 Welcome back to our lecture series Math 42-20, Abstract Algebra 1 for students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Missildine. Lecture 26 in our series starts chapter 9 of Tom Judson's Abstract Algebra textbook, which seemingly seems like a short chapter with only two sections. We'll actually spend four lectures talking about the topic of isomorphisms. It's a pretty big deal right here, isomorphism. Well, when you look at the word in Greek itself, iso means same, morph means shape, and so an isomorphism is going to be a way of saying when two groups have the same shape. And there are, of course, analogs to the notion of isomorphism for other algebraic structures like rings, fields, modules, vector spaces, etc. So what exactly do we mean by an isomorphism? What does it specifically mean? So imagine we have two groups in play right here. So we have a group G and we have a group H. Now our usual notion here of a group, the usual way we express a group is just describe it as a set. So we have a set G and a set H, but that can be a little bit misleading because a group is a set equipped with a binary operation and that binary operation needs to satisfy conditions of associativity, identities, and inverses. But as the notation can cumbersome, sometimes we often omit the operation in play here. Despite that convention with this definition, I want to emphasize not just the set but the operation in play as well because the two operations associated to the groups G and H do not have to necessarily have anything in common like these don't have to both be multiplication, they don't have to both be addition. They're just two operations which are whatevs, right? They're just group operations. We have these two groups. So we say that the groups G and H are isomorphic. If there exists a bijection wrapped around the line there, I'm going to write it right here. So we have a bijection fee which goes from G to H. So what does it mean to be a bijection again? So this map needs to be both injective and surjective, one-to-one and onto. It has a function inverse. But in addition to be bijective, this function needs to preserve the operations. What does it mean to preserve? All right. So preserve the group operation here means the following very important identity. This is often called the homomorphic identity. Later on in this series when we broaden our functions in play here from isomorphisms to homomorphisms, a homomorphism will essentially be functions between groups that have this homomorphic property and the bijective part will be dropped. So the homomorphic property says that if we operate on the group G, so group G has the star operation, we operate on the group G and then we hit the with the function. This is equal to the images operated here. So that is you have the operation inside the function. This acts the same way as the operation outside the function. So if you take a product of two elements in G and you map it over, this is going to map to phi of A operate phi of B. So basically a product will map to a product in the two respective groups. And again, the operation could be very different. The star operation here, the circle operation here, that might not seem like have anything to do with each other, but isomorphism will preserve the operation, this homomorphic property by sending the product to a product where product might change based upon the two groups we're in right here. Now if G is isomorphic to H, then we'll denote this as G isomorphic to H, this little symbol here, a little bit of a textbook moment here. If you want to write the isomorphic symbol, this is in geometry, this is the congruent symbol. And so in latex, this is backslash c-o-n-g. I also want to mention that if you want to write the function phi, the usual command for latex, if you do backslash phi, this actually gives you the symbol that looks like this, which is perfectly fine. But this symbol right here, which is more common in group theory, this is actually a variant of the traditional phi symbol. So this is actually called var phi. And this will produce this more handwritten phi symbol in case you care to distinguish between the two. So if G is isomorphic to H, then we say that the function phi is an isomorphism. I want to show you some examples of what this means in just a second. But again, I want you to drill into your minds right here the following. When two groups are isomorphic, they are essentially the same group, only their outward appearance is what differs. That we're talking about labeling, we're talking about cosmetic things, things that really do not matter in terms of the algebra. Now, of course, the symbol for isomorphism here is an equivalence relation, which I'll leave it up to an exercise to the viewer to prove here. So let's look at some examples. So the first example is one I really like a lot because this is not something you would expect to be isomorphic. I want you to consider the two groups associated to real numbers. We're going to take the real numbers under addition. So this will include all real numbers, positive, negative and zero. And we want to compare this to the real numbers with respect to multiplication, for which we're only going to consider the positive real numbers. So the real numbers with respect to addition is a group and the positive real numbers with respect to multiplication is likewise a group. We are going to define an isomorphism between the two, we'll call it phi for the moment. And this is really just going to be exponentiation. That is, we're going to use the natural exponential. We're going to send the real number x to the exponential e to the x. And I do want you to commit yourself that this is going to be a well-defined map. Notice that for every x, there's only one e to the x we can associate to it. And really the part we have to focus on here is that for any real number x, e to the x is actually a positive number. And so in particular, e to the x will land inside of h plus. So this gives us a well-defined function between the real numbers and the positive real numbers. Because again, it doesn't matter what real number you shove in there, a positive, zero or negative e to the x is defined for any real number. So we have a function defined between the two groups, the reals and real plus. Is it a bijection? Well, it turns out the natural exponential map is a bijection because it has an inverse. To show that something is bijective, you could show it's one to one and onto or you could just produce a function inverse. And so the