# 06 Groups (quotient groups)

- Channel: [Berlin Ethereum Meetup](https://streameth.org/berlin-ethereum-meetup)
- Date: 2024-10-07
- Duration: 2:00:26
- Watch: https://streameth.org/watch/yt-TCrpAegQIDk
- YouTube: https://www.youtube.com/watch?v=TCrpAegQIDk

## Description

Lecture notes to be found here: https://drive.google.com/file/d/18KRG6J8YA-UxWhZmk1aE47S8iAdTanDE/view?usp=sharing

## Transcript

and um a given a subgroup age of g i I forgot to say this is Den noted with this smaller equal relation and given a subgroup age of g and an element X in G the left cette of H with respect to X is the set it's denoted x. H or just ex age and this is the set of all expressions of the form x * y such that Y is an element in H okay so the cette you can think of it as um as a shift of the of the subgroup so um example we take G to be um the the set of um of a the real plane this is the the set of tels XY maybe we do it AB such that A and B are real numbers and we have a natural operation um of adding two vectors I mean two two elements here and let say a prime B Prime this is defined to be A+ a prime B plus b Prime and and the the neutral element is a 0 so this is a group it's very easy to check and as a subgroup uh we take H just pick um for um I guess say do M and Alpha two real numbers um I can consider um H to be the set of solutions X comma y or maybe we said a such that um B is equal to M A + Alpha so this set of solution is a is a line in the I mean for a given M and Alpha H is simply a line that passes actually Alpha needs to be alha needs to be zero I'm sorry so it it has to be a line that passes through the origin because for for age to be a subgroup it needs to contain the the neutral element so the picture is is like this um this is AG and age is indeed a subgroup here by the way I mean um if this is Alpha then m is going to be I think tangents tangents Alpha ah uh yeah okay Alp is um so so a is a subgroup of G and it's very easy to see because if you take two elements um if you take you take two elements in age say a b and a prime B Prime um you add them two together you just get another another guy that s i mean then you know that you have this equation satisfied and you have also the other equation satisfied and so when you sum the equations you get that b + B Prime is equal to M A plus a prime and this means this is equivalent say that a A plus a prime B plus b Prime is an element in h right an element in h is something that satisfy this equation is this okay is it clear why H is a subgroup of a of R2 okay and now if I do the the cetes now I pick um I pick P to be or maybe we do it a capital x to be an element in uh in R2 and I want to consider what is the cette X with of age with respect to X but note that we are for this group G we are using additive notation right instead of multiplicative notation that we did in in the general case so with additive notation um the cette of AG with respect to X is denoted x + H and what is it x + H it's um let's say that X is some um s comma T then this is all s comma t B plus a comma B such that um a comma B is an element in R2 in sorry in h and but if you look at it this is just of course I can write it simply as s+ a t + B such that um a comma B is in age but for a comma B to be in age I know what has to happen I need that m b is equal to ma a right this this is the condition to be in age = M because m is definition of yeah no it's s comma T was picking was picked not in h s comma p s comma T is is an element in the the ambient group in G yeah X is s comma t X is s comma T remember a cette of of AG is you pick an element in the the ambient group in the the big group and then you take the the coet of of the subgroup with respect to this element so I pick a general element in the group this it just a point s comma t with no restrict friction and now I look at the coet of this element with respect to the or the CET of age with respect to this element what is it again I write Plus instead of I mean in the general case we we wrote instead of plus we wrote a multiplication but this is just the binary operation of the the group the binary operation here is better WR in in additive notation and X Plus H by definition is you take X and you take all the X operation with an element in h right this H I write I write again here in in purple xh in in the multiplicative notation it was all the X comma sorry X operation age where AG is in age right so if if I do this if I do this for for our case this is X operation a comma B where a comma B is in is in h right this a comma B now takes the role of of AG in the the purple art is this clear yeah okay so now but what is it um we know what it is to um I mean I have this Vector um say say it is like this this is X x is s comma T I can I can picture it as a vector and this is H and now I need to um I need to add every two every Vector in a AG I need to add to X how do you add two vectors in a in the XY plane you do a parallelogram so I I picture a par parallelogram is fortunately not not very precise but this I hope you get the idea and the the corner of the parallelogram is the result Vector of and I can do this parallelogram for all the all the arrow I mean all the points of of age where I consider them as arrows so what I get is is this um this