A maximal abelian subalgebra that is not a Cartan subalgebra

The one-dimensional subalgebra $\mathbb{C}Y$ is abelian and maximal abelian: no abelian subalgebra can strictly contain it, because $\mathfrak{g}$ itself is not abelian, as $[X,Y]=Y\mathrm{\ne }0$. It is also a nonzero ideal, since both $[X,Y]$ and $[Y,Y]$ lie in $\mathbb{C}Y$. Being abelian it is solvable, so $0\mathrm{\ne }\mathbb{C}Y\subseteq \mathrm{rad}(\mathfrak{g})$; consequently $\mathfrak{g}$ is not semisimple.

However $N_{\mathfrak{g}}(\mathbb{C}Y)=\mathfrak{g}$: for $aX+bY$ one has $[aX+bY,Y]=aY\in \mathbb{C}Y$ and $[Y,Y]=0$, so every element of $\mathfrak{g}$ normalizes $\mathbb{C}Y$, while $\mathbb{C}Y\mathrm{\ne }\mathfrak{g}$. Hence $\mathbb{C}Y$ is not a Cartan subalgebra, since a Cartan subalgebra must equal its normalizer.

The notion is therefore strictly finer than maximal abelianness in this nonsemisimple algebra: nonsemisimplicity was proved in step 1.1, and $\mathbb{C}X$ is a Cartan subalgebra, as $[aX+bY,X]=-bY\in \mathbb{C}X$ forces $b=0$, so $N_{\mathfrak{g}}(\mathbb{C}X)=\mathbb{C}X$ and $\mathbb{C}X$ is abelian, whereas $\mathbb{C}Y$ is maximal abelian but not Cartan. This witnesses the failure of the claimed equivalence.