 In this video, I wanna talk about the automorphism groups of abelian groups. So we define the automorphism group in the previous videos. So remember, odd of G, this is the collection of all automorphisms of G. These are gonna be isomorphisms, phi from G back into itself. So they're bijective, they're permutations. This is what we have as an automorphism. Well, can we describe automorphisms for abelian groups? The simplest example would be to consider a cyclic group. So just for a concrete example, let's consider for a moment, the cyclic group of order 12. Now, as we learned about in math 4,2,20, abstract algebra one, we've learned that every homomorphism, so I'm not even considering automorphisms necessarily at the moment, but if you take any homomorphism from a cyclic group, this is actually uniquely determined by what the generator is doing. So if we have any math, sigma from Z12 to any group whatsoever, if you know where one goes, then you actually know what everything does, right? So let's say, for example, this group G is actually some other cyclic group, because again, our ultimate goal is gonna be Z12, right? But if we know what the identity does, excuse me, not the identity, if we know what the generator does, then we know what everything does. So let's say that sigma of one is equal to K for some K inside of Z12. Because again, we're trying to consider automorphisms, and this is a property that's true for any homomorphism of a cyclic group. We know what the generator does. Well, if you take any other element in Z12, so say N for example, what's sigma of N? Well, N is gonna be one added together in times, okay? That's exactly what we mean here. So we take like this N dot one here. We're not necessarily talking about the ring multiplication that Z12 has. We're saying N times one is just one added together in times. Now, of course, because of the distributive property in Z12 viewed as a ring, these two notions are equivalent to each other. So it doesn't really matter which one you fixate, but we can describe it from a purely group theoretic point of view. N one just means you add together one N times. And so as a homomorphism, sigma here, if you have a sum of ones, this will then become a sum of sigma ones for which each of those sigma ones is a K. And so you've added together K in times. So you get back in times K. And so this is what we were saying earlier that if I know what the generator of a cyclic group is under a homomorphism, I actually know what the homomorphic image of every element is. We can use the homomorphic property to do this. And this will in fact be a homomorphism, right? Sigma of N plus M here. Well, because of the previous observation, sigma is just multiplication by K. It doesn't matter what the input is. You're just gonna get K times the input there. And so sigma of N plus M is gonna be K times N plus M there for which this is multiplication, the usual sense it distributes. So we get K in and KM, which becomes sigma of N and sigma of M. So in fact, if you choose any K in Z12, sigma will be a homomorphism from Z12 back to Z12. So that's what we're considering right here. Sigma is this map from Z12 into Z12. And we define it to be multiplication by K where K is some element in Z12. And so this actually gives you all of the so-called endomorphisms. Remember an endomorphism is a homomorphism from the group back into itself. But is it an automorphism? Well, an automorphism is gonna be an endomorphism which is bijective. That is an automorphism is an endomorphism, which is also an isomorphism. So that's what we have to investigate right here. Under what conditions will sigma be bijective? Well, it probably has to do with this element K because K uniquely determines this map sigma right here. So if sigma belongs to the automorphism group there, another way of thinking about is that the generator one has to map to another generator because in particularly the order of the element one has to be preserved by an automorphism. Isomorphism is preserved order. The element one has order 12. So the only chance would be to send one to some other element of order 12. For which in Z12, we have four elements of order 12. You got the generator one, but you have the other generators five, seven and 11 as well. So in particular, we wanna choose our element K so it belongs to Z12 star, all right? If K belongs to Z12 star, so this is the group of elements which have multiplicative inverses of Z12. If you pick an element there, then K will have an inverse element, call it D. And then the composition of those two would then give you the identity map. So if we take sigma inverse composed with sigma of N, well, sigma of N is going to give you KN, but sigma inverse is associated to the inverse of K there. You're gonna end up with this element D, DKN for which this becomes DKN for which this would then reduce to be one times N because after all, we're working mod 12 here and which K since you just give back an N, this is the identity map. So if we choose K so that has a multiplicative inverse because after all, sigma is just multiplication by K, then it'll have an inverse, it'll be a bijection. And in fact, every automorphism must have this form, must be invertible. So the automorphisms of Z12 coincide to multiplication by these multiplicatively invertible elements, 1, 5, 7, 11. And so if we look at them in collection, the automorphisms of Z12 coincide exactly with the group Z12 star. And this actually establishes the isomorphism, the automorphism group of Z12 star is, it's isomorphic to, excuse me, the automorphism group of Z12 is isomorphic to Z12 star. And nothing, as nothing in this example here really requires mod 12. Sure, specific details about who is invertible, mind you, but this argument didn't depend on 12 whatsoever. We can actually lift this argument and say something about all cyclic groups. In fact, each homomorphism from ZN to ZN, which of course that means these are gonna be endomorphisms, each endomorphism from ZN to ZN is determined by multiplication by some integer K mod N. We've mentioned that. And therefore, if we restrict our attention to the automorphisms on ZN, these are gonna be exactly those elements of ZN which are multiplicatively invertible. So these are gonna be elements which are co-prime, relatively prime to N. And so this establishes a very important property here that when it comes to cyclic groups, the automorphism group of a cyclic group is just those elements co-prime to it. The automorphism group of ZN is isomorphic to ZN star. And this is true for any N whatsoever. This is actually even true for the infinite cyclic group. If you take the automorphism group of Z here, the infinite cyclic group, this is gonna be isomorphic to Z star, which we typically define