# Left translation on Lie group of a discrete subgroup is properly discontinuous

This question has been asked before here and there but has not received answers which make clear my difficulties understanding this argument. I am quite rusty in both group theory and topology, and I suppose this is why it's getting the better of me. The proposition (and the arguments I fail to comprehend) comes from Boothby's "An Introduction to Differentiable Manifolds and Riemannian Geometry" on page 96.

It says the following. Suppose $$M$$ is a Lie group and $$G$$ a discrete subgroup of $$M$$. Then for $$g\in G$$ the left translation $$g:M\ni x\mapsto gx\in M$$ is properly discontinuous.

A group action $$g$$ is properly discontinuous if (i) and (ii) hold:

(i) There is a neighbourhood $$U\subset M$$ of every $$x\in M$$ making the set $$\{g\in G : gU\cap U \neq \emptyset \}$$ finite, (ii) If $$Gx\neq Gy$$ then there are respective neighbourhoods $$U,V\subset M$$ of $$x,y$$ such that their intersection $$gU\cap V = \emptyset$$ for all $$g\in G$$.

Specifically I am having trouble showing (ii). I could give more detail as to up to which point I am stuck, but perhaps it is better to simply leave it as is since the argument as a whole is quite short. I should mention that it makes use of the fact found in there and on request I can outline what parts I have understood so far.

Fact: Let $$G$$ be a topological group. For any neighbourhood $$U$$ of $$1$$, there exists a (possibly smaller) open neighbourhood $$V\ni 1$$ such that $$V=V^{-1}$$ and $$V^2\subseteq U$$.
Proof: Let $$V=U_1\cap U_2\cap U_1^{-1}\cap U_2^{-1}$$, where $$U_1\times U_2\subseteq m^{-1}(U)$$ is a neighbourhood of $$(1,1)$$, $$m\colon G\times G\to G$$ is the multiplication in $$G$$.
Now for your question, note that WLOG we may assume $$x=1$$. Take the open neighbourhood $$U=M-Gy$$ (note $$Gy$$ is closed since $$M$$ is Hausdorff) of $$1$$ and obtain $$V\ni 1$$ with $$V^2\subseteq U$$ and $$V=V^{-1}$$. We claim the neighbourhoods $$V$$ of $$x$$ and $$yV$$ of $$y$$ works. Indeed, if $$z\in gV\cap yV$$, then $$g^{-1}z\in V$$ and $$y^{-1}z\in V$$, so $$g^{-1}y=(g^{-1}z)(y^{-1}z)^{-1}\in VV^{-1}=V^2$$, contradicting the construction $$V^2\subseteq M-Gy$$.