 So, let us continue from where we left off yesterday. So, we were looking at this as p power m and we saw that finally we could come down to this structure clearly as a vector suppose. We could write down the feed like this and also give you a construction. So, this we came from the abstract idea. So, we started with a field of size p power m. So, we actually we started with a finite field and we said there has to be a p and m. So, that it is a size p power m and then we said it has to have this form it should be a vector space. So, it is of this form this is from the abstract angle and then I gave you a concrete construction. So, what was this? So, this is from this is from the abstract and then we have a construction right? What was the construction? What do I do? Get out of all this madness. What did I do something? I mean is it some multimedia? Multimedia. Multimedia I do not want any such thing. Oh my goodness. See that is the problem with this connector you know I mean this is also not very nice if I take it on this side. How do you make sure that that does not happen? I tap my hand on something right? There is a way to get out of that. I should do this. It is not the lock. Yeah, I should type the task there. How do I do this? It is a complicated operation to have properties. So, we also had a construction. We said f p power m basically set of all polynomials right? So, this is where it happened ok. So, set of all polynomials once again with the a i coming from f p. But the degree is restricted to n minus 1 ok. So, degree is less than or equal to n minus 1. Why does that happen now? Sorry how does this go away? Ok. So, I have to say somebody has to put some path into how to use these things you know. I think people who do not use it at all build it. I think that is what makes sense to me. I cannot understand how they can build something like this and market it and sell it actually. Anyway, so let us forget about it. So, we have polynomials of degree less than or equal to n minus 1 and the coefficients come from f p ok. So, that is fine. So, that much is easy to see and then addition is quite easy ok. So, it is a polynomial addition that for multiplication we needed a irreducible polynomial right. So, we said there is a power of alpha which is a degree m irreducible polynomial. And I gave you kind of an argument saying such polynomials always exist for every m and every p there is an irreducible polynomial f degree m and that is quite important. So, how did we multiply? If you have a of alpha b of alpha n f p power m a of alpha times b of alpha. So, this multiplication in f p power m if you are to multiply in f p power m you can do the regular multiplication, but you do modulo phi of alpha ok. So, that was the idea in the construction and they gave you a very simple argument for why this should be a field ok. So, addition is very trivial only thing you have to check is multiplicative inverse ok. So, multiplicative calculator and all everything is trivial only the inverse has to be checked and with I said this is the same proof like you did for z p you can repeat for this same exact proof you take the element one element one polynomial for which you want to find the inverse you multiply it by all the elements of the multiplicative group you cannot have a repetition because power alpha is irreducible you cannot have a repetition. So, each of them is distinct and you have exactly as many as the elements of the multiplicative group. So, one of them should be equal to 1 and that gives you the inverse ok. So, that is the idea you can quickly show it ok. So, let us see an example of this construction before we proceed ok. So, the similar to the first we have this f 9 right. So, the 0 1 2 let me see. So, all the polynomials I am going to write it down explicitly just to see how it works ok. So, you have alpha plus 1 ok. So, let me just write alpha 2 alpha alpha plus 1 alpha plus 2 2 alpha plus 1 2 alpha plus 1 ok. So, those are all the polynomials the coefficients from f 3 right. So, remember the 0 1 2 these guys are from z 3 ok. So, everything is modulated 3 then I have to think of a irreducible power formula degree 2 n in f 3 ok. So, we can let for instance power alpha 3 ok something else happen now something else is going to start now I apologize in advance for it. So, power alpha is alpha square plus 1 ok. So, we could do that and then let me show you how to multiply something. So, let us say if you want to multiply alpha plus 1 by 2 alpha plus 2 ok. You can go ahead and multiply like you would multiply any normal polynomial except that coefficients you have to do modulo 3 ok. So, if you do quickly multiplication you will get 2 alpha square plus 4 alpha which will become alpha itself plus 2 ok this is what you get. So, clearly this is not an f 9 immediately then what we should do should divide by alpha square plus 1 and take the reminder ok. So, that is the same as using alpha square equals minus 1 ok in this expression both of them are exactly equal ok. When you divide by