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I am a bit confused about semisimple Lie algebras.

For the sake of simplicity, let's take $\mathfrak{g}=M_n(k)$ where $k=\bar{k}$. According to Wiki, $M_n(k)$ is solvable if the radical of $M_n(k)$ is zero. But the set of all upper triangular matrices $\mathfrak{b}_n$ is a subalgebra of $M_n(k)$, which is solvable. Why would that not imply that $M_n(k)$ is solvable?

$\mathit{Edit}$: Here is a better question. Suppose $\mathfrak{h}$ is a subalgebra of a Lie algebra $\mathfrak{g}$. Suppose $\mathfrak{h}$ is solvable.

$\mathbf{first \; question}$: When is it the case that solvability cannot be induced to $\mathfrak{g}$?

$\mathbf{second \; question}$: When is it the case that solvability can be induced to $\mathfrak{g}$?

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    Thanks for the additional comment, Qiaochu!2012-06-28

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