 Lagrange's theorem guarantees that the order of a subgroup divides the order of a group, but can we always find a subgroup with a given order? Well, let's explore, because what's the worst that could happen? Since we need a normal subgroup, let's start with abellion groups so that all of our subgroups are normal. So suppose G is abellion with the order of G equal to PQ, where P and Q are both prime. What orders must be present? So let's keep in mind two important ideas. The powers of any element form a subgroup, and the order of a subgroup must divide the order of a group. So consider some element A of G. If A is a generator, then we know that A to power PQ must be the identity of the group, and because the exponents are counts, all of our standard rules apply. So that means A to power P raised to power Q must also be the identity. So A to power P generates a subgroup of order Q. Or does it? You should prove that this is actually the case. Now similarly, A to power Q will generate a subgroup of order P. So what this means is that if A is a generator, then G has subgroups of order P and Q. So suppose A is not a generator, but has order P. Then G mod the group generated by A has order Q. So let H be the subgroup generated by A. G mod H has to have order Q, and if B is not in H, then the coset BH in G mod H must have order Q. And so B to power Q must be an element of H. Which means that B to power Q is a power of A. And so we can say that for any B not in the subgroup generated by A, B to power Q is a power of A, and maybe it's the identity. If B to power Q is the identity, then B has order Q. Otherwise suppose B to power Q is A to power R for some R that's less than P. Since A has order P, we know that A to the R to the P must be the identity. So B to the Q to the P must be the identity. And again, exponents are counts, so we can switch those exponents. B to power P raised to the Q must be the identity. So the order of B to the P must be a divisor of Q. But since Q is prime, so B to power P has order Q. And what this means is that if we find a generator, then we can produce subgroups of order P and Q. And if we don't find a generator, we can still produce subgroups of order P and Q. And so if the order of G is a product of primes P and Q, then G itself must have subgroups of order P and Q. Now, since we assume nothing about G beyond its being a bellion and that P and Q were arbitrary primes, we prove the following. Suppose the order of a bellion group G is P, Q, a product of primes. Then G must have elements of order P and Q. But wait, we can go further. Notice we didn't even require that P and Q be distinct primes. But if they are, then we know that the group G has an element A of order P, and it also has an element B of order Q. And in a bellion group, if A has order P and B has order Q, where the greatest common divisor of P and Q is 1, then AB has order equal to the least common multiple of P and Q. So if P and Q are distinct primes, then their greatest common divisor is 1, and their least common multiple will be the product. So AB will have order P, Q. And consequently, if G is a bellion with order P, Q, a product of distinct primes, then G is cyclic. So for example, let's find subgroups of all possible orders of the multiplicative group of integers mod 23. So since 23 is prime, we note that the multiplicative group of integers has order 23 minus 1, that's 22, or 2 times 11, a product of distinct primes. And so it must have subgroups of order 2 and 11. So let's pick a random element, about 22. So we find the subgroup generated by 22, which will be... And since this has two elements, we'll let H, our normal subgroup, be the subgroup generated by 22. So Z23 mod H must have order 11. So again, we'll pick a random element, say 10H. So we find the coset 10H has order 11, but 10 to the 11th is congruent to 22, so 10 does not have order 11. But since 10 to the 11th must be an element of our subgroup H, and H has order 2, we know that 10 to the 11th to the 2nd must be congruent to 1. And again, since the exponents are counts, we can switch the order 10 to the 2nd to the 11th is still congruent to 1. And so 10 to the 2nd must have order 11. Now we can actually compute that we find 10 to the 2nd is congruent to 8, and so 8 has order 11. And since 22 has order 2 and 8 has order 11, then 22 times 8, 176 or 15, must have order 2 times 11, 22. And that means that 15 is a generator of our group. So we can do this with two primes, and so now the natural question to ask is, could we do this if we have a product of three or more primes? Well, let's find out.