 Last time, starting with a family of groups, we took the disjoint union of these groups as our alphabets. Over these alphabets, we constructed what are known as free words, which are nothing but finite sequences of alphabets. From the free words, we introduced an equivalence relation and that made it into a group, namely concatenation of two sequences, two words, gives you a free word. Along with the empty word, it makes it a semi-group, associative binary operation with a two-sided identity, all right. But it is not a group. To not make it a group, we took equivalence classes of these, namely equivalence relation was defined by a chain of elementary collapsing or its inverse. Finite chain of such a thing, if two words are related, then they are said to be equivalent. So that equivalence classes were denoted by the bracket W. These equivalence classes becomes a semi-group under the obvious inverse operation, namely take a sequence x1, x2, xn, take inverse formally as inverses of each xi written in the opposite order. These inverses make sense because they are elements of various groups that we have taken. This on the right hand side, these inverses are genuine inverses of group elements. On the left hand side, this inverse is a notation. Now you can verify that when you go to the classes, equivalence classes, that SS inverse is equivalent empty set, therefore on the classes, this is the inverse. So this much we had seen. So today we shall complete the proof of the existence by showing that this bracket W is the one which satisfies the required properties, namely the universal property. So we introduce a terminology here. A free word is called a reduced word if either it is empty, empty, empty word is a reduced word or the following conditions hold. None of the entries of this free word should be an identity in any of these groups at the first condition. The second condition is no two consecutive elements should belong to the same group. However under these two conditions, we can collapse. Whenever such a condition is there, whenever any element is an identity element, we delete it. So that is for the first collapse. Whenever two elements are consecutive and belong to the same group, again we collapse by writing them together as one single element in the corresponding group. Therefore these two operations, if performed finitely many times, any given word will be equivalent to a reduced word. After all you start with a finite sequence. So in finitely many words, all these things is possible because each time you perform this, the length is reduced. Therefore what we have just observed is that in the equivalence class of free words, there is always a reduced word. It is always a reduced word. I mean it may be empty also. Empty word is by definition reduced. So the point is, if there is always a reduced word, if I say that, that is weaker than saying that every free word in finitely many steps can be reduced to a reduced word, can be collapsed to a reduced word. Actually collapsing thing else, one single way. Of course collapsing, collapsing, collapsing, that is the equivalence class. What is not clear is that if we have, suppose you have a very huge word, collapsing can take place in different ways in different directions, different order, then it is not at all clear that you will get the same, two people, if they start working or two different people or two different kind of algorithms are written down, whether you will get the same reduced word, whether there may be two different reduced words representing the same class. So this is not clear, okay. So what is fundamental importance to us is that the equivalence class of every word contains a unique reduced word. So this is what we want to see and this is more or less key to the universal property of this, the group that you have constructed, okay, they go simultaneously, okay. So let us study this one carefully. So here is a technically superior method. This method should be learned, it can be useful elsewhere also, okay. It is actually used in linear algebra quite often, linear algebra of, you know, functional analysis, exterior algebra and so on. So what we will do is we shall denote the set W hat, subset of W hat consisting of all reduced words by removing that hat, let us say reduced words. So here is a subset, actually it is there, each element there in each equivalence class is below, namely square W, so W surjects on to square W, okay. So what you would like to say is this square W, the equivalence class is in a projection with W, okay. So for seeing this one, what we do is take the permutation group of W, W is now reduced words, not the whole of W hat. The permutation group is a very huge group of words, that is a key, okay. Let this pW denote the group of all permutations of W. Fix a group one of the GIs, concentrate on one of the GIs, we shall define a subgroup LGI of this pW, okay. And I will denote for each G belongs to GI, GI six set, each is corresponding LG, these are the elements of pW, so that will be a correspondence between G going to LG, that is going to be a monomorphism, under that we can get a group LGI, okay, a subgroup of LGI which is isomorphic to GI. You see the first you know starting point of this universal property starting group, okay. So if G is identity, take LG to be identity, because after all you want it to have a monomorphism, okay. Identity of what? Identity, endomorphism, identity permutation of W. To each G which is not identity, let LG be defined as follows. So you have to have patience, because we want to take a reduced word to a reduced word. If you have to define it W hat there