 The easiest groups to work with are abelian, and the simplest abelian groups are cyclic. And the integers mod n under addition is cyclic with order n, so it's isomorphic to any cyclic group with order n. So let's consider direct products of the group of integers mod n under addition. And one quick note, if we are using addition, we typically designate the group of integers mod n under addition as Zn without specifying plus. So let's consider G, Z2 cross Z2, and H, Z4 are these two groups isomorphic. So the elements of Z2 are 0 and 1, and so G, the direct product of Z2 with itself, will have elements. Meanwhile, the elements of C4, the integers mod 4 under addition, will be, and we note that both of these groups are abelian, both have four elements. But every non-identity element of G has order 2, while H has an order 4 element, and so G is not isomorphic to H. Which we should have suspected since we know there are two abelian groups of order 4. How about Z2 cross Z3 and Z6? Now since both groups are cyclic with six elements, then Z2 cross Z3 is isomorphic to Z6. The fact that Z6 is isomorphic to Z2 cross Z3 illustrates an important result. Let G be a cyclic group with order P, Q, where P and Q are relatively prime, then G is going to be isomorphic to ZP cross ZQ. Add the proof this is part of your homework. And to suggest the following, suppose H is abelian with the order of H equal to N, since every whole number N can be expressed as a product of distinct primes, might H be isomorphic to a corresponding direct product of the integers mod P to power N. And the answer is no. In fact, we've already seen that Z4 is not isomorphic to Z2 cross Z2, since Z4 has an element of order 4, and every non-identity element of Z2 cross Z2 has order 2. While we can't guarantee we can express H as a direct product this way, we can still make use of the fact that it might be possible. For example, let's try to find two groups of order 12 and prove that they're actually different. So we note that 12 is 2 squared times 3, so one group of order 12 is Z4 cross Z3. And since 4 and 3 are relatively prime, then 1, 1 in G has order 4 times 3 or 12. However, in Z2 cross Z2 cross Z3, no element has order greater than 6, and so G is not isomorphic to H. And so here we have two distinct groups, both of order 12. As our example shows, there may be more than one abelian group of order K depending on how we split the factors. But in fact, the fundamental theorem of finite abelian groups. With H being an abelian group with order N, there exist not necessarily distinct prime factors P1, P2, and so on, and exponents N1, N1, and so on, where N is the product of these primes to their powers. And H is isomorphic to the direct product of the groups of integers mod P to the power N. Now the statement is a little bit daunting, so somewhat less formally. If H is a finite abelian group, then H is isomorphic to some direct product for some numbers A, B, C, and so on, where the order of H is the product of the moduli. So we might consider how many distinct abelian groups of order 12 exist. So again, since 12 is 2 squared times 3, then we can form an abelian group by considering different partitions of the factors. So we might lump all the factors into one group, C12, or we might factor 12 as 6 times 2, giving us one direct product, or we could factor 12 as 4 times 3, giving us another direct product, or we could factor 12 as 2 times 2 times 3, giving us another direct product. And so there are 4 abelian groups of order 12. Or are there? So remember that if P and Q are relatively prime, ZP cross ZQ is isomorphic to ZPQ. Since 4 and 3 are relatively prime, Z4 cross Z3 is isomorphic to Z12. And so G1 is isomorphic to G3. So since 2 and 3 are relatively prime, then Z2 cross Z3 is isomorphic to Z6, which means Z2 cross Z2 cross Z3 is isomorphic to Z2 cross Z6. But Z2 cross Z6 is isomorphic to Z6 cross Z2. And so Z2 cross Z2 cross Z3 is isomorphic to Z6 cross Z2. And so G2 is isomorphic to G4. So this means there are only two distinct groups, G1, the integers mod 12 under addition, and G2, Z6 cross Z2.