Algebra(02)_160229_01

© Art Traynor 2011
Algebra
Subgroup
Group
Subgroup
A Subset “ H ”
of a Group “ G ”
is a Subgroup of “ G ”
if it satisfies the following Axioms:
Artin Section 2.2, ( Pg. 44 )
Definition 2.1
Closure:
Identity:
Inverses:
∀ a, b ∊ ℤ , if a ∊ H ∧ b ∊ H then a · b ∊ H
1 ∊ H
“ H ” contains the Additive Identity
Element which is Unity or “ 1 ”
∀ a ∊ ℤ , if a ∊ H then a – 1 ∊ H

© Art Traynor 2011
Algebra
Proof:
1a
Subgroups of ℤ +
Subgroup
Additive Group ℤ + of Integers
Gross: “ Abstract Algebra ”
Lecture 2 @ 36:30
Proposition 2.4
Proposition:
 ∃! b ∊ ℤ , b ·ℤ ⊂ ℤ + and
For any integer “ b ” the product
subset bℤ is a subgroup of bℤ+
 ∀ H ⊂ ℤ + , ∃! b | H = b ·ℤ Every Subgroup H of ℤ+ is of the
type H = bℤ for some integer “ b ”
Artin Section 2.2, ( Pg. 47)
Lemma 2.8
∃ G ∊ ℤ 1 ≤ |G ||:
Existential Quantification
( Instantiation ) of a Set/Group
over the Field ℤ ( EQ / EI )
Any “Quantification” is a
species of “Instantiation”
The Set/Group “ H ” is
not empty and contains at
least two elements
∃ x ∊ G x 0 = 1|:1b
Zero Property of Exponentiation
b0 = 1 If | 1 | is in the Set/Group
then zero must be as well
∴ 0, 1 ∊ G1c
1bBy Implication and1a
0 + 1 = 0
G includes the Identity Element of Zero
with LOC of Addition
1d
We will Demonstrate that
G is a Group
( 1 + 0 ) + 1 = 1 + ( 0 + 1 ) = 01e The LOC for G Associative

© Art Traynor 2011
Algebra
Proof:
Lecture 2 @ 36:30
Proposition 2.4
Proposition: ∃! b ∊ ℤ , b ·ℤ ⊂ ℤ + and ∀ H ⊂ ℤ + , ∃! b | H = b ·ℤ
Lemma 2.8
Subgroups of ℤ +
Subgroup
∴ – n ∊ G if n ∊ G Additive Inverses are defined
for all elements of “ G ”
1f
1g
We will Demonstrate that
G is a Group
for x 0 = 1, ∃ n ∊ G
x n · x – n = x n + ( – n ) ,
x n – n = x 0 = 1
|: By Implication 1c
G is a Group

© Art Traynor 2011
Algebra
Proof:
Lecture 2 @ 36:30
Proposition 2.4
Lemma 2.8
Subgroups of ℤ +
Subgroup
We will next demonstrate that
there is a Subset of “ G ”
namely “ H ” forming a Group
included within “ G ” ( i.e. a
Subgroup ) which is identical
to ℤ +
∀ m , n ∊ H 0 ≤ |m, n |
∃ ( x m = 1 ∧ x n = 1 )
x m + n = ( x m · x n ) = 1
By Introduction, G.A.
Let m, n = 0
or Iff m = n = 0
∴ m + n ∊ H if m , n ∊ H “ H ” is Closed under Addition2c
2a ∃ H ∊ G H ⊂ G|: By Introduction, G.A.: EQ / EI
2b
Let 1 ≤ |H | 1 ∊ H|:2d
1bBy Implication and1a
“ H ” Contains the
Additive Identity Element, Unity
By Implication 1g
Additive Inverses are defined
for all elements of “ H ”
∃ n ∊ H – n ∊ H if n ∊ H|:2e
“ H ” is a Subgroup
of “ G ”
|:

© Art Traynor 2011
Algebra
Proof:
Lecture 2 @ 36:30
Proposition 2.4
Subgroups of ℤ +
Subgroup
3a
∴ ∃ k ∊ H k · m ∊ H|:
∃ b ∊ H b = m
∴ bk = b1 + b2 +…+ bk – 1 + bk
if m + n ∊ H ∧ n = m
then m + m ∊ H
3b By Implication , and Induction3a
By Implication 2c
|: Summation Property of Repeated Addition
a1 + a2 +…+ an – 1 + an = na
Establishing that any “k”
multiple of “b” is in “H” is
important for us to
demonstrate the two trivial
subgroups of “H”: the empty
set and the Additive Identity,
Unity
3c
3d if k = 1
then bk ∊ H
∴ b ( k + 1 ) ∊ H
By Induction

© Art Traynor 2011
Algebra
Proof:
Lecture 2 @ 36:30
Proposition 2.4
Subgroups of ℤ +
Subgroup
4
5
if H = { 0 }
then b = 0 H = 0 ·G = Ø|:
Trivial Subgroup #1 ( smallest )
H = { 0 , + }
if H = { 1 }
then b = 1 H = G|:
Trivial Subgroup #2 ( largest )
H = { 1 , + }
“ H ” consists of a single
element, zero, with the
LOC of Addition, so that H
is the empty set
“ H ” consists of a single
element, Unity or the
Identity Element, with the
LOC of Addition, H is thus
G itself!
6a ∃ r ∊ H n = bk + r
∧ 0 ≤ r < b
|: By Introduction, G.A.: EQ / EI
We can multiply then add
any integer to “ b ”
∴ r = n – b · k ∊ H
∧ n – r = b · k ∊ H
Blitzer, Section 2.1,
pg. 115
Addition Property of Equality
if a = b then a + c = b + c
6c
We situated G in ℤ by 1a,
restricting the cardinality of G
to something equal to or
greater than one, then defined
H as a Subgroup of G by 2a
b has been defined in H and
therefor may only assume
integer values such that the
modulus inequality establishes
b as the smallest positive
integer
Let 1 ≤ |b |6b G.A.: Domain Restriction

© Art Traynor 2011
Algebra
Proof:
Lecture 2 @ 36:30
Proposition 2.4
Subgroups of ℤ +
Subgroup
0
7
n = b(1)
k = 1
n = b(2)
k = 2
n = b(3)
k = 3
n
bk
n
∊ ℤ
n = b(4)
n + r
bk
∉ ℤ
r
If b is a fixed integer 0 ≤ r < b
then bk will not divide
any integer sum n + r for r ≠ 0
Another way of saying that we
cannot have Rationals in a
Field of Integers
a – b
Congruence Property of Integers
n
∊ ℤ a ≡ b ( mod n ) = 0
Rationals need not apply!
Geometric Proof – Number Line

© Art Traynor 2011
Algebra
Proof:
Lecture 2 @ 36:30
Proposition 2.4
Subgroups of ℤ +
Subgroup
8a
∀ r = 0 , n = bk + r = bk
∧ bk ∊ H
∴ if H = b ·ℤ
then H ⊂ ℤ +
∀ k = 1 ,
bk = b
∧ ∃! b ∊ ℤ b ·ℤ ⊂ ℤ +
By Implication 7
3d
By Implication , , &7 6a
8b
8c
a ≡ b ( mod n ) = 0
|:

© Art Traynor 2011
Algebra
Lecture 2 @ 45:50
Cyclic Subgroups
Subgroup
Cyclic Subgroup
The Cyclic Group
is a natural Subgroup “ H ”
containing the Identity
and all powers of an arbitrary element “ g ” ∊ G
H = { … , g – n , g 1 – n , … , g – k , g 1 – k , … , g – 1 , g 0 , g 1 , … , g k , g k + 1 , … , g n – 1 , g n , … }
The Cyclic Group
is the smallest Group
containing its generating element “ g ”
The index value which returns
an element to Identity is the
Order of the Cyclic Subgroup
“ generated by ” that element
and which also divides the
cardinality of the parent Group
H ⊂ G g ∊ H|:

© Art Traynor 2011
Algebra
Lecture 2 @ 45:50
Cyclic Subgroups
Subgroup
Cyclic Subgroup
Cyclic Group
H = { … , g – n , g 1 – n , … , g – k , g 1 – k , … , g – 1 , g 0 , g 1 , … , g k , g k + 1 , … , g n – 1 , g n , … }
A Subgroup “ H ” of a Group “ G ”
The Group of least Cardinality “ g ” ∊ H , H ⊂ G g ∊ H
Inclusions:
Identity
All indices of the generating element “ g ”
Order:
The index which transforms the element into Identity
Divides the Cardinality of the parent Group
The smallest positive integer “n” such that gn = 1
∃ n ∊ ℤ 1 ≤ |n | ∧ gn = 1 ∀ g ∊ H|:
If there is no such “n” then the
Cyclic Subgroup is of infinite
Order
Lemma 2.8
|:

© Art Traynor 2011
Algebra
Lecture 3 @ 0:30
Examples
Group
Group
A Group is a Set G
together with a Law Of Composition ( LOC )
which is Associative
and possesses an Identity Element
and such that each element of G
possesses a unique Inverse Element
GLn (ℝ ) – General Linear Group
The set of all n x n ( square ) invertible matrices
Sn

The Group of all permutations on a set of “ n ” integers
Of Dimension “n” over the
Real Numbers
The Integers with LOC of addition
{ ℤ , + }

© Art Traynor 2011
Algebra
Lecture 3 @ 0:30
Examples
Group
Group ( Examples )
Each of these Groups arises as the consequence of a “Structure
Preserving” bijective transformations on their respective Domain sets
GLn (ℝ ) – General Linear Group
The set of all n x n ( square ) invertible matrices
Sn

The Group of all permutations on a set of “ n ” integers
Of Dimension “n” over the
Real Numbers
The Integers with LOC of addition
{ ℤ , + }

© Art Traynor 2011
Algebra
Lecture 3 @ 0:30
Examples
Group
Group ( Examples )
Each of these Groups arises as the consequence of a “Structure
Preserving” bijective transformations on their respective Domain sets
GLn (ℝ )
Sn
{ ℤ , + }
Group Structure Preserved under Transformation
Linearity
Permutations ( Set Structure )
Symmetries of Geometric Objects
( e.g. a vector Translated in one of two
directions along the Real Number Line )

© Art Traynor 2011
Algebra
Lecture 3 @ 0:30
Subgroups
Group
Group/Subgroup ( Examples )
Each of the Groups/Subgroups considered this far into the prior two lectures will
reveal the structural enrichments that later lectures will explore in great detail
Groups
GLn (ℝ ) , Sn , { ℤ , + }
Subgroups
H ⊂ G g ∊ H|:
General Linear
Symmetric
Integers with Addition
Cyclic Group
Subgroups of ℤ +
b ·ℤ ⊂ ℤ + Additive Group ℤ + of Integers
A structure-invariant
mapping between
mathematical structures
Morphisms
1. Homomorphism:
A structure preserving map
between two algebraic
structures (i.e. Groups,
Rings, or Vector Spaces)
2. Isomorphism:
A Bijective Homomorphism
including an element such that
composition the composition
of an inverse yields Identity
3. Automorphism:
An Isomorphism with image
set precisely equal to its
domain set – a formalism of
symmetry
Hom ( ℝ n , ℝ n ) has a
Vector Space structure
Cyclic Subgroups
b ·ℤ ⊂ ℤ + All Powers of an element of a
Group ( e.g. A Generator )

© Art Traynor 2011
Algebra
Lecture 3 @ 0:30
Subgroups
Group
Groups
GLn (ℝ ) , Sn , { ℤ , + }
Subgroups
H ⊂ G g ∊ H|:
General Linear
Symmetric
Integers with Addition
Cyclic Group
b ·ℤ ⊂ ℤ + Additive Group ℤ + of Integers
Cyclic Subgroups
1
0
1
1
⊂ GL2 (ℝ )
1
0
1
1
n
→
The Cyclic Group
and all powers of an
arbitrary element “ g ” ∊ G1
0
n
1
→
Order = ∞

© Art Traynor 2011
Algebra
Lecture 3 @ 0:30
Subgroups
Group
Groups
Subgroups
Cyclic Subgroups
1
0
1
1
⊂ GL2 (ℝ )
1
0
1
1
n
→
The Cyclic Group
and all powers of an
arbitrary element “ g ” ∊ G1
0
n
1
→
Order = ∞
A note on Order
– 1
0
0
– 1
1
0
0
1
2
→
Order = 2
The negation of the identity matrix
squared returns the identity matrix
therefore the order this cyclic
subgroup is two
– e → ( – e ) 2 = e

© Art Traynor 2011
Algebra
Morphism
Homomorphism
A Morphism f : X → Y constitutes an Isomorphism if there
exists a Morphism g : Y → X such that f ○ g = Iy
and g ○ f = Ix , implying that Ix = Iy
I.e.: f has an inverse, the
compositions of which return
the Identity element for the
respective variables
n Isomorphism
I.e.: if a set and another
compose to an identity, then
those sets are Isomorphic to
one another
Homomorphism
Isomorphism
A structure-invariant mapping of one mathematical structure to another
( i.e. a domain set to a corresponding image set) , featuring special cases such as:

Example 1:
G1 = { ± 1 , ± i } ⊂ ℂ * = ( ℂ { 0 } , x ) I.e.: A Group populated by
non-zero complex elements
with a multiplicative LOC
G1 = { i k , k = 0 … 3 }
I.e.: A four element Group
populated by k = 0, k = 1,
k = 2, and k = 3

© Art Traynor 2011
Algebra
Morphism
Homomorphism
A Morphism f : X → Y constitutes an Isomorphism if there
exists a Morphism g : Y → X such that f ○ g = Iy
and g ○ f = Ix , implying that Ix = Iy
I.e.: f has an inverse, the
compositions of which return
the Identity element for the
respective variables
n Isomorphism
I.e.: if a set and another
compose to an identity, then
those sets are Isomorphic to
one another
Homomorphism
Isomorphism

Example 2:
G2 = S4 Cyclic Subgroup
⍴ =
1 1
3 3
2 2
4 4
⍴2 ≠ e , ⍴
⍴3 ≠ e , ⍴ , ⍴ 2
⍴4 = e
Basically a σ sigma permutation

© Art Traynor 2011
Algebra
Morphism
Homomorphism
n Isomorphism
Homomorphism
Isomorphism

Example 2:
⍴ =
1 1
3 3
2 2
4 4
⍴2 ≠ e , ⍴
⍴ 3 ≠ e , ⍴ , ⍴ 2
⍴4 = e
e
e
MultiplicationTable
e
⍴
⍴
LOC
( x )
⍴ 2
⍴3
⍴ 2 ⍴3
⍴3
⍴ 2⍴
⍴ ⍴ 2 ⍴3
⍴
⍴ 2
⍴ 2
⍴3
e
e⍴3
e ⍴
S4 shares the same multiplication table as “ i ” → i 4 = 1
Therefore S4 is really the same group as “ i ” , e.g. i 4 = Unity
G2 = S4 Cyclic Subgroup Basically a σ
sigma permutation

© Art Traynor 2011
Algebra
Morphism
Homomorphism
n Isomorphism
Homomorphism
Isomorphism

