 So now we've introduced the roots of unity in the complex number system. So we previously remember we had these symbols zeta sub n, which we define to be e to the two pi i over n. And we argued that zeta, whoops, zeta to the n raised to the k, these numbers right here are the primitive, they're the primitive nth roots of unity whenever the GCD of k and n is equal to one. And these are gonna be our primitive nth roots of unity. Remember the roots of unity are those numbers which would raise to the nth power, give you one. The primitive ones are those where no smaller power than n will give you one earlier than that. So we are gonna define a set, we're gonna call it zn. Notice that we're using a different font than the usual zn we're gonna be doing, where this zn with the blackboard font right here, this is representing the congruence classes of integers modulo n. While in this situation zn, which is a regular z, we're viewing this as a subset of the complex number system. Zn is gonna be the set of all nth roots of unity inside of C star. And I should mention that this is not a universal notation. There really isn't one for this group, but we're gonna go for that right now. What we claim is that zn is in fact a cyclic subgroup of C star right here. And the generators of that cyclic subgroup are going to be the primitive nth roots of unity, all right? So let's consider some primitive nth root of unity zeta inside of C star. I mean, that could be zeta n, that could be one of these guys zeta into the k. I'm not specifying which one it is, but we have some primitive nth root of zeta. And then we're gonna take the cyclic group, the cyclic subgroup generated by zeta, and so g can be viewed as a subgroup of C star. Zeta is in C star and so the cyclic subgroup generated by zeta will be inside of C star as well. All right, so by exponent rule, some things we should mention here that if you take zeta to the m power, then you take that to the nth. By exponent rules, you can actually commute the exponents. And so this becomes zeta to the n to the m, which means you're taking one to the m, which is equal to one. And so what this tells us here is that if you take zeta, all of the powers of zeta themselves are gonna be nth roots of unity. So we're basically saying that zeta to the m is an nth, I guess I have a lower case in there, nth root of unity. And as such, zeta to the m belongs to the set z sub n. Of course, zeta belong to that set as well. So because all the powers of zeta belong to zn, this tells us that g is a subset of zn. Now, as g is, of course, a group, you could say that g is a subgroup of zn, but the problem is we haven't yet established that z is itself a group, so it doesn't really make sense to talk about subgroups. So we do see that in terms of set containment, the cyclic subgroup g is contained inside of zn. It's gonna be our goal to prove equality right here, right? We wanna show that g is equal to zn, and as z is the cyclic subgroup generated by zeta, that would improve that zn is a cyclic subgroup of c star right here. So how do we proceed next in this regard? So notice here that in terms of order, zeta is an element of order n. Now you have to be a little bit careful with the notation right here, because depending on context, am I talking about the modulus of the complex number, which would be one, or am I talking about the order of the group? It can be a little bit confusing, but as viewing zeta as a member of this group, it has finite order, and since it's a primitive intrude of unity, its order in that group is going to be n. And so the cyclic subgroup generated by zeta will also be a group of order n. Now notice we have a set of size n inside of another set. Well, how big is that set? Does zn contain n many elements? Because if both of these sets have size n, then the containment would actually force equality right here because they both have order n. So zn is all of the nth roots of unity total. And so how many roots, how many nth roots of unity are there? Well, notice that every nth root of unity is a root to the polynomial x to the n minus one. So if we took like a polynomial f of x equals x to the n minus one, notice that f of zeta is equal to zeta to the n minus one, which is one minus one, which is zero. And so the roots of the polynomial x to the n minus one are exactly the nth roots of unity. And because the other thing to mention here is that if you were looking for the roots of this polynomial, we're looking for things where x to the n equals one, right? So the roots of x to the n minus one are exactly the nth roots of unity. Now by the fundamental theorem of algebra, the number of roots to this polynomial will exactly be its degree if you count multiplicities. And therefore there's at most nth roots of unity. And so then when we look at the containment here, notice that g contains n elements. It is a subset of zn. So in terms of their cardinality, the cardinality of g is less than or equal to the cardinality of zn. But by we mentioned here with FTA, zn itself has a cardinality of n. And so that then forces, that should be an equal sign right there. I guess I take that back because the fundamental theorem of algebra doesn't actually tell us that it has exactly n. It just says with multiplicity, you have at most n. So because n equals n right here, you're gonna force equality in both of these situations here. So g then equals z to the n. And as it was a cyclic group, that then tells us that z to the n is a cyclic subgroup of c star. Well, what about this other idea about primitive roots, right? Well, the generators of a cyclic group are gonna be those elements whose order is equal to the order of the group. That is we're looking for elements whose multiplicative order is n. And that's exactly what the primitive roots are. No smaller power does it. All right. And then related to this idea. Oh, and so let me actually say that with the zn here, so for every n possible, you get a cyclic group of order n inside of c star. And these, of course, are gonna be distinct. If