How that the Lie algebra of the normalizer of a Lie group equal the normalizer of the Lie algebra

Solution 1:

The derivative of $\mathrm{Ad}(\mathrm{exp}_G(tx))$ is not in $\mathrm{GL}(\mathfrak{g})$ since as you note it is $\mathrm{ad}(tx)$, which is in general not invertible. However it does preserve $E$ which is seen for example by pulling $z$ inside the derivative as you suggested:

$$ \left(\frac{d}{dt}\Bigg|_{t=0} \mathrm{Ad}(\mathrm{exp}_G(tx))\right)z = \left(\lim_{t\to 0} \frac{\mathrm{Ad}(\mathrm{exp}_G(tx)) - I}{t} \right)z = \lim_{t\to 0}\frac{\mathrm{Ad}(\mathrm{exp}_G(tx))z - z}{t} = \gamma'(0) $$

Where $\gamma(t) = \mathrm{Ad}(\mathrm{exp}_G(tx))z$, this gives you that $[x, z] \in E$ since $\gamma(t) \in E$.

You still need to prove that every element of $\mathfrak{n}_\mathfrak{g}$ is an element of $L(N_G(E))$. To prove that you once again use that $\mathrm{Ad}(\mathrm{exp}_G(tx)) = e^{\mathrm{ad}(tx)}$ and note that since $\mathrm{ad}(x)$ preserves $E$ then $\mathrm{ad}(tx)$ does too and also $e^{\mathrm{ad}(tx)}(E) \subset E$, you can see this for example by writing the matrix exponential as an infinite sum. All in all this gives that $\mathrm{exp}_G(tx) \in N_G(E)$ and hence $x \in L(N_G(E))$.

I hope this answers your question.