S4 shares the same multiplication table as “ i ” → i 4 = 1
Therefore S4 is really the same group as “ i ” ,
e.g. i 4 = Unity
Thus G1 & G2 are Isomorphic i
i
MultiplicationTable
– 1
i2
i2
LOC
( x )
i3
i3
1– i
– i 1 i
1 – 1i
e is eliminated in this table
as its operation is trivial
A map between two Groups f : G1 → G2o
Bijectiveo
Preserves LOC : f ( x · y ) = f ( x ) · f ( y )o
G2
G1

© Art Traynor 2011
Algebra
Morphism
Homomorphism
n Isomorphism
Homomorphism
Isomorphism

ik ⟼ ⍴ k
f( ik ) = ⍴ k
Any two cyclic groups of Order “ n ” are Isomorphico

© Art Traynor 2011
Algebra
Morphism
Homomorphism
n Isomorphism
Homomorphism
Isomorphism

G1 = (ℝ , + ) and G2 = ( 0 < ℝ , x )
Example:
Are they Isomorphic?
Consider a Function “ f ” such that:
f : G1 → G2
f ( x ) = ex
Note that ex supplies just such an Isomorphism
bm+n = bm · bn
An Isomorphism between
Groups that are neither
Finite nor Cyclic

© Art Traynor 2011
Algebra
Morphism
Homomorphism
n Isomorphism
Homomorphism
Isomorphism

Klein 4-Groupo
Aka: Vierergruppe
Denoted “ V ” or “ k4 ”
Direct product of two copies of the Cyclic Group of Order Two
Smallest non-cyclic Group
Abelian
Isomorphic to Dihedral Group, Order Four

© Art Traynor 2011
Algebra
Morphism
Homomorphism
n Isomorphism
Homomorphism
Isomorphism

Example : Klein 4-Group
G1 = { e, τ1 , τ2 , τ1 τ2 , τ2 τ1 }
2 2
1 1
4 4
3 3
τ1 =
2 2
1 1
4 4
3 3
τ2 =
2 2
1 1
4 4
3 3
τ1 · τ2 = = τ2 · τ1
Identitieso
τ1
2 = e
τ2
2 = e
( τ1 τ2 ) 2 = e
Any product of two Order
Two elements yields the
third element
Any element multiplied by
itself yields the Identity

© Art Traynor 2011
Algebra
Morphism
Homomorphism
n Isomorphism
Homomorphism
Isomorphism

Example : Klein 4-Group
G1 = { e, τ1 , τ2 , τ1 τ2 , τ2 τ1 }
2 2
1 1
4 4
3 3
τ1 =
2 2
1 1
4 4
3 3
τ2 =
2 2
1 1
4 4
3 3
τ1 · τ2 = = τ2 · τ1
– 1
0
0
1 ,
1
0
0
– 1
– 1
0
0
– 1
G1 =
,
τ1
e is eliminated in this
Group membership set
as its operation is trivial
τ2 τ1 τ2
1
0
0
1

© Art Traynor 2011
Algebra
Morphism
Homomorphism
n Isomorphism
Homomorphism
Isomorphism

Counter-Example : is the Klein 4-Group isomorphic to
G1 = { ± 1 , ± i } ?
A: No. Because k4 has no elements of Order 4 or |k4 | ≠ |G1 |

© Art Traynor 2011
Algebra
Morphism
Homomorphism
n Isomorphism
Homomorphism
Isomorphism

Tests for Isomorphism:
Same Order: |G1 | = |G2 |o
G1 Abelian ⟺ G2 Abeliano
G1 & G2 have the same number of elements of every ordero
Subgroups can be discovered by examining symmetrieso
Necessarily so as to be
Isomorphic a Bijection
must exist between the
two Groups

© Art Traynor 2011
Algebra
Morphism
Homomorphism
n Automorphism
Homomorphism
Automorphism

An Automorphism can be
regarded as a species of
Isomorphism, inheriting
much of its structure
Alternatively referred to as
an Endomorphism
Inverses are Automorphismso
An Isomorphism with equivalent Domain & Codomain sets
as mapped by a structure invariant Transformation
o
Aut ( G ) = Automorphism Group or Symmetry Group of the setso

© Art Traynor 2011
Algebra
Morphism
Homomorphism
n Isomorphism
Homomorphism
Isomorphism

Composition of Isomorphic Transformations
Compositions of Isomorphisms are Isomorphic to each othero
With the resulting
composition being
necessarily Bijective
f ○ g ( x · y ) = f( g( x ) · g( y ) )
f ○ g ( x · y ) = f · g( x ) · f · g ( y )
fg ( x · y ) = f( g( x ) · g( y ) )
fg ( x · y ) = f g( x ) · f g ( y )

© Art Traynor 2011
Algebra
Morphism
Morphism
Homomorphism

Homomorphism
A map which operates between two algebraic structures such as
Groups, Rings, or Vector Spaces preserving structure between these
constituent mathematical objects under transformation
det ( G ) : GLn (ℝ ) → ℝ x
= { ℝ { 0 } , x )
det ( AB ) = det ( A ) · det ( B )
What are the indicia that this is not an Isomorphism?
n One is Abelian ( RHS ) whereas the other is not ( LHS )
n Mapping is not One-to-One ( fails Injection ? )
n Contra: Group structure is preserved
n Contra: Bijections exist ( have the same Cardinality )

© Art Traynor 2011
Algebra
Morphism

Homomorphism
n Structure invariant map operating between two algebraic structures
( e.g. Groups, Rings, or Vector Spaces )
n Not necessarily Bijective
n f : G1 → G2 such that
f ( x · y ) = f ( x ) · f( y )
n There is always a Trivial Homomorphism
f : G1 → G2 such that
f ( x ) = e
The Identity ( not the
natural exponent )
Morphism
Homomorphism

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Example : Symmetric Group Sn for 3 ≤ n
f : S3 → Sn
Recall that S3 is the first Set
( beyond S2 ) exhibiting
non-Abelian permutation
transformation??
n The maps for Sn for 3 ≤ n are
merely Injective…they are not Isomorphisms
I.e.: An Injective
Homomorphism
Morphism
Homomorphism

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Example :
f : ℤ → S2
Even S ⟼ e
Odd S ⟼ τ
Morphism
Homomorphism

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Example :
f : G1 → G2
‘ G2 ’‘ G1 ’
The “ Image ” is that
part of G2“hit” by G1
i.e. the “ Column Space ”
of a Matrix
Image = { f ( x ) ∊ G2 , x ∊ G1 }
As “x” ranges over G 1
Morphism
Homomorphism

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Example : G → G′
f
n A map from one group to another
n A map of Sets
f( g · h ) = f ( g ) · f ( h )
Proposition 4.1
G → G′
f
in G in G ′
A linear map between
vector spaces
f( e ) = e′ → e · e = e
The identity is the only
element with this property
f( e ) · f( e ) = f ( e · e ) = f( e )
f ( g – 1 ) = f ( g ) – 1
Morphism
Homomorphism

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Example : G → G′ → G″
f h
The composition of two
Homomorphisms yields a third
Morphism
Homomorphism
f ○ h
Image = { g′ = f ( g ) in G′ } ⊂ G′
Kernel = { g : f ( g ) = e′ } ⊂ G
Both are Subgroups
If Image = G′
and Kernel = { e } f is an Isomorphism
If G = G′ and f is an Isomorphism
then f is an Automorphism

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Conjugation
n Kernel : A special type of Subgroup
a Normal Subgroup H ⨞ G H is a Normal Subgroup of G
∀ g ∊ G
g · h ·g – 1 The Conjugate
( g H g – 1 ) · ( g H′ g – 1 )
The inner “ g ” terms will cancel
(i.e. evaluate to Unity)
g ( h h′ ) g – 1 Closed under Multiplication
∀ g ∊ G
g · H ·g – 1 = H
Normal Subgroups are closed
under Conjugation
True of a Kernel, but not so for
an Ordinary Subgroup
A map including a conjugate
codomain element is an
Automorphism
Proposition 4.1
I.e. The conjugate of
“ h by g ”

