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 is the set , where for . We denote the Cartesian product by

.

Fact: The Cartesian product forms a group under the binary operation .

Proof: 1. Closure: Note that G is closed. This is true because, since each is a group, each

is closed and for any . Hence since

2. Associativity: Let . Then , , and where . It can be show that both and equal

. Hence, is associative.

3. Identity: The identity is given by , where each is the identity for the group . Note that for , we have

.

Similarly, we can show .

4. Inverse. For each , since is a group. Hence, the inverse of is . Note that

Similarly, .

Hence, by definition, is a group. █Example 1: List out the elements of the groups and .

Solution:

1

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Fact: The group has elements.

Exponential Notation

In multiplicative form, , where .

In additive form, , where .

Other Exponent Facts

1. , where .

2. if the binary operation is in multiplicative form.

3. if the binary operation is in additive from.

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Cyclic Groups

Theorem 1: Let G be a group and let . Then the set

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

1. Closure: Since , and for Then

since

Thus, the closure property holds.

2. Inverse: If , it follows that since

Since , it follows that

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 . Then for some Since K is a subgroup and , the closure property says that . Thus, every element of H is in K and hence , 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 (written as for an additive operation) is called the cyclic subgroup of G generated by a and is denoted by

Definition 3: Let G be a group. An element is said to be a cyclic generator of G if . A group G is cyclic if there is some element that generates G.

Example 2: Determine if the group is cyclic. If so, name all of its cyclic generators.

Solution:

3

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Example 3: Determine if the group is cyclic. If so, name all of its cyclic generators.

Solution:

4

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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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Example 5: Determine if the group is cyclic. If so, name all of its cyclic generators.

Solution:

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Example 6: Determine if the group < , + > is cyclic.

Solution:

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Example 7: Describe the elements of the cyclic group under addition.

Solution:

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Example 8: Describe the elements of the cyclic group under multiplication.

Solution: Since we can see that , ,

, , , , , the elements of this cyclic

group have the form

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Example 9: Describe the elements of the cyclic subgroup of generated by the

matrix under multiplication.

Solution:

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Order of an Element of a Group

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Definition 4: Let . We say the element a has finite order if there exists where ( if the operation is additive). The order of the element a, denoted as

, is the smallest positive integer n where . We say that . An element a has infinite order if for every positive integer k.

Note: If is a finite group with cyclic generator of a, then .

Example 10: For the cyclic group , find the order of the elements 4 and 5.

Solution:

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Example 11: For the cyclic group , find the order of the element 1.

Solution:

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Example 12: Find the order of the elements (1, 2) and (1, 1) in

Solution: Note that . For the element (1, 2), we have

Thus, (1, 2) is of order 2 and generates the subgroup of ,

For the element (1, 1), we have

Thus, (1, 1) is of order 4 and generates the subgroup of ,

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Example 12: Find the order of the elements 2 and 3 in

Solution:

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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 , then H is cyclic ( for any integer k). Suppose . Let . Since H is a subgroup of G (and hence a subset), for some since the parent group G is cyclic. Let m be the smallest positive integer such that . We claim that is a cyclic generator for the subgroup H, that is .

Again, let . Since G is cyclic, for some . If we take , the division algorithm says there exist unique integers q (called the quotient) and r (called the remainder) such that

.

Then

or.

Hence,

or.

Since both , by the closure property. This implies . Since m is the smallest positive integer where and , .Thus, and hence

Since x is an arbitrary element of H, this shows that any element of H can be generated using . Thus, and H is cyclic. █The Multiplicative Cyclic Group for a prime p

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An important cyclic group is the set of non-zero integers modulo p, designated by , 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 is cyclic.

For example, in the group , the element 3 is a cyclic generator since

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 since

The question we want to ask is are we guaranteed to have cyclic generators for the group . The following theorem guarantees the possibility that a cyclic generator will exist.

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Theorem 4: (Fermat’s Little Theorem) Let p be a prime number, a an integer where . Then

.

Fermat’s Little Theorem guarantees that for any , . However, the theorem does not guarantee that is the smallest exponent that a base 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 that . However, we saw that . In contrast, 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 , which we define next.

Definition 4: An element is said to be primitive if of G if and is the smallest exponent that a can be raised to obtain a result of 1. That is, a will be

a cyclic generator of .

Here are a couple of notable items.

Facts

1. For , a primitive element will always exist, that is , is always cyclic.2. If is not primitive, it will generate a cyclic subgroup of . The order of this

cyclic subgroup is guarantee by Lagrange’s Theorem to divide . 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 are primitive. However for each , one does not have to compute in the worst case scenario

to verify a is primitive.

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Method for Determining if is Primitive

i.) Factor into prime factors.ii.) For each distinct prime factor, say , of , if for , if

for all ,

then a will be a primitive element of .

Example 12: Determine whether the elements 2 and 5 are primitive in .

Solution:

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Example 12: Determine whether the elements 3 and 5 are primitive in .

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Solution: For , and the prime factorization of is . Hence, the distinct prime factors of 36 is and . Using the

formula

we can test the elements 3 and 5 as follows.

Element 3:

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 (note there is no need to check the result for

). The element 3 will generate a cyclic subgroup of

Element 5:

Since in either case we did not get a result of 1, 5 will be a primitive element and hence a cyclic generator of .

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Exercises

1. Perform the following computations in the in following groups.a. in b. in c. in d. in e. in f. in

2. Determine the generators of the following cyclic groups.a.b.c.d. .e.f.g.

3. For the group of eight elements, perform the following.a. Compute the cyclic subgroups , , , , , ,

, and of .b. Which elements are generators of from part a.

4. Find the order of the cyclic subgroup of the given group generated by the indicated element.a. The subgroup of generated by 2.b. The subgroup of generated by 3.c. The subgroup of generated by 4.d. The subgroup of generated by 4.e. The subgroup of generated by 5.f. The subgroup of generated by 6.g. The subgroup of generated by 3.

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5. Find the order of the cyclic subgroup of the given group generated by the indicated element.a. The subgroup of generated by (0, 1).b. The subgroup of generated by (2, 6).c. The subgroup of generated by (2, 3).d. The subgroup of generated by 3.e. The subgroup of generated by 7.f. The subgroup of generated by 2.g. The subgroup of generated by 6.

6. Describe all of the elements of the cyclic subgroup of generated by the given matrices.

a.

b.

c.

d.

7. Determine if the following elements are primitive elements of the given groups.a. The element 2 in .b. The element 4 in .c. The element 3 in .d. The element 7 in .e. The element 2 in .f. The element 6 in .

8. Determine all of the primitive elements in the following groups.a.b.c.d.

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.

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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 , where p and q are prime numbers. Show that every proper subgroup (every subgroup except G itself) is cyclic.

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