Section 6.4

Section 6.4 SubgroupsSection 6.4 SubgroupsSection 6.4 SubgroupsSection 6.4 Subgroups 629

Section 6.4 Subgroups : Groups Inside a GroupSection 6.4 Subgroups : Groups Inside a GroupSection 6.4 Subgroups : Groups Inside a GroupSection 6.4 Subgroups : Groups Inside a Group

Purpose of SectionPurpose of SectionPurpose of SectionPurpose of Section To introduce the concept of a subgroupsubgroupsubgroupsubgroup and find the

subgroups of various symmetry groups.

IntroductionIntroductionIntroductionIntroduction

Recall the six symmetries of an equilateral triangle; the identity map, three

flips about the midlines through the vertices of the triangle, and two

(counterclockwise) rotations of 120 and 240 degrees.

Symmetries of an equilateral triangle

Figure 1

This set, along with the group operation of composition, forms a self-

contained algebraic system called a group. It is distinguished by the fact the

group operation is closed and the group contains an identity (do nothing

operation), and every element in the group has an inverse. But this group is

only the outside of the shell, inside there may be smaller groups. For

example, in the dihedral group 3

D of six symmetries of an equilateral triangle,

consider the subset of three rotational symmetries, the identity map e and the

two rotations of 120 and 240 degrees. The Cayley table for these symmetries

{ }120 240, ,e R R is drawn in Figure 2, which can easily be verified to form a

group. The group operation is closed (i.e. the product of two elements

belongs to the group), e is the identity, and each element has an inverse.


Subgroup of rotations of symmetries of an equilateral triangle

Figure 2

This motivates the following definition of “groups within groups,” or

subgroups.

Definition:Definition:Definition:Definition: Let ( ),G ∗ be a group with operation ∗ . If a subset H G⊆ itself

forms a group with the same operation ∗ , then H is called a subgroupsubgroupsubgroupsubgroup of G .

If H is neither the identity { }e nor the entire group G , which are groups

called trivial subgroups trivial subgroups trivial subgroups trivial subgroups of G , then H is called a proper subgroupproper subgroupproper subgroupproper subgroup of .G

Example 1 (SubgroupExample 1 (SubgroupExample 1 (SubgroupExample 1 (Subgroups of Symmetries of an Equilateral Triangle)s of Symmetries of an Equilateral Triangle)s of Symmetries of an Equilateral Triangle)s of Symmetries of an Equilateral Triangle)

Find the proper subgroups of the dihedral group 3

D the symmetries of

an equilateral triangle.

Solution:Solution:Solution:Solution: The Cayley table for the dihedral group3

D of symmetries of an

equilateral triangle and its proper subgroups are displayed in Figure 3. There

are four proper subgroups of3

D ; the rotational subgroup { }120 240, ,e R R of order

3 and three “flip” subgroups { } { } { }, , , , ,v ne nw

e F e F e F , each of order 2.


∗ e 120R

240R

vF

neF

nwF

e e 120R

240R

vF

neF

nwF

120R

120R

240R e ne

F nw

F v

F

240R

240R e 120

R nw

F v

F ne

F

vF

vF

nwF

neF e 240

R 120

R

neF

neF

vF

nwF

120R e 240

R

nwF

nwF

neF

vF

240R

120R e

∗ e 120R

240R

vF

neF

nwF

e e 120R

240R

vF

neF

nwF

120R

120R

240R e ne

F nw

F v

F

240R

240R e 120

R nw

F v

F ne

F

vF

vF

nwF

neF e 240

R 120

R

neF

neF

vF

nwF

120R e 240

R

nwF

nwF

neF

vF

240R

120R e

{ }1,

vH e F= Flip around vertical axis { }2

,nw

H e F= Flip around the northwest axis

∗ e 120R

240R

vF

neF

nwF

e e 120R

240R

vF

neF

nwF

120R

120R

240R e ne

F nw

F v

F

240R

240R e 120

R nw

F v

F ne

F

vF

vF

nwF

neF e 240

R 120

R

neF

neF

vF

nwF

120R e 240

R

nwF

nwF

neF

vF

240R

120R e

∗ e 120R

240R

vF

neF

nwF

e e 120R

240R

vF

neF

nwF

120R

120R

240R e ne

F nw

F v

F

240R

240R e 120

R nw

F v

F ne

F

vF

vF

nwF

neF e 240

R 120

R

neF

neF

vF

nwF

120R e 240

R

nwF

nwF

neF

vF

240R

120R e

{ }3,

neH e F= Flip around the northeast axis. { }4 120 240

, ,H e R R= Identity and two rotations

Four Proper Subgroups of Symmetries of an Equilateral Triangle

Figure 3

We let the reader verify that each of these subgroups satisfy the necessary

requirements to be groups. See Problem 1.

