 Welcome back we are in the middle of the proof actually we are proving that un is cyclic whenever n is power of an odd prime. We have proved the case up square being cyclic in a separate result and then we started this proof of un being cyclic when n is p power e where p is odd and we have just proved that up cube is cyclic that was to give you a glimpse of the general proof. I told you that the general proof is by induction and now we have the proof. So, we begin with the assumption that so the assumption is as follows let a in n be primitive u p square and we show that a is primitive u p cube u p power e for we proved that it is actually primitive in p power 3 but we will show that it remains primitive in all the higher p power u p power is. So, let us assume the induction hypothesis that is a is primitive in u p square u p cube so on in fact it will remain primitive in u p also up to p power e and then that a is primitive in u p power e plus 1. This is our plan we will start with the induction as hypothesis that our element remains primitive for all these groups and we will prove that it remains primitive in the next group as well. So, since a is primitive in u p power e minus 1 and u p power e we get the following statements. First of all you will have that p power e minus 1 will divide a power p power e minus 2 into p minus 1 minus 1. Here the Euler-Phoi function of p power e minus 1 is 1 less power of p into p minus 1 and similarly we will have that p power e divides a power p power e minus 1 into p minus 1. This is because the Euler-Phoi function of p power e is p power e minus 1 into p minus 1. Moreover we have that p power e does not divide the earlier thing as a is primitive in u p power e. These are the three things that we are going to need. So, we have these three things that p power e divides this quantity, p power e minus 1 will divide it but p power e will not divide. You will check that these two are the same p power a power p power e minus 2 p minus 1 and a power p power e minus 2 into p minus 1. Further when we look at the group u p power e plus 1 the order of a in u p power e plus 1 is either p power e into p minus 1 or p power e minus 1 into p minus 1 as the Euler-Phoi function of p power e should divide the order should divide order a and further order a should divide the Euler-Phoi function of p power e plus 1. So, we have these possibilities and the difference the ratio of phi of p power e plus 1 upon phi of p power e is p. Therefore, these are the only two possibilities. If we show that this is not the order of the element a in u p power e plus 1 then we are done then this will have to be the order of a in u p power e plus 1. So, for that we go one step back and notice that a raise to p raise to e minus 2 p minus 1 minus 1 because of what we have observed here is p power e minus 2 into some element alpha which is co-prime to p. This is because the thing on the left hand side is divisible by p power e minus 2 but not by p power e minus 1. So, therefore, we got that one second we have to make a correction here that this 2 is 1. So, we see that this left hand side is divisible by p power e minus 1 but not by p power e therefore, alpha is co-prime to p. And now we just raise both sides to the power p. So, we look at a power p power e minus 2 into p minus 1 to the power p. This is 1 plus alpha p power e minus 1 to the power p and then apply the binomial theorem. So, 1 plus alpha p power e minus 1 to the power p will give you 1 plus p alpha p power e minus 1 plus summation i from 2 to p p choose i 1 will remain as it is and we will have alpha p power e minus 1 to the power i. Now, depending on what our p power e is, we will see this quite easily that these terms are all divisible by p power e plus 1. And moreover on the left hand side, we have this to be a power p power e minus 1 into p minus 1. And what we have is that this quantity is divisible by p power e but not by p power e plus 1. So, it tells us that a power p power e minus 1 into p minus 1 is not congruent to 1 modulo p power e plus 1 which is to say that the order of a in u p e plus 1 is not this quantity which is p power e minus 1 into p minus 1. Hence, a is primitive in u p e plus 1. This completes our long proof which has been going on over several last lectures. So, this proves that when p is an odd prime, the group of units modulo n is a cyclic group. And the only remaining case now is where n is 2 times p power e. But before that, let us look at this particular example where we have p equal to 5 and a equal to 2. I hope you will remember that 2 is not just a generator for u 5 but it is also a generator for u 25. And now we have to only check or verify that the order of 2 modulo the next prime power, next power of 5 which is 125 is correct number or not. So, e is 2 primitive modulo 125. This is the question. So, we observe that 2 power 4 is not congruent to 1 modulo 25. And this by the proof that we have done will tell you that 2 power 20 is not going to be congruent to 1 modulo 125. And therefore, the order of 2 in u 125 is equal to 100. Remember that the order of 2 in u 125 could have been 20 or 5 into 20 which is 100. And therefore, what we have is that 2 also generates u 125. The next example that we had seen in our last lecture was where our a was 7. And we observed that 7 is a generator, it is a primitive element for u 5. But we saw that it is not a primitive element for u 25. And clearly 7 is not going to be primitive in u 125. This is because to have something primitive modulo 125 we should have the element to be primitive modulo 25 to begin with. Now, the only case remaining is u 2 into p power e where p is an odd prime. And I have promised you that we will show that this group is also a cyclic group. So, let me remain let me remark here again that p remains an odd prime. So, observe if p is