 I just went on to homomorphism very quickly and there were some questions at the end. So, I thought let me clarify homomorphism. So, the first thing is it is a map between two groups we wrote this between group G 1 and G 2 we can write a map and then we look for. So, you have a map between G 1 to G 2 and then you want to find phi G which is going to be identity. So, identity I called it as capital E this is an identity element belonging to G 2 right and this set. So, set of elements in G 1. So, there is a set of elements let me call it as set G which satisfies the condition that phi of G equal to E then we say that this set whatever is the set we call it it includes the set will also include the identity element because the map should be. So, G 1 of the element will be G equal to E will also satisfy phi of E equal to E you can have phi of G 1 equal to E phi of G 2 equal to E suppose you have this then you say that E G 1 G 2 is the kernel of the homomorphism. If only this is the set which you find which is map to identity element then you call it to be the kernel of the homomorphism. If we find only E only E satisfying phi of E equal to E and no other element no other element if the identity of G 1 is map to identity of G 2 and no other element then we say that G 1 is isomorphic G 2 ok. So, this we formally write it as no other elements. Once this is satisfied those things will also be not satisfied because of the group properties you can disprove it ok. So, this is a very stringent condition once I put this condition all those 2 to 1 mapping and all will never be possible ok. And for this case we try to look at a simple example which was symmetric where I took the K which is an invariant subgroup or a normal subgroup that one had E B and B squared this has an invariant subgroup which is this. And then we looked at a factor group which is G mod K and this G mod K was just E and A they are all with me and then I said that the factor group should always be done with the invariant subgroup and then somebody pointed out that I could have factored it by this group and that is also fine. Why is it fine in this case? Can somebody tell me? It turns out that the group G divided by a group E A the symmetric group let us take. Why this is also a group? Why does it satisfy the group properties? You know why? This group has only one non-trivial element there is no point in talking about left coset and right coset right. Invariant subgroup has this property that left coset is same as right coset. If the group itself has only one element which means that you have only 2 cosets to start with right one coset with. So, you have only 2 cosets to start with. So, there is no point in talking about left coset and right coset. So, that is why accidentally the G by E A turned out to be your you know there will be only 2 cosets for this group and you find that it turns out to be also a group, but in general for an abstract group G if you find only an invariant subgroup the factor will be having the group properties ok. So, in this particular example it is satisfying this. So, that is one thing which was an accident here because it is a 2 element group one is identity other one is a non-trivial element, but this will not happen in general. Factor group will always be a group factor group will always be a group provided the factoring on the group is done by an invariant subgroup. If the group has only 2 elements subgroup has only 2 elements then your cosets are only 2 cosets. So, that is trivially going to be the factor will also have a group properties. It is also a factor group, but this factor group also has left coset equal to right coset trivially. See any group which will allow left coset being right coset will always the factor will always be a group. In this case it is non-trivial in the 2 element groups order 2 groups if you have that becomes a very trivial statement that is why it happened that that one is also a factor someone was pointing it out at the end of the lecture ok. So, this is the non-trivial statement for an arbitrary group, but this particular case it also turns out that it has the group properties because it is just a 2 element. So, whenever you have a 2 element subgroup then the number of cosets will be only 2 right right for a subgroup you can list out the cosets. Suppose the subgroup is EA the cosets are going to be HE and HE that is it and if you do the left or the right HE is same as EH. So, which means HE should be same as AH also. So, this is why this property was satisfied left coset equal to right coset for the 2 ordered to group that is why this factoring factor group was also good ok. So, this is one thing which I want you to remember. So, I have put in here is a subset of elements of one group map to an identity element of the other this is what we call it as the image of the map is identity element yeah I agree with you sorry I what I wrote here is so, what you are saying is that the H itself is EA that is correct. So, there will be 3. So, then my argument is not right thank you this is what you are saying and HB is it same as BH is the question. If the left coset and right coset are same then we can use