 Okay. So, let me start by reminding you what we did at the end of Wednesday, which was a discussion of cyclic groups. Cyclic groups is one with the property that G is cyclic in case there exists what we call a generator for G, meaning G is cyclic. There's some element, there is, and I don't care how many, at least one, be with the property that the group G looks subgroup generated by A. In other words, that every element of G can be expressed as a power means you do the binary operation on A as many times as is necessary, or you include the identity element, or you include the powers of the inverse of A. That's what a cyclic group is. And what we looked at last time are some examples of cyclic groups. We made a statement about what cyclic groups look like, and then we made some comment about what subgroups of cyclic groups look like. Okay. And so from last time, the reminder first, any cyclic group is abelian. We prove that in the notes, and that actually would be a quiz for you on Wednesday. It implies G abelian done last time, but not necessarily conversely, there are a lot of examples of abelian groups that are not cyclic. Secondly, these groups are cyclic. The group of integers under addition, of course, and any group that looks like Z sub n, one of these arithmetic addition mod n groups are cyclic. And what we noticed at the end of Wednesday, or at least was stated, and in fact, these are all the cyclic groups. Meaning if somebody hands you a cyclic group, i.e., if G is cyclic and the number of elements in G is n, where n is some positive integer, or you can include infinity here, then G is isomorphic to Zn. Same structure. That statement, two-byte cyclic groups. And finally, the statement that we made at the end of last time, which I'm not going to prove this semester, is this, if H is a subgroup of G, and G is cyclic, then 2 also H is cyclic. In other words, here's the way we'd phrase this in mathematics. Any subgroup of a cyclic group is cyclic, and H is cyclic also. H is cyclic. Which means, of course, that if someone asks you to write down all of the subgroups of a cyclic group, it's actually doable because what this says is, if you handed a subgroup of a cyclic group, it's necessarily cyclic. Meaning that you can find an element in there with the property that when you look at all of its powers, you get the entire subgroup. So in order to find all of the subgroups of a cyclic group, you simply look at each of the elements, you write down the cyclic subgroup that each of them generates, and you look at all the subgroups that arise in that way. And what this proposition says is that that's necessarily all the subgroups. And in general, that won't be the case. If you hand me a group, even an abelian group, it's certainly the case that there might be subgroups that are not cyclic. So this is an important and powerful property of cyclic groups. What we're going to do today is look more closely at the subgroups of this form, whether or not the subgroup of the form actually gives the entire group or not. So today, we consider subgroups of the form, cyclic subgroup generated by A for elements in a group. Again, in a cyclic group, there's necessarily at least one element for which this is the entire group. But if I simply hand you some element in a group and ask you to look at a subgroup of this form, we may get the whole group, we may not get the entire group. And so what we'll do is we'll look at how the powers of a given element actually play out when you keep taking an element and raising it to higher and higher powers. And what we saw last Wednesday was if we looked at a specific example. So for example, when we looked in, let's say, G equals Z, let's make a slightly different one. Inside Z sub 10, if I ask you to find the cyclic subgroup of Z10 generated by 2, let's see, by definition of cyclic subgroup, you necessarily put the identity element of the group in. You put the element that you've asked me to generate the cyclic subgroup by. Here it happens to be 2. You then do the operation on this element as many times as is necessary and you see what comes out. Well, the operation on 2 is 2 plus 2 and then 2 plus 2 plus 2 and then 2 plus 2 plus 2 plus 2 and then 2 plus 2 plus 2 plus 2 plus 2, that's 10. In other words, it's two five times, but 10 in Z sub 10 is zero, mod 10. Thinking, all right, so I've got to repeat. In other words, I don't want to write zero down. But wait a minute. If I keep going, if I do two six times, well, that's the same as doing two five times and then two more. So that's just going to be two five times where there that is plus two. In other words, I just get two. And if I do two plus two plus two seven times, in fact, I get this and eight times I get this and nine times I get this and 10 times I get this and 11 and 12 and 13. And what we get is a pattern. In fact, we get a cycle and this is why these are called cyclic subgroups. So if we continue, we see here that, let's see, 2 plus 2 plus 2 plus 2 plus 2 is the identity of the group five times. And what we're going to do in general is give this number a name. But let's notice how the pattern plays out. And moreover, continuing, this is just two. This is if we do two plus two plus two six times, this is the same as zero is the same as two. In other words, it's two to the first. If I do two plus two plus two seven times, this is the same as two plus two. Keep going. If I do two plus two plus two eight times, I get the same as two plus two plus two. In other words, three times. In other words, I get six, et cetera. And what's going to happen is we're going to take the element in question. We're going to do the binary operation to it as many times as is necessary so that we see the identity element. Now, sometimes that won't happen. There are situations where we can start with elements where the identity element won't result. When that happens, somehow this number, the number of times that you do the binary operation on an element to itself before you see the identity is going to play a special role. And the reason it's going to be special is once we see the identity, if we then continue the process of taking this element and combining it with itself according to the binary operation, what we're going to wind up getting is simply the original elements in the order that they were presented