PSE Lectures - Ep 2 - Groups (1/4)

Transcript

Vivian did you hear that Carl's gonna join today or maybe not I don't know I don't think so and I think so okay okay great um yeah glad he's on vacation day okay so um welcome back uh and I before before I start I I just want to make a organizational uh a a consideration I mean

uh I hope that so I cannot always look at the chat and I will try but uh generally speaking just try to speak up or if there is a possibility in Zoom to to ping when someone raises a hand or something like that that would be a ideal and then then I could I could take questions but they otherwise just a just speak up um Okay so

this time we we talked about a a several things but the first thing in order is a correction so when I when I Define the natural numbers uh I said that the natural number N is a the Union of uh all it's a

previous natural numbers but this this is not this is not true what should be is a it's the set of all previously defined natural and the confusion I I mean another another way to to Define n is by saying that it is n minus 1 Union the set containing n minus one

so this is consistent with the way we defined um zero to be the M the empty set one to be well it is the empty set Union the set containing the empty set but any set Union with the empty set is just the set itself so it is and this and then two is going to be reset the set containing the empty set which is

a this and this set so this is not a this is not something we are going to rely on but still I want to be precise about it the other things um we talked about injective Maps a injective amount this is a function from A to B such that for any a different than a prime in a F of a

is different than F of a prime and subjective is a is kind of a dual dual notion it means that for any B in B there exists some A and A F of a is equal to B and similarly

you can say that bijective functions is just a injective and subjective now um we had a we had I think the last thing we did is a theorem or proposition that a if from A to B is bijective there exists a function

which we denote as F minus 1. from B to a F minus 1 composed f is the identity function on a and F composed F minus 1 is the identity function only so this this uh this function f minus 1 is called an inverse to f

um but note that the converse is also true that is a if there exists F minus 1 from B to a F minus 1 composed f is the identity of F composed F minus 1 is the identity of

B then F must be f is is bijective how how do we see this uh this is very easy so suppose so we need to show that f is bijective that is to say we need to show that it is injective and that it is subjective so how to show that f is injective if I take two

different elements a and a prime in a I need to show that F of a is different but what I can do um well I can I can take F minus 1 of f of a this because F minus 1 f is a is the identity this must must be equal to a minus 1 of f of a

Prime must be equal to a prime so it cannot be that A F of a is equal to F of a prime because F minus 1 is a function if these two purple elements would be equal then since f f minus 1 is a function it it would it would have to

associate the a to associate to them the same the same output but we know that a is not equal to a prime so a F of a cannot be equal to F of a prime similarly I guess we can we can put it as an exercise that in in this case f is is also subjective one thing that uh is kind of uh

[Music] a meta principle I guess you could say um Notions of I mean Notions of types of functions like like injective and subjective what is called closed under composition

and what I mean by that is that if if you give me two functions F from A to B and G from B to C um and both f and g r injective or both of them are subjective G composed f

is also of this of this type and a how to see this it's very easy um suppose so suppose we go for a um the injective case so both both functions are injective and I want to show that the composition is uh so I take two different elements a

and a prime in a is different than F of a prime since f is injective and it's G of f of a must be different than G of f of a prime since G is injective so what we get is that g composed f

is injective and the similar argument again you could you could try it as an exercise and shows that a composition of subjective Maps is a subjective and of course if you combine a the two statements about injective and subjective you get that the composition of bijective maps is bijected before we we move on to a to new material I want to

give it a try and tell you about a country's Theory I guess I forgot to to give you the updated notes so you should see it in the chat um so this is this is an aside which is not something we are going to

use but I think you will find it interesting and this is called the country's Theory and what it says is a quite simply there is no subjective map from the natural numbers to the real numbers before I go to the proof are there any

on what what I said so far Okay so how does one go about proving that there is no subjective map this is usually done in a in method I mean approved by contradiction so uh suppose by contradiction or towards towards contradiction there is

some function if from the natural numbers to the real a m and that is rejective a real number I can represent I I I can write in a decimal representation well let's denote F of n the image of the

natural number n under f as x n then I can write x n to be X1 index n dot X2 index n X3 index n and so on where X1 is some integer and all the the rest of the the x i n so for any I bigger than one

x i n is a digit okay so if this is the case now I can order the I can I can basically write f is is an infinite table as follows so I get one is mapped to and X1 we said we denote it as X what x 1 1 dot x 2 1.

and here possibly a infinite sequence in in fact we can we can always say it's in infinite sequence that may end in a A Tale Of A Tale of zeros so the same goes for two two maps two X2 which is X 2 1 X 2.