Proper Proper Proper Proper SuSuSuSubgroups of the Klein 4bgroups of the Klein 4bgroups of the Klein 4bgroups of the Klein 4----GroupGroupGroupGroup

Recall from Section 6.1 that the group of (rotational and reflective)

symmetries of a rectangle form a group, called the Klein 4-group, with

elements { }180, , ,G e R H V= , where as always " "e denotes the group identity,

180R a rotation of 180 degrees, and ,H V flips around the horizontal and

vertical midlines, respectively. Figure 4 shows the Cayley table of the

symmetries of a rectangle and its three proper subgroups, all of order 2.

Note how the order of the subgroups always divides the order of the group.

We will not prove it here but this is a fundamental property was one of the

first fundamental theorems proven in group theory and is called Lagrange’s

theorem, after the great French/Italian mathematician Joseph-Louis Lagrange

(1736-1813).


∗ e 180R H V

e e 180R H V

180R

180R e V H

H H V e 180R

V V H 180R e

∗ e 180R H V

e e 180R H V

180R

180R e V H

H H V e 180R

V V H 180R e

Group { }180, , ,G e R V H= of symmetries of

a rectangle.

Subgroup of symmetres { },H e H= about

the horizontal midline.

∗ e 180R H V

e e 180R H V

180R

180R e V H

H H V e 180R

V V H 180R e

∗ e 180R H V

e e 180R H V

180R

180R e V H

H H V e 180R

V V H 180R e

Subgroup of symmetries { },H e V= about

the horizontal midline.

Subgroup of rotational symmetries

{ }180,H e R= .

Symmetry Group of a Rectangle and Three Subgroups

Figure 4

Test of SubTest of SubTest of SubTest of Subgroupsgroupsgroupsgroups

Although a subset H of a groupG is a group only if it satisfies the four

axioms of a group; i.e.. Closure, Associativity, Identity, Inverse, the fact that

H is a subset of G , it is only necessary to verify that the group operation ∗

is closed in H and that every element of H has an inverse in H . There is no

need to show the existence of an identity; the identity in the larger group G is

also an identity in the subgroup H , This result is summarized in the following

theorem.


Theorem Theorem Theorem Theorem 1111 (Conditions for Being a Subgroup) (Conditions for Being a Subgroup) (Conditions for Being a Subgroup) (Conditions for Being a Subgroup) Let ( ),G ∗ be a group with

operation ∗ and H a nonempty subset of G . The set H with operation ∗ is

a subgroupsubgroupsubgroupsubgroup ( ),H ∗ of ( ),G ∗ if the following two conditions hold:

i) H is closedclosedclosedclosed under ∗ . That is, ,x y H x y H∀ ∈ ⇒ ∗ ∈ .

ii) Every element in H has an inverseinverseinverseinverse in H . That is

( )( )( )1 1 1h H h H h h h h e

− − −∀ ∈ ∃ ∈ ∗ = ∗ = .

where " "e is the identity element in G .

Proof: Proof: Proof: Proof:

Since ∗ is a binary operation on G , it is also a binary operation on the

subset H , and by assumption i) we know ∗ maps H H× into H . Next, the

associative law ( ) ( )a b c a b c∗ ∗ = ∗ ∗ holds for all , ,a b c H∈ since H is a

subset of G and we know it holds for all , ,a b c G∈ . We now ask if the identity

e G∈ also belongs to H and is the identity of H ? The answer is yes since

by picking an h H∈ we know by hypothesis ii) there exists a 1h H−∈ , and by

closure 1h h e H−∗ = ∈ . Hence, we have verified the four properties required

for a group: closure, associativity, identity, and inverse. Hence H is a group.

▌

Example 2 (Test of Subgroup)Example 2 (Test of Subgroup)Example 2 (Test of Subgroup)Example 2 (Test of Subgroup) Let { }0, 1, 2,...G = = ± ±� be the group of

integers with the binary operation of addition + . Show the even integers

{ }2 0, 2, 4,...= ± ±� is a subgroup of G .

SolutionSolutionSolutionSolution

We observe that + is closed binary operation in 2� since if

1 22 , 2m k n k= = are even integers, so is their sum ( )1 2

2 2m n k k+ = + ∈ � .

