Exercise 3.1.34 ( $D_{2n}/\langle r^k \rangle \simeq D_{2k}$)

Proof. Let

(a)

Put $N=⟨r^k⟩$, where $k>0,k\mid n$.

By Exercise 29, since $G=⟨r,s⟩$ and $N=⟨r^k⟩$, it suffices to verify $r{r}^k{r}^{-1}\in N$ and $s{r}^k{s}^{-1}\in N$:

$$\begin{array}{llll}\hfill r{r}^k{r}^{-1}& =r^k\in N,& \hfill & \\ \hfill s{r}^k{s}^{-1}& =r^{-k}\in N.& \hfill & \end{array}$$

Therefore

$$⟨r^k⟩⊴D_{2n}.$$

(b)

Since $k\mid n$ ( $k>0$), the order of $r^k$ is $|r|=n∕k$, so $|⟨r^k⟩|=n∕k$. Therefore $$|D_{2n}∕⟨r^k⟩|=2n∕(n∕k)=2k.$$

Let $\bar{x}=\mathit{𝑥𝐻}$ denote the elements of $D_{2n}∕⟨r^k⟩$. Then $D_{2n}∕⟨r^k⟩=⟨\bar{r},\bar{s}⟩$ and

$${\bar{r}}^k={\bar{s}}^2=\bar{1},\bar{r}\bar{s}=\bar{s}{\bar{r}}^{-1}.$$

Since $D_{2k}=⟨\rho ,\sigma \mid {\rho }^k={\sigma }^2=1,\mathit{𝜌𝜎}=\sigma {\rho }^{-1}⟩$, by van Dyck’s Theorem, there is surjective homomorphism $\phi :D_{2k}\to D_{2n}∕⟨r^k⟩$ such that $\phi (\rho )=\bar{r}$ and $\phi (\sigma )=\bar{s}$. Moreover, $|D_{2k}|=2k=|D_{2n}∕⟨r^k⟩|$, so $\phi$ is an isomorphism, and

$$D_{2n}∕⟨r^k⟩\simeq D_{2k}.$$