 So we have the fundamental theorem of finite abelian groups. Essentially, that any finite abelian group can be decomposed into a direct product of the integers mod n under addition. So let's see if we can decompose some groups. So let's set up with z31 under multiplication. Since z31 under multiplication, 31 is prime, so this has 31 minus 1, or 30, or 2, times 3, times 5 elements. Now remember, if g is a cyclic group with order p, q, where p and q are relatively prime, then g is isomorphic to zp cross zq. And so that means z31 under multiplication, which is cyclic with 30 elements, is going to be isomorphic to z2 cross z3 cross z5. So if I want to express the integers mod 1800 multiplication, well, since 18 is prime, then our group has 18 minus 117 elements. Wait, 18 is composite. Well, since 18 is composite, the integers mod 18 under multiplication will have phi of 18 elements, where we can calculate phi of 18 to be 6. And so the integers mod 18 under multiplication are isomorphic to z2 cross z3. How about the integers mod 16 under multiplication? Since 16 is composite, then the integers mod 16 under multiplication will have phi of 16 elements, which will be, now, since 8 is 2 to the 3rd, that means our integers under multiplication mod 16 could be isomorphic to one of the groups z8, z4 cross z2, or z2 cross z2 cross z2. So to decide, let's look for a feature of the integers mod 16 under multiplication that is not present in one of these products. And the easiest feature to compare is the order of an element. z8 has an element of order 8. The highest possible order of an element in z4 cross z2 is 4. And every non-identity element in z2 cross z2 cross z2 has order 2. So we can choose a random element and find its order. For example, let's pick the element 3. So the powers of 3 mod 16 are, this means we have an order 4 element, and so the integers mod 16 under multiplication can't be isomorphic to z2 cross z2 cross z2. This does mean we have to decide between z4 cross z2 and z8. And if we try other elements of our group, we find no element of the integers mod 16 under multiplication has order 8. So we can't be isomorphic to z8, and so we must be isomorphic to z4 cross z2. So sure, it's easy to prove there's an isomorphism, but what is that isomorphism really? So let's try and find an isomorphism. And since isomorphisms are bijections, it really doesn't matter which group we start with for a variety of reasons. It's easier to go from z4 cross z2 into the integers mod 16 under multiplication. So there are 8 elements of the integers mod 16 under multiplication, and they are. And to begin the construction of the isomorphism, remember that if a cross b is isomorphic to g, then the isomorphism has to map identities to identities. And so the identity in the integers mod 16 under multiplication must be the image of 00. Next, remember that isomorphisms must map elements of order n to elements of order n. So we found that 3 has order 4 in our group of integers mod 16 under multiplication, and the order 4 elements in z4 cross z2 are... And so the question is, which one gets mapped to 3? And the answer is, let's try it and see what happens. So suppose phi of 1, 0 gets mapped to 3. Then the powers of 1, 0 in z4 cross z2 must map to the powers of 3 in our group of integers mod 16 under multiplication. But remember the powers in z4 cross z2 are actually repeated sums. So these powers are... And these have to be assigned to the powers of 3, which are 3, 9, 11, and 1. We also notice that 15 has order 2, so it must be the image for an order 2 element in z4 cross z2. And these elements are... So picking one of these at random, how about 01, let's assign 01 to 15. Since we want phi to be a homomorphism, then we must preserve the binary operation. So we have phi of 01 equals 15, phi of 1, 0 equals 3, and if phi of something is 5, then we must have, since 15 is 3 times 5, phi of 01 equals phi of 1, 0 times phi of x, y. And since we have a homomorphism, we can combine the arguments. And since we want this to be a 1 to 1 function, then the argument 1 plus x, 0 plus y, must be the same as 01. And so comparing the terms component-wise, we have... And in fact, there is a unique solution, x equals 3, y equals 1, which means that phi must end 3, 1, to 5. And continuing, we might find phi of 3, 1, phi of 3, 1, phi of 3, 1. Well, that's really the same as phi of 9, 3, or phi of 1, 1. Over on the left-hand side, that's 5 by 5 by 5, or 13, and that gives us a new assignment. And at this point, 7 is the only element that doesn't have a preimage, and this leaves the last element, 2, 1, must be mapped to 7, which completes our isomorphism.