Can adding a single element to a Lie group make it infinite-dimensional?

Question

Suppose that I have a finite-dimensional Lie group, $F\subset H=\text{Diff}(\mathcal{M})$, of diffeomorphisms on some smooth finite-dimensional manifold $\mathcal{M}$. Suppose that there is some element in the larger group, $g\in H$, which is not in the smaller group $g\notin F$. Let $G\subset H$ be the smallest Lie subgroup of $H$ which contains both $F$ and $g$.

$G:= \text{Closure}(\{g\}\cup F)\subset H$

I am interested in bounding the size of $G$ (i.e., the dimensionality of $G$) in terms of the dimensionalities of $F$ and of $\mathcal{M}$. For instance, one might conjecture that adding a single element $g$ to a Lie group $F$ will never increase its dimension by more than a factor of $\text{dim}(\mathcal{M})$. Namely, one might conjecture $\text{dim}(G)\leq\text{dim}(\mathcal{M}) * \text{dim}(F)$. At the end of the day, all I need is proof that $G$ is finite-dimensional (even if $\text{dim}(G) < \infty$ is arbitrarily large). Is it possible that adding a single element $g$ to a finite-dimensional Lie group $F$ would make it infinite dimensional?

(* Edited to emphasize the problem of interest with $F\subset H=\text{Diff}(\mathcal{M})$ being diffeomorphisms).

Answer 1

There is no bound on the dimensionality of $G$. In fact, a single element can generate a Lie group of arbitrarily large dimension. Thus the $n$-dimensional torus $\mathbb T^n$ is the closure of the group generated by a single element $(e^{2\pi i \omega_1}. \ldots, e^{2\pi i\omega_n})$ if $1$ and the $\omega_j$ are linearly independent over the rationals.

Answer 2

It's probably easier to work with Lie algebras rather than Lie groups here. If we take $M = \mathbb{R}^2$ then $Lie(Diff(\mathbb{R}^2)$ is the Lie algebra of smooth vector fields on $\mathbb{R}^2$.

Let's take say $y^2 \frac{d}{dx}$ and $x^2 \frac{d}{dy}$. Either one of these generates a finite (one) dimensional Lie algebra, but together they generate something infinite dimensional.