Permutations of an Initial Segment Form a Group under Composition

Let $n$ be a natural number, let $[n]$ be the initial segment determined by $n$, that is, the set $\{1,\dots,n\}$, and let $S_{n}$ be the set of permutations of $[n]$, that is, the set of bijections from $[n]$ to $[n]$. For maps $\sigma,\tau:[n]\to[n]$ let $\sigma\circ\tau$ be the map with $(\sigma\circ\tau)(k)=\sigma(\tau(k))$, and let $\mathrm{id}$ be the map with $\mathrm{id}(k)=k$.

Then the following hold.

1. (Identity) $\mathrm{id}\in S_{n}$, and $\sigma\circ\mathrm{id}=\mathrm{id}\circ\sigma=\sigma$ for every $\sigma\in S_{n}$.

2. (Composition) If $\sigma,\tau\in S_{n}$ then $\sigma\circ\tau\in S_{n}$; and $(\rho\circ\sigma)\circ\tau=\rho\circ(\sigma\circ\tau)$ for all $\rho,\sigma,\tau\in S_{n}$.

3. (Inverses) For every $\sigma\in S_{n}$ there is exactly one map $\sigma^{-1}:[n]\to[n]$ with $\sigma^{-1}\circ\sigma=\sigma\circ\sigma^{-1}=\mathrm{id}$; it belongs to $S_{n}$, and $(\sigma^{-1})^{-1}=\sigma$.

4. (Translation and inversion) Let $\tau\in S_{n}$. The map $R_{\tau}:S_{n}\to S_{n}$ with $R_{\tau}(\sigma)=\sigma\circ\tau$ is a bijection from $S_{n}$ onto $S_{n}$, and so is the map $\sigma\mapsto\sigma^{-1}$.