 Welcome back to our lecture series Math 42-20, Abstract Algebra 1 for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Missildine. We're going to continue in lecture 27 to talk about isomorphisms like some of the topics found from section 9.1 of Tom Judson's Abstract Algebra Textbook. Our focus for lecture 27 is going to be on Cayley's theorem. We'll actually prove Cayley's theorem in the second video of this lecture. In this first video, I want to make mention that when we define isomorphisms last time, it should go without saying that isomorphisms form an equivalence relation on the class of groups. It suffices when we consider groups, we really only have to consider groups up to isomorphism. If we want to classify all the groups in the world, we just have to classify them up to isomorphism, because different representations of the same group should really be considered the same thing. There are some types of groups where it's easy to classify them. The family of cyclic groups is very easy to classify, and it turns out that cyclic groups will be unique up to their order. This is a statement I made in the previous lecture, but we'll actually want to prove this right now. What we're going to prove here is what we'll call theorem 9.1.7. Two cyclic groups are considered isomorphic if and only if they have the same order. In particular, the order of a cyclic group determines the group up to isomorphism. Now, one direction is very, very clear. We proved previously that if two groups are isomorphic, then they have the same order. So if we assume the two groups are isomorphic, that means we'll have the same order. So that direction is pretty straightforward from what we've already seen previously. So in order to prove the converse, so we want to prove now the other direction, suppose we have two cyclic groups with the same order, we want to prove that these cyclic groups are actually isomorphic. Now, since isomorphism is an equivalence relationship, it actually is transitive, right? So it suffices that if a cyclic group has order N, that it's isomorphic to ZN. Because if we can prove, let's say we have two cyclic groups, both H and G, whose orders are N, well, if we can prove that G is isomorphic to ZN, and we can prove that H is isomorphic to ZN, then by symmetry, this second statement would imply that ZN is isomorphic to H, and then by transitivity, we get that G is isomorphic to H. So I guess we need transitivity and symmetry here. But because we have an equivalence relationship, it suffices to prove that a cyclic group of order N is congruent to our isomorphic to ZN. So let's take a cyclic group. So, okay, it's cyclic G, it has a generator, which we call little G, and let's suppose that the order of G is N, and for the current situation, let's assume this is a finite cyclic group. We'll deal with the infinite cyclic group in just a moment. Well, in this situation, we're gonna define a map phi, where phi is gonna be a map from, and I'm gonna write this to the side, so it's a little bit easier to see. So we take the cyclic group as the domain, we take our group in question as the co-domain, and we're gonna map phi of K to G to the K, like so. And that's gonna be our map. Now, we have to be a little bit concerned because as the cyclic group ZN here is defined via congruence classes, this and this function I defined using an integer as opposed to a congruence class, I have to make sure it's well-defined. So suppose we have two integers L and K, which are congruent mod N. Well, if the two integers are congruent mod N, that means there's some integer A such that K is equal to L plus A to the N. So we can actually make an equality of integers there. And so let's look what this function does here. So we have phi of K, well, by the formula above, phi of K will map to G to the K. But by this identity right here, K equals L plus AN, we make that substitution G to the L plus AN. Then by usual exponent rules, which hold in any group, we're gonna get G to the L times G to the N to the A. Now, since the order of the group is N, every element raised to the Nth power is one, and particularly the generator also is an element of order N. So G to the N is equal to the identity of the group, which if you raise that to the eighth power, that's still the identity. And so we see that G to the K is actually equal to G to the L, which is what phi of L will do. This proves that phi here is a well-defined function. So we define the function based upon representatives of congruence classes. We have to make sure this is well-defined. We double-check that. That two different integers that are congruent to each other mod N will give you the same image. So this map does not depend on the integer K that we choose here, up to congruence. So now, let's imagine we take two integers, K and L in ZN. And this time I don't suppose initially that the two are congruent to each other. So I'm dropping the assumption we had before. In a computer program, we'd say we're overloading these numbers here. So we have K and L, but we are gonna assume this time that phi of K is equal to phi of