Cayley’s Theorem

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

Proof: Let $$G$$ be a finite group 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
\[{f_a}\left( x \right)ax\,\,\,\,\,\forall x \in G\]

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

The function $${f_a}$$ is also onto because if $$x$$ is any element of  then there exists an element $${a^{ – 1}}x$$ such that
\[{f_a}\left( {{a^{ – 1}}x} \right) = a\left( {{a^{ – 1}}x} \right) = \left( {a{a^{ – 1}}} \right)x = ex = x\]

Thus $${f_a}$$ is one-one from $$G$$ onto $$G$$. Therefore, $${f_a}$$ is a permutation on $$G$$. Let $$G’$$ denote 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
\[\begin{gathered} \left( {{f_a} \circ {f_b}} \right)x = {f_a}\left[ {{f_b}\left( x \right)} \right] = {f_a}\left( {bx} \right) = a\left( {bx} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {ab} \right)x = {f_{ab}}\left( x \right)\,\,\,\forall x \in G \\ \end{gathered} \]

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
\[\begin{gathered} {f_a} \circ \left( {{f_b} \circ {f_c}} \right) = {f_a} \circ {f_{bc}} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {f_a}\left( {bc} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = f\left( {ab} \right)c \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {f_{ab}} \circ {f_c} = \left( {{f_a} \circ {f_b}} \right) \circ {f_c} \\ \end{gathered} \]

The 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 the 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 $$a,b \in G$$.
\[g\left( a \right) = g\left( b \right) \Rightarrow {f_a} = {f_b} \Rightarrow {f_a}\left( x \right) = {f_b}\left( x \right)\]
\[ \Rightarrow ax = bx \Rightarrow a = b,\,\,\,\forall x \in G\]
$$g$$ 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
\[g\left( {ab} \right) = {f_{ab}} = {f_a} \circ {f_b} = g\left( a \right) \circ g\left( b \right)\]

Hence $$G \cong G’$$.