Proof of Permutations of an Initial Segment Form a Group under Composition

Claim 1. For every $k\in[n]$ the element $k$ is the one and only element of $[n]$ whose image under $\mathrm{id}$ is $k$, so $\mathrm{id}$ is a bijection and lies in $S_{n}$ by Permutation of the Set $\{1,\dots,r\}$. The identities $\sigma\circ\mathrm{id}=\mathrm{id}\circ\sigma=\sigma$ hold because both sides send $k$ to $\sigma(k)$.

Claim 2. If $\sigma$ and $\tau$ are bijections from $[n]$ to $[n]$, then $\sigma\circ\tau$ is a bijection by claim 2 of Injectivity, Composition, and Restriction of Bijections, hence lies in $S_{n}$. Associativity holds because both $(\rho\circ\sigma)\circ\tau$ and $\rho\circ(\sigma\circ\tau)$ send $k$ to $\rho\bigl(\sigma(\tau(k))\bigr)$.

Claim 3. Let $\sigma\in S_{n}$. By claims 1 and 2 of Inverse of a Bijection, applied with $X=Y=[n]$, there is exactly one map $\sigma^{-1}:[n]\to[n]$ with $\sigma^{-1}\circ\sigma=\mathrm{id}$ and $\sigma\circ\sigma^{-1}=\mathrm{id}$, this map is a bijection, so it lies in $S_{n}$, and $(\sigma^{-1})^{-1}=\sigma$.

Claim 4. By claims 2 and 3 the maps $R_{\tau}(\sigma)=\sigma\circ\tau$ and $R_{\tau^{-1}}(\rho)=\rho\circ\tau^{-1}$ do send $S_{n}$ into $S_{n}$. For $\sigma\in S_{n}$, using associativity, claim 3 and claim 1,

and symmetrically $R_{\tau}\bigl(R_{\tau^{-1}}(\rho)\bigr)=\rho$ for every $\rho\in S_{n}$. So $R_{\tau}$ and $R_{\tau^{-1}}$ are two-sided inverses of one another as maps of $S_{n}$, and claim 3 of Inverse of a Bijection shows that $R_{\tau}$ is a bijection from $S_{n}$ onto $S_{n}$.

The map $\iota:S_{n}\to S_{n}$ with $\iota(\sigma)=\sigma^{-1}$ takes values in $S_{n}$ by claim 3, and $\iota(\iota(\sigma))=(\sigma^{-1})^{-1}=\sigma$ by claim 3 as well, so $\iota$ is a two-sided inverse of itself; claim 3 of Inverse of a Bijection again shows that $\iota$ is a bijection from $S_{n}$ onto $S_{n}$.