something then take the reminder you basically take the modulus means the divisor you can equate to 0 it is the same operation ok. So, that is a quick way in which you can do the division here it is not really hard. So, you see when you do this it becomes alpha mod alpha square ok. So, you put alpha square equals minus 1 what happens minus 2 plus alpha plus 2. So, minus 2 and plus 2 go away it becomes alpha ok. So, that is how we do the multiplication in these fields ok in this construction that is how we do it all right. So, there is there appears to be there appears to be a gap between this abstract idea that we have come up with in the construction. As far as addition is concerned both of these are identical right the way you do addition is that the addition in the abstracting we also do the same thing here in the construction ok. The multiplication seems a bit more fancy seems to come from snow up ok in the construction and just come up with the multiplication. But, it turns out even from the abstract idea you can come down to that multiplication you can show that nothing else can happen ok. So, which is what I am going to do next and the ideas are a bit more complicated the multiplicative group is slightly more complicated that well the proofs are slightly more complicated than the addition but eventually you will see it is very simple ok. So, that is the idea all right. So, we are going to begin by looking at the multiplicative group ok. So, that is denoted f p and star ok right that is the multiplicative group. So, star is without the 0 that is the multiplicative group. So, it will have how many elements p power m minus 1 elements ok. So, there is there is someone who gave us this finite field right. So, we can ask that there is some more questions to figure out what the multiplicative group looks like ok. So, if in the entire field we have 0 and 1 and we could add 1 to itself and then figure out something about that we can prove ok that is what we did. Now, in the multiplicative group what can you do? So, you do not know anything else you have 1, but then if you multiply 1 with itself you will just keep getting 1 you will not get anything new ok. So, you have to say something like you take an element of f p and star and then multiply it with itself ok. So, if you keep multiplying it with itself what can happen? Suppose I say p power is an element of f p and star p power and star and then I look at this p power p power squared p power 3 so on what should happen? Once again it has to repeat right it cannot go on and on and on forever. So, it has to repeat at some point and then once again you can show that first place it repeats it will become 1 ok. So, it is also not very hard. So, basically in this sequence there exists r such that there exists minimal r such that beta power r is 1 ok what do I mean by minimal r? Beta power 1 comma beta power so on till beta power r minus 1 those days will never be 1 none of them will be 1, but beta power r is 1. First time one appears in this sequence is this minimal r such that beta power r is 1. So, this minimal r has a name it is called the multiplicative order of beta ok alright. So, this this guy is called the multiplicative order of beta ok. So, the minimal r so, let us define multiplicative order ok let us do something here I do not know why why does it do that but once in a while it does not matter. Do you guys know why this is happening? When I go close it changes it to star or something I do not know why. So, I have to go to the next page of this ok. So, multiplicative order of beta is the minimal r beta power r equals 1 ok. So, this is an interesting and important quantity ok we can show a lot of interesting relations in properties for this beta power r ok. So, the first thing I am going to show is so ok. So, you may ask about additive order the element also has an additive order right. So, if you take the element and add repeatedly added eventually it will go to 0 right, but then the additive order in finite fields is not very interesting ok. So, if you have a prime characteristic and all of them have to be added a prime number of times to get you get you to 0. So, additive order is really not interesting. So, when people say order of an element in a finite field usually it is the multiplicative order ok additive order is not so meaningful in finite fields it is not so varied or interesting. So, usually it is the multiplicative order ok. So, so often times I will drop the multiplicative order and simply say order you have to interpret this multiplicative order this. Yeah in general this is a general definition in any group if you have an identity element how many times you have to operate beta by itself to get one multiplicative order. But in a general group so, there might be no order it is an infinite group there may not be any order defined ok. So, but in finite groups it will always be there no, no, no as in different for group as in. For each element. For each element it is not it is not ok there is no order for the group