2 and so on then on the nth element we have x n which is X and one dot X and two and here it goes on let's say I can I can continue it to X and n and of course it continues after that so just for a for symmetry here I also have X

2. n and here I have X I Define some element Y in r as follows so why it has a decimal representation let's say y1 dot Y2 here it goes up to y n for example and

it continues y1 I'm going to take it to be x 1 1 plus 1. Y2 is going to be X 2. well I could do a I could do cases or I could do modulus but let's say um it's going to be x 2 2. plus one x22

is different than nine and otherwise I could take a x 2 2 minus one if X2 is equal to 9. and and the same goes for any for any n so y n is going to be equal to X and n plus 1 X and n and x and N minus 1 if x and n

so the picture and this is why it is called a the diagonal argument a I have this this infinite table and I look at the diagonal and what what I'm doing is I Define a number y a by tweaking each of the the entries in the diagonal of this infinite table and so um note Y is not equal to x n

for for any end I guess say it like this and for any n because the nth digit of Y is different by definition from the nth digit of x n thus f is not subjective

and that's that's a contradiction any questions about the proof hello I'm not sure you're I'm not sure I can hear you oh can't hear you ah okay so this is a bit of an advanced topic but I thought you could digest it and it has an interesting application

uh for a computer science Computer Sciences and and this is the following example um I choose choose a programming language programming and consider and and let

P bar a a all computer programs in this language I can I can look at a sub collection of p p bar be the set of all programs print

a decimal representation for example you can have the program that print a 0.3 or let's say it it prints zero zero point and then it it prints in a loop let's say while while true

um print three so this is a computer program a legitimate it will compile the issue is that it does not stop but we can consider for such a program p in

curly key the theoretical and print output is an element in r right I mean as if you you would in this example the theoretical output would be

0.33 and so on which is of course is is just a third music a real number and now I claim that there exists uh real number X that is not obtained as a theoretical output is a uh

a theoretical output of any program and in fact there are many many such real numbers um so if you think of it it tells you something about the the impossibility of uh of uh through a randomness and it how do you prove it I can give you a sketch because I get a feeling

you're not a you're not completely a a completely interested but it the the the way to prove it is is you say um if um it it wasn't it wasn't the case then for for any real number we could a construct write a program who is a theoretical or eventual a printing output is uh is this number and this means they would exist a

let's call it f from the set of all computer programs that print a decimal representation this we called Curly p into the real numbers uh by by assigning a for each computer program the theoretical output so p goes to the the output and and what what it implies in fact I can

simply Define a let's say F Bar from P bar to R m um by saying that F Bar of some p is let's say zero does not belong to this uh F Bar of P is zero if p does not belong

to curly p and it is the output theoretical output of P P does belong and this this map is also subjective because the previous map F was rejected uh no um P bar furthermore

we can order the set of all computer programs in a given language in a lexicographic order and this means that I get a

let's call it G bar I get a map from the natural numbers to the set of all computer programs in my my given language this for example um you could do that g of of one is is the program that is simply the string a um up to what is it g over 24 26 I forgot this would be Z and then Geo 20 25 would be a a

and this this map is clearly subjective and and all of this is to say that now I would get a um I have a subjective map from n to P bar this I call G bar and I have a subjective map from P bar we called it F Bar of course as we said the composition of subjective Maps is rejective

so the map F Bar composed G bar from n to R and that that is a contradiction to uh to counter the ethereum I feel that uh you are not completely a the same page are there any questions or I think it's very good interesting

thank you yeah I I wasn't sure if if that would would be of interest to you I thought it would uh I mean you always hear that true Randomness is is not is not possible but how do you actually prove it right so this this is an application of this uh subjective and injective maps um okay unless we have you have an other uh the next topic I want to discuss is the is groups which is uh certainly

relevant to almost anything we we do later on a group is a set G together with binary operation

which I call at the moment I call it mu binary operation simply means a function from the Cartesian product so the Cartesian product of G with itself 2G and you you write it as you take a pair I mean the Cartesian product is the collection of all pairs ordered pairs and you you map it mu of X Y here if I wanted to be formal I would mu double brackets