was a very obvious, but here you want to take a reduced word to a reduced word. By the very notation LG is going to be left multiplication in some sense. So what we are going to do, we are going to do a left multiplication on W, but then we have to do reduction also, reduced word. So G is a non-trivial element, okay. Then LG of S where S is a reduced word will be defined as follows. If S is empty, LG is just the sequence little G that namely one single element, okay. If S is let us say one single element S1, okay, the first one only one element and K is equal to 1. So S is a singleton sequence S1, but this S1 is G inverse. Then when you multiply by G, it will not be reduced, right? So you combine them, G, G inverse identity, just check it out, what you get is empty sentence, empty word. So LG of S, namely LG of G inverse, which defined to be empty word. Next suppose it is now K is bigger than 1, then all that you have to do is, S1 is G inverse, so you cut it out, right? It has S2 up to SK, remember this is S1, S2, SK was a reduced word. If you cut it short at one part, okay, either initial part or end part, it will be still reduced. If you cut out somewhere middle, then it may not be, okay. So I have dropped out S1 here, this is still a reduced word, okay, take this as definition. Now up to these three cases are over, the next case is S1 is not an element of given at all, then just put the G extra, so S1, S2, SK as it is, G will come on the left. So this is L of S1, SK, L of S1, LG of S is this one, okay, if S1 is not in GYI at all. The last condition is suppose S1 is in GI, but it is not the inverse of G, then you just combine G and S1, G S1 is not identity, so this itself is a reduced word, okay, so that is a complete definition of what is the meaning of LG of S. LG take a reduced word, I have produced another reduced word, therefore LG for each G is a function from W to W. We have to show that it is a bijection, then only it is a permutation, right. So when it is checked that LG of W is contained inside W, contained inside W means what W is a reduced word is also, LG of W is also reduced word, okay. The bijection part is taken care very easily by just checking that LG inverse, composite LG is identity, similarly LG, composite LG inverse, okay. LG inverse of LG, go through this definition at each step, you get back the original sequences here. So this is actually an element of PWF. It is also clear that if LG could LH, then G must be equal to H because you operate it on, these are two permutations, right, operate it on empty set, empty set is a reduced word. LG of empty set by definition is the singleton G, LH is of singleton H, right, therefore singleton G equal to singleton H, therefore G equal to H. Therefore this L is what is injective, okay. So simply taking value of unencumbered set, finally we need to verify that L is a homomorphism. See L itself is injective, each LG is a permutation, okay that is different than saying L is injective, L itself is injective and now I have to verify that L is a homomorphism, okay. That is L of GH equal to LG, composite LH for G and H belong to GI. This is also easy but you have to have patience and verify everything properly, okay. So in order to show this, what you have to show, LG of a reduced word S is equal to LG of LH of S, okay. So what I want to say, suppose S is of length more than 1, then this will follow by just, suppose first I have proved it for length 1 elements, suppose this is formula is true for length 1 elements, then for any element of length more than 1, the S2 to SK part remember is never disturbed in the definition. So you have to just put back those S2 to SK part after getting the formula for K equal to 1, alright. Therefore you have to verify it only for length 1 elements, okay. So LG H of S equal to LG composite LH of S is what you have to verify and this is very easy. Again you have to go through this list, okay. For example, suppose first case namely S is empty that is also length is actually length 0, then it is very clear LG of what? LG LH of empty set is LG of operative 1 H, G and H are in the same group. So LG of that one is LG H, so LG H operative upon operative upon empty set is also G H. So like that you can verify the entire thing is verified for any one. So what we have done so far is for each I GI to permutation of W's that is a monomorphism given by for each GI left multiplication by GI, it is a monomorphism. So I can identify GI with this subgroup L of GI inside Pw, okay. So this I can do for all I, alright. So therefore I should need a different notation I will call them as Li. So this was earlier I had just L because I had fixed GI. So let us call these as Li, okay. Now what you have is they are not necessary because they are different L is Li, right Li. Now what you have what is union of all LGIs, they are subgroups, each of them is subgroup, they will generate a subgroup, okay. Union is not a subgroup, the union will generate a subgroup, okay. That subgroup is isomorphist, oh this bracket W is what we want to show. So let us have notation, this is final notation, union of all GI generates a subgroup of Pw and let us take that as G, okay. So our definition of G is this one now. What we are going to show is that this subgroup of Pw is actually isomorphic to this bracket W, the equivalence classes of three words and we are also going to show that it is biject to correspond to a set of reduced words. So various ways of looking at this group that will help to consolidate the entire thing, alright. So this bracket W we know it is a group, ordinary W reduced words we do not know it is a