How to prove that the adjoint group is a Lie subgroup of $Gl(\mathfrak{g})$

Question

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. Let $Ad: G \rightarrow GL(\mathfrak{g})$ be the Adjoint representation. I want to prove that $Ad(G)$ is a Lie subgroup of $GL(\mathfrak{g})$.

Here is what I tried: From Cartan's Theorem, it suffices to show that $Ag(G)$ is closed in $GL(\mathfrak{g})$. So take $x \in Ad(G)$ and $(Ad(g_n))_n$ a sequence in $Ad(G)$ that converges to $x$, with $g_n \in G$. I can't continue from here. It may be possible that some additional other hypothesis are required.

Thanks!

Answer 1

User proposition 7.1 in San Martin book. http://www.ime.unicamp.br/~smartin/cursos/grupolie-2013/gruplie0.pdf

Answer 2

Hilgert, Neeb: The Structure and Geometry of Lie Groups, Exercise 18.1.2

Pick $\theta\in{\mathbb R}$ irrational and define $G := {\mathbb C}^2\rtimes_\alpha{\mathbb R}$, where $\alpha: {\mathbb R}\to {\text{GL}}({\mathbb C}^2)$ is given by $$\alpha(t) := \begin{pmatrix} e^{it} & 0 \\ 0 & e^{i\theta t}\end{pmatrix}.$$ Then $\text{Ad}(G)$ is claimed not to be closed in $\text{GL}({\mathfrak g})$, though I haven't had the time to check.