 One important idea is that when you have subgroups of a group, you also have what are called cosets of that subgroup. So for example, say I have my subgroup of g, and for convenience we'll assume that our subgroup is not equal to the entire group, then I know that there's something in g that is not going to be in the subgroup. So what can you do with that? Well, one of the nice things about groups is that there's not a lot you can answer to the question of what can you do with that. If I have two elements, the only thing I can really do is I can multiply them. So what I can do is I can take this element b that's not in h and multiply it by every element of h. And because multiplication in a group is not necessarily commutative, I do want to be careful about what that order is. So I'll write b h to indicate the set consisting of the elements b times every element of h. Here I'm multiplying my subgroup h on the left by b, and so this is going to be called the left coset. And likewise, we can form h times b, which is going to be called the right coset. So for example, let's let g be the group generated by i under multiplication. That's our complex number i. And let's consider the subgroup generated by minus 1, and we'll find the left coset i h. Now if you want to think about things like a mathematician would, one of the things he'll do is to ask and answer the question that isn't there. I've talked about a subgroup generated by something I've found a subgroup. I haven't actually done too much about what the group is, so let's actually find the group as well. So since i is the generator of our group, I'm going to find the elements of g by repeatedly multiplying i with itself. So that's going to start with i, i multiplied by itself. That's i squared is negative 1, i multiplied by itself 3 times i times i times i is going to be negative i, and i multiplied by itself 4 times 1, 2, 3, 4 is going to be equal to 1, and there's our multiplicative identity. And so here's our group g consists of these four elements. Now further powers of i are just going to reproduce this sequence i negative 1, negative i1, you should prove this. And our subgroup that's generated by negative 1 is going to be negative 1 and 1. Now again further products of minus 1 will just reproduce those elements. So I have my group, I have my subgroup, and I can form my cosets by multiplying the subgroup by any element of the group. So ih in this case will be the left coset that's going to be formed by multiplying every element of h by i on the left. So that's going to be i times negative 1, that's going to be i times 1. And so those are my group elements and they simplify to minus i and i. Now we can have more than one coset, so let's consider two cosets a, h, and b, h, and let's assume that a is not equal to b. And the first question of mathematician would ask given two things, are they the same or are they different? And so the answer to this question is, well let's get on the coset bus and see where it takes us. Again an important idea in mathematics is I can never prove anything from an example. However that doesn't mean examples aren't useful. A concrete example can help us guide our thinking about a concept or an idea. So let's take a look at that. We found our group g and our group h, a subgroup h, so we can now find some cosets. So ih, well that's going to be i multiplied by every element of h, that'll give us minus i and i, negative 1 times h, well that's negative 1 times every element of h. Negative i times h, negative i times h gives us this, and 1 times h is going to give us this. And let's take a look at this and it seems that when I form the cosets of h we get one of two different things. They are either identical to h, remember the order doesn't matter. They are either the same thing as h is, or they are completely different. They have nothing in common with h. In other words the cosets are disjoint. Well one piece of evidence is not enough for us to decide something is probably true, but we can collect more data in class and see that there may be something here. So let's compare two cosets ah and bh, and our conclusion was they're either identical or they are disjoint. So let's suppose that they are not disjoint, that they don't have nothing in common, then there's got to be some element in ah, that's also in bh. And so the fact that these two cosets do have something in common, that their intersection is non-empty, says that they have to have some element in common. Something in ah, ahi, has to be equal to bhj. So since h is a subgroup, then it's a group and every element has an inverse. And that's an important idea. Also we cannot assume commutativity. We need to choose which side we're going to multiply by, and we'll use the right hand side. So I know that ahi is equal to bhj. There's some element in common to the two cosets. So I know that something in ah is equal to something in bh. So because there's an inverse, I can multiply on the right by the inverse. And that gives me b is equal to a times hi times hj inverse. Well, h is a subgroup, so I know this product here, hihj inverse, is another element of h. So I know that b is equal to ah times ahk. And it's worth noting that what this means, because b is a times some element of h, then b has to be in this coset ah. Well, let's go a little bit farther. If b is ahk, then when I take a look at the coset bh, well, that's really ahk times h. So bh is going to look like b multiplied by every element of h. And that's ahk multiplied by every element of h. So that's going to be ahk and so on. And again, this works because h is a group. So any time I multiply ahk times any of these, I get another element of the group. And this set looks a lot like ah, because it's a multiplied by some of the elements of h. Now there is an important hidden assumption here. While we know that we've multiplied a by many of the elements of h, I don't know that we've included all of the elements of h. So if nothing else, this is at least a subset of ah. But we don't know if it includes the entire thing. So let's check that out. So a quick summary of where we are. If my two cosets have one element in common, it doesn't really matter which one it is. If they have one element in common, then I know that b is going to be in ah. Now what I want to do is I want to show that everything in ah is also in bh. So I know that h looks like this. I know that ah looks like this. And what I want to do is I want to show that everything in here is some element of bh. So how do we prove that? Well here's a useful approach to prove. We can start where we want to end. And in this particular case, we want to show that ah i is in bh. And that means our ending point is to show that ah i is bh m. That a times h is b times something else in h. So let's go ahead and end there. And we'll work our way backwards. So what do we know? Well I do know that b is equal to ah k. And because that's an equality, I have no objections to replacing this b with an ah k in the line preceding it. So let's go ahead and do that. Because I have a group and I can multiply on the left by a inverse and get rid of that a. And I know that again because I have an inverse, I can multiply on the left by h k. And so h m is this value h k inverse times h i. But because h is a subgroup and i, h i and h j are both in h. So is the inverse and so is the product. And so likewise. And so that means that this thing, h m, has to be in h. Which means that this has to be in h. Which means that this is an element of h, multiplied by b, has to be in bh. And so ah has to be a subset of bh. And we can do almost the same thing to prove that bh is a subset of ah. You should go about and prove it. And because these are subsets of each other, we know the two sets have to be equal. And so this gives us a very important lemma, disjoint cosets. If I have two cosets, they're either identical or they have to have an empty intersection. They must be completely disjoint.