 Now, we've already defined a group action on a particular subset of G, given H a subgroup of G, and A an element of G. The coset AH consists of all elements of H left multiplied by A. While we've considered H to be a group, remember it's also a subset. So the cosets correspond to a group action of G on a quotient group, well, set, anyway, AH, where A is an element of G. And this allows us to define a group action from G cross P to P, where P is a set of subsets of G. Now, since the cardinality of AH is the same as the cardinality of H, then P must be the set of all subsets of G with K elements. Where K is some specific number. So if F from G cross P to P is a group action on the subsets of G with K elements, where F of GK is GK, left multiplication, we can define the orbit of K to be the set of cosets we can get to by left multiplying by G. And likewise, the stabilizer of K is the elements of G, which acting on K just return K. So for example, let G be our multiplicative group of integers mod 31 and K be, again, not a group, but this set, and let's find the stabilizer. And so we note that 1 is always in the stabilizer. The identity always reproduces the set. But after some effort, we also discover that 5 will also reproduce the set. 5K is the same as the set we started with. And so the stabilizer consists of the elements 1 and 5. At this point, you might notice that our set K was actually the union of the subgroup generated by 5 and the coset produced by multiplying this group by 2. In other words, this is a union of the cosets of a subgroup of G. Well, since it happened once, it must always happen, but let's prove it. So to begin with, if K is actually a subset of G, then the stabilizer of K includes, at the very least, the identity of G. Remember, the stabilizer must be a group. So let H be the stabilizer of our set K. Suppose A is in K. For G in H, since it's a stabilizer, G applied to K must be K itself. Now, remember the elements of K are elements of G, and so GA is a thing. And we know that must be an element of K. But now let's think about this. Remember, G is an element of a subgroup. It's the element of H. And so that means that GA is actually an element of HA, a right coset of H. And remember, when we introduced left and right cosets, we said that right cosets existed, but then we pretty much never did anything with them until now. So GA is an element of this right coset of H, and that means that the right coset HA is a subset of K. Now, if K was a group, we could directly appeal to Lagrange's theorem, but K isn't a group. So let's check it out. Suppose I have two of these cosets, HA and HB, and their intersection is non-empty. So that means H1A must be H2B for some elements H1 and H2 in a group. And because they're in a group, we know that H2 inverse exists, and so we can solve for B. Now let's consider some other element of HB. Well, we know that C must be H3 times B, but we know what B is, and so that means that C is an element of HA. And so the coset HB must be a subset of the coset HA, and we could make the same argument that the coset HA must be a subset of HB. So if these two cosets have a non-empty intersection, they must be identical. And consequently, we have the following theorem. Suppose G is a group and K is a subset of G. The cosets of the stabilizer of K form a partition of our subset. And remember, all of these cosets have the same size, and so consequently the order of the stabilizer is a divisor of the cardinality of our subset K. So let's take a look at our multiplicative group of integers mod 31 and again this random subset 15810, and let's find the stabilizer of K. So note that K has four elements, so the stabilizer of K must be a divisor, and so it must have one, two, or four elements. But remember, the stabilizer is also a subgroup, and since the stabilizer of K is a subgroup of Z mod 31, which has 30 elements, then its order is either one, two, three, or some number that's higher than four. And so the stabilizer of K is either the identity or a subgroup with two elements. So since the order of the stabilizer of K must be a divisor of the order of K, then the stabilizer of K must have one or two elements. But the only subgroup of Z31 with two elements is the subgroup generated by 30. But 30 is not an element of the stabilizer of K, we can verify that directly. And so the stabilizer of K is just the identity.