1
Cyclic Groups
HW # 1-13 p. 16 at the end of the notes
In this section, we discuss the basics of cyclic groups and subgroups. First, discuss the
basics of another type of group.
Definition 1: The Cartesian product of the groups G1 , G2 , , Gn is the set
(a1 , a2 , , an ) , where ai  Gi for i  1, 2,  , n . We denote the Cartesian product by
n
G1  G2    Gn   Gi .
i 1
Fact: The Cartesian product G1  G2    Gn forms a group under the binary operation
(a1 , a2 , , an )(b1 , b2 , , bn )  (a1b1 , a2b2 , , an bn ), ai , bi  Gi .
Proof:
1. Closure: Note that G is closed. This is true because, since each Gi is a group, each
Gi is closed and ai bi  Gi for any ai , bi  Gi . Hence
(a1b1 , a2b2 , , an bn )  G1  G2    Gn since ai bi  Gi
2. Associativity: Let x, y, z  G1  G2    Gn . Then x  (a1 , a2 , , an ) , y  (b1 , b2 , , bn ) ,
and z  (c1 , c2 , , cn ) where ai , bi , ci  Gi . It can be show that both ( xy) z and x( yz )
equal (a1b1c1 , a2b2 c2 , , an bn cn ) . Hence, G1  G2    Gn is associative.
3. Identity: The identity is given by e  (e1 , e2 , , en ) , where each ei is the identity for
the group Gi . Note that for x  (a1 , a2 , , an ) , we have
xe  (a1 , a2 , , an )(e1 , e2 , , en )  (a1e1 , a2 e2 , , an en )  (a1 , a2 , , an )  x .
Similarly, we can show ex  x .
4. Inverse. For each ai  Gi , ai1  Gi since Gi is a group. Hence, the inverse of
x  (a1 , a2 , , an ) is x 1  (a11 , a21 , , an1 ) . Note that
xx1  (a1 , a2 , , an )( a11 , a21 , , an1 )  (a1a11 , a2 a21 , , an an1 )  (e1 , e2 , , en )  e
Similarly, x 1 x  e .
Hence, by definition, G1  G2    Gn is a group.
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2
Example 1: List out the elements of the groups Z 2  Z 3 and Z 3  Z 4 .
Solution:
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Fact: The group Z m  Z n has m n elements.
Exponential Notation
In multiplicative form, a n  a
a
 a

a , where n  Z  .
n times
a


a , where n  Z  .
In additive form, na  a


n times
Other Exponent Facts
1. a m  a n  a m  n , where m, n  Z .
1
1
1
1

a 
 a

a
2. a  n  a
if the binary operation is in multiplicative form.


n times
a) 
 (

a
) 

( 
a ) if the binary operation is in additive from.
3.  na  (

n times
3
Cyclic Groups
Theorem 1: Let G be a group and let a  G . Then the set
H  {a n | n  Z }
is a subgroup of G and is the smallest subgroup of G that contains a, that is, every
subgroup that contains the element a will contain H.
Proof: To prove that H is a subgroup of G, let x, y  H .
1. Closure: Since x, y  H , x  a r and y  a s for r, s  Z. Then
xy  a r a s  a r  s  H since r  s  Z .
Thus, the closure property holds.
2. Inverse: If x  a r , it follows that x 1  a  r since
a r a r  a r  (r)  a 0  e
Since  r  Z , it follows that x 1  a  r  H
Thus, the closure property holds
Hence, H is a subgroup of the group G. To show H is the smallest subgroup of G
containing a, suppose K is a smallest subgroup of G that contains a. Let x  H . Then
x  a r for some r  Z . Since K is a subgroup and a  K , the closure property says that
x  a r  K . Thus, every element of H is in K and hence H  K , which contradicts the
assertion that K is the smallest subgroup of G containing a. Hence, H is the smallest
subgroup of G containing a.
Definition 2: Let G be a group. Then the subgroup {a n | n  Z} (written as {na | n  Z }
for an additive operation) is called the cyclic subgroup of G generated by a and is
denoted by  a 
Definition 3: Let G be a group. An element a  G is said to be a cyclic generator of G if
G  a  . A group G is cyclic if there is some element a  G that generates G.
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4
Example 2: Determine if the group Z 6 is cyclic. If so, name all of its cyclic generators.
Solution:
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5
Example 3: Determine if the group Z is cyclic. If so, name all of its cyclic generators.
Solution:
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Example 4: Determine if the group < R, + > is cyclic. If so, name all of its cyclic
generators.
Solution:
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6
Example 5: Determine if the group Z 2  Z 3 is cyclic. If so, name all of its cyclic
generators.
Solution:
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7
Example 6: Determine if the group < Z n , + > is cyclic.
Solution:
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Example 7: Describe the elements of the cyclic group 5Z under addition.
Solution:
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8
 1  n