X Y because X Y comes with brackets as is and our official Syntax for writing a function is that we do the function brackets and the the argument but we abuse notation to avoid a multiple a brackets we we simply write mu x y and this will happen many times later on WE omit a repetition of brackets to to get a clearer more intuitive notation okay so a group is a set G together with

the binary operation a chosen element e the following conditions are permit one is that or any X in G if I take the binary operation of X and the The Element e I guess you could you could call it a

unit this is often called Unit or a neutral element and if I take a um the binary operation of X with the unit element in other side either side I get simply X so this is equal to Mu of e x this this condition is called the unitarity you need charity and the second condition is that for any

X Y and Z in G mu of I guess I need more space of mu x y Z is equal to Mu sorry to yes to Mu of x mu of y z and this is called a associativity

and a the less condition e is is the existence of of an inverse so for any X in G some element which we denote is x minus 1 mu of x minus 1 is equal to the unit Element e but also if you do it on the other side mu x minus 1 X okay and this is existence

of inverse all right so let us try to digest this definition I think you you you probably saw a definition of a group but maybe not in this way it's it's completely equivalent to the way you uh to the the other definitions you saw um but it's a bit more abstract because I Define it in in terms of this this function mu but really what what there is here is a um meta principle

algebraic Notions I guess you could you could even say mathematical Notions am can be described by first giving giving data this data is usually is basically from the axiomatic point of view is just a

collection of sets and two conditions that this data need to satisfy if you think if you think about this a meta principle in in the definition I gave for a group

you could say that the the purple part is the data and the yellow part is the conditions and this will this will a this will repeat in almost all uh definitions that uh that we give um so of course this we discussed last

time I mean here I say that um new is a is a function but a function is just it can be viewed as a set this condition it seems like it may be a a condition that is that is data but in fact it is you can say that this is simply a the existence of you could say that

um I have another function which I call I from g to g and this function is a is bijective a and this function needs to need to satisfy mu of a um of X with with I of I of x yeah I'll try to write it here it's like it's going to be mu of x I of x to e and similarly on the other side

so so I of X we denote as a this we can denote as X goes to x minus 1. okay now the convention any question about the the definition so the conventions in the notations um I am not going to write a mu it's it's too it's too cumbersome to to write

a me all the time so we write Dot instead of of or mu but so it means that dot is like is again is like a a function from the Cartesian product of D of G with 2G and we we map the pair x y we write it as x dot Y and sometimes we even omit the DOT so this is the dot notation but

of course it may be that the binary operation comes from a usual addition or multiplication of and if it is the case we will often denote the the binary operation is a is additional multiplication and the unit element if the binary operation was a multiplication of numbers then usually the the unit element is one so we will write one instead of E and if if the unit if the the binary

operation is is addition of numbers and then we will write 0 for the unit element because this is usually the uh the case this is all a bit of a warning but um we will soon see examples okay so but the main but if if I don't assume anything on the group then I have this a DOT notation instead of mu so the axioms are one x times e is equal to e to X which is equal to e

times x is for all X and G two that for all x y z in G x y times Z to X y z this is what we call the associativity and three for every X there exists x minus one every X in G to exist x minus 1 in G x times x minus 1 is e

which is also x minus 1 times x famous groups one I take a take an object a then

you take G as a set it's just the element e and the binary operation it's the the only the only uh the only possible function that you can Define I mean e e binary operation with itself is is e right note that a okay let me wait for the the next example so this is called the trivial trivial group of course the empty set is not a group

because it does not contain any element in particular it cannot contain an that will be the the unit element a second example is you take the integers with a addition and a the zero element now I need to tell you what is the inverse so inverse of e x in Z is simply minus X

so you see when we when we deal with the integers it is more natural to denote the binary operation as as Plus the unit element is zero and the inverse of X is -6 so you can think of it like it's it's the same it's the same principle that is both applicable to multiplication and addition I mean you have a neutral element with respect to the binary operation

and the inverse of this element is an element such that if you do the binary operation with the the so-called inverse you get the neutral element so if you have real numbers with multiplication then the unit element or the neutral element with respect to multiplication is one and the inverse of an element is of an element X is one over X or what we denote is X to the power of minus one now a note

now I take the natural numbers it also has it has addition and it has a an element let's say I take the natural numbers with with zero and I take addition and I I treat zero as the neutral this is not a group can anybody say why white is not a group because there are no inverse elements

yes exactly they it does not contain the inverse of of of elements basically of of any any non-zero element does not admit an inverse the third example is a one that you you you know I mean I take the integers module n this is you can view it as the set of numbers from 0 up to n minus one but the binary operation is

