Symmetric group on a finite set is a Coxeter group

This article states and proves a fact about a particular group, or kind of group, (i.e., Symmetric group (?)) having a particular presentation (or kind of presentation), i.e., where the generic name for such presentations or groups having such presentations is: Coxeter presentation (?)
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Statement

Suppose $n$ is a nonnegative integer and $G$ is the symmetric group on the set $\{ 1,2,3, \dots, n+1 \}$. Then, $G$ is isomorphic to a Coxeter group with $n$ generators $s_i, 1 \le i \le n$, where $m_{i(i+1)} = 3$ and $m_{ij} = 2$ for $i$ and $j$ differing by more than one.

In other words:

$G \cong \langle s_i \mid s_i^2 = 1, (s_is_{i+1})^3 = 1, (s_is_j)^2 = 1 \ \forall \ |i - j| > 1 \rangle$.

The isomorphism identifies $s_i$ with the transposition $(i,i+1)$.

Facts used

1. Transpositions of adjacent elements generate the symmetric group on a finite set

Proof

The Coxeter group described admits a surjective homomorphism to the symmetric group

Consider a map:

$\langle s_i \mid s_i^2 = 1, (s_is_{i+1})^3 = 1, (s_is_j)^2 = 1 \ \forall \ |i - j| > 1 \rangle \to G$

given by:

$s_i \mapsto (i,i+1)$.

To check that the map is well-defined, we need to check that all the Coxeter relations are satisfied by the images of the $s_i$. This is direct: the square of any transposition is the identity element, the product of two adjacent transpositions has order three, and the product of two disjoint transpositions has order two.

Next, the map is surjective because by fact (1), elements of the form $(i, i+1)$ generate $G$.

The size of the Coxeter group is at most $(n+1)!$

Let $C_n$ be the Coxeter group for $n$ letters.

First observe that the subgroup $H$ of $C_n$ generated by $s_1,s_2, \dots, s_{n-1}$ satisfies all the relations for the Coxeter group for $n-1$ generators, hence is a homomorphic image of $C_{n-1}$. Hence, by induction, $|H| \le |C_{n-1}| \le n!$.

We now show that $H$ has at most $n+1$ left cosets in $C_n$. In fact, we can show that the every left coset of $H$ in $C_n$ has a representative among these:

$s_1s_2 \dots s_n, s_2 \dots s_n, s_3 \dots s_n, \dots s_{n-1}s_n, s_n$.

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Thus, $[C_n:H] \le n + 1$. Using Lagrange's theorem, we get $|C_n| = |H|[C_n:H] \le (n+1)n! = (n+1)!$.