The following three steps suffice.
The morphisms L and R are injective.
Each group is isomorphic with a permutation group.
The kernel of the morphism C is the
center of G.
If
Lg = Lh, then, in
particular,
Lg(e)
= Lh(e). But
Lg (e) = g e = g and
Lh (e) = h e = h. Therefore
g = h. It follows
that the morphism L : G -> Sym(G) is injective.
The proof for R is similar.
If
Cg = Ce, then, for
all x 
G, we have
Lg(x)
= Le(x), so
gxg-1
= x.
Thus, g belongs to the kernel of C if and only if, for all
x G,
we have gx
= xg, that is, if and only if g Z(G).
G, we have
Lg(x)
= Le(x), so
gxg-1
= x.
Thus, g belongs to the kernel of C if and only if, for all
x G,
we have gx
= xg, that is, if and only if g Z(G).
As L is injective, the group G is isomorphic with
its image {Lg | g G}
in Sym(G). The image is a permutation group.
G}
in Sym(G). The image is a permutation group.