Lie algebra of a Normalizer

Lie algebra of a Normalizer

Let $G$ be a Lie group and let $\mathfrak{g}$ be its Lie algbera. Let $E$
be a subspace of $\mathfrak{g}$. Define the normalizer of $E$ in $G$ as:
$$ N_{G}(E) = \{ g \in G \;\; | \;\; Ad(g)E = E\}$$
Let the normalizer of $E$ in $\mathfrak{g}$ be defined as:
$$ \mathfrak{n}_{\mathfrak{g}} = \{ x \in \mathfrak{g} \;\; | \;\; [x,E]
\subseteq E \}$$
Then I must show that $N_{G}(E)$ is closed and its Lie algebra is
$\mathfrak{n}_{\mathfrak{g}}$.



I could prove that it is closed as the adjoint representation is
continuous. For the Lie algbera, this is what I did:
For $x \in L(N_{G}(E))$ (the Lie algebra of $N_{G}(E)$), and $z \in E$, we
have
$$ [x,z] = ad(x)(z) = \left(\frac{d}{dt}\bigg|_{t=0}e^{ad(tx)}\right)(z) =
\left(\frac{d}{dt}\bigg|_{t=0}Ad(exp_{G}(tx)\right)z$$
where $e^{x}$ is the matrix exponentiation and $exp_{G}$ is the Lie groups
exponential map. Now I know that $Ad(exp_{G}(tx))z \in E$ by various
definitions, but I do not know how to pull the $z$ inside the derivative.
Any ideas about the above?