Cayley's Theorem

Cayley’s Theorem:
Every group is isomorphic to a permutation group.

Proof: Let $G$ be a finite groups of order $n$. If $a \in G$, then $\forall \,\,x \in G$, $ax \in G$. Now consider a function from $G$ into $G$ defined by

For $x,y \in G,\,{f_a}\left( x \right) = {f_a}\left( y \right) \Rightarrow ax = ay \Rightarrow x = y$ Therefore, function ${f_a}$ is one-one.

The function ${f_a}$ is also onto because if $x$ is any element of  then there exist an element ${a^{ - 1}}x$ such that

Thus ${f_a}$ is one-one from $G$ onto$G$. Therefore, ${f_a}$ is a permutation on $G$. Let $G'$ denotes the set of all such one-to-one functions defined on $G$ corresponding to every element of $G$, i.e. $G' = \left\{ {{f_a}:a \in G} \right\}$

Now, we show that $G'$ is a group with respect to the product of functions.

(i) Closure Axiom: Let ${f_a},{f_b} \in G'$ where$a,b \in G$, then

Since $ab \in G$, therefore ${f_{ab}} \in G'$ and thus $G'$ is closed under the product of functions.

(ii) Associative Axiom: Let ${f_a},{f_b},{f_c} \in G'$ where $a,b,c \in G$, then

Product of functions is associative in $G'$.

(iii) Identity Axiom: If $e$ is the identity element in $G$, then ${f_e}$ is the identity of $G'$ because $\forall \,\,{f_x} \in G'$ we have ${f_e} \circ {f_x} = {f_{ex}} = {f_x}$ and ${f_x} \circ {f_e} = {f_{xe}} = {f_x}$.

(iv) Inverse Element: If ${a^{ - 1}}$ is the inverse of $a$ in $G$, then ${f_{{a^{ - 1}}}}$ is the inverse of ${f_a}$ in $G'$ because ${f_{{a^{ - 1}}}} \circ {f_a} = {f_{{a^{ - 1}}a}} = {f_e}$ and ${f_a} \circ {f_{{a^{ - 1}}}} = {f_{a{a^{ - 1}}}} = {f_e}$

Hence $G'$ is a group with respect to composite of functions denoted by the symbol $\circ$.

Now consider the function $g$ and $G$ into $G'$ defined by $g\left( a \right) = {f_a}\,\,\,\forall a \in G$.
$g$ is one-one because for $a,b \in G$.

$g$ is onto because if ${f_a} \in G'$ then for $a \in G$, we have $g\left( a \right) = {f_a}$
$g$ preserves composition in $G$ and $G'$ because if  $a,b \in G$ then

Hence $G \cong G'$.