Abelian Group or Commutative Group

"A group (G,O) is said to be an abelian group or a commutative group if (a O b) = (b O a)  a, b ᗴ G." 
OR

"A non-empty set 'G' together with a binary operation 'O' is said to be a group, if the following laws holds for all the operations 'O'."


Group holds the following given five laws :
  
(a) Clouser Law : 'G' is cloused under the binary operation 'O' that is ( a O b ) = ( b O a ) ∀ a, b ᗴ G.

(b) Associative Law : The associative law for the operation holds that is  a O ( b O c ) = ( a O b ) O c ∀ a, b, c ᗴ G.

(c) Existence of Identity Element : The identity element exists in 'G' that is there exists an element e ᗴ G such that a ᗴ  e = e ᗴ a = a ∀ a ᗴ G.

(d) Existence of Inverse Element : The inverse element of all the elements of 'G' exists that is for every a ᗴ G there exists an element b ᗴ G such that a O b = b O a = e , where 'e' bring the identity element.

(e) Commutative Law : 'G' is commutative under binary operation 'O' that is a O ( b O c ) = ( a O b ) O c  ∀ a, b, c ᗴ G.


Prove that the 4th roots of unity forms Abelian group with respect to ordinary multiplication.

Since, G = { -1, 1, i, -i }; where i = √-1.

 If set 'G' follows the five laws i.e. clouser law, commutative law, associative law, existance of identity element and existence of inverse element then only we can say that 'G' is an abelian group.

Testing for clouser law

Multiplicaion table

X -1 1 i -i
-1 1 -1 -i i
1 -1 1 i -i
i -i i -1 1
-i i -i 1 -1


From the above table, we see that clouser law holds good for the operation ordinary multiplication


Testing for associative law

   Since, 'G' is a set of complex number and we know that associative law holds good for ordinary multiplication in set of complex number. Hence, it also holds good in 'G'.


Testing for existence of identity element

 Since, identity element exists in 'G' and which is 1; because

          a * 1 = 1 * a = a ∀ a ᗴ G.

Testing for existence of inverse element

Inverse of  -1 :  ( -1 ) * ( -1 ) = 1
Inverse of   1 :  ( 1 ) * ( 1 ) = 1
Inverse of    i :  ( i ) * ( -i ) = 1
Inverse of   -i :  ( -i ) * ( i ) = 1

Thus, we see that all elements have an inverse element.


Testing of commutative law

              Since, 'G' is a sub-set of complex number and we know that commutative law holds good for ordinary multiplication in set of complex number, so it also holds good in 'G'.


Since, 'G' follows all the five laws. Hence; 'G' is an abelian group with respect to ordinary multiplication.



Prove that the cube roots of unity forms Abelian group with respect to ordinary multiplication.



Since, G = { 1, w, w2 }; where w3 = 1.

 If set 'G' follows the five laws i.e. clouser law, commutative law, associative law, existance of identity element and existence of inverse element then only we can say that 'G' is an abelian group.

Testing for clouser law

Multiplicaion table

X1ww2
11ww2
www21
w2w21w


From the above table, we see that clouser law holds good for the operation ordinary multiplication


Testing for associative law

   Since, 'G' is a set of complex number and we know that associative law holds good for ordinary multiplication in set of complex number. Hence, it also holds good in 'G'.


Testing for existence of identity element

 Since, identity element exists in 'G' and which is 1; because

          a * 1 = 1 * a = a ∀ a ᗴ G.

Testing for existence of inverse element

Inverse of  1   :  ( 1 ) * ( -1 ) = 1
Inverse of  w  :  ( w ) * ( w2 ) = 1
Inverse of  w2 :  ( w2 ) * ( w ) = 1


Thus, we see that all elements have an inverse element.


Testing of commutative law

              Since, 'G' is a sub-set of complex number and we know that commutative law holds good for ordinary multiplication in set of complex number, so it also holds good in 'G'.


Since, 'G' follows all the five laws. Hence; 'G' is an abelian group with respect to ordinary multiplication.




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