ISRO2016-4

Question

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

• A. Commutative semi group
• B. Abelian group
• C. Non-abelian group
• D. None of these

Comments

which set ?

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Can someone please tell me how to prove G is closed?

Answer 1

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.

Comments

@LeenSharma how did you arrive at  Equation (1) ?

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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.

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@Leensharma I have not seen this property in any of the video lectures of group theory. In which standard textbook is this theorem there?

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

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

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

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

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

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

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https://math.mit.edu/~gs/linearalgebra/ila0205.pdf

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 ∴(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.

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Don't we have to check other properties for Abelian group like closed, Identity, Inverse and associativity?

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Its already given that (G,.) is a group, so Properties like closed, Identity, Inverse and associativity are implicitly valid.

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can anyone please tell how
a-1b-1   = b-1 a-1

we can also write it as

ab=ba

how we can prove ab=ba from above eqn

Answer 2

B abelian group

Answer 3

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.