group, okay. But if there is a, there is at least each equivalence class, there is a reduced word here. We want to show that all these are isomorphic to each other. There is a group structure here, there is a group structure G here, we want to say these two are equivalent, they are same isomorphic, okay. So how do you proceed? Start with a three word S inside W hat, that is an element of W hat, okay. Now you define a map, a function from this W hat into Pw. What is that? In fact, you define a map inside G which you land into G. What is that? Look at S1, S2, S2, S1, S1, S1 will be in some Li, Li1, Li2 and so on. Sorry, Gi1, Gi2 and so on. Accordingly, you take those homomorphism, monomorphism Li1, Li2 and so on. Take Li1, S1, LiK, SK, compose them in the same order, S1, S2, S3, same order. These are all permutations of W, remember that, okay. So you compose them, but they are all inside the union of all LiG. Therefore, this element is inside Captain G, the subgroup generated by this thing. So what we have got is for each element W hat, we have got an element of G. So that is a function, okay. This function is clearly a homomorphism of semi-group into a group, right. How do you compose here, S1, S2, SK and then SK plus 1, etc., next one, next sequence. The same way you are composing here also. So this is a homomorphism of semi-groups, okay. Of course the codomain is a group, but it is homomorphism groups. One more thing you can say that, suppose S collapses to S prime, yes, simple elementary collapse. Then PSYS will be equal to PSYS prime. Why? There are two cases when this collapses. One of the SI's here may be identity. Then LiJ whatever of SJ whatever you are taking there, that will be identity. So the two compositions will be the same. Next suppose two of them say S1 and S2 are consecutive things which are in the same group. Then I am taking S1, S2. Then but here I will get the same Li, Li of S1 and Li of S2. But Li is the homomorphism. So Li of S1 into Li of S2 is Li of S1, S2. Therefore under elementary collapsing the values of PSYS and PSYS prime are the same. What does this mean? This means that this PSY map actually factors down to give you a map PSY bar from equivalence classes of W2G. This map is going to be an isomorphism, okay. Let us see how. So given an element G by definition what is G? G is finite combination, finite sequence of elements, product of elements. GI belonging to GI actually of LGI's. So G is actually LG1, L1G1, LKGK where I assume GI's are inside corresponding GI's. This is just a free notation. I should strictly speaking I should write Li1G1, Li2G2, LiKGK. Every element looks like that, okay. Take S to be just this G1G2GK belonging to W hat. This is a free word, okay. Then PSY of S will be clearly this one by a very definition which implies PSY bar of S when you go down in the class it is also G. So PSY is very induced by PSY. What we have shown in just now is that PSY bar is subjective because PSY itself is subjective. Now we want to show that this is injective. PSY bar of S remember this is already homomorphism. So PSY bar of S equal to PSY bar of S prime we must show that S must be equal to S prime. Not S equal to S prime but the classes. So this is where we have to go to the classes. S may not be equal to S prime but their equivalence classes they are the same. So all that you have to do pick up reduced words which represent them. We have seen that in each class there is a reduced word. Take the reduced word representing them, okay. Then PSY S is equal to PSY S prime because when you go down this is the equation that I have. Therefore what is PSY S of empty word it must be equal to PSY S prime on a empty word because these are endomorphisms. If they are equal their value on the empty word must be the same. But by definition what is PSY of PSY S of this one empty word, okay. Suppose for any reduced word S is equal to SK we have PSY because I have chosen S and S prime reduced words PSY S of this one empty set it will be just S1, S2, SK, right. Remember what is it? It is last LIK of SK acting on a empty word that will produce SK. Next one LIK minus 1 acting upon SK minus 1. But SK minus 1 and SK cannot cancel out neither they are trivial elements. Therefore when you multiply this one it will become again SK minus 1 into SK. Inductively like this you will get S1, S2, SK, okay the sequence S1, S2, SK. Therefore on S prime also we will get S1, S2, S1, SK prime. If these two are equal it must be put S1. You can directly do this by saying that suppose PSY S prime is identity then show that S is collapsible. If S is reduced word not identity it cannot be identity that is what you can you can directly show that also, okay. There are different ways of putting this one. Idea is the same. What we have done so far is that PSY bar is an isomorphism from the equivalence classes of words three words to this group which is a subgroup of the permutation groups of reduced words. Already we have already hidden thing here namely in this proof I wrote down this one not for fun. This already tells you that in each equivalence class there is only one reduced word. Can you see that? This tells you that the two reduced words if they are in same equivalence class then they must be equal. So let us run through that proof again because this is so important, okay. Suppose S1, S2, S2 reduced word in the same equivalence class that is I am assuming S equal to S prime then S which is PSY S of empty this much what we have seen because S is reduced PSY S of empty word is just S. But by definition this is PSY bar of S operating on empty word PSY bar of S equal to PSY bar of S