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.