 So we defined a group as an algebraic structure with the following properties, let g be a set and star, some binary operation where g is closed under star. Star is associative, there is an identity element e, and every element of g has an inverse, and in that case we say that g star is a group. So we go through our laundry list, let's prove or possibly disprove that z plus is a group where z is the set of integers and plus is ordinary addition. And so definitions are the whole of mathematics, all else is commentary, we need to check our group requirements. The set of integers z with operation plus is, well let's see, is it closed? Well if p and q are integers then p plus q is an integer, so g is closed under plus. Are we associative? Since plus is ordinary addition, and we know ordinary addition is associative, then plus is associative. Now strictly speaking we should prove that it's associative, but that requires building up a theory of the integers, and we'll do that in a different course. We need to find an identity element, we need an element e, where for any integer p we have p plus e equals what we started with, and also e plus p gives us p. And let's think, thinking, thinking, thinking, the integer 0 satisfies this property, and so our set of integers has identity, 0. We also need an inverse, and so for any integer p we need an inverse q where p plus q is equal to 0, it's a right inverse, and it also has to serve as a left inverse q plus p has to give us 0. And the integer minus p satisfies this property. And since we've met all the requirements, z with operation plus is a group. Now remember commutativity is not a requirement of our operation. If we are lucky though our binary operation is commutative, stop jumping your q. And this leads to the following definition. A group is this if its binary operation is commutative. Now different people will pronounce this word differently, and that's because it's a person's name. This is actually named after a Norwegian mathematician by the name of, as nearly as I can pronounce it, Niels Henrik Abel. And so this word should probably be pronounced abelian. But most English speakers will pronounce this as abelian. In any case, this also gives the obligatory mouth joke what's purple and commutes. It's an abelian grape. And now you know why my career as a comedian never really took off. So let's find a set of matrices that form a group under matrix multiplication. And let's find out if m times is an abelian group. So let's think about that. We need closure. So given a matrix A, well at the very least we have to be able to multiply A by itself. So when can we find the product of a matrix with itself? In order for that to happen, our matrices must be square matrices. Now if A and B are differently sized square matrices, we can't multiply them. And since closure is a requirement for a group, then in order for the set m to be closed under matrix multiplication, all matrices must be square and the same size. So we'll make m the set of 23 by 23 matrices. We do need to check our other group requirements. So there is an identity matrix. It's a 23 by 23 identity matrix. And matrix multiplication is associative, so that meets some of our group requirements. Now the important thing to remember is that in order for m to be a group, every element must have an inverse. But not all matrices are invertible, so we'll limit ourselves to invertible matrices. And while we could just write that requirement into the set, let's go ahead and show that we've taken some linear algebra. This is the same as requiring the determinant b non-zero. And finally we want to know whether we're dealing with an abelian group. And since ab is not generally equal to ba, at least not when we're dealing with matrix multiplication, then m times is not commutative. Or let's find a set of numbers of the form a plus b square root of 2 that form a group under ordinary multiplication where a and b are integers. So first let's check closure. If we have a plus b square root of 2 and c plus d square root of 2, their product, and since ab, c, and d are integers, so is ac plus 2bd and bc plus ad. And so this number is also the form integer plus integer times square root of 2, and so we have closure. Since this is ordinary multiplication, we also have associativity. Now the identity for ordinary multiplication is 1, but we're dealing with numbers of the form a plus b square root of 2. So we have to make sure that the identity is actually in this set. Fortunately, it is. And so we only need to make sure that any element a plus b square root of 2 has an inverse. So we need a plus b square root of 2 times x plus y square root of 2 to be our identity 1 plus 0 square root of 2. Now if we expand out the left-hand side, and this means we can find an inverse by solving the system ax plus 2by equals 1, our term not multiplied by the square root of 2, and bx plus ay equals 0, the term that is multiplied by the square root of 2. Now we can solve this in a number of different ways, but if we multiply the first equation by b and the second by minus a and ad, that will get rid of our x terms. And so we can find y by dividing by 2b squared minus a squared. Or can we? We do have to make sure that this isn't 0, and so we'll require that 2b squared minus a squared not be equal to 0. Now it is possible there may be additional requirements, and we've only solved for y. So let's see if we can solve for x. And here we notice that if we multiply our first equation by a and the second by minus 2b and ad, this eliminates our y terms, and so we can solve it for x. And again we can evaluate this as long as a squared minus 2b squared is not equal to 0. And this gives us a second requirement. Well, not really. We are already requiring that 2b squared minus a squared not be 0, and that will also give us this a squared minus 2b squared not equal to 0. And so this single requirement is enough, and then a plus b squared 2 will have an inverse under multiplication.