The symmetric group has the Coxeter presentation

Fix $n\ge 2$ and let $G_n$ be the group given by the displayed generators and relations. The adjacent transpositions ${\tau }_i=(ii+1)$ satisfy ${\tau }_i^2=1$. Transpositions with disjoint supports commute, and the two length-three words in ${\tau }_i,{\tau }_{i+1}$ both exchange $i$ with $i+2$ and fix $i+1$. Thus every relator holds in $S_n$. By [F2] there is a homomorphism ${\pi }_n:G_n\to S_n$ sending $s_i$ to ${\tau }_i$, and [F1] makes it surjective.

Put $H=⟨s_1,\dots ,s_{n-2}⟩\le G_n$, with $H=\{1\}$ when $n=2$. For $1\le j<n$ set $r_j=s_{n-1}{s}_{n-2}\cdots s_j$, and set $r_n=1$. The defining relations give the following right-multiplication rules for $1\le i<n$: if $i\le j-2$, then $r_j{s}_i=s_i{r}_j$ by commuting $s_i$ past every factor of $r_j$; if $i=j-1$, then $r_j{s}_i=r_{j-1}$; if $i=j$, then $r_j{s}_i=r_{j+1}$ by $s_j^2=1$; and if $i>j$, then $r_j{s}_i=s_{i-1}{r}_j$. For the last rule commute the final $s_i$ left past $s_j,\dots ,s_{i-2}$, replace $s_i{s}_{i-1}{s}_i$ by $s_{i-1}{s}_i{s}_{i-1}$, and commute the leading $s_{i-1}$ left past $s_{i+1},\dots ,s_{n-1}$. For $j=n$, the rules are $r_n{s}_i=s_i{r}_n$ when $i\le n-2$ and $r_n{s}_{n-1}=r_{n-1}$. Every coefficient on the left of a resulting $r_k$ lies in $H$.

The union $T={\bigcup }_{j=1}^{n}H{r}_j$ contains $1=r_n$ and, by step 1.2, is stable under right multiplication by every generator $s_i$. Since each $s_i$ is its own inverse, every word in the generators lies in $T$, so $G_n=T$ and $\mathrm{\mid }{G}_n\mathrm{\mid }\le n\mathrm{\mid }H\mathrm{\mid }$. The defining relations of the presented group $G_{n-1}$ hold among $s_1,\dots ,s_{n-2}$, hence [F2] gives a surjection $G_{n-1}\twoheadrightarrow H$; this includes $G_1=\{1\}$ when $n=2$. Induction yields $\mathrm{\mid }{G}_n\mathrm{\mid }\le n!$. Independently, a bijection of an $n$-element set has $n$ choices for the first image, then $n-1$ for the second, and so on, hence $\mathrm{\mid }{S}_n\mathrm{\mid }=n!$ by [F3]. The surjection ${\pi }_n$ of step 1.1 is therefore a bijection and an isomorphism. Transport it through [L1] to obtain exactly the stated presentation on the library's underlying set.

If $n=0$ or $1$, there are no adjacent transpositions and the only bijection of the underlying set is the identity. Its group is presented by the empty set of generators and relators.