The concept of homomorphism helps to understand the structural similarity between two given algebraic structures.

*Let (S , *) and (T , D ) be any two semigroups.*

*A mapping g : S ® T such that for any two elements a, b S,*

g (a * b) = g (a) D g (b) is called

g (a * b) = g (a) D g (b) is called

**semigroup homomorphism**.

Remark:

Remark:

Homomorphism essentially preserves the behavior of the operations on the mapped elements.

*.*

A semigroup homomorphism is called a semigroup

whether the mapping is "one - one", "onto", or "one - one and onto " respectively

A semigroup homomorphism is called a semigroup

**monomorphism, epimorphism or isomorphism**depending onwhether the mapping is "one - one", "onto", or "one - one and onto " respectively

*A semigroup homomorphism*

onto itself is called.

onto itself is called

**endomorphism**

Two semigroups (S , *) and (T , D ) are said to be

S onto T.

Two semigroups (S , *) and (T , D ) are said to be

**isomorphic**if there exists a semigroup isomorphism mapping fromS onto T.

Example 1:

Example 1:

Consider the semigroup (Z , ·) and (N, ·). Define f : Z ® N by f(x) = |x|.

Then, for all, x, y Z , f(x , y) = |x · y| = |x| · |y| = f(x). f(y).Thus f is homomorphism.

f is also onto, since for every n Î N, there exists –n, n Î Z such that f(n) = |n| = n and f(-n) = |-n| = n.

But, f is not one-one for, f(3) = f(-3) = 3, but 3 ¹ -3.

**Example 2:**

Define g : (Z , ·) ®(N

_{0}, ·) by g(x) = 0. Then g is homomorphism.

For, g(x · y) = g(xy) = 0 = 0 · 0 = g(x) · g(y), for all x , y Z.

Example 3:

Example 3:

Let (N

_{0}, +) be the semigroup of natural numbers and (S , *) be the semigroup on S ={e, 0, 1} with the operation *

given by the following table.

A mapping g : N

_{0}® S given by g(0) = 1 and g( j ) = 0 for j ¹ 0, is homomorphism,

for, .

Therefore from the table of (S , * ) , we have g(a + b) = g(a) * g(b).

Thus, g is a semigroup homomorphism. But g does not map the identity element 0 (zero) of the semigroup (N , +) onto the

identity element e of (S , *).

From the Example 3 of semigroup homomorphism, one can observe that semigroup homomorphism not necessarily always

maps an identity element onto identity element. Thus, we have the following definition for the monoid homomorphism.

*Let, (M , * , e*

_{M}) and (T , D , e_{T}) be any two monoids.*A mapping*g : M ® T

*such that for any two elements,*a, b Î M

*,*

g(a * b) = g(a) D g(b) and g(e

_{M }) = e

_{T }

*is called a*

**monoid homomorphism**.Example 1 of semigroup homomorphism is also monoid homomorphism, since it maps the identity element 1 of (Z , ·) onto

the identify element 1 of (N , ·).

Theorem 2.2.1 :

Theorem 2.2.1 :

Let (S , *) be a semigroup. Then H, the set of all endomorphism on S is a semigroup with respect to the operation composition

of mapping.

Proof :

Proof :

Let H be the set of all endomorphism on a semigroup (S , *).

Let f, g Î H and let x, y Î S.

Then, (f g) (x * y) = f (g(x * y))

= f(g(x) * g(y))

= f(g(x)) * f(g(y))

= (f g)(x) * (f g)(y).

Thus, f g Î H.

Hence (H , ) is a semigroup, since the composition of mapping '' is associative.

**Theorem 2.2.2 :**

Let (S , *) be a given semigroup. There exits a homomorphism g : S ® S

^{S}, where (S

^{S}, ) is a semigroup of mappings from S

into S under the operation of composition of mappings.

Proof :

Proof :

Let (S , *) be a given semigroup. For each a Î S, define g : S ® S

^{S}by g(a) = f

_{a }, where f

_{a}(x) = a * x, for all x Î S.

Then, f

_{a * b }(x) = (a * b) * x

= a * ( b * x)

= f

_{a}( b * x)

= f

_{a}(f

_{b}(x))

= f

_{a}f

_{b}(x), for all x Î S.

Thus, f

_{a }

_{* b}= f

_{a}f

_{b}.

Now, g(a * b) = f

_{a*b}= f

_{a}f

_{b}= g(a) g(b).

Hence, g is a semigroup homomorphism from S into S

^{S}.

Remark 1:

Remark 1:

Let g(S) denote the image set of S under the mapping g. Therefore by the Theorem 2.2.2, g (S) Í S

^{S}. That is, S can be

identified as a substructure of S

^{S}. Therefore every semigroup can be visualized as a substructure of S

^{S}.

Theorem 2.2.3:

Theorem 2.2.3:

Let (M , * , e) be a monoid. Then there exists a subset T Í M

^{M}such that (M , * , e) is isomorphic to the monoid T.

Proof:

Proof:

For any element a M , define g (a) = f

_{a}, where f

_{a}(x) = a * x, for all x Î M.

Then f

_{a}Î M

^{M}.

Let T = {f

_{a }Î

_{ }M

^{M}

_{ }/ f

_{a}=a * x , for all a Î M and for all x Î M }.

Then by the Theorem 2.2.1, g is a homomorphism from M into T.

Also, for each f

_{a}Î T, there exist a Î M such that g(a) = f

_{a}.

Thus, g is onto.

Further, if a ¹ b, then g(a) ¹ g(b). For, suppose g(a) = g(b), then a * x = b * x, for all x Î M, in particular for x Î e also.