Example 8: Describe the elements of the cyclic group   | n  Z  under
 3 

multiplication.
1
Solution: Since we can see that  
 3
1
0
1
1
1
1
   1  3  3 ,    1 ,
 3
 3
3
cyclic group have the form
3

1
3
1
3
1
1
   ,
3
 3
1
 3  27 ,  
 3
2
3
2
1
1
   ,
9
 3

1
3
2
 32  9 ,
3
1
1
, the elements of this
  
27
 3
1 1 1 1


, , , , , 1, 3, 9, 27, 81,
 81 27 9 3

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Example 9: Describe the elements of the cyclic subgroup of GL( 2, R ) generated by the
 0 1
matrix 
 under multiplication.
  1 0
Solution:
9
Order of an Element of a Group
Definition 4: Let a  G . We say the element a has finite order if there exists k  Z 
where a k  e ( ka  e if the operation is additive). The order of the element a, denoted as
| a | , is the smallest positive integer n where a n  e . We say that | a | n . An element a
has infinite order if a k  e for every positive integer k.
Note: If G  a  is a finite group with cyclic generator of a, then | a || G || a | .
Example 10: For the cyclic group Z 6  {0, 1, 2, 3, 4, 5} , find the order of the elements 4
and 5.
Solution:
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Example 11: For the cyclic group Z , find the order of the element 1.
Solution:
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10
Example 12: Find the order of the elements (1, 2) and (1, 1) in Z 2  Z 4
Solution: Note that Z 2  Z 4  {( 0,0), (0,1), (0,2), (0,3), (1,0), (1,1), (1,2), (1,3)} . For the
element (1, 2), we have
1  (1, 2)  (1, 2)
2  (1, 2)  (1, 2)  (1, 2)  ( 2 mod 2, 4 mod 4)  (0, 0)
Thus, (1, 2) is of order 2 and generates the subgroup {( 0,0), (1,2)} of Z 2  Z 4 ,
For the element (1, 1), we have
1  (1, 1)  (1, 1)
2  (1, 1)  (1, 1)  (1, 1)  ( 2 mod 2, 2 mod 4)  (0, 2)
3  (1, 1)  (1, 1)  (1, 1)  (1, 1)  (0, 2)  (1, 1)  (1 mod 2, 3 mod 4)  (1, 3)
4  (1, 1)  (1, 1)  (1, 1)  (1, 1)  (1, 1)  (1, 3)  (1,1)  ( 2 mod 2, 4 mod 4)  (0, 0)
Thus, (1, 1) is of order 4 and generates the subgroup {( 0,0), (1, 1), (0, 2), (1, 3)} of Z 2  Z 4 ,
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Example 12: Find the order of the elements 2 and 3 in U 7
Solution:
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11
We next examine two important theorems concerning cyclic groups.
Theorem 2: Every cyclic group is abelian.
Proof:
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Theorem 3: A subgroup of a cyclic group is cyclic.
Proof: Let H be a subgroup of a cyclic group G generated by a. If H  {e} , then H is
cyclic ( e k  e for any integer k). Suppose H  {e} . Let x  H . Since H is a
subgroup of G (and hence a subset), x  a n for some n  Z  since the parent group
G is cyclic. Let m be the smallest positive integer such that a m  H . We claim that
a m is a cyclic generator for the subgroup H, that is H  a m  .
Again, let x  H . Since G is cyclic, x  a n for some n  Z  . If we take n  m , the
division algorithm says there exist unique integers q (called the quotient) and r (called
the remainder) such that
n  qm  r, 0  r  m .
Then
a n  a qm  r
or
a n  a qm  a r .
Hence,
a r  a  qm  a n
or
a r  (a m )  q  a n .
Since both a m , a n  H , (a m ) q  a n  H by the closure property. This implies
a r  H . Since m is the smallest positive integer where a r  H and 0  r  m , r  0 .
