For-all-a-b-G-where-G-is-a-group-with-respect-to-operation-o-Prove-that-aob-1-b-1-o-a-1-

Question Number 14353 by tawa tawa last updated on 30/May/17

For all a, b \in G, where G is a group with respect to operation^{'} o^{'}

Prove that {(aob)}^{- 1} = b^{- 1} o a^{- 1}

Commented by Tinkutara last updated on 31/May/17

Sorry, I want to say that are a and b

functions on the set G ?

Also, is a o b a composite function of b

and a ?

Commented by tawa tawa last updated on 31/May/17

Yes sir

Commented by Tinkutara last updated on 01/Jun/17

We have to show that (a o b) o (b^{- 1} {o a}^{- 1})

= I_{G} = (b^{- 1} {o a}^{- 1}) (a o b)

\Rightarrow (a o b) o (b^{- 1} {o a}^{- 1}) = a o ({b o b}^{- 1}) {o a}^{- 1}

= (a o I_{G}) {o a}^{- 1} = {a o a}^{- 1} = I_{G}

Similarly you can prove the other

part .

[Here it is not needed to prove that

{a o a}^{- 1} = {b o b}^{- 1} = I_{G}, identity functions

on G]