PSE Lectures - Ep 4 - Groups (3/4)

Transcript

4.2 and in the notes um what what we're going to talk about well you you know about groups and uh you know suppose I have two groups one with uh operation dot g and the other with operation dot age

and I would like to to be able to move between the two groups so to move between sets is just to have a function but to move between groups it needs to be something that respects the additional structure of uh of the groups so this is this is the following definition a G and H B groups um a function

if from G to H is called uh oromorphism if for any x and x Prime in G f of x times x Prime is equal to f of x times f of x Prime

foreign with sub G and sub H but here the multiplication is is done in G and here the multiplication is done in H okay um so well one one immediate question you you can ask multiplication is not uh is not the entire structure of a group right I mean we have this unit element and we also have inverses

but in fact from G to H is homomorphism then immediately uh automatically F of the unit element in G

has to go to the unit element in age why why is that so well you take F of we take for any X and G you can take f of x but but X is also e times x so since f is a is a homomorphism I get that this is f of E times f of x and now

f of x is an element in age and age has inverses for all elements so I can multiply this equation on the right by f of x minus one and then I get that e with e of H is equal to F of E that's one thing and the other thing if I take any any element in G and then f of x

inverse must be equal to f of x and inverse here again the inverse on the left hand side is done in G and the inverse on the right hand side is done in H why is that I can write e of H as we saw this is f of e of G but e of G is equal to f x minus 1.

and since f is a homomorphism I get that this is f of x f of x minus 1. now I multiply this equation on the on the left by f of x to the power of -1 and I get that f of x minus 1 is equal to well f of x minus 1 f of x minus one so

all together I get it is simply f of x Apple sorry so the the inverse of f of x is f of the inverse of x any questions about this a definition and and remark so here we see a principle that happens a lot in mathematics

you are given a definition and then there are some some properties that follow automatically from the the and so this can be used in um in two different ways I mean if someone gives you a map and and ask you to to say whether it's a homomorphism or not then the easiest thing for you to do is to check whether this definition uh this one condition let's call it one is satisfied on the other hand if someone a

tells you that a certain map is a homomorphism then you can immediately deduce uh the two other properties let's call it two and three okay so this is a this is typically uh what is happening what will happen with the a lot of the Notions we we Define of course equivalently I could just say I could have said that the homomorphism is a function that satisfy condition one two and three but that's that's really a matter of taste

example we take the map from the integers to the integers module n and what I do is I take some integer X and I map it to each module n well that's a that's that's a homomorphism uh how do we see this well I need to

check that f of x of X Plus y that's now the the binary operation in in the integers is plus and the binary operation in the integers module to n is plus modulo and so I need to check that f of x plus y which I know is equal to X Plus y module n is equal to a f of x plus F of Y but what is f of x plus F of Y

you do f of x a so here the the Plus is modulo n but f of x is simply X module n F of Y sorry here should be y and F of Y is y module n and of course if I do modulo n to all of that

and then this question mark is in fact a true equality so here okay that should should have been an easy example are slightly less easy example I take two natural numbers m and n

and I I consider the function from integers modulu n times m to integers modulo n and this function is a is actually like like we had before is equal to X multiple l okay and now we have a

slightly more involved a condition to check because the the Plus in z modulo n times m is is already a modulus um to see is a homomorphism we write

X as n times q1 plus R1 this I can always do for any integer it's like doing the integer a long division with n so I get some quotient and I get some some remainder and similarly I can I can write y is equal to nq2 plus R2 and I also do it for X Plus y

X Plus Y is some multiple of n plus some remainder then if I do f of x then this is by definition it's R1 F of Y is R2 and f of x plus y is R3 and what I need to check is that a f of x plus y

is a f of x plus F of Y but f of x plus F of Y a r R1 plus R2 R3 and from these equations what I have is that I can I can