when we looked at the powers of the original element cycling through the first time, cycling through before we saw the identity element. So the idea is going to be idea that once the identity of the group appears in the list, let's see, a, a star, a, a star, a, star, a, et cetera, et cetera, et cetera, then this list, the list starts repeating. And so here then is the definition. Can I, is this okay to write over here? Okay. Here's the definition. We give an importance to the number of times that you have to combine the element with itself before you see the identity element. So let G be a group. Pick any element in the group. Let A be an element of G. Then the order of A of A in G is the smallest positive integer. Let's call it N, having the property that when you do A star A star A with itself N times, you get the identity of the group. It's the smallest number so that when you combine A with itself according to the binary operation, you get the identity element of the group. So just as a quick example, if we look in the group Z10 and we look at the particular element A equals 2, well, 2 itself, namely A, is not the identity. A star A is not the identity. A star A star A is not. A star A star A star A is not. But when I do A star A star A star A star A, I finally get the identity element of the group. So in that particular case, example, in the previous case, the order of the element 2 inside the group Z10, so the order of the element inside the group turns out to be 5. And that's done simply by banging out this computation. Many remarks, remark one is, well, there are certain situations where the element inside the group doesn't have an order. Because there are situations where you can start with an element of the group and you continue to combine it with itself according to the binary operation. You might not see the identity element. So it turns out if no such n exists, we say that the element A has infinite order in G. And a good example of an element with infinite order, for example, if I look at, it doesn't matter which one we start with, how about the element 2 in Z? Well, 2 had order 5. Well, 2 had order 5 in the context of the group Z10. But if I look at the number 2, viewed as an element of the group Z, of course, is Z with plus. And if I do 2 and then 2 plus 2 and then 2 plus 2 plus 2, in other words, if I keep combining 2 with itself, I certainly never get the identity element of this group on that list. Because the identity element of this group is 0. And if you keep adding 2 to itself, you keep getting larger and larger numbers. So there is no integer for which this particular sequence eventually spits out the identity element of the group. So we say that the element in the group. In fact, it's pretty easy to see that if you hand me any element in Z, except for the identity element itself, then you're going to get an element of infinite order. So there's the notion of the order of an element in a group. It's the smallest positive integer with the property that when you take the element in the group and you start it with itself, you do the binary operation with itself that many times that you see the identity element. And again, keep in mind that this number need not exist. In other words, the order might be infinite. All right. What will happen is that the order of the element, the number of times that you combine it with itself before you see the identity element, closely connected with the cyclic subgroup generated by that element. And in fact, what this observation shows is if you start grinding out the elements that would appear in the cyclic subgroup generated by this thing, once you see the identity element for the first time, then what this computation indicates is that if you keep combining that element with itself, you're certainly free to do that, but you're not going to get anything new. You're not going to get anything that you hadn't seen in the first pass through when you generated the powers of the original element a. So the observation and this isn't something I'm going to prove technically for you, but the punchline turns to be this, turns out, turns out, following is true. If a is an element of g and the order of a is finite, the order of a is some finite number is, let's call it n, then this number is precisely the number of elements in the subgroup generated by a. So the observation is because this element had order 5 inside z10, in other words, because I, when I was combining two with itself, because it took me five times, or two to the fifth, to get back to the identity element, if I then go back and count up how many different elements I see along the way, I get precisely that number. So what that means, of course, is that, well, let's see, we saw how things were cycling through here. What it means is that there's actually no repeats along the way. So in particular, moreover, or more generally, if a has order n and g in the group g, and we have two powers of a, let's say a to the power i. Remember what power means? It means a star a star a with itself, i times, if i is positive, or a inverse star inverse with itself, if i is negative. If i is negative. If I have two powers that are the same, well, then the powers might not be the same. I can have two powers of a equal without the powers being equal. For example, what we showed here is, for example, if we do two plus two, we get four. But if I do two plus two plus two, seven times, I also get four. In other words, two, quote unquote, squared is the same as two, quote unquote, to the seventh. But there's a relationship between two and seven. Specifically then, i is congruent to j mod the power, mod n. Okay, and conversely. So what happens with these powers is you get some sort of modular arithmetic, regardless of whether or not the original group has anything to do with mod n arithmetic, with modular arithmetic. You can decide when two powers of a given element in the group give the same element. That happens precisely when the two powers are congruent, mod, whatever the order of the element is. So let's give another example. Example, if we look at the element i inside the group of non-zero complex numbers, and I'll emphasize that the operation there is multiplication. And let's see if we can compute the order of i, the order of i in this group. Well, what do you do? There's really no