 Welcome back to our lecture series, Math 42-20, Abstract Algebra I for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misildine. In this video, the first video for lecture 33, we're going to start chapter 11 in Tom Judson's Abstract Algebra Textbook, which is kind of a curious thing. A lot of businesses end in chapter 11, but we still have several more chapters in this series, plus another semester after this one. So we'll keep on going past chapter 11. But in chapter 11, we're going to talk about group homomorphism, which we've talked about what a homomorphism was in the previous lecture, lecture 32. In this one, we want to talk about the idea of a kernel, a very important aspect associated to homomorphisms here. So in this lecture, we're talking about, of course, group homomorphisms, but analogous things can be said for rings, module homomorphisms, etc. So let phi be a group homomorphism between two groups G and H. We define the kernel of phi, which is typically denoted K-E-R of phi, which honestly, this is actually a built-in latex command, assuming you use the AMS math packages, just backslash cur, and then you can use backslash var phi if you want to get that symbol right there. All right, so the kernel of phi is defined to be the set of all elements G inside the domain, such that the image maps to the identity of H. So we're looking for those elements which map to the identity under this group homomorphisms. Now, based upon what we proved previously, we can actually see very quickly that the kernel of a homomorphism is a normal subgroup of G. And to see that, notice that the kernel of phi is just the pre-image of the identity of the group H there. And in fact, it's the pre-image of the trivial subgroup of the co-domain H right there. And so as we've proven previously, the pre-image of a subgroup of the co-domain will be a subgroup of the domain. And if that subgroup is normal, so the trivial subgroup is always normal, right? If the pre-image of it, if you take the pre-image of a normal subgroup, that also will give you a normal subgroup. We've proven this previously in this series. And so the kernel will always be a normal subgroup of G. And I mean, you can prove this directly, just showing the fact that kernels are closed under conjugations and things like that. I actually think it's a great exercise for the viewer to try to do that on their own. But the machine that we have already shows us that the kernel of any group homomorphism will be a normal subgroup of the domain. And this is actually a very convenient way of creating normal subgroups for domains. And we'll actually see a little bit later in this video that essentially, normal subgroups and kernels are one of the same thing. But let's look at some examples first. So we saw previously that the determinant is actually a group homomorphism from the general linear group to the group of non-zero numbers over a field. And so for this example, let's just consider the field of real numbers right here. So if we take, remember GLN of R, this would be all the N by N non-singular real matrices. That is those matrices which have a matrix inverse. And then R star right here, this is the multiplicative group of non-zero real numbers. So if you take the determinant of a N by N matrix, you'll get back a number. If it was non-singular originally, then the determinant will be zero. A matrix is singular if and only if it's determinant is zero, in fact. And so we prove previously that this is a group homomorphism. Determinant has the property. Take the determinant of a times b. This is the same thing as the determinant of a times the determinant of b. That's exactly the homomorphic property right there. So we have this homomorphism. What would be the kernel of this homomorphism? Well, looking at the co-domain, the identity of the co-domain will be the number one. This is a multiplicative group after all. So the kernel is gonna be all of those non-singular matrices whose determinant is one. And so the kernel of this map, the kernel of the determinant map is what we call the special linear group, SLN of R. And then of course R could be replaced with any field that you want here. But SLN of R, the special linear group, it's the set of all matrices, N by N matrices whose determinant is equal to one. And since it's the kernel of the determinant map, this implies that the special linear group is a normal subgroup of the general linear group. Another example we see before is the trace map. The trace, remember, you take it in by N matrix, which remember this notation here. GLN of R, this is the multiplicative set of non-singular matrices. That is the group operation in play is matrix multiplication. R to the N by N here, this represents the vector space of N by N real matrices with respect to addition. So this is an additive group. And so we could take the map from R to the N by N to R. So this would be the map from N by N matrices just to a single number, kind of like the determinant here. But the trace is when we take the sum of the diagonals. You have your N by N matrix, you take the sum of all the diagonal entries, you add those together and get the trace, okay? And so this will turn a matrix into a number. We proved previously that the trace is in fact a group homomorphism, right? Because for the trace here, you wanna take the sum of two matrices. So the trace of A plus B equals the