Question Number 76328 by arkanmath7@gmail.com last updated on 26/Dec/19
Commented by arkanmath7@gmail.com last updated on 26/Dec/19
$${sol}.\:{plz} \\ $$
Commented by kaivan.ahmadi last updated on 26/Dec/19
$$\left.\mathrm{1}\right)\:{aa}^{−\mathrm{1}} =\mathrm{1}\in{H}\Rightarrow{a}\equiv{a}^{−\mathrm{1}} \\ $$$$\left.\mathrm{2}\right){if}\:{a}\equiv{b}\Rightarrow{ab}^{−\mathrm{1}} \in{H}\Rightarrow \\ $$$${ba}^{−\mathrm{1}} {ab}^{−\mathrm{1}} =\mathrm{1}\in{H}\Rightarrow{ba}^{−\mathrm{1}} \in{H}\Rightarrow{b}\equiv{a} \\ $$$$\left.\mathrm{3}\right){if}\:{a}\equiv{b}\:{and}\:{b}\equiv{c}\Rightarrow{ab}^{−\mathrm{1}} \in{H}\:{and}\:{bc}^{−\mathrm{1}} \in{H}\Rightarrow \\ $$$$ \\ $$$${ac}^{−\mathrm{1}} ={ab}^{−\mathrm{1}} {bc}^{−\mathrm{1}} \in{H}\Rightarrow{a}\equiv{c} \\ $$$$ \\ $$