this properties what I am thinking, but maybe you will have a proof for a ordered to group thank you this is not what I wrote is not right. H itself is EA. So, you have to multiply on the right or on the left. So, this is not clear now. So, this is what you will get as left coset and right coset most probably left coset and right coset are not same, but still we find this to be a group right this one is nothing, but EBB squared which is a group why is it happening is a question can you give a proof maybe you can think about it. I think it is the ordered to group, but it may be something more ok thank you fine ok. So, just to recap so, I have already done groups subgroups then coset of a subgroup coset is always associated with a subgroup and generators are all possible powers of the generators will constitute the group elements. We discussed this in the last lecture also told that for every group you can write a multiplication table along row and column it will be some permutations of the group elements, but all the elements will be there along a row or along a column. Normal or invariant subgroup in this case left coset is same as right coset conjugate subgroups remember that we said that for the symmetric group you could write the conjugate subgroups which were the conjugate subgroups conjugate subgroups are take EA as a subgroup and then do the conjugation right and then you will end up getting H 1 will be some element B H B inverse right H 2 will be AB H AB inverse I do not know I am just trying to write it like this elements which does not belong to that set. So, if you do this then you know that H 1 is conjugate to H 2 and H 1 is conjugate to H ok. So, both are conjugate to H according to this equations and from these two you can also say that H 1 is conjugate to ok. So, this statements you can say. So, I said that you start with the subgroup and generate other groups if it is an invariant subgroup can you find a conjugate to an invariant subgroup? Invariant subgroup definition is that arbitrary element of a group A multiplying that invariant subgroup is same as invariant subgroup time say. So, from here you can see that the. So, invariant subgroup is always self conjugate. So, these are couple of points with a particular example in mind. So, that you know you do not get confused ok. So, this also I have elaborated now factor group K is an invariant subgroup. So, this is the most general statement, but arbitrary groups whether it will the factor will be a group or not is something which one could try to prove even the simple case ok. Homomorphism is a map that also I have said kernel of phi will be the invariant subgroup K if the kernel is only identity element then the map is an isomorphism is that ok. So, so far for the group structures subgroup structures now let us get on to concrete you know I want to do permutations of objects and for that as I have already said anything related by similarity transformation has no new physics. So, you write G and G prime by a similarity transformation where A also belongs to the group G and G prime also belongs to the group then this such a relation will tell you the element G is conjugate to G prime ok. Till now I was talking about groups generating conjugate subgroups, but now I am going to confine to conjugates of the group elements then G and G prime are conjugate elements ok. Suppose the element G had certain order like say that if the element G is like A, order of A was A was 2 right. So, if A had 2 then the conjugate element which you find G prime in case you find that then that order should also be 2. So, in this particular case H 1 had a set of elements will again be 2 elements right let me not write what that element is this one is E and another element A had order A squared equal to identity and how can you find G? G will be conjugate to A G prime will also be conjugate to A. So, both G and G prime should also satisfy G squared equal to E G prime squared equal to E of G also will dictate this also you can prove it use this conjugation and prove it, but as of now let me just give it as a statement if G has certain order then its conjugate elements will also have the same orders that is what happened I do not know whether you recall we had H 1 one was E and A B, other one was someone A squared B or A B squared you can check A B, A B and A B squared will be conjugate to A ok. Why it is that? You can show that A B is A A B if you do it you can get what? B A was A B squared A squared cancel B squared A will become again A B it will become B squared I think I do something wrong A B squared. So, this is what you will get. So, A is conjugate to A B squared A B squared A B squared if A squared is identity you can do it twice here you can show that A B squared the whole squared is also identity. Is that clear? Order of the group element which is conjugate to another group element should share the same order that is why H 1, H 2, H 3 each one had order 2 element. So, these 2 will have order 2 element ok, yeah A B inverse did I make a mistake? Yeah it should be inverse. Inverse I was just thinking it is an order 2 group so it is the same, but technically I have to put it