I can deduce from Star R1 Plus nq1 plus R2 Plus nq2 Plus n q three and and this means that R1 plus R2 are equal to R3 module n and so so I get so I get

the quality I wanted f of x that's F of Y for importing Expressway okay is is this clear any any questions so yes I see I see a question in in the

chair a model operation is an easy way to construct group homomorphisms this is is is the thumb rule it is a it is true but you need to be careful uh let me show you another example suppose I I do the same thing but from integers modulo NM plus 1 to integers modulo m so I take I take an element X

and I I send it to X module n now I will not be able to uh to use the argument I did before because well I mean because I have this this plus one so I guess exercise show this is not homorphism

okay so so you need to be a a bit more careful here's another another uh example so let me just say in in a terminology um we always have for any groups G and age we have the the trivial that takes

an element X in G and send it to the unit element this it's very easy to check that this is a homomorphism because if I have x and x Prime then well x times x Prime must go to E of H and also X maps to e of H and X Prime maps to e of H so when you multiply when you multiply the results you get

again little bit so this this is a claim there is no non-trivial or morphism from the integers modulo n to the integers if there was

let's call it F from integers modulo n two integers then for any non-zero element in the integers modulo n X Plus itself n times which is usually denoted as n times x

right this is equal to zero in in Zim in z modulu n and since f is a homomorphism I get that F of X Plus itself n times to f of x plus itself n times this is like a a repeated application of

the homomorphism property I do it n times if you want you can uh I mean you can write of course this the sum X Plus X Plus X and times you can write it as X Plus and plus the sum of the X's n minus 1 and and apply the the homomorphism property this is f of zero and F must must send the unit element or the neutral element to the neutral element so this is zero

so we get that a f of x n times which is a at the end n times f of x must be equal to zero in the integers and there is only one one number that satisfied this property namely f of x must be zero F must be must be the trivial so you see there is a a break of symmetry if you will I mean we have

constructed an easy map from the integers to the integers modulus n this was like modulo n but conversely it's a I mean apart from the the trivial homomorphism it's impossible to find any any way back I have a question yes and is it sufficient to say uh or to check if the target group is smaller than the yeah then the other group it it is sufficient for what uh that there can't be any homomorphism no no that's uh that's not sufficient

uh generally speaking you can have a homomorphism which is subjective and the homomorphism which is injective and they I mean both both existers so you know what let me give you an example I suppose G is a billion and age is a subgroup then

I can Define what is called the quotient mark I denote it as Q This goes from Gmod h sorry from g to g mode h in Gmod H I remind you is the the set of all cosets exh such that X is in G and Q is simply given by Q of x xh this is a well-defined function

and and in fact this is a group Q is a homomorphism well what do I need to check I need to check that Q of x times x Prime to Q of x times Q of X Prime but Q of x times x Prime is x times x h and in the the right hand side is simply

times x Prime H which by definition this is x times x Prime h so indeed we have inequality we have a and to answer your question of course if if G is finite and age is not is not the the trivial group consisting of only the unit element then g modulo h is is strictly smaller than G but still you always have a homomorphism and my question was if it's now reversed

basically so from from a smaller group into a bigger group okay A a note also that the inclusion age I denote it like this this you can view as a function age is is a subset of of G so you can Define the function let's call it maybe y why in age goes to Y because Y is is a subset of G

and this is also a group homomorphism this is the trivial one right this is not trivial this is not trivial you could say it's simple but there are other examples so so basically you know I could well we haven't talked about isomorphisms uh but there could be an isomorphic copy of age in G

and I could map Age into into this isomorphic copy uh okay for another example maybe this will answer your question take G times h G and age are some groups you have G times the unit element in H remember G times H is the set of all X Y such that X is in G Y is in H

and G times H this is again this is the Cartesian product of a set with a set in with one element so this is by definition the set of all ordered pairs X comma e h such that X is in G okay then a then of course G times a d g times e is it can be included into a G times H right so this

