Question Number 14353 by tawa tawa last updated on 30/May/17
$$\mathrm{For}\:\mathrm{all}\:\mathrm{a},\:\mathrm{b}\:\in\:\mathrm{G}\:,\:\mathrm{where}\:\mathrm{G}\:\mathrm{is}\:\mathrm{a}\:\mathrm{group}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{operation}\:\:'\mathrm{o}' \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\:\left(\mathrm{aob}\right)^{−\mathrm{1}} \:=\:\mathrm{b}^{−\mathrm{1}} \:\mathrm{o}\:\:\mathrm{a}^{−\mathrm{1}} \\ $$
Commented by Tinkutara last updated on 31/May/17
$$\mathrm{Sorry},\:\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{say}\:\mathrm{that}\:\mathrm{are}\:{a}\:\mathrm{and}\:{b} \\ $$$$\mathrm{functions}\:\mathrm{on}\:\mathrm{the}\:\mathrm{set}\:{G}? \\ $$$$\mathrm{Also},\:\mathrm{is}\:{aob}\:\mathrm{a}\:\mathrm{composite}\:\mathrm{function}\:\mathrm{of}\:{b} \\ $$$$\mathrm{and}\:{a}? \\ $$
Commented by tawa tawa last updated on 31/May/17
$$\mathrm{Yes}\:\mathrm{sir} \\ $$
Commented by Tinkutara last updated on 01/Jun/17
![We have to show that (aob)o(b^(−1) oa^(−1) ) = I_G = (b^(−1) oa^(−1) )(aob) ⇒ (aob)o(b^(−1) oa^(−1) ) = ao(bob^(−1) )oa^(−1) = (aoI_G )oa^(−1) = aoa^(−1) = I_G Similarly you can prove the other part. [Here it is not needed to prove that aoa^(−1) = bob^(−1) = I_G , identity functions on G]](https://www.tinkutara.com/question/Q14466.png)
$$\mathrm{We}\:\mathrm{have}\:\mathrm{to}\:\mathrm{show}\:\mathrm{that}\:\left({aob}\right){o}\left({b}^{−\mathrm{1}} {oa}^{−\mathrm{1}} \right) \\ $$$$=\:\mathrm{I}_{{G}} \:=\:\left({b}^{−\mathrm{1}} {oa}^{−\mathrm{1}} \right)\left({aob}\right) \\ $$$$\Rightarrow\:\left({aob}\right){o}\left({b}^{−\mathrm{1}} {oa}^{−\mathrm{1}} \right)\:=\:{ao}\left({bob}^{−\mathrm{1}} \right){oa}^{−\mathrm{1}} \\ $$$$=\:\left({ao}\mathrm{I}_{{G}} \right){oa}^{−\mathrm{1}} \:=\:{aoa}^{−\mathrm{1}} \:=\:\mathrm{I}_{{G}} \\ $$$$\mathrm{Similarly}\:\mathrm{you}\:\mathrm{can}\:\mathrm{prove}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{part}. \\ $$$$\left[\mathrm{Here}\:\mathrm{it}\:\mathrm{is}\:\mathrm{not}\:\mathrm{needed}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that}\right. \\ $$$${aoa}^{−\mathrm{1}} \:=\:{bob}^{−\mathrm{1}} \:=\:\mathrm{I}_{{G}} ,\:\mathrm{identity}\:\mathrm{functions} \\ $$$$\left.\mathrm{on}\:{G}\right] \\ $$