(a) There exists a finite-dimensional extension $\rho$ of $\rho'$ to $g$ whose largest nilpotency ideal contains that of $\rho'$.

(b) If for all $x \in h$ the restriction to $g'$ of $\operatorname{ad}_gx$ is nilpotent, $\rho$ can be chosen such that moreover the largest nilpotency ideal of $\rho$ contains $h$.

Let $U'$ be the enveloping algebra of $g'$. Suppose that $U'$ and ${U'}^*$ have the $g$-module structures defined at the beginning of this no.

We now show that $S$ is finite-dimensional over $K$. Let $V'$ be the space on which $\rho'$ operates and $V' = V'_0 \supset V'_1 \supset \cdots \supset V'_d = \{0\}$ a Jordan-Hölder series of the $g'$-module $V'$. Let $\rho'_i$ be the representation of $g'$ on $V'_{i-1}/V'_i$ derived from $\rho'$ ($1 \leq i \leq d$). Let $I' \subset U'$ be the intersection of the kernels of the $\rho'_i$ (identified with representations of $U'$). Then

and $I' \cap g'$ is the largest nilpotency ideal of $\rho'$. By § 2, no. 6, Corollary to Proposition 6, ${I'}^d$ is of finite codimension in $U'$. If $x \in h$, the derivation $u \mapsto xu - ux$ of $U'$ maps $g'$ into $[h, g'] \subset I'$, hence $U'$ into $I'$ and hence ${I'}^d$ into ${I'}^d$. On the other hand, clearly ${I'}^d$ is a sub-$g'$-module of $U'$. Hence ${I'}^d$ is a sub-$g$-module of $U'$. The orthogonal of ${I'}^d$ in ${U'}^*$ is a finite-dimensional sub-$g$-module which contains $C(\rho')$ and therefore $S$. This shows that $S$ is finite-dimensional over $K$. For $x \in I' \cap g'$, $x^d$ is obviously contained in the annihilator of the $g$-module $U'/{I'}^d$ and hence also in the annihilator of the $g$-module $S$.

We saw in no. 1 that the $g'$-module $V'$ is isomorphic to a sub-$g'$-module of a product $(C(\rho'))^n$. Hence the $g$-module $S^n$ provides a finite-dimensional extension $\rho$ of $\rho'$ to $g$. Moreover, $\rho(x)$ is nilpotent for $x \in I' \cap g'$; as $I' \cap g'$ is an ideal of $g$ (for it contains $[h, g']$ by hypothesis), we see that $I' \cap g'$ is contained in the largest nilpotency ideal of $\rho$. Thus (a) is proved.

Suppose finally that for all $x \in h$ the restriction to $g'$ of $\mathrm{ad}_{\psi}x$ is nilpotent. As the elements of $h$ operate on $U'$ by derivations, there exists for all $u \in U'$ and all $x \in h$ an integer $e$ such that $(x_{U'})^e.u = 0$; the endomorphisms derived from $x_{U'}$ on $U'/{I'}^d$ and on $S$ (which are finite-dimensional spaces) are therefore nilpotent. Thus $\rho(x)$ is nilpotent for all $x \in h$. We have seen earlier that this is also true for $x \in I' \cap g'$. As $I' \cap g'$ is an ideal of $g'$ containing $[h, g']$, the sum $h + (I' \cap g')$ is also an ideal of $g$. Assertion (b) of Theorem 1 then results from the following lemma:

Passing to the quotient by the kernel of $\sigma$, $\sigma$ may be assumed to be faithful. Then $g'$ and $h$ are nilpotent and hence $g$, which is an extension of a quotient of $h$ by $g'$, is solvable. Then $h$ and $g'$ are contained in the largest nilpotency ideal of $\sigma$ (§ 5, no. 3, Corollary 6 to Theorem 1).

For an improvement of Theorem 1, cf. Exercise 4.

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.

More precisely:

The 1-dimensional Lie algebra $K$ admits finite-dimensional faithful representations $\tau$ such that every element of $\tau(K)$ is nilpotent, for example the representation

It is easily deduced that the centre $c$ of $g$, which is a product of 1-dimensional algebras, admits a finite-dimensional faithful representation $\sigma$ such that every element of $\sigma(c)$ is nilpotent. Let $\sigma_1$ be a finite-dimensional extension of $\sigma$ to $g$ such that every element of $\sigma_1(n)$ is nilpotent (Proposition 1); if $k$ denotes the kernel of $\sigma_1$, then $k \cap c = \{0\}$. On the other hand, let $\sigma_2$ be the adjoint representation of $g$, whose kernel is $c$; every element of $\sigma_2(n)$ is nilpotent. The direct sum $\rho$ of $\sigma_1$ and $\sigma_2$ is finite-dimensional; every element of $\rho(n)$ is nilpotent; and the kernel of $\rho$, contained in $k$ and in $c$, is zero, so that $\rho$ is faithful.