issue in trying to compute the order. You simply start writing out the elements, the element itself, and then what you get when you combine the element with itself. So we get i, we get i squared, which happens to be minus one. We get i cubed, which happens to be minus i. We get i to the fourth. Oh, but i to the fourth is one. i times i times, it's the same as i squared, which is minus one squared. Oh, so what happens is after four tries, we see the identity element of the group, because this group is non-zero complex numbers under multiplication. So in the end, the order, the order of i in c star is four. And it might be a good way just to make the idea more concrete in your mind to do a few more computations. In other words, let's compute what you'd get if you take i to the fifth. Well, let's see what this result says is if you have two things that are congruent, mod whatever the order is, and we've just found this is order four, then this should be the same as i to the first. In other words, i, because five and one are congruent mod four. Is it sure if you take i times i times i times i times i? Well, that's the same as i to the fourth times i, but i to the fourth is already one. So we get one times i, which is i. i to the sixth is the same as i squared. i to the seventh is the same as i cubed. You know, et cetera, et cetera, et cetera. If you want, I could write down something like i to the hundredth is the same as, well, let's see what's 100 congruent to mod four is the same as i to the fourth, if you want, which is one. So you can compute any power of i that you'd like. Don't go ahead and do all the computations. Simply ask what is the power congruent to modulo whatever the order of the element is. So that's the notion of the order. The notation for order, and this turns out to be pretty handy, is this is the letter small o, and I'll try to make it as clean as possible. So this is o standing for order. So order of a denotes the order of a, denotes the order of a. And so in effect, once we've got this notation, then there's really three things that give the same quantity. The order of a, that's the smallest positive integer for which a to that power is the identity element of the group. That's the same as the smallest positive integer. Let's call it n with an equals e in the group. And that's the same thing as asking how many elements are there when you take the cyclic subgroup generated by the element a. So these three are equal. These are. Now if the question is how do I figure out the order of an element in general? The answer is if you handed the group and you handed the element, it's the brute force approach. Typically you just take the element and start beating on it a, a star a, etc, etc, etc. Of course the one easy case is if somebody hands you the identity element of the group, its order is one. How many times do you have to do the operation to itself to see the identity element? You already see the identity element, so that's a non-issue. But in general, if you're asking for the order of an element like here, well we just had to start pounding out the appropriate powers of the given element in the identity. There is one situation where it's relatively straightforward to figure out what the order of an element is in a group. And that one situation is this. Suppose you've already done the hard work required to figure out what the order of an element in the group is. Like we've done the brute force computation to figure out the order of i in C star is four. We did the brute force computation to figure out that the order of two inside z ten is five. Once you've done that, the question is what if then somebody comes back and hands you a power of the given element whose order you already know? Is there some way of computing what the order of that new element is without reinventing the wheel? So question, can we easily compute? Compute the orders of elements in a group. The answer is in general no. Usually, no. We've got to just do the brute force method. Which typically isn't too terrible. You just keep beating on it. But if you already know, you already know what the order of some element is. It's called the order of a. In other words, if you've already done the hard work to compute this and someone then asks, then asks you to compute, not the order of some arbitrary other element in the group, but rather asks you to compute the order of some power of a. For some a, I'm sorry, some power i. The question is, can you compute this number without reinventing the wheel without simply taking whatever this element is and beating on it and seeing how long it takes to get back to the identity element? And the answer turns out to be yes. Asks you to compute this. Can we easily do it? And the answer turns out to be yes. Yes. Here's the proposition. I'm simply going to hand you this formula. We could prove it to take about a half hour for those of you that saw the number theory course. The proof that we would give here is virtually identical to the one we used in the number theory course. When the specific situation happened to just be the Zn groups, it turns out the same formula will hold in general. If the order of a is something that we'll call, let's call it k, then the order of a to the i is k divided by the GCD, the greatest common divisor of i and k. In other words, if you know what the order of the original element a is, you can easily write down what the order of any power of a is. It's the order of the original element divided by the greatest common divisor of the order, this thing called k, and the power here. For example, let's see, in C star, what did we find? We found that the order of i is 4, so now the question is, compute the order of i squared. Well, I give this example because it's trivial to figure out what the order of i squared is. i squared, we already know, is minus 1. Question, what's the order of minus 1? Well, you take the element minus 1 and you start raising it to powers. That's how many wax you have to take before you get to the identity element. Well, hey, if you take minus 1 and you square it, in other words, minus 1 star minus 1, you already get 1. So this should be order 2. Let's make sure that the formula actually gives us this. This formula says you take what the order of the original element was, you take the order of i, which turned out to be 4, and you divide through by the GCD of whatever that number i was, whatever the number that appears as the exponent is, it happens to be 2, and whatever the original order of i was, that's the number k. So this is 4 over the greatest common divisor of 2 and 4 is 2, and we get that the order of i squared is 2. Another quick example, if we look at in Z sub 10, find the order of 6. Well, here too, folks, if we wanted to, we could simply do this brute force. Here's an element, here's the group. Find its order, just start adding it to itself. 