Lemma 6

For every $x\in X$, the group N acts faithfully and transitively on the fibre $\overset{-1}{p}(x)$.

Since $p$ is surjective, the fibre $\overset{-1}{p}(x)$ is not empty. Let $y, y'\in$ $\overset{-1}{p}(x)$; let us prove that there exists a unique element $n\in$ N such that $n\cdot y=y'$. Let $g, h\in$ F and let $u, v\in$ M be such that $y=\pi (g, u)$ and $y'=\pi (h, v)$. Put $s=\varphi (g^{-1}h)$. Since $x=\varphi (g)\cdot u=\varphi (h)\cdot v$, we have $u=s\cdot v$, whence $s\in S$. It follows that $\varphi (h) =\varphi (gx_s)$, so that there exists $n\in N$ such that $h=ngx_s$. Then,

Let $n'$ be an element of N such that $n'\cdot y=y'$. We have $n'\cdot \pi (g, u) =n\cdot \pi (g, u)$, whence $\pi (n'g, u) =\pi (ng, u)$. Consequently, there exists $t\in S$ such that $n'g=$ $ngx_t$; this relation implies that $\varphi (x_t) =e$, whence $t=e$ and $n=n'$.