarily abelian, we do have to specifically prove that the operation for this cyclic subgroup is a commutative operation. Hence, that cyclic subgroups are always abelian, even if the overgroup is itself not abelian. So to see that H and K commute, because again, H and K are still arbitrarily chosen right here, we're going to get H times K, which as we saw a moment ago, this is G to the N plus M. But in terms of the exponent laws that we prove for any group, this again, this is derived solely from associativity, that G to the N plus M becomes G to the N plus N, which then can be factored as G to the N. That's a typo there. Sorry, everyone. You'll only even notice it. And then G to the M times G to the N is equal to K times H. So this shows us that powers of G will commute with each other. And so the cyclic subgroup will be, it'll be an abelian subgroup, assuming that even is a subgroup, we haven't finished that fact yet. To finish showing that it's a subgroup, we do need it has the identity, which we can see very clearly right here by definition G to the zero is equal to the identity of the group, which we're calling it one right here, as it's a multiplicative group. So H contains the identity. What about N versus? Well, that's also pretty easy, because if you take any element H inside of the set H, it can be written as G to the N. But because we take all integer powers of G, we will also contain G to the negative N. And by exponent rules, G to the N times G to the negative N, you know, that's just going to give the identity. G to the N is just G inverse N times, right? And so the inverse of H is contained inside of capital H. This now shows us that H is closed under the inverse operation. And in fact, it is an abelian subgroup of G. So this cyclic subgroup generated by G, we've now established is, it's always an abelian subgroup. Now to show that H is the smallest subgroup of G containing this little element G, we have to kind of first make sense, what does it mean to be small, right? This of course means that if we take any other, any other subgroup, any other subgroup K that's inside of G, and little G belongs to it. So if this happens, little G is inside of a subgroup of G, then we weren't going to show that H is a subgroup of K. So this is what we mean by H is the smallest subgroup. And to show that H is a subgroup of K, we really don't have to show it's a subgroup again, we just have to show that it's a subset. Because the subgroup relationship is transitive, a subgroup of a subgroup, you know, if H is a subgroup of K and K is a subgroup of G, then H will be a subgroup of G because if H is a subgroup, is a group inside of K and K is a group inside of G, then H will be a group inside of G, right? So it suffices just to show that H is a subset of K, that's what we mean by H is the smallest subgroup that contains little G. All right, so with that, with that clarification in mind here, let's go about showing that here. Suppose that K is a arbitrary subgroup of G that contains little G. Now, since K is a subgroup of G, it contains the identity, right? And this is going to act like a base case because this argument is going to follow by induction. Because after all, the cyclic subgroup is going to be integer powers of little G. So we're going to play induction on the power of G. So K contains the zero with power. For our inductive hypothesis, we're going to assume, so this is our inductive hypothesis right now, we're going to assume that the nth power of G belongs to K, all right? And again, this isn't too much of a stretch, right? Because if you have the identity, we also have G by assumption, you're going to have one times G, which is G, you're going to have G times G, which is G squared, you're going to have G squared times G, which is G cubed. This is the train of thought we're going down right now. So suppose we've now argued that G to the n is inside of K. Well, since K is a subgroup, it's closed under inverses. If it contains G, then it'll contain G inverse. Likewise, if it has G, so that's one thing to be aware of. Let's see. Likewise, it'll contain the inverse of G to the n, which G to the n inverse is G to the negative n, since it's closed under inverses, it has that. And likewise, our group K is closed under multiplication. So it'll contain G times G to the n, but that's just the same thing as G to the n plus one. And likewise, it'll contain the element G inverse times G to the negative n, and that's just the inverse of G to the negative one. That is, it's G to the negative n plus one. And so then we see using our inductive hypothesis that G to the n was in there, that we get all integer powers of G. And so therefore, because K is a subgroup and it contains G, it'll contain all integer powers of G. Well, as the integer powers of G are just that's what H is defined to be, this then shows us that H is a subset of K. And like we mentioned earlier, this then proves that every subgroup of G that contains little g will contain its entire cyclically generated subgroup. In this case, we would call it H.