6. 6 plus 6, is it the identity yet? No, 6 plus 6 plus 6, is it the identity yet? No, because it's 18. All right, so that's one way we could use brute force, but notice here, since 6 is 2 plus 2 plus 2 inside Z10, which is, and you know, it scares me to use this notation, but this is the notation that we'd get from general group theory. This says, do the operation to the element three times, the operation happens to be plus, then what this formula says in the proposition is that the order of 6 is whatever the order of 2 is, and I'll remind you what that is, divided by the GCD of whatever the power is, here the power on the element 6 is 3, and the order of 2. So all I've done here is sort of use the long term or the long hand notation for this thing called K. If you prefer the short hand notation, defining it as some number called K, or if you prefer to use the order of notation, it's all the same, it gets you to the same place. The order of 2 we saw was 5 by previous work, divided by the GCD of 3 and 5. This is 5 divided by, let's see, the GCD of 3 and 5 is 1, 5 divided by 1, which is 5. So it turns out that the order of 6 is the same as the order of 2, and both of them happen to be 5. Again, I'm going to omit the proof of this result. It's a relatively straightforward one to do, but it would take us 20 minutes or so and sort of standard caveat. I think our 20 minutes might be better spent doing something. A little bit different. So what we're now going to do is move away from this notion of taking groups that we already know and analyzing what's going on inside them. That was the whole idea of looking at subgroups and then looking at cyclic subgroups. We're going to back up again and we're going to look at a completely different type of group, one that we haven't seen before. We've seen lots of different flavors of groups. There have been groups of integers and groups of complex numbers and groups of matrices and groups of functions. You saw one of those for homework, et cetera, et cetera, et cetera. What we're now going to look at is what's typically referred to as permutation groups. This is section 6. Permutation groups will provide us with a number of very interesting examples of groups. It will also give us a context for many of the types of groups that we'll wind up looking at later on so let me just do a brief recap of what a permutation is. Remember the definition of a permutation is the following. Let A be any set, any set. Now, I'm not assuming there's any sort of structure on A. There's no binary operations necessarily on A. It's simply a set. Maybe it's the letters A, B and C. Maybe it's the numbers 2, 3 and 4. It doesn't really matter. The definition of a permutation is this, a function F from A to A. It means a function whose domain is A and whose range is A is a permutation of A. Of A, in case, is pretty easy. F is simply 1 to 1 and onto. 1 to 1 meaning that there are no two different things that produce the same output. That's 1 to 1. Onto means that every element of A appears. The words 1 to 1 and onto are often replaced by the single word is a bijection. A permutation of a set A is nothing more than a bijection from the set to itself. For example, here's the underlying set. A is the set A, B, C. There are lots of different permutations of this set. Here's 1. Here's A permutation. Well, permutation is a function. In order to tell you what the permutation is, I have to tell you what the function does. I have to tell you what the function does to each of the input values. There's three different input values. Let's call the permutation F. F of A equals B. F of B equals A and F of C equals C. Here is a permutation of the set A. I've described a function from A to itself. The function is 1 to 1. I don't see the same output value coming up or occurring from different input values, and I see each of the three elements of capital A appearing as output. Here's an example of a permutation. In fact, let's see, this is a consequence of something that you learned back in the discrete math course, in the math 215 course. If the set A has size n, then the number of permutations of a set with n elements of A is n factorial. In fact, because I started in this example by giving you a set having three elements, it would be possible to write down three factorial or six permutations of the set A. All I've done here is given you one of them. Here's the notation. If I hand you a set capital A, we write the letter capital S with the letter capital A in its lower right-hand corner to denote the set, the set of all permutations of the set A of A. So it's a set of functions corresponding to an underlying set A. Please again, notice there's no structure that's coherent or implied on the set capital A. The set capital A might just be letters. It might be numbers. It could be anything you want. It could be symbols. It could be circle, square, solid, et cetera. So simply rephrasing the comment that I've made here, this result from math 215. So the number of permutations on the set A is n factorial. So why is this of interest? The answer is it turns out this set of permutations, not the underlying set capital A necessarily, but the set S of A, the collection of permutations of A, actually has a binary operation on it that turns it into a group. So that's our interest in this. But wait, let's see. If F and G are permutations of the set A, in other words, are one-to-one and onto functions from A to itself, then this composition makes sense. Is what? Is a function from... Well, let's see. G is a function from A to A. In turn, F is a function from A to A. So when I compose them, I get a function that starts at A and ends at A. In fact, if this is one-to-one and this is one-to-one, then the composition is one-to-one. If this is onto and this is