 So here's an obligatory riddle. How many mathematicians does it take to screw in a light bulb? 1. They give it a 3-visigoth, thereby reducing it to the previous riddle. This riddle illustrates a few important points. First, it explains why I'm still teaching mathematics and don't have a successful career as a stand-up comic. But the other thing it illustrates is that mathematicians like to transform a problem into a problem they've already solved. So let's talk about Koshy's theorem again. We found that if G is a bellion with the order of G equal to PQ, a product of primes, then we found that G has subgroups of order P and Q, and that also G is cyclic. But what if the order of G is the product of 3 primes? Well, let's see what happens. Suppose the order of G is the product of 3 primes, including P. Can we create a subgroup of order P? So the important thing to remember is that any element generates a cyclic subgroup. So we can take any element at random and see what happens. Now if we choose an element A and it generates a subgroup with order P, we're done. Otherwise, G mod H has order PQ, a product of 2 primes, and we already know what to do in that case. G mod H has a subgroup of order P, but does this mean that G has a subgroup of order P? What we need is to show that an element of G mod H of order P corresponds to an element of G of order P. So suppose the coset BH in G mod H has order P, then we know that B to the P must be an element of H. Now if B to the P is the identity, we're done because then B generates a subgroup of order H. But if B to the P is not the identity, then since B to the P is an H, we know that B to the P to the K is the identity for some K that divides the order of H. And we can rearrange our exponents. So remember, the exponents are counts, so all of our standard rules of working with exponents still hold true. So B to the P to the K must be the identity, and we can swap the order of the exponents, B to the K to the P must be the identity. And so the order of B to the K must be a divisor of P, but P is a prime number, and so that means this must have an order of P. And so if B to the P is not the identity, there is some K for which B to power K has order P. So for example, let's try to find a subgroup of the multiplicative group of integers mod 31 with order 5. So remember, if you can get to your destination despite any roadblocks, you can always get to your destination. So every time we run across an easy way of finding that element of order 5, let's not take the easy way and see if we can still find an element of order 5. And what that means is we want to choose an element that doesn't have order 5 and can't be used to create an element of order 5. So we can begin by choosing an element at random, say 2, and we find the cyclic subgroup generated by 2 will be, which has order 5. So we want to throw up a roadblock here, so we don't choose 2. What about 3? We find that the cyclic subgroup generated by 3 is, in fact, the integers mod 31 under multiplication, which means that 3 is a generator. And since this means 3 to the 30th is congruent to 1, then 3 to the 6th to the 5th is congruent to 1, and 3 to the 6th will have order 5. And so that means 3 to the 6th or 16 has order 5. So here's a case where our choice allowed us to compute an element with order 5. So let's throw up a roadblock and don't choose 3. So, well, what if we choose 15? We find that 15 has order 10. Well, this means that 15 to power 10 is congruent to 1. And again, we can rearrange our exponents. 15 to the 2nd to the 5th is congruent to 1. So 15 to the 2nd has order 5. So 15 to the 2nd or 8 has order 5. And again, our initial choice allows us to compute an element of order 5. And so we don't choose 15. So we choose 25, and we find that 25 has order 3. And we can't use the power of 25 to create an order 5 element. So now let's consider the subgroup generated by 25. Since h has order 3, then our quotient group will have order 10, which is the product of two primes. So we're guaranteed it has elements of order 2 and 5. And we find that the coset for h has order 5. But since 4 to the 5th is congruent to 1, then 4 has order 5. And this would be the easy way out, so we don't use it. Instead, we find that 9h has order 5. And in this case, 9 to the 5th is congruent to 25 is not congruent to 1. So 9 does not have order 5. Now, since 9 to the 5th must be an element of h, which has order 3, then we know that 9 to the 5th to the 3rd is congruent to 1. Again, the exponents are counts, so we know that 9 to the 3rd to the 5th is congruent to 1. And so 9 to the 3rd has order 5. And so we find that 9 to the 3rd or 16 has order 5. And even though we did our best to avoid the order 5 element, there was no way why he could avoid it forever, and we found our subgroup of order 5. Now, if g is a Boolean with order equal to the product of 3 primes including p, then it has a subgroup of order p. And so the natural question to ask is, what if g's order is the product of 4 primes? We can approach it in the same way, but what we're proving is an infinite ordered list of statements. If g is a Boolean and the order of g is the product of 2 primes including p, then g has a subgroup of order p. If g is a Boolean and g has the product of 3 primes including p, then g has a subgroup of order p. If g is a Boolean and the order of g is the product of 4 primes including p, then g has a subgroup of order p, and so on. And this suggests we should look for a proof by induction. So let's take a look at that next.