 So I'll talk a little bit about why cosets are important, and so we prove that if h is a subgroup of g, then the cosets a h and b h, they're either identical or they are disjoint. And so this suggests the following possibility. I could take any group g, and I'll find a subgroup h. And if h is not the entire group, I can find an element that is not in h, but is an element of the group, and I can find the coset a h. And if possible, I now have two sets, the set that makes up the subgroup and the coset. I might be able to find another element that's in neither, and this allows me to form another coset, and I will lather, rinse, repeat. And if I do this, eventually every element of g is going to end up in one of the cosets, and because cosets are disjoint, every element will end up in exactly one coset. And so this gives me a partition of my group. Now remember the order of a group or a subgroup is written this way, might think about that as absolute g. It's the number of elements in the group or a subgroup, and let's say I have a subgroup of g with h not equal to g. And I'll go ahead and find my coset. Now because a is not in h, what that tells me is my coset and my subgroup have to be completely disjoint. And you might want to think about why that happens. A mathematician would then ask, well, how does the size of a h, how does the number of elements in a h compare to the number of elements in h? And we can go through a couple of things. First of all, because a h is formed by multiplying every element of h by a, then we know that it can't get larger. We can't have more elements in a h than we started with. So it might be equal, or maybe it's less. Well, suppose it's less. By the pigeonhole principle, what that means is that we have to multiply a by two different elements, but get the same thing. So there has to be some pair that are equal when the h values are not. So, well, I know that a is in a group, so a inverse is also in the group. So if I have a hi equal a hj, I can multiply on the left by the inverse of a, and the terms drop out over on the left hand side of hi. On the right hand side, I have hj, and those two must be equal, except I assume that they worked. So this is a contradiction. You got on the bus that says these two are equal. We came up with something that we know isn't true, so don't get on this bus. So a hi hj have to be different for all values. And so that tells me that these two actually have to have the same size. The order of the coset must be the same as the order of the original subgroup. And this leads to a very important result known as Lagrange's theorem. So let g be a group, and let h not equal to g be some sort of subgroup of g. And what I can do is, again, if h isn't the entire thing, I can find some element that's not in my subgroup, and I can form the coset, and they are disjoint. And because a group and its coset are the same size, then I know that the size of the union of the two is just going to be twice the size of the subgroup. Now, if that union happens to be the entire group, then I know that twice the size of the subgroup is the size of the group, so two is g over h. If, on the other hand, they aren't the same size, if they don't include the entire group, then I know there's some other element in the group that's not in h and not in the coset, and so I can have a rinse repeat, and I can form another coset. All three of these are now disjoint, and their union is going to be three times the size of h. And as before, if they happen to be the whole thing, then I know that three is the size of g divided by h, otherwise I can do the same thing over and over again. And what we find here is that the order of a subgroup always divides the order of a group. If a subgroup and its coset give you the entire group, the order of the group divided by the order of the subgroup is going to be two. If the subgroup, coset, and another coset give you the entire group, then my group is three times the size of the subgroup, and so on regardless of how big or small that subgroup is. Now one term that we use for that, if h is a subgroup of g, then this quantity, size of g divided by size of h, this quantity is called the index of the subgroup, and it's usually written in some variation of this. So in this case, if my subgroup and coset give you the whole thing, then my subgroup has index two. If my subgroup, coset, and another coset give you the whole thing, then my subgroup has index three, and so on. Now it's always useful to think about what this tells us, so one of the important things that we can get out of this is suppose my group has a prime order and I have a subgroup. Well then one of two things happens because the order of the subgroup has to divide the order of the group, well the only divisors of a prime number are the prime itself, the order of the group has to be p, well that means h has p elements, g has p elements, they have to be the same thing, the subgroup has to be the whole thing. The other possibility, the other divisor of the prime number is one, so the order of the subgroup has to be one, and since every subgroup has to contain the identity element, my subgroup has to be the identity only, and what that tells us is this very nice corollary. If the order of a group is prime, then g has no proper subgroups. Every subgroup of g is going to either be just the identity or it's going to be the whole thing. So let's continue writing this bus and see what happens. Suppose the order of a group is a prime number. Well let's take any element of that group other than the identity, which wouldn't be very interesting. We know that the sequence of powers of this element g, g times g, g times g times g, and so on, the sequence is going to form an abelian subgroup of g, but because the order of g is prime, we know there's no proper subgroups, which means that this sequence has to include the entire group. This sequence of powers must be, as a subgroup, must be g itself, and what that tells us is that if p is a prime number and the order of my group is prime, then g has to be abelian, because this set is the entire group and this set is abelian. And we can in fact go one step further, because this is true for any element of g other than the identity. We also know that any element of the group other than the identity is going to be a generator of the group. Well, this raises a new question. Suppose p is a prime number, and so I'll let the order of the group be p, and I'll take g and g other than the identity, and so g is a generator and a group consists of the elements g, g squared, and so on all the way up to g to power p minus one. g to power p is going to take us back to the identity element, and let's take a different group. Let's let the order of h be the same with h and h, and h not equal to the identity, and h is completely different from g, whatever that means. Then I can pick some element to be the generator, and our group is going to consist of the elements h, h squared, h cubed, and so on all the way up to h to power p, which is going to be the same as the identity. So our group consists of these elements, which are completely different from these elements. Now, some observations we might make, g is a bilion, and any element other than e is of the form g to power k, and that element has order p. On the other hand, h is a bilion, and any element other than the identity is of the form h to power k, and that element has order p. Well, remember, if it quacks like a duck, swims like a duck, and flies like a duck, then it's probably a duck. But we said that h and g were completely different, so these two can't be the same even though they look an awful lot like the same thing. So maybe they are the same, but this is kind of a mystery of mathematics. Well, there really aren't any mysteries of mathematics. We'll look at this problem in some detail later on.