 Suppose H is a subgroup of G, then the set of cosets H, A, H, B, H, and so on is a set, and so the question is, could we make this into a group? Now remember, in order to do that, we need a binary operation, and the obvious choice is the following. Let H be a subgroup of G, and let A, H, and B, H be cosets. Then let's define A, H by B, H to be A, B, H. So let's consider, say, H being the identity and the transposition 1, 2, a subgroup of S3. Let's find the cosets of H, and then form a Cayley table for the cosets. So we can form the distinct cosets by choosing elements not already in a coset. So the cosets are the subgroup H being the identity and the transposition 1, 2. And we know, I don't know, how about the transposition 1, 3 is not in a coset. So we find 1, 3, H, and the transposition 2, 3 is not in a coset. So we find the coset 2, 3, H, and since S3 only has six elements, this is all of the cosets. And so our set of cosets is the subgroup H, 1, 3, H, and 2, 3, H. And so our Cayley table is set up as... Now, while we've defined a binary operation, we still need to verify that we actually have a group, which means we have to have an identity. So since the subgroup itself can be written as E, H, then H times BH, well that's really EH times BH, which is EB times H, which is just BH. And so H is the identity for the group under coset multiplication. And so we can form our products. So because H is the identity, that gives us the first row and first column of our Cayley table. And using our definition, we can find the remaining products. 1, 3, H by 1, 3, H. Well that's the product 1, 3, 1, 3, H. And 1, 3, 1, 3 is the identity, and so that just gives us H. We can find 1, 3, H by 2, 3, H, which gives us 1, 3 by 2, 3, H. And this product 1, 3 by 2, 3 is 1, 3, 2, H. And it appears we don't have closure. But wait, remember that if A is an element of the coset BH, then the coset AH is the same as the coset BH. And we notice that this element 1, 3, 2 is an element of the coset 2, 3, H. And so that says the coset 1, 3, 2, H is actually the same as the coset 2, 3, H. So our product 1, 3, 2, H is just 2, 3, H. And we can complete the third row of our Cayley table. 2, 3, H by 1, 3, H. Well that's 2, 3, 1, 3, H, which is 1, 2, 3, H. And since 1, 2, 3 is an element of 1, 3, H, then this product is 1, 3, H. And finally our last product 2, 3, H by 2, 3, H gives us. And so we've produced the Cayley table and we observe that it satisfies the Latin square property and we have the identity. And so it appears that H, 1, 3, H, and 2, 3, H could be the elements of a group, except our operation is based on the class representative. And we have to make sure that the result of the operation is not affected by the representative we choose. In other words, our operation must be well-defined. No, no, no, we can't engage in wishful thinking. We actually have to verify that our operation is well-defined. Now earlier we found that 1, 3, H by 1, 3, H was H itself. But remember that any element of a coset can be used as the generator of the coset. So since 1, 3, H has the element 1, 2, 3, then 1, 3, H is the same as 1, 3, H. And so we could find the product 1, 3, H by 1, 3, H as the product 1, 2, 3, H by 1, 2, 3, H. And if we find this product we get 1, 3, 2, H. And since 1, 3, 2 is in 2, 3, H, then this product is 2, 3, H. And since we get different results depending on which representative we use, coset multiplication is not well-defined. So at this point we have a choice. To form a group from the cosets of a subgroup, we can either use our coset multiplication A H B H equals A B H on a limited type of subgroup, namely whatever subgroups we need to make sure that it's well-defined, or we can define the product H B H in some other way. Now, we could do either, but since the definition A H B H equals A B H seems more natural, it's better to limit the types of subgroups we can use. And so we'll define the following, let H be a subgroup of G. We say that H is a normal subgroup if, for any A in our group, A H equals H A. The left and right cosets are the same. And if we do that, then we have the following theorem. If H is a normal subgroup, then A H B H equals A B H is well-defined. So when do we get normal subgroups? Well, the boring case is that if G is a bellion, and in that case every subgroup of a bellion group is normal, how about from a non-bellion group? For example, the symmetric group on three elements. Let's show that the subgroup generated by 1, 2, 3 is normal. So, first of all, we find our subgroup, which we'll include, and we need to verify that for all G in our group, the left and right cosets are the same. And so we see that 1, 2, H is, meanwhile the right coset H 1, 2 is, and these are the only cosets that we can have. And so G H is H G for all elements, and H is a normal subgroup.