Is the point group of a discrete group also discrete? Every isometry $g$ of the Euclidean plane $\mathbb{E}^2$ is uniquely written as a rotation $\rho_\theta$ by angle $\theta (g)$ about the origin followed by a translation $t_v$ by a vector $v (g) \in \mathbb R^2$.
For a group $G$ of isometries of $\mathbb{E}^2$, define its point group $\overline G = \{\rho_{\theta (g)} | g \in G\}$.
In Artin's Algebra (Second Edition, Section 6.5, 'Discrete Groups of Isometries') I feel it is implicitly assumed that for a discrete group $G$, its point group $\overline G$ is also discrete. Can anyone provide a proof of this?
Just to be sure, my definition of discrete is this:
The set of isometries is equal to $\mathbb R^2 \times S^1$ (as a set) and so is given the topology arising as this product. $G$ is a discrete group if it is discrete in the subspace topology. In other words there is an $\varepsilon > 0$ such that whenever $g \in G$ is a nontrivial element, we have $\lVert v (g) \rVert^2 + \lvert \theta (g) \rvert^2 \ge \varepsilon$.
 A: I'll write an answer which makes use of a general theorem of group theory:

For any group $G$, and any normal subgroup $N < G$ with quotient $G/N = Q$, there is a natural homomorphism $Q \mapsto \text{Out}(N)$ defined as follows: pick an element of $Q$, pick a representative $g \in G$, let $i_g : G \to G$ be the inner automorphism $h \mapsto g^{-1}hg$, and restrict $i_g$ to $N$ to get an automorphism of $N$

The key to this theorem is well-definedness: if you replace $g$ by another element of its coset modulo $N$ then that will change the restricted automorphism of $N$ but it will not change the outer automorphism class of that restriction.
So, let me denote the translation subgroup of $G$ as $T < G$, which is a discrete group because $G$ is discrete. $T$ is a normal subgroup of $G$, and it is the kernel of the quotient homomorphism $G \mapsto \overline{G}$, so the previous theorem applies and we get a homomorphism $$\mathcal{A} : \overline G \mapsto \text{Out}(T) \approx \text{Aut}(T)
$$
The last equation follows because $T$ is abelian. 
So now given a rotation $\rho \in \overline G$ with image $\mathcal{A}\rho \in \text{Aut}(T)$, and given a translation $\tau \in T$, lets consider the translation $\mathcal{A}\rho(\tau)$. Let's represent $\tau$ by its translation vector $v_\tau = \tau(0) - 0$, so $\tau(x) = v_\tau + x$. If you trace through the definitions, you will discover that
$$v_{\mathcal{A}(\rho)(\tau)} = \rho(v_\tau)
$$
So if we fix one $\tau \in T$ and its translation vector $v_\tau$, it follows that every vector in the set 
$$\{\mathcal{A}\rho(v_\tau) \mid \rho \in \overline G\} = \{\rho(v_\tau) \mid \rho \in \overline G\}
$$
has the same length. But since $T$ is discrete, this set of vectors must be finite. 
This is enough, with a bit more argument, to let you conclude that $\overline G$ is finite (hence discrete). 
