Automorphism group of a lie algebra as a lie subgroup of $GL(\frak g)$ Let $G$ be a lie group with lie algebra $\frak{g}$. Let $Aut(\frak g)$ be the automorphism group of $\frak{g}$. 
Its clear to me that $Aut(\frak{g})$ $\subset GL(\frak{g})$ since any automorphism of $\frak{g}$ is also a linear transformation. Now i have several questions:


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*How do i prove that $Aut(\frak{g})$ is actually a lie subgroup of $GL(\frak{g})$? I've tried playing with actions to realize $Aut(\frak g)$ as an orbit or an isotropy subgroup but nothing worked out...

*Say $H \subset G$ is an abstract subgroup (not necessarily lie). Are the following statements equivalent?
$$\{T_e H \text{  is closed under the lie bracket of } \frak{g} \}$$ 
$$ \{H \text{is a lie subgroup of } G\}$$


*How does the lie algebra of $Aut(\frak g)$ look like? When does $Aut(\frak g)$ $= G$? what does $Aut(\frak g)$ tell me about $G$?

 A: *

*You are almost there: The group $Aut(\mathfrak{g})$ of automorphisms of a Lie algebra $\mathfrak{g}$ is closed in the group $End(\mathfrak{g})^×$ of vector space automorphisms, hence it is a Lie group. 

*No, only one direction is true in general: If $H$ is a Lie subgroup of $G$, then $\mathfrak{h}\simeq T_eH\subset T_eG\simeq \mathfrak{g}$ is a Lie subalgebra. Conversely we only know that if $\mathfrak{h}\subset \mathfrak{g}$ is a Lie subalgebra, there exists a unique connected Lie subgroup
$H \subset G$ with Lie algebra $\mathfrak{h}$. 

*The group $Aut(\mathfrak{g})$ is a Lie subgroup of $GL(\mathfrak{g})$ ( and not of $G$). Its Lie algebra is given by $Der(\mathfrak{g})$, the algebra of derivations. It is an important invariant for $\mathfrak{g}$ resp. $G$. It can be "small", i.e., $Der(\mathfrak{g})=Inn(\mathfrak{g})$, for example for semisimple Lie groups, or "big", i.e., $Der(\mathfrak{g})=End(\mathfrak{g})$, for example for abelian Lie groups.
Analoguous results hold for the automorphism group. We have the exponential mapping $\exp : \mathfrak{g}\rightarrow G$.

