 So, we found that if f from g to h is a homomorphism, then the image of g is a subgroup of h. Now, since f is a function, it could have other properties, so it might be 1 to 1. f of g1 is f of g2 if and only if g1 is equal to g2. Or, it might also be onto for any h in h. There is some g where f applied to g gives us h. And the fact that the function could be 1 to 1 and onto suggests we might be interested in the following. A homomorphism is an isomorphism if f is 1 to 1 and onto. Now, I like to use those terms 1 to 1 and onto, but it's worth remembering that a 1 to 1 and onto function is called a bijection or as an adjective bijective. And so, while my preference is to use the more descriptive phrases 1 to 1 and onto, you will hear people describe an isomorphism as a bijective homomorphism. Well, let's think about that. Suppose f mapping g to h is an isomorphism. Well, let's consider some of the properties of elements of g. So, for example, let a in g have order n. Then, well, we know that a to the n is the identity in g. And a to the x is not the identity for any x less than n. So, now suppose b is the image of a. Then we know that the nth power of that image, because it's a homomorphism, we know that that's f of a to the n, which is f of the identity in g. But remember, a homomorphism will map the identity in g to the identity in h. And so this f applied to the identity in g is the identity in h. And so that means that the order of f of a divides n. But if the order is k strictly less than n, then f of a to the k, again, because it's also a homomorphism, is f of a to the k. And so the identity in h is the image of a to the k. But since f is an isomorphism, it's one to one. And since we know the identity in g maps to the identity in h, then the pre-image of the identity in h, a to the k, must be the identity in g. But that would say that a has order less than n, which isn't true. And so f of a in h must also have order n. And so if we put this together, that means if f is an isomorphism that maps g to h, then if a in g has order n, then the image of a in h also has order n. How about some other properties? For example, suppose a and b are elements of g and ab is ba. In other words, a and b happen to commute. So we know that f of ab is f of a times f of b. That's because we're assuming that f is an isomorphism and therefore a homomorphism. We also have f of ba is f of b times f of a. But since ab is equal to ba, then we know that f of ab and f of ba must be the same thing. So f of a times f of b must be f of b times f of a. And so that tells us that if a and b commute, then the image of a and b will also commute. Well, what if they don't commute? Suppose a and b are in g and ab is not ba. Once again, because we do have a homomorphism, we know that f of ab is the product of f a times f b and f of ba is f b times f a. So what if these commute? Well, let me know that f of a f b is f b f a. And again, because we're dealing with a homomorphism, we know that f of ab must be f of ba. But since f is a bijection, it's one to one and on to, we know that f of ab equals f of ba requires that the arguments be the same. Ab must be ba, which isn't true. And so we know that if a and b and g don't commute, neither do the images of a and b in h. And the preceding lead to an important idea. If f from g to h is an isomorphism, then any properties of a in g are shared by the image of a in h. And it's worth keeping in mind, if it walks like a duck, swims like a duck, and quacks like a duck, it's a duck. In other words, an isomorphism just renames the elements of a group. And so this leads to the following definition. Suppose f from g to h is an isomorphism. We say that g is isomorphic to h and write it this way. And essentially the idea is that if two groups are isomorphic, they're really the same groups and all we've done is we've renamed the elements. And we'll take a look at some of the implications of this renaming next.