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

{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, 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

{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 ontoG. 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' wherea,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}

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\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'.