Sub-group Test

2-Step Test subgroups:
H subset of group G is subgroup if:
1. H is non-empty
(check: identity of G ∈ H)
2. $Latex a.b^{-1} \in H$

Prove Subset not a subgroup:

1. For infinite Group: sufficient to prove subset doesn't contain e (identity).

2. For finite group: sufficient to prove subset not closed.
H is subgroup of G
$latex \iff a*b^{-1} \in H, \forall a, b \in H$

Group Problem

Let G be a group.
∀x,y ∈ G
Prove that xy and yx have the same order n ?
ie. $latex (xy)^{n}= (yx)^{n} = e$

Proof:
Let
$latex (xy)^{n}=e$
(xy)(xy).....(xy)(xy) = e
[(xy) n times]
By associativity,
x.(yx)(yx)...(yx).y=e
$latex (yx)(yx)...(yx).y = x^{-1}$
$latex (yx)^{n-1}= x^{-1}.y{^-1}$
$latex (yx)^{n-1}= (y.x)^{-1}$
$latex (yx) ^{n-1}.(yx) = e$
$latex (yx)^{n} = e$
ie yx has order n.
[QED]

Galois Theory Simplified

Galois discovered Quintic Equation has no radical (expressed with +,-,*,/, square root) solutions, but his new Math "Group Theory" also explains:
$latex x^{5} - 1 = 0 \text { has radical solution}$
but
$latex x^{5} -x -1 = 0 \text{ has no radical solution}$

Why ?

$latex x^{5} - 1 = 0 \text { has 5 solutions: } \displaystyle x = e^{\frac{ik\pi}{5}}$
$latex \text{where k } \in \{0,1,2,3,4\}$
which can be expressed in
$latex x= cos \frac{k\pi}{5} + i.sin \frac{k\pi}{5} $
hence in {+,-,*,/, √ }
ie
$latex x_0 = e^{\frac{i.0\pi}{5}}=1$
$latex x_1 = e^{\frac{i\pi}{5}}$
$latex x_2 = e^{\frac{2i\pi}{5}}$
$latex x_3 = e^{\frac{3i\pi}{5}}$
$latex x_4 = e^{\frac{4i\pi}{5}}$
$latex x_5 = e^{\frac{5i\pi}{5}}=1=x_0$

=>
$latex \text {Permutation of solutions }{x_j} \text { forms a Cyclic Group: }
\{x_0,x_1,x_2,x_3,x_4\} $

Theorem: All Cyclic Groups are Solvable
=>
$latex x^{5} -1 = 0 \text { has radical solutions.}$

However,
$latex x^{5} -x -1 = 0 \text{ has no radical solution } $
because the permutation of solutions is A5 (Alternating Group) which is Simple
ie no Normal Subgroup
=> no Symmetry!

Group Action: Orbit

Orbit demystified
Let G= {rotation around center O }
Let X= point {x,y } ∈ R²
Orbit (X) = subset of points X on the concentric circles of different radius, centered at point O.
In fact,
Orbit = Coset
Hence is equivalent relation
=> can partition set X
=> X = Uj Orbit (Xj)

French Flag Colors

National Flag of France has 3 color stripes: red,white, blue = (r,w,b)
If we permute the 3 colors, how many flags are there ?

Eg. (w,b,r) = (r,b,w) since they are same if flip over the flag.
Eg. (b,b,b),(r,w,r)... repeat color allowed.
[Ans: 18]
Hint: Apply Group action on the Set (r,w,b)


French Flag:
Let Group G = {g1,g2 }
g1={e}= {(1)(2)(3)}
g2={(1 3)(2)}
|G|=order = 2
Apply Group Action on 3 colors set X {r,w,b}:
g1.X=e.{r,w,b} => Fix 3 stripes each with 3 colors
=> |Fix(g1)|= 3³ ways

(1 3)= Fix 1st & 3rd stripe as 1 stripe, x 3 colors => 3 ways
(2) => Fix 2nd stripe x 3 colors = 3 ways

g2.X=(1 3)(2).X=> 3x3
=> |Fix(g2)|= 3²

By Counting Theorem:
Total Flags = Orbit(X)
= ∑|Fix(g) | /|G|
= (3³ + 3² ) / 2
= 18



Exercise:
Chessboard (8x8=64) colored arbitrarily in black or white.
How many different patterns are there ?