Thus, a n  a qm and hence
x  a n  (a m ) q
Since x is an arbitrary element of H, this shows that any element of H can be generated
using a m . Thus, H  a m  and H is cyclic.
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12
The Multiplicative Cyclic Group Z *p for a prime p
An important cyclic group is the set of non-zero integers modulo p, designated by
Z *p  {1, 2, 3, , p  1} , where p is a prime. We have already said that this set represents
a group under multiplication. We next state this basic fact without proof.
Fact: The multiplicative group Z *p is cyclic.
For example, in the group Z 7*  {1, 2, 3, 4, 5, 6} , the element 3 is a cyclic generator
since
31 mod 7  3
32 mod 7  3 mod 7  2
33 mod 7  (32  3) mod 7  ( 2  3) mod 7  6 mod 7  6
34 mod 7  (33  3) mod 7  (6  3) mod 7  18 mod 7  4
35 mod 7  (34  3) mod 7  ( 4  3) mod 7  12 mod 7  5
36 mod 7  (35  3) mod 7  (5  3) mod 7  15 mod 7  1
However, it is easy to see the element 4 is not a cyclic generator but only generates a
the cyclic subgroup {1, 2, 4} of Z 7* since
41 mod 7  4
4 2 mod 7  16 mod 7  2
4 3 mod 7  (4 2  4) mod 7  (2  4) mod 7  1
The question we want to ask is are we guaranteed to have cyclic generators for the
group Z *p . The following theorem guarantees the possibility that a cyclic generator
will exist.
13
Theorem 4: (Fermat’s Little Theorem) Let p be a prime number, a an integer where
p | a . Then
a p1  1(mod p) .
Fermat’s Little Theorem guarantees that for any a  Z *p , a p1  1(mod p) . However, the
theorem does not guarantee that p  1 is the smallest exponent that a base a  Z *p can be
raised to get a result of 1 in modulo p arithmetic. For example, Fermat’s Little Theorem
guarantees for the element 4 in Z 7*  {1, 2, 3, 4, 5, 6} that 4 6 mod 7  1. However, we saw
that 4 3 mod 7  1 . In contrast, 36 mod 7  1 and 6 was the smallest exponent where the
element 3 raised to a power gives 1. In this case, 3 is a primitive element for Z 7* , which we
define next.
Definition 4: An element a  Z *p is said to be primitive if of G if a p1  1(mod p) and
p  1 is the smallest exponent that a can be raised to obtain a result of 1. That is, a will be
a cyclic generator of Z *p .
Here are a couple of notable items.
Facts
1. For Z *p , a primitive element will always exist, that is , Z *p is always cyclic.
2. If a  Z *p is not primitive, it will generate a cyclic subgroup of Z *p . The order of this
cyclic subgroup is guarantee by Lagrange’s Theorem to divide | Z *p | p  1 . Note
that Lagrange’s Theorem states that if H is a subgroup of a finite group G, then the
order of H must divide the order of its parent group G.
3. There is no known way to generate which elements of Z *p are primitive. However
for each a  Z *p , one does not have to compute in the worst case scenario
a, a 2 , a 3 ,, a p 1  1 to verify a is primitive.
14
Method for Determining if a  Z *p is Primitive
i.) Factor p  1 into prime factors.
ii.) For each distinct prime factor, say q1 , q2 ,, qr , of p  1 , if for a  Z *p , if
a ( p 1) / qi mod p  1 for all i  1, 2,, r ,
then a will be a primitive element of Z *p .
*
Example 12: Determine whether the elements 2 and 5 are primitive in Z11
.
Solution:
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15
