 Let's talk about a few important groups and some important group classifications. So one very, very, very, very important set of groups is known as glnr. What does that mean? Well, let's take the part gl is shorthand for general linear, and that doesn't really mean a whole lot, until you know that n refers to the fact that we're dealing with n by n matrices. And in fact, we can go a little bit farther. This r means that the entries of those n by n matrices are real. So this is a group that's based on n by n real matrices. And well, that's a set. In order for glnr to be a group, we need to make sure that we have some sort of defined operation. And in this particular case, we're going to use the operation of matrix multiplication. We need an identity. Well, we'll leave that as an exercise for the viewer. We also need to make sure that we're dealing with a set where all of our inverses are included. And so that means glnr has to include only those matrices that are invertible. And here's a quick review from linear algebra. Make sure that you can determine whether a matrix is invertible. What do you look at? Now, the glnr groups are useful because we know that matrix multiplication is not in general commutative. And again, as a reminder in general for a group, you cannot assume that the group operation is commutative. You cannot assume that A star B is the same as B star A. Again, it is one of the most common mistakes in proofs and abstract algebras to assume commutativity when it doesn't exist. Now, in very, very, very, very, very, very rare cases, we do actually have this commutativity that if I pick two elements in the group, A star B and B star A are going to be the same. Importantly, this is for any two elements of the group. Some group elements will commute. The identity commutes with everything. The inverse commutes with whatever it's the inverse of. But in general, if I pick two arbitrary elements of a group, I can't guarantee commutativity. On the other hand, I may actually have a group in a very, very, very rare case. I may actually have a group where commutativity exists, in which case I have this type of group, which is named after a Norwegian mathematician by this name. Now, the name of the mathematician is Nils Henrik Abel. But everybody calls these Abelian groups. And just a historical footnote, you might notice here, Abel didn't live that long. And this is actually true for many of the chief founders of abstract algebra. Abstract algebra is hazardous to your health. Well, that's not really true. Most of those who developed abstract algebra lived long, healthy, happy, and productive lives. Abel and a few others died at extremely young ages. And of course, here is an important thing to know as a mathematician. Here's an obligatory math joke. What's purple and commutes? That would be an Abelian group. All right, so let's talk a little bit more about these Abelian groups. Is GL2R an Abelian group? And if not, we want to find a subset that's Abelian. So first of all, GL2R is a group. Prove it. Make sure that you can prove it. So we only need to determine whether or not it's an Abelian group. And what that means is we want to determine whether it is true for any elements A and B, that A times B is the same as B times A. And we'll start off with a bad proof. Here's a bad proof that it's Abelian. So I'll pick two matrices. This is two by two matrices with real entries. We'll have about this matrix and this matrix. I'll multiply them together this way, and I'll get this product. And then I'll switch the order of multiplication. So I'll switch that order. And I end up with, lo and behold, it's the same product. And so now I have a proof that A times B is the same as B times A. Well, this is actually a bad proof. And we might go into some detail about why it's a bad proof. Proves in general should give us general results, not specific ones. All we've actually proven here is that this matrix times this matrix gives us the same results, no matter which way we multiply them. However, a bad proof can still be useful if it gives us some insight. And the most important thing here is that if you want to start off with an example, don't end there. And so here I might take a look at a different example. What if I change the matrices? So here I have this matrix times this matrix. I run the matrix multiplication and I get that. If I swap the order, I get, oh wait, that's not the same thing. And so that tells me that A times B is not equal to B times A. Now you might wonder why we can use this example to prove this statement. And the reason is that when I make this claim, I'm saying that there is some matrix A and B where the commutativity does not hold. If I claim they're equal, then I'm saying that for any matrix the product is the same. If I claim they're unequal, all I'm saying is that I can find at least one matrix for which this is not true. And because we can do that, we can say that our group GL2R is not abelian. Now it's helpful to start with an example, but here's the important idea. You should try as hard as possible to find something that disproves your claim. And the reason you want to do that is that if you don't, someone else will. Now we do have this problem. We do want to find a subset of GL2R that is an abelian group. And again, here's where it's useful. We actually found an example that does commute. So we started out with these two matrices and found that the product of these two matrices is the same regardless of the order that we multiplied them in. And so we might want to look closely at this example where we do have commutativity and compare it to this example where we don't. And the one striking feature about the two examples is that in the case where we did have commutativity, our matrices had a further special property, they were diagonal matrices. So it appears that a diagonal matrix times another diagonal matrix is going to give you the same result regardless of the order that you do it in. Well unless you're a politician, you're not going to generalize from one example. So if you're a person who deals with facts and evidence, you might want to look for other examples where this is true. So let's find a couple of diagonal matrices and multiply them and swap the order and I get the same result. And maybe here's another pair of diagonal matrices and I multiply them and I multiply them in the other order and again I get the same result. And again, take another pair of diagonal matrices, multiply them, get the same result either way, and it suggests that if I look at the diagonal matrices in GL2R, I do get an abelian group. But again, examples do not constitute a proof. Because we've only seen four examples, it's conceivable that I can find a pair of diagonal matrices that do not commute. On the other hand, our evidence suggests that they do in general commute, so at this point it's worth looking for a proof. Wait a minute. Now let's think about this in a bigger picture. We found a bunch of examples where we had something that seemed to be true and now we're going to look for a proof. We're not even going to try to find a proof unless you already believe something to be true. And so this raises the question, why bother with proof in the first place? And one of the important things about proof is every proof requires you to review what you know or find out what you should know. So proof is a good way to study mathematics. So in this case, we're looking at invertible 2x2 diagonal matrices. So to show that this is a abelian group, we'll take two invertible 2x2 diagonal matrices and show that the product is the same regardless of the order that we perform the multiplication. Because we want to do this in general, we're going to take two generic invertible 2x2 matrices and multiply them in both ways. So here's a generic 2x2 diagonal matrix. Well, actually here is a generic 2x2 diagonal matrix. And here's another one. And when I multiply them, I get this product. If I swap the order, I get this product. And well, since A, E, T, and F are real numbers, then A, E, and E, A are the same. D, F, and F, D are the same. So these two products are going to be the same. And so the product of the two matrices is the same regardless of the order of multiplication. And so the set of invertible diagonal matrices under matrix multiplication does give us an abelian group.