05 Subgroups and cosets

the no that's okay that's uh sorted okay so I think we're we're ready so recall that a group is a set G with um a specified Element e in G and the binary operation let me write it as as star for now this is like a a function from the cartisian product of G with itself into G such that first um we want that the specified Element e would be a neutral with with respect to the binary operation so for any X in g e star X is equal to to X and is also equal to X star two we want um associativity so for any X Y and Z in g x star y star Z is equal to X star y star Z and three for any X and G there exist an element which we denote as X one such that X star x - one is equal to e and is also equal to x - one star X okay so um but but as a notation convention we use usually replace star either by um dot when we talk about um structures with mul with multiplication um or we we can denote we can denote start by plus when we talk about structures that have a a natural addition in them and correspondingly we denote the unit element in the the plus case we denote the unit element is as zero and the unit or the neutral element in the multiplication case we denote as one the this is a this is the convention and and in fact um I I don't want to drag the notation star I I just wanted to stress the fact that this is um kind of a a vanilla a binary operation that you could think of as as multiplication and and and as addition depending on the the context but um when nothing is known on G we [Music] um we use um g dot one now a group G is a billion if for any X and Y in g x y is equal to YX okay and this um a bilan groups or also sometimes called commutative um our our particular kind of groups that um we will we will H ER Focus mostly on in the course but um I I think it's it's a good good um it's a good pedagogic approach to not um not dive straight into Aidan groups and and and do the theory for General possibly non ailan groups um as a notation in that case we use uh we write G in additive notation I mean again if we we only know that g is in a billion group then we naturally write the the binary operation as addition and this is a conventional of course we could have used the multiplication still in this case h if we just take the analogy from a from numbers then both multiplication and addition are a commutative operations so er there is no apparent reason to prefer one over the other um but of course we um we may run into a an ailan group that um whose binary operation is naturally multiplication and example for it h let N be some natural number and um G as a set is the set of n roots of unity over the complex numbers this I remind you this is the set of solutions to the equation so this is all X in C maybe Z in C such that Z to the N is equal to one or in other words it's the set of of roots of the polom Z to the N minus one and we saw that this this H um set comes naturally with um with a structure where you take the binary operation to be multiplication of complex numbers and the unit element is the complex number one which is 1 + 0 I so this is indeed a a group and in fact an a billion group but it's natural to write this ailan group as in a multiplicative notation okay because we do multiplication of complex numbers [Music] um let me give you an example of another group so let let N it be a natural number and the symmetri group on N letters uh this is Al also called a permutation group um is the set SN is uh the set of all functions from n that is n is the set of U numbers one up to n to itself s such that a sigma is projective okay let me you know what let me just write it again so for for a natural number N I write this bracket n to be the set of numbers one up to n and the symmetric group on N letters is the set SN the set of all functions Sigma from n to n such that Sigma is bjective and on the set I have a natural operation um in if Sigma and town are two elements in SN that is two functions from n to n then Sigma operation to maybe we do it like this start to is is defined to be Sigma composed with to okay and um I have to say something because um if I just compose two functions I mean I get a function but I need to know that I got a bjective function note that Sigma composed to is bjective as a composition of such okay we had uh we had a a proposition when we talked about injective subjective and bjective functions that if you compose two injective functions you get an injective function and if compose two subjective functions you get a subjective function so the same is true for sorry I have to take this hello you uh sorry so um so we get that we have um a binary operation um the identity map identity function e n from n to n satisfies Sigma composed identity is equal to Sigma which is equal to Identity composed Sigma for any Sigma any Sigma in s in fact for for any for any Sigma from n to n this equation will be true so we have a binary operation and we have a a candidate for a neutral element and what what we need to check well we need to check associativity but associativity of the binary operation comes from associativity of composition okay um star or or rather circle is associative since a composition of functions is associative and the last axom we need to check is