 So far we've found that if G is a Boolean and the order of G is the product of two or three primes including P, then G has a subgroup of order P. Since there's an infinite but ordered list of cases, we can try an induction proof of Koji's theorem for finite a Boolean groups, let G be a Boolean group, then for any prime P that divides the order of G, there is a subgroup of G of order P. Now remember that any element generates a subgroup and so it will be sufficient to show that there is an element of order P because that will give us our subgroup of order P. We've proven the result if the order of G is a product of two or three primes, so let's use our standard induction step. Let's assume the result holds if the order of G is a product of K primes. So let G be a group whose order is a product of K plus one primes including P. So remember any element generates a subgroup and so if A in G does not generate a subgroup H of order P, then the order of G mod H is a product of K primes or E is it? Well, let's think about this. Suppose the order of G is the product of four primes P, Q, R, S and an element A could produce a subgroup H with order Q, R and quotient group G mod H with order P, S. However, our induction assumption only tells us what happens if the order is a product of three primes. It doesn't tell us what happens in these two cases. This means we need to use strong induction. So let's assume the result holds if the order of G is a product of K or fewer primes. So let the order of G be a product of K plus one primes and let A in G generate a subgroup H. So both the order of H and the order of G mod H will be a product of K or fewer primes. So our induction assumption does tell us something about these cases. Now if P divides the order of H, then since the order of H is the product of K or fewer primes, our induction assumption tells us that there is an element in H of order P. On the other hand, if P divides the order of G mod H, then since the order of G mod H is the product of K or fewer primes, our induction assumption tells us there is an element BH of order P. So there's some B not in H, where B to power P is an element of our subgroup H. Now it's tempting to proceed as follows. Let the order of H be N. If B to the P is the identity, then B generates our subgroup of order P. If B to the P is not the identity, then since the order of H is N, we know that B to the P to the N must be the identity, and so we can swap our exponents, so B to the N has order P. Or does it? And the problem is, we don't know the order of B. And in particular, if B to the N is the identity, then we haven't found an element of order P. Well let's find out. Suppose B to the P is an element of H, and the order of B is Q. So B to the Q is the identity. Then we know that B to power Px plus Qy is also an element of H for any integers x and y. Now by Bayes-Zou's theorem, there are integers x and y, where Px plus Qy is the greatest common divisor of P and Q. But since P is prime, the greatest common divisor is either 1 or P. But if it's 1, then B to the 1 is B, which is an element of H, but we assume that wasn't the case. And so this means the greatest common divisor of P and Q is P, which means that P divides Q. So B to the PR is the identity. So B to the R to the P is the identity. And so that means the order of B is a multiple of P, so for some R, B to the R has order P. So for example, let's prove that the multiplicative group of integers mod 79 is cyclic. So we note that 79 is prime, so the number of elements in this multiplicative group is V of 79, that's 79 minus 1, 78, or 2 times 3 times 13. Now by Cauchy's theorem, there is an element A of order 2, an element B of order 3, and an element C of order 13. And since 2, 3, and 13 are relatively prime, then the product A, B, C has order 2 times 3 times 13, 78, which is the size of our group. And so the multiplicative group of integers mod 79 is generated by this element A, B, C. And at this point, we'll point out something useful. Notice that our proof that this multiplicative group of integers showed that A, B, C, and the product A, B, C existed, but they didn't tell us how to find them. So we say that the result that this group is cyclic is an existence theorem with a non-constructive proof. We know that it has a generator A, B, C, but the proof doesn't tell you how to find it. What we'd like is an algorithm, a sequence of steps that is guaranteed to produce a desired object. For that, we need a constructive proof. Let's take a look at that next.