onto, then the composition is onto. So in fact, when I do composition of two permutations from A to A is another element. I'll have S sub A. The composition of two permutations of a set A yields another permutation of the set A. For example, let's look at the same set A as we did before and I'll haul out the same permutation F that I just did. If I define a new permutation, maybe G from A to A. And again, A here is the same A, B and C. A, B and C, A, B and C. Let's define it this way. How about G of A equals C, G of B equals B and G of C equals A? This is a permutation. Each element of A appears and appears exactly once it's an output. Let's look at F circle G. Well, this presumably is also a permutation of A. So it makes sense to ask, what would come out if I plug A into this permutation? Well, that's by definition of composition. It's simply F of G of A. Let's see, G of A is C. That's the given information about G. But F of C is, we go back to the definition of F of C turns out to be C. So in the end, when I do the composition F circle G of A, I get C. If I do F circle G of B, let's see what comes out when I plug B in, that's F of G of B. Let's see, what is G of B? Oh, it's given information, that's B. So that's F of B. But the information about F of B is that that's A. And finally, if we do F circle G of C, we get F of G of C. F of G of C, let's see, that's F of, what is G of C? Given information about G of C, that's A, and F of A turns out to be B. So we get a new permutation, the permutation where when you plug in A you get C, when you plug in B you get A, when you plug in C you get B. Simply another permutation of the set A, or rephrase simply another element of the set, capital S of A. All right, so what we're about to do is phrase this observation, which is nothing more than a discrete math result, a math 215 result. If you take two permutations or take two bijections from a set to itself and you compose them, you get another one. So, it turns out, composition of functions, which is denoted by little circle, as usual, is a binary operation, operation on the set S of A. On the collection of permutations of A, not on A itself, it might not make any sense to compose elements of A. The elements of A might just be letters or numbers or something like that, but it makes perfect sense to compose bijections from A to itself. And in fact, given the context of this course, so you are expecting the following statement. Proposition, if we look at the set of permutations on whatever set you want to start with, I don't really care what A is, look at the collection of permutations together with function composition gives a group. So, here's a new group. Here's a group of a different flavor. It's a group where the elements of the group are functions, they are permutations corresponding to a given underlying set. The example that I gave happened to be the underlying set A was the three letters little A, little B, and little C. It doesn't really matter at all what the underlying set is. In fact, the underlying set could be infinite, but the examples that we'll wind up looking at for this semester typically will involve the underlying set being finite. All right, so how to prove this proposition? Well, we know what to do, prove to show something's a group, well, look folks, I'm not claiming this is a subgroup of something that's already known to be a group. This is a new beast, so when we're trying to verify that something is a group, something that we haven't confronted before, we can't just say, well, use the subgroup there and we have to actually go back to basics. So what do we have to do? Well, we have to check that we actually have a binary operation, check that the given operation really is a binary operation on S of A, and we've done that on S of A done previously, basically done in the last three minutes. So that's step zero, we actually get this thing off the ground. We have to check that composition of functions is associative, it is, that's okay, a discrete math result. Let's see, we have to check then that what? We have to check that there's a special element inside the set, not the underlying set, but the set that we're interested in showing is a group, namely the collection of permutations of A that behaves as the identity element. So we have to find a special function, because that's what the elements of S of A are, with the property that when you compose that special function with any other function, that yields the original function. Well, we know how to do that. Check the existence of an identity element in a binary operation. Identity element in S of A with composition. Well, here's one, but for any set A, define, and this will be standard notation, this is, you can either think of it as the letter I without a dot on it, or the Greek letter iota from A to A. So I'm defining a function from the set A to itself, even though I have no idea what this might look like. The definition of the function is, whatever goes in comes out. This is the identity function, and that's a good thing to call it, because it'll turn out to be precisely the identity element in this group. It's the function that doesn't change anything. So, for instance, the identity function on this set A, B, C is simply the function that inputs A, outputs A, inputs B, outputs B, inputs C, outputs C. Well, what do we have to check? We have to check not only that this thing exists, but we have to check that it's in the correct set. In other words, we have to check that it really is a one-to-one and on-to function. That it really is an element of this, but it is. It's clear that i is one-to-one and on-to. It's just the identity function. Heck, if you hand me something, that's what comes out. So you certainly can't send two different things to the same thing. Similarly, does everything come out? Yeah, because whatever comes out is what went in. In other words, the identity function is in S of A and this function clearly has the property that for any other function, let's call it f, for any other permutation, if I do f circle i, I get the same as i circle f. The identity function doesn't change anything. If you compose with the identity function, it's going to keep the original function the same. So this particular set has an identity element. Final. See what do we have to do? It has an inverse. In other words, for