you choose different n's, you're gonna get different cyclic groups. And so this gives actually an infinite family of cyclic subgroups inside of c star. Now, what we're gonna do next is basically put all of these things together in a manner of speaking. We're gonna take the set s1 to be the set of all complex numbers whose modulus is equal to one. And I claim that this set s1 is a subgroup of c star. And in fact, this is a subgroup that contains every, it contains all of these cyclic subgroups of roots of unity. Zn is contained in each and every one of these things. Now, I wanna mention, before we talk about the proof here, let's talk about the notation. Why is it s1 right here? This is actually borrowing from topology because this is what's referred to as the circle group. The reason why I call it s1 is that s1 is the one sphere. A circle is just a one dimensional sphere where two sphere, the way we usually think of a sphere as a two sphere and then you have higher spheres and higher dimensions as well. So this object right here, the set s1, if we think of the complex plane, as we think of the geometry of this thing, then the circle group, those complex numbers whose modulus is one, we're looking for complex numbers which are exactly one unit away from the origin in the complex plane. Well, this will actually create for us the unit circle, s1 right here, this is gonna be the unit circle in the complex plane. And so that's why we call this the circle group. And so let's prove that the circle group is in fact a group. It will be an infinite group and I should mention that this is actually an uncountable group inside of C star, but we're gonna prove that it's a group. So take two elements of the circle group, say mu and nu. Well, because these belong to the circle group, their modulus is one. So consider the modulus of their product. Because the modulus has this multiplicative property, the norm of mu nu is equal to the norm of mu times the norm of nu, which is one times one, which is one. So this shows that the circle group is closed under multiplication. The multiplicative identity in the complex numbers is the number one, its modulus is equal to one. So one is inside of the circle group. So the last thing to do is to show that it is closed under inverses. And so if we take the multiplicative inverse of mu, so it's reciprocal mu inverse, their products can equal one because that's what just happens when you multiply things together. And so as you investigate the modulus of mu inverse, this is the same thing as mu inverse times the modulus of mu because the modulus of mu is one. But the multiplicative principle says you can put these things together, their product is one and the modulus of one is one. And so this tells you that the modulus of mu inverse is itself one. In general, if you have a multiplicative norm, what you can see here is if you have some number raised to the negative one, then its norm will be one over the norm of A. That happens in general. So the reciprocal of one of course is one. Put that back on the screen. Oops. And then so that shows that it's a group. So S one is a subgroup of C star. I should also mention that it is an abelian subgroup but that's because C star is an abelian group itself. And so what about the primitive roots of unity? Well, let's take zeta to be e to the two pi i over n. So this is the principle nth root of unity. It's primitive, it's the first one if you think about it. Because one should mention that when it comes to the roots of unity, you have like, let's say we think of like fifth roots of unity, for example. So you're gonna have one that's always there. But then the next one, if you rotate this thing counterclockwise, the next one on the unit circle, that's gonna be our zeta, okay it's called like zeta five right here. Then the next one is gonna be zeta five squared. Then the next one is gonna be zeta five cubed. Then the next one is gonna be zeta five to the four. Then the next one's gonna be one again because one is the same thing as zeta to the fifth. And so one thing I should mention here is when it comes to these primitive roots of unity, if we orient them in the complex plane, they actually form a regular in-gone. So if you take the fifth roots of unity, this will form a regular pentagon, which my picture's not drawn perfectly scale. You'll excuse me for that though. And this in-gone will have the real axis as a line of symmetry. One will always be on there. If you reflect across the real axis, you actually get the complex conjugate. So the conjugate of a root of unity is always, it's root of unity, it's also a root of unity as well. Just kind of visualizing these things a little bit. But in particular, the in-throats of unity are gonna be complex numbers whose modulus is equal to one. And as such, every root of unity will be belonging to S1. So the entire subgroup Zn will belong to S1 as well. And so I just wanna kinda introduce you to these important groups inside of C star. There's a circle group, but also we can visualize primitive, well not just primitive, but complex roots of unity can be formed to make cyclic groups. And I wanna throw this out here because this is the first taste of a branch of group theory that's called representation theory. A representation theory, as the name kinda suggests, is about representing abstract groups using concrete representations. And so we can take the idea of a cyclic group which we've learned about recently, but we can actually represent the group not as like these abstract notions floating in space, but we can represent them as specific sets of complex numbers, or in more generality, you represent groups as groups of matrices or groups of permutations or things like that. And so this is the first taste of a branch of abstract algebra called representation theory, which is a very, very important part of algebra. And the topic we might go into a little bit in the future, but what this lecture tells us is that every cyclic group can be represented as a group of complex roots of unity.