Secondly, every even integer 2 2k ∈ � has an inverse, namely

( )2 2 2k k− = − ∈ � . ▌

Note:Note:Note:Note: The order of any subgroup of a group is a divisor of the group, and if the

order of the subgroup is a prime number then there will be a subgroup of that order.

Hence, there is not a subgroup of order 9 of a subgroup of order 30, and there

might be a subgroup of order 15, and there is a subgroup of order 5.


Example Example Example Example 3333 (Group of Infinite Order) (Group of Infinite Order) (Group of Infinite Order) (Group of Infinite Order) Let ( ),G ∗ be the group of points in the

plane 2� where the group operation :+ × →� � � is coordinate wise addition

of points ( ) ( ) ( ), , ,a b c d a c b d+ = + + . We leave it to the reader to show ( )2, +�

is a group. Show that the x -axis ( ){ },0 :H x x= ∈� is a subgroup of ( )2, +� .


The x -axis is a subset of the plane and the operation + is closed in H

since

( ) ( ) ( )1 2 1 2,0 , ,0 ,0x H x H x x H∈ ∈ ⇒ + ∈

Also every ( )1,0x H∈ has an inverse ( )1

,0x H− ∈ , i.e. ( ) ( ) ( )1 1,0 ,0 0,0x x+ − = ,

which is the group identity in 2� ▌

In general it is not a simple task to find all subgroups of a group, but for

cyclic groups it is an easy task.

Example Example Example Example 4444 (S (S (S (Subgroups of ubgroups of ubgroups of ubgroups of the Dihedral Groupthe Dihedral Groupthe Dihedral Groupthe Dihedral Group 4D )))) Figure 5 shows the dihedral

group 4

D of eight symmetries of a square, also called the octic octic octic octic group.

a) Is the octic group commutative? Hint: Compare products 270 ne

R F

and270

ne

F R .

b) There are several subsets of the eight symmetries that form a group

in their own right. These are called subgroups of the octic group. Can you

find all ten of them?


a) The reader can check but 270 270ne ne

R F F R≠ . Hence, the octic group is

not commutative.

b) The 9 subgroups of the octic group are

{ } { } { } { } { } { } { } { }180 180 180, , , , , , , , , , , , , , , , , ,

nw ne nw nee e V e H e F e F e R e R V H e R F F


MotionMotionMotionMotion SymbolSymbolSymbolSymbol First and Final PositionsFirst and Final PositionsFirst and Final PositionsFirst and Final Positions

No motion 0e R=

Rotate 90�

Counterclockwise 90R

Rotate 180�


Rotate 270�


Horizontal flip H

Vertical flip V

Northeast flip neF

Northwest flip nwF


Symmetries of a Square

Figure 5

The nine subgroups of the symmetry group of a square form a partially

ordered set with ordering set inclusion, which can be illustrated in a Hasse

diagram as shown in Figure 6. The group 4

D is itself a subgroup of the group

4S of permutations of four elements.

Hasse Diagram for the Subgroups of the Octic Group4

D

Figure 6

Note:Note:Note:Note: The two groups and � � are subsets of � and under the same operation

of addition, hence both are subgroups of � .

Subgroups of Cyclic Groups Subgroups of Cyclic Groups Subgroups of Cyclic Groups Subgroups of Cyclic Groups

We have seen that the finite cyclic group n

Z of order n is a group

generated by a single element in the group. That is, there exists a g Z∈

such that


{ }2 3 1, , , ,...,

n

ng e g g g g Z

−≡ = .

To find the subgroups of n

Z we start with an element n

g Z∈ and compute the

set g generated by g . This set may or may not be all of n

Z , but it will be a

subgroup of n

Z . We then move on to a newn

h Z∈ that is not in g and

compute the set h generated by h . Continuing in this manner we will

eventually obtain all subgroups of n

Z . For example to apply this technique to

the group

{ }120,1, 2,3, 4,5,6,7,8,9,10,11Z =

where the group operation is addition modulo 12, where the group operation is

addition modulo 12. If we start taking “powers” of 1g = , we get (remember

powers are really adding 1)

{ }1 1,2,3,4,5,6,7,8,9,10,11,0=

which has generated the entire group 12� . On the other hand the element

2g = generates the subgroup { }2 0, 2, 4,6,8 G= ⊆ . Figure 7 shows the

subgroups generated by 1, 2,3, 4g = . Do you see why 12

5 = � and

{ }6 0,6= .