odd, then the Euler 5 function of 2 times p power e is the Euler 5 function of p power e. So, the number of elements in u n where n is 2 power 2 times p power e is same as the number of elements in u n by 2. But that does not mean that the numbers are the elements themselves are same because to belong to u n where n is 2 into p power e, each number will have to be an odd number which is not the case when you were looking at u p power e. So, let me do one basic example where p power e is 3 square which is 9. So, let p be equal to 3 and e be equal to 2. So, you have n to be 18 and n by 2 is 9. Let us observe that u 18 has elements 1, 2 will not come because 2 is even. So, we should look at the next odd numbers only, 3 will not come, but 5 will come, 7 comes, 9 does not come, but 11 and 13 come, 15 does not come, but 17 will come. So, u 18 has exactly 6 elements because 5 18 is 5 9 which is 6, u 9 has also 6 elements and these are the elements 1, 2, 4, 5 and 7, 8. So, note that for every even element here, there is an element which is just obtained by adding 1 times 9 to it to get the next element. So, 8 plus 9 is actually 17 and this is how we get this number. So, 1, 5 and 7, these are the elements which remained as they were and for all the even elements in u 9, we simply had to add 9 ones to get the next element. So, this is how we see that the order of these, the orders of these 2 groups are same and in fact, we will prove that the both the groups are actually isomorphic. So, the proof would be as follows, let a in n be a primitive root in u p power e because whenever we have a divisor of n which is 2 times p power e and we are looking for primitive roots modulo n, it should remain a primitive root modulo u p power e as well. So, what it says is that p power e is going to divide the element a raised to the correct order which is phi of p power e minus 1. Now, if a is an odd number, then a raised to phi p power e minus 1 is even. So, 2 divides this quantity. You have that 2 divides one particular number, you have that p power e also divides the same number and 2 comma p power e these are co-prime because p is odd. So, 2 times p power e will have to divide this number a to the power phi p power e minus 1 and then we are done because then a is primitive in u 2 times p power e as well. So, the only situation where we will not get a to be a primitive root would be when a is not odd. If a is even, then this number that we get here will not be even. This will actually be odd and you will not have to dividing a raised to phi p power e minus 1. So, if a is even, then we have to get a separate proof. Let b be a plus p power e here, b is an element in u 2 times p power e and b is congruent to a mod p power e. So, all the earlier discussion now holds for b, a is even and you have added an odd prime power to it. So, b is odd. So, all the earlier discussion now gives us that b is primitive in u 2 times p power e. And so, whenever n is 2 times power of an odd prime, then u n is a cyclic group. So, we have computed all u n which are cyclic groups and the just to recall this for you these are where n is 2, n is 4, n is power of an odd prime or n is 2 times p power e where p is an odd prime. These are all the cases where our groups u n are cyclic and in all the remaining cases we have seen that the groups u n cannot be cyclic. So, as far as getting the cyclic group structure on u n is concerned, we have solved the problem. However, we cannot say that we have understood all u n because we do not understand the structure of the other u n which are not yet cyclic. So, that is something that we would want to do. We are yet to determine the structure of the remaining u n which are not cyclic. If you remember these were the n for which n had 2 or more prime factors or n was of the form 4 into m where m is odd or it was of the form 8 into anything. So, 8 divided n or 4 divided n and the factor is an odd element bigger than one of course. And the other third possibility was that n is a product of 2 or more distinct primes. So, powers. So, it has 2 or more distinct prime factors. These were the cases where the groups u n are not cyclic and we want to understand the structure of these groups. So, once again we will begin by looking at the groups of the form u n where n is a power of 2. The very basic such case would be where we would look at u 8. So, we consider the case 2 power e equal to 8. If you remember u 2 and u 4 are already cyclic. And we also saw that u 8 is isomorphic to c 2 cross c 2. So, the next case we should look at is 16. u 16 has 8 elements all the elements which are odd numbers up to 16. And here I will give you one particular subgroup. The subgroup generated by 5 is 5 phi square which is 9 modulo 16, phi cube is 45 and so modulo 16 this is 13 and then 5 raise to 4 which is 13 into 5 which is 65. So, that is 1 modulo 16. So, we get this group to be 5, 9, 13 and 1. And we observe that this therefore is plus minus 5 because we have these elements of 5, 9, 13 and 1. These are already there and their negatives are these elements. So, negative 5 is 11, negative 9 is 7, negative 13 is 3 and negative 1 is 15. So, we get that our group is actually plus minus 5 plus minus 5 square plus minus 5 cube and plus minus 5 raise to 4. Thus u 16 is c 2 cross c 4. So, the structure of the group u 2 power e where 2 power e is 8 or above is that it will have a direct factor which is c 2. In both these cases 8 as well as 16 we got a direct factor which is c 2 and then there is one more cyclic subgroup which together with c 2 gives us the u 2 power e completely. So, this is the way we are going to prove our result finally that the group u n where n is 2 power e bigger than or equal to 8 then u n is product of two cyclic groups, one of them being of order 2. So, I hope to see you in the next lecture to study this. Thank you.