anything I choose here, I can find some other thing in the set so that when I combine it with that other thing in the set via the given binary operation, that I get to whatever the identity element is. We've already identified the identity element. It's the identity function. But the point is this, if I take any function in S of A, then by definition it means that this thing is one-to-one and on-to. So because the elements of this set are by definition one-to-one and on-to, then f inverse exists. Again, just a math 215 result. If you have a one-to-one and on-to function, then its inverse exists. You simply do things backwards. And not only does it exist, f inverse is also one-to-one and on-to. It's not special. It's just math 215 stuff. And by definition, when I do f circle f inverse, I get the identity element and f inverse circle f is the identity element. So check. This collection, S sub A, where A is any set, the collection of permutations on the set A actually gives a group structure where the appropriate operation is composition of functions. Now I've said it a couple of times already and I'll say it again because this seems to be where students have some difficulty understanding what the things are and with the elements in the underlying set capillary really play no role here. The things that are playing the role are functions from the set A to itself. So if you want to describe what those functions do, okay, then maybe the element in the underlying set will be written down. But the structure of the underlying set is not what's an issue. What's an issue here is the functions from the underlying set to itself that happen to be one-to-one and on-to. All right, example. Let's see. Well, we've already seen some. If I look at the set A equals ABC, then the group S sub A well, consists of, I'm going to use this result that we hold out. We know how many permutations there are on a set having an element. Three-factor equals six elements. We've listed out a few of them already. Let's see. We have seen three of them already. And what we'll do is we will see three more momentarily, but in a slightly different guise. The idea is this, because the structure of the underlying set doesn't play any role in the structure of the permutations of the set. In other words, I could just as easily have started with the set square circle triangle and asked you to tell me something about the permutations of those three things, or I could just as easily have started with the numbers one, two, and three and asked you to tell me about the permutations of those three things. That's the context of the following comment, because the structure of the underlying set really doesn't play a role. Only the number of elements in the underlying set plays a role. Set doesn't play a role other than the number of elements in the set. Number in the set. What we usually do is, we usually denote the group S of A by the notation S sub order of A. If we were to talk about S sub A, where A is the set, ABC, because the set A has three elements in it, this would be denoted by S with a three on it. Why? Because there's three elements in the underlying set. The understanding is that if you've given me some set, if you've called it ABC or square circle triangle or one, two, three, the structure of the permutations on that set are going to be identical. How you move them one to the other is completely irrelevant or independent whether you happen to have called them letters or symbols or numbers. So in general, what we will wind up talking about are permutations groups that are referred to as or denoted by S sub N. Now, what's of interest here is the following. We've seen some examples of non-abelian groups, of groups that have the property that when you combine elements in different orders, they might produce different outcomes. But we haven't yet seen an example of non-abelian groups having only finitely many elements. So non-abelian finite groups, all the examples of non-abelian groups that we've given so far are these infinite groups of matrices, the GLN groups. Proposition is this, if you hand me any integer bigger than or equal to three, then the group of permutations on the underlying set, having that many elements, so maybe S sub three or S sub four or S sub 100, I don't really care, is not a beauty. And what we'll do is we'll give prove momentarily, momentarily, what I want to do first is give notation, first notation. So it'll be easier to describe what this thing is doing if we have a little bit of notation under our belts. All right, the advantage of describing the underlying set as numbers. And look, folks, if you hand me the numbers one, two, three or the numbers 18, 19 or 20, both are sets of three numbers, so either one is perfectly good but typically we use the numbers one through N, that's a more natural collection to view. And what we'll do is we use some parentheses. They look like matrices, but parentheses to describe what permutations do, permutations do do the elements in the underlying set. Well, the underlying set we've agreed to simply take as the numbers from one to N. For example, here's a permutation, let's call it, and I'm going to start using the standard notation when we talk about permutations, typically we'll use Greek letters. So I've used up until now the letters F and G just because that's comfortable as far as your function notation goes. But when we talk about permutations, typically the letters sigma or kappa or tau will be used to describe those. Here's a permutation from the set 1, 2, 3 to the set 1, 2, 3. In other words, what I'm about to do is give you an element of S3. It's the element that takes the following, sigma of 1 is 2, sigma of 2 is 3 and sigma of 3 is 1. So the equation sure it is. Each element goes to a different element and each of the image elements appears exactly once. Rather than calling this sigma and writing it out in a line this way, what I'm going to do is describe it this way. I'm going to give you the elements of the underlying set, which happened to be 1, 2, 3. And then I'm simply going to ask you to read down and indicate what the output is for each of the specific inputs. So sigma of 2 is 3, sigma of 3 is 1. So that's the description that we will use. So sigma of 3 is 1, 2, 3. And that's the end form of sigma. So here is sigma written in its long