So far, I have shown that any symmetry of $\mathbb{R}^2$ which fixes the origin is either identity or a reflection about the line passing through the origin or a rotation about the origin. Further, I got that any symmetry of $\mathbb{R}^2$(say $g$) can be written in the form: $g=\tau \circ f$, where $\tau$ is a translation and $f$ is the symmetry about the origin(which would certainly be a linear transformation).
My claim is simple.
Claim: $f$ has order $n$ iff $g$ has order $n$, provided f is not identity.
This is quite intuitive to see(eg: a rotation of $\frac{\pi}{3}$ about the point $(1,1)$, would be simply $\tau\circ f$, where f is a rotation of $\frac{\pi}{3}$ about the point $(0,0)$ and $\tau$ is a translation which takes the origin to $(1,1)$). However, I am facing some difficulty proving it. This is what I have done.
(say $\tau(x)=x+v$)
If $g^n=I$, then $(\tau\circ f)^n=I$ $\Rightarrow$ $(\tau\circ f)^n(x)=x$
$\Rightarrow f^n(x)+f^{n-1}(v)+f^{n-2}(v)+\cdots +f(v)+v=x$.
I'm trying to prove that the above polynomial of degree $n-1$ in $v$ is $0$, so that $f$ also has degree $n$. I'm stuck here, can someone help me out?
Best Answer
This
is not quite right. Rotation about the point $(1,1)$ fixes the point $(1,1)$. So what you actually need is $\tau \circ f \circ \tau^{-1}$. That is, drag the point $(1,1)$ to the origin, rotate, then drag it back to $(1,1)$.
Of course, it should be easy to see that $$ (\tau \circ f \circ \tau^{-1})^n = \tau \circ f^n \circ \tau^{-1} \text{,} $$ by observing all the cancellation.