# 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}

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 $$4$$.
\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.