 In this video, I wanna talk about some homomorphic images of cyclic groups. And it turns out it's very easy to describe when you have a function, I have a homomorphism between cyclic groups and when you do you not, right? So imagine you have an integer N for which it has a divisor D that you see right here. So with a divisor D in hand, we can define a homomorphism from ZD to ZN. So ZD is the cyclic group of order D, ZN is the cyclic group of order N. And we'll think of this as additive notation. So we're working mod D and mod N respectively. So since D is a divisor of N, we're gonna define a function from ZD to ZN by the rule that the integer M will map to the integer MN divided by D. Well, since D of course divides N, that implies that MN divided by D is of course, well, I should say that N over D is an integer, right? So therefore M times N divided by D would be an integer right here. So that makes sense. Now we have to of course be concerned, is this map well-defined? Because after all ZD here is the set of representatives of congruence classes. So if we choose a different representative from the same congruence class, do we get the same element? That's always has to be of concern when you define a function based upon representatives of some equivalence relation. So if we were to consider that, take phi of M plus KD, right? Well, what does phi do? It just times the integer you started with by N divided by D. For instance, this is just multiplication of integers you can distribute here. You're gonna get MN divided by D and then you're gonna get KD times ND. So the D that actually cancel and you're gonna get KN, which means that MN divided by D plus KN is congruent to MN divided by D mod N, right? Which is where the co-domain lives, right? And so therefore these two elements are congruence to each other. And so this would just equal phi of M. Whoops, that didn't look right. Phi of M. So if two numbers are congruent mod D, then their images will be congruent mod N. So this is in fact a well-defined map. Is it homomorphic? Well, let's take the sum and send it through the function and see what happens. So if we take two integers mod D, we add them together, A plus B. We take phi of A plus B. You're gonna get A plus B times ND, which again, this is an integer. We distribute it, because this is multiplication of integers right now. You're gonna get A N divided by D plus B times N divided by D, which this first number right here is just phi of A. And then the second number of course will just be phi of B. And so we do in fact see that this is homomorphic. So we do get the homomorphic property. In fact, this is an example of a monomorphism. That is it's a one-to-one homomorphism. How we know that this thing is one-to-one, right? So if you take phi of A and you set it equal to phi of B, then this would tell us that A times N divided by D is equal to B times N divided by D, right? For which, which admittedly this is up to congruence, right? So you're gonna have some type of KN going on over here. That's what you would see there. But another word, but that's not such a big deal really because when you have this statement right here, we can cancel, we can cancel the N plus D going on right here. But let's try to do it a little bit more properly right here. So you have this. And so the reason this is of advantage to us is we just factor this, right? Just times it by D over D. And so it's in the right-hand side, you can factor out the ND. So we're gonna get KDN times N over D. And as this is just multiplication of integers, we can cancel that. And so we're gonna get that A is equal to B plus KD times N, which tells us that A and B are congruent to each other, mod D, like we were looking for. So this is in fact a one-to-one function. And what we're gonna see, again, this is something we'll explore much more in the future, but note that the image of fee is actually equal to N divided by D, right? That's what's going on here. And that's isomorphic to ZD, which was the domain, right? Now, I'm not saying these sets are equal to each other, but they're isomorphics. It's the same group up to isomorphism. We can actually generalize the previous example. Take any group you want, any, any, any, any, any group, and then pick your favorite element inside of the group, G, what we'll call a little G. So then I can define a homomorphism from the infinite cyclic group, right? So this is the integer with respect to addition. We're gonna define a map where fee of N maps to G to the N. Where again, little G was our favorite element we picked here. Now the usual exponent rules are gonna justify that this is a homomorphism. I'll leave it up to you to verify this. Some other things I can mention here is that this map will be a monomorphism if and only if G has infinite order. Because if G has a finite order, let's say like the G has order three like that, then you're gonna get the following, like zero will map to the identity of one. One will map to G, two will map to G squared, three would map to G cubed, which of course is equal to the identity. So this maps with here right here. So it wouldn't be one to one in that case. And that's basically the argument. But we can also do the following, right? Similarly, we can define ZD to G by the rule, right? Phi of N equals GN. And this is gonna be well-defined if and only if D is a divisor of the order of G. And so again, I'm gonna leave it up to the viewer here to prove this. But if D divides the order of G, this will be a well-defined map and it's well-defined if and only if, right? Likewise, this will be injective only if D is equal to the order of G. And so I'll let you try to convince yourself so those things. But essentially, this then determines the only types of homomorphisms between groups, right? When it comes to a homomorphism from a, when it comes, well, actually we'll just say it right here. When it comes to a homomorphism from a cyclic group into a group, this is where the cyclic group is in the domain. These are the only options, right? You have to pick your favorite element in the group, but its order has to be divisible by the element, the order of the original cyclic group, whether the infant cyclic group or the finite cyclic group right here. Let me give you one more map in this video here, kind of related. If we look at the real numbers versus the circle group, right? I say this is kind of related because the circle group actually, this would be, this right here would be the group of all complex numbers of modules one, right? For which the circle group contains all of the roots, the complex roots of unity. So actually the circle group contains every finite cyclic group, up to isomorphism, it's kind of cool little thing there. And so we're gonna define a map from the real numbers to the circle group where phi, excuse me, theta maps to e to the i theta, which every complex number of modules one could be written of this form, e to the i theta, somewhat of a consequence of, you know, the exponentials of imaginary numbers and things like that. And so this map right here is, again, gonna be a homomorphism by the usual exponent rules. I'll let you convince yourself of that, but it has to do with addition of exponents and things like that. This is also surjective because like I said, every complex number of modules one has this form e to the i theta. So this is an example of a surjective homomorphism, a so-called epimorphism, right? Now this one also has sort of a fun little geometric or topological feel to it. Because after all, why do we call it the circle group? Because in the complex plane, the circle group is the real McCoy unit circle, right? It's a circle if we look at it from above. And now the real numbers, we can think of it geometrically as a line. Now, if we add a little bit of a coil to our line, what we can kind of think is the following idea, right? That if we take the real numbers and we kind of spin it around like it's this infinite long spring, and as it lives above the circle group, right? This is our real numbers right here. Oh boy, let me try that one more time. Here's our real numbers right here. And so there's a nice little like topological identification here that the real numbers essentially cover the circle group here for which this epimorphism preserves the multiplication, but also it's gonna be continuous, right? It's not a notion we really worry about in abstract algebra, but as the real numbers in the circle are topological sets, we could ask, is this map between them, does it preserve topology whatsoever? So we actually have a continuous epimorphism. And that's because both of these things are examples of so-called Lee groups, which are groups which also have some type of geometry attached to them. In particular, these are groups for which the multiplication inversion maps are both continuous and differentiable. Not something we're gonna worry so much about in this lecture, so you just wanna kind of throw out those buzzwords for us here. But this homomorphism here, it's not saying that the real numbers in the circle are the same, but basically it's saying that we can cover, because it's an epimorphism, we can cover the circle with just a line. It's kind of, it'll just wrap around, right? So it just wraps around and around and around and around and around and around, right? You're basically working mod two pi in this situation here. And so this algebra os is actually compatible with the geometry. This is a continuous epimorphism.