trace of A plus the trace of B, right? This is a very simple property. You see here, so it's a group homomorphism. So we could ask, what is its kernel? The kernel is gonna be the set of the so-called traceless matrices that is the matrices whose trace is equal to zero. So those matrices that disappear without a trace. And so as the identity of R, speaking of it as now an additive group, the additive identity is gonna be zero. So we're looking for those matrices whose trace is equal to zero. And so this is gonna form a normal subgroup of R and by N. As this is an Abelian group, every subgroup is normal, so it doesn't say so much. But in fact, since R to the N by N is actually a subspace, you can actually make an argument that this is, excuse me, that R to the N by N is actually a vector space. It's an N by N, an N squared dimensional real vector space. You can actually show that the kernel of this is gonna be a subspace as well. It's closed under a scalar multiplication. In fact, you could, this is in fact, a null space, which is always a subspace. So that gives us a little bit more linear algebra than I really wanna worry about right now. But the rank nullity theorem, we can actually show that this thing is a N squared minus one dimensional subspace. The idea here is that R to the N by N is an N squared. That is N times N dimensional subspace vector spaces because when you take your matrices here, you have a choice for the first position, a choice for the second position, and choice for the next row. And then if you continue on, you're gonna get N squared possibilities, N squared degrees of freedom when you choose a matrix with no restriction on whatsoever. When it comes to traceless matrices, again, you can use the rank nullity theorem. That is the rank and nullity will always add up to be the dimension in that situation. But another way to do it is sort of like the following. Look at the, let's skip the diagonal for a moment. You can choose the things in the off diagonal be whatever you want, because they don't affect the trace. And you can set the things on the diagonal to be whatever you want with the exception of the very last term right there. The very last term will be determined by the other terms, right? So if you call those entries like A11 plus A22 plus A33 all the way down to ANN, if this is traceless, it should have to be zero. And so solving for ANN, you get negative A11 minus A22 all the way down to AN minus one, N minus one. And so you see that the last entry on the diagonal actually is a dependent variable. It doesn't have freedom compared to the other ones. And so in some respect, if your matrix is traceless, the first N minus one bits along the diagonal are kind of like information bits, but then that last bit in the last digit or whatever, the last number along the diagonal, it kind of like a check bit in some regards. If it's gonna be traceless, you gotta satisfy this condition. So that's why you get N minus one degrees of freedom. You can't choose all of the diagonals. You can choose all of the off diagonal entries and all but one of the diagonal entries and that gives you a traceless matrix there. Some other examples, let's consider the homomorphisms we saw previously. Let's take the sigma map. Remember, this is the map here, sigma of N. Well, sorry, sigma of sigma here. This will give you plus one or minus one exactly in the situations where sigma is even. Whoops. When sigma is even, it'll give you plus one and when sigma is odd, it'll give you a minus one. So we proved previously this is a homomorphism. And so remember this map right here one in, all right, excuse me, this group one and negative one. This is just the group of second roots of unity inside of the complex numbers, right? Viewed as a subgroup, a multiplicative subgroup of C star right here. We can identify this with the cyclic group of order two because there's just the two elements and it's a cyclic group, of course. But what would be the kernel of the sigma map, right? Well, for Z two, this is a multiplicative group, right? The identity is gonna be the number one, right? One times anything will give you back that number. And so if we're looking for the kernel of the sigma map, then we're looking for those permutations which map to one, which is exactly the even permutations. We refer to this as the alternating group. And so A N is the kernel of this sigma map. And we've seen previously that the alternating group is normal. That's an easy thing to see when you look at the index, the index of the alternating group is two. But we can also see it as it's the kernel of the signum homomorphism, it's also a normal subgroup. Okay, some other examples. Just trying, I'm just grabbing some homomorphisms we saw in lecture 32 here. So with this one, consider the map from the real numbers to the circle group. So remember the circle group, this is the set of all complex numbers whose modulus is equal to one. And so this is gonna be naturally a subgroup of C star, okay? And so we're gonna take the real numbers, map to the circle group, and it's gonna follow the formula right here. Theta, which is just gonna be a generic real number, will map to E to the i theta, like so. And so like I mentioned before, there's like a geometric interpretation of this. If you have your circle group S one, then this is the group where we're basically kind of wrapping. We kind of take