 Another important collection of groups is known as the cyclic or cyclic groups. So, let's start with any group at all. Let G be any group, and let's take any element in G. Because G is a group, we know the following. The product of A with itself has to be in G. We have closure, and, well, we also know that the product of A with itself three times is going to be in G. And, in fact, the product of A with itself four times is going to be G, and so on. Now, by analogy with the notation, but very importantly, not the meaning. The notation, not the meaning, of basic arithmetic, we're going to define A star A to be A to the second. And likewise, A star A star A, we're going to call that A3. And A star A star A star A, we're going to call that A4 and so on. So, this notation A2, A3, A4 just means we're going to take this many copies of A and apply our group operation to them. Now, let's consider this sequence A, A squared, A cubed, A to the fourth, and so on. Well, there's two possibilities. Again, a fairly standard thing to consider in mathematics as you develop your mathematical sophistication. Always look at what the possibilities are. And there's two possibilities here. Remember, all of these are elements of some group, and so one possibility is this sequence contains all elements of G of our group. In that case, G is said to be a cyclic group, and A is a generator. A generates all elements of G, and there's a cyclic nature to it that we'll talk about in a moment. The other possibility, of course, is that this does not contain all the elements of G. In that case, well, we'll see what happens. Most of mathematics really centers around this question. Let's see what happens. But for right now, we'll focus on the case where the sequence produces all elements of G, where A is a generator of the group. Well, I'll suppose that that sequence does actually contain all elements of our group. Well, let's consider two elements. Let B and C be two elements of our group. And again, because this sequence contains all elements, then B is A to some power, C is A to some power. Well, let's see what happens. B star C, well, that's going to be A to the M star A to the N. By our definition of what these are, A to the M is the star of M copies of A. A to the N is the star of N copies of A. And so when I multiply them together, I have this mass of things. Now, if I run these all together, I have M plus N copies of A. So this product here is A starts with itself M plus N times. Now, it's an important thing to remember. This is a count. I count the number of As here. I have M of them. So all together, I have M plus N As. This count has nothing to do with arithmetic. Well, it has everything to do with arithmetic, but it has very little to do with what we think about as arithmetic. It's just counting. Now, that also means that if I have M plus N As here, I can break them up this way. So now I'll put N of those here. I'll put M of those there. I'm allowed to do that because, remember, G is a group, so I have associativity. I can combine these in any way that I choose to. And by my definition, this is A to the N. This is A to the M. And again, by assumption, B is A to the M. C is A to the N. This is C star B. And now I started, I got on this bus, the B times C bus, and I found that it was actually the same as the C times B bus. And what that means is the following. Remember that in general for groups, I can't assume commutativity. Star does not have to be commutative. But if I have a cyclic group, if this sequence contains all elements of G, what this proves is that if G with operation star is a cyclic group, then it's commutative that B star C is the same as C star B. So let's talk about birds. The pigeonhole principle is one of the most powerful tools in mathematics, and it emerges as follows. Suppose I have M cages and N greater than M pigeons. Well then, surprise, at least one cage will have more than one pigeon. Well, let's say, well, that's kind of obvious. What can we do with it? Well, here's an important thing we can do with it. Suppose I have a set G with M elements. And I'll consider this sequence of powers A, A squared, A cubed, and so on all the way up to A to the N, where N is, say, anything greater than M, doesn't really matter what it is, as long as it's larger. Because there's M pigeonholes, those are the possible products, because again, G contains M elements, every one of these products has to be one of those M elements. But there's N greater than M pigeons, these are the powers of A, because I have more pigeons than pigeonholes, then at least two pigeons have to be assigned to one pigeonhole. Two powers of A have to be assigned to the same product. And so that tells me they have to be equal to the same group element B, and so I have A to K equal B, and A to the I equal to B. And I'll assume that K is greater than I. You might pause a minute and think about why we're allowed to do this. But if that's true, then I have A to the K equal A to the I. Now, because K is greater than I, I can use my associativity that my group has. I can split off Ki factors of A, and then half what's left over. And because G is a group, so G is a group, it has an inverse. Every element, A to the I in particular, has an inverse. And notationally, we'll write that as A to the power minus I. So I'll multiply on the left by A to the power minus I. And so that gives me multiplying on the left, multiplying on the left. Because it's the inverse and because it's a group, I have associativity. I can multiply these two things together because if it's an inverse, the product of these two is the identity. The product of these two is the identity. And so I get A to the power K minus I equal to the identity element. And this proves an important result. If I have G with operation star as a finite group and I have some element in G, then A to some power is going to be the identity for some P among the natural numbers. And this is an important result in abstract algebra and later on number theory.