Commutator subgroup of a dihedral group. I have a few questions concerning an example of the commutator subgroups in the dihedral group. This example is found on pg.171 of Abstract Algebra by Dummit and Foote. 

Let $D_{2n}=\langle r,s |r^n=s^2=1, s^{-1}rs=r^{-1}\rangle$. Since $[r,s]=r^{-2}$ we have that $\langle r^{-2} \rangle = \langle r^2 \rangle \le D'_{2n}$. Furthermore, $\langle r^2\rangle \trianglelefteq D_{2n}$ and the images of $r$ and $s$ in $D_{2n} / \langle r^2 \rangle$ generate this quotient. 

What exactly is meant by the image of $r$ and $s$?

They (is this referring to $r$ and $s$?) are commuting elements of order $\le 2$ (I know $s$ is of order $2$ but $r$ is of order $n$??) so the quotient is abelian and $D'_{2n} \le \langle r^2 \rangle$. (I thought the quotient was abelian due to the already established properties of $\langle r^2 \rangle$ i.e it is normal and a subgroup of the commutator subgroup.)

 A: When D&F say "The image of $g$" in such contexts, they mean its image under the canonical homomorphism that takes $g$ to the coset $gH$. (Here the subgroup $H$ is $\left\langle r^{2}\right\rangle $.) 
The next sentence is about the orders of the cosets $r\left\langle r^{2}\right\rangle ,s\left\langle r^{2}\right\rangle $ in the quotient group $D_{2n}/\left\langle r^{2}\right\rangle$, which are indeed $\leq 2$. (Not the orders of $r,s$ in the dihedral group.) 
Now, by the important defining relation of the dihedral group $srs=r^{-1}$, one can directly verify that these generators commute. Since they are also of order  $\leq 2$ , this implies that $D_{2n}/\left\langle r^{2}\right\rangle$  is abelian. Now, using Proposition 7(4) on page 169, ('the commutator subgroup is the largest abelian quotient'), we conclude $D_{2n}' \leq \left\langle r^{2}\right\rangle$. Therefore, $\left\langle r^{2}\right\rangle =D_{2n}'$
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A: I think you have confused the commutator subgroup with the quotient. The quotient is (isomorphic to the group) formed by adding the relation $g=1$ for every generator of the subgroup. In this case, the only generator is $r^2$, so we get:
$$<r, s | r^2=s^2=r^n=1, s^{-1}rs=r^{-1}>$$
$$=<r, s | s^2=r^{n \text{mod} 2}=rsrs=1>$$