the Axiom of inverse we need existence of inverse for every every function but um if Sigma is some element in um in SN then since Sigma is bjective it has an inverse as an inverse as as a function and this inverse is typically denoted as Sigma minus one this is again a function from n to n so this is a function such that Sigma composed Sigma minus one is the identity function and sigma minus one composed Sigma is also the identity function here the the the domain and the range of the functions is the same so um so in both cases we get the identity function on n h so to sum up SN together with the binary operation of composition of functions and and the neutral element being the identity function on N is a group and I claim that this group is not a billion generally speaking composition of functions is a non obedient operation you can think of it it's like a a function is like an operator and if I do one operator and then I apply another operator there's no reason that it's going to be the same as doing it in in the opposite order but so um ER SN is not an a bilan group so to see why let me represent a a permutation in in SN I can represent it a for example let's take n equal n = 3 but this this is true for for general for General and um I represent a permutation by this kind of Matrix that says that I um I map where do I map one I map one into H one let's say and I M two into three and three into okay and I take to this is Sigma to so this is going to be Sigma and now I take to to be a some something that um moves one so let's say one moves to two two moves to three and three moves to one I think already here it's not going to be a video um I mean the composition of these two what you what you get I mean you need to see where one goes so one remember composition we start from here when we start from the the the the right hand side ER function so the element one first goes to two and then the element two in in Sigma goes to three so the element one eventually goes to three the element two goes to three and then the element three goes to two so two goes to two and lastly a three goes to one and one goes to one so three goes to one now if I do this in the opposite direction in the opposite order so this is tow right I do one two three three sorry 2 three one this is Tow and I do one 2 3 1 3 2 this is sigma and now you can already see that one goes first to one and then to two so one goes to two one goes to two two goes to three and then three goes to one so two goes to one and three goes to two and two goes to three so three goes to three okay and this um this this is a different permutation than the the composition in the other order uh if you want as an exercise you you can in fact do what I did here for General end and if you want to show that for any n the permutation group is not an Aban group you need to find two elements that do not commute you can I think you can do the trick I did here for General right but do we need to because we already found one counter example for nals 2 so oh but it could be that for n equals 2 it's not opinion and for it could be that above a certain integer it's in a begin yeah you know hypothetically yes um so maybe exercise and prove it for General end now we will not we will not do a lot with the permutation group but at some point later on we will so I am ER I don't I don't going I'm not going to spend the um too much time on it now um for the next example I need a definition really at the moment it's it's a notation and let FP be the set zero up to P minus one I'm sorry before we get cared along actually you were saying prove it for n greater than two it's actually n greater than three because for two I think it actually is the Milli yes you're right yeah yeah for General n greater I didn't say but let's say greater than four but in fact we we proved it for three so you need four right yeah yeah um so I take a a a prime p and I denote by FP the set of elements I mean integers zero up to P minus one together with um a two binary operations um one of them is plus modu p and the other is multiplication modul P okay now I claim this is an important L for any X and Y in FP if you take x + y to the power of P to the N so this is for any x x y and p in FP and for any natural number n if you take x + y to the power of P to the N you get x to the P to the N plus y to the P to the n okay this is like the this is called the The Freshman freshman L because uh freshman I like to ER just ignore the brackets and do the operation on on each of the terms so this is the this is the case in which it actually it's actually true ER how do we how do we see this so we are going to prove it by induction the induction is going to be very very simple so um we start with the case of n equal Nal one I write x + y to the P according to the binomial formula so the binomial formula says that you sum over all um terms of the form x to the i y to the P minus I where the coefficient I goes from zero up to p and the coefficient is what is called the binomial efficient this is a a bomal coefficient um this is by definition this is p factorial divided by I factorial times P minus I factorial this is something that is usually er done in school so I'm I'm not going to prove this formula but you