[Math] Prove $D_{2n}/\langle r^k\rangle$ is isomorphic to $D_{2k}$

Question

Let $D_{2n} =\langle r,s\ |\quad r^n = s^2 = 1, rs = sr^{-1}\rangle$ be the usual presentation of the dihedral group of order $2n$ and let $k$ be a positive integer dividing $n$.

Prove that $D_{2n}/\langle r^k\rangle$ is isomorphic to $D_{2k}$,
where $\langle r^k\rangle$ is a normal subgroup of $D_{2n}$.

My attempt: Suppose $f: D_{2n}/\langle r^k\rangle\ \longrightarrow\ D_{2k}$ is a homomorphism of groups. Then recall the first isomorphism theorem, if $f: G \rightarrow H$ is a homomorphism of groups, then $\ker f$ is a normal subgroup of $G$ and $G/\ker f$ is isomorphic to $\operatorname{im}f$.

Then I need to show $\ker f =\langle r^k\rangle$ and $\operatorname{im}f= D_{2k}$. So we can conclude by the first isomorphism theorem that $D_{2n}/\langle r^k\rangle$ is isomorphic to $D_{2k}$..
Can someone please help me? I am stuck.Thank you

Answer 1

You mention all the right ingredients, but you don't seem to put them together in a coherent way. Try the following approach:

1. Show that for any positive integer $k$ dividing $n$ the map defined by $$\varphi_k:\ D_{2n}\ \longrightarrow\ D_{2k}:\ \begin{array}{c}s\ \longmapsto\ s'\\ r\ \longmapsto\ r'\end{array},$$ is a group homomorphism, where \begin{eqnarray*} D_{2n}&=&\langle r,s\ \mid\quad r^n=s^2=1,\ rs=sr^{-1}\rangle\\ D_{2k}&=&\langle r',s'\ \mid\quad r'^k=s'^2=1,\ r's'=s'r'^{-1}\rangle \end{eqnarray*}
2. Show that $\varphi_k$ is surjective.
3. Show that $\ker\varphi_k=\langle r^k\rangle$.
4. Apply the first isomorphism theorem.