An ideal that contains the commutator of a solvable Lie algebra

Question

Let $\mathfrak g$ be a real solvable Lie algebra and $\mathfrak n$ be an ideal of $\mathfrak g$ such that the commutator algebra $\mathfrak g'$ is contained in $\mathfrak n$. Now, let $\mathfrak m$ be any Lie subalgebra of $\mathfrak g$ such that $\mathfrak m\lhd \mathfrak n$. Then, is $\mathfrak m\lhd \mathfrak g$?

Answer 1

The answer is no. Consider the Heisenberg algebra $\mathfrak{h}$ with generators $x,y,z$ and bracket relations $$ [x,y]=z,\, [y,z]=[z,x]=0,$$ and take $\mathfrak{n}=\mathrm{Span}\{z,x\}$ and $\mathfrak{m}=\mathrm{Span}\{x\}$.