Download Examples, Binary Structures, Isomorphisms

Chapter 1
Groups and Subgroups
1.1
Introduction and Examples
1.1.1
Revisit some examples in brief
Ex 1.1. (R, +).
Ex 1.2. (Z, +).
Ex 1.3. (R+ , ·).
Ex 1.4. (V, +) where V is a vector space.
Ex 1.5. Rotations on a plane, with composition.
Ex 1.6. The set of bijections from R to R, with function composition.
Question: What are in common?
A set together with an operation. The operation acts “closely” and
“associatively” in the set. The set has an identity. Every element in the set
has an “inverse”.
These are the examples of an algebraic structure called group.
1.1.2
Complex numbers
• Real numbers are numbers on a line.
• Complex numbers are numbers on a plane (with x-axis
and yi-axis). The
√
2
number i satisfies that i = −1. We may write i = −1. The set of
complex number is C = {a + bi | a, b ∈ R}.
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8
CHAPTER 1. GROUPS AND SUBGROUPS
A complex number can be expressed in two ways: (by graphs)
1. a + bi, for a, b ∈ R.
2. Euler’s formula states that eiθ = cos θ + i sin θ. So we may write
z = |z|(cos θ + i sin θ) = |z|eiθ , |z| ≥ 0 and θ ∈ R.
p
|a + bi| = a2 + b2 .
Multiplication of complex numbers: (by graphs)
1. (a + bi)(c + di) = ac + adi + bci + (bi)(di) = (ac − bd) + (ad + bc)i.
2. aeiα · beiβ = abei(α+β) .
Below are some examples of isomorphisms:
1.1.3
Algebra on Circles
• Addition modulo 2π on R2π .
• Multiplication on unit circle U .
• Let φ : U → R2π be the bijection given by eiθ ↔ θ. We see that:
if eiθ1 ↔ θ1 and eiθ2 ↔ θ2 ,
then eiθ1 · eiθ2 ↔ (θ1 +2π θ2 ).
This is called an isomorphism.
1.1.4
Roots of Unity
The elements of Un = {z ∈ C | z n = 1} is called the n-th roots of unity.
Using Euler’s formula, we see that
2π
2π
Un =
cos m
+ i sin m
| m = 0, 1, 2, · · · , n − 1
n
n
2π
2π
0 1 2
n−1
= {1 = ξ , ξ , ξ , · · · , ξ
},
ξ = cos
+ i sin
.
n
n
Let Zn = {0, 1, 2, · · · , n − 1} be a subset of Rn . There is an isomorphism
Un ∼ Zn in the way that
if ξ i ↔ i and ξ j ↔ j,
1.1.5
I-1, p19
Homework
8, 17, 38
then ξ i · ξ j ↔ (i +n j).
1.2. BINARY OPERATIONS
1.2
1.2.1
9
Binary Operations
Definitions and Examples
Def 1.7. A binary operation ∗ on a set S is a function mapping S × S to
S. We write a ∗ b instead of ∗((a, b)).
In other words, for every a, b ∈ S, there is exactly one element c ∈ S
with c = a ∗ b.
Ex 1.8 (Ex 2.2, p21). The usual addition “+” is a binary operation on
R, Z = the set of all integers, C. The usual multiplication “·” is a binary
operation on R, C, Z, R+ , respectively.
Ex 1.9 (Ex 2.5, p21). Let R∗ = R−{0} = the set of nonzero real numbers.
Then “·” is closed on R∗ and hence “·” is a binary operation on R∗ . However, “+” is not closed on R∗ (e.g. −2 + 2 = 0 ∈
/ R∗ ). So “+” is NOT a
∗
binary operation on R .
Ex 1.10 (Ex 2.8, p22, projection). Given a set S, define a ∗0 b = a for
(a, b) ∈ S × S. Then ∗0 is a binary operation.
Let ∗ be a binary operation ∗ on a set S.
Def 1.11. The ∗ is commutative iff a ∗ b = b ∗ a for all a, b ∈ S.
Revisit the previous three examples.
Def 1.12. The ∗ is associative iff (a ∗ b) ∗ c = a ∗ (b ∗ c) for all a, b, c ∈ S.
Thm 1.13. Let S be a set. Let f, g, h be functions mapping S into S. Then
(f ◦ g) ◦ h = f ◦ (g ◦ h).
Ex 1.14 (Linear Algebra). If A, B, C, are three n × n matrices, then
(AB)C = A(BC). The proof is not easy. However, you may check some
examples for n = 2.