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
To verify that an element h ∊ kernel
Confirm that the element conjugate
g · h ·g – 1 is in the kernel
o
Apply “ f ” to the element conjugateo
f ( g · h ·g – 1 ) = f( g ) · f ( h ) · f ( g – 1 )
This is a property of a
Homomorphism
= f( g ) · e′ · f ( g – 1 ) = e′

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
Compare to a Subgroup that is not Normal
Can’t be Abeliano Are all Abelian Subgroups Normal?
“ Cancellation ” of the conjugate is
obvious ( i.e. evaluation to unity )
Consider S3o
The simplest non-abelian Group

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
Consider S3o
The simplest non-abelian Group
G = S3
H = 〈 e , τ 〉 τ tau
1 1
3 → 3
2 2
Inspection alone reveals this
not to form a Normal Group
We must identify an element in the Group
that takes the subgroup out of itself…

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
Consider conjugation by tau prime τ ′o
1 → 1
3 3
2 2
τ ′
τ ′ · τ · ( τ ′ ) – 1 ≠ τ = τ ″
τ ′ · τ · ( τ ′ ) ( 3 ) τ ′ of 3 is 2

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
1 → 1
3 3
2 2
τ ′
τ ′ · τ · ( τ ′ ) – 1 ≠ τ = τ ″
τ ′ · τ · ( τ ′ ) – 1 ( 3 ) ( τ ′ ) – 1 of 3 is 2
τ ′ · τ ( 2 ) τ of 2 is 1
1 1
3 → 3
2 2
τ

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
1 → 1
3 3
2 2
τ ′
τ ′ · τ · ( τ ′ ) – 1 ≠ τ = τ ″
τ ′ · τ · ( τ ′ ) – 1 ( 3 ) ( τ ′ ) – 1 of 3 is 2
τ ′ · τ ( 2 ) τ of 2 is 1
1 1
3 → 3
2 2
τ
τ ′ ( 1 ) τ ′ of 1 is 1

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
1 → 1
3 3
2 2
τ ′
τ ′ · τ · ( τ ′ ) – 1 ≠ τ = τ ″
τ ′ · τ · ( τ ′ ) – 1 ( 3 ) ( τ ′ ) – 1 of 3 is 2
τ ′ · τ ( 2 ) τ of 2 is 1
1 1
3 → 3
2 2
τ
τ ′ ( 1 ) τ ′ of 1 is 1
So the composed permutation sends 3 to 1

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
1 → 1
3 3
2 2
τ ′
τ ′ · τ · ( τ ′ ) – 1 ≠ τ = τ ″
τ ′ · τ · ( τ ′ ) – 1 ( 3 ) ( τ ′ ) – 1 of 3 is 2
τ ′ · τ ( 2 ) τ of 2 is 1
1 1
3 → 3
2 2
τ
τ ′ ( 1 ) τ ′ of 1 is 1
So the composed permutation sends 3 to 1
1 1
3 3
2 2
τ ″
Which is what H τ ″
does for a living!

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
1 → 1
3 3
2 2
τ ′
τ ′ · τ · ( τ ′ ) – 1 ≠ τ = τ ″
τ ′ · τ · ( τ ′ ) – 1 ( 3 ) ( τ ′ ) – 1 of 3 is 2
τ ′ · τ ( 2 ) τ of 2 is 1
1 1
3 → 3
2 2
τ
τ ′ ( 1 ) τ ′ of 1 is 1
1 1
3 3
2 2
τ ″
τ ′ · τ · ( τ ′ ) – 1 = 〈 e , τ ″ 〉 This Subgroup is not the
Kernel of a Homomorphism

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
The Kernel of a Homomorphism is a Normal Subgroupo
Given any Normal Subgroup, there exists a Homomorphism
the Kernel of which is that Subgroup
o
The notion of a Homomorphism is thus more “ elemental ”
than even that of a Group
o
e.g. Category Theory – not about the Categories themselves but
more concerned with the Morphisms that exist between them…

Every Normal Subgroup is
the Kernel of some
Homomorphism

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
G = GLn (ℝ ) → G′ ℝ x
= GL1 (ℝ )1
f
det ( AB ) = det ( A ) · det ( B )
f( A ) = det ( A )
Multiplication
in GL n ( ℝ )
Multiplication in GL 1 ( ℝ )
A determinant is a unique mapo
This map has an Image ( f ) = ℝ x
o
We can obtain a matrix with any determinanto
A determinant is an area, or
more generally a Manifold

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
G = GLn (ℝ ) → G′ ℝ x
= GL1 (ℝ )1
f
det ( AB ) = det ( A ) · det ( B )
f( A ) = det ( A )
Multiplication
in GL n ( ℝ )
Multiplication in GL 1 ( ℝ )
We can obtain a matrix with any determinanto
A determinant is an area, or
more generally a Manifold
If we want to “ hit ” some real number, we need a
matrix with that determinant
o

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
λ
0
0
λ
det
.
.
.
.
.
.
.
.
.
. . .
. . .
= λ
kernel ( f ) is the matrices which have det = 1o
kernel ( f ) = { A : det ( A ) = 1 } = SLn (ℝ )o
Any matrix of det = 1 when conjugated with some arbitrary
matrix will yield a det = 1 resultant matrix
o

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
Example :
det ( B · A ·B – 1 ) = det ( B ) · det ( A ) · det ( B – 1 ) = det ( A ) This is a property of a
Homomorphism
This must be Abelian ( i.e. Commutative )o
Both elements of det = 1 are stable under Conjugationo
All elements of fixed determinant are stable under Conjugation
– a necessary condition of a Normal Subgroup
o

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
kernel ( f ) = { A : det ( A ) = 1 } = SLn (ℝ ) ⨞ GLn (ℝ ) GL n ( ℝ ) is a Normal
Subgroup of GL n ( ℝ )
f : Sn → GL1 (ℝ )2 A consequential
Homomorphism
f ( σ ) = Aσ : the permutation matrix associated to σ
0
.
.
.
. . .
. . .
1
.
.
.
. . .
. . .
0
0
0
.
.
.
.
.
.
jth
A =
The “1” indicates to where
in the j
th
column vector the
σ permutation takes j

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
Recall S3o
G = S3 1 1
3 3
2 2 σ sigma
0
1
0
0
0
1
1
0
0
Cj Ri
One goes to Two
Two goes
to Three
Three goes
to One
= Aσ

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
How to confirm that σ is a Homomorphism ? Apply “ f ” to a composition of
permutations and evaluate its
equivalence to the multiplication
of matrices representing those
permutationsf ( σ · τ ) = Aσ · Aτ
Composition of
Permutations
Matrix
Multiplication
The determinant of any permutation matrix is ± 1o
det = ( f ( σ ) ) = ± 1 ∀ σ ∊ Sno
In ( f ) = Permutation Matriceso
ker ( f ) = { e }o

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
τ tau
1 1
3 → 3
2 2
.
.
.
.
.
.
n → n
0
.
.
.
. . .
. . .0
1
.
.
.
A =
1
0
0
.
.
.
.
.
. 1
0
0
1
.
.
.
0 0
Identity Block
det = 1
“ A ” Block
det = – 1
det = ( f ( e ) ) = det ( I ) = 1o
det = ( f ( τ ) ) = – 1o

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Kernel
Now we consider the composition of homomorphisms &1 2
Sn → GLn (ℝ ) → ℝ x
f det
No kernel Surjective
(all of the Group is in
the Image)
image = 〈 ± 1 〉 ⊂ ℝ x
kernel = { σ : det f ( σ ) = ± 1 }
An even permutation
The sign of a permutation is
the sign of the determinant
An ⨞ Sn which has order of one-half that of SLn (ℝ ) The alternating Group is a Normal
Subgroup of the Symmetric Group