*
Example 13: Determine whether the elements 3 and 5 are primitive in Z 37
.
*
Solution: For Z 37
 {1, 2, 3, ,35, 36} , p  37 and the prime factorization of p  1  36
is 36  4  9  2 2  32 . Hence, the distinct prime factors of 36 is q1  2 and q2  3 . Using
the formula
a ( p 1) / qi mod p
we can test the elements 3 and 5 as follows.
Element 3:
q1  2 :
a  3  336 / 2 mod 37  318 mod 37  1
Since we have a result of 1 for the element 3, the element 3 will not be primitive element
*
and will not be a cyclic generator of Z 37
(note there is no need to check the result for
*
q2  3 ). The element 3 will generate a cyclic subgroup of Z 37
Element 5:
q1  2 :
a  5  536 / 2 mod 37  518 mod 37  36  1
q1  3 :
a  5  536 / 3 mod 37  512 mod 37  10  1
Since in either case we did not get a result of 1, 5 will be a primitive element and hence a
*
cyclic generator of Z 37
.
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16
Exercises
1. Perform the following computations in the in following groups.
a. (1, 1)  (1, 2) in Z 2  Z 3
b. (1, 2)  (0, 2) in Z 2  Z 3
c. ( 2, 4)  (3, 3) in Z 4  Z 5
d. ( 4, 7)  (3, 6) in Z 5  Z 8
e. (7, 10)  (8, 1) in Z10  Z11
f. (17, 25)  (14,28) in Z 25  Z 50
2. Determine the generators of the following cyclic groups.
a. 2 Z
b. 10Z
c. Z 5
d. Z 6 .
e. Z 2  Z 3
f. U 8
g. Z 7*
3. For the group Z 8 of eight elements, perform the following.
a. Compute the cyclic subgroups  0  ,  1  ,  2  ,  3  ,  4  ,  5  ,  6  , and
 7  of Z 8 .
b. Which elements are generators of Z 8 from part a.
4. Find the order of the cyclic subgroup of the given group generated by the indicated
element.
a. The subgroup of Z 4 generated by 2.
b. The subgroup of Z 4 generated by 3.
c. The subgroup of Z 5 generated by 4.
d. The subgroup of Z 6 generated by 4.
e. The subgroup of Z 6 generated by 5.
f. The subgroup of Z10 generated by 6.
g. The subgroup of Z12 generated by 3.
17
5. Find the order of the cyclic subgroup of the given group generated by the indicated
element.
a. The subgroup of Z 2  Z 3 generated by (0, 1).
b. The subgroup of Z 4  Z12 generated by (2, 6).
c. The subgroup of Z 6  Z15 generated by (2, 3).
d. The subgroup of U 8 generated by 3.
e. The subgroup of U10 generated by 7.
f. The subgroup of Z 7* generated by 2.
*
g. The subgroup of Z11
generated by 6.
6. Describe all of the elements of the cyclic subgroup of GL( 2, R ) generated by the given
2  2 matrices.
 0  1
a. 

 1 0 
1
b. 
0
3
c. 
0
1
1
0
2
 0  2
d. 

 2 0 
7. Determine if the following elements are primitive elements of the given groups.
a. The element 2 in Z 5* .
b. The element 4 in Z 5* .
*
c. The element 3 in Z11
.
*
d. The element 7 in Z11
.
*
e. The element 2 in Z 41
.
*
f. The element 6 in Z 41
.
8. Determine all of the primitive elements in the following groups.
a. Z 5*
b. Z 7*
*
c. Z11
*
d. Z13
18
9. Prove that a cyclic group with only one generator and have at most 2 elements.
10. Show that a group with no proper nontrivial subgroups is cyclic.
11. Prove that a direct product of abelian groups is abelian.
12. Show that every group G of prime order must be cyclic.
13. Let G be a group of order p  q , where p and q are prime numbers. Show that every
proper subgroup (every subgroup except G itself) is cyclic.