 In the last lecture, we saw the structure of groups. Here we will see some examples. So there are these are some of the examples. We have the integers, the set of all integers. So remember this denotes the set of all integers minus 1, 0, 1, 2 and so on. So these are the integers. These are the rationals. You may call them rational numbers. These are the real numbers. These are the complex numbers and these are our residue classes modulo n. All these form groups under addition and this is something that we have seen already or if you have not seen then maybe you should go back and see these things. These are all groups under addition. There are also groups under multiplication. So these are the group. In general what we have for all this is that this is the set plus minus 1. These 3, these are all the nonzero elements in the corresponding sets. And remember these are all the elements i where the GCD of i and n is 1. So these are all invertible elements with respect to the product in the sets Z, Q, R, C, Z and which we have seen earlier which are groups under addition. And here we have that these corresponding invertible elements form a group under multiplication which also tells you that the elements which we have seen earlier the examples also have one more structure than addition and that is the structure of multiplication and this is what takes us to our next concept of a ring. So we have seen the definition of group in the last lecture perhaps in a hurried manner but we are not going to be able to spend more time on these things. A ring is something which is equipped with two binary operation and traditionally one of them is denoted by plus and the other is denoted by product. What we require is that the ring or the set R be a group with respect to plus and there should be some more properties with respect to the product. The properties with respect to product R first of all that the set be closed undertaking product the closure property. We also demand that our rings often be associative there are of course examples of non associative rings that people study but we are going to look at only associativity. We do not ask for the identity element to be there and of course we do not ask for inverse of every element. As you see this set of z does not have the inverse property the multiplicative inverse of 2 is not there in z. So identity element may be there and in that case we say that our rings are rings with identity. Identity element may not be there. So we have two structures plus and product with respect to plus we have that it should be a group with respect to product. We demand that there be two properties closure and associativity but we are not going to put these two structures in isolation. We are going to have a condition which will tell you how these two structures can be combined. Whenever you have a set and you are putting some structures randomly you would demand that the different structures have something to do with the first structure only then there is some meaning to the whole structure. So here we demand that the product and addition be tied up with a rule which is called distributivity which says that a into b plus c should be a b plus a c. So you have b and c these are the two elements of the set R, b plus c is yet another element of the set R and you are multiplying to the b plus c by a. So whether you take the addition first and then multiply by a or you take the multiplications first which is a b and a c and then add you should get the same answer. This is called the first distributivity law. The second law of distributivity is that you have a plus b into c should be equal to a c plus b c. So there is a multiplication on the left hand side and there is a multiplication on the right hand side also. So a dot b plus c this is the multiplication on the left hand side a plus b dot c or a plus b multiplied to c is the multiplication on the right hand side. These give you two distributivity laws. This all constitutes a ring. Once again with respect to plus it should be a group there is one more important condition that it should be an abelian group which is to say that a plus b is always equal to b plus a. So with respect to plus it should be an abelian group with respect to product there should be two properties closure and associativity and product should distribute over the sum both on the left hand side as well as on the right hand side. And the examples that we see here are the examples of commutative rings with identity. If you see very carefully the examples q, r and c have the property that every nonzero element is invertible. Whereas in z we have the element 2 here which is not invertible although it is nonzero. And here if n is composite then we get nonzero elements which are not invertible. Composite just means that it is not a prime. Whenever we have a prime p dividing n and p is less than n then p itself is an element in your ring. So p for instance will give you such an example. So for us what we are going to do is to consider only the groups or rings which are finite. We have been looking at these things so far in our lectures and this is what we are going to study. So we are going to be dealing with finite groups finite rings Moreover we are going to look at commutative rings where the product is also commutative. We demand that under addition the group should be abelian but here we demand that here we have that the product is also abelian that is 1 plus thing and we also have that all these rings have identity in them. So what we are looking at are finite commutative rings with identity. We are going to require some more concepts of group theory and ring theory. So we are going to study these groups called the unit groups of Zn. These are denoted by Zn