this is maybe a the same the same as the example here we have an inclusion of a of a subgroup this G times e h is a subgroup but I I can also Define a map from g into G times e h given by X going to X comma e or X comma e h and the composition of these Maps is again a homomorphism so we will we will soon prove it yeah you know what maybe let's let's

prove it so observation a f from G to H and G from age to K are group homomorphisms G composed f which is of course which is a function from g into k

is a group homomorphism why is that it's very easy if I take two elements x and x Prime and I apply the function G composed f on this uh the product of X and X Prime this is by definition taking G of F applied on the product X times x Prime

but f is a homomorphism so it is like saying taking G of f of x Prime and G is also a homomorphism so this is like taking a g of f of x times G of f of X Prime and G of f of x is by definition and G of f of of X Prime is by

definition G composed f okay so we get the left hand side equal to the right hand side that's exactly the homomorphism property so now just going back to this this you you can define a homomorphism in blue by taking X to X comma the unit Element e h and this is a homomorphism it's very easy to see uh well the the multiplication in G

times e h is the one induced from G times H which means you multiply coordinate Wise by Co but in the second coordinate you don't do anything so it is like multiplying in G so this this blue map is a is clearly compatible with the the binary operational on the on the left and on the right so it's a homomorphism and on the other hand the the purple map which is just the inclusion of a subgroup is also a group of homomorphism so the composition

is is a group homomorphism so you you get a group homomorphism from G into G times h so here is something small in here something big is this answering your question yes okay so this this is all uh rather uh simple examples let me remind you that last time

we took we talked about the quotient group and uh we took the the integers module I mean sorry we took the integers and within the integers we we could identify a subgroup which we denote as NZ this is all the multiples of a of n so it's like n a a and z and then we can look at the at the quotient and we identify the quotient just by definition it's the set of all

cosets so it's like X Plus n z where X is in in Z but there is a lot of repetition here so to find a a a set of Representatives without repetition you could you could take zero plus n z 1 plus NZ up to n minus 1. thus NZ and the operation in in the quotient is well you take let's say X Plus NZ

and you you want to add it to Y plus NZ then you simply do X Plus y modulo n plus NZ and so what we see here this this is like this is like z modulo n right I mean I have n elements they are denoted slightly differently I mean instead of zero up to n minus one I have this uh

this tale of a I mean 0 plus and Z One Plus NZ and so on but the operation is is essentially the same I mean that's what you have here in blue is really the operation in in Z in the integers module n but they are not equal as six so so what can I say well this this is the notion of an isomorphism or the groups

F from G to H and isomorphism it is rejected Okay so

a a to get the intuition uh suppose I have a an isomorphism from G to H and a g is a set of three elements say X1 X2 X3 you can think of X1 as as the unit element but it doesn't really matter then the the data of of the group G can be depicted as as a multiplication table

so I can I can do X1 X2 X3 and also vertically X3 X2

X3 X3 since a a f is an isomorphism so so suppose f is in isomorphism this is denoted by the way a by by saying a by putting a equality with the tilde over the arrow or you could write a g is isomorphic to H like this and since so if we assume that f is an isomorphism then I can then F reset is bijective so

I can enumerate age as F of X1 F of X2 F of X3 and I can write the multiplication table for this F so f of x one and here what I have is f of X1

f of x one and so on but f of x one times F of X1 because f is a this is equal to f of X1 times X1 and similarly in all the other cells so I have like f of X1 X2 F of X1 X3 F of X2 X1 and from X2 X2