1g = generates the entire group

{ }121 0,1,2,...,11= =�

2g = generates the subgroup

{ }2 0,2,4,6,8,10=



{ }3 0,3,6,9=


{ }4 0,4,8=

Four typical subgroups generated by elements of the group

Figure 7

Table 1 shows the subgroups generated by each element of the group and the

order of the subgroup generated by the generator.

GeneratorGeneratorGeneratorGenerator Order of the GeneratorOrder of the GeneratorOrder of the GeneratorOrder of the Generator

121 = � 12 ( )12

1 0=

{ }2 0,2,4,6,8,10= 6 ( )62 0=

{ }3 0,3,6,9= 4 ( )43 0=

{ }4 0,4,8= 3 ( )34 0=

125 = � 12 ( )12

5 0=

{ }6 0,6= 2 ( )26 0=

127 = � 12 ( )12

7 0=

{ }8 0,4,8= 3 ( )38 0=

{ }9 0,3,6,9= 4 ( )49 0=

{ }10 0, 2, 4,6,8,10= 6 ( )610 0=

1211 = � 12 ( )12

11 0=

Generators of Subsets of 12�

Table 1


Note:Note:Note:Note: You may have noticed that the order of the subgroups seems to

always divide the order of the group. This is not a coincidence. The order of

a subgroup always divides the order of a group. For example a group of order

11 will only have the trivial subgroups of the group itself and the identity

subgroup. On the other hand the groups of order 6 we have seen (cyclic

group of order six and the dihedral group 3

D of symmetries of an equilateral

triangle both have subgroups of order 2 and 3.

Example Example Example Example 5555 (Subgroup Generated by (Subgroup Generated by (Subgroup Generated by (Subgroup Generated by 120

R ))))

Find the subgroup of the dihedral group 3

D of symmetries of an

equilateral triangle generated by 120

R .


Starting with 120

R and the identity 0

e R= we form the set { }0 120,R R after

which we compute 2

120 240R R= . Since this is not in { }0 120

,R R we include it,

getting{ }0 120 240, ,R R R . We now compute the next power 3

120 0R R= in which case

we stop, getting the subgroup { }120 0 120 240, ,R R R R= of rotations of

3D .

Example Example Example Example 6666 (Subgroups of a Cyclic Group) (Subgroups of a Cyclic Group) (Subgroups of a Cyclic Group) (Subgroups of a Cyclic Group) Find the subgroups of 8

Z


Systematically trying different generators, we find the 8 subgroups.

{ }

{ }

{ }

{ }

{ }

8

8

8

1 0,1, 2,3,4,5,6,7 (order 8)

2 0,2,4,6 (order 4)

3 0,3,6,1, 4,7, 2,5 (order 8)

4 0,4 (order 2)

5 0,5, 2,7,4,1,6,3 (order 8)

= =

=

= =

=

= =

�

�

�

{ }

{ }

{ }

8

6 0,6, 4, 2 (order 4)

7 0,7,6,5, 4,3, 2,1 (order 8)

8 0 (order 1)

=

= =

=

�

Hence, the four subgroups of 8

Z are { } { } { } { }{ }0 0 4 0 2 4 6 0 1 2 3 4 5 6 7, , , , , , , , , , , , , ,

under the same addition and multiplication mod 8 as 8

Z .


ProblemsProblemsProblemsProblems, Section 6.4, Subgroups, Section 6.4, Subgroups, Section 6.4, Subgroups, Section 6.4, Subgroups

1. (True or False)(True or False)(True or False)(True or False)

a) The order of any subgroup always divides the order of the group.

Ans: Ans: Ans: Ans: true

b) Every subgroup of a group must contain the identity element of the

group.

AnsAnsAnsAns: true

c) Some groups do not have any subgroups.

AnsAnsAnsAns: false, the identity alone is a subgroup

d) � is a subgroup of � under the operation of addition.

Ans:Ans:Ans:Ans: true

e) The symmetric group 2

S has two subgroups.

Ans:Ans:Ans:Ans: true, the identity and the group itself are both subgroups of 2

S

a) There are some groups where every subset is a subgroup.

Ans:Ans:Ans:Ans: false, the subsets must contain the identity element and some subsets fo

not.

g) The set { },e h is a subgroup of the group of symmetries of a square,

where e denotes the identity map, and h is the horizontal flip.

