Lemma 2

Let $\mathfrak{n}$ be a subalgebra of $\mathfrak{g}$ such that, for all $n\in \mathfrak{n}$, ad$_{\mathfrak{g}}(n)$ is nilpotent. Let $h\in \mathfrak{g}$ be such that $[h,\mathfrak{n}] =\mathfrak{n}$. Then $e^{ad_{\mathfrak{g}}\mathfrak{n}}.h=h+\mathfrak{n}$.

It is clear that $e^{ad_{\mathfrak{g}}(\mathfrak{n})}.h\subset h+\mathfrak{n}$. We shall prove that, if $v\in \mathfrak{n}$, then $h+v\in e^{ad_{\mathfrak{g}}(\mathfrak{n})}.h$. It suffices to prove that $h+v\in e^{ad_{\mathfrak{g}}(\mathfrak{n})}.h+\mathscr{C}0\mathfrak{n}$ for all $p\geq 1$ (since $\mathscr{C}^p\mathfrak{n}= 0$ for sufficiently large $p$). This is clear for $p= 1$ since $\mathscr{C}^1\mathfrak{n}=\mathfrak{n}$. Assume now that we have proved the existence of $y_p\in \mathfrak{n}$ and $z_p\in \mathscr{C}^p\mathfrak{n}$ such that $h+v=e^{ad_{\mathfrak{g}}y_p}.h+z_p$. Since (ad$_{\mathfrak{g}}h$)$(\mathfrak{n}) =\mathfrak{n}$, (ad$_{\mathfrak{g}}h$)$|\mathfrak{n}$ is a bijection from $\mathfrak{n}$ to $\mathfrak{n}$, hence its restriction to $\mathscr{C}^p\mathfrak{n}$, which leaves $\mathscr{C}^p\mathfrak{n}$ stable, is also bijective; consequently, there exists $z\in \mathscr{C}^p\mathfrak{n}$ such that $z_p= [z, h]$. Then

so

which establishes our assertion by induction on $p$.