can prove this formula in induction it's a it's I mean just think of a this it's a it does not depend on the the fact that P is prime you can put instead of P you can put any integer K it just think of the the if P was a was two then when you do x + y SAR you have um XY I mean you have a term with XY the coefficient of XY is two and then you have a term X squ and you have a term x y s and the coefficient of each of them is one okay so it's kind of a combinatorial um problem to to do it for a general General integer but but you can count you do a counting argument and you you get into this binomial formula okay so now P clearly divides P factorial and P does not divide I factorial times pus I factorial this is for um for I am greater than zero and smaller than p p always divide a p factorial but P does not divide I factorial times P minus I factorial because a is prime I mean I factorial is the pro is the product of all numbers up to I and P minus I factorial is the product of all numbers up to P minus I if I mean if p is prime then this is a product of numbers that each of them is smaller than P so um none of them divides I mean p does not divide either of the the factors of the product so it does not divide the hence the coefficient p over I this is often is it's it's called P choose I and for I bigger than zero and smaller than p p divides this coefficient because it divides the the numerator and it does not divide the the denominator right so there's there could be no cancellation there is a factor of p in the numerator and there is no factor of p in the denominator which implies that x + y to the p is equal to a I mean modul modul p or in FP right in FP these coefficients um are equal to zero so it x + y to the P becomes just the first term that is X um to the P plus the last term y to the P right the first term in fact the first term is when I is equal to zero so this is just technically this is a well okay this is the the last term is X to the p and the first term is y to the P right okay so we we proved ER we proved the claim for n equal one and now for n = 2 m X+ y to the p² is by definition x + y to the p and all that to the P but by the Nal one case this is equal to x + y to the p and applying the Nal 1 case again I get that this is equal to x to the P plus y p um I want to no I had a sorry applying the Nal one case once I get that this is X to the P plus y to the P to the P but the the the claim was true for for any X and Y so I can apply the anyal one case for x to the p and Y to the P so this is X to the P to the P that's y to the P to the P which is X to the p² that's y to the p² okay and and ER for n bigger than two this is the same argument um the same repeated application of the N equal one case this is what mathematicians usually call a trivial induction I mean formally if I wanted to to write a completely formal proof I would I would do it by induction but you can see that all I'm doing here is just applying the Nal one case again and again any questions on this LMA the so if I know that the coefficient here I mean x + y to the p is equal by I mean the binomial formula to the this this sum in blue now if I know that the coefficients in this sum almost all the coefficients are zero mod not all of them so which one the ones so this is for any I that is not zero and not P that is the first coefficient is is not zero and the last coefficient is not zero if you think of it the first coefficient when I is equal to zero this is p over I what is p over I it's a it's one because this is a I mean p over over zero is p factorial divided by 0 factorial which is 1 time P factorial okay so the coefficient the first coefficient is one and also the last coefficient is one now what is the the monomial corresponding to the the first coefficient that is what is this X and Y term when I is equal to zero when I is equal to Z this is X to the 0 times y to the here this is not P minus one this is p minus I okay so this is the the ter is is y to the p and the last the last one is is going to be x to the P okay so I get I get that x + y to the p is X to the P plus y to the p uh any other questions uh maybe from online okay so um with this LMA we can do a we can do something nice and this is called um if for m a little theor and for my theorem says that if p is Prime and a is some number between one and P minus one then a to the P minus one well let me do it like this a to the a to the p is equal to a Okay so proof oh um so by by induction on a if a is equal to one this is clear suppose um a to the p is equal to a and now we do the a + one case then a + 1 to the p is equal to a to the p plus one to the P by The Freshman lar and by the induction hypothesis this is equal to a + one by induction hypothesis a to the p is equal to a so I get that a a + 1 to the p is equal to a + 1 okay that's that's the whole proof question I thought of a different proof but I don't think it works but what if we say uh a minus I know what it say it doesn't work out okay forget it sorry okay so using Forma a little theorem we we get into this a um important example let FP star the FP with the zero element removed and only multiplication module P okay so now I take only the elements one up to P minus one and I um I only