equivalence relation proof

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September 3rd 2011, 04:39 AM
shelford

equivalence relation proof

Hi, I am stuck and need some help.

Let G be a group. For $x,y \in G$ , $x \sim y$ iff $x=y^{\pm 1}$ . Prove that $\sim$ is an equivalence relation on G.

Thanks for any help in advance.
September 3rd 2011, 05:18 AM
FernandoRevilla

Re: equivalence relation proof

Quote:

Originally Posted by shelford

Let G be a group. For $x,y \in G$ , $x \sim y$ iff $x=y^{\pm 1}$ . Prove that $\sim$ is an equivalence relation on G.

I suppose you meant $x\sim y\Leftrightarrow (x=y)\;\vee \;(x=y^{-1})$ . For example : Symmetric: If $x\sim y$ then, $x=y$ or $x=y^{-1}$ . (i) If $x=y$ then $y=x$ i.e. $y\sim x$ (ii) If $x=y^{-1}$ then, taking inverses $y=x^{-1}$ i.e. also $y\sim x$ . Try the Reflexive and Transitive.

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