1. Cyclic Groups A Cyclic Group is a group which can be generated by one of its elements. That is, for some a in G, G={a n | n is an element of Z} Or, in addition notation, G={na |n is an element of Z} This element a (which need not be unique) is called a generator of G. Alternatively, we may write G=. Examples: (Z,.+) is generated by 1 or -1. Z n, the integers mod n under modular addition, is generated by 1 or by any element k in Z n which is relatively prime to n. Non-Examples: Q* is not a cyclic group, although it contains an infinite number of cyclic subgroups. U(8) is not a cyclic group. D n is not a cyclic group although it contains a cyclic subgroup http://www.math.csusb.edu/faculty/susan/modular/modular.html

2. Properties of Cyclic Groups: Criterion for a i = a j For |a| = n, a i = a j iff n divides (i-j) (alternatively, if i=j mod n.) Or, in additive notation, ia = ja iff i=j mod n. For example, in Z 5, 2x4 = 7x4 = 3 because 2=7 mod 5 Corollaries: 1. |a|=| | that is, the order of an element is equal to the order of the cyclic group generated by that element. 2. If a k =e then the order of a divides k. For example... in Z 10, |2|=5 and ={2,4,6,8,0} Caution: This is why it is an error to say, that the order of an element is the power that you need to raise the element to, to get e. A correct statement is, the order of an element is the smallest positive power you need to raise the element to, to get e. math cartoons from http://www.math.kent.edu/~sather/ugcolloq.html

3. Properties of Cyclic Groups For |a|=n, = and |a k |= n / gcd(n,k) For example... In Z 30, let a=1. Then |a|=30. Since Z 30 uses modular addition, a 26 is the same as 26. What is the order of 26? Since gcd (26,30)=2, and gcd(2,30)=2, it follows that |26|= |2| = gcd(2,30)=15. Thus we expect that =, and, in fact, this is {0,2,4,6,8...24,26,28}. So we see that | |=| | = 30 / gcd(2,30) = 30 / 2 = 15. In words, this reads as: If the order of a is n, then the cyclic group generated by a to the k power is the same as the cyclic group generated by a to the power of the greatest common divisor of n and k. Also, the order of a to the k power is equal to the order of a divided by the greatest common divisor of k and the order of a. ( Exercise: Try verbalizing a similar statement for additive notation!) “You may have to do something like this in a stressful situation” - Dr. Englund

4. Properties of Cyclic Groups For |a|=n, = and |a k |= n / gcd(n,k) Corollaries.... This gives us an easy way to specify the generators of a group, the generators of its subgroups, and to tell how these are related. Corollary 1: When are cyclic subgroups equal to one another? Let |a|=n. Then = iff gcd(n,i)=gcd(n,j) For example... In Z 30, let a=1. Again, consider a 2 and a 26. Since gcd (26,30)=2, and gcd(2,30)=2, it follows that = = gcd(2,30)=15. Thus we expect that =, and, in fact, this is {0,2,4,6,8...24,26,28}. So we see that | |=| | = 30 / gcd(2,30) = 30 / 2 = 15. On the other hand, ≠ because gcd(30,3) = 3 while gcd (30,2)=2. And in fact, ={0,3,6...24,27} and | |=30/3 = 10. However, =. Do you see why?

5. Properties of Cyclic Groups For |a|=n, = and |a k |= n / gcd(n,k) Corollaries.... For example... In Z 10, let a=1, so that Z 10 =. Other generators for Z 10 are a k for each k less than 10 and relatively prime to 10. So the other generators are 3,7,and 9. Corollary 2: Generators of Cyclic Groups In any cyclic group G= with order n, the generators are a k for each k relatively prime to n. This gives an easy way to find all of the generators of a cyclic group. Corollary 3 specifies this for Z n, the integers mod n under modular addition. Since any Z n is a cyclic group of order n, its generators would be the positive integers less than n and relatively prime to n. gauss stamp http://webpages.math.luc.e du/~ajs/courses/322spring 2004/worksheets/ws5.html

6. Properties of Cyclic Groups: The Fundamental Theorem of Cyclic Groups Let G= be a cyclic group of order n. Then... 1. Every subgroup of a cyclic group is also cyclic. 2. The order of each subgroup divides the order of the group. 3. For each divisor k of n, there is exactly one subgroup of order k, that is, 3. For each divisor k of n, there is exactly one subgroup of order k, that is, For example consider Z 10 = with |1|=10. Let a=1. Every subgroup of Z 10 is also cyclic. The divisors of 10 are 1, 2, 5, and 10. For each of these divisors we have exactly one subgroup of Z 10, that is,, the group itself, with order 10/1=10, the group itself, with order 10/1=10 ={0,2,4,6,8} with order 10/2 = 5 ={0,2,4,6,8} with order 10/2 = 5 ={0,5} with order 10/5 = 2 ={0,5} with order 10/5 = 2 ={0} with order 10/10=1 ={0} with order 10/10=1 The order of each of these subgroups is a divisor of the order of the group, 10. So the generators of Z 10 would be 1, and the remaining elements: 3, 7, and 9. symmetry 6 ceiling art http://architecture-buildingconstruction.blogspot.com/2006_03_01_archive.html

7. Properties of Cyclic Groups: The Fundamental Theorem of Cyclic Groups - Corollary For example consider Z 10 = with |1|=10. Let a=1. Every subgroup of Z 10 is also cyclic. The divisors of 10 are 1, 2, 5, and 10. For each of these divisors we have exactly one subgroup of Z 10, that is,, the group itself, with order 10/1=10 ={0,2,4,6,8} with order 10/2 = 5 ={0,5} with order 10/5 = 2 ={0} with order 10/10=1 The order of each of these subgroups is a divisor of the order of the group, 10. So the generators of Z 10 would be 1, and the remaining elements: 3, 7, and 9.. Corollary -- Subgroups of Z n : For each positive divisor k of n, the set is the unique subgroup of Z n of order k. These are the only subgroups of Z n.

8. Number of Elements of Each Order in a Cyclic Group Let G be a cyclic group of order n. Then, if d is a positive divisor of n, then the number of elements of order d is φ (d) where φ is the Euler Phi function φ (d) is defined as the number of positive integers less than d and relatively prime to d. The first few values of φ (d) are: d 1 2 3 4 5 6 7 8 9 10 11 12 φ(d) 1 1 2 2 4 2 6 4 6 4 10 4 In non-cyclic groups, if d is a divisor of the order of the group, then the number of elements of order d is a multiple of φ (d), For example, consider Z 12... Z 12 = {0,1,2,3,4,5,6,7,8,9,10,11} We have 1 element of order 2 = {6} because φ (2)=1. We have 2 elements of order 3 = {4,8} because φ (3)=2. We have 2 elements of order 4 = {3,9} because φ (4)=2. And 2 elements of order 6 = {2,10} because φ (6)=2 euler totient equation http://en.wikipedia.org/wiki/List_of_formulae_involving_%CF%80 euler totient graph http://www.123exp-math.com/t/01704079357/