consider multiplication model p no addition then if P star together with multiplication and the element one is a group in fact an a billion group so um unitary and associativity axom are immediate I mean obviously one is a neutral element with respect to multiplication modle p and also multiplication model p is an associative operation whatever you put into it you can move the brackets either from the I mean in the beginning or to the end so neutral elements neutral element a plus associativity and [Music] axioms are immediate what is a a a what requires some some argument is the inverse ation so for the inverse let a be an element in FP star i e a is a number between one and P minus one then byma um a to the p is equal to a which which means that um a times a to the P minus1 - 1 is equal to 0 modu P right this is modu p which is equivalent I mean two two numbers that are um equal to zero must must satisfy that one of them is equal to zero so a is not equal to zero so it must be that a to the P minus one minus one is equal to Z mod P ER which which which means that a to the p minus1 is equal to 1 mod P which is the same as saying that a time a to the P minus 2 isal to one module P so the inverse of a is a to the P minus minus 2 okay as as desired uh so so you see um we have a we have a collection of examples of a a group groups some of them I mean most of the groups we talked about are a billion but the exception was the symmetric group and some M bilan groups may come naturally with multiplication some the binary operation can be naturally viewed as addition ER and if as I said earlier by convention if we don't know anything about the biner operation but we only know that it is a billion we write it as a as a plus oh if we don't know that if we don't know yeah if we know if we know the the the details I mean this you know if we have it like here FP star we know the binary operation is a a a multiplication modu so we are not going to write it as addition this is going going to cause a lot of confusion um last H let thing before the break let G and and AG the groups the the group G * H is as a said it is just the cartisian product of g& H this is all X comma y such that X is in G and Y is in H and I need to tell you what is the binary operation um we Define star this needs to be a binary operation from G * H to itself so this is it takes as an input two pairs in G * H so I take let's say x comma Y and I want to multiply it by x x Prime comma y Prime so I do the the obvious thing this is going to be defined as x times maybe I I I do it like this G comes with operation star subg and a you a neutral element one subg and AG has a similar notation and then in the first coordinate I do multiplication in G and in the second coordinate I do multiplication in h so sorry this needs to be X maybe I WR like this is X multiplication in g x Prime y multiplication in ag y Prime then a a exercise Maybe this this set G * H with a a the operation Dot and the unit element or the neutral element 1 G comma 1 h is a group okay it's very easy to check um all axioms follow essentially because they they are valid in in both G and H any questions before we take a break yes is it a group or a build if this you can add to the exercise if G and H are both a billion then it is going to be an a billion group but if G if one of them is not a billion then it's not going to be an a bilon okay so uh let's break here and see you in 10 minutes e okay um let me just remark um something about this St in blue here I mean in the assuming we know by Forma that a a to the p is equal to a then we can definitely write we can definitely write it as a * a pus1 -1 = to z modu p but how do we know that if the multiplication is zero module P then the the right hand side factor of the multiplication has to be zero modle P if it was a zero on the nose then of course we know that two integers that multiply into zero one of them has to be zero but this is zero mod P so but but we know that a is smaller than P minus one and bigger than one so if the the product is equal to zero modu in P it means that P divides the product if P divides the product if any number divides a product it has to divide well actually if if the number is prime it has to divide either the first factor of the product or the second factor of of the product and since a is smaller than P minus one it means that P cannot divide a so I will add here P does not divide a hence it has to divide a to the P minus right one okay so this is just H to to be more uh more complete on the proof but we will use so if you if you notice what we have here this FP H the set of of numbers zero up to P minus one with addition module p and multiplication module p h it um it is what is called the field that is we have a a a structure of a a a aent group with respect to multiplication and a new a new sorry a billian group with respect to addition with the neutral element being zero and if we remove zero we have an a bilent group structure with a a multiplication and neutral element being one and in in a field we we will get to it later on but in the field you have additional axom that requires compatibility of addition and multiplication this is called the distri distributivity so you want that x times