 Now that we know about subgroups, let's move on to a new type of group called a cyclic group, cyclic cyclic cyclic cyclic group. Don't confuse that with cyclic subgroups, you know, they're also cyclic groups because a subgroup is also a group and depending on how this is explained to you, sometimes the explanation will involve the subgroup, sometimes it'll just involve a proper group. So just get your mind around those. So what is a cyclic group? Well, first of all, it is a group. So, you know, we must have a group and it must have a set and it must be a binary operation between all the elements of that set with our four properties. Let's just say our four properties there, and we know that's closure, associativity, unique identity element and every element has its unique inverse in that. So, what is a cyclic group? A cyclic group is then if I take one of the elements in my set and under that binary operation, it generates all the other elements in that set and say we take this element A, it's an element of, I'm going to say G so that it incorporates both the set and its binary operation. Then A, then A is a generator of the group and we write usually also in these bracket notation. It's a generator of that group, G, such, and we write it so that any multiple under the binary operation of that generator, such that N is an element of the integers. What does that mean? You know, what does that notation mean? So it says A to the power N, but you know, that normally would indicate to us it's multiplicative, but we also actually just want it to include in A, you know, so that it's additive as well. So this would be really for this binary operation being multiplication. This will be for this binary operation to be addition, but most textbooks and most explanations just, you know, leave that out and assume that no matter what the binary operation is, you know, there's more than just addition and multiplication, this this means repeated ones of that. So, you know, if my binary operation is just this generic this, you know, A on itself, that will just be A1, A that, that would be written as A squared, but it's not really squared because if this was addition, that means A plus A and that was 2A and not A squared, but you understand what we mean by this square, this power notation here, so that if we have A A, that's A3 and this A3 is also an element of the set. If that is so, that every element in the set can be generated by multiples under the binary operation of that element, then that is a cyclic group. So let's look at an example, let's look at an example of a cyclic group. Well, one cyclic group would be the integers, the integers under addition because it has a generator which is just one. So any multiple of, integer multiple of one, so and remember n is that multiple and that's an integer, so from all the negative to zero to all the positive integers, that all this whole set, this infinite set can be generated by one. Okay, so which makes it also this infinite set being generated by one, some uniqueness there, but we might get into that a bit later. So let's move on just to stating the very obvious here, that cyclic groups are abelian. So that cyclic groups are abelian, in other words, they commute. Not all abelian groups are cyclic, but cyclic groups do commute. And let's prove that very quickly. Ah, don't you just love proofs. So let's just prove that very quickly. It's not a typical proof. We're going to say let g1 and g2 be elements of my cyclic group, g. So g is a cyclic group, g. That means it has a generator, you know, let the generator, let the generator, let, let's say a element of g be the generator of that cyclic group because that's how we defined it. In other words, we can write we have, we have two elements, say r and s and they're both elements of z. They can be both of them. They're all n. So that means I can write g1 as a r because g1, remember, is generated by a. So it's just some multiple of a and it's an r multiple of a and it is in there. And let g2 equals a s. If I now have the binary, this g1, g2, that is going to equal a r, a s. And remember, that's just a r plus s. And since r and s are both integers, they commute. So that is a s plus r. And that equals a s, a r, and that equals g2, g1. So g1, g2 equals g2, g1 for any, for any g1, g2 elements of a cyclic group. So cyclic groups do commute. I'm going to clean the board and then we'll carry on. Now let's consider, clean board, let's consider the integers of mod 6. What were they? Well, that's a group of 6 elements, 0, 1, these equivalence classes, and 5. Let's see if any one of them are generators of the whole group. Can they be generators of the whole group? Remember, we have modular addition here. So in other words, 5 plus 4 is 9. 