 So it turns out that isomorphisms preserve more than just multiplication. We saw previously that if we have two groups G and H, we say they're isomorphic if they're some function phi, which will send G to H, it needs to be bijective to be an isomorphism. And it has to also be true that if we take phi times AB, this is equal to phi of A times phi of B, that the operation in the domain is preserved. That is products are mapped to products is so-called homomorphic property. So that's what one means by an isomorphism. So it turns out isomorphism, they also prove the following things are also true for isomorphisms that if we have an isomorphism between groups G and H, the order of G is equal to the order of H. That argument is actually fairly simple to see because every isomorphism is a bijection. It's bijective. And so the order of a group is just its cardinality as a set and every bijection preserves cardinality. So this is a medi consequence that we have a bijection between the two sets. In particular, the two groups will have to have the same order if they're isomorphic. An infinite group cannot be isomorphic to a finite group and things like that. All right. Next, it's also true that if G is abelian, that H is likewise abelian. If two groups are isomorphic, they're either both abelian or they're both non-abelian. And what's the argument there? Let's suppose that the domain G is abelian. Well, let's take two elements in H and we're gonna prove that they commute, right? So you take two elements in H. Well, since phi here is a isomorphism bijective by the surjective principle, there will exist elements X and Y inside of G so that phi of X equals A and phi of Y equals B. So there's something that maps onto these two elements. So now let's prove that the product A times B commutes. So what we're gonna do is we have this product inside of H. By the isomorphism, we're gonna pull it back to the group G, right? So I want you to kind of pay attention to this. This is a product right here, these elements, they're inside of H. But the elements X and Y are inside of G. So we basically were pulling back our product AB. We kind of like, whoop, we pull it back. We pull it back from some product X, Y over here. Well, okay, so you have A times B, this equals phi of X times phi of Y. But because this is a homomorphism by the homomorphic property, a product on the outside equals a product on the inside. So phi of X, phi of Y is equal to phi of X, Y. So this is what I mean here, X, Y. This is an element that moves inside of G. Now inside of G, we're a billion, right? So X, Y is equal to Y, X. So X, Y is equal to Y, X. We can commute inside of G. And then we're gonna send it back over, right? We're gonna send it back over. So then phi of Y, X becomes phi of Y and phi of X. But like we said, phi of Y was B, phi of X is A. So this is gonna go back to B times A. And basically we can force equality by going in this sort of like round robin exchange right here. So if the domain is abelian, then the co-domain has to be abelian as well. Now I want you to remember here that, well, not one, you should know how to spell abelian correctly. But more importantly, what I want you to remember is that if two groups have an isomorphism between it, the inverse of the isomorphism itself is an isomorphism. So there exists some inverse map, phi inverse that sends H to G here, like so, which would also be an isomorphism. So that means this statement's reversible. So if G is abelian, then H is abelian. But if H is abelian, then G is abelian as well. So we can actually improve this statement and say the following that we can improve this, that G is abelian if and only if H is abelian. Two groups, if they're isomorphic, they're either both abelian or they're both non-abelian. That's gonna have to be the case. And so similar proofs can be provided to prove statement C and D right here. If G is cyclic, then H is cyclic. This is left as an exercise to the viewer here. But basically the argument is we have to produce a generator for H, but G has a generator, right? G is equal to the cyclic subgroup generated by G. Well, look at phi of G and see what happens. You're gonna argue that H is now a cyclic group. If G has a subgroup of order H, let's excuse me, a subgroup of order N, let's say that group is, we already have an H, let's call it K, then H will likewise have a subgroup of order N. Well, how do we produce it? Basically we look at phi of K, right? And we have to argue that's a subgroup of H whose order is gonna be N still. All right, so I'm gonna let the viewer here finish up the details of that. But like I mentioned over here with property B, this principle is actually an if and only if statement. Property C here, we can actually restate it as G is cyclic if and only if H is cyclic. And likewise, G has a subgroup of order N if and only if H has a subgroup of order N. And then you can push that a little bit further using cyclic groups that G has an element of order N if and only if H has an element of order N. And you can actually start counting them like G will have say A mini elements of order N if and only if H has A mini elements of order N. The number of elements of order N have to be the same. The number of subgroups of order N have to be the same because again, essentially other than the cosmetic differences between the two groups, the two groups are gonna be the same and all these properties are gonna be the same about them. And like I said, these some of these additional properties I mentioned I will provide with, I'll state with no proof. I think it's a great exercise for the viewer to complete the proofs of those statements.