Hi, I am stuck and need some help.
Let G be a group. For $\displaystyle x,y \in G$, $\displaystyle x \sim y$ iff $\displaystyle x=y^{\pm 1}$. Prove that $\displaystyle \sim$ is an equivalence relation on G.
Thanks for any help in advance.
Hi, I am stuck and need some help.
Let G be a group. For $\displaystyle x,y \in G$, $\displaystyle x \sim y$ iff $\displaystyle x=y^{\pm 1}$. Prove that $\displaystyle \sim$ is an equivalence relation on G.
Thanks for any help in advance.
I suppose you meant $\displaystyle x\sim y\Leftrightarrow (x=y)\;\vee \;(x=y^{-1})$ . For example : Symmetric: If $\displaystyle x\sim y$ then, $\displaystyle x=y$ or $\displaystyle x=y^{-1}$ . (i) If $\displaystyle x=y$ then $\displaystyle y=x$ i.e. $\displaystyle y\sim x$ (ii) If $\displaystyle x=y^{-1}$ then, taking inverses $\displaystyle y=x^{-1}$ i.e. also $\displaystyle y\sim x$ . Try the Reflexive and Transitive.