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- Exercise 2.8 (Solutions)
- $G$ is group so $a^{-1}\in G$ such that \begin{align}&{{a}^{-1}}*\left( a*x \right)={{a}^{-1}}*b \\ \R... ={{a}^{-1}}*b\,\,\,\, \text{by identity law.}\end{align} And for \begin{align} & x*a=b \\ \Rightarrow \,\,\,\,& \left( x*a \right)*{{a}^{-1}}=b*{{a}^{-1}} \te... =b*{{a}^{-1}} \,\,\,\,\text{by identity law.}\end{align} </panel> <panel> **Question # 7** Show that th