# Subgroups of isometries on Euclidean space

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.

The fact that $O(2,\mathbb{R})$ (which I will henceforth write as $O(2)$) is a subgroup is simpler than you're making it. Given that $Iso(\mathbb{R}^2)$ is a group, you only need to check that the group axioms are satisfied for the subset so described.

1. Is the identity in $O(2)$? That is, does the identity fix the origin?
2. Does the composition of two elements that fix the origin also fix the origin?
3. If $g$ fixes the origin, does $g^{-1}$ fix the origin?

So this should be a straightforward check based on the definitions.

Now, as for the fact that $O(2)$ consists only of rotations and reflections through the origin, since you have a list of all possible types of isometries, you should be able to cross many of those off of your list of things that can be in $O(2)$. For example, are non-trivial translations in $O(2)$?

As for your first question as to whether or not you need to argue that $\mathcal{T}_v$ (although I'm not sure what this is exactly, since it doesn't match your previous notation) is an isometry, you certainly can use the fact that you already know that $Iso(\mathbb{R}^2)$ is an isometry. I suppose it depends on the structure of the course and the preferences of your instructor, but odds are that you can always use proven facts from your class. If you're not sure though, ask your prof or TA, and they will be the ultimate arbiter.

• $T_v$ basically means $T$ on $v$ so $T_v = T(v)$ Commented Oct 27, 2014 at 13:48