Unit-4

Algebraic systems

Q1) Define algebraic system.

A1)

Binary operations-

A binary operation * is a set A is a function from A × A to A.

If * is a binary operation in a set A then than for the * image of the ordered pair (a, b) ∈ A×A, we write a*b.

For example:

Addition + is a binary operation in the set of natural number N, integer Z and real number R.

Multiplication is a binary operation in N, Q, Z, R and C.

General properties-

Propery-1: Let A be any set. A binary operation * A × A →A is said to be commutative if for every a, a, b ∈A.

a* b = b * a

Property-2:Let A be a non-empty set. A binary operation *; A × A →A is said to be associative if

a* b) * c = a * (b * c) for every a, b, c ∈A.

Property-3: Let * be a binary operation on a non-empty set A. If there exists an element e ∈A such that e * a = a * e = a for every a ∈A, then the element e is called identity with respect to * in A.

Property-4: Let * be a binary operation on a non-empty set A and e be the identity element in A with respect the operation *. An element a ∈A is said to be invertible if there exists an element b ∈A such that

a* b = b * a = e

In which case a andb are inverses of each other. For the operation * if b is the inverses of a ∈A then we can write b =

Q2) Define semi-groups.

A2)

Semi-group-

Let S be a non-empty set and * be a binary operation on S. The algebraic (S, *) is called a semi-group if the operation * is associative. In other words, the groupoid is a semi-group if

(a* b) * c = a * (b * c) for all a, b, c ∈S

Thus, a semi-group requires the following:

(i) A sets.

(ii) A binary operation * defined on the elements of S.

(iii) Closure, a * b whenever a, b ∈S.

(iv) Associativity i(a * b) * c = a * (b * c) for all a, b, c ∈S.

Example: Let N be the set of natural numbers. Then (N, +) and (N, *) are semi-groups.

Q3) What is homomorphism and Isomorphism of Semi-groups.

A3)

Homomorphism of semi-groups-

Let (S, *) and (T, 0) be any two semi-groups. A mapping f: S →T such that for any two elements a, b ∈S

f(a * b) = f (a) o f(b) is called a semi-group homomorphism.

Q4) Define Monoids.

A4)

A semi-group (M, *) with an identity element with respect to the binary operation * is called a monoid. In other words, an algebraic system (M, *) is called a monoid if:

(i) (a * b) * c = a * (b * c) ∀a, b, c ∈M.

(ii) There exists an element e ∈M such that e * a = a * e = a ∀a ∈M.

Example: Let Z be the set of integers (Z, +) is a monoid 0 is the identity element in Z with respect

to +.

Q5) Define groups.

A5)

A group is an algebraic structure (G, *) in which the binary operation * on G satisfiesthe following conditions:

Condition-1: for all a, b, c, ∈G

a* (b * c) = (a * b) * c (associativity)

Condition-2:there exists an elements e ∈G such that for any a ∈G

a* e= e * a = a (existence of identity)

Condition-3: for every a ∈G, there exists an element denoted by in G such that

a* = * a = e

is called the inverse of a in G.

Example: (Z, +) is a groupwhere Z denote the set of integers.

Example: (R, +) is a group where R denote the set of real numbers.

Q6) If G = {1, -1, i, -i} where i= , then show that G is an abelian group with respect to multiplication as a binary operation.

A6)

First we will construct a composition table-

. | 1 | -1 | i | -i |

1 | 1 | -1 | i | -i |

-1 | -1 | 1 | -i | i |

i | i | -i | -1 | 1 |

-i | -i | i | 1 | -1 |

It is clear from the above table that algebraic structure (G, .) is closed and satisfies the following conditions.

Associativity- For any three elements a, b, c ∈G (a ⋅b) ⋅c = a ⋅(b ⋅c)

Since

1 ⋅(−1 ⋅i) = 1 ⋅−i= −i

(1 ⋅−1) ⋅i= −1 ⋅i= −i

⇒1 ⋅(−1 ⋅i) = (1 ⋅−1) i

Similarly with any other three elements of G the properties holds.

∴Associative law holds in (G, ⋅)

Existence of identity: 1 is the identity element (G, ⋅) such that 1 ⋅a = a = a ⋅1 ∀a ∈G

Existence of inverse: 1 ⋅1 = 1 = 1 ⋅1 ⇒1 is inverse of 1

(−1) ⋅(−1) = 1 = (−1) ⋅(−1) ⇒–1 is the inverse of (–1)

i⋅(−i) = 1 = −i⋅i⇒–iis the inverse of iin G.

−i⋅i= 1 = i⋅(−i) ⇒iis the inverse of –iin G.

Hence inverse of every element in G exists.

Thus all the axioms of a group are satisfied.

Commutativity: a ⋅b = b ⋅a ∀a, b ∈G hold in G

1 ⋅1 = 1 = 1 ⋅1, −1 ⋅1 = −1 = 1 ⋅−1

i⋅1 = i= 1 ⋅i; i⋅−i= −i⋅i= 1 = 1 etc.

Commutative law is satisfied

Hence (G, ⋅) is an abelian group.

Q7) Prove that the set Z of all integers with binary operation * defined by a * b = a + b + 1 ∀a, b ∈G is an abelian group.

A7)

Sum of two integers is again an integer; therefore a +b ∈Z ∀a, b ∈Z

⇒a +b + 1 ⋅∈Z ∀a, b ∈Z

⇒Z is called with respect to *

Associative law for all a, b, a, b ∈G we have (a * b) * c = a * (b * c) as

(a* b) * c = (a + b + 1) * c

= a + b + 1 + c + 1

= a + b + c + 2

Also

a* (b * c) = a * (b + c + 1)

= a + b + c + 1 + 1

= a + b + c + 2

Hence (a * b) * c = a * (b * c) ∈a, b ∈Z.

Q8) If (G, *) is a group and H ≤G, then (H, *) is a sub-group of (G, *) if and only if

(i) a, b ∈H ⇒a * b ∈H;

(ii) a∈H ⇒∈H

A8)

If (H, *) is a sub-group of (G, *), then both the conditions are obviously satisfied.

We, therefore prove now that if conditions (i) and (ii) are satisfied then (H, *) is a sub-group of (G, *).

To prove that (H, *) is a sub-group of (G, *) all that we are required to prove is : * is associative in

H and identity e ∈H.

That * is associative in H follows from the fact that * is associative in G.

Also,

a∈H ⇒∈H by (ii) and e ∈H and ∈H ⇒a * = e ∈H by (i)

Hence, H is a sub-group of G

Q9) If (H, *) is a sub-group (G, *), then a * H = H if and only if a ∈H.

A9)

Let a * H = H

Since e ∈H then a = a * e ∈a * H

Hence a ∈H

Conversely

Let a ∈H then a * H ⊆H

(H, *) is a sub-group.

∴a∈H, h ∈H ⇒* h ∈H.

Now h ∈H

⇒h = a * ( * h) ∈a * H

∴h∈H ⇒h ∈a * H

⇒H ⊆a * H

Hence a * H = H

Q10) Let G be (Z, +) i.e., the group of integers under addition and let f: G →G defined by

∅(x) = 3x ∀x ∈G. Prove that f is homomorphism, determine its Kernel.

A10)

We have ∅(x) = 3x ∀x ∈G

∀x, y ∈G ⇒x + y ∈G (∴G is a group under addition)

Now

f(x + y) = 3 (x + y)

= 3x + 3y

= f (x) + f (y)

Hence f is homomorphism.

Kernel of homomorphism consists of half of zero i.e., the integers whose double is zero.

Thus K = {0}