Let $K$ be a field of characteristic zero, let $\mathfrak{g}$ be a nilpotent Lie algebra over $K$, and let $\mathfrak{h}$ be a self-centralising abelian ideal of $\mathfrak{g}$, and we assume that $\mathfrak{g}/\mathfrak{h}$ is also abelian.
Given an element $x\in\mathfrak{g}$ and a subset $Y$ of $\mathfrak{h}$, we denote:
$C_{\mathfrak{h}}(x):=\{u\in\mathfrak{h}:[x,u]=0\}$,
$C_{\mathfrak{g}}(Y):=\{v\in\mathfrak{g}:[y,v]=0$ for all $y\in Y\}$.
My question is, can we find an element $x\in\mathfrak{g}$ such that $x\notin\mathfrak{h}$, and if $\mathcal{Z}$ is the centre of $\mathfrak{g}$ then:
\begin{equation} \dim_{K}\mathfrak{g}\geq \dim_K C_{\mathfrak{h}}(x)+\dim_K{C_{\mathfrak{g}}(C_{\mathfrak{h}}(x))}-\dim_K\mathcal{Z}. \end{equation}
Any ideas or counterexamples would be welcome.
If $\mathfrak{g}$ is Abelian, then there is no such element, since $\mathfrak{h}=\mathfrak{g}$ in this case. So the answer is no for Abelian Lie algebras.
Now let $g'\neq 0$. From the conditions we get $\mathcal{Z},\mathfrak{g}'\subseteq \mathfrak{h}$, and $C_\mathfrak{h}(x)$ is an Abelian ideal for all $x\in \mathfrak{g}$. Our statement then reads $\dim \mathfrak{g}/C_\mathfrak{h}(x) \geq \dim C_\mathfrak{g}(C_\mathfrak{h}(x))/\mathcal{Z}$ which is certainly true, as $Z\subseteq C_\mathfrak{h}(x)$. Since this is true for any element, the only remaining question is, whether there is such an ideal $\mathfrak{h}\subsetneq \mathfrak{g}$ at all.