I'm currently working out the equivalence of the following definitions, when the ambient space $\mathfrak{g}$ is a complex, semi-simple Lie algebra.
(1) $\mathfrak{h}$ is called a Cartan-1 subalgebra when $\mathfrak{h}$ is maximally abelian (not contained in a larger abelian subalgebra) and $ad$-diagonalizable, that is, $ad_X$ is diagonalizable for any $X \in \mathfrak{h}$.
(2) $\mathfrak{h}$ is called Cartan-2 subalgebra when $\mathfrak{h}$ is its own normalizer, $N(\mathfrak{h}) = \mathfrak{h}$ and $\mathfrak{h}$ is nilpotent.
Here, the normalizer of $\mathfrak{h}$ in $\mathfrak{g}$ is $N(\mathfrak{h}) := \{ X \in \mathfrak{g} \; | \; ad_X(\mathfrak{h}) \subset \mathfrak{h} \}$. For clarity, I'm trying to prove
Theorem - Let $\mathfrak{g}$ be a complex semi-simple Lie algebra. Then a sub-algebra $\mathfrak{h}$ is Cartan 1 iff it is Cartan 2.
So far, I've been able to deal with one direction. I showed Cartan 1 implies Cartan 2. For the converse Cartan 2 implies Cartan 1, I would like to proceed as follows:
pf: C2 -> C1: First, we can show any $\mathfrak{h}$ that is Cartan 2 will be maximally nilpotent. So, it suffices to show that $\mathfrak{h}$ is abelian and diagonalizable, as we will then get maximally abelian for free. On the other hand, we actually need only show diagonalizability of $\mathfrak{h}$. For then take any $X \in \mathfrak{h}$ and $ad_X$ is diagonalizable, acts nilpotently on $\mathfrak{h}$ by Engel's Theorem. We conclude that $ad_X|_{\mathfrak{h}} \equiv 0$ so that $\mathfrak{h}$ is abelian. Now, on to proving diagonalizability. Here, I want to use the Jordan-Chevalley decomposition to write $ad_X = S +N$ for some $S,N \in \mathfrak{gl}(\mathfrak{g})$. Doing some clever work, we can show that $S, N$ are actually derivations on $\mathfrak{g}$. Since $\mathfrak{g}$ is semi-simple, we see that $S,N$ are inner derivations. That is, $S = ad_Y, N = ad_Z$ for some $Y, Z \in \mathfrak{g}$. But then since $S, N$ are polynomials in $ad_X$ by Jordan-Chevalley, we see that they preserve the subalgebra $\mathfrak{h}$. Hence, $Y, Z \in N(\mathfrak{h}) = \mathfrak{h}$, by definition. Ok, my suspicion is that we can finish the proof by using the non-degeneracy of the killing form to show that $Z =0$ and hence $ad_X = S$ is diagonalizable. I have been unable to put the pieces together yet, though.
I'd very much appreciate someone who can help me finish this proof or offer guidance on a simpler way to prove Cartan-2 implies Cartan 1.