blue line the collection of all points of so this BL line is going to be x + H yeah could you just scroll a tiny bit scroll up a tiny bit uh right so my question is in the case of X belonging to H as well you don't have a coet at all you you have a coet it's going to be H so H is a coet of H we I mean our terminology was a coet of age with respect to so so coet is always a a coet is like a shi that's what I'm this is what I'm trying to demonstrate in this example that a coet is like a shift of the original subgroup right by an element right you see or instead of a shift you can say translation this I translated X Plus H is a translation of age and I mean in the direction of X yeah okay and what I did here for X was General right I didn't assume anything on X so you can do it for any for any element in in any Vector ER and so what you get is a collection I mean you do it for all vectors you get a collection of lines that covers all the plane right okay um yeah I have a question but I don't know what's coming next so maybe it's not the right time to to ask it okay I'll try yeah um if we have if we look at a different model uh that would be a circle well okay more exactly the operation we take the the plane yeah the usual plane R square yes and we instead of having um yeah like the let's say you have ah yeah but we need right now actually it's one operation that would be both scaling and uh rotating let me give you an example I think that is maybe is something similar to what you had in mind I take the circle the circle is is denoted as S1 yeah I I treat it as um as a subset of the complex numbers okay okay now I view the the plane as the the complex plane instead of the and um this is a subgroup with respect to complex multiplication because complex multiplic I mean if I have a um Zed and Z Prime in the circle then I can just multiply them H the the the norm the complex Norm of Z * Z Prime is simply is going to be equal to the complex Norm of Z times the complex Norm of Z Prime complex Norm is simply the um the distance from the origin okay and if if they are on S1 on the circle the circle I mean the unit circle the distance of every point in the unit circle from the origin is one so this is going to be 1 * 1 which is one so you see that Z * Z Prime is also of distance one from the from the origin and hence it's again an element in the circle and in fact what happens if Z is a this vector and Z Prime is this Vector then you have two angles say Alpha and Alpha Prime and the multiplication of Z and Z Prime is simply adding these two angles this is z * Z Prime the resulting angle is Alpha plus Alp Prim and we okay so this so you see I mean this this group is written multiplicatively but what is really going on is addition of angles okay okay now within this group I have the group of n roots of unity um this this was called MN C this is the set of solutions to the equation the complex equation Z to the N equals 1 okay so if you take n equals a three for example what you do is you divide the circle into three equal parts be like this I gu and you call these guys z0 Z1 and Z2 this is M m3c and it is a subgroup it is a subgroup of S1 y right now I can consider I can I take an element in S1 let's say let's say um e and I consider the cette of X with respect to M3 C what is is it this is the this is a this m m3c i mean I will write as H okay but this is by definition x times Z where Z is an element in the sub group in h M3 C but what does it mean we said we know how to do what how to interpret geometrically um multiplication of complex numbers especially on the on the circle on the circle is just adding the two angles so this set this Co set x times m3c it's first of all we know that this can be a set of at most three elements because m3c is has three elements I cannot I cannot get more than three elements but in fact it's it's just a shift so the cette will be um I mean let's say that this is Alpha then I shift z0 Z1 and Z2 by Alpha so it will be now this is going to be x times z0 is simply X right because I shifted z0 by Alpha z0 has has angle angle zero so I get so I get this Vector then I shift Z1 by Alpha so I get this Vector this is X Z1 and then I shift Z2 by by Alpha I get this Vector X Z2 okay so you you see I the coet of the the subgroup m3c with respect to some general element X is just rotating it's just shifting so these cetes are just shifts of the of the group um yeah yes understood I mean okay maybe someone has a question my question was more instead of multiplying that that was actually to take the other variable thing which was r instead of the the alpha the scaling the so still multiplication by a uh by what is your group r with respect to plus R2 with respect to plus uh basically it's you still take the the group which is the the unit circle so S1 yes yes but and you still have this uh complex multiplication M but you multiply by a series of numbers that are by real numbers so you just scale the the tree sorry the tree the circle uh that would create coets too right yeah okay so so your example that's maybe it's good also to to picture it so no no no we can do uh no it's a CL yeah okay so the group is a circle ccle ah right no so what you can you can make it work but you need to be a bit more careful you can look at h