itself ok. Yeah. Usually when people say order of the group they mean the size of the group. So, it is all you are all terminologies ok. And you say order for every element of a group there is a different order finite group definitely yes ok. All right. So, so multiplicative order the minimum as a step will at R as 1. So, the first property you can show is R will divide the size of us people around stuff ok. This is in general proof for any group in any group any finite group the order of any element will divide the size of the group ok or order of the group ok. So, that is the general property. So, for any beta. So, see I am writing R here it means you have to think of R as order of any element of f p m star. You do not do not think that this one only this R will divide ok. So, order of any element of f p m star will divide this ok. So, there are couple of ways of proving it ok. So, which is what d power m minus 1 ok. There are couple of ways of proving it and the first thing first way is well I will appeal to some notion that I defined before. So, you can but you can define something as a subgroup generated by beta ok subgroup of f p m star generated by beta what is that basically it is beta, beta squared, subgroup beta power R minus 1 and then beta power R which is actually 1 ok. This actually is a subgroup subgroup of the multiplicative group ok. It is very easy to check it has the identity and you multiply any two of these elements in the domain in this element in this subgroup 1 ok. If you multiply beta power i times beta power j what do you get? beta power i plus j again belongs to this subgroup ok. So, this is clearly a subgroup. What is the size of this subgroup? R ok. What do we know about sizes of subgroups? Those are the result I mentioned in the course of decomposition. What should happen to the size of a subgroup? It should divide the size of the group which is what the result is ok. So, R has to divide p power m minus 1. So, when we see we are proving it if you want a more laborious way there is also other proofs but I will just pick that ok. So, there is more more literal proofs of this fact but anyway. So, R divides p power m minus 1. Is that ok? All right. So, that is the second property. Second property that is interesting is if beta power a equals 1 when R divides a ok. So, in fact if R divides p power m minus 1 what would mean? What would be beta power p power m minus 1? 1 ok. So, this is an interesting fact if in case you did not know before if you have a finite group you take any element raise it to the power of the size of the group you will get 1 ok. So, that is all this ok. And then what is slightly more interesting about the minimality is if I give you some number a and say tell you that beta power a is 1 then the other has to divide a ok that has to happen it cannot be in either way ok. So, how do you prove this? . Yeah. So, what you have to do is use the division theorem here. So, you take a and divide by R you get q times R plus some reminder ok right. Now, when you raise when the reminder you know is less than R ok. Once then you show that in case if the reminder is not 0 then you will get a lower element than R for which the power will be 1 and that will violate the minimality of ok. So, that is the idea. So, what you do is divide ok. So, proving this divide a by R a by R ok. So, you get a equals q times R plus some q prime and this q prime is strictly less than R a can be 0 also but it is strictly less than R ok. And then what we do? We look at this beta power a ok the beta power q R plus q prime which is beta power q R times beta power q prime what is beta power q R that is 1. So, this is equal to beta power q prime ok. So, now, when we know about a beta power a is 1. So, beta power q prime is also 1 and q prime is . . So, it cannot be anything from 1 to R minus 1. So, it has to be 0 so that implies q prime is 0 ok and that is the end of the proof q prime is 0 R divides. So, these two properties are quite implement ok. The order multiplied the order of element element of fpm star divides p power m minus 1 and if anything else is such that beta power a is 1 then the multiplicative order should divide a also. And there are more properties based on this minimality. So, I can write them down and try to prove them that it is bit laborious. So, whenever we need it I will invoke it and I may not prove it completely but you will see that I will give it as an exercise for you, try to prove it. It is not very hard to prove these properties. Usually some LCM and CCD and all will be involved. So, for instance one question that is often asked is beta has order r what is the order of beta power i. So, what will be the order of let us say beta squared what will be the order of that. So, you have to look at some LCM of 2 and r, if r is even then the other here will go by 2, r is r then what will happen? It will be the same. So, basically it will be r divided by LCM of 2 comma r that is a general fact order of beta power i will be r divided by LCM of i and