y + z is equal to x x * y + x * Z but this is also true for addition and multiplication mod P so FP is a field with P elements yeah's the confusion is the group over this field does not zero but this proof over Prime field so because they like assume the result of modifcation is zero so remember that in a group we we really require that every element has an in so if I take FP FP is all the all the numbers zero up to P minus one and I take the operation of multiplication module P if I take all FP that is I include zero then not every element will have an inverse yeah what I how I understand this is that uh so this proof is over FP so these operations in this are over but multiplication of in the group group operation is something different it's different multiplication no I mean in F from the set FP let me write it here we take a a the set zero one up to P minus one this I called FP and I call um well FP star is the set one up to P minus one so here if I add addition module p and I take take the neutral element to be zero then this is a group and here I can take multiplication module p and neutral element being one and this is again a a group both are a groups and so okay we will get to definition of a field later on but this will be something that has both addition and multiplication and addition makes makes it in a bilent group and multiplication if you remove the zero makes it another so it's like two compatible aident group structures okay so let G and I remind you um for a general group we write um G multiplication and one and a subset age in G is called a subgroup if the following conditions are at one is that the neutral element has to be in in the candidate to be a subgroup two for any X and Y in h x * Y is also in H a prior all we know is that x * Y is in G but it the condition for a subgroup requires that if you multiply any two elements in the subset it remains an element in the subset and three for any X in h the inverse of x is also in AG in other words a subset age in G is a subgroup if it's if and only if in um age is a group with respect to the operation from G the binary operation star from G okay now we saw a few examples so let's H let's check this notion with um the examples we had so I take the complex numbers and I remove zero and I consider it as a group with the multiplication of complex numbers and the unit element or the neutral element being the complex number one which is 1 + 0 I this is a group this you can call G and within this group I have um H the the group of n roots of unity mu n of C this I remind you this is the set of all complex numbers Z such that Z to the N is equal to one and here again I take multiplication and a a unit element I should I should remark the the notation if H is indeed a subgroup of G we write h smaller equal to G instead of just inclusion right we write a a smaller equal to G to denote that AG is a subgroup of G so here H is a subgroup of G um second example G is the group of integers mod 10 with addition right and AG is um The subgroup Well it's going to be a sub group but it's as a subset it's the set of the elements zero and five zero and five is indeed a subgroup of integers mod 10 because if I add two elements in the set H I get I mean the only non-trivial thing that I could I could do is to add five + 5 5 + 5 is 0o modle 10 so I stay with within the with in the subset Edge if I want what is the inverse of of five in Z modle 10 it's five so the inverse of five is still it's still in set and um yeah that's it right and the UN the the neutral element zero is of [Music] course yes this is also yes yeah that's fine we I mean all all we care about I mean we want that the third for example the third condition will be met the third condition means for any EX in age the inverse of X is in age again here the inverse I I forgot to say it in the notational convention but when we when we use addition the notation for the inverse is instead of to the power one the notation for the inverse is minus the element because the inverse needs to be inverse with respect to the the operation so the inverse of zero is zero the inverse of five is five and it follows that H is a subgroup of G however if I take the set H Prime to be 02 for example then it is not a subr because well for many reasons but um one reason would be that if you take two plus two then you get four and and four is not in the it's not in the set so you need to I mean a subgroup needs to be closed under the the operation of the the ambient group so I write it like this not a subgroup of two okay um the set consider G to be the integers and age is the subset 0 up to n minus one of course when I say G is is the integers I really mean G of being the integers with addition and the zero element and I pick some some natural number n this is this is what we refer to as Z module n but this is not a subgroup of z z modu n as a set is not a subgroup of G so you see each of them independently is a group but there are not I mean the smaller one is not a subgroup of the bigger one and the reason is that the the operation I mean the operations that make z modu n a group the binary operation is is radically different than the binary operation that makes a g as a group I mean addition mod n is not the