Let $G=\mathbb Z / n \mathbb Z$ for $n > 2$ and let $G$ act on $\mathbb R^2$ linearly and effectively. Let $T_g (v)$ denote the element $gv$ where $v \in \mathbb R^2$. Assume that $\det T_g > 0$ and that $\langle T_g v, T_g v'\rangle = \langle v,v'\rangle$.
I want to show that there exists exactly one $g \in G$ such that $T_g$ is a rotation of angle ${2 \pi \over n}$.
Please could someone tell me if what I did is correct?
That $G$ acts linearly means that $T_g$ is a linear transformation, that is, a $2 \times 2$ matrix. (use that matrices and linear transformations are in bijection without proof)
That $\langle T_g v, T_g v'\rangle = \langle v,v'\rangle$ therefore means that $T_g$ is a linear isometry of $\mathbb R^2$. Hence its determinant is either $1$ or $-1$ but since we have $\det T_g>0$ it must be $1$. (use that linear isometries have determinant $\pm 1$ without proof)
Also without proof I want to use that linear isometries of $\mathbb R^n$ are either rotations or reflections but since $\det =1$ this one is a rotation.
Edit
To simplify notation we use the complex representation of the problem: $\mathbb R^2 = \mathbb C^2$ and rotation by $\theta$ is multiplication by $e^{i\theta}$.
We have $T_1^n = I$ (because the action is effective) that is, $(e^{in \theta}) = 1 = e^{2 \pi i k}$. Taking the logarithm we get $\theta = {2\pi k\over n}$ where $k \in \mathbb Z$.
Therefore $T_1$ is rotation by $\theta = {2\pi k \over n}$. Since $1$ generates $G$, $T_1$ must generate the group $T_1, \dots, T_n$.
Do I have to prove this claim? (by providing a group isomorphism from $G$ to $\{T_k\}_k$)
Now I have two things left to do: I have to argue that $k$ has to be coprime to $n$ (otherwise it wouldn't generate the group) and finally I somehow need to argue why $k=1$ (at the moment I really don't see how this is possible)
$\color{grey}{\text{ Finally, consider $1 \in \mathbb Z / n \mathbb Z$. Note that $1^n = id$. Hence $T_1^n = I$. This is true for the matrix that is rotation by ${2\pi / n}$. Hence there exists $g\in G$ such that $T_g$ is rotation by ${2\pi / n}$.}}$
$\color{grey}{\text{ Now assume $T_g$ was another rotation such that $T_g^n=I$. It's not clear how to argue with more mathematical rigor but it is clear}}$ $\color{grey}{\text{ that the only rotations that achieve the identity when applied $n$ times are rotations by ${2\pi \over n}+k 2\pi$. }}$
There is a problem in your conclusion. Let me denote by $k(\theta)$ the $2 \times 2$ rotation matrix with rotation angle $\theta$. Your argument beginning with "Finally, consider $1$ ..." isn't really complete. Essentially what you've shown there is that $T_1 \in \{k(2\pi m/n) : m = 0,1,\ldots,n\}$ (although you should probably justify why these are the only rotation matrices with order dividing $n$), and it very well may be the case that $T_1 = k(2\pi m/n)$ where $m \neq 1$.
So what can you do to show that $T_g = k(2\pi/n)$ for some $g$? My hint would be to look at the other information about the group action you haven't used yet, i.e., that it's effective. This will also be useful (critical, really) in showing that $g \in G$ with $T_g = k(2\pi/n)$ is unique. I've provided a bigger hint in a spoiler tag below, but try doing this yourself before looking at it.
Edit (in response to your new argument and comments):
A few points:
Minor correction here (it's probably just a typo), but $\mathbb R^2 \cong \mathbb C$ as real vector spaces, not $\mathbb C^2$.
Although your intuition that you may identify $\mathbb R^2$ and $\mathbb C$ is correct, in any formal setting [that is, one outside the realm of math.SE
:)] there are definitely some things to justify about why it's ok to make this substitution here. First, what is the $\mathbb R$-linear isomorphism $\varphi: \mathbb R^2 \to \mathbb C$ you're using? Second, and more significantly here, the map $\varphi$ enables us to identify these two vectors spaces, but we're concerned with the linear transformations on these spaces, not the spaces themselves. How does $\varphi$ affect linear transformations? Moreover, given a linear transformation $T: \mathbb R^2 \to \mathbb R^2$, the map $\varphi$ should identify $T$ with some other transformation $T': \mathbb C \to \mathbb C$. Given $T$, what does $T'$ look like? (A common notation for this new linear map is $\varphi^*(T)$.) For instance, if I begin with the rotation-by-$\theta$ mapping $T = k(\theta)$, what does $\varphi$ "send" $k(\theta)$ to? interact with the group action we're looking at? This is totally possible (and a cool way to rethink the problem), but I thought I'd provide some food for thought.:)To avoid going through the work of identifying $\mathbb R^2$ and $\mathbb C$, you can use a more elementary argument to conclude that $T^n = I$ for $T$ a rotation matrix implies $T = k(2\pi m/n)$ for some $m$. Write $T = k(\theta)$ for some $\theta \in [0,2\pi)$. Then $T^n = k(n\theta) = I$, and the only real numbers $\psi$ (think $\psi = n\theta$) for which $k(\psi) = I$ are $\psi = 2\pi m$ for $m \in \mathbb Z$ (show this by explicitly writing out the components of $k(\psi)$ if this is unclear to you). Hence $n\theta = 2\pi m$ for some integer $m$, and therefore $\theta = 2\pi m/n$. The assumption that $0 \leq \theta < 2\pi$ then implies that $m \in \{0,1,\ldots,n-1\}$.
Here is a quick way to reach the desired conclusion, given what we already know. From this, it follows that $T_1 = k(2\pi m/n)$ for some $m \in \{0,1,\ldots,n-1\}$. In fact, since $n\cdot g = \text{Id}$ for all $g \in G = \mathbb Z/n\mathbb Z$, the same argument (beginning with the observation $T_g^n = I$) implies that for each $g \in G$ there exists $m_g \in \{0,1,\ldots,n-1\}$ such that $T_g = k(2\pi m_g/n)$. Since the action is effective, we know that $T_g \neq T_{g'}$ for all $g,g'\in G$. Thus $\{T_g : g \in G\} \subseteq \{k(2\pi m/n) : m = 0,1,\ldots,n-1\}$ and
$$ \#\{T_g : g \in G\} = \# G = n = \#\{k(2\pi m/n) : m = 0,1,\ldots,n-1\}, $$
from which it follows that $\{T_g : g \in G\} = \{k(2\pi m/n) : m = 0,1,\ldots,n-1\}$. Therefore $T_g = k(2\pi/n)$ must hold for some $g \in G$.