Let S be set of all bijective functions on the set {a, b, c} . Recall that S is a goup under function composition.
That is, < S. . > is a group. Describe all six elements of the group < S, "> and their inverses.
Alsois the group abelian? Justify your answer.

Detailed Stepwise Solution is given below as is it theoretical answer type in nature 

Step-by-step explanation

 

Given

     X=  {   a,b,c  }   and    S be the set of all bijective function on the set 'X'

 

 Step-1        It is given also we know  S is a group under function composition  <S,?> and have Six elements in it 

                   All the six elements 

 

               X={ a,b,c }                   fi?:X→X     where fi? is the bijective function on set X  for   i=1,2,3,4,5,6

 

                Elements

 

 

           f1?:???abc?abc????  

   

  

          f2?:???abc?acb????     

 

 

          f3?:???abc?cba????

 

        

          f4?:???abc?bac????

   

 

          f5?:???abc?bca????

 

      

         f6?:???abc?cab????

 

 

Step-2       we know inverse of bijective function from set X is the  inverse image of element  f(x)=y

                    such that   f−1(y)=x

 

            Inverse elements 

 

  

       f1−1?:???abc?abc????

 

 

       f2−1?:???abc?acb????

 

 

       f3−1?:???abc?cba????

 

 

        f4−1?:???abc?bac????

 

 

        f5−1?:???abc?cab????

 

 

        f6−1?:???abc?bca????

 

 

  Step-3    Verifying is Group abelian or not 

                  We know group G is abelian 

                 if  for all     a,b?G    

                  ⇒ab=ba

  

              if there exist any two elements a,b   in group G s.that 

                ⇒ab?=ba

             Then we say the group is non-Abelian 

        

               here the  group    (S,o)       where "o"  is function composition    Is Non-Abelian 

 

       Justification

    

      f2?of3?=???abc?bca????                                            

      ⇒f2?of3?=f5?

 

Also  

       ⇒f3?of2?=???abc?cab????

       ⇒  f3?of2?=f6?       

 

       ⇒f2?of3??=f3?of2?

       ⇒    there exist elements in   S     s.that   they do not follow  ab?=ba

          hence group S is non - abaelian.