# equivalence relation proof

• Sep 3rd 2011, 04:39 AM
shelford
equivalence relation proof
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.
• Sep 3rd 2011, 05:18 AM
FernandoRevilla
Re: equivalence relation proof
Quote:

Originally Posted by shelford
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.

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.