Dimension of Lie Group and Lie Algebra generated by underlying manifold

Question

I am a beginner to Lie groups and Lie algebra. Recently, I learnt that the dimension of the Lie algebra generated by a Lie group is the same as that of the Lie group.

I have known for a long time that the space of vector fields on a manifold (even when not considered a Lie group), forms a Lie algebra.

Is there any conclusion that can be made about the dimension of this Lie algebra?

It seems to me that the Lie algebra generated a Lie Group (space of left-invariant vector fields) is the subalgebra of the Lie algebra generated by the underlying manifold (space of all vector fields), so their dimensions should not be the same.

Any enlightenment is greatly appreciated!

Answer 1

A nice exercise to prove is the following. Let $G$ be a Lie group and $\mathfrak{g}$ be the Lie algebra of $G$. Show that $\mathfrak{g}$ is isomorphic (as a vector space) to $T_eG$, where $e\in G$ is the identity. Hint below.

Use the differential of left translation by elements of $G$, i.e., $L_g:G\to G$ defined by $L_g(h)=gh$ and use differential.

Hence, $G$ and $\mathfrak{g}$ have the same dimension.