SECTION 6 Cyclic Groups
Recall: If G is a group and aG, then H={an
|n Z} is a subgroup of G.
This group is the cyclic subgroup a of G generated by a.
Also, given a group G and an element a in G, if G ={an
|n Z} , then a is
a generator of G and the group G= a is cyclic
Let a be an element of a group G. If the cyclic subgroup a is finite,
then the order of a is the order | a | of this cyclic subgroup.
Otherwise, we say that a is of infinite order.
Elementary Properties of Cyclic Groups
Theorem
Every cyclic group is abelian.
Proof: Let G be a cyclic group and let a be a generator of G so that
G = a ={an
|n Z}.
If g1 and g2 are any two elements of G, there exists integers r and s
such that g1=ar
and g2=as
. Then
g1g2= ar
as
= ar+s
= as+r
= as
ar
= g2g1,
So G is abelian.
Division Algorithm for Z
Division Algorithm for Z
If m is a positive integer and n is any integer, then there exist unique
integers q and r such that
n = m q + r and 0 r < m
Here we regard q as the quotient and r as the nonnegative remainder
when n is divided by m.
Example:
Find the quotient q and remainder r when 38 is divided by 7.
q=5, r=3
Find the quotient q and remainder r when -38 is divided by 7.
q=-6, r=4
Theorem
Theorem
A subgroup of a cyclic group is cyclic.
Proof: by the division algorithm.
Corollary
The subgroups of Z under addition are precisely the groups nZ under
addition for nZ.
Greatest common divisor
Let r and s be two positive integers. The positive generator d of the
cyclic group
H={ nr + ms |n, m Z}
Under addition is the greatest common divisor (gcd) of r and s. W write
d = gcd (r, s).
Note that d=nr+ms for some integers n and m. Every integer dividing
both r and s divides the right-hand side of the equation, and hence
must be a divisor of d also. Thus d must be the largest number
dividing both r and s.
Example: Find the gcd of 42 andf 72.
6
Relatively Prime
Two positive integers are relatively prime if their gcd is 1.
Fact
If r and s are relatively prime and if r divides sm, then r must divide m.
The structure of Cyclic Groups
We can now describe all cyclic groups, up to an isomorphism.
Theorem
Let G be a cyclic group with generator a. If the order of G is infinite,
then G is isomorphic to  Z, + . If G has finite order n, then G is
isomorphic to  Zn, +n .
Subgroups of Finite Cyclic Groups
Theorem
Let G be a cyclic group with n elements and generated by a. Let bG
and let b=as
. Then b generates a cyclic subgroup H of G containing
n/d elements, where d = gcd (n, s).
Also  as
=  ar
 if and only gcd (s, n) = gcd (t, n).
Example: using additive notation, consider in Z12, with the generator
a=1.
• 3 = 31, gcd(3, 12)=3, so  3  has 12/3=4 elements.  3 ={0, 3, 6, 9}
Furthermore,  3 =  9  since gcd(3, 12)=gcd(9, 12).
• 8= 81, gcd (8, 12)=4, so  8  has 12/4=3 elements.  8 ={0, 4, 8}
• 5= 51, gcd (5, 12)=12, so  5  has 12 elements.  5 =Z12.
Subgroup Diagram of Z18
Corollary
If a is a generator of a finite cyclic group G of order n, then the other
generators of G are the elements of the form ar
, where r is relatively
prime to n.
Example: Find all subgroups of Z18 and give their subgroup diagram.
• All subgroups are cyclic
• By Corollary, 1 is the generator of Z18, so is 5, 7, 11, 13, and 17.
• Starting with 2,  2  ={0, 2, 4, 6, 8, 10, 12, 14, 16 }is of order 9, and
gcd(2, 18)=2=gcd(k, 18) where k is 2, 4, 8, 10, 14, and 16. Thus 2,
4, 8, 10, 14, and 16 are all generators of 2.
• 3={0, 3, 6, 9, 12, 15} is of order 6, and gcd(3, 18)=3=gcd(k, 18)
where k=15
 6={0, 6, 12} is of order 3, so is 12
 9={0, 9} is of order 2
Subgroup diagram of Z18
1
2 3
6 9
0

SECTION 6 Cyclic Groups.ppt notes impor

  • 1.