same as addition now on the other hand if I take G to be the integers I take a natural number n so fix a natural number n and consider AG to be and this this is denoted as n Z this is all the multiples of n so this is n time x where X is an integer then H is indeed a subgroup of G why h if n * X and n * y are two elements in h then N X Plus n y is equal to n * x + y hence it belongs to it is a multiple of n again so the the set age is closed under under the the operation of the the ambient Group G um obviously the the the neutral element zero is equal to n * Z so it belongs to NZ and um now I I want to see that the inverse um stays in the in I mean the inverse of an element in age is again an element in age but this this is easy if NX is an element in NZ then n Time - x which is an element in Z sorry in n z is the inverse and satisfy N X Plus n * - x = z okay so NZ all the all the multiples of of a given integer is a subgroup of the integers um we talked about the the uh the product of two groups um let G and H be any groups any two groups and and consider well then I claim um G * H or the group um has a a two canonical sub groups one so I remind you g * H is the set of elements X Y such that X is in G and Y is in h within this set I can I can take um G cartisian product with um the the set that has one element being the the neutral element of H this cartisian product is simply the set of all pairs X comma 1 h X is in G and obviously this is a a subset of the whole cartisian product G * H and on the other hand I can take the unit the neutral element of G I take the the the set with one element being the the neutral element of G and I take the cartisian product with h this is a specifically this is the set of all pairs 1 G comma Y where Y is in h and again this is a subset of G * h and I claim that these two subsets are in fact subgroups both are Subs yeah I leave it as an exercise uh what is the LW or the yeah what is the operation for this uh if I claim that they are subgroups then the operation has to be induced from the operation of the ambient of the big group right that's my question what is the the operation is coordinate wise you take X XY * x x Prime y Prime you do x * X Prime y * y Prime the times in each coordinate you do the operation in the relevant group the first coordinate the operation G the second okay so let me perform a main construction that uh we will consider in groups so the G be a group um and X an element in G for a sub group AG in G the left of set or of H with respect to X is a a the set xh this is um all the product all the elements of the form x * h where AG is in age sometimes this is denoted like x. H so what we do here with cetes we kind of um abuse notation and we use the the the binary operation Dot to apply to to be an operation between an element and a subset okay ER this is a this is just for a for getting notation to be intuitively understandable what it means is that I take all the products I take X and I I take all the product of X with elements in H now observe just a few you know a few simple things I mean if I take e h the cette e not e and one one h then this is simply H right I take one times every element in h one this one is one the one in G I take the cent of one is simply H uh sorry I have a termin terminology question you say the left Co oh left C of H with respect to X okay yeah okay of course I could have done a a right cette right cette would be age X it's all the products age times x if we don't assume that that g is a billion then left cetes could be different than right cetes but we are very soon we are going to assume that g is a billion and it then it's not going to matter so you know you we might as well just choose one either left or right and and stick to it um and um if x so this the first observation is this the second if x is in ag then the CET of X AG is again AG right why H you take xh I mean X has an inverse in h x is in h then the inverse of x xus one has to be in h because H is a subr so um the the neutral element one which is which can be written as X time xus one it has to be in xh right because it is x times an element in h you see this and okay so one is an element in xh um and now given um some Y in h so I guess to show equality of sets I need to show um inclusion of both sides so um if um T is equal to xh an element in a in xh then clearly e belongs to H because it is a product of two elements in h okay so this means that xh is contained in ag and if um H is an element in age then I can write I want to write age as a then I can write age is x * x -1 * H and this is this is an element in inh because it is a product of X and some other element in age why is this other element in age xus one is in ag because X is in ag and H is a subgroup little H is assumed to be an element in h and since H is a subgroup product of two elements in h are in h so I can write a a small a as x times another element in h and hence small H is in fact an element of the cette the left cette is H this means I started with a general element in age this means that um AG is contained in xh which of course um means the that H is equal to e AG okay so it's a funny it's a funny operation I mean kind of swallows everything everything that you put into