Theorem 1 (Zassenhaus)

Let $g = g' + h$ be a Lie algebra which is the direct sum of an ideal $g'$ and a subalgebra $h$ and $\rho'$ a finite-dimensional representation of $g'$ whose largest nilpotency ideal contains $[h, g']$.

(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: