Math 3121Abstract Algebra I

Lecture 7:Finish Section 7

Sections 8

Finish Section 7

• Examples in class of Cayley Digraphs

Cayley Diagraph

• For each generating set of a finite group G, we can draw a graph whose vertices are elements of G and whose arcs represent right multiplication by a generator. Each arc is labeled according to the generator.

• Examples in class: Z6

Properties of Cayley Diagraphs1. Can get to any vertex from any other by a path. Reason: Every equation

g x = h has a solution in G and each member of G can be written as a product of generators and their inverses.

2. At most one arc goes from any vertex g to a vertex h. Reason: The solution of g x = h is unique.

3. Each vertex g has exactly one arc of each type starting at g and exactly one arc of each type ending at g. Reason: It is constructed this way.

4. If two different sequences of arc types go from vertex g to vertex h, then these two sequences applied to any other vertex will go to the same vertex.

Note: These four properties characterize Cayley diagraphs.

Examples

• Examples: Write out table of group described by Cayley

diagraph on page 72. Note the inner and outer squares have different directions.

Try this with triangles - note the directions of inner and outer triangles have same direction on page 71. Now do them with opposite directions.

What about pentagons?

HW for Section 7

• Don’t hand in:Pages 72-73: 1, 3, 5, 9

• Hand in (Due Tues, Oct 28):page 73: 12, 14, 16

Section 8

• Section 8: Groups of Permutations– Definition and Notation of Permutation– Theorem: Permutations on a set form a group

with composition as binary operation.– Definition: Symmetric Group on n letters– Definition: Dihedral group– Cayley’s Theorem

Permutations

Definition: A permutation of a set A is a function from A to A that is one-to-one and onto.

Examples:1) Let A = {a, b, c} , and let f: A A such that

f(a) = bf(b) = cf(c) = a

2) Let A = the set of real numbers, and let f: A A such that f(x) = 2 x. Does this work if A is the set of integers?

Permutation Groups

Theorem: Let A be a nonempty set, and let Perm[A] be the set of permutations of A. Then Perm[A] is a group with composition as the binary operation.

Proof: Composition is a well defined binary operation on Perm[A]. It satisfies: 1) It is associative because composition is associative.2) The identity map from A to itself acts as an identity for composition. Hence Perm[A] has an identity. 3) Every permutation is one-to-one onto, and thus has an inverse function. The inverse is also a 1-1 function from A onto itself. Hence Perm[A] is closed under inverses.

Permutation Notation

Write

for f: {x1, x2, …, xn} {x1, x2, …, xn}

Note: Most of the time, the set will be {1, 2, …,n}

)(...)()(

...

21

21

n

n

xfxfxf

xxx

Composition in Permutation Notation

))((...))(())((

...

)(...)()(

...

)(...)()(

...

21

21

21

21

21

21

n

n

n

n

n

n

xfgxfgxfg

xxx

xfxfxf

xxx

xgxgxg

xxx

• Composition in permutation notation is represented by multiplicatively. Note the order is right to left in this textbook (and hence in this class).

Composition in Permutation Notation

))((...))(())((

...

)(...)()(

...

)(...)()(

...

21

21

21

21

21

21

n

n

n

n

n

n

xfgxfgxfg

xxx

xfxfxf

xxx

xgxgxg

xxx

Procedure: Fill in each column at a time. For each column of the result, the top row determines a column of the right system. In that column, the entry in the second row determines a column of the left system. In that column, the entry in the second column determines the entry of the selected column of the result.

Example

• Try

4321

2143

4321

4132

4321

The Symmetric Group on n Letters

• Sn denotes the permutation group on the set {1, 2, …, n} and is called the symmetric group on n letters.

• Note: Sn has n! elements. Why?

