Reductive Lie algebra Does it exist a Lie algebra $\mathfrak{g}$ that is reductive but if we consider the inclusion of Lie agebras $\mathfrak{g} \subset \mathfrak{h}$ then $\mathfrak{g}$ is not reductive in $\mathfrak{h}?$
 A: As remarked by Jim Humphreys in a comment to my answer to a previous question, the notion of reductive for a Lie algebra (in characteristic zero) has no intrinsic interest, which means that the answer to this question has to be positive.
Here is one possible construction.  Let $\mathfrak{g} = \mathfrak{s} \oplus \mathfrak{z}$ be a reductive Lie algebra, where $\mathfrak{s}$ is semisimple and $\mathfrak{z}$ is the centre of $\mathfrak{g}$.  Consider a representation $V$ of $\mathfrak{g}$ where $\mathfrak{z}$ does not act semisimply.  Now let $\mathfrak{h}$ be the semidirect product $\mathfrak{g} \ltimes V$, with $V$ abelian.
A: A reductive Lie algebra $L$ is the direct sum of a semisimple
Lie algebra $L_1$ and an abelian Lie algebra $L_2$. Let's consider
the case where $L_2$ is one-dimensional.
We can embed $L$ into a larger Lie algebra $L=L_1\oplus L_2'$
by embedding $L_2$ into $L_2'$. Let $L_2'$ be the two-dimensional
Lie subalgebra
$$\left(\begin{array}{cc}
*& *\\\
0& 0
\end{array}
\right)$$
of $\mathfrak{gl}(\mathbf{C})$
and
$$L_2=\left(\begin{array}{cc}
0& *\\\
0& 0
\end{array}
\right).$$
Then $L_2$ does not act semisimply on $L_2'$, so $L$ does not
act semisimply on $L'$.