as A B inverse is so it is same right. A B already I have a definition that A B is A B is B squared A. So, I am just saying that the inverse is also B squared A is that ok. Anyway this is a nice simple example to play around and to clarify all your notations. So, that is why I given one non trivial non abelian group ok. For the symmetric group now I want you to we did various things for symmetric group we wrote it as a disjoint union of cosets we also looked at what are the subgroups we looked at what whether it has an invariant subgroup and so on. So, we did all these things for the symmetric group. I want you to find out the grouping of conjugate elements can you do that now we have done it already. So, it is not very difficult can you tell me what are the conjugate elements? We have already shown there that E E has any conjugate element E is self conjugate union 1 conjugacy class what are the conjugacy class what are the conjugate elements union what is left out. We will see that this bracketed one is the set of elements conjugate to each other. By doing some similarity transformation you can show that B A B inverse will give you one of the elements in this set only ok. So, this is what we call it as a conjugacy class. Similarly, this is another conjugacy class this is always a lone element and which we call it as one conjugacy class ok. So, essentially your symmetric group can be having this conjugacy class decomposition with three conjugacy class ok. You will have three conjugacy class. So, this we have already checked a couple of elements and you will not be able to find any overlap between conjugacy class. Conjugacy classes are distinct ok. So, you cannot make B and A to become conjugate and so on. So, B and B squared are conjugate to each other they will be related by some similarity transmission ok. Any G B G inverse will only give you either B or it can give you B squared. It gives B it is a trivial statement for all G you have to span it. If you do that you will find B and B squared are the only elements in that conjugacy class. Similarly, take one element here G inverse do this for all G. It can either give you A it may an either give you A B or it can give you A B square. I have given some assignment problems which you have to submit and you have to see the conjugacy classes ok. Is this clear? So, this is what I have put it in here that the symmetric group which I was discussing which is an order 6 group can be treated like a disjoint union of conjugacy classes. The reason why I am stressing on conjugacy classes that we can work with one candidate within the conjugacy class. So, you do not need to work with all the 6 elements ok. Instead we can take one one candidate from every conjugacy class and the number of conjugacy class is very important. Different groups will have if it is an abelian group how many conjugacy classes will be there? If it is an abelian group what happens to the statement? All elements are self conjugates. So, the number of conjugacy classes is also the number order of the group. The number of conjugacy class will only change if you are looking at non abelian groups. A couple of things which is interesting about this group is that no complex numbers at times looks very unphysical when you put an i factor and then you are used to i squared equal to minus 1 right. So, here there is something called as a Quotinian group where you have i j k are like your imaginary numbers s is negative 1 roughly as a mapping you can give which satisfies properties like squares i squared equal to j squared equal to k squared and equal to minus 1 and then s squared is identity ok. So, these are some properties and then I have asked you to find the group elements of such a group those are generators and decompose the group elements into conjugacy class ok. So, take it as an exercise and try to do whether you can find the conjugacy classes here ok. The other thing which we will find you know it will appear when I do the point groups of molecules is you can have additional symmetry ok. So, this the terminology of dihedral group will come back later these just like I wrote cyclic group C subscript n D subscript n has two n elements ok. It has two generators one generator R order of that generator is order of the generator is R has what order n order of the generator is n the second generator you have order of that generator s is 2 ok. Now, the question is given this to constraint ok. I also give you one more constraint S R S R inverse S using these conditions just like we played around with a squared equal to b cube equal to identity then a b equal to b squared a we made a multiplication table at least try and do it for some specific n n equal to 4 to get some clarity n equal to 4 will have 8 elements and then you try and find out subgroups normal subgroups conjugacy classes first non-trivial example where you can start playing around to get more clarity on what I was saying ok. So, there are some crucial points here n being odd and n being even could play some different roles I want you to figure out doing for D 3 and D 4 in general can you say how many conjugacy classes are there for a D n ok is that fine.