Cardinality of a coxeter group Let ${G}$ be a Coxeter group with the next presentation 
\begin{equation}
G = \left\langle s_1,s_2,\cdots,s_{n-1} : (s_is_{i+1})^3=1 , \ (s_is_j)^2=1 \ ,\ |i-j| > 1 \right\rangle
\end{equation}
Where $S=\{s_1,\cdots,s_{n-1}\}$ denote the set of generators of the group
I need to prove that $G \cong S_n $.
For that, I define a function $\phi : S \longrightarrow S_n $ by $\phi(s_i)=(i,i+1)$. it is clear that the images of this function satisfy the relations of the group G, so I can extend $\phi$ to a homomorphism $\widetilde\phi:G \longrightarrow S_n$. This homomorphism is surjective, hence $|G| \geq n!$, so if i can show that $|G|\leq n!$ then $\widetilde\phi$ is a isomorphism. 
My idea is:
I denote by $S_k=\{s_1,\cdots,s_{k} \}$ and $G_k=\left\langle S_k : (s_is_{i+1})^3=1 , \ (s_is_j)^2=1 \ ,\ |i-j| > 1   \right\rangle$. 
I want to prove that $|G_k|\leq (k+1)!$
Some hint?
 A: Since I happen to have a proof of this result in some lecture notes, I will cut and paste it here. You will see that it uses Proposition ???, so you will need to figure out what that says! Note also that permutations are composed from left to right.
Theorem Let $ G_n := \langle X \mid R_1 \cup R_2 \cup R_3 \rangle $, where
\begin{align*}
X & := \{x_1,\ldots,x_{n-1}\} \\
R_1 & := \left\{ x_i^2 = 1 : 1 \leq i \leq n-1 \right\} \\
R_2 & := \left\{ (x_ix_{i+1})^3 = 1 : 1 \leq i \leq n-2 \right\} \\
R_3 & := \left\{ (x_ix_j)^2 = 1 : 1 \leq i < j-1 \leq n-2 \right\}
\end{align*}
Then $ G_n \cong S_n $ by $ x_i \mapsto \tau_i =(i,i+1)$.
Proof By the fundamental theorem, $ x_i \mapsto \tau_i $ for $ 1 \leq i \leq n-1 $ extends to a homomorphism $ \theta : G_n \to S_n $, and in fact $ \theta $ is an epimorphism as $ S_n $ is generated by the $ \tau_i $, so $ |G_n| \geq |S_n| = n! $.
To show that $ |G_n| \leq |S_n| $, we use induction on $ n $. We have $ G_1 =
\langle \varnothing \mid \varnothing \rangle $, so $ G_1 = \{1\} $ and $ |G_1|
= 1 $. For $ n = 2 $, $ G_2 = \langle x_1 \mid x_1^2 \rangle \cong C_2 $, so $
|G_2| = 2 $, which is fine.
Note that $ G_3 = D(2,2,3) \cong D_{6} \cong S_3 $. Now we assume that $ |G_n|
\leq n! $, and aim to show that $ |G_{n+1}| \leq (n+1)! $.
Let $ H := \langle x_2,\ldots,x_n \rangle \leq G_{n+1} $. Since the relations
of $ G_n $ on generators of $ H $ are all in $ R_1 \cup R_2 \cup R_3 $, we may
apply induction to get $ |H| \leq n! $. We now aim to express $ G_{n+1} $ as
a union of $ n+1 $ cosets of $ H $ to get $ |G_{n+1}| \leq (n+1)! $.
Note that the stabiliser of 1 in $S_{n+1}$ is
$ \langle \tau_2,\ldots,\tau_n \rangle $ By the Orbit-Stabiliser
Theorem, the $n+1$ coset representatives of this subgroup in $S_{n+1}$
map 1 to $1,2,\ldots,n+1$.  Since $ 1_{S_{n+1}} : 1 \mapsto 1 $, $ \tau_1 : 1
\mapsto 2 $, $ \tau_1\tau_2 : 1 \mapsto 3 $ etc, we see that $
\{1,\tau_1,\tau_1\tau_2,\ldots,\tau_1\cdots\tau_n\} $ is a set of coset
representatives. So we will try to prove the result by showing that
$\{1,x_1,x_1x_2,\ldots,x_1\cdots x_n\} $ is a set of coset representatives
of $H$ in $G_{n+1}$.
To do that, we will use Proposition ??? with
$ S = Hg_0 \cup Hg_1 \cup \cdots \cup H g_n, $
where $ g_0 = 1 $, $ g_1 = x_1 ,\ldots, g_i = x_1 \cdots x_i $ for
$ 1 \leq i \leq n $. We need to prove that $ g_ix_j \in S $
for $0 \leq i \leq n$ and $1 \leq j \leq n$.
Then Proposition ??? gives $ G = S $, so $ |G| =
|S| \leq (n+1)|H| \leq (n+1)!$, and we are done. There are four cases:


*

*[$ j > i+1 $]. For $k\le i$, we have
$ (x_kx_j)^2 = 1 \Rightarrow x_j^{-1}x_k^{-1} = x_kx_j $,
and so $ x_jx_k = x_kx_j $.
Hence $ g_ix_j = x_1 \cdots x_ix_j = x_jx_1 \cdots x_i = x_jg_i $.
But $ j \geq 2 $, so $ x_j \in H $. Hence $ x_jg_i \in Hg_i \subseteq S $.

*[$ j = i+1 $] We have $ g_ix_j = g_{i+1} \in S $.

*[$ i = j $] This implies that
$ g_ix_i = x_1 \cdots x_{i-1}x_i^2 = x_1 \cdots x_{i-1} = g_{i-1} $ and
$ g_{i-1} \in S $.

*[$ i > j $] Since $g_0=1$, we can assume that $j \ge 1$.
Note that $ (x_jx_{j+1})^3=1 $, so $ x_jx_{j+1}x_j = x_{j+1}x_jx_{j+1} $,
since $ x_j^2 = x_{j+1}^2 = 1 $, so
\begin{align*}
 g_ix_j & = x_1 \cdots x_ix_j \\
 & = x_1 \cdots x_jx_{j+1}x_{j+2} \cdots x_ix_j \\
 & = x_1 \cdots x_jx_{j+1}x_jx_{j+2} \cdots x_i \\
 & = x_1 \cdots x_{j-1}x_{j+1}x_jx_{j+1}x_{j+2} \cdots x_i \\
 & = x_{j+1}x_1 \cdots x_{j-1}x_jx_{j+1} \cdots x_i
 \end{align*}
Since $ j+1 \geq 2 $, $ x_{j+1} \in H $, so $ g_ix_j \in Hg_i \subseteq S $.
So $ g_ix_j \in S$ for all $ 0 \leq i \leq n $ and $1 \leq j \leq n $,
and we are done.
