 Today we shall prepare ourselves with some algebra needed to understand even the statement of one compounds theorem, this algebra of groups, so some little group theory we have to do. Let us start with an indexing set lambda by a diagram, this is a technical term now, so it consists of it is not just it does have a ordinate sense of a diagram also, but now I am going to define the diagram of groups and homomorphisms. Index over lambda, so this consists of a number of you know data here, a collection of groups GI, see the I is inside lambda these are groups, a collection of homomorphisms is from GI to G, so there is a unique group G here without any suffix here, alpha is GI to G and a collection of homomorphisms eta ij, so these are now double indexed from G ij to G i, so G i's are also some groups such that alpha i composes eta ij equal to alpha j composes eta ij, so you can come from G ij to G i and then go to G in two different ways they should commute whenever i is not equal to G, so these all these data is there, such a thing is called a diagram of groups. You can say all of them defining over G, G is the one which is fixed, these things are varying here G i, G ij and so on and lambda itself may be varying. Such a diagram G, G i, alpha i, G ij, eta ij these are all these, is called a pushup diagram, so it has to certify some more condition, namely if for each diagram suppose you take another diagram like this the same indexing set, but groups are and groups and homomorphisms are different, G prime, G i, G i prime, alpha i prime, G ij, G eta ij, then there exists a unique homomorphism gamma from G to G prime such that gamma composite alpha i could alpha i from R, so this is similar to the pushup diagrams but it looks like more data is there and arrows are all reversed here, so that was a pullback diagram and this is push out, in categorical language this push out diagram is an initial object, for every other G prime there is a map from G to G prime making all these diagrams committed. Let me look at this case carefully again, so let us take a simpler case which you may be able to understand easily, namely suppose all the G ij's are the trivial group, then this data is obvious, all eta ij's have to be trivial, this condition is also trivial, so C goes away only the collection of G i with homomorphism from G i to G remain, that is a special case, then the push out diagram will be called a free product, free product of this family, so free product of a family corresponds to a group G along with homomorphism from G i to G satisfying some universal property, for every other such collection this G will admit a map homomorphism from here to the new G, G primes making all those various diagrams commutative, so that is the definition of free product, for a free product we will have a neat notation, in other words a push out diagram is an initial object in an appropriate category, since category I have not emphasized, you can just keep it in your memory box itself, given a collection of groups and homomorphisms eta ij the problem is to determine the existence and uniqueness of this push out, suppose you are given only this data, can you construct the G and alpha ij on this way, which will satisfy this push out, okay, uniqueness is something which is easy, memory if there is a G like this and G prime like this, you will get a unique map like this, but reverse this one, you will get a unique map this way, the composite will be an identity because G to G identity map will be there already, which satisfies it, by uniqueness of this one, this composite this back to this one must be identity, so this becomes an isomorphism, such that the diagrams commutative, that is the uniqueness of the push out, uniqueness of every kind of this one which is defined by this kind of universal properties is built in the definition itself and it's easy to verify that, just like in the case of fiber product we have seen that this is, but the harder part is the existence, so you will see that the existence of this one is much much harder than the pullback, the pullback was an easy case, an important observation is that the defining homomorphism alpha j from gj, gj to g, in the case of free product they all become monomorphisms, in the case of free product means what, again I recall that this part is absent, this is trivial group here, so you don't have to write anything, what I want to say is when star of g i that exists, let us assume that exists, the homomorphisms will be already there in the structure, alpha i's and alpha j's from each g i's, these will be all monomorphisms that means they are injective homomorphisms, you see you don't know what this g is, how this g looks like and so on, but you can determine quite a few things by just by the definition it has to be like this, it must be like this, it must satisfy like this and so on and that helps, those things help finally to locate what exactly it is, namely to construct this one, so let us see why these alpha i's are injective, so to see this fix an index j take g prime equal to gj and alpha i prime equal from g i to gj I have to define to be identity homomorphism if i is equal to j and the trivial homomorphism i is not equal to g, so what I am doing, I am forcing some extra information from this one by choosing this one to be one of the gj put this one here and take this as identity map, take all these for i j not equal to i i not equal to j, j is fixed all these varying this map to be homomorphism to be trivial homomorphism obviously with this data and there is no this is identity here this part identity okay once you have this there must be unique map here unique homomorphism the unique homomorphism here such that this composite this is identity means what this must be injective over if there is homomorphism here let us say this composite this is identity map that is what I have taken identity map now then this must be injective in fact this becomes a retract this is stronger than an ingest injectivity that is a map from g to here back here g prime this g prime is gj yesterday this composite identity retract okay to this observation helps a little bit in guessing how to construct a free free product so I told him this free product is in some sense to minimal thing it should contain all these groups the given groups and it should have this property of extending any kind of homomorphism defined on all these