 So, earlier we found what we might call Lagrange-Lite. The order of a subgroup generated by an element is a divisor of the order of the group. But what if our subgroup isn't generated by an element? The proof for the multiplicative group and the subgroup generated by an element relied on the subgroup generated by the element, having k terms, the product of each element of A by an element not in A, and then proving that BA also had k elements, and proving that if C wasn't in A or BA, then CA produced another distinct set of k elements. Then Lather, Rinse, Repeat. So can we do this even if our subgroup isn't a cyclic group? So by analogy we define the following. Let G be a group and H a subgroup. Let A be an element of our group G. The set AH, which we'll define as the set of products AH where H is an element of our subgroup, is called a left coset of H. And similarly we can multiply on the right HA where we take every element of our subgroup and multiply it on the right by some element, and in a fit of creativity, mathematicians call this a right coset of H. So for example, let's find a subgroup of S3, the symmetric group on three elements, and a left coset of the subgroup. So remember we can form a subgroup by finding the powers of any element, say how about the cycle 1, 2, 3. So we'll find the powers of our cycle, that'll be sigma 1, 2, 3, sigma squared, sigma cubed, and since we're back at the identity, that completes our subgroup. We can form a coset by picking any element of S3, say 1, 3, and so we find that 1, 3 times the identity is 1, 3, and 1, 3 times the other two elements of the subgroup give us, and so the left coset 1, 3, H will be. Now with the multiplicative group, we saw the cosets were either disjoint or identical. What about a general subgroup? Suppose we have a coset A, H, and B not an element of A, H. Could the cosets A, H, and B, H have an element in common? Well, let's compare this to what we did with the multiplicative group. Suppose A, H1 is B, H2, where H1 and H2 are elements of our subgroup. Then if that's true, then I can write B as A times H1 times H2 inverse. Now since H is also a group, H2 inverse is an element of H, and since H is a group, and H1, H2 inverse are in H, so is their product. And we'll call this something, and in a fit of creativity, we'll call it H prime. And so this says that B is A, H prime, which means that B is an element of the left coset A, H, but it can't be. And so the two cosets can't have a common element, and this gives us a lemma. Let H be a subgroup of G and A, H a coset. If there's some other element B not in A, H, then the two cosets have to be disjoint. The other important result was that with the multiplicative group, we showed the cosets were all the same size. Now since A, H consists of the elements of H multiplied by A, we can get a coset collapse if any two products are the same. So suppose A, H1 is the same as A, H2, then that stuff happens. And you should be able to complete the proof on your own. And we put all these pieces together, we get what's called Lagrange's theorem. Given a finite group with subgroup H, the subgroup H has order K, and every coset also has order K. And if our set of cosets doesn't include all of G, we can form another disjoint coset with K elements. And so the order of G is a multiple of K, or again, equivalently, the order of a subgroup is a divisor of the order of a group. So for example, what sized subgroups can S4 have? So we note that S4, the set of permutations on four elements, has four factorial 24 elements, so it might have subgroups of order 1, 2, 3, 4, 6, 8, and 12, all of the divisors of 24. Now the important question to ask at this point is, will it have subgroups of these orders? And the answer to that is, watch this space, because we'll take a look at that problem later.