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

$$\mathrm{For}\mathrm{all}\mathrm{a},\mathrm{b}\in \mathrm{G},\mathrm{where}\mathrm{G}\mathrm{is}\mathrm{a}\mathrm{group}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{operation}{}^{\prime }{\mathrm{o}}^{\prime }$$$$\mathrm{Prove}\mathrm{that}{\left(\mathrm{aob}\right)}^{-1}={\mathrm{b}}^{-1}\mathrm{o}{\mathrm{a}}^{-1}$$

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

$$\mathrm{We}\mathrm{have}\mathrm{to}\mathrm{show}\mathrm{that}\left(aob\right)o\left(b^{-1}{oa}^{-1}\right)$$$$={\mathrm{I}}_G=\left(b^{-1}{oa}^{-1}\right)\left(aob\right)$$$$\Rightarrow \left(aob\right)o\left(b^{-1}{oa}^{-1}\right)=ao\left({bob}^{-1}\right){oa}^{-1}$$$$=\left(ao{\mathrm{I}}_G\right){oa}^{-1}={aoa}^{-1}={\mathrm{I}}_G$$$$\mathrm{Similarly}\mathrm{you}\mathrm{can}\mathrm{prove}\mathrm{the}\mathrm{other}$$$$\mathrm{part}.$$$$[\mathrm{Here}\mathrm{it}\mathrm{is}\mathrm{not}\mathrm{needed}\mathrm{to}\mathrm{prove}\mathrm{that}$$$${aoa}^{-1}={bob}^{-1}={\mathrm{I}}_G,\mathrm{identity}\mathrm{functions}$$$$\mathrm{on}G]$$

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