 Welcome back. In the last lecture we proved that U p has primitive roots. So, what are U n in general? Let us have a very quick recall. So, these are the groups U n denotes the groups g n cross. These are the groups that we are studying and these are denoted by U n where n is a natural number and we want to study when exactly these groups are cyclic. So, there is one thing which we should have noticed which is that cardinality of U n which is also denoted by this symbol sometimes. This is nothing but the Euler phi function evaluated on n. So, to say that this group is cyclic means to say that there is an element whose order is exactly equal to the order of the group. So, when we show in the next slide that the group U p have primitive elements that means we will actually produce one element of order p minus 1 and if you remember this proof what we proved was that whenever there was any element of order m one such element will generate a cyclic group of order m. Further all these elements will have property that they are equal to 1 when you raise them to the power m and x power m minus 1 over z p cannot have more than m solutions. So, once you have one or element of order m you have a subgroup of order m generated by these. These will contain exactly phi m many elements of order m. If there was anything outside of this which is also of order m in other words if you had any elements of order m more than these phi m elements then we get a contradiction. So, for each m you will have only phi m elements of order m or perhaps you will have no elements. But when you take summation of phi d d divides p minus 1 it should give you p minus 1. Therefore by counting orders of all elements in the group U p we see that for every divisor of p minus 1 there has to be at least one element of that order and in particular there is an element of order p minus 1. This is the proof that we had to show that U p is cyclic for every prime p or in other words that U p have primitive elements. We now study the U n slightly more in detail where n is a composite number. Let us compute these things for some small values of n. So, first of all we look at U 6. 6 is the smallest number which is divisible by 2 different primes. So, we consider U 6. U 6 has elements which are co-prime to 6 which are co-prime to 2 and 3. So, you get only 2 elements here. U 6 has exactly 2 elements and we know that every group of order 2 is isomorphic to the cyclic group of order 2. So, U 6 is isomorphic to C 2. This is the complete information about the group U 6. U 7 is a cyclic group because 7 is a prime number U 8. So, we see that U 8 has only the odd elements because all the even elements are not going to be congruent to are not going to be co-prime to 8. They will have the GCD equal to 2 at least. So, these are all the elements further. We notice that the squares of all the non-trivial elements are equal to 1 modulo 8. Thus, although you have a group of order 4 the all 3 non-trivial elements are of order 2 and so U 8 is C 2 cross C 2. You actually have 1 C 2 which is 1, 3. This is the subgroup and another C 2 which is 1, 5 and note that 5 into 3 is equal to 7 in U 8. So, there is a natural map from the product of these two subgroups to the whole group U 8 which gives you that U 8 is an internal direct product of these two subgroups. So, U 8 is C 2 cross C 2. U 9 we of course, forgot 1 composite number 4 but that is something that we will be looking at later. So, this is the next composite number U 9. When we write U 9 we will have to be careful that no multiple of 3 should come. These are all the elements in U 9. Let us try to find orders of elements in the groups in the group U 9. So, suppose we start with 2, 2 square is 4, 2 cube is 8, 2 raise to 4 is 16 but you are modulo 9. So, this is 7, 2 raise to 5 is now 14 because it is 7 into 2 but modulo 9, 14 is 5 and then we get 2 raise to 6 which is 10. This is also 1. So, 2 square, 2 cube, 2 raise to 4, 2 raise to 5 and 2 raise to 6 are all distinct elements. In other words, the order of 2 in this group is 6. So, U 9 is actually a cyclic group of order 6. So, we have that U 8 was C 2 cross C 2, U 6 was cyclic, U 7 was cyclic and U 9 is also cyclic. Let us go one step ahead and look at U 10. So, in U 10 we should not because 2 divides 10. So, we should not write the even elements but we should also make sure that 5 does not come. So, the only elements you have are 1, 3, 7 and 9. So, note first of all that cardinality of U 10 is equal to the cardinality of U 8 which is 4. This is one small observation because of course, we have that 5, 10 is equal to 5, 8 which is equal to 4 and here now let us try to compute orders of 3. So, 3 square is 9, 3 cube is 27 which is 7 because we are working modulo 10 and then 3 raise to 4 is 7 into 3, this is 21 and therefore this is 1. So, order of 3 in this group is 4 and so we get that U 10 is isomorphic to C 4. So, this is one major thing because U 8 has exactly as many elements as there are in U 10. If you were looking at just the number of elements you would have no way to distinguish U 8 and U 10 and what we find here is that U 8 is isomorphic to C 2 cross C 2 but U 10 is cyclic it is isomorphic to C 4 and as a small corollary we get that U 10 is not isomorphic to the group U 8. This is why we are looking at the group structure on these sets. If you