Why not $D_8/\langle r^2 \rangle = M?$

Question

Source: Dummit and Foote book.

Page number: $85$.

It is written that

$D_8/\langle r^2 \rangle = \{ x \{ 1, r^2\} : x \in D_8 \}=\{ \{1,r^2\} , \{ r,r^3\} , \{s, sr^2\} , \{sr, sr^3\}\} $

My confusion: we know that elements of $D_8$ are $1,r,r^2,r^3,s,sr,sr^2$ and $sr^3$

I think it should be like this

$D_8/\langle r^2 \rangle = \{ x \{ 1, r^2\} : x \in D_8 \}=\{ \{1,r^2\} , \{ r,r^3\} , \{r^2, r^4\} , \{r^3,r^5\},\{s,sr^2\},\{sr,sr^3\},\{sr^2,sr^4 \},\{sr^3,sr^5\}\} $

Here I take $M=\{ \{1,r^2\} , \{ r,r^3\} , \{r^2, r^4\} , \{r^3,r^5\},\{s,sr^2\},\{sr,sr^3\},\{sr^2,sr^4 \},\{sr^3,sr^5\}\} $

My question: Why not $D_8/\langle r^2 \rangle = \{ \{1,r^2\} , \{ r,r^3\} , \{r^2, r^4\} , \{r^3,r^5\},\{s,sr^2\},\{sr,sr^3\},\{sr^2,sr^4 \},\{sr^3,sr^5\}\} ?$

Answer 1

There are repeating elements there. For instance, since $r^4=1$, $\{r^2,r^4\}=\{1,r^2\}$.

Besides, if your suggestion was correct, $|D_8/\langle r^2\rangle|=8$. But$$|D_8/\langle r^2\rangle|=|D_8|/|\langle r^2\rangle|=8/2=4.$$