[Consider 2 patterns as the same if a rotation (90, 180, 270 degrees) takes one to another ?]

Group Representation

Group Representation
Faithful: Isomorphic (1 to 1)
Unfaithful: homomorphic (m to 1)

Group of Equilateral Triangle
Permute the 3 vertices.
Isomorphic to S3 (permute 1 2 3)
Faithful representation by matrices:
e = $latex
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
$

C₃= $latex
\begin{pmatrix}
-\frac {1}{2} & \frac{\sqrt{3}}{4} \\
-\frac{\sqrt{3}}{4} & -\frac {1}{2}
\end{pmatrix}
$

C₃² = $latex
\begin{pmatrix}
-\frac {1}{2} & -\frac{\sqrt{3}}{4} \\
\frac{\sqrt{3}}{4} & -\frac {1}{2}
\end{pmatrix}
$

σ₁ = $latex
\begin{pmatrix}
-1 & 0 \\
0 & 1
\end{pmatrix}
$

σ₂ = $latex
\begin{pmatrix}
\frac {1}{2} & \frac{\sqrt{3}}{4} \\
\frac{\sqrt{3}}{4} & -\frac {1}{2}
\end{pmatrix}
$


σ₃ = $latex
\begin{pmatrix}
\frac {1}{2} & -\frac{\sqrt{3}}{4} \\
-\frac{\sqrt{3}}{4} & -\frac {1}{2}
\end{pmatrix}
$

Group Theorems: Lagrange, Sylow, Cauchy

1. Lagrange Theorem:
Order of subgroup H divides order of Group G

Converse false:
having h | g does not imply there exists a subgroup H of order h.
Example: Z3 = {0,1,2} is not subgroup of Z6
although o(Z3)= 3 which divides o(Z6)= 6

However,
if h = p (prime number),
=>
2. Cauchy Theorem: if p | g
then G contains an element x (so a subgroup) of order p.
ie.
$latex x^{p} = e $ ∀x∈ G

3. Sylow Theorem :
for p prime,
if p^n | g
=> G has a subgroup H of order p^n:
$latex h= p^{n}$

Conclusion: h | g
Lagrange (h) => Sylow (h=p^n) => Cauchy (h= p, n=1)


Trick to Remember:

g = kh (god =kind holy)
=> h | g
g : order of group G
h : order of subgroup H of G
k : index

Note:
Prime order Group is cyclic
(Z/pZ, +) order p is cyclic & commutative.

Order 4: Z4 not isomorphic to Z2xZ2

Order 6: only Z6 isomorphic Z2xZ3.
Z6 non-commutative

S3 = {1 2 3} ≈ D3 Not Abelian
(1 2)(1 3) = (1 3 2)

(1 3)(1 2) =(1 2 3)

Lagrange: |G|=6
=> order of subgroups in G = 1,2,3,6
6= 2x3
Cauchy : 2|6, 3|6 (2,3 prime)
=> order of elements in G
= 2, 3

Abelian Group

This interesting example is like solving simultaneous equations in Group, using only one Group property tool (Inverse => cancellation law)

Let Group G, ∀a,b ∈G,
for any 3 consecutive integers i,
$latex (a.b)^{i}= a^{i}.b^{i} $
Prove：G is abelian?
[Herstein: i, i+1,i+2]

Proof:
(a.b)ⁿ= aⁿ.bⁿ ...(1)
(a.b)ⁿ⁺¹= aⁿ⁺¹.bⁿ⁺¹ ...(2)
(a.b)ⁿ⁺² = aⁿ⁺².bⁿ⁺² ...(3)
Take inverse (1):
(a.b)⁻ⁿ = (aⁿ.bⁿ)⁻¹ = b⁻ⁿ.a⁻ⁿ ...(4)

Left * (4) to (2)
(a.b)ⁿ⁺¹(a.b)⁻ⁿ =ab

{Right *} (4) to (2):

ab = (aⁿ⁺¹.bⁿ⁺¹).(b⁻ⁿ.a⁻ⁿ) = aⁿ⁺¹.b.a⁻ⁿ
{Left *} a⁻¹
=> a⁻¹(ab) = b = (a⁻¹aⁿ⁺¹).b.a⁻ⁿ = aⁿ.ba⁻ⁿ
Right x aⁿ
=> b.aⁿ = aⁿ.b(a⁻ⁿaⁿ) = aⁿ.b ...(5)
Take inverse of (2):
(a.b)⁻ⁿ⁻¹= b⁻ⁿ⁻¹.a⁻ⁿ⁻¹ ...(6)
Right x (6) to (3)
(a.b) = aⁿ⁺².bⁿ⁺².(b⁻ⁿ⁻¹.a⁻ⁿ⁻¹)
ab = aⁿ⁺².b.a⁻ⁿ⁻¹ ...(7)
Right x aⁿ⁺¹
abaⁿ⁺¹ = aⁿ⁺².b
Left x a⁻¹
baⁿ⁺¹ = aⁿ⁺¹.b ...(8)
(b.aⁿ).a = (aⁿ.b).a from (5) b.aⁿ = aⁿ.b
aⁿ.b.a = aⁿ⁺¹.b
cancellation law:
=> ba= ab
=> G is abelian [QED]

Note:

Trick is inverse (1) then x to (2):

b.aⁿ = aⁿ.b ...(5)
Similarly inverse (2) then x (3):
baⁿ⁺¹ = aⁿ⁺¹.b ...(8)
Solve (5) & (8): by cancellation law
aⁿ.b.a = aⁿ⁺¹.b
=> ba= ab

Group is Symmetry

Landau's book "Symmetry" explains it as follow:

Automorphism = Congruence= 叠合 has
1). Proper 真叠合 (symmetry: left= left, right = right)
2). Improper 非真叠合 (non-symmetry: reflection: left changed to right, vice-versa).
Congruence => preserve size / length
=> Movement 运动 (translation 平移, rotation about O )
= Proper congruence (Symmetry)

In Space S, the Automorphism that preserves the structure of S forms a Group Aut(G).
=> Group Aut(G) describes the Symmetry of Space S.

Hence Group is the language to describe Symmetry.

 

Song of Simple Group

http://www.youtube.com/watch?v=lJhmJJ_q6Go&NR=1

The lyrics to the song can be found in "Contemporary Abstract Algebra" (Seventh Edition) by Joseph A. Gallian

(ISBN 978-0-547-16509-7) pages 422 & 423.

Group: Fermat Little Theorem

Use Group to prove Fermat Little Theorem:
For any prime p,
Let Group (Zp,*mod p) = {1,2,3....p-1};    [*mod p= multiply modulo p]
For any non-zero b in Zp,
$latex b^{p-1} = 1 \: \mbox {in Zp }$
Since Zp isomorphic~ to the ring of cosets  of the form a+pZ   (eg. Z2 ~ {0+2Z, 1+2Z}
For any m in Z not in the coset {0+pZ}
ie m ≠0 (mod p)
or p not divisible by a
$latex m^{p-1} = 1 (mod \: p)$
(x m both sides)
$latex m^{p} = m (mod \: p)$

Automorphism = Symmetry

Automorphism of a Set is an expression of its SYMMETRY.
1. Geometry figure (e.g. triangle) under certain transformations (reflection, rotation, ...), it is mapped upon itself, certain properties (distance, angle, relative location) are preserved.
=> the figure admits certain automorphism relative to its properties.
2. Automorphism of an arbitrary Set (with arbitrary relations between its elements) form an Automorphism Group of the set.

Group Definition

Memorize Trick for Group Definition

C.A.N. I. ?
C= Closure
A= Associative: (ab)c = a(bc)
N= Neutral element (e): ae=ea=a
I= Inverse: a $Latex ^ {-1} $ = e

If only 50% (C.A.)=> Semi-Group

If Semi-Group + Neutral = (C.A.N.) = MoNoid

Note: No Inverse => not a Group

Arthur Cayley (UK, 1821-1895): first gave an abstract definition of Group @1854 while being a lawyer for 14 yrs, couldn't find a teaching job. His definition was ignored for 25 years until 1882 by Walter Van Dyck who gave the final Axiomatic definition of Abstract Group. ie above [C.A.N.I.]