 So suppose H is a normal subgroup of G. We saw that we could define coset multiplication A H B H equals A B H. Consequently, the cosets of H have an associative binary operation with an identity and an inverse. And this motivates the following definition. Let G be a group and H a normal subgroup. The quotient group G mod H is the group of cosets of H under coset multiplication. So for example, suppose G is a group with 30 elements and H is a subgroup with 5 elements. Let's find the order of G mod H, and let's consider what are the possible orders of an element of G mod H. Since H has 5 elements and G has 30, then there must be 30 divided by 5 or 6 distinct cosets. Because remember, given a subgroup H of G, every element of G is in exactly one of the cosets of H, and every coset of H has the same number of elements. And so that means that the quotient group G mod H has order 6, and so an element A H of G mod H must have order 1, 2, 3, or 6, what are the divisors of the order of the group? Now Lagrange's theorem says that if a subgroup exists, then its order must be a divisor of the order of the group. But what about the converse? If a number is a divisor of the order of a group, then a subgroup of that order exists? Well, maybe. So here's an idea that we might consider in the investigation to follow. If you can reach your destination despite any roadblocks, you can reach your destination. So let's see if we can find a subgroup of the multiplicative group of integers mod 11 of order 5. So remember, every element of a group generates a subgroup, and the order of a subgroup is a divisor of the order of the group. So to begin with, we note that the multiplicative group of integers mod 11 has 11 minus 1, 10 elements. So its subgroups could have order 1, 2, 5, or 10. So we can find a subgroup by considering powers of any element. Well, let's consider the powers of 3 mod 11. And we find these are, and we're done. We found a subgroup of order 5. And so if we pick an element of order 5, we're done. But we don't want to really rely on being so lucky as to have picked an element of order 5. Remember, if you can reach your destination despite any roadblocks, you can reach your destination. So let's put a roadblock. Let's suppose we don't pick an element of order 5. So again, we can form a subgroup by considering powers of any element. So let's consider the powers of 2 mod 11. And these are, and we see that the powers of 2 actually generate all of the elements of the group. Now, since 2 to the 10th is congruent to 1, then 2 to the 2nd to the 5th is also congruent to 1. And so that means 2 squared 4 has order 5. And so if we pick a generator, we're done. We can find that element of order 5. And correspondingly, the subgroup of order 5. But again, if you can reach your destination despite any roadblocks, you can reach your destination. So suppose we don't pick a generator. Let's close off that possibility. So again, if we pick an element of order 5 or a generator, we're done. So suppose we pick neither and generate a subgroup H. Well, the only other possible order is 2. And so we find that the element 10 generates a subgroup, 1, 10, until it has order 2. And since the subgroup has order 2, that means there have to be 5 cosets. And so our quotient group has to have order 5. And so that means any non-identity element of our quotient group must also have order 5. So A H in G mod H must have order 5. And so we'll pick a random coset, about 4, 10. And we find that 4, 10 has order 5. But 4 itself has order 5, and so we don't want it. Because stop me if you've heard this. If you can reach your destination despite any roadblocks, you can reach your destination. So in this case when we found our coset of order 5, it turned out that the element that produced that coset also has order 5. So let's throw up a roadblock that keeps us from taking that path. So we find a different coset, about 2, 10. And we find that 2, 10 has order 5, but 2 itself does not have order 5. 2 to the 5th is not congruent to 1. But now let's think about this. 2H to the 5th is 2 to the 5th H, which is H itself, because again 2H has order 5. And so that means 2 to the 5th is an element of H. And that's because A H is equal to H, if and only if A itself is in H. So we'll give it a name, and in a spate of creativity, we'll say that 2 to the 5th is, how about H? And we can actually compute it, it turns out to be 10 in fact. But wait, there's more. Since the subgroup H has two elements, then every element must have order 2. And so 2 to the 5th to the 2nd, well that's really the same as 2 to the 2nd to the 5th, must be the identity element. And so we've overcome all of our roadblocks. Since A H has to have order 5, then if A itself has order 5, we're done. So suppose we don't pick an element of order 5, say 2. Then 2 to the 5th is an element of H, and 2 to the 5th to the 2nd is congruent to 1. But because the exponents are counts, we can rearrange them in our usual way. 2 to the 5th to the 2nd is the same as 2 to the 2nd to the 5th. And so 2 to the 2nd or 4 is our element with order 5. So even though we close off every possible road to an element of order 5, we still found one. And so the question is, can we do this in general? Well let's find out.