Commutator of two Lie subalgebras Let $\mathfrak{g}$ be a Lie algebra, and $\mathfrak{h},\mathfrak{m}$ two Lie subalgebras. Is it true that $[\mathfrak{h},\mathfrak{m}]$ is a Lie algebra in itself?. I have tried to develop a commutator of the form $[[a,b],[a',b']]$ (where $a,a'\in\mathfrak{h}$, and $b,b'\in\mathfrak{m}$) by applying repeatedly Jacobi's rule, but it seems that one of the subalgebras must be an ideal.
 A: No. Let $\mathfrak{g}$ be $\mathfrak{gl}_3$. Inside, consider the subalgebra $\mathfrak{h}\simeq\mathfrak{gl}_2$ consisting of the upper left $2\times 2$ block and the subalgebra $\mathfrak{m}\simeq\mathfrak{gl}_2$ consisting of the lower right $2\times 2$ block. Then $[\mathfrak{h},\mathfrak{m}]$ is the 6-dimensional subspace of all matrices with zero diagonal. This is not a subalgebra.
When you say "one of the subalgebras must be an ideal", you probably mean that the answer is indeed yes if one of the subalgebras is an ideal [edit: which actually also fails, see Jendrik Stelzner's comment, pointing to this question]. It holds when both $\mathfrak{h}$ and $\mathfrak{m}$ are ideals, but of course this is not a necessary condition.

Added: Here's a generalization of the linked counterexample: let $\mathfrak{g}$ be an arbitrary split semisimple Lie algebra (in char. 0), with a Cartan grading, so $\mathfrak{h}=\mathfrak{g}_0$ is a split Cartan subalgebra. Then $[\mathfrak{g},\mathfrak{h}]=\bigoplus_{\alpha\neq 0}\mathfrak{g}_\alpha$, which is not a subalgebra if $\mathfrak{g}\neq\{0\}$.
