Quotient group of the dihedral group by $\langle r^2 \rangle.$

Question

Show that $G/H$ is abelian, where $G$ is the dihedral group $$ G={\langle r,\, f \mid r^n=f^2=1,\, rf=fr^{-1}\rangle}$$
and $H$ is the subgroup $\langle r^2 \rangle.$

I've tried showing that for $a,b \in G$ then $ar^{2i}br^{2j}=br^{2i}ar^{2j}$ but im sure if the steps I used are true.
$$\begin{align}
ar^{2i}br^{2j} &=r^{2i}a^{-1}br^{2j}\\
&=r^{2i}b^{-1}ar^{2j} \\
&=br^{2i}ar^{2j}.
\end{align}$$

Answer 1

Hint If $n$ is odd, $\mid\langle r^2\rangle\mid=\mid\langle r\rangle \mid=n$. Hence the index is $2$. That is, $D_{2n}/\langle r^2\rangle=C_2$.

If $n$ is even, $\mid \langle r^2\rangle \mid=n/2$. Thus the index is $4$. Thus $D_{2n}/\langle r^2\rangle $ has order $4$.