Cyclic Groups

A A Cyclic GroupCyclic Group is a group is a group which can be generated by which can be generated by

one of its elements. one of its elements.

That is, for some That is, for some aa in in GG, , GG={={aann | | n n is an element of is an element of ZZ}}

Or, in addition notation, Or, in addition notation, GG={={nana | |nn is an element of is an element of ZZ}}

This element This element aa (which need not be unique) is called a (which need not be unique) is called a

generatorgenerator of of GG..

Alternatively, we may write Alternatively, we may write GG==<a><a>..

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 Zn

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

<R(360/n)

>http://www.math.csusb.edu/faculty/susan/modular/modular.html

Properties of Cyclic Groups:

Criterion for ai = aj

For |a| = n, ai = aj 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 Z5,

2x4 = 7x4 = 3 because 2=7 mod 5

Corollaries:1. |a|=|<a>| that is, the order of an element is equal to the order of the cyclic group generated by that element.

2. If ak=e then the order of a divides k.

For example...in Z

10, |2|=5 and

<2>={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 fromhttp://www.math.kent.edu/~sather/ugcolloq.html

Properties of Cyclic Groups

For |a|=n, <ak>=<agcd(n,k)>and |ak|=n/

gcd(n,k)

For example ...In Z

30, let a=1. Then |a|=30.

Since Z30

uses modular addition, a26 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 <26>=<2>, and, in fact, this is {0,2,4,6,8...24,26,28}.

So we see that |<2>|=|<26>|=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

Properties of Cyclic Groups

For |a|=n, <ak>=<agcd(n,k)>and |ak|=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 <ai>=<aj> iff gcd(n,i)=gcd(n,j)

For example ...In Z

30, let a=1. Again, consider a2 and a26.

Since gcd (26,30)=2, and gcd(2,30)=2, it follows that <|26|>= <|2|> =

gcd(2,30)=15.

Thus we expect that <26>=<2>, and, in fact, this is {0,2,4,6,8...24,26,28}.

So we see that |<2>|=|<26>|=30/

gcd(2,30) = 30/

2 = 15.

On the other hand, <3> ≠<2> because gcd(30,3) = 3 while gcd (30,2)=2.

And in fact, <3>={0,3,6...24,27} and |<3>|=30/3 = 10.

However, <3>=<9>. Do you see why?

Properties of Cyclic Groups

For |a|=n, <ak>=<agcd(n,k)>and |ak|=n/

gcd(n,k) Corollaries....

For example ...In Z

10, let a=1,

so that Z10

= <a>.Other generators for Z

10 are

ak 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 GroupsIn any cyclic group G=<a> with order n,the generators are ak 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 Zn is a cyclic group of order n,

its generators would be the positive integers less than n and relatively prime to n.

gauss stamphttp://webpages.math.luc.edu/~ajs/courses/322spring2004/worksheets/ws5.html

Properties of Cyclic Groups:

The Fundamental Theorem of Cyclic Groups

Let Let GG=<=<aa> be a cyclic > be a cyclic group of order group of order nn. . Then ...Then ...

1. 1. Every subgroup of a Every subgroup of a cyclic group is also cyclic group is also cyclic.cyclic.

2. 2. The order of each The order of each subgroup divides the subgroup divides the order of the group.order of the group.

3. 3. For each divisor For each divisor kk of of nn, there is , there is exactly oneexactly one subgroup of ordersubgroup of order k k, that , that

is, is, <<aan/n/kk>>

For example For example considerconsider ZZ1010 = <1> with |1|=10. = <1> with |1|=10.

Let Let aa=1.=1.Every subgroup of Every subgroup of ZZ1010 is also cyclic. is also cyclic.

The divisors of 10 are 1, 2, 5, and 10.The divisors of 10 are 1, 2, 5, and 10.

For each of these divisors we have For each of these divisors we have exactly one subgroup of exactly one subgroup of ZZ1010, that is,, that is,

<1>, the group itself, with order 10/1=10<1>, the group itself, with order 10/1=10<2>={0,2,4,6,8} with order 10/2 = 5<2>={0,2,4,6,8} with order 10/2 = 5

<5>={0,5} with order 10/5 = 2<5>={0,5} with order 10/5 = 2<10>={0} with order 10/10=1<10>={0} with order 10/10=1

The order of each of these subgroups is a The order of each of these subgroups is a divisor of the order of the group, 10.divisor of the order of the group, 10.

So the So the generatorsgenerators of of Z Z1010 would be 1, and would be 1, and the remaining elements: 3, 7, and 9.the remaining elements: 3, 7, and 9.

symmetry 6 ceiling arthttp://architecture-buildingconstruction.blogspot.com/2006_03_01_archive.html

Properties of Cyclic Groups:

The Fundamental Theorem of Cyclic Groups - Corollary

For example consider Z

10 = <1> 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,

<1>, the group itself, with order 10/1=10<2>={0,2,4,6,8} with order 10/2 = 5

<5>={0,5} with order 10/5 = 2<10>={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 Z10

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 <n/k>

is the unique subgroup of Z

n

of order k. These are the

only subgroups of Z

n.

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 12φ(d) 1 1 2 2 4 2 6 4 6 4 10 4

In non-cyclic groupsIn non-cyclic groups, if , if dd is a divisor of the is a divisor of the order of the group, thenorder of the group, thenthe number of elements of order the number of elements of order dd is is a a multiplemultiple of of φφ((dd), ),

For example, consider Z12

...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 equationhttp://en.wikipedia.org/wiki/List_of_formulae_involving_%CF%80

euler totient graphhttp://www.123exp-math.com/t/01704079357/