Therefore, a * e = b * e, that is, a = b, a contradiction.

Thus, g (a) ¹ g (b).

Hence, g is one - one.

So, g is isomorphism from M to T.

Hence, M T.

Theorem 2.2.4:

Theorem 2.2.4:

Let X be a set containing n elements, let X* denote free semigroup generated by X, and let (S , Å ) be any other semigroup

generated by any n generators. Then there exists a homomorphism g : X

^{*}® S.

Proof:

Proof:

Let Y be the set of n generators of S, consisting of y

_{1},y

_{2}, … ,y

_{n}and let x

_{1}, x

_{2}, … ,x

_{n}be the n elements of X.

Define g(x

_{i}) = y

_{i, }for i = 1,2, … ,n.

Then, for any string a = x

_{1}, x

_{2}… x

_{n}of X*, we define g(a ) = g(x

_{1}) Å g(x

_{2}) Å … Å g(x

_{m}) .

Then, for any b = … , we have

g (a b ) = g (x

_{1}x

_{2}… x

_{m}… )

= g(x

_{1}) Å g(x

_{2}) Å … Å g(x

_{m}) Å g() Å g() Å … Å g()

= g(a ) Å g(b ) .

Thus, g is a homomorphism.

Hence the theorem.

*Let (X ,*

*) be an algebraic system in which*

*is a binary operation on X. An equivalence relation E on X is said*

to have the

to have the

**substitution property**with respect to the operation*for any x*

(x

_{1 }, x_{2 }Î_{ }X.(x

_{1}E ) and (x_{2}*E ) then (x*

_{1}*x*

_{2}) E (*), where , Î X.*

That is, if x

_{1}is related to and x

_{2}is related to , then the result of x

_{1}and x

_{2}by the binary operation , i.e., x

_{1}x

_{2 },

is related to , the result of the binary operation on the related elements and. More precisely, x

_{1}

^{ }is related

to and x

_{2 }is related to

^{ }then x

_{1}x

_{2 }can be equivalently substituted for .

Let (X ,

Let (X ,

*) be an algebraic system and E be an equivalence relation on X. The relation E is called a*

**congruence**

relationon (X ,relation

*) if E satisfies the substitution property with respect to the operation ‘*’

*.*

Theorem 2.2.5:

Theorem 2.2.5:

Let (S , *) and (T , D ) be two semigroups and g be a semigroup homomorphism from (S , *) to (T , D ). Corresponding to

the homomorphism g, there exists a congruence relation R on (S , *) defined by xRy, if g(x) = g(y) for x, y Î S.

**Since g(x) = g(x), for all x Î S, xRx, for all x Î S.**

Proof:

Proof:

The relation R defined above is an equivalence relation:

The relation R defined above is an equivalence relation:

Therefore, R is reflexive.

Also, if g(x) = g(y), then g(y) = g(x).

That is, if xRy then yRx .

Therefore, R is symmetry.

Further, if g(x) = g(y) and g(y) = g(z), then g(x) = g(y) = g(z), that is, g(x) = g(z) .

That is, if xRy and yRz then xRz.

Therefore, R is transitive.

Thus, R is an equivalence relation .

Let x

_{1}, x

_{2 }, and Î S such that xR and x

_{2}R.

Then g(x

_{1}* x

_{2}) = g(x

_{1}) D g(x

_{2}) (since g is a homomorphism.)

= g() g() (since x

_{1}R and x

_{2}R, so g(x

_{1}) = g() and g(x

_{2}) = g()]

= g( * ) (since g is a homomorphism.).

Therefore, by definition of R, x

_{1}* x

_{2}R *

*.*

Thus, R is congruence relation on (S , *) .

Theorem 2.2.6:

Theorem 2.2.6:

Let (S , *) be a semigroup and R be a congruence relation on (S , *) on the quotient set

S/R = {[a] / a Î S}, define [a] Å [b] = [a * b], for all [a], [b] Î S/R . Then (S/R , Å ) is a semigroup. Further, there exists a

homomorphism from (S , *) onto (S/R , Å ) called natural homomorphism.

Proof:

Proof:

Let (S, *) be given semigroup and let [a] denote an equivalence class of a corresponding to the congruence relation R. By the

Theorem 2.2.5 such congruence relation exists.

Claim 1 :

Claim 1 :

*The operation Å is well – defined.*

For, if [a] = [a'] and [a] = [b'], then aRa' and bRb' .

Since R is congruence relation, we have, a * b R a' * b' .

Thus, [a * b] = [a' * b'] .

Consequently, [a] Å [b] = [a * b] = [a' * b'] = [a'] Å [b'].

Hence the Claim 1.

**Claim 2 :**

*(S/R ,Å ) is a semigroup.*

For all [a], [b], [c] Î S/R, we have

[a] Å ([b] Å [c] ) = [a] Å [ b * c]

= [a * ( b * c)]

= [ (a * b) * c ]

= [ a * b] Å [c]

= ( [a] Å [b] ) Å [c].

That is, Å is associative.

Hence (S/R , Å ) is a semigroup.

Hence the Claim 2.

**:**

Claim 3

Claim 3

*There exists a natural homomorphism.*

Define a mapping g : S ® S/R by g(a) = [a] ,for every a Î S .

Then, g(a* b) = [a * b] = [a] Å [b] = g(a) Å g(b) ,for all a , b Î S.

Thus, g is a homomorphism.

Further, for each [a] Î S/R there exist a Î S such that g(a) = [a].

Thus, g is a homomorphism from (S , *) onto (S/R , Å ) .

Hence the Claim 3.

Hence the theorem.

## No comments:

Post a Comment