prove that cayley's theoram
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If g is any element of a group G with operation ∗, consider the function fg : G → G, defined by fg(x) = g ∗ x. By the existence of inverses, this function has a two-sided inverse, {\displaystyle f_{g^{-1}}} f_{{g^{{-1}}}}. So multiplication by g acts as a bijective function. Thus, fg is a permutation of G, and so is a member of Sym(G).
The set K = {fg : g ∈ G} is a subgroup of Sym(G) that is isomorphic to G. The fastest way to establish this is to consider the function T : G → Sym(G) with T(g) = fg for every g in G. T is a group homomorphism because (using · to denote composition in Sym(G)):
{\displaystyle (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),} (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),
for all x in G, and hence:
{\displaystyle T(g)\cdot T(h)=f_{g}\cdot f_{h}=f_{g*h}=T(g*h).} T(g)\cdot T(h)=f_{g}\cdot f_{h}=f_{{g*h}}=T(g*h).
The homomorphism T is injective since T(g) = idG (the identity element of Sym(G)) implies that g ∗ x = x for all x in G, and taking x to be the identity element e of G yields g = g ∗ e = e, i.e. the kernel is trivial. Alternatively, T is also injective since g ∗ x = g′ ∗ x implies that 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.
The set K = {fg : g ∈ G} is a subgroup of Sym(G) that is isomorphic to G. The fastest way to establish this is to consider the function T : G → Sym(G) with T(g) = fg for every g in G. T is a group homomorphism because (using · to denote composition in Sym(G)):
{\displaystyle (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),} (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),
for all x in G, and hence:
{\displaystyle T(g)\cdot T(h)=f_{g}\cdot f_{h}=f_{g*h}=T(g*h).} T(g)\cdot T(h)=f_{g}\cdot f_{h}=f_{{g*h}}=T(g*h).
The homomorphism T is injective since T(g) = idG (the identity element of Sym(G)) implies that g ∗ x = x for all x in G, and taking x to be the identity element e of G yields g = g ∗ e = e, i.e. the kernel is trivial. Alternatively, T is also injective since g ∗ x = g′ ∗ x implies that 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.
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