Some Examples of Groups

Example 1: The set S = {0}, under the operation of useal addition of integers, is an abelian group of order one.

Sol.  Closure Property: For all a, b ∈ S.

                 a = 0, b = 0, a + b ∈ S as 0 + 0  = 0 ∈ S.

          Thus closure property holds in S.

         Associativity: We know addition of integers is associative

         i.e., (a + b) + c = a + (b + c) ∀ a, b, c ∈ S.

         Here a = 0, b = 0, c = 0 and (0 + 0) + 0 = 0 + (0 + 0)

         Thus associative property holds in S.

         Existence of identity: ∀ a ∈ S, there exist 0 ∈ S such that

                               a + 0 = a = 0 + a

        Here a = 0 and 0 + 0 = 0 = 0 + 0.

        Thus 0 ∈ S works for the identity element of S.

       Existence of inverse: ∀ a ∈ S, there exist -a ∈ S such that

                              a + -(a) = 0 = (-a) + a

        Here a = 0 and -a = -0 = 0.

        Thus every element of S has inverse.

        Since all the axioms of a group are satisfied. Hence 
<S, +> is a group.

        Also addition of integers is commutative

       i.e.,        a + b = b + a ∀ a, b ∈ S.

       Here     0 + 0 = 0 + 0.   S contains only one element.

       Therefore, <S, +> is ab abelian group of order one.

Example 2: The set S = {-1, 1} under the operation of usual multiplication of integers is an abelian group of order two.

Sol: Closure Property: We know

                                        1 . 1 = 1, 1 . (-1) = -1, (-1) . 1 = -1

        and                         (-1) . (-1) = 1.

        ^∴    ∀ a, b ∈ S ⇒ a . b ∈ S.

       Thus closure property holds in S.

       Associativity: We know multiplication of integers is associative

       i.e.,                   (a . b) . c = a . (b . c) ∀ a, b, c ∈ S.

      Thus associative property holds in S.

       Existence of identity: Since 1 . 1 = 1 = 1 . 1 and (-1) . 1 = -1 = 1 . (-1)

       ∴ ∀ a, b ∈ S, ∃ 1 ∈ S such that a . 1 = a = 1 . a

       The element 1 ∈ S works for identity element in S.

      Existence of inverse: Since 1 . 1 = 1 = 1 . 1

                                            and  (-1) . (-1) = 1 = (-1) . (-1)

       i.e., inverse of 1 is 1 and inverse of -1 is -1.

       ∴    every element of S has inverse in S.

      Therefore all the axioms of a group are satisfied. Hence <S, .> is a group.

      Moreover, multiplication is commutative

      i.e.,   a . b = b . a ∀ a, b ∈ S

      and S contains only two elements.

      Thus <S, .> is an abelian group of order two.

Example 3: Give an example of a semi-group, which is not a group.

Sol: Let ℤ' = Set of all non-negative integers.

         The operation is the usual addition of integers.

         Closure Property:  We know the sum of two non-negative integers is a non-negative                integer

         i.e., ∀ a, b ∈ ℤ', a + b ∈ ℤ'

         ∴ closure property holds in ℤ'.

       

         Associativity: We know addition of non-negative integers is associative

          i.e., ∀ a, b ∈ ℤ', (a + b) + c = a + (b + c)

         Thus associative property holds in ℤ'.

         Therefore <ℤ' , +> is semi-group.

          

         Existence of identity: Since ∀ a ∈ ℤ',  there exist 0 ∈ ℤ' such that

                               a + 0 = a = 0 + a.

          Thus 0 ∈ ℤ' works for the identity element.

         Non-Existence of inverse: For any a ≠ 0 ∈ ℤ', there exists no non-negative integers                b  ∈ ℤ' such that

                                      a + b = 0

         Thus inverse of a does not exist.

          ∴    <ℤ', +> is not a group.

Example 4: The set Q* of all non-zero rational numbers forms an infinite abelian group under the operation of multiplication of rational numbers.

Sol: Closure Property: We know that the product of two non-zero rational numbers is also a non-zero rational number.

        i.e.,            a . b ∈ Q*, ∀ a, b ∈ Q*.

     

        Assiciativity: We know that multiplication of rational numbers is an associative operation.

        ∴          (a . b) . c = a . (b . c) ∀ a, b, c ∈ Q*.

      Thus associative property holds in Q*.

      Existence of identity: There exist rational number 1 ∈ Q* such that

                    a . 1 = a = 1 . a ∀ a ∈ Q*.

      Thus, The element 1 ∈ Q* works for identity element in Q*.

      Existence of inverse: For all a ∈ Q* , ∃ (1/a) ∈ Q* such that

                                            a . (1/a) = 1 = (1/a) . a.

       Thus, the element 1/a is the inverse of the element a ∈ Q*.

       Therefore all the axioms of a group are satisfied. Hence <Q*, .> is a group.

      Moreover, multiplication is commutative

      i.e.,   a . b = b . a ∀ a, b ∈ Q*.

      and Q* contains an infinite number of elements.

      Thus <Q*, .> is an infinite abelian group.