I have the following exercise:
Consider a Lie algebra $\mathfrak{g}$. Decompose $\mathfrak{g}$ using the Levi decomposition, so $\mathfrak{g}=\mathfrak{s}\oplus \mathfrak{r}$. Let $\mathfrak{a}$ be a maximal torus inside the radical. Is it always true that $[\mathfrak{s},\mathfrak{a}]=0$? Discuss which hypotesis are needed to obtain such a result.
Now I started assuming that $\mathfrak{g}$ is algebraic (to avoid exceptional or strange cases, if any), and I procede as suggested, so I get
$$\mathfrak{g}=\mathfrak{s}\oplus \mathfrak{a}\oplus\mathfrak{m}$$
for some complement $\mathfrak{m}$ of $\mathfrak{a}$ in the radical $\mathfrak{r}$. For sure $[\mathfrak{s},\mathfrak{a}]\subset \mathfrak{r}$. But then I don't have any idea, nor I can prove it is true always (I guess it is false).
For the second question, I can think only of "trivial" answers: it is true for solvable Lie algebras and for semi-simple Lie algebras.
For characterisitic zero, it is true for reductive Lie algebra, because the radical is the center, so it is abelian and commute with everythings.
Suggestions?
Best Answer
Hint: Any element $x$ of a Lie algebra (over a field $k$ say) is contained in the abelian Lie subalgebra $k \cdot x$. Hence by your definition of maximal torus, by a "finite dimension" argument, every element of $\mathfrak r$ is contained in some maximal torus $\mathfrak a$ inside $\mathfrak r$.
That means that for $\mathfrak s$ to commute with all such maximal tori $\mathfrak a$, it has to commute with ...