 So we spend an enormous amount of time on cosets so we better put this knowledge of ours cosets in the ground just set him to some use and the first place that we're going to use it is to develop our understanding of what is called the quotient group the quotient group So what I want you to imagine is that we have a group G now it contains some set and some Group operation and I'm going to have a normal subgroup now Just to confuse things sometimes in textbooks you'll see they change from H Which we had in the discussion before to N So we now going to have this special subgroup called N of G and we want N You know we don't want it to be one of the trivial cases So that we don't want the cardinality of G the set that makes up G to be a prime number we know what that is all about now and So so that we can make this normal subgroup and remember the normal subgroup is is Stable under conjugation. What does that mean? Well if I have G element of G and I have N element of N I have this fact that G composed with N composed with G inverse still an element of N Stable under conjugation And we know we can also have these the other way around because it'll just be the inverse of the inverse So that is what I want us to start with As this as this very normal normal subgroup now we want to To have N now the reason why we do this at least is is the following is That we now have the fact that our left cosets in our right cosets are equal You know such that this N is an element of N and this is to find this A o N And this is to find this in a binary operation in an N binary operation with a for all N elements of it So I have this left and right cosets and we saw from Lagrange's theorem that if this is our large group G and Say for instance, this is N now that we have all of these We have all of these cosets Imagine they are just three just just for argument's sake. So it'll be a N and B N and C N Imagine imagine that was true What we want to show now is If I take any element of The the I want to define this as Something new I want to bring in a new just some new terminology I just want to to call this instead of calling it in all these elements and all these elements Left coset or I coset of a left to right coset with respect to B with C I'm going to call them just I'm just going to give them another name and in many textbooks You'll see that this normal subgroup is called E with a bar on it And the name is taken is that remember E would be the identity element would be one of the elements here But there'll still be many other elements in Inside of this and there'll be many elements the same number of elements here same number of elements here And I'm going to take one of them and one of them and we have shown is that a will be One of them so I'm going to rewrite rename this coset as a bar I'm going to write this coset is B bar and this coset is C bar and if there were more I'm just going to write them and From this I'm going to form this new quotient group But first we just going to look at some set and usually it's written as G over N It's a division. It's a fraction. It's a quotient No, we're talking about division here the division that's sort of what is going on We're dividing this group into different subgroups Which is the normal subgroup and cosets that that the unique cosets that follow from that And I'm going to define this as a set first of all and the set is going to be this set B and C bar and if there were more I'm talking finite groups here and I just mentioned that in passing. It's actually quite important. Anyway, so I'm just rewriting this And I'm so this is a set in itself. This is a set of sets. This is a set This is a set. This is a set. This is a set because there's many elements There's many elements in each of these and what I'm trying to suggest is that this this These sets are mathematical objects in their own right that I can do Mathematics with that object. So I needn't just view them as individual remember integers mod n. I mean we we viewed them as you know that that Class as this as a separate mathematical entity with which I can do I can I can look at them as as that This combination of all of this or as the individual elements the same is going to happen here So I'm saying this is a mathematical object and I can do mathematics with these objects So much so that I'm going to define remember We still have this binary operation that I'm going to define the binary operation of Two of these I'm seeing that as an object and that is an object not necessarily now as an element of each of these And I have to define this and I'm going to state that this is if I just take two arbitrary Elements in there and I take that binary operation that composition and that is going to land up in somewhere else That that element that I get now I have to go find that element and it's going to be inside of Now one of these and I'm just going to call that because I don't know for now what it is because this is an arbitrary Arbitrary, I'm just going to call it that In other words, it is one of these it is one of these and My claim here is that no matter which element I take in here and which of the elements I take in here not necessarily the A or the B You know any of the elements in there that if I do the binary operation between them. I will always land up in the same One of these it's going to be an element and I'm going to go look for that element But it will always be in the same if it wasn't see it will always be in C No matter what a I what element and I choose what element and B I choose this a binary operation B if I go look for that element for that element I'm saying it's always going to be in C. Okay, so