inverse here, which will reverse the direction, it'll send a positive real number back to a real number. This is actually done by the natural logarithm. So take psi of x to be the natural log of x. We can see that when we compose these two functions together, if you take phi of psi of x, this would look like e to the natural log of x, which is equal to x. So remember, what is the natural log of x doing? I mean, you can just take it as the inverse of e to the x here. But the natural log of x is supposed to be the power, it's the number which if e is raised to that power produces the given number. So the natural log of x is the power of e that produces x. Well, if I take e to the power of the power of e that produces x, then of course you're going to produce x. They cancel out. On the other hand, of course, if you take psi of phi of x, you get the natural log of e to the x. Well, the natural log, like I said, it's asking what power of e will give you this number? Well, the x power will produce the e to the x right here. So these things are inverse operations, logs and exponentials. That's something we probably don't need to go into much more here. What really is important to mention here is the homomorphic property, which you've seen before in previous algebraic settings, but maybe didn't realize actually that when you were seeing was a homomorphism. So take, for example, phi of x plus y. So the operation of the domain is addition. So I'm going to take two arbitrary elements and add them together and put that inside the function. So we have phi of x plus y. Well, what does the function phi do? It's the natural exponential. So this then becomes e to the x plus y. Now here's the kicker, right? If you get e to the x plus y, this is an exponential function. When you take a sum of exponents, you can actually factor your exponential as e to the x times e to the y. Notice how addition turned into multiplication by this exponent law. We get e to the x times e to the y. Well, e to the x is just phi of x and e to the y is just phi of y. This right here is the homomorphic property. This says that the natural exponential turned addition into multiplication. Now, I'll also leave it as an exercise for the viewer here to prove that the inverse of an isomorphism itself is an isomorphism. I'll demonstrate this just for this example right here. Clearly, if a function is bijective, its inverse will also be bijective. But the natural log also has this homomorphic property. Notice if we take the natural log of x times y, you also know from calculus, college, algebra type settings that if you have a product inside of the natural log, this turns into the natural log of x plus the natural log of y. So notice what happened here. A product inside of a natural log turns into a sum outside of the natural log. This is again an example of the homomorphic property. Now I don't have to prove the homomorphic property for the natural log because honestly if it works for, if its inverse has the homomorphic property, it will have the homomorphic property as well. That's what I mean by proving that the inverse of an isomorphism is an isomorphism. And so this then establishes that the two groups, the real numbers under addition and the positive real numbers under multiplication are isomorphic. So essentially from an algebraic point of view, there is no difference between these two groups. Everything true about the first group is true also about the second group, at least with respect of algebra. Now these sets, the real numbers, the positive real numbers, they have more structure to them than just algebra. It's like there's a linear order attached to the real numbers. That leads to some geometric and topological consequences as well as well, right. And so I'm not saying that the geometry of the real numbers is the same as the geometry of the positive real numbers. I'm not not saying it either. I'm actually no statement about the geometry. What I'm saying here is that with respect to the algebraic structure as groups. So the group of addition with real numbers is the same as the multiplication of positive real numbers. Now I don't want you to take this too far. I am not saying that this is the same thing as the real numbers with respect to addition with respect to multiplication because there's course of issues there. First of all, this is not a group. So a group cannot be isomorphic to a not group. The problem is that the zero element doesn't have an inverse. But even if you restrict to non-zero real numbers, I'm also not saying that these two groups are isomorphic. There actually is a difference between these groups that any homomorphism would have to detect. Again, topic for a later conversation here. But we are saying that in terms of addition, the real numbers are the same in terms of multiplication if you restrict to the positive. It's a fascinating concept that the addition and multiplication of real numbers is essentially the same operation. Let me give you an example that might be easier to digest here. Let's consider the two groups which we're going to take the cyclic group of order four. And by this, of course, I mean by Z four, we mean the group where you have congruence classes mod four. So this is a group of order four. And so by things like the congruence class of order four, we mean all multiples of four, where K is any integer. What we mean by the congruence class containing one, we're looking for integers of the form 4K plus one, where K is an arbitrary integer, et cetera. That's what the other ones mean as well. So this is how we define the cyclic group of order four. Now we often abbreviated this as zero, one, two, and three. And this is really just an abusive notation here, right? And we just use zero to represent the congruence class that contains zero. And we would use one to represent the congruence class that contains one. And so that way we can get away with saying things like one is congruent to five, mod four, is because both one and five, both one and five belong to the congruence class one, right? And so we might sometimes drop the brackets and say one just to represent this congruence class, right? That's an important thing to mention here. So we can take the, we take z four, take the multiplicative group generated by the complex unit i. So we're clear here. We're talking about the group one, i, negative one, and i cubed, which of course is just negative i. And this is in fact a cyclic group as well, right? Because you get one, you get i, you get i squared, which is equal to