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Let $\mathfrak{g}$ be a finite dimensional Lie algebra and $\mathfrak{h}$ a Lie subalgebra. If I know that $\mathfrak{g}/\mathfrak{h}$ is abelian, does it follow that $\mathfrak{g}/\mathfrak{h} \subset \mathfrak{g}/[\mathfrak{g}, \mathfrak{g}]$?

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The correct statement is that $\mathfrak{g}/\mathfrak{h}$ is a quotient, not a subalgebra. –  Qiaochu Yuan Mar 26 '12 at 4:22
Er, I'm not sure what you mean. –  Shayla Mar 26 '12 at 4:36
$\mathfrak{g}/\mathfrak{h}$ is a quotient, not a subalgebra, of $\mathfrak{g}/[\mathfrak{g}, \mathfrak{g}]$. –  Ted Mar 26 '12 at 5:27
g/h doesn't make sense as a Lie algebra unless h is an ideal, not just a subalgebra –  mt_ Mar 26 '12 at 7:29
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up vote 1 down vote accepted

Assuming $\mathfrak{h}$ is indeed an ideal (to make sense of the statement, as noted by Qiaochu), saying that $\mathfrak{g}/\mathfrak{h}$ is abelian means that for any $x,y \in \mathfrak{g}$, $[x,y]\in\mathfrak{h}$, hence $[\mathfrak{g},\mathfrak{g}]\subset\mathfrak{h}$. Now consider the application $$ \pi :\mathfrak{g}/[\mathfrak{g},\mathfrak{g}] \rightarrow \mathfrak{g}/\mathfrak{h}\\ \pi:x+[\mathfrak{g},\mathfrak{g}] \mapsto x+\mathfrak{h} $$ It is well-defined because $[\mathfrak{g},\mathfrak{g}]\subset\mathfrak{h}$, and it is obiously linear, and commutes with the bracket. Now by elementary commutative algebra, $\mathfrak{g}/\mathfrak{h}$ is isomorpic to the quotient of $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$ by the kernel of $\pi$, which is precisely what you want (following the remark of Ted).

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This is analogous to the situation in group theory. $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$ is the largest quotient of $\mathfrak{g}$ which is abelian. Just as $G/G'$ (where $G'=[G,G]$) is the largest abelian quotient of a group $G$. –  Bill Cook Mar 26 '12 at 13:18
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