X 31. so I under under this a a under this map f the two multiplication tables are are identified I mean up to a change of symbols in one table I call it x 1 x 2 and X3 and in the other table I call it X Y I mean f of x one f of x two and f of x three the multiplication tables are identical I just I just apply F on the on the symbols so you can think intuitively on

an isomorphism it really tells you that the two groups are the same up to a change of symbols um and of course a a the same principle holds also if G and age are infinite I mean then I would not have this uh this easy drawing a a a alternative but you can you can still think abstractly the the if GN group and G and H are infinite groups that are isomorphic then

really the the whole data of G is the same as the as the data of age up to a change of symbols which is given by the by uh by the isomorphism any questions about this this notion of so the the example uh we we actually basically talked about is a a defined the map from

the integers modulo NZ into the the integers module n and this is given by a X Plus NZ maps to X then this is this is a homomorphism uh simply because

what you do on the left when you add to elements it is simply as we said it's simply X Plus y so this is sent into a X Plus y module n and of course this is equal to x modulo that's why

so this is a homomorphism but it's also it's very clear that this home this function is bijective because what I have here is zero plus NZ up 2. n minus 1 plus NZ and here on the on the left sorry on the right I have 0 up 2. n minus 1. so this is a one-to-one correspondence I mean I take X Plus and Z into X f is clearly

adjective so this means that the integers modulu and Z is isomorphic to the integers Model N Let's uh let's take a uh maybe a more interesting example so recall

a I have I have the circle in in the complex numbers and on the circle I can look for a what we called mu n of C this is the a the collection of all n roots of unity if n is equal to 3 for example then you can talk about a Omega 0

Omegle one Omega 2. Omega 0 will be of course one mu mu n of C is the collection of all X such that x to the n is equal to 1. or X in the complex numbers you see here I have a I have an equal division of the circle so this is

120 degrees this angle is 240 degrees a 120 is like pi divided by three times two and this is a two times this so a 2 pi times 2 divided by 3.

so in fact Omega I is going to be equal to cosinus 2 pi maybe let's do it k Omega K is equal to cosines 2 pi k over n plus I sinus 2 pi okay okay this is generally true I mean for any n so this is a this is the

uh the collection of all and roots of unity I can write mu n of C Omega 0 up to Omega n minus 1. okay and Omega k has this this formula now we said that when you multiply two two elements in this two elements in the circle using the the complex number multiplication what you in fact do is you simply add the two angles

if I take K and K Prime between one or sorry from zero to n minus one and I multiply the two omegas Omega K Omega K plus a k Prime then what I get is Omega of K plus K let let me just write it down so I get of course a kiss K Over N

times kiss of 2 pi K Prime over n and as I said you multiply two two elements in the circle you simply add their their angles so this is case a 2 pi K plus K Prime okay so this is a this is a nice group it's a it's a multiplicative group and now I I can Define from

to the group of n roots of unity given by K maps to Omega k and I claim that this map is in fact an isomorphism of course of groups

I think now would be a good time to uh to pause if you have any questions and if not we can we can take a short break yes sir okay so see you here in uh in 10 minutes all right so I see another a set of questions in the chat um the first is whether the identity function is always a homomorphism and the answer is of course yes it is in

fact an isomorphism so remark the identity function of G from g to g is an isomorphism and this is always true because well the identity function is clearly bijective and what you need to check is the the homomorphism property but if you multiply x and x Prime this is simply map to x times x Prime and so

which is of course the identity G applied on x times the identity Gene applied on X Prime so that's that's easy now the second Point yes but why is that subjective well take a take X here the the element X is mapped to X so it is a source I mean I see now ah okay yeah okay so the identity map is injective and subjective alternatively we we saw

that a characterization of bijective map between sets is that you can go back but here you of course you can go back by simply the identity right so let's there's another way to to see that this is bijecting now the second question is a says isomorphisms are a special type of homomorphisms they chose the two groups a pretty much have the same structure yeah that's that's uh that's a reformulation of what I was trying to say I mean with this multiplication

tables I mean right it's a it tells you that the the multiplication table is is the same up to a change of a of symbols and the multiplication table is all the structure of the of the group I mean of course multiplication table I can write for for finite groups for infinite groups I I need to uh to talk more abstractly um but this is uh this is the the notion of an isomorphism and then the third question is a well homomorphisms as a weaker claim that shows that there is a