hand form. Let's give another example of a permutation of the set 1, 2, 3 without describing it this way. Here's another one. Another permutation. It's the permutation tau. It's the permutation that takes 1 to 2, 2 to 1 and leaves 3 long. So if you want to write this thing out using function notation, you'd write tau of 1 equals 2, tau of 2 equals 1, tau of 3 equals 3. Here's another permutation. In order to indicate whether or not or to decide whether or not the function you've written down really is a permutation, all we're doing is making sure that what's coming out here includes each of the elements of the underlying set exactly once. So as soon as you've seen each element of A appear exactly once, then you know you've written down a permutation. That is, the permutation that you get if you do sigma, then tau. All right. Well, remember, function composition reads from right to left. If you do sigma of tau of 1, and we've already done an example of this already, so I'll run through this just for one or two of the examples. Sigma circle tau of 1. What comes out when you plug in 1? Well, it's sigma of tau of 1. So that's sigma of, let's see, what's tau of 1? Let's get off here. Tau of 1 is 2. And sigma of 2, I can read off from this. Let's see, sigma of 2 is 3. So if I want to write out the composition, sigma circle tau in this long-hand form, what I've just shown is when I do sigma circle tau that when I plug in 1, that 3 comes out. Let's see what else we can say about sigma circle tau. Sigma circle tau of 2. Sigma of tau of 2. Oh wait, that's sigma of, I know what tau of 2 is. Tau of 2 is 1. Tau of 2 is 2. So I get 2 goes to 2. And now rather than doing the computation because we know we're going to get a permutation, in other words, each of the three elements in the set, 1, 2, 3 have to appear along the bottom. By default, sigma circle tau of 3 has to be 1. If you want to check that, that might be a good computation, a good exercise for you to do. So there's a long-hand version of sigma circle tau. Now, here's the beauty. But if we try to take these same two permutations and ask what do we get if we do the composition tau circle sigma in the other order, let's see what we get. Well, rather than pounding things out this way, what I'd like you to do is start thinking about taking the long-hand version of these and seeing if we can read that information from the long-hand version. So let's see, what was tau? I'll write it out in this long-hand version. 1, 2, 2, 1, 3, 3. I'll write out sigma in this long-hand version. 1, 2, 2, 3, 3, 1. Now, keep in mind how we compose functions. Functions are composed from right to left. It means you first do this one, then you do this one. Well, let's follow through and see what happens. What happens to 1? Well, when I run it through sigma, 1 goes to 2. But then when I run 2 through the function tau, 2 goes back to 1. So the net result is sigma takes 1 to 2. In turn, tau takes 2 to 1. And so we get sigma takes 1 to 1. Next one. What happens to 2? Well, if I run 2 through the function sigma, 2 gets taken to 3. In turn, 3 gets, oh, taken back to 3. So the net result is that 2 gets taken to 3. And finally, what happens to input value 3? If I input 3 into this composition of functions, 3 goes to 1. And in turn, 1 goes to 2. Quick sniff test here, yeah, that's fine. It is a permutation. And so here in the end, turns out to be the permutation tau circle sigma. And notice what's just happened. If I compute sigma circle tau, I get this. If I compute tau circle sigma using the same tau and sigma, I get this. And these obviously are different. So these are not equal. Not equal. In other words, sigma circle tau tau circle sigma. At least for these two functions in S sub 3. So what we've proved already is the particular group S sub 3 is not a billion because I've produced two specific elements inside S sub 3 that have the property that when you combine them in the two possible orders, that you get different outcomes. And in fact, more generally, this shows not only that the group S3 is not a billion, and shows more. Using the same idea, same idea, we get the result of the proposition. Proposition. I'd say, well, how do you know just because inside S sub 3 you have two things that don't commute that you can build at least two things inside S sub 4 or S sub M that don't commute. That's what I want you to do. Inside S sub 4, describe the permutation that does these things but then takes 4 to 4. And similarly, tau takes 4 to 4. In other words, define the permutation that somehow extends what's going on for tau and sigma to S sub 4, S sub 5, simply by doing the identity element on anything else that appears. And what'll happen is, folks, that when you look at what happens when you plug in 4 or 5 or anything up to N, because what's going to happen in the first three slots of sigma and tau will still give you two different things and the first three slots of sigma circle tau, you'll see that, and the first three slots of tau circle sigma, you'll see something different. So we get in general that the collection of permutations on a finite set, and in fact the same is true on infinite sets, but as I mentioned, we're not going to really concern ourselves too much on the permutation groups for infinite sets, that these things give examples on the permutation groups. Let's see, how am I doing on time here? Well, I'm at about an hour. I'll go about, I don't know, five, seven more minutes and then we'll call it a day. All right, so to end today, what I want to do is give you some additional notation surrounding permutations. The longhand notation, this parentheses notation for permutations is actually pretty useful, but it turns out it gets a little bit cumbersome when you have to continue to write out, one, two, three, you write out one, two, three, four, and then you tell me what one, two, three, four, go to, et cetera. There's a somewhat more streamlined notation that has to do what we call, do with what we call the orbits of a permutation. The orbits, the orbits of an element permutation. And I'll describe this just by giving an example. So here's an example. We'll look at the element sigma in S sub, how about S sub nine? And at this stage, I can describe what permutations do by simply listing out the elements in the underlying set. Presumably it's any set