Proposition 6

Let G be a group, E a homogeneous G-set, $a \in E$, H the stabilizer of a and K a subgroup of G contained in H. There exists one and only one G-morphism f of G/K into E such that $f(e.K) = a$. If K = H, f is an isomorphism.

If f is a solution, then $f(x.K) = x.a$ for all x in G, whence the uniqueness; we show the existence. The orbital mapping defined by a is compatible with the equivalence relation $y \in xK$ on G. For, if $y = xk, k \in K$, then

A mapping f is thus derived of G/K into H which satisfies $f(x.K) = x.a$ for all x in G. This mapping is a G-morphism and $f(K) = a$. This mapping is surjective for its image is a non-empty stable subset of E. Suppose now that K = H and let us show that f is injective. If $f(x.H) = f(y.H)$, then $x.a = y.a$, whence $x^{-1}y.a = a$ and $x^{-1}y \in H$, whence $x.H = y.H$.