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Group
Order
An ⨞ Sn which has order of one-half that of SLn (ℝ ) The alternating Group is a
Normal Subgroup of the
Symmetric Group
∀ n : 2 ≤ n
ord ( An ) =
n!
2
Examples :
Ord ( S3 ) = 6 , Ord ( A3 ) = 3o
Even Permutations = { e , σ , σ ′ = σ 2 }o
Odd Permutations = {τ , τ ′, τ ″ }o

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Structure invariant map operating between two algebraic structures
n Center of a Group : Artin Section 2.4, ( Pg. 52 )
Proposition 4.10
Z ( G ) = Center of G = { Z ∊ G : Z · g = g · Z ∀ g }
That element which commutes with every other element of a Groupo
For an Abelian Group : Z ( G ) = Go
Z ( G ) = { e , … , G }o
The Center is the whole
Group
The Center can assume
values from e up to G
For G = GLn (ℝ ) , Z ( G ) = { λ · I , λ ∊ ℝ x
}o Where “ I ” represents the
set of all diagonal matrices
Given a non-diagonal matrix , an element that does not
commute with it can be identified

Group
Center

© Art Traynor 2011
Algebra
Morphism

Homomorphism
n Center of a Group : Artin Section 2.4, ( Pg. 52 )
Proposition 4.10
Z ( G ) = Center of G = { Z ∊ G : Z · g = g · Z ∀ g }
Z ( G ) = Center of G , is a an Abelian Groupo
Z ( G ) = Center of G , is a Normal Groupo
Every/Any Group will feature a Center, if only a trivial centero
e.g. Sn has Abelian Subgroups ∴ Z ( G ) is trivial
Group
Center

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Automorphism
Automorphism:
symmetryn Automorphism
G → Aut ( G ) : The set of all Isomorphisms h : G → Gf
Those Isomorphisms which
map G to “ itself “ under
Composition ( i.e. to a
subdomain Group identical
to the domain Group )Examples :
f ( g ) · ( h ) = g · h Is this an Isomorphism?
f ( g · g′ ) = f( g ) · f( g′ )
No: it is a Set Isomorphism not
however a Group Isomorphism
f ( g · g′ ) · h = f( g ) · f( g′ ) · ( h )

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Automorphism
Automorphism:
f ( g ) · ( h ) = g · h
f ( g ) · ( h · h′ ) = g · h · h′
f ( g ) · ( h ) ○ f ( g ) · ( h′ ) = g · h · g · h′
There is nothing to
suggest that these
two must be equal

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Automorphism
Automorphism:
For ( g ) · ( h ) · ( g ) – 1 = ( g ) · ( h′ ) · ( g ) – 1
the products are Automorphic only if h = h′
Conjugation by “ g ” : f ( g ) · ( h ) = g · h · g – 1
To constitute an Automorphism every
element of the Automorphic Group must be
capable of being rendered in this fashion

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Automorphism
Automorphism:
f ( g ) · ( h · h′ ) = g · h · h′ · g – 1
= g · h · g – 1 · g · h′ · g′
= f ( g ) · ( h ) · f ( g ) · ( h′ )
These two RHS expansions
don’t seem quite right ??
This RHS expansion seems
to be the correct one…

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Automorphism
Automorphism:
f ( g · g′ ) = f( g ) · f( g′ )
f ( g · g′ ) · h = ( g · g′ ) · h · ( g · g′ ) – 1
= g · g′ · h · g′ – 1 · g – 1
f ( g ) ○ f ( g ′ ) · ( h ) = f( g ) · ( g′ · h · g′ – 1 ) = g · ( g′ · h · g′ – 1 ) · g – 1

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Group
Kernel
Those elements of a Group G such that when we Conjugate by
them we yield the trivial Automorphism
Examples :
( g ) · ( h ) · ( g ) – 1 = h ∀ h ∊ G
This condition must be satisfied
for an element “ h ” to be
included in the kernel of G
ker ( G ) = Z ( G ) The kernel of G is exactly equal
to the Center of G !
“ g ” commutes ∀ h : ( g ) · ( h ) · ( g ) – 1 = h
Z ( G ) is the kernel ( a normal subgroup ) of a Homomorphism

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Automorphism
Automorphism:
to the domain Group )
Examples :
Do we get every Automorphism by Conjugation ? i.e. as a Image Set
Consider K4 Klein 4-Group
± 1
0
0
± 1
G = = { e, τ , τ ′, τ ″ } The product of any two permutations
yields the third
τ · τ ′ = τ ″

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Automorphism
Automorphism:
symmetry
n Automorphism
– 1
0
0
1
1
0
0
– 1
– 1
0
0
– 1
τ τ′ τ″
Examples :
± 1
0
0
± 1
G = = { e, τ , τ ′, τ ″ }

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Automorphism
Automorphism:
symmetry
n Automorphism
Examples :
± 1
0
0
± 1
G = = { e, τ , τ ′, τ ″ }
Z ( G ) = G , the Center is the entire Groupo
An Abelian Group ord ( G ) = 4o
Img ( G ) in Aut ( G ) is “ e ”o
e.g. Conjugation by an arbitrary element of G is a trivial Automorphism

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Automorphism
n Automorphism
Examples :
± 1
0
0
± 1
G = = { e, τ , τ ′, τ ″ }
K4 features a non-trivial Automorphism Groupo
e.g. Aut ( K4 ) ≃ S3 S3 = { e, τ, τ′, τ″, σ , σ ′ }
1 → 1
3 3
2 2
1 1
3 → 3
2 2
1 1
3 3
2 2
1 1
3 3
2 2
1 1
3 3
2 2
τ τ′ τ″ σ σ ′

© Art Traynor 2011
Algebra
Morphism

Homomorphism
Homomorphism
Automorphism
n Automorphism
Examples :
± 1
0
0
± 1
G = = { e, τ , τ ′, τ ″ }
K4 features a non-trivial Automorphism Groupo
e.g. Aut ( K4 ) ≃ S3 S3 = { e, τ, τ′, τ″, σ , σ ′ }
Note Also That : the non-identity elements in K4
must be permuted by an Automorphism !
o
Professor Gross uses the
“ Asymptotically Equal To ”
symbol to express the
relation between the Auto-
morphism Group of K 4 and
the Symmetric Group S 3

© Art Traynor 2011
Algebra
Morphism
 Hom ( ℝ n , ℝ n ) has a
Homomorphism
Homomorphism
Automorphism
Automorphism:
symmetry
n Automorphism Those Isomorphisms which
I think it’s appropriate to
adopt Professor Gross’s
use of the “ Asymptotically
Equal To ” symbol here with
these three permutations
he’s identified in S 3 as
representing the Auto-
morphism Group of K 4
Examples : Consider K4 Klein 4-Group
If a : G → G represents an Automorphism then we
can associate “a” to a permutation of S3 ≃ { τ, τ′, τ″ }
oo
Aut ( K4 ) features a “ full ” Image of S3 with trivial kerneloo
oo G → Aut ( G ) : Need not be an Injection nor a Surjectionf
oo Img ( f ) = Inn ( G ) = { a ( h ) = g ·h· g – 1 } for some g ∊ G
( where Inn ( G ) is the Inner Automorphism Group of G )

© Art Traynor 2011
Algebra
Equivalence Relation on a Set
A partition of a set into disjoint subsets
Sets
Relations
The Union of the partitions thus (re-)constitutes the referent Set
S = { s1 ∪…∪ sn }
s1
s2
si …si +1 sn – 1
sn
The centipede of subsets
a
b
a ∼ b if two distinct elements share
the same subset inclusion

Properties
a ∼ a Reflective
a ∼ b → b ∼ a Symmetric
a ∼ b ∧ b ∼ c → a ∼ c Transitive
Order Invariant,
Commutative,
Abelian
An Equivalence Relation can also be regarded as a Subset
of the product of S x S = { a , b : a ∼ b }
 Of S with itself !

