The Cartan subalgebras of a reductive Lie algebra are abelian.
Are there non-reductive Lie algebras with abelian Cartan subalgebras?
In fact, the elements of a Cartan subalgebra of a reductive Lie algebra are semisimple, so a weaker question is:
Are there non-reductive Lie algebras with abelian Cartan subalgebras all of whose elements are semisimple?
N.B.: I asked this earlier at math.SE.
The subalgebra of diagonal matrices in the (non-reductive) standard Borel subalgebra ${\mathfrak b}$ of ${\mathfrak gl}_2$ is an abelian Cartan subalgebra of ${\mathfrak b}$, since it is abelian (hence nilpotent) and its own normalizer (directly calculate the bracket of an element $X$ of ${\mathfrak b}$ with a diagonal matrix $A$ to see that the result is diagonal only when $X$ was in fact diagonal or $A$ was in the center).
Maybe it should be emphasized that these questions deal with finite dimensional Lie algebras over a field of characteristic 0; in prime characteristic there are further complications. BR has addressed the first question and observed that a subalgebra of a reductive Lie algebra containing a given Borel subalgebra (in other words a parabolic subalgebra) will have the same type of abelian Cartan subalgebras found in the big algebra.
The follow-up question is more problematic, since the notion of "semisimple" element in a Lie algebra isn't always intrinsically defined. In fact, there is an intrinsic Jordan decomposition precisely when the Lie algebra equals its derived algebra (which for instance excludes nonzero abelian Lie algebras): this is proved by building on the classical theorem for semisimple Lie algebras, in a paper by Cagliero and Szechtman here. (The intrinsic nature of the Jordan decomposition involves good behavior under linear representations. Leaving that aside, parabolic subalgebras of a given semisimple Lie algebra do contain "semisimple" and "nilpotent" elements relative to the big algebra but not relative to their own representations taken in isolation.)
Anyway, I'm not sure right away how much can be said about Cartan subalgebras in a non-reductive Lie algebra which equals its derived algebra. Examples?
If $H$ is a Cartan subalgebra of a Lie algebra $L$, and $A$ is an associative commutative algebra, then $H \otimes A$ is a Cartan subalgebra of $L \otimes A$. Specializing this to the case $L$ classical simple, we get another positive answer to the first question (but not to the second one). This example can be, probably, varied, by taking subalgebras of $L \otimes A$, or adding "tails" of derivations, etc.