different groups to a unique homomorphism so this is the kind of criteria what we have observed in the space of vector spaces free abelian groups and so on if you remember vector space is definitely remember therefore we would like to have this one think of this one as some kind of group generated by all these areas groups a group relations within each of these groups should be maintained and nothing more no further relations that is why the word free product comes in this is a general idea but this is hardly a proof this can't be a proof because we know what is the meaning of generation only when we have a group right so this part we have to take care of it and that is where the free words come into picture so what we do is the special case is simpler namely the free product case we will first prove the existence there okay then with a little bit of work we will complete the existence of pushouts in general case okay the proof of the existence of pushout will be presented in two stages first stage is you take up simpler case of the existence of free products then we should derive the existence of pushouts without much effort okay so let us restate the definition of free product a little more carefully because all these eta ij's are there they were not necessary so let us make it what you have is a collection of groups indexed by an indexing side the free product of this collection is a group together with the collection of homomorphisms I am now denoting it as eta i because I know finally they are going to be inclusion maps okay of some kind monomorphism g i to g such that I am not assuming that they are inclusion maps because I don't know what g is eta i from g to g such that the following universal property holds for every group h and a collection of homomorphism f i from g i to h there exists a unique homomorphism free from g to h such that when you compose it with eta i phi comes with eta i is given f i that is a meaning of that this f this phi is an extension of all the f i's so this is a definition okay so let us look at existence uniqueness you have already seen let us look at existence and existence is by again construction not just there exists something say by construction so go through these new concepts carefully let us take any set s okay you may just for the sake of definiteness let us we can assume that s is non-empty by a free wordiness we just mean a finite sequence of elements of s okay each element of s is identified with a singleton sequence this is just convention so that when I write s belongs to s it also belongs to the sequence in that of length 1 elements of s will be called alphabets and these sequences will be called free words okay therefore an alphabet is also a word a word of what length 1 you can say this k is the length of s okay the set of all free words in s let us have a notation we will have w hat of s okay w for words but w hat is for free words we include the empty sequence also just for sake of carefully s could have been empty then sequence if there are no sequences of elements there empty sequence also it should be allowed okay and this empty sequence becomes the hero we will see how okay given two free words s 1 s 2 s k s 1 prime s 2 prime s l prime so these s 1s s i s and s primes are elements of s these are sequences so length k and length l something we can define a sequence of length k plus l by just juxtaposing the two sequences s s prime is the new sequence of length k plus l the first k elements of s later on it is s primes so this k plus 1 element is s 1 prime s 2 prime and so on s l primes okay so this is the definition of composition so i am defining a binary operation on w hat two finite sequences are there juxtaposing you see another finite sequence of course the length will interfere no problem okay i want to complete the definition what is the juxtaposing a empty sequence s empty is empty s s empty s is also empty s s this is true for all s where s is a sequence therefore you see for this binary operation this empty sequence becomes a two sided identity so this is very important for us this makes the w hat into a semi group what is a semi group a set with an associative binary operation together with an element which is a two sided identity that is all okay the associativity of juxtaposing is obvious you can put the bracket here and then write another one or first put the bracket this way and and take the left part here so that that part is okay it is completely obvious just like the associativity of composition of functions so we shall write when i say s is a sequence s j will be the its j th entry okay this is just a temporary convention okay again i want to tell you that an alphabet which is an element of this capital s is also a word what is a what is it it is a word of length one namely s equal to the sequences bracket if you put that bracket okay okay now suppose we are given a family of groups gi this was all general thing about any set s now i am coming to the construction of free product start with a family of group gi just take s to be the disjoint union of all these gi's okay so that is my alphabet now look at w hat of that so this time i will not write s at all here this s is understood here what is it says it's a disjoint union of gi's okay take all sequences of binar sequences in this letter they are alphabets and it has a binary operation okay its two sided identity is the empty sequence all right now we go ahead we don't have a group stress area it is only semi groups inverses don't make sense area so we want to introduce an equivalence relation here okay inside w hat so before that i have to go step by step so i define what is the meaning of elementary collapsing we say s prime which is another sequence is obtained by s by an elementary collapsing which i write it as s collapses to s prime if s prime is obtained by one of the following two operations what are these two operations there is some j such that the jth entry is the identity element of one of the groups remember s is union of all gi's the identity element in each group has a different element after all in this group these are disjoint union but i am just writing just one for all of them okay i should write one comma i or something index by i to show that it is in the gi group gi okay whatever it is suppose the jth element happens to be one okay in the group operations this one has no role to play