© Art Traynor 2011
Algebra
Sets
Relations
S = { s1 ∪…∪ sn }
s1
s2
si …si +1 sn – 1
sn
a
b

S
S
Properties
Order Invariant,
Commutative,
Abelian
①
①
②
②
① The first property a ∼ a requires that the
subset contain the diagonal ( x = y )
② The second property a ∼ b → b ∼ a requires
that the subset be stable under reflection
③ The third property is harder to describe…

© Art Traynor 2011
Algebra
Sets
Relations
S = { s1 ∪…∪ sn }
s1
s2
si …si +1 sn – 1
sn
a
b

Properties
Order Invariant,
Commutative,
Abelian
n a ∼ a requires that any (arbitrary ) element of a Set S
populate at least one Subset
Note That :
n The transitive property requires that the subsets be disjoint

© Art Traynor 2011
Algebra
Sets
Relations
S = { s1 ∪…∪ sn }
s1
s2
si …si +1 sn – 1
sn
a
b

Properties
Note That :
si ∩ s¬i = 0o
If si ∩ s¬i ≠ 0 , then ∃ c ∊ s¬i but c ∊ si
and si ≠ s¬i else S = { si , s¬i } are not disjoint
o |:
c

© Art Traynor 2011
Algebra
Sets
Relations
S = { s1 ∪…∪ sn }
s1
s2
si …si +1 sn – 1
sn
a
b

Properties
Note That :
These disjoint Subsets form distinct Equivalent Classeso
c
If si ∩ s¬i ≠ 0 , then ∃ c ∊ s¬i but c ∊ si
and si ≠ s¬i else S = { si , s¬i } are not disjoint
o |:

© Art Traynor 2011
Algebra
Sets
Relations
S = { s1 ∪…∪ sn }
s1
s2
si …si +1 sn – 1
sn
a
b
S = Set
S = { Equivalence Classes in S }
S → S There exists a map that takes an
element of S into S ( a subset, the
equivalence class ) containing “ a ”
a ⟼ a The Equivalence Class containing “ a ”
Note That :
Every element in S arises from S because every Equivalence Class
contains something ( i.e. is non-zero )

Not an Injection of Sets
It is a Surjection of Sets
An Equivalence Relation gives a Map of Sets
A Map of Sets always determines an Equivalence Relation

© Art Traynor 2011
Algebra
Sets
Relations
S ,T = Sets
f : S → T A Map of Sets yields an Equivalence
Relation ( i.e. a Partition ) on S
a ∼ b ⟷ f( a ) = f( b ) if they have the same
Image in T Image of “ f ”
T = Codomain of fS = Domain of f
f
Equivalence
Classes
t
t′
Note That :
The equivalence classes in S which map to T
are also referred to as the Fibres above t

In this case we posit a map
which is not Surjective
( i.e. not all of T is covered by S )
Img ( f ) < Cod ( f )
The function is Injective : every element in S maps to some element in T ( in the Image )
Fibre
Fibre
Those elements mapping to “ t ” have no intersection with those mapping to t′
All of the Image is covered by S however the Fibres are distinct
The map yields an equivalence relation

© Art Traynor 2011
Algebra
Sets
Relations
S ,T = Sets
f : S → T A Map of Sets yields an Equivalence
Relation ( i.e. a Partition ) on S
a ∼ b ⟷ f( a ) = f( b ) if they have the same
Image in T Image of “ f ”
T = Codomain of fS = Domain of f
f
Equivalence
Classes
t
t′
Note That :
The Set S , the Set of Equivalence Classes ,
can be identified with the Image of “ f ”

In this case we posit a map
which is not Surjective
( i.e. not all of T is covered by S )
Img ( f ) < Cod ( f )
Fibre
Fibre
Because the disparate points in the Image “ index ” the distinct
Equivalence Classes in S


© Art Traynor 2011
Algebra
Sets
Relations
Examples : From Algebraic Topology
0 1 2– 1– 2
1
dy 2 πi t
ef =
dy 2 πi t
ef = Consider the map
A Group Homomorphism between :
Complex
Unit Circle
n The additive Group of Real Numbers
( i.e. the real number line ) and…
n The multiplicative Group of numbers
on the complex Unit Circle
ℝ = S
ℂ = S1 = T = S
f ( a + b ) = f ( a ) · f( b )
Additive Multiplicative
A property of the natural
exponential function
f – 1 ( 1 )
The Fibre over the
point “1” on the
Complex Circle
A Surjective Map
The Set of Equivalence Classes of the Map :
n Is equal to the Image
n Consists of only the real line points 0 ≤ x ≤ 1
n Any point on the unit circle has a pre-image
on the real line
n The Real Line endpoints are adjoined to identify
the Equivalence Classes with the unit circle

© Art Traynor 2011
Algebra
Homomorphism
Relations
Given a Map of Sets “ f ” which forms a Group Homomorphism
f : G → G′
f : G → G′
S → T
H = f – 1 ( e′ ) = { a ∊ G : f( a ) = f( e ) = e′ }
∃ H ∊ G′ ker ( f ) = H ⨞ G : A Normal Subgroup of G′ |:
∃ H ∼ G′ : An Equivalence Relation
dom ( f ) = { S } ( G = the set S )
cod ( f ) = { T } ( G′ = the set T )
T = { ti ∪ ti+1 ∪… ∪ tn – 1 ∪ tn } : Equivalence Classes of G′
Proposition:
ti ⊂ T ⊆ G ∧ H ⊆ T
H = e – 1 : The Inverse of the Identity
Whenever we encounter a
Group Homomorphism we
are also visited by a
partition of G into pieces,
one of which is the Kernel
of the Homomorphism
Those elements in G that
map to e ′
Kernel
I have bollixed up some of the
notation here…the lecture
was a bit ambiguous here ??
“H” is the Kernel
One of the Equivalence
Class partitions of G ′
A normal Subgroup of G ′
The Inverse of the Identity

© Art Traynor 2011
Algebra
Homomorphism
Relations
Given a Map of Sets “ f ” which forms a Group Homomorphism f : G → G′
S → T
Proposition:
What are the other Equivalence Classes of the form :
∃ H ∊ G′ ker ( f ) = H ⨞ G |:
T = { ti ∪ ti+1 ∪… ∪ tn – 1 ∪ tn } : Equivalence Classes of G′
ti ⊂ T ⊆ G ∧ H ⊆ T
a · H = { a · h h ∊ H } , for some a ∊ G|:
The Kernel of a Homomorphism supplies a good description
of the Equivalence Classes of Group

n Those elements mapping to a fixed point in G′

© Art Traynor 2011
Algebra
Homomorphism
Relations
a · H = { a · h h ∊ H } , for some a ∊ G′|:
Proof:
Let f( a ) = f( b ) ∊ G′ ( i.e. a ∼ b )
Equivalence Classes of G′ : T = { ti ∪ ti+1 ∪… ∪ tn – 1 ∪ tn }
As it must be
if f ( a ) = f ( b )
Then f( a – 1 · b ) = e′
Because this expression is equivalent to: f( a ) – 1 · f( b ) = e′
This is valid because we
already have that G ′ is a
Group Homomorphism
we’ve already established that f( b ) = f( a )
we know full well that f( a ) – 1 · f( a ) = e′
∴ f( a ) – 1 · f( b ) = e′
And :

© Art Traynor 2011
Algebra
Homomorphism
Relations
Proof:
Let f( a ) = f( b ) ∊ G′ ( i.e. a ∼ b )
As it must be
if f ( a ) = f ( b )
Then f( a – 1 · b ) = e′
And a – 1 · b ∊ H
An element multiplied by its
inverse maps to the Identity
a – 1 · b = h ∊ H
This product of a Group
element and it’s inverse can
then be equated to an
element of H ( the Kernel )
∴ it is included in the Kernel
a · ( a – 1 · b = h )
( a · a – 1 )· b = a · h
b = a · h