cross. We put this cross on the head to signify that we are going to look at elements which are invertible with respect to the multiplication. These are the groups unit groups that we are going to study. They are going to be called Un where U stands for units. We have already seen what the cardinality of these groups are going to be. But after that we want to find the exact structure of these groups. So we want to see how they behave if you are looking at the group structure of these. This will help us in distinguishing various Un. Later on we will see examples of these Un and you will see what I mean by distinguishing the Un by the group structure that we put on them. So once again I think it would be good if you brushed up with your knowledge of basic group theory and the concepts that you should look for are subgroups, cyclic groups and direct product. If this seems to be too much for you then what I suggest is the following. While we go on proving the later results we are going to use some particular concepts in groups and I will of course have to mention that while I give you the proofs. So when I mention that you note all these concepts down and go back and check the definitions that will help you understand these proofs in a better way. If of course there are ways to do these proofs without using the group theory which is to say that we will do the same operations but in more detail and we will not use the word group cyclic group direct product or so on but then our proofs will become very lengthy and such proofs are not inspiring. What happens is that there are same methods applied in many proofs and therefore you combine these methods and give them some name. So the group theory is one such method which we are going to use in number theory. It has been our experience that the primes are well behaved whenever we study any structure associated with a general integer n. So we start by looking at the group of units modulo a prime p. We call this to be up. Of course the z n star is un. So whenever your n is the prime number p we will call them up. What we are going to prove first is that all these groups are cyclic. This is to say that there is an element in z p star whose powers will give you all elements. So there is an element a in z p star such that a a square a cube a raise to 4 comma dot dot dot will list all the elements in up. This is what we want to prove. Perhaps the result is true in more generality. So we make one definition here an element a in general un where n need not be a prime and element a in un is called a primitive root modulo n. So we will work with the concept primitive root but it is understood that whenever we are talking about primitive root there is an n which is there in the background. So we call this element a to be a primitive root if the order of a is equal to phi n. What it means to say is that a power phi n is equal to 1 in un and a power m is not 1 for any m strictly less than phi n. So when you list all these elements a a square a cube and so on you do not hit 1. Once you hit 1 after taking power for some times the next number that you will take will again be equal to a. If you have that a power 10 is 1 then a power 11 which is a power 10 into a and if a power 10 is 1 then a power 11 is going to be a. So you have a a square a cube up to a power 9 and a power 10 which is 1 then a power 11 will be a and you are going to get the same loop again. So here what we demand is that of course we are asking for the order to be phi n but not smaller than phi n. Therefore the distinct elements that you get are phi n in number and we also know that cardinality of un is equal to the Euler phi function phi n. So whenever you have a primitive root modulo a number n you have that the corresponding group is a cyclic group un will then be a cyclic group. So we are going to be studying these structures of these un's trying to see whether they are cyclic or not by trying to verify whether there exists a primitive element modulo that n or not. So to begin with where we want to prove that our group up is cyclic this is to say that each up has a primitive element this is what we will have to prove. So the theorem says that the groups up have primitive elements. So we observe first of all that for a prime p the cardinality of up is p minus 1 and we already know that up being gp cross is a group under multiplication and if you take any element then the p minus 1th power will always give you 1. The question is to find an element a for which no smaller power gives you 1 this is what we want to show. Furthermore using group theory if a if the order so this is where I am now going to use some concept from group theory which is to talk about order of a. Furthermore if the order is m then we also know that m has to divide p minus 1. So if you take all the elements in the group up and separate them out by their orders then they would be in the sets which are indexed by divisors of p minus 1. So thus we have a m which are elements a in up with the property that the order is m give disjoint union of a m where you have m dividing p minus 1 is your up. The disjoint union of all these m, a m cannot have any intersection with a different a l because in a m the order of the element is m in a l the order of the element is going to be l and if you have any element in common it will say that the order of the element is both m and l but that cannot happen if m is not equal to l. So this is the disjoint union and what we have observed before is that the summation of their cardinalities will give you p minus 1. This is something that is quite nice. So what does it tell you? It tells you that when you take the summation of the cardinalities of a m. So for instance here you should check that a 1 is only the