# Discrete Isometries

• Nov 22nd 2011, 02:16 PM
monomoco
Discrete Isometries
Let G be a discrete subgroup of Iso(R2). Show that every subgroup of G is also discrete.

Isn't this true simply because it's a subgroup? So the elements of the subgroup are also in G?
• Nov 22nd 2011, 07:39 PM
Drexel28
Re: Discrete Isometries
Quote:

Originally Posted by monomoco
Let G be a discrete subgroup of Iso(R2). Show that every subgroup of G is also discrete.

Isn't this true simply because it's a subgroup? So the elements of the subgroup are also in G?

I'm confused, if you are talking about the fact that a subgroup of a discrete subgroup is discrete, this really is just the (more general statement) that a subspace of a discrete space is discrete. In particular, you want to show that each element of $\displaystyle H\leqslant G$ is open, but for each $\displaystyle h\in H$ you know (via the fact that $\displaystyle G$ is discrete) that there exists some open set $\displaystyle O\subseteq \mathbb{R}^2$ such that $\displaystyle O\cap G=\{h\}$. Clearly though $\displaystyle O\cap H=\{h\}$ and so $\displaystyle \{h\}$ is open in $\displaystyle H$. Since $\displaystyle h$ was arbitrary the conclusion follows. Make sense?
• Nov 22nd 2011, 07:44 PM
monomoco
Re: Discrete Isometries
We're talking about a subgroup of Iso (R2) so by discrete I have the definition:
G< Iso (R2) is discrete there exists an E such that

for all translation t_a in G, a < E
for all rotations in r_o in G, o<E

so the motions cannot be arbitrarily small. I understand what you mean, but I don't know how to apply it to this example.
• Nov 22nd 2011, 07:49 PM
Drexel28
Re: Discrete Isometries
Quote:

Originally Posted by monomoco
We're talking about a subgroup of Iso (R2) so by discrete I have the definition:
G< Iso (R2) is discrete there exists an E such that

for all translation t_a in G, a < E
for all rotations in r_o in G, o<E

so the motions cannot be arbitrarily small. I understand what you mean, but I don't know how to apply it to this example.

Oh, well then in that case (the two are equivalent concepts though, which I think you noticed) you must merely note that the same $\displaystyle \varepsilon$ that works for $\displaystyle G$ works for $\displaystyle H$. So yes, I think it's as simple as you indicated in the initial post.
• Nov 22nd 2011, 07:52 PM
monomoco
Re: Discrete Isometries
I thought I must be missing something. Thanks for your help!