 So I like to go very fast with these videos. I hate it when I learn something online And I start the playlist or I start the video and I see it's in a 40 minutes long Difficult, so I like to keep things short, but sometimes you have to slow down a little bit And so what I want to do here is to just slow it down so we can just look again at homomorphisms We've looked at homomorphisms isomorphisms before but we didn't really have enough to build to where we really want to go And that is to look at the theorems of isomorphisms to get a very deep understanding of what is really going on here at the Moment we still set with that problem. I mentioned right at the beginning you have this box and you have to fill up Keep on filling up this box. You don't know where these things fit in you don't really know why but That's the thing about abstract algebra. You've got to build these You've got to build all this knowledge put things in the box put things in the box Pull them out when you need it and two things come together into something greater But that goes in the box again until you can pull it out and that makes it very difficult because sometimes to know this You first have to learn this but to learn this you need it to know something about it goes all over the show Anyway, let's just make sure that we understand slow down a little bit and understand what what is happening here I have a group G it consists of a set of elements and a binary operation I have a group H and it consists of a set of perhaps different elements and perhaps a different binary operation and I've got this homomorphism F and that maps G to H Many textbooks you're gonna see it's Phi and it maps a G to G prime All sorts of things it's all over the show textbooks lectures all over the show Doesn't matter what is written on the board or in the book. It's a mapping of one group to another group Maintains the structure this hope this homomorphism maintains the structure Now what we're interested in this isomorphisms and we know for an isomorphism We have to have that this mapping is bijective. So it's got to be injective and surjective Now we want to stick to and you'll see most books It's going to stick to the fact that we only need to look initially at surjective homomorphisms And it's almost we can state that any homomorphism we have we can reduce it to a surjective Homomorphism so doesn't matter what the homomorphism is We can we can reduce it to a surjective one so that we only need to look at it being injective before we say It's an isomorphism so can we really prove that we can do this because remember if I had h then and it has a lot of elements G then and it has elements here What we can say is that for for surjective more than one of the elements here can map to the same element The same element there and it might very well be that some of these are not mapped to Imagine if I have a function or old-fashioned function e you know old-fashioned meaning You know what you learned at school or the beginning of College of University e to the power x. I'm mapping r So f of x equals mapping r to r real numbers to real numbers This is the domain of all values x and this is the co-domain all the real values there on the y-axis But I'm not using up all of them. I'm only using up the positive ones here And that is called the image of this co-domain So I have a domain I have a co-domain, but only the ones that be mapped to and I'm suggesting that whenever I'm You know busy with this Homomorphism if you're between these two groups that I can discard all of these and I'm and I only need to consider the image It's called the image of f So of this whole the set this is the set of elements that make up H I only need to consider those elements that are that make up the image of F. So the image of F Really is just the set of all elements the f of G Such that G is an element of G. So there might be elements here in H You know that is not mapped to by an element in G and I'm suggesting I can discard them So the only thing I'm left with is this Surjective homomorphism and no matter what H was before I'm claiming that I can always do that How do I prove that the first thing that I really need to do? I've got to show that F maps identity element in G to the identity element in H Because in that image, I've got to have the identity element. Otherwise, this doesn't make up a group anymore So let's just let's just have a look at how we can prove that The way that we go about it that you'll see in most book take X which an element of G and state It has the following property that if I take the binary operation Between X in itself and I land up with X again The only way that that can be is let's write binary operation with the inverse So I'm gonna have X binary operation X inverse which just leaves me with the identity that is going to be This and X has got to be the identity element in this case in G. So let's just remember that fact Let's just remember that fact. So let's take the F of Let's take the identity element in G Binary operation the identity element in G and by definition of homomorphisms that is going to be the F of the identity element of G binary operation different binary operation now if of that Okay In group G binary operation between the two identity the identity element with itself. Well, that's this F of the identity element in G Okay Looks in G looks a lot like that. Let's just rewrite that and we put this first and then this second So that's binary operation equals Okay, so I've just swung it around