    SECTION 6 CyclicGroups Recall: If G is a group and aG, then H={an |n Z} is a subgroup of G. This group is the cyclic subgroup a of G generated by a. Also, given a group G and an element a in G, if G ={an |n Z} , then a is a generator of G and the group G= a is cyclic Let a be an element of a group G. If the cyclic subgroup a is finite, then the order of a is the order | a | of this cyclic subgroup. Otherwise, we say that a is of infinite order.
  • 2.
    Elementary Properties ofCyclic Groups Theorem Every cyclic group is abelian. Proof: Let G be a cyclic group and let a be a generator of G so that G = a ={an |n Z}. If g1 and g2 are any two elements of G, there exists integers r and s such that g1=ar and g2=as . Then g1g2= ar as = ar+s = as+r = as ar = g2g1, So G is abelian.
  • 3.
    Division Algorithm forZ Division Algorithm for Z If m is a positive integer and n is any integer, then there exist unique integers q and r such that n = m q + r and 0 r < m Here we regard q as the quotient and r as the nonnegative remainder when n is divided by m. Example: Find the quotient q and remainder r when 38 is divided by 7. q=5, r=3 Find the quotient q and remainder r when -38 is divided by 7. q=-6, r=4
  • 4.
    Theorem Theorem A subgroup ofa cyclic group is cyclic. Proof: by the division algorithm. Corollary The subgroups of Z under addition are precisely the groups nZ under addition for nZ.
  • 5.
    Greatest common divisor Letr and s be two positive integers. The positive generator d of the cyclic group H={ nr + ms |n, m Z} Under addition is the greatest common divisor (gcd) of r and s. W write d = gcd (r, s). Note that d=nr+ms for some integers n and m. Every integer dividing both r and s divides the right-hand side of the equation, and hence must be a divisor of d also. Thus d must be the largest number dividing both r and s. Example: Find the gcd of 42 andf 72. 6
  • 6.
    Relatively Prime Two positiveintegers are relatively prime if their gcd is 1. Fact If r and s are relatively prime and if r divides sm, then r must divide m.
  • 7.
    The structure ofCyclic Groups We can now describe all cyclic groups, up to an isomorphism. Theorem Let G be a cyclic group with generator a. If the order of G is infinite, then G is isomorphic to  Z, + . If G has finite order n, then G is isomorphic to  Zn, +n .
  • 8.
    Subgroups of FiniteCyclic Groups Theorem Let G be a cyclic group with n elements and generated by a. Let bG and let b=as . Then b generates a cyclic subgroup H of G containing n/d elements, where d = gcd (n, s). Also  as =  ar  if and only gcd (s, n) = gcd (t, n). Example: using additive notation, consider in Z12, with the generator a=1. • 3 = 31, gcd(3, 12)=3, so  3  has 12/3=4 elements.  3 ={0, 3, 6, 9} Furthermore,  3 =  9  since gcd(3, 12)=gcd(9, 12). • 8= 81, gcd (8, 12)=4, so  8  has 12/4=3 elements.  8 ={0, 4, 8} • 5= 51, gcd (5, 12)=12, so  5  has 12 elements.  5 =Z12.
  • 9.
    Subgroup Diagram ofZ18 Corollary If a is a generator of a finite cyclic group G of order n, then the other generators of G are the elements of the form ar , where r is relatively prime to n. Example: Find all subgroups of Z18 and give their subgroup diagram. • All subgroups are cyclic • By Corollary, 1 is the generator of Z18, so is 5, 7, 11, 13, and 17. • Starting with 2,  2  ={0, 2, 4, 6, 8, 10, 12, 14, 16 }is of order 9, and gcd(2, 18)=2=gcd(k, 18) where k is 2, 4, 8, 10, 14, and 16. Thus 2, 4, 8, 10, 14, and 16 are all generators of 2. • 3={0, 3, 6, 9, 12, 15} is of order 6, and gcd(3, 18)=3=gcd(k, 18) where k=15  6={0, 6, 12} is of order 3, so is 12  9={0, 9} is of order 2
  • 10.
    Subgroup diagram ofZ18 1 2 3 6 9 0