> [!NOTE] Theorem (Group Presentation of Dihedral Group) > Let $n\geq 3$. Then [[Dihedral Group|dihedral group]] $D_{2n}$ has the [[Group Presentation|group presentation]] $\langle r,s \mid r^n = 1, s^2 = 1, srs = r^{-1} \rangle.$ ###### Proof of Group Presentation of Dihedral Group \[MA268\] By definition, the $D_{2n}$ is the group of symmetries of the regular $n$-gon. Label its vertices $1$ through $n$. Let $r\in D_{2n}$ denote the anticlockwise rotation through $2\pi/n$ and $s\in D_{2n}$ denote the reflection in the line of symmetry through vertex $1$. The first two relations are clear. NTS the last one. Note that $s^{-1}=s.$ Thus $srs=srs^{-1}.$ Since we know that $R$ is normal in $D_{2n}$ and $r\in R,$ we have that $srs\in R.$ Hence $srs=r^k$ for some natural number $k.$ Let $n\geq3.$ Recall that $D_2n$ has elements $r,s$ that satisfy $r^n=s^2=\mathrm{id},\quad srs=r^{-1}.$ If you've forgotten these facts see once again Theorem V.1.1 and Lemma V.1.4. For now let Moreover, (V.16) $D_{2n}=\{\mathrm{id},r,r^2,\ldots,r^{n-1}\}\cup\{s,sr,sr^2,\ldots,sr^{n-1}\}.$ $G=\langle a,\:b\mid a^n=b^2=\mathrm{id},\:bab=a^{-1}\rangle.$ We will show that $G$ is in fact isomorphic to $D_2n.$ Define $\phi:G\to D_{2n},\quad\phi(a)=r,\quad\phi(b)=s.$ Since $r,s$ satisfy (V.16), the fundamental theorem of group presentations (Theorem V.4.1) tells us that $\phi$ is a well-defined homomorphism Moreover, $r,s\in\operatorname{Im}(\dot{\phi}).$ As Im$(\phi)$ is a subgroup we find that $D_{2n}=\{\mathrm{id},r,r^2,\ldots,r^{n-1}\}\cup\{s,sr,sr^2,\ldots,sr^{n-1}\}\subseteq\mathrm{Im}(\phi).$ However, Im$(\phi)\subseteq D_2n.$ Hence Im$(\phi)=D_2n$, so $\phi$ is surjective. In particular, $\#G\:\geq\:\#\operatorname{Im}(\phi)\:=\:\#D_{2n}\:=\:2n.$ Now think about the elements of $G.$ An element of $W\in G$ is a word (i.e. a fnite product) of $a,a^-1,b,b^{-1}$ in any order. As $a^-1=a^{n-1}$, $b^{-1}=b$ we may suppose that each exponent of $a,b$ in the word $W$ is non-negative. $\bar{\text{Since }bab^{-1}}=a^{-1}$ we have $ba=a^-1b=a^{n-1}b.$ Thus we may swap powers of $a$ and powers of $b$ in $W$ so that all $a$ appear on the left and all $b$ appear on the right. Thus $W=a^ub^v$ where $u,v\geq0.$ By division with remainder $u=nk+r,\quad v=2\ell+s,\quad0\leq r\leq n-1,\quad0\leq s\leq1.$ Hence $W=(a^{n})^{k}\cdot a^{r}\cdot(b^{2})^{\ell}\cdot b^{s}=a^{s}b^{s},\quad0\leq r\leq n-1,\quad0\leq s\leq1.$ There are at most $2n$ possibilities for $W\in G.$ Hence $\#G\leq2n.$ But we already noted that $\#G\leq2n$ so $\#G=2n.$ Since $\phi:G\to D_2n$ is surjective, and $\#G=\#D_{2n}=2n$ we see that $\phi$ is injective. Hence $\phi$ is an isomorphism of $G$ to $D_{2n}$ that takes $a$ to $r$ and $b$ to $s.$ Now that we've established this isomorphism we simply identify $D_2n$ with $G,a$ with $r$ and $b$ with $s$ and just write $D_{2n}=\langle r,\:s\:|r^n=s^2=\mathrm{id},\:srs=r^{-1}\rangle.$ You should memorise this presentation for $D_{2n}.$ It's very useful.