# Left Translation Action on Lie Group is Proper

Let $$G$$ be a Lie group with left translation action:

$$L: G \times G \rightarrow G$$ $$\quad$$ $$(g,h) \rightarrow L_g(h) := gh$$

I want to show that this action is transitive,free, and proper.

Transitive and free follow from definition more or less.

However, I am having trouble showing it is proper.

This is the definition of a proper action that I am given:

The action $$\phi:G \times M \rightarrow M$$ of a group $$G$$ on a manifold $$M$$ is proper if, whenever the sequences $$\{x_n\}$$ and $$\{g_nx_n\}$$ converge in $$M$$, the sequence $$\{g_n\}$$ has a convergent subsequence in $$G$$.

It is clear to me that if $$G$$ is a compact Lie group, then the action is proper. This is so because in a compact group, every sequence $$\{g_n\}$$ has a convergent subsequence. But we are not given that the group $$G$$ is compact. My question is, how do we show this for a generic Lie group $$G$$?

Any hints/suggestions are most welcomed.

Hint : if $$x_n\to x, g_nx_n \to y$$, then $$g_n = g_nx_nx_n^{-1}\to ?$$
• if $x_n\to x, g_nx_n \to y$, then $g_n = g_nx_nx_n^{-1}\to yx^{-1}$. But ok, can we just conclude that since we've found a convergent subsequence $g_n$ that works, we're done? Apr 29 '19 at 18:17