L, right? We wanna argue that actually, that these things are congruent to each other. Because we're trying to show that the function's one-to-one right now. After all, we have to form an isomorphism. So if phi of K equals phi of L, that means G to the K is equal to G to the L. Which if you then take that equation, G to the K is equal to G to the L, I'm gonna times on the right by G to the negative L. G to the negative L. And that gives us that G to the K minus L is equal to one. Well, then by Lagrange's theorem, that since K minus L, that power of G gives you the identity, this tells us that N must divide K minus L. Only multiples of N will raise an element to the identity. Where N was the order of G. Because after all, capital G was a cyclic group generated by little G in the order of N. That's the order of the elements as well. Okay, so N divides K minus L. So that tells us that K and L are congruent to each other, mod N. And so you'll see this a lot. That when you define a map, when you define a map using equivalence relations, then you have to prove it's well-defined. But if you wanna show the map as one-to-one, oftentimes the two arguments are kind of duels of each other. Notice what we did here. We first assumed we had two different representatives of the same congruence class. And then we showed that the images were the same. That gave us well-defined. On the other hand, we now showed that we assumed, I should say, that the images were the same. We then proved that that actually means that we represent the same congruence class. So up to congruence, this is the same element. This then proves that we're one-to-one. So there's sort of like a dual argument that goes on there. You see that a lot. That being well-defined is sort of like the dual argument to being one-to-one. Well, to show that it's a bijection, so we have a one-to-one function now. Well-defined meaning it's a function one-to-one we just proved. Let's show it's surjective. This is actually a very easy claim, right? Because elements of g will look like some power of little g. So if we take an arbitrary element, this will look like g to the k. Well, phi of k will map to g to the k, so surjectivity is easy, right? And so now we have a bijective function, it's a bijection. To show that it's isomorphism, we have to also show that it satisfies the homomorphic property. We have to show that phi of k plus l is equal to phi of k times phi of l. Because the cyclic group generated by g we're assuming is a multiplicative group there. All right, so how does that work? Well, phi of k plus l, addition is the operation of zn here. Phi of k plus l means g to the k plus l, but by exponent properties, this becomes g to the k times g to the l, which gives you phi of k times phi of l. And so the homomorphic property is immediate, right? It's almost natural in a manner of speaking. And this proves that phi is an isomorphism, therefore, z sub n is isomorphic to g, where g was a cyclic group of order n. Well, okay, what if we had an infinite group, right? What if the order of g was infinite? Well, the only part of the above proof that needs to be adapted really is the injectivity argument. Because notice here that in terms of being well-defined, we really didn't need to use, we didn't need to use, well, honestly, if you're infinite, you don't need to make a well-defined argument whatsoever because in that situation, you're comparing to the infinite cyclic group z. There's no congruence classes and therefore this map has no worries about being well-defined, great. The other thing about being surjective, right? If you're cyclic, right? If g's cyclic here, then everything still looks like g to the k and so surjectivity will be the same. So it's the injective argument that potentially could change. But when you look at that situation, if you take these elements k and l, such that phi of k equals phi of l, will you still get k g to the k equals g to the l? Or in other words, you're gonna get that g to the k minus l is equal to one. But now since we're the infinite cyclic group, what that means is there actually is no finite power, there's no positive power of g that'll give you the identity. G to the zero is the only thing that can happen. And therefore you would have to conclude instead that k minus l is equal to zero or k equals l. So in that case, again, there's no congruence going in play there. We actually get that authentically k equals l. And so that would show that for the infinite cyclic group, you are gonna get isomorphism there as well. Now I should mention that when it comes to the infinite cyclic group, there's only one infinite cyclic group that the cardinality has to be countable. There's no such thing as an uncountable infinite cyclic group. Because if you look at the elements of the cyclic group, you notice here that we can put them in bijective correspondence with the set of integers, which is a countable set. So there's only one infinite cyclic group. It's the integers up to isomorphism. And then every other, every finite cyclic group is isomorphic to Zn. Again, up to isomorphism there.