Assume $k|n$, show $D_{2n}/\langle r_n^k \rangle \cong D_{2k}$

Question

Definition of Dihedral Group: $D_{2n}=\langle r,s|r^n=s^2=1, sr=r^{-1}s \rangle $

I need to show that $D_{2n}/\langle r_n^k \rangle \cong D_{2k}$, assuming $k|n$.

I have already proven that $\langle r_n^k \rangle \trianglelefteq D_{2n}$ so we know that $D_{2n}/\langle r_n^k \rangle$ is a quotient group. Also I know that $|\langle r_n^k \rangle| = \frac{n}{k}$, so by Lagrange's Thm

$$|D_{2n}/\langle r_n^k \rangle| = \frac{|D_{2n}|}{|\langle r_n^k \rangle|}= \frac{2n}{\frac{n}{k}}=2k=|D_{2k}|$$

So everything looks good, but I am just not quite understanding how I show that these two groups are isomorphic. I feel there is one step missing, or I need to use one of the isomorphism Theorems maybe?

Answer 1

Here is a rigorous argument, using the very definition of the dihedral groups via ther group presentations and the universal property of free groups and factor groups.

Let $G = ⟨R, S⟩$ be the free group generated by two elements, $R, S$. Let $n, k ∈ ℕ$ with $k \mid n$.

Let $N_i = (R^i, S^2, SRSR) ⊆ G$ be the generated normal subgroup for $i = k, n$. Now we have $N_k ⊇ N_n$ and $N_k/N_n = (r^k)$ in $D_{2n}$. So in short, by the third isomorphism theorem, we have $$\frac {D_{2n}} {(r^k)} = \frac {G/N_n} {N_k/N_n} \cong \frac G {N_k} = D_{2k}.$$

That’s it. Here is an elaboration:

Let $H$ be any group with $r, s ∈ H$. Then we have an equivalence of equations: $$r^k = s^2 = srsr = 1 \iff r^n = s^2 = srsr = 1~\text{and}~ r^k = 1.$$ Also, the left side still implies the right side when we leave out “$r^k = 1$”. We can use this to proceed as follows: Again let $N_k$ and $N_n$ be the normal subgroups from above and let $$ψ \colon G → D_{2k},~R ↦ r,~S ↦ s.$$ Then – by definition – $ψ$ is surjective with $\ker ψ = N_k$. But $N_k ⊇ N_n$, so the map $ψ$ factorizes via $$\overline ψ\colon D_{2n} = G/N_n → D_{2k}.$$ Because the projection $G → G/N_n$ is surjective, the kernel of $\overline ψ$ is exactly the image of $N_k$ under the projection, hence $$\ker \overline ψ = N_k/N_n = (r^k).$$ Hence, by the first isomorphism theorem, $D_{2k} \cong D_{2n}/(r^k)$.

It may be worth noting that this recipe is making rigorous the argument “The defining equations are equivalent, so the groups must be isomorphic”, so in future you may use this line of thinking.