it that is already in the in the CLE um exactle I take a g to be the integers and age to be integers which are multiple of a given integer here again I I fix an in I fix n okay so clearly AG is a subgroup of G this this is what we talked about in the examples what is the cette so remember the the the the way I wrote left cetes I wrote it with multiplication because for a general our notational convention for a general group is that we use multiplication as the the binary operation but here in this in this example the binary operation is plus so for a a x in in G the left cette is written as um x + n z right this is instead of product I put a plus and instead of AG I put LC so X Plus n z if you think of it it's just the the set of all integers integers of the form x + n a for a in z note so I take for example 1+ NZ this I claim that it's the same cette it's it's the same the same set as n + one plus n and it's the same as 2 n + one plus n and so on right you see why because um let me do the first um the first equality and then you get the the principle um 1 + n z or or let me do n + one plus n z is the set of of numbers of the form n + 1 plus um n a where a is in Z but I can write it as the set of numbers of the form 1 plus n * a + 1 or a is in Z and this is equal to the set of numbers of the form 1 + n let's call it a prime where a prime is in Z right I mean if a is allowed to vary over all integers then a + one also varies over all integers and this last set is what we call 1 + NZ okay um so you see we have a lot of different cetes that are equal in fact how many distinct distinct coets should we expect in this case we fix an N we look at the group of integers and the subgroup NZ how many distinct cets should we expect to have I mean how many X Plus n z how many different different sets will will we obtain when we do X Plus n z where X runs over all Z it should be n right so um I I don't want to prove it now because we are uh getting close to the end um but let me put it as put something as observation I want to know I mean in order to realize how many distinct cetes I have I want to know when um x + n z is equal as a set to y + n z i I want to to have a a a a condition depending only on X and Y that will be that will guarantee me that that if this condition is met then these two coets are equal division by yes um I claim that these two cets are equal if and only if x is equal to Y mod n and the way to see it is is as follows I mean if x is equal to Y module n then um X minus y or rather um yeah I guess I could do x X - Y is equal to n a i e x is equal to a y plus n a maybe instead of a I put P but then when I do y + n z this is the set of all y + n a a in Z then I I might as well write oh sorry when I do X X Plus NZ this is x + na a where a is in C but X is equal to y + NT so I can write it as y + n t plus n a where a RS over all elements in in Z and this is equal to y + n t + a where a varies over all Z but T is fixed so if a varies over all T so over all Z then a plus T also varies overall Z so this is y + n a prime for a prime is allowed to vary over all Z and this is by definition the set Y A Plus n z conversely um if I know that x + n z is equal to y + n z then a for a given a a given element in z um x + n a which is an element in x + NZ it must be an element in y + n z so there exist an element a prime in Z such that x + na a is equal to y I + n a prime but this means that x - Y is is equal to n a prime minus a which is the same as saying that x - Y is equal to Z module n okay so you see um we know actually when two coets here are equal they are equal if um X X and Y are equal mod all right um I guess it's it's good to stop here uh any questions about what we did today this it was a I think it was quite a lot of new things um yeah have a question but we can talk about this next time what would be an example of the coent that is not using numbers Theory I mean like modular modular arithmetic well a lot I mean think of the ER the group of any roots of unit okay right this is a subgroup of the complex numbers right so take any complex number yeah and look at the cette of the complex number with I mean of the subgroup of n roots of unity with respect to this complex number but that's yeah okay I mean eventually yes because the the ends of unities is is isomorphic to but it's not I mean you know one is addition n one is multiplication of complex numbers right and okay so one example would be because this is also isomorphic to rotations somehow I don't I don't want to answer that too much because we didn't talk about isomorphism right but we will we will definitely see many examples of cetes and eventually I mean what we will do with these cets is we we will Define question Books Okay so okay but I I don't want to jump ahead that um maybe from the chats yeah yeah I'm I was hoping someone would ask any anything today from chat nazari Craig ER you have maybe requests for next time because I I didn't hear um any question from you today yeah and Nazar are you trying to speak okay okay so um it's good but it would be good I would be happy to hear questions that you might have er er in generally in the sessions so uh feel invited and I guess otherwise we um we meet again next week right bye-bye e