this to be plus or minus one. That is those integers which have multiplicative inverses, only one and negative one have reciprocals which are also integers themselves. So this theorem, 11-4-4 on the screen completely characterizes the automorphism group of a cyclic group. In general, automorphism groups can be a little bit more complicated to describe, but there's one other Abelian group which has an easy to describe automorphism group that's worth mentioning. And this is the idea of what we call an elementary Abelian group. So an elementary Abelian group, often denoted as E sub Q, where Q is itself often abbreviation, Q is gonna be a power of a prime. So P is some prime number, it could be two, it could be an odd prime, doesn't matter. They often abbreviate as Q, like I said. So you have E to the QN here. And so just for the details here, what is E sub P to the N? This is gonna be the direct product of ZP N times. So you take the exact same cyclic group ZP, and this is necessarily a cyclic group of prime order, and you take N copies, N direct copies of it with respect to the direct product here. So E P N is just the direct product of ZP to the N power. Sometimes this notation is used, sometimes this notation is used. These are probably the most common notations for elementary Abelian groups. And so besides cyclic groups, elementary Abelian groups are some of the simplest, easiest Abelian groups to study, because after all, something that's interesting about elementary Abelian groups is that the exponent of an elementary Abelian group is always just P. That is, if you raise every element to the P power, you'll get back the identity, okay? Now one reason that elementary Abelian groups are so simple, other than the fact, of course, that their exponent is very small. It's the minimum exponent there, is that we can actually study elementary Abelian groups using linear algebra, because we can view each element of E Q as in fact a column vector, whose scalars come from ZP, right? Cause after all these elements here, you have some type of like X1, some X2, all the way down to XN. In which case these XIs here, XI will just say it that way, XI belongs to ZP. So, and we'll think of these as column vectors, so we'll take the transpose of this thing. So each element of ZPN can be viewed as a column vector whose scalars come from ZP. And this is something we'll talk about later on in this lecture series, but ZP is one of the simplest examples of a so-called finite field. It is in fact a field that is for ZP, you have well-defined addition, multiplication, subtraction and division with the exception divided by zero there. And it satisfies all the actions you want, associativity, commutivity, distributed properties. It's super good. And because such you can do linear algebra when you have scalars that come from a field. Again, these are all topics we'll see later on in our lecture series. And so because of this observation, you can actually view, you can actually view EPN as a vector space with ZP coordinates. Okay? And then it turns out that every group homomorphism between elementary and building groups can then be viewed as a linear transformation that is a vector space homomorphism because scalar multiplication is essentially automatic in this situation. Because scalar multiplication, if you times a vector by N, this just means you take X plus X and you do this N times. And so scalar multiplication in this finite setting is just repeated addition that follows from the distributed property for which a group homomorphism will preserve this automatically. And so every elementary building group can be viewed as a finite vector space over a finite field. And the group homomorphisms are exactly just linear transformations. So in particular, the linear endomorphisms are just the group endomorphisms. And so then if we get to this idea right here, the linear isomorphisms, that is, this is a bijective linear transformation. This is identical just to the group homomorphism, the group automorphisms of this set right here. And so with that perspective in mind, we can then describe the automorphism group of the elementary building group using the general linear group. There you go. Because the general linear group of N by N matrices whose scalars come from Zp, this then gives you all of the linear automorphisms from a vector space back into itself. And as these linear automorphisms are identical to the group automorphisms for elementary building groups, this gives us this isomorphism. The automorphisms of an elementary building group is just the general linear group over the appropriate finite field of scalars. So let's look at an example real quick. So let's take the Klein-4 group. The Klein-4 group can be identified as Z2 crosses Z2. And therefore the Klein-4 group is the elementary abelian group of order four. And so it's automorphism group would then be identical to GL2 Z2. So this would be the collection of all two by two non-singular binary matrices. So we only have as coordinates in the matrix zero and one. And we'll of course arithmetic will be mod two as usual. So in particular, one plus one is zero in this situation. Now we only want non-singular matrices. So since you have four coordinates to fill in here, you have two options for the first, two options, two options, two options. So you get two times two times two times two. There are 16 such matrices, two by two matrices with Z2 coordinates. But there's only six of them that are going to be non-singular. And you can think of this in terms of the determinant, right? This has determinant one, this one has determinant of negative one, which mod two is just one again. So we're really just looking for those matrices which have determinant one when you're working with Z2 coordinates right here. And you can work through all of these. These are the six matrices which have a determinant that's non-zero, which in this case would have to be one. And that gives us the automorphism group for the Klein-4 group. Now, previously we have observed that the Klein-4 group's automorphism group is actually just S3. You can look at any permutation on the elements one, zero and zero, one and one, one. Any permutation of these three elements gives you an automorphism of Z4. And that actually agrees with our observation right here. And so in general, though, if you're studying these elementary deli groups, its automorphism group will just be this general linear group. So this gives us an example of how employed linear algebra can actually help us study groups in a deeper, deeper way. This is the basic idea behind something called representation theory. We can strengthen our study of linear algebra by representing the group maybe as a vector space or something related in that regard.