r. So, those are easy things to prove you can just repeatedly use these results and argue the minimality that has to be. So, I will use such results to write along then we try to prove the two multiplicative structure. So, now the subgroup generated the data is very interesting. So, first thing I want to condense you is such a group that is generated the only one element is very easy to work with. It would be a dream. When you want to multiply 2 numbers, 2 elements from some abstract group and you know that it is generated the only one element it is much much better. If you are just of beta, beta, the only thing you have to know is the exponent. You simply add the exponent like modulo r it becomes just a binary like an integer operation nothing more you have to worry about. Otherwise you have to multiply polynomials, do division all kinds of complicated stuff. So, when the our entire FPM star be generated by a single element, when can I say there will be some single element beta maybe this beta that I picked? Yeah, so when r equals when the order of an element is equal to theta r minus 1. So, if I can show that there is some element in this FPM star whose other will not just divide theta r minus 1, but it will be equal to theta r minus 1 then what does that mean? I have one element of my FPM star that generates the entire multiplicative group. It turns out that is true for this FPM star we can prove. So, which is what I am going to prove next. I am going to show that you have all these elements which have different multiplicative orders it looks like we do not know what the different multiplicative orders are. We know that they have to divide theta r minus 1 we know some properties, but we do not know anything. You know that what we are going to show next is there is one element in this multiplicative group for any finite field whose order will be equal to theta r minus 1. So, now we can ask our friend to give us a finite group for that element and you will know exactly how to multiply also. So, multiply into elements with that that generated by the same thing very easy. So, just like addition was easy once you know the structure multiplication also becomes. So, that is the next result and it is a bit of a long proof. So, you store with me I will skip some steps that I will try to give you the most important and essential steps ok. All right. So, this is the fact I am going to try and prove that exists theta n of p star theta equals p power m minus 1 ok. So, this is the main result you can think of it as a theorem if you like. This is quite an important statement about finite fields. It simplifies finite fields considerably yes this is one of the main results which makes finite field structure so much easy ok. Every finite field has 0 and then it has just one other element which generates the entire group. So, what will the FPM look like? 0, theta, beta squared or the way to beta power r theta by minus 2 and then beta power theta by minus 1 which is 1 ok. So, this is how every finite field looks like ok. So, contrast this with this picture ok based on the additive quantity we have that picture for the finite field right oh my goodness am I doing this ok. So, contrast with this picture right ok. So, from the additive structure we came up with this picture right you said any finite field has to look like this ok. So, they are looking at the multiplicative structure they are able to come up with another picture any finite field has to look like that ok. So, you have two different pictures which we will use rest of the time again and again in doing addition and multiplication. So, this picture is useful for addition this is useful for addition this picture is useful for that is the main idea. So, let us try to prove it proof is not it is not it is not elementary, but it is a bit involved or some idea ok. So, first thing that I am going to do is I have theta m minus 1 elements in my FPM stop. Let us say we go through each element and pick that element which has the maximum order ok. If there are more than one you pick any one ok take any one element of maximum order ok. So, I have to show that how it is the primitive element what can I do I can try to show that the element with maximum order that ever has to be equal to theta m minus 1. So, we will do the strategy ok. So, we will first pick theta m FPM star order of theta it shows equal to r greater than or equal to order of theta prime for all theta prime in FPM star. Our strategy is to show that r equals theta m minus 1 ok. So, it is the greatest order among all the elements of FPM star order and you show that that equals theta m minus that is the status ok. So, the first thing we know is r is r divides theta m minus 1. So, r has to be less than or equal to theta m minus 1 ok. So, that is the first reason which is very easy is r is less than or equal to theta m minus 1 that we already had ok. So, the only thing that we show is r is greater than or equal to theta m minus 1 we show both then it has to be equal to theta m. So, we will now