Proposition 5

With the above notation, G/H is a homogeneous G-set. The mapping $\phi$ induces an isomorphism of $(N/H)^0$ onto the group of automorphisms of the G-set G/H.

The orbit in $G/H$ of the element $e = H$ is $G/H$, whence the first assertion. We now prove the second. If $n \in N$ defines by right translation the identity mapping on $G/H$, then $e.n = e$, that is $H.n = H$, whence $n \in H$. Therefore

N/H operates faithfully on G/H on the right and $\phi$ is injective. The left operation of G and right operation of N/H on G/H commute and hence the operators of N/H define G-morphisms of G/H into itself, which are necessarily G-automorphisms since they are bijective. Therefore $\phi$ takes its values in the group $\Phi$ of G-automorphisms of G/H. We show that the image of $\phi$ is $\Phi$. Let $f \in \Phi$. By transporting the structure, the stabilizer of $f(\dot{e})$ in G is equal to the stabilizer of $\dot{e}$ and hence to H. Let $n \in G$ be such that $f(\dot{e}) = n\dot{e}$. The stabilizer of $n\dot{e}$ in G is $nHn^{-1}$ (no. 2, Proposition 2), whence $nHn^{-1} = H$ and $n \in N$. For every element $xH$ of G/H, $f(xH) = f(x.\dot{e}) = x.f(\dot{e}) = xnH = xHn$ and $f$ coincides with the mapping defined by $n$.