let's get a Representative of this one and that is going to be a binary operation with N And I'm going to get a null and that's binary operation with a representative of this Let's make this one. Let's like make this one That's B. Oh, and I've B. I don't know why we so but anyway binary operation And I've just got to make sure that this is not necessarily the same end remember for the cosets I have to run through all the ends element of N And now I want to be slightly sneaky about all of this Because I'm going to move this in here I'm going to move that in here and inside here. I'm going to put a B and a B inverse No, that's just identity element. So I've done nothing. I haven't changed anything and by the Societivity remember this is all part of a group associativity. I'm going to put these two together And remember this is a normal group. So that is an element of N. Let's call this whole thing Let's make it in star star Because it is one of the elements of that and binary operation in star But now remember both in star star and Insta they both elements of N So if I take their binary operation that is also going to be an element of N So what I have here If I do a associativity is just this Let's make those three stars because those two these two together a system element of N And let's just choose an arbitrary element of N. So what do I have here? I basically have this coset This left coset and remember it can also be a right coset because this is a normal group I have just one of the cosets but irrespective. This was arbitrary. This was arbitrary and It will always then end up in the same. I hope you can see that this will always end up in the same coset So if I have this and this I'm going to land up in a coset And I've just proven that it will always be irrespective of which one I chose an arbitrary one a binary operation in I chose a arbitrary one for B bar, which was B Binary operation with the N star. I'm always going to land up in the same coset. So that's very important To show that that if I have the binary operation between any two of these I'm always going to and I take any element inside of any one of those sets and the binary operation with them will always land up in The same irrespective of inside of that which one I chose. So I hope that's very clear Now the only thing I want to do is to show that this this is actually a group So this is actually a group. So I'm going to have my my Elements here and I'm sticking with three But you do understand that they came in more and I still have this binary operation. I need to show that Now if I want to show that that's a group I must have the group properties the first of which is closure and is there closure well, I have to find this as Does that so I take the binary operation between any of them one of those two now since A is an element of my group B is an element of my group the binary operation between them better also be Under the laws of property of closure a part of the group So by definition, I really have closure by the way that I've defined the binary operation between any of these elements in that set Do I have a sociativity? Sociativity yes by definition really I have because what I'm trying to show here is if I have these two that it's exactly the same as this B and C and Remember, you know, how did I define this well? I took an arbitrary element there and an arbitrary element there That they're binary operation with that and then see and then on this side. It was the same thing I had boc, which is now its own element and it falls in one of these it falls in one of these sets and You know still by definition This would be taking that one and that one and finding it and the same the goes for there This just follows from the definition how I define this. So associativity is no problem. What about the identity element? Do I have an identity element? So what I want to want to show if I take a And binary operation E that the identity element that will be the identity element binary operations a and that's just gonna be a And how do I define this? Well, that's this a oh e And that is this e binary operation a forgive me for saying oh all the time But anyway, and what what is this? Well by definition, that's an element of a group That's an element of our group G the binary operation that is just going to be a cause a So we know that there is this identity element One of them is going to be the identity element and four is then inverse is there inverse for every element as every one of these elements The normal set and all the cosets do they all have an inverse and for that You know what I really want to show is that if I have this that there is some Inverse that one of these is the inverse of this one and that is going to give me the identity element Well since we you know we can choose an a element a element of this one We might as well choose the fact that our inverse now is going to be this a inverse is You know in a inverse is going to be an element of of that and By that I just have a a inverse and I take that and that is going to give me that is just e So quite easy to see that you know every element every element does have its inverse And I'm going to find that inverse somewhere and that is going to be inside of one of these In one of those and that will give me the binary operation no matter what arbitrary element I take what arbitrary element I take inside of these I'm going to get the identity element So let me so so I've shown that this is indeed with this set This is indeed a group easy to understand and very easy to understand why I if I do the binary operation Define it as such I will always land up in exactly the same coset