$D_n\cong ⟨r,s\mid r^n,s^2,sr{s}^{-1}r⟩$ for the dihedral group $D_n=⟨\{\rho ,\sigma \}⟩\le \mathrm{Sym}(\mathbb{Z}\mathrm{/}n)$, $n\ge 3$

Direct substitution on $\mathbb{Z}\mathrm{/}n$ gives ${\rho }^n=\mathrm{id}$, ${\sigma }^2=\mathrm{id}$, and $\sigma \rho {\sigma }^{-1}\rho =\mathrm{id}$ because $\sigma \rho \sigma (k)=k-1$; thus the displayed relators hold.

In $P$, the relators give $s^{-1}=s$, $sr=r^{-1}s$, $r^n=e$, and $s^2=e$; moving every $s$ to the right and reducing exponents therefore writes every element as $r^i{s}^{\epsilon }$ with $0\le i<n$ and $\epsilon \in \{0,1\}$.

The permutations ${\rho }^i$ are distinct by their values ${\rho }^i(0)=i$, and the permutations ${\rho }^i\sigma$ are likewise distinct; the two families are disjoint because equality would first force the same $i$ from the value at $0$ and then force $i+1=i-1$ in $\mathbb{Z}\mathrm{/}n$ from the value at $1$, impossible for $n\ge 3$. Step 1.1 gives the same relations among $\rho$ and $\sigma$, so the set $\{{\rho }^i{\sigma }^{\epsilon }:0\le i<n,\epsilon \in \{0,1\}\}$ is closed under products and inverses and contains $\rho$ and $\sigma$; being a subgroup containing $\{\rho ,\sigma \}$, it equals $D_n$ by the minimality in [F1]. Hence [L1] and [L3] give exactly $2n$ elements of $D_n$.

By [L2], construct a homomorphism from the displayed presentation $P$ to $D_n$, sending $r$ to $\rho$ and $s$ to $\sigma$; its image is a subgroup containing $\rho$ and $\sigma$, so [F1] makes it all of $D_n$ and the homomorphism surjective.

Step 1.2 gives at most the same $2n$ normal forms in $P$, and step 2.1 shows that their images under the surjection of step 2.2 are all distinct; therefore that homomorphism is bijective and hence an isomorphism.

The target is the subgroup $D_n=⟨\{\rho ,\sigma \}⟩$ of $\mathrm{Sym}(\mathbb{Z}\mathrm{/}n)$ specified in the Example, which step 2.1 shows has order $2n$; so the displayed isomorphism holds with the stated conventions and boundary $n\ge 3$.