Interpreting a Quotient Group ($D_8/\langle r^2\rangle$) in 2 Distinct Ways

Question

I seem to have a misunderstanding about quotient groups. Let $D_{2n}$ denote the group of symmetries of an $n$-gon and let $V_4$ denote the Klein-4 group. On one hand, if we identify $r^2$ with $1$, we have $$D_8/\langle r^2\rangle=\lbrace 1,r,s,sr\rbrace$$ which is $V_4$.

On the other hand, using the definition of the quotient group as the set of left cosets, $$D_8/\langle r^2\rangle=\Big\lbrace g\lbrace 1,r^2\rbrace :g\in D_8 \Big\rbrace=\Big\lbrace \lbrace 1,r^2\rbrace, \lbrace r,r^3\rbrace,\lbrace s,sr^2\rbrace, \lbrace sr,sr^3\rbrace \Big\rbrace$$

In the latter set, I am not identifying $r^2$ with $1$ since the definition of cosets or quotient spaces does not require me to (by definition, $G/K=\lbrace gK:g\in G\rbrace$). However, none of these elements appears to be the identity, and the second entry squared, $\lbrace r^2,r^2\rbrace$, doesn't even belong to the set. Where am I going wrong? Is one allowed to replace $r^2$ with 1 in the latter approach?

EDIT: $D_8=\langle r,s:r^4=s^2=1, rs=sr^3\rangle$, so $r$ denotes the rotation and $s$ denotes the reflection.

Answer 1

when multiplying two complexes (subsets) $A,B \subset G$ the result is: $$ AB=\{ab|a \in A,b \in B\} $$ so for your problem case: $$ \{r,r^3\}^2 = \{rr,rr^3,r^3r,r^3r^3\} =\{r^2,1,1,r^2\}=\{1,r^2\} $$

Answer 2

The subgroup $\langle r^2\rangle$ is normal in $D_8$ because $r^2$ is in the center:

For an element $g \in G,\; gr^2g^{-1}=gg^{-1}r^{-2}=r^{-2}=r^2 \in \langle r^2\rangle$.

Since the subgroup is normal, the set of cosets becomes a group with the following operation:

$$a\langle r^2\rangle b\langle r^2\rangle = ab \langle r^2\rangle \; \forall a,b \in D_8$$

This group has identity $\langle r^2\rangle$ because $\langle r^2\rangle=1\cdot\langle r^2\rangle$, so $$a\langle r^2\rangle\cdot \langle r^2\rangle=a\cdot 1\langle r^2\rangle = a\langle r^2\rangle$$

for any coset $a\langle r^2\rangle \in D_8/\langle r^2\rangle$.

I'm not sure of how exactly you derived the isomorphism with $V_4$ in your first interpretation, since $r^2$ is not an element of $D_8/\langle r^2\rangle$, but the isomorphism can be justified with the group in the second interpretation.

The easiest way to see the isomorphism is to show every nonidentity element has order 2:

$$\{r,r^3\}^2=\langle r^2\rangle \text{ (David Holden's answer shows this)}\\ \{s,sr^2\}^2=(s\langle r^2\rangle)^2=s^2\langle r^2\rangle=\langle r^2\rangle \\ \{sr,sr^3\}^2=(sr\langle r^2\rangle)^2=(sr)^2\langle r^2\rangle=srsr\langle r^2\rangle=ssr^{3}r\langle r^2\rangle=\langle r^2\rangle$$