I saw a question which was about a 3-dimensional Lie algebra with a 2-dimensional derived subalgebra, and it was asserted that the derived subalgebra must be abelian. If $\mathfrak{g}$ is 3-dimensional with 2-dimensional derived subalgebra $[\mathfrak{g},\mathfrak{g}]$, is the derived subalgebra nessecarily abelian.
Could someone show me a proof of this as I've been unable to prove it myself. Thanks very much!
A $2$-dimensional subalgebra is solvable, and abelian if nilpotent.
It follows from that that if $g$ is a $3$-dimensional Lie algebra with $\dim g'=2$, then $g$ is solvable. Indeed, there is then a short exact sequence $0\to g'\to g\to g/g'\to0$ and both the ideal $g'$ and the quotient $g/g'$ are solvable.
Now a solvable Lie algebra has nilpotent derived algebra, so $g'$ is $2$-dimensional and nilpotent. Our opening observation, then, implies that $g'$ must be abelian.