negative one. You get i cubed, which is equal to negative i. And then the next element, i to the fourth is one again, right? Which you don't need to mention it twice there. So we have these two cyclic groups of order four. I claim that they're isomorphic to each other. And so how is this accomplished? So we're going to define the function phi from z four to i, the cyclic group generated by i. As phi of n is equal to i to the n. Now you'll notice here that I'm defining a function on the representatives of z four, not the classes. I didn't define, I didn't say something like, oh, phi of the phi of n bracket here is equal to whatever, right? I didn't say that at all. Instead, I said phi of n is equal to i to the n, right? That's a critical difference here. I'm not defining the function on the congruence classes, which is what z four is, it's a set of congruence classes. I'm instead defining it by a representative. And so it draws into question, is this map even well-defined? What if I pick two different representatives of the same function, right? What happens in that situation? I mean, we have to be, this is a real thing to be cautious about because notice if I were to do something slightly different. Let's take like for example, z five, and we define and we send this to the cyclic group generated by i. And if you do the exact same map, phi of n is equal to i to the n. You're going to see some problems here because for example, if you take phi of one, that's equal to i. On the other hand, if I take phi of say six, which is congruence to one mod five, this would be e to the six, which reduces just to be negative one, right? That's not the same number, right? And so this would this would be the example of something that's not well-defined. It's not actually a function because the definition depends on the representative, not on the congruence class. So how do we know the same problem isn't occurring here as we take z four to the cyclic subgroup generated by i? Well, so what we have to do is we have to take two representatives of the same congruence class. So let's take m and n, which are congruent mod four. Well, what that means is if m and n are congruent, that means there's some integer k such that m is equal to n plus 4k as integers. Now let's look at the images of these two, these two elements individually. So if we look at phi of m, phi of m by definition should be i to the m, but by this identity right here, m is equal to n plus 4k, which by usual exponent rules, because this is a complex number, you get i to the n times i to the four to the k power. Now notice i to the fourth is equal to one, the order, the multiplicative order of i is equal to one. And if you take one to the k power, you're just going to get a one. So this will simplify just to be i to the n, which is equal to phi of n. So as opposed to our map Z5, which maps to this one right here, which was not well-defined, shame on you, this one is well-defined. And there of course, there is a significance of this. Notice that the order of i was equal to four, right? And we're doing the cyclic group of order four as opposed to the cyclic group of order five. That's really the kicker here in this argument, but this is a well-defined map. So it's well-defined. Is it bijective? Well, we're going to show this one is bijective by showing it's one to one and onto, right? So let's suppose we have two integers here, Z4 and Z4, so phi of m and phi of n. I'm no longer assuming that they're congruent mod four, right? We're just saying their images are the same. That would tell us that i to the m is equal to i to the n, right? And then if you divide both sides by i to the n, you get i to the m minus n is equal to one. Well, if a power of i is equal to one by Lagrange's theorem, that means the order of i, which is four, must divide m minus n, because again, the order of i is four right here. That's the consequence of Lagrange's theorem. And if four divides m minus n, then actually suggests that m and n are congruent mod four. So the only way the two images are the same is if m was congruent to n, which in Z4 means they are the same number, so to speak. And so this, it's kind of similar to how we showed it was well defined already. This would show that the function is one to one. How about onto? Well, let's pick our favorite element of here. I should say let's just pick an arbitrary element. So every element of the subgroup generated by i here will look like i to the n. Well, this is pretty easy. Phi of n will map to i to the n. So surjectivity, boom, drop the mic and walk away. This shows us that we have a bijection. That's pretty nice. What about the homomorphic property? Does it preserve the operation, which again, the operation of Z4 is addition mod four. And the operation of this cyclic group, this is what we called earlier, the cyclic group Z4. And we saw this as a subgroup of Z star, the non-zero complex numbers with respect to addition. Multiplication, excuse me. In fact, we can see it as a subgroup of S1, the circle group. So the operation of Z4, this is Z4, not the blackboard Z4, that's multiplication. On the other hand, the other Z4, which we see right here, its operation is addition mod four. Are these the same operation? Does it have the homomorphic property? Well, if we take phi of a sum, implicit. Again, we have no assumptions about m and n, except they belong to Z4. If we take that, well, this will map it to i to the n plus n, which again, by usual exponent rules of complex numbers, you get i to the m times i to the n, which looks like phi of m times phi of n. And so again, using exponent rules, we turn addition into multiplication and we see that these two groups are in fact isomorphic to each other. And this principle works actually in more generality. We could do this for arbitrary orders. The group Zn, which is addition mod n, will be isomorphic to the cyclic group generated multiplicatively by a primitive and through immunity. So you could take e to the 2 pi i over n, that's the group we called Zn. And you'll notice that although it's a different font, why did we call both of these groups Zn? And this is because this was a foreshadowing to the observation that these two groups are in fact isomorphic. Integers under addition mod n are isomorphic as a group to the nth roots of unity under multiplication. That although the labels are different, the operations are different, they behave in the same way. And so with for all intents and purposes, we should treat these two groups as if they are the same group.