9 minus 6 is 3. So 4 plus 5 would be 3. So let's look at just the generator 0. What is 0 going to get us? Well, that is just 0, very uninteresting. Does 1 generate? Well, it will have 0. 1, 1 is 1. 2 is 2. 3 is 3. 3 1s, 4 1s is 4. 5 1s is 5. 6 1s. I'm just carrying on with my integers. 6 1s under this modular addition is 6, but that is mod 6 is 0 plus another one, which will give me 7, which is, but 7 mod 6 is just 1. So I'm just going to repeat as n grows bigger and bigger and bigger. I just repeat this modular addition. So look at that. 1 is a generator of the whole group. 1 is a generator of the whole group. Let's go to 2. Well, that's going to give me a 0. It's going to be 2. It's going to give me 4. When it gets to 6, it'll be back to 0. When it gets to 8, that's 4. And so on. So that's actually all I have. That doesn't matter how big n grows. That mod 6 will be one of those. Let's look at 3. Well, that is going to be 0. And that's going to be 3. And 3 plus 3 is 6, which is just back to 0. So that's all you're going to get. Let's go to 4. What is 4 going to give us? Well, that's just going to give us a 0. It's going to give us a 4. 4 and 4 is 8. And 8 minus 6 is 2. And 4 and 4 and 4 is 12. And that's just back to 0. And 4 and 4 and 4 is 16 minus 6 minus 12 is back to 4. So I'm just going to cycle around all of those. 5 is going to be interesting again. Okay, so let's just do this. 0, 0, 5 is 0. 1, 5 is 5. So that's, you know, just a 5. 5 plus another 5, because remember this is addition. 5 plus 5 is 10. Mod 6 is 4, because 10 minus 6 is 4. 5 plus 5 plus another 5, that's 15. Minus 6 is 9. Minus another 6 is 3. So we have mod 3 there. 5 plus 5 plus 5 plus 5. Now we have 20. And 6 and 6 and 3 is 18. So I can subtract 6 from it 3 times. And that leaves me with mod 2. And the next one's going to be 25 minus 24. That's mod 1. And just look at that. I have 0, 1, 2, 3, 4, 5. I have the whole z6 again. So look at this. There were two generators inside of this that gave me the whole set. So this is, number 1, it is a cyclic group. The integer's mod 6, because there's always going to be one that is going to, with modular arithmetic additive addition, is going to give me one of those. But there's more than just one. In this instance, it was 5 that gave me this. 5 that generated the whole group for me. So don't get these confused. I'm just looking at these elements as generators. And of course, we can check each of these if they are cyclic. We can check each of these if they are cyclic. And we have to check each of these to be a group, because remember a cyclic group has to be both a group and then have to be cyclic. So don't get all of this confused. So we first have to check if this is a group, if this is a group, and then check if it's cyclic. That's not what I'm doing here. I'm just showing this is cyclic, because this element and this element generates the whole, this whole set. Let's clean the board. Okay, let's move on to the integers. I'm doing it the other way around now. Let's move to the integers mod 12. And I want to look at one of its subgroups. And I claim that 0, 3, 6 and 9 is a subgroup. It's one of these subgroups. So as I say, don't get these things confused. Cyclic groups, subgroups, cyclic subgroups. So just let's just concentrate on what we're trying to achieve here. I'm looking at Z12. So remember that is going to be 0, 1 all the way. And these are all congruence. These are all congruence classes or equivalence classes up to 11. And so I'm saying I'm taking some of its elements, 0, 3, 6 and 9. And I'm asking, I'm not going to put these square brackets in every time I'm asking, is that a subgroup? Remember, in the previous lecture, put the link up there. If I remember, is that it has to be, there has to be closure. And then every element must have its inverse. So there are only two things we need. So let's have modular addition. Sometimes we just put plus there. But remember, this is modular addition. And we have 0, 0, 3, 6 and 9. 0, 3, 6 and 9. 0 and 0, 0. And 3 and 6, and 9. And 3 and 6 and 9. 3 and 3 is 6. 6 and 3 is 9. 9 and 3 is 12. Mod 12 is 0. 6 and 3, 6, 7, 8, 9. 9 is there. 6 and 6 is 12. Minus 12 is 0. 9 and 6 is 15. Minus 12 is 3. 9 and 3 is 12. Mod 12 is 0. 6 and 9 is 15. That's mod 3. 9 and 9 is 18. And we have 6 times 3. So that's mod 0. So just have a look at that. So we have closure. Definitely we have closure because no matter what the modular addition is between any of these, it remains inside of my set. So we have closure. Do we have unique inverses for each? What is the inverse for 0? Well, if I do mod 0, 0, modular addition with 0, I get 0. So it's its own. What do I have to do with 3 to get 0? Well, its inverse is 9. 