Mathematical Foundation of Computer Science

(Chris Devlin) #1
DHARM

ALGEBRAIC STRUCTURES 83


  1. From the result of equality (1), for any x ∈ X, we have the inverse element x–1, where x × x–1
    = ê; and also y × y–1 = ê for any y ∈ X. Since, operation × is closed so x × y ∈ X,
    Let u = x × y and w = y–1 × x–1, then
    u × w = (x × y) × (y–1 × x–1)
    = x × (y × y–1) × x–1 (by Associativity)
    = x × ê × x–1 (since y × y–1 = ê)
    = x × x–1
    = ê (identity element)
    Thus we have,
    (x × y) × (y–1 × x–1) = ê = (y–1 × x–1) × (x × y) also.
    ⇒ (x × y) = (y–1 × x–1)–1, and due to symmetricity
    (x × y)–1 = (y–1 × x–1). (Proved)
    Example 4.5. Let W = {1, ω, ω^2 }, where ω is cube root of unity i.e. ω^3 = 1 and 1 + ω + ω^2 = 0,
    then show that algebraic system (W, ×) is a group.
    Sol. Since,

    1. Operation × is closed for W, i.e.
      1 × ω = ω ∈ W; ω × ω^2 = ω^3 = 1 ∈ W; 1 × ω^2 = ω^2 ∈ W; and others.

    2. Operation × is associative over W, viz.
      ω^2 × (1 × ω) = (ω^2 × 1) × ω = ω^3 ; and others.

    3. Existence of an identity element i.e.,
      ω × ê = x = x × ê for any x ∈ W
      if x = ω then ê = 1, i.e. 1 × ω = ω × 1 = ω;
      if x = ω^2 then ê = 1, i.e. 1 × ω^2 = ω^2 × 1 = ω^2 ;

    4. Existence of unique inverse element for every element of W, i.e.,
      x × y = y × x = ê where x, y ∈ W and y is inverse of x.
      if x = 1 then y = 1, so 1 × 1 = ê;
      if x = ω then y = ω^2 , so ω × ω^2 = ω^3 = 1 = ê;
      if x = ω^2 then y = ω, so ω^2 × ω = ω^3 = 1 = ê;
      Therefore, Algebraic system (W, ×) is a group.




4.3 Semi Subgroup...................................................................................................................


Let algebraic system (X, ) is a semi group and let Y ⊆ X, then (Y, ) is said to be a semi
subgroup if, (Y, ⊗) is itself a semigroup, i.e.
(Y, ⊗) ⊆ (X, ⊗)
For example, (I+, +) is semigroup, where binary operation + is defines over set of posi-
tive integers I+. Now consider two subsets of I+ i.e., (1) positive integers that are all even (J)
and (2) positive integers that are all odd (K). Thus,
J ⊆ I+ and K ⊆ I+
Now test whether (J, +) is a semigroup or (K, +) is a semigroup. Former, is a semigroup
because operation + is closed and associative over J but latter is not a semigroup due to violation

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