© Art Traynor 2011
Algebra
Homomorphism
Relations
Conclusions :
Let f( a ) = f( b ) ∊ G′ ( i.e. a ∼ b )
As it must be
if f ( a ) = f ( b )
Then b = a · h and Therefore:
n Any element of G′ equivalent to “ a ” lies inside the set { a · H }
Because if we apply “ f ” to
such an element “ a · h ”
we arrive at f ( a ) · f ( h )
where f ( h ) = e
n Conversely any b ∊ { a · H } is equivalent to “ a ”
as f( b ) = f( a · h )
As f ( a ) · f ( h ) = f ( a ) · e
n So the elements equivalent to “ a ” are exactly those
elements included in the Subset { a · H } of G′ ,
a Left Coset of H in G′ for some a ∊ G′

© Art Traynor 2011
Algebra
Homomorphism
Relations
Conclusions :
Let f( a ) = f( b ) ∊ G′ ( i.e. a ∼ b )
As it must be
if f ( a ) = f ( b )
n The Equivalence Classes of a Group Homomorphism are precisely
the collection of Left Cosets of the Kernel
The Kernel is one of the Left Cosets of G′ (?)o
Another Left Coset of G′ is “ h ” ( the Identity )o

© Art Traynor 2011
Algebra
Homomorphism
Relations
Conclusions :
Let f( a ) = f( b ) ∊ G′ ( i.e. a ∼ b )
As it must be
if f ( a ) = f ( b )
n Every Coset has the same Cardinality ( Order )
If |H | is finite ( |H | < ∞ )o
then |H | = |a · H | for all “ a ”

© Art Traynor 2011
Algebra
Homomorphism
Relations
Conclusions :
Let f( a ) = f( b ) ∊ G′ ( i.e. a ∼ b )
As it must be
if f ( a ) = f ( b )
What’s so good about that?
The map h ⟼ a · h renders a bijection of Setso
H ⥴ a · H
Notation: Symbol for
Bijective Map
Test for Bijectiono
Injection: a · h = a · h ′ iff h = h ′ ( multiplying by a – 1 )
Surjection: Clearly because the elements populating { a · h } are
merely product elements of “ a ” and an element in “ H ”


© Art Traynor 2011
Algebra
Homomorphism
Relations
Conclusions :
Let f( a ) = f( b ) ∊ G′ ( i.e. a ∼ b )
As it must be
if f ( a ) = f ( b )
For a Group Homomorphism
n All Cosets are in Set-Theoretic Bijection to “ H ”
n All Equivalency Classes are of the form:
{ a · H } where H = ker ( f )o

© Art Traynor 2011
Algebra
S = { s1 ∪…∪ sn } ⊆ G
s1 = H
si
si +1
sn – 1
sn
The centipede of Cosets
Homomorphism
Relations
Image of “ f ”
T = Codomain of f
f ( a )
S = Domain of f
f
a · H = { a · h h ∊ H } , for some a ∊ G|:
Conclusions :
Equivalence Classes of G:
S = { si ∪ si+1 ∪… ∪ sn – 1 ∪ sn },
i.e. The Cosets of G

s2 = aH
G′
( a )
Cosets of G = { a · H } where H = ker ( f )
Cosets of G are Bijective and share the same Cardinality
This condition does not hold for a Map of Sets S
2
3
T
n Because all the different pieces into which G is
partitioned in a Group Homomorphism contain
the same number of elements
Where “S” is
populated with
Equivalence
Classes

© Art Traynor 2011
Algebra
Homomorphism
Relations
Corollary:
Assuming G is finite and
f : G → G′ is a homomorphism with kernel “ H ”
then |G | = |H | · |img ( f ) | splits the Cardinality
Because we “ exhaust ” G by adding up the elements in the Cosets
( with each Coset containing the same number of elements )
and the number of Cosets is the number of points in the image
|G | = | ker ( f ) | · |img ( f ) |

© Art Traynor 2011
Algebra
Homomorphism
Relations
Corollary:
 |G | = | ker ( f ) | · |img ( f ) |
 If T : V → W is a linear map,
then the dimension of V
is the dimension of ker ( T ) plus the dimension of the Image of “ T ”
dim ( V ) = dim ( ker ( T ) ) + dim ( img ( T ) )
so |G | = | ker ( f ) | · |img ( f ) | splits the Cardinality
Example :
|Sn | = n!
for 2 ≤ n , |An | =
n!
2

© Art Traynor 2011
Algebra
Homomorphism
Cardinality
Corollary:
 |G | = | ker ( f ) | · |img ( f ) |
Example :
|Sn | = n!
for 2 ≤ n , |An | = n!
2
Proof:
f : Sn → 〈 ± 1 〉
A Homomorphism
Given by the sign of a permutation
Surjective for 2 ≤ n
Kernel = An
|Sn | = n!
|img ( Sn ) | = 2
|A3 | = 3
|A4 | = 12
|A5 | = 60

© Art Traynor 2011
Algebra
Homomorphism
Structure
Corollary:
 |G | = | ker ( f ) | · |img ( f ) |
But have we used all of our Hypotheses?
If we have a Homomorphism, we get a Normal Subgroup but nowhere
have we used the notion that H is Normal in G
Corollary( more Generally ):
Let H ⊂ G “ H ” is any Subgroup ( not necessarily Normal )
“ H ” therefore does not arise from the Kernel of a Homomorphism
The Left Coset of a ∊ G is denoted as:
a · H = { a · h h ∊ H }|:

© Art Traynor 2011
Algebra
Homomorphism
Structure
But have we used all of our Hypotheses?
The Left Coset of a ∊ G is denoted as:
a · H = { a · h h ∊ H }|:
If we have a Homomorphism
we get a Normal Subgroup “
H ” , but we have yet to
declare the restriction that “
H ” is Normal in “ G ”
Proposition:
The Subsets { a · h } are:
Disjoint and partition G
In Set Theoretic Bijection with “ H ” i.e.: The same Cardinality
or Order

© Art Traynor 2011
Algebra
Homomorphism
Structure
The Left Coset of a ∊ G , a · H = { a · h h ∊ H }|:
If we have a Homomorphism
we get a Normal Subgroup “
H ” , but we have yet to
declare the restriction that “
H ” is Normal in “ G ”
Proposition:
In Set Theoretic Bijection with “ H ”
i.e.: The same Cardinality
or Order
Yet we have not used all of
our Hypotheses…
So the following arise in consequence:
An Equivalence Relation
n Not originating from a Homomorphism
n Originating from any Subgroup
Which would of course
supply a Normal Subgroup
A “ breaking down ” of the Group into the
distinct ( same Cardinality ) Cosets of “ H ”

n The same map takes an element in “H”
to an element in “ a · h ”
All of which are in Set
Theoretic Bijection with “ H ”
Supplies a new
Equivalence Relation :
The Index

© Art Traynor 2011
Algebra
Homomorphism
Structure
Proposition:
So the new Equivalence Relation arising in consequence thereof is :
The Index of “ H ”
Which might be infinite…
n The number of distinct Left Cosets
n The number of Equivalence Classes
Denoted:
An arbitrary Subgroup of “ G ”
Supergroup
Index of “ H ” in “ G ”[ G : H ]

© Art Traynor 2011
Algebra
Homomorphism
Structure
Proposition:
 |G | = | H | · [ G : H ]
Just the Index ( not the
Cardinality of the Index)
As the equation suggests, the
Index as well as the Cardinality of
“ H ” divide the Cardinality of “ G ”
Represents the number of Cosets
in the Normal Group which is the
same as the Image
This is because we’ve divided
“G” into equal parts, the number
of parts being the Index
The Order (Cardinality) of “ G ”
is the Order of “ H ” times the
number of parts, i.e.: The Index
G = { si }
s1
s2
si …si +1 sn – 1
sn
# of Parts = The Index
[ G : H ]
∪i = 1
n
H a ·h b ·h … n – 1 ·h n ·h
The number of
elements in each
“part” is the
Cardinality of “H”

© Art Traynor 2011
Algebra
1736 – 1813
Joseph-Louis Lagrange
Kingdom of Sardinia
City of Turin
Homomorphism
Structure
G = { si }
s1
s2
si …si +1 sn – 1
sn
[ G : H ]
∪i = 1
n
H a ·h b ·h … n – 1 ·h n ·h
The number of
elements in each
“part” is the

|G | = | H | · [ G : H ]
This relationship was first noted by Lagrange at
the conclusion of the 18th century
 It constitutes the first formula in Group Theory
 No formal education at age 17 when he began a
correspondence with Euler in St. Petersberg
Read books in his rich Uncle’s library
In his third letter he set out all that had
then been known about Calculus

Euler arranged for his formal education

© Art Traynor 2011
Algebra
1736 – 1813
Kingdom of Sardinia
City of Turin
Homomorphism
Structure
G = { si }
s1
s2
si …si +1 sn – 1
sn
[ G : H ]
∪i = 1
n
H a ·h b ·h … n – 1 ·h n ·h
The number of
elements in each
“part” is the
|G | = | H | · [ G : H ]
Lagrange’s Theorem
If |G | is finite, and g ∊ G then | g | divides |G |
 i.e.: the smallest power m g m = e|:
Proof: Let “ H ” be the Subgroup generated by “ g ”
which contains all the elements { e, gi ,…, gm – 1 } :
H = 〈 g 〉 = 〈 e , g1, g2,…, gm – 1 〉
| H | = m = Cardinality of “ g ”
And since the Cardinality of any Subgroup of
a Finite Group divides the Cardinality of the
Group (by considering the decomposition into
Cosets ) it is certainly true that | H | divides
the Order of the Group , which is the
Cardinality of the Element

© Art Traynor 2011
Algebra
1736 – 1813
Kingdom of Sardinia
City of Turin
Homomorphism
Structure
G = { si }
s1
s2
si …si +1 sn – 1
sn
[ G : H ]
∪i = 1
n
H a ·h b ·h … n – 1 ·h n ·h
The number of elements in each
“part” is the Cardinality of “H”
|G | = | H | · [ G : H ]
Lagrange’s Theorem
If |G | is finite, and g ∊ G then | g | divides |G |
 i.e.: the smallest power m g m = e|:
Proof: Let “ H ” be the Subgroup generated by “ g ”
which contains all the elements { e, gi ,…, gm – 1 } :
H = 〈 g 〉 = 〈 e , g1, g2,…, gm – 1 〉
| H | = m = Cardinality of “ g ”
 Note that the inverses of the generating element
are also in the Cyclic Subgroup
H = 〈 g 〉 = 〈 e , g1, g2,…, g – 2,( gm – 1 = g– 1) 〉

© Art Traynor 2011
Algebra
1736 – 1813
Kingdom of Sardinia
City of Turin
Homomorphism
Structure
G = { si }
s1
s2
si …si +1 sn – 1
sn
[ G : H ]
∪i = 1
n
H a ·h b ·h … n – 1 ·h n ·h
|G | = | H | · [ G : H ]
Lagrange’s Theorem ( Corollary )
Let “ G ” be a finite group, with |G | = p “ p ” is prime number…
i.e. a Group of Prime Order
Then “ G ” is a Cyclic Group, generated by
any g ∊ G with G ≠ e
Furthermore, the only Subgroups of “ G ” are G & e
So if we have a Group of prime order then the only
Subgroups are the entire Group and the Group
If not of Prime Order, the Group can have
a plethora of Subgroups
This is a nexus of Group Theory with Number Theory
because the Orders of Groups being certain numbers will
provide useful, more generalizable information about the
Group structure itself…

© Art Traynor 2011
Algebra
1736 – 1813
Kingdom of Sardinia
City of Turin
Homomorphism
Structure
G = { si }
s1
s2
si …si +1 sn – 1
sn
[ G : H ]
∪i = 1
n
H a ·h b ·h … n – 1 ·h n ·h
|G | = | H | · [ G : H ]
Let “ G ” be a finite group, with |G | = p “ p ” is prime number…
i.e. a Group of Prime Order
Then “ G ” is a Cyclic Group, generated by
any g ∊ G with G ≠ e
Furthermore, the only Subgroups of “ G ” are G & e
Proof:
Let G ≠ e in “ G”
The order of “ g ” divides “ p ” , and is not “ 1 ”
( since only “ e ” has an order = 1 )
Hence if “ p ” is prime, its only divisors are “ p ” & “ 1 ”
and the Order of “ g ” is thus p ( i.e. | g | = p )

© Art Traynor 2011
Algebra
1736 – 1813
Kingdom of Sardinia
City of Turin
Homomorphism
Structure
G = { si }
s1
s2
si …si +1 sn – 1
sn
[ G : H ]
∪i = 1
n
H a ·h b ·h … n – 1 ·h n ·h
|G | = | H | · [ G : H ]
“ p ” is prime number…
i.e. a Group of Prime OrderProof:
Let G ≠ e in “ G”
The order of “ g ” divides “ p ” , and is not “ 1 ”
( since only “ e ” has an order = 1 )
Hence if “ p ” is prime, its only divisors are “ p ” & “ 1 ”
and the Order of “ g ” is thus p ( i.e. | g | = p )
〈 g 〉 ⊂ G
Order = “ p ”
The Subgroup
generated by “ g ”
If we have Finite Sets
of the same Cardinality
then they are equal !
i.e. 〈 g 〉 = G

© Art Traynor 2011
Algebra
Homomorphism
Structure
Proposition:
A Group can be Cyclic without being generated by any of its constituent elements
G is a Cyclic Group generated by any non-trivial element
( i.e. not the Identity )

e.g.: The Cyclic Group of Order 4
n Contains an element of Order 2 And two is not a generator of this Group
So if | H | = 1 , H is the Identity
if | H | = p , H is the entire Group “ G ”
Note also that if H “ consumes ” the Group “ G ”
then the Complement Set is the Null Set !

Can we show this is a strong theorem
by exhibiting non-cyclic groups of Order:

n p2
n p · q

© Art Traynor 2011
Algebra
Homomorphism
Structure
Proposition:
A Group can be Cyclic without being generated by any of its constituent elements
Can we show this is a strong theorem
by exhibiting non-cyclic groups of Order:

n p2
n p · q
Yes: G = K4 ( the Klein Group Four )
| G | = 4 not cyclic – no included elements are of Order 4
n K4 only exhibits Orders = 1 and 2 for its elements
n e.g. for G = S3 | G | = 6 = 2, 3 and is not Abelian
It will be shown that there are in fact non-cyclic groups of Order p2
n But all Groups of Order p2 are Abelian
i.e. : we can’t construct a non-Abelian Group of Order 4…
the simplest Non-Abelian Group we can construct is of Order 6

© Art Traynor 2011
Algebra
Homomorphism
Structure
Definition:
A Finite Group is Simple if its only Normal Subgroups “ H ” are “ e ” and “ G ”
If there are only two Normal Subgroups, that Group is Simple
n Which means the group cannot be broken into ( 2 or more ? ) smaller Groups
Any Group of Prime Order is cyclico
The only Abelian Simple Groups
An is Simple for 5 ≤ no
Any finite, non-Abelian, Simple Group has Even Ordero

Algebra(02)_160229_01

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