 So I want us to cement this understanding of homomorphism, so let's today do a few examples. I'm given two groups, G and H. This is the set of integers under the binary operation of addition, and group H is the set of the roots of 1, which is 1, negative 1, i, the imaginary number, and minus i, under the binary operation of multiplication. My function maps G to H such that any element x is mapped to i to the power x. So any element x and g is mapped to an element i to the power x in H. Is this a homomorphism? So let's consider, for a moment, let's consider a, b and c elements of G such that a plus b equals c. I know that it's a group, so there must be a closure. So if a plus b plus c is an element of G, then under closure this must hold. My binary operation is addition, so there's no problem there. Let's map f of a, well that's going to be i to the power a, f of b, well that's going to be mapped to i to the power b, and f of c is going to be mapped to i to the power c. And this is going to be for all if a, fb, fc, or i to the power a, i to the power b, i to the power b elements of H. The binary operation is multiplication, so what we're really saying here is i to the power a times i to the power b is going to equal i to the power c. i to the power c, because there must be closure, and this is the way I set it up. And under the properties of, properties of, of exponents, so exponents, powers here, i to the power a times i to the power b equals i to the power a plus b. But I know what a plus b is, that c, that equals i to the power c. I have i to the power c, i to the power c, I have proven that under f, or f at least is a homomorphism for these groups, g and h, if f is this mapping. Clean the board, let's do another example. So let's do this very good example. I'm given two groups, g and h. G is the set of all real numbers, the binary operation is addition, h is the set of all elements z, and they are elements of the set of complex numbers, such that the norm, the length of that complex number equals one. So only ones that would have a norm of one. Under the binary operation of multiplication, and we told that f maps g to h, such that every element x and g is mapped to e to the power x. So e to the power x will then be an element of h. First things though, let's not take this for granted. Let's prove that h is indeed, is h a group? Let's answer that question. Let's just look at the closure. So what I'm suggesting here is that z, there's one z and another z, and if I multiply them by each other, I have got to have another z, let's make that k, and that z must also be an element of the group. So I've got to show that this, for any two of these z's, z i and z sub j, if I multiply them, it's under multiplication. The resultant element must also be, now just let's remember the laws of the multiplication of if I multiply two complex numbers, multiply two complex numbers. If I multiply two complex numbers, what happens to the argument, the principal argument? The principal argument is added, and what happens to the norms? Well with the norms, they are just multiplied by each other. They are multiplied by each other. So if I multiply two of these, I can do, you know, if I have a plus b i and c plus d i, I can do that multiplication of the two terms there, but what I could also do is work out the argument and the norm, and I can just add the two arguments, the one vector and the argon plane plus the other vector, I'll add those two to get another one, and the length of that vector then will be the multiplication of the norm. But I know the norm of this is one, I know the norm of that is one, so one times one is one, which means the norm of this one is also going to be, sorry, one, so that it is a part of that set. So the closure property really does hold, and if you think about the associative property, it doesn't matter if I multiply two and then one or one and then two, it doesn't matter what I do there, the norms are always going to be one, so it's still going to be the same. So I still, associativity is going to hold, is there an identity element? Yeah, of course there is, if I write z equals a plus b i, such that a equals one and b equals zero, I have that z equals one, the norm of one is one, one times any of these z's is going to give me that same z back, z times one is going to give me that z back, so I have this identity element, one plus zero i, there's my identity element, and is there a unique inverse for any of these? Well, remember how we do the norm, the norm of z is going to be a square plus b squared, square root of that, and that's going to equal one, well, you know, if I square that, that's still one, so under our special circumstance here, b squared, a squared plus b squared equals one, so I'm squaring the a and I'm squaring the b, so they'll both be positive. Now think about this, think about the complex conjugate of any element, that's going to be a plus b i and a minus b i, and if I multiply that I get a squared plus b squared, and that equals just one, because a squared plus b squared must be one for these, because I'm taking this to be both of these to be elements, and what I have here is that every element, and remember that is my identity element here, every element has its unique identity and that is going to be this complex conjugate of that, so I have proven that this is indeed a group, so let's have a comma b comma c, you know how this is going to go now, those are all elements of g such that a plus b equals c, it all holds because that is my binary operation, the f of a is going to be e to the power a i, f of b, that's going to be e to the power i b, and f of c is going to be e to the power i c, and remember this is multiplication, so I'm going to have that e to the power a i times e to the power i b, it's going to equal e to the power i c, and I know this because I am told that it is a group, I've shown that it is a group, but let's just say I don't have to do that, I'm told that it is a group, so I know this must hold, no issues there, and by the properties of exponents or powers here, we have e to the power a a plus b equals e to the power i c, but a plus b equals c, so e to the i c equals e to the i c, that is not a contradiction that holds, in other words, f is a homomorphism for these two groups, clean the board, one more example, so here's our last example, I have a group onto itself, so I have g and h, that's exactly the same thing, my set is the set of real or real numbers, my binary operation is addition, and I'm mapping one, I'm actually mapping here the group to itself, and such that any element in x is mapped to x squared, so is this a homomorphism, does it maintain this structure of the group mapped to itself, so let's let a, b and c be elements of g, such that a is under addition plus b equals c, it is a group, there is closure, so that is all going to help, all going to hold, so let's check what the f of a is, well that's going to be a squared, the f of b is going to be b squared, and f of c is going to be c squared, so such that this is also under addition, so I have to have that a squared plus b squared is going to equal c squared, and you know under the Pythagorean theorem what's going on here, and you know this is not going to hold, because c is a plus b, so that equals a plus b squared, and that is definitely a squared plus b squared, but there's this 2ab term here, and a squared plus b squared is not equal to that, that's a contradiction, that's a contradiction, therefore f is not a homomorphism, so there we have an example of a mapping from a set to itself that is definitely not a homomorphism.