try to show r is greater than or equal to theta m minus 1 ok. So, the strategy there is as follows ok. So, you know that r equals omega minus what did I do know? I went some three pages in advance oh my god ok. So, what we are going to do is we are going to show these modern devices are here extremely smart or extremely dumb. Anyway, so r is less than or equal to theta m minus 1 or I want to show that r is greater than or equal to theta m minus 1. So, my strategy there will be to show the following. First thing we will show is we will show if you have theta prime on FPM star with order of theta prime equal to r prime. Suppose I give you that this is the claim I am going to claim then r prime divides r ok. So, this is the slightly strong claim. So, I am going to say r is my maximum order among all the elements and then I am saying see remember when I decide maximum I only require that r is greater than or equal to r prime for every other element. Now, once I pick that maximum element and then I will prove that any other elements order will have to not only be less than or equal to this r, but should also divide the r ok. So, that is all next proof ok. So, once you prove this claim you will have made some progress ok and this proof is a little bit more twisted than you can imagine ok. So, I will try to write it down ok. Let us say there is a prime number pi ok. Let us say we have a prime number pi that divides both r and r prime ok. So, that divides where does this want to keep coming up I thought I keep there sorry what did I do? So, I can move to the top how do I do that? Unlock it and then how do you grab? Can I click? I do not even have to auto write it know if it is in the top ok man that is not bad it is a good idea ok. So, so pi is a prime number ok prime number. So, that divides divides r ok. So, let us pick a prime number that divides r remember. So, if we integer can that has a prime characterization. So, there will be a prime number that divides r let us say ok. So, prime number type that divides r and let us write r as pi power of a times r1 gcd of r1 from a pi is 1 ok. So, what do I mean by this? What do I mean by that? So, I have extracted all the pi's that are in r ok and there are a pi power a as them ok. So, a as them are there. So, I have extracted all of them and then I have r1 remaining r1 does not have any factor as pi ok pi does not divide r1 ok. So, the gcd of r1 and pi is ok. So, a is the largest power of pi that divides r ok r1 does not have any pi power ok. So, now r prime will be some pi power b times r2 right. So, some b this will also be 2 b could be 0 I do not know ok. So, some some some b this will be 2 right. Again once again we will suppose what r2 from a pi is ok. So, this notation is basically gcd ok. What is gcd? Greater common divisor ok for both no there is no common divisor. When I say gcd is 1 it is very easy gcd 1 means they are relatively prime they do not have any common factor ok. So, r prime will also be like that. We can show that b has to be less than or equal to a ok. So, again this is like claim star claim prime claim prime is that b is less than or equal to a ok. So, if I show this I have showed that r prime has to divide up ok do you agree with that ok taken an arbitrary prime pi and I have looked at the factor of pi that is there in r that is a then I am showing the same pi the power that is in r prime has to be quickly smaller ok which means every prime this is true which means the prime factorization of r prime will be such that it divides the prime factorization of r and that shows r prime divides. So, that is the idea and the proof ok. So, once I show this claim prime I am done it is less than or equal to b. So, the proof for this will basically construct. So, we have to look at whether this beta prime power r 2 times beta power pi power a ok alright. So, I know order of beta prime is r prime order of beta is r what will be the order of beta power pi power a r 1 what will be the order of beta prime power r 2 pi power b and pi and r r r 1 are relatively prime. So, if I multiply these two the order will also multiply ok. So, that is another result which I have not shown ok. But you can show this you have two elements there are two orders product of those two if you take the order of the you will look LCM of the respective orders that also something you can show at least for this kind of groups you can show very easily ok. So, finite group is easy. So, this has to have order equal to power b times r 1 ok and that has to be definitely less than r right order of any element has to be less than r less than or equal to r and that proves my result ok. So, this is less than or equal to r which equals pi power a times r ok. So, that shows b is less than or equal to b. So, what I have done is I have taken the element beta which has maximal order r then I am looking at an element beta prime which has some order r prime and then I want to show r prime the words are. How do I do that? I take every prime factor of r and look at its power