element 1 just to give you an example of 1 a m. This is the only element whose order is 1 there is no other element which to the first power is equal to 1. So 1 is the only element of order 1 then you will look at other divisors of p minus 1 and find the sets a m. It may happen that some particular a m are empty it may happen that you are looking that say 3 is a divisor of p minus 1 but there is no element of order 3. So that in that case a 3 will be empty and then the cardinality of a 3 will be 0 that is quite okay. But what we certainly have is that every element in U p has to have some order dividing p minus 1. Therefore every element of U p is in sum a m and clearly a m are subsets of U p. So you have the equality that the disjoint union of a m is U p which when you take the cardinalities gives you that the summation of cardinalities a m for m dividing p minus 1 is p minus 1. Now what we have further is the following thing. So each note that each a m cardinality is bigger than or equal to phi m. Why do I say this? This is because when you take a particular m and you take an element of order m we are looking at so this is true whenever a m is non-empty. So start with an element a of order m we are assuming that a m is non-empty. So there is an element a of order m then look at the cyclic subgroup generated by a. So this will be the element a a square a cube so on all the way up to a power m minus 1 and then a power m gives you 1. So this is the cyclic subgroup of order m of the group U p. Now we want to know how many generators are there of this cyclic subgroup. So a is one generator but there could be some more generators and the basic fact in cyclic groups tells you that a power i is a generator of this cyclic group generated by a precisely when i is co-prime to m. So the group generated by a has phi m generators these are all distinct elements. So whenever you have an order element of order m you are going to take one of them it will have phi m generators since they generate the same subgroup they will also be of order m. Therefore once you have one element of order m you have at least phi m elements of order m. So we have two equations here our equation number one tells you that these cardinalities give you p minus 1 but then whenever this is non-empty each such cardinality is bigger than or equal to phi m. Let us put these together to get the following nice result then p minus 1 which is summation of cardinality a m where we have that m divides p minus 1 this is also clearly summation of the cardinality is a m where you have that m divides p minus 1 and a m is non-empty. The only difference between these two sums the only difference between this and this is that here you are taking all the m's dividing p minus 1 whether the a m is empty or not and here taking you are taking all those m's dividing p minus 1 where a m is non-empty. So the only difference between these two is the cardinality of a m which are empty but those are anyway 0. So these two equalities are the same that tells us that we have this equality and this is now bigger than or equal to m dividing p minus 1 phi m of course with the condition that a m is non-empty but if I have my a m to be non-empty and I take the summation over m dividing p minus 1 I get phi of summation of phi m where m dividing p minus 1 whereas on this side I also have summation over phi m where m divides p minus 1 this is something that we have proved right after studying the Euler phi function that summation phi d where d divides n has to be n. So what we now have is summation phi m m dividing p minus 1 is of course bigger than or equal to summation phi m where m divides p minus 1 and a m is non-empty. We want to say that these two equations are the same and therefore we want to say that whenever a m is non-empty we should get phi m and nothing more than that. So here we have when we have this inequality and if you are not getting all phi m then some phi m should come with more multiplicity. Further each a in a m gives a cyclic subgroup order m in particular we get roots x rest to m minus 1 in Zp. But once we have roots of this polynomial it should immediately strike to you that polynomial of degree m cannot have more than m roots. So what we get is that the cardinality of a m is also less than or equal to phi m. This is because once you have one element generating a cyclic group of order m in that cyclic subgroup you will have phi m elements of order m. If you have any more elements of order m it should give you at least one more element outside this cyclic group and then the polynomial x to the m minus 1 will have more than m solutions. This is something which cannot happen. So whenever we have that the cardinality of a m is non-zero you will actually have that the cardinality of a m is exactly equal to phi m. And now we go back to the previous slide which tells us that summation phi m where m divides p minus 1 is p minus 1 which is also summation phi m where m divides p minus 1. But here the condition is that a m is non-empty. Since these two terms are equal you should have that a m is non-empty for every m dividing p minus 1 which tells you that for every m, for every divisor of p minus 1 there is an element of order m and therefore we get that in particular u p has a primitive root. So we have proved that the group of units modulo p is a cyclic group of order p minus 1. Go through this proof it uses some concepts from basic group theory which are not very difficult and you will get familiar with these when you study these things more. In the next lecture we are going to look at structure of un in general and see whether they are cyclic and if they are not cyclic then we would like to find the exact structure of these unit groups un. Thank you.