Let $X$ be a non-empty set and $W$ be a set of permutations of $X$. Assume we are given a set $\mathscr{R}$ of equivalence relations on $X$, a fixed element $x_0\in X$ and a map $\varphi:\mathscr{R}\to W$, $H\mapsto s_H$. Define $$ \mathscr{R}_0 = \{H\in\mathscr{R} \, | \, s_H(x_0)\equiv x_0 (\operatorname{mod} H') \text{ for all } H'\in \mathscr{R}\setminus\{H\}\}, $$ and let $S_0 = \{s_H \, | \, H\in\mathscr{R}_0\}$. We also assume that
- For each $H\in\mathscr{R}$ there are two equivalence classes modulo $H$ permuted by $s_H$ and $s_H^2=1$.
- For each $H\in\mathscr{R}$ and each $w\in W$ we have that $w(H)\in \mathscr{R}$ and $s_{w(H)} = ws_Hw^{-1}$.
- For each $w\neq 1$ in $W$, the set $\{H\in\mathscr{R}\, | \, w(x_0)\not\equiv x_0 (\operatorname{mod} H)\}$ is finite and intersects $\mathscr{R}_0$.
The aim is to show that $(W,S_0)$ is a Coxeter system. For this the first thing I need to prove is that $W$ is a group and that it is generated by $S_0$. Once we know this, we have to use (as Bourbaki's book suggests) Proposition 6 of no.7 (in Section 1 of Chapter IV). For this I defined $$ P_H = \{w\in W \, | \, w(x_0) \equiv x_0 (\operatorname{mod} H)\}, \quad \text{for each } H\in\mathscr{R}_0, $$ and showed that it satisfies conditions (A'), (B') and (C) of the proposition mentioned above. Condition (A') is obvious, condition (B') follows from 3. and condition (C) I believe follows from 2. My first question is if this is the correct approach for solving the exercise.
The real problem for me is I don't know how to prove that $W$ is a group and that it is generated by $S_0$, so my second question is on how to show that $W$ is the group generated by $S_0$.
EDIT: It is not possible to show that $W$ is a group from the information given, as if $W$ is a set of permutations satisfying the conditions, then so is $W\setminus\{1\}$. Thus we need to assume that $W$ is a group. Now, let $w\in W$ and write $$ \mathscr{R}(w) = \{ H\in\mathscr{R} \, | \, w(x_0)\not\equiv x_0 (\operatorname{mod} H) \}. $$ Let $n(w) = |\mathscr{R}(w)|$. Note that if $w=1$ then $n(w)=0$. I was suggested to show by induction on $n(w)$ that $w$ can be written as a product of elements in $S_0$. The case $n(w)=0$ is trivial, so I only need to care for the inductive step. It is easy to see (using condition 2.) that if $H\in \mathscr{R}(w)\cap \mathscr{R}_0$ (such $H$ exists by condition 3.) then the map $H'\mapsto s_H(H')$ defines a bijection $$ \mathscr{R}(w)\setminus\{H\} \cong \mathscr{R}(s_Hw) \setminus\{s_H(H)\}. $$ Thus in order to complete the induction the only thing I need to prove is that $s_H(H)\not\in \mathscr{R}(s_Hw)$ for at least one $H\in \mathscr{R}(w)\cap \mathscr{R}_0$. How can we prove this claim? Is there another way to solve this?