I will write like this plus mod n and the unit the unit or the neutral element is is zero so all this stuff is is a group so so if a is is an element n minus a is is going to be the inverse of a

right since if you take the binary operation and you apply it on a and N minus a you get well n mod n which is equal to zero and so so n minus a is the inverse of of a associativity you need to check it um but taking module n is is an associative operation I will just write the the equation that that you need to verify um so let's say you have a b and c

in Z mode n then you need to verify that if you take a plus b and you do mod n and then you add C and you do more than you get the same thing as you take a you add to it B plus c mod n and then you do more than this this is really the the equation so

I guess we can put it as an exercise or a b and c I think it's a good time to take a break the the next example I want to talk about is kind of a geometric uh but how about we we take 10 minutes break

yes okay so see you soon okay so the next example I mean I want to to spend some time on it because it gives you an illustration of a a the type of object we eventually are going to study a a namely elliptic curves they have we said that this is a geometric object

a like a solution of a certain equation and it also has an algebraic structure of an abedin group so um consider S1 which is oh you know what I will I will simply start with the complex numbers the complex numbers I'm going to write as X Plus i y such that X and Y are real numbers and you can multiply a two complex

numbers let's say Z is equal to X Plus i and Z Prime is X prime plus I prime y sorry i y Prime Z multiplied by Z Prime is you just do the the usual multiplication what you get when you assemble everything is X x Prime minus y y Prime Plus X Y Prime

X Prime y okay and this is another another complex but a a complex numbers you can represent is if I have Z this is the real axis this is the image the so-called imaginary axis and Zed has this this quantity is well if Z is X Plus i y then this quantity is X and this

quantity is y right this is a it's the same as you can draw kind of a vector that it takes a a goes from the origin to the point Z and and this Vector will create an angle with the Positive a part of the red axis or of the x-axis if you want um and moreover this this uh this Vector will have a length

and call it r so I could write in fact Z is equal to R times cosinosteta plus I the serum stata because here if you do the the usual sine and cosine a definition you get that Y is Arsenal status

and X is our cosine of theta and in in shortened notation we write it as R kiss Theta if you you examine the the multiplication I just defined and you use some trigonometric identities you will get that so if I have Z is equal to are keys and Z Prime is R Prime

kiss Theta Prime then when you multiply the way to multiply is you multiply the the radai r and r Prime and you add the angles so in slightly more expanded the writing this is r r Prime Times cosinos Theta as Theta Prime plus I sinus theta plus data Prime now inside the complex numbers

the so-called unit circle which is the set of complex numbers Z such that it um the the radius of Z is equal to R keys data and r is equal to 1. usually we denote a so if Z is is Arc is Theta we will denote by absolute value of R of Z to be the the real number r

so I could say simply the circle is the set of all Z such that the absolute value of Z is R is is one if I take two points on the circle and is Z Prime each of them creates it has it has an angle say Theta and Theta Prime

and when I multiply them and of course the radius of both of them is one when when I multiply Z and Z Prime what I get is well it's a point that has a radius one so it it must lie again on the circle and the way the way I compute it is I mean I simply add the the angle stata and Theta Prime so this would be something like these were the angle here is Theta

theta plus Theta Prime this is z um and I claim observe the multiplication of complex numbers and a a the element one one is is here the purple element this is one which is a simply one plus zero I

this I claim that this is a group well associativity associativity of a of the binary operation it follows from associativity of multiplication of complex numbers which in turn if you think of it it simply follows from associativity of of multiplication of real numbers of course the the unit element one it angle angle zero

so if I multiply the unit element with any other element I I get this other element so so the unit element is neutral indeed and then I need to tell you what is the inverse can anybody say what is the inverse if um Z is equal to R Keys Theta what is the the inverse we are going to denote the inverse is z minus one because we are in multiplicative notation

um but what is what is the description of a z minus one I mean here Z is equal to our key stata but R is equal to one so I just delete r can anybody say what is a z z minus one is it negative data yes but what is negative Theta oh says of negatives Theta yes but what

is but what is negative Theta I mean we the way we defined it Theta must be positive okay then um three sixty minus yes yes and and I usually use a what is called Radiance so 360 I call it 2 pi so Pi is one 180 and this a a and if you you take 2 pi minus Theta and you you multiply it with with Z so you you take Z minus one

you get a CIS of a m 2 pi and this is this is indeed the simply so it's kind of uh it's a funny it's a funny uh uh case because we the operation is is a formally speaking I mean the operation is multiplication of complex numbers but in fact what the only thing we do is we we add the the angles uh so this this is an example of a a a a

group which is which is also a geometric object a geometric really means that it well it can have different meanings but for example you can look at it as a this set of solutions to the equation e x squared plus y squared minus 1. this is the circle so okay all the examples we did are a a what is called a