map between groups that preserves the group operation so yeah so you could say that homomorphism is a weaker notion than an isomorphism in that every every isomorphism is in particular a homomorphism but conversely is not true so not not every homomorphism is necessarily an and you can think about the homomorphism as something that is compatible with the group structure but not necessarily mapping the multiplication tables in one-to-one correspondence and maybe here is a good time to revisit an

you take we took integers modulu n times M into integers module to n and this this goes as X goes to x modulo now to check that this is a homomorphism I need to check that X Plus y under f a f of x plus F of Y here the plus is in fact Plus modulo

NM and here the plus is a module n so what I really need to check well I take X Plus y to do X Plus Y modulu and M I write X Plus y as n m times n Q3 plus some remainder and I write X as n times Q one plus some remainder and also Y is n times Q2 and then when I apply f

of X Plus y this is simply taking well I first I first do X Plus y module NM so this is f of x plus mod n m Y is simply taking this R3 so F of R3 but luckily because n divides n times m is also it's also I mean a

R1 plus n q1 of course F of R3 or I should say okay uh indeed this is going to be of course f of x is going to be R1 is going to be R2 and we know that R1 plus R2 is equal to R3 modulu so when I do F of R3

this is like taking R3 modulo module n but this is this is the same as doing a f of x plus F of Y a modulo n so this is the homomorphism this all this all argument would not work if you would take in the same map but from N M plus 1 into integers module n taking X

this is is not because it is not compatible with I mean if you you do a the binary operation on the domain this is taking modulo n plus n plus one and then the whole argument that relies on the fact that when I did I got something that is a multiple of n plus some reminder and now it will be a multiple of n times

n plus one and and this and this would not be compatible with the with the module n operation and you can do you can take a a small values of N and M to to really play with the a concrete example is this answering your questions you will be hello yes yes we had a claim before the break that

this map that I defined from the integers module and to the the group of n roots of unity taking K into Omega K is a group of and in the interest of time I will leave it as an exercise it's not hard you need to show basically that well this map is clearly bijective well you you can you can really see that uh it Maps it one it's a one-to-one correspondence between the the elements

of the two sets so all you need to check is that if of K plus K Prime is equal well here the the the slightly a potentially confusing issue is that the operation on the left hand side is plus well plus modulo and the operation on the right hand side is multiplication so you need to check that X that F of K plus K Prime is equal to F of K

times F of K Prime but in fact we had uh we had a claim here and where is it so if you multiply Omega K and Omega K Prime you get this in fact the exercises is almost solved using using this star identity we talked about isomorphisms there is a an easy but important observation

maybe let's make it a proposition is an ISO then so is F minus 1 H to G so I remind you an isomorphism is a bijective map

and a four bijective Maps we we were we defined a the inverse map of course the the inverse map does not does not exist generally but if you if you have a bijective map it always exists and this is in fact an equivalent condition to uh to the to the fact that f is bijective um the definition or F minus 1

well if Y is in age then since there exists a unique X in G is equal to y uh and then we Define f minus 1 of Y to bd6 now we saw

F minus 1 is bijective so what is left to check is that it is a group homomorphism F minus one is a group and so so

let Y and Y Prime B2 elements in H um there exist unique X f of x is equal to y which of course by the definition of f minus 1 it means that X is is equal to F minus 1 of Y there exists a unique X Prime in G is equal to Y Prime

and of course then X Prime is going to be equal to F of Y Prime sorry F minus 1. wipe and now since f is a homomorphism I can take f of x times x Prime and this is going to be equal to f of x but f of x is y and f of x Prime is y

Prime so I get y times y Prime and this really means by the definition of f minus 1 that F minus 1 of Y y Prime is going to be equal to x times x Prime but here I use the the blue brackets this is equal to F minus 1.