in r pi power a ok and then I say the same type will have power b in r prime and b has to be less than or equal to b. How do I show the less than or equal to b? I use the maximality of the order of r in this twisted way. I construct an element whose order is pi power b times r 1 and that has to be less than or equal to pi power a times r 1 because r is the maximal order and you cancel everything and you show p is less than or equal to b. So, that is QED for the claim ok. So, you have shown that if you have any element if this element beta which has this has maximal order r is such that the order of any other element has to divide that out ok. So, r prime if it is the order of any other element it has to divide out ok. So, it is a fairly strong result ok. Is that clear? This proof is a little bit abstract, but I am just good to see it once since no other way to prove ok. So, r prime divides r ok. So, now what I am going to look at is this. So, we are very close we are now going to show that r equals theta r minus 1 ok. So, about what we do is we are going to look at this polynomial x bar r minus 1 ok. So, how many roots can it have in f p p power m? You can have at most r roots ok. So, it cannot have more than r roots right. So, but now I am sure p power m minus 1 roots for it ok. So, whatever p power m minus 1 roots roots of all elements of f p m star do you agree? Every element of this f p m star will be a root of x bar r minus 1 right because I know the order of every element divides r. So, you raise it to the power r it is going to go to 1. So, x bar r minus 1 will have roots as every element of f p m star ok. So, that implies r is greater than or equal to p power m minus 1. So, I know it is a field it cannot have more than degree roots. So, it is distinct roots that is p power m minus 1 right. So, r is greater than or equal to p power m minus 1 and that proves other side ok. So, that is our Q d d further and towards that there is an element in my finite field whose multiplicator order is equal to p power m minus 1 whose already had r less than or equal to p m p power m minus 1. So, how it is related ok. So, the argument for the multiplication relies on so many other facts that you might have picked up here and there ultimately it gives a very very simple characterization of the multiplicator ok. There is one element beta which will generate the entire multiplicator ok. Such an element beta is called the primitive element of the finite field ok. So, that is the definition which you should know ok primitive element of f p bar m is theta and f p bar m such that order of f theta equals p power m minus 1 ok. So, there can be more than one primitive element in a finite field it is not unique or anything, but there is at least no. You know that every element has at least one every finite field has at least one primitive element ok right yes. So, this polynomial is not true for a completely see up to the previous thing everything is true maximum model divides and all is fine. Then polynomial is not true you know see I can come up with the polynomial in say z 8 for instance it is only degree 2 by less over roots it is easy to come up say x square minus 1. So, you look at x square minus 1 in z 8 ok multiplicator group model one of additive group model 8 multiplicator yes you do that it is well maybe it is not a group, but I guess for groups also you can come up with some examples like that. So, this polynomial thing is not there that is the main thing. So, this polynomial identity the algebraic theorem it says number of groups is less than e to the power m it is true only for fields only for fields this is work. There are very many groups which are not generated by one element ok that is the main problems in problem in group here. But fields they are not that much well ok. So, let us see some examples this is worth looking at a few examples. So, let us begin by looking at simple examples in first look at f 2 what is the primitive element of f 2? This is just 1 and 1 is the primitive element ok. So, there is no problem not a word as 3 there can be only one primitive element here also this is 2 who is the primitive element not a word as 5 0 1 2 3 4 what is the primitive element? 2 is primitive 2 2 square is 1 4 2 power 3 is 3 then 2 power 4 is 1 ok. So, what else is primitive? 3 is also primitive what about 4? 