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If $(G , .)$ is a group such that $(ab)^{-1}=a^{-1}b^{-1},\forall a,b \in G,$ then $G$ is a/an

  1. Commutative semi group
  2. Abelian group
  3. Non-abelian group
  4. None of these
asked in Graph Theory by Veteran (37.5k points)   | 1.6k views
which set ?

3 Answers

+7 votes
Best answer

In a group (G , .) is said to be abelian if 

(a*b) =(b*a) ∀a,b ∈G

 (ab)-1 = (b--1a-1)............(1)

 

Given ,(ab)-1= a-1b-1     ..........(2)

from (1) and (2) we can write 

 a-1b-1   b-1 a-1

we can also write it as

ab=ba

 

Hence,Option(B)Abelian Group is the correct choice.

answered by Veteran (32.2k points)  
selected by
@LeenSharma how did you arrive at  Equation (1) ?
eq(1) is a property of inverse.don't be confuse here eq(1) is property and eq(2) is given in the question which i already mentioned in my solution.
@Leensharma I have not seen this property in any of the video lectures of group theory. In which standard textbook is this theorem there?
$(AB)(AB)^{-1} = I$

$(A^ {-1}AB) (AB)^{-1} = A ^{-1} I$

$(I B) (AB)^{-1} = A ^{-1}$

$B (AB)^{-1} = A ^{-1}$

$B ^{-1} B (AB)^{-1} = B ^{-1} A ^{-1}$

$(AB)^{-1} = B ^{-1} A ^{-1}$

 ∴(ab)-1 = (b--1a-1)............(1)

This property is applicable only for Matrix not for any variables and in this Q matrix is not mention btw.

 

+1 vote
B abelian group
answered by (145 points)  
0 votes

A group is called as abelian group if and only if it fallow commutative rule properly.

Example as::

a*b=b*a

Now According to given problem

let:

(x*y)^1=(y*x)^-1

x^-1y^-=y^-1x^-1

so xy=yx that's why it abelian group.

answered by Boss (5.4k points)  


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