times F minus 1. of Y Prime these two purple left hand side and right hand side are exactly the condition I need to check to see that f is a f minus 1 is a okay so this this finishes the proof this justification this is a justification for

a a this notation G isomorphic to H which is symmetric right I mean so so G is isomorphic to age you can equivalently say that there exists an isomorphism between G and H are that they exist in isomorphism between H and G any questions maybe an exercise

take the integers modulu 10. and take the subgroup consisting of the elements 0 and 5. and now you take the the quotient integers module 10 modulu let's call it age and a I want you to show that this is isomorphic

to the integers modulo 5. H should be intuitively easy what the the the main point of the exercise is to to practice the formal next a suppose so let F from G to H would be the kernel

is a the following set it's denoted as care of f it's all the elements X in G is equal to e h this is one the image of if and this is this is denoted is

in f is the set of all f of x such that X is in G so of course the kernel of f is a subset of G and the image of f is a subset of H now we have uh we have uh an easy proposition maybe Lemma the kernel of f

is a subgroup of G and the image of f so what do we need to check in order to show that something is a is a subgroup well we need to check that the unit element is is in the the candidate for a we need to check that the product or the

binary operation of two elements in the are again an element in the subgroup and we need to check that the inverse of an element in the subgroup is also an element in the sub of course a e I mean the the neutral element is an is an element in the kernel because F of is is e a and

if if X and Y are in the kernel f of x is equal to to E F of Y is also equal to E thus f of x times F of Y is equal to E but since f is a homomorphism this left-hand side is equal to f of x times y and it follows that X y

is in the kernel a let's think if x is in G f of x minus 1 we we talked about it f of the inverse of X is simply f of x inverse if x is in the kernel minus 1

is going to be e to the minus 1 which is e x minus 1 is also in the Curve so this this proves uh that the kernel is a subgroup of the domain G for the second claim a suppose Alpha and beta

are elements in the image there exists X and Y in June Alpha is equal to f of x and beta is equal to F of Y of course note that this X and Y they don't need to be unique uh but nevertheless

when I do alpha times beta then it's the same as doing f x times FY and since f is a homomorphism this is f of x times y Alpha times beta also belongs to the image of f of course a trivially the unit element is in the image of Earth because it is f of the unit element in G which is some f f of x is in the image

Alpha minus one is f of x to the minus one and and we showed that this is f of x minus one so Alpha minus 1 belongs to the image okay any question about this uh this this Lemma the next thing to do is what is called

the first Theory and this goes as follows uh let F from G to H B a group over billion groups

and the note K to be the kernel of f then there there is an isomorphism between the quotient G module okay the image of reference okay is is the statement clear I mean we are now a

when I say of course the image of f I said that it is a subgroup it's a subgroup of age but as you you might have noticed if something is a subgroup then it is a it is a group in its own you can multiply two elements you can take the inverse and and you have the the neutral element so so it is a group in its own right I can I can re regard it as such and say uh well and I can claim that it is isomorphic to another group and the statement of the

theorem says that is the quotient of of the domain Group G by the kernel okay so we Define a map we call it f Prime g modulu k to the image of f and this map well

what is g module Okay g module okay is all the X K such that and so I Define f Prime of an element x k to be simply K clearly f of x is in the image of f okay so this this is the definition

as as always or as often in the quotient group we need to check that a a what we what we want to Define is indeed well defined so to check F Prime is well defined suppose well suppose you have two a two representatives of the same element namely

that x k is equal to y k then we saw that two cosets are equal if and only if X Y minus 1 is equal is an element in the in the group K and K I remind you is is simply the kernel well okay so f of x y minus 1 must be equal to E this is equivalent since f is a

homomorphism to say that f of x is equal but let me write it a bit more detail this this says that f of x times F of Y minus 1 is equal to E which is equivalent to say that f of x is equal to F of Y a a thus that the f of x is equal to F of Y is the same as saying that F Prime of x k is equal to F Prime

