Quotient group of the dihedral group by $\langle r^2 \rangle.$ 
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. 
$$\begin{align}
ar^{2i}br^{2j} &=r^{2i}a^{-1}br^{2j}\\
&=r^{2i}b^{-1}ar^{2j} \\
&=br^{2i}ar^{2j}.
\end{align}$$
 A: 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 $4$.
A: One way to look at this is to note that
$$\begin{align}
G/H &\cong\langle r, f\mid r^2=1, r^n=f^2=1, rf=fr^{-1}\rangle \\
&\cong\langle r, f\mid r^{\gcd(2, n)}=f^2=1, \underbrace{rf=fr}_{r^2=1}\rangle \\
&\cong \Bbb Z_{\gcd(2,n)}\times\Bbb Z_2,
\end{align}$$
where the first isomorphism$^\dagger$ holds because $G/H$ is, essentially, just introducing the relation $r^2=1$ to $G$; the second isomorphism is a consequence of manipulating the relations in the first presentation above (and a standard result on the $\gcd$ of powers of the same generator of a presentation); and the latter is, again, standard, since one just has to recognise a typical presentation for the direct product cyclic groups.

$\dagger$ A presentation is not a group, technically speaking. What I am referring to is the group defined by the presentation.
