# Lie algebra of the automorphism group of an arbitrary Lie Algebra?

I'm trying to prove the Lie algebra of $$Aut(\mathfrak{g})$$ is $$Der(\mathfrak{g})$$, the set of all derivations over $$\mathfrak{g}$$. Where $$\mathfrak{g}$$ is an arbitrary finite lie algebra.

Due to $$\mathfrak{g}$$ is finite Lie algebra, I am using that $$Lie(Aut(\mathfrak{g}))=\{X \in M(n, \mathbb{C}) / e^{tX}\in Aut(\mathfrak{g})\}$$.

How can I prove if $$\delta \in Der(\mathfrak{g})$$ then $$e^{t\delta}([X,Y])=[e^{t\delta}(X),e^{t\delta}(Y)]$$?

And if $$\delta \in Lie(Aut(\mathfrak{g}))$$ then $$\delta \in Der(\mathfrak{g})$$?

Let $$\varphi(t)$$ be a smooth path of automorphisms with $$\varphi(0)=\mathrm{id}$$. Then for $$X,Y\in\mathfrak g$$ you've got:

$$0=\frac d{dt}[X,Y]\lvert_{t=0}= \frac d{dt}[\varphi(t) X,\varphi(t)Y]\lvert_{t=0}=[\varphi'(0)X, Y]+[X,\varphi'(0)Y]$$ and $$\varphi'(0)$$ is a derivation. Hence the elements of Lie algebra of the automorphism group are derivations.

And the other direction? Let $$\delta$$ be a derivation, then:

$$[\delta^n X,\delta^m Y]= (-1)^m[\delta^{n+m}X,Y]$$

and

$$[\sum_n\frac{(t\delta)^n}{n!}X, \sum_m \frac{(t\delta)^m}{m!}Y]=\sum_{n,m}\frac{(-1)^mt^{n+m}}{n!\,m!}[\delta^{n+m}X,Y]$$ do a substitution of the sums, let $$k=n+m$$ and multiply by $$\frac{k!}{k!}$$ and $$1^{k-m}$$. Then the sum becomes:

$$\sum_k\frac{t^k}{k!}\sum_{m≤k}\frac{(-1)^m 1^{k-m}k!}{m!\,(k-m)!} [\delta^{k}X,Y] = \sum_k \frac{t^k}{k!}\sum_{m≤k}(1+(-1))^k[\delta^kX,Y]$$

Now only the $$k=0$$ term survives, so you are left with $$[X,Y]$$.