were looking at only the orders of these sets then we would have that U 10 and U 8 have the same number of elements however as groups these two are distinct. Just as a curiosity let us also look at U 12. U 12 has number of elements so 1, 2 will not come, 3 will not come, 4 will not come but 5 comes, 6 will not come and then 7 comes, 8, 9, 10 these 3 will not come. So, next one is 11 and of course 12 will not come. So, if I now want to compute the orders of these elements let us again look at the order of 5. So, you have 5 and you have 5 square which is 25 but 25 is 1 modulo 12. Then you will look at 7 and 7 square is 49 which is also 1 modulo 12 and of course 11 square is going to be 1 because 11 is minus 1 modulo 12. So, these 3 tell you that U 12 is isomorphic to C 2 cross C 2. Therefore, we have this nice result that U 10 is also not isomorphic to U 12 but we do have that as groups U 12 and U 8 are isomorphic both being isomorphic to C 2 cross C 2. In general what we would want to do is to find the structure of all U n. We want to understand all U n for every natural number n. We have made a step, we have began the study of these structures of U n by looking at U p where p is a prime. Our experience has been that U p are much more tractable the prime numbers things related to prime numbers are more tractable than things related to natural numbers. So, we would want to see whether the same thing holds in general. However, when you are looking at a group U n and you wonder whether that is a cyclic group. We notice that in a cyclic group there are unique subgroups of any order which divides the order of the group. So, what I mean is that if C n has a cyclic group then it is the unique such subgroup. Meaning if I take a cyclic group of order n here that was C n then first of all for every divisor of n there is a cyclic subgroup of that order. But suppose we have a cyclic subgroup of C n of order L then there is only one such subgroup. You cannot have more than one such subgroups. Therefore, when we are looking at U n for n higher up. So, we have looked at 2, 4 and the primes and so on. Then we notice that whenever n is not equal to 2 then 2 has to divide phi n. So, there are 2 things that we have to notice here. First of all the cardinality of each U n is phi n and each phi n for n bigger than 2 is even. So, 2 is going to divide the cardinality of U n and you may ask how many elements are there which have order 2. Is there a single such subgroup? So, whenever the number of square roots of 1 is more than 2 then the group U n cannot be a cyclic group. So, what we are going to do is that we will look at all U n. We will take away the U n's which are never going to be cyclic because we have computed the number of square roots of 1 in all z by nz. We have their exact number depending on the number of prime factors and whether 2 divides it, 4 divides it, 8 divides it. We have the exact description of the number of square roots of 1. So, using that we will be able to tell exactly when these groups cannot be cyclic. So, the groups which can be cyclic that number is though those groups are among the remaining ones and we will study them one by one that is our plan. So, now let us look at the U n which cannot be cyclic and therefore we will have to recall the number of square roots of 1 that we have already computed. So, these are the number of square roots of 1 in z n and if you remember this was the exact description. So, whenever we had k distinct prime factors we had the numbers to be these 3 numbers. Now, whenever k whenever these numbers are more than 2 then our U n cannot be a cyclic group. So, here we want to look at these to be more than 2 that is the thing that we have to find. So, k because you will have at least 1 prime factor. So, whenever you have 2 or more factors then this will tell you if there are 2 or more odd factors let us say because the even factors are divided in 2 cases here. So, whenever you have 2 or more odd prime factors then the number of square roots of 1 is going to be 2 power k which is bigger than 2. So, this is bigger than 2 if this number k is bigger equal to. So, whenever you have 2 or more odd prime factors the group U n cannot be cyclic. So, for instance if you were looking at 3 into 5 and therefore you would look at U 15 then U 15 is never going to be a cyclic group that is because the number of square roots of 1 in U 15 or in Z mod 15 Z which is the same thing is equal to 4. There are 2 coming from 3 and 2 coming from 5. So, once you have more than or equal to 2 odd prime factors then gone U n cannot be cyclic. Let us now look at the possibility where 1 of the factors can be 2. So, if 2 is the only prime power of 2 which divides the group and no higher prime powers divide then you are in this situation. So, 2 divides but 4 does not divide that is the situation that we are looking at and here once again we have 2 power k minus 1 among the k minus 1, 1 k is already equal to 2 the 1 of the primes that you have is already equal to 2. So, you are looking at all other odd primes and therefore if you have k minus 1 odd prime factors together with 2 where that k minus 1 is 2 or more which means to say that k is 3 or more the total number of prime factors including 2 is 3 or more and 4 does not divide the number then you cannot have the group U n to be cyclic. And