• Look at S1, S2 , S3

S3 = Symmetric Group on 3 Letters

ρ0 ρ1 ρ2 μ1 μ2 μ3

ρ0 ρ0 ρ1 ρ2 μ1 μ2 μ3

ρ1 ρ1 ρ2 ρ0 μ3 μ1 μ2

ρ2 ρ2 ρ0 ρ1 μ2 μ3 μ1

μ1 μ1 μ2 μ3 ρ0 ρ1 ρ2

μ2 μ2 μ3 μ1 ρ2 ρ0 ρ1

μ3 μ3 μ1 μ2 ρ1 ρ2 ρ0

312

321

123

321

231

321

213

321

132

321

321

321

3

2

1

2

1

0

Symmetries of the Equilateral Triangle

ρ01

3

2

μ31

3

2μ2

1

3

2μ1

1

3

2

ρ11

3

2ρ2

1

3

2

S3=D3 = 3rd Dihedral Group

ρ0 ρ1 ρ2 μ1 μ2 μ3

ρ0 ρ0 ρ1 ρ2 μ1 μ2 μ3

ρ1 ρ1 ρ2 ρ0 μ3 μ1 μ2

ρ2 ρ2 ρ0 ρ1 μ2 μ3 μ1

μ1 μ1 μ2 μ3 ρ0 ρ1 ρ2

μ2 μ2 μ3 μ1 ρ2 ρ0 ρ1

μ3 μ3 μ1 μ2 ρ1 ρ2 ρ0

312

321

123

321

231

321

213

321

132

321

321

321

3

2

1

2

1

0

1

3

2

S4=D4 = Dihedral Group

ρ0 ρ1 ρ2 ρ3 μ1 μ2 δ1 δ2

ρ0 ρ0 ρ1 ρ2 ρ3 μ1 μ2 δ1 δ2

ρ1 ρ1 ρ2 ρ3 ρ0 δ1 δ2 μ2 μ1

ρ2 ρ2 ρ3 ρ0 ρ1 μ2 μ1 δ2 δ1

ρ3 ρ3 ρ0 ρ1 ρ2 δ2 δ1 μ1 μ2

μ1 μ1 δ2 μ2 δ1 ρ0 ρ2 ρ3 ρ1

μ2 μ2 δ1 μ1 δ2 ρ2 ρ0 ρ1 ρ3

δ1 δ1 μ1 δ2 μ2 ρ1 ρ3 ρ0 ρ2

δ2 δ2 μ2 δ1 μ1 ρ3 ρ1 ρ2 ρ0

2341

4321

4123

4321

1234

4321

3412

4321

3214

4321

2143

4321

1432

4321

4321

4321

2

1

2

1

3

2

1

0

1

3

2

4

Symmetries of the Square

3ρ04

21

3ρ14

21

3ρ24

21

3ρ34

21

3μ14

21

3μ24

21

3δ14

21

3δ24

21

Cayley’s TheoremTheorem (Cayley’s Theorem): Every group is isomorphic to a group of

permutations.Proof: Let G be a group. We show that G is isomorphic to a subgroup of the

permutations on the set G. For each a in G, let ρa be the map from G to itself given by left multiplication by a. That is, ρa(g) = a g. Then ρa is one-to-one and onto. In fact, it has an inverse. So ρa is a permutation of the set G. The map ρ that takes a to ρa is a homomorphism, because

ρa b(g) = a b g = a ρb(g) = ρa (ρb(g)) = (ρa ρb)(g) ρ is one-to-one and it is onto its image f[G]. It is straightforward to show that f[G] is a subgroup of Perm[G]. Thus f induces an isomorphism between G and f[G].

Right and Left Regular Representations

• ρx(g) = g x as in the theorem defined the left regular representation f of G.

• Multiplication on the right gives the property that is not homomorphism: μx y = μy μx. This is sometimes called the antihomomorphism property. The inverse map reverses the order. So

• μx(g) = g x-1 defines a map that has the correct order to be a homomorphism. This is called the right regular representation f of G.

Examples

• Find the left and right representations of Z2 , Z3 , Z4, S3

HW for Section 8

• Don’t hand in: pages 83-84: 1, 3, 5, 7, 9, 11, 13• Hand in (Due, Tues, Oct 28): Page 84: 18