Why does rotation $r^2$ by $\pi$ give an order $2$ subgroup in $D_8?$

Question

Find a subgroup of $D_8$ of order $2.$

My attempt :

A presentation of $D_{2n}$ is $\langle r,s\mid r^n=s^2=\mathrm{id},srs=r^{-1}\rangle$, the reflections are $s,sr,sr^2,\ldots,sr^{n-1}$

Put $ n=4$ then $D_{8}$ is $\langle r,s\mid r^4=s^2=\mathrm{id},srs=r^{-1}\rangle.$

The element in $D_8$ of order $2$ are the reflection $s,sr,s^2$ and $sr^3$ because they all are identity symmetry

But there are $5$ distinct subgroup of order $2$

Rotation $r^2$ by $\pi$ also give order $2$

Here im not getting why rotation $r^2$ by $\pi$ give order $2$ subgroup in $D_8?$

Answer 1

Doing it twice gets you back to where you started (a rotation of $2\pi$), so the element has order $2$, and hence generates a subgroup of order $2$.