finally, if you have n congruent to 0 mod 8 whatever number of prime factors you have 2 power k plus n is always going to be bigger than or equal to 2. So, here you always have this to be bigger equal to this is something that you have to study later. So, we will say sometimes. So, these are the 3 major observations which we write down here. If n has 2 or more odd prime factors then U n is not cyclic. This is once again recalling that the number of prime factors being k will give you that there are 2 power k square roots of 1 and therefore once k is bigger equal to then U n cannot be cyclic. All right. So, whenever you have 2 or more odd prime factors this has to be prime then U n cannot be cyclic. So, we next come to dividing the number n and assume that 8 divides. So, we will look at the cases where 2 divides but 4 does not divide 4 divides but 8 does not divide and 8 divides these are the 3 cases. So, whenever you have n congruent to 0 mod 8 then U n is not cyclic. So, here the number of square roots of 1 was 2 power k plus 1 which is always bigger than 2 because you have at least one prime which is 2 dividing it. So, k is bigger than or equal to 1. So, this group is also not a cyclic group. So, now the cases are as follows you will have that 4 divides your number n and then there can be one more odd factor. If there are 2 or more odd factors then we will refer to the first statement and U n cannot be cyclic. So, now we are going to deal with the cases where n is p power e where p is odd, 2 into p power e where p is odd or 4 into p power e where p is odd. Those are the cases that we have to deal with. So, this is the last case 4 n is 4 m where m is odd then again U n is not cyclic. This is because in this case if you were looking at the square roots modulo n there is a square root modulo 4 namely 3 the square root of 1 and m being odd will give you at least one more square root and together with the element 1 you have more than 2 square roots of 1 and so U n is not going to be cyclic and the final thing that we have to now notice is where 2 is the only factor or you have that n is equal to 4 because when you are looking at here n is 4 m, m is odd and we are going to take m to be strictly bigger than 1. This is the case when U n is not cyclic and so the final thing is where we have U 2 and U 4 these are cyclic. This is something that we can easily see. So, U 2 for instance is the trivial group and therefore it is isomorphic to C 1 and U 4 has only two elements 1, 3 and therefore it is isomorphic to C 2. So, both U 2 and U 4 are cyclic so the only remaining cases now are where you have 2 divides the number n no higher power of 2 divides n and the only other factor the only other prime factor is a single odd prime. So, the only remaining cases are n equal to p power e or 2 times p power e and what we are now going to show is that in both these cases the groups U n are cyclic and this is going to be done by producing a primitive element in each of these U n. So, primitive element just to recall is an element whose order is exactly equal to phi n. This is what we are going to do we will produce a primitive element in each of these U n. So, the very first stage we actually have to look at p power e and 2 p power e but we will start with the case where n is p square and remember this is something that I will not repeat that p is always an odd prime. So, we are going to prove that whenever n is p square then U n is a cyclic group. So, one basic statement here that we are going to use is the following. If a in the natural numbers is a primitive root modulo p square then a is also a primitive root modulo p this is a basic statement. So, what we observe here is that when we are starting with a in n being a primitive root modulo p square we will start with the assumption that the order of a which is d modulo p square. So, in U p square is p into p minus 1 this is the order we are starting with this statement. This implies that the order of a which we can call d 1 in U p is p minus 1. So, what we will have to show in this case is that if you have some so if p divides a power d 1 minus 1 then p square divides a power d 1 p minus 1 this is something that we will have to show. So, this will imply that if you had any smaller d 1 inside the group z by p z as your order of the element a in U p then the order of a in U p square will have to be smaller than p into p minus 1. So, with this exercise we have this observation that a primitive root modulo p square is also a primitive root modulo p. Here we will want to show that U n is cyclic U p square is cyclic. So, we will need to produce a primitive root modulo p square and when you look at it modulo p that should already be a primitive root modulo p we know that there are primitive roots modulo p. So, we will pick one of those we will see whether that is already a primitive root modulo p square and if not then we will add a suitable multiple of p to get a primitive root modulo p square keeping the same modulo class modulo p the same residue class modulo p. Since this proof is going to involve few more slides and some more time I think we should begin this proof in our next lecture. Please remember what we have done and the plan of the proof is also something that I have told you just now. So, we will start with this proof in our next lecture. Thank you.