 In this video, we're going to prove that every subgroup of a cyclic group is itself cyclic. So let's start off with some cyclic group in mind. Let's take our group G and let's say it's a generator is given by little g and let H be then a subgroup of our cyclic group G. So we need to prove that H itself is a cyclic group. Now because every element in G is a power of little g, every element in H will likewise be a power of little g. So using the well ordering principle, we want to choose the smallest power of G, the least power of G that we can find inside of H. Let's call this element G to the K and we're going to call it H for short. So because H, the subgroup H contains the little H, that means it'll the subgroup H will contain the cyclic subgroup, the cyclic subgroup generated by little H. So what we need to do is we need to show that H is itself a subset of the cyclic subgroup generated by little H. So how are we going to do that? Well, in this regard, we're going to let G of N be some element in H. Now of course, every element in H looks like some G to the N where K is the smallest one. So the minimality of the choice of K is going to be the critical element right here. Take the integer, so we just take an arbitrary element of H. It's going to be G to the N. Take the exponent of G in that situation. N is an integer, K is an integer. We can run the division algorithm with N and K. Therefore, by the division algorithm, there exists a unique quotient and remainder such that N is equal to QK plus R, where R itself is either 0 or R is strictly less than K. This is where the minimality condition is going to get into play here. Because notice the following. If GN is inside of H and of GK is inside of H, then some things we know are the following. Because H is a subgroup, GK to the Q power will be inside of H, but this itself is just G to the QK, right? And then we also know that G to the negative QK, the inverse of that element is inside of H. And so then here's the main kicker. We're going to get that G to the N times G to the negative QK, which when you combine those exponents together, you're going to get G to the R. This is an element of H. And so this is where the minimality condition comes into play here. K was chosen to be the minimal positive power of G that's inside of H. R is a smaller number than K, but G to the R is inside of H. So because K is the smallest positive power, it means that R cannot be positive. But by the division algorithm, if R is not positive, then it actually must be 0. So we get R to the 0 right here. Well, if R equals 0, that means that N equals Q times R. That is, N is a multiple of K, K divides N in this situation. So G to the N equals G to the QK, which is just G to the K to the Q, which is H to the Q, which would belong to the cyclic subgroup generated by H. So this, in fact, does show that the subgroup H is contained inside of the cyclic subgroup generated by H, thus proving equality. The subgroup, this arbitrary subgroup of G was itself a cyclic subgroup. Now let's come back to our known cyclic groups. We know that the integers under addition is an infinite cyclic group. And we've explored that NZ are cyclic subgroups of Z. But since Z is itself a cyclic group, the previous theorem that implies that every subgroup of Z must also be cyclic. And so therefore, these are the only subgroups of Z. The only subgroups of Z are gonna be NZ. And by similar reasoning, if we look at the cyclic subgroup ZN, which we've already established, clearly MZN are cyclic subgroups. This is the subgroup generated by M. But the only subgroups of a cyclic group are cyclic subgroups. And therefore, these are the only subgroups of ZN. So I want you to kind of consider that for a moment. Previously, we had drawn the following picture, right? We had previously drawn the following picture. We had the group Z6. Well, there's the trivial group, which itself will write it as the cyclic subgroup generated by zero. We have the cyclic subgroup generated by one, but that itself is one. We have the cyclic subgroup generated by two, which looks like zero, two, and four. The cyclic subgroup generated by zero, of course, is just the trivial subgroup. We have the cyclic subgroup generated by three, which was just zero and three. We have the cyclic subgroup generated by four, which, of course, is just the same thing as the cyclic subgroup generated by, it's inverse two. And then there's the cyclic subgroup generated by five, which as five and one are inverses, we get this picture right here. And so we had previously drawn this HAZI diagram, not with any reference to cyclic subgroups. But I had mentioned that this diamond shape was all of the subgroups of Z6. And this is the argument that supplies that. That for our cyclic group Z6, every subgroup has to be a cyclic subgroup. And if we go through all six elements and think of all the cyclic subgroups generated by that, we only get four options. The subgroup generated by zero, the subgroup generated by two, the subgroup generated by three, and the subgroup generated by one.