I've played around with different isometries in $\mathbb{R}^2$ and it if $f$ is any isometry with a flip (like a flip followed by a rotation), its order is $2$. This makes sense geomtrically, how do I prove this?
If $\tau$ denotes a flip, and $\sigma$ denotes a rotation, then $f = \sigma ^a \tau$, where $a \in \mathbb{R}$. How can I prove that $\tau \sigma \tau = \sigma^{-1}$? Is this a fact for isometries in $\mathbb{R}^2$ ? If so, why?
Isometries in $\mathbb{R}^2$ form a group, O(2), the orthogonal group. The map $f \mapsto determinant(f)$ is a group morphism from $O(2)$ into the multiplicative group $(\{ -1, 1\}, \times) $.
$\sigma \in O(2)$ is a rotation if and only if $det(\sigma)=1$ (in other works, the subset of rotations, $SO(2)$, is the kernel of $determinant$); $\tau$ is a flip if and only if $det(\tau)=-1$. Besides, pour any flip $\tau, \tau^2 = id$.
$\tau \sigma$ is a flip for $det(\tau \sigma)= det( \tau) det(\sigma)= -1 \times 1 = -1$. Hence $(\tau \sigma)^2 = id$, i.e. $\tau \sigma \tau \sigma = id \iff \tau \sigma \tau = \sigma^{-1}$