and what do I what do I have here something that looks very similar to this The only way that this can be so if I then let me do the full thing for you on the board. So I've got this I've got this I take the binary operation with the inverse of this and that equals okay These two if I take this element and it's inverse that leaves me with identity element So I'm left left with you with this Equals and I have this is an element of H. This is an element of H It's inverse of this one that leaves me with the identity element in H And that shows that this homomorphism E of G is going to be here. It is always going to map to Identity element in H. So I'm suggesting here that under the image now this G is This it is going to map to that. So it's always going to be in the image of F So we know that identity element in H is always going to is always going to be there I'm going to clean the board because the next one that I want to show you is this That remember a is mapped to a f of a a inverse that's mapped to the f of a inverse Now curious factors these two and G the inverses of each other are these two inverses of each other In other words is the f of a inverse is that equal to the f of a inverse That's what I've got to show now. So remember a is also an element of G and it has been mapped to something so that is going to be in the image and This is in G. So it's also going to be mapped to so it's also in the image This is also element of the image of F and That is so that would mean that all the inverses Really all the inverses, you know, so for every element f of a in here It's and it's inverse is also there and the only way that I can show that is if these two are really equal to each other So let's prove that Okay, now we need to show these inverses now So if I have the f of a and I have the f of a inverse That is by definition This and that is nothing other than the f of the identity element in G and that is nothing we've just shown to be the identity element and the only way that The binary operation between these two can give me the identity element if this one and this this one are inverses of each other In other words this one and it's inverse Equals that in other words. This is exactly this is exactly the same thing so I could just left binary operation here with with F is inverse with F is inverse and Let's do that. So I have F a it's inverse binary operation with F of a and binary operation with the F of a inverse So they and that has got to be the F of a Inverse binary operation with E of H So I've just taken that first one and on this side The inverse of that the inverse of that and it that gives me the identity element of H So on this side I'm just left with F a inverse and that has got to equal the binary operation with this and its identity element is just this F of a inverse This is an element of G. This is an element of G. They both be mapped to the image So what we really have now is we we have actually shown that this image of F This image of F is a subgroup of H because we've shown that it will have the identity element We've shown that all the inverses will be there So this is a subset. Remember, we just took this as a subset of Of each so we are going to inherit The associativity property there and the only thing that we really still need to show that this is a group in itself Is if there is closure So if I have two elements H and H this element of the image of F I want to show that the binary operation between these two So these are two different elements there must be an element of the image of F I've still got to show that I've got to show that that's so let's just take elements G and G star Just arbitrary elements of G. So I am going to have the F of G and let that be H I'm in going to have the F of G Star let's make that this element Quite legal to do these are arbitrary elements and if I take the binary operation between these two So I'm going to take the F of G at the binary operation and the F of G star Remember that's going to equal the F of G binary operation G star That's in G. This is nothing other than H This is nothing other than H star And this is F of G and G star But remember G is a group so the binary operation between two elements of that is still an element of G So I am mapping an element of G An element of G to something and in other words, and that's how I define the image So these two must be an element of the image of F Because that exists in G The mapping is mapping to something inside of the image of F one of the elements This is that so the binary operation with these two must also be there So we've shown that we can really take any isomorphism And we can reduce it to a surjective homomorphism So we needn't worry about that at all. You know just reduce it. It's always reducible I've shown you that it is reducible just reduce it to you know If need be throw some of the elements of HOA so that you only have a surjective homomorphism Now we only need to show that it is injective if we want to show that this homomorphism is indeed an isomorphism a specific kind of homomorphism called an isomorphism Because we can always just Just reduce a homomorphism to a surjective homomorphism. We don't have to worry about that anymore All the all these mappings that we see that maintain structure now if or phi or whatever It's named in your textbook or by your lecture We can always just reduce it to a surjective no issues there and now we can move forward and just they now test for You know which ones really are injective as well