Test by Bikram | Mock GATE | Test 4 | Question: 21

Question

$\left ( G, . \right )$ is a group such that $\left ( x,y \right )^{-1} = x^{-1}y^{-1}, \forall \left ( x,y \right ) \in G$.
Here, $G$ is a:

• A.  Monoid
• B.  Commutative semi group
• C.  Abelian group
• D.  Semi group

Comments

It's already stated that G is a group, so A and D are trivially true. I guess we have to find the most appropriate answer.

 

We know $(xy)^{-1}=y^{-1}x^{-1}$

Given: $(xy)^{-1}=x^{-1}y^{-1}$

 

Hence, $y^{-1}x^{-1}=x^{-1}y^{-1}$

So, this is commutative.

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Also, we know: $(y^{-1}x^{-1})^{-1}=xy$   //Just do step by step. It's easy.

By what's given: $(y^{-1}x^{-1})^{-1}=yx$

 

So, $xy=yx$

This is definitely commutative. :P

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A group that's commutative is an Abelian Group.

Answer 1

In a group $\left ( G , . \right )$ is said to be abelian if  $\left ( a\ast b \right ) = \left ( b\ast a \right ) a,b $∀$\epsilon G$

one of the well known property for inverse is given below:

 (ab)^-1 = (b^-1a^-1)............$(1)$

Given ,$\left ( a,b \right )^{-1}$ $=$ $a^{-1}b^{-1}$     

from $(1)$ and $(2)$ we can write 

$a^{-1}b^{-1}$  = $b^{-1}a^{-1}$     

One more property of groups: "In a group (G, *) if inverse(x) = x   for all x ϵ G then G is abelian group "

$a.b=b.a$

Hence, Option (C) Abelian Group