Statement:

  Let \((G,\circ )\) be a group and \(S\) be a non-empty set and \(S\) be a G-Set. Then the action of \(G\) on \(S\) induces a homomorphism from \(G\) onto \(A(S)\), where \(A(S)\) is the group of all permutations of \(S\).

  Proof:
  Given that \((G,\circ )\) is a group and \(S\) is a non-empty set and \(S\) is a G-Set and \(A(S)\) is the group of all permutations of \(S\).
  Let \(\cdot :G\times S \to S\) be an action of \(G\) on \(S\).
  Let \(g\in G\) and let us define a mapping \(\tau_{g}:S\to S\) such that \(\tau_{g}(a)=g\cdot a,~a\in S\).

• To prove \(\tau_{g}\) is well-defined
  Let \(a,b\in S\) such that \begin{align*} & a=b \\ \implies & g\cdot a = g\cdot b~ \big[\because \cdot\text{ is well-defined} \big]\\ \implies & \tau_{g}(a)=\tau_{g}(b) \end{align*} \(\therefore\) \(\tau_{g}\) is well-defined.

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• To prove \(\tau_{g}\) is injective
  Let \(a,b\in S\) such that \begin{align*} & \tau_{g}(a)=\tau_{g}(b) \\ \implies & g\cdot a = g\cdot b\\ \implies & g^{-1}\cdot\big(g\cdot a\big) = g^{-1}\cdot\big(g\cdot b\big)\\ \implies & \big(g^{-1}\circ g\big)\cdot a = \big(g^{-1}\circ g\big)\cdot b\\ \implies & e\cdot a = e\cdot b~ \big[\text{ where e is the identity element of G} \big]\\ \implies & a=b \end{align*} \(\therefore\) \(\tau_{g}\) is injective.

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• To prove \(\tau_{g}\) is surjective
  Let \(c\in S\) then \begin{align*} & c=e\cdot c\\ \implies & c= \big(g\circ g^{-1}\big)\cdot c\\ \implies & c= g\cdot \big(g^{-1} \cdot c\big)\\ \implies & c= \tau_{g} \big(g^{-1} \cdot c\big)\\ \end{align*} \(\therefore g^{-1} \cdot c\) is the pre-image of \(c \) under \(\tau_{g}\). Since \(c \) is an arbitrary element of \(S\) then each member of \(S\) has a pre-image in \(S\), \(\therefore\) \(\tau_{g}\) is surjective.

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  \(\therefore\) \(\tau_{g}\) is bijective \(\implies \tau_{g}=A(S)\).
  Now let us define a mapping \(\psi:G\to A(S) \) such that \(\psi(g)=\tau_{g}~,~g\in G\).

• To prove \(\psi\) is well-defined.
  Let \(g_{1},g_{2} \in G\) such that \( g_{1}= g_{2} \). Let \(s\in S \) \begin{align*} & \tau_{g_{1}}(s)=g_{1}\cdot s~\forall s\in S \\ \implies & \tau_{g_{1}}(s)=g_{2}\cdot s~\forall s\in S \\ \implies & \tau_{g_{1}}(s)=\tau_{g_{2}} (s)~\forall s\in S\\ \implies & \tau_{g_{1}}=\tau_{g_{2}} \\ \implies & \psi(g_{1})=\psi(g_{2}) \end{align*} \(\therefore\) \(\psi\) is well-defined.

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• To prove \(\psi\) is surjective.
  Let \(\tau_{g}\in A(S)\) for some \(g\in G\) then \(\psi(g)=\tau_{g} \). Therefore \(\tau_{g}\) has a pre-image \(g\in G\). Since \(\tau_{g}\) is an arbitrary element of \(A(S)\), then each member of \(A(S)\) has a pre-image in \(G\).
  \(\therefore\) \(\psi\) is surjective.

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• To prove \(\psi\) is homomorphism.
  Let \(g_{1},g_{2} \in G\) then \begin{align*} \psi\big(g_{1}\circ g_{2} \big)=\tau_{g_{1}\circ g_{2}} \end{align*} Let \(s\in S\) then \begin{align*} &\tau_{g_{1}\circ g_{2}}(s)=\big(g_{1}\circ g_{2} \big)\cdot s~\forall~s\in S\\ \implies & \tau_{g_{1}\circ g_{2}}(s)=g_{1}\cdot \big(g_{2} \cdot s\big)~\forall~s\in S\\ \implies & \tau_{g_{1}\circ g_{2}}(s)=\tau_{g_{1}} \big(g_{2} \cdot s\big)~\forall~s\in S\\ \implies & \tau_{g_{1}\circ g_{2}}(s)=\tau_{g_{1}} \big(\tau_{g_{2}}( s)\big)~\forall~s\in S\\ \implies & \tau_{g_{1}\circ g_{2}}(s)=\big(\tau_{g_{1}}\circ \tau_{g_{2}}\big)( s)~\forall~s\in S\\ \implies & \tau_{g_{1}\circ g_{2}}=\tau_{g_{1}}\circ \tau_{g_{2}}\\ \implies & \psi\big(g_{1}\circ g_{2} \big)=\psi\big(g_{1} \big)\circ \psi\big(g_{2} \big)\\ \end{align*} \(\therefore\) \(\psi\) is homomorphism.

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  Hence the theorem is proved.