Does semisimple element of a semisimple complex Lie algebra belong to some Cartan subalgebra? Main question: Suppose that $\mathfrak g$ is a complex Lie algebra over $\mathbb C$ and let $J$ be a semisimple element of $\mathfrak g$ (meaning that $\mathrm{ad}_J$ is a diagonalizable operator on $\mathfrak g$). Is there a Cartan subalgebra $\mathfrak h \subseteq \mathfrak g$ which contains $J$?
Some motivation: for any Cartan subalgebra $\mathfrak h \subseteq \mathfrak g$ all elements of $\mathfrak h$ are semisimple. Moreover there exists a condition on an element $J$ of $\mathfrak g$ stronger than semisimplicity, called regularity (cf. Serre's book on semisimple Lie algebras) which guarantees existence of a unique Cartan subalgebra $\mathfrak h \subseteq \mathfrak g$ such that that $J \in \mathfrak h$. Here I ask if existence (but with no uniqueness) can be inferred under weaker assumptions. In fact, if the answer to the main question is negative, I would like to ask a slightly more general question.
Generalization: Let $\mathfrak g$ be as above and let $S$ be the union of all Cartan subalgebras of $\mathfrak g$. Is it possible to describe explicitly the set $S$? Let me just mention that $S$ is clearly dense, as it contains the non-empty, Zariski open set of all regular elements of $\mathfrak g$.
 A: Yes, and a more general statement is true even over general fields of characteristic $0$, according to Bourbaki's take on Cartan subalgebras (in book VII, §2 of the volume on Lie Groups and Lie Algebras). Namely, proposition 10 says that for an abelian subalgebra $\mathfrak{a} \subseteq \mathfrak{g}$ consisting of semisimple elements,
$$\lbrace \text{Cartan subalgebras of } \mathfrak{g} \text{ containing } \mathfrak{a} \rbrace = \lbrace \text{Cartan subalgebras of } \mathfrak{z}_{\mathfrak{g}}(\mathfrak{a}) \rbrace $$
($\mathfrak{z}_{\mathfrak g} =$ centraliser). But every Lie algebra has Cartan subalgebras (see e.g. corollary 1 to theorem 1 loc. cit.), in particular so does $\mathfrak{z}_{\mathfrak{g}}(\mathbb{C} J)$ in your question.
A: Oh, I just saw that in your question title, you restrict to semisimple Lie algebras. Well there's a much easier argument for this case:
Cartan subalgebras of semisimple Lie algebras can equivalently be characterised as the maximal toral ones, where a subalgebra is called "toral" iff it is abelian and consists of semisimple elements. (E.g. Humphreys actually uses this as definition; cf. my answer to Are there common inequivalent definitions of Cartan subalgebra of a real Lie algebra?; a proof for equivalence of these definitions is an exercise in Bourbaki, and done in prop. 3.1.5 of my thesis.) With that, it's just a usual maximality argument in finite dimension.
Again, this works over any field of characteristic $0$ (just that "semisimple" does not necessarily mean diagonalisable, but, well, semisimple, i.e. diagonalisable over an algebraic closure).
