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- Apr 14, 2013

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Let $\alpha\in \mathbb{R}$. We define the following maps:

(a) For $p\in \mathbb{R}^2$ let $\delta_{p,\alpha}=\tau_p\circ \delta_{\alpha}\circ\tau_p^{-1}$.

(b) For $p\in \mathbb{R}f_{\alpha}$ let $\sigma_{p,\alpha}=\tau_p\circ \sigma_{\alpha}\circ\tau_p^{-1}$.

(c) For $p\in \mathbb{R}f_{\alpha}$, $q\in \mathbb{R}e_{\alpha}$ let $\gamma_{q,p,\alpha}=\tau_q\circ \sigma_{p,\alpha}$.

Let $\beta\in \text{Isom}(\mathbb{R}^2)$. Then one of the following statements is true.

(i) $\beta=\text{id}$

(ii) $\beta=r_v$, $0\neq v\in \mathbb{R}^2$.

(iii) $\beta=\delta_{p,\alpha}$,$0<\alpha<2\pi$

(iv) $\beta=\sigma_{p,\alpha}$, $\alpha\in \mathbb{R}$, $p\in \mathbb{R}f_{\alpha}$

(v) $\beta=\gamma_{q,p,\alpha}$, $\alpha\in \mathbb{R}$, $p\in \mathbb{R}f_{\alpha}$ , $q\in \mathbb{R}e_{\alpha}$ , $0\neq q$

So we have to check which of these cases for $\beta$ we have an isometry, or not?

The identity function is in $\text{Isom}(\mathbb{R}^2)$, isn't it?

For the other ones do we have to checkif these maps are bijective?