6's inverse is just its own inverse. And 9's inverse is going to be 3. So 0, 3, 6, 9. Each of them has a unique inverse. So indeed, this is a subgroup. It is a subgroup. Let's see if it's cyclic. One of its elements must generate all of them. Let's choose 3 for very obvious reasons. You can guess what's going to be 3. Because what does that give me? Multiples of that. No multiple letters 3. One multiple of that is 3. 3 and 3 is 6. And 3 and 3 and 3 is 9. And 3 and 3 and 3 and 3 is 12. But that's mod 0. And then I'll carry on. No matter how many 3's I add with modular addition, I'll just get back to that. Remember, we are talking in these congruence classes. Okay? So not only is it a subgroup but it's also a cyclic subgroup. So I've shown both things. So don't get those mixed up. So I've shown that it's a subgroup and I've shown now that it's a cyclic subgroup. So that leaves me with one more topic for, that leaves me for one more topic for this. Let's just look at Z12. Z12. Which of its generators is going to generate the whole group? So I'm just walking away now from cyclic groups. But this is usually added to cyclic groups. Which of its generators, so if I have a 0 and I have 1, you know, elements I should say, not generators. Which one? 4, 5, 6, 7, 8, 9, 10, 11. Which ones of those will generate? Well, I'm going to give you the answer now. We won't look at a proof. But it is the coprimes. The coprimes, just remember that always. The coprimes of 12. The coprimes of 12. What is a coprime, you ask? Something we haven't put in the box yet. Not in this video series also. Let's put that in the box. Let's take the numbers. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 in this instance. 12. Let's just see what all the factors of 1 is. What times what gives me 1? Just 1 times 1. 1 and 2. Let's do this. So 1 and 2, because that is 1 and 3. 1, 2 and 4. Because 4 times 1 is 4. 2 times 2 is 4. 1 times 4 is 4. 1 and 5. So that's prime. 1, 2, 3 and 6. 6 times 1 is 6. 3 times 2 is 6. 2 times 3 is 6. 1 times 6. 6 times 1. 7 is prime again. 8 is 1, 2, 4, 8. 9 is 1, 3 and 9. 10 is 1, 2, 5, 10. 11 is prime again. 12 is 1, 2, 3, 4, 6. 12. So it's all the factors that you can factor these values into. And co-primes, what a co-prime is, if you take that number, the two numbers, to decide whether they're a co-prime, the only factor that they can have in common is 1. If they have any other factors in common, they are not co-prime. So each of the two numbers does not have to be prime. That's not what co-primes. It's not both primes. Definitely 12 is not prime. But what I want between those two numbers is that the only factor that they have in common is 1. So let's look. 1 is definitely co-prime with 12, because the only factor they have in common is 1. 2 is not co-prime because there's a 2 and there's a 2. It's not only 1. 3 is definitely not co-prime. 4, a 2 and a 4. It's not co-prime. 5 is definitely co-prime. 6 is not going to be co-prime. 7. Yes, it's only the 1. There's no 7 there. So 7 is co-prime. 8 is not. 9 is not. 10 is not. 11 definitely is co-prime. So the generator is 1. The generator 5. 7 is a generator. Element 7 is a generator. And element 11 is a generator. And that is going to give us the integers mod 12. The whole of integers mod 12. That is going to give us the integers mod 12. And that is going to give us the integers mod 12 again. So just to look at which of the elements are generators of this. Remember, this is always going to be a cyclic group because definitely one is going to always with the modular arithmetic integers mod n is always going to give you generate the whole group for you. But there will be others. And the others are the ones, are the elements that are co-prime to mod n, whatever that n is. So what we've looked at here is what a cyclic group is. You see that cyclic group? Catches is we always write this. But remember, this just depends on what our binary operation is. But we always write it like we never do addition. And we always just keep to this same thing. And remember, when we talk about subgroups or cyclic subgroups, first show that it is a subgroup. And then you can show that it is cyclic. So you can take any one of these subgroups. First show that it is a subgroup. And now an interesting thing are all the subgroups. Cyclic. Look at that. All right. And then we looked at which of the elements in this integers modulo n is going to generate for us the whole group. So there's different things that's usually discussed together in the cyclic group. Please remember, a cyclic group is a group. So it has to be a group first. So a set and its binary operations with the full properties. And if one, at least one, just one, it can be more as we shown. If at least one of them, not there, that's not what you showed. If at least one of them generates by just multiples of itself, generates all the elements in that group, then that becomes a cyclic group.