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CHAPTER 1. GROUPS AND SUBGROUPS
1.2.2
Tables
When S is a finite set, a binary operation ∗ can be described by a table.
Ex 1.15. Suppose S = {a, b, c}, then a binary operation ∗ on S may be
represented by
∗
a
b
c
a a∗a a∗b a∗c
b b∗a b∗b b∗c
c c∗a c∗b c∗c
Ex 1.16 (Ex 2.14, p24). It is not commutative, not associative.
The ∗ is commutative iff the table is symmetric.
1.2.3
Warning
To be a binary operation, ∗ should be well-defined and closed on S.
Ex 1.17 (Ex 2.19, 2.20, 2.21, p25).
1.2.4
I-2, p25
1.3
Homework
3, 4, 8, 26, 36
Isomorphic Binary Structures
Ex 1.18. Compare the tables of binary operations (“+3 ”) on S1 = {1, 2, 0}
and on S2 = {one, two, zero} respectively:
+
1
2
0
1
2
0
1
2
0
1
2
0
1
2
0
and
+
one
two
zero
one
two
zero
one
two
zero
one
two
If we make a one-to-one and onto correspondence:
1 ↔ one,
2 ↔ two,
then S1 and S2 have the same structure.
Ex 1.19 (3.1, 3.2 Tables, p29).
0 ↔ zero,
zero
one
two
zero
1.3. ISOMORPHIC BINARY STRUCTURES
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Def 1.20. A binary algebraic structure (S, ∗) is a set S together with a
binary operation ∗ on S.
Def 1.21 (isomorphism). Let (S, ∗) and (S 0 , ∗0 ) be algebraic structures.
An isomorphism of S with S 0 is a one-to-one function φ mapping S onto S 0
such that (There is a misprinted on the book.)
φ(x ∗ y) = φ(x) ∗0 φ(y)
for all x, y ∈ S
(homomorphism property)
If such a map φ exists, then S and S 0 are isomorphic binary structures,
which we denote by S ' S 0 .
? Process To show that (S, ∗) and (S 0 , ∗0 ) are isomorphic:
1. Define a function φ : S → S 0 that gives an isomorphism of S with S 0 .
2. Show that φ is a one-to-one function. That is, prove that φ(x) = φ(y)
iff x = y.
3. Show that φ is onto S 0 . That is, every element s0 of S 0 can be written
as φ(s) for some s ∈ S.
4. Show that φ(x ∗ y) = φ(x) ∗0 φ(y) for all x, y ∈ S.
Ex 1.22 (Ex 3.8, p30). Show that (R, +) and (R+ , ·) are isomorphic.
Ex 1.23 (Top of p18). Show that (Rc , +c ) and (Rd , +d ) are isomorphic.
There is no universal way to prove that two binary algebraic structures
(S, ∗) and (S 0 , ∗0 ) are not isomorphic. However, if S and S 0 have different
cardinalities, then surely there is no one-to-one and onto map between S
and S 0 , and thus S and S 0 are not isomorphic.
Ex 1.24. (Zn , +n ) and (Zm , +m ) are not isomorphic (But the previous
example says that (Rn , +n ) and (Rm , +m ) are isomorphic).
Def 1.25. A structural property of a binary structure is one that must be
shared by any isomorphic structure.
Ex 1.26 (Ex 3.11, p32). (Z, ·) and (Z+ , ·) are not isomorphic.
Ex 1.27 (p 32). Examples of possible structural properties and possible
not structural properties.
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CHAPTER 1. GROUPS AND SUBGROUPS
Def 1.28. Let (S, ∗) be a binary structure, an element e of S is an identity
element for ∗ if e ∗ s = s ∗ e = s for all s ∈ S.
Thm 1.29. A binary structure (S, ∗) has at most one identity element.
The existence of an identity element is a structural property.
Thm 1.30. Suppose φ : S → S 0 is an isomorphism between binary algebraic
structure (S, ∗) and (S 0 , ∗0 ). If S has an identity element e for ∗, then φ(e)
is the identity element for ∗0 in S 0 .
Ex 1.31 (Ex 3.15, 3.16, 3.17, p 33). If two binary algebraic sets have
some different structural properties, then they are not isomorphic. This
provides one efficient way to prove that two binary algebraic sets are not
isomorphic.
1.3.1
Homework (I-3, p34-p36)
3, 4, 5, 26, 27, 30.
(Optional:) Some challenging problems: 18, 24, 25, 34