Editing Cayley's theorem (section)

== Proof of the theorem ==
If ''g'' is any element of a group ''G'' with operation ∗, consider the function {{nowrap|''f''<sub>''g''</sub> : ''G'' → ''G''}}, defined by {{nowrap|1=''f''<sub>''g''</sub>(''x'') = ''g'' ∗ ''x''}}. By the existence of inverses, this function has also an inverse, <math>f_{g^{-1}}</math>. So multiplication by ''g'' acts as a [[bijective]] function. Thus, ''f''<sub>''g''</sub> is a permutation of ''G'', and so is a member of Sym(''G'').

The set {{nowrap|1=''K'' = {''f''<sub>''g''</sub> : ''g'' ∈ ''G''} }} is a subgroup of Sym(''G'') that is isomorphic to ''G''. The fastest way to establish this is to consider the function {{nowrap|''T'' : ''G'' → Sym(''G'')}} with {{nowrap|1=''T''(''g'') = ''f''<sub>''g''</sub>}} for every ''g'' in ''G''. ''T'' is a [[group homomorphism]] because (using · to denote composition in Sym(''G'')):

:<math> (f_g \cdot f_h)(x) = f_g(f_h(x)) = f_g(h*x) = g*(h*x) = (g*h)*x = f_{g*h}(x) ,</math>
for all ''x'' in ''G'', and hence:
:<math> T(g) \cdot T(h) = f_g \cdot f_h = f_{g*h} = T(g*h) .</math>
The homomorphism ''T'' is [[injective]] since {{nowrap|1=''T''(''g'') = id<sub>''G''</sub>}} (the identity element of Sym(''G'')) implies that {{nowrap|1=''g'' ∗ ''x'' = ''x''}} for all ''x'' in ''G'', and taking ''x'' to be the identity element ''e'' of ''G'' yields {{nowrap|1=''g'' = ''g'' ∗ ''e'' = ''e''}}, i.e. the kernel is trivial. Alternatively, ''T'' is also [[injective]] since {{nowrap|1=''g'' ∗ ''x'' = ''g''′ ∗ ''x''}} implies that {{nowrap|1=''g'' = ''g''′}} (because every group is [[cancellative]]).

Thus ''G'' is isomorphic to the image of ''T'', which is the subgroup ''K''.

''T'' is sometimes called the ''[[regular representation]] of'' ''G''.

=== Alternative setting of proof ===
An alternative setting uses the language of [[Group action (mathematics)|group action]]s. We consider the group <math>G</math> as acting on itself by left multiplication, i.e. <math>g \cdot x = gx</math>, which has a permutation representation, say <math>\phi : G \to \mathrm{Sym}(G)</math>.

The representation is faithful if <math>\phi</math> is injective, that is, if the kernel of <math>\phi</math> is trivial. Suppose <math>g\in\ker\phi</math>. Then, <math>g = ge = g\cdot e = e</math>. Thus, <math>\ker\phi</math> is trivial. The result follows by use of the [[first isomorphism theorem]], from which we get <math>\mathrm{Im}\, \phi \cong G</math>.