Statement:

  Let \((G,\circ )\) be a group and \(S\) be a non-empty set and \(S\) be a G-Set. Then prove that the relation \(\rho\) on \(S\) defined by \(\forall~a,b \in S \), \(a\rho b\iff g\cdot a=b \) for some \(g\in G\) is an equivalence relation.

  Proof:
  Given that \((G,\circ )\) is a group and \(S\) is a non-empty set and \(S\) is a G-Set.
  Let \(\cdot :G\times S \to S\) be an action of \(G\) on \(S\).
  Let \(\rho\) be a relation on \(S\) defined by \(\forall~a,b \in S \), \(a\rho b\iff g\cdot a=b \) for some \(g\in G\).
  To prove that \(\rho\) is an equivalence relation.

• To prove \(\rho\) is reflexive.
  Let \(a\in S\).
  And \(e\in G\) be the identity element. Then we have \(e\cdot a=a\).
  \(\implies a \rho a \)
  \(\therefore\) \(\rho\) is reflexive.

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• To prove \(\rho\) is symmetric.
  Let \(a,b\in S\) such that \( a \rho b \).
  Then there exists an element \(g\in G\) such that \begin{align*} & b=g\cdot a\\ \implies & g^{-1}\cdot b=g^{-1}\cdot\big(g\cdot a\big)\\ \implies & g^{-1}\cdot b=\big(g^{-1}\circ g\big)\cdot a\\ \implies & g^{-1}\cdot b=e\cdot a\\ \implies & g^{-1}\cdot b=a\\ \implies & b\rho a ~\big[\because g^{-1}\in G] \end{align*} \(\therefore\) \(\rho\) is symmetric.

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• To prove \(\rho\) is transitive.
  Let \(a,b,c\in S\) such that \( a \rho b \) and \( b \rho c \).
  Then there exists an element \(g_{1},g_{2}\in G\) such that \(b=g_{1}\cdot a\) and \(c=g_{2}\cdot b\). Now \begin{align*} & c=g_{2}\cdot b\\ \implies & c=g_{2}\cdot \big( g_{1}\cdot a \big)\\ \implies & c=\big( g_{2}\circ g_{1}\big) \cdot a \\ \implies & a\rho c ~\big[\because g_{2}\circ g_{1}\in G\big] \end{align*} \(\therefore\) \(\rho\) is transitive.

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  Hence \(\rho\) is an equivalence relation.