What is an example of a noncompact Lie group $ G $ and two finite order elements $ a,b \in G $ such that the closure (in the manifold topology) of the subgroup $ <a,b> $ is all of $ G $?
Ideally the example would be a matrix group $ G $ and $ a,b $ would be given explicitly as finite order matrices.
It turns out that such a pair of finite order elements always exists if $ G $ is compact semisimple
For example $$ a=\frac{i}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} $$ $$ b=\begin{bmatrix} \overline{\zeta_{16}} & 0 \\ 0 & \zeta_{16} \end{bmatrix} $$ generate a dense subgroup of $ SU_2 $.
Here is how to find such examples in, say, $G=PSL(2, {\mathbb R})$. First of all, if $a, b$ are noncommuting elements of finite order $\ge 3$, they generate a Zariski dense subgroup (this is a nice exercise in understanding Lie subgroups of $G$). Now, you just have to choose $a, b$ so that they generate a nondiscrete subgroup. For instance, take them both the Zassenhaus neighborhood of $1\in G$. Or, if you prefer, make them violate the Kazhdan-Margulis lemma: Fix the order and make sure that the fixed points in the hyperbolic plane are close enough. Or make $a, b$ violate the Jorgensen inequality...
With more thought, you will realize that such elements exist in any complex semisimple Lie group $G$. The key is to observe that compact elements of the Lie algebra form a Zariski dense subset. Then you use the fact that if $a, b\in G$ generate a nondiscrete subgroup and $\log(a), \log(b)$ generate the Lie algebra of $G$, then $a, b$ generate a dense subgroup of $G$.