g to be the complex numbers without the origin right without okay right because that was going to be my question precisely that's what I was trying to get and and and then this is a group with respect to complex multiplication okay okay and now S1 is a subgroup is a subgroup of G with respect to multiplication now I take an element X so this is H I take an element X in C minus the origin and um and I look at the cette xh xh is like X S1 and now you need to understand what does it mean to do multiplication of complex numbers geometrically if you want to but doing multiplication of complex numbers geometrically is you multiply the the radius and you add the angles okay so um if x is uh like this H then um I mean since I I do x times S1 for all S1 then let's say the radius of X is bigger than one then I get whatever this this R then I take the unit circle and I stretch it by R and that's that's [Music] myly that's my coet so that's a coet but it's not a group right because it doesn't have a neutral elements it's not a group no it has a neutral it has a neutral element ah yeah it it doesn't because it's not it's not one it's not a group for several reasons yes also when you multiply two elements here this is like X S1 but the radius the radius here is like um it's basically it's basically actually it's r i i i i drew it wrongly and now I take the the unit circle the the circle with radius R you see and this this is not closed under multiplication because you take two elements you multiply them the radius is being multiplied so you don't get a new element on this circle right right but nevertheless I took it for all X I can cover all the plane with this cets yeah okay so and this is really my my next proposition so just before we go uh yeah so you could cover all the plane except the zero that was that's what prompted me to C to to bring this example but I didn't start with zero right so you can't said zero if I if I included zero then then it wouldn't be a group with respect to multiplication yeah okay cool sorry okay so so er to really pin down the phenomena we have the next proposition um for Group G and a subgroup h the set of cetes of all cets um I write it like this x h such that X is in G or I guess maybe a notation you're are more used to is like this okay let me just I'll write the the the two notations usually H it's more handy to write it like this X and G but this this notation really means just x h such that X is in G okay so just to um abbreviate this uh this notation and we have we have this one and I claim this this is a set of sets right it's a set of sets or a set a set of subsets of G and um I claim that this um forms a partition of g i e what is a partition a partition it means that first the union of all the guys in the set that is the union of all X age such that X is in G this is a union of subset of G and I claim that the union is G itself that is this collection of subset covers G and the second the second condition for being a partition is that um if x age and Y age are are two two elements in the the candidate for being a a partian then either xh is equal to y or x h intersection y h is the empty set okay so in words in plain words um it means that um let's write it like this X Y and G then either x x is X AG is equal to Y age or X age intersection y is this joint is the empty set and and so a partition in words is is a collection of subsets of the ambient set that covers the ambient set so that the union of all the all the subsets is the the big set and every two sets in in this collection are disjointed or I mean either equal or District okay now note that um there is a lot of repetition when I write it like this x age or or like this all the X age where X is in G this is a collection of subsets of G but it has a lot of repetitions we actually um I'm not sure if we proved it last time but um well okay we can leave it later but but we um for example we did see that um if for example um X is in age then X age is simply equal to age because I mean this this is very easy manipulation I mean X age this cette of age with respect to X is X AG aging AG but um I can write I can write it simply as um x * x -1 age we age in age or yeah and this this is simply and it's all the age that age you see it's like um for example if x is the unit element the neutral element one then clearly this is H so the the cette X age may be equal to age I mean if x is in age there will be many cetes X age that are simply equal to AG all the cetes I mean with respect to to elements in age are are age again so in other words here you have a lot of repetitions but remember set a set we we consider a set as as something that ignores repetition so it's handy to write it like this x age X is in G and do not worry about repetition and because I write it as a set it means that I I omit all the repetitions okay so back to the proposition let's H let's see how it goes so so so first we have condition a right condition a we want to show that the union of all each age is G the X in that's very easy because um um for one well for any one is always an element in age so x * 1 so for any X in g x * 1 belongs to the CET x h right so so so the union the union of all X age is already d right because I got I took a general element X and I can express it as an element in some CET named the cette of age with respect to X and this means that um I can get all the elements of of G as elements as as as belonging to at least one of the cetes is this clear this this should be the easy part but I don't want to move on so okay so to show equality of sets we need to show double inclusion but one inclusion is obvious this is obvious because it's a union of subset of G okay a union of a union of subsets of anything is a subset of this of this thing okay so it's a union of subsets of G so it must be a subset of G so now I need to show this inclusion so I take an element here x and I want to show that it belongs to the Union what does it mean to belong to the union it means that it belongs to at least one of the of the factors of the Union okay so I I took X and I X always belongs to x * H to the CET of H with respect to X because X can be written as x * 1 and one is an element in h h is a subgroup right so X which is x * 1 always belongs to age and hence it belongs to the union okay Yan how do you feel about it more yeah I think I I got some of it I'm fine with the the first obvious one it's more like the other one yeah okay so let me repeat I mean I take and how do you prove inclusion of sets you take an element in the the small set and you show that this this this element is in the big set we want to show now this the blue direction right so I take an element so I can write it like this let let X be in G this is an element in the small set and since one is an element in ede X which can be written as x * 1 is an element in X age because X age I remind you is just x times AG where age is taken arbitrarily from the subgroup age in particular I can take small age to be one so ex so the the bottom line is that X is an element in the cette X AG okay so if it is an element in one of the group in one of the sets of the Union then it is an element in the union union of of set so X is in the union of all X Ed I guess in here is a bit confusing because um and maybe we do it like this all th with T's in G okay it just uh okay so this is One Direction This is the easy um now we want to show that for any X when Y in G either X age is equal to y or xh intersection y age is the empty set so there could there could not be only partial intersection so suppose xh intersection suppose um xh intersection y contains an element Alpha and then I can write Alpha in two ways right I can write Alpha as xh there exist some age in h Prime such that Alpha on the one hand is equal to xh and on the other hand it is equal to y h Prime right that's what it means to be a cette um and then I can write X as y h Prime hus one and I just multiplied this equation by H minus one on the right and age AG Prime H minus one is in h right this guy is in h because H H minus one I mean small H to the minus one is obviously an element in h because H is a sub and then again because H is a sub group you take H Prime Times H to the minus one you get an element in ag because AG is closed under multiplication so what I'm saying is that this guy y h Prime H minus one is an element in y AG the CET of AG with respect to Y because I can express it as as y times an element in ag now I do the same trick for y I can express y as x h h Prime the minus one so I I conclude that Y is an element in the CET of xh now um now show that in this case X age is in fact equal to Y so the claim I I just reformulated the claim that I need to prove I mean if I have one element Alpha in that is common to the two cetes then they are equal that's that's an equivalent claim to what we want so I have this Alpha and I want to show that xh is equal to y h again equality of sets you prove by a double inclusion and so here is One Direction of inclusion H let a equals um X um I don't know Xs be an element in xh okay then X I have a formula to write X as an element in y AG right so I use this formula I do um X sorry a a is equal to a um y h Prime H -1 this this is x times s but by associativity I can write it as y h Prime hus one time s and the whole thing is in h right when I say A is equal to XS it means that s is in h so Al together I get that a is an element in YH I start you see I started with General element in xh and I I conclude that it is in y now I do the same this is symmetric um so y age is contained in X age and I take B to be and um YT a general element in y right so this I really mean that t is an element in ag and I have a formula for y what I have a formula for y here so instead so I plug in this formula and then B is equal to x h h Prime minus one times T but this whole thing this whole thing is an element in age because it's a multiplication of elements elements in age so this guy is an element in X Edge okay so now this proves the the inclusion in the other direction so hence xh is equal to Y and this this ends the questions before we take a break all right let's meet in uh 10 minutes e oh okay so the picture for the the L theorem Let me Give an example we had um you pick um you pick an integer and and you consider even let's pick a a positive positive integer and you consider the subgroup of all multiples of N and we saw that this is a subgroup of the the group of integers group of integers with with addition and so what is claim the the proposition we we proved before the break it says that if you take if you take the collection of all cetes x + n z where you run over all elements in Z you get the whole thing but I I remind you we said before the break that this collection of cetes has a lot of repetitions and we would like to pin down exactly when there is a repetition this is the fing LMA um for AG and G is above I mean G is a group and H is a subgroup and where and X and Y some elements in G xh is equal y h if and only if x y -1 belongs to H and this is is the same as saying that x -1 y belongs to H going to be equivalent okay so this this LMA characterizes when two cetes are equal two cetes are equal if the representative this is called a representative if um when you multiply the two Representatives where one representative you take the inverse you get an element in h okay so how how to prove this um suppose remember we have an if and only if so um One Direction is if xh is equal to y h then I want to prove that X Yus one is in h so I assume the two cetes are equal since y can be expressed as y * 1 it it is in y h and since the two cetes are equal then why must be an element in xh IE there exist some AG in age such that Y is equal to xh right that's what it means to be in xh and but then um we want X so y x -1 is equal to H and H is an element in h so YX - one is an element in h and note also if you want to do it in the other direction X is equal to x * 1 so it is X is an always an element in the CET xh and the CET xh is assumed to be equal to the cette Y age so X must be an element in the CET y h IE there exists some H Prime such that X is equal to y h Prime and then X well y -1 x is equal to H Prime which is an element in h okay so this is it's the same the whole the whole claim is anyway symmetric so you can do either X x y - one or x - one y doesn't matter and this this proves the the first Direction um conversely suppose X Y -1 is an element in H and we want to show that the two cets are equal and we we to show equality of sets again we do double inclusion so um Y is equal to x x -1 y and uh okay so let's suppose let's suppose x - one y is in h then Y is equal to x * x -1 Y and this is this guy is assumed to be in ag so y belongs to xh and um if y belongs to X age then it means that X age must be equal to Y AG [Music] why I claim that now now that we know that y belongs to x h to the CET of AG with respect to X I can immediately deduce that X age is equal to Y age well because the partitioning theorem we saw before yes two coets are either equal or disjoint if one element is is in both sets then they must be equal okay so this proves the LMA so LMA again tells us um that in order for two cetes to be equal if you multiply their representatives and you do inverse on one of them then it must be in the in the Su this an equivalent condition of when the two coets are ER and just as a remark um if Alpha is is equal to xh well if Alpha belongs to the CET xh the and Alpha is equal to xh or some age but I claim that there is a unique for a unique age age in age so you can express Alpha uniquely as as a product of X with an element in H means the coet is not in the case we were talking about uh you know like with the lines of the plane if the if the base group or I don't know how you call it the the group The ambient or the sub uh the sub group that we used to create all the coet this one is also considered as a coet in that case it can not be uniquely because if the X is not unique but once you choose X so if you fix an X ah okay um if you fix an X then Alpha is equal so um or fixed X if you get an element in the cette then it is uniquely expressed as x * said why um if you if you can write Alpha as X AG on one hand and Alpha as X AG Prime on the other hand then I can multiply all this equation on the on the left by xus one and what I get is that x -1 x h Prime is equal to x -1 x h this is the same as saying that H Prime is equal to [Music] H okay so you cannot do I mean this um there exist an age but it this this age unique now fix X and Y and Define a function f from the cette of age with respect to x to the cette of Y with of at age with respect to y by what do I do well f i I plug in some xh and I mop it into y okay this is this function is well defined because of the remark because I I I I know that every every element in the cette xh has a unique representation as as E's small age and I take this representation and I I mop it into another unique representation y now I claim okay f is clearly subjective right because every element in the cette y h is is being hit by by some I mean by an output of the function f right I mean the function f covers all the all the all the range all the target y okay take an element in in the the target target set y h how does it look like it looks like y * small H okay if beta is an element in YH then there exist some AG in ag such that beta is equal to YH and then um f of x h with this particular age will be y h which is better okay so this is clearly subjective I claim that this is also injective over if is injective what do I need to show in order to prove that f is injective I take two elements in the domain the domain set two elements that are different and I need to show that they that F Maps them to different outputs different elements in the in the [Music] Target H if I have um let's say Alpha and Alpha Prime alpha alpha Prime in the domain H with let's say Alpha is is xh and Alpha Prime is xh prime then F of alpha well let's put it like this if F of alpha is equal to F of alpha Prime then I just by definition of f it means that X sorry y h is equal to y h Prime right because F Alpha is X H so F Marx xh to Y and an alpha Prime is xh prime so F Maps Alpha Prime into YH Prime if these two are equal then it means that YH is equal to y h Prime but then I multiply all this equation on the on the left by Yus one I get that H is equal to H Prime and therefore Alpha is equal to Alpha Prime the representation is unique okay so what I get is sum up the function f from X age to Y age that takes an element Alpha and Maps it into say F of alha is bjective this is without any assumption that Ag and G are finite this is is always bjective function but in case G and H are finite bjective means that the two sets have the same number of elements right so um so now suppose G is finite then this this discussion says that the number of elements in xh is equal to the number of element in y AG for any X Y in G right this is what we just proved and so G can be partitioned into a a um sets of equal size how many of them I don't know because there is a repetition and I have a condition to pin down when a repetition happens but I don't know exactly how to found how many repetitions but either way I get a famous theorum by the r this L and L L gr says if G is a finite group and H is a subgroup then the number of elements in h must divide the number of elements in G and the proof is basically the discussion we had CU if I can if I can partition g into a a sets of equal size what is the size of xh the size of xh for any X for any X is the size of H age is also in this partition maybe I add it into the equal size and this the size is equal to for example the cette 1 * H the cette 1 time H is simply the the the subgroup age okay so the proof is what we I mean m g is equal to the unit of all X age G and as a partition and um the number of elements in X age for any EX the number of elements in X age is has to be equal to the number of elements in the cette 1 time H which is the number of elements in h so I get that e x is partitioned to sets of equal size each of them is of size the size of edge okay so the number of elements in G is is just a just a multiple of of the number ag yeah maybe it's me over interpreting the complexity of uh the last not the like but everything I came before but just the fact that we have um you know for every Alpha there was there will be a unique H in in h in Capital H to Define it doesn't that already say it's a bje and therefore every don't you get that more like faster you get it more faster if I mean yeah I was just uh I mean if you I mean I was explaining stuff that are usually referred to as trivan details right right but again it depends on your background I mean if you if you didn't see the the material before what I call trivial is not immediate right so I just explain all the details okay yeah I mean it's it's in hindsight for yeah I mean but to you know I mean to show that something is bjective you need I mean I I break it up to all the pieces you need to show subjective you need to show injective ER injective basically I mean injectivity is is completely equivalent to the unique represent ation yeah subjectivity is immediate okay so you get bjective yeah right right yeah okay okay so this is this is a one um one dividend we can draw from the this this proposition the fact that we have a partition but there is another another thing that we can conclude from the dep partition I remind you that when we talked about equivalence relations we said that one S I mean one equivalent way to to show I mean to talk about equivalence relation is to talk about partitions if I have a set weall or set s um a relation Ta on S is an equivalence relation if if and only if we can partition s to the equivalence classes of the elements in s so I write it like this Union of all brackets s and s is in s and just to to give you one example I mean you take um you take the equivalence relation on S equal z um you take the relation F to be um a is equivalent to B if and only if a is equal to B mod n this is an equivalence relation right and what is the equivalence class of of a the equivalence class of a is um by definition it's a a is the set of all let's say X in in Z such that a is equivalent to X right the equivalence class is all all the guys that are equivalent to youu and to write it more conely in our case it means all the X in Z such that X is equal to a modu and if you so this is the equivalence class of a if you now run over all a in Z you get a partition because I mean the equivalence actually you have the equivalence class of zero this would be 0 n and minus n also 2 n- 2 N and so on you have the equivalence class of one it will be 1 n + one - n + 1 and so on and you have the equivalence class up to the equivalence class of n minus one because the equivalence class of n is going to be equal to the equivalence class of of zero so you see this is a partition of the integers I mean the collection of all equivalence classes or if you wish it's the same as the collection of all equivalence classes where I don't I mean I run over all elements in Z this is a partition of Z right I mean I take z z is is all the dots on the the line of numbers and uh here I will have let's say zero and here um so it would be zero and let's say n is equal to 3 so I have in the same equivalence CL class of zero I have three and some other guys and then a second equivalence class would be the equivalence class of one it would be one four and so on and the third would be in the orange guy and two so you you take all equivalence classes you get a partition of the set this is this is true for any equivalence relation it's in fact it's it's an equivalent definition of an equivalence relation I just um as you give me a partition I mean if um let's say s is equal to s i i is in some index set and Si is a partition of s that is a remind you um for any I and J in the index set um either SI I is equal to SJ or Si intersection SJ is the empty set if I have such a partition I can define an equivalence relation till that on S by um setting X equivalent to Y if and on if there exist some I such that X and Y are elements in si okay so this this is the other the other direction you give me a partition of Z of the integers for example then I I def I I can define an equivalence relation by this partition so if you take this partition for example you take this partition of the integers then I say the two elements are equivalent if they belong to one one set of the partition and this this would be the same as saying that two elements are equivalent if if they are equal mod but coming back to the theorem I mean we had the partition and so for um two sub groups we have a partition of g h by The Collection um of Coes x h x is in G and this partition defines an equivalence relation sorry just you so the equivalence relation is um I take Alpha and beta in G and I say that Alpha is equivalent to Beta if and only if Alpha and beta belong to the same CET or so of course if they belong to one cette they cannot belong to another cette because the cetes are either equal or distraint and so if I have an equivalence relation I can take the quotient the quotient is taking all the um all the equivalence classes of element the equivalent class of alpha in this case the equivalent class of alpha which I remind you it's it's all the all the beta in G such that Alpha is equivalent to Beta but I defined Alpha equivalent to Beta if and only if they belong to the same CET so the equivalence class of alpha is simply the cette xh okay then G modul remember when we talked about equivalence relation we talked about the quotient of a set by an equivalence relation this by definition it's the set of all equivalence classes but in our case it's simply the set of all cets definition the quotient of G by H is um descri G mod h i define it to be the quotient of G by the equivalence relation coming from the cetes all the cetes of H IE this I just this is just the quent on in plain plain words is is simply the the collection of all coets okay and but I I wanted to explain it from the view of equivalent relations to justify the terminology quotient really means here that we quotient by an equivalence relation that is canonically obtained from from the sub group AG I take all the translations of of the subgroup AG by elements this is the cetes and this this gives me a partition hence an equivalence relation and I mod out uh I think this this is a good place to stop I mean of course I can define a I mean the the next thing would be to define a group structure on the quotient set but just to wrap it ER wrap it in examples in example um let's say G is the integers and H is more multiples of a particular natural number n then g mode H is um it's the collection of all cets so this is the CET of zero CET of one up to the CET of n minus one of course um I can I can write cetes of other integers but this would be a repeating the cets written in this list okay so you see there is a group structure a natural group structure that we know on this quotient because I can I mean if you want to um the the operation here is addition right so I just in the in the quotient group in the quotient set the operation I can take is addition mod n so zero or I don't know a plus b would be the coet of a + b modu n okay and in fact this is generally true if H is a billion so this this is a a generally and one can Define canonical group structure canonical I mean really induced from G from g& H um on on the quotient set G modu age if AG is a billion or if if G is a billion okay this this uh something you can look in the notes and of course we will cover it eventually if we get to it but it's a good place to stop I think here all right uh questions on this um yes it's a pretty stupid as a question but why do you say g modu h instead of g quoti h this say modu and quo for me is syn is it uh to mod out is to right right yes yes okay okay um so I hope we can H ER get more people I will write something on Discord H maybe if if some of you if any of you know other people that want to be I mean might be interested in joining ER either let me know directly or some post something on Discord we'll try to collect a [Music] group sorry there is a meet at some point soon yeah okay I'll just end the call e e