Proposition 1

Let $g$ be a Lie algebra, $n$ its largest nilpotent ideal, $a$ a nilpotent ideal of $g$ and $\rho$ a finite-dimensional representation of $a$ such that every element of $\rho(a)$ is nilpotent. Then $\rho$ admits a finite-dimensional extension $\sigma$ to $g$ such that every element of $\sigma(n)$ is nilpotent.

Let $a = n_0 \subset n_1 \subset \cdots \subset n_p = n$ be a sequence of subalgebras of $n$ such that $n_{i-1}$ is an ideal of $n_i$ of codimension 1 for $1 \leq i \leq p$ (§ 4, no. 1, Proposition 1 (e)). The algebra $n_i$ is therefore the direct sum of $n_{i-1}$ and a 1-dimensional subalgebra. As $\mathrm{ad}_{n}x$ is nilpotent for all $x \in n$, it is possible (Theorem 1) to find one by one finite-dimensional extensions $\rho_1, \rho_2, \ldots, \rho_p = \rho'$ of $\rho$ to $n_1, n_2, \ldots, n_p = n$ such that every element of $\rho'(n)$ is nilpotent.

Let $r$ be the radical of $g$ and let $n = r_0 \subset r_1 \subset \cdots \subset r_q = r$ be a sequence of subalgebras of $r$ such that $r_{i-1}$ is an ideal of $r_i$ of codimension 1 for $1 \leq i \leq q$ (§ 5, no. 1, Proposition 2 (d)). The algebra $r_i$ is thus the direct sum of $r_{i-1}$ and a 1-dimensional subalgebra. As $[r, r] \subset n$, it is possible (Theorem 1) to find one by one finite-dimensional extensions $\rho'_1, \rho'_2, \ldots, \rho'_q = \rho''$ of $\rho'$ to $r_1, r_2, \ldots, r_q = r$ such that every element of $\rho''(n)$ is nilpotent.

Finally $g$ is the direct sum of $r$ and a subalgebra $s$ (§ 6, no. 8, Theorem 5). As $[s, r] \subset n$, it is possible (Theorem 1) to find a finite-dimensional extension $\sigma$ of $\rho''$ to $g$ such that every element of $\sigma(n)$ is nilpotent.