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.

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- Sep 3rd 2011, 04:39 AMshelfordequivalence 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 AMFernandoRevillaRe: equivalence relation proof
I suppose you meant $\displaystyle x\sim y\Leftrightarrow (x=y)\;\vee \;(x=y^{-1})$ . For example :

: 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.__Symmetric__