4 is not primitive 2 and 3 are primitive ok. So, 2 and 3 I will just put a star on top of the primitive element ok 2 and 3 are primitive. So, you can keep on answering and I think there are results saying in f p more if you just have prime p there are some characterizations for when 2 will be a primitive element. I think there is also a result which says either 2 or 3 or some other number has to be primitive always or some such there are some results. There are f p there are ways to find primitive element ok you can pick up some number theory book and you will see that there are a lot of kinds of interesting methods for finding primitive element ok. It is quite important and f p sometimes see a lot of character operations use this as p case you should know something suppose how this thing works but it is possible to find primitive element I think MATLAB I am sure will have a function which is primitive element ok. So, this is quite easy. So, let us move on to other more interesting examples. So, if you do not if you just look at m equals 1 it is obtained as a basic field right ok. So, base field ok. So, these are called base fields ok. So, let me say base field if m is greater than 1 you have what is called an extension field ok. So, this is the terminology ok. So, every extension field will contain a base field right. So, the base field will not contain any other field. So, that way the extension fields are not as basic as the base fields. So, base fields are the most important ok. So, mostly extension fields are also interesting to us and let us look at f 4 ok. So, the more I say f 4 you already know that the base field is f 2 so many other things you can quickly figure out. You know what f 4 is ok alpha star of alpha plus 1. So, this is a model of 2 ok what is the primitive element? Alpha is the primitive element right. So, you do 1 alpha alpha alpha star what about 1 plus alpha? That is also primitive. So, 1 plus alpha is also primitive ok both of them have order 3. You do 1 plus alpha 1 plus alpha squared is 1 plus alpha star is alpha and alpha that squared gain will give you the ok. So, let us look at f 9 that we had. So, the problem will get more interesting ok. So, let me do it quickly ok. So, let us say f 9 did I miss anything? Yeah. No, did I miss any element here? Is it 9? No. 2 elements ok. Alright. So, what are the primitive elements? So, remember I should tell you what the rule is. Let us say the rule is alpha squared 1 is 0 ok. Of course, model of 3 right from the thing is model of 3 addition ok. So, I want an element of order what? Order 8 ok. So, if you have an element of order 8 ok. So, the next see remember so 8 I know is the maximum possible order. Order of any element has to divide 8 ok. So, possible orders for the elements of f 9 is what? 1, 2, 4 and 8 ok. So, you can quickly check those things ok. 1 is not is an order only for 1 right it is very easy to check that. And quickly you have to eliminate 2 and 4 itself. So, you have to pretty much look at only 4 if an element ratio of alpha 4 is not 1 then it has to have order 8 ok. There can be no other order in the thing ok. So, those are quick ways of checking it ok. So, if you get see an element you first square it right see if it is 1. So, it is not very square it again see if it is 1. If it is not 1 that should be primitive ok. So, it takes some effort to prove this it is not quite easy. Alpha is not a primitive element ok where alpha 4 will be 1. Alpha square is minus 1 alpha power 4 will be 1 ok. So, alpha is not a primitive element you guys you are saying that alpha power 1 alpha plus 2 are primitive ok. There can be only 2 primitive elements I think or maybe more than 2. I think let us say so this was going to be at least 2 ok. So, these 2 are primitive maybe the other ones are also primitive I am not sure ok. You can check that ok. Let us see again. So, in general signing primitive element is a bit more involved in these extension feeds. It is not so maybe it is not so easy that you have to check ok. So, you can check for instance let us look at alpha plus 1 power 4 what is this guy ok. So, you have alpha power 4 right what is alpha power 4? It can be just 1 it is 2 is it ok that is to be 2 ok. So, let us see and 3 times alpha that will be just alpha less than 6 times alpha square right. 6 times alpha 6 will go away right and then 3 times alpha power 3 which is what ok. Let me write it like that no 4 no did I get this right I think it should be an alpha square right. This is fine no and then there is a plus 1 yeah. So, then alpha power 3 and then alpha power 4 right ok. 6 will go away because 6 mod 3 is 0 ok they are running out of time. So, this will reduce to what? It is may be minus 1 alpha power 1 no 1 plus 1 is 2 whatever 4 alpha plus 4 yeah that goes to 0. If it is 4 alpha out you get 1 plus alpha square so that goes to 0 this evaluates to 2 so alpha plus 1 is 2 ok. So, we will stop here for now. So, we have to be multiplicative group of primitive and we are meeting again tomorrow right. Actually tomorrow there is no club because I am not in town we will meet again on Monday ok. When is the quiz it is in the first week of September? So, we will have to do something about the quiz. So, if you go to my web page and click on e512 ok we will go to another page in this there will be some assignments. There will be at least 2 assignments which are relevant for what I have done up to now ok. One is called linear block codes another is called string algebra or something the first one ok. So, those 2 assignments I think actually I have like at least some 30-40 problems. If you make sure to practice those problems and we try to discuss that sometime next week. Ok. So, that is it.