I am solving the following exercise:
Let $\mathcal{T}(E) := \{ T_v \mid v \in \mathbb{R^2} \}$ be the set of all translations of E and $\mathcal{O}(2,\mathbb{R}) := \{g \in \DeclareMathOperator{\Iso}{Iso}\Iso (E) \mid g(0) = 0 \}$.
Show that $\mathcal{T}(E)$ and $\mathcal{O}(2,\mathbb{R})$ are subgroups of $\Iso(E)$
Further show that every element of $\mathcal{O}(2,\mathbb{R})$ is either a rotation around $0$ or a reflection with respect to a line which intersects $0$.
Previous knowledge: In our lectures we cover in our second chapter the topic of isometries of a plane. Therefore we defined a list of orientation-preserving isometries such as: translation and rotation and orientation reversing isometries such as: reflections and glides. Further we stated a theorem which says:
Every isometry is element of the given list. The set of orientation-preserving isometries is a subgroup.
My Attempt: After having a look at the theorem we stated in our lectures one can see that the subgroup of orientation-preserving isometries is a subgroup of all isometries. Therefore the subgroup which contains only the translations is also a subgroup of all isometries. For $$T_1(v), T_2(v) \in \mathcal{T}(E)$$ we can say: $$T_1(v) \circ T_2(v) = \overbrace{T_1 \circ T_2}^{\in \ \mathcal{T}} \ (v)$$ and $$ T_1 \circ T_1^{-1} (v) = \DeclareMathOperator{\Id}{Id} \Id(v) = v$$ My Question at this point is, whether we have to prove if $T_v \in \mathcal{T}(E)$ is an isometry or is it sufficient to argue with the theorem?
Further we need to show that: $\mathcal{O}(2,\mathbb{R})$ is a subgroup. By the condition that $\mathcal{O}(2,\mathbb{R}) := \{g \in \Iso (E) \mid g(0) = 0 \}$ we know that $0$ is always going to map on $0$. This is the indicator that only rotations around $0$ and reflections with respect to a line which intersects $0$ are possible isometries for this subgroup.
From a logical imaginational point of view it is clear that a combination of rotation and reflection always can be recalled by doing only a reflection or rotation or even both. My problem here is to state it mathematically correct. $$ d_{p,\theta}\ ,\ S_{g_0} \in \mathcal{O}(2,\mathbb{R}) \ , \ p \in \mathbb{R^2} $$ where $S_{g_0}$ is the reflection with respect to a line which intersects $0$ and $g_0$ the straight line. Now one can show that:
\begin{align}(d_{\theta_1} \circ d_{\theta_2})(p) &= d_{\theta_3}(p) \in \mathcal{O}(2,\mathbb{R}) \\ (d_{\theta_1} \circ d_{\theta_1}^{-1})(p) &= \Id(p) \in \mathcal{O}(2,\mathbb{R}) \\ (S_{g_{0_1}} \circ S_{g_{0_2}}) &= S_{g_{0_3}}(p) \in \mathcal{O}(2,\mathbb{R}) \\ (S_{g_{0_1}} \circ S_{g_{0_1}}^{-1}) &= \Id(p) \in \mathcal{O}(2,\mathbb{R}) \end{align} Further we show that the combination is also closed: $$ (d_{\theta_1} \circ S_{g_{0_1}})(p) = d_{\theta_3}(p) \in \mathcal{O}(2,\mathbb{R})$$ I think I got the problem but I am not very sure if I proved it sufficiently.
I appreciate every hint and correction to my attempt.