one into g into one so we want to just drop it that is what we are okay identity element of some g then we delete this entry from s to obtain s prime when you delete it the length of sequence becomes one less so this is the operation of collapsing okay there is another way i can collapse namely suppose for some j the jth entry and the j plus one entry are in the same group one of the groups namely gi one gi two gi whatever i don't know which one but they must be in the same group then i combine them using the group law inside that group they are two elements of same group they are occurring consecutive so you combine them so two entries become one single entry combine and put it there so these collapses elementary collapses each time they reduce the length of the word okay so these two operations i have defined now out of this i cook up a equivalence relation how these operations are non symmetric you see right they are not even reflexive also so you have to just make them reflexive and and symmetry by inverting them so we say s is equal to s prime if there exists a finite sequence say s is s one s two s n these are not elements now they are all sequences okay the last sequence is s prime okay so what is the relation these sequence of sequences s is the first sequence of words s two is another one what is the relation s two can be got by s one by an elementary collapsing or s one can be got by s two by elementary collapsing collapsing could be this way or this way okay so that is the meaning sk plus one is collapse to sk or sk is collapsed to sk plus one this should happen for each consecutive consecutive elements here then we say s c is equivalent to s prime now it is easy to see that this is an equivalence relation okay here are some examples of what how how is it like suppose there is a sequence like this one one singleton one that one could be one of any group any of the group gis what do i do i drop it so i get a empty sequence suppose g and g prime both g and g prime are the same group g i then i combine them and get g g prime this is length two this length one okay suppose i 11111 this should be equivalent to empty sequence because go on doing collapsing collapsing collapsing and times finally you will be left with empty sequence suppose i have a b b inverse a inverse what do i do i have to combine b and b inverse by by this notation they are in the same group right b and b inverse are in the same group so combine them what i get b b inverse is identity of that group drop it then what do i have a and a inverse again i have to combine them they may be in a different group does not matter again they produce one of that group so drop it so finally i get empty so these are some easy examples of what we are doing with these words three words they are no longer you know when you go model this one there are the three words but but they are more restricted okay so you have cancelled out certain things that is the point here we shall denote the equivalence classes of s for example 11111 is the same thing as empty sequence a b a inverse b inverse same thing as empty sequences right like that equivalence classes of this one putting a bracket take a sequence put a bracket means it is the equivalence class all of them are in one that with a simplification notation we do not want to put any bracket for the empty sequence okay empty word is just represented like that the set of these equivalence classes of words will be denoted by bracket w remember what was w hat you started with unions disjoint union of all the groups then we took all the three words in it then we took equivalence classes and that is this bracket it is also clear that if s collapses to s prime you multiply multiply means what juxtaposing it by a sequence t then s t collapses to s prime t because collapsing is happening inside s similarly t s will collapse to t s prime okay so juxtaposing either on the left or right preserves this collapsing therefore it preserves the equivalence relation also therefore the binary operation gets goes down to bracket w under equivalence relations the binary operation can be defined now because it respects the equivalence these equivalence classes i can define bracket s into bracket s prime to be bracket of s s prime you choose any representative representatives may be different here the light and side will be well defined s s prime is not well defined bracket s s prime is well defined okay and you can check that the empty sequence box with s the same thing as the box of the the equivalence class of just s right on either side therefore the empty sequence plays the role of two sided identity on this binary operation on bracket w okay so in particular bracket w will be also a semi group what we want to say is there is more here now what is that now let us do that suppose you have sequences which you can write as x 1, x 2, x k i have just changed the notation for here you can put s 1, s 2, s k it does not matter the inverse word i will define s 1 inverse s inverse is the inverses of this one taken in the opposite order x k inverse etc x 1 inverse what are these inverses remember each g i is a group each x i is in some group some g j okay so inverse makes sense there and that inverse will be another element so it is another alphabet so this is a sequence again and this is a an element of w hat all right and if you combine s and s inverse follow s by s inverse this is not a empty sequence okay but it is equivalent to empty sequence why because look at this one s 1, s 2, s k followed by s k inverse so i am canceling out s k and s k inverse because i have to combine these two get identity drop it next combine identity drop it and so on you get this one similarly if i put s inverse okay therefore when you take the brackets bracket s bracket s prime is nothing but the empty sequence so what we have done is we have actually shown that bracket w is a group under this operations it was not if you take the free words it was not after equivalence relation it becomes a group okay therefore w together with ever multiplication becomes a group construction of free groups is over however it takes some more time and effort to see that this is actually the free group that we are looking for namely it satisfies what it satisfies the universal property that we have to prove we have just constructed something okay that will take some time which we will do next time