Regarding proof of Theorem 3.3 in Humphreys (Lie alg. and Rep.)

Theorem 3.3. of Humphreys goes something like this: given a subalgebra $\mathfrak{g}$ of $\mathfrak{gl}(V)$ where $V$ is nonzero, finite-dimensional and $\mathfrak g$ consists of nilpotent endomorphisms, then there is a nonzero vector $v \in V$ such that ${\frak g}.v = 0$. Assuming I understand the first part of the proof I understand the rest, however I am not confident that is the case. I paraphrase the first part of the proof:

By induction on $\dim \frak g.$Suppose $\frak h \subset \frak g$ a proper subalgebra. $\frak h$ acts via ad as a Lie algebra of nilpotent linear transformations on the vector space $\frak g$, hence on $\frak g / \frak h$. As the dimension of $\frak h$ is strictly smaller than that of $\frak g$, the induction hypothesis ensures a nonzero $x + \frak h$ killed by the image of $\frak h$ in ${\frak gl}(\frak g /\frak h)$.

Breaking this down, I want to confirm that this is the situation: we have a composite $$\frak h \to {\frak gl}(\frak g) \to {\frak gl}(\frak g / \frak h)$$ the first map being $y \mapsto \text{ad}_y(-)$, the latter being the canonical projection. For dimensional reasons the theorem is in this particular situation assumed true (by the inductive hypothesis), and nilpotency of $\text{ad}_y(-)$ follows from nilpotency of $y$, so we are in the situation of the theorem. Here I am starting to lose confidence; where is the subspace of $\frak gl(\frak g / \frak h)$? Why is the theorem valid for this composite?

The induction hypothesis is applied for the vector space $V=\mathfrak{g}/\mathfrak{h}$, which is of lower dimension, since $\mathfrak{h}$ is a proper subalgebra. So you find a vector $x+\mathfrak{h}\neq \mathfrak{h}$ in $V$ killed by the image of $\mathfrak{h}$ in $\mathfrak{gl}(V)$. It follows that $\mathfrak{h}$ is properly contained in its normalizer $N_{\mathfrak{g}}(\mathfrak{h})$, and the proof can go on. I do not see why you need composite maps.
• The induction hypothesis explicitly says: there exists $\overline{x}\neq \overline{0}$ in $V=\mathfrak{g}/\mathfrak{h}$ with $(\mathfrak{g}/\mathfrak{h}). \overline{x}=\overline{0}$. Now remember how the residue classes of $\mathfrak{g}/\mathfrak{h}$ look like. – Dietrich Burde Jun 30 '15 at 9:40
• Assuming here $({\frak g / \frak h})\cdot \bar{x} = 0$ means $\text{ad}_{\frak g / h}(\bar{x}) = 0$, otherwise the statement does not make sense. So, $V = \frak g / h$, the subalgebra consisting of nilpotent endomorphisms is $\text{ad}_{\frak g / h}$. Then the induction hypothesis guarantees $x + \frak h$ such that $\text{ad}_{\frak g / h}(x + {\frak h}) = 0$, or said differently, $[{\frak g / h}, x + {\frak h}] = 0$ in the quotient $\frak g / h$, hence necessarily in $\frak h$. As $x + \frak h$ is nonzero, this implies that the bracket $[y, x] \in \frak h$, where $x \not\in \frak h$. Correct? – Andrew Thompson Jun 30 '15 at 9:59
• Sorry for being silly about this. Will leave comments for future reference. If the reader should have a brain similar to mine; write $\bar{ad}$ for the composition $\pi \circ \text{ad} \colon \frak h \to \frak gl(g / h)$. Then $\bar{ad}_{\frak h}$ is a subalgebra of $\frak gl(g / h)$ consisting of nilpotent endomorphisms, and the induction hypothesis gives a nonzero $x + \frak h$ such that $\bar{ad}_{\frak h}(x + {\frak h}) = 0$, so $[{\frak h}, x] \in \frak h$ and as Dietrich pointed out, the proof may go on. – Andrew Thompson Jun 30 '15 at 10:14