of y k and and this means that F Prime is well I claim that f is is a homomorphism it's actually it's very easy to see

you see we have from G module okay I take an element x k and I map it but if I now take two elements x k and X Prime k the first is mapped to f of x the second is mapped to f of x Prime and their product is mapped

f of x times x Prime but f is a homomorphism so the product is mapped to the product of f of x times f of x Prime so essentially F Prime is a is a homomorphism because f is such and now the the statement of the theorem is that F Prime is or that there is an isomorphism and F Prime is actually the candidate to be the isomorphism so we need to show that that f f Prime is

bijective um subjective subjective is easy if Alpha is in image f a alpha is equal to f of x or some eggs

if I take the cassette x k and I apply F Prime on it then I get f of x which is Alpha so f is rejecting sorry F Prime a to show that F Prime is injective F Prime of x k then what this really says is that f of x

but this really says that f of x y minus 1 which is equivalent to say that X Y is in the kernel of f but the kernel of f is what we denoted as K and the condition that X Y minus 1 belongs to K is equivalent to the condition that the cosets XK and y k are equal so if I had two elements

x k and y k that are mapped by F Prime to the same the same element in in image of f then they in fact there they must be equal so f is f Prime is injected and this finishes the proof any question about the theorem do you have some examples with a

non-trivial kernel uh sure so uh let's let's do two examples one I take a from the the integers to the integers what is the kernel of f the kernel of f is all the X in z

such that X modulu this of course this map is a x goes to X it's all the X's such that if you do module one you get the neutral element in this case it's zero so in other words this is what we called right this is all the multiples of of n a a the image of f can anybody say

this it should be easy uh but I I want to see that you we are on the same page what what is the image of f I mean I remind you the definition this is all the f x X is in z come on you can do it

I have do I have a do I have one in image of f would it be the whole Z in exactly okay so what I get automatically by the the first isomorphism theorem is that if I take

which is NZ it is isomorphic to z modulo n another example from Z modulo 10 to Z modulo 2 b a f of x is X module 2.

what is the kernel I remind you the kernel of f is all the x's in Z modulo 10 x modulo 2 is equal to zero so what is it it's it's all the even numbers

we need 0 2 4 6 8 and well ten I I do not have the image of f what is the image of f more it's modulus right I thought you you are comfortable with it isn't it that two again

it is Z2 again okay so what I get is that if I do Z module 10 modulo this kernel let's call it k I'll do K I get something isomorphic Z module two now in fact K of course K is a subgroup of Z modulus 10.

I claim and that K is isomorphic to Z modulus five well if you think of it what is the multiplication table the multiplication table is just is the same up to a change of symbols I mean I can define a map for example a let's call it maybe G from Z modulo 5 to K by setting a g of x to be equal to 2X and if you look at it

it's a this map is compatible with the group operation this is this is a it's a one-to-one map it's it's a projective map and it is a group homomorphism because what is a what is the operation in in k I mean the operation in K is Plus modulo 10. but the the numbers in K are simply a a multiples of two I mean there there are double then the the numbers in in Z module five and instead of doing module five you do the double you do a modulo 10.

so so it is in fact the same operation I mean up to a a change of uh of symbols you can check that g is an isomorphism okay so so what we have here it's like we have Z module 10 not exactly Z module 5 but rather an isomorphic a subgroup of Z Model modulo well a subgroup of Z module 10 that is isomorphic to Z module five

and then I get something that is isomorphic to Z modulo 2. maybe another example so this was two you take the quotient map

from G to G times h here I assume that age is a subgroup of G what is the kernel of Q the kernel of Q is all the elements X in such that Q X to the zero element in the in the quotient which is a e h and Q is taking X into so what I get

is that kernel of Q is simply h let me pause for for a couple of minutes to to let you ask more questions but I think this this would be a good a good place to stop all right then let's let's stop here and

see you on the next week I hope yeah thank you thanks