up, we think of our real numbers as this like infinite coil, right? We wrap the real numbers in this coil and then we project it down. The shadow of the real line is where we're projecting to this right now. And so what would the kernel of this map be? What things map to the identity? Well, the identity of the circle group is just the same as the identity of C star right here. It's gonna be the number one. So we're looking for those complex numbers or those real numbers, excuse me, such that E to the i theta, which is the complex number maps to one. Well, given sort of this oiler's identity from complex analysis here, complex variables, we see the important thing to realize here is that E to the pi i is equal to one. And this comes from the fact that E to the i theta can be written as cosine of theta plus i sine of theta, like so. We're hearing of one exactly when theta is a multiple of two pi because that makes cosine equal to one and sine equal to zero, like so. So it turns out that the kernel of this map is going to be the cyclic subgroup generated by the number two pi. And this is an additive group. So we're looking at integer multiples of two pi. So like zero, two pi, four pi, six pi, negative two pi, those type of things. That's what this cyclic group is right here. This would be the kernel, which it is a normal subgroup. Again, normal doesn't really mean too much here because after all R is an abelian group with respect to addition. But it, so that would be the kernel here, the cyclic subgroup generated by two pi. And I should mention that this is going to be an infinite cyclic group. There's no multiple of two pi that gets you back to zero. And so this is an infinite cyclic group. So the cyclic subgroup generated by two pi is actually isomorphic to the infinite cyclic group and the integers. This will be important as we see in just a second. Well, I should say it in the next video when we talk about the first isomorphism theorem. Let's do one more example. And I think that'll be enough for this video here. So let phi be the homomorphism from the infinite cyclic group Z just to any group whatsoever, right? The group G could be anything, don't really care. So we take the homomorphism Z to G here where the integer M is gonna map to the Mth power of G where G is just our favorite element. Little G is just our favorite element in our group here. And so we saw previously that this always gives a group homomorphism, right? No restriction whatsoever. And if the order of G is finite, let's say that it's order N, then this is what we need to know what the kernel is gonna be because the kernel depends on the order of G. If the order of G turned out to be infinite, we'll actually see this maps one to one. That's actually written down here. But we'll get to that detail in just a second. So let's first suppose that little G has finite order. Let's say it's order N. Well, that means if I take phi of K to the N, that'll map to G to the K to the N, which by exponent root laws, which apply to this associative group, we're gonna get G to the NK, which G to the N is the identity since N is the order of G. So you get E to the K, which is just the identity E right there. This is just consequences of Lagrange's theorem. And so we see that any multiple of N is going to give you some, it's gonna give the identity. So that's what the kernel is. The kernel of phi right here is gonna be NZ, right? This right here is the cyclic subgroup of the infinite cyclic group here generated by N, right? So we get NZ right there as the kernel. That's if the finite was, excuse me, if the order was finite. If the order of G of course is infinite, then there's actually nothing that'll map to the identity other than the identity itself. And so I guess since we're talking about, we should probably say like, oh, this is just zero because we're talking about the group of integers right here. If you want, this is actually zero Z. You could say that by multiples of zero, right? And so we prove previously, in fact that when this homomorphism is gonna be one to one, that is it's injective if and only if the order of the group is infinite. That is when the order of this element, excuse me, let me say that again, we prove that this homomorphism Z to G is a one to one map exactly when the element little G has infinite order. And it won't be one to one if it has finite order. So the order of the element we're mapping onto here makes a big difference on whether it's one to one or not. And so in this example, you'll see that the kernel of phi is actually just the identity, right? And that proved that the function, the homomorphism is one to one in that situation. And so I'm gonna leave it as an exercise to the viewer here to prove the more general statement. This is actually a very useful statement that we'll use throughout this, whenever we talk about homomorphisms in the future, that a homomorphism is injective, that is it's one to one, if and only if the kernel is trivial. We always know that the identity maps to the identity. So the identity of G will map onto the identity of H. So therefore the kernel will always contain the identity. But the thing is this map is one to one, if and only if the only thing that maps the identity is the identity itself. And so I'll leave it as you leave the viewer to prove this fact, but we'll be using this the future. It's really, really useful.