= Solution
Use the same nonsemisimple matrix from part (i) and the one-dimensional <Lie subalgebra>
$$
\mathfrak g=\mathbb C X\subseteq\mathfrak{gl}_2,\qquad X=\begin{pmatrix}2&1\\-1&0\end{pmatrix}.
$$
It is a <Lie subalgebra> because $[aX,bX]=0$ for all scalars $a,b$. Its <Jordan–Chevalley decomposition> has $X_s=I_2$ and $X_n=X-I_2\ne0$. If $I_2=cX$, comparing off-diagonal entries forces $c=0$, which contradicts the diagonal entries. If $X_n=cX$, taking traces gives $0=2c$, again forcing $c=0$ although $X_n\ne0$. Thus
$$
\boxed{X_s\notin\mathfrak g,\qquad X_n\notin\mathfrak g.}
$$
The semisimplicity assumption on the <Lie algebra> in part (ii) is therefore essential.
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