Ans:Ans:Ans:Ans: yes

h) There are 5 subgroups of order 2 of the group of symmetries of a square.

AnsAnsAnsAns: yes, and you should be able to envision them

2. (Subgroups of (Subgroups of (Subgroups of (Subgroups of 6� ) ) ) ) List all subgroups of { }6

0 1 2 3 4 5, , , , ,=� generated by the

elements of the group. What is the order of each generator?

Ans:Ans:Ans:Ans:


{ }

{ }

{ }

6

6

1 order 6

2 0 2 4 order 3

3 0 3 order 2

4 0 2 4 order 3

5 order 6

, ,

,

, ,

=

=

=

=

=

�

�

3. Find the Cayley table for the subgroup { }180, , ,e R v h of the group of

symmetries of a square.

Ans:Ans:Ans:Ans:

4. Show that the group defined by the following Cayley table is a subgroup of

3S .

∗ ( ) ( )123 ( )132

( ) ( ) ( )123 ( )132

( )123 ( )123 ( )132 ( )

( )132 ( )132 ( ) ( )123

Ans:Ans:Ans:Ans: By Theorem 1 all we need to do is verify that the subset

( ) ( ) ( ){ }123 132, ,H = is closed under the operation *, and that each member of

H has an inverse. Clearly the operation * is closed since the table consists of

these members. Also, ( ) ( )123 and 132 are inverses of each other, and of

course the inverse of the identity ( ) is itself.

5555. (Subgroup Geneated by . (Subgroup Geneated by . (Subgroup Geneated by . (Subgroup Geneated by 240

R ) ) ) ) Find the subgroup of the dihedral group 3

D

of symmetries of an equilateral triangle generated by 240

R .

Ans:Ans:Ans:Ans: { }240 120 240, ,R e R R=

* e 180R v h

e e 180R v h

180R

180R e h v

v v h e 180R

h h v 180R e


6. (Generated Groups of Symmetries of a Rectangle)(Generated Groups of Symmetries of a Rectangle)(Generated Groups of Symmetries of a Rectangle)(Generated Groups of Symmetries of a Rectangle) In the Klein 4-group

{ }180, , ,e R v h of symmetries of a rectangle, find the subgroups generated by

each element in the group. What is the order of each member?

Ans:Ans:Ans:Ans:

{ }

{ }

{ }

180 180 180 has order 2

has order 2

has order 2

,

,

,

R e R R

v e v v

h e h h

=

=

=

7. (Center of a Group)(Center of a Group)(Center of a Group)(Center of a Group) The center ( )Z G of a group G consists of all

elements of the group that commute with all elements of the group. That is

( ) { }: for all Z G g G gx xg x G= ∈ = ∈

It can be shown that the center of any group is a subgroup of the group. Find

the center of the group of symmetries of a rectangle. Note: The center of a

group is never empty since the identity element of a group always commutes

with every element of the group. The question is, are there other elements

that commute with every element of the group.

Ans:Ans:Ans:Ans: The center of the Klein 4-group is { }180,e R

8. (Hasse Diagram)(Hasse Diagram)(Hasse Diagram)(Hasse Diagram) Draw a Hasse diagram for the subgroups of symmetries

of a rectangle; i.e. the Klein 4-group.

Ans: Ans: Ans: Ans:


9. (Subgroups of (Subgroups of (Subgroups of (Subgroups of 8� )))) Find the subgroups of the cyclic group

8� .

Ans:Ans:Ans:Ans: Systematically trying difference generators of 8� we find

{ }

{ }

{ }

{ }

{ }

{ }

{ }

8

8

8

8

1 1,2,3,4,5,6,7,0

2 2, 4,6,0

3 3,6,1, 4,7, 2,5

4 4,0

5 5, 2,7,4,1,6,3,0

6 6, 4, 2,0

7 7,6,5, 4,3, 2,1,0

= =

=

= =

=

= =

=

= =

�

�

�

�

which gives us four subgroups { } { } { } 80 0 4 0 2 4 6 and , , , , , , , � .

10. 11. (Subgroups(Subgroups(Subgroups(Subgroups of of of of 11� )))) Find the subgroups of the cyclic group

11� .


Ans:Ans:Ans:Ans: Systematically trying difference generators of 11� the only subgroups

are the trivial group { }0 and 11� . Also, the order of the subgroup divides the

order of the group and since 11� has order 11 and the only two numbers that

divide 11 are 1 and 11.

Section 6.4

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Transcript of Section 6.4