Corollary 2

Assume in addition that $\mathfrak{g}$ is solvable. Then:

(i) The subgroup of Aut($\mathfrak{g}$) consisting of the $e^{adx}, x\in \mathscr{C}^{\infty}\mathfrak{g}$ (cf. §3, no. 4), operates transitively on $\mathscr{T}_1$.

(ii) If $\mathfrak{t}\in \mathscr{T}_1,\mathfrak{g}$ is the semi-direct product of $\mathfrak{t}$ and $\mathfrak{n}_V(\mathfrak{g})$.

Assertion (i) follows from the fact that the group of the $e^{adx},x\in \mathscr{C}^{\infty}\mathfrak{g}$, operates transitively on $\mathscr{H}($§3, no. 4, Th. 3).

We prove (ii). Let $\mathfrak{t}\in \mathscr{T}_1$, and let $\mathfrak{h}=\psi (\mathfrak{t})$ be the corresponding Cartan subalgebra of $\mathfrak{g}$. In view of Prop. $5,\mathfrak{h}=\mathfrak{t}+\mathfrak{n}_V(\mathfrak{h})\subset \mathfrak{t}+\mathfrak{n}_V(\mathfrak{g})$. On the other hand, $\mathfrak{g}=\mathfrak{h}+ [\mathfrak{g},\mathfrak{g}] ($§2, no. 1, Cor. 3 of Prop. 4) and $[\mathfrak{g},\mathfrak{g}]\subset \mathfrak{n}_V(\mathfrak{g})$, so $\mathfrak{g}=\mathfrak{t}+\mathfrak{n}_V(\mathfrak{g})$. But it is clear that $\mathfrak{t}\cap \mathfrak{n}_V(\mathfrak{g}) =\{0\}$. The algebra $\mathfrak{g}$ is thus the semi-direct product of $\mathfrak{t}$ and the ideal $\mathfrak{n}_V(\mathfrak{g})$.