For a nilpotent Lie subalgebra, $\mathfrak{h}$, is $ad(\mathfrak{h})$ simultaneously diagonalizable if each $ad(H)$ is diagonalizable? Let $\mathfrak{g}$ be a Lie algebra and $\mathfrak{h}\subseteq \mathfrak{g}$ be a nilpotent subalgebra such that for every $H \in \mathfrak{h}$, the adjoint map $ad(H): \mathfrak{g} \rightarrow \mathfrak{g}$ is diagonalizable.  Does it follow that the set of endomorphisms $ad(\mathfrak{h})$ is simultaneously diagonalizable?
 A: Yes, and you don't even need to assume that $\mathfrak h$ is nilpotent.
Lemma: Let $\mathfrak h$ be a subalgebra of a Lie algebra $\mathfrak g$ (finite dimensional over a field $K$) such that for each $h \in \mathfrak h$, $ad(h) : \mathfrak g \rightarrow \mathfrak g$ is diagonalisable. Then $\mathfrak h$ is abelian. (Consequently, all its elements are simultaneously diagonalisable.)
Proof (Here I gave section 8.1 of Humphreys' Introduction to Lie Algebras and Representation Theory as reference for this. I do not have that reference handy right now, but I think the following works):
Let $x \in \mathfrak h$. Since $\mathfrak h$ is invariant under $ad(x)$, the restriction $ad_{\mathfrak h}(x)$ is also diagonalisable, i.e. $\mathfrak h$ has a basis $y_1, ..., y_n$ such that for each $1 \le i \le n$, there exists $a_i \in K$ with $[x,y_i] = a_i \,y_i$. Then for each $i$, the subspace $\mathfrak h_i :=\mathrm{span}(x,y_i)$ is also invariant under $ad(y_i)$, which by assumption is diagonalisable, hence again its restriction to $\mathfrak h_i$ is diagonalisable as well. But if any $a_i \neq 0$, then $x,y_i$ would be a basis of $\mathfrak h_i$, and the matrix of $ad_{\mathfrak h_i}(y_i)$ in that basis would be $\pmatrix {0 & 0\\-a_i&0}$, contradiction. So all $a_i =0$.

Note: If we already know that $\mathfrak h$ is nilpotent, we can shorten the above proof by saying that -- not $ad(x)$, but -- $ad_{\color{red}{\mathfrak h}}(x)$ is diagonalisable (as above, because it's the restriction of a diagonalisable map to an invariant subspace) and at the same time nilpotent (by Engel's theorem on the Lie algebra $\mathfrak h$, as alluded to in comments), which forces it to be zero.
Note also that the lemma becomes false if we try to generalize from $ad$-diagonalisable to $ad$-semisimple elements. I blundered about this in my very first MSE question, an easy counterexample is $\mathfrak g = \mathfrak h = \mathfrak{su}_2$ (or any compact simple real Lie algebra).
