 In order for G to be isomorphic to H, the cardinalities of the two groups have to be the same. G and H are either both abelian, or both non-abelian, and if G has element of order K, H does as well, and conversely. So, let's start by seeing if we can classify abelian groups. The simplest abelian groups are cyclic groups, so let's start with them. Suppose G has generator A and H has generator B. We can find an isomorphism, F from G to H, where F of A is equal to B. And in general, if G and H are cyclic groups of the same size, they're isomorphic, with one isomorphism, F from G to H, mapping a generator of G to a generator of H. Now, if G is abelian with order equal to a product of distinct primes, it's cyclic, that's something we proved earlier. And consequently, if G and H are abelian groups, where the order of G is the order of H, and the order is a product of distinct primes, then G is isomorphic to H. And so we now know everything there is to know about all cyclic groups. So the question you've got to ask yourself is, could we have an abelian group that is not cyclic? This would require its order include a repeated prime factor. So let's consider the smallest possible group. Let's consider groups G and H with four elements. So let G and H be groups with four elements. Will G be isomorphic to H? Now, since four is equal to two times two, then Cayley's theorem guarantees there is an element of order two. And so the question we might ask is, could all elements have order two? So there's four elements, but the group must have an identity E. So let's throw down an element A with order two, and so we can complete a row and column of our Cayley table. We also have another element, and we'll assume that also has order two. So we can complete the row and column that has the identity, and B times B is also the identity. Now we have to figure out what's ab. Now ab can't be the identity since A and B are already their own inverses. So ab has to be equal to C, another element, and you'll notice that's actually the fourth element of the group. So we fill out the row with the identity and the column. Now, while we've figured out what ab is, what's ba? Well, ba can't be E since A and B are their own inverses. Ba can't be B since A would be the identity, and similarly ba can't be A, and so that means ba must be C, the fourth element. What's that? Question from the back? Isn't the group a billion? Oh, yeah, that would make this a lot easier, wouldn't it? Well, remember Gauss's comment, we should try to prove things any way that we can, and then later on find another proof. In this case, since we're assuming G is abelian, we'd also know that ab is equal to ba, which is equal to C. And since G is a group, its Cayley table will satisfy the Latin square property. So we can complete the Cayley table, and we see that a squared equals b squared equals c squared equals the identity, so every element has order two. Now, we also know there's a cyclic group of order four, namely the integers mod four under addition. Since this group is cyclic, this group has an element of order four, and so this group G cannot be isomorphic to H. And so there are at least two distinct abelian groups of order four. And in fact, there are only two ways we can complete a Cayley table on four elements, so there are only two groups of order four. And in fact, if you go back to the way we computed ba, that also takes care of the case if our group is not assumed to be abelian, we find out that it must in fact be abelian at that point. Which is a good example of why Gauss made the suggestion try to prove things as many ways as you can, because sometimes you find things that you weren't expecting. In this case, all groups of order four must be abelian. Now, we can make similar arguments for abelian groups of order, say, 12, which is two to the second times three, but how do we know such a group can be non-cyclic? We'll need to introduce a new idea.