with nine elements, but we're going to choose the particular set that consists of the numbers one through nine. And we'll look at this. One, two, four, three, nine, five, six, let's see, eight, seven, and what number do I need to make this permutation one? So this is the permutation that takes one to two, two to four, three to three, et cetera, and simply gives all the other values as are indicated. So here is a permutation. But watch, what happens to one? What happens to one? Well, let's just follow it through. Let's see, first one goes to two. All right, that's cool. What happens to two? Four goes to nine. And nine in turn goes back to one. So here's what sigma does. It takes one to two. In turn, it takes two to four. In turn, it takes four to nine. It takes nine to one. And so after one, two, three, four steps, we wind up back where we started. We're going to call this the orbit of one under the permutation sigma. It's the elements inside the underlying set that eventually get hit when you look at what happens when you start with the particular element here, one. So what's the orbit of the element two? Well, two goes to four, goes to nine, goes to one, goes to two. So once you've got the orbit of an element and you've got some other elements that appear in that orbit, you, in effect, already know then what the orbits of the other elements that appear are. The orbit of two, again, is two, four, nine, one. Back to two. The orbit of four. What the orbit of two is, what the orbit of four is, and what the orbit of nine is. Let's look at some other orbits. How about the orbit of three? Orbit of three? Well, this says three goes to three. So the orbit of three consists just of itself. I already know what the orbit of four is that appeared here. The orbit of five goes to itself. It's just five. The orbit of six is just six. How about the orbit of seven? Well, this says seven goes to eight. Seven goes to eight. But in turn, eight goes back to seven. So the orbit of seven is the collection seven and eight. Sometimes we'll list out the orbit of an element as some sort of sequence. Other times we'll simply list it out as the set. So if you want, the orbit of any of the elements that appeared here, one, two, four, nine, can be written as this set. Here's the final comment and then we'll call it a day. It's this. In general, if I hand you two permutations, two permutations may or may not commute. Sometimes you're lucky. Sometimes sigma, circle tau is tau, circle sigma. But in general, you're not necessarily guaranteed that. Here's the primary examples. What we can do is always take any permutation and write it somehow as permutations that are made up by what happens to each of the orbits that make up the original permutation. So that if I talk about something like, let's call it sigma sub one, is the permutation that takes one to two, let's just look at what this permutation does on this orbit of one, takes two to four, takes four to nine, and takes nine back to one. So what I've just done is I've written out a piece of a permutation that tells me exactly what's happening to the four things inside the orbit of one. And now what I'm going to do is simply fill in the rest of the slots with the identity. Six, six, seven, seven, eight, eight. I'm going to call that sigma sub one. Let's see if I do the permutation that corresponds to the orbit of three. Well, the orbit of three consists only of three. So if I'm going to write out what happens to three, three goes to three. And then if I'm going to fill in everything else with the identity, this permutation corresponding to an element with one orbit is just going to be the identity. And I'm going to then look at another permutation that somehow describes what's happening on this second chunk. Seven goes to eight. Eight goes to seven. And then I'm going to fill in all the rest of the entries with the identity. Five, five, four, four, two, two, one, one. And the punchline is this. Then the original permutation, sigma, can be viewed as made up as the composition of these two permutations, where individually what these two permutations are doing is simply telling you what's happening to the orbit of the original element one, and then in the second case of the original element seven. All right, this is a good place to quit. What we will do on Wednesday is start with this notion of taking a permutation, breaking it up somehow as a composition of smaller or easier or more manageable permutations, the permutations corresponding to what's happening on each of the orbits of the original permutation. And then we'll make the observation that in general, while in general permutations need not commute, the permutations that arise under this construction do commute. And we'll be able to say something relatively intelligent about the result. Okay, so let me give you a homework assignment. Let's see, homework assignment, yeah, will be this. I'll post this on the web as well, so if you don't get all the details here, not to worry. Homework assignment is this, so this is assigned on Monday, September 10th, so it's due a week and a half from now, Wednesday, September 19th. And it looks like the following. In section six, someone should do problem 17 through 21, turn in 17 and 20. Section eight, problems one through nine, 16 through 26, and 44 through 46, so I want you to turn in problems five and eight. In section nine, problems seven through 13 and 24, I want you to turn in problems 12 and 13. And then finally, and I want you to turn this one in as well, E and S five, define the following subset. H is the collection of elements I'll call them sigma, with the property that sigma of three is three. In other words, I want you to look at the permutations of S five, but not all of them, only the ones that have the property that when you plug in three that three comes out. Prove that H is a subgroup of S five, S five, and then answer the question to what known group, so turn in, and then to what known group is H isomorphic? That's a turning question. And finally, two days from now, Wednesday, September 12th, quiz two, and quiz two is prove that every cyclic group is abelian, and then that's one. And question two will be give an example, example, an abelian group which is not cyclic. You should refer to your notes from last Wednesday in order to do. Okay, that's it.