G is a billion if for any two elements you can do x times y and this is equal to Y times x and if you think of it all the examples we saw are a a billion groups in all the examples we saw the binary operation is commutative foreign let

n be the set of of all numbers 1 up to n SN I take all functions from n to n such that f is bijective this is called the symmetric group um now I need to tell you what is the

operation so maybe we can call it a a instead of f we call it Sigma Sigma operation tau is defined to be simply decomposition Sigma composed Tau of course well we have this this condition that the elements of the group need to be a bijective Maps but we had if you remember in the first uh the

first part of this session we said that composition of bijective maps is bijected so in fact this is this operation is well defined I mean um you get another element of SN and associativity if you have a sigma Tau in in row and you do it like this this by definition it's Sigma composed

Tau composed row but since composition of functions is associative you get Sigma composed Tau compose row which by by the definition of the binary operation this is Sigma operation Tau in operation no so this is associativity and the unit element is the identity function identity

and when of course if I compose any any function with the identity function I get the the other function and existence of inverse follows from the the proposition we said that a bijective map to say that a map is dejective is actually equivalent to say that it has an inverse inverse in the sense so of Sigma is is what we denoted we also

denoted it as Sigma minus one exist since Sigma is bijected this operation is not commutative it's because composition of functions usually of course it could by some coincidence it could be commutative but it usually isn't so a exercise

find Sigma and Tau choose choose an n Sigma operation tau is not equal to Tau operations a any questions on that for the next example a a I would need a

theorem which is called firma a little Theory and it goes as follows a p is a prime a is some number between 1 and P minus

a to the P minus 1 is equal to one mod P so here I do I do multiplication mod P the proof I need the Lemma I need the Lemma so dilemma says consider

what we denoted FP is the set of all numbers from zero up to P minus one uh with a plus mod P multiplication won't be for any n and any X Y in f p

X Plus y to the p to x to the p plus y to the p month p okay this is a it's like the heaven of a he equations for for beginners I mean you don't have to worry about the brackets you just do you just raise each variable to the power of a of P there is the so-called a binomial

formula that is if I take x y times x y times and I do it a in our case P times but this this formula of course is is valid for any any power you take of X Plus y then this is equal to the following expression I write sum I equals 0 to p P choose I

x to the I y to the P minus I this formula is really a a follows from combinatorics it is valid in any in any place you have a multiplication in addition I mean basically because what did I do here I mean to to a I mean if I open these brackets completely well I I will have an expression for example of the form x times x times x to the p but a a generic expression will be you

choose either X or Y here then you choose either X or Y here and you multiply it all together but what are the options I have I mean basically once I chose a I I places of of X let's say I chose three places of X and that's all the X's I chose the rest of them need to be y so I have a a now how many options do I have to choose a three X's out of this P it's a p choose

a choose three and for each such option the expression I would get is X to the I a times a the remainder I I have to have a y to the P minus I so this is a sketch a sketch of the proof of the the binomial binomial formula but I think many people uh get it in the school now this P to the I sorry p p choose I is a p factorial over I factorial times P minus I factorial since p is prime

a p does not divide I factorial because I factorial is is a a multiplication of numbers which are all smaller than p so none of them can really uh I mean p p cannot divide it and and also does not divide P minus I factorial because again it's a multiplication of numbers that are all smaller than p P does divide

P factorial and what I get is that a a as a result P must divide this expression a p over I foreign but this is true of course if if I is not in the edges so I need to be bigger than than zero and smaller than P because if I for example if I is equal to p

well I mean a a if I is equal to P then I get a p p does divide I factorial right for example this is all to say um or all the inner Expressions here I get it is equal to well now I need to take the sum or only with the edges so I equals 0

and I equals P when I equals 0 I get x to the 0 times y to the p a I should say if P divides p a a p over I this means that p over I is equal to zero mod P so so all these expressions here in blue where I is a is is not 0 or P they all have a coefficient which is zero modulo

