 It's often easier to find when two groups are not isomorphic, so let's consider a couple of cases. Suppose the order of G is not equal to the order of H. Is G isomorphic to H? And the answer is No, since there could be no bijective homomorphism from G to H. Remember an isomorphism has to be 1 to 1. But what if the order of G is the order of H? Then maybe there's an isomorphism. So we've already seen a couple of important properties of isomorphisms. Suppose F from G to H is an isomorphism. We saw that the order of A in G must be the same as the order of F of A in H. And also we saw that if A and B commute, then F of A and F of B must also commute. And this allows us to find when two groups are not isomorphic. So for example, let's consider the integers mod 6 under addition and the symmetric group with three elements. Now both of these groups have three elements, but we note that S3 is not a baleen. So there are elements sigma and tau in S3, where sigma tau is not tau sigma. But if we have an isomorphism from S3 into Z6, then we have F of sigma equals S and F of tau equals T. Now because F is an isomorphism, it's also a homomorphism, and remember the fundamental property of a homomorphism is that our function applied to a product is the product of the function values. But wait, there's more. Since our function values live in Z6, which is a baleen, then we can rearrange the order. So F of tau F of sigma is equal to F of sigma F of tau. But again, because F is an isomorphism, it's also a homomorphism, and that equality works in both directions. So if I have a product of function values, I can combine them into the function value of the product. But remember F is an isomorphism, so it's got to be one to one. So if F of a is equal to F of b, we must have a equal to b. And so here we have F of sigma tau equal to F of tau sigma, and so we must have sigma tau equal to tau sigma, which is a contradiction. And so the two groups cannot be isomorphic. And more generally what this says is that a non-a baleen group cannot be isomorphic to an a baleen group. So we also saw that if a in G has order n, then if F from G to H is an isomorphism, F of a in H must also have order n. For example, let's consider two groups of symmetries. The rotation symmetries of a square and the symmetries of a non-square rectangle. Let's find the two groups and decide if G is isomorphic to H. So for the square, our group elements are the identity, do nothing, a 90-degree counterclockwise rotation, we'll call that R, a 180-degree counterclockwise rotation, we'll call that R squared because that really is two 90-degree rotations put together. The 270-degree counterclockwise rotation are cubed. And so the rotation symmetries of a square can be described as E, R, R squared, R cubed. What about the rectangle? So here our group elements are the identity, do nothing. The reflection across a vertical line, we'll call that V. The reflection across a horizontal line, we'll call that H, and the 180-degree rotation, R. And so the symmetries of a non-square rectangle are the set of four elements, E, V, H, R. So while G and H have the same number of elements, G has an element R with order four. If you rotate four times, you'll get back to where you started. But every element of H has order two, and this means that G cannot be isomorphic to H. Now, Gauss once said that mathematicians should prove their results as quickly as possible using whatever method they need it, and then they should try to prove their results many more times trying to improve the proof and possibly extend the result. And Gauss proved the fundamental theorem of algebra four times and the law of quadratic reciprocity six times. Euler had a similar record with the Euler-Firma theorem giving four proofs over the course of his career. What this means is that any time we prove a result or solve a problem, it's worth going back to the problem and asking, how can we have done that differently? So taking another look at our two groups, we might note that the rotation symmetries of a square form an abelian group, while the symmetries of our rectangle are not abelian. And we know that an abelian group and a non-abelian group can't be isomorphic. And so G cannot be isomorphic to H.