Is every semisimple element of a Borel subalgebra contained in a Cartan subalebra of the Borel subalgebra? Let $\mathfrak{g}$ be a semisimple Lie algebra (over $\mathbb{C}$), $\mathfrak{b}$ a Borel subalgebra of $\mathfrak{g}$, and $L$ a semisimple element contained in $\mathfrak{b}$. 
I know that $L$ is contained in a Cartan subalgebra (CSA) of $\mathfrak{g}$. Is it true that $L$ is contained in a CSA of $\mathfrak{b}$? If so, why?
 A: Here is another answer, using only Lie algebra theory. After applying ${\rm ad}_{\mathfrak{g}}$ we can assume that $\mathfrak{g}$ is a semisimple subalgebra of some $\mathfrak{gl}_k(V)$. As such, $\mathfrak{g}$ is almost algebraic, as defined in Jacobson's book, page 98 (called "decomposable" in the English translation of Bourbaki, Lie VII, or "scindable" in the original French version), i.e., it contains the semisimple and nilpotent components of all its elements. The Borel subalgebra $\mathfrak{b}$ is almost algebraic too. This follows for example from Bourbaki, Lie VII, section 5.2, Cor 1: The almost algebraic envelope $e(\mathfrak{b})$ of $\mathfrak{b}$ is a solvable subalgebra of $\mathfrak{g}$ because $\mathfrak{g}$ is almost algebraic, hence equals $\mathfrak{b}$, because $\mathfrak{b}$ is a maximal solvable subalgebra of $\mathfrak{g}$. Hence $\mathfrak{b}$ is an almost algebraic subalgebra of $\mathfrak{gl}_k(V)$.   
The element $L$ is semisimple. It is therefore contained in an abelian subalgebra $\mathfrak{t}$ of $\mathfrak{b}$ consisting of semisimple elements (called a toral subalgebra in Humphreys' book). By reasons of dimensions we can assume that $\mathfrak{t}$ is a maximal such subalgebra. Now Bourbaki, Lie VII, section 5.3, Proposition 6 says that the centralizer $\mathfrak{c}$ of $\mathfrak{t}$ in $\mathfrak{b}$ is a Cartan subalgebra of $\mathfrak{b}$. Since $\mathfrak{t}$ is abelian, we have $\mathfrak{c} \supset \mathfrak{t} \ni L$. 
