Lemma 3

The group G is generated by S.

Let H be the subgroup of G generated by S. Let $U = H\cdot M$; observe that U is the union of the subsets of the form $h\cdot (S\cdot M)$, for $h\in H$. Since $S\cdot M$ is a neighbourhood of M (lemma 2), the set U is a neighbourhood of $H\cdot M = U$; it follows that U is open in X. Let $g, g'\in G$ be such that $g\cdot U\cap g'\cdot U=\not\emptyset$; let $h, h'\in H$ and $x, x'\in M$ be such that $gh\cdot x=g'h'\cdot x'$; then $h^{-1}g^{-1}g'h'\cdot x'=x$, so that $h^{-1}g^{-1}g'h'\in S$; in particular, $g^{-1}g'\in H$. When $g$ runs through a system of representatives of the left classes modulo H, the sets $g\cdot U$ are thus pairwise disjoint; since $G\cdot M = X$, they cover X. As X is connected, it follows that (G : H) = 1, whence H = G.

Let T be the set of pairs $(s, t)$ of elements of G such that $M\cap s\cdot$ $M\cap st\cdot M=\not\emptyset$; if $(s, t)\in T$, then $s,t$ and $st$ belong to S. Let F be the group $F(S,\mathbf{r})$ defined by the generating set S and by the set $\mathbf{r}$ of relators $str^{-1}$ for $r, s, t\in S$ such that $(s, t)\in T$ and $r=st$; let us denote by $\varepsilon$ its identity element and by $\varphi : F\rightarrow G$ the canonical homomorphism (A, I, p. 86, prop. 9). If $s\in S$, we shall denote by $x_s$ the element of F, image of $s$ under the canonical mapping of S into F; for every $s\in S$, one has $\varphi (x_s) =s$. The homomorphism $\varphi$ is therefore surjective (lemma 3). For $(s, t)\in T$ and $r=st$, one has $x_r=x_sx_t$; hence $x_e=\varepsilon$, since $(e, e)\in T$; for every $s\in S$, one has also $x_{s^{-1}}=x^{-1}_s$ since $(s, s^{-1})\in T$.

Endow the set F with the discrete topology and let Z denote the topological space $F\times M$, endowed with the operation of F given by $(\gamma ,(g, x))\mapsto$ $(\gamma g, x)$, for $\gamma$ and $g\in$ F and $u\in$ M. Let $(g_1, u_1)$ and $(g_2, u_2)$ be elements of Z; we shall say that $(g_1, u_1)$ is congruent to $(g_2, u_2)$ if there exists $s\in S$ such that $g_2=g_1\cdot x_s$ and $s\cdot u_2=u_1$.