Let $\frak{g}$ be a finite dimensional Lie algebra over an algebraically closed field, $k$, of characteristic $0$.
Suppose that $I \triangleleft \frak{g}$ is an ideal of co-dimension $1$, and that $\{0\} \neq I \cap Z(\frak{g})$.
Moreover, define $cen(I) = \{g \in \frak{g} : \forall$ $x \in I, [x,g] =0\}$.
Prove that $[\frak{g}$$,cen(I)] \neq cen(I)$.
Where I'm at:
First, it can be easily shown that $cen(I)$ is an ideal itself, and hence we get that $[\frak{g}$$,cen(I)] \subset cen(I)$. So we must show that the inclusion is strict.
Since $I$ is of co-dimension $1$ we have that $[\frak{g},g] \subset$ $I$. Therefore $[\frak{g}$$,cen(I)] \subset I$.
Clearly, if $Z(\frak{g})$ isn't contained in $I$ we can write that $\frak{g} =$ $kx_0 \oplus I$ for $x_0 \in Z(\frak{g})$. This gives $[\frak{g}$$,cen(I)] = [kx_0 \oplus I, cen(I)] = 0$ and we'd be done, since $I \cap Z(\frak{g})$ is non-zero and is contained in $cen(I)$.
Thus we can assume: $\{0\} \neq Z(\frak{g}) \subset$ $I$.
Here I got stuck; I was trying to use Lie's lemma, and find a functional that is non zero on $cen(I)$, and has a common eigenvector. This would give the required inequality. However it is not clear to me how I may do so.
Moreover, I'm having trouble interpreting the assumption that $\{0\} \neq I \cap Z(\frak{g})$
Any hints?
Let $x$ an element which is not in $I$, $[g,cen(I)]=[kx\oplus I,cen(I)]=[kx,cen(I)]$, this implies that $[g,cen(I)]$ is the image of the restriction of $ad_x$ defined by $ad_x(y)=[x,y]$ to $cen(I)$, but the kernel of the restriction of $ad_x$ to $cen(I)$ is not trivial since it contains $Z(I)$, thus $ad_x$ is not surjective and $[g,cen(I)]\neq cen(I)$.