prime because the classes are the same PSY bar is very different that what we have seen. But PSY bar S prime of empty set is PSY S prime of this one because S prime is reduced but this is quote S prime, okay. Therefore in each equivalence class there is only one reduced word that is exactly one reduced word. From that if you take PSY bar it will become a bijection onto the group G reduced words themselves defining a group operation was bit awkward. We have avoided that. Now you can put back that definition of group structure on G onto reduced words. Finally if you want to just work with reduced words what you have to do take a reduced word, take another reduced word put one after the other it may not be reduced but keep collapsing you will go to a unique reduced word that is the definition of the composition law on reduced words. You see why we needed to have unique reduced words otherwise definition of you know alpha composite beta does not make sense. Even after this proving associativity is a hell of a problem we avoided all this cleverly by going to the permutation group of all reduced words, okay. Also we have shown that there are monomorphisms from E GI to this W now, okay. This W is reduced word which is in bijection with G. So I can replace this W by G if you want. So these eta i's along with this G this is going to be our free product. So we have to verify the universal property, okay. Given any homomorphism F i from GI to H a sequence of homomorphisms, okay. We define F from W hat namely free words to H by the formula F of S1, S2, SK equal to F i, F i1 of S1, F i, K of SK just take the corresponding F i's and their compositions in the correct same order whereas J's are in GIJ's. This IJ's which F i have to take depends upon where S1 is, where S2 is, where S3 is and so on. Take this product. There is no ambiguity in the free word defining anything there is no ambiguity, okay. Verify that this is a homomorphism just like in the case of our L this is a homomorphism is also obvious homomorphism of what of the semi group into H, okay. The next thing is just like we have done for L i's verify that it takes same values on equivalence classes namely if S reduces to S prime if S is collapses to S prime we have to verify then the value of F is the same. This is also again the same way the way we have done it for L i because if S1 is identity then this is really this is this can be checked out. If S1 and S2 are the same element then F i1, F i2 are the same F i1 of S1 compose it S2 equal to F i1 of S2 because F i's are homomorphisms the same argument. Therefore, there is a well defined homomorphism F from what is this W these are now reduced words but they are equivalence classes that is a better way of thinking to think of them as equivalence classes because 1 W is representing the whole equivalence class and that is a uniqueness that is why I can put it as W okay. So, there is a well defined map from W to H okay which clearly has the property that if you take eta i of g eta i's are how we define eta i of g is just singleton g that is a reduced word okay. So, this F i operating upon this precisely that if this sequence is a singleton 1 then wherever it is that coming from g i it is F i of that. So, eta F compose eta i is F i of g okay for every g inside g i. So, it satisfies these conditions only thing is why F is unique F is already agreeing on all g i's and g i's generate the capital G if you have homomorphism okay of a growth it is completely determined once you know its values on a set of generators therefore F is unique. So, this completes the proof of that this eta i, g, eta i this family is the free product of the family g i okay. So, we have this notation for free product g i's are families of groups on indexing such lambda then you can have this one. So, we summarize this one whatever you have done namely start with a collection of groups then the set W of all reduced words in the union of this g i now for reduce words you can check away the identity there is no identity does not play anywhere you have to cut it anyway right whenever. So, just reduce words in these letters of course reduction you have to still do the disjointing of this one this forms a free product product of g i under the usual of composition concatenation and reduction just concatenate it will not give you a reduced words. So, we have not directly proved this we have proved it indirectly okay yeah we have verified that this definition is the same as the other definition but we have not verified that this definition is associative left multiple get wrapped identity right identity those things we have avoided. Now, here is an easy if you have understood this construction and so on easy result which is very useful namely suppose you have a sequence of or a collection of homomorphism from g i to h i form the free product here form the free product here then is all these alphas will extend to a single map alpha homomorphism from the on the free product of all these first family of eyebrows the free product of all these h i's in such a way that on subgroups alphas g i's they are equal it is equal to alpha that means alpha is an extension of all these remember g i's and h i's are subgroups here okay if you follow this construction then this is obvious all that you have to do is look at the free words and define there there it's obvious then all that you have to do is keep ready reductions reduction process that's all okay so let us do all other constructions like we have to push outs and various things we have just done one single memory construction of free product that we will do next time thank you.