P so they all vanish and then I'm left with the the coefficients in the in I mean with the expression the the terms in the edges which are x x to the 0 y to the P plus x to the p y to the zero so this is x to the p y to the P but of course this is only true if I do moment p any questions about the Dilemma dilemma

is going to be a central in in a in calculus modulo p okay and now I'm going to prove now let's prove a remember I want to prove

a to the P minus 1 is equal to one mod P so we want or any a for any a between well for any a in in a in the set one up to P minus 1. but equivalently I could just say that a to the p is equal to a mod P so I I will prove we will prove this a a

purple equivalent statement and we we prove by induction induction on a so a equals one a a equals one is clear 1 to the power of p is one modulo p even even not modulo p a to the p

and now I want to prove it for a plus a but a plus one by ferma is a to the P plus 1 to the p is Perma a and a by induction a to the p is simply a so I get and 1 to the P of course is one this is mod P okay and this this finishes the proof I

mean uh we showed by induction that a to the p is equal to a mod P uh any questions about these two these two statements now the next example uses these

statements I take a I call it FP star FB star this is all the numbers one up to P minus one with a a the binary operation being multiplication mode p and I claim that this is a group what okay the the the the obvious the obvious conditions to check is that it

is this binary operation is associative that's very clear because we we said that multiplication mode p is associative the unit element is is one because we are like in multiplication of numbers one is going to be the neutral element to the the binary operation but but what what we did the non-trivial condition to check is that for any any a in FP star

there exists what is called a minus 1 well a times a minus 1 is equal to 1. can anybody say what what is going to be a minus 1 hello this is something you should definitely be able to answer and it uses the previous uh in the previous propositions we we discussed

is it a to the P minus two exactly exactly a minus 1 to be a to the P minus 2. a times a minus 1 is a to the P minus 1 which is 1. all right so this is a a this is the example of a the what is called the multiplicative group of the Prime field FB we we will

of course discuss a Fields extensively later on but for now you can you can you can see the the the non-trivial the non-trivial part is the existence of inverses and for that you you really use the the fact that P is prime so everything here in terms of operations could have been defined for a some any any a natural number n instead of P but you will not have this property uh here in in in blue

rectangle this will this property will not will not be true if P if p is not a prime and essentially it because um this argument that all the all the binomial coefficients vanish module p is not true for a a natural number that is not Prime so the whole the whole argument collapses in a and therefore we we really have to have a a prime number in order for the uh for for example for for this collection to be a group

any questions okay so suppose H and G are groups I guess I I want to specify the unit elements so a g operation and I will I will write operation G to denote that it is from G

and unit element sub G and the same I do for H this is just to to disambiguate the the two uh in the two terms and so suppose I am given two groups G and H each of them it has a unit element uh e in in one case e of G in the other e of H and it has a binary operation there then I can consider the Cartesian product G times h

which I remind you is the set of all X comma y all ordered pairs such that X is in in G and Y is in H and on this set G times h i can define a uh so or X and X Prime y Prime in G times h

operation X Prime y Prime to be X operation in G X Prime y operation in H the unit Element e of G times h I simply Define as you take e the unit element of G and the unit element of H in the second

coordinate this is of course an element in the Cartesian product the inverse of course if I want to take the inverse of x x comma y what I do is simply the inverse of the element X in G and the inverse of the element Y in in H a exercised

G times h with a in these duration is a group if you want you can think of it as now we are obstructing one level that is with a group it has it has like a it's something with the binary operation so in a group I can I can take two

elements and apply apply the binary operation to get a new element but you could also View the collection of all groups I will write it this is a bit informal so the collection of all groups you take two groups G and H and you give it a new group I mean G times H so in fact you can view this star I mean the Cartesian product

is a binary operation on the collection of all loops and a note uh the the number of elements in G times H is simply the number of elements in G multiplied by the number of elements in H okay so uh I think this is a good place to to take

questions and and finish the session um are there any uh questions or comments was it too slow for me it was fine yeah it's fine okay I I mean I don't want to to uh I mean I don't want to go too slow because then uh people people feel uh it's a waste of time uh but I I also don't want to rush I I

think it's good to to lay down the foundations carefully and then it's better in the in the continuation yes I agree I think the same this is that they okay so if there are no other questions or comments uh let's let's finish here we see each other next week