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I have two problems involving equivalence relations. I will only ask the first for now, in the hopes of figuring out the second one after this one is clear.

Let $X$ be a set, and $G$ be a subgroup $G \subset S_X$. Consider the relation on $X$ given by $x \equiv_G y$ if there's an element $g \in G$ s.t. $g(x)=y$. Prove that this is an equivalence relation on $X$.

Attempt: Let $E$ be an equivalence relation on $X$. Then, we have:

  1. $(x,y)\in E \Rightarrow (y,y)\in E \Rightarrow (g(x),g(x))\in E$ since $g$ is bijective. So $G$ is reflexive on $X$.

  2. Assume $(g(x),g(y))\in E \Rightarrow (y,x)\in E \Rightarrow (x,y)\in E \Rightarrow (g(y),g(x))\in E$. So $G$ is symmetric.

  3. To prove transitivity, I think I need a second fact such as $y \equiv_G z$. I am not sure.

I am convinced that I have erred in notation at some point. The notation I see varies greatly in this area of equivalence.

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2 Answers 2

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I'm a little confused by your three properties here. Reflexive would mean $$ (x,x)\in E\iff \exists g\in G:g\cdot x=x\\ $$ Take $g=e_G$ for that one. Symmetric: $$ (x,y)\in E\implies\exists g:g\cdot x=y\implies x=g^{-1}\cdot y\iff(y,x)\in E. $$ Transitive: $$ (x,y),(y,z)\in E\implies \exists g,h:g\cdot x=y,h\cdot y=z\implies h\cdot(g\cdot x)=z=(h\cdot g)\cdot x\iff(x,z)\in E. $$

Sorry this answer was so wordless, but is it clear?

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  • $\begingroup$ Yes, this clears up my confusion of looking at bijections as a group. My other issue involves the wording of an equivalence relation "on" something. $\endgroup$
    – user113525
    Feb 14, 2014 at 4:37
  • $\begingroup$ Actually, I have a question about your notation here. Why is it appropriate to multiply $g\in G$ by $x\in X$, when $g$ is a bijection but $x$ is just a set element? $\endgroup$
    – user113525
    Feb 14, 2014 at 4:45
  • $\begingroup$ Since $g$ is an automorphism of $X$, we let $g\cdot x=g(x)$. $\endgroup$
    – Ian Coley
    Feb 14, 2014 at 4:58
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There is a natural action of $G$ on $X$. So $x$ and $y$ are equivalent if and only if $Orbit_G(x)=Orbit_G(y)$. see Ian's answer for the remaining argument.

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