 So let's look at a new topic. It's not really new. We're just going to create something a bit more specific and we'll call it the kernel. The kernel of a homomorphism. So the kernel of a homomorphism, if our homomorphism was f that maps g to h, we call the kernel of f. If we usually write kernel of f this might be fine in your textbook, then that'll be fine. This is the kernel of the homomorphism and we define the kernel of the homomorphism. So the kernel of the homomorphism, let's stick to f here, that is all the elements g in g, g element of g, such that the f of g equals the identity element in h. So very easily all the elements here in g that maps to the identity element in this group. So some of them are going to map to it, some are going to map to other things there. Remember, we could reduce it always to a surject of homomorphism, but it's all these elements that that that under the homomorphism goes to the identity element. So a couple of things that we need to show, really, we need to show that this is a subgroup. To do that, we're going to have closure, we have to show closure, so closure, and one way to show closure is I'm talking about take any two elements a and b elements of g, so that the f of a, remember, so that is going to map to this and f of b, that is also this, so a maps to the identity element, b goes to the identity element, but what about that? So by definition of a homomorphism, this is really easy proof because it's just by whatever the, you know, by definition of a homomorphism. So this must hold true because this is a homomorphism, that's the identity element, that's the identity element, so this is the identity element. So for all a and b, if they individually map to that, the binary operation under g, remember g would have that binary operation, I'm just showing that they might have two different binary operations that that their binary operation would also map to the identity element, so that just really falls from from the definition of what we have here. The inverses, let's have a look, let's just clean the board and we can have a look at the inverses. Okay, so clean board, now we want to show the inverses, so we're going to let a, there exists an a element of g such that f of a equals the identity element in h, we already know that the identity element in g is the identity, maps to the identity element in h, we know that. The identity element in g we can write as, we can write as a times a inverse, a binary operation a inverse, that is that, so that's got to equal this. By definition though, this is nothing other than the f of a binary operation with the f of a inverse. This f of a, we know that that's this and we know that this is this is the inverse, we've already shown that the f of the inverse is f of a the inverse of that. This is still the identity element though, this is still the identity element though, and the only way, remember we've shown that before if I have x equals x binary operation x, the only way that that's going to be is if this is all that. So this really, this or if you want to write it, f of a is the whole inverse, that is just e of h as well. So that's just e of h, the identity element in h as well, I suppose we don't have to go through that step, but it shows to us that if a is in the kernel then its inverse must also be in the kernel because it also maps to the identity element in h and that really is just, that is our definition. The last thing that we're going to show is that this is a normal subgroup. This isn't actually this kernel, this kernel of the homomorphism is a normal subgroup. So remember what we say there, if we take an element g and g inverse and the elements of g, that I have this that f of g and h and g inverse, that is going to equal, that must also still be in e of n equal to the identity element in h, that is what I need to show. So, you know how do we write this if by the definition of a homomorphism, that's just going to be the f of g, binary operation f of h, binary operation f of g inverse. Now, so we're going to suggest that then this h must be, h must be an element of the kernel of f and I want to show back if I conjugate it, that it's still for any element of g and h that it still stays within that. So, what we have here is the f of g and we have the f of h which is actually the identity element that is in the kernel and we have the f of inverse and that is just the f of g, binary operation with the f of g inverse and that is the f of g, oh it's terrible and remember we've already shown that this would be g inverse and that is just the identity element in h, an element times its inverse, that's just there. So, under conjugation this stays in the kernel, it stays in the kernel, in other ways we've shown that this kernel is a normal subgroup of, a normal subgroup of g and it has this property, it's all the elements in g that map to the identity element in h, if it is my homomorphism from g to h. So, pretty easy stuff, this actually the kernel but it's going to give us some very deep insight that's going to lead us to the first theorem of isomorphisms.