**Group Theory**

**2.1. Definition of group**** **

A nonempty set of elements G is said to form a *group* if in G there is defined a binary operation, called the product and denoted by such that

1) a, b Є G implies that a . b Є G (closed)

2) a, b, c Є G implies that a . (b . c)= (a . b). c (associative law)

3) There exists an element e Є G such that a. e = e . a = a for all a Є G (the existence of an identity element in G).

4) For every a Є G there exists an element a* ^{ -1}* Є G such that a . a

^{ -1}*= a*

^{ }

^{ -1}*.a = e (the existence of inverses in G)*

^{ }** Definition**** of Abelian**

A group G is said to be *abelian* (or *commutative*) if for every a,b ЄG, a.b = b.a.

A group which is not abelian is called, naturally enough, non-abelian.

Another characteristic of a group G is the number of elements it contains. We call this the order of G and denote it by °(G).

**2.2. some example of group**

**Example 2.2.1**

Let G consist of the integer 0, ±1, ±2, . . . Where mean by a.b for a,bЄG the usual sum of integers, that is, a.b = a+b. G is an infinite abelian group in which 0 plays the role of e and –a that of a^{–}^{1}

**Ex****ample 2.2.2 **

Let G consist of the real number 1,-1 under the multiplication of real numbers. G is then an abelian group of real numbes. G is an group of order 2.

**Example 2.2.3 **

Let G = S3, the group of all 1-1 mappings of the set {x_{1},x_{2},x_{3}} onto itself, under the product which we defined in chapter 1. G is a group of order 6. we digress a little before returning to S3

For …notation, not just in S3, but in any group G, let us define for any aЄG, a^{0} = e, a^{1}=a, a^{2}=a.a, a^{3}=a.a^{2}, . . . A^{k} = a.a^{k-1}, and a^{-2}=(a^{-1})^{2},a^{-3}=(a^{-1})^{3}, etc. The reader may verify that the usual rules of exponents prevail; namely, for any two integers (positive, negative, or zero)m,n.

A^{m}.a^{n} = a ^{m+n}

(a^{m})^{n} = a^{mn}

(It is worthwhile nothing that, in this notation, if G is the group of example 2.2.1, a^{n} means the integer na) with this notation at our disposal let us examine S3 more closely. Consider the mapping φ defined on the set x_{1},x_{2},x_{3} by

x_{1} → x_{2}

x_{2} → x_{1}

x_{3 } → x_{3 }

And the mapping

x_{1}_{ }→ x_{2}

x_{2 }_{ }→ x_{3}

x_{3}_{ }→ x_{1}

Checking, we ready see that

φ^{2}^{ }=e, ψ^{3} = e and that

x_{1} → x_{3}

x_{2} → x_{2}

x_{3} → x_{1}

Whereas

x_{1} → x_{1}

x_{2} → x_{3}

x_{3} → x_{2}

**Example 2.2.4**

Let n be any integer. We construct a group of order n as follows: G will consist of all symbols a^{ }^{i} , i = 0, 1, 2, …, n-1 where we insist that a^{ }^{0} = a^{ }^{n} = e, a^{ }^{i} . a^{ }^{j} =a^{ }^{i+ j} if i+ j ≤ n and a^{ }^{i} .a^{ }^{j} = a^{ }^{I +j-n} if I + j>n. It is called a cyclic group of order n.

**Example 2.2.5**

Let S be the set of integers and, as usual, let A(S) be the set of all one to one mappings of S onto itself. Let G be the set of all elements in A(S) which move only a finite number of elements of S; that is, σ Є G if and only if the number of x in S such that x σ≠ x is finite.

**Example 2.2.6**

Let G be the set of all 2×2 matrices

Where a,b,c,d are real numbers, such that ad-bc ≠ 0. for the operation in G we use the multiplication of matrices; that is.

=

This is the required relation on the entries of a matrix which puts it in G we merely must show that (aw+by) (cx+dz)- (ax+bz)(cw+dy) ≠ 0

**Example 2.2.7**

Let G be the set of all 2×2 matrices,

Where a,b,c,d are real number such that ad-bc =1. define that ad-bc =1. Define the operation in G, as we did in example 2.2.6, via the multiplication of metrices.

**2.3. Some preliminary Lemmas**

**Lemma 2.3.1 **

If G is a group, then

- The identity element of G is unique
- Every aЄG has a unique inverse in G
- For every aЄG, (a
^{-1})^{-1}=a - For all a,b ЄG, (a.b)
^{-1}=b^{-1}.a^{-1 }

**Lemma 2.3.2**

Given a,b in the group G, then the equation a.x = b and y.a = b have unique solutions for x and y in G. In perticular, the two cancellation laws

a°u = a°w implies u = w, and

u°a= w°a implies u = w

Hold in G.

Discussion:

1. What is the meaning of binary operation?

Binary operation is a mathematical operation in which to element are combined to yield a single result.

Examples include the familiar arithmetic operation of addition, subtraction, multiplication, and division.

More precisely, the binary operation on a set S is a binary function from S and S to S.

Suppose X, Y, and Z are sets, suppose that given any element x of X and y of Y, f(x, y) is a unique element of that. Then f is a binary function from X and Y to Z.

To explain the concept and to solve the problem in group theory in post graduate, we don’t need to explain all steps that we need because we suggest that we already have mastered it.

2. Does the equation of (a. b)^{ n} =a^{n}.b^{n} in an Abelian group satisfy for all operation?

Yes, because this operation is a character of Abelian group. This make all operation can be solved with this equation.

2.5 A Counting Principle

If H, K are two subgroups of G, let HK = {x}

Example:

In S_{3} let H={e, } , K={e, }

HK={e, }, ^{2}=e, H, K subgroups

^{2}=

○ (HK) = 4

4 is not a divisor of 6, the order of S_{3}

HK is not a subgroup of S_{3} (Lagrange’s Theorem)

LEMMA 2.5.1 HK is a subgroup of G if and only if HK=KH.

Proof

Suppose that HK=KH

If h and k then hk=k_{1}h_{1}, k_{1}, h_{1}

It will be proven that HK is a subgroup

Suppose x=hk and y=h’k’

Then xy=hkh’k’, kh’

kh’=h_{2}k_{2} , h_{2} k_{2}

xy=h(h_{2}k_{2})k’

=(hh_{2})(k_{2}k’)

HK is closed.

x^{-1}=(hk)^{-1}

=k^{-1}h^{-1}

So x^{-1}

HK is a subgroup of G.

Then for any h, k

h^{-1}k^{-1} and kh=(h^{-1}k^{-1})^{-1}

KHHK

xHK

if x^{-1}=hkHK

so x=(x^{-1})^{-1}=(hk)^{-1}=k^{-1}h^{-1} KH

HKKH

HK=KH

COROLLARY If H, K are subgroups of the abelian group G, then HK is a subgroup of G

THEOREM 2.5.1 If H and K are finite subgroups of G of orders ○ (H) and ○ (K), respectively, then

Proof

H(e)

It will be proven that

hh_{1}H, hk=h_{1}k_{1}

h_{1}^{-1}h=k_{1}k^{-1} , h_{1}H, h_{1}^{-1}H

h_{1}^{-1}h H

k_{1}K, k_{1}^{-1}K

k_{1}^{-1}k K

h_{1}^{-1}h H=(e)

h_{1}^{-1}h=e

h=h_{1}

Contradiction with hh_{1